Page 102 - Solutions Manual to accompany Electric Machinery Fundamentals

 
nP 3570 r/min 2

f  m   59.5 Hz
fl1
120 120
The no-load frequency of generator 2 corresponds to a frequency of
 
nP 1800 r/min 4

f  m   60.00 Hz
nl2
120 120
The full-load frequency of generator 2 corresponds to a frequency of
 

nP 1780 r/min 4
f  m   59.333 Hz
fl2
120 120
(a) The speed droop of generator 1 is given by
n  n 3650 r/min 3570 r/min

SD  nl fl  100%   100% 2.24%

1
n fl 3570 r/min
The speed droop of generator 2 is given by
n  n 1800 r/min 1780 r/min

SD  nl fl  100%   100% 1.12%

2
n fl 1780 r/min
(b) The full load power supplied by generator 1 is

P 1,full load  S  cos  250 kVA 0.8  200 kW
The maximum power supplied by generator 2 is
P 2,full load  S  cos  250 kVA 0.85  212.5 kW

The power supplied by generator 1 as a function of frequency is given by

P  s  f  f 
1 P 1 nl1 sys
and the power supplied by generator 2 as a function of frequency is given by
P  2 s P 2  f nl2  f sys 

The power curve’s slope for generator 1 is
P 0.2 MW
s  P 1   0.15 MW/Hz
f 

nl f fl 60.833 Hz 59.5 Hz
The power curve’s slope for generator 1 is

s  P  0.2125 MW  0.319 MW/Hz
P 2

f  nl f fl 60.00 Hz 59.333 Hz
The no-load frequency of generator 1 is 60.833 Hz and the no-load frequency of generator 2 is 60 Hz.
The total power that they must supply is 300 kW, so the system frequency can be found from the
equations
P LOAD  P  1 P 2

P LOAD  s P 1  f nl1  f sys  s P 2  f nl2  f sys 

300 kW   0.15 MW/Hz 60.833 Hz   f sys  0.319 MW/Hz 60.0 Hz   f sys 

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