Nepal Engineering Council · Electrical Engineering · Chapter 3
Power Plants Engineering
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180 questions in 6 syllabus topics.
3.1 Hydroelectric power plants
31 questions · AElE0301
1. A hydropower plant that uses the natural river flow with little or no storage, so its output varies with the river discharge, is called a:
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Think about whether water is held back for later use.
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Answer: C. Run-of-river plant
A run-of-river plant diverts the available river flow directly through the turbines with at most small daily pondage, so its output follows the river flow.
2. In hydropower plant classification, 'pondage' refers to:
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It is about short-term, not seasonal, regulation.
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Answer: B. Small storage that holds water for a few hours to meet daily load variations
Pondage is limited storage (hours to a day) that lets a run-of-river plant follow daily load changes; seasonal storage is a reservoir.
3. For a high-head site (several hundred metres) with relatively small discharge, the most suitable turbine is the:
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High head means a high-velocity jet is available.
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Answer: C. Pelton wheel
Pelton (impulse) wheels suit high head and low discharge; Kaplan, propeller and bulb turbines suit low head and large discharge.
4. A Kaplan turbine is best described as a:
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Its blades can be turned like a ship's propeller.
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Answer: C. Low-head, axial-flow reaction turbine with adjustable runner blades
The Kaplan runner is an axial-flow reaction runner whose blade angle can be changed, giving good efficiency at part load on low heads.
5. A Francis turbine is a:
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Consider both the flow direction and whether pressure changes in the runner.
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Answer: A. Mixed (radial-inward) flow reaction turbine
In a Francis runner water enters radially inward and leaves axially, and pressure drops across the runner, so it is a mixed-flow reaction turbine.
6. Which statement about an impulse turbine is correct?
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Where is the head converted to kinetic energy?
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Answer: A. The pressure of water remains atmospheric as it passes over the runner
In an impulse (Pelton) turbine the whole head is converted to jet velocity in the nozzle; the buckets work at atmospheric pressure, so no draft tube is needed.
7. The main function of a draft tube in a reaction turbine is to:
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It is located at the outlet of the runner.
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Answer: A. Recover kinetic energy at runner exit and allow the turbine to be set above tailwater level
The diverging draft tube converts exit velocity head into pressure (creating suction at runner exit) and lets the runner sit above tailwater without losing head.
8. The main purpose of a surge tank in a hydroelectric scheme is to:
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Think of sudden closing of turbine gates.
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Answer: B. Reduce water hammer pressure in the conduit during sudden load changes
A surge tank absorbs pressure rise on load rejection and supplies water on load acceptance, protecting the headrace/pressure tunnel from water hammer.
9. The pipe that carries water under pressure from the forebay or surge tank to the turbine is the:
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It is usually a steel pipe running down the hillside.
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Answer: A. Penstock
The penstock is the pressure pipe feeding the turbine; the tailrace carries water away and the spillway passes surplus flood water.
10. A trash rack is provided at the intake of a hydro plant to:
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It is a screen.
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Answer: A. Prevent floating debris from entering the water conductor system
Trash racks are screens of steel bars that stop logs, leaves and other debris from reaching the penstock and turbine.
11. Cavitation in hydraulic turbines is most likely to occur:
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Look for the point of lowest pressure.
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Answer: A. At the runner blade outlet and draft tube inlet of reaction turbines, where pressure is lowest
Cavitation happens where local pressure falls below vapour pressure; in reaction turbines this is the runner exit/draft tube inlet region, which is why Thoma's cavitation factor limits the setting height.
12. Thoma's cavitation factor (σ) is used mainly to decide:
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It relates to suction head at the runner exit.
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Answer: D. The safe setting height of a reaction turbine above tailwater
σ compares the net positive suction head to the net head; keeping the plant σ above the critical value fixes the maximum height of the runner above tailwater.
13. A turbine develops 10,000 kW at 500 rpm under a net head of 100 m. Its specific speed (N√P/H^1.25, P in kW) is about:
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100^1.25 = 316.2.
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Answer: B. 158
Ns = 500 × √10000 / 100^1.25 = 500 × 100 / 316.2 ≈ 158, which lies in the Francis range.
14. A turbine with a specific speed of about 25 (metric, P in kW) would most likely be a:
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Low specific speed goes with high head.
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Answer: A. Pelton wheel
Low specific speeds (roughly below 50–60 for a single jet) correspond to Pelton wheels; Francis and Kaplan have progressively higher Ns.
15. Water flows at 10 m³/s through a turbine under a net head of 100 m. With an overall efficiency of 85%, the electrical output is about:
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Use P = ρgQHη with ρ = 1000 kg/m³.
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Answer: B. 8.34 MW
P = ρgQHη = 1000 × 9.81 × 10 × 100 × 0.85 ≈ 8.34 × 10⁶ W = 8.34 MW.
16. A small hydro plant must give 1 MW with a discharge of 2 m³/s and overall efficiency 80%. The net head required is about:
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Rearrange P = ρgQHη for H.
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Answer: B. 63.7 m
H = P/(ρgQη) = 10⁶/(1000 × 9.81 × 2 × 0.8) ≈ 63.7 m.
17. A plant has a gross head of 120 m and penstock losses of 8 m. With Q = 5 m³/s and overall efficiency 90%, its output is about:
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Use net head, not gross head.
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Answer: B. 4.94 MW
Net head = 120 − 8 = 112 m; P = 9.81 × 5 × 112 × 0.9 ≈ 4,944 kW ≈ 4.94 MW.
18. The jet velocity of a Pelton wheel working under a net head of 400 m with a nozzle velocity coefficient of 0.98 is about:
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V = Cv√(2gH).
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Answer: C. 86.8 m/s
V = Cv√(2gH) = 0.98 × √(2 × 9.81 × 400) = 0.98 × 88.6 ≈ 86.8 m/s.
19. For maximum hydraulic efficiency of an ideal Pelton wheel, the bucket (peripheral) speed should be about:
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The jet should leave the bucket with almost no absolute velocity.
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Answer: C. Half of the jet velocity
For an ideal Pelton wheel, efficiency is maximum when bucket speed u = V/2; in practice u ≈ 0.45–0.47 V.
20. A hydro generator is to be directly coupled to a turbine running at 375 rpm on a 50 Hz system. The number of poles required is:
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Ns = 120f/P.
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Answer: C. 16
P = 120f/N = 120 × 50 / 375 = 16 poles.
21. A 5 MW hydro plant operates with an annual capacity factor of 0.6. Its annual energy generation is about:
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A year has 8760 hours.
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Answer: B. 26.3 GWh
E = 5 MW × 8760 h × 0.6 = 26,280 MWh ≈ 26.3 GWh.
