Obtaining Data to Plot Circle Diagram

Obtaining Data to Plot Circle Diagram

The data required to draw the circle diagram is obtained by conducting two tests which are,

1. No load test or open circuit test

2. Blocked rotor test or short circuit test

Let us discuss these tests in detail. 

 

1. No load test

In this test, the motor is made to run without any load i.e. no load condition. The speed of the motor is very close to the synchronous speed but less than the synchronous speed. The rated voltage is applied to the stator. The input line current and total input power is measured. The two wattmeter method is used to measure the total input power. The circuit diagram for the test is shown in the Fig. 6.4.1.

As the motor is on no load, the power factor is very low which is less than 0.5 and one of the two wattmeters reads negative. It is necessary to reverse the current coil or pressure coil connections of such a wattmeter to get the positive reading. This reading must be taken negative for the further calculations.

The total power input W₀ is the algebraic sum of the two wattmeter readings. The observation table is,

The calculations are,

This is no load power factor.

Thus we are now in a position to obtain magnitude and phase angle of no load current I₀, which is required for the circle diagram.

From the knowledge of I₀ and ϕ₀, the parameters of the equivalent circuit can be obtained as,

I_c = I₀ cos ϕ₀ = active component of no load current

I_m = I₀ sin ϕ₀ = magnetising component of no load current

R₀ = V₀ (per phase) / I_c (per phase) = no load branch resistance

X₀= V₀ (per phase) / I_m (per phase) = no load branch reactance

The power input W₀ consists of following losses,

1. Stator copper loss i.e. 3I₀² R₁ where I₀ is no load per phase current and R₁ is stator resistance per phase.

2. Stator core loss i.e. iron loss.

3. Friction and windage loss.

The no load rotor current is very small and hence rotor copper loss is negligibly small. The rotor frequency is s times supply frequency and on no load it is very small. Rotor iron losses are proportional to this frequency and hence are negligibly small.

Under no load condition, I₀ is also very small and in many practical cases it is also neglected. Thus W₀ consists of stator iron loss and friction and windage loss which are constants for all load conditions. Hence W₀ is said to give fixed losses of the motor.

W₀ = no load power input

W₀  = ... Neglecting stator copper loss

a. Separating No Load Losses

The no load losses are the constant losses which include core loss and friction and windage loss. The separation between the two can be carried out by the no load test conducted from variable voltage, rated frequency supply.

When the voltage is decreased below the rated value, the core loss reduces as nearly square of voltage. The slip does not increase significantly the friction and windage loss almost remains constant.

The voltage is continuously decreased till the machine slip suddenly begins to increase and the motor tends to stall. At no load, this takes place at a sufficiently reduced voltage. The graph showing no load losses PN.L. versus V as shown in the Fig. 6.4.2 is extrapolated to V = 0 which gives friction and windage loss as iron or core loss is zero at zero voltage.

 

2. Blocked Rotor Test

In this test, the rotor is locked and it is not allowed to rotate. Thus the slip s = 1 and R_L= R₂ (1 - s)/s is zero. If the motor is slip ring induction motor then the windings are short circuited at the slip rings. 

The situation is exactly similar to the short circuit test on transformer. If under short circuit condition, if primary is excited with rated voltage, a large short circuit current can flow which is dangerous from the windings point of view. So similar to the transformer short circuit test, the reduced voltage (about 10 to 15 % of rated voltage) just enough such that stator carries rated current is applied. Now the applied voltage V_sc, the input power W_sc and a short circuit current I_sc are measured.

As R_L = 0, the equivalent circuit is exactly similar to that of a transformer and hence the calculations are similar to that of short circuit test on a transformer.

V_sc = Short circuit reduced voltage (line value)

I_sc = Short circuit current (line value)

W_sc = Short circuit input power

Now W_sc = √3 V_sc I_sc  … Line values

Cos ϕ_sc = W_sc / √3 V_sc I_sc  

This gives us short circuit power factor of a motor.

Now the equivalent circuit as shown in the Fig.6.4.3.

This is equivalent resistance referred to stator.

During this test, the stator carries rated current hence the stator copper loss is also dominant. Similarly the rotor also carriers short circuit current to produce dominant rotor copper loss. As the voltage is reduced, the iron loss which is proportional to voltage is negligibly small. The motor is at standstill hence mechanical loss i.e. friction and windage loss is absent. Hence we can write,

W_sc = Stator copper loss + Rotor copper loss 

But it is necessary to obtain short circuit current when normal voltage is applied to the motor. This is practically not possible. But the reduced voltage test results can be used to find current ISN which is short circuit current if normal voltage is applied.

If V_L = Normal rated voltage (line value)

V_sc = Reduced short circuit voltage (line value)

where          I_sc = Short circuit current at reduced voltage

Thus, I_SN = Short circuit current at normal voltage

Now power input is proportional to square of the current

So     W_SN = Short circuit input power at normal voltage

This can be obtained as,

But at normal voltage core loss can not be negligible hence,

W_SN = Core loss + Stator and rotor copper loss

Review Questions

1. Which tests are required to be performed to obtain the data for the circle diagram ? How these tests are performed?

2. What data is obtained by conducting the no load test and short circuit test on three phase induction motor ?