83 questions — 59 objective (1 mark), 14 short (5 marks), 10 long (10 marks). Every answer is checked against the 2014 BEE guidebook and carries its book section reference plus an explanation. ▶ Practice this chapter interactively (timer, read-aloud, progress saving).
Objective questions (1 mark) — 59
📖 §9.1 Diesel engine cycle
1. AI PRACTICE: In a diesel engine (compression ignition), what is the typical compression ratio range as per the guidebook?
6:1 to 10:1
14:1 to 25:1
25:1 to 40:1
4:1 to 8:1
Answer: B) 14:1 to 25:1
Confirmed vs Book-3 §9.1 — In the diesel engine air is drawn into the cylinder and compressed to a high ratio of 14:1 to 25:1, heating the air to 700-900 degC, so the metered diesel then injected ignites spontaneously; hence 'compression ignition (CI) engine'. (a) and (d) are spark-ignition (petrol/gas) ranges; the book's own objective question offers 10:1 to 15:1 as the trap. The high compression ratio is why CI engines are more efficient than spark-ignition engines.
Source: AI practice
📖 §9.4 Specific fuel consumption (lit/kWh), Table 9.7
2. AI PRACTICE: A 1500 kVA DG set operates at 0.8 power factor with a specific fuel consumption of 0.24 L/kWh. What is the fuel consumption rate?
360 L/hr
288 L/hr
450 L/hr
240 L/hr
Answer: B) 288 L/hr
Confirmed vs Book-3 §9.4 — Convert first: kW = kVA x PF = 1500 x 0.8 = 1200 kW. Fuel rate = kW x SFC = 1200 x 0.24 = 288 litres/hr. (a) 360 L/hr is the trap of applying the SFC to 1500 kVA without applying the power factor; (d) 240 uses 1000 kW. Table 9.7 monitors DG sets on exactly this specific fuel consumption in lit/kWh (field values typically 0.29-0.36).
Source: AI practice
📖 §9.5 Solved example (a) – BHP to kW
3. AI PRACTICE: An engine is rated at 200 BHP. What is its equivalent shaft power in kW?
268.1 kW
147.0 kW
149.2 kW
200.0 kW
Answer: C) 149.2 kW
Confirmed vs Book-3 §9.5 solved example, which uses 1 BHP = 0.746 kW: 200 x 0.746 = 149.2 kW. (a) 268.1 kW comes from dividing by 0.746 instead of multiplying. (b) 147.0 kW uses the metric horsepower factor 0.735, which the guidebook does not use — always take 0.746.
Source: AI practice
📖 §9.5 Solved example (a) – maximum power factor
4. AI PRACTICE: A 180 kVA alternator is driven by a 210 BHP engine with alternator plus exciter losses of 5.66 kW. What is the maximum power factor at which the set can run without overloading the engine?
0.80
0.84
0.87
0.90
Answer: B) 0.84
Confirmed vs Book-3 §9.5 solved example (a), identical figures — Engine rated power = 210 x 0.746 = 156.66 kW; power available for the alternator = 156.66 - 5.66 = 151 kW; maximum PF = 151/180 = 0.84. (a) 0.80 is the alternator's RATED power factor, the standard trap. (c) 0.87 comes from forgetting to deduct the 5.66 kW alternator plus exciter losses (156.66/180).
Source: AI practice
📖 §9.4 Energy performance assessment (1 kWh = 860 kcal)
5. AI PRACTICE: In the overall efficiency formula for a DG set, 1 kWh of electrical output is equivalent to how many kilocalories of heat?
1000 kcal
632 kcal
860 kcal
746 kcal
Answer: C) 860 kcal
Confirmed vs Book-3 §9.4 — 1 kWh is equivalent to 860 kcal. It converts electrical output into heat units, e.g. DG overall efficiency = (units generated x 860)/(fuel litres x density x GCV) x 100, and the book's cooling-water question: 12.9 m3/hr x 1000 x 1 x 10 degC = 1,29,000 kcal/hr, / 860 = 150 kW. (a) 1000 kcal is a rounding trap, (b) 632 kcal is one horsepower-hour and (d) 746 is the BHP-to-Watt factor, not a heat equivalent.
