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# Solved Problems: Electrical Engineering and Instrumentation - DC Machines

Electrical Engineering and Instrumentation - DC Machines - Solved Problems: Electrical Engineering and Instrumentation - DC Machines

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A 4 pole generator with wave wound armature has 51 slots each having 24 conductors. The flux per pole is 10 mWb. At what speed must the armature rotate to give an induced emf of 0.24 kV. What will be the voltage developed, if the winding is lap connected and the armature rotates at the same speed?

Given data

P = 4

No.of slots = 51

No.of conductors/slot = 20

Eg= 0.24 Kv = 240 V

Φ = 10 mW= 10/1000 Web

Find N &Eg at same N?

Solution

Total no. of conductors, Z = 51x20 = 1224

Wave winding, A=2

From EMF equation,

N= Eg60A / ΦZP = (240x60x2)/(10/1000x1224x4) = 612.75 rpm

Lap winding, A=P=4

Eg = PΦZN/60A = (4x10/1000x1224x612.75)/(60x4) = 0.125 kV

2. A 250 volt DC shunt motor has armature resistance of 0.25 ohm on load it takes an armature current of 50A and runs at 750rpm. If the flux of the motor is reduced by 10% without changing the load torque, find the new speed of the motor.

Given data

V = 250

Ra = 0.25

Ia = 50

N1 = 750

Φ2 = 90%Φ1

Find N2?

Solution Eb1 = V-Ia1Ra  = 250-(50x0.25) = 237.5V

Eb2 = V-Ia2Ra

Ta1 = Ta2

Φ1Ia1 = Φ2Ia2

Ia2 = 55.55A

Eb2 = 250-55.55X0.25 = 236.12V

N2 = 828 rpm

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Electrical Engineering and Instrumentation : DC Machines : Solved Problems: Electrical Engineering and Instrumentation - DC Machines |