Question 6 of 7: Parallel Synchronous Generators — Governor Droop
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: Professional Engineers Ontario — 07-Elec-A6 Power Systems and Machines, Spring 2016. Closed book (formula sheets supplied); do any five of seven questions, all of equal value (20 marks each). Three hours. All ac quantities are rms; three-phase voltages are line-to-line and power is total real power unless noted. All seven questions are solved here as a complete study resource.
Reference texts. S. J. Chapman, Electric Machinery Fundamentals, 5th ed. (DC machines ch. 8, transformers ch. 2, induction motors ch. 7, synchronous machines ch. 4–5, parallel operation ch. 5); J. D. Glover et al., Power System Analysis and Design, 6th ed. (per-phase and power-factor analysis ch. 2); T. Wildi, Electrical Machines, Drives, and Power Systems. Canadian frame: EGBC/PEO national exam; SI units.
Find. (a) governor characteristics with the operating point; (b) the shifted operating condition after −400 kW; (c) the new frequency and each machine's load share.
Figure 6 — governor (frequency–load) characteristics: the stiffer 2% machine (A) and the softer 5% machine (B) intersect at the 57 Hz / 500 kW operating point; the dashed line marks the new 57.57 Hz condition after the 400 kW load drop.
Approach. Convert each droop into a frequency–power slope (Hz/kW), fix each straight-line characteristic through the known 500 kW / 57 Hz point to get the no-load frequencies, then impose the new total power at a common frequency to solve for that frequency and the split.
Governor slopes. Droop $=\dfrac{f_{nl}-f_{fl}}{f_{fl}}$; taking the nominal 60 Hz over the 600 kW rated range, $$m_A=\frac{0.02(60)}{600}=0.002\ \tfrac{\text{Hz}}{\text{kW}},\qquad m_B=\frac{0.05(60)}{600}=0.005\ \tfrac{\text{Hz}}{\text{kW}}.$$ The 2% machine has the flatter (stiffer) characteristic.
No-load frequency settings. Each characteristic is $f=f_{nl}-mP$ and passes through (500 kW, 57 Hz): $$f_{nl,A}=57+0.002(500)=58.0\text{ Hz},\qquad f_{nl,B}=57+0.005(500)=59.5\text{ Hz}.$$
New common frequency (c). After the drop the bus carries 600 kW: $P_A+P_B=\dfrac{f_{nl,A}-f}{m_A}+\dfrac{f_{nl,B}-f}{m_B}=600.$ Substituting, $500(58-f)+200(59.5-f)=600$, which gives $$\boxed{f=\frac{500(58)+200(59.5)-600}{700}=57.57\text{ Hz}.}$$
New load distribution (c). $P_A=\dfrac{58.0-57.57}{0.002}=214\text{ kW}$ and $P_B=\dfrac{59.5-57.57}{0.005}=386\text{ kW}$ (sum 600 kW). Equivalently, the load change divides inversely with the slopes: $\Delta P_A:\Delta P_B=m_B:m_A=2.5:1$, so the stiffer machine A sheds the larger share (−286 kW) while B sheds only −114 kW.
Final results — Question 6
Quantity
Initial (1000 kW)
After −400 kW (600 kW)
Bus frequency
57.0 Hz
57.57 Hz
Gen A output (2% droop)
500 kW
214 kW
Gen B output (5% droop)
500 kW
386 kW
No-load set frequencies
fnl,A = 58.0 Hz, fnl,B = 59.5 Hz
Check: The droop percentages are referred to the nominal 60 Hz over the 600 kW rating to fix the slopes; the resulting load split (214/386 kW) depends only on the slope ratio and is independent of that reference, while the new frequency (57.57 Hz) shifts by only ±0.03 Hz if 57 Hz is used as the reference instead.
