induction Machine report

8/18/2019 induction Machine report

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EE212 – SYNCHRONOUS AND INDUCTION MACHINES

LABORATORY

MACHINE REPORT

Roll No. 107114093

EEE

IV Semester



Aim:

To perform no load and blocked rotor tests on a three phase induction motor to determine

its equivalent circuit parameters.

To generate theoretical performance characteristics from the equivalent parameters.

To perform load test on the three phase induction motor and to obtain the experimental

performance characteristics and to compare it with the theoretical performance

characteristics.

To obtain theoretical performance characteristics for a given different operating voltage,

frequency, rotor resistance and for a constant v/f ratio.

MACHINE NAME PLATE DETAILS

Motor:

Rated Voltage : 415 VRated Current : 7.5 A

Rated Power : 3.7 kWRated Speed : 1430 rpm

NO LOAD AND BLOCKED ROTOR TEST

Apparatus Required:

1. Ammeter 0-10 A (MI) 10-5 A (MC) 1

2. Voltmeter 0-600 V (MI) 10-300 V (MI) 1

0-50 V (MC) 13. Wattmeter 500 V, 10 A (LPF) 1

250 V, 10 A (UPF) 14. Rheostat 230 Ω, 5 A 1

Precautions:

1. Appropriate fuse wires should be fixed.

2. The TPST switch should be kept in open position.

3. Autotransformer should be kept at minimum voltage position.

4. Initially the motor should be in no load condition.



Fig 1. CIRCUIT DIAGRAM – NO-LOAD TEST ON THREE-PHASE INDUCTION MOTOR

R

B

3-Phase400V

50 Hz

AC

Supply

Y

B

EC

C

C

B

B

E

E

Three-phase

Induction

Motor

Double element wattmeter

M L

CV

M L

CV

R

YB

T

PS

T

S

V

A

VL

IL

3Ф

Autotransformer

A1

A2

B1B2

C1

C2

PANEL BOARD



Fig 2. CIRCUIT DIAGRAM – BLOCKED ROTOR TEST ON THREE-PHASE INDUCTION MOTOR

Brake

Drum

S2

R

B

3-Phase400V

50 HzAC

Supply

Y

B

EC

C

C

B

B

E

E

Three-phaseInduction

Motor


M L

CV

M L

CV

R

YB

#

T

P

S

T

S

V

A

VL

IL

3Ф

Autotransformer

A1

A2

B1B2

C1

C2

S1

PANEL BOARD



FORMULAE TO BE USED

I. NO LOAD TEST

1. Woc = 3 VocIoc COS oc (W)

2.

COSϕoc =ococ

oc

I V W

3

3. Iw = ococ I

cos3

(A)

4. Im = ococ I

sin3

(A)

5. Xo =m

oc

I

V (Ω)

6. R o =

Iw

V oc (Ω)

II. BLOCKED ROTOR TEST

7. Z01 = )( phase per SC

phase per SC

I

V

8. R 01 =

phase per SC

phase per SC

I

W

2

(Ω)

9. X01 =2

01

2

01 R Z (Ω)

10. R 2' = (R 01 – R eff ) (Ω)

11.

R eff = R mean 1.6 (Ω)

III. CALCULATION OF LOSSES AND EFFICIENCY

12. Cos NL = W NL / (3 V NL I NL)

13. Cos sc = W sc / (3 V sc I sc)

14. I sc at rated voltage (ISN) = (V NL / V sc) I sc (A)

15.

W sc at rated voltage (WSN) = (V NL / V sc) W sc (W)16. Stator copper loss = 3 I2

NL R 1 (W)

17. Rotor copper loss = WSN – (3 I2 NL R 1) (W)

18. Percentage efficiency = (Output power / Input power) * 100



19. Input power (stator input) = (Output power + RotorCopper loss +

Stator Copperloss +Fixed losses)(W)

20. Rotor input = (Stator input – Stator Copperloss)(W)

21. Slip = Rotor copper loss / Rotor input

22. Multiplication Factor =deflection scale full

I V cos3

where,

Woc - Power consumed under no load (W).