22. A 20 MW hydro unit has a governor speed droop of 4%. If system frequency falls by 0.5 Hz from 50 Hz, the increase in its output (from the governor) is:
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Droop = per-unit frequency change for 100% change in output.
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Answer: C. 5 MW
ΔP/Prated = (Δf/f)/R = (0.5/50)/0.04 = 0.25, so ΔP = 0.25 × 20 = 5 MW.
23. The governor of a Francis turbine controls the water flow mainly by adjusting the:
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Francis turbines have no nozzle.
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Answer: C. Guide vanes (wicket gates)
In Francis turbines, the governor servomotor moves the guide vanes to vary flow; the spear and deflector belong to Pelton wheels.
24. Why is a jet deflector used along with the spear valve in a Pelton turbine governing system?
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Closing a nozzle quickly on a long penstock is dangerous.
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Answer: D. To divert the jet quickly on load rejection while the spear closes slowly to avoid water hammer
Sudden closing of the nozzle would cause dangerous pressure rise in the penstock; the deflector instantly turns the jet away from the buckets while the spear valve closes gradually.
25. 'Double regulation' in a Kaplan turbine means that the governor adjusts:
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Kaplan blades are adjustable.
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Answer: D. Both the guide vane opening and the runner blade angle
A Kaplan governor moves the wicket gates and, through a cam relationship, the runner blade angle, keeping efficiency high over a wide load range.
26. Which factor is LEAST important when selecting a site for a hydroelectric plant?
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A hydro plant burns no fuel.
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Answer: B. Nearness to a coal or oil supply
Hydro plants need water, head, storage and sound geology; fuel supply is a consideration for thermal and diesel plants, not hydro.
27. The flow duration curve of a river is mainly used in hydropower planning to:
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It shows flow versus percentage of time exceeded.
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Answer: D. Determine firm (dependable) flow and the plant's design discharge
A flow duration curve shows the percentage of time a given flow is equalled or exceeded, from which firm power and design discharge are chosen.
28. Which of the following is an auxiliary system of a hydro power plant?
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Exclude equipment that belongs to a steam plant.
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Answer: A. Governor oil pressure unit
Hydro auxiliaries include the oil pressure unit for governor servomotors, cooling water, compressed air, drainage/dewatering pumps and station service supply; the others belong to steam plants.
29. Compressed air in a hydro power station is commonly used for:
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Think of brakes and accumulators.
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Answer: D. Applying generator brakes during stopping and charging the governor oil pressure tank
Station compressed air systems supply the mechanical brakes used to stop the unit and keep the air cushion in the governor pressure accumulator.
30. Compared with thermal plants, a key operating advantage of hydro plants is that they:
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Consider start-up time.
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Answer: D. Can be started and loaded quickly, making them suitable for peak load and frequency control
Hydro units can start and pick up load within minutes and have very low running cost, though their capital cost is high and output depends on water availability.
31. The penstock is usually fitted with a main inlet valve (butterfly or spherical) just before the turbine mainly to:
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It is an on/off valve, not a control valve.
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Answer: D. Isolate the turbine for maintenance and act as an emergency shut-off
The main inlet valve is fully open or fully closed; it isolates the turbine and closes in emergencies, while speed is regulated by guide vanes or spear valve.
3.2 Diesel electric power plants
30 questions · AElE0302
32. Diesel engines used in power plants work on the principle of:
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There is no spark plug.
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Answer: B. Compression ignition
In a diesel engine air is compressed to a high temperature and the injected fuel ignites by itself; no spark plug is used.
33. Which of the following is the most common role of diesel power plants in an interconnected grid?
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Compare start-up time with fuel cost.
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Answer: A. Standby, emergency and peak-load supply
Diesel plants start quickly but have high fuel cost, so they are used for standby/emergency, peak load and supply to isolated areas rather than base load.
34. Which is NOT a normal consideration in selecting a site for a diesel power plant?
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One option belongs to a different type of plant.
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Answer: D. Large catchment area and high head
Catchment area and head are hydro site factors; diesel sites need fuel access, nearness to load, cooling water and a firm foundation.
35. A diesel plant site is preferably kept somewhat away from densely populated areas mainly because of:
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Think about what a running engine gives off.
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Answer: D. Noise, vibration and exhaust emissions
Diesel engines produce noise, vibration and exhaust gases, so they are located away from residential areas where possible while staying close to the load.
36. The fuel system of a diesel power plant includes storage tanks, transfer pumps, filters, injection pumps and:
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It holds about one day's fuel near the engine.
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Answer: D. A day (service) tank
Fuel is pumped from bulk storage to a smaller day tank near the engines, then through filters to injection pumps and injectors.
37. The purpose of a muffler (silencer) in the exhaust system of a diesel plant is to:
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It works on the exhaust side.
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Answer: A. Reduce the noise of the exhaust gases
The exhaust silencer damps the pressure pulsations of exhaust gas leaving the engine, reducing noise.
38. Supercharging a diesel engine means:
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More air per stroke allows more fuel.
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Answer: A. Supplying air at higher than atmospheric pressure to increase power output
A supercharger or turbocharger forces more air into the cylinders, allowing more fuel to be burnt and more power from the same engine size.
39. A turbocharger on a diesel engine is driven by:
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'Turbo' refers to a turbine using waste energy.
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Answer: B. The exhaust gases of the engine
A turbocharger uses an exhaust-gas turbine to drive the intake air compressor, recovering some of the exhaust energy.
40. Large diesel engines in power plants are usually started by:
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Large plants keep air receivers for this purpose.
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Answer: D. Compressed air admitted to the cylinders
Large engines are started by compressed air from receivers; small engines use battery-powered electric starter motors.
41. Which component in the cooling system of a diesel plant removes heat from the jacket water?
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Heat must go to the atmosphere somehow.
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Answer: A. Radiator or heat exchanger with cooling tower
Jacket cooling water is circulated through a radiator or heat exchanger (with raw water/cooling tower) to reject engine heat.
42. Which of the following is NOT a function of the lubrication system of a diesel engine?
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Pick the one that has nothing to do with oil.
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Answer: D. Raising the compression temperature
Lubricating oil reduces friction, cools, cleans and seals; it does not raise the compression temperature.
43. The air intake system of a diesel plant includes an air filter mainly to:
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Dust is abrasive.
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Answer: B. Remove dust that would cause wear of cylinders and pistons
Dust entering the cylinder acts as an abrasive; the air filter removes it to protect liners, pistons and rings.
44. The governor of a diesel generating set keeps the speed (frequency) nearly constant by controlling the:
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What sets the engine's power each cycle?
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Answer: A. Quantity of fuel injected
Diesel governors act on the fuel rack of the injection pumps, changing the fuel delivered per stroke as load changes.
