Two generating units rated for 250 MW and 400 MW have governor speed regulations of 6% and 6.4%, respectively, from no load to full load. Both the generating units are operating in parallel to share a load of 500 MW. Assuming free governor action, the load shared in MW, by the 250 MW generating unit is ______. (round off to nearest integer)

The correct answer is 200.

Step-by-step Explanation:

Let us define the parameters for the two generating units operating in parallel:

For Unit 1 (250 MW unit):
Rated capacity: $P_{R1}=250\text{MW}$
Governor speed regulation: $R_1=6\%=0.06$

For Unit 2 (400 MW unit):
Rated capacity: $P_{R2}=400\text{MW}$
Governor speed regulation: $R_2=6.4\%=0.064$

Let the load shared by Unit 1 be $P_1$ (in MW) and the load shared by Unit 2 be $P_2$ (in MW).

The total load shared by both units is 500 MW:
$P_1+P_2=500$

Assuming free governor action, both units operate at the same system frequency. Therefore, the change in frequency (or percentage drop in speed/frequency from no load to the operating load) must be identical for both generators:

$R_1\cdot \frac{P_1}{P_{R1}}=R_2\cdot \frac{P_2}{P_{R2}}$

Substituting the given values into the equation:
$0.06\cdot \frac{P_1}{250}=0.064\cdot \frac{P_2}{400}$

Simplifying the relation:
$\frac{6}{25000}{P}_1=\frac{6.4}{40000}{P}_2$
$\frac{6}{25000}{P}_1=\frac{4}{25000}{P}_2$
$6P_1=4P_2$
$P_2=1.5P_1$

Now, substitute this relationship into the total load equation:
$P_1+1.5P_1=500$
$2.5P_1=500$
$P_1=\frac{500}{2.5}=200\text{MW}$

Thus, the load shared by the 250 MW generating unit is 200 MW.