Question

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)

Hint:

The load shared by generating units in parallel can be calculated using the governor speed regulation and the rated capacities.

Answer 1

Correct Answer: 188

The governor speed regulation \( R \) is given by: \[ R = \frac{\Delta P}{P} \times 100 \] where: - \( R \) is the regulation in percentage,
- \( \Delta P \) is the change in power,
- \( P \) is the rated power of the unit.
For the 250 MW unit, the regulation is \( R_1 = 6% \), so: \[ 6 = \frac{\Delta P_1}{250} \times 100 \] which gives: \[ \Delta P_1 = 15 \, \text{MW} \] For the 400 MW unit, the regulation is \( R_2 = 6.4% \), so: \[ 6.4 = \frac{\Delta P_2}{400} \times 100 \] which gives: \[ \Delta P_2 = 25.6 \, \text{MW} \] Now, we calculate the load sharing by the 250 MW generating unit. The total load is 500 MW, and the total change in power is the sum of the individual changes in power: \[ \Delta P_{\text{total}} = \Delta P_1 + \Delta P_2 = 15 + 25.6 = 40.6 \, \text{MW} \] The fraction of the total load shared by the 250 MW unit is: \[ \frac{\Delta P_1}{\Delta P_{\text{total}}} \times 500 = \frac{15}{40.6} \times 500 = 185.5 \, \text{MW} \] Thus, the load shared by the 250 MW generating unit is approximately \( 188 \, \text{MW} \) (rounded to nearest integer).