# Testing Of Induction Motor

Like other machines induction motors are tested to get the necessary information to know parameters, losses, efficiency, speed, torque, current, P.f., etc. The following tests are made on induction motors:

- Direct test
- No-load test
- Blocked (or locked) rotor test
## Direct or Load Test

Another machine like a generator, mechanically coupled with the shaft of the motor acts as a load. The load on the motor is changed by changing the electric load on the generator. The output of the generator and input of the motor are determined by means of various meters connected as shown in Figure 1.

**Figure 1: Direct or load test on 3 Phase induction motor**Let the generator is of DC type. Generator output = V

_{2}I_{2}, where V_{2}, I_{2}are output voltage and current:Efficiency of generator,

**η**_{g}= output of generator / Input of generator = V_{2}I_{2}/ input of generatoror

Input of generator = V

_{2}I_{2}/**η**_{g}But input of generator = output of motor = V

_{2}I_{2}/**η**_{g}The efficiency of the motor, ηm = output of motor / Input of motor = (V

_{2}I_{2}/**η**_{g}) / W_{1}+ W_{2}where W_{1}+ W_{2}= input power to the motor.The direct test has the disadvantage of large consumption of power during testing, Also the availability of a load connected to the motor is necessary.

## Indirect Or No-Load Test

The no-load test of an induction motor determines the friction and windage losses and the magnetization current. The no-load test is carried out with different values of applied voltage. The power input is measured by two wattmeters, no-load current "I

_{o}" by ammeter, and the input voltage "V" by voltmeter which are included in the circuit shown in Figure 2(a). The readings of the total power input "W_{o}", "I_{o}" and voltage "V" are plotted as shown in Figure 2(b). If we extend the curve for "W_{o}" it cuts the vertical axis at point A. "OA" represents losses due to friction and windage. If we subtract the loss corresponding to "OA" from "W_{o}" then we get the no-load electrical and magnetic losses in the machine.

**Figure 2: (a) No-Load Test (b) Curves obtained from No-Load test**"OB" represents normal voltage. Hence at normal voltage, the no-load input power "W

_{o}" consists of:(i) Small stator cu. Loss

(ii) Core loss

(iii) Loss due to friction and windageThese losses can be found from the curve by drawing a vertical line from B.

BD = loss due to friction and windage

DE = stator cu. Loss

EF = Core lossKnowing the core loss, G

_{o}, B_{o,}and Y_{o}can be determined asG

_{o}= Core loss / 3V_{2}

Y_{o}= I_{o}/V

B_{o}= √( Y_{o}^{2}– G_{o}^{2})Input power factor angle can also be found from input power:

W

_{o}= √3 V_{L}Io Cos Ø_{o}

CosØ_{o}=√3 V_{L}I_{o}

Where

V_{L}= Line voltage.## Blocked Rotor Test

This test is also known as the locked-rotor or short-circuit test. This test is used to find:

- Short circuit current with normal voltage applied to the stator
- P.f. on short circuit
- Total Cu. Losses
- Equivalent circuit elements i.e., total leakage reactance "X
_{01}” referred to the stator, and total resistance "R_{01}" referred to the stator.

In this test rotor is locked i.e. it is held stationary by some means. The rotor is short-circuited at slip rings if the motor has a wound rotor. Just as in the case of a short circuit test on a transformer, a reduced voltage (up to 20 percent of normal value) is applied to the stator terminals. This voltage is adjusted so that full load current flows in the stator. As here slip, equal to one, the equivalent circuit of the motor is exactly like a transformer having short-circuited secondary. The values of current, voltage, and power on the short circuit are measured by different meters connected to the circuit.

The following are found from this test:

**Short circuit current I**_{SN}obtainable with normal voltage "V"Let:

Z

_{s}= Short circuit impedance

I_{s}= Short circuit current with voltage "V_{s}"

Short circuit current on normal voltage can be found as I_{sN}= IsV / V_{s}**Power factor on the short circuit:**W

_{s}=√3 V_{L}I_{SL}CosØs

CosØ_{s}= W_{s}/ √3 V_{SL}I_{SL}where

W

_{s}= short circuit input power

V_{SL }= line voltage on short circuit

I_{SL}= line current on short circuit.**Cu. Loss.**Motor input on short circuits consists of mainly stator and rotor Cu. Losses and core losses:

W

_{s}= C_{cu}+ W_{CL}

where Wcu = Total copper losses

W_{CL}= Core losses

Or

W_{cu}= W_{s}- W_{CL}**R**_{01}, X_{01}, Z_{01},From an equivalent circuit diagram of the induction motor

Z

_{01}= V_{s}/I_{s}

And from total Cu. Losses

W_{cu}= 3I_{s}^{2}R_{01}

Or

R_{01}= W_{cu}/3Is^{2}

X_{01}= Root (Z_{01}^{2}– R_{01}^{2})In the case of the squirrel cage rotor, R

_{1}is determined as usual and after allowing for Skin effect is subtracted from R_{01}to give the effective rotor resistance R_{2}as referred to the stator.R

_{2}= R_{01}- R_{1}