45. The typical compression ratio of a diesel engine is in the range of about:
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It must be higher than a petrol engine.
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Answer: C. 14 to 22
Diesel engines need high compression (roughly 14–22) to reach the self-ignition temperature of the fuel; petrol engines work at about 6–10.
46. Which is an advantage of a diesel power plant compared with a steam power plant of similar small rating?
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Think about start-up time and plant size.
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Answer: A. It can be started quickly and needs less space
Diesel plants are compact and quick to start, but have higher fuel and maintenance costs and are limited to relatively small unit sizes.
47. A heavy flywheel is fitted to a diesel engine-alternator set mainly to:
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Power strokes come in pulses.
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Answer: A. Smooth out the cyclic fluctuation of engine torque and speed
Torque from a reciprocating engine is pulsating; the flywheel's inertia reduces speed fluctuation within each cycle, which keeps the alternator frequency and voltage steady.
48. In the layout of a diesel power plant, the engines are usually placed:
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Maintenance access is a priority.
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Answer: B. Side by side in the engine room with space for maintenance and an overhead crane
Engine-generator sets are arranged in parallel rows in the engine hall with access space and an overhead crane for overhauling; fuel bulk storage is kept separate for fire safety.
49. A diesel plant runs at 500 kW for 10 hours with a specific fuel consumption of 0.25 kg/kWh. The fuel used is:
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Fuel = energy × SFC.
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Answer: C. 1250 kg
Energy = 500 × 10 = 5000 kWh; fuel = 5000 × 0.25 = 1250 kg.
50. A diesel set has a specific fuel consumption of 0.25 kg/kWh, and the fuel's calorific value is 42,000 kJ/kg. Its overall efficiency is about:
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1 kWh = 3600 kJ.
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Answer: C. 34.3%
η = 3600/(SFC × CV) = 3600/(0.25 × 42000) = 0.343 = 34.3%.
51. An engine has an indicated power of 600 kW and mechanical efficiency of 85%. If the alternator efficiency is 95%, the electrical output is about:
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Multiply the efficiencies in sequence.
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Answer: B. 484.5 kW
BP = 600 × 0.85 = 510 kW; electrical output = 510 × 0.95 = 484.5 kW.
52. A diesel engine runs at 750 rpm driving an alternator for a 50 Hz supply. The alternator must have:
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N = 120f/P.
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Answer: C. 8 poles
P = 120f/N = 120 × 50 / 750 = 8 poles.
53. A diesel set burns 20 kg/h of fuel (CV 44,000 kJ/kg) and delivers 80 kW. Its overall thermal efficiency is about:
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Convert kJ/h to kW by dividing by 3600.
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Answer: C. 32.7%
Input = 20 × 44000/3600 = 244.4 kW; η = 80/244.4 = 0.327 = 32.7%.
54. A 1 MW diesel plant runs at full load 8 h/day for 30 days with SFC 0.22 kg/kWh. If fuel density is 0.85 kg/L, the fuel storage needed is about:
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Find mass first, then divide by density.
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Answer: C. 62,100 L
Energy = 1000 × 8 × 30 = 240,000 kWh; fuel = 240,000 × 0.22 = 52,800 kg; volume = 52,800/0.85 ≈ 62,100 L.
55. The jacket cooling water must remove 300 kW of heat with a temperature rise of 10 °C (c = 4.186 kJ/kg·°C). The water flow required is about:
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Q = m·c·ΔT.
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Answer: B. 7.2 kg/s
m = Q/(cΔT) = 300/(4.186 × 10) ≈ 7.17 kg/s.
56. A diesel station has a maximum demand of 800 kW and generates 9600 kWh in a day. Its daily load factor is:
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Load factor = average load / maximum demand.
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Answer: B. 50%
Average load = 9600/24 = 400 kW; load factor = 400/800 = 50%.
57. In a four-stroke single-cylinder diesel engine running at 1500 rpm, the number of power strokes per minute is:
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How many revolutions per cycle in a four-stroke engine?
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Answer: B. 750
A four-stroke engine has one power stroke every two revolutions: 1500/2 = 750 per minute.
58. In a typical heat balance of a diesel engine, roughly what fraction of the fuel energy appears as useful shaft work?
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The rest goes to exhaust and cooling.
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Answer: A. About one third
Typically about 30–40% of fuel energy becomes brake power, with the rest lost to cooling water, exhaust and radiation.
59. Waste heat from diesel engine exhaust in a power plant can be usefully recovered by:
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Exhaust is hot gas.
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Answer: D. A waste-heat boiler producing steam or hot water
Hot exhaust gas can pass through a waste-heat recovery boiler to make steam or hot water, raising overall plant efficiency.
60. Why are diesel plants generally not used for large-scale base-load generation in a grid with hydro and thermal resources?
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Look at the running cost.
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Answer: D. High fuel and maintenance cost per kWh
Diesel generation has high running cost (imported oil, lubricants, maintenance), so it is economical only for short-duration, standby or remote use.
61. The fuel injector of a diesel engine is designed to:
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Diesel engines have no carburettor.
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Answer: C. Atomise the fuel into a fine spray inside the combustion chamber
The injector sprays high-pressure fuel as fine droplets into the hot compressed air for rapid evaporation and self-ignition.
3.3 Non-conventional power generation
30 questions · AElE0303
62. A photovoltaic (solar) cell converts:
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No moving parts or heat engine are involved.
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Answer: A. Light energy directly into electrical energy
A PV cell is a p-n junction in which absorbed photons create electron-hole pairs that are separated by the junction field, giving a direct current.
63. The most widely used semiconductor material for commercial solar cells is:
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It is the same material used in most ICs.
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Answer: C. Silicon
Crystalline silicon (mono and poly) dominates the PV market; its band gap (about 1.1 eV) suits the solar spectrum.
64. Standard Test Conditions (STC) for rating PV modules are:
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1367 W/m² is the extraterrestrial solar constant.
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Answer: A. 1000 W/m² irradiance, 25 °C cell temperature, air mass 1.5
Module nameplate (peak watt) ratings are given at STC: 1000 W/m², 25 °C cell temperature and AM1.5 spectrum.
65. As the cell temperature of a silicon PV module rises, its:
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Hot panels perform worse.
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Answer: A. Open-circuit voltage decreases, reducing output power
Voc of silicon cells drops by about 2 mV/°C per cell; the small rise in Isc does not compensate, so power falls with temperature.
66. The short-circuit current of a PV cell is approximately proportional to the:
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More photons, more carriers.
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Answer: C. Incident solar irradiance
Photocurrent, and hence Isc, is nearly proportional to irradiance; voltage changes only logarithmically with irradiance.
