Swinburne's Test or No Load Test

 Swinburne's Test or No Load Test

AU May-04.05,06,08.16.17. Dec.-04,06,08,16

• This is indirect method of testing d.c. motors in which flux remains practically constant i.e. specially in case of shunt and compound motors. Without actually loading the motor the losses and hence efficiency at different loads can be found out.

 • The motor is run on no load at its rated voltage. At the starting some resistance is connected in series with the armature which is cut when motor attains sufficient speed. Now the speed of the motor is adjusted to the rated speed with the help of shunt field rheostat as shown in the Fig. 5.5.1.

• The no load armature current I_a is measured by ammeter A₁ whereas the shunt current is measured by ammeter A₂. om air to bac

• If V is the supply voltage then motor input at no load will be,

Power input at no load = V (I_a+I_sh) watts

There will be Cu loss in the field winding which will be given as,

Field copper loss = V × I_sh

Let R_a be the resistance of armature,

Armature copper loss = I_a² R_a

• Thus the stray losses which includes iron, friction and windage losses can be obtained as,

Stray losses = Input at no load - Field copper losses - No load armature copper losses

 Stray losses =V (I_a+I_sh) - (V × I_sh) - (I_a² R_a) = W_a

• In the field and armature windings there will be copper loss due to flow of current which will increase the temperature of the field and armature winding when the motor is loaded. This increase in temperature will affect their resistances.

• Thus the new value of field resistance R'_sh and that of armature R'a can be found by considering that rise in temperature as about 40 °C.

If α₁ = Resistance temperature coefficient of copper at room temperature

R'_a = R_a (1+α₁×40)

At room temperature the shunt field winding resistance will be,

Now if we want to find the efficiency of the motor at say ¼ th full load. It can be calculated as follows,

Let I_F.L = Full load current of motor

W_F =Field copper loss

W = Stray losses

This is the efficiency of motor when the load on motor is ¼  th of full load which can be found without loading the motor. The efficiencies at other loads can be calculated similarly.

 

1. Advantages

1. Since constant losses are known, the efficiency can be estimated at any load.

2. The method is convenient and economical as less power is required for testing even a large motor i.e. only no load power is to be supplied. gorb

3. The motor is not required to be loaded i.e. only test to be carried out is the no load test.

 

2. Disadvantages

1. In this method, the iron losses are assumed to be constant which is not the true case as they change from no load to full load. Due to armature reaction at full load there will be distortion in flux which will increase the iron loss.

2. The only test which is carried out is the no load test. Hence it is difficult to know whether there will be satisfactory commutation at full load.

3. We have assumed that there is rise in temperature of 40 °C at full load which cannot be checked actually as we are not actually loading the motor.

4. As it is a no load test it cannot be performed on a series motor.

Efficiency as a motor :

Input = V I, Armature copper loss = 1_a² R_a

W = Stray losses, W_F = Field copper loss

Efficiency as a generator:

Output = V_tI_L, Armature copper loss = I_a²R_a

W = Stray losses, W_F = Field copper loss

Ex. 5.5.1 The no load test of a 44.76 kW, 220 V d.c. shunt motor gave the following results: Input current = 13.25 A, Field current = 2.55 A, Resistance armature at 75 °C = 0.032 Ω, Brush drop = 2 V Estimate the full load current and efficiency.AU May-16, Marks 16

Sol. :

Sol : No load current I = 13.25 A,

Full load power output = 44.76 kW = 44760 W

No load power input = V.I. = 220 × 13.25 = 2915 W

In case of shunt motors, I = I_sh + I_a

I_a = I - I_sh = 13.25 - 2.55 = 10.7 A

No load armature copper loss = I_a² . R_a = (10.7)² (0.032) = 3.664 W

Loss due to brush drop = 2 × 10.7 = 21.4 W

Constant losses = No load input - No load armature

Cu loss - Loss due to brush drop

= 2915 - 3.664 - 21.4 = 2889.936 W

Now let I_a be the full load current in armature.

