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?

Question

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 [latex] G_{2}[/latex], for the system frequency to be 50 Hz? What would be the load sharing now?

Accepted Answer

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, [latex] \left(x_{1}+x_{2} ight)[/latex]

AC =[latex] x_{1}[/latex] , , load supplied by G1

BC = [latex] x_{2}[/latex], load supplied by G2

OC = system frequency

(a) Substituting values.

[latex] G_{1}[/latex] [latex] y=-x_{1}+51.8 ext { (i) }[/latex] m = – 1 Hz/MW

[latex] G_{2}[/latex] [latex] y=-x_{2}+51 [/latex] m = – 1 Hz/MW

It follows

[latex] -x_{1}+51.8=-x_{2}+51[/latex]

Or [latex] \left(x_{1}-x_{2} ight)=0.8[/latex] (iii)

Solving [latex] x_{1}+x_{2}=2.8 MW[/latex] (total load, AB)

Load supplied by [latex] G_{1}[/latex] ; [latex] x_{1}=1.8 MW[/latex]

Load supplied by [latex] G_{2}[/latex] ; [latex] x_{2}=1 MW[/latex]

System frequency. [latex]f=y=-1.8+51.8=50 Hz[/latex]

(b) Load increased by 1 MW

[latex] x_{1}+x_{2}=2.8+1=3.8 MW[/latex] (iv)

Solving Eqs (iii) and (iv)

Load [latex] G_{1}[/latex] : [latex] x_{1} = 2.3 MW[/latex]; increase 2.3 – 1.8 = 0.5 MW

Load [latex] G_{2}[/latex] : [latex] x_{2} = 1.5 MW[/latex]; increase 1.5 – 1 = 0.5 MW

Note: Additional load equally divided as the two characteristics have the same slope

system frequency [latex] f=y=-2.3+51.8=49.5 Hz[/latex]

(c) The [latex] G_{1}[/latex] can supply the load of 1.8 MW at the frequency of 50 Hz. The remaining power must be shared by [latex] G_{2}[/latex]. So the new load of [latex] G_{2}[/latex] is 2 MW. For the new load, the set point of [latex] G_{2}[/latex] must be

[latex] 50=-1 imes 2+[/latex] set frequency

Set frequency = 52 Hz

Question 8.48: Figure 8.115 shows speed (frequency)-load characteristics of...

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