67. The purpose of Maximum Power Point Tracking (MPPT) in a PV system is to:
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It is an electrical, not mechanical, tracker.
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Answer: A. Operate the array at the voltage and current that give maximum power as conditions change
MPPT is a DC-DC converter control that keeps the array at the knee of its I-V curve where V × I is maximum, as irradiance and temperature vary.
68. Bypass diodes are connected across groups of cells in a PV module to:
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Think about partial shading.
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Answer: D. Limit power loss and hot spots when some cells are shaded
A shaded cell becomes reverse-biased and heats up; the bypass diode lets the string current flow around it, reducing loss and hot-spot damage.
69. In a stand-alone PV system with battery storage, a blocking diode (or charge controller) prevents:
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What happens when there is no sunlight?
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Answer: D. The battery from discharging back through the PV array at night
At night the array voltage falls below the battery voltage; a blocking diode or the charge controller stops reverse current from the battery into the array.
70. The device that converts the DC output of a PV array to AC for feeding the grid is a(n):
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Rectifiers do the opposite.
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Answer: B. Inverter
An inverter converts DC to AC; grid-tied inverters also synchronise with the grid and usually include MPPT.
71. Fill factor of a solar cell is defined as:
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It compares the actual maximum rectangle with the ideal one.
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Answer: D. Maximum power divided by the product of open-circuit voltage and short-circuit current
FF = Pmax/(Voc × Isc) = (Vmp × Imp)/(Voc × Isc); it measures the 'squareness' of the I-V curve.
72. A PV module has Voc = 21 V, Isc = 5.5 A, Vmp = 17 V and Imp = 5 A. Its fill factor is about:
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FF = (Vmp·Imp)/(Voc·Isc).
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Answer: B. 0.74
FF = (17 × 5)/(21 × 5.5) = 85/115.5 ≈ 0.74.
73. A 250 W PV module has an area of 1.6 m². At STC (1000 W/m²) its efficiency is about:
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Input power = irradiance × area.
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Answer: C. 15.6%
η = 250/(1000 × 1.6) = 0.156 = 15.6%.
74. A 3 kWp rooftop PV system at a site with 5 peak sun hours per day and a performance ratio of 0.8 produces about:
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Energy = peak power × peak sun hours × performance ratio.
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Answer: B. 12 kWh per day
Daily energy = 3 kW × 5 h × 0.8 = 12 kWh.
75. A stand-alone PV array must supply 48 V and 20 A using modules rated 24 V, 5 A each. The arrangement required is:
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Series adds voltage, parallel adds current.
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Answer: A. 2 modules in series per string, 4 strings in parallel
Series modules per string = 48/24 = 2; parallel strings = 20/5 = 4; total 8 modules.
76. A concentrating solar thermal power plant differs from a PV plant in that it:
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Concentration means focusing.
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Answer: A. Uses mirrors to focus sunlight to heat a fluid that drives a heat engine
CSP systems (parabolic trough, central tower, dish) concentrate direct beam radiation to produce high-temperature heat for a steam or Stirling engine.
77. The power available in the wind passing through a rotor of swept area A at speed v is:
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Mass flow rate × kinetic energy per kg.
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Answer: D. ½ρAv³
Kinetic energy per second of air mass flow ρAv is ½(ρAv)v² = ½ρAv³.
78. If the wind speed doubles, the power available in the wind becomes:
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Power depends on the cube of speed.
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Answer: C. 8 times
Power ∝ v³, so doubling v gives 2³ = 8 times the power.
79. If the rotor diameter of a wind turbine is doubled (same wind speed), the available power becomes:
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Swept area depends on diameter squared.
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Answer: C. 4 times
Power ∝ swept area ∝ D², so doubling D gives 4 times the power.
80. The Betz limit gives the theoretical maximum power coefficient of an ideal wind turbine as about:
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It is 16/27.
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Answer: B. 59.3%
Betz showed the maximum fraction of wind power that can be extracted is 16/27 ≈ 0.593, reached when the downstream velocity is one third of the upstream.
81. A wind turbine has a rotor diameter of 40 m. With air density 1.225 kg/m³ and wind speed 10 m/s, the power available in the wind is about:
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Use the radius, 20 m, for the swept area.
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Answer: B. 770 kW
A = π × 20² = 1256.6 m²; P = 0.5 × 1.225 × 1256.6 × 10³ ≈ 769.7 kW.
82. For the same 40 m rotor at 10 m/s (ρ = 1.225 kg/m³), if the power coefficient is 0.4, the turbine's mechanical output is about:
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Multiply the available wind power by Cp.
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Answer: B. 308 kW
P = Cp × ½ρAv³ = 0.4 × 769.7 ≈ 308 kW.
83. A wind turbine with rotor radius 20 m rotates at 30 rpm in a 10 m/s wind. Its tip speed ratio is about:
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TSR = ωR/v, with ω in rad/s.
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Answer: C. 6.3
ω = 2π × 30/60 = 3.14 rad/s; tip speed = 3.14 × 20 = 62.8 m/s; λ = 62.8/10 ≈ 6.3.
84. A 2 MW wind turbine generates 5256 MWh in a year. Its capacity factor is:
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Divide actual energy by rated power × 8760 h.
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Answer: B. 30%
Maximum possible = 2 × 8760 = 17,520 MWh; CF = 5256/17,520 = 0.30.
85. The wind speed at which a wind turbine is shut down to protect it from damage is called the:
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It is the upper limit of operation.
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Answer: C. Cut-out speed
Below cut-in the turbine produces no power; between rated and cut-out it is held at rated power; above cut-out it is stopped and feathered.
86. Pitch control in a large wind turbine is used mainly to:
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It acts on the blades, not the nacelle.
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Answer: D. Limit the rotor power above rated wind speed by changing blade angle
Pitch control rotates the blades about their own axis to reduce angle of attack and hold power at rated value in high winds; yaw control turns the nacelle.
87. The yaw mechanism of a horizontal-axis wind turbine:
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It acts on the nacelle's direction.
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Answer: A. Turns the nacelle and rotor to face the wind direction
The yaw drive rotates the nacelle on top of the tower so the rotor axis follows the wind direction.
88. Which of the following is a vertical-axis wind turbine?
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It looks like an eggbeater.
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Answer: D. Darrieus turbine
Darrieus (eggbeater) and Savonius rotors are vertical-axis designs and accept wind from any direction without yawing.
89. A gearbox is used in many wind turbines because:
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Compare rotor rpm with generator rpm.
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Answer: B. The rotor speed is low while conventional generators need high speed
Large rotors turn at roughly 10–30 rpm, so a step-up gearbox raises the speed for a standard 4- or 6-pole generator; direct-drive designs use multi-pole generators instead.
