electrical machines-i laboratory manualsmec.ac.in/sites/default/files/lab1/em-i lab manual.pdf ·...

ST.MARTIN’S ENGINEERING COLLEGE ELECTRICAL MACHINES-I LAB MANUAL

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

ELECTRICAL MACHINES-ILABORATORY MANUAL

FOR YEAR 2015-2016

BY

DEPARTMENT OF ELECTRICAL AND ELECTRON-ICS ENGINEERING

DHULLAPALLY, KOMPALLYSECUNDERABAD-500014



230 V DC Supply

DPST Switch

(0-500)V

(0-2)A MC

− F

FF

V

+400Ω/ 1.7A

−

A

F

M

F

+DPST Switch

Fuse

3 point starterFL A

A

Fuse

A

G

A−

A

+

−

DC Excit-

er

+

1. Magnetization Characteristics of DC Shunt Generator

Aim: To conduct an experiment on a D.C shunt generator and draw the magnetization characteristics (OCC) and to determine the critical field resistance and critical speed.

Apparatus:

Name plate details: (To be noted Down from the Machine)

Circuit diagram:

S. No Apparatus Type Range Qty1 Voltmeter M.C 0-250/500V 1

2 Ammeter M.C 0-1/2A 1

3 Rheostats Wire wound

400/1.7A 1

4 Tachometer Digital 0-9999 1



Shunt Field Resistance (Rsh):-

Theory: Open circuit characteristics or magnetization curve is the graph be-tween the generated emf and field current of a dc shunt generator. For field cur-rent is equal to zero there will be residual voltage of 10 to 12V because of the residual magnetism present in the machine .If this is absent there the machine can not build up voltage to obtain residual magnetism the machine is separately excited by a dc source from OCC we can get critical field resistance and critical speed.

Critical field resistance: It is the resistance above which the machine cannot build up emf.

Critical speed: It is the speed below which the machine cannot build up emf.

Procedure:

1. Connections are made as per the circuit diagram.2. Start the motor and bring it to rated speed..3. The switch SPST is opened and If=04. For the different values of excitations (If) the generated voltage (Eg)from

the voltmeter is taken at rated speed, with increasing and decreasing orders.5. Calculate average Eg from increasing and decreasing orders.6. A graph is drawn between Avg Eg & If. From the graph (OCC) Critical

field resistance and critical speed are calculated.

400Ω/1.7A (0-2A)

(0-250V)

A

V

F

FF

+

+

−

−

230 V DC

Sup-ply

+

−

DPST FUSE



Tabular column:

Graph:

S.NOField current

IfGenerated Voltage (Eg)

Average EgIncreasing Decreasing

Q

A

CO

Eg (V)

If (A)

Rf

P

R



Critical field resistance (Rc) = OA/OC

Field resistance (Rf) = OR

The maximum voltage the Generator can induce With this field resistance. = OM

Critical Speed = PQ/PR * N

Result:



(0-2)A MC

−

L

(0-250)V MC−

V

+400Ω/ 1.7A

−

A

FM

FF

230 V DC Sup-ply

+DPST Switch

Fuse

3 point starterFA

A

A

(0-20)A MC

S1 S2

A

+

2. BRAKE TEST ON DC SHUNT MOTOR. DETERMI-NATION OF ITS PERFORMANCE CURVES

Aim: To conduct brake test on DC Shunt motor. And to determine its perfor-mance curves.

Apparatus:

S. No Equipment Range Type Qty1. Voltmeter 0-250V M.C. 1

2. Ammeter 0-20A M.C 1

3 Ammeter 0-1/2A M.C 14 Rheostat 400/1.7A Wire wound 15. Tachometer Digital type 1

6. Connecting wires


Circuit diagram:



Theory: When if is required to determine directly efficiency if comparatively small motors, the motor is loaded directly by means of Mechanical Break. Hence in the case of shunt motor there is no drastic change in speed. The TorqueT = (S1 ~ S2) g. r – Nm. where S1 S2 is the spring balance reading, r = Break drum Radius and g=9.81.P = Power developed. Efficiency of DC motor = Po/ Pi x 100

Procedure:-

01. Make Connections as per the circuit diagram.02. Start the motor with the help of the starter.03. Then bring the motor to rated speed by adjusting field rheostat.04. Put the mechanical load on the motor in steps and note down correspond-

ing readings of all meters.05. Do calculations accordingly.