Source: AI practice
📖 §9.3 Waste heat recovery in DG sets
6. AI PRACTICE: In waste heat recovery from DG exhaust, why must the exit flue gas temperature not be allowed to fall below 180 degC?
To increase steam pressure
To avoid acid (sulphur) dew-point corrosion
To raise the turbocharger speed
To reduce the back pressure
Answer: B) To avoid acid (sulphur) dew-point corrosion
Confirmed vs Book-3 §9.3 — The WHR potential formula carries the term (t_gas - 180 degC) because the limiting exit gas temperature cannot be less than 180 degC, to avoid acid dew point corrosion of the cold-end heat transfer surfaces. (d) back pressure is a different WHR constraint (the book allows about 250-300 mm WC maximum) and is not the reason for the 180 degC floor. Gross under-loading of the set also drives gas temperatures down towards cold-end corrosion.
Source: AI practice
📖 §9.3 Waste heat recovery potential formula
7. AI PRACTICE: A DG set generating 900 kW has an exhaust gas temperature of 480 degC. Using 8 kg gas/kWh, specific heat 0.25 kcal/kg degC and a limiting exit temperature of 180 degC, what is the waste heat recovery potential?
4,32,000 kcal/hr
5,40,000 kcal/hr
6,00,000 kcal/hr
3,60,000 kcal/hr
Answer: B) 5,40,000 kcal/hr
Confirmed vs Book-3 §9.3 — Potential WHR = (kWh output/hr) x 8 kg gas/kWh x 0.25 kcal/kg degC x (t_gas - 180) = 900 x 8 x 0.25 x (480 - 180) = 1800 x 300 = 5,40,000 kcal/hr. (a) 4,32,000 comes from using a 240 degC drop instead of 300 degC. Cross-check with the book's own case: 800 kW at 480 degC gives 4,80,000 kcal/hr.
Source: AI practice
📖 §9.1 Table 9.1 Comparison of captive power plants
8. AI PRACTICE: As per the guidebook, what is the typical thermal efficiency range of a diesel engine power plant?
33-36%
40-46%
43-45%
25-30%
Answer: C) 43-45%
Confirmed vs Book-3 §9.1, Table 9.1 — Thermal efficiency of the diesel engine power plant is 43-45%. (b) 40-46% is the combined GT and ST figure and (a) 33-36% the conventional steam plant — both are in the same table, which is why they are such effective distractors. The same table is why the book says captive diesel wins on efficiency, capital cost (Rs 7,500-9,000/kW) and plant load factor (7200-7500 kWh/kW).
Source: AI practice
📖 §9.3 Waste heat recovery / Table 9.5 Energy balance
9. AI PRACTICE: Waste heat from a DG set is tapped mainly from the exhaust gas and from the engine jacket cooling water. Which stream does the guidebook call the MORE VERSATILE (more attractive) source for waste heat recovery, because of its much higher temperature?
Jacket cooling water
Radiation to atmosphere
Exhaust (flue) gas
Lubricating oil heat
Answer: C) Exhaust (flue) gas
Rewritten to be book-grounded (answer unchanged, c) — Book-3 §9.3: engine exhaust and cooling water each provide about half of the useful thermal energy, but the exhaust is at around 450 degC against about 100 degC for the jacket water and is 'hence more versatile' — only the exhaust can raise steam in a waste heat recovery boiler. Trap: (a) jacket water is actually the LARGER stream by quantity in Table 9.5 for a 500 kW diesel set (32% vs 24% exhaust), so 'biggest loss' and 'best for recovery' are not the same thing; the original wording of this item wrongly assumed they were. (b) radiated heat (9%) and (d) lube-oil heat are minor and low-grade.
Source: AI practice
📖 §9.2 Maximum single load on DG set
10. AI PRACTICE: The largest motor that can normally be started Direct-On-Line (DOL) on a DG set is approximately what fraction of the set kVA rating?
