· web viewoperation of a polyphase induction motor ( 94 – 117 ) type: single cage, double cage,...

ELECTRICAL AND ELECTRONICS DIVISION

HIGHER NATIONALDIPLOMA

YEAR 2

UTILIZATION OF ELECTRICAL ENERGY

EET 442

SEMESTER 1 2012-2013

BTI

BTI

BTI

BA

HR

AIN

TR

AIN

ING

INST

ITU

TE

Outcomes1. Operation Of Power Transformers ( 1 – 41 )

Construction: shell and core typesOperating principles: derive the equivalent circuit for an ideal transformer on load, phasor diagram for an ideal transformer on load, identify the no-load losses, derive the equivalent circuit to represent no-load losses, leakage reactance, winding impedance, derive the complete equivalent circuit, components of the equivalent circuit referred to one winding, phasor diagram for the loaded transformer, voltage regulation, approximate formula for voltage regulation, calculation of voltage regulation, losses on load, efficiency of transformer, calculation of efficiency under load conditions, effects of load changes on losses, load conditions for maximum efficiency, calculation of maximum efficiency.Connections: star-stare, delta-star, delta-zigzag

2. Circuit protection ( 42 - 68 )Over-current protection devices: basic construction of oil, vacuum and airblast circuit breakers, high rupture capacity fuse, overcurrent relay and miniature circuit breakerOperating principles: characteristics and circuit positions of over-current relays, high rupture capacity fuse, and miniature circuit breaker, calculations of ‘time to clear’ over-current faults, discriminationsEarth fault protection devices: basic construction of earth fault relay and residual current circuit breaker, performance requirements of earth fault protection, principle of operation and characteristics of earth fault relays and residual current circuit breaker, position in circuit, calculation of ‘time to clear’ earth faults, discrimination

3. Design of a simple lighting system ( 69 – 93 )Common lamp types: low pressure mercury, high pressure mercury, low pressure sodium, high pressure sodium, fluorescent and halogenLighting design: quality of light, control of glare, luminance distribution, consistency of lighting levels, interior lighting design codes, lighting for visual tasks, emergency lighting Light scheme: produce a scheme for one of the following developments or equivalent, given the appropriate plans: small commercial development to involve roads, tunnel, pedestrian areas and car parks; small supermarket; administration office of a college, including computer stations.

4. Operation of a polyphase induction motor ( 94 – 117 )Type: single cage, double cage, wound rotorOperating principles: production of a rotating magnetic field in the stator; synchronous speed; rotor resistance, reactance, and induced voltage; standstill conditions; slip speed; the effect of rotor speed on rotor resistance and reactance; torque equations; for a three-phase induction motor; torque/speed characteristics, stator and rotor losses, efficiency calculationsStarting methods: direct on-line, stator voltage reduction, rotor resistance methodSpeed control: change of stator voltage and frequency

5. Energy management and tariffs ( 118 –126 )Tariff structures: domestic, Domestic Economy, Domestic Smart, business (eg Economy all-purpose, Economy combined premises, evening and weekend), restricted hour, methods of controlling maximum demand, metering arrangementsEnergy consumption: load scheduling, power factor correction techniques, calculation of apparent power rating of a capacitor to improve power factor of a load, location of power factor correction capacitors, efficient control of heating and lighting systems, recycling heat from heating and lighting systems Cost of energy: cost of running a system using the different tariffs available, selection of appropriate tariff for a given installation and set of circumstances.

BTI - Electrical & Electronics Division a Elbaz Bader AyadYOSRY IDREES

Outcomes and assessment criteria

Outcomes Assessment criteriaTo achieve each outcome a student must demonstrate the ability to:

1. Investigate the operation of power transformers

Describe the construction of different types of power transformer

Identify the operating principle of a power transformer under no-load and load conditions

Describe the modes of connection for polyphase transformers.

2. Investigate circuit protection for distribution and installation systems

Describe the construction of over-current protection devices

Determine the operating principles of circuit over-current protection devices

Determine the operating principles of earth fault protection devices

3. Investigate the design of a simple lighting system

Describe the construction and associated circuit of common lamp type

Determine the operation of common lamp types Determine the principles of good lighting design Plan a simple light scheme

4. Investigate the operation of a polyphase induction motor

Describe the types and construction of induction motors

Determine the operating principles and methods of starting induction motor

Determine the methods of speed control of induction motors.

5. Investigate energy management and tariffs

Determine the factors governing tariff structures Analyse methods for reducing energy

consumption Determine cost of energy used in a system

Prepared By: Yosry Idrees

BTI - Electrical & Electronics Division b Elbaz Bader AyadYOSRY IDREES

BAHRAIN TRAINING INSTITUTE - 77585Electrical and Electronics Division

BTEC Level 5 HND Diploma in Electrical and Electronics Engineering (XN040) – Year 2Module Name: Utilization Of Electrical Energy

Module Code (BTEC/BTI): 21576D/HEE432

ASSESSMENT MAPPING SHEET 2012-2013(Semester1)

LearningOutcome Ref.

Assessment

Criteria Ref.

Assessment Criteria Wk

8

Wk

(10-14)

Wk

15

Wk

16

Wk

18

LO1: Understand the operation of power transformers

1.1 Explain the construction of different types of power transformer Ph1 S/OP

1.2 Identify the operating principles of a power transformer under no-load and load conditions Ph1 Exam

1.3 Discuss the modes of connection for polyphase transformers Ph1 S/OP

LO2: Understand the applications of circuit protection for distribution and installation systems

2.1** Explain the construction of over-current protection devices Ph1 S/OP

2.2 Explain the operating principles of circuit over current protection devices Ph1 Exam

2.3 Explain the operating principles of earth fault protection devices Ph1 S/OP

LO3: Understand the design and construction of lighting systems

3.1 Explain the construction, operation and associated circuitry of common lamp types A Exam3.2 Explain the principles of good lighting design A S/OP3.3 Plan a light scheme A S/OP

LO4: Be able to determine the cost of energy used in a system in order to be energy efficient

4.1 Discuss the factors governing tariff structures Ph2 S/OP4.2 Analyse methods for reducing energy consumption Ph2 S/OP4.3 Determine the cost of energy used in a system Ph2 Exam

LO5: Understand the operation of a poly phase induction motor

5.1 Describe the types and explain the construction of induction motors Ph2 S/OP5.2 Explain the operating principles and methods of starting induction motors Ph2 Exam5.3 Analyse the methods of speed control of induction motors. Ph2 S/OP

M1** Identify and apply strategies to find appropriate solutions (Assessment Criteria Ref. 4.3) Ph2M2 Select/design and apply appropriate techniques (Assessment Criteria Ref. 2.2) ExamM3 Present and communicate appropriate findings (Assessment Criteria Ref. 3.1) A

D1 Use critical reflection to evaluate own work and justify valid conclusions (Assessment Criteria Ref. 5.3) Ph2

D2 Take responsibility for the management and organisation of activities (Assessment Criteria Ref. 5.2) Exam

D3** Demonstrate convergent/lateral/creative thinking (Assessment Criteria Ref. 1.3) Ph1

** Assessed for U, S, G and EA= Assignment; LB= Lab; PA= Practical Assignment; CS= Case Study; Ph= Phase Test; Pr= Project; Qz= Quiz; S/OP= Second Opportunity; Exam= End of Semester Exam

BTEC – HND Year 2 - Utilization of Electrical Energy Ideal Transformer

1. The Ideal Transformer

The transformer is one of the most useful electrical device.. It can raise or lower the voltage or current in an ac circuit, it can isolate circuits from each other, and it can increase or decrease the apparent value of a capacitor, an inductor, or a resistor. Furthermore, the transformer enables us to transmit electrical energy over great distances and to distribute it safely in factories and homes.

We will study some of the basic properties of transformers in this chapter. It will help us understand not only the commercial transformers covered in later, but also the basic operating principle of induction motors, alternators, and synchronous motors. All these devices are based upon the laws of electromagnetic induction. Consequently, we encourage the reader to pay particular attention to the subject matter covered here.

1.1 Voltage induced in a coil:Consider the coil of Fig. 1.l-a, which surrounds (or links) a variable flux . The flux alternates sinusoidal at a frequency f; periodically reaching positive and negative peaks max . The alternating flux induces a sinusoidal ac voltage in the coil, whose effective value is given by

E = 4.44 f N max (1.1)

where: E = effective voltage induced [V] f = frequency of the flux [Hz] N = number of turns on the coil

max = peak value of the flux [Wb] 4.44 = a constant [exact value = 2 / 2]

(a) (b)

Fig. 1.1: (a) A voltage is induced in a coil when it links a variable flux. (b) A sinusoidal flux induced a sinusoidal voltage.

It does not matter where the ac flux comes from: It may be created by a moving magnet, a nearby a coil, or even by an ac current that flows in the coil itself.Equation 1 is derived from Faraday's law equation e = N(/t) in which /t is the rate of change of flux and e is the instantaneous induced voltage. Thus, in Fig. 1-b when /t is positive, the flux is increasing with time and the voltage is positive. Then, when /t is negative, the induced voltage e is negative. Finally, when /t is momentarily zero, the voltage is zero.

BTI - Electrical & Electronics Division 1 Elbaz Bader AyadYOSRY IDREES

N turns

I

+max

E

time

-max

T/2


Example 1.1: The coil in Fig.1 possesses 4000 turns and links an ac flux having a peak value of 2 mWb. If the frequency is 60Hz, calculate the effective value and frequency of the induced voltage E.

Solution:E = 4.44 f N max (1.1) = 4.44 60 4000 0.002 = 2131 V

The induced voltage has an effective or RMS value of 2131 V and a frequency of 60 Hz. The peak voltage is 2131 2 = 3013 V.

1.2 Applied voltage and induced voltage:

Fig. 1.2-a shows a coil of N turns connected to a sinusoidal ac source Eg. The coil has a reactance Xm and draws a current Im . If the resistance of the coil is negligible, the current is given by:

Im = Eg / Xm

(a) (b)

Fig. 1.2: (a) The voltage E induced in a coil is equal to the applied voltage Eg

(b) Phasor relationships between Eg, E, Im and

As in any inductive circuit, Im lags 90° behind Eg and is in phase with the current (Fig. 1.2-b).

The detailed behaviour of the circuit can be explained as follows:The sinusoidal current Im produces a sinusoidal mmf (N Im) which in turn creates a sinusoidal flux . The peak value of this ac flux is max. The flux induces an effective voltage E across the terminals of the coil, whose value is given by Eq. 1.1. On the other hand, the applied voltage Eg and the induced voltage E must be identical because they appear between the same pair of conductors. Because Eg = E, we can write

Eg = 4.44 f N max

so,max = Eg / 4.44 f N (1.2)

This equation shows that for a given frequency and a given number of turns, max varies in proportion to the applied voltage Eg . This means that if Eg is kept constant, the peak flux must remain constant.For example, suppose we gradually insert an iron core into the coil while keeping Eg fixed (Fig. 1.3). The peak value of the ac flux will remain absolutely constant during this operation, retaining its original value max even when the core is completely inside the coil. In effect, if the flux increased (as we would expect), the induced voltage E would also increase. But this is impossible because E = Eg at every instant and, as we said, Eg is kept fixed.

For a given supply voltage Eg, the ac flux in Figs. 1.2 and 1.3 is therefore the same. However, the magnetizing current Im is much smaller when the iron core is inside the coil. In


N turns

Im

Eg E

+

Eg, E

Im


effect, to produce the same flux, a smaller magnetomotive force is needed with an iron core than with an air core. Consequently, the magnetizing current in Fig. 1.3 is much smaller than in Fig. 1.2

(a) (b)

Fig. 1.3: (a) The flux in the coil remain constant so long as Eg is constant. (b) Phasor relationships.

Example 1.2:A coil having 90 turns is connected to a 120 V, 60 Hz source. If the effective value of the magnetizing current is 4 A, calculate the following:

a. The peak value of fluxb. The peak value of the mmfc. The inductive reactance of the coild. The inductance of the coil

Solution:a- max = Eg / (4.44 f N) = 120/(4.44 60 90) = 0.005 = 5 mWbb- The peak current is: Im(peak) = 2 I = 1.41 4 = 5.64 A

The peak mmf (U) is: U = N Im = 90 5.64 = 507.6 AT The flux is equal to 5 mWb at the instant when the coil mmf is 507.6 ampere-turns.

c- The inductive reactance is: Xm = Eg / Im = 120 / 4 = 30 d- The inductance is: L = X m / 2 f = 30 / (2 x 60) = 0.0796 = 79.6 mH

1.3 Elementary transformer:

In Fig. 1.4, a coil having an air core is excited by an ac source Eg. The resulting current Im produces a total flux , which is dispersed in the space around the coil. If we bring a second coil close to the first, it will surround a portion m1 of the total flux. An ac voltage E2 is therefore induced in the second coil and its value can be measured with a voltmeter. The combination of the two coils is called a transformer.

Fig. 1.4: Voltage induced in a secondary winding. Mutual flux is m1 , leakage flux is f1

The coil connected to the source is called the primary winding (or primary) and the other one is called the secondary winding (or secondary).A voltage exists only between primary terminals 1-2 and secondary terminals 3-4, respectively. No voltage exists between primary terminal 1 and secondary terminal 3. The secondary is therefore electrically isolated from the primary.


Eg, E

Im

Im

Eg

1

2

3

4

E2

m1

f1

N turns

Im

Eg E

+


The flux created by the primary can be broken up into two parts: 1- a mutual flux m1, which links the turns of both coils; and 2- a leakage flux f1, which links only the turns of the primary.

If the coils are far apart, the mutual flux is very small compared to the total flux ; we then say that the coupling between the two coils is weak. We can obtain a better coupling (and a higher secondary voltage E2) by bringing the two coils closer together. However, even if we bring the secondary right up to the primary so that the two coils touch, the mutual flux will still be small compared to the total flux . When the coupling is weak, voltage E2 is relatively small and, worse still, it collapses almost completely when a load is connected across the secondary terminals.

In most industrial transformers, the primary and secondary windings are wound on top of each other to improve the coupling between them.

1.4 Polarity of a transformer:

In Fig. 1.4 fluxes f1 and m1 are both produced by magnetizing current Im. Consequently, the fluxes are in phase, both reaching their peak values at the same instant. They also pass through zero at the same instant. It follows that voltage E2 will reach its peak value at the same instant as Eg does. Suppose, during one of these peak moments, that primary terminal 1 is positive with respect to primary terminal 2 and that secondary terminal 3 is positive with respect to secondary terminal 4 (Fig. 1.5).

Terminals 1 and 3 are then said to possess the same polarity. This sameness can be shown by placing a large dot beside primary terminal 1 and another large dot beside secondary terminal 3. The dots are called polarity marks.The polarity marks in Fig. 1.5 could equally well be placed beside terminals 2 and 4 because, as the voltage alternates, they too, become simultaneously positive, every half-cycle. Consequently, the polarity marks may be shown beside terminals 1 and 3 or beside terminals 2 and 4.

Fig. 1.5: Terminals having the same instantaneous polarity are marked with a dot.

1.5 Properties of polarity marks:

A transformer is usually installed in a metal enclosure and so only the primary and secondary terminals are accessible, together with their polarity marks. But although the transformer may not be visible, the following rules always apply to polarity marks:

1. A current entering a polarity-marked terminal produces a mmf that acts in a "positive" direction. As a result, it produces a flux in the "positive" direction (Fig. 1.6). Conversely, a current flowing out of a polarity-marked terminal produces a mmf and flux in the "negative" direction. Thus, currents that respectively flow into and out of polarity-marked terminals of two coils produce magnetomotive forces that buck each other.

2. If one polarity-marked terminal is momentarily positive, then the other polarity-marked terminals is momentarily positive (each with respect to its other terminal). This rule


Im

Eg

1

2

3

4

E2

m1

f1

Polarity mark


enables us to relate the phasor voltage on the secondary side with the phasor voltage on the primary side. For example, in Fig. 1.7, phasor Edc is in phase with phasor Eab.

Fig. 1.6: A current entering a polarity-marked terminal produce a flux in a “positive” direction

(a) (b)

Fig. 1.7: (a) Instantaneous polarities when the magnetizing current is increasing (b) Phasor relationship.

1.6 Polarity tests:

To carry out he polarity test, we proceed as follows (Fig. 1.8):1. Connect the high-voltage winding to an ac source Eg.1. Connect a jumper J between any two adjacent HV and LV terminals.2. Connect a voltmeter Ex between the other two adjacent HV and LV terminals. 4. Connect another voltmeter Ep across the HV winding. If Ex gives a higher reading than Ep,

the polarity is additive. On the other hand, if Ex gives a lower reading than Ep , the polarity is subtractive.

Fig. 1.8: Determining the polarity of a transformer using an ac source.

In this polarity test, jumper J effectively connects the secondary voltage Es in series with the primary voltage Ep. Consequently, Es either adds to or subtracts from Ep. In other words,: Ex = Ep + Es or Ex = Ep - Es, depending on the polarity.

In making the polarity test, an ordinary 120 V, 60 Hz source can be connected to the HV winding, even though its nominal voltage may be several hundred kilovolts.


Edc Eab

Im current

Increasing

Eg

a

b

c

d+

+

-

-

Im

Eg

1

2

3

4

Polarity mark

+

Transformer

Eg

H1

H2

X1

X2

Ep V

EX

V

J


Example 1.3:During a polarity test on a 500 kVA, 69 kV/600 V transformer (Fig. 1.8), the following readings were obtained: Ep = 118 V, Ex = 119 V. Determine the polarity markings of the terminals.

Solution:The polarity is additive because Ex is greater than Ep. Consequently, the HV and LV terminals connected by the jumper must respectively be labelled H1 and X2 (or H2 and X1).

Fig. 1.9 shows another circuit that may be used to determine the polarity of a transformer. A dc source, in series with an open switch, is connected to the LV winding of the transformer. The transformer terminal connected to the positive side of the source is marked Xl. A dc voltmeter is connected across the HV terminals. When the switch is closed, a voltage is momentarily induced in the HV winding. If, at this moment, the pointer of the voltmeter moves upscale, the transformer terminal connected to the (+) terminal of the voltmeter is marked H1 and the other is marked H2.

Fig. 1.9: Determining the polarity of a transformer using a dc source.

1.7 Ideal transformer at no-load; voltage ratio:

Before undertaking the study of practical, commercial transformers, we shall examine the properties of the so-called ideal transformer. By definition, an ideal transformer has no losses and its core is infinitely permeable. Furthermore, any flux produced by the primary is completely linked by the secondary, and vice versa. Consequently, an ideal transformer has no leakage flux of any kind.

(a) (b)

Fig. 1.10: (a) The ideal transformer at no-load. (b) Phasor relationship at no-load.

Figure 1.10-a shows an ideal transformer in which the primary and secondary respectively possess N1 and N2 turns. The primary is connected to a sinusoidal source Eg and the magnetizing current Im creates a flux m. The flux is completely linked by the primary and secondary windings and, consequently, it is a mutual flux. The flux varies sinusoidally, and reaches a peak value m . According to Eq. 1.1, we can therefore write:

E1 = 4.44 f N1 m (1.3)


+

Im

Eg N1 N2E2

m

E1

+E2 Eg, E1

Im

m

H1

H2

X1

X2

V

+

-

+

-


andE2 = 4.44 f N2 m (1.4)

From these equations, we deduce the expression for the voltage ratio of an ideal transformer:

E1 / E2 = N1 / N2 (1.5)where:

E1 = voltage induced in the primary [V]E2 = voltage induced in the secondary [V] N1 = numbers of turns on the primaryN2 = numbers of turns on the secondary

This equation shows that the ratio of the primary and secondary voltages is equal to the ratio of the number of turns. Furthermore, because the primary and secondary voltages are induced by the same mutual m, they are necessarily in phase.The phasor diagram at no load is given in Fig. 1.10-b. Phasor E2 is in phase with phasor E1 (and not 180° out of phase) as indicated by the polarity marks. If the transformer has fewer turns on the secondary than on the primary, phasor E2 is shorter than phasor E1. As in any inductor, current Im lags 90° behind applied voltage Eg . The phasor representing flux m is obviously in phase with magnetizing current Im which produces it.

However, because this is an ideal transformer, the magnetic circuit is infinitely permeable and so no magnetizing current is required to produce the flux m. Thus, under no-load conditions, the phasor diagram of such a transformer is identical to Fig. 1.10-b except that phasor Im is infinitesimally small.

Example 1.4:A not quite ideal transformer having 90 turns on the primary and 2250 turns on the secondary, is connected to a 120 V, 60 Hz source. The coupling between the primary and secondary is perfect, but the magnetizing current is 4 A.Calculate:

a. The effective voltage across the secondary terminalsb. The peak voltage across the secondary terminals

c. The instantaneous voltage across the secondary when the instantaneous voltage across the primary is 37 V

Solution: a. The turns ratio is: N2 / N1 = 2250 / 90 = 25

The secondary voltage is therefore 25 times greater than the primary voltage because the secondary has 25 times more turns. Consequently:

E2 = 25 x E1 = 25 x 120 = 3000 VInstead of reasoning as above, we can apply Eq. 1.5:

E1 / E2 = N1 / N2 = 120 / E2 = 90 / 2250 V which again yields E2 = 3000 V

b. The voltage varies sinusoidally; consequently, the peak secondary voltage is:E2(peak) = 2 E = 1.414 x 3000 = 4242 V

c. The secondary voltage is 25 times greater than E1 at every instant. Consequently, when e1 = 37 V e 2 =25 X 37 = 925 V

1.8 Ideal transformer under load; current ratio:

Pursuing our analysis, let us connect a load Z across the secondary of the ideal transformer (Fig. 1.11). A secondary current I2 will immediately flow, given by: I2 = E2 / ZDoes E2 change when we connect the load? To answer this question, we must recall two facts.



(a) (b)

Fig. 1.11: (a) Ideal transformer under load. (b) Phasor relationships under load.

First, in an ideal transformer the primary and secondary windings are linked by a mutual flux m, and by no other flux. In other words, an ideal transformer, by definition, has no leakage flux. Consequently, the voltage ratio under load is the same as at no-load, namely:

E1 / E2 = N1 / N2

Second, if the supply voltage Eg is kept fixed, then the primary induced voltage E1 remains fixed. Consequently, mutual flux m also remains fixed. It follows that E2 also remains fixed. We conclude that E2 remains fixed whether a load is connected or not.

Let us now examine the magnetomotive forces created by the primary and secondary windings. Current I2 produces a secondary mmf (N2 I2). If it acted alone, this mmf would produce a profound change in the mutual flux m. But we just saw that Him does not change under load. We conclude that flux m can only remain fixed if the primary develops a mmf which exactly counterbalances (N2 I2 )at every instant. Thus, a primary current I1 must flow so that:

N1 I1 = N2 I2 (1.6)

To obtain the required instant-to-instant bucking effect, currents I1 and I2 must increase and decrease at the same time. Thus, when I2 goes through zero, I1 goes through zero, and when I2 is maximum (+) I1 is maximum (+). In other words, the currents must be in phase. Furthermore, in order to produce the bucking effect, when I1 flows into a polarity mark on the primary side, I2 must flow out of the polarity mark on the secondary side (see Fig. 1.11-a).

Using these facts, we can now draw the phasor diagram of an ideal transformer under load (Fig. 1.11-b). Assuming a resistive-inductive load, current I2 lags behind E2 by an angle , Flux m lags 90° behind Eg, but no magnetizing current Im is needed to produce this flux because this is an ideal transformer. Finally, the primary and secondary currents are in phase. According to Eq. 1.6, they are related by the equation:

I1 / I2 = N2 / N1 (1.7)where:

I1 = primary current [A]I2 = secondary current [A]N1 = number of turns on the primaryN2 = number of turns on the secondary

Comparing Eq. 1.5 and Eq. 1.7, we see that the transformer current ratio is the inverse of the voltage ratio. In effect, what we gain in voltage, we lose in current and vice versa. This is consistent with the requirement that the instantaneous power input E1 I1 to the primary must equal the instantaneous power output E2 I2 of the secondary. If the power inputs and outputs were not identical, it would mean that the transformer itself absorbs power. By definition, this is impossible in an ideal transformer.

Example 1.5:


+

I1

Eg N1 N2E2

m

E1

+I2

Z

E2 Eg, E1

m

I1

I2


An ideal transformer having 90 turns on the primary and 2250 turns on the secondary is connected to a 200 V, 50 Hz source. The load across the secondary draws a current of 2 A at a power factor of 80 percent lagging (Fig. 1.l2-a).

Calculate:a) The effective value of the primary currentb) The instantaneous current in the primary when the instantaneous current in the

secondary is 100mA c) The peak flux linked by the secondary winding d) Draw the phasor diagram

Solution: a- The turns ratio is: a = N1/ N2 = 90 / 2250 = 1 / 25

The current ratio is therefore 25 and because the primary has fewer turns, the primary current is 25 times greater than the secondary current. Consequently:

I1 = 25 x 2 = 50A

Instead of reasoning as above, we can calculate the current by means of Eq. 1.6.N1 I1 = N2 I2 = 90 I1 = 2250 2 I1 = 50 A

b- The instantaneous current in the primary is always 25 times greater than the instantaneous current in the secondary. Therefore when I2 = 100 mA,I1 is:I1 instantaneous = 25 I2 instantaneous = 25 x 0.1 = 2.5A

c- In an ideal transformer, the flux linking the secondary is the same as that linking the primary. The peak flux in the secondary is:

m = Eg / (4.44 f N1) = 200 / (4.44 x 50 90) = 0.01 = 10 mWb

(a) (b)

Fig. 1.12: (a) See Example 4 (b) Phasor relationship.

d. To draw the phasor diagram, we reason as follows: Secondary voltage is:E2 = 25 x E1 = 25 x 200 = 5000 V

E2 is in phase with E1 indicated by the polarity marks. For the same reason, I1 is in phase with I2. Phase angle between E2 and I2 is: power factor = cos = 0.8. The phase angle between E1 and I1 is also 36.9°. The mutual flux lags 90° behind Eg (Fig. 1.12-b).

1.9 Circuit symbol for an ideal transformer:

We simply as shown in Fig. 1.13 by drawing the primary and secondary windings and the mutual flux m. Polarity marks are added, enabling us to indicate the direction of current flow as well as the polarities of voltages E1 and E2 . For example, a current I1 flowing into one polarity-marked terminal is always accompanied by a current I2 flowing out of the other polarity-marked terminal. Consequently, I1 and I2 are always in phase .


Eg, =E1= 200V E2= 5000V

m =

10 mWb

I2= 2A

I1= 50A

= 36.9+

I1

EgN1

90

N2

2250E2

m

E1

+I2 = 2A

Z Cos = 0.8

200 V60 HZ


Fig. 1.13: (a) Symbol for an ideal trans. & phasor diagram using sign notation. (b) Symbol for an ideal trans. & phasor diagram using double-subscript notation.

Furthermore, if we let the ratio of transformation N1 / N2 = a , we obtain E1 = a E2 and I1 = I2 / a

In an ideal transformer, and specifically referring to (Fig. 1.13-a), E1 and E2 are always in phase, and so are I1 and I2

If the double-subscript notation is used (Fig. 1.13-b), Eab and Ecd are always in phase and so are I1 and I2.

1.10 Impedance ratio:

Although a transformer is generally used to transform a voltage or current, it also has the important ability to transform an impedance. Consider, for example, (Fig. 1.14-a) in which an ideal transformer is connected between a source Eg and a load Z. The ratio of transformation is a, and so we can write E1 / E2 = a and I1 / I2 = 1/a

As far as the source is concerned, it sees an impedance ZX between the primary terminals given by: ZX = E1 / I1

On the other hand, the secondary sees an impedance Z given by: Z = E2 / I2

However, ZX can be expressed in another way:

Z X E

1I1

a E

2I / a

a2 E

2I2

a2 Z2

Consequently, ZX = a2 Z (1.8)

This means that the impedance seen by the source is a2 times the real impedance (Fig. 1.14-b). Thus, an ideal transformer has the amazing ability to increase or decrease the value of an impedance. In effect, the impedance seen across the primary terminals is identical to the actual impedance across the secondary terminals multiplied by the square of the turns ratio.


I1 +

E1

I2+

E2

N1 N2

a = N1/N2

Ideal Trans.

I1

E1

N1

a = N1/N2

Ideal Trans.

Eg

I2

N2

E2 Z

I1

E1 Zx = a2 ZEg

I1 a

E1

I2 c

E2

N1 N2

a = N1/N2

Ideal Trans.

b d

I1

I2

E2 E1

Any angle

(a)

Ecd Eab

Any angle

(b)

I1

I2


(a) (b)Fig. 1.14: (a) Impedance transformation using a transformer. (b) The impedance seen by the source differs from Z

The impedance transformation is real, and not illusory like the image produced by a magnifying glass. An ideal transformer can modify the value of any component, be it a resistor, capacitor, or inductor.

For example, if a 1000 resistor is placed across the secondary of a transformer having a primary to secondary turns ratio of 1:5, it will appear across the primary as if it had a resistance of 1000 (1/5)2 = 40. Similarly, if a capacitor having a reactance of 1000 is connected to the secondary, it appears as a 40 capacitive reactance across the primary. However, because the reactance of a capacitor is inversely proportional to its capacitance (XC = 1 / 2 f C), the apparent capacitance between the primary terminals is 25 times greater than its actual value. We can therefore artificially increase (or decrease) the microfarad value of a capacitor by means of a transformer.

1.11 Shifting impedances from secondary to primary and vice versa:

As a further illustration of the impedance-changing properties of an ideal transformer, consider the circuits of Fig. 1.15. It is composed of a source Eg, a transformer T, and four impedances Z 1 to Z4. The transformer has a turns ratio a. As the impedances are shifted, the circuit configuration remains the same, but the shifted impedance values are multiplied by a2.The reason is that the voltage E across each element in the secondary side becomes aE when the element is shifted to the primary side. Similarly, the current I in each element in the secondary side becomes I / a when the element is shifted to the primary side.On account of this relationship, it is easy to solve a real circuit such as the one shown in Fig. 1.15-a. We simply reduce it to the equivalent form shown in Fig. 1.15-e and solve for all the voltages and currents. These values are then respectively multiplied by 1/a and by a, which yields the actual voltages and currents of each element in the secondary side.

(a)


E3

E2

I1

a = 1

Ideal Trans.

Eg

I3

Z1 Z3

Z2I2

E4Z4

I4

E4

E3

aE2

I1

a = 1

Ideal Trans.

Eg

Z1

I3

Z3

Z4

I4

a2Z2I2/a


(b)

(c)

(d)

(e)

Fig. 1.15:a) The actual circuit showing the actual voltages and currents.b) Impedance Z2 is shifted to the primary side.c) Impedance Z3 is shifted to the primary side..d) Impedance Z4 is shifted to the primary side.e) All the impedances are now transferred to the primary side the transformer is no longer

needed.

To illustrate, suppose that the real voltage across Z4 in Fig. 1.16 is E4 volts and that the real current through it is I4 amperes. Then, in the equivalent circuit, the voltage across the a2 Z4 impedance is equal to (E4 a ) volts. On the other hand, the current through the impedance is equal to ( I4 a ) amperes (Fig. 1.17). In other words, whenever an impedance is transferred to the primary side, the real voltage across the impedance multiplying by a factor a, while the real current dividing by the factor a.


aE4a2Z2I2/a

I1

Eg

Z1 a2Z3

a2Z4I3/a

I2/a

aE3

I1

a = 1

Ideal Trans.

Eg

Z1

I3/a

a2Z3

E4Z4

I4

aE2a2Z2

aE4

a = 1

Ideal Trans.