Tabular columns :

S.No Voltage(V)

Current(I)

Speed(N)

Spring Balance Readings

Torque=9.8 1(S1 ~ S2) .r -Nm

Pout = 2 nT/60 -Watts

Pin = Vi -Watts

Eff = op/ip x100.

S1 S2



EXPECTED GRAPHS:

Ta vs Ia.

N

Ia0 Y

X

N vs Ia

0

TshT

X

Y

Ia

Ta



Vs O/P

Result:

η

O/P

X

Y

N

N vs T

Y

X

T



(0-2)A MC

−

L

(0-250)V MC−

V

+400Ω/ 1.7A

−

AA

FM

FF

230 V DC Sup-ply

+DPST Switch

Fuse

3 point starterFA

A

A

(0-20)A MC

S1 S2

A

+Y

YY

3. BRAKE TEST ON DC COMPOUND MOTOR DE-TERMINATION OF PERFORMANCE CURVES

Aim: To conduct brake test on dc compound motor.

Apparatus:

S. No Equipment Range Type Quantity1 Voltmeter (0-250V) M.C. 1 No2 Ammeter (0-20A) M.C 1 No.3 Rheostat 400/1.7A Wire wound 14 Tachometer digital 0-9999 1No5 Connecting wires

Name plate details (To be noted Down from the Machine)

Circuit diagram:

Cumulative:



(0-2)A MC

−

L

(0-250)V MC−

V

+400Ω/ 1.7A

−

AA

FM

FF

230 V DC Sup-ply

+DPST Switch

Fuse

3 point starterFA

A

A

(0-20)A MC

S1 S2

A

+YY

Y

+ −

Differential:

Theory:A Compound motor has a shunt field winding as well as series field winding. If the series field mmf and shunt field mmf help each other it is a cumulative compoundmotor. If the series and shunt fields appose each other it is a differentially com-pound motor. The operation of differential compound motor is unstable In a cumu-lative compounded motor the fluxes are add each other at light loads the shunt field is stronger than series field so motor behaves shunt motor. At high loads series field is stronger than shunt field so the characteristics like nearly to series motor.

Procedure:1. Connections are made as per the circuit diagram.2. Start the motor with the help of the starter.3. Then bring the motor to rated speed by adjusting field rheostat.4. Put the mechanical load on the motor in steps and note down all the meter read-ings.



Tabular columns:

s.no Voltage(V)

Current(I)

Speed(N)

Spring balance readings

Torque=9.81xS1~S2xr -Nm

Output power2πNT/60-Watts

Input powerVI-Watts

EfficiencyPout/P in x 100.

S1 S2

Graph: Draw graphs O/P Vs Speed, Current, Torque, Efficiency.

N vs Ia

0

CumDiff

X

Y

T

N



N

N vs T

Y

X

T

Cum

Diff

T

Ia0 Y

X

T vs Ia

Cum

Diff



Vs O/P

Result:

η

O/P

X

Y

Cum

Diff



(0-5A)

+ −

Resistive Load

(0-250v)

F

−

F

400Ω/ 1.7A

FF

230 V DC Sup-ply

+DPST Switch

Fuse

3 point starterAL

AA

M

A

400Ω/ 1.7A

A

AA

G

A

A

V

+

−

(0-20A)

+ −DPST Switch

F

FF

4. LOAD TEST ON DC SHUNT GENERATOR.

Aim: To conduct a load test on the given DC Shunt generator and to obtain the performance characteristics.

Apparatus required:


Circuit diagram:

1 Ammeter 0-20A, MC 1 0-1A, MC 1

0-5A MC 12 Voltmeter 0-250V, MC 1

0-30V, MC 13 Rheostat 400/1.7A 14 Rheostat 100/5A 15 Load 3 Kw / 220V 16 Tachometer 15 Connecting wires



Armature Resistance (Ra):-

Theory: . By conducting load test on DC shunt generator we can get load characteris-tics i.e, Internal & External characteristics. By exciting the m/c, the field current increases and voltage build up. After the machine has attained 220V the rated load is switched on. With increase in load, the voltage will be droppedProcedure:

1. Connections are made as per the circuit diagram.2. Start the machine with the help of starter and bring to rated speed by vary-

ing field rheostat of motor, then by varying field rheostat of the generator set the rated voltage of the generator.. Then close the DPST switch of the load and increase the load by step 0.125Kw, up to full load of the generator.

