Page 204 - Electrician - TT (Volume 1)

ELECTRICIAN - CITS

Eg
I
Efficiency of dc generator C Rf Ra RL
VI
η = Output / Input
Power developed in the armature B EgIa
Mechanical Efficiency
Mechanical powerinput B EgIa

the

develope
Power
General Eg electrical in d power armature input A BHP 746
Mechanical Efficiency Efficiency I Eg C VI

Electrical
I
Power Mechanical
armature
developed
RL
Rf Rf Ra Ra RL in powerinput
Genera
GeneralVIC VI electrical l electrical powe input r input A BHP 746
power
C
Electrical Efficiency Efficiency

Overall

power in d
EgIa
input
Mechanical

EgIa
B B Power develope armature
EgIa
General electrical power input
B
Overall Efficiency B EgIa
BHP power
A A Mechanical 746 input
746
BHP
C C VI VI

BHP 746P
A A BH 746
Condition for maximum efficiency
As I <<< I , so neglect shunt current.
SH L
Therefore
Generator Input = V I + I R + W
2
c
L
L
a
η = Output / Input
= V I / V I + I R + Wc
2
L
L
L
a
= V / [V + I Ra + Wc / I ]
L
L
For maximum efficiency, denominator should be minimum. Take the derivative of denominator with respect to IL
(variable parameter) and equate to zero.
d / d I [V + I Ra + Wc / I ] = 0
L
L
L
[0 + Ra + ( -W / I ) = 0
2
L
c
I Ra =Wc
2
L
Variable losses = Constant losses
VOLTAGE REGULATION (VR)
Change voltage from no-load voltage to full-load voltage, expressed as percentage of no-load voltage, at
constant speed
V - V
NL
FL
% Voltage regulation = x 100
V FL
Lower the regulation, better the output.
Problem # 1
A 20 KW dc shunt generator has the following losses at full-load:
Mechanical losses = 300 W; Iron losses = 400 W; Shunt Cu. loss = 120 W, Armature copper loss = 600 W
Calculate the efficiency at (i) No-load (ii) 25 % load.
191
CITS : Power - Electrician & Wireman - Lesson 30-37