I2/a

aE3

I1

Eg

Z1

I3/a

a2Z3

a2Z4I4/a

I = 0

aE2a2Z2

a = 1

Ideal Trans.

Eg

Z1 Z3

Z2

E4Z4

I4

A

V


Fig. 1.16: Actual voltage & current in impedance Z4

Fig. 1.17: Equivalent voltage & current in Z4 referring to the primary side

In some cases it is useful to shift impedances in the opposite way, that is from the primary side to the secondary side (Figs. 1.18). The procedure is the same, but all impedances so transferred are now divided by a2 .

(a) (b)

(c) (d)

Fig. 1.18: (a) The actual circuit, showing the real voltages and currents on the primary side.(b) Impedance Z1 is transferred to the secondary side. Note the corresponding change

in E1 and I1.(c) The source is transferred to the secondary side. Note the corresponding change in

Eg. Note also that the currents in T are zero.(d) All the impedances and even the source are now on the secondary side. The

transformer is no longer needed because its currents are zero.

Example 1.6:Calculate voltage E and current I in the circuit of Fig. 1.19, knowing that ideal transformer has a primary to secondary turns ratio of 1:100.

Solution:The easiest way to solve this problem is to shift all the impedances to the primary side of the transformer. Because the primary has 100 times fewer turns than the secondary, the impedance values are divided by 1002 , or 10 000.


aE4a2Z2I2/a

I1

Eg

Z1 a2Z3

a2Z4I3/a

E1

I1

a = 1

Ideal Trans.

Eg

Z1 Z3

Z2 Z4

a = 1

Ideal Trans.

Eg

Z3

Z2 Z4

E1/a

aI1

Z1/a2

a = 1

Ideal Trans.

Eg/a

Z3

Z2 Z4

E1/a

aI1

Z1/a2

I = 0 I = 0

Eg/a

Z3

Z2 Z4

Z1/a2


Voltage E becomes E/100, but current I remains unchanged because it is already on the primary side (Fig. 1.18).The impedance of the circuit in Fig. 1.20 is:

Z = R2 + (XL - XC )2 = 42 + (5 - 2)2 = 5

The current in the circuit is: I = E / Z = 10 / 5 = 2 A

The voltage across the resistor is: E /100 = IR = 2 4 = 8 V

The actual voltage E is, therefore, E = 8 100 = 800 V

Fig. 1.19: See Example 5

Fig. 1.20: Equivalent circuit of Fig. 17


10V 4

2

I1

5

E/100

E

a = 1/100

Ideal Trans.

Eg

20K

40K

I1

5

10 V


1.12 Questions and Problems:

1. The coil in Fig. 1.2-a has 500 turns and a reactance of 60 but negligible resistance. If it is connected to a 120 V, 60 Hz source Eg, calculate the following:a. The effective value of the magnetizing current Im.b. The peak value of Im.c. The peak and mmf produced by the coil.d. The peak flux max.

2. In Problem 1, if the voltage Eg is reduced to 40 V, calculate the new mmf developed by the coil and the peak flux max.

3. What is meant by mutual flux? by leakage flux?

4. The ideal transformer in Fig. 1.11 has 500 turns on the primary and 300 turns on the secondary. The source produces a voltage Eg of 600 V, and the load Z is a resistance of 12.

Calculate the following:a. The voltage E2 b. The current I2

c. The current I1 d. The power delivered to the primarye. The power output from the secondary

5. In Problem 4, what is the impedance seen by the source Eg?

6. In Fig. 1.19, calculate the voltage across the capacitor and the current flowing through it.

7. The nameplate on a 50 kVA transformer shows a primary voltage of 480 V and a sec-ondary voltage of 120 V. We wish to determine the approximate number of turns on the primary and secondary windings. Toward this end, three turns of wire are wound around the external winding, and a voltmeter is connected across this 3-turn coil. A voltage of 76 V is then applied to the 120 V winding, and the voltage across the 3-turn winding is found to be 0.93 V. How many turns are there on the 480 V and 120 V windings (approximately)?

8. A coil with an air core has a resistance of 14.7. When it is connected to a 42 V, 60 Hz ac source, it draws a current of 1.24 A. Calculate the following:

a) The impedance of the coil.b) The reactance of the coil, and its inductance.c) The phase angle between the applied voltage.

9. Two coils are set up as shown in Fig. 1.4. Their respective resistances are small and may be neglected. The coil having terminals 1, 2 has 320 turns while the coil having terminals 3, 4 has 160 turns. It is found that when a 56 V, 60 Hz voltage is applied to terminals 1-2, the voltage across terminals 3-4 is 22 V. Calculate the peak values of , f1 and m1.

10. A 40 F, 600 V paper capacitor is available, but we need one having a rating of about300 F. It is proposed to use a transformer to modify the 40 F so that it appears as 300 F. The following transformer ratios are available: 120 V/330 V; 60 V/450V; 480 V/150 V. Which transformer is the most appropriate and what is the reflected value of the 40 F capacitance? To which side of the transformer should the 40 F capacitor be connected?


BTEC – HND Year 2 - Utilization of Electrical Energy Practical Transformer

2. Practical Transformers

We studied the ideal transformer and discovered its basic properties. However, in the real world transformers are not ideal and so our simple analysis must be modified to take this into account. Thus, the windings of practical transformers have resistance and the cores are not infinitely permeable. Furthermore, the flux produced by the primary is not completely captured by the secondary. Consequently the leakage flux must be taken into account. And finally, the iron cores produce eddy-current and hysteresis losses, which contribute to the temperature rise of the transformer.The properties of a practical transformer can be described by an equivalent circuit comprising an ideal transformer and resistances and reactances. The equivalent circuit is developed from fundamental concepts. This enables us to calculate such characteristics as voltage regulation and the of transformers that are connected in parallel. The per-unit method is also used to illustrate its mode of application.

2.1. Ideal transformer with an iron core:

The ideal transformer studied before had an infinitely permeable core. We can replace the ideal transformer core by an iron core having hysteresis and eddy-current losses and a circuit with two elements Rm and Xm in parallel with the primary terminals of the ideal transformer (Fig. 2.l-a). The primary is excited by a source Eg that develops a voltage E1 .

The resistance Rm the iron losses and the resulting heat they produce. To furnish these losses a small current IF is drawn from the line. This current is in phase with E1 (Fig. 2.l-b).The magnetizing reactance Xm is a measure of the permeability of the transformer core. Thus if the permeability is low, Xm is relatively low. The current Im flowing through Xm represents the magnetizing current needed to create the flux m in the core. This current lags 90° behind E1.

Note:In practice, the normal procedure ensure the ES is in phase with EP (secondary and primary terminal voltages) also I2 and I1 (secondary and primary currents). However, if ES and EP were drown in phase with one another Fig. 2.1-b, the diagram would become cluttered and therefore, for convenience, it is usual to show EP and ES in phase opposition with one another and also E1

and E2. This should be remembered that the manner of drawing is for convenience only.

(a) (b)

Fig. 2.1: (a) An imperfect core represented by a reactance Xm and a resistance Rm. (b) Phasor diagram of a practical transformer at no-load.

The values of the impedances Rm and Xm can be found experimentally by connecting the trans-former to an ac source and measuring the active power and reactive power absorbed. The following equations then apply:

Rm = E12 / Pm (2.1)

Xm = E12 /Qm (2.2)


Io

E1 Industrial application:

34. A transformer has a rating 200 kVA, 14 400 V/277 V. The high

voltage winding has a resistance of 62 . What is the approximate resistance of the 277 V winding?

35. The primary winding of the transformer in Problem34 is wound with No. 11 gauge AWG wire. Calculate the approximate cross section (in square millimeters) of the conductors in the secondary winding.

36. An oil

filled distribution transformer rated at 10 kVA weighs 118 kg, whereas a 100 kVA transformer of the same kind weighs 445 kg. Calculate the power output in watts per kilogram in each case.

37. The transformer shown in Fig. 13 has a

rating of 40 kVA. If 80 V is applied between terminals X1

, and X2

, what voltage will appear between terminals 3 and 4? If a single load is applied between terminals 3 and 4 what is the maximum allowable current that can be drawn?

1

Eg

I2= 0

E2RmXmIf

Im

I1 = 0

Im

If Eg, E1E2

m

Io


where:Rm = resistance representing the iron losses []Xm = magnetizing reactance of the primary winding []E1 = primary induced voltage [V]Pm = iron losses [W]Qm = reactive power needed to set up the mutual flux m [var]

The total current needed to produce the flux m in an imperfect core is equal to the phasor sum of If and Im. It is called the exciting current lo. The phasor diagram at no-load for the transformer is shown in Fig. l-b. The peak value of the mutual flux m is again given by :

m = E1 / (4.44 f N1) (1.3)

Example 2.1:A large transformer operating at no-load draws an exciting current Io of 5 A when the primary is connected to a 120 V, 60 Hz source (Fig. 2.2-a). From a wattmeter test it is known that the iron losses are equal to 180 W.

Calculate:a. The reactive power absorbed by the coreb. The value of Rm and Xm

c. The value of If and Im

(a) (b)

Fig. 2.2: (a) See Example 1. (b) Phasor diagram.

Solution:a. The apparent power supplied to the core is: Sm = E1 Io = 120 5 = 600 VA

The iron losses are: Pm = 180WThe reactive power absorbed by the core is:

Qm = (Sm 2 - Pm 2) = 600 2 - 1802 = 572 var

b. The impedance corresponding to the iron losses is:Rm = E1

2 / Pm = 1202 / 180 = 80The magnetizing reactance is: Xm = Em

2 / Qm = 1202 / 572 = 25.2

c. The current needed to supply the iron losses is: If = E1 / Rm = 120 / 80 = 1.5AThe magnetizing current is: Im = E1 / Xm = 120 / 25.2 = 4.8 AThe exciting current Io is: Io = (If

2 + Im 2) = 1.52 + 4.82 = 5 A

The phasor diagram is given in Fig. 2-b.

2.2 Ideal transformer with loose coupling:


E1Eg

I2= 0

E2120 V60 HZ

5 A

Pm = 180 W

Im= 4.8A

If = 1.5A120 V

m

Io= 5A


We have just seen how an ideal transformer behaves when it has an imperfect core. We now assume a transformer having a perfect core but rather loose coupling between its primary and secondary windings. We also assume that the primary and secondary windings have negligible resistance and the turns are N1, N2.

Let us now connect a load Z across the secondary, keeping the source voltage EP fixed ( see Fig. 2.3 ). The operation sets as the following list:

1. Currents I1 and I2 immediately begin to flow in the primary and secondary windings. They are related by the ideal-transformer equation:

I1 /I2 = N2 / N1, hence N1 I1 = N2 I2 2. I2 produces an mmf (N2 I2) while I1 produces an mmf (N1 I1) . These

magnetomotive forces are equal and in direct opposition because when I1 flows into the polarity-marked terminal 1, I2 flows out of polarity-marked terminal 3.

3. The mmf N2 I2 produces a total ac flux 2 . A portion of 2 (m2) links with the primary winding while another portion (f2) does not. Flux f2 is called the secondary leakage flux.

Fig. 2.3: Mutual fluxes and leakage fluxes produced by a transformer under load. The leakage fluxes are due to the imperfect coupling between the coils.

4. Similarly, the mmf N1 I1 produces a total ac flux 1 . A portion of 1 (m1 ) links with the secondary winding, while another portion (f1)) does not. Flux f1 is called the primary leakage flux.

5. We combine m1 and m2 into a single mutual flux m (Fig. 2.4). This mutual flux is created by the joint action of the primary and secondary mmfs.

Fig. 2.4: A transformer possesses two leakage fluxes and a mutual flux.

6. We note that the primary leakage flux f1 is created by N1 I1 , while the secondary leakage f2 created by N2 I2. Consequently, leakage flux f1 is in phase with I1 and leakage flux f2 is in phase with I2.

7. The voltage Es induced in the secondary is actually composed of two parts:a. A voltage Ef2 induced by leakage flux f2 and given by:

Ef2 = 4.44 f N2 f2 (2.3)

b. A voltage E2 induced by mutual flux m and given by: E2 = 4.44 f N2 m (2.4)


m2

2

m1 11

f1 f2

I2I1

Es

3

4

1

2

EpEg N1 N2 Z

m

f1 f2

I2I1

Es

3

4

1

2

EpEg N1 N2 Z


In general, Ef2 and E2 are not in phase. Similarly, the voltage Ep induced in the primary is composed of two parts:

a. A voltage Efl induced by leakage flux f1 and given byEf1 = 4.44 f N1 f1 (2.5)

b. A voltage E1 induced by mutual flux m and given by:E1 = 4.44 f N1 m (2.6)

8. Induced voltage Ep = applied voltage Eg.Using these eight basic facts, we now proceed to develop the equivalent circuit of the

transformer.

2.3 Primary and secondary leakage reactance:

We can better identify the four induced voltages E1 , E2 , Ef1 and Ef2 by rearranging the transformer circuit as shown in Fig. 2.5. Thus, the secondary winding is drawn twice to show even more clearly that the N2 turns are linked by two fluxes, f2 and m.

Fig. 2.5: Separating the various induced voltages due to the mutual flux and the leakage fluxes.

This rearrangement does not change the value of the induced voltages, but it does make each voltage stand out by itself. Thus, it becomes clear that Ef2 is really a voltage drop across a re-actance. This secondary leakage reactance Xf2 is given by:

Xf2 = Ef2 / I2 (2.7)

The primary winding is also shown twice, to separate E1 from Ef1. Again, it is clear that Ef1 is simply a voltage drop across a reactance. This primary leakage reactance Xf1 is given by:

Xf1 = Ef1 / I1 (2.8)

The primary and secondary leakage reactances are shown in Fig. 2.6. We have also added the primary and secondary winding resistances R1 and R2, which, of course, act in series with the respective windings.

Fig. 2.6: Resistance and leakage reactance of the primary and secondary windings.2.4 Equivalent circuit of a practical transformer:

The circuit of Fig. 2.6 is composed of resistive and inductive elements (R1 , R2 , Xf1 , Xf2 , Z) coupled together by a mutual flux m which links the primary and secondary windings. The leakage-free magnetic coupling enclosed in the dotted oval is actually an ideal transformer. It


I1R1 Xf1

1

2

EpEg E1

N1

Xf2 I2

Es

3

4

Z

R2

E2

N2

m

m

N1 N2

I1

f11

2

N1

Ef1EpEg E1

I2

Es

3

4

ZE2

f2

N2

Ef2


possesses the same properties and obeys the same rules as the ideal transformer discussed before. For example, we can shift impedances to the primary side by multiplying their values by (NI / N2)2, as we did before.

If we add circuit elements Xm and Rm to represent a practical core, we obtain the complete equivalent circuit of a practical transformer (Fig. 2.7-a). In this circuit T is an ideal transformer, but only the primary and secondary terminals 1-2 and 3-4 are accessible; all other components are "buried" inside the transformer itself. However, by appropriate tests we can find the values of all the circuit elements that make up a practical transformer.

Fig. 2.7-b shows the phasor diagram of a practical transformer. It is clear that the primary current IP equal to the phasor sum of I1 and Io .

where :Io = no-load current.I1 = secondary current (I2) referred to the primary side. [I1 = I2 / a = I2 (N2 / N1)]

(a)

Class work 2.2:The secondary winding of a transformer possesses 180 turns. When the transformer is under load, the secondary current has an effective value of 18 A, 60 Hz. Furthermore, the mutual m

has a peak value of 20 mWb. The secondary leakage flux f2 has a peak value of 3 mWb.Calculate:

a) The voltage induced in the secondary winding by its leakage flux b) The value of the secondary leakage reactancec) The value of E2 induced by the mutual flux m

2.5 Transformers and the per-unit system:

The per-unit system of measurement is particularly useful when dealing with transformer. The nominal power rating SB of the transformer is selected as the base power, and the nominal


IpR1 Xf1

1

2

EpEg E1

N1

Xf2 I2

Es

3

4

Z

R2

E2

N2

Io

Rm

If

Xm

Im

I1

ES , E2

EP , E1

IP

Io

Im m

I1

I2

If

S

P

o

(b)

Fig. 2.7: a) Complete equivalent circuit of a practical transformer.

b) Phasor diagram of a practical transformer


voltage EB is selected as the base voltage. It follows that the base current In is equal to SB / EB, which is the same as the nominal current.

The base impedance ZB is therefore equal to EB / IB. Note that EB and IB are the nominal voltage and nominal current on either the primary or secondary side. The base impedance may also be calculated by the equation:

ZB = EB2 / SB (2.9)

where:ZB = base impedance of the transformer []SB = power rating of the transformer [VA]EB = nominal voltage of the transformer on either the primary or secondary side [V]

Because the primary and secondary voltages are usually different, it follows that a transformer has two base impedances, one for the primary and one for the secondary. Depending upon the respective voltages, the two impedances may be vastly different.

Example 2.3:Calculate the base impedance of a transformer rated 250 kVA, 4160 V / 480 V, 60 Hz, on the

a) primary and b) secondary side.

Solution:a. Base impedance on the primary side is: ZBp = Ep

2 / SB = 41602 / 250 000 = 69

b. Base impedance on the secondary side is: ZBs = Es2 / SB = 4802 / 250000 = 0.92

2.6 Per-unit impedances:

We can get a better idea of the relative magnitude of the winding resistance, leakage reactance, etc., of a transformer by comparing these impedances with the base impedance of the transformer. In making the comparison, circuit elements located on the primary side are compared with the primary base impedance. Similarly, circuit elements on the secondary side are compared with the secondary base impedance. The comparison can be made either on a percentage or on a per-unit basis; we shall use the latter. Typical per-unit values are listed in Table 2-A, for transformers ranging from 3 kVA to 100 MVA. For example, the per-unit resistance of the primary winding of a transformer ranges from 0.009 to 0.002 for all power rat-ings between 3 kVA and 100 MVA. Over this tremendous power range, the per-unit resistance of the primary or secondary windings varies only from 0.009 to 0.002 of the base impedance of the transformer. Knowing the base impedance of either the primary or the secondary winding, we can readily estimate the order of magnitude of the real values of the transformer impedances. Table 2-A is, therefore, a useful source of information.

Fig. 2.8: Equivalent circuit of a transformer.

Circuit elementTypical per-unit values

3 KVA to 250 KVA 1 MVA to 100 MVAR1 or R2 0.009 - 0.005 0.005 - 0.002Xf1 or Xf2 0.008 - 0.025 0.03 - 0.06


IpR1 Xf1

1

2

EpEg E1

Xf2 I2

Es

3

4

Z

R2

E2

N2

Io

Rm

If

Xm

Im

I1

N1


Xm 20 - 30 50 - 200Rm 20 - 50 100 - 500Io 0.05 - 0.03 0.02 - 0.005

Table 2-A: Typical per-unit values of transformers

Example 2.4:Using the information given in Table 2-A, calculate the approximate real values of the impedances of a 250 kVA, 4160 V/480 V, 60 Hz distribution transformer.

Solution:We first determine the base impedances on the primary and secondary side. From the results of Example 3, we have: ZBp = 69 ZBs = 0.92 We now calculate the real impedance by multiplying ZBp and ZBs by the per-unit values given in Table 2-A. Thus, referring to Fig. 2-8 we find the following:

R1 = 0.005 69 = 0.35 R2 = 0.005 0.92 = 4.6 mXf1 = 0.025 69 = 1.7 Xf2 = 0.025 0.92 = 23 mXm = 30 69 = 2 K Rm = 50 69 = 3.5 K

Fig. 2.9: See Example 2.4

This example shows the usefulness of the per-unit method of estimating impedances. The equivalent circuit of the 250 kVA transformer is shown in Fig. 2.9. The true values may be 20 to 50 percent higher or lower than those shown in the figure. The reason is that the per-unit values given in Table 2-A are broad estimates covering a wide range of transformers.

2.7 Simplifying the equivalent circuit:

The equivalent circuit of a transformer (Figs. 2.7 and 2.8) is relatively simple, but it gives far more detail than is needed in most practical problems. Consequently, we can simplify it to make the calculations easier. Let us try to simplify the circuit when the transformer operates:

1. at no-load and 2. at full-load.

1. At no-load: (Fig. 2.10) I2 is zero and so is I1 because T is an ideal transformer.Consequently, only the exciting current Io flows in Rl and Xfl. These impedances are so small that the voltage drop across them is negligible. Furthermore, the current in R2 and Xf2 is zero. We can, therefore, neglect these four impedances, giving us the much simpler circuit of Fig. 2.11.


0.924160VEg

N1

480V Z

N2Rm

3.5KXm

2K

R1

0.35

Xf1

1.7

Xf2

23

R2

4.6m

Z

a = 4160/480 = 8.87

Ip R1 Xf1

EpEg

N1

Xf2I2

Es

R2

N2

Io

RmXm

X1


Fig. 2.10: Complete equivalent circuit of a transformer at no-load.

Fig. 2.11: Simplified circuit at no-load

The turns ratio, a = N1 / N2, is obviously equal to the ratio of the primary to secondary voltages Ep / Es measured across the terminals.

2. At full-load: Ip is at least 20 times larger than Io. Consequently, we can neglect Io and the corresponding magnetizing branch. The resulting circuit is shown in Fig. 2.12. This simplified circuit may be used even when the load is only 10 percent of the rated capacity of the trans-former.

Fig. 2.12: Simplified equivalent circuit of a transformer at full-load.

Fig. 2.13: Equivalent circuit with impedances shifted to the primary side.

We can further simplify the circuit by shifting everything to the primary side, thus eliminating transformer T (Fig. 2.13). Then, by summing the respective resistances and reactances, we obtain the circuit of Fig. 2.14. In this circuit:

Rp = R1 + a2R2 (2.10)

Xp = Xfl + a2 Xf2 (2.11)where:Rp = total transformer resistance referred to the primary side

Xp = total transformer leakage reactance referred to the side

Fig. 2.14: The internal impedance of a large transformer is mainly reactive.The combination of Rp and Xp constitutes the total transformer impedance Zp referred to the pri-mary side. From the impedance triangle we have:

Zp = (Rp2 + Xp

2 ) (2.12)

Impedance Zp is usually expressed as either a percent or per-unit of the base primary impedance Zn, mentioned before. The per-unit impedance is always given on the nameplate of the transformer.


E1

Io

EpEg

N1

Es

N2

RmXmE2

Ip

R1 Xf1

EpEg

N1 N2

Xf2

Es

R2

I2

Z

Ip

R1Xf1

EpEg

a2Xf2

aEs

a2R2

a2Z

a2Z

Ip

EpEg

XpRp

aEsZp


Transformers above 500 kVA possess a leakage reactance Xp at least five times greater than Rp. In such transformers we can neglect Rp , as far as voltages and currents are concerned. The equivalent circuit is thus reduced to a simple reactance Xp between the source and the load (Fig. 2.15). It is quite remarkable that the relatively complex circuit of Fig. 2-8 can be reduced to a simple reactance in series with the load.

Fig. 2.15: The internal impedance of a large transformer is mainly reactive.

2.8 Voltage Regulation of a Transformer:

The voltage regulation of a transformer is defined as the variation of the secondary voltage between no-load E2 and full-load VS , expressed as either a per-unit or a percentage of the no-load voltage E2 , the primary voltage being assumed constant.

Percentage Regulation =No load voltage Full load voltage

No load voltage

100

= (E2 - ES) 100 / E2

Per-unit Regulation = (E2 - ES) / E2 (2.13)

To calculate voltage regulation, the primary winding resistance and reactance are referred to the secondary; the equivalent values are used to construct the secondary phasor diagram in which the load current is with a lagging phase angle S (See Fig. 2.16 & Fig. 2.17)

Fig. 2.16: Equivalent Circuit Referred to Secondary Side

In a resistance, the voltage and current are in-phase therefore the (Is. Rs) is in-phase with IS. This is added to Es as shown in Fig. 2.17.

The voltage developed across a pure inductance lead the current by 90 so that the voltage (IsXs) leads Is by 90.

Finally add (Is Rs) and (Is Xs) to give (Is Zs) and the later will add to Es to gives E2 . For the leading power factor, the load voltage Es is more than E2 .


EpEg

Xp

aEs a2ZIp

Ep = E1

N1

Is

LoadN2

EsE2

RS XS

M

B

E2

I 2 Zs

I 2 Xs

I 2 Rs

s

s

S

A

I2

sO

N

C

ES


Fig. 2.17: Phasor diagram

Transformer are highly efficient and losses are small. Thus, the impedance triangle will be small compared to the voltages (Es & E2 ).

The angle will be so small that it can be said that :

OC = ON E2 - Es = OC - OA = ON - OA

= AN = AM + MN

= Is Rs Cos s IS XS Sin s

= I2 ( RS Cos s + XS Sin s ) at lagging power factor= I2 ( RS Cos s - XS Sin s ) at leading power factor

Therefore, the fractional regulation can be given by :

(2.14)

Example 2.5:A single-phase transformer that is rated 3000 kVA, 69 kV/4.16 kV,60 Hz has an impedance of 8 percent.

Calculate:a. The total impedance of the transformer referred to the primary sideb. The voltage regulation from no-load to full-load for a 2000 kW resistive load, knowing that the

primary supply voltage is fixed at 69 kVc. The primary and secondary currents if the secondary is accidentally short-circuited.

Solution:a. The base impedance on the primary side is:

ZBp = EB2 / SB = (69000)2 / 3 000 000 = 1587

The transformer impedance referred to the primary side is:Zp = Z(pu) ZBp = 0.08 1587 = 127

Because the transformer exceeds 500 kVA, the windings have negligible resistance compared to their leakage reactance; we can therefore write: Zp = Xp = 127


EpEg

Xp= 127

aEs a2Z2380

Ip

69 KV


Fig. 2.18: See Example 5.

b. Approximate impedance Z of the 2000 kW load: Z = Es

2 / P = 41602 / 2 000 000 = 8.65

Impedance referred to primary side: a2 Z = (69 / 4.16)2 8.65 = 2380

Referring to Fig. 2.18 we have:Ip = 69 000 / 1272 + 23802

= 28.95 A

a Es = (a2 Z) Ip = 2380 28.95 = 68 902 V

Es = 68 902 (4.16 / 69) = 4154 V

Secondary no-load voltage = 4160 V Voltage regulation in percent is:

=

c. Referring again to Fig. 2.18, if the secondary is accidentally short-circuited, we find:

Ip = Ep / Xp = 69 000 / 127 = 543 A

The corresponding current Is on the secondary side:Is = a Ip = (69 / 4.16) 543 = 9006 A

The short-circuit currents in the primary and secondary windings are 12.5 times greater than the rated values. The I2R losses are, therefore, 12.52 or 156 times greater than normal. The circuit-breaker or fuse protecting the transformer must open immediately to prevent overheating. Very powerful electromagnetic forces are also set up. They, too, are 156 times greater than normal and, unless the winding are firmly braced and supported, they may be damaged or torn apart.

2.9 Measuring transformer impedances:

For a given transformer, we can determine the actual values of all the impedances shown in Fig. 2.8 by means of an open-circuit test and a short-circuit test.During the open-circuit test, rated voltage is applied to the primary winding and current Io , voltage Ep, and active power Pm are measured (Fig. 2.19). The secondary open-circuit voltage Es is also measured. These test results give us the following information:active power absorbed by core = Pm apparent power absorbed by core = Sm = Ep Io reactive power absorbed by core = Qm



where : Qm = Resistance Rm corresponding to the core loss is: Rm = Ep

2 / Pm (2.1)Magnetizing reactance Xm is: Xm = Ep

2 / Qm (2.2)Turns ratio a is: a = N1 / N2 = Ep / Es

Fig. 2.19: Open-circuit test and determination of Rm, Xm, and turns ratio.

During the short-circuit test, the secondary winding is short-circuited and a voltage Eg much lower than normal (usually less than 5 percent of rated voltage) is applied to the primary. The primary current ISc should be less than its nominal value to prevent over-heating and, particularly, to prevent a rapid change in winding resistance while the test is being made.

The voltage Esc, current Isc , and power Psc are measured on the primary side (Fig. 2.20) and the following calculations made:

Total transformer impedance referred to the primary side is:Zp = Esc / Isc (2.15)

Total transformer resistance referred to the primary side is:Rp = Psc / Isc

2 (2.16)Total transformer leakage reactance referred to the primary side is:

Xp = Z2p - R2

p (2.12)

Fig. 2.20: Short-circuit test to determine leakage reactance and winding resistance.

Example 2.6:During a short-circuit test on a transformer rated 500kVA, 69 kV/4.16 kV, 60 Hz, the following voltage current, and power measurements were made.Terminals X1 , X2 were in short-circuit (see Fig. 2.20):

Esc = 2600 V Isc = 4 A Psc = 2400 WCalculate: the value of the reactance and resistance of the transformer, referred to the HV side, as well as the percent impedances.

Solution :Referring to the equivalent circuit of the transformer under short-circuit conditions (Fig. 2.21), we find the following values:

Transformer impedance referred to the primary is: ZP = Esc / Isc = 2600 / 4 = 650 Resistance referred to the primary is: Rp = Psc / Isc

2 = 2400 /1 6 = l50 Leakage reactance referred to the primary is: Xp = 6502 - 1502 = 632 Nominal primary voltage is: Ep = 69 kV Nominal primary current is: Ip = 500 kVA / 69 kV = 7.25 A Base impedance referred to the primary is: ZBP = Ep / Ip = 69 000 / 7.25 = 9517


Io

E g

W A

Es

Pm

V

X1

X2H2

H1

Ep V

Isc

E g

W A

Psc

V

X1

X2H2

H1

Esc

Isc= 4 AEsc

XpRp

2600 V


Fig. 2.21 : See Example 2.6.

Per-unit impedance = Zp / ZBP = 650 / 9517 = 0.0682 (Percent impedance = 6.82%)Per-unit reactance = Xp / ZBP = 632 / 9517 = 0.0664 (Percent reactance = 6.64%)Per-unit resistance = Rp / ZBP = 150 / 9517 = 0.0158 (Percent resistance = 1.58%)

2.10 Efficiency of a Transformer:

The losses which occur in a transformer on load can divided into two groups:1. I2 R losses in primary and secondary windings, namely I1

2 R1 + I22 R2 ;

2. Core losses due to hysteresis and eddy currents.

Since the maximum value of the flux in a normal transformer does not vary by more than about 2 per cent between no-load and full-load, it is usual to assume the core loss constant at all loads.Hence, if Pm = total core loss, total losses in transformer are:

Pm + I12 R1 + I2

2 R2

Efficiency = output power / input power = output power / (output power + losses)

= (2.17)

Greater accuracy is possible by expressing the efficiency thus:

Efficiency = output power / input power = input power - losses / input power

= 1 - ( losses / input power) (2.18)

Example 2.7:An open-circuit test was conducted on the transformer given in Example 2.6. The following results were obtained when the low-voltage winding was excited. (In some cases, such as in a repair shop, a 69 kV voltage may not be available and the open-circuit test has to be done by exciting the LV winding.)

Es = 4160 V Io = 2 A Pm = 5000 WUsing this information and the transformer characteristics found in Example 2.6, calculate:

a) the efficiency and b) the voltage regulation of the transformer when it supplies a 250 kVA, 80

percent power factor (lagging) load.

