Efficiency and engine BHP of 48.75 kW shunt generator

Efficiency is output divided by input, where input = output + all losses. The driving engine must supply this input, so its B.H.P. is the input power in horsepower.

Given: $P_{out} = 48.75$ kW, $V = 250$ V, $R_a = 0.02\ \Omega$, $R_{sh} = 50\ \Omega$, no-load (iron + friction) loss = 950 W.

Assumption: the 950 W no-load loss is the iron and mechanical (stray) loss; shunt field copper loss is added separately.

Step 1: Currents

$$\begin{aligned} I_L &= \frac{48750}{250} = 195\ \text{A} \\ I_{sh} &= \frac{250}{50} = 5\ \text{A} \\ I_a &= I_L + I_{sh} = 200\ \text{A} \end{aligned}$$

Step 2: Losses

$$\begin{aligned} P_{cu,a} &= I_a^2 R_a = 200^2 \times 0.02 = 800\ \text{W} \\ P_{cu,sh} &= V I_{sh} = 250 \times 5 = 1250\ \text{W} \\ P_{stray} &= 950\ \text{W} \\ P_{loss} &= 800 + 1250 + 950 = 3000\ \text{W} \end{aligned}$$

Step 3: Efficiency

$$\begin{aligned} P_{in} &= 48750 + 3000 = 51750\ \text{W} \\ \eta &= \frac{48750}{51750} \times 100 = 94.20\% \end{aligned}$$

Step 4: B.H.P. of the engine

Taking 1 hp = 746 W:

$$\text{B.H.P.} = \frac{51750}{746} = 69.37\ \text{hp}$$

(With metric hp, 735.5 W, this is 70.36 hp.)

Answer: Efficiency $\eta = 94.2\%$; B.H.P. of driving engine $\approx 69.37$ hp.