R oc - No load resistance (Ω)

Ioc - No load current (A)

Xoc - No load reactance (Ω)

Voc - Rated voltage (V)

Wsc - Power consumed when the Rotor is blocked (W)

R 01 - Blocked rotor resistance (Ω)

X01 - Blocked rotor reactance (Ω)

Z01 - Blocked rotor impedance (Ω)

W NL - Power consumed under no load (W)V NL - No load supply voltage (V)

I NL - No load current (A)

Wsc - Power consumed when rotor is blocked (W)

Vsc - Blocked rotor voltage (V)

Isc - Blocked rotor current (A)

R 1 - Stator resistance (Ω)

NL - No load power factor

sc - Blocked rotor power factor



TABULATION

Multiplication factor : 4

Terminalvoltage,

Voc (V)

Line current,Ioc (A)

WOC

Observed (div) Actual (W)

415 4.3 122 488

Table 1. NO LOAD TEST ON THREE-PHASE INDUCTION MOTOR


Terminal voltage

Vsc (V)

Line Current

Isc (A)

Wsc

Observed

(div)

Actual

(W)

82 7.5 265 530

Table 2. BLOCKED ROTOR TEST ON THREE PHASE INDUCTION MOTOR

Sl.No Voltage, V (V) Current, I (A) Resistance, R=V/I (Ω)

1. 4 0.9 4.45

2. 6 1.35 4.45

3. 8 1.75 4.57

4. 10 2.2 4.54

5. 12 2.65 4.53

Table 3. TABLE MEASUREMENT OF STATOR RESISTANCE R eff

R mean = 4.508 Ω R eff = R mean*1.6 = 7.213 Ω



36 VDC Supply

+

_

DPST

I

VV

A+

+

_

_

A1

A2

Fig 3. CIRCUIT DIAGRAM – MEASUREMENT OF STATOR RESISTANCE ( R eff )

Fig 4. PER PHASE EQUIVALENT CIRCUIT OF THREE-PHASE INDUCTION

MOTOR

R 'L

X01

R 0

R 01

X0

I1

> >

>>

I2

Io

IW Iμ AC

Supply Voltage

>

N

P

Machine parameters obtained:

Iw = 0.392 A Xo = 169.39 Ω X01 = 16.419 Ω

Im = 2.45 A R o = 1058.67 Ω R eff = 7.213 Ω Ioc = 4.3 A R 01 = 9.42 Ω

Woc = 488 W R 2’ = 2.207 Ω



LOAD TEST

Apparatus required:

1. Ammeter 0-10 A (MI) 12. Voltmeter 0-600 V (MI) 1

3. Wattmeter 500 V, 10 A (UPF) 1

Formulae to be used:

1. Torque, T = 9.81× (S1~S2) ×R (N-m)

2. Output power, P0 =60

2 NT (W)

3. Input power, Pin = 3 L LV I cos (W)

4. % efficiency, = 100

in

O

P

P

5. Power Factor, cos = L L

in

I V

P

3

6. Synchronous speed, Ns = 120 f

p

(rpm)

7. Percentage slip, %s = 100 s

s

N N

N

8. Multiplication Factor =deflection scale full

I V cos3

where S1 and S2 are the readings of spring balance (Kg)

R is the radius of the brake drum (m)

N is the speed of the motor (rpm)

Pin is the input power (W)P0 is the output power (W)

VL is the line voltage (V)

IL is the line current (A)

f is the frequency of the input supply (Hz)

p is the number of poles



Fig 5. CIRCUIT DIAGRAM – LOAD TEST ON THREE-PHASE INDUCTION MOTOR

Brake

Drum

S2

R

B

3-Phase

400V

50 Hz

AC

Supply

Y

B

EC

C

C

B

B

E

E

Three-phase

Induction

Motor


M L

CV

M L

CV

R

YB

#

TP

S

T

S

V

A

VL

IL

3Ф Autotransformer

A1

A2

B1B2

C1

C2

S1

PANEL BOARD




Sl.No

LineVoltage

VL (V)

LineCurrent

IL (A)

Speed

N

(rpm)

Spring Balance

Readings (kg) Torque

T

(Nm)

Power

factor

Cos ф

Input

Power Pin

Output

Power

Po

(Watts)

%Slip

%Efficiency

η S1

(kg)

S2

(kg)

S1 ~ S2

(kg)

Obser

ved

(div)

Actual

(W)

1. 415 4.5 1497 0 0 0 0 0.062 100 200 0 0.2 0

2. 415 4.6 1491 4.7 1 3.7 3.147 0.278 460 920 491.36 0.6 53.41

3. 415 4.7 1486 7 1 6 5.103 0.37 625 1250 744.09 0.93 63.53

4.