Solution:The equivalent circuit of the transformer and its load is represented by Fig. 2.22. The values of Rp end Xp are already known, and so we only have to add the magnetizing branch. To simplify the calculations,


N1 N2

Rm

5000WXm

I1=3.62A

Rp

150

Xp

632I2=60A

4160V250KVAcos = 0.869KV

1

2

3

4 6

5


Fig. 2.22: See Example 2.7.

we shifted Xm and Rm from terminals 3, 4 to the input terminals 1, 2. This change is justified because these impedances are much greater than Xp and Rp. Let us assume that the voltage across the load is 4160 V. We now calculate the efficiency of the transformer.

a. The load current is: I2 = S / Es = 250 000 / 4160 = 60 AThe current in the primary is: I1 = I2 / a = 60 4.16 / 69= 3.62 AThe total copper loss (primary and secondary) is:

Pcopper = I12 Rp = 3.622 150 = 1966 W

The iron loss is the same as that measured on the LV side of the transformer. Piron = 5000 W

Total losses are: Plosses = 5000 + 1966 = 6966 W = 7 kW The active power delivered by the transformer to the load is:

Po/p = S cos = 250 0.8 = 200 kWThe active power received by the transformer is:

Pi/p = Po + Plosses = 200 + 7 = 207 kW The efficiency is: = Po/p / Pi/p = 200 / 207 = 0.966 or 96.6%

Note that in making the above calculations, we only consider the active power. The reactive power of the transformer and its load does not enter into efficiency calculation. Let us now determine the voltage regulation.

b. In examining Fig. 2.22, it is obvious that the presence of the magnetizing branch does not affect the voltage drop across RP and Xp. Consequently, the magnetizing branch does not affect the voltage regulation.

To determine the voltage regulation, we will use the per-unit method. All voltages, imped-ances, and currents are referred to the HV (69 kV) side. We assume the voltage between terminals 1, 2 is 69 kV, and that it remains fixed. The base power PB is 500 kVA The base voltage EB is 69 kV. Consequently, the base current is:

IB = PB / EB = 500 000 / 69 000 = 7.25 A and the base impedance is:

ZB = EB / IB = 69 000 / 7.25 = 9517 The per-unit value of RP is: RP(PU) = 150 / 9517 = 0.0158 The per-unit value of Xp is: Xp(pu) = 632 / 9517 = 0.0664 The per-unit value of voltage El2 is: El2(pu) = 69 000 / 69 kV = 1.0 The per-unit value of the apparent power absorbed by the load is:

S(pu) = 250 kVA / 500 kVA = 0.5

The per-unit value of the active power absorbed by the load is:P(pu)= S(pu) cos = 0.5 X 0.8 = 0.4

The per-unit value of the reactive power absorbed by the load is:Q(pu) = S2

(pu) - P2(pu) = 0.52 - 0.42 = 0.3

The per-unit load resistance RL corresponding to P is:RL(PU) = E2

(PU) / P(PU) = 12 / 0.4 = 2.50The per-unit load reactance XL corresponding to Q is:

XL(PU) = E2(PU) / Q(PU) = 12 / 0.3 = 3.333


E(pu)1.0

Eg RL(pu)2.5

XL(pu)j3.33

I1

Rp(pu)0.058

Xp

j0.0664

1

2

3

4


Fig. 2.23: Per-unit equivalent circuit of a 500 kVA transformerfeeding a 250 kVA load.

We now draw the equivalent per-unit circuit shown in Fig. 2.23. The magnetizing branch is not shown because it does not enter into the calculations. Note that the load appears across the primary terminals 3 ,4 of the circuit shown in Fig. 2.22.

The per-unit impedance between terminals 3, 4 is:

The per-unit impedance between terminals 1, 2 is:Z12(PU) = 0.0158 + 1.6 + j(1.2 + 0.0664) = 1.616 + j 1.266 = 2.053 38.07

The per-unit current I1 is:I1(pu) = E12(PU) / Z12(PU) = 1.0 / 2.053 38.07 =0.4872 -38.07

The per-unit voltage E34 across the load is:E34(pu) = Il(pu) x Z34(pU) = (0.4872 -38.07°) (2 36.87°) = 0.9744 -1.20°

The per-unit voltage regulation is:= ( E34(pu) at no-load - E34(pu) at full-load) / E34(pu) at no-load

= (1.0 - 0.9744 ) / 1 = 0.0256

If we wish, we can calculate the actual values of the voltage and current as follows:a. Voltage across terminals 3, 4 is:

E34 = E34(pU) EB == 0.9744 x 69 000 = 67.23 kV

b. Actual voltage across the load is: E56 = E34 x (4160 / 69 000) = 4053 V

c. Actual line current is: I1 = Il(pu) x IB = 0.4872 7.246 = 3.53 A

2.11 Condition for Maximum Efficiency of a Transformer:

If RS is the equivalent resistance of the transformer referred to the secondary circuit,

RS = R1(N2 / N1 )2 + R2 (2.10)

= a constant for a given transformer

Hence for any load current I2, total losses I2 R = I22 RS , and

Efficiency =

=

For a normal transformer ES is approximately constant, hence for a load of given power factor the efficiency is a maximum when the denominator of the last equation is minimum, i.e. when:



therefore

or I22 RS = Pm (2.19)

Now it is clear that the efficiency is a maximum when the variable I2R (copper losses) is equal the constant core losses



2.12 Questions and Problems:

Practical level:1. Name the principal parts of a transformer.2. Explain how a voltage is induced in the secondary winding of a transformer.

3. The secondary winding of a transformer has twice as many turns as the primary. Is the secondary voltage higher or lower than the primary voltage?

4. Which winding is connected to the load: the primary or secondary?5. State the voltage and current relationships between the primary and secondary windings of

a transformer under load. The primary and secondary windings have N1 and N2 turns, respectively.

6. Name the losses produced in a transformer.7. What purpose does the no-load current of a transformer serve?8. Name three conditions that must be met in order to connect two transformers in parallel.9. What is the purpose of taps on a transformer?10. Name three methods used to cool transformers.11. The primary of a transformer is connected : to a 600 V, 60 Hz source. If the primary has

1200 turns and the secondary has 240, calculate the secondary voltage. [120V]12 The windings of a transformer respectively have 300 and 7500 turns. If the low-voltage

winding is excited by a 2400 V source, calculate the voltage across the HV winding. [60 KV]

13. A 6.9 kV transmission line is connected to a transformer having 1500 turns on the primary and 24 turns on the secondary. If the load across the secondary has an impedance of 5 . calculate the following: a. The secondary voltage b. The primary and secondary currents

[110.4V, 0.353A, 22.08A]14. The primary of a transformer has twice as many turns as the secondary. The primary

voltage is 220 V and a 5 load is connected across the secondary. Calculate the power delivered by the transformer, as well as the primary and secondary currents.

[2.42KW, 11A, 22A]15. A 3000 kVA transformer has a ratio of 60 kV to 2.4 kV. Calculate the nominal current of

each winding. [50A, 1250A]

Intermediate level:16. In Problem 11, calculate the peak value of the flux in the core.17. Explain why the peak flux in a 60 Hz transformer remains fixed as long as the ac supply

voltage is fixed.18. The transformer in Fig. 2.24 is excited by a 120 V, 60 Hz source and draws a no-load

current Io of 3 A. The primary and secondary respectively possess 200 and 600 turns. If 40 percent of the primary flux is linked by the secondary, calculate the following:a. The voltage indicated by the voltmeter [144V]b. The peak value of flux [2.25mwb]c. The peak value of m [0.9mwb]d. Draw the phasor diagram showing El, E2,, Io ,m and f1

Fig. 2.24 See Problem 18.



19. Explain why the secondary voltage of a practical transformer decreases with increasing resistive load.

20. What is meant by the following terms: a. Transformer impedance b. Percent impedance of a transformer

21. The transformer in Problem 15 has an impedance of 6 percent. Calculate the impedance referred to:a. The 60 kV primary [72]b. The 2.4 kV secondary [0.115]

22. A 66.7 MVA transformer has an efficiency of 99.3 percent when it delivers full power to a load having a power factor of 100 percent.

a. Calculate the losses in the transformer under these conditions. [466.9 KW]b. Calculate the losses and efficiency when the transformer delivers 66.7 MVA to a load

having a power factor of 80 percent. [466.9Kw, 99.1%]

Advanced level:23. The impedance of a transformer increases as the coupling is reduced between the primary

and secondary windings. Explain.

24. The following information is given for the transformer circuit of Fig. 2.12.Rl = 18 Ep = 14.4 kV (nominal)R2 = 0.005 Es = 240 V (nominal)Xfl = 40 Xf2 = 0.01 If the transformer has a nominal rating of 75kVA, calculate the following:

a. The transformer impedance [] referred to the primary side [84.1]b. The percent impedance of the transformer [3.04%]c. The impedance [] referred to the secondary side [23.36m]d. The percent impedance referred to the secondary side [3.04%]e. The total copper losses at full load [976 W]f. The percent resistance and percent reactance of the transformer [1.3%, 2.75%]

25. During a short-circuit test on a 10 MVA, 66 kV/7.2 kV transformer, the following results were obtained:Eg = 2640 V ISc = 72 A Psc = 9.85 kWCalculate the following:

a. The total resistance and the total leakage reactance referred to the 66 kV primary side. [1.9, 36.6]

b. The nominal impedance of the transformer referred to the primary side. [453.6]c. The percent impedance of the transformer. [8.4%]


BTEC – HND Year 2 - Utilization of Electrical Energy Three Phase Transformer

3. Three Phase Transformers

In order to transmit and distribute an electrical power efficiently and economically, the voltages must be at appropriate levels. These levels ((13.8 kV to 765 kV) depend upon the amount of power that has to be transmitted and the distance it has to be carried. Another aspect is the appropriate voltage levels used in factories and homes. These are fairly uniform, ranging from 120/240 V single-phase systems to 600 V 3-phase systems. Clearly, this required the use of 3-phase transformers to transform the voltages from one level to another.

The transformers may be inherently 3-phase, having three primary windings and three secondary windings mounted on a 3-legged core. However, the same result can be achieved by using three single-phase transformers connected together to form a 3-phase transformer bank.

3.1 Basic properties of 3-phase transformer banks:

When three single-phase transformers are used to transform a 3-phase voltage, the windings can be connected in several ways. Thus, the primaries may be connected in delta and the secondaries in star (wye), or vice versa. As a result, the ratio of the 3-phase input voltage to the 3-phase output voltage depends not only upon the turns ratio of the transformers, but also upon how they are connected.A 3-phase transformer bank can also produce a phase shift between the 3-phase input voltage and the 3-phase output voltage. The amount of phase shift depends again upon the turns ratio of the transformers, and on how the primaries and secondaries are connected.

Furthermore, the phase shift feature enable us to change the number of phases. Thus, a 3-phase system can be converted into a 2-phase, a 6-phase, or a 12-phase system. Indeed, if there were a practical application for it, we could even convert a 3-phase system into a 5-phase system by an appropriate choice of single-phase transformers and interconnections.

In making the various connections, it is important to observe transformer polarities. An error in polarity may produce a short-circuit or unbalance the line voltages and currents.

The basic behavior of balanced 3-phase transformer banks can be understood by making the following simplifying assumptions:

1. The exciting currents are negligible.2. The transformer impedances, due to the resistance and leakage reactance of the

windings, are negligible.3. The total apparent input power to the transformer bank is equal to the total apparent

output power.

Furthermore, when single-phase transformers are connected into a 3-phase system, they retain all their basic single-phase properties, such as current ratio, voltage ratio, and flux in the core. Given the polarity marks X1, X2 and H1, H2, the phase shift between primary and secondary is zero, in the sense that EX1X2 is in phase with EH1H2

3.2 Delta-Delta Connection:

The three single-phase transformers P, Q, and R of Fig. 3.1 transform the voltage of the incoming transmission line A, B, C to a level appropriate for the outgoing transmission line 1, 2, 3. The incoming line is connected to the source, and the outgoing line is connected to the load.



The transformers are connected in delta-delta. Terminal H1 of each transformer is connected to terminal H2 of the next transformer. Similarly, terminals X1 and X2 of successive transformers are connected together. The actual physical layout of the transformers is shown in Fig. 3.1. The corresponding schematic diagram is given in Fig. 3.2. The schematic diagram is drawn in such a way to show not only the connections, but also the phasor relationship between the primary and secondary voltages.

Fig. 3.1: Delta-delta connection of three single-phase transformers. The incoming lines (source) are A, B, C and the outgoing lines (load) are 1, 2, 3.

Fig. 3.2: Schematic diagram of a delta-delta connection and associated phasor diagram.

Thus, each secondary winding is drawn parallel to the corresponding primary winding to which it is coupled. Furthermore, if source G produces voltages EAB, EBC, ECA according to the indicated phasor diagram, the primary windings are oriented the same way, phase by phase. For example, the primary of transformer P between lines A and B is oriented horizontally, in the same direction as phasor EAB. Because the primary and secondary voltages EH1H2 and EX1X2 of a given transformer must be in phase, it follows that E12 (secondary voltage of transformer P) must be in phase with EAB (primary of the same transformer). Similarly, E23 is in phase with EBC and E31

with ECA.In such a delta-delta connection, the voltages between the respective incoming and outgoing transmission lines are in phase.


H1 X1

H2 X2

H1 X1

H2 X2

H1

X2H2

A

B

C

1

2

3

L

O

A

D


If a balanced load is connected to lines 1-2-3, the resulting line currents are equal in magnitude. This produces balanced line currents in the incoming lines A-B-C. As in any delta connection, the line currents are ,3 times greater than the respective currents Ip and Is flowing in the primary and secondary windings (Fig. 3.2). The power rating of the transformer bank is three times the rating of a single transformer.

Example 3.1.Three single-phase transformers are connected in delta-delta to step down a line voltage of 138 kV to 4160 V to supply power to a manufacturing plant. The plant draws 21 MW at a lagging power factor of 86 percent.

Calculate:a. The apparent power drawn by the plant b. The apparent power furnished by the HV linec. The current in the HV linesd. The current in the LV linese. The currents in the primary and secondary windings of each transformerf. The load carried by each transformer

Solution:a. The apparent power drawn by the plant is:

S = P / cos = 21 / 0.86 = 24.4 MVA

b. The transformer bank itself absorbs a negligible amount of active and reactive power because the I2R losses and the reactive power associated with the mutual flux and the leakage fluxes are small. It follows that the apparent power furnished by the HV line is also 24.4 MVA.

c. The current in each HV line is:I1 = S / (3 E) = (24.4 106) / (3 138 000) = 102 A

d. The current in the LV lines is:I2 = S / (3 E) = (24.4 106) / (3 4160) = 3384 A

e. Referring to Fig. 3.2, the current in each primary winding is:Ip = 102 / 3 = 58.9 A

The current in each secondary winding is:Is = 3384 / 3 = 1954 A

f. Because the plant load is balanced, each transformer carries one-third of the total load, or 24.4 / 3 = 8.13 MVA.

The individual transformer load can also be obtained by multiplying the primary voltage times the primary current:

S = Ep Ip = 138 000 58.9 = 8.13 MVA

3.3 Delta-Star Connection:

When the transformers are connected in delta-star, the three primary windings are connected the same way as in Fig. 3.1. However, the secondary windings are connected so that all the X2

terminals are joined together, creating a common neutral N (Fig. 3.3). In such a delta-star



connection, the voltage across each primary winding is equal to the incoming line voltage. However, the outgoing line voltage is 3 times the secondary voltage across each transformer.

Fig. 3.3: Delta-Star connection of three single-phase transform

The relative values of the currents in the transformers windings and transmission lines given in Fig. 3.4. Thus, the line currents in phases A, B, and C are 3 times the currents in the primary windings. The line currents in phases 1, 2, 3 are the same as the currents in the secondary windings.

Fig. 3.4: Schematic diagram of a Delta-Star connection and associated phasor diagram.

A Delta-Star connection produces a 30° phase shift between the line voltages of the incoming and outgoing transmission lines. Thus, outgoing line voltage E12 is 30° ahead of incoming line voltage EAB as can be seen from the phasor diagram. If the outgoing line feeds an isolated group of loads, the phase shift creates no problem. But, if the outgoing line has to be connected in parallel with a line coming from another source, the 30° shift may make such a parallel connection impossible, even if the line voltages are otherwise identical.



One of the important advantages of the Star connection is that it reduces the amount of insulation needed inside the transformer. The HV winding has to be insulated for only 1/3 or 58 percent of the line voltage.

Example 3.2:Three single-phase step-up transformers rated at 40 MVA,13.2 kV / 80 kV are connected in Delta-Star on a 13.2 kV transmission line (Fig. 3.5). If they feed a 90 MVA load, calculate the following:

a. The secondary line voltageb. The currents in the transformer windingsc. The incoming and outgoing transmission line currents

Fig. 3.5: See Example 2.

Solution:

The easiest Star to solve this problem is to consider the windings of only one transformer, say, transformer P.

a. The voltage across the primary winding is obviously 13.2 kV.

The voltage across the secondary is, therefore, 80 kV. The voltage between the outgoing lines 1, 2, and 3 is: ES = 803 = 138 kV

b. The load carried by each transformer is: S = 90 / 3 = 30 MVA

The phase current in the primary winding is: Ip = 30 MVA / 13.2 kV = 2272 A

The phase current in the secondary winding is: Is = 30 MVA / 80 kV = 375 A

c. The current in each incoming line A, B, C is: I = 22723 = 3932 A

The current in each outgoing line 1, 2, 3 is: I = 375 A

3.4 Star-Delta Connection:

The currents and voltages in a Star-Delta connection are identical to those in the Delta-Star connection of Section 3.3. The primary and secondary connections are simply interchanged. In other words, the H2 terminals are connected together to create a neutral, and the X1, X2 terminals are connected in delta. Again, there results a 30° phase shift between the voltages of the incoming and outgoing lines.

3.5 Star-Star Connection:



When transformers are connected in Star-Star, special precautions have to be taken to prevent severe distortion of the line-to-neutral voltages. One way to prevent the distortion is to connect the neutral of the primary to the neutral of the source, usually by way of the ground (Fig. 3.6). Another way is to provide each transformer with a third winding, called tertiary winding. The tertiary windings of the three transformers are connected in delta (Fig. 3.7). They often provide the substation service voltage where the transformers are installed.Note that there is no phase shift between the incoming and outgoing transmission line voltages of a Star-Star connected transformer.

Fig. 3.6: Star-Star connection with neutral of the primary to the neutral of the source.

Fig. 3.7: Star-Star connection using a tertiary winding.

3.6 Open-Delta Connection:

It is possible to transform the voltage of a 3-phase system by using only 2 transformers, connected in open-delta. The Open-Delta arrangement is identical to a delta-delta connection, except that one transformer is absent (Fig. 3.8). However, the Open-Delta connection is seldom used because the load capacity of the transformer bank is only 86.6 percent of the installed transformer capacity.

For example, if two 50 kVA transformers are connected in open-delta, the installed capacity of the transformer bank is obviously 2 x 50 = 100 kVA. But, strange as it may seem, it can only deliver 86.6 kVA before the transformers begin to overheat.

The Open-Delta connection is mainly used in emergency situations. Thus, if three transformers are connected in delta-delta and one of them becomes defective and has to be removed, it is possible to feed the load on a temporary basis with the two remaining transformers.



(a)

(b)Fig. 3.8 (a): Open-Delta connection.

(b): Associated schematic & phasorExample 3.3:Two single-phase 150 kVA, 7200 V/600 V transformers are connected in Open-Delta. Calculate the maximum 3-phase load they can carry.

Solution:Although each transformer has a rating of 150 kVA, the two together cannot carry a load of 300 kVA.The nominal secondary current of each transformer is:

Is = 150 kVA / 600 V = 250 A

The current Is in lines 1, 2, 3 cannot, therefore, exceed 250A (Fig. 3.8-b). Consequently, the maximum load that the transformers can carry is:

S = 3 E I = 1.73 600 250 = 259 500 VA = 260 kVA

Thus, the ratio = maximum load / installed transformer rating = 260 kVA / 300 kVA = 0.866, or 86.6%



3.7 Questions and Problems

Practical level:1. Assuming that the transformer terminals have polarity marks H1, H2, X1, X2, make schematic

drawings of the following connections:a. Delta-Starb. Open-delta

2. Three single-phase transformers rated at 250 kVA, 7200 V/600 V, 60 Hz, are connected in Star-delta on a 12 470 V, 3-phase line. If the load is 450 kVA, calculate the following currents:

a. In the incoming and outgoing transmission lines [20.8A, 433A]b. In the primary and secondary windings [20.8A, 250A]

3. A transformer has a rating of 36 MVA, 13.8 kV/320 V. Calculate the nominal currents in the primary and secondary lines. [1506A, 65A]

Intermediate level: 4. The transformers in Problem 2 are used to raise the voltage of a 3-phase 600V line to 7.2kV.

a. How must they be connected? a. Calculate the line currents for a 600 kVA load. [577A, 48.1A]b. Calculate the corresponding primary and secondary currents. [333A, 27.8A]

5. In order to meet an emergency, three single-phase transformers rated 100 kVA, 13.2 kV/2.4 kV are connected in wye-delta on a 3-phase 18 kV line. a. What is the maximum load that can be connected to the transformer bank? [236KVA]b. What is the outgoing line voltage? [1.89KV]

6. transformers rated at 250 kVA, 2.4 kV/600 V are connected in open-delta to supply a load of 400 kVA. a. Are the transformers overloaded? b. What is the maximum load the bank can carry on a continuous basis? [433KVA]

7. Referring to Figs. 3.3 and 3.4, the line voltage between phases A-B-C is 6.9 kV and the voltage between lines 1, 2, and 3 is balanced and equal to 600 V. Then, in a similar installation the secondary of transformer P are by mistake connected inverse.

a. Determine the voltages measured between lines 1-2, 2-3, and 3-1. [347, 600, 347V] b. Draw the new phasor diagram.

Industrial application: 8. Three 150 kVA, 480 V/4000 V, 60 Hz single-phase transformers are to be installed on a 4000

V, 3-phase line. The exciting current has a value of 0.02 put. Calculate the line current when the transformers are operating at no-load. [1.3A]

9. Three single-phase transformers rated at 15 kVA, 480 V/120 V, 60 Hz are connected in delta to function as auto-transformer on a 600 V 3-phase line. The H1, H2, X1, X2 polarity marks appear on the metal housing. a. Show how the transformers should be connected. b. Determine the 3-phase voltage output of the transformer. [465V]c. Determine the phase shift between the 3-phase voltage output and the 600 V, 3-phase

input. [12.9]


BTEC – HND Year 2 - Utilization of Electrical Energy Protective Relay

Protective Relays

1. Operating principles and Construction Protective relays are classified into the following three categories, depending on the technologies they use for their construction and operation.

1. Electromagnetic and electrothermal2. Static; using semiconductor devices like ICs, transistor, diodes, logic gates, etc.3. Microprocessor based

There are various types of protective relays in each category, depending on the operating principle and application.

1. Electromagnetic Relays

The following are the important types of construction of electromagnetic Relays.(i) Attracted armature(ii) Induction disc(iii) Printed disc dynamometer(iv) Permanent magnet(v) Moving-coil(vi) Polarized moving-iron

1.1 Attracted Armature Relay:

Hinged armature and plunger type construction comes under this group of relays. Fig. (1.1-a) shows a hinged armature type construction. The coil is energised by an operating quantity proportional to the system current or voltage. The operating quantity produces a magnetic flux which in turn produces an electromagnetic force. The electromagnetic force is proportional to the square of the flux in the air gap or the square of the current. The attractive force increases as the armature approaches the pole of the electromagnet. This type of a relay is used for the protection of small machines, equipment, etc. It is also used for auxiliary relays, such as indicating flags, slave relays, alarm relays, annunciators, semaphores, etc.

The actuating quantity of the relay may be either ac or dc in dc relays, the electromagnetic force of attraction is constant. In the case of ac relays, sinusoidal current flows through the coil and hence the force of attraction is given by:

F = KI2

= K( Imax sin t)2 = ½ KI2

max ( 1 - cos 2 t) = ½ K(I2

max - I2max cos 2 t )

From the above expression, it is evident that the electromagnetic force consists of two components. One component is constant and is equal to ½KI2

max . The other component is time dependent and pulsates at double the frequency of the applied ac quantity. Its magnitude is ½ KI2

max cos 2 t. The total force is a double frequency pulsating force. This may cause the armature to vibrate at double the frequency. Consequently, the relay produces a humming sound and becomes noisy. This difficulty can be overcome by making the pole of the electromagnet of shaded construction. Alternatively, the electromagnet may be provided with two coils. One coil is energised with the actuating quantity. The other coil gets its supply through a phase shifting circuit.



The restraining force is provided by a spring. The reset to pick-up ratio for attracted armature type relays is 0.5 to 0.9. For this type of a relay, the ratio for ac relays is higher as compared to dc relays. The VA burden is low, which is 0.08 W at pick-up for the relay with one contact, 0.2 W for the relay with four contacts. The relay is an instantaneous relay. The operating speed is very high. For a modern relay, the operating time is about 5 ms. It is faster than the induction disc and cup type relays.

Attracted armature relays are compact, robust and reliable. They are affected by transients as they are fast and operate on both dc and ac The fault current contains a dc component in the beginning for a few cycles. Due to the presence of dc transient, the relay may operate though the steady state value of the fault current may be less than its pick-up. A modified constructions as shown in Fig. (1.1-b) reduces the effect of dc transients.

(a) Hinged armature type relay (b) Modified hinged armature type relay

Fig. 1.1

1.2 Induction Disc Relay:

There are two types of construction of induction disc relays, namely the shaded pole type, as shown in Fig. 1.2; and watt hour meter type, as shown in Fig. 1.3.

Fig. (1.2-a) shows a simple theoretical figure, whereas Fig. (1.2-b) shows the construction which is actually used in practice. The rotating disc is made of aluminium. In the shaded pole type construction, a C-shaped electromagnet is used. One half of each pole of the electromagnet is surrounded by a copper band known as the shading ring. The shaded portion of the pole produces a flux which is displaced in space and time with respect to the flux produced by the unshaded portion of the pole. Thus two alternating fluxes displaced in space and time cut the disc and produce eddy currents in it. Torques are produced by the interaction of each flux with the eddy current produced by the other flux. The resultant torque causes the disc to rotate.In wattmetric type of construction, two electromagnets are used: upper and lower one. Each magnet produces an alternating flux which cuts the disc. To obtain a phase displacement between two fluxes produced by upper and lower electromagnets, their coils may be energised by two different sources. If they are energised by the same source, the resistances and



reactances of the two circuits are made different so that there will be sufficient phase difference between the two fluxes.Induction disc type construction is robust and reliable. It is used for over-current protection. Disc type units give an inverse time current characteristic and are slow compared to the induction cup and attracted armature type relays. The induction disc type is used for slow-speed relays. Its operating time is adjustable and is employed where a time-delay is required. Its reset/pick-up ratio is high, above 95% because its operation does not involve any change in the air gap. The VA burden depends on its application, and is generally of the order of 2.5 VA. The torque is proportional to the square of the actuating current if single actuating quantity is used.

A spring is used to supply the resetting torque. A permanent magnet is employed to produce eddy current braking to the disc. The magnets should remain stable with age so that its accuracy will not be affected. Magnets of high coercive force are used for the purpose. The braking torque is proportional to the speed of the disc. When the operating current exceeds pick-up value, driving torque is produced and the disc accelerates to a speed where the braking torque balances the driving torque. The disc rotates at a speed proportional to the driving torque.

Fig. 1.2: Shaded pole type induction disc relay

It rotates at a constant speed for a given current. The disc inertia should be as small as possible, so that it should stop rotating as soon as the fault current disappears when circuit breaker operates at any other location or fault current is for a short moment (i.e.. transient in nature). After the cessation of the fault current, the disc will travel to some distance due to inertia. This distance should be minimum. It is called the over-run of the disc. A brake magnet is used to minimise over-run. The over-run is usually not more than 2 cycles on the interruption of a current which is 20 times the current setting.



At a current below pick-up value, the disc remains stationary by the tension of the control spring acting against the normal direction of disc rotation. The disc rests against a backstop. The position of the backstop is adjustable and therefore, the distance by which the moving contact of the relay travels before it closes contacts, can be varied. The distance of travel is adjusted for the time setting of the relay.

The rotor disc carries an arm which is attached to its spindle. The spindle is supported by jewelled bearings. The arm bridges the relay contacts. In earlier constructions, there were two contacts which were bridged when the relay operated. In modern units however, there is a single contact with a flexible lead-in.

Current setting:In disc type units, there are a number of tappings provided on coil to select the desired pick-up value of the current. These tappings are shown in Fig. 1.3. This will be discussed in more detail in the next chapter.

Fig. 1.3: Wattmetric type induction-disc relay

Time setting:The distance which the disc travels before it closes the relay contact can be adjusted by adjusting the position of the backstop. If the backstop is advanced in the normal direction of rotation, the distance of travel is reduced, resulting in a shorter operating time of the relay. More details on time-setting will be discussed in the next chapter.

1.3. Reverse Power or Directional Relay

Fig. (1.4-a) shows an electromagnetic directional relay. A directional relay is energised by two quantities, namely voltage and current. Fluxes 1 and 2 are set up by voltage and current, respectively. Eddy currents induced in the disc by 1 interact with 2 and produce a torque. Similarly, 2 also induces eddy currents in the disc, which interact with 1 and produce a



torque. The resultant torque rotates the disc. The torque is proportional to VI cos where is the phase angle between V and I. The torque is maximum when voltage and current are in phase. To produce maximum torque during the fault condition, when the power factor is very poor, a compensating winding and shading are provided, as shown in Fig. (1.4-a).

Earlier it has been mentioned that the torque produced by an induction relay is given by:

T = 1 2 sin I1 I2 sin

where 1 and 2 are fluxes produced by I1 and I2 respectively. The angle between 1 and 2 or I1

and I2 is . If one of the actuating quantities is voltage, the current flowing in the voltage coil lags behind voltage by approximately 90. Assume this current to be I2. The load current I (say I1) lags V by . Then the angle between I1 and I2 is equal to (90- ), as shown in Fig. (1.4-b).

(a) Construction (b) Phasor diagram

Fig. 1.4: Induction disc type directional relay

T = I1 I2 sin (9O - ) I1 I2 cos VI cos

Torque produced is positive when cos is positive, i.e. is less than 90. When is more than 90 (between 90 and 180), the torque is negative. At a particular relay location, when power flows in the normal direction, the relay is connected to producing negative torque. The angle between the actuating quantities supplied to the relays is kept (180 - ) to produce negative torque. If due to any reason, the power flows in the reverse direction, the relay produces a positive torque and it operates.

In this condition, the angle between the actuating quantities is kept less than 90 to produce a positive torque. This is shown in Fig. (1.5-a). For normal flow of power, the relay is supplied with V and (-I). For reverse flow, the actuating quantities become V and I. Torque becomes VI cos , i.e. positive. This can be achieved easily by reversing the current coil, as shown in Fig. (1.5-b).

Relaying units supplied with single actuating quantity discussed earlier are non-directional over-current relays. Non-directional relays are simple and less expensive than directional relays.



(a) (b)

Fig. 1.5 (a): Phasor diagram for directional relay (b): Connection of current coil for reverse power relay

2. Static Relays

The expansion and growing complexity of power systems have brought a need for protective relays with a higher performance and more sophisticated characteristics. These have been made possible by the development of semiconductors and other associated components, and many types of static relay are commercially available. All duties performed by electromagnetic relays can now be undertaken by static devices, in most cases with performance and economic advantages.

The emergence of the static relay has posed a new set of problems for the relay designer. The requirement of a d.c. supply for measurement purposes and the necessity to protect static devices from high voltage surge are two which have been thrown up in the development of this kind of equipment.

As the development of static relays has progressed, certain basic circuits have emerged which go to make up the required relaying characteristic. The use of semiconductors has allowed greater flexibility in the design process and has presented an opportunity to optimise the parameters of each particular element.

