Power Systems m m the due date: 0-1 A 1.1 kW, 100-V DC shunt generator has an ar

Question

Power Systems

m m the due date: 0-1 A 1.1 kW, 100-V DC shunt generator has an armature resistance of 1.0 and fieldcircuit resistance of 100. The total voltage drop on the commutator brushes is about 2.5V. Neglect the effects of the armature reaction, the core losses, and the mechanical and stray losses. With the generator operating at rated voltage and at rated speed of 1050 rpm, determine the followings: a- The developed mechanical power and developed mechanical torque. b- The full load efficiency of the generator. c- The half load efficiency of the generator. d- The voltage regulation of the generator. Q-2 A 5.0kW 220V series DC mot is 25A and the ra or has an armature resistance of 0.25. At full load, the input current ted speed is 1200 rpm. The mechanical and stray losses are constant and equal 200W age drop on the brushes, and the effect of the armature reaction, all are heglected, calculate the following: a- The resistance of the series field windings. b- The efficiency of the motor at full load. c- The starting current of the motor and the starting torque of the motor d- The speed of the motor at half load,

Explanation / Answer

Answer:- Max output power = 1.1 kW at 100 V, Ra = 1 ohm, Rf = 100 ohm, Vbrush = 2.5 V.

at full load, load current IL = (1.1 x 1000)/100 A = 11 A

field current = 100/Rf = 100/100 = 1A

Thus armature current, Ia = 11 + 1 A = 12 A.

Thus loss power loss on brush for full load = 12 x 2.5 W = 30 W. Armature loss = 122 x 1 W = 12 W

a) Since rest losses are neglected here, so total mechanical power developed = input power = output power + lost power = 1100 + 30 + 12 W = 1142 W.
Mechanical torque = (1142 x 60)/(2 x pi x 1050) Nm = 10.39 Nm

b) Full load efficiency(in %) = output power / input power = (1100/1142) x 100 = 96.32 %

c) For Half load-

Load current = (1100/2)/100 A = 5.5 A, Field current is same as full load case i.e 100/100 = 1 A.

Thus armature current = 5.5 + 1 A = 6.5 A. Hence brush loss = 6.5 x 2.5 W = 16.25 W. Armature Loss = 42.25 W.

So input power = 550 W + 16.25 + 42.25 W = 608.5 W, and hence efficiency = (550/608.5) x 100 % = 90.4 %.

d) Voltage generated(full load) = load voltage + brush voltage + armature drop = 100 + 2.5 + 12 V = 114.5 V.

Hence voltage regulation = ((114.5 - 100)/100) x 100 % = 14.5 %

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