Power plant engineering: board exam questions with solutions
On the REE exam under Electrical Engineering Professional Subjects. Try each question before you open its solution, then practise with fresh numbers below: 2 computed question types on this topic.
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Questions
1. The load factor of a power plant over a period is defined as:
- Average load divided by peak (maximum) load
- Peak load divided by installed capacity
- Energy generated divided by installed capacity × hours
- Maximum demand divided by connected load
Show the answer and solution
Answer: A. Average load divided by peak (maximum) load
Load factor = average load/peak load for the same period. Energy over capacity × hours is the capacity factor, and maximum demand over connected load is the demand factor.
2. A plant generates 4800 kWh in a day and its peak load that day is 400 kW. What is the daily load factor?
- 83%
- 50%
- 12%
- 200%
Show the answer and solution
Answer: B. 50%
Average load = 4800/24 = 200 kW. Load factor = 200/400 = 0.50 = 50%.
3. A 100 MW plant produces 438 000 MWh in a (non-leap) year. What is its annual capacity factor?
- 44%
- 60%
- 50%
- 25%
Show the answer and solution
Answer: C. 50%
CF = energy/(capacity × hours) = 438 000/(100 × 8760) = 0.50 = 50%.
4. A thermal plant has a net heat rate of 10 000 kJ/kWh. What is its overall thermal efficiency?
- 10%
- 28%
- 100%
- 36%
Show the answer and solution
Answer: D. 36%
1 kWh = 3600 kJ, so η = 3600/heat rate = 3600/10 000 = 0.36 = 36%.
5. A 50 MW coal plant operates at full load with a heat rate of 11 000 kJ/kWh. The coal has a heating value of 25 000 kJ/kg. How much coal does it burn per hour?
- 22 t/h
- 114 t/h
- 7.2 t/h
- 2.2 t/h
Show the answer and solution
Answer: A. 22 t/h
Heat input = 50 000 kWh/h × 11 000 kJ/kWh = 5.5 × 10⁸ kJ/h. Coal = 5.5 × 10⁸/25 000 = 22000 kg/h = 22 t/h.
6. Which type of plant is best suited to serve short daily peak loads because it can start and reach full load quickly?
- Nuclear plant
- Open-cycle gas turbine plant
- Coal-fired steam plant
- Geothermal plant
Show the answer and solution
Answer: B. Open-cycle gas turbine plant
Open-cycle gas turbines start in minutes and have low capital cost, making them suitable for peaking duty. Nuclear, coal and geothermal plants are run as base load.
7. A building has a connected load of 500 kW and a measured maximum demand of 300 kW. What is the demand factor?
- 1.67
- 0.40
- 0.60
- 0.30
Show the answer and solution
Answer: C. 0.60
Demand factor = maximum demand/connected load = 300/500 = 0.60.
8. A hydro plant has a net head of 50 m, a flow of 20 m³/s, and an overall efficiency of 85%. What is its electrical output?
- 9.81 MW
- 7.09 MW
- 83.38 MW
- 8.34 MW
Show the answer and solution
Answer: D. 8.34 MW
P = ρ g Q H η = 1000 × 9.81 × 20 × 50 × 0.85 = 8338500 W ≈ 8.34 MW.
Worked examples with computed numbers
These numbers are fixed on this page; in practice they change every time.
9. A plant generates 6480 MWh in a 30-day month and its peak demand is 20 MW. What is its load factor?
- 45%
- 56.3%
- 22.2%
- 3.7%
Show the answer and solution
Answer: A. 45%
Average demand = 6480 MWh/(30 × 24 h) = 9 MW; load factor = average/peak = 9/20 = 45%.
10. A 100 W appliance runs 6 hours a day for 30 days at ₱11.5/kWh. What does it cost?
- ₱207
- ₱29.57
- ₱6.9
- ₱2070
Show the answer and solution
Answer: A. ₱207
100 W × 6 h × 30 days = 18 kWh; × ₱11.5 = ₱207.
Other topics in the same subjects
DC circuits · AC circuits · Three-phase systems · Transients (RC, RL, RLC) · Semiconductors and diodes · Power electronics · Instrumentation and measurement · Control systems · Transformers · DC machines · Induction motors · Synchronous machines · Power systems and transmission · Protection: fuses, breakers, relays · Illumination · PEC: branch circuits, ampacity and protection
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