2.1 Over-current definite time relay:

As an example, Fig. (1.6-a) shows the elements used in a single-phase definite time over-current relay. Alternating current is converted to a proportional direct voltage and compared with a fixed d.c. level. When it exceeds the reference level a timer is initiated. After the set time delay the second level detector operates to activate the output element.



The input circuit consists of a current transformer, the secondary current of which is rectified and fed into a resistive shunt. The current setting of the relay can be changed by means of primary taps on the CT or by varying the value of the secondary shunt. A time delay setting can be obtained by varying the resistance value of RC delay circuit with a calibrated potentiometer. Instantaneous operation above a set level is achieved by by-passing the time delay element to a further level detector.

Fig. 1.6: Definite time over-current static relay.

Fig. (1.6-b) shows the circuit of a simplified level detector which also includes an output stage driving the coil of an attracted armature relay. In this circuit all transistors are biased off until the input voltage exceeds that at the emitter of TR1, set by potentiometer chain R1 and R2. When this voltage is exceeds both transistors turn on and the output relay is energized. The circuit has the advantages of drawing no current when in the non-operated state and giving a drop-out level of almost 100% of the pick-up level.

Timing can be performed by connecting an RC time delay network, shown dashed in Fig. (1.6-b), at the input to the second level detector monitoring the voltage across the capacitor.

2.2 Advantages of Static Relays:

The main advantages of static relays over electromagnetic relays are:1- C.T. burden is about one tenth, thereby a smaller C.T. can be employed.2- The space required for a single-phase relay is half and that for a three-phase relay is

about one third. Consequently, the panel space and overall cost of installation are reduced. This helps in miniaturization of control equipment.

3- Instantaneous reset can easily be achieved. This allows the application of automatic reclosing of circuit breaker.

4- Accuracy in time-current characteristics.5- Fast operation, absence of mechanical inertia and bouncing of contacts.6- Long life and less maintenance, immunity to vibration, dust and polluted atmosphere.


BTI – HND Year 2 – Utilization of Electrical Energy Protective Relay

2. Over Current Protection

A protective relay which operates when the load current exceeds a preset value, is called an over-current relay. The value of the preset current above which the relay operates is known as its pick-up value. An over-current relay is used for the protection of distribution lines, large motors, power equipment, etc. A scheme which incorporates over-current relays for the protection of an element of a power system, is known as an over-current scheme or over-current protection. An over-current scheme may include one or more over-current relays.At present electromagnetic relays are widely used for over-current protection. The induction disc type construction, as shown in Fig. (1.2-b) is commonly used.

1 Time-current Characteristics

A wide variety of time-current characteristics is available for over-current relays. The name assigned to an over-current relay indicates its time-current characteristic as described below.

1.1 Definite-time Over-current Relay

A definite-time over-current relay operates after a predetermined time when the current exceeds its pick-up value. Curve (a) of Fig. 2.1 shows the time-current characteristic for this type of relay. The operating time is constant, irrespective of the magnitude of the current above the pick-up value. The desired definite operating time can be set with the help of an intentional time-delay mechanism provided in the relaying unit.

1.2 Instantaneous Over-current Relay

An instantaneous relay operates in a definite time when the current exceeds its pick-up value. The operating time is constant, irrespective of the magnitude of the current, as shown by the curve (a) of Fig. 2.1. There is no intentional time delay. It operates in 0.1 s or less. Sometimes the term like “high set” or “high speed” is used for very fast relays having operating times less than 0.1 s.

Fig. 2.1: Definite-time and inverse-time characteristics of over-current relays



1.3 Inverse-time Over-current Relay

An inverse-time over-current relay operates when the current exceeds its pickup value. The operating time depends on the magnitude of the operating current. The operating time decreases as the current increases. Curve (b) of Fig. 2.1 shows the inverse time-current characteristic of this type of relays.

1.4 Inverse Definite Minimum Time Over-current (I.D.M. T.) Relay

This type of a relay gives an inverse-time current characteristic at lower values of the fault current and definite-time characteristic at higher values of the fault current. Generally, an inverse-time characteristic is obtained if the value of the plug setting multiplier is below 10. For values of plug setting multiplier between 10 and 20, the characteristic tends to become a straight line, i.e. towards the definite time characteristic. Fig. 2.2 shows the characteristic of an I.D.M.T. relay along with other characteristics. I.D.M.T. relays are widely used for the protection of distribution lines. Such relays have a provision for current and time settings which will be discussed later on.

1.5 Very Inverse-time Over-current Relay

A very inverse-time over-current relay gives more inverse characteristic than that of the I.D.M.T. relay. Its time-current characteristic lies between an I.D.M.T. characteristic and extremely inverse characteristic, as shown in Fig. 2.2. The very inverse characteristic gives better selectivity than the I.D.MT characteristic. Hence, it can be used where an I.D.M.T. relay fails to achieve good selectivity. Very inverse time-current relays are recommended for the cases where there is a substantial reduction of fault current as the distance from the power source increases. They are particularly effective with ground faults because of their steep characteristic.

Fig. 2.2: IDMT, very inverse-time and extremely inverse-time characteristics



1.6 Extremely Inverse-time Over-current Relay

An extremely inverse time over-current relay gives a time-current characteristic more inverse than that of the very inverse and I.D.M.T. relays, as shown in Fig. 2.2. When I.D.M.T. and very inverse relays fall in selectivity, extremely inverse relays are employed. I.D.M.T. relays are not suitable to be graded with fuses.

The electromagnetic relay which gives the steepest time-current characteristic is an extremely inverse relay. The time-current characteristic of an extremely inverse relay is not good enough to be graded with fuses.

An extremely inverse relay is very suitable for the protection of machines against overheating. Hence, this type of relays are used for the protection of alternators, power transformers, earthing transformers, expensive cables, railways trolley wires, etc.

The rotors of large alternators may be overheated if an unbalanced load or fault remains for a longer period on the system. In such a case, an extremely inverse relay, in conjunction with a negative sequence network is used. By adjusting the time and current settings, a suitable characteristic of the relay is obtained for a particular machine to be protected.

A relay should not operate on momentary overloads. But it must operate on sustained short circuit current. For such a situation, it is difficult to set I.D.M.T. relays. An extremely inverse relay is quite suitable for such a situation. This relay is used for the protection of alternators against overloads and internal faults. It is also used for reclosing distribution circuits after a long outage. After long outages, when the circuit breaker is reclosed there is a heavy inrush current which is comparable to a fault current.

An I.D.M.T. relay is not able to distinguish between the rapidly decaying inrush current of the load and the persistent high current of a fault. Hence, an I.D.M.T. relay trips again after reclosing. But an extremely inverse relay is able to distinguish between a fault current and inrush current due to its steep time-current characteristic. Therefore an extremely inverse relay is quite suitable for the load restoration purpose.

2. Current Setting

The current above which an over-current relay should operate can be set. Suppose that a relay is set at 5A. It will then operate if the current exceeds 5A. Below 5A, the relay will not operate. There are a number of tappings on the current coil, available for current setting, as shown in Fig. 1.2 and Fig. 1.3. An over-current relay which is used for phase to phase fault protection, can be set at 50% to 200% of the rated current in steps of 25%.

The usual current rating of this relay is 5A. So it can be set. at 2.5 A, 3.75 A, 5 A,..., 10 A. When a relay is set at 2.5 A, it will operate when current exceeds 2.5 A. When the relay is set at 10 A, it will operate when current exceeds 10 A. The relay which is used for protection against ground faults (earth-fault relay) has settings 20% to 80% of the rated current in steps of 10%. The current rating of an earth-fault relay is usually 1 A.

If time-current curves are drawn, taking current in amperes on the X - axis, there will be one graph for each setting of the relay. To avoid this complex situation, the plug setting multipliers are taken on the X - axis. The actual r.m.s. current flowing in the relay expressed as a multiple of the setting current (pickup current) is known as the plug setting multiplier (PSM).



Suppose, the rating of a relay is 5A and it is set at 200%, i.e. at 10 A. If the current flowing through the relay is 100 A, then the plug setting multiplier will be 10. The PSM = 4 means 40 A of current is flowing, PSM = 6 means 60 A of current is flowing and so on.

Fig 2.3: Standard I.D.M.T. characteristic

If the same relay is set at 50%, i.e. at 2.5 A, the PSM = 4 means 10 A; PSM = 6 means l5A;PSM= l0 means 25A and so on. Hence, PSM can be expressed as:

PSM = Secondary current / Relay current setting = Primary current during fault / (Relay current setting CT. Ratio)

While plotting the time-current characteristic, if PSM is taken on the X-axis, there will be only one curve for all the settings of the relay. Fig. 2.3 shows a time-current characteristic with PSM on the X-axis. The curve is generally plotted on log/log graph. Only this curve will give the operating time for different settings of the relay. Suppose the relay is set at 5 A. The operating times for different currents are shown in Table 2.1.

If the same relay is set at 10 A, the corresponding operating times for different currents are shown in Table 3.2, using the same curve of Fig. 2.3.

Table 2.1

Current in Amperes

5 10 20 50

PSM 1 2 4 10Operating time in Sec No

operation10 5 3

Table 2.2Current in Amperes 5 10 20 40 100

PSM less than 1 1 2 4 10Operating time in

secondsRelay will

not operate

No operation 10 5 3



3. Time Setting

The operating time of the relay can be set at a desired value. In induction disc type relay, the angular distance by which the moving part of the relay travels for closing the contacts can be adjusted to get different operating time. There are 10 steps in which time can be set. The term time setting multiplier (TSM) is used for these steps of time settings. The values of TSM are 0.1,0.2,..., 0.9, 1. Suppose that at a particular value of the current or plug setting multiplier (PSM), the operating time is 4 s with TSM = 1. The operating time for the same current with TSM = 0.5 will be 4 0.5 = 2 s. The operating time with TSM=0.2 will be 4 0.2=0.8 s.

Fig. (2.4-a) shows time-current characteristics for different values of TSM. The characteristic at TSM = 1 can also be presented in the form shown in Fig. (2.4-b)

Fig. (2.4-a): Time-current characteristics for different values of TSM

Fig. (2.4-b): Logarithmic scale for IDMT relay at TSM = 1


1 2 2.5 31.3 4 6 10 20

30 1015 8 6 5 4 33.5

14

2.2

Plug Setting Multiplier

Time in Seconds


Example:The current rating of a relay is 5 A. Rating current setting = 1.5, TSM = 0.4, C.T. ratio = 400/5, fault current = 6000A. Determine the operating time of the relay. At TSM = 1, operating time at various PSM are indicated in the table below.

PSM 2 4 5 8 10 20Operating time in second 10 5 4 3 2.8 2.4

Solution: C.T. ratio = 400 / 5 = 80Relay current setting =5 1.5 = 7.5 A

PSM = Secondary current / Relay current setting = Primary current (fault current) / (Relay current setting CT. Ratio) = 6000 / (7.5 80) = 10

Operating time from the given table at PSM = 10 is 2.8 Sec. This time is for TSM = 1. The operating time for TSM = 0.4 will be equal to 2.8 x 0.4 = 1.12 Sec..

4. Over-current Protective Schemes

Over-current protective schemes are widely used for the protection of distribution lines. Protection against excess current was naturally the earliest protective system to evolve. From this basic principle has been evolved the graded over-current system, a discriminative fault protection. This should not be confused with ‘overload’ protection, which normally makes use of relays that operate in a time related in some degree to the thermal capability of the plant to be protected.

Over-current protection, on the other hand, is directed entirely to the clearance of faults, although with the settings usually adopted some measure of overload protection is obtained.

4.1 CO-Ordination Procedure

Correct current relay application requires a knowledge of the fault current that can flow in each part of the network. Since large scale tests are normally impracticable, system analysis must be used. It is generally sufficient to use machine transient reactance X and to work on the instantaneous symmetrical currents. The data required for a relay setting study are:

a. A one-line diagram of the power system involved, showing the type and rating of the protective devices and their associated current transformers.

b. The impedances in ohms, per cent or per unit, of all power transformers, rotating machines and feeder circuits.

c. The maximum and minimum values of short circuit currents that are expected to flow through each protective device.

d. The starting current requirements of motors and the starting and stalling times of induction motors.

e. The maximum peak load current through protective devices.

f. Decrement curves showing the rate of decay of the fault current supplied by the generators.

g. Performance curves of the current transformers.

The relay settings are first determined so as to give the shortest operating times at maximum fault levels and then checked to see if operation will also be satisfactory at the minimum fault



current expected. It is always advisable to plot the curves of relays and other protective devices, such as fuses, that are to operate in series, on a common scale. It is usually more convenient to use a scale corresponding to the current expected at the lowest voltage base or to use the predominant voltage base.

The alternatives are a common MVA base or a separate current scale for each system voltage. The basic rules for correct relay co-ordination can generally be stated as follows:

i. Whenever possible, use relays with the same operating characteristic in series with each other.

ii. Make sure that the relay farthest from the source has current settings equal to or less than the relays behind it, that is, that the primary current required to operate the relay in front is always equal to or less than the primary current required to operate the relay behind it.

4.2 Principles of Time/Current Grading

Among the various possible methods used to achieve correct relay co-ordination are those using either time or over-current or a combination of both time and over-current. The common aim of all three methods is to give correct discrimination. That is to say, each one must select and isolate only the faulty section of the power system network, leaving the rest of the system undisturbed.

4.3 Discrimination by time

In this method an appropriate time interval is given by each of the relays controlling the circuit breakers in a power system to ensure that the breaker nearest to the fault opens first. A simple radial distribution system is shown in Fig. 2.5, to illustrate the principles.

Fig. 2.5: Radial system with time discrimination.

Circuit breaker protection is provided at B, C, D and E, that is, at the infeed end of each section of the power system. Each protection unit comprises a definite time delay over-current relay in which the operation of the current sensitive element simply initiates the time delay element. Provided the setting of the current element is below the fault current value this element plays no part in the achievement of discrimination. For this reason, the relay is sometimes described as an ‘independent definite time delay relay’ since its operating time is for practical purposes independent of the level of over-current.

It is the time delay element, therefore, which provides the means of discrimination. The relay at B is set at the shortest time delay permissible to allow a fuse to blow for a fault on the secondary side of transformer A. Typically, a time delay of 0.25s is adequate.

If a fault occurs at F, the relay at B will operate in 0.25s and the subsequent operation of the circuit breaker at B will clear the fault before the relays at C, D and E have time to operate. The main disadvantage of this method of discrimination is that the longest fault clearance time occurs for faults in the section closest to the power source, where the fault level (MVA) is highest.

4.4 Discrimination by current

Discrimination by current relies on the fact that the fault current varies with the position of the fault, because of the difference in impedance values between the source and the fault. Hence, typically, the relays controlling the various circuit breakers are set to operate at suitably tapered values such that only the relay nearest to the fault trips its breaker. Fig. 2.6 illustrates the method.



Fig. 2.6: Radial system with current discrimination

For a fault at F1, the system short-circuit current is given by:

I = 6350 / (ZS + ZL1)Where: ZS = source impedance = 112 / 250 = 0.485 ohms ZL1 = cable impedance between C and B = 0.24 ohmsHence I = 6350 / 0.725 = 8800 ASo a relay controlling the circuit breaker at C and set to operate at a fault current of 8800 A

would in simple theory protect the whole of the cable section between C and B.

Assuming a fault at F4, the short-circuit current is given by:

I = 6350 / (ZS + ZL1 + ZL2 + ZT)

Where: ZS = source impedance = 112 / 250 = 0.485 ohms ZL1 = cable impedance between C and B = 0.24 ohms ZL2 = cable impedance between B and 4 MVA transformer = 0.04 ohms ZT = transformer impedance = 0.07 (112 / 4) = 2.12 ohms

Hence I = 6350 / 2.885 = 2200 A

For this reason, a relay controlling the circuit breaker at B and set to operate at a current of 2200A plus a safety margin would not operate for a fault at F4 and would thus discriminate with the relay at A. Assuming a safety margin of 20% to allow for relay errors and a further 10% for variations in the system impedance values, it is reasonable to choose a relay setting of 1.3 x 2200, that is, 2860 A for the relay at B. Now, assuming a fault at F3, that is, at the end of the 11 kV cable feeding the 4 MVA transformer, the short-circuit current is given by:

I = 6350 / (ZS + ZL1 + ZL2)

Thus, assuming a 250 MVA source fault level: I = 6350 / (0.485 + 0.24 + 0.04) = 8300 A

Alternatively, assuming a source fault level of 130 MVA: I = 6350 / (0.93 + 0.24 + 0.04) = 5250 A

In other words, for either value of source level, the relay at B would operate correctly for faults anywhere on the 11 kV cable feeding the transformer.

4.5 Discrimination by both time and current

Each of the two methods described so far has a fundamental disadvantage. In the case of discrimination by time alone, the disadvantage is due to the fact that the more severe faults are cleared in the longest operating time. Discrimination by current can only be applied where there is an appreciable impedance between the two circuit breakers concerned.

It is because of the limitations imposed by the independent use of either time or current co-ordination that the inverse time over-current relay characteristic has evolved. With this characteristic, the time of operation is inversely proportional to the fault current level and the actual characteristic is a function of both ‘time’ and ‘current’ settings. The advantage of this method of relay co-ordination may be best illustrated by the system shown in Fig. 2.7, which is identical to that shown in Fig. 2.5 except that typical system parameters have been added.



Fig. 2.7 : Time and Current Grading

In order to carry out a system analysis, before a relay co-ordination study of the system shown in Fig. 2.7, it is necessary to refer all the system impedances to a common base and thus, using 10 MVA as the reference base, we have:

4 MVA transformer percentage impedance on 10 MVA base = 7 (10 / 4) = 17.5%11 kV cable between B and A percentage impedance on 10 MVA base:

= 0.04 100 10 / 112 = 0.33 %11 kV cable between C and B Percentage impedance on 10MVA base:

= 0.24 100 10 / 112 =1.98 %30 MVA transformer percentage impedance on10 MVA base:

= 22.5 10 / 30 = 7.5%132 kV overhead line percentage impedance on 10MVA base:

= )6.2 100 10( / (132)2 = 0.36 %132 kV source percentage impedance on 10 MVA base:

= (100 10) / 3500 = 0.29 %



The graph in Fig. 2.7 illustrates the use of ‘discrimination curves’, which are an important aid to satisfactory protection co-ordination. In this example, a voltage base of 3.3 kV has been chosen and the first curve plotted is that of the 200A fuse, which is assumed to protect the largest outgoing 3.3 kV circuit. Once the operating characteristic of the highest rated 3.3 kV fuse has been plotted, the grading of the over-current relays at the various substations of the radial system is carried out as follows:

Substation B:CT ratio 250/5ARelay over-current characteristic assumed to be extremely inverse relay. This relay must discriminate with the 200A fuse at fault levels up to:

(10 x 100) / (17.5 + 0.33 + 1.98 + 7.5 + 0.36 + 0.29) = 35.7 MVA

that is, 6260A at 3.3 kV or 1880A at 11 kV. The operating characteristics of the relay show that at a plug setting of 100%, that is, 250 A and 4.76 MVA at 11 kV, and at a time multiplier setting of 0.2, suitable discrimination with the 200 A fuse is achieved.

Substation C:CT ratio 500/5ARelay over-current characteristic assumed to be extremely inverse relay. This relay must discriminate with the relay in substation B at fault levels up to:

(10 x 100) / (1.98 + 7.5 + 0.36 + 0.29) = 98.7 MVA

that is, 17,280A at 3.3kV or 5180A at 11 kV. The operating characteristics of the relay show that at a plug setting of 100%, that is, 500 A and 9.52 MVA at 11 kV, and at a time multiplier setting of 0.7, suitable discrimination with the relay at substation B is achieved.

Substation D:CT ratio 150/1 ARelay over-current characteristic assumed to be extremely inverse relay. This relay must discriminate with the relay in substation C at fault levels up to:

(10 x 100) / (7.5 + 0.36 + 0.29) = 123 MVA

that is, 21,500 A at 3.3 kV or 538 A at 132 kV. The operating characteristics of the relay show that at a plug setting of 100%, that is, 150 A and 34.2 MVA at 132 kV and at a time multiplier setting of 0.25, suitable discrimination with the relay at substation C is achieved.

Substation E:CT ratio 500/1 ARelay over-current characteristic assumed to be extremely inverse relay. This relay must discriminate with the relay in substation D at fault levels up to:

(10 x 100) / ( 0.36 + 0.29) = 1540 MVA

that is, 270,000 A at 3.3 kV or 6750 A at 132 kV. The operating characteristics of the relay show that at a plug setting of 100%, that is, 500A and 114 MVA at 1 32 kV, and at a time multiplier setting of 0.9, suitable discrimination with the relay at substation D is achieved.

4.6. Protection of Parallel Feeders

Fig. 2.8 shows an over-current protective scheme for parallel feeders. At the sending end of the feeders (at A and B), non-directional relays are required. The symbol ( ↔ ) indicates a non-directional relay. At the other end of feeders (at C and D), directional over-current relays are required. The arrow mark ( ) for directional relays placed at C and D indicate that the relay will operate if current flows in the direction shown by the arrow. If a fault occurs at F, the directional relay at D trips, as the direction of the current is reversed. The relay at C does not trip, as the current flows in the normal direction. The relay at B trips for a fault at F after relay D. Thus, the faulty feeder is isolated and the supply of the healthy feeder is maintained.



Fig. 2.8: Protective scheme for parallel feeder

It is not desired to place non-directional relays at C and D to clear a fault at F. Due to this very reason relays at C and D are directional over-current relays. For faults at feeders, the direction of current at A and B does not change and hence relays used at A and B are non-directional.

4.7. Protection of Ring Mains

Fig. (2.9-a) shows an over-current scheme for the protection of a ring feeder. Fig. (2.9-b) is another way of drawing the same scheme. Compared with radial feeders, the protection of ring feeders is costly and complex. Each feeder requires two relays. A non-directional relay is required at one end and a directional relay at the other end. The operating times for relays are determined by considering the grading, first in one direction and then in the other direction, as shown in Fig. 2.9. If a fault occurs at F1 as shown in Fig. (2.9-a) , the relays at C’ and D’ will trip to isolate the faulty feeder. The relay at C will not trip as the fault current is not flowing in its tripping direction though its operating time is the same as that of C’. Similarly, the relays at B and D will not trip as the fault currents are not in their tripping direction, though their operating time is less than the operating time of B’ and D’ respectively.

Fig. 2.9: Protection of Ring Feeder.

Fig. (2.9-c) shows a scheme involving an even greater number of feeders.



3. Earth Protection

1. Earth Fault and Phase Fault Protection

A fault which involves ground is called an earth fault. Examples are:i) Single line to ground (L-G) fault and ii) Double line to ground (2L-G) fault.

Faults which do not involve ground are called phase faults. The protective scheme used for the protection of an element of a power system against earth faults is known as earth fault protection. Similarly, the scheme used for the protection against phase faults is known as phase fault protection.

1.1 Earth Fault Relay and Over-current Relay

Relays which are used for the protection of a section (or an element) of the power system against earth faults are called earth fault relays. Similarly, relays used for the protection of a section of the power system against phase faults are called phase fault relays or over-current relays. The operating principles and constructional features of earth fault relays and phase fault relays are the same. They differ only in the current levels of their operation.

The plug setting for earth fault relays varies from 20% to 80% of the C.T. secondary rating in steps of 10%. Earth fault relays are more sensitive than the relays used for phase faults. The plug setting for phase fault relays varies from 50% to 200% of the C.T. secondary rating in steps of 25%. The name phase fault relay or phase relays is not common. The common name for such relays is over-current relay. One should not confuse this term with the general meaning of over-current relay.

In a general sense, a relay which operates when the current exceeds its pick-up value is called an over-current relay. But in the context under consideration, i.e. phase fault protection and earth fault protection, the relays which are used for the protection of the system against phase faults are called over-current relays.

1.2 Earth Fault Protective Schemes

An earth fault relay may be energised by a residual current as shown in Fig. (3.1-a) ia ,ib and ic are currents in the secondary of C.T.s of different phases. The sum (ia + ib + ic) is called residual current. Under normal conditions the residual current is zero .

When an earth fault occurs, the residual current is non-zero. When it exceeds pick-up value, the earth fault relay operates. In this scheme, the relay operates only for earth faults. During balanced load conditions, the earth fault relay carries no current hence theoretically its current setting may be any value greater than zero. But in practice, it is not true as ideal conditions do not exist in the system. Usually, the minimum plug setting is made at 20% or 30%.

The manufacturer provides a range of plug settings for earth fault relay from 20% to 80% of the C.T. secondary rating in steps of 10%.



Fig. 3.1: Various earth fault protection schemes

The magnitude of the earth current depends on the fault impedance. In case of an earth fault, the fault impedance depends on the system parameter and also on the type of neutral earthing. The neutral may be solidly grounded, through resistance or reactance. The fault impedance for earth faults is much higher than that for phase faults. Hence, the earth fault current is low compared to the phase fault currents. An earth fault relay is set independent of load current. Its setting is below normal load current. When an earth fault relay is set at lower values, its ohmic impedance is high, resulting in a high C.T. burden.

Fig. (3.1-b) and 3.1 (c) show an earth fault relay used for the protection of transformer and an alternator, respectively. When an earth fault occurs, zero sequence current flows through the neutral. It actuates earth fault relay. Fig. (3.1-d) shows the connection of an earth fault relay using a special type of C.T. known as a core-balance C.T., which encircles the three-phase conductors.



2. Combined Earth Fault and Phase Fault Protective Scheme

Fig. 3.2 shows two over-current relays (phase to phase fault relays) and one earth fault relay. When an earth fault occurs, the burden on the active C.T. is that of an over-current relay (phase fault relay) and the earth fault relay in series. Thus, the C.T. burden becomes high and may cause saturation.

Fig. 3.2: Two over-current and one earth fault relays

3. Phase Fault Protective Scheme

Fig. 3.3 shows three over-current relays for the protection of a three-phase system. This scheme is mainly for the protection of the system against phase faults. If there is no separate scheme for earth fault protection, the over-current relays used in this scheme will also sense earth faults but they will be less sensitive.

Fig. 3.3: Three over-current relays


BTEC - HND Year 2 - Utilization of Electrical Energy Circuit Breakers

4. Circuit Breakers Operation and Construction

1. Principles of Operation

In interrupting a circuit, the circuit breaker actually makes a physical separation in the current-carrying or conducting element by inserting an insulating medium sufficient to prevent current from continuing to flow.

The circuit is usually opened by drawing out an arc between contacts until the arc can no longer support itself. The arc is formed when the contacts of a circuit breaker move apart to interrupt a circuit. The arc is a conductor made up of ionized particles of the insulating material. With the advent of higher voltage power systems, the length of arc that would extinguish itself in air requires breakers of such physical size as to be impractical. Hence, whenever voltages and currents are going to be large, other forms of insulation are used in place of air to extinguish the arc as quickly as possible. The magnetic fields set up and the heat generated will persist as long as current continues to flow through the contacts and the arc. Should this current be “fault current,” the enormous magnetic fields and the great amount of heat will tend to cause damage in terms of their own duration.

The most commonly used insulation is oil. For higher voltages and larger capacities, the insulating medium may be a vacuum or an inert gas, such as Sulphur hexafluoride (SF6).Other means are also employed to limit and extinguish the arc as quickly as possible as well as to reduce the magnitude and duration of the fault current, both of which are factors in the size, reliability, and cost of circuit breakers.

2. Types of Breakers Categorized by Insulation Medium

The classification of circuit breakers has been made on the basis of insulation medium employed in the circuit breaker to extinguish the arc. Depending on the arc quenching medium employed, the following are important types of circuit breakers:

1. Air circuit Breakers2. Oil Circuit Breakers3. Vacuum Circuit Breakers.4. Sulphur Hexafluoride (SF6) Circuit Breakers.

2.1 Air Circuit Breaker:

An air circuit breaker employs air as the interrupting insulation medium. Of all the insulating media mentioned, air is the most easily ionized and, hence, arcs formed in air tend to be severe and persistent.

The switching elements, for an air current breaker, therefore, may often consist of main and auxiliary contacts. The auxiliary contacts open before the main contacts do, and the arc is drawn on them, thereby avoiding severe pitting of the main contacts (see Fig. 4.1). As the contacts separate, a puff of air is sometimes used to blow the arc into “arc chutes.” Moreover, the current itself is made to flow through coils that set up magnetic fields that force the arc into these arc chutes. An electric arc is essentially an electric conductor with a surrounding magnetic field; hence it can be attracted or repulsed by another magnetic field.

The arc chutes are usually made of corrugated insulating material (such as porcelain or slate). Their function is to stretch out the arc until it can no longer sustain itself. In simple form, it is an enclosed channel in which the arc is dissipated and kept from striking surrounding objects.



Beakers of this type are usually employed in electric systems where fault currents are relatively small. Their advantages are simplicity of construction, low cost, and relatively low maintenance requirements.As a means of extending the short-circuit duty of this type of breaker, a blast of air trained on the arc from a reservoir of compressed air enables a heavier arc to be extinguished.

(a) (b)Fig. 4.1: (a) Air circuit breaker, and

(b) schematic showing main and auxiliary contacts and arcing arrangement

2.2 Oil Circuit Breaker:

Oil circuit breakers have their contacts immersed in insulating oil. They are used to open and close high-voltage circuits carrying relatively large currents in situations where air circuit breakers would be impractical because of the danger of the exposed arcs that might be formed (see Fig. 4.2).

When the contacts are drawn apart, the oil covering them tends to quench the arc by its cooling effect and by the gases thereby generated, which tend to “blow out” the arc. At the instant the contacts part, the arc formed at each contact not only displaces the oil but decomposes it, creating gas and a carbon residue. If these carbon particles were to remain in place, as a conductor they would tend to sustain the arc formed. However, the violence of the gas and the resulting turbulence of the oil disperse these particles, and they eventually settle to the bottom of the tank. The rate of expansion and the volume of gas generated depend to a great extent on the energy to be interrupted, the speed of the separation of the contacts, the cooling effect, and the quantity of oil used as the insulating medium.

The use of oil as insulation also permits smaller distances between live parts and between live parts and ground (the structural and mechanical support parts of the circuit breaker) because of its better insulating qualities (higher dielectric strength) in comparison to air.

The insulating oil normally used has a dielectric strength of around 30 kV per one-tenth of an inch (compared to a similar value of 1 kV for air). Oil is also an effective cooling medium.



Fig. 4.2: Oil circuit breaker

2.3 Vacuum Circuit Breaker:

In this type of breaker, the contacts are drawn apart in a chamber from which air has been evacuated. (see Fig. 4.3). The electric arc is essentially an electric conductor made up of ionized air. Thus, if there is no air, theoretically the arc cannot form. In practice, however, a perfect vacuum is not likely to be obtained. The small residual amount of air that may exist permits only a small arc to be formed and one of only a very short duration. The same vacuum, however, will not dissipate the heat generated as readily as other insulating media. Although this type of breaker has certain advantages in terms of its size and simplicity, its interrupting ratings are not comparable to those of oil circuit breakers. Furthermore, maintenance procedures are more complex and hence more costly.

Fig. 4.3: Vacuum circuit breaker



2.4 Sulphur Hexafluoride (SF6) Breaker:

This type of breaker (see Fig. 4.4) is similar to the vacuum type of breaker except that the vacuum is replaced by an inert, non-toxic, odourless gas Sulphur Hexafluoride (SF6). This gas, under pressures that may vary from 45 to 240 lb/sq in., extinguishes the arc so rapidly as almost to prevent its formation. It also has excellent heat-dissipating characteristics, and its dielectric strength is very much greater than that of oil.