415 4.75 1484 8 1.4 6.6 5.613 0.395 675 1350 872.28 1.07 64.61

5. 415 4.8 1483 10 1.5 8.5 7.229 0.435 775 1550 1122.66 1.13 72.43

Table 4. LOAD TEST ON THREE-PHASE INDUCTION MOTOR



Comparison of theoretical & experimental performance curves

Fig 5. Torque – slip characteristics

Fig 6. Output power vs Torque

It is seen from Fig 5. that the torque increases steeply with slight increase in slip in the operatingregion. The experimental and the theoretical curves almost coincide.

From Fig 6. It is seen that output power and torque follow linear relationship and this holds good

even under practical conditions.



Fig 7. Output power vs Line current

Fig 8. Output power vs Speed

From Fig 7. It is seen that the line current drawn experimentally is less than expected. This isdue to the experimental loading conditions which were far less than the rated load.

From Fig 8. It is seen that the experimental rotor speed is a bit less than the expected values.This is because of the mechanical losses which are unaccounted in the theoretical curve. It isalso seen that the speed reduces only by a few rpm.



Fig 9. Output power vs % slip

Fig 10. Output power vs efficiency

It is seen in Fig 9. that the experimental slip is a bit more than the expected values. This is because of the the same reason as of speed, i.e the mechanical losses which are unaccounted in

the theoretical curve.



Fig 11. Output power vs Power Factor

Fig 12. Torque vs slip for varying supply voltage

The power factor increases with loading (Fig 11.) because, the mechanical load on the motor isseen as an equivalent resistive load. Thus, as the load increases, the motor becomes more and

more resistive, thus the power factor increases.



Fig 13. Torque vs slip for varying rotor resistance

Fig 14. Torque vs slip for varying frequency



Fig 15. Torque vs slip for constant v/f

The efficiency of the motor (Fig 10.) is low initially because under no load and small loadconditions, comparatively more power is dissipated in magnetization and losses than on theload. This is why efficiency is small initially. On loading, the power used on the load (i.e the

output power) increases, and the output power is larger than the rotor copper losses. Thus theefficiency increases. But after a limit, the losses start increasing and the curve gets flatter.

In Fig 12. It is seen that torque decreases with decrease in line voltage. This is because torque

is proportional to the square of the line voltage.

So, as voltage is decreased, the shaft torque also decreases

It is seen from Fig 13 that higher rotor resistance provides higher starting torque (atslip=1). The critical torque for all cases is same because it is proportional to (E2

2)/(2X2) which

remains constant. The slip increases with increase in rotor resistance because increased rotorresistance causes an increase in the rotor copper losses. This results in decrease in efficiency of

the machine.

From Fig 14 it is observed that the torque increases with decrease in frequency of thesupply. This is because of the fact that torque is inversely proportional to the synchronous

speed Ns as given by the above expression. As frequency is decreased, Ns also decreases,causing an increase in torque. But the machine must be designed to withstand high torques.

Since V = 4.44ΦKTf , we can say that, V/f = 4.44ΦKT. (K is constant, T is constant underrunning conditions) Flux decreases with decrease in Voltage but flux increases with decreasein frequency. Thus, when the v/f ratio is kept constant, the flux remains constant, thus the

critical torque is constant. But the speed of the machine decreases because the synchronousspeed decreases with decrease in supply frequency, thus the slip increases with decrease in

frequency.





Fig 17. Power vs slip

75% Rated Voltage – Since torque is proportional to the square of voltage, the torque

decreases with decrease in applied voltage. Thus, speed decreases or slip increases with

decrease in voltage.

33 Hz - Torque increases with decrease in frequency because torque is inversely proportional

to Ns. This increase in torque causes the speed to improve, and the slip to decrease, as seen in

the graph.

33 Hz, constant v/f – when v/f is constant, but both V and f are decreased, the torque decreases

because torque is proportional to the square to voltage but inversely proportional to frequency.

But the stalling torque remains the same. The decreased torque causes the speed to reduce, or

the slip to increase which is reflected in the above graph.