90. A doubly fed induction generator (DFIG) is popular in variable-speed wind turbines mainly because:
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Compare it with a full-rated converter.
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Answer: A. Its rotor-side converter handles only a fraction of rated power
In a DFIG the stator connects directly to the grid and a back-to-back converter in the rotor circuit, rated roughly 25–30% of turbine power, allows a limited variable-speed range.
91. Wind speed generally increases with height above ground because:
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Think of friction at the surface.
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Answer: D. Surface friction slows the air close to the ground
Ground roughness slows air near the surface (wind shear), so taller towers reach faster winds and more energy.
3.4 Energy storages
29 questions · AElE0304
92. A pumped storage plant normally pumps water to the upper reservoir:
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When is energy cheapest?
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Answer: A. During off-peak hours using cheap surplus energy
Pumped storage uses low-value off-peak energy to pump water up, and generates during peak hours, flattening the load curve.
93. The main benefit of a pumped storage plant to a power system is that it:
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No storage system creates energy.
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Answer: A. Shifts energy from off-peak to peak periods and improves the load factor of base-load plants
Pumped storage is a net consumer of energy (losses ~20–30%), but it fills valleys and shaves peaks, letting base plants run steadily and supplying fast reserve.
94. Modern pumped storage plants commonly use reversible pump-turbines of the:
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The same runner must work as a centrifugal pump.
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Answer: B. Francis type coupled to a motor-generator
A reversible Francis-type pump-turbine works as a pump in one direction and a turbine in the other, coupled to a synchronous motor-generator.
95. A pumped storage scheme has a pumping efficiency of 88% and a generating efficiency of 90%. Ignoring other losses, its round-trip efficiency is about:
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The efficiencies multiply.
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Answer: B. 79.2%
Round-trip efficiency = 0.88 × 0.90 = 0.792 = 79.2%.
96. An upper reservoir stores 1 × 10⁶ m³ of water at an average head of 300 m. The potential energy stored is about:
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1 MWh = 3.6 × 10⁹ J.
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Answer: C. 817 MWh
E = ρgVH = 1000 × 9.81 × 10⁶ × 300 = 2.943 × 10¹² J = 2.943 × 10¹²/3.6 × 10⁹ ≈ 817 MWh.
97. A pumped storage plant must deliver 1200 MWh during peak hours. If its overall cycle efficiency is 75%, the energy needed for pumping is:
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Input is larger than output.
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Answer: D. 1600 MWh
Pumping energy = output/efficiency = 1200/0.75 = 1600 MWh.
98. In a pumped storage plant the electrical machine works as a:
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Energy flows in opposite directions in the two modes.
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Answer: B. Motor during pumping and generator during generation
The synchronous machine runs as a motor to drive the pump and as a generator when water flows back through the turbine.
99. The nominal voltage of one lead-acid cell is about:
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A 12 V car battery has six cells.
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Answer: C. 2.0 V
A lead-acid cell has a nominal EMF of about 2 V; NiCd/NiMH cells are 1.2 V, alkaline 1.5 V and Li-ion about 3.6–3.7 V.
100. During discharge of a lead-acid battery, the specific gravity of the electrolyte:
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Acid is used up in the reaction.
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Answer: B. Decreases because sulphuric acid is consumed and water is formed
On discharge both plates form PbSO₄ and water is produced, diluting the acid, so specific gravity falls; this is used to check state of charge.
101. A battery bank rated 48 V, 200 Ah can store about:
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Wh = V × Ah.
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Answer: C. 9.6 kWh
Energy = V × Ah = 48 × 200 = 9600 Wh = 9.6 kWh.
102. A 12 V, 100 Ah battery supplies a 120 W load. If only 80% depth of discharge is allowed, the backup time is about:
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Usable energy = rated energy × DoD.
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Answer: B. 8 hours
Usable energy = 12 × 100 × 0.8 = 960 Wh; time = 960/120 = 8 h.
103. A 100 Ah battery is discharged at a rate of 0.5C. The discharge current is:
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1C would discharge it in one hour.
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Answer: C. 50 A
C-rate × capacity = 0.5 × 100 = 50 A, so the battery would last about 2 h.
104. Four 12 V, 100 Ah batteries are connected in series. The resulting bank is rated:
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Series adds voltage only.
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Answer: D. 48 V, 100 Ah
Series connection adds voltages while Ah capacity stays that of one battery: 48 V, 100 Ah (4.8 kWh).
105. Increasing the depth of discharge (DoD) used in each cycle of a lead-acid battery generally:
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Deeper cycles cause more wear.
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Answer: A. Reduces its cycle life
Deeper cycling stresses the plates more, so the number of cycles to end of life falls as DoD increases.
106. In a grid-scale Battery Energy Storage System (BESS), the power conversion system (PCS) is needed because:
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Compare the battery's current type with the grid's.
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Answer: B. Batteries store DC while the grid operates on AC
The PCS is a bidirectional inverter/rectifier that charges the DC battery from the AC grid and discharges it back as AC.
107. Compared with lead-acid batteries, lithium-ion batteries generally offer:
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They are used in phones and EVs for a reason.
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Answer: A. Higher energy density and longer cycle life
Li-ion cells have higher energy density (Wh/kg), higher cell voltage and more cycles, but need a BMS for safety (overcharge, temperature).
108. The main function of a Battery Management System (BMS) is to:
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It is the battery's protection and monitoring brain.
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Answer: D. Monitor cell voltage, current and temperature and protect cells from unsafe operation
A BMS monitors each cell, balances charge, estimates state of charge and disconnects the battery under over-voltage, under-voltage or over-temperature.
109. In a compressed air energy storage (CAES) plant, air is compressed during off-peak hours and stored in:
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Very large volumes are needed.
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Answer: C. Underground caverns such as salt caverns or depleted mines
Large CAES plants such as Huntorf (Germany) and McIntosh (USA) store air in solution-mined salt caverns.
110. In a conventional (diabatic) CAES plant, the stored compressed air is used to:
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Normally about two thirds of a gas turbine's work drives its own compressor.
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Answer: A. Supply the combustion chamber of a gas turbine, so the turbine needs no compressor during generation
The stored air is heated by burning fuel and expanded in a gas turbine; because compression was done earlier, nearly all turbine output goes to the generator.
111. A main source of inefficiency in a diabatic CAES plant is:
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Compressing air makes it hot.
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Answer: D. Loss of the heat produced during air compression
Compression heat is rejected to the atmosphere, so fuel must be burnt on expansion; adiabatic CAES stores this heat to raise efficiency.
112. The energy stored in a flywheel of moment of inertia I rotating at angular speed ω is:
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Compare with ½mv².
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Answer: D. ½Iω²
Rotational kinetic energy is E = ½Iω², analogous to ½mv².
