Question

(a) With reference to an a.c. generator used in marine practice derive an expression for the frequency of the generated e.m.f. in terms of the speed of the machine and the number of poles. (6)

(b) A 200V, long-shunt compound-wound generator has a full-load output of 20KW. The various resistances are as follows: armature (including brush contact) 0.15 Ω, series field 0.025 Ω, interpole field 0.028 Ω, shunt field (including the field-regulating resistance) 115 Ω. The iron losses at full load are 780W, and the friction and windage losses 590W. Calculate the efficiency at full load. (10)

Answer 1

Part (b)

Given:

$$Output=20000W$$

$$Iron\:loss=780W$$

$$Friction\:loss\:\left(mech\operatorname{loss}\right)=590W$$

$$R_{a}=0.15\Omega$$

$$R_{se}=0.025\Omega$$

$$R_{int}=0.028\Omega$$

$$R_{sh}=115\Omega$$

$$I_{sh}=\frac{V}{R_{sh}}=\frac{200}{115}$$

$$I_{sh}=1.74A$$

$$Copper\:loss\:in\:stator=I^2R$$

$$C_{S}=1.74^2\times115$$

$$C_{S}=348W$$

$$Gen\:output\:=\:V\times I_{L}=20000W\:\left(given\right)$$

$$200\times I_{L}=20000$$

$$I_{L}=100A$$

$$I_{a}=I_{sh}+I_{L}$$

$$=1.74+100$$

$$I_{a}=101.74A$$

$$Copper\:loss\:in\:stator=I^2R=I_{a}^2\left(R_{se}+R_{a}+R_{int}\right)$$

$$C_{R}=101.74^2\times\left(0.025+0.15+0.028\right)$$

$$C_{R}=2101W$$

$$Total\:Copper\:loss=C_{S}+C_{R}$$

$$=348+2101$$

$$=2449W$$

$$\eta=\frac{Output}{Input}=\frac{Output}{Output+losses}$$

$$=\frac{20000}{20000+780+590+2449}$$

$$\eta=83.96\%$$

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