4 R 2 – Since the rotor resistance is increased four times the actual value, the I2R losses in the rotor become so high that the torque reduces drastically. Due to increased losses, speed also decreases.



Fig 18. Power vs Speed

75% Rated Voltage – Since Speed and torque are directly proportional to voltage, speed dips as

voltage is decreased.

33 Hz – Here, Ns becomes 990 since frequency is changed. So, the speed is very less compared to

rated operation. But the speed does not dip quickly. It is seen that the slop of the curve depends

on the voltage and not frequency.

33 Hz, constant v/f – Here also, Ns becomes 990 since frequency is changed. But, voltage is also

decreased, so the decrease in the speed is more compared to the previous case and it is observed in

the graph.

4 R 2 – The drastic increase in rotor resistance causes rotor copper losses to increase, and hence the speto decrease.



Fig 19. Power vs power factor

75% Rated Voltage – When voltage is decreased, the flux produced in the motor decreases,

thus the reactive power supplied to the motor decreases, causing an increase in power factor.

33 Hz – With decrease in frequency, the flux produced in the machine increases, i.e the

reactive power required for the machine will increase. This increase in reactive power will

cause a decrease in power factor.

33 Hz, constant v/f – a decrease in voltage and decrease in frequency such that v/f ratio

remains constant will result in a constant flux, thus the reactive power does not vary. Thus, the

curve coincides with that of the rated operating conditions

.

4 R 2 – the increase in rotor resistance causes the motor to be more resistive, thus increasing the powerfactor.



Fig 20. Power vs Efficiency

75% Rated Voltage – When voltage is decreased, the flux produced in the motor decreases, thetorque also decreases but the losses remain the same. Thus, efficiency decreases when voltage is

decreased.

33 Hz – With decrease in frequency, the flux produced in the machine increases, thus

increasing the torque. Also, the decreased frequency also decreases the inductive reactance

losses, hence resulting in an improved efficiency.

33 Hz, constant v/f – Since voltage is decreased torque (which is proportional to the square of

voltage) also decreases. Even though frequency is decreased, the effect of decrease in voltage

dominates, resulting in a decreased efficiency.

4 R 2 – the increase in rotor resistance results in increased rotor copper losses, thus the efficiencyof the machine decreases due to increased losses.



Fig 21. Power vs Line current

75% Rated Voltage – When voltage is decreased, to provide a required torque, the motor needs

more current than while operating at rated voltage. Thus, for a particular output power, the line

current drawn increases.

33 Hz – Since frequency is decreased, torque increases, and thus, to provide a given output

power, the machine requires less current than that of the rated operating conditions. Moreover,

more flux is produced with less current when frequency is reduced.

33 Hz, constant v/f – Since voltage is decreased, it results in increased line current for a given

load condition. But due to the effect of decreased frequency, the machine draws less current

compared to the 75% of rated voltage operating condition.

4 R 2 – The increase in rotor resistance will result in more losses, thus more current will berequired for a particular load, compared to the rated operating conditions.



INFERENCES:

By analysing the machine performance characteristics at different operating voltages, we candecide the method of speed control to be used for different necessities.

Since speed is proportional to square of applied voltage, voltage can be varied to control speedover a large range.

v/f speed control is used because changing frequency alone may cause the cores to saturatequickly and the no load current of the motor will increase. By keeping flux constant this iseliminated.

Since increase in rotor resistance increases starting torque, this method is used in starting ofinduction motors. An external resistance is included in series with the rotor resistance whilestarting to improve the starting torque. Care should be taken so that the starting torque does not

exceed the breakdown torque of the machine. The external resistance is then removed after themotor has started.

It is seen that increased frequency results in decreased torque and speed. This shows the effect of

harmonics on the motor. Higher frequency harmonics cause decrease in torque and even crawlingin some cases.

Since power factor is poor at low loads, measures must be taken (such as using capacitor banks)while operating at low loads.

RESULT:

No load and blocked rotor tests were performed on a three phase induction motor to

determine its equivalent circuit parameters. Theoretical performance characteristics were generated

from the equivalent parameters. Load test was performed on the three phase induction motor and the

experimental performance characteristics were obtained and compared with the theoretical

performance characteristics. Theoretical performance characteristics were obtained for a given

different operating voltage, frequency, rotor resistance and for a constant v/f ratio and compared

with the rated performance characteristics.

induction Machine report

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