Question

(a) Explain how drooping characteristics cater for stable operation when running in parallel. (6)

(b) Two shunt generators X and Y work in parallel. Their external characteristics may be assumed to be linear over their normal working range, the terminal voltage of X falls 265V on no-load 230V when delivering 350A to the busbars, while the voltage of Y falls from 270V on no-load to 240V when delivering 400A to the busbar. Calculate the current with each machine delivers when they share a common load of 500A. What is the busbar voltage under this condition and the power delivered by each machine (10)

Answer 1

Part (a)

How drooping characteristics cater for stable operation when running in parallel:

A drooping (falling) external characteristic means the terminal voltage falls as the load current increases. When two generators run in parallel, they share the load automatically because of the droop.
If one generator momentarily takes more load, its voltage falls (due to the droop), which reduces the current it supplies, while the other generator, seeing a slightly higher voltage, takes more load. This self-balancing action keeps the load shared stably.
Without droop (a flat characteristic), a small difference in the e.m.f.s would cause one machine to take all the load or even motor the other, making stable parallel operation impossible.
The droop also allows the load to be shared in proportion to the ratings of the machines, and the governor/voltage regulator settings can be adjusted to control the sharing.

Part (b)

Two shunt generators X and Y in parallel. X: 265 V no-load to 230 V at 350 A. Y: 270 V no-load to 240 V at 400 A. Common load 500 A.

Droop of X: voltage drop = 265 - 230 = 35 V over 350 A, so drop per ampere = 35/350 = 0.1 V/A. V_X = 265 - 0.1 I_X.
Droop of Y: voltage drop = 270 - 240 = 30 V over 400 A, so drop per ampere = 30/400 = 0.075 V/A. V_Y = 270 - 0.075 I_Y.
In parallel, the terminal voltages are equal: 265 - 0.1 I_X = 270 - 0.075 I_Y.
Also I_X + I_Y = 500 A.
From the voltage equation: 0.1 I_X - 0.075 I_Y = -5.
Substituting I_Y = 500 - I_X: 0.1 I_X - 0.075(500 - I_X) = -5 -> 0.1 I_X - 37.5 + 0.075 I_X = -5 -> 0.175 I_X = 32.5 -> I_X = 185.7 A.
I_Y = 500 - 185.7 = 314.3 A.
Busbar voltage V = 265 - 0.1 x 185.7 = 265 - 18.57 = 246.4 V. (Check: 270 - 0.075 x 314.3 = 270 - 23.57 = 246.4 V.)
Power delivered: P_X = V x I_X = 246.4 x 185.7 = 45,760 W = 45.8 kW. P_Y = 246.4 x 314.3 = 77,440 W = 77.4 kW.

So X delivers 185.7 A (45.8 kW) and Y delivers 314.3 A (77.4 kW), at a busbar voltage of 246.4 V.

✓ Verified Model Answer (Text Solution)

How drooping characteristics cater for stable operation when running in parallel:

Two shunt generators X and Y in parallel. X: 265 V no-load to 230 V at 350 A. Y: 270 V no-load to 240 V at 400 A. Common load 500 A.