Fig. 4.4: Sulphur Hexafluoride Circuit Breaker, 34.5 to 69 KV

The breakers are constructed in modules capable of operation at voltages from 34.5 kV at gas pressures of 45 psi to 362 kV at 240 psi. (see Fig. 4.5). By connecting two or three such modules in series, breakers capable of operating at 800 kV at 240 psi can be constructed, with two- to three-cycle interrupting time. Installation and maintenance requirements of these breakers are less costly than for other types.

Fig. 4.5: Features of Sulphur Hexafluoride Interrupting Module

2.5 Types of SF5 Circuit Breakers:



2.5.1 Double-pressure SF6 circuit breakers:

This is the early design of SF6 circuit breakers. Its operating principle is similar to that of air-blast circuit breakers. In this type of a circuit breaker, the gas from a high-pressure compartment is released to the low pressure compartment to extinguish the arc. Because of its complicated design and construction, and its need for various auxiliaries such as gas compressors, filters and control devices, this type of circuit breakers have become obsolete.

2.5.2 Puffer-type SF6 circuit breakers:

This type of circuit breakers are also sometimes called single-pressure or impulse type SF6

circuit breakers. In this type of breakers, gas is compressed by a moving cylinder system and is released through a nozzle to quench the arc. This type is available in the voltage range 3.6kV to 765 kV.

Fig. (4.6-a) shows a puffer-type breaker in closed position. The moving cylinder and the moving contact are coupled together. When the contacts separate and the moving cylinder moves, the trapped gas is compressed. The trapped gas is released through a nozzle and flows axially to quench the arc as shown in Fig. (4.6-b). There are two types of tank designs. Live tank design and dead tank design. In live tank design, interrupters are supported on porcelain insulators. In the dead tank design, interrupters are placed in SF6 filled-tank which is at earth potential. Live tank design is preferred for outdoor substations.

A number of interrupters (connected in series) on insulating supports are employed for EHV systems up to 765 kV. Two interrupters are used in a 420 kV system. Breaking time of 2 to 3 cycles can be achieved. In the circuit breaker the steady pressure of the gas is kept at 5kg/cm2. The gas pressure in the interrupter compartment increases rapidly to a level much above its steady value to quench the arc.

Fig. 4.6: Puffer-Type SF6 Circuit Breaker



3. Selection of Circuit Breakers

Table 4.1 shows the summary of various types of circuit breakers, their voltage ranges and arc quenching medium they employ.

Table 4.1 Types Of Circuit Breaker

Type Arc Quenching Medium Voltage Range &Breaking Capacity

Miniature C.B. Air at atmospheric pressure 400V; for small current rating

Air-Break C.B. Air at atmospheric pressure 400V-11KV; 5-750MVA

Minimum Oil C.B. Transformer Oil 3.3KV-220KV;150-25000MVA

Vacuum C.B. Vacuum 3.3KV-33KV; 250-2000MVA

SF6 C.B. SF6 at 5Kg/cm2 pressure 3.3-765KV;1000-50,000MVA

Air Blast C.B. Comp. Air at high presser (20-30Kg/cm2)

66-1100KV; 2500-60,000MVA

Table 4.2 shows the choice of circuit breakers for various voltage ranges.

Table 4.2: Selection of Circuit Breakers

Rated Voltage Choice of C.B. Remark

Below 1KV Air-Break C.B.

3.3KV-33KV Vacuum; SF6; Minimum Oil C.B.`s Vacuum Preferred

132KV-220KV SF6; Air Blast; Minimum Oil C.B.`s

SF6 Preferred

400KV-760KV SF6; Air Blast C.B.`s SF6 is Preferred

Earlier oil circuit breakers were preferred in the voltage range of 3.3 kV-66kV. Between 132kV and 220 kV, either oil circuit breakers or air blast circuit breakers were recommended. For voltages 400 kV and above, air blast circuit breakers were preferred. The present trend



is to recommend vacuum or SF6 circuit breakers in the voltage range 3.3 kV-33 kV. For 132kV and above, SF6 circuit breakers are preferred. Up to 1 kV, air break circuit breakers are used. Air blast circuit breakers are becoming obsolete and oil circuit breakers are being superseded by SF6 and vacuum circuit breakers.


BTEC - HND Year 2 - Utilization of Electrical Energy Lighting System

Lighting System

1 Types of Lamps

The whole of interior lighting design revolves around the lamp, and development and competition within the industry has resulted in an enormous range of lamps now being available to the lighting designers. The proliferation of lamp types is a mixed blessing because lamp development can often make an existing installation obsolete both in appearance and efficiency before its time. However, a surfeit of choice is not to be scorned. It means that the lighting designer must appreciate the range of lamps at his disposal and keep himself well informed on new developments.

1. Filament lamps

In a tungsten filament lamp the passage of electricity through the filament raises the temperature of the molecules within the filament to the point where they begin to give off light, or they incandesce.

1.1 The general lighting service lamp (GLS):

A lamp designed for general lighting service (GLS) is illustrated in Fig. 1.

Fig. 1: General lighting service - Filament Lamp (GLS)

Filaments: These are constructed of drawn tungsten wire which is then coiled, and in the lower-wattage gas-filled lamps the coils can be coiled again to reduce convection losses (coiled-coil).

Gas filling: The rate of filament evaporation can be reduced by raising the vapour pressure in the lamp. This can only be done by introducing gases which are chemically inactive with hot tungsten. The most common gas filling is a mixture of argon and nitrogen, and this can only be used with coiled filaments where the heat losses are much less than for a straight-wire filament.

Lamp caps: The British GLS lamps normally have Bayonet Caps (BC) up to 150 W. Edison Screw (ES) caps are used for 200 W lamps and Goliath Edison Screw (GES) caps are used for 300-1500 W lamps.



1.2 Tungsten-halogen lamps:

The most important factor deciding the light output of a lamp is the filament temperature, and this determines the life of the lamp.The addition of a halogen vapour to the gas filling effectively reduces the evaporation rate of the filament. This allows the use of a higher filament temperature and, hence, promotes greater efficacy without the detriment of short lamp life.

Fig. 2: Tungsten-halogen linear lamp

Construction: The bulb must be capable of operating continuously above 250 °C and a form of fused silica called quartz is used. These lamps operate with higher filament temperatures, resulting in increased efficacy and an increased lamp life, generally 2000 h. A typical linear lamp is illustrated in Fig. 2.

Operation: There are several operating conditions which must be observed for satisfactory performance.

1. The quartz bulb wall must be maintained above the condensation temperature of the halide (approx. 250 °C).

2. Linear lamps must be operated within 4° of horizontal. Outside this range, the halogen vapour will migrate to the lower end of the lamp, resulting in early lamp failure of the upper part of the filament.

3. Contamination of the outside surface of the lamp wall must be avoided. For example, a deposit of grease by handling will cause the quartz envelope to fail at high temperatures as the grease will cause the surface to develop fine cracks. Before use, any such contamination should be cleaned off with a suitable solvent such as methylated spirit.



2. Discharge lamps:

The method of light production previously discussed is concerned with the passage of electricity through a solid. Lamps which produce light due to the passage of electricity through a gas are termed discharge lamps.

Ionisation: Initially when a voltage is applied to a discharge lamp the gas inside the lamp is in an insulating state and will not conduct electricity. Ionisation can occur within the lamp in two ways. Initially the application of a high voltage pulse to the lamp will pull many electrons from their orbits. However, once the lamp is running, charged particles moving along the lamp will collide with gas atoms dislodging electrons in the process.

High- and low-pressure discharges: In the low-pressure discharge the charged particles move along the lamp a relatively long distance between collisions and will easily build up sufficient speed to cause excitation to the first possible orbit. As the gas pressure is increased the distance between collisions decreases and so the energy gained by the particle during its acceleration time is proportionally lower.

Fluorescence: The low-pressure mercury discharge produces radiant energy within the ultra-violet part of the spectrum. By coating the inside of the arc-tube with a fluorescent coating or phosphor, this UV radiation is converted to visible radiation.

Gas fillings in discharge lamps: The discharge envelope is filled with a mixture of gases and vapours. The main gas or vapour is the one responsible for the emission of light. This may be in solid state at room temperature, so a further gas is needed to initiate the discharge. When the lamp is running there may be loss of energy due to the current electrons not colliding with the main gas or vapour, and even passing out of the envelope. To increase the probability of collision a buffer gas or vapour is included. Examples of these three elements are given in Table 1.1.

Table 1.1 Typical gas and vapour fillings

Lamp type Main filling Starting BufferFluorescent tube Mercury Argon ArgonLow-pressure sodium Sodium NeonHigh-pressure sodium Sodium Xenon MercuryHigh-pressure mercury Mercury Argon and Nitrogen

3. Types of discharge lamps and typical control circuits

There are two kinds of discharge lamp, one is called the cold-cathode type and the other hot-cathode type. The former is normally referred to as a neon lamp, which required a very high voltage to initiate the discharge.

Five of these will now be discussed.

1. Low-pressure mercury vapour lamps (MCF).2. High-pressure mercury vapour lamps (MBF)3. Metal halide lamps (HBI)4. Low-pressure sodium vapour lamps (SOX).5. High-pressure sodium vapour lamps (SON)

3.1 Low-pressure Mercury Vapour Lamps (MCF):BTI - Electrical & Electronics Division 93 Elbaz Bader Ayad

YOSRY IDREES


The low-pressure mercury vapour lamp (it is coded MCF) can have a relatively high light output and efficacy provided the ultra-violet radiation generated in the lamp is converted into light by means of the fluorescent coating. This type of lamp is one of the most common lamps manufactured today. It is used extensively in industrial and commercial premises and is beginning to be used in the home. Its general construction is shown in Fig. 3

Fig. 3: Tubular fluorescent lamp (MCF)

Electrodes: The electrodes, which are at the ends of the tube, act as both anode and cathode alternately when connected to alternating current but are normally called cathodes.

Electrons are emitted at the cathode and collected at the anode. For an electron to leave the cathode, it has to be given sufficient energy to overcome the forces holding it to the atoms of the cathode material. Heating the electrodes by passing a current through them before the arc is struck makes sure that a large number of free electrons exist around the electrodes, making the striking of the lamp easier. During operation the electrodes are kept hot by the passage of the discharge current.Control gear: In fluorescent lamp circuits operated on a.c. supplies, the ballast is usually a choke and is connected in series with the lamp, as shown in Fig. 4. There are many more complex circuits used to control the lamp current and to assist in striking the lamp, and the term ballast is used to cover all forms of control gear.

Fig.4: Switch-start circuit


Fig. 5: Glow type starter switch


Switch-start circuits: Glow type starter switch This switch, shown in Fig. 5, has two switch contacts, one of which is bimetal strip in a small envelope containing argon.

Starter capacitor: The small capacitor which is connected across the starter terminals (i.e. in parallel with the switch contacts) reduces emission by the lamp at radio frequencies and also improves the waveform of the starting voltage pulse.

Circuit Operation: Circuit operation on starting having considered the sequence of events that takes place in the starter switch, this must be related to the lamp and choke during this time. First there is a current flowing through the choke, cathodes, and starter switch. This current is about 1.5 times the running current; it heats the cathodes and causes free electrons to be emitted and ionisation to take place near the electrodes.

When the starter switch opens, the choke opposes the change by producing a voltage which is in the same direction as the supply voltage and therefore adds to it. A high-voltage surge across the lamp is enough to cause an arc between the electrodes of the lamp.

Electronic starter switch: A switch is available, at extra cost, which uses a thyristor trigger. This sets up a pre-heat circuit and a high voltage is generated. Starting is near instant and, in the event of lamp failure, the electrode current ceases. This overcomes damage to the starter and ballast.

Starterless circuits: Although the switch-start circuit is reliable and is more suitable for low-temperature applications, the starter switch is the one item in the circuit with moving parts liable to fail. Also, in some applications its operation can be a nuisance due to flicker until the lamp has struck - especially with old lamps.Starterless circuits operate without a starter switch and so reduce the maintenance required. They have disadvantages, however: they are more complex and expensive; ballast losses are higher than those of the simple choke; and they are more susceptible to difficulties in starting at temperatures below 5°C. The circuit is shown in Fin 6.

Fig. 6: Starterless transformer circuit

The major change from the switch-start circuit is that a transformer is used to provide a low voltage across each electrode, which is contained within the ballast canister together with the choke.

When the circuit is first switched on, nearly all the supply voltage appears across the lamp and the transformer. Current flows through the electrodes from the two transformer windings, one heating each electrode and producing emission of electrons, until there are enough free electrons to enable the lamp to strike with only the supply voltage across the lamp. Once the arc has struck the current through the choke increases and therefore the voltage drop



previously across the transformer now appears across the choke, reducing the lamp and transformer voltage so that the lamp current is limited to its correct value. The voltage across the electrodes falls to about half.

This circuit is more complex and expensive than the switch start but is the only circuit suitable for use with dimming control. To dim fluorescent tubes the lamp voltage and filament temperatures must be maintained while the lamp current is reduced. A typical circuit is shown in Fig. 7.

Fig. 7: Circuit for dimming a fluorescent tube

Semi-resonant start circuit: This comes within the category of starterless circuits, but is simpler and has lower gear losses than the transformer type. It comprises a double wound choke with two equal coils wound in opposition and a series capacitor. The term resonant is used since the sub-circuit of the lamp, coil B. and the capacitor (Fig. 8) produces a high lamp voltage and filament current for starting (the capacitor and coil combination have minimum impedance). When the lamp is running, the coil B becomes ineffective and the circuit is coil A, lamp, and capacitor. This circuit will operate at low temperatures.

Fig. 8: Semi-resonant start circuit

Electronic control circuits: Raising the electrical frequency to the lamp from 50 Hz to a high frequency of around 30 kHz brings a number of significant improvements to performance. These include:

1. reduction in ballast losses, and weight; 2. elimination of flicker and noise; 3. improved lamp efficacy and life; 4. instant lamp start.

The electronic element of the circuit is the frequency conversion. Figure 9 shows an example of this type of circuit.



Fig. 9: Typical circuit for a high frequency electronic ballast

The HF system offers particular advantages in twin-lamp luminaires, because a single ballast drives two lamps instead of the two separate ballasts in conventional luminaires.3.2. High-pressure mercury lamp (MBF):

Increasing the vapour pressure of a discharge has the effect of changing the spectral power distribution by increasing the width of the spectral lines (bands) produced, and by changing the relative intensities. Absorption of the resonance radiation also takes place.

Lamp construction: The basic high-pressure mercury vapour lamp shown in Fig. 10 consists of a glass outer envelope containing the quartz arc tube, its supports, and the lead-in wires. This outer envelope contains nitrogen to dissipate the heat produced and maintain the arc tube at the correct temperature.Inside the arc tube; there are small quantity of mercury, argon (which acts as a starting gas), and the tungsten electrodes. There are usually three or four electrodes - two of which are main electrodes and a smaller auxiliary electrode (or electrodes) used to initiate the arc. The auxiliary electrode is very close to a main electrode and is connected through a resistor to the electrode at the other end of the lamp.

Lamp operation: When the lamp is connected to the supply, there will initially be no current flow and mains voltage will exist between the two main electrodes and between a main electrode and its auxiliary electrode via the internal resistor.

This voltage will produce a local arc in the argon gas with the current limited to a safe value by a high resistance of a few thousand ohms connected in series with the auxiliary electrode as shown on Fig. 10. The initial arc produces ionization, and an argon arc is soon struck between the two main electrodes.

This produces further ionisation until sufficient mercury ions exist for the mercury arc to be formed. At first a low-pressure arc exists and very little light is produced; but gradually, as the lamp heats up, the mercury vapour pressure rises and a high pressure arc is formed and more light is emitted. Once the main arc is established, its resistance is lower than in the auxiliary electrode circuit and this auxiliary electrode ceases to function. The time taken for the lamp to completely 'run up' - that is, to reach full light output - is approximately 5 min.This type of lamp is also widely used for street lighting in residential areas where its colour rendering may be preferred to low-pressure sodium.



Fig. 10: High-pressure mercury fluorescent lamp (MBF)

Lamp circuits: The circuit is relatively simple as the lamp incorporates its own starter. The circuit is shown in Fig. 11 and consists of a choke and power factor capacitor. A series resistance in the form of a tungsten filament can be used instead of the choke. These are called MBT-type lamps; they have improved colour rendering but only about half the luminous efficacy.

Fig. 11: Control circuit for an MBF lamp3.3. Metal halide lamps (MBI):

This lamp, known as the mercury halide or metal halide lamp, is similar in construction to the MB lamp but has a much shorter arc tube. Strictly speaking, they are not mercury lamps but are a series of lamps combined within a single arc tube with output characteristics of different mixtures of elements.

One disadvantage of these lamps is that a high starting voltage is necessary. This is usually provided by an electronic igniter which generates very short duration high-voltage pulses is connected across the lamp terminals. This type of starter enables the lamp to be restruck within one minute of being switched off.

Due to its high efficacy and good colour rendering properties, the MBI lamp is suited to both high bay interior and high mast exterior installations, and is often used in sports stadiums, especially where colour television transmission is expected. A typical lamp is shown in Fig. 12.

Fig. 12: 365 W metal halide lamp (MBI)

Lamp circuits: The lamp requires an igniter, choke, and power factor capacitor. A typical circuit is shown in Fig. 13.



Fig. 13: Control circuit with ignitor for an MBI lamp3.4 Low-pressure sodium vapour lamp (SOX):

Light is produced by the action of the excited electrons returning to lower energy levels. The sodium vapour arc produces a number of spectral lines but most of the visible radiation (95 %) is concentrated in a very narrow region of the spectrum at 589.0 and 589.6 nm (known as the sodium 'D' lines). The light output is therefore monochromatic (single coloured) and the lamp is easily recognised by its yellow appearance.

Although this yellow light has the disadvantage that colours are greatly distorted, it has the advantage that, as the relative response of the eye, it has a potential for a high lumen output, depending upon how much energy is actually put into the lamp.

Lamp construction and performance: There are a number of different types of lamp. The two main types are SOX and SL1. SOX is a single-ended lamp with an integral outer jacket coated with a heated reflector (see Fig. 14). The inner discharge tube is made of 2-layer glass, the outer being soft glass and the inner a thin film of sodium-resistant class.

Fig. 14: Single-ended low-pressure sodium lamp (SOX)

SL1 is a double-ended single-line discharge tube. This lamp is of particular value in floodlights which require a narrow linear source.

The electrical characteristics are similar to those of the fluorescent tube. The lamps are not temperature sensitive as they operate within a vacuum jacket.

Lamp circuits: At room temperature the metallic sodium solid, and in order to vaporise the sodium, neon and argon are added. Initially a neon-argon arc is formed and the heat produced gradually vaporises the sodium. The lamp has a run-up period of about 10 min before maximum light output is reached.

A transformer is used in the sodium lamp circuit because of the 'higher-than-mains' ignition voltage required, i.e. approximately 470 to 490 V. If a normal transformer is used, a choke would also be needed to limit the current after the discharge has struck because of the negative resistance characteristic of the arc. Such a circuit is, however, expensive in material



usage, when one considers that the transformer output voltage is about 500 V while the normal operating lamp voltage required is 100 V.A lot of material can be saved by, first of all, using an auto-transformer and, secondly, by making it an auto-leak-transformer which removes the need for a separate choke (see Fig. 15).

This leak transformer type of circuit has a low power factor (0.3) and a capacitor is connected across the supply to increase the power factor to an acceptable value.The linear lamp can be operated on circuits similar to fluorescent lamps.

Fig. 15: 'Leakage-flux' transformer circuit

3.5 High-pressure sodium lamp (SON, SON-L, SON-R):

The low-pressure sodium arc emits all of its radiation in the two yellow. The colour rendering is, as stated earlier, very unsuitable for most applications other than road lighting and security lighting.

Lamp construction and performance: The high temperature (1300 °C) necessary to produce the very high vapour pressure required in this lamp would melt any glass arc tube, and at this temperature quartz also is readily attacked by hot sodium vapour. A 400 W high-pressure SON sodium lamp is shown in Fig. 16.

The lamp consists of an evacuated outer envelope containing the arc tube and its support-similar to the MB lamp except that the arc tube is longer and narrower than the arc tube of the MB lamp and no auxiliary electrode is possible. The arc tube contains an amalgam of mercury and sodium together with a small quantity of argon or xenon to assist in striking the arc. The mercury is used to reduce the lamp current for a given wattage.

A high voltage is required across the lamp in order to strike the arc. This high voltage may be provided, as in the MBI circuit, by an electronic igniter which generates pulses of a few thousand volts for a few microseconds duration and is connected across the lamp terminals. The igniter continues to pulse until the lamp is struck.



Fig. 16: High-pressure sodium lamp

Some manufacturers include in the outer envelope of the lamp a bimetal starting switch and small heating element. The switch and element are in series and are connected across the electrodes. This type of lamp is manufactured in many sizes from 125 W (76 lm/W) to 1 kW (130 lm/W) and the outer envelope may be either elliptical with a diffusing coating on the inside, or tubular clear glass.



2. Design of lighting

This should be the culmination of all the knowledge gained from the previous chapters. Unfortunately, this is where many people responsible for the design of lighting start and finish. Good design is the application of the best or most appropriate equipment in an economical but effective manner.This chapter covers the normal design processes for establishing a satisfactory illuminance on the working plane, provided in a glare controlled manner.

1. Amount of light Required

Standard values of illuminance (lux) are specified in the CIBSE 1984 Lighting Code for all the usual locations and occupations. These values are necessary to achieve:

(1) satisfactory illuminance of the task;(2) an agreeable general appearance of the interior.

They are quoted as 'standard service illuminances', normally on a horizontal working plane. A 'service' illuminance is the mean illuminance throughout the life of the lighting system and averaged over the relevant area. 'Standard' implies a value suitable for many locations, unless there are special requirements.

Table 3.1 shows the illuminances relevant to a range of different visual tasks. This table is expanded in the Guide to quote recommended standard service illuminance values covering nearly 150 activities or types of interiors.The illuminance values serve as a guide to good practice. They are not mandatory although some authorities may adopt them as part of their standard requirements.

It is also recognised that there can be circumstances for a given situation where the standard service value should be modified to give a design service value, i.e. one applicable to the specific circumstances.

2. Designing a general layout

The first stage is to decide on the type of luminaire to be used. This decision may seem premature, but, unless using computer programs, it is difficult to achieve anything without immediate access to specific photometric data. Alternative approaches can be used where all the design requirements are established, and luminaires are then found to fit these requirements. However, it is easier to explain the design process by the first method.

Next, the total number of lamp lumens is calculated to achieve the recommended illuminance, and finally the layout is planned.

2.1 Determination of lumens required:A commonly accepted method involves the use of the 'lumen formula' which is:

Lamp lumens required = E A / UF LLFwhere: E = standard service illuminance (Ix)

A = area of working plane (m2)UF = utilization factorLLF = light loss factor



Table 2.1 Examples of activities/interiors appropriate for each standard service illuminance (reproduced by permission of the Chartered Institution of Building Services Engineers. CIBSE 1984 Interior Lighting Code).

StandardIllumin.

(lx)

Characteristics of the activity/interior Representative activities/interiors

50Interiors visited rarely with visual tasks confined to movement and casual seeing without perception of detail.

Cable tunnels, indoor storage tanks, walkways

100Interiors visited occasionally with visual tasks confined to movement and casual seeing calling for only limited perception of detail.

Corridors, changing rooms, bulk

150

Interiors visited occasionally with visual tasks requiring some perception of detail or involving some risk to people, plant or product.

Loading bays, medical stores, switch-rooms.

200Continuously occupied interiors, visual tasks not requiring any perception or detail.

Monitoring automatic processes in manufacture, casting concrete, turbine halls.

300Continuously occupied interiors, visual tasks moderately easy, i.e. large details > 10 min arc and/or high contrast.

Packing goods, rough core making infoundries, rough sawing.

500

Visual tasks moderately difficult, i.e. details to be seen are of moderate Size (5-10 min arc) and may be of low contrast. Also colour judgement may be required.

General of fires, engine assembly, painting and spraying.

750

Visual tasks difficult, i.e. details to be seen are small (3-5 min arc) and of low contrast, also good colour judgements may be required.

Drawing of rice’s, ceramic decoration, meat inspection.

1000

Visual tasks very difficult, i.e. details to be seen are very small (2-3 min arc) and can be of very low contrast. Also accurate colour judgements may be required.

Electronic component assembly, gauge and tool rooms, retouching paintwork.

1500Visual tasks extremely difficult, i.e. details to be seen extremely small (1-2 min arc) and of low contrast. Visual aids may be of advantage.

Inspection of graphic reproduction, hand tailoring, fine die sinking.

2000

Visual tasks exceptionally difficult, i.e. details to be seen exceptionally small ( < I min arc) with very low contrasts. Visual aids will be of advantage.

Assembly of minute mechanisms, finished fabric inspection.



2.2 Utilization factor:This is a measure of the efficiency of the lighting scheme and is the proportion of lamp flux that reaches the working plane. Some will come direct and some will come after reflection from other room surfaces.Table 2.2 shows a typical set of data, and to determine the precise value involves details of the room dimensions and surface reflectance.

The room dimensions are required in terms of the room index, where

Room index= (L W) / [(L + W) H] where: L = Room Length

W = Room WidthH = Height of lamp above working plane

Table 2.2: A typical set of data, and to determine the precise value involves details of the room dimensions and surface reflectance.

Room reflectance Room IndexC W F 0.75 1.0 1.25 1.50 2.00 2.50 3.00 4.00 5.00

0.70.50 0.20 0.41 0.47 0.52 0.55 0.60 0.63 0.66 0.69 0.710.30 0.36 0.42 0.47 0.50 0.56 0.59 0.62 0.66 0.680.10 0.32 0.38 0.43 0.47 0.52 0.56 0.59 0.63 0.66

0.50.50 0.20 0.37 0.42 0.46 0.49 0.53 0.55 0.57 0.60 0.610.30 0.33 0.38 0.42 0.45 0.49 0.52 0.55 0.57 0.590.10 0.29 0.34 0.39 0.42 0.47 0.50 0.52 0.56 0.58

0.30.50 0.20 0.33 0.37 0.40 0.43 0.46 0.48 0.49 0.51 0.530.30 0.29 0.34 0.37 0.40 0.43 0.46 0.48 0.50 0.510.10 0.27 0.31 0.35 0.38 0.41 0.42 0.46 0.48 0.50

0.00 0.00 0.00 0.23 0.26 0.28 0.30 0.33 0.35 0.36 0.38 0.39

2.3 Light loss factor and maintenance factor

In previous lighting codes the effect of hours of operation on the lumen output from lamps has been allowed for by using lighting design lumens rather than initial lumens, and the effect of dirt depreciation has been allowed for by using a maintenance factor.

The maintenance factor is the ratio of the illuminance provided when the installation is in an average condition of dirtiness to the illuminance provided when it is clean.

This system has been abandoned because with the range of light sources now available it has become unrealistic to consider the illuminance provided at 2000 h as an accurate description of the average through-life illuminance provided by the installation. Dealing with maintenance by means of the light loss factor allows for a more comprehensive examination of the effect of the operating conditions and maintenance procedures on the illuminance provided.

2.4 Determination of light loss factor:



Light loss factor is the product of three other factors:Light loss factor = Lamp lumen maintenance factor

Luminaire maintenance factor Room surface maintenance factor

Lamp lumen maintenance factor estimates the decline in light output of the light source over a specified time. Luminaire maintenance factor estimates the effect of dirt deposited on or in the luminaire over a set time on the light output of the luminaire. Room surface maintenance factor estimates the effect of dirt deposited on the room surfaces over a set time on the illuminance produced by the installation.

By calculating the light loss factor for different times, and taking into account the proposed maintenance schedule, it is possible to predict the pattern of illuminance produced by the installation over time. This pattern can be used to assess the suitability of any proposed maintenance schedule. It can also be used to estimate the average illuminance provided by the installation over time, and hence to determine whether the installation is likely to meet the appropriate design service illuminance criterion recommended.

As a guide, the lumen output of discharge lamps after 2000 h of use is about 0.92 of their initial lumens. Typical depreciation curves for luminaires are shown in Fig. 1. This also includes data on light loss due to dirt on the room surfaces.

Fig. 1: Loss of light due to dirt on luminaires and room surfaces

Example:What is the light loss factor due to open base fluorescent reflector luminaires assuming a 6 month cleaning interval based on lamp lumen output after 2000 h?

Solution:Lamp lumen maintenance factor = 0.92 (from lamp data)

Average 3 month depreciation = 0.90 (from Fig. 1)

Light loss factor = 0.92 0.90 = 0.83

It must be stressed that this is an over simplification, and that to obtain an accurate factor the Guide must be used.

3. Lighting Fittings Layout


Elapsed time months0 2 4 6 8 10 12 14

20

40

60

80

100 abcd

a. Ventilated reflectorb. Open base reflectorc. Non-ventilated

reflectord. Indirect cornice

%Lightoutput


The previous section has shown how to calculate the number of lumens necessary for efficient lighting. This section will consider how the fittings chosen should be disposed to give the best results. It should first be appreciated that the type of fitting to be used must be chosen before the lumen calculation is completed, since this effect the coefficient of utilization. The choice may be altered in the light of figures obtained. In most applications the type of fitting used will often be dictated by architectural or financial considerations. In many cases, the pattern used will be dictated because the ceiling is divided into bays or panels. The example which follow shows the methods to be adopted, but the following points should always be in mind.

1) Always arrange the fittings symmetrically in straight or staggered rows, unless there is something (such as a special architectural feature in the ceiling) to prevent it .

2) If the calculated number of fittings will not give a suitable pattern, increase the number to be used. A decrease will result in a lower level of illumination than that planed.

3) Allow even spaces between fittings, and “half-spaces” between fittings and walls.The fittings may need to be nearer the wall if they are in line with work-benches.

4) Make sure that the space / height ratio is reasonable. The correct ratio depends on the type of fitting used, but as a rough guide, the distance between the fittings should not exceed 1.5 the mounting height.

5) Glare must be prevented by shielding lamps where necessary. This is of particular importance where lamps become visible by only a slight elevation of the angle of vision, as in low-ceiling or very large rooms.

The required uniformity should be achieved if the layout conforms to the manufacturers quoted maximum spacing to height ratio (S / hm). The values of S and hm are illustrated in Fig. 2. Where manufacturers' data are not available, the BZ classification gives a guide to the ratio.

(a) (b)

Fig. 2: Space to mounting height ratio

If linear luminaires are used in continues rows, then the manufacturer may quote a maximum transverse spacing to height ratio. This indicates the maximum spacing between parallel rows of luminaires (Normally from 1 to 1.5).

Example:A lighting scheme is required for a small drawing office, 6 6 m , with a 3 m ceiling height.


S S/2

hm

Working plane

S/2

S/2 S


The office is air conditioned (LLF = 0.85) and the room reflectance are 0.7 ceiling, 0.3 walls, and 0.2 floor and the disk height is 0.8m.

Solution:Step 1. Select the mounting height. In this case the ceiling is relatively low and the luminaires would be mounted direct to the ceiling and the height is 0.8 m.Mounting height hm = Ceiling height - Desk height = 3.0 - 0.8 = 2.2 m

Step 2. Determine the illuminance and, hence, the lumens required. The illuminance would normally be selected from the Table 3.1. However, a drawing office is an example given in that table and the value is 750 Ix. Values of UF and LLF must be determined. The room index is given by: RI = (6 6) / 12 2.2 = 1.4From Table 2.2 UF = 0.49 and as the office is air conditioned, an LLF value of 0.85 will be assumed. Lamp lumens =(750 6 6) / (0.49 0.85) = 64 826 Im

Step 3. Determine the minimum number of lighting points. This may not produce a satisfactory layout but the golden rule is that this number can be increased but not reduced.

