3.-electrotecnia_unit3 corrente trifásica -

5/28/2018 3.-Electrotecnia_Unit3 Corrente trifsica -

1/11

11

Alternating current

three-phasecircuits

Unit 3. AC THREE-PHASE CIRCUITS

22


Three-phase systems characteristics

Generation of three-phase voltages

Three-phase loads

-Y and Y- transformation Instantaneous power

Three-phase power: S, P and Q

Power measurement. Aaron connection

Power factor improvement

Electrical measurements

Exercises

CONTENTS:

33


THREE-PHASE SYSTEM CHARACTERISTICS

44



The electricity grid is made up of four main components:

GENERATION: production of electricity from energy sources

such as coal, natural gas, hydropower, wind and solar.

TRANSMISSION: the transmission system carries the electricpower from power plants over long distances to a distribution

system.

DISTRIBUTION: the distribution system brings the power to the

customers.

COSTUMERS: these are the consumers of electric power

(industry, service sector and residential uses).


2/11

55



Instantaneous electric power has a sinusoidal shape with double

frequency than voltage or current.

SINGLE-PHASE AC CIRCUITS: instantaneous electric power is

negative twice a period (power flows from the load to thegenerator) and positive twice a period, falling to zero.

BALANCED THREE-PHASE AC CIRCUITS: instantaneous

electric power is constant. Three-phase power never fal ls to

zero.

Three-phase electric motors perform better than single-phase AC

motors.

Three-phase power systems allow two voltage levels (L-L, L-N).

When electric power is transmitted, three-phase AC systems

require 25% less Cu/Al than single-phase AC systems.

66


GENERATION OF THREE-PHASE VOLTAGES

Three-phase generators contain three sinusoidal voltage sources

with voltages of the same frequency but a 120-phase shift with

respect to each other.

This is achieved by positioning three coils separated by 120angles. There is only one rotor.

Amplitudes of the three phases are also equal.

The generator is then balanced.

77


INTRODUCTION

88


INTRODUCTION

R

S

T

N

N: neutral point

R S T (or A B C) direct sequence or sequence RST

VRS, VST, VTR: line voltages or line-to-line voltages

VRN, VSN, VTN: line-to-neutral voltages

Vline= Vline-to-neutral3

R

ST

N

V

V V

V

V

V

120

TN SN

RSTR

ST

RN

vRN

(t) = V0

cos(t + 90) VvSN(t) = V0cos(t - 30) VvTN(t) = V0cos(t +210) V


3/11

99


INTRODUCTION

50 Hz Usual system

Vphase = Vline-to-neutral230 volt

Vline 400 volt

Frequency 50 Hz

phase

Ntolineline .V3V

R

ST

N

V

V V

V

V

V

120

TN SN

RSTR

ST

RN

1010


THREE-PHASE LOAD CLASSIFICATION

WYE(two voltages)

Balanced

3-wires

4-wires

Unbalanced

3-wires

4-wires

DELTA(one voltage)

Balanced

Unbalanced

Z Z

Z

R

N

S

T

O=N

IR

IN

IT

ISZ Z

Z

R

S

T

O=N

R

S

T

Z

Z

Z

RSTR

S

R

T

ST

I

I

I

I

I

I

R

S

T

R

S

T

Z

Z

Z

RSTR

S

R

T

ST

I

I

I

I

I

I

R

S

T

RS

ST

TR

1111


BALANCED WYE-CONNECTED LOAD

The wye or star connection is made by connecting one end of each of

the three-phase loads together.

The voltage measured across a single load or phase is known as the

phase voltage.

The voltage measured between the lines is known as the line-to-line

voltage or the line voltage.

In a wye-connected system, the line voltage is higher than the load

phase voltage by a factor of the square root of 3.

In a wye-connected system, the phase current and line current are the

same.

1212


BALANCED DELTA-CONNECTED LOAD

This connection received its name from the fact that a schematic

diagram of it resembles the Greek letter delta ().

In the delta connection, the line voltage and phase voltage in the load

are the same.

The line current of a delta connection is higher than the phase currentby a factor of the square root of 3.


