Question

Two generators rated 200 mw and 400 mw are operating in parallel. The droop characteristics of their governors are 4% and 5%, respectively from no load to full load. Assuming that the generators are operating at 50 hz at no load, assume free governor operation, how would a load of 600 mw be shared between them? 230, 370 229, 371 231, 369 (d) 270, 330

Answer 1

Answer:

For generator 1:

P(rated) = 200 MW and the drooping characteristics = 4%

For generator 2:

P(rated) = 400 MW and the drooping characteristics = 5%

Both the generators are operating at 50 HZ at no load

A load of 600 MW has to be shared between generator 1 and generator 2

So, let the load on generator 1 be “x MW” and on generator 2 be “(600 – x) MW”.

Now,  

Reduction in frequency, ∆f for generator 1 = [0.04 * 50 / 200] * x ….. (i)

and,

Reduction in frequency, ∆f for generator 2 = [0.05 * 50 / 400] * (600 - x) ….. (ii)

Equating the value for ∆f from (i) & (ii), we get

[0.04 * 50 / 200] * x   = [0.05 * 50 / 400]  * (600 – x)      

Or, 0.0002 x = (0.000125 * 600) – 0.000125 x

Or, 0.000325 x = 0.075

Or, x = 230.76 MW ≈ 231 MW = load on generator 1

∴ load on generator 2 = (600 – x) = 369 MW

Thus, we can clearly see that the generator 1 gets overloaded due to a load of 231 MW and generator 2 gets under-loaded due to a load of 369 MW.