25%
50%
75%
100%
Answer: B) 50%
Confirmed vs Book-3 §9.2 — 'In general, the HP of the largest motor that can be started with direct on line starting is about 50% of the kVA rating of the generating set.' (c) 75% is the trap — that is the limit reached only when starting is changed to star-delta or auto-transformer. Related book limit: the starting current must not exceed 200% of the alternator full load capacity, even though DOL starting current is about six times rated current.
Source: AI practice
📖 §9.2 Sizing of a genset
11. AI PRACTICE: A captive plant has a connected load of 650 kW, diversity factor 0.54, expected loading 70%, and operates at 0.8 PF. What is the required set rating in kVA?
500 kVA
560 kVA
625 kVA
700 kVA
Answer: C) 625 kVA
Confirmed vs Book-3 §9.2 sizing example, identical figures — Max demand = connected load x diversity factor = 650 x 0.54 = 351, taken as 350 kW; set rating = 350/0.7 = 500 kW; at 0.8 PF, rating = 500/0.8 = 625 kVA. (a) 500 kVA is the trap of stopping at the kW rating and forgetting to divide by the power factor. Note the diversity factor here is defined as demand/connected load (<1), so it MULTIPLIES the connected load.
Source: AI practice
📖 §9.3 Load characteristics – alternator losses
12. AI PRACTICE: Copper (I-squared-R) losses in a DG set alternator are proportional to what?
The current delivered by the alternator
The square of the current delivered by the alternator
The square root of the current delivered by the alternator
The frequency of the alternator output
Answer: B) The square of the current delivered by the alternator
Confirmed vs Book-3 §9.3 (the book's own objective question 5) — Alternator copper losses are I2R losses and therefore vary as the SQUARE of the current delivered. (a) assumes a linear relation, which is why it is the popular wrong answer. This is why a low power factor, which raises current for the same kW, increases losses and heating; the book's remedy is power factor improvement capacitors rather than over-sizing the AC generator.
Source: AI practice
📖 §9.2 Site condition effects on performance derating — Table 9.3 altitude & intake-temperature corrections
13. How does altitude affect the performance of a DG set?
Increases power output
Reduces fuel consumption
Reduces power output
No change in fuel consumption
Answer: C) Reduces power output
Confirmed vs Book-3 §9.2 — Table 9.3 gives altitude correction factors on engine output that fall steadily with height (e.g. 0.980 at 610 m, 0.855 at 1525 m, 0.494 at 4880 m for a non-supercharged engine; supercharged engines derate less). Thinner air means less oxygen mass per stroke, so output is derated. Options (a)/(b)/(d) are wrong: altitude neither raises output nor lowers fuel needed per kWh.
Source: Sep 2024
📖 §9.4 Energy performance assessment — specific fuel consumption (L/kWh)
14. In a D G set, the generator is generating 1000kVA at 0.7PF. If the specific fuel consumption of this D G set is 0.25 lits per kWh, then how much fuel in litres will be consumed while delivering generated power for one hour?
230
250
175
225
Answer: C) 175
Confirmed vs Book-3 §9.4 — Fuel is consumed against real power (kW = kVA × PF), not kVA. 1000 kVA × 0.7 = 700 kW → 700 kWh in one hour × 0.25 L/kWh = 175 litres. Option (b) 250 L wrongly multiplies the kVA (1000 × 0.25); (a) and (d) do not follow from the data.
Source: Sep 2024
📖 §9.3 Waste heat recovery in DG sets — Table 9.5 energy balance for reciprocating engine
15. In a DG set, the component causing maximum energy loss is:
Coolant loss
Alternator loss
Radiation loss
Flue gas loss
Answer: A) Coolant loss
Corrected (was d) — Book-3 §9.3: Table 9.5 energy balance for a 500-kW diesel engine generator: electric power 35%, jacket (coolant) water 32%, exhaust heat 24%, radiated 9% (natural-gas engine: jacket 38%, exhaust 24%). So the jacket-cooling-water (coolant) stream is the single largest loss, ahead of flue gas. Flue gas (d) is tempting because the book calls the exhaust "more versatile" — it is at ~450 °C vs ~100 °C for cooling water — but it carries less energy (24% vs 32%). Alternator and radiation losses are small.
