Problem: A Separately excited d.c motor has an armature resistance of 0.5 ohm. It runs off a 250 V d.c supply drawing an armature current of 20 A at 1,500 rpm. The torque developed for an armature current of 10 A will be______for the same field current.
Solution: Vt = 250 V; Ra = 0.5 Ω; N = 1500 rpm; Ia = 20 A
Induced EMF, Ea = Vt - Ia Ra = 250 - 20*0.5 = 240 V
Developed Power, P = Ea Ia = 240*20 = 4800 W
Developed Torque, T = P / Ꞷ = 4800 / (2𝜋*1500 / 60) = 30.557 N-m
Assuming flux remains constant, T ∝ Ia
T2 / T1 = Ia2 / Ia1
T2 / 30.557 = 10 / 20
T2 = 0.5 * 30.557 = 15.278 N-m
Problem: A 5 kw, 2000 V d.c shunt motor has an armature resistance of 1 ohm and shunt field resistance of 100 ohms. At no-load, the motor draws 6 A from a 200 V supply and runs at 1000 rpm. The rotational loss of the machine is ____W and the no load torque is _____N-m.
Solution: Under No-Load conditions there is no shaft output of the DC Machine and hence the developed power will be used to supply the Rotational Losses.
If = Vt / Rf = 200 / 100 = 2 A
Ia = IL - If = 6 - 2 = 4 A
Eb = Vt - Ia Ra = 200 - 4*1 = 196 V
Developed Power, P = Eb Ia = 196*4 = 784 W
Rotational Losses = 784 W
No-Load Torque, T = P / Ꞷ = 784 / (2𝜋 * 1000 / 60) = 7.486 N-m
Problem: The armature resistance of permanent magnet dc motor is 0.8 ohm. At no load, the motor draws 1.5 A from a supply voltage of 25 V and runs at 1500 rpm. The efficiency of the motor while it is operating on load drawing a current of 3.5 A from the same source will be?
Solution: At No-Load Developed Power is same as Rotational Losses.
At no load developed power, P = Eb Ia
Eb = Vt - Ia Ra =25 - 1.5*0.8 = 23.8 V
P = Eb Ia = 23.8 * 1.5 = 35.7 W
∴ friction and windage losses = 35.7 W
Under loaded condition Armature Cu loss = (3.5^2)*0.8 = 9.8 W
∴ Total losses = 35.7 + 9.8 = 45.5 W
Input to motor = V*Ia = 25*3.5 = 87.5 W
Efficiency, 𝜂 = [(input - losses) / input] * 100 %
= [(87.5 - 45.5) / 87.5] * 100 %
= 48 %
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