113. A flywheel with moment of inertia 50 kg·m² rotates at 3000 rpm. The energy stored is about:
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Convert rpm to rad/s first.
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Answer: B. 2.47 MJ
ω = 2π × 3000/60 = 314.2 rad/s; E = 0.5 × 50 × 314.2² ≈ 2.47 × 10⁶ J.
114. If the speed of a flywheel is doubled, the energy it stores becomes:
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Energy depends on speed squared.
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Answer: D. 4 times
E ∝ ω², so doubling speed gives 4 times the energy.
115. A flywheel slows from 3000 rpm to 2000 rpm while supplying a load. The fraction of its initial stored energy delivered is about:
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Remaining energy fraction = (N2/N1)².
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Answer: C. 55.6%
E ∝ N²: remaining = (2000/3000)² = 0.444; delivered = 1 − 0.444 = 0.556 ≈ 55.6%.
116. Modern high-speed flywheel storage systems use vacuum enclosures and magnetic bearings mainly to:
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What causes a spinning wheel to slow down by itself?
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Answer: C. Reduce windage and friction losses
Air drag and bearing friction cause self-discharge; vacuum and magnetic bearings minimise these losses.
117. Flywheel energy storage is best suited for:
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Consider self-discharge and energy capacity.
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Answer: A. Short-duration, high-power applications such as frequency regulation and ride-through
Flywheels respond very fast and tolerate many cycles but store relatively little energy and self-discharge, so they suit seconds-to-minutes applications.
118. High-strength composite (e.g., carbon fibre) rotors are used in modern flywheels because they:
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Energy depends on speed more strongly than on mass.
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Answer: D. Allow much higher rim speeds, and stored energy rises with speed squared
Maximum speed is limited by rotor tensile strength; strong, light composites allow very high speeds, which gives high energy per unit mass.
119. Globally, the energy storage technology with the largest installed capacity is:
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It is the oldest large-scale method.
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Answer: A. Pumped hydro storage
Pumped hydro accounts for the large majority of the world's grid energy storage capacity, though battery installations are growing fast.
120. Which storage technology is generally limited by geography because it needs two reservoirs at different elevations?
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Think of head.
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Answer: A. Pumped storage
Pumped storage needs suitable topography and water for an upper and lower reservoir with a good head difference.
3.5 Excitation systems
30 questions · AElE0305
121. The main function of an excitation system of a synchronous generator is to:
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The field winding of a synchronous machine needs DC.
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Answer: C. Supply DC current to the rotor field winding and control it
The exciter provides and regulates the DC field current; by varying it, terminal voltage and reactive power output are controlled.
122. In a conventional DC excitation system, the main exciter is:
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The name of the system tells you the type of machine.
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Answer: D. A DC generator usually mounted on the same shaft as the alternator
Older systems used a shaft-driven DC generator (often with a pilot exciter) whose output fed the alternator field through slip rings.
123. In a DC excitation system, the pilot exciter is used to:
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It is the smaller stage of a two-stage exciter.
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Answer: A. Supply the field of the main exciter
The pilot exciter (a small DC generator) excites the main exciter's field, which in turn feeds the alternator field.
124. A major disadvantage of DC exciters for large generators is:
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Think about the commutator.
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Answer: B. Commutator and brush problems at high current and speed
DC machines need commutators and brushes that are hard to maintain at large ratings and high speeds, and their response is relatively slow.
125. In an AC excitation system with stationary rectifiers, the alternator field is fed through:
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The rectifier does not rotate.
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Answer: D. Slip rings and brushes from a rectified AC exciter output
With stationary rectifiers the AC exciter output is rectified outside the rotor, so slip rings are still needed to reach the field winding.
126. In a brushless excitation system, the AC exciter has:
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Everything that feeds the main field must rotate with it.
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Answer: B. A stationary field and rotating armature, with rotating diode rectifiers on the shaft
The exciter's armature rotates with the shaft; its AC output is rectified by diodes on the shaft and fed directly to the main field, eliminating brushes and slip rings.
127. The main advantage of a brushless excitation system is:
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The name says what is removed.
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Answer: D. Elimination of slip rings, brushes and commutators, reducing maintenance
Brushless systems avoid brush wear and sparking; a drawback is that field current and voltage cannot easily be measured or quickly forced to zero.
128. A limitation of brushless excitation compared with static excitation is that:
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Can you put a breaker on a rotating circuit?
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Answer: B. The main field cannot be de-excited quickly because there is no direct access to it
Since the rotating diodes connect straight to the field, a field breaker cannot be inserted; de-excitation relies on the exciter field and the slow field time constant.
129. In a static excitation system, the field current is supplied by:
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No rotating exciter machine is involved.
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Answer: B. A thyristor rectifier fed from an excitation transformer, through slip rings
A static exciter takes power from the generator terminals (or auxiliary bus) via an excitation transformer, rectifies it with thyristors and feeds the field through slip rings.
130. The main advantage of a static excitation system is:
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Thyristors can switch very quickly.
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Answer: D. Very fast response to voltage changes
Thyristor control can change the field voltage almost instantly, including negative forcing, which helps transient stability.
131. A self-excited static excitation system (fed from the generator terminals) needs 'field flashing' at start-up because:
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Where does the exciter's power come from before the voltage builds up?
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Answer: A. The generator has almost no terminal voltage until the field is energised
At start the residual voltage is too low to fire the thyristors usefully, so a station battery or auxiliary supply momentarily energises the field to build up voltage.
132. A three-phase fully controlled thyristor bridge in a static exciter is fed at 400 V line-to-line. At a firing angle of 30°, the DC output voltage is about:
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Vdc = 1.35 VLL cos α.
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Answer: C. 468 V
Vdc = 1.35 × VLL × cos α = 1.35 × 400 × cos 30° ≈ 468 V.
133. A brushless exciter uses a three-phase diode bridge on the shaft fed with 300 V line-to-line. The DC voltage applied to the main field is about:
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A diode bridge acts like a thyristor bridge with α = 0.
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Answer: C. 405 V
For a 3-phase diode bridge Vdc ≈ 1.35 × VLL = 1.35 × 300 = 405 V.
134. A generator field winding of resistance 1.25 Ω is fed at 250 V by the exciter. The field current and excitation power are:
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I = V/R and P = VI.
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Answer: B. 200 A and 50 kW
If = 250/1.25 = 200 A; P = 250 × 200 = 50,000 W = 50 kW.
135. A field winding has an inductance of 2 H and a resistance of 0.5 Ω. Its time constant is:
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τ = L/R.
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Answer: D. 4 s
τ = L/R = 2/0.5 = 4 s, which is why field forcing (ceiling voltage) is used to speed up response.
