EMF and efficiency of 20 kW compound generator

Given: $P_{out} = 20$ kW, $V = 220$ V, $R_a = 0.02\ \Omega$, $R_{se} = 0.04\ \Omega$, $R_{sh} = 100\ \Omega$, iron + friction loss = 250 W.

Assumption: the type of compound connection is not stated, so the long-shunt connection is taken (the usual assumption). The short-shunt result is given at the end for comparison.

     +--Ra--Rse--+-----------+
     |           |           |
    (Eg)        Rsh        Load 220 V
     |           |           |
     +-----------+-----------+

Step 1: Currents

$$\begin{aligned} I_L &= \frac{20000}{220} = 90.909\ \text{A} \\ I_{sh} &= \frac{220}{100} = 2.2\ \text{A} \\ I_a &= I_{se} = I_L + I_{sh} = 93.109\ \text{A} \end{aligned}$$

Step 2: Generated emf

$$\begin{aligned} E_g &= V + I_a (R_a + R_{se}) \\ &= 220 + 93.109 \times 0.06 \\ &= 220 + 5.587 = 225.59\ \text{V} \end{aligned}$$

Step 3: Losses

$$\begin{aligned} P_{cu,a} &= I_a^2 R_a = 93.109^2 \times 0.02 = 173.39\ \text{W} \\ P_{cu,se} &= I_a^2 R_{se} = 93.109^2 \times 0.04 = 346.77\ \text{W} \\ P_{cu,sh} &= V I_{sh} = 220 \times 2.2 = 484\ \text{W} \\ P_{loss} &= 173.39 + 346.77 + 484 + 250 = 1254.16\ \text{W} \end{aligned}$$

Step 4: Efficiency

$$\begin{aligned} P_{in} &= 20000 + 1254.16 = 21254.16\ \text{W} \\ \eta &= \frac{20000}{21254.16} \times 100 = 94.10\% \end{aligned}$$

Answer: $E_g \approx 225.59$ V, efficiency $\eta \approx 94.1\%$.

(If short shunt is taken: $V_{sh} = 223.64$ V, $I_{sh} = 2.236$ A, $I_a = 93.15$ A, $E_g \approx 225.50$ V, total loss $\approx 1254.2$ W, $\eta \approx 94.1\%$.)