3. Note down all the meter readings at every step.4. Do necessary calculations.

Observations:

S no

IL, in amps

If , in amps

Ia= IL+ifin amps

Vt in volts

Ia Ra involts

EG = Vt + IaRa

in volts

A

M V

A

AA

+

+

−

−

230 V DC

Supply

+

−

DPST FUSE

100Ω/5A (0-5A)

(0-30V)



Graph:

E&V vs I

Result:

External

E& vs I

E&

I

X

Y

Internal

P



Y

Resistive Load

F

−

F

400Ω/ 1.7A

FF

230 V DC Sup-ply

+DPST Switch

Fuse

3 point starterAL

AA

M

A

AA

G

A

A

(0-250v)V

+

−

(0-20A)

+ −DPST Switch

YY

5. LOAD TEST ON DC SERIES GENERATOR

Aim: To conduct load test on the given DC series generator and to obtain its per-formance characteristics.

Apparatus required:

S.NO Equipment Range Type Qty1 Ammeter. 0-20A M.C. 1

0-5A MC 12 Voltmeter. 0-250V M.C 1

0-30V MC 13 Rheostat 400/1.7A Wire wound 1

100/5A Wire wound 14 Load 5,Kw 15 Tachometer 0-9999 Digital 1

Nameplate Details: (To be noted Down from the Machine)

Circuit diagram:-




Series Field Resistance (Rse):-

Theory: The load characteristics curve of DC series generator shows the relation

b/w its terminal voltage and load current. The characteristics are rising in nature and excitation increases with load. At large values of load current, the terminal voltage must be start decreasing owing to the saturation of the machine iron & rapidly increasing voltage drop of armature and armature resistance.

100Ω/5A(0-5A)

(0-30V)

A

V

Y

YY

+

+

−

−

230 V DC

Sup-ply

+

−

DPST FUSE

AA

100Ω/5A(0-5A)

(0-30V)

A

M V

A

+

+

−

−

230 V DC

Sup-ply

+

−

DPST FUSE



Procedure: 1. Make connections as per the circuit diagram.2. Adjust the speed of the motor to its rated value using field rheostat.(motor).3. Connect the load to generator with the help of load box, and increase the

load 0.125Kw at every step and note the corresponding readings.4. Plot the graph b/w terminal voltage Vs current and generated voltage Vs

armature current.

Observations:Speed of the motor, N =

Sno Terminal volt-age, in volts

Load current, IL=Ia=Ise in amps

IaRaIn volts

IaRseIn volts

Eg=V+IaRa+IaRseIn volts

Graph: Plot the graph b/w terminal voltage and load current by taking ‘V’ on Y-

axis and ‘IL’ on X-axis, and Eg on Y axis and Ia on X axis.

Result:

Eg & V vs Ia= Ise

Eg & V

Ia = Ise

X

Y

Internal

OCC

External



6. SPEED CONTROL OF DC SHUNT MOTOR

Aim: To conduct speed controls on DC shunt motor. The methods are1. Armature voltage control method 2. Flux control method

Apparatus:

S.No Equipment Range Type Qty1 Ammeter 0-5A

0-2AMCMC

1No1No

2 Voltmeter 0-250V MC 1No3 Rheostats 100/5A Wire wound 1NO

400/1.7A Wire wound 1No4 Tachometer 0-2000rpm Digital 1No5 Connecting Wires LS

Nameplate Details (To be noted Down from the Machine)

Circuit diagram:

A

M

A

AA

230 V DC

Supply

+

−

DPST FUSE

100Ω/5A

+

−(0-5A)

V

+

−

(0-250V)

L A F

A

400Ω/1.7A

+

−(0-2A)

F

FF




Theory:i) Armature voltage control method:

For a load of constant Torque, the speed is proportional to the applied to the arma-ture. Therefore speed voltage characteristic is linear and is a straight line. As the voltage is decrease across the armature the speed falls. This method gives speeds less than rated speeds.

Eb α ΦN Eb α NV-Ia(Ra+R) α NAs the voltage is decreased speed decreases.

ii) Flux Control Method:With rated voltage applied to the motor, the field resistance is increased i.e field current is decreased. I t is observed that speed increases.