4/11

1313


-Y TRANSFORMATION

R

S

T

Z

Z

Z

1

23

13

23

12

Z

Z Z

1

32

231312

1312

1

ZZZ

ZZZ

231312

2312

2

ZZZ

ZZZ

231312

2313

3

ZZZ

ZZZ

Balanced loads: ZY = Z/3

)(

)(

)1( 231312

231312

21ZZZ

ZZZ

ZZ

Z between nodes 1 and 2:

)(

)()2(

231312

23121331

ZZZ

ZZZZZ

)(

)()3(

231312

13122332

ZZZ

ZZZZZ

From expressions (1), (2) and (3) the result is:

)(

)(

231312

2313122,1

ZZZ

ZZZZ

212,1 ZZZY Y:

Z nodes 1-2:

Z nodes 1-3:

Z nodes 2-3:

(1) + (2) - (3) (1) - (2) + (3) -(1) + (2) + (3)

1414


BALANCED THREE/FOUR-WIRE WYE-CONNECTED LOAD

Z Z

Z

R

N

S

T

O=N

IR

IN

IT

IS

Z Z

Z

R

S

T

O=N

0)III(I

Z

VI

Z

VI

Z

VI

TSRN

TNT

SNS

RNR

Q

lnialnia

P

lnialnia

lnialniaRNfasetotal

.sin.I.V3.cos.I.V3

.I.V3I.V3.S3.S *R

j

The three currents are balanced.

Thus the sum of them is always zero.

Since the neutral current of a balanced, Y-connected three-phase load is always zero,

the neutral conductor may be removed with

no change in the results.

1515


BALANCED THREE/FOUR-WIRE WYE-CONNECTED LOAD

Z Z

Z

R

S

T

O=N

Example A three-phase, RST system (400 V, 50 Hz), has a

three-wire Y-connected load for which Z = 1030 Obtain theline currents and the complex power consumption

A3

40

10

3400/

Z

VI

60

30

90

RNR

A3

40

10

3400/

Z

VI

60

30

30SN

S

A3

40

10

3400/

Z

VI

180

30

210

TNT

(VAr)j8000(watt).4138561VA00061

)3)(40/33(400/IV3S3S

30

*6090

RNphasetotal*R

1watt4.1385630cos3

404003cosIV3P lltotal

VAr800030sin3

404003sinIV3Q lltotal

VA160003

40

4003IV3S lltotal

1616


UNBALANCED FOUR-WIRE WYE-CONNECTED LOAD

Z Z

Z

R

N

S

T

O=N

R

S T

IR

IN

IS

IT

0)III(I

Z

VI

Z

VI

Z

VI

TSRN

T

TNT

S

SNS

R

RNR

*

T

*

S

*

R IVIVIVS TNSNRNtotal


5/11

1717


UNBALANCED THREE-WIRE WYE-CONNECTED LOAD

Z Z

Z

R

S

T

O=N

R

S T

IR

IS

IT

0III TSR

TSR

TTNSSNRRN

TSR

T

TN

S

SN

R

RN

ON

YYY

YVYVYV

Z

1

Z

1

Z

1

ZV

ZV

ZV

V

0Z

V

Z

V

Z

V

T

TO

S

SO

R

RO

0Z

VV

Z

VV

Z

VV

T

ONTN

S

ONSN

R

ONRN

1818



Z Z

Z

R

S

T

O=N

R

S T

IR

IS

IT

T

ONTN

T

TO

T

S

ONSN

S

SOS

R

ONRN

R

ROR

Z

VV

Z

V

I

Z

VV

Z

VI

Z

VV

Z

VI2)

*

T

*

S

*

RIVIVIVS3) TOSOROtotal

TSR

T

TN

S

SN

R

RN

ON

Z

1

Z

1

Z

1

Z

V

Z

V

Z

V

V)1

1919



Example A three-phase, RST system (400 V, 50 Hz), has a three-wire

unbalanced Y-connected load for which ZR= 100 , ZS= 100 and ZT= 1030

Obtain the line currents and the total complex power consumption.

O N

R

S

T

IR

IS

IT

ZR

ZS

ZTV40.93

1.01.01.0

0.12300.12300.1230

YYY

YVYVYVV

114.89

3000

30-210030-090

TSR

TTNSSNRRNON

VRO= VRNVON= 23090 - 40.93114.89 = 193.6484.90 V

VSO= VSNVON= 230-30 - 40.93114.89 = 264,54-35.10 V

VTO= VTNVON= 230210 - 40.93114.89 = 237,18-140.10 V

IR= VRO/ZR= 193.6484.90/100 = 19,3684.90A

IS= VSO/ZS= 264.54-35.10/100 = 26,45-35.10A

IT= VTO/ZT= 237.18-140.10/1030 = 23.72-170.10A

Stot= VROIR* + VSOIS* + VTOIT* = 15619.56 W + j2812.72 VAr

1)

2)

3)