The luminaire has a quoted maximum S / hm ratio of 1.4. If hm = 2.2 m, then S = 2.2 1.4 = 3 m So the minimum number of points is 2 2, i.e. 4.

Step 4. Can this number achieve 750 Ix with a sensible lamping? The available lamp sizes are shown in Table 2.3 and range from a 40 W to a 125 W fluorescent lamp. For a layout of four points:

Lumens per point = 64 826 / 4 = 16206 lm

No single lamp approaches this initial lumen value. The nearest is the 125 W white tube with 9500 Im or triple tube of 75 W with 6050 lm . A possible solution is to use twin or triple lamp luminaires, but a new set of calculations is required. As it is a drawing office it would be more advantageous to increase the number of points to eight or nine to reduce the risk of shadow ( try to do it). A suggested layout is shown in Fig. 3.

Fig. 3: Suggest layout for the above example.

Table 2.3 : Range of typical lamps and their lumen outputs


1.5 m

3.0 m

1.5 m

1.5 m

3.0 m

1.5 m


Lamp type Wattage Average initialLumens (100 h)

Lighting designLumens (2000 h)

Tungsten filamentGLS 60 610 -

150 2 060 -500 8 200 -

Tungsten Halogen K 750 15 000 -1 500 32 000 -

Fluorescent tubesWhite

Normal

125 9 500 8 50075 6 050 5 75065 5 100 4 75040 3 050 2 80075 4 400 4 00065 3 700 3 40040 2 300 2 100

Krypton filled 26 mmTriphosphor

Compact lamps

70 6 550 6 30058 5 400 5 10036 3 450 3 20016 1 050 1 80024 1 800 1 690

Low-pressure sodiumSOX

SLI

35 4 600 4 500180 32 000 31 50060 6 000 5 700200 20 500 20 000

High-pressure sodiumSON

SON Deluxe

70 6 000 5 500400 47 000 45 000150 12 500 11 500400 38 5000 35 500

High-pressure mercuryMBF

MBIF

50 1 900 1 750125 6 200 5 700

1 000 58 000 54 000250 19 000 16 000

1 000 92 000 85 000

Example :



A workshop has a floor area of 10m 15m and is to be illuminated at 100 (lux) by using single-tube 4-ft, 40 W fluorescent fittings. The tube output through life averages 2150 (lm) and the estimated values for the coefficient of utilization factor is 0.4 and the light loss/maintenance factor is 0.75. Find the number of fittings required and show on a drawing how they could be arranged.

Solution: E A

UF MF lm

Lumens required 100 ( 10 15 )

0.4 0.75 50 000 lm

Number of fittings required = 50 0002150

= 23.26 24 fittings

Since it is impossible to have a part of a fitting, at least twenty four will be needed. In a rectangular space 10m 15m it is easy to arrange 24-fittings symmetrically, so 4-rows of 6-fittings is suggested arrangement.

To space 4-fittings over 10m with a half-space at the wall on Y axis, spacing will be:10 4 = 2.5 m.

To space 6-fittings over 15m with a half-space at the wall on X axis, spacing will be:15 6 = 2.5 m.

The proposed layout is shown in Fig. 4.

It is important to appreciate that this solution is one of many. Particularly in workshops, where appearance may not be too important, a symmetrical pattern must be sacrificed to provide even illumination for benches and machinery.

Fig. 4: Layout of the example above

Class work :

A heavy engineering assemble bay 120m 40m is to be illuminated by twin mercury-vapour/tungsten fittings mounted 15m above floor level, at which the illumination must be 250 lux. Each fitting comprises 1-250W h.p.m.v. lamp emitting 9250 lm, and 1-800W tungsten filament lamp emitting 4650 lm. If the coefficient of utilization is 0.25 and the maintenance factor is 0.6, find the number of fittings required and suggest a suitable layout. 4. Calculation of glareBTI - Electrical & Electronics Division 109 Elbaz Bader Ayad

YOSRY IDREES

1.25m

1.25m

2.5m

2.5m

2.5m

1.25 m

2.5 m

2.5 m

2.5 m

2.5 m

2.5 m

1.25 m


In both daylight and electric light there is always a need to restrict the luminance within the normal field of view, otherwise an unacceptable degree of glare will be experienced.

4.1 Disability glare

This can be defined as: Glare which impairs the ability to see detail without necessarily causing visual discomfort. As it is a definite reduction in the ability to see, it can be measured in terms of visual performance, and its effect can be expressed as a shift in the adaptation level of the eye. If the adaptation level is raised by a light source, such as an unshaded lamp or window, then the eye becomes less sensitive to small differences in brightness.

4.2 Discomfort glare

This can be defined as: Glare which causes visual discomfort without necessarily impairing the vision. The degree of discomfort will depend on the type of location and angle of view; for instance, people will tolerate a much brighter installation in a supermarket, where they are continually on the move and looking in all directions, than in a classroom, where the direction of view is fixed, the visual task more exacting, and the occupants have more time to become aware of the lighting. This is very much a subjective assessment; it cannot be measured in terms of performance, but only in relative degrees of discomfort.

To calculate the glare from each light point for a range of room positions would be too time-consuming. The method can be simplified by considering a number of standard conditions and making adjustments as necessary. The standard conditions and assumptions are:

1. Glare is additive; therefore, for a complete installation, the glare from all the luminaires would be the same as for a single luminaire of the same luminance.

2. The direction of view is horizontal, straight across the room, and 1.2 m above floor level.3. The room is rectangular and has a regular array of luminaires.

The calculations and corrections depend on the tabular data being used. A list of typical limiting glare indices is given in Table 2.4.

Table 2.4 CIBSE recommended limiting glare indices for different classes of visual task

Class of visual task Examples Limiting glare index

Critical, often with a fixed direction of view

Drawing offices; very fine inspection

16

Critical, but general direction of view

Offices, libraries; computer buildings

19

Ordinary task with general direction of view

Kitchens; reception areas; fine assembly work

22

Large task or limited viewing time

Stock rooms; medium assembly work

25

No specific visual task or direction of view

Rough industrial work; indoor parking areas

28

Table 2.5 shows a typical glare data sheet for prismatic enclosure type luminaire.



Glare indicesCeiling reflectance 0.7 0.7 0.5 0.5 0.3 0.7 0.7 0.5 0.5 0.3Wall reflectance 0.5 0.3 0.5 0.3 0.3 0.5 0.3 0.5 0.3 0.3

Floor reflectance0.14 0.14 0.14 0.14 0.14 0.14 0.14 0.14 0.14

0.14

Room dimension Viewed crosswise Viewed endwiseX Y

2H

2H 7.3 8.5 8.4 9.7 11.1 6.7 8.0 7.8 9.1 10.63H 9.4 10.5 10.5 11.6 13.1 8.4 9.5 9.5 10.7 12.24H 10.4 11.4 11.5 12.6 14.1 9.1 10.2 10.3 11.4 12.96H 11.5 12.5 12.6 13.6 15.2 9.8 10.7 10.9 11.9 13.58H 12.1 13.0 13.3 14.2 15.8 10.0 10.9 11.1 12.1 13.7

12H 12.8 13.7 13.9 14.9 16.4 10.1 11.0 11.2 12.2 13.7

4H

2H 8.1 9.1 9.2 15.3 11.8 7.6 8.7 8.8 9.8 11.43H 10.5 11.4 11.6 12.6 14.1 9.6 10.5 10.8 11.7 13.24H 11.7 12.5 12.8 13.7 15.3 10.5 11.3 11.7 12.5 14.16H 13.0 13.8 14.2 15.0 16.6 11.4 12.1 12.6 13.3 14.98H 13.8 14.5 15.0 15.7 17.3 11.7 12.4 12.9 13.6 15.2

12H 14.6 15.2 15.8 16.4 18.1 11.9 12.5 13.1 13.8 15.4

8H

4H 12.3 13.0 13.5 14.2 15.8 11.3 12.0 12.5 13.2 14.86H 13.9 14.5 15.2 15.7 17.4 12.5 13.1 13.7 14.3 15.98H 14.9 15.4 16.1 16.6 18.2 13.0 13.5 14.2 14.7 16.4

12H 15.9 16.4 17.2 17.6 19.3 13.4 13.8 14.6 15.1 16.7

12H

4H 12.4 13.0 13.6 14.2 15.8 11.6 12.2 12.8 13.4 15.06H 14.1 14.7 15.4 15.9 17.5 12.9 13.4 14.1 14.6 16.28H 15.2 15.7 16.5 16.9 18.6 13.5 14.0 14.7 15.2 16.9

12H 16.1 16.5 17.4 17.8 19.4 13.7 14.1 15.0 15.4 17.0

Glare index conversion:

Luminaire length (mm) 600 1200 1500 1800 2400Conversion +3.82 +1.41 +0.63 0 -1.0

The glare data sheet is based on a number of assumptions:

1. the luminaire is 2 m above eye level;2. the lamp emits 1000 lumens;3. the spacing to height ratio is 1 : 1.

Table 3.6 gives corrections for factors (1) and (2) . No corrections are available for (3).

Initial glare indices are tabulated according to room dimensions and reflectances. Fig..5 shows the method of specifying the room dimensions. The Y dimension is always parallel to the line of sight and the X dimension is perpendicular to the line of sight. They are both expressed as multiples of the mounting height above eye level.

One view of the room will show the ends of the luminaires (endwise view) and the other view will show the sides (crosswise view).



Fig. 5 : Room dimensions for glare calculation

Table 2.6: Glare Correction data

Initial lampLumens

Correction term

Height H above1.2 m eye level

Correction tem

100 -6.0 1 -1.2150 -4.9 1.5 -0.520 -4.2

300 -3.1 2 0.0500 -1.8700 -0.9 2.5 +0.4

1 000 0.0 3 +0.71 500 +1.1 3.5 +1.02 000 +1.83 000 +2.9 4 +1.25 000 +4.2 5 +1.67 000 +5.1 6 +1.910 000 +6.015 000 +7.1 8 +2.420 000 +7.8 10 +2.830 000 +8.9 12 +3.150 000 +10.2

When the initial glare index has found it must be corrected for:1. mounting height above 1.2 m eye level if this differs from 3 m. (table 2.6)2. total downward luminous flux if this differs from 1000 lm. (table 2.6)3. extra correction sizes or lamp types. (table 2.5)

These correction terms are added to (or subtract from) the initial glare index to give the final glare index of the installation.


Y

XDirection of viewPlan of room

H

1.2 m

Eye level

Elevation


Example:Referring to the scheme designed earlier in Fig 3, what is the glare index?

Solution:The room is square, so X = Y = 6m. The ceiling height is 3m and the height of eye level for a seated man is 1.2m above the floor.

H = 3 - 1.2 = 1.8 m X = 6 m = 6 H / 1.8 = 3.3 H = Y

The reflectances are 0.7 ceiling, 0.3 wall and 0.2 floor. In this case the worst glare situation is when the luminaires are viewed crosswise. From Table 2.5, the initial glare index is found by interpolation:

X Y Y = 3.3H X = 3.3 H 2H 3H 10.5

10.8 4H 11.4

11.44H 3H 11.4

11.73 4H 12.5

Applying correction from Table 2.6 :

Initial lamp lumens = 6050 lm Correction = + 4.8

Height above eye level is 1.8 m Correction = - 0.2

There is a further glare index conversion of + 0.63 (see Table 2.5) for 1.5 m lamp.

Final glare index = 11.4 + 4.8 - 0.2 + 0.63 = 16.23

As the limiting glare index for a drawing office is 16, this installation barely meets the glare limit requirements. It does suggest that glare could be a problem, and another type of luminaire should be selected.



3. Emergency lighting

It is an increasing requirement that commercial, industrial, and public buildings are provided with some form of emergency lighting. Requirements vary for different types of buildings and even their geographical location. General guidance is given in BS 5266: 1975 The emergency lighting of premises.

The following is a summary of the design requirements. The main classifications are:

1. Emergency lighting: lighting provided for use when the normal lighting fails.

2. Escape lighting: that part of emergency lighting which is provided to ensure that the means of escape can be safely and effectively used at all material times.

3. Standby lighting: that part of emergency lighting which may be provided to enable normal activities to continue.

The basic recommendations are that, during the period of use, the horizontal illuminance measured on the centre line of any escape route should never fall below 0.2 /x and in halls and corridors it should be measured at floor level. The emergency system should be in operation within 5 seconds of the failure of the normal lighting installation.

The ratio of the minimum illuminance to the maximum illuminance along the escape route should not exceed 1: 40 and care should be taken to avoid abrupt alternations of excessive dark and light areas on the floor.

There are points in these recommendations that require some elaboration.

1. The illumination level quoted, of 0.2 lx, is roughly equivalent to a moonlit night. It is the intention of the BS that this should be an absolute minimum, i.e. with an aged lamp and battery, soiled luminaire and at the end of the period of duration. Such a situation would suggest a desirable illumination level of between one-tenth and one-fiftieth of the normal lighting level with a minimum value of 1.0 Ix.

2. The BS allowance of 5 sec. before the emergency lighting must be in operation (it can be 15 sec. for some areas at the local authority's discretion) is based upon practical tests that indicated that after 15 sec. of darkness people began to move irrationally. Fortunately many of the specialist emergency luminaires are capable of virtually instantaneous operation and the 5 sec. requirement will be redundant in such cases.

The BS goes on to deal with the problems of specific areas and provides useful comment on the factors that should be considered in the design of an emergency lighting installation. In particular, the merits of the various possible systems are dealt with.

1. Planning the Layouts:

Positioning of luminaires: The correct positioning of emergency luminaires is essential in order to provide a system that not only complies with the various legislative requirements but provides a safe and effective way of evacuating a building in the event of a mains failure. Therefore, apart from achieving the minimum light levels, various obstructions, hazards and routes must be covered.



Luminaires and signs should be positioned:

(a) To show exit routes and final exits from premises clearly. Signs should be illuminated;

(b) To ensure exterior areas of final exits are lit to at least the same level as the area immediately inside the exit to enable people to move away from the exit to areas of safety;

(c) Near each intersection of corridors (less than 2 m horizontally);

(d) Near each change of direction (less than 2 m horizontally);

(e) Near each staircase so that each flight of stairs receives direct light (less than 2 m horizontally);

(f) Near any other change of floor level which may constitute a hazard (less than 2 m horizontally);

(g) To illuminate fire alarm call points and fire fighting equipment at all material times;

(h) To ensure normal pedestrian escape routes from covered car parks are illuminated to the same standard as escape routes within buildings;

(i) In plant, switch and control rooms;

(I) Within passenger lift cars, (only self-contained emergency luminaires are suitable for this application);

(k) In toilets exceeding 8 m2.

In order to meet the illuminance levels manufacturers should provide a table of recommended spacing. A typical example is shown in Table 3.1.

Table 3.1 Spacing for a 4W fluorescent surface mounted luminaire

Ceiling Lighting Transverse Transverse Axial Axialmounting level to wall spacing spacing to wallheight directly(m) under (m) (m) (m) (m)

luminaires

2.5 1.8 lux 3.8 10.2 9.2 3.44.0 0.8 lux 4.0 11.2 10.2 3.66.0 0.4 1ux 3.2 11.2 10.2 3.0

Spacing Chart for Briklite Luminaire (dimensions in metros).

2 Undefined Escape Routes:

In an open area, such as a hall or shop, it is not possible to define a precise escape route. In this case an average illuminance of 1 Ix is required, again with a uniformity of 40: 1. This



may be achieved with separate luminaires or emergency packs in some of the normal luminaires. The lumen method is used to calculate the lamps required for 1 Ix and a spacing to height ratio of 2.5: 1 can give a guide to the minimum number of points.

Calculation:

N = (E L W) / (UFO SF ELDL K)

where:N = number of luminaires

UFO= utilization factor at zero reflectance. This is provided by the luminaire manufacturer for a given room index. When using conversion sets in standard mains luminaires the UFO is taken for the mains luminaire

L = room length (in metros)W = room width (in metros)E = average illuminance required, i.e. 1.0 lxSF = service correction factor to cover effects of lamp ageing, dust build up, etc,

assumed to be 0.8 for normal environments.ELDL= emergency lighting design lumens at nominal volts. For self-contained

luminaires the ELDL is given in the specification for each luminaire. For conversion units, multiply the design lumen output of the lamp by the ballast lumen factor.

K = a factor to cover the reduction in light output at end of discharge, or 5 seconds after mains failure, whichever is the lowest.

Values of ELDL and K depend on the type of lamp, system and luminaire. It would be misleading to include typical tables in the text. The values should be provided by the supplier of the luminaire. These factors can in some cases effectively reduce the lighting design lumens to less than 10% of the initial lamp lumens.


BTEC - HND Year 2 - Utilization of Electrical Energy Three Phase Induction Motors

Three-Phase Induction Motors

Three-phase induction motors are the motors most frequently encountered in industry. They are simple, rugged, low-priced, and easy to maintain. They run at essentially constant speed from zero to full-load. The speed is frequency-dependent and, consequently, these motors are not easily adapted to speed control. However, variable frequency electronic drives are being used more and more to control the speed of commercial induction motors.

In this chapter we cover the basic principles of the 3-phase induction motor and develop the fundamental equations describing its behavior. We then discuss its general construction and the way the windings are made.

Squirrel-cage, wound-rotor, and linear induction motors ranging from a few horsepower to several thousand horsepower permit the reader to see that they all operate on the same basic principles.

1. Principal components

A 3-phase induction motor (Fig.1) has two main parts: a stationary stator and a revolving rotor. The rotor is separated from the stator by a small air gap that ranges from 0.4 mm to 4 mm, depending on the power of the motor.

The stator consists of a steel frame that supports a hollow, cylindrical core made up of stacked laminations. A number of evenly spaced slots, punched out of the internal circumference of the laminations, provide the space for the stator winding.

The rotor is also composed of punched laminations. These are carefully stacked to create a series of rotor slots to provide space for the rotor winding. We use two- types of rotor windings: a) conventional 3-phase windings made of insulated wire, called wound-rotor induction

motors, andb) squirrel-cage windings, called cage-rotor induction motors.

Fig. 1: Exploded view of the cage motor

a) A squirrel-cage rotor is composed of bare copper bars, slightly longer than the rotor, which are pushed into the slots. The opposite ends are welded to two copper end-rings, so that all the bars are short-circuited together. The entire construction (bars and end-rings) resembles a squirrel cage, from which the name is derived. In small and medium-size motors, the bars and end-rings are made of die-cast aluminum, molded to form an integral block Fig. 2.



Fig. 2: Squirrel-cage rotor with integral cooling fan

b) A wound rotor has a 3-phase winding, similar to the one on the stator. The winding is uniformly distributed in the slots and is usually connected in wye. The terminals are connected to three slip-rings, which turn with the rotor (Fig. 3). The revolving slip-rings and associated stationary brushes enable us to connect external resistors in series with the rotor winding. The external resistors are mainly used during the start-up period; under normal running conditions, the three brushes are short-circuited.

Fig. 3: Exploded view of wound-rotor induction motor



2. Principle of operation

The operation of a 3-phase induction motor is based upon the application of Faraday's Law and the Lorentz force on a conductor early studied. To understand the principle of operation, the rotating magnetic field and it's direction of rotation should be in consideration.

2.1 The rotating field:Consider a simple stator having 6 salient poles, each of which carries a coil having 2 turns (Fig. 4). Coils that are diametrically opposite are connected in series by means of three jumpers that respectively connect terminals a-a, b-b, and c-c. This creates three identical sets of windings AN, BN, CN, that are mechanically spaced at 120° to each other. The two coils in each winding produce magnetomotive forces that act in the same direction.The three sets of windings are connected in wye (Star), thus forming a common neutral N. Owing to the perfectly symmetrical arrangement, the line to neutral impedances are identical. In other words, as regards terminals A, B. C, the windings constitute a balanced 3-phase system.If we connect a 3-phase source to terminals A, B. C, alternating currents la, Ib , and Ic will flow in the windings. The currents will have the same value but will be displaced in time by an angle of 120°. These currents produce magnetomotive forces which, in turn, create a magnetic flux. It is this flux we are interested in.

Fig. 4: Elementary stator having terminals A, B, C connected to a 3-phase source

As time goes by, we can determine the instantaneous value and direction of the current in each winding and thereby establish the successive flux patterns. Thus, referring to Fig. 5 for each of the successive instants 1. 2, 3, 4, 5, 6, and 7, separated by intervals of 1/6 cycle, we find that the magnetic field makes one complete turn during one cycle.



Fig. 5: Instantaneous value of current and position of the rotating flux

The rotational speed of the field depends, therefore, upon the duration of one cycle, which in turn depends on the frequency of the source. If the frequency is 60 Hz, the resulting field makes one turn in 1/60 s, that is, 3600 revolutions per minute. On the other hand, if the frequency were 5 Hz, the field would make one turn in 1/5 s, giving a speed of only 300 r/min. Because the speed of the rotating field is necessarily synchronized with the frequency of the source, it is called synchronous speed.

2.2 Direction of rotationThe positive crests of the currents in (Fig. 5), follow each other in the order A-B-C. This phase sequence produces a field that rotates clockwise. If we interchange any two of the lines connected to the stator, the new phase sequence will be A-C-B. By following the same line of reasoning explained before, we find that the field now revolves at synchronous speed in the opposite, or counterclockwise direction. Interchanging any two lines of a 3-phase motor will, therefore, reverse its direction of rotation.

3. Synchronous speed

Soon after the invention of the induction motor, it was found that the speed of the revolving flux could be reduced by increasing the number of poles.

The speed of a rotating field depends therefore upon the frequency of the source and the number of poles on the stator. Using the same reasoning as above, we can prove that the synchronous speed is always given by:

Ns = (120 f) / pwhere:

Ns = synchronous speed (r/min)f = frequency of the source (HZ)p = number of poles per phase.



4. Starting characteristics of a squirrel-cage motor

Let us connect the stator of an induction motor to a 3-phase source, with the rotor locked. The revolving field created by the stator cuts a the rotor bars and induces an a.c. voltage in all of them.

The frequency of the voltage depends upon the number of N and S poles that sweep across a conductor per second; when the rotor is at rest, it is always equal to the frequency of the source. Since the rotor bars are short-circuited by the end-rings, the induced voltage causes a large current to flow usually several hundred amperes per bar in machines of medium power. The current-carrying conductors are in the path of the flux created by the stator, consequently, they all experience a strong mechanical force. These forces tend to drag the rotor along with the revolving field. In summary:

1. A revolving magnetic field is set up when a 3-phase voltage is applied to the stator of an induction motor.

2. The revolving field induces a voltage in the rotor bars.3. The induced voltage creates large circulating current which flow in the rotor bars and end-rings.4. The current-carrying rotor bars are immersed in the magnetic field created by the stator; they are

therefore subjected to a strong mechanical force.5. The sum of the mechanical forces on all the rotor bars produces a torque which tends to drag the

rotor along in the same direction as revolving field.

5. Motor under load

Suppose the motor is initially running at no-load. If we apply a mechanical load to the shaft, the motor will begin to slow down and the revolving field will cut the rotor bars at a higher and higher rate. The induced voltage and the resulting current in the bars will increase progressively, producing a greater and greater motor torque. The motor and the mechanical load will reach a state of equilibrium when the motor torque is exactly equal to the load torque. When this state is reached, the speed will cease to drop and the motor will turn at a constant rate. It is very important to understand that a motor only turns at constant speed when its torque is exactly equal to the torque exerted by the mechanical load. The moment this state of equilibrium is upset, the motor speed will start to change.

Under normal loads, induction motors run very close to synchronous speed. Thus, at full-load, the slip for large motors (1000 kW and more) rarely exceeds 0.5% of synchronous speed, and for small machines (10 kW and less), it seldom exceeds 3%. That is why induction motors are considered to be constant speed machines. However, because they never actually turn at synchronous speed, they are sometimes called asynchronous machines.

6. Slip

The slip s of an induction motor is the difference between the synchronous speed and the rotor speed, expressed as a percent (or per-unit) of synchronous speed. The per-unit slip is given by the equation

S = (Ns - N) / Ns where:

S = SlipNs = Synchronous speed (r/min)N = Rotor speed (r/min)

The slip is practically zero at no-load and is equal to 1 (or 100%) when the rotor is locked.



7. Voltage and frequency induced in the rotor

The voltage and frequency induced in the rotor both depend upon the slip. They are given by the following equations:

fr = s fEr = s EO (approx.)

where:fr = frequency of the voltage and current in the rotor [Hz]f = frequency of the source connected to the stator [Hz]s = slip

Er = voltage induced in the rotor at slip s Eo = open-circuit voltage induced in the rotor when at rest [V]

In a cage motor, the open-circuit voltage Eo is the voltage that would be induced in the rotor bars if the bars were disconnected from the end-rings. In the case of a wound-rotor motor the open-circuit voltage is 1/3 times the voltage between the open circuit slip-rings. It should be noted that fr (in the Equ. Above) always holds true, but Er (in the Equ. above) is valid only if the revolving flux (expressed in webers) remains absolutely constant. However, between zero and full-load the actual value of Er is only slightly less than the value given by the equation.

Classwork: A 6-pole wound-rotor induction motor is excited by a 3-phase 60 Hz source. If the full-load speed is 1140 r/min calculate the frequency of the rotor current under the following conditions:a. At standstillb. Motor turning at 500 r/min in the same direction as the revolving field.c. Motor turning at 500 r/min in the opposite direction to the revolving field d. Motor turning at 2000 r/min in the same direction as the revolving field

8. Characteristics of Squirrel cage Induction Motors

Table 1 lists the typical properties of squirrel-cage induction motors in the power range between 1 kW and 20 000 kW. Note that the current and torque are expressed in per-unit values.

Table 1 Typical Characteristics of cage-rotor induction motors

Loading I (per unit) T (per unit) S (per unit) P.F.

Motor Size Small Big Small Big Small Big Small Big Small BigFuul-load 1 1 1 1 0.03 0.004 0.7 0.96 0.8 0.87

to to to to0.9 0.98 0.85 0.9

No-load 0.5 0.3 0 0 ≈ 0 ≈ 0 0 0 0.2 0.05

Locked rotor 5to6

4to6

1.5to3

0.5to1

1 1 0 0 0.4 0.1

Small means under 11 KW (15 hp); Big means over 1000 KW (~ 1350 hp)

The following explanations will clarify the meaning of the values given in the table.

8.1 Motor at no-load.



When the motor runs at no load, the stator current lies between 0.5 and 0.3 pu (of full-load current). The no-load current is similar to the exciting current in a transformer. Thus, it is composed of a magnetizing component that creates the revolving flux m and a small active component that supplies the windage and friction losses in the rotor plus the iron losses in the stator. The flux m links both the stator and the rotor; consequently it is similar to the mutual flux in a transformer (Fig. 6).

Considerable reactive power is needed to create the revolving field and, in order to keep it within acceptable limits, the air gap is made as short as mechanical tolerances will permit. The power factor at no-load is therefore low; it ranges from 0.2 (or 20%) for small machines to 0.05 for large machines. The efficiency is zero because the output power is zero.

Fig. 6: Flux m at No-load

8.2 Motor under load. When the motor is under load, the current in the rotor produces a mmf which tends to change the mutual flux m.. This sets up an opposing current flow in the stator. The opposing mmfs of the rotor and stator are very similar to the opposing mmfs of the secondary and primary in a transformer. As a result, leakage fluxes f1 , and f2 are created, in addition to the mutual flux m (Fig. 7).

The total reactive power needed to produce these three fluxes is slightly greater than when the motor is operating at no-load. However, the active power (kW) absorbed by the motor increases in almost direct proportion to the mechanical load. It follows that the power factor of the motor improves dramatically as the mechanical load increases. At full-load it ranges from 0.80 for small machines to 0.90 for large machines. The efficiency at full-load is particularly high; it can attain 98% for very large machines.

Fig. 7: Fluxes at full-load8.3 Locked-rotor characteristics. The locked-rotor current is 5 to 6 times the full-load current, making the I2R losses 25 to 36 times higher than normal. The rotor must therefore never remain locked for more than a few seconds.



Although the mechanical power at standstill is zero, the motor develops a strong torque. The power factor is low because considerable reactive power is needed to produce the leakage flux in the rotor and stator windings. These leakage fluxes are much larger than in a transformer because the stator and the rotor windings are not as tightly coupled.

9. Active power flow

Voltages, currents, and phasor diagrams enable us to understand the detailed behavior of an induction motor. However, it is easier to see how electrical energy is converted into mechanical energy by following the active power as it flows through the machine. Thus, referring to Fig. 8, active power Pe

flows from the line into the 3-phase stator. Due to the stator copper losses, a portion Pjs is dissipated as heat in the windings. Another portion Pf is dissipated as heat in the stator core, owing to the iron losses. The remaining active power Pr is carried across the air gap and transferred to the rotor by electromagnetic induction.

Due to the I2R losses in the rotor, a third portion Pjr is dissipated as heat, and the remainder is finally available in the form of mechanical power Pm. By subtracting a small fourth portion Pv, representing windage and bearing-friction losses, we finally obtain PL, the mechanical power available at the shaft to drive the load.

Fig. 8: Active power flow in a 3-phase induction motor.

The power flow diagram of Fig. 8 enables us to identify and to calculate three important properties of the induction motor: (1) its efficiency, (2) its rotor copper losses, (3) its power, and (4) its torque.

9.1 Efficiency. By definition, the efficiency of a motor is the ratio of the output power to the input power:

Efficiency () = PL / Pe

9.2 I2 R losses in the rotor. It can be shown that the rotor I2 R losses Pjr are related to the rotor input power Pr by the equation

Pjr = s Pr

where:Pjr = rotor I2 R losses [W]s = slipPr = power transmitted to the rotor [W]

The equation above shows that as the slip increases, the rotor I2 R losses consume a larger and larger proportion of the power Pr transmitted across the air gap to the rotor. A rotor turning at half synchronous speed (s = 0.5) dissipates in the form of heat 50 percent of the active power it receives. When the rotor is locked (s = I ), all the power transmitted to the rotor is dissipated as heat.

9.3 Mechanical power.



The mechanical power Pm developed by the motor is equal to the power transmitted to the rotor minus its I2 R losses. Thus, Pm = Pr - Pjr

= Pr - s Pr whence Pm = (1 - s) Pr The actual mechanical power available to drive the load is slightly less than Pm, due to the power needed to overcome the windage and friction losses. In most calculations we can neglect this small loss.

9.4 Motor torque. The torque Tm developed by the motor at any speed is given by

Tm = 9.55 Pm / N = [9.55(1 - s)Pr] / Ns(1 - s) = 9.55 Pr / Ns

therefore, Tm = 9.55 Pr / Ns

where:Tm = torque developed by the motor at any speed [N.m] Pr = power transmitted to the rotor [W]Ns = synchronous speed [r/min]9.55 = multiplier to take care of units [exact value: 60/2]

The actual torque TL available at the shaft is slightly less than Tm, due to the torque required to overcome the windage and friction losses. However, in most calculations we can neglect this small difference.The torque equations show that the torque is directly proportional to the active power transmitted to the rotor. Thus, to develop a high locked-rotor torque, the rotor must absorb a large amount of active power. The latter is dissipated in the form of heat, consequently, the temperature of the rotor rises very rapidly.

Classwork: 1A 3-phase induction motor having a synchronous speed of 1200 r/min draws 80 kW from a 3-phase feeder. The copper losses and iron losses in the stator amount to 5 kW. If the motor runs at 1152 r/min, calculate the following:

a. The active power transmitted to the rotorb. The rotor I2R losses c. The mechanical power developed d. The mechanical power delivered to the load, knowing that the windage and friction losses are

equal to 2 kWe. The efficiency of the motor.