Eb/Φ α N N α Eb/If

The characteristics If Vs N is inverse (or) if it is hyperbola.Procedure:i) Armature Voltage Control Method

1) Make connections as per the circuit diagram.2) Show the connections to the lab instructor.3) Keeping both rheostats at minimum, Start the motor with the help of starter and

by adjusting field rheostat bring the motor to rated speed.4) By increasing armature circuit rheostat in steps note down voltage, Ia and speed

at every step.5) The corresponding graph is draw between armature Voltage Vs speed.

AA

100Ω/5A(0-5A)

(0-30V)

A

M V

A

+

+

−

−

230 V DC

Sup-ply

+

−

DPST FUSE



ii) Flux Control method:

1) The machine run at its rated speed and rated voltage obtained.2) The voltage is kept constant and for different values of field current the

speed are noted.

Tabular Column:Armature Voltage Control Method:

S.No Armature Voltage in volts

Armaturecurrent=Ia in amps

Speed in RPM

Eb=V-IaRa in volts

Flux Control Method:

S.No Field Current in amps Speed in RPM

Expected graphs:-

OX

Y

If(I)

N

N Vs If



N vs Va(Armature voltage)

Result:

OX

Y

Va(V)

N

N Vs Va



6. SWINBURNE’ S TEST ON DC SHUNT MACHINE PREDETERMINATION OF EFFICIENCIES

Aim: To perform no load test on dc motor and to predetermine the efficiencies of the machine acting as a motor and generator.

Equipment:

S.No Apparatus Type Range qty

1 Voltmeter MC 0-250v 1

2 Voltmeter MC 0-30V 1

3 Ammeter MC 0-5A 1

4 Ammeter MC 0-2A 1

5 Rheostats Wire wound 400/1.7A 1Wire wound 100/5A 1


Circuit diagram:

(0-250)V MC

(0-5)A MC

(0-2)A MC

L

−−

+400Ω/ 1.7A

−

AA

FM

FF

230 V DC

Supply

+DPST Switch

Fuse

3 point starterFA

A

A

+

A

V

+ −



−

(0-5)A MC

100Ω/5A

−

AA

M

230 V DC Sup-

ply

+DPST Switch

Fuse

A

A

(0-30)V MC

−

V

+

+

Circuit diagram to find out Ra:

Theory: It is simple indirect method in which losses are measured separately and the efficiency at any desired load can be predetermined. This test applicable to those machines in which flux is practically constant i.e. shunt and compound wound machines. The no load power input to armature consist iron losses in core, friction loss, windage loss and armature copper loss. It is convenient and economi-cal because power required to test a large machine is small i.e. only no load power. But no account is taken the change in iron losses from no load to full load due to armature reaction flux is distorted which increases the iron losses in some cases by as 50%

Procedure:

1. Make connections as per the circuit diagram.2. Show the connections to the lab instructor.3. Keeping both rheostats at minimum, Start the motor with the help of starter and by adjusting field rheostat bring the motor to rated speed.4. Note down all the meter readings at no load..5. Do necessary calculations and find out the efficiency of the Machine as a motor and as a generator.6. Draw the graphs between output Vs efficiency of the Machine as a generator and as a motor.



Observations:

IL IF IA V N

For Ra

Expected graphs:-

Efficiency Vs Output

S.NO V I Ra=V/I

η (%)

Motor

Generator

O Out-



Tabular Column to find out efficiency:

GENERATOR:

Motor:

S.No Voltagein volts

Load Current in amps

ArmatureCurrent Ia =(IL+If)

Armature Cu loss=Ia XIaXRa

Total lossesWt=Wc+

IaXIaXRa

Input-VxIL

Output-Input-total losses=VxIL-Wt

=OutputInput.

S.No Voltagein volts

Load Current in amps

ArmatureCurrent Ia =(IL-If)

Armature Cu loss=Ia XIaXRa

Total lossesWt=Wc+

IaXIaXRa

Output=VxIL

Input=output+total loss-es=VxIL+Wt

=OutputInput.



Model calculations:

No load input=V IL

No load armature copper losses =Ia 2 Ra =(Il –If)2 Ra

Constant losses Wc=V l–(Il-If )2 Ra

Efficiency as a motor:

I= Assumed load currentMotor i/p=VIIa=IL-IfMotor armature losses=I2a .Ra Total losses=I2

a Ra+ WcEfficiency of motor= VI- I2

a Ra+ Wc / VI x 100

Efficiency as generator:

I=assumed load currentGenerator O/P =VI

Generator armature cu. Losses= I2a .RaTotal losses= I2

a Ra+ WcEfficiency of generator=VI / VI+ I2

a Ra+ Wc

Results:



(0-2A)

+ −

Resistive Load

(0-250v)

F

−

F

400Ω/ 1.7A

FF

230 V DC Sup-ply

+DPST Switch

Fuse

3 point starterAL

AA

M

A

400Ω/ 1.7A

A

AA

G

A

A

V

+

−

(0-20A)

+ −DPST Switch

F

FF

Y

YY

7. LOAD TEST ON COMPOUND GENERATOR

Aim: To conduct load test on DC compound generator and to determine its characteristics.

