**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 (**

**η**

_{m})

η

_{m}= B/A = Total watts generated in armature/ Mechanical power supplied
= E

_{g}I_{a}/Output of driving engine**(b) Electrical efficiency (**

**η**

_{e})

η

_{e}= C/B = Watts available in load circuit/Total watts generated
= VI/E

_{g}I_{a}

**(c) Overall or Commercial efficiency (**

**η**

_{c})

η

_{c}= C/A = Watts available in load circuit/Mechanical power supplied
= VI/Output of driving engine

Overall efficiency can also be found by

η

_{c}= η_{m}x η_{e}**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 = I

_{sh}^{2}R_{sh}
The variable losses occur in the armature winding and series field winding and are equal to

I

^{2}_{a}R_{a}+ I^{2}_{se}R_{se}
Therefore,

Generator Input = Output + Variable losses + Constant losses

= VI + I

^{2}_{a}R_{a}+ I^{2}_{se}R_{se}+ W_{C}
η = Output/Input

= VI/( VI + I

^{2}_{a}R_{a}+ I^{2}_{se}R_{se}+ W_{C})
Since shunt field current (I

_{sh}) is supplied by armature. I_{a}> 1 but in practice the shunt field current is negligible as compared to line current. Therefore I_{a}= I (approx).
η = Output/Input = VI/(VI + I

^{2}R_{a}+ W_{C})
Dividing numerator and denominator by VI,

η = 1/{1 + (IR

_{a}/V + W_{C}/VI)}
Efficiency shall be maximum when the denominator is minimum. Differentiating the denominator with respect to I and equating to zero.

d/dI(IR

_{a}/V + W_{C}/VI) = 0
We get,

R

_{a}/V – W_{C}/VI^{2}= 0
Or R

_{a}/V = W_{C}/VI^{2}
Or I

^{2}R_{a}= W_{C}
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,

I

^{2}R_{a}= W_{C}
Or I = √(W

_{C}/R_{a})
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