Parallel Operation of Two Generators

When two generators are connected in parallel and are jointly supplying the demand in a small power system, the load is shared according to the set points of their governors. The best way to illustrate this sharing mechanism and the resulting system frequency is by means of an example.

In the small power system of Figure 3.5(a) two generators A and B rated at 50 and 100 MW respectively supply a load of 100MW. Both generators are fitted with governors having a droop of 4% and a no-load set point of 52Hz. Lines aa and bb in Figure 3.5(b) show the frequency-power (f-P) characteristics of the two generators.

The division of load between the generators when they are supplying 100MW will now be determined. Let line cc located Af below 52 Hz represent the frequency at which this sharing takes place.

(a) Using trigonometry PA/Af = 50/2 and PB/Af = 100/2. Therefore PA/PB = 1/2, but as Pa + PB = 100MW, then PA = 33.3MW and PB = 66.6MW In this case the demand is shared in proportion to the generator ratings

(b) To determine the system frequency, from PA/Af = 50/2 Af = 33.3 x 2/50 = 1.33 and hence the frequency at which line cc intercepts the f axis is f = (52 – 1.33) = 50.67Hz

(c) Suppose that generator A is more modern and therefore more efficient in terms of fuel used per generated kWh. The set point adjustment of the governors could be used to arrive at a new more economic load-sharing condition that maximizes the contribution from generator A.

Parallel Operation of Two Generators

One way to implement this is to shift the set point of generator B so that its characteristic is displaced from bb to BB, with the result that each generator is supplying 50MW. The system frequency is now 50 Hz and the new set point of generator B can be found from f/50 = 2/100. Hence f = 1 and the new set point is 51 Hz

Updated: September 26, 2015 — 4:50 am