Apparatus:


Circuit diagram for cumulative compound generator:

S. No Equipment Range Type Qty

1. Voltmeter 0-250 V M.C. 1

2. Ammeter 0-2A0-20A

M.C.M.C.

11

3. Rheostats 400/1.7A Wire wound 2

4. Tachometer Digital 1

5 Connecting wires



(0-2A)

+ −

Resistive Load

(0-250v)

F

−

F

400Ω/ 1.7A

FF

230 V DC Sup-ply

+DPST Switch

Fuse

3 point starterAL

AA

M

A

400Ω/ 1.7A

A

AA

G

A

A

V

+

−

(0-20A)

+ −DPST Switch

F

FF

YY

Y

Circuit diagram for differential compound generator:



100Ω/5A(0-5A)

(0-30V)

A

V

Y

YY

+

+

−

−

230 V DC

Sup-ply

+

−

DPST FUSE

AA

100Ω/5A(0-5A)

(0-30V)

A

M V

A

+

+

−

−

230 V DC

Sup-ply

+

−

DPST FUSE



Theory: D.C. Compound generator consists of both series and shunt field wind-ings. The shunt and series fields can be connected in two ways.

1. Short shunt.

2. Long shunt.

When the MMF of series field opposes the MMF of shunt field, the gener-

ator is differentially compound. The terminal voltage decreases sharply with in-

creasing load current. Evidently this connection is not used.

In cumulative compound the connections of the two fields are such that

their MMF’s added and help each other. If the series field is very strong, the termi-

nal voltage may increase as the load current increases and it is called over com-

pounding. When terminal voltage on full load and no load are equal, it is known as

flat compounded generator. If the series field is not strong, the terminal voltage

will decreases with increase in load current (under compound)

Procedure:

1. Connections are made as per the circuit diagram.

2. The machine is run at rated speed and the rated voltage is obtained by vary-

ing field excitation

3. There the switch is closed so that load is connected across the generator.

4. Increase the load step by step with 0.125Kw and note down all the meter

readings and calculations are made accordingly and the characteristics are

obtained.

5. Plot graph for internal external characteristics.



Tabular Column:-

Model graphs:-

Internal characteristics

S.No IL , Amps VL Volts If , Amps Eg = vl+IA (rn+rsc)

OX

Y

Differential

Cumulative

Ia (A)

E (V)

E (V) Vs Ia



External characteristics

Result:-

OX

Y

Differential

Cumula-

Ia (A)

V

V Vs Ia



(0-250v)

(0-20A)

+ −

Resistive Load

(0-250v)

−

A

AA

230 V DC Sup-ply

+DPST Switch

Fuse

AL

M

AA

G

A

A

V

+

−

(0-20A)

+ −DPST Switch

Y

YY

A

V

Y

YY+

−

2 point starter

8. FIELDS TEST ON TWO IDENTICAL DC SERIES MACHINES

Aim: To determination the efficiency of two mechanically coupled series ma-chines by conducting field’s test.

Apparatus:

Name Plate Details (To be noted Down from the Machine)

Circuit diagram:

S.No Equipment Range Type Qty

1. Voltmeter 0-250V2-30V

M.C.MC

31

2 Ammeter 0-20A0-5A

M.C.M.C.

21

3. Resistive load 5Kw4. Connecting wires





Theory: This test is applicable for two series machines which are coupled mechani-cally. Series machines cannot be tested on no load conditions due to dangerous high speeds. One machine normally run as motor and drives generator whose out put is wasted in a variable load R. The fields of two machines are connected in se-ries in order to make iron losses of both the machines equal.