2020


BALANCED DELTA-CONNECTED LOAD

R

S

T

Z

Z

Z

RSTR

S

R

T

ST

I

I

I

I

I

I

R

S

T

Z

VI

Z

VI

Z

VI

TRTR

STST

RSRS

Q

lnialnia

P

lnialnia

lnialniaRSfasetotal

sinIV3cosIV3

IV3IV3S3S *RS

j

STTRT

RSSTS

TRRSR

III

III

III


6/11

2121


UNBALANCED DELTA-CONNECTED LOAD

R

S

T

Z

Z

Z

RSTR

S

R

T

ST

I

I

I

I

I

I

R

S

T

RS

ST

TR

TR

TRTR

ST

STST

RS

RSRS

Z

VI

Z

VI

Z

VI

STTRT

RSSTS

TRRSR

III

III

III

*TR

*ST

*RS

IVIVIVS TRSTRStotal

2222


UNBALANCED THREE-WIRE -CONNECTED LOAD

ExampleA three-phase, RST system (400 V, 50 Hz), has an unbalanced -connected load for which ZRS= 10

0 , ZST= 1030 i ZTR= 10-30 Obtainthe line currents and the complex power consumption

R

S

T

Z

Z

Z

RSTR

S

R

T

ST

I

I

I

I

I

I

R

S

T

RS

ST

TRA4010

400

Z

VI

A4010

400

Z

VI

A40

10

400

Z

VI

90

30

120

TR

TRTR

30

30

0

ST

STST

120

0

120

RS

RSRS

A40.00IIIA77.29III

A77.29III

150STTRT

45RSSTS

105TRRSR

43712.81VA

(VAr)j0)43712.81(WIVIVIVS *TR

*

ST

*

RS TRSTRStotal

R

ST

N

V

V V

V

V

V

120

TN SN

RSTR

ST

RN

2323


POWER MEASURMENT. Four-wire load

Unbalanced wye-connected, four-wire load

Z Z

Z

R

S

T

O=N

N

W

R

S TW

W

R

S

T

WR = VRNIRcos( VRN - IR)

WS= VSNIScos( VSN - IS)

WT= VTNITcos( VTN - IT)

Ptotal = WR + WS + WT

Z Z

Z

R

S

T

O=N

N

W

Balanced wye-connected, four-wire load

Ptotal = 3W

W = VRNIRcos( VRN - IR)

2424


POWER MEASURMENT. ARON CONNECTION

General 3-wire load. Two-wattmeter method (ARON connectio n)

R

ST

N

V

V V

V

V

V

120

TN SN

RS

TR

ST

RNVRT

VST

VSR

)30cos(VI)cos(IVW

)30cos(VI)cos(IVW

SST

RRT

IVSST2

IVRRT1

cosVI3)]30cos()30cos(VI[WWP 21TOTAL

,

LOAD

inVI)]cos()cos(VI[]W[WQ 21TOTAL s3303033

Demonstration done for a balanced 3-wire load


7/11

2525


POWER MEASURMENT. BALANCED LOAD

,

LOAD

)WW

WW3arctg(

21

21

)W(W3Q

WWP

21total

21total

Balanced load, g eneral (Y/D, 3/4 wires). Two-wattmeter method

(ARON connection)

,

LOAD

Unbalanced wye/delta-connected, three-wire load

Ptotal = W1 + W2

BALANCED

UNBALANCED

2626

Aron cycl ic permutat ions


POWER MEASURMENT: THE TWO-WATTMETER METHOD

W3QTOT

W1 W2

V I V I

RT R ST S

SR S TR T

TS T RS R

WV I

ST R

TR S

RS T

Q measurement: cyclic permutations

3inIV)os(90-IV)cos(IVW TOTALlinelinelinelineRST

QscIRVST

2727


INSTANTANEOUS THREE-PHASE POWER

Single-phase load: cosAcosB = 05[cos(A+B) + cos(A-B)]

p(t) = v(t)i(t) = V0cos(wt + V)I0cos(wt + )

p(t) =1/2V0I0cos(V-I) + 1/2V0I0cos(2wt + V+ I) watt

Constant Oscillates at twice the mains frequency!

