The following readings are obtained when doing a load test on a d. c. shunt motor using a brake drum:

Spring balance reading 10 kg and 35 kg, Diameter of drum 40 cm, Speed of motor 950 rpm, Applied voltage 200 V, Line current 30 A. Calculate efficiency of motor.

This question was previously asked in

PTCUL AE E&M 2017 Official Paper (Set B)

Option 4 : None of these

**Concept of Brake test:**

- It is a direct method and consists of applying a brake to a water-cooled pulley mounted on the motor shaft as shown in fig.
- The brake band is fixed with the help of wooden blocks gripping the pulley.
- One end of the band is fixed to earth via a spring balance S and the other is connected to a suspended weight W1.
- The motor is running and the load on the motor is adjusted till it carries its full load current.

Let W_{1} = suspended weight in kg

W_{2} = reading on spring balance in kg-wt

The net pull on the band due to friction at the pulley is

= (W_{1} - W_{2}) kg. wt. or 9.81 (W_{1} - W_{2}) N.

If R = radius of the pulley in meter

and N = motor or pulley speed in r.p.s.

Then, shaft torque Tsh developed by the motor

= (W_{1} - W_{2}) R kg-m = 9.81 (W_{1} - W_{2}) R N-m

Motor output power = Tsh × 2π N watt

= 2π × 9.81 N (W_{1} - W_{2}) R watt

= 61.68 N (W_{1} - W_{2}) R watt

Let V = supply voltage; I = full-load current taken by the motor.

Then, input power = VI watt

\(\eta = \frac{{\rm Output}}{{\rm Input}} = \left( {\frac{{61.68\left( {{W_1}\ -\ {W_2}} \right)R}}{{{V_1}}}} \right)\)

**Calculation:**

Given: W_{1} = 35 kg wt, W_{2} = 10 kg wt, V = 200 V, I = 30 A, N = 950 rpm, Diameter of drum = 40 cm

Force on the drum surface,

F = (35 - 10) = 25 kg wt

= 25 × 9.8 N

Drum Radius R = 40/2 = 20 cm = 0.2 m

Torque (T_{sh}) = F × R

T_{sh} = 25 × 9.8 × 0.2

= 49 N

Speed (N) = 950 rpm = 950/ 60 rps = 95/6 rps

\(ω = \frac{{2\pi f}}{{60}}\)= \(ω = 2\pi \left( {\frac{{95}}{6}} \right)\) = 99.5 rad/sec

Motor output = T_{sh} × ω Watt

= 49 × 99.5

= 4876 Watt

Motor input = 200 × 30 = 6000 Watt

\(\eta = \frac{{4876}}{{6000}} = 0.813 = 81.3\;\% \)