Model calculations:Generator output = V3.I2 Watts. - - - - - - - - - (1)Total input = V1.I1 Watts. - - - - - - - - (2)Total losses Pt of both machines = ( V1.I1)-(V3.I2) Watts - - (3) Motor Field Cu loss = I1.I1.Rsem - - - - - - - - - - (4)Motor Armature Cu loss = I1.I1.Ram - - - - - - - - - - -(5)

100Ω/5A(0-5A)

(0-30V)

A

V

Y

YY

+

+

−

−

230 V DC

Sup-ply

+

−

DPST FUSE

AA

100Ω/5A(0-5A)

(0-30V)

A

M V

A

+

+

−

−

230 V DC

Sup-ply

+

−

DPST FUSE



Generator Field Cu loss = I1.I1.Rseg - - - - - - - - - - (6)Generator Armature Cu loss = I2.I2.Rag - - - - - - - - - - - -(7)P2,Total Ra and Se ,Cu losses Of both machines = (4)+(5)+(6)+(7)- - - - -- - --(8)Ps, Stray losses of both m/c = (Pt-P2) Watts - - - - - - - - -(9)Ps/2, Stray losses of each m/c = (Pt-P2)/2 Watts- - - - - - - - -(10)

Efficiency calculations for Generator:-Output of Generator =V3 . I2 Watts---------------(11)Stray losses of Generator = (Pt-P2)/2 Watts ----------(12)Field Cu loss of Generator. = I1 . I1 .Rseg----------------(13)Armature Cu loss of Generator. = I2 .I2 .Rag------------------(14)Input to Generator. = (11)+(12)+(13)+(14)-------(15)Efficiency of Generator =Output/Input =(11)/(15)Efficiency calculations for Generator:-Motor in put . =V2 .I1 Watts----------------(16)Motor Field Cu loss = I1 .I1.Rsem-----------------(17)Motor Armature Cu loss =I1 . I1 .Ram-----------------(18)Motor Stray losses =(Pt-P2)/2 Watts------------(19)Motor Output =(16)-(17)+(18)+(19)------(20)Efficiency of Motor =Output/Input=(20)/(16)

Model graphs:-

Result: By conducting the field’s test we found the efficiency of series machines (motor and generator).

η (%)

Motor

Genera-

O Out-



230 V

DC Sup-

(0-2)A MC

F

FF

A

AA

(0-230)V MC

L

−V

+

−

AA

M

+DPST Switch

Fuse

FA

A

400Ω/ 1.7A

+

V

A

M

A

400Ω/ 1.7A

F

FF

3 point start-

−

(0-500)V MC

A

(0-2A) MC

+

−

+

−

(0-20)A MC

+ −

9. REGENERATIVE (OR) HOPKINSON’S TEST

Aim : To conduct a Hopkinson’s test on a two similar D.C shunt machines and find out the efficiency.

Apparatus Required:

S.no Equipment Range Type Qty1 Volt meter 0-250V M.C. 12 Ammeter 0-20A

0-2AM.CM.C

22

3 Rheostat 400/1.7A Wire wound 24 Connecting wires


Circuit Diagram:




Theory:

Hopkinson’s test is also called as regenerative test or back-to-back test. It is an indirect test or full test, which is used to determine the efficiency of the two identical shunt machines. The two machines are mechanically coupled and are also adjusted electrically that are of then run as motor and other as a generator. The two-shunt machines are connected in parallel. The power input from the mains is only that needed for supplying the losses of the two machines. The two machines can be tested under full load conditions (for determining the efficiency and maxi-mum temperature rise).

Procedure:

1. Connected the circuit as per the circuit diagram.

2. Keep the field regulator minimum resistance position and start the motor by

using starter, Keeping S.P.S.T switch open.

3. Adjust the regulator on generator side until the rated voltage equal to both

in magnitude and polarity as that of main supply. i. e; voltmeter reads zero.

4. The S. P. S.T switch is closed to parallel the machines, by adjusting the re-

spective field regulators, any load can how be thrown on to machines.

