Question

A 100kW, 230V, shunt generator has an armature resistance and field resistance of 0.05Ω
and 57.5Ω. If the generator operates at rated voltage. SPL is 1.8kW, calculate
a) At what load does the generator achieve maximum efficiency?
b) What is the value of this maximum efficiency?

Answer 1

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Answer 2

Given:

Output power, P₀ = 100 kW = 100 × 1000 W = 100000 W

Potential difference, V = 230 V

Armature resistance, R₁ = 0.05 Ω

Field resistance, R₂ = 57.5 Ω

SPL = 1.8 kW  = 1800 W

Find:

a) Load at which the generator achieve maximum efficiency.

b) The value of this maximum efficiency.

Solution and explanation:

Field current, $I_f$ = V/ R₂

                          = 230/57.5

                          = 4 A

Load current, $I_L$ = $P_0/V$

                           = 100000/230

                           = 434.8 A

Armature current, $I_a$ = $I_L + I_f$ = 434.8 + 4 = 438.8 A

Induced emf, E = V + $I_aR_1$ = 230 + 438.8 $\times$ 0.05

                          = 230 + 21.94 = 251.94 V

For maximum efficiency, armature losses should be equal to the constant losses, i.e.,

$P_c = I_a^2R_1$

$I_a = \sqrt{\frac{P_c}{R_1} }$

Constant losses, $P_c$ = $I_f^2R_2 + SPL$

                           $P_c = 4^2 \times 57.5 + 1800\\P_c = 16 \times 57.5 + 1800\\P_c = 920 + 1800 = 2720 W$

But at maximum efficiency, armature current is also maximum

$I_a = I_{a_{max}}$

$I_{a_{max}} = \sqrt{\frac{2720}{0.05} }$

$I_{a_{max}} = \sqrt{54400} = 233.24 \ A$    

a)

Load current for maximum efficiency,    

$I_L = I_{a_{max}} - I_f\\I_L = 233.24 - 4 = 229.24 \ A$

b)

Output power, $P_0$ = 100000 W

Input power, $P_i$ = $P_o + P_c$

                         = 100000 + 2720

                         = 102720 W

Hence, Maximum efficiency, $\eta_{max}$ = $\frac{P_o}{P_i} \times 100$

                                                        = $\frac{100000}{102720} \times 100$

                                                        = $0.9735 \times 100$ %

                                                        = 97.35 %

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