Classwork:2



A 3-phase induction motor having a nominal rating of 100 hp (~75 kW) and a synchronous speed of 1800 r/min is connected to a 600 V source

The two-wattmeter method shows a total power consumption of 70 kW, and an ammeter indicates a line current of 78 A. Precise measurements give a rotor speed of 1763 r/min. In addition, the following characteristics are known about the motor:stator iron losses Pf = 2 kW windage and friction losses Pv = 1.2 kW resistance between two stator terminals = 0.34 Calculate:a. Power supplied to the rotorb. Rotor I2R lossesc. Mechanical power supplied to the load, in horsepowerd. Efficiencye. Torque developed at 1763 r/min

10. Torque versus speed curve

The torque developed by a motor depends upon its speed, but the relationship between the two cannot be expressed by a simple equation. Consequently, we prefer to show the relationship in the form of a curve. Fig. 9 shows the torque-speed curve of a conventional 3-phase induction motor whose nominal full-load torque is T. The starting torque is 1.5 T and the maximum torque (called breakdown torque) is 2.5 T.

At full-load the motor runs at a speed n. If the mechanical load increases slightly, the speed will drop until the motor torque is again equal to the load torque. As soon as the two torques are in balance, the motor will turn at a constant but slightly lower speed. However, if the load torque exceeds 2.5 T (the breakdown torque), the motor will quickly stop.Small motors (10 kW and less) develop Their breakdown torque at a speed rid of about 80% of synchronous speed. Big motors (1000 kW and more) attain their breakdown torque at about 98% of synchronous speed.The rotor resistance of a squirrel-cage rotor is essentially constant, except chat it increases with temperature. Thus, the resistance increases with increasing load because the temperature rises.In designing a squirrel-cage motor, the rotor resistance can be set over a wide range by using copper, aluminum, or ocher metals in the rotor bars and end-rings. The torque-speed curve is greatly affected by such a change in resistance. The only characteristic that remains unchanged is the breakdown torque. The following example illustrates the changes chat occur.

Fig. 9: Typical torque-speed curve of a 3-phase cage-rotor motor

11. Wound-rotor motor



We explained the basic difference between a squirrel-cage motor and a wound-rotor motor previously. Although a wound-rotor motor costs more than a squirrel-cage motor, it offers the following advantages:1. The locked-rotor current can be drastically reduced by inserting three external resistors in series

with the rotor. Nevertheless, the locked -rotor torque will still be as high as of a squirrel-cage motor.

2. The speed can be varied by varying the external rotor resistors.3. The motor is ideally suited to accelerate high-inertia loads, which require a long time to bring up to

speed.

Fig. 10 is a diagram of the circuit used to start a wound-rotor motor. The rotor windings are connected to three Y-connected external resistors by means of a set of slip-rings and brushes. Under locked-rotor (LR) conditions, The variable resistors are set to Their highest value. As the motor speeds up, the resistance is gradually reduced until full-load speed is reached, whereupon the brushes are short-circuited. By properly selecting the resistance values, we can produce a high accelerating torque with a stator current chat never exceeds twice full-load current.

Fig. 10: External resistors connected to the three slip-rings of a wound-rotor induction motor.

We can also regulate the speed of a wound-rotor motor by varying the resistance of the rheostat. As we increase the resistance, the speed will drop. This method of speed control has the disadvantage that a lot of heat is dissipated in the resistors; the efficiency is therefore low. Furthermore, for a given rheostat setting, the speed varies considerably if the mechanical load varies.

12. Equivalent Circuit of the Induction Motor

Studying of the principle of operation and characteristics of the three-phase have shown that we can describe the important properties of squirrel-cage and wound-rotor induction motors without using a circuit diagram. However, if we want to gain even a better understanding of the properties of the motor, an equivalent circuit diagram is absolutely essentially. In this part we develop the equivalent circuit from basic principles. We then analyze the characteristics of a low-power and high-power motor and observe their basic differences.Finally, we develop the equivalent circuit of an asynchronous generator and determine its proper ties under load.

12.1 The wound-rotor Induction Motor



A 3-phase wound-rotor induction motor is very similar in construction to a 3-phase transformer. Thus, the motor has 3 identical primary windings and 3 identical secondary windings one set for each phase. On account of the perfect symmetry, we can consider a single primary winding and a single secondary winding in analyzing the behavior of the motor.

When the motor is at standstill, it acts exactly like a conventional transformer, and so its equivalent circuit (Fig. 14) is the same as a transformer equivalent circuit.

We assume a wye connection for the stator and the rotor, and a turns ratio of 1:1. The circuit para-meters, per phase, are identified as follows:

Eg = source voltage, line to neutralr1 = stator winding resistancer2 = rotor winding resistancex1 = stator leakage reactancex2 = rotor leakage reactanceRx = external resistance, effectively connected between one slip-ring and the neutral of the

rotorXm = magnetizing reactanceRm = resistance corresponding to the iron losses and windage and friction lossesT = ideal transformer having a turns ratio of 1:1

Fig. 14:

Equivalent circuit of a wound-rotor induction motor at standstill

In the case of a conventional 3-phase transformer, we would be justified in removing the magnetizing branch composed of jXm and Rm because the exciting current Io is negligible compared to the load current Ip. However, in a motor this is no longer true: Io may be as high as 40 percent of Ip because of the air gap. Consequently, we cannot eliminate the magnetizing branch. However, for motors exceeding 2 hp. we can shift it to the input terminals, as shown in Fig. 15.

Fig. 15: Approximation of the equivalent circuit is acceptable for motors above 5 hp.


Ip

r1 jX1

jX2 r2

Io

RmjXm

I1

RXEg

3

42

1

T

1:1

I2

E2=E1

Ipr1 jX1 jX2 r2

Rm

I1

RXEg

3

42

1 I2

jXm

E1

T

1:1


Suppose the motor runs at a slip s, meaning that the rotor speed is NS (1 - s), where NS is the syn-chronous speed. This will modify the values of E1, I1 and E2, I2 on the primary and secondary side of the ideal transformer T. Furthermore, the frequency in the secondary winding will become sf where f is the frequency of the source Eg. Fig. 16 shows these new operating conditions.

Directing our attention to the secondary side, the amplitude of the induced voltage E2 would be equal to E1 (the turns ratio is 1:1) if the motor were stationary. But because the slip is s, the actual voltage induced is

E2 = s E1

The frequency is sf and this changes the impedance of the secondary leakage reactance from jx2 to jsX2. Because resistors are not frequency-sensitive, the values of r2 and Rx remain the same. Let us lump the two together to form a single secondary resistance R2, given by;

R2 = r2 + Rx

Fig. 16:


E2=E1

Ip

r1

Stator

resistance

jX1 jsX2 r2

Rotor

resistance

Rm

I1

RX External

resistance

Eg

3

42

1 I2

jXm

E1

T

N1=N2

Frequency = f

Frequency = sf


Equivalent circuit of a wound-rotor motor when it is running at a slip s. The frequency of the voltages and currents in the stator is f . But the frequency of the voltages and current in the rotor is sf

The details of the secondary circuit are shown in Fig. 4-a, and the resulting current I2 is:

I2 = sE1 / (R2 + jsx2 ) = sE1- / [ (R2 )2 + (sX2)2]where:

= arctan sX2 / R2

The corresponding phasor diagram is shown in Fig. 17-b.

(a) (b)

Fig. 17- a) Equivalent circuit of the rotor; E2 and I2 have a frequency sf

b) Phasor diagram showing the current lagging behind the voltage by angle .

It is important to realize that this phasor diagram relates to the frequency sf. Consequently, it cannot be integrated into the phasor diagram on the primary side, where the frequency is f. Nevertheless, there is a direct relationship between I2 (frequency sf) in the rotor and I1 (frequency f) in the stator. In effect, the absolute value of I1 is exactly the same as that of I2 . Furthermore, the phase angle , between E1 and I1 is exactly the same as that between E2 and I2. This enables us to draw the phasor diagram for E1 and I1 as shown in Fig. 18.

Fig. 18: The voltage and current in the stator are separated by the same phase angle even though the frequency different.

To summarize:1. The effective value of I1 is equal to the effective value of I2, even though their frequencies are

different.

2. The effective value of E1 is equal to the effective value of E2 divided by the slip s.

3. The phase angle between E1 and I1 is the same as that between E2 and I2.

Thus, on the primary side we can write:


sE1

I2

frequency = sf

E1

I1

frequency = f

I1 absolute = sE1/ (R2)2 + (sx2)2

= arctan sx2/R2

E2=sE1R2 Total resistance

of rotor circuit

jsX23

4

I2

Frequency = sf


I1 = I2 = sE1 / (R2 + jsx2 ) = E1 / [(R2 /s) + jx2] = E1 / Z2

The impedance Z2 seen between the primary terminals 1, 2 of the ideal transformer is, therefore,Z2 = E1 / I1 = (R2 / s) + jx2

As a result, we can simplify the circuit of Fig. 16 to that shown in Fig. 19. The leakage reactances jx1, jx2 can now be lumped together to create a single total leakage reactance jx. It is equal to the total leakage reactance of the motor referred to the stator side.

Fig. 19: Equivalent circuit of a wound-rotor motor referred

to the primary (stator) side.

The final equivalent circuit of the wound-rotor induction motor is shown in Fig. 20. In this diagram, the circuit elements are fixed, except for the resistance R2 /s. Its value depends upon the slip and hence upon the speed of the motor. Thus, the value of R2 /s will vary from R2 to infinity as the motor goes from start-up (s = 1) to synchronous speed (s = 0).

This equivalent circuit of a wound-rotor induction motor is so similar to that of a transformer that it is not surprising that the wound-rotor induction motor is sometimes called a rotary transformer.The equivalent circuit of a squirrel-cage induction motor is the same, except that R2 is then equal to the equivalent resistance r2 of the rotor alone referred to the stator, there being no external resistor.

Fig. 20: The primary and secondary leakage reactances x1 and x2 are combined to form an equivalent total leakage reactance x.

12.2 Power relationshipsThe equivalent circuit enables us to arrive at some basic electromechanical power relationships for the 3-phase induction motor. The following equations can be deduced by visual inspection of the equiva-lent circuit of the wound-rotor motor (Fig. 20):

1. Active power absorbed by the motor is: P = [(E g) 2 / Rm ] + [(I1)2 r1 ] + [(I1)2 (R2 /s)]2. Reactive power absorbed by the motor is: Q = [(Eg)2 / xm ] + [(I1)2 x]3. Apparent power absorbed by the motor is: S = P2 + Q2

4. Power factor of the motor is: cos = P / S5. Line current is: Ip = S / Eg 6. Active power supplied to the rotor is: Pr = (I1)2 (R2 / s)7. Power dissipated as I2 R losses in the rotor circuit is: Pjr = (I1)2 R2 = sPr


Ipr1 jX1

Rm

I1

Eg

2

1

jXm

Io

jX2

R2/s

Ipr1 jX

Rm

I1

Eg jXm

Io

R2/s

Im

If

Pr


8. Mechanical power developed by the motor is: Pm = Pr - Pjr = Pr (1 - s)9. Torque developed by the motor is:

T = 9.55 Pm / N = [9.55 Pr (1 -s)] / Ns ( 1 - s) = 9.55 Pr / Ns

10. Efficiency of the motor is: = Pm / P

13. Starting Method For Cage-Rotor Machines : 13.1 Direct On-line Starting :Induction motors when direct-switched by this method of starting, the three-phase supply is supplied at full voltage to the stator windings employing a circuit breaker or contactor.

The motors starting current is large being typically between five and seven times the normal full load value. Such large currents can cause problems of volt drop in supply lines and affect other electrical equipment connected to the same areas of the system .

Direct on-line starting method is acceptable for example a small motor of rating to 20 KW in a factory which is fed from its own, probably large, supply transformer.

When the motor is direct-switched onto normal voltage, then the slip is 100%, the rotor induced voltage is maximum voltage, the starting current is the short-circuit current Isc and the starting torque Tst is about 1.5 to 2.5 times their full-load torque Tf. Therefore:

Tst / Tf = (Ist / If)2 S = a2 SWhere:

S = Full-load motor slip

a = (Ist/If)Suppose in a case, Ist = 6.If , S = 5% then:Tst / Tf = 62 0.05 = 1.8

Therefore Starting torque Tst = 1.8 full-load currentHence, we find that with a current as great as six times the full current, the motor develops a staring torque which in only 1.8 times the full-load value.

13.2 Star- Delta Starting :

The motor windings must be such that all six coil ends are available for connection.The motor is started with the windings in a star connection effectively connecting two coils in series between the supply line voltages. This results in a reduced voltage across each winding of VL/3. Since the torque proportional to the square of the voltage, the torque developed becomes 1/3 of that which would have been developed if motor were directly connected in delta and consequently a reduced current to 1/3 . Tst / Tf = (Ist/ If)2 S = (Ist/ 3)2 S = 1/3 (Ist/If) S = 1/3 a2 S

This method is cheap and effective provided the starting torque is required not to be more than 1.5 times the full-load torque.

Fig. 27 shows, at the beginning the main contactor and the star contactor are closed which provides a star point.When the rotor speed is approximately full speed, star contactor will open and delta contactor will close which joins stator windings, so creating the delta connection.



Class work:

A 3-phase,6-pole,50HZ induction motor takes 60A at full-load speed of 960 rpm, and develops a torque of 15 N.m. The staring current at rated voltage is 300A. What is the starting torque?. If a star/delta starter is used, determine the starting torque and starting current.

13.3 Auto-transformer Starting :

The supply to the motor is applied via a star-connected auto-transformer which a voltage tapping.


Fig. 27: Power Circuit Diagram

Z YX

WVU

3-Motor

L1

L2

L3

Fuses

Contactors

Overload Relay

Main Delta Star


On starting, the supply is taken from the tap connecting 50, 65 or 80 percent of the supply voltage to the motor windings thus reducing the staring current.When the rotor speed approaches to full speed, the full supply voltage is connected to the motor (See Fig. 28 )

With this method of starting, the motor remains connected delta at all times while the reduction in voltage is achieved by the use of the transformer. The power input to the motor is reduced in the same proportion as is the starting torque.

This method is therefore only suitable for off load starting of induction motors

(a)

(b)

Fig. 28: a) the case when the motor is DOL b) the case when the motor is connected to auto-transformer

13.4 static voltage controllerThe speed of a 3-phase squirrel-cage induction motor can be varied by simply varying the stator voltage. This method of speed control is particularly useful for a motor driving a blower


Stator

Ist = 5If

V

Auto-Transformer

I1

V

Stator

Ist = 5If

V


or centrifugal pump. At rated voltage, the torque-speed characteristic of the motor is given by curve 1 of Fig. 29. If we apply half the rated voltage, we obtain curve 2. Because torque is proportional to the square of the applied voltage, the torques in curve 2 are only 1/4 of the corresponding torques in curve 1. For example, the breakdown torque drops from 184% to 46%. Similarly, the torque at 60 percent speed drops from 175% to 43.75%.

Fig. 29: Torque speed carve1) at rated voltage

2) at 50 % of rated voltage

The load torque varies nearly as the square of the speed. This typical characteristic, shown by curve 3, is superimposed on the motor torque-speed curves. Thus, at rated voltage, the intersection of curves 1 and 3 shows that the motor runs at 90 percent of synchronous speed. On the other hand, at half rated voltage, the blower rotates at only 60 percent of synchronous speed. By varying the voltage this way, we can control the speed.The variable-voltage autotransformer shown in Fig. 28 can be replaced by three sets of thyristors connected back-to-back, as shown in Fig. 30. The sets are called valves. To produce rated voltage across the motor, the respective thyristors are fired with a delay equal to the phase angle lag that would exist if the motor were directly connected to the line.

Fig. 30: Variable-voltage speed control of cage-rotor Induction motor

Fig. 31 shows the resulting current and line-to-neutral voltage for phase A. The valves in phases B and C are triggered the same way, except for an additional delay of 120 and 240, respectively.



Fig. 31:Wave shapes at rated voltage. Fig. 32:Wave shapes at 50% rated voltage

To reduce the voltage across the motor, the firing angle is delayed still more. For example, to obtain 50 percent rated voltage, all the pulses are delayed by about 100. The resulting distorted voltage and current wave shape for phase A are pictured very approximately in Fig. 32.

The distortion increases the losses in the motor compared to the autotransformer method. Furthermore, the power factor is considerably lower because of the large phase angle lag . Nevertheless, to a first approximation, the torque-speed characteristics shown in Fig. 29 still apply.

Due to the considerable I2R losses and lower power factor, this type of electronic speed control is only feasible for motors rated below 20 hp. Small hoists are also suited to this type of control, because they operate intermittently. Consequently, they can cool off during the idle and light-load periods.

13.5 Self-commutated inverters for cage motors

The cage-rotor induction motor can be driven using a self-commutated inverter (also called force commutated inverter). It operates quite differently from a line-commutated inverter.

First, it can generate its own frequency, determined by the frequency of the pulses applied to the gates.

Second, it can either absorb or deliver reactive power. The reactive power generated or absorbed depends upon the nature of the load and the switching action of the power semiconductors. They may be IGBTs, power MOSFETs, GTOs, or ordinary thyristors.

In the latter case, the thyristors are arranged in a conventional 3-phase bridge circuit. However, each thyristor is surrounded by an array of capacitors, inductors, diodes, and auxiliary thyristors. The purpose of these auxiliary components is to force some power thyristors to conduct when normally they would not, and to force other thyristors to stop con-ducting before their “natural” time. It is precisely this forced switching action that enables these converters to generate and absorb reactive power.

Because of the variety of the switching circuits used, we show the self-commutated inverter as a simple 5-terminal device having two dc input terminals and three ac output terminals to provide 3-phase power to the motor. The basic type of inverters: current-source inverters is shown in Fig. 33.



Fig. 33: Current-fed frequency converter.



14. Tutorial Questions and Problems

14.1 Practical level:1. Name the principal components of an induction motor.

2. Explain how a revolving field is set up in a 3-phase induction motor.

3. If we double the number of poles on the stator of an induction motor, will its synchronous speed also double?

4. The rotor of an induction should never be locked while full voltage is being applied to the stator. Explain.

5. Why does the rotor of an induction motor turn slower than the revolving field?

6. What happens to the rotor speed and rotor current when the mechanical load on an induction motor increases?

7. Would you recommend using a 50 hp induction motor to drive a 10 hp load? Explain.

8. Give two advantages of a wound-rotor motor over a squirrel-cage motor.

9. Both the voltage and frequency induced in the rotor of an induction motor decrease as the rotor speeds up. Explain.

10. A 3-phase, 20-pole induction motor is connected to a 600 V, 60 Hz source.a. What is the synchronous speed? [360 rpm]h If the voltage is reduced to 300 V, will the synchronous speed change?c. How many groups are there, per phase? [20]

11. Make a drawing of the magnetic field created by a 3-phase, 12-pole induction motor.

12. How can we change the direction of rotation of a 3-phase induction motor?

14.2 Intermediate level:13-a. Calculate the synchronous speed of a 3-phase, 12-pole induction motor that is excited by a 60 Hz

source. [600 rpm] b. What is the nominal speed if the slip at full load is 6 percent? [564 rpm]

14. A 3-phase 6-pole induction motor is connected to a 60 Hz supply. The voltage induced in the rotor bars is 4 V when the rotor is locked. If the motor turns in the same direction as the flux, calculate the approximate voltage induced and its frequency: a. At 300 r/min [3V, 45 Hz]b. At 1000 r/min [0.67V, 10 Hz]c. At 1500 r/min [1V, 15 Hz]

15. A 3-phase, 75 hp. 440 V induction motor has a full-load efficiency of 91 percent and a power factor of 83 percent. Calculate the nominal current per phase. [97.2A]

16. An open-circuit voltage of 240 V appears across the slip-rings of a wound-rotor induction motor when the rotor is locked. The stator has 6 poles and is excited by a 60 Hz source. If the rotor is driven by a variable-speed dc motor, calculate the open-circuit voltage and frequency across the slip-rings if the dc motor turns



a. At 600 r/min, in the same direction as the rotating field [120V,30Hz]b. At 900 r/min, in the same direction as the rotating field [60V,15Hz]c. At 3600 r/min, opposite to the rotating field [960V, 240Hz]

17. A large 3-phase, 4000 V, 60 Hz squirrel-cage induction motor draws a current of 385 A and a total active power of 2344 kW when operating at full-load. The corresponding speed is accurately measured and is found to be 709.2 r/min. The stator is connected in wye and the resistance between two stator terminals is 0.10 . The total iron losses are 23.4 kW and the windage and friction losses are 12 kW.

Calculate the following:a. The power factor at full-load [0.879]b. The active power supplied to the rotor [2298Kw]e. The total I2R losses in the rotor [24.5Kw]d. The load mechanical power [kW], torque [kN m], and efficiency [%]

[2251Kw, 30.3KNm, 96%]

18. If we slightly increase the rotor resistance of an induction motor, what effect does this have (increase or decrease) upon a. Starting torque b. Starting current c. Full- load speed d. Efficiencye. Power factorf. Temperature rise of the motor at its rated power output

14.3 Advanced level:19. A 3-phase, 5000 hp. 6000 V, 60 Hz 12 poles wound-rotor induction motor turns at 594 r/min.

The motor has the following characteristics:a) dc resistance between stator terminals = 0.112 at 17°C

2. dc resistance between rotor slip-rings = 0.0073 3. open-circuit voltage induced between slip rings with rotor locked = 1600 V4. Iine – to- line stator voltage = 6000 V5. no-load stator current, per phase = 100 A6. active power supplied to motor at no-load = 91 kW7. windage and friction losses = 51 kW8. iron losses in the stator = 39 kW9. Locked-rotor current at 6000 V = 1800 A

10. active power to stator with rotor locked = 2207 kW

Calculate:a) Rotor and stator resistance per phase at 75°C (assume a wye connection)

[4.49m, 66.9m]b) Voltage and frequency induced in the rotor when it turns at 200 r/min and at

594 r/min [1067V, 40Hz, 16V, 0.6Hz]c. Reactive power absorbed by the motor to create the revolving field, at no-load

[1035 KVAr]d. I2R losses in the stator when the motor runs at no-load (winding temperature 75°C)

[2.07 Kw]e. Active power supplied to the rotor at no-load [50Kw]

20. Referring to the motor described in Problem (19) above, calculate under full voltage LR (locked-rotor) conditions:

a. Reactive power absorbed by the motor [18.6 MVAr]b. I2R losses in the stator [670 Kw]c. Active power supplied to the rotor [1498 Kw]



d. Mechanical power output [ zero ]e. Torque developed by the rotor [23.8 KNm]

21. We wish to control the speed of the motor given in Problem (19) by inserting resistance in series with the rotor (see Fig. 10). If the motor has to develop a torque of 20 kN.m at a speed of 450 r/min, calculate the following:

a. Voltage between the slip rings [400 V]b. Rotor resistance (per phase) and the total power dissipated [508 m, 314 Kw]c. Approximate rotor current, per phase [455 A]

23. A 3-phase, 300 kW, 2300 V, 60 Hz,1780 r/min induction motor is used to drive a compressor. The motor has a full-load efficiency and power factor of 92% and 86%, respectively. If the terminal voltage rises to 2760 V while the motor operates at full-load, determine the effect (increase or decrease) upon

a. Mechanical power delivered by the motorb. Motor torquec. Rotational speedd. Full-load currente. Power factor and efficiencyf. Starting torqueg. Starting currenth. Breakdown torquei. Motor temperature risej. Flux per polek. Exciting currentl. Iron losses


BTEC - HND Year 2 - Utilization of Electrical Energy Energy Management and tariffs


BTEC - HND Year 2 - Utilization of Electrical Energy Experiments

Experiment 1Transformer Polarity

OBJECTIVES• To determine the polarity of transformer windings.

• To learn how to connect transformer windings in series aiding.

• To learn how to connect transformer windings in series opposing.

DISCUSSION

When the primary winding of a transformer is energized by an ac source, an alternating magnetic flux is established in the transformer core. This alternating flux links the turn of each winding on the transformer, thereby inducing ac voltages in them. Consider the circuit shown in Figure 1.

Figure 1

By definition an ac voltage is continually changing its magnitude and its polarity, therefore, the voltage across the primary winding (terminals 1 and 2) keeps changing the polarity of terminal 1 with respect to terminal 2. Terminals 1 and 2 can never have the same polarity. Terminal 1 must always be positive or negative with respect to terminal 2. Thus, the alternating magnetic flux induces voltages in all of the other windings causing an ac voltage to be produced across each pair of terminals. The terminals of each winding also change polarity with respect to each other.

When we speak of "the polarity" of transformer windings we are identifying all of the terminals that are the same polarity (positive or negative) at any instant of time. Polarity marks are employed to identify these terminals. These marks may be black dots, crosses, numerals, letters or any other convenient means of showing which terminals are of the same polarity. For example, in Figure 1, we have used black dots. These black dots, "polarity marks", indicate that for a given instant in time,

1 is positive with respect to 2,3 is positive with respect to 4,6 is positive with respect to 5,7 is positive with respect to 8,10 is positive with respect to 9.It should be noted that a terminal cannot be positive by itself. It can only be positive with respect to some other terminal. Therefore, at any given instant of time, terminals 1, 3, 6, 7 and 10 are all positive with respect to terminals 2, 4, 5, 8 and 9.

When batteries (or cells) are connected in series, to obtain a higher output voltage, the positive terminal of one battery must be connected to the negative terminal of the other battery. Connected in this manner, their individual voltages will add. Similarly, if transformer windings are to be connected in series, so that their



individual voltages add, the "polarity mark" terminal of one winding must connect to the "unmarked" terminal of the other winding.

INSTRUMENTS AND COMPONENTS

DESCRIPTION MODEL

Single-Phase Transformer 8341DC Voltmeter/Ammeter 8412AC Voltmeter 8426Power Supply 8821Connection Leads 8941

PROCEDUREWarning: High voltages are present in this Laboratory Experiment! Do notmake any connections with the power on! The power should be turned offafter completing each individual measurement!

1. a) Connect the 0-40 V meter of the DC Voltmeter/Ammeter across the variable dc output of your Power Supply, terminals 7 and N.

b) Turn on the Power Supply and slowly adjust the voltage to 20 V dc.

c) Without touching the voltage control knob, turn off the Power Supply and disconnect your DC Voltmeter/Ammeter.

d) Using the Single-Phase Transformer, Power Supply and DC Voltmeter/Ammeter, connect the circuit shown in Figure 2. Note that the 40 V dc meter is connected across terminals 3 and 4.

Figure 2

e) Note the deflection of the DC Voltmeter/Ammeter upon turning on the Power Supply switch. If the Voltmeter pointer momentarily deflects to the right, then terminals 1 and 3 have the same polarity mark. (Terminal 1 is connected to the positive side of the dc supply and terminal 3 is connected to the positive side of the Voltmeter).

f) Which two terminals are positive in windings 1 , 2 and 3 , 4?

________________________________________________________________________

g) Disconnect the DC Voltmeter from winding 3 , 4 and connect it across winding 5 , 6. Repeat (e).

h) Which two terminals are positive in windings 1 , 2 or 5 , 6?

________________________________________________________________________



i) Return the voltage to zero and turn off the Power Supply.

2. In this Procedure step you will see the effect of connecting two windings of a transformer in series and the importance of polarity.

a) Using your AC Voltmeter, connect the circuit shown in Figure 3. Note that terminals 1 , 5 are connected together.

b) Turn on the Power Supply and adjust for exactly 208 V ac (one-half of the rated voltage of winding 3 , 4).

Figure 3

c) Measure and record the voltages across the following terminals:

E1 to 2 = ____________ Vac

E5 to 6 = ____________ Vac

E2 to 6 = ____________ Vac

d) Return the voltage to zero and turn off the Power Supply,

e) Remove the connection between terminals 1 and 5. Connect terminals 1 and 6 together and connect the Voltmeter across terminals 2 and 5 as shown in Figure 4.



Figure 4f) Turn on the Power Supply and adjust for exactly 208 V ac

g) Measure and record the voltages across the following terminals:

E1 , 2 = ____________ Vac

E5 , 6 = ____________ Vac

E2 , 6 = ____________ Vac

h) Return the voltage to zero and turn off the Power Supply.

i) Explain why the voltage with the two windings in series is approximately zero in one case and nearly 240 V ac in the other.

___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

j) Which terminals have the same polarity?

________________________________________________________________________

3. a) Consider the circuit shown in Figure 5. Note that winding 3 , 4 is connected to a 208 V ac power source. Do not connect the circuit at this time!

b) What would be the induced voltage across winding 1to 2?

E12 = _____________ V ac

Figure 5

c) If winding 1 , 2 is connected in series with winding 3 , 4 what three possible output voltages can be obtained?

E = ______________ Vac E = ____________ Vac E = _____________ Vac

d) Connect the circuit shown in Figure 5 and place the windings in series, joining terminals 1 and 3.

e) Turn on the Power Supply and adjust for 208 V ac. Measure and record the voltage between terminals 2 and 4.

E2 4 = —————— V ac

f) Return the voltage to zero and turn off the Power Supply.

g) Remove the connection between terminals 1 and 3 and join terminals 1 and 4.

h) Turn on the Power Supply and adjust for 208 V ac. Measure and record the voltage between terminals 2 to 3 and 1 to 2.



E23 = —————— V ac E12 = —————— V ac

i) Return the voltage to zero and turn off the Power Supply.

j) Do the results of (e) and (h) meet with your prediction in (c)?

Yes No

Explain __________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

k) Which terminals have the same polarity?

_____________________________________________________________________________________________________________________________________________________________________________________________________________________

TEST YOUR KNOWLEDGE

1. Assuming you have a 240 V ac power source and that all of the windings on your transformer module develop their rated voltage, show in the spaces provided, how you would connect the windings to obtain the following voltages.

a) 480V:

b) 176V:

c) 360V:

d) 186V:



Experiment 2Transformer Regulation

OBJECTIVES

• To study the voltage regulation of the transformer with varying loads.

• To study transformer regulation with inductive and capacitive loading.

DISCUSSION

The load on a large power transformer in a sub-station will vary from a very small value in the early hours of the morning to a very high value during the heavy peaks of maximum industrial and commercial activity. The transformer secondary voltage will vary somewhat with the load, and because motors and incandescent lamps and heating devices are all quite sensitive to voltage changes, transformer regulation is of considerable importance. The secondary voltage is also dependent upon whether the power factor of the load is leading, lagging, or unity. Therefore, it should be known how the transformer will behave when it is loaded with a capacitive, aninductive, or a resistive load.If a transformer were perfect (ideal) its windings would have no resistance. Furthermore, it would require no reactive power (vars) to set up the magnetic field within it. Such a transformer would have perfect regulation under all load conditions and the secondary voltage would remain absolutely constant. But, practical transformers do have winding resistance and they do require reactive power to produce their magnetic fields. The primary and secondary windings possess, therefore, an overall resistance R and an overall reactance X. The equivalent circuit of a powertransformer having a turns ratio of 1 to 1, can be approximated by the circuit shown in Figure 1. The actual transformer terminals are P1P2 on the primary side and S1S2 on the secondary.

Figure 1

In between these terminals we have shown the transformer as being composed of a perfect (ideal) transformer in series with an impedance consisting of R and X, which represents its imperfections. It is clear that if the primary voltage is held constant, then the secondary voltage will vary with loading because of R and X.An interesting phenomenon occurs with a capacitive load. Partial resonance is set up between the capacitance and the reactance X. Secondary voltage E; may actually tend to rise as the capacitive load value increases.


DESCRIPTION MODEL

Variable Resistance 8311Variable Inductance 8321Variable Capacitance 8331Single-Phase Transformer 8341



AC Ammeter 8425AC Voltmeter 8426Power Supply 8821Connection Leads 8941

PROCEDURE

Warning: High voltages are present in this Laboratory Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!

