Posted On : 30.04.2017 07:03 pm

## Chapter: __Electrical Engineering and Instrumentation : DC Machines__

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

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

**PROBLEMS**

** 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

R_{a}
= 0.25

I_{a}
= 50

N_{1}
= 750

Φ_{2}
= 90%Φ_{1}

**Find N**_{2}?

**Solution**

E_{b1}
= V-I_{a1}R_{a} = 250-(50x0.25)
= 237.5V

E_{b2}
= V-I_{a2}R_{a}

Load
torque is constant

Ta1 = Ta2

Φ1Ia1 =
Φ2Ia2

I_{a2}
= 55.55A

E_{b2}
= 250-55.55X0.25 = 236.12V

**N**_{2} = 828 rpm

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