Short shunt compound generator with heater load

The heater load takes $I_L = 175$ A, so the terminal voltage follows from Ohm's law. In a short shunt machine the series field carries the load current and the shunt field is across the armature.

       Ia                 Rse   IL
  +--->---+----------+--[~~~]--->--+--o +
  |       |          |            |
 (E,Ra)   |        [Rsh]       Heaters
  |       |          |            |
  +-------+----------+------------+--o -

Terminal voltage

$$V = I_L R_L = 175 \times 1.4 = 245.0\ \text{V}$$

Shunt and armature currents

$$V_{sh} = V + I_LR_{se} = 245.0 + 175(0.1) = 262.5\ \text{V}$$ $$I_{sh} = \frac{V_{sh}}{R_{sh}} = \frac{262.5}{100} = 2.625\ \text{A}, \qquad I_a = I_L + I_{sh} = 177.625\ \text{A}$$

Generated emf

Brush drop $= 2\ \text{V}\times 2 = 4\ \text{V}$.

$$E = V_{sh} + I_aR_a + V_{brush} = 262.5 + (177.625)(15) + 4 = 2930.9\ \text{V}$$

Copper losses

$$\begin{aligned} P_a &= I_a^2R_a = (177.625)^2(15) = 473259.6\ \text{W} \\ P_{se} &= I_L^2R_{se} = 175^2(0.1) = 3062.5\ \text{W} \\ P_{sh} &= I_{sh}^2R_{sh} = (2.625)^2(100) = 689.1\ \text{W} \\ P_{cu} &= 473259.6 + 3062.5 + 689.1 = 477011.2\ \text{W} \end{aligned}$$

Brush contact loss $= V_{brush}I_a = 4 \times 177.625 = 710.5\ \text{W}$, so the total electrical loss is $477011.2 + 710.5 = 477721.7\ \text{W}$.

Electrical efficiency

$$\text{Output} = VI_L = 245.0 \times 175 = 42875\ \text{W}$$ $$\text{Electrical power generated} = EI_a = 2930.9 \times 177.625 = 520596.7\ \text{W} \;(= 42875 + 477721.7)$$ $$\eta_{elec} = \frac{VI_L}{EI_a} = \frac{42875}{520596.7}\times 100 = 8.24\%$$

Answer: $E = 2930.9\ \text{V}$; copper loss $= 477011.2\ \text{W}$ (plus $710.5\ \text{W}$ brush contact loss); electrical efficiency $= 8.24\%$. (The armature resistance of $15\ \Omega$ is used as given. It is almost certainly a misprint, since a practical armature resistance is a few hundredths of an ohm, which is why the emf is so high and the efficiency so low.)