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Saturday, 28 December 2019

Power Flow Diagram of DC Generator

POWER FLOW DIAGRAM OF DC GENERATOR


Power flow diagrams give the stages of power in the case of a dc generators showing clearly the losses that occur at various stages of generation of power. The input to a generator is mechanical power a hydel, steam, diesel or nuclear source in the form of rotational power. The power flow diagram is shown in figure below.

EFFICIENCY OF DC GENERATOR:

Efficiency, η = Output/Input

In the case of a dc generator the output is a known factor from which input can be found. Thus the formula gets modified as,

Efficiency, η = Output/[Output +  Losses]
= Output/[Output +  Copper loss + Stray losses]

The following three types of efficiencies can be determined for a DC generator.

(a) Mechanical efficiency (ηm)

ηm = B/A = Total watts generated in armature/ Mechanical power supplied
= EgIa/Output of driving engine

(b) Electrical efficiency (ηe)

ηe = C/B = Watts available in load circuit/Total watts generated
= VI/EgIa

(c) Overall or Commercial efficiency (ηc)

ηc = C/A = Watts available in load circuit/Mechanical power supplied
= VI/Output of driving engine

Overall efficiency can also be found by

ηc = ηm x ηe


CONDITIONS FOR MAXIMUM EFFICIENCY:

Let the generator,

Output = VI watts
Input = Output + Losses

The losses comprises of constant losses and variable losses.

Constant Losses (WC) = Stray losses + Shunt Field copper losses

In the case of shunt field,
Field copper losses = Ish2 Rsh

The variable losses occur in the armature winding and series field winding and are equal to
I2aRa + I2seRse

Therefore,

Generator Input = Output + Variable losses + Constant losses
= VI + I2aRa + I2seRse + WC

η = Output/Input
= VI/( VI + I2aRa + I2seRse + WC)

Since shunt field current (Ish) is supplied by armature. Ia > 1 but in practice the shunt field current is negligible as compared to line current. Therefore Ia = I (approx).

η = Output/Input = VI/(VI + I2Ra + WC)

Dividing numerator and denominator by VI,

η = 1/{1 + (IRa/V + WC/VI)}

Efficiency shall be maximum when the denominator is minimum. Differentiating the denominator with respect to I and equating to zero.

d/dI(IRa/V + WC/VI) = 0

We get,

Ra/V – WC/VI2 = 0   
Or Ra/V = WC/VI2
Or I2Ra = WC 

That is, when the variable losses is equal to constant losses, the efficiency of a DC generator is maximum. The load current corresponding to maximum efficiency is given by,

I2Ra = WC 

Or I = (WC/Ra)

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