.. Full load motor input current,

I = I_a + I_sh =  (I_a + 2.55) A

Full load motor input = V.I. = 220 (I_a + 2.55) W

I = I_a + I_sh = (I_a + 2.55) A

Full load motor input = V.I.= 220 (I_a + 2.55) W

Motor input = Motor output + Total Losses

= Motor output + constant loss + brush loss + armature Cu loss

220 (I_a + 2.55) = 44760 + 2889.936 + 2 I_a + I_a² R_a

= 44760 + 2889.936 + 2 I_a + I_a² (0.032)

0.032 I_a² + 2 I_a - 220 I_a + 44760+ 2889.936

- (220) (2.55) = 0

0.032 I_a² - 218 I_a + 47088.936 = 0 i.e. I_a = 223.325 A

Full load motor input current = 223.325 + 2.55

= 225.875 A

Full load motor input = V . I = (220) (225.875)

= 49692.5 W

Full load motor efficiency = Output / Input × 100

= 44760 / 49692.5 × 100 = 90. 074%

Ex. 5.5.2 A 50 kW, 440 V, shunt generator having an armature circuit resistance including the interpole winding of 0.15 at normal working temperature was run as a shunt motor on no load at its rated voltage and speed. The total current drawn by the motor was 5 A, including shunt field current of 1.5 A. Calculate the efficiency of the shunt generator at i) Full load ii) 3/4th load and iii) Half load.

Sol. :

Ex. 5.5.3 A 500 V, D.C. shunt motor takes a total current of 5 A when running unloaded. The resistance of armature circuit is 0.25 Ω and the field resistance is 125Ω. Calculate the efficiency and output when the motor is loaded and draws a current of 100 A.AU Dec.-06, Marks 6

Sol. :

Ex. 5.5.4 A 220 V, D.C. shunt motor takes 4 A at no-load when running at 700 r.p.m. The field resistance is 100 ohm. The resistance of armature at standstill gives a drop of 6 volts across armature terminals when 10 A pass through it.

Calculate i) Speed on load ii) Torque in N-m and iii) Efficiency: The normal input of the motor is 8 kW. AU: Dec.-08, Marks 16

Sol. :

Review Questions

1. With neat circuit diagram explain how will you conduct Swinburne's test. Also show how to calculate the efficiency when the machine is running a) as a motor b) as a generator. Also state the merits and demerits of this method. AU: Dec.-04, 16, May-04, 05, 08, 17, Marks 16

2. A 440 V d.c. shunt motor takes a no load current of 2.5 A. The resistance of the shunt field and the armature are 550 Ω and 1.2 Ω respectively. The full load line current is 32 A. Find the full load output and the efficiency of the motor. 

(Ans.: 83.91 %)

3. A 80 h.p., 250 V d.c. shunt motor runs on full load with a speed of 1200 r.p.m. The armature resistance is 0.04 Ω and shunt field resistance is 250 Ω. Determine the mechanical power developed on full load in kW. Assume full load efficiency of the motor as 80 %.

If now the load on the motor is completely removed and it is made to run light, find its no load speed and no load current drawn from the supply. Assume stray losses constant on all load conditions. 

(Ans.: 70.812 kW, 1250.9122 r.p.m.)

 4. A 250 V, d.c. shunt motor takes a full load current of 67 A. Its armature resistance is 0.15 Ω and shunt field resistance is 166.67 Ω. When running on no load, the current drawn by the motor is 6.5 A and the speed as 1280 r.p.m. Determine i) Full load speed ii) Speed regulation iii) H.P. rating of the machine iv) Full load efficiency. 

(Ans. 1233.396 r.p.m., 3.78%, 19.42 h.p., 86.47%)

5. A 400 V shunt motor takes 4.1 A when running light at 950 r.p.m. The armature resistance is 0.8 Ω, the total brush drop is 2 V and R_sh = 250 Ω. Determine full load efficiency, full load regulation and h.p. output when full load current is 50 A. Assume 5 % armature reaction. 

(Ans.: 82.03%, 5.26 %, 22.3 h.p.)

6. A 10 kW, 250 V, d.c. shunt motor with an armature resistance of 0.8 Ω and a field resistance of 275 Ω takes 3.91 A, when running light at rated voltage and rated speed.

What conclusions can you draw from the above data regarding machine losses?

Calculate the machine efficiency as a generator when delivering an output of 10 kW at rated voltage and speed and as a motor drawing an input of 10 kW. What assumption if any do you have to make in this computation. 

(Ans. ng = 81.24%, = 78.07 %)

7. The full load rating of 240 V, d.c. shunt motor running at 1000 r.p.m. is 8 kW. The armature and shunt field resistance are 0.4 Ω and 240 Ω respectively. The efficiency at full Determine,

i) No load current ii) Speed when motor takes 12 A iii) The armature current when the torque developed is 60 N-m.

(Ans. : I_L = 2.2814 A, N₁ = 1045.95 r. p. m, I_a1 = 27.89 A)

8. A 230 V d.c. shunt motor is taking 5 A when running light. The armature resistance is 0.2 Ω and field circuit resistance is 115 Ω. For an input current of 72 A, calculate the shaft output and efficiency. Also calculate the armature current at wo which the efficiency is maximum. 

(Ans. : η= 87.14%, I_a = 75.76 A)