No Load Test on Induction Motor / Procedure & Application


No Load Test on Induction Motor

There are three types of tests to be performed for checking the performance of an induction motor

1.No load test or open circuit test.
2.Blocked rotor test or short circuit test or locked rotor test.
3.Stator winding resistance test.
Now let's see each test in detail:

No load test:

Main objective of conducting the no load test is:

1. To find the constant losses of the induction machine.
2. To divide constant losses into iron losses and mechanical losses.
3. To find shunt branch elements of the equivalent circuit.
4. To find no load power factor of the induction machine.

Condition to conduct no load test:

1.To conduct the no load test we need to apply rated voltage and rated frequency to stator under no load condition and also the air gap flux is to be maintained constant.
2. If the power supply given to the induction machine is constant then one watt meter is enough in the circuit if not two watt meter's are required.

Circuit diagram to conduct no load test:



Procedure to conduct no load test:

1. Connections are to be done as per the circuit diagram given above.
2. load connected to the motor is to be removed.
3. Now we need to supply rated voltage at rated frequency to the stator and also maintain constant air gap flux.
4. Input power can be measured with the help of two watt meters connected in the circuit.
5. Ammeter and voltmeter connected in the circuit gives no load voltage and current.
6. As said there is no load given to the motor the total input power given to the motor is equal to the sum of constant iron loss, mechanical losses like friction and windage losses and copper loss but it is in small amount.
7. Copper losses at rotor side can be neglected in no load test because the slip is small at no load.

Experiment values table:  

               P1
            P2
           ILINE
           VLINE









Equivalent circuit:

at no load as slip is very less we can take slip approximately equal to zero.
R2'( 1/s -1) ≈ ∞  since no load I2r' = 0
so the equivalent circuit now gets modified as follows:

                        
And the current flowing through xm is Iu  and R0 is I2.

Calculations:

V = Line to line voltage.
I = Average value of line currents flowing throw ammeters.
WNL  = P1  + Pwhich is input power or three phase power.
All the readings we get from the circuit are line values.

With star connected stator winding:

VNLperphase = V/√3 here v is the voltmeter reading in the circuit.
I0 perphase =  I here I is the ammeter reading in the circuit.
cosᴓ0 =  WNL / 3×VNLperphase×I0perphase
I1perphase = I0perphase × cosᴓ0.
Iuperphase = I0perphase × sinᴓ0.
R0 = VNLperphase / I2perphase.
Xm = VNLperphase / Iu perphase.

With delta connected stator winding:

VNL  = V = vlotlmeter reading of the circuit.
I0 perphase = I/√3 I is the ammeter reading.
cosᴓ0 =  WNL / 3×VNLperphase×I0perphase.
I1perphase = I0perphase × cosᴓ0.
Iuperphase = I0perphase × sinᴓ0.
R0 = VNLperphase / I2perphase.
Xm = VNLperphase / Iu perphase.

Power factor:

Based on watt meter readings we can find power factor approximately
If  P1 = P2  then power factor (lagging) = Unity.
if  P1 = 2Pthen power factor(lagging) = 0.866.
if  P2= 0 , P1 = total 3 phase power then power factor(lagging) = 0.5
if P1 = - P2 then power factor(lagging) =  Zero.

Finding out constant losses:

At no load input power supplied WNL = iron losses of stator + mechanical losses + stator copper losses.
1.Here the core losses of rotor are not considered as there is no load losses at rotor side are negligible.
2.And as there is no load mechanical power output is zero.
3.For induction machine mechanical losses are almost constant irrespective of load because of less speed variations.
Therefore iron loss of stator + mechanical losses = Constant losses.
WNL = Constant losses + stator copper loss at no load .
constant losses = WNL - stator copper losses at no load.
constant losses =  WNL - 3I02R1.
where R1 is the per phase resistance of stator winding.
R1 can be measured by using bridges.

Dividing constant losses into Iron losses and Mechanical losses:

1. We need to conduct the no lo load test using variable voltage at constant frequency.
2. By changing the applied voltage note down the readings of watt meter.
3. Now find constant losses using, constant losses = WNL - 3I02R1.

Calculating iron loss:

As frequency is constant  iron losses α V12.
V1 is the varying voltage.

Calculating mechanical loss:

1.As we said that speed is almost constant in induction motor at constant frequency the mechanical losses are constant.
2. By adding load to the induction motor and decreasing the voltage at constant frequency the speed of induction motor falls to maintain load torque constant so mechanical losses will decrease. If applied voltage becomes zero iron losses will become zero.
Now draw the graph and from graph we can calculate the values of mechanical loss and iron loss.
Test has to be conducted from Vrated to Vmin.

In this way we can find out Iron losses and mechanical losses.
Now  all the objectives of No load test are fulfilled and hence the performance of machine can be known.