16. In a 4-stroke diesel engine, fuel is injected during:
Induction stroke
Compression stroke
Ignition and Power stroke
Exhaust stroke
Answer: C) Ignition and Power stroke
Confirmed vs Book-3 §9.1 — The book names the third stroke "ignition and power stroke: fuel is injected while the valves are closed (fuel injection actually starts at the end of the previous stroke), the fuel ignites spontaneously and the piston is forced downwards". Only air is drawn in on the induction stroke and only air is compressed on the compression stroke (CI engine), so (a)/(b) are wrong; nothing is injected during exhaust.
Source: Sep 2025
📖 §9.4 Energy performance assessment — specific fuel consumption (L/kWh)
17. A DG set operates at 1250 kVA, 0.8 PF, with specific fuel consumption of 0.23 L/kWh. Quantity of fuel used is_____________.
175 L/h
230 L/h
250 L/h
300 L/h
Answer: B) 230 L/h
Confirmed vs Book-3 §9.4 — Real power = 1250 kVA × 0.8 = 1000 kW; fuel = 1000 kW × 0.23 L/kWh = 230 L/h. Option (c) 250 L/h would need SFC 0.25; (a) 175 L/h is the 1000 kVA × 0.7 PF × 0.25 case from another exam. Always convert kVA to kW with PF before applying SFC.
Source: Sep 2025
📖 §9.1 Diesel engine cycle — compression ratio & CI principle
18. The compression ratio in diesel engines is in the range of:
10:1 to 15:1
14:1 to 25:1
5:1 to 10:1
1:2 to 3:1
Answer: B) 14:1 to 25:1
Confirmed vs Book-3 §9.1 — "air is drawn into the cylinder and is compressed to a high ratio (14:1 to 25:1). During this compression the air is heated to 700–900 °C" so the injected diesel ignites spontaneously (compression ignition). 10:1–15:1 (a) is a spark-ignition-type range; (c)/(d) are far too low to reach auto-ignition temperature.
Source: Book EOC
📖 §9.1 Diesel engine power plant developments — specific fuel consumption trend
19. Present specific fuel consumption value of DG sets in industries is about
220 g/kWh
100 g/kWh
160 g/kWh
50 g/kWh
Answer: C) 160 g/kWh
Corrected (was a) — Book-3 §9.1: "The specific fuel consumption has come down from a value of 220 g/kWh in the 1970's to a value of around 160 g/kWh in present times." 220 g/kWh (a) is the tempting distractor because it is the historical (1970s) figure printed in the same sentence; 100 and 50 g/kWh are unrealistically low.
Source: Book EOC
📖 §9.2 Selection and installation factors — sizing of a genset (diversity factor, % loading, PF)
20. The rating required for a DG set with 500 kW connected load and diversity factor of 1.5, 80% loading and 0.8 power factor is
520 kVA
600 kVA
625 kVA
500 kVA
Answer: A) 520 kVA
Confirmed vs Book-3 §9.2 — Sizing chain from the book example: Max demand = connected load ÷ diversity factor = 500/1.5 = 333.3 kW; set rating (kW) = demand ÷ % loading = 333.3/0.8 = 416.7 kW; at 0.8 PF, kVA = 416.7/0.8 = 520.8 ≈ 520 kVA. Option (c) 625 kVA is the tempting distractor because it is the answer of the book's own worked example (650 kW, DF 0.54, 70% loading) and also equals 500/0.8 if diversity and loading are ignored.
