EMF and efficiency of short shunt compound generator (50 A, 200 V)

In a short-shunt compound generator the shunt field is connected directly across the armature, and the series field is in series with the load.

       Ia            Rse = 0.05     IL = 50 A
   +---->----+-------/\/\/\-----+--->---o +
   |         |                  |
 ( E )     [Rsh]                       V = 200 V
 Ra=0.08   100 ohm                load
   |         |                  |
   +---------+------------------+-------o -

Step 1: Voltage across the shunt field

$$V_{sh} = V + I_LR_{se} = 200 + 50 \times 0.05 = 202.5\ \text{V}$$

Step 2: Shunt field and armature currents

$$\begin{aligned} I_{sh} &= \frac{V_{sh}}{R_{sh}} = \frac{202.5}{100} = 2.025\ \text{A} \\ I_a &= I_L + I_{sh} = 50 + 2.025 = 52.025\ \text{A} \end{aligned}$$

Step 3: Generated EMF

$$\begin{aligned} E_g &= V_{sh} + I_aR_a = 202.5 + 52.025 \times 0.08 \\ &= 202.5 + 4.162 = 206.66\ \text{V} \end{aligned}$$

Step 4: Losses

The no-load (rotational: iron + friction) loss of 300 W is taken as the stray loss; the field and armature copper losses are added separately.

$$\begin{aligned} \text{Armature Cu loss} &= I_a^2R_a = 52.025^2 \times 0.08 = 216.53\ \text{W} \\ \text{Series field loss} &= I_L^2R_{se} = 50^2 \times 0.05 = 125\ \text{W} \\ \text{Shunt field loss} &= V_{sh}I_{sh} = 202.5 \times 2.025 = 410.06\ \text{W} \\ \text{Stray (no-load) loss} &= 300\ \text{W} \\ \text{Total loss} &= 216.53 + 125 + 410.06 + 300 = 1051.59\ \text{W} \end{aligned}$$

Check: $E_gI_a = 206.66 \times 52.025 = 10751.6$ W = output + copper losses ($10000 + 751.6$).

Step 5: Efficiency

$$\begin{aligned} \text{Output} &= VI_L = 200 \times 50 = 10000\ \text{W} \\ \eta &= \frac{10000}{10000 + 1051.59}\times 100 = 90.48\% \end{aligned}$$

Answer: Generated EMF $E_g =$ 206.66 V; efficiency $\approx$ 90.48%.