Three-phase wye balanced load:p(t)= vRN(t)iR(t) + vSN(t)iS(t) + vTN(t)iT(t) =

=2Vpcos(wt +V)2Ipcos(wt +)

+2Vpcos(wt -120+V)2Ipcos(wt -120+)

+2Vpcos(wt +120+V)2Ipcos(wt +120+)

=

VpIpcos(V-I) + VpIpcos(2wt +V+I)

+ VpIpcos(V-I) + VpIpcos(2wt -240+V+I)+ VpIpcos(V-I) + VpIpcos(2wt +240+V+I)

= 3/2VpIpcos(V-I) = 3/2VpIpcos = constant! 2828


INSTANTANEOUS POWER: SINGLE-PHASE LOAD

v(t)

i(t)

p(t)Average p ower = P


8/11

2929


INSTANTANEOUS THREE-PHASE POWER

v(t)

i(t)

p(t)

pTOTAL = pR(t) + pS(t) + pT(t)

3030


POWER LOSSES: THREE-PHASE/SINGLE PHASE

Single-phase lineR1

R

N

LOAD

R1

Three-phase line

Supposing same losses1p3p

3p1p

21 S2

1S

SS2R2R ll

Single-phase line: 2 conductors of length land section S1p

Three-phase line: 3 conductors of length l and section S3p= 1/2S1p

As a result: weight3p-cables= 3/4weight1p-cables

22

2

load1

2

1losses

cosV

P2RI2RP V

V

22

2

load2222

2

load2

2

2losses

cosV

PR

cos.V)3(

P3RI3RP

cosV

PI load

cosV3

PI load

3131


Example 1

A

Balanced load 1capacitive

C

K

CC

UU1

ZL

ZL

ZL

R

S

T A1

W1

Balanced load 2inductive

A2

W2

Three-phase balanced RST system for which A1= 1.633 A, A2= 5.773 A, W1= 6928.2

W, W2= 12000 W, U = 6000 V and Z line=4+j3 a) Obtain the complex power in theloads, as well as the ammeter A and the voltmeter U1 readings. b) Obtain the value of

Cto improve the loads PF to 1, assuming U = 6000 V.

48000VArsinUI3Q

12.53

cosUI3W36000W3P

W12000cosUI3P

45

sinUI3VAr12000Q

VAr12000W3Q

222

2

2222

111

1

111

11

VA6000060003j48000)Qj(Q)P(PS 36.87212121

A774.5773.5633.1III 85.369012.5390459021

TOTAL

6050VU3 UV97.34923

6000)34(774.5UZ.IU phase1,1

90

90

85.3690

phaseL,1

jphase

F1.06C).50.C1/(2

(6000)-r-36000/3VAQ/3Q

2

,1

C

R

ST

3232


Example 2

Three-phase 50 Hz system for which V = 400 V, W1= -8569.24 W, W2= -5286.36 W, AS= 21.56 A. Obtain a) the value of R. b) the reading of A R.c) the value of the inductance

L

22

11

W'W

and

W'W

23.095R:inresultingR

4002W13855.6'W'WPa)

2

21otalt

A32.17095.23

400

R

VIA32.17

095.23

400

R

VIb) 60

60

RTRT

6060

SRSR

A30I-II:resultsIt90

SRRTR

L

90

L

180

60

L

TS60TSSRS

X400jj1566.8

X40032.17

jXV32.17IIIc)

H0.2684L:isresultThe

50L284.308X)X

40015(8.66A21.56I L

2

L

2

S

ARR

S

T

W1

R L

W2

R

AS

AT

V

R

S

T

R

ST

W1 W2V I V I

RT R ST S

SR S TR T

TS T RS R


9/11

3333


Example 3

Varley phase-sequence indicator. Calculate the voltage in each element and deduce

the practical consequences Three-phase 400 V/50 Hz system

V171.55

5290/15290/13183/1

/5290230/5290230/3183230

YYY

YVYVYVV

171.31

0090

0210030-90-90

TSR

TTNSSNRRNON

VRO= VRNVON= 23090 - 171.55171.31 = 265.3450.28 V

VSO= VSNVON= 230-30 - 171.55171.31 = 394.78-20.91 V

VTO= VTNVON= 230210 - 171.55171.31 = 144.00-101.86 V

Conductor R is s ituated where the capac itor is p laced, conductor S

is si tuated where the brigh ter bu lb is p laced and T is the remaining

conductor.

C = 1F XC= 3183

R2bulbs= 2V2/P = 22302/20 = 5290

R

ST

0V0N

VR0

VT0 VS0

3434


Example 4

A 400 V and 50 Hz three-phase line feeds two balanced loads through a line which has

an internal impedance of ZL=0.5 + j1 The-connected load has phase impedanceswhose values are 45+j30 , whereas the Y-connected load has phase impedances of

15j30 Determine: a) the reading of the ammeter A, b) the reading of the voltmeter

V and c) the readings of watt-meters W1and W2.