5. Calculate efficiency by applying load.(changing excitation)

Observations:

S. NoInput Volt-age in volts

Input Cur-rent=I1

Gen Ia=I2Gen If=I3 Motor If=I4

AA

100Ω/5A(0-5A)

(0-30V)

A

M V

A

+

+

−

−

230 V DC

Sup-ply

+

−

DPST FUSE



To find out efficiency:Motor:S. No Motor

inputMotorArmatureCu loss

MotorField Cu loss

Strayloss

Total lossesof mo-tor

Out putOf Mo-tor

of Mo-tor

Generator:S. No

Generatoroutput

GeneratorArmatureCu loss

GeneratorField Cu loss

Strayloss

Total losses of generator

InputOf generator

of generator

Calculations:Armature Resistance of each machine =RaGenerator Armature cu loss =I1 x I2 x Ra Watts.Motor Armature cu loss =(I1 + I2)(I1 + I2)Ra Watts.Armature power input to the set. =VL x I1 Watts. Ps, Stray losses of both machines = VL x I1 Armature Cu loss of (Gen +Motor) Stray losses of each machine = Ps/2

Efficiency of Generator: Generator output =VL . I2 WattsGenerator Losses Pg = V.I3 + I2 x I2+(Ps/2)Efficiency of Generator =( VL .I2 )/ (VL .I2 +Pg)

Efficiency of Motor :

Motor in put: =VL (I1 +I2 +I4).Motor losses Pm =(I1 +I2)(I1 +I2)Ra +VxI4 +Ps/2 . Motor Efficiency = VL (I1 +I2+I4)-Pm/ VL (I1+I2+I4)

Plot the following graphs:-(a) Output Vs Efficiency for Generator.(b) Output Vs Efficiency for Motor.



Model graphs:-

Result:

η (%)

Motor

Genera-

O Output



10.SEPARATION OF STRAY LOSSES IN A DC MACHINE

Aim :To conduct the No-load Test at various speeds at two different excitations on a DC shunt machine and to determine following losses:-(a) Hysteresis Loss (b) Eddy Current Loss (c) Mechanical Loss.

Apparatus Required:

S.no Equipment Range Type Qty1 Volt meter 0-250V M.C. 1

0-30V M C 12 Ammeter 0-5A

0-2AM.CM.C

22

3 Rheostat 400/1.7A Wire wound 1100/ 5A Wire wound 1

4 Connecting wires5 Tachometer digital 0-9999 1


Circuit Diagram:

A

M

A

AA

230 V DC

Supply

+

−

DPST FUSE

100Ω/5A

+

−(0-5A)

V

+

−

(0-250V)

L A F

A

400Ω/1.7A

+

−(0-2A)

F

FF



Circuit diagram to find out Ra:

Procedure:

1. Connected the circuit as per the circuit diagram.

2. Keep both field and armature rheostats at minimum position and start the

motor by using starter, and bring to rated speed by adjusting field rheostat.

3. Note down all the meter readings, repeat this by varying armature rheostat.

The field current to be kept constant.

4. Adjust the field to another suitable value and repeat step -3

5. Find the armature resistance by conducting the experiment.

Observations: Field current (if)=

S. No

ArmatureVoltage=Va

ArmatureCurrent=Ia

SpeedN

Back EMFEb

Armatureinput

ArmatureCu loss Stray loss

A

M V

A

AA

+

+

−

−

230 V DC

Supply

+

−

DPST FUSE

100Ω/5A (0-5A)

(0-30V)



Evaluation of friction, Hysterisis and Eddy current losses(Different speeds)

Calculations:-

Stray losses (Ps)= Mechanical loss + Eddy current loss + Hysterisis lossAt constant normal excitation:Ps=AN+BN2 +CN+DN2 ----------------------------- (1)

At constant reduced excitation(Ps/N)=(A+C1)+(B+D1)N --------------------------------- (2)

Plot the graph between speeds Vs Ps/NFrom the graph at two different speeds determine the values of Ps/N, for normal and reduced excitations and find the values (A+C),(B+D),(A+C1) and (B+D1)And from these values calculate the values of C-C1,D-D1.

The co-efficient of hysteresis loss C is proportional to 1.6, and the co-efficient of eddy current loss D is proportional to 2. If and ’ are the fluxes corresponding to the normal and reduced excitation ,the:-

(C’/C) = (’/)1.6

(D’/D) = (’/)2 at the same speed Also,(’/) = (E’,E) at any speed.

