**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:**

P
_{1} |
P
_{2} |
I
_{LINE} |
V
_{LINE} |

**Equivalent circuit:**

at no load
as slip is very less we can take slip approximately equal to zero.

R

_{2}'( 1/s -1) ≈ ∞ since no load I_{2r}' = 0
so the equivalent
circuit now gets modified as follows:

And the current flowing through x

_{m }is I_{u }and R_{0 }is I_{2}.**Calculations:**

V = Line to line voltage.

I = Average value of line currents flowing throw ammeters.

W

_{NL }= P_{1 }+ P_{2 }which is input power or three phase power.
All the
readings we get from the circuit are line values.

**With star connected stator winding**:

V

_{NLperphase}= V/√3 here v is the voltmeter reading in the circuit.
I

_{0 perphase}= I here I is the ammeter reading in the circuit.
cosᴓ

_{0 }= W_{NL }/ 3×V_{NLperphase}×I_{0}_{perphase}
I

_{1perphase}= I_{0perphase}× cosᴓ_{0}.
I

_{uperphase}= I_{0perphase }× sinᴓ_{0}.
R

_{0 }= V_{NLperphase}/ I_{2perphase}.
X

_{m }= V_{NLperphase }/ I_{u perphase}.**With delta connected stator winding:**

V

_{NL }= V = vlotlmeter reading of the circuit.
I

_{0 perphase }= I/√3 I is the ammeter reading.
cosᴓ

_{0 }= W_{NL }/ 3×V_{NLperphase}×I_{0}_{perphase}.
I

_{1perphase}= I_{0perphase}× cosᴓ_{0}.
I

_{uperphase}= I_{0perphase }× sinᴓ_{0}.
R

_{0 }= V_{NLperphase}/ I_{2perphase}.
X

_{m }= V_{NLperphase }/ I_{u perphase.}_{}**Power factor:**

Based on
watt meter readings we can find power factor approximately

If P

_{1 }= P_{2 }then power factor (lagging) = Unity.
if P

_{1 }= 2P_{2 }then power factor(lagging) = 0.866.
if P

_{2}= 0 , P_{1}= total 3 phase power then power factor(lagging) = 0.5
if P

_{1}= - P_{2}then power factor(lagging) = Zero.**Finding out constant losses:**

At no load
input power supplied W

_{NL }= 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.

W

_{NL}= Constant losses + stator copper loss at no load .
constant
losses = W

_{NL}- stator copper losses at no load.
constant
losses = W

_{NL }- 3I_{0}^{2}R_{1}.
where R

_{1}is the per phase resistance of stator winding.
R

_{1}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 = W

_{NL }- 3I_{0}^{2}R_{1.}_{}**Calculating iron loss:**

As frequency
is constant iron losses α V

_{1}^{2}.
V

_{1}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 V

_{rated }to V_{min}.
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.