A dc shunt generator delivers 195 Amp at terminal voltage of
250V. The armature resistance and shunt field resistances are
0.02Ω and 50Ω respectively. The iron and friction losses are
950W. Calculate: 8
i) emf generated
ii) total copper losses
iii) input to the generator
iv) electrical efficiency of generator
Answers
Answer:
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A dc shunt generator delivers 195 Amp at terminal voltage of 250V. The armature resistance and shunt field resistances are. 0.02Ω and 50Ω respectively. The iron and friction losses are
i) emf generated
answer - In nature, emf is generated when magnetic field fluctuations occur through a surface. For example, the shifting of the Earth's magnetic field during a geomagnetic storm induces currents in an electrical grid as the lines of the magnetic field are shifted about and cut across the conductor.
ii) total copper losses
answer -The total copper loss (I 2R) loss in transformer is load dependent. The copper losses are proportional to square of the RMS current flowing in the winding and also proportional to the resistance of winding. The resistance of the conductor varies with the rise in temperature
iii) input to the generator
answer-dont know
iv) electrical efficiency of generator
answer-The efficiency of a generator is determined by the power of the load circuit and the total watts produced by the generator. It is expressed as a percentage since you are dividing units of power by units of power. For most commercial electrical generators, this ratio can be upwards of 95 percent.
Concept:
The generated EMF of a DC shunt generator can be calculated by the equation, E = V + IaRa
Given:
Current of the DC shunt generator = 195 A
Terminal voltage = 250V
Armature Resistance = 0.02 ohm
Shunt field resistance = 50 ohm
Iron and friction losses = 950W
Find:
We need to determine:
i) Generated EMF
ii) Total Copper loss
iii) Input to the generator
iv) Electrical efficiency of generator
Solution:
The output power of the generator can then be calculated by the following relation- Current of the DC shunt generator × Terminal voltage
= 250 × 195
= 48,750 W
The field current is, therefore, Terminal voltage/Shunt field resistance
Field current = 250/50 = 5A
Hence armature current = 195 + 5 = 200A
Then, the field copper loss = Terminal voltage × field current
= 250 × 5 = 1250 Watt
While the armature copper loss is equivalent to product of square of armature current with armature resistance
Therefore, the armature copper loss = 200² × 0.02 = 800 Watts
It is given to us that the iron and friction losses are 950W.
i) The EMF generated can be calculated by the formula- E = V + IaRa where V is the terminal voltage, Ia is the armature current and Ra is the armature resistance
Therefore, E = 250 + 200×0.02
E = 254 V
ii) The Total copper loss can be calculated by = the field copper loss + Armature Copper loss
Total copper loss = 1250 + 800
= 2050 Watts
iii) Prime mover output serves as the generator's input.
Therefore, Input to generator = The output power of the generator + Total copper loss + Iron and Friction Loss
Input to generator = 48750 + 2050 + 950
Input to generator = 51,750 Watts
iv) While the electrical efficiency of generator = Output to generator/ Input to generator
Therefore, electrical efficiency of generator = 48,750/51750 × 100
electrical efficiency of generator = 94.20%
Thus, the emf generated is 254V, the total copper loss is 2050 Watts, the input to the generator is 51,750 Watts and the electrical efficiency of the generator is 94.20%.
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