Question

Two generators rated 200MW and 400MW are operating in parallel. The droop characteristics of their governors are 4% and 5% respectively from no load to full load. The speed changers are so set that the generators operate at 50 Hz sharing the full load of 600 MW in the ratio of their ratings. If the load reduces to 400MW how will it be shared among the generators and what will be the system frequency be? Assume free governor operation
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Answer 1

The frequency of the system is f = 47.69 Hz

Solution:

• Load On 200 MW generator = P
• Then the load on a 400 MW generator = (600 − P) MW
• Let the reduction in the frequency = Δf

Now,

Δf ⁄ P = 0.04 × 50  ⁄ 200

Δf ⁄ P = 0.01 ———(1)

Also,

Δf ⁄ (600 − P) = 0.05 × 50  ⁄ 400

Δf ⁄ (600 − P) = 0.006 ———-(2)

Now equating Δf in (i) and (ii), we get

P = 231 MW

600 − P = 369 MW

The frequency of the system is

f = fo - Δf / P(rated) x P1/P2

After putting the values

f = 50 - 0.04 x x 50 / 200 x 231

f = 47.69 Hz