136. An exciter has a rated field voltage of 250 V and a ceiling voltage of 500 V. The ceiling voltage ratio is:
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Divide the maximum by the rated value.
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Answer: C. 2.0
Ceiling ratio = ceiling voltage/rated field voltage = 500/250 = 2.
137. A generator produces 10 kV at a field current of 200 A, operating in the linear (unsaturated) region. The field current needed for 11 kV at the same speed is about:
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EMF is proportional to field current when unsaturated.
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Answer: C. 220 A
In the linear region E ∝ If, so If = 200 × 11/10 = 220 A.
138. An 8-pole brushless AC exciter is mounted on the shaft of a 1500 rpm generator. The frequency of the exciter's armature voltage is:
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f = PN/120.
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Answer: C. 100 Hz
f = PN/120 = 8 × 1500/120 = 100 Hz.
139. An alternator's terminal voltage is 11.55 kV on no load and 11 kV on full load at the same excitation and speed. Its voltage regulation is:
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Divide by the full-load voltage.
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Answer: C. 5%
Regulation = (Vnl − Vfl)/Vfl = (11.55 − 11)/11 = 0.05 = 5%.
140. The basic function of an Automatic Voltage Regulator (AVR) is to:
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Voltage depends on field current.
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Answer: D. Keep the generator terminal voltage constant by adjusting the excitation
The AVR compares the measured terminal voltage with a reference and changes exciter output to correct the error.
141. In an AVR, the terminal voltage of the generator is measured using a:
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The quantity being measured is voltage.
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Answer: A. Potential (voltage) transformer and rectifier/transducer
A PT steps down the terminal voltage; it is rectified and filtered (or digitally measured) and compared with the reference setting.
142. Which correctly lists the basic chain of an AVR?
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Identify the loop that ends at the field winding.
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Answer: A. Voltage sensing → comparator with reference → amplifier → exciter → generator field
Error between sensed and reference voltage is amplified and drives the exciter, which changes field current and thus terminal voltage, closing the loop.
143. An AVR uses proportional control with a loop gain of 99. The steady-state voltage error is about:
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Error = 1/(1 + K).
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Answer: B. 1%
Steady-state error of a type-0 loop = 1/(1 + K) = 1/100 = 1%.
144. Increasing the gain of an AVR loop generally:
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Think of the trade-off in feedback control.
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Answer: A. Reduces steady-state error but may reduce stability
Higher gain lowers steady-state error, but too much gain causes oscillation; stabilising feedback or a power system stabiliser is added.
145. A generator connected to an infinite bus is over-excited. It will:
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Excitation controls reactive, not active, power on an infinite bus.
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Answer: A. Deliver lagging reactive power (VAr) to the system
Increasing excitation above the normal level makes the machine supply lagging VArs; real power is set by the prime mover, not by excitation.
146. When the excitation of a generator connected to a large grid is reduced too much, the risk is:
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Pmax = EV/X.
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Answer: A. Loss of synchronism due to reduced steady-state stability margin
Lower excitation reduces internal EMF and maximum power (EV/X), so the machine operates at a larger load angle and may pull out of step; under-excitation limiters prevent this.
147. The purpose of a field discharge resistor connected across the generator field when the field breaker opens is to:
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The field winding is a large inductor.
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Answer: B. Dissipate the stored magnetic energy and limit the induced voltage across the field
Opening an inductive circuit causes high voltage; the discharge resistor provides a path for field current to decay safely.
148. A power system stabiliser (PSS) adds a supplementary signal to the AVR in order to:
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It deals with oscillations of the rotor.
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Answer: A. Damp low-frequency electromechanical (rotor) oscillations
A fast high-gain AVR can reduce damping; a PSS uses speed, frequency or power deviation to add damping torque.
149. Which of the following excitation systems does NOT need slip rings?
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Where is the rectifier located?
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Answer: B. Brushless excitation system
Only the brushless system rectifies on the rotor and feeds the field directly; all others feed the field from outside through slip rings.
150. The response ratio of an excitation system indicates:
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It concerns speed of response.
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Answer: D. How fast the exciter output voltage rises when forced
Response ratio is the rate of rise of exciter voltage (over the first half second) per unit of rated field voltage; higher values mean faster excitation.
3.6 Starting of generators
30 questions · AElE0306
151. Before starting a hydro generating unit, which auxiliary must be confirmed healthy so that the turbine gates can be operated?
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What powers the gate servomotors?
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Answer: B. Governor oil pressure unit at normal pressure
Guide vanes or spear valves are moved by governor servomotors, which need the oil pressure unit at working pressure before start.
152. Before a hydro unit is started, the generator brakes must be:
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The rotor must be free to turn.
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Answer: C. Released (off) and the brake air pressure removed
Brakes are used only to stop the unit at low speed; they must be released before the guide vanes open, otherwise the brake pads and rotor will be damaged.
153. Before opening the main inlet valve of a hydro turbine, the bypass valve is opened in order to:
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Large valves should not open against a big pressure difference.
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Answer: A. Fill the spiral casing and equalise pressure across the inlet valve
Opening a large inlet valve against full penstock pressure would need huge force and cause surges; the bypass first fills the casing and balances pressure on both sides.
154. On a hydro unit with large thrust bearings, a high-pressure oil lifting (jacking) pump is run during starting to:
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Oil films form only above some speed.
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Answer: B. Form an oil film under the thrust bearing pads at low speed
At standstill and low speed no hydrodynamic oil film exists; high-pressure oil lifts the rotor slightly to prevent metal-to-metal contact.
155. Which check is part of the pre-start routine of a hydro generating unit?
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Pick the item that belongs to a hydro unit.
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Answer: D. Cooling water flow to the bearing and generator air coolers is available
Hydro units need cooling water for bearing oil coolers and generator air coolers; fuel and boilers relate to thermal or diesel units.
156. When starting a hydro generator, the field breaker is normally closed (excitation applied):
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Consider the V/Hz ratio at low speed.
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Answer: A. When the unit reaches about 90–95% of rated speed
Excitation is applied near rated speed so that voltage builds up at nearly correct frequency, avoiding over-fluxing (high V/Hz) at low speed.
157. Applying full excitation to a generator at low speed should be avoided mainly because it causes:
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Flux ∝ V/f.
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Answer: A. Over-fluxing of the generator and connected transformer (high V/Hz)
Flux is proportional to V/f; rated voltage at low frequency saturates the iron and overheats it, so V/Hz limiters are provided.
158. The 'creep detector' on a large hydro unit is used to detect:
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It concerns a unit that should be at standstill.
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Answer: C. Slow rotation of the stopped unit due to leakage through closed guide vanes
Leakage past closed wicket gates can slowly turn the rotor, damaging bearings without an oil film; the creep detector initiates brakes or oil lift.