R

ST

A30.13365.17

3/400

Z

VIa) 875.82

125.7

90

TOT

RNR

386.33V.7706)(13.30163)(IZ3Vb) //

1.154313.30)W(W3Q

16.731313.30WWPc)

2

21LOAD

2

21LOAD

The Aron connection results in:

W1 = 4616,1 W, W2 = 4262,5 W

17.365j1.154)(16.731j1)(0.5Z)Z//Z(ZZ7.125

//LYLTOT ZY

W1 W2

V I V I

RT R ST SSR S TR T

TS T RS R

W1

W2ZL

ZL

ZL

AV

R

S

T

3535


Example 5

Three-phase 400 V-50 Hz line. When switch K2is closed , WA= 4000 W. When K1and

K3are closed, WA= 28352.6 W and WB= -11647.4 W. Determine: a) R2,b) R1and c) AT.

R

ST

j34.64A6010

400

10

400

R

V

R

VIII

60180

1

RT

1

TSRTTST1

R

S

T

N

WA

AT

WB

K1

C1 R1

R1 R2

K2 K3

AS

a) K2 closed: WA = VSTIScos(VST -IS)

4000 = 400IScos(0+30) IS = 11.55 A

R2 = VSN/IS = (400/3)/11.55 = 20

b) K1 and K3 closed:

PTOT = W1 + W2 = -WB + WA = 40000 W = 24002/R1 + 400

2/R2 R1 = 10

c) K1 and K3 closed:

j0A2020

400

R

VI180

2

TST2

A87.18j34.64-80j0)(-20j34.64)60(III -156.6T2T1Ttotal

This results in AT = 87.18 A

W1 W2

V I V I

RT R ST S

SR S TR T

TS T RS R

36


Question 1

An electrical lineman is connecting three single-phase transformers in a Y(primary)-

Y(secondary) configuration, for a businesss power service. Draw the connecting

wires necessary between the transformer windings, and those required between the

transformer terminals and the lines. Note: fuses have been omitted from this

illustration for simplicity.


10/11

37


Question 2

Identify the primary-secondary connection configuration of these pole-mounted

power transformers (i.e. Y-Y, Y-Delta, Delta-Y, etc.).

HV

LV

R ST

These transformers are connected in a Yyconfiguration.

38


Question 3



These transformers are connected in a Ydconfiguration.

39


Question 3



These transformers areconnected in open-delta

configuration.

Three single-phase transformers are not normally used because this is more

expensive than using one three-phase transformer.

However, there is an advantageous method called the open-Delta or V-

connection

It functions as follows: a defective single-phase transformer in a Dd three-phase

bank can be removed for repair. Partial service can be restored using the open-Delta configuration until a replacement transformer is obtained.

Three-phase is still obtained with two transformers, but at 57.7% of the original

power.

This is a very practical transformer application for emergency conditions.

40


Question 4

One of the conductors connecting the secondary of a three-phase power distribution

transformer to a large office building fails when open. Upon inspection, the source of

the failure is obvious: the wire overheated at a point of contact with a terminal block,

until it physically separated from the terminal. What is strange is that the overheated

wire is the neutral conductor, not any one of the line conductors. Based on this

observation, what do you think caused the failure?

After repairing the wire, what would you do to verify the cause of the failure?

Heres a hint (pista): if you were to repair the neutral wire and take current

measurements with a digital instrument (using a clamp-on current probe, for safety), you

would find that the predominant frequency of the current is 150 Hz, rather than 50 Hz.

This scenario is all too common in modern power systems, as non-linear loads such asswitching power supplies and electronic power controls have become more prevalent.

Special instruments exist to measure harmonics in power systems, but a simple DMM

(digital multimeter) may be used as well to make crude assessments.


11/11

4141


POWER MEASURMENT. ARON CONNECTION

General 3-wire load. Two-wattmeter method (ARON connection)

p(t) = vRN(t)iR(t) + vSN(t)iS(t) + vTN(t)iT(t)

p(t) = vRN(t)iR(t) + vSN(t)iS(t) + vTN(t)[-iR(t) - iS(t)]

p(t) = iR(t)[vRN(t) - vTN(t)] + iS(t)[vSN(t) - vTN(t)] = vRT(t)iR(t) + vST(t)iS(t)

Mean value

Ptotal= W1+ W2= VRTIRcos(VRT-IR) + VSTIScos(VST-IS)

,

LOAD

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