Question

(a) Explain the purpose of interpoles and state their magnetic polarity relative to the main poles of both generators and motors. (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 ohm, series field 0.025ohm, interpole field 0.028ohm, shunt field (including the field-regulator resistance) 115ohm. 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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