159. Before starting a diesel generating set, which check is essential?
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Pick the item specific to a diesel engine.
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Answer: B. Adequate fuel in the day tank and lubricating oil at the correct level
A diesel set needs fuel in the service tank, lube oil level, cooling water and a healthy starting system before cranking.
160. Pre-lubrication of a large diesel engine before starting is done to:
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The engine-driven pump gives no pressure at standstill.
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Answer: C. Supply oil to bearings so that the first revolutions do not cause dry wear
A pre-lube pump fills the oil galleries and bearings before cranking, preventing wear during the first turns when the engine-driven pump gives little pressure.
161. Jacket water preheaters on standby diesel generators are used to:
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Cold engines start poorly.
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Answer: D. Keep the engine warm so that it can start and accept load quickly
A warm engine starts reliably and can take load quickly with less wear, which is vital for emergency and black-start sets.
162. For an air-started diesel engine, before starting the operator must check that:
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Large engines are turned over by stored air.
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Answer: B. The starting air receivers are at the required pressure
Air-start engines rely on stored compressed air; low receiver pressure may not turn the engine enough to start.
163. A diesel engine's starter battery is 24 V and the starter draws 800 A for 6 seconds per attempt. The charge consumed by three attempts is about:
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1 Ah = 3600 A·s.
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Answer: C. 4 Ah
Charge = 800 A × 6 s × 3 = 14,400 A·s = 14,400/3600 = 4 Ah.
164. After a diesel set is started, it is usually allowed to run unloaded or lightly loaded for a short time before full load mainly to:
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Engines like to warm up.
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Answer: D. Let oil pressure and temperatures stabilise
A short warm-up lets lube oil pressure build and cylinder temperatures rise, reducing thermal stress and wear before heavy loading.
165. Which condition is NOT required for synchronising an alternator with the bus bars?
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Ratings of machines in parallel can differ.
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Answer: A. Equal kVA ratings of the incoming machine and the bus
Synchronising needs matching voltage, frequency, phase sequence and phase angle; machine ratings need not be equal.
166. An incoming generator runs at 50.2 Hz while the bus is at 50.0 Hz. The synchroscope pointer completes one revolution in:
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The pointer turns at the slip frequency.
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Answer: C. 5 seconds
The pointer rotates at the slip frequency 0.2 Hz, so one revolution takes 1/0.2 = 5 s (in the 'fast' direction).
167. When the synchroscope pointer rotates in the 'fast' direction, it means the incoming machine:
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The synchroscope compares phase and frequency only.
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Answer: A. Has a higher frequency than the bus
The direction of rotation shows whether incoming frequency is above (fast) or below (slow) the bus frequency; voltage is checked on voltmeters.
168. In the 'dark lamp' method of synchronisation, the correct moment to close the breaker is when:
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Zero voltage across the lamps means in phase.
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Answer: A. All the lamps are dark
With lamps connected across corresponding phases, all lamps go dark when the voltages are in phase; uniform flicker of all lamps also confirms correct phase sequence.
169. An 8-pole hydro generator runs at 760 rpm while being brought up for synchronising to a 50 Hz grid. Its frequency is about:
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f = PN/120.
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Answer: C. 50.67 Hz
f = PN/120 = 8 × 760/120 ≈ 50.67 Hz; the governor must lower the speed slightly to 750 rpm for synchronising.
170. Two 11 kV (line) systems are 180° out of phase at the instant of closing a breaker. The voltage across each breaker pole (phase) is about:
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Compare phase voltages, not line voltages.
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Answer: C. 12.7 kV
Phase voltage = 11/√3 = 6.35 kV; at 180° the difference is 2 × 6.35 = 12.7 kV.
171. When synchronising an 11 kV generator with equal voltages but a phase angle error of 10°, the voltage difference across each breaker pole is about:
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ΔV = 2V sin(δ/2) using phase voltage.
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Answer: B. 1.11 kV
ΔV = 2Vph sin(δ/2) = 2 × 6351 × sin 5° ≈ 1107 V.
172. Closing the generator breaker with a large phase angle error during synchronisation results in:
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A voltage difference drives current through a low impedance.
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Answer: B. Large circulating current and severe mechanical shock to the shaft
The voltage difference drives a large transient current, producing high torque that stresses windings, shaft and couplings.
173. An automatic synchroniser normally adjusts which two quantities of the incoming unit before closing the breaker?
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Frequency is controlled by speed.
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Answer: B. Speed (through the governor) and voltage (through the AVR)
The auto-synchroniser raises/lowers governor speed reference and AVR voltage reference, then issues a closing command with advance angle compensation.
174. A 'black start' in a power system refers to:
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There is no grid supply available.
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Answer: D. Restoring a power station or grid after a total blackout without external supply
Black start is the process of starting generation and re-energising the network after a total or partial shutdown, using units that need no grid supply.
175. Which type of unit is generally most suitable as a black-start unit?
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Look for small auxiliary power demand and fast starting.
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Answer: A. Hydro unit or diesel generator with its own auxiliary supply
Hydro and diesel units need little auxiliary power and can start quickly; large thermal and nuclear units need substantial off-site power for auxiliaries.
176. A small diesel generator is often installed in a hydro power station for black-start purposes in order to:
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Main units need auxiliaries before they can start.
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Answer: D. Supply station auxiliaries such as oil pumps, compressors and battery chargers when the grid is dead
With no grid, the station needs power for governor oil pumps, cooling water, compressors and controls; the DG set provides this so the main units can start.
177. Why is the station DC battery system critical for black start of a power station?
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What still works when all AC is lost?
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Answer: A. It supplies control, protection, breaker tripping/closing and field flashing when no AC is available
The DC battery keeps controls, relays, breaker coils and emergency lighting alive and can flash the field of a static exciter when no AC supply exists.
178. During grid restoration after a blackout, black-start units are typically first used to:
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Restoration is done step by step.
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Answer: D. Energise a portion of the transmission network and supply auxiliaries of other power stations
Restoration proceeds in stages: black-start units energise cranking paths, start other stations and pick up load in small blocks, forming islands that are later synchronised.
179. When a black-start unit energises a long unloaded transmission line, a main concern is:
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Unloaded lines behave capacitively.
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Answer: D. Overvoltage due to line charging (Ferranti effect) and generator self-excitation
An unloaded line draws leading charging current, raising receiving-end voltage; a small generator absorbing much leading VAr may self-excite, so line lengths and reactors are managed carefully.
180. When picking up load during black start, load is restored in small blocks mainly to:
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An isolated small generator has little inertia.
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Answer: B. Limit frequency dips caused by sudden load addition
An isolated black-start unit has limited inertia and governor response; large load steps would drop the frequency below safe limits and trip it.