Figure 2

1. Using your Single-Phase Transformer, Power Supply, Variable Resistance and AC Voltmeter and AC Ammeter, connect the circuit shown in Figure 2

2. a) Place all of the Variable Resistance module switches in their open position for zero load current.

b) Turn on the Power Supply and adjust for exactly 240 V ac as indicated by Voltmeter E1.

c) Measure and record in Table 1 the input current l1, the output current I2, and the output voltage E2 .

d) Adjust the load resistance ZL to 4800 Ω. Make sure that the input voltage remains at exactly 240 V ac. Measure and record I1 I2 and E2 .

e) Repeat (d) for each of the listed values in Table 1.

f) Return the voltage to zero and turn off the Power Supply.

ZL I2 E2 I1

Ω mA Vac mA∞

4800240016001200960

Table 1

3. a) Calculate the transformer regulation using the no-load and full-load output voltages from Table 1.



____________________________________________________________________

____________________________________________________________________

____________________________________________________________________

___________________________________________Regulation_______________%

b) Does the primary winding VA equal the secondary winding VA for everyvalue of load resistance in the Table?

Yes NoExplain ____________________________________________________________

____________________________________________________________________

____________________________________________________________________

4. a) Repeat Procedure step 2 using the Variable Inductance in place of the resistance load.

b) Record your measurements in Table 2.

ZL I2 E2 I1

Ω mA Vac mA∞

4800240016001200960

Table 2

5. a) Repeat Procedure step 2 using the Variable Capacitance in place of the resistance load.

b) Record your measurements in Table 3.

ZL I2 E2 I1

Ω mA Vac mA∞

4800240016001200960

Table 3



6. You will now plot a regulation curve of the output voltage E2 vs output current I2 regulation curve for each type of transformer load.

a) Plot your recorded values of E2 (at each value of I2 listed in Table 3-1) on the graph of Figure 3.

b) Draw a smooth curve through your plotted points. Label this curve "resistive load".

c) Repeat (a) for the inductive (Table 2) and capacitive (Table 3) loads. Label these curves "inductive load" and "capacitive load".

E2

240

230

220

210

200

19050 100 150 200 I2 mA

Figure 3TEST YOUR KNOWLEDGE

1.Explain why the output voltage increases when capacitance loading is used.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

2.A transformer has a very low impedance (small R and X)

a) What effect does this have on the regulation?________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

b) What effect does this have on short-circuit current?________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

3.Very large transformers are sometimes designed not to have optimum regulation properties in order for the associated circuit breakers to be within reasonable size. Explain,

________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

4.Will transformer heating be approximately the same for resistive, inductive or capacitive loads of the same VA rating?

Yes NoExplain _____________________________________________________________________________________________________________________________________________________



__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________



Experiment 3Transformers in Parallel

OBJECTIVES

• To learn how to connect transformers in parallel.

• To determine the efficiency of parallel-connected transformers.

DISCUSSION

Transformers may be connected in parallel to furnish load currents greater than therated current of each transformer. There are two precautions to be observed whenconnecting transformers in parallel.

1 The windings to be paralleled must have identical output voltage ratings.

2 The windings to be paralleled must have identical polarities.

Very large short-circuit currents can be developed if these rules are not followed. Infact, transformers, circuit breakers and associated circuitry can be severely dam-aged, or may even explode, if these short-circuit currents are large enough.

The efficiency of any machine or electrical device is given by the ratio of outputpower to input power. (Apparent power and reactive power are not used in calcula-ting transformer efficiency). The equation for percent efficiency is:

efficiency = [ active power out / active power in ] x 100%


DESCRIPTION MODEL

Variable Resistance 8311Single-Phase Transformer 8341AC Ammeter 8425AC Voltmeter 8426Single-Phase Wattmeter 8431Power Supply 8821Connection Leads 8941

PROCEDURE

Warning: High voltages are present in this Laboratory Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!

1. Using your Single-Phase Transformer, Power Supply, Variable Resistance, AC Voltmeter and AC Ammeter, connect the circuit shown in Figure5-1. Note that the two transformers are connected in parallel. The primary windings (1 to 2) are connected together to the 240 V ac power source. The wattmeter will indicate the input power. Each secondary winding (3 to 4) is connected in parallel to the load RL. Ammeters are inserted to measure load current IL and transformer secondary currents I1 and I2



Figure 1

2. Place all the resistance switches in their open positions for zero load current. Note that the windings are connected for voltage step-up operation (240 V primary to 415 V secondary).

3. Have your circuit wiring approved by the instructor before proceeding.

4. a) Turn on the Power Supply and slowly turn the voltage output control knob while observing the transformer secondary current meters l1 and I2 and the load current meter IL. If the windings are properly phased, no load or secondary currents should be flowing.

b) Adjust the Power Supply voltage to 240 V ac as indicated by the voltmeter connected across the wattmeter.

5. a) Gradually increase the load RL until the load current IL equals approximately 250 mA. Check to see that the input voltage is exactly 240 V ac.

b) Measure and record the load voltage, load current, transformer secondary currents and the input power.

EL = ___________ V ac

IL = ____________ A ac

li = ____________ A ac

Iz = ____________ A ac

Pin = ____________ W

c) Return the voltage to zero and turn off the Power Supply.

6. a) Calculate the load power.

EL_________________ x IL_________________ = _________W

b) Calculate the circuit efficiency.

Pout____________ / Pin____________ X 100 = _________ %

c) Calculate the transformer losses.



Pin ____________ _ Pout _______________ = _________ wd) Calculate the power delivered by transformer 1.

I1________________ x EL ________________ = ___________ W

e) Calculate the power delivered by transformer 2.I2 ________________ x EL ________________ = ___________ W

7. Is the load reasonably distributed between the two transformers?

Yes No

Explain

TEST YOUR KNOWLEDGE

1 . Show how you would parallel connect the transformers to the source and the load in Figure 5-2. Windings 1 to 2 and 3 to 4 are rated for 1,1 kV ac and windings 5 to 6 and 7 to 8 are rated for 415 V ac.

Figure 5-2

2. The efficiency of a transformer which supplies a pure capacitive load is zero.Explain.

3. Name the losses which cause a transformer to heat up.

4. How does the efficiency of your Single-Phase Transformer compare to the efficiency of your DC Motor? Explain



Experiment 4Distribution Transformer

OBJECTIVE• To understand the standard distribution transformer with a 240/415 V secondary winding.

DISCUSSIONSome distribution transformers which supply homes and stores with power have one primary high voltage winding. The secondary winding furnishes 240 V for lighting and the operation of small appliances and also 415 V for electric stoves, water heaters, clothes dryers, etc. The secondary may be a centre tapped single-winding or two separate windings connected in series.

This Laboratory Experiment will show how such a transformer reacts under various load conditions.


DESCRIPTION MODEL

Variable Resistance 8311Variable Inductance 8321Single-Phase Transformer 8341AC Amrneter 8425AC Voltmeter 8426Power Supply 8821Connection Leads 8941

PROCEDUREWarning: High voltages are present in this Laboratory Experiment! Do not make any connections with

the power on! The power should be turned off after completing each individual measurement!

1. Using your Single-Phase Transformer, Variable Resistance, Power Supply, AC Voltmeter, and AC Ammeter, connect the circuit shown in Figure 1. Note that the primary winding (3 , 4) is connected to the 0-415 V ac output of the Power Supply, terminals 4 and 5. The transformer secondary windings 1 , 2 and 5 , 6 are connected in series to obtain 480 V ac between points A and B. R1 and R2 are two Variable Resistance modules.

2. a) Place all of the Variable Resistance module switches in their open positions.

b) Turn on the Power Supply and adjust for 415 V ac as indicated by the Power Supply voltmeter.

c) Measure and record in Table 1 the transformer total output voltage ET, the voltages across each of the loads E1 and E2, the line currents l1 and I2, and the current in the neutral line IN.



Figure 13. a) Place 1200 Ω in each load circuit by closing the appropriate switches.

b) Measure and record all quantities in Table 1.

c) Why is the neutral line current zero?

PROCEDURE STEP R1 R2 I1 I2 IN E1 E2 ET

Ω Ω2c ∞ ∞3b 1200 12004b 1200 48005c 1200 48006c 1600 1600

Table 14. a) Place 4800 Ω in the R2 load while leaving 1200 Ω in load R1.

b) Measure and record all quantities.


d) Is the neutral line current equal to the difference of the line currents?Yes No

5. a) Disconnect the neutral line from the transformer by removing the connection between the transformer and the neutral current meter IN.

b) Turn on the Power Supply and adjust for 415 V ac as indicated by the Power Supply voltmeter.

c) Measure and record all quantities.

d) Return the voltage to zero and turn off the Power Supply.

e) If the load R1 and R2 were incandescent lamps in a home, what would be noticeable?____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

6. a) Reconnect the neutral line from the transformer to the neutral current meter IN.

b) Replace the load R2 with the Variable Inductance.

c) Adjust R1 for a resistance of 1600 Ω.

d) Adjust R2 for an inductive reactance Xi. of 1600 Ω.

e) Turn on the Power Supply and adjust for 415 V ac.

f) Measure and record all quantities.

g) Return the voltage to zero and turn off the Power Supply.

h) Is the current in the neutral line equal to the arithmetic difference of the line currents?

Yes No

Explain ____________________________________________________________________________



______________________________________________________________________________________________________________________________________________________________

TEST YOUR KNOWLEDGE

1. A home is equipped with a 240/415 V ac power system and the following electri-cal loads are in operation:

Line 1 to Neutral

7 ea lamps 60 W1 ea lamp 100W1 ea motor (2,5 A ac)

Line 2 to Neutral

1 ea television 200 W1 ea toaster 1200 W4 ea lamps 40 W

Line 1 to Line 2

1 ea dryer 2 kW1 ea stove 1 kW

a) Calculate the currents in Line 1, Line 2 and the Neutral (assume a 100 %power factor for all appliances).

____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Line 1 = _____ A ac____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Line 2 = ______ A ac____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Neutral = ______ A ac

b) If the neutral wire opened, which lamps would get brighter and which oneswould get dimmer?

2. A 2400 V to 240/415 V distribution transformer has a capacity of 60 kVA.a) What is the rated secondary (415 V) line current?

____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

b) If the load is all placed on one side (line to neutral, 240 V) what is the maxi-mum load that the transformer will carry without overheating?



_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________



Experiment 5Prime Mover and Torque Measurement

OBJECTIVES

• To learn how to connect a split-phase induction motor.

• To learn how to connect the Electrodynamometer.

• To learn how to use the Prony Brake.

DISCUSSION

The split-phase induction motor is the simplest alternating current single-phase motor. It is the least expensive type, and for that reason, the most frequently used. Its speed varies very little from no load to full load.

Even though the study of the split-phase motor is not the main object of this Laboratory Experiment, you will learn how to connect this motor as a driver for the Electrodynamometer and the Prony Brake. The complete study of its characteristics will be taken up in later Laboratory Experiments.

You will note that the motor is started with two windings and that one of them, the auxiliary (starting) winding, is disconnected from the supply by a centrifugal switch when the motor has reached approximately 75% of its normal running speed. You will also note that when a split-phase induction motor is severely overloaded, its speed decreases rapidly until the centrifugal switch closes again and reenergizes the auxiliary winding. When such a condition occurs, the motor should be disconnected immediately from the power source in order to prevent it from burning.

The load imposed on a motor can be measured by two different means; these two torque measuring devices are the Prony Brake and the Electrodynamometer. The Prony Brake is an entirely passive device (no electrical power is required) while the Electrodynamometer requires external power.

The Prony Brake is a friction brake which is used to act as a load for any type of motor and to measure the torque developed by these motors. This brake is entirely mechanical, and consist of a spring balance mounted in a standard full size module. It is built to accurately measure the torque developed by any rotating machine placed on its left-hand side.

A self-cooling friction wheel is slipped over the shaft of the machine under test, and attached to its output pulley by means of two screws. The friction belt of the Prony Brake is then removed from inside the module and slipped over the friction wheel. The braking torque applied to the machine can be varied by turning the knurled wheel (LOAD) in the upper left corner of the module. A second knurled wheel (TORQUE PRESET) in the upper centre allows to bring the spring balance back into equilibrium by aligning the red line in the right-hand window (ZERO) with the blackline; the torque can then be read directly on the 0-3,3 N-m, 360° circular scale, in steps of 0,1 N-m.The accuracy is better than 2% and the torque is continuously adjustable over the full range from no load to locked rotor. When one wants to apply a known torque to the machine, the TORQUE PRESET wheel must first be set so that the calibrated circular scale reads exactly the desired torque value and the LOAD wheel must then be turned in such a way that the red line in the ZERO window is aligned with the black line.

The Electrodynamometer is a device used to accurately measure the torque developed by motors of all kinds. It is actually an electrical brake in which the braking force can be varied electrically rather than by mechanical friction. The Electrodynamometer is a more stable, easier to adjust, device than the mechanical friction brakes.

The Electrodynamometer consists of a stator and a squirrel-cage rotor. The stator, unlike other electromechanical devices, is free to turn, but its motion is restricted by a helical spring.

In normal operation, dc current is applied to the stator winding. This sets up a magnetic field which passes through both the stator and the rotor. As the rotor turns (being belt-coupled to the driving motor), a voltage is induced in the rotor bars, and the resulting eddy currents react with the magnetic field causing the stator to turn in the same direction as the rotor.



The stator rotation is limited by the helical spring and the amount that it turns is marked off on a scale attached to the external stator housing.

The Electrodynamometer is calibrated from -0,3 to 3 N-m which is more than adequate for the testing of 0,2 kW motors even when they are tested at overload conditions.

The power output of a motor depends on its speed and the torque it develops, This relationship is given by the following equation:

P = [ 2 π x N x T ] / 60where

P = power output (W)N = speed (r/min)T = torque (N.m)


DESCRIPTION MODEL

Capacitor-Start Motor 8251AC Ammeter 8425Power Supply 8821Electrodynamometer 8911or Prony Brake 8913Hand Tachometer 8920Connection Leads 8941Timing Belt 8942

PROCEDURE

Warning: High voltages are present in this Laboratory Experiment! Do notmake any connections with the power on! The power should be turned offafter completing each individual measurement!

The Split-Phase Induction Motor

1. Examine the front panel of the Capacitor-Start Motor. (The entire Capacitor-Start Motor will be fully described in a later Laboratory Experiment).

a) Note the two separate windings. The main winding is rated 2,5 A while the auxiliary winding is marked "intermittent". A circuit breaker protecting the auxiliary winding will trip if the winding is left connected to the input line (240 V) for longer than a few seconds.

b) Identify the centrifugal switch connected between terminals 6 and 7. The centrifugal switch should always be connected in series with the auxiliary winding.

c) Note also the capacitor connected between terminals 4 and 5. The capacitor is not used in the split-phase induction motor wiring; it will be used in the capacitor-start motor wiring.

2. Using the Power Supply, Capacitor-Start Motor and AC Ammeter, connect the circuit shown in Figure 1.

Terminals 1 and N on the Power Supply provide fixed single-phase 240 V power to the main winding in parallel with the series connection of the centrifugal switch and the auxiliary winding. The capacitor connected between terminals 4 and 5 is not used when this module is operated as a split-phase motor.



Figure 1

a) Have your instructor check your completed circuit.

b) Turn on the Power Supply, The motor should start running immediately and you should hear the click of the centrifugal switch as it opens.

c) c) Note the indication of the current meter.I = _____________ A ac

d) d) Note whether the direction of rotation is clockwise or counter clockwise.Rotation = ____________

e) e) Measure with your Hand Tachometer the speed of the motor without load.

Speed without load = __________ r/min

f) Turn off the Power Supply. You should hear the closing of the centrifugal switch as the motor slows down and comes to a rest.

g) Disconnect the motor.

Note: If you are not using the Electrodynamometer in this experiment, proceed to Procedure 12.

3. a) Examine the construction of the Electrodynamometer.

b) Note the cradle construction for the Electrodynamometer housing. (This is also called turning mounting).

c) Note the helical spring at the rear of the machine. This spring has been accurately calibrated against the graduations marked on the front of the housing.

d) Note the mechanical stops that limit the rotational travel of the stator housing.

e) Identify the stator winding attached to the inside of the housing. (This winding carries dc current).

f) Identify the two wire leads that carry dc current to the stator winding. (They enter the housing through the centre of the helical spring).

g) Identify the bridge rectifier located at the rear of the module. (This bridge furnishes dc power for the stator magnetic field).



h) Identify the variable autotransformer mounted on the front panel of the module. The braking effect of the Electrodynamometer is controlled by the strength of the stator magnetic field, which is proportional to the dc output of the bridge rectifier. The dc output is varied by the variable autotransformer.

i) Identify the two ac connections terminals mounted on the front panel of the module.

4. a) Connect the Electrodynamometer to the fixed ac output of the power source by connecting the two input terminals of the Electrodynamometer to terminals 1 and N of the Power Supply.

DO NOT APPLY POWER AT THIS TIME!

b) Set the Electrodynamometer variable transformer control knob to its mid-position.

c) Lower the front panel of the module so that you may turn the pulley by hand.

5. a) Turn on the Power Supply.

b) Keeping one hand in your pocket, for safety reasons, carefully reach in and try to turn the pulley.

Warning: Caution is advised because there are several live terminals exposedwhen the front panel is dropped.

Do you feel a drag when you turn the pulley?

Yes NoDoes the stator housing tend to turn in the same direction as the pulley?

Yes No

6. a) Remove your hand from inside the module and advance the control knob, thereby increasing the magnetic stator field.

b) Carefully reach in and turn the pulley. Did the drag increase?Yes No

c) Repeat (a) but this time reduce the stator magnetic field.

d) Carefully reach in and turn the pulley. Did the drag increase?Yes No

e) Turn off the Power Supply.

7. a) Couple the Capacitor-Start Motor to the Electrodynamometer with the Timing Belt.

b) Connect the motor as shown in Figure 7-1.

c) Connect the Electrodynamometer to terminals 1 and N of the Power Supply. (There should now be two connection leads at these terminals of one to the Electrodynamometer).

d) Set the Electrodynamometer control knob at its full ccw position (to provide minimum starting load for the motor).

8. Apply power and note if the motor revolves in a cw direction. If not, reverse its rotation by interchanging the connections to one of the motor windings. (The Electrodynamometer torque can only be measured for cw rotation).

9. Increase the load on the motor (the Electrodynamometer braking action) by varying the control knob on the Electrodynamometer until the scale marked on the stator housing indicates 1 N-m. (The numeral 1 should be directly beneath the red vertical line on the window beneath the pulley).

10. a) Measure and record the motor current with a 1 N-m load on the motor.



I = __________ A ac

b) Measure and record the motor speed with a 1 N-m load.Speed with load = ______________ r/min

11. Gradually increase the load on the motor, by advancing the variable auto-transformer control knob on the Electrodynamometer, until the motor stalls. Quickly note the maximum value measured by the Electrodynamometer and immediately turn off the Power Supply.

"Break-down" torque = ____________ N-m

Prony Brake

Note: If you are not using a Prony Brake in this experiment, proceed to the TEST YOURKNOWLEDGE

12, a) Examine the construction of the Prony Brake.

b) Remove the friction wheel from inside the module and note the holes in the web of the pulley; they are for cooling purposes. Slip the friction wheel over the shaft of the motor and secure it firmly to the motor pulley by tightening the two screws in the grooves of the motor pulley.

c) Note the floating plate inside the module. In operation, this plate is pulled in one direction by the friction belt and in the other direction by a spring which is connected to the fixed frame by means of a rack gear. A pinion gear, driven by the rack, turns the front plastic disc which has a red line to read the torque. The TORQUE PRESET wheel effectively increases the spring tension and so, the force which tends to turn the floating plate.

d) The LOAD wheel effectively increases the tension on the friction belt,and tends to turn the floating plate in the opposite direction.

e) Note the friction belt attached at one end to the floating plate and at the other end to the LOAD wheel screw.

f) The complete system constitutes a spring balance.

13. a) Turn the LOAD wheel downwards to release the tension on the friction belt, and slip the belt over the friction pulley mounted on the motor. Leave the belt loose.

b) Connect the motor as shown in Figure 7-1.

c) Turn on the Power Supply and note if the motor revolves in a cw direction. If not, reverse its rotation by interchanging the connections to one of the motor windings. (The Prony Brake can only measure torque for cw rotation).

d) Measure the motor current.

I = __________ A ac

e) Vary the TORQUE PRESET wheel until the circular scale indicates 1 N-m. This does not impose any torque on the motor and the current should not change. You have just preset the balance to the indicated torque.

f) Turn slowly the LOAD wheel upwards to tighten the belt over the friction pulley. Note the gradual increase in the motor current. Keep turning the LOAD wheel until the hairlines in the right-hand window are aligned. The balance is now in equilibrium and is imposing a torque of 1 N-m on the motor.

g) Measure and record the motor current with a 1 N-m load.

I = _______________ A ac



h) Measure and record the motor speed with a 1 N-m load.Speed with load = ____________ r/min

14. a) Gradually increase the load on the motor by turning the LOAD wheel Then bring the balance into equilibrium. Repeat successively these last two steps until the motor stalls. Immediately turn off the Power Supply.

b) Record the value of the "break-down" torque as indicated by the calibrated dial.

"Break-down" torque = ____________ N-m

TEST YOUR KNOWLEDGE

1 How can your reverse the direction of rotation of a split-phase induction motor?____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

2. Calculate the developed motor power P in Procedure 10 or 13 (h).____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

P = ____________W

3. Which torque measuring device is easier to use, the Electrodynamometer or theProny Brake?

____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

4. Where is the power (heat) dissipated in the Electrodynamometer?

____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

5. Where is the power (heat) dissipated in the Prony Brake?

____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________



Experiment No. 6 The Wound-Rotor Induction Motor-part I

Objectives:

To examine the construction of the three-phase wound-rotor induction motor.

To understand synchronous speed and slip in a three-phase induction motor.

To observe the effect of the revolving field and rotor speed on the voltage induced in the rotor.

Discussing :

The wound rotor consists of a rotor core with three windings instead of conducting bars of the squirrel cage rotor. In this case, currents are induced in the windings just as they would be in short-circuited turns. However. the advantage of using windings is that the wires can be brought out through slip rings so that resistance, and therefore the current through the rotor windings, can be controlled

The rotating stator field induces an alternating voltage in each winding of the rotor. When the rotor is at a standstill, the frequency of the induced rotor voltage is equal to that of the power source. If the rotor is now rotated by some external means, in the same direction as the rotating stator field, the rate at which the magnetic flux cuts the rotor windings will diminish. The induced voltage and its frequency will drop.

When the rotor revolves at the same speed and in the same direction as the rotating stator field, the induced voltage, as well as its frequency, will drop to zero. (The rotor is now at synchronous speed).

Conversely, if the rotor is driven at synchronous speed, but in the opposite direction to the rotating stator field, the induced voltage, as well as its frequency, will be twice what they were when the rotor was at a standstill (although the rotor will be driven by an external motor in this case), it should be noted that for a given rotor speed the induced voltage value and its frequency will be the same even if the rotor were turning by itself.

Instruments And Components:

DC Motor/Generator Three-phase Wound Rotor-Induction Motor AC Ammeter AC Voltmeter Three-Phase Wattmeter Power Supply Hand Tachometer Connection Leads Timing Belt

Procedure:



WarningHigh voltages are present in this laboratory Experiment! Do not make any connections with power on! The power should be turned off after completing each individual measurement!

1. Using your Power Supply Module, Three-Phase Wound-Rotor Induction Motor, DC Motor/Generator, AC Voltmeter and Three-Phase Wattmeter, connect the circuit in Fig. 1. The field rheostat should be turned to its full cw position for minimum resistance.

Fig. 12. a) Couple the DC Motor/Generator and the Three-Phase Wound-Rotor

Induction Motor using the Timing Belt.

b) Make sure the variable output control knob is at zero. Turn on the Power supply. The dc motor should not be turning.

c) Measure and record the following:

E1 = ------- Vac E2 = --------- V ac P = --------- W

I1 = -------- A ac I2 = ---------- Aac I3 = --------- Aac

d) Turn off the Power Supply.3. a) Turn on the Power Supply and gradually adjust the variable dc output

voltage for a motor speed of 750 r/min.



If the value of E2 gets higher than in Procedure step 4, turn off the Power Supply and interchange any two of the three stator leads.

b) Measure and record the following:

E1 = --------- V E2 = ----------- V P = ------------ W

I1 = --------- A I2 = ----------- A I3 = ---------- A

4. a) Increase the variable dc output voltage to 240 V and adjust the field rheostat for a motor speed of exactly 1500 r/min.


E1 = ------------ V E2 = --------- V P = -------- W

I1 = ----------- A I2 = ---------- A I3 = -------- A.


5. a) Interchange your dc armature connections in order to reverse the motor direction. Turn the field rheostat to its full cw position.

b) Turn on the Power Supply and adjust the dc output voltage for a motor speed of 750 r/min.

c) Measure and record the following:

E1 = ------------ V E 2 = ---------------- V

I1 = ------- A I2 = -------A and I3 = --------- A.

W1 = ---------------- W W2 = ------------- W

6. a) Increase the variable dc output voltage to 240 V dc and adjust the field rheostat for a motor speed of 1500 r/min.


E1 = --------- V E2 = --------------- V

I1 = --------- A I2 = -------- A and I3 = -------- A .

W1 = ------------ W W2 = ------------- W




HND2 – Utilization of Electrical EnergyThe Wound-Rotor Induction Motor Experiment 1

Testing Knowledge

Name: _______________________ No.: __________ Date: / / Group:

1. Refer to procedure 2-c Calculate the following: [15-marks]

a) Apparent power S = ----------------------------------------- = -------- VA

b) Power factor Cos = ---------------------------------------- = ---------

c) Reactive power Q = ------------------------------------ = --------- VAr

2. In Procedure steps 3 and 4, is the rotor being driven with or against the rotating field? Explain? [10-marks]

-------------------------------------------------------------------------------------------------------

3. In Procedures 5 and 6 is the rotor being turned with or against the rotating stator field? Explain? [10-marks]

----------------------------------------------------------------------------------------------------

4. Knowing that the voltage induced in the rotor winding is zero when it is turning at synchronous speed, what is the synchronous speed of your motor? [5-marks]----------------------------------------------------------------------------------------------------

-----------------------------------------------------------------------------------------------------

5. Determine the number of poles in your motor. [5-marks]

-----------------------------------------------------------------------------------------------------

6. Calculate the rotor slip in Procedure steps 2, 3, 4, 5 and 6. [30-marks]

Step 2: -----------------------------------------------------------------------------------------

Step 3: ------------------------------------------------------------------------------------------

Step 4: ------------------------------------------------------------------------------------------



Step 5: ------------------------------------------------------------------------------------------

Step 6: ------------------------------------------------------------------------------------------

7. How much reactive power (Q) is needed to produce the magnetic field in your motor? [5-marks]

----------------------------------------------------------------------------------------------------

8. How much active power (P) is needed to supply the losses associated with the production of the magnetic field? [5-marks]

9. Plot the rotor speed against the rotor voltage on the graph paper below and write your conclusion. [15-marks]

Conclusion:--------------------------------------------------------------------------------

-----------------------------------------------------------------------------------------------------



----------------------------------------------------------------------------------------------------

Experiment No. 8

The Wound-Rotor Induction Motor-part II

Objectives

• To observe the rotor and stator currents and the torque values at different motor speeds.

• To determine the starting characteristics of the wound-rotor induction motor.

Discussion

In the previous Laboratory Experiment we saw that a considerable voltage appears across the rotor windings on open circuit, and that this voltage varies linearly with rotor slip in r/min, becoming zero at synchronous speed.

If the rotor windings are short-circuited, the induced voltage will cause large circulating currents in the windings. To supply this rotor current, the stator current must increase in value above its ordinary exciting current level. The power consumed (VA) in the rotor windings (and associated circuitry) must be supplied by the stator windings. Therefore, we should expect the following:

a) At standstill, or at low speed, the rotor currents, stator currents, and torque will be high.

b) At synchronous speed, the rotor current and torque will be zero, and the stator will only carry the exciting current.

c) At any other motor speed, the currents will be between the above extremes.

INSTRUMENTS AND COMPONENTSDescriptions Model

Three-Phase Wound-Rotor Induction Motor 8231 AC Ammeter 8425 AC Voltmeter 8426 Electrodynamometer 8911 Power Supply 8821 Connection Leads 8941 Timing Belt 8942



PROCEDURE

Fig. 1

1. Using your Power Supply, AC Ammeter, AC Voltmeter, and Electrody-namometer, connect the circuit shown in Fig. 1. Note that the three stator windings are connected to the variable 3-phase output of the Power Supply (terminals 4, 5 and 6).

2. a) Couple the Electrodynamometer to the motor with the Timing Belt.

b) Connect the input terminals of the Electrodynamometer to the fixed 240 V ac output of the Power Supply, terminals 1 and N.

c) Set the Electrodynamometer control knob at its full cw position (to provide a maximum starting load for the motor).

3. a) Turn on the Power Supply and adjust E1 at 208 V ac (50% of the rated

voltage). The motor should be turning slowly.

b) Measure and record the three rotor currents and the developed torque.

I1 = -------- A ac I2 = ------A ac


Caution:

High voltages are present in this Laboratory Experiment! Do not make any connections with the power on! The power should be

turned off after completing each individual measurement


I3 = -------- A ac Torque = -------- Nm

4. a) Gradually reduce the load on the motor by slowly adjusting the Electrodynamometer control knob. As the load is reduced the motor speed will increase.

b) Do the three rotor currents remain approximately equal?

Yes ------- OR No ------------

c) Do the three rotor currents decrease as the motor speeds up?

Yes -------- OR No --------

d) Measure and record the rotor currents at a torque of 0,2 Nm.I 1 = _______ A ac I2 = _______ A ac, I3 = _______ A ac

e) Return the voltage to zero and turn off the Power Supply.

5. a) Connect the circuit shown in Fig. 2. Note that the fixed 3-phase output of the Power Supply, terminals 1, 2 and 3 are now being used.

Fig. 2

b) Set the Electrodynamometer control knob at its full cw position (to provide a maximum starting load for the motor).

6. Turn on the Power Supply and quickly measure E1 , I1, I2 and the developed starting torque. Turn off the Power Supply.

I1= --------- Aac I2 = ---------- Aac

E1= -------- Vac Torque = --------- Nm



HND2 – Utilization of Electrical EnergyThe Wound-Rotor Induction Motor Experiment 2

Testing Knowledge

Name: _______________________ No.: __________ Date: / / Group:

TEST YOUR KNOWLEDGE

1. Assuming the full load (175 W) motor speed is 1315 r/min, calculate the value of the full load torque using the formula for output power:

P = 2NT / 60 = ----------------------------------------------------------------------------[8 marks]

2. Calculate the ratio of starting torque to full load torque:

-------------------------------------------------------------------------------------- [7 marks]

3. Assuming that the full load stator current is 0,48 A per phase, calculate the ratio of starting current to full load operating current.

---------------------------------------------------------------------------------------[7 marks]

4. In Procedure step 6 Calculate the apparent power (S) to the

motor at starting torque. Apparent power (S) = -------------------------------------- = -------------- VA

[7 marks]

5. If the stator voltage of a wound rotor motor is reduced by approximately 50% of the rated value:

a) By how much is the starting current reduced?

----------------------------------------------------------------------------------[7 marks]

b) By how much is the apparent power reduced?

-----------------------------------------------------------------------------------[7 marks]

c) By how much is the starting torque reduced?

-----------------------------------------------------------------------------------

[7 marks]


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