From the above equations evaluate the equations the constants A,B,C&D. Hence evaluate the friction, Hysteresis and Eddy current losses at various speeds up to the rated speed and tabulate the results in the table:-

S. NoSpeedN

Friction lossAN+BN2

Hysterisis LossCM

Eddy current lossDN2



Model graphs:-

Result

OX

Y

3/4 Excitation

Normal Excitation

Speed (N)

PS / N

Ps/N Vs N



L

(0-250)V MC

−

VA

+

−

AA

M

230 V DC

Supply

+DPST Switch

Fuse

2 point starter

A

A

A

(0-20)A MC

S1 S2

Y

YY

+ −

11. Brake Test On DC Series Motor

AIM: To draw the performance characteristics of DC series motor by performing Brake

APPARATUS:


2 Ammeter 0-20A M.C 13 Connecting wires4 Tachometer 0-9999 digital 1


Circuit Diagram:



THEORY : DC series motor is having high starting torque and its speed will be decreases by increasing of load .series motor runs on load only. It implies that the motor starts only when the load is applied on it. If S1, S2 are spring balance read-ing force

T= (S1-S2)*G*r

r- brake drum radiuso/p power P=T*W

= 2πNT/60Input power Pin =VILefficiency η = Pout/Pin*100

SPECIFICATION RATINGS OF DC SERIES MOTOR :

PROCEDURE:

1) Construct the circuit as shown in the figure 2) Apply some load and then switch on DPST switch3) Take down the readings of N,S1,S2,IL4) Calculate the efficiency under different loads5) Plot the graph between o/p and i/p

Efficiency vs o/p Torque vs IL

Speed vs IL

Speed vs T

PRECAUTIONS:

1) See that before switching on DPST whether some load is applied or not. If notapply some load

2) Pour water on brake drum whenever you are changing the load



TABLE:

Model Graphs

S.NO. VOLTAGE LOAD CURR ENT

SPEED S1 S2 TORQUE P=2ΠNT/60 (out put power)

Input power

η = Pout/Pin

1) - - - -

-

- - - -

2) - - - -

-

- - - -

3) - - - -

-

- - - -

0

Tsh

T

X

Y

Ia

Ta



N

N vs T

Y

X

Ta

N

Ia0 Y

X

N vs Ia



RESULT:

vs O/P

η

O/P

X

Y

η



(0-250V)MC

A

(0-20A)MC

+ −

(0-20A)MC

+ −L F A

MC

A F

400Ω/1.7A 1.7A

−

F

FF

230 V

DC

+DPST

Fuse

3 point starter

A A

M

A

V

V

G2G1M

230 V DC

Supply

Fuse

A

Field REV Switch

F

F

(0-250V)MC

+ −

OFF

S

DPDT

A

AA

A

AA

400Ω/1.7A 1.7A

F

FF

F

(0-20A)MC

+

−+−

Resistive load

Machine Machine

12. Parallel Operation of Two DC Shunt Generators



12. Parallel Operation of Two DC Shunt Generators

AIM: To run two DC shunts generators in parallel and study the load sharing.

Name Plate Details (To be noted down from the machine):

Apparatus:


2 Ammeter 0-20A M.C 33 Rheostat 400/1.7A Wire wound 24 Resistive Load 5 KW 15 DPDT Switch 26 SPST Switch 14 Connecting wires5 Tachometer digital 1

PROCEDURE:

1. Ensure that the paralleling switch S1 is ‘OFF’ positions .open and the change over switch S IS IN

2. Start machine NO1 and adjust the field excitation so that it generates the rated voltage and record the reading.

3. Put switch ‘S’ in the positon-1 and the gradually increase in the load in the steps.

4. Note the load current of machine-1 and its terminal voltage.5. Repeat the step [d] till the machine one is fully loaded.6. Bring the load to zero and the stop the machine-1.7. Put change over switch in ‘OFF’ position. Now start machine-2 and adjust the

voltage to rated value and repeat the steps done for machine-1.8. Stop the machine and put the change over switch in ‘OFF’ position.9. Run both machine keeping parallel switches S1 open.10. Adjust the voltage each machine to its rated value and if the polarity is correct the

parallel volt meter V2 will read zero if not reverse the polarity of any one machine. when parallel volt meter reads zero , close the parallel switch S1 by keeping the



change over switch in either voltmeter reads zero , close the parallel switch S1 by keeping the change over switch in either position 1or 2. Load the machine and note down the individual machine load current, the total load current and the busbar voltage.

11. Change the excitation of one of machine and observe the changes in ammeter readings of each machine.

Observation Table:

S.no Generator 1 Generator 2 Gen1 & Gen 2 Parallel Total Current

C, Bus Bar vol

Voltage Current Voltage Current Load CurrentGen 1

Load CurrentGen 2

Result:

electrical machines-i laboratory manualsmec.ac.in/sites/default/files/lab1/em-i lab manual.pdf ·...

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