POWER FLOW, MAXIMUM POWER OUTPUT and MAXIMUM EFFICIENCY IN A DC MACHINE

Power flow

The flow of power from mechanical to electrical domain in a generator and electrical to mechanical domain in a motor occurs as shown in the diagram below:

Power Flow Diagram

For Series and Separately excited machines: IL = Ia
For Shunt Generator: IL = Ia - If
For Shunt Motor:  IL = Ia + If  

NOTE:- The amount of conductor and iron materials required for a Machine of given rating ∝ 1 / speed of machine     


Maximum Power Output

To achieve maximum power output in a DC Machine the following condition must be satisfied.

Motor

Developed Power, Pa = EaIa = (V - IaRa)Ia = VIa - Ia^2 Ra 
For maximum output,

dPa/dIa = V - 2IaRa = 0

Ia = V / 2Ra

And  Ea = V - 2IaRa

Ea = V/2

Efficiency, 𝜂 = Pa / VIa = EaIa / VIa = 1/2 = 50%

Suppose, Ra = 0.03pu
for maximum output

Ia = 100% / (2*3%) ≈ 17 times full load

Due to such high current, windings will be burnt immediately. Hence, maximum power output is undesirable condition.

Generator

Output Power, Pa = VaIa
                                      = (Ea - IaRa)Ia
                                      = EaIa - Ia^2 Ra

For maximum output power,  dPa / dIa = Ea - 2IaRa = 0

Ia = Ea / 2Ra

Va = Ea - IaRa = Ea/2

Efficiency, 𝜂 = VaIa / EaIa = 1/2 = 50%

Here, also current will be high due to small Armature Resistance and it can damage the windings.

So, the desirable condition is maximum efficiency so that maximum part of Input Power is converted to Output Power.


Maximum Efficiency

Assuming a Short Shunt Machine with the following parameters:

VBD = Brush Drop Voltage
Pk = No-Load Rotational Losses
Ra = Armature Resistance
Rf = Shunt Field Resistance
Rse = Series Field Resistance

If Machine operates as a generator,
Ia = IL + If

Where IL is current delivered to load
 If is current in Shunt Field Winding
Ia is Armature Current

 𝜂 = Vt IL / (Vt Ia + Ia^2 Ra + VBD Ia + Pk + IL^2 Rse + If^2 Rf)

For Maximum Efficiency, d𝜂/dIL = 0

Vt(Vt Ia + Ia^2 Ra + VBD Ia + Pk + IL^2 Rse + If^2 Rf) - Vt IL(Vt + 2IaRa + VBD + 2ILRse) = 0

Substitute Ia = IL + If

We get, IL^2 (Rse + Ra) = If^2 Rf + Pk + VBDIf + If^2 Ra

which implies, Constant Losses = Variable Cu Losses

This can be applied to Shunt Machines by equating Rse to zero and for Series Machines by equating If and Rf to zero.