Figure 8.115 shows speed (frequency)-load characteristics of two generators supplying in parallel a load of 2.8 MW at 0.8 pf lagging:
(a) At what frequency is the system operating and what is the load supplied by each generator?
(b) If the load is now increased by 1 MW, what will be the frequency and the load sharing?
(c) In part (b) which should be the set point of G_{2}, for the system frequency to be 50 Hz? What would be the load sharing now?
The speed-load characteristics of the two generators are drawn in Fig. 8.116 on the sides of the load axis. These being linear can the represented by the equation of a line as
y = mx + c
where
x = generator load, it is taken positive on both sides
y = system frequency, same for both generators
c = set frequency (on frequency axis)
m = slope, Hz/MW
On the figure
AB = total load, \left(x_{1}+x_{2}\right)
AC = x_{1} , , load supplied by G1
BC = x_{2}, load supplied by G2
OC = system frequency
(a) Substituting values.
G_{1} y=-x_{1}+51.8 \text { (i) } m = – 1 Hz/MW
G_{2} y=-x_{2}+51 m = – 1 Hz/MW
It follows
-x_{1}+51.8=-x_{2}+51Or \left(x_{1}-x_{2}\right)=0.8 (iii)
Solving x_{1}+x_{2}=2.8 MW (total load, AB)
Load supplied by G_{1} ; x_{1}=1.8 MW
Load supplied by G_{2} ; x_{2}=1 MW
System frequency. f=y=-1.8+51.8=50 Hz
(b) Load increased by 1 MW
x_{1}+x_{2}=2.8+1=3.8 MW (iv)
Solving Eqs (iii) and (iv)
Load G_{1} : x_{1} = 2.3 MW; increase 2.3 – 1.8 = 0.5 MW
Load G_{2} : x_{2} = 1.5 MW; increase 1.5 – 1 = 0.5 MW
Note: Additional load equally divided as the two characteristics have the same slope
system frequency f=y=-2.3+51.8=49.5 Hz
(c) The G_{1} can supply the load of 1.8 MW at the frequency of 50 Hz. The remaining power must be shared by G_{2}. So the new load of G_{2} is 2 MW. For the new load, the set point of G_{2} must be
50=-1 \times 2+ set frequency
Set frequency = 52 Hz