Source: Book EOC
📖 §9.3 Waste heat recovery in DG sets — potential WHR formula
21. The waste heat potential for an 1100 kVA set at 800 kW loading and with 480 C exhaust gas temperature is
4.8 lakh kcal/hr
3.5 lakh kcal/hr
3 lakh kcal/hr
2 lakh kcal/hr
Answer: A) 4.8 lakh kcal/hr
Confirmed vs Book-3 §9.3 — Book formula: Potential WHR = (kWh output/hr) × 8 kg gas/kWh × 0.25 kcal/kg °C × (t_g − 180 °C), the 180 °C floor avoiding acid dew-point corrosion. The book works this exact case: 800 × 8 × 0.25 × (480 − 180) = 4,80,000 kcal/hr = 4.8 lakh kcal/hr. Using a lower gas rate or a 120 °C exit temperature gives the smaller distractor values.
Source: Book EOC
📖 §9.3 Load characteristics — power factor and alternator losses
22. For a DG set, the copper losses in the alternator are proportional to the:
current delivered by the alternator
square of the current delivered by the alternator
square root of the current delivered by the alternator
none of the above
Answer: B) square of the current delivered by the alternator
Confirmed vs Book-3 §9.3 — Copper (I²R) loss in the alternator windings varies with the square of the load current; that is why the book says lower power factor (more current for the same kW) "demands higher excitation currents and results in increased losses". Losses are not linear (a) or square-root (c) in current.
Source: Book EOC
📖 §9.2 Maximum single load on DG set
23. The capacity of the largest motor that can be started on a given DG set is
25%
50% kVA rating of DG set
75%
100%
Answer: B) 50% kVA rating of DG set
Confirmed vs Book-3 §9.2 — "the HP of the largest motor that can be started with direct on line starting is about 50% of the kVA rating of the generating set", because DOL starting current is ~6× rated and should not exceed 200% of alternator full-load capacity. 75% (c) applies only with star-delta or auto-transformer starting; 25%/100% are not book figures.
24. The jacket cooling water in a diesel engine flows at 12.9 m3/hr with a range of 10 C and accounts for 30% of the engine input energy. The power loss in cooling water is
430 kW
500 kW
387 kW
150 kW
Answer: D) 150 kW
Confirmed vs Book-3 §9.3 — Heat carried by jacket water = 12.9 m³/hr × 1000 kg/m³ × 1 kcal/kg °C × 10 °C = 1,29,000 kcal/hr; ÷ 860 kcal/kWh = 150 kW. The "30% of input" figure is not needed for this part (it gives input = 430,000 kcal/hr → 43 kg/hr of 10,000 kcal/kg fuel, the sister question). 430 kW (a) is the tempting wrong pick — it is the fuel INPUT in kcal/hr ÷ 1000, not the cooling-water loss.
Source: Book EOC
📖 §9.3 Trigeneration technology
25. The operating economics of trigeneration primarily depends on the
capacity of generator
capacity of waste heat boiler
capacity of VAM
cost of fuel
Answer: D) cost of fuel
Confirmed vs Book-3 §9.3 — Trigeneration (simultaneous electricity, heat and cooling with WHR boiler + VAM on jacket water) is developed "to further optimize fuel utilization"; its operating economics therefore hinge on the cost of fuel, since all three outputs come from the one fuel stream. Equipment capacities (a)–(c) set capital cost and output, not the running economics.
26. The efficiency of a Diesel Generator power plant ranges between:
20 - 25%
0 - 20%
40 - 45%
60 - 70%
Answer: C) 40 - 45%
Confirmed vs Book-3 §9.1 — Advantages of diesel power plants include "higher efficiency (as high as 43–45%)", and Table 9.1 lists diesel engine plant thermal efficiency 43–45% (vs 33–36% conventional steam, 40–46% combined cycle). 20–25% (a) understates it; 60–70% (d) is beyond any single prime mover.
Source: Book EOC
📖 §9.1 Introduction — spark ignition vs compression ignition engines
27. A spark ignition engine is used for firing which of the following fuels?
high speed diesel
light diesel oil
natural gas
furnace oil
Answer: C) natural gas
Confirmed vs Book-3 §9.1 — "Spark ignition engines use a spark ... Typical fuels for such engines are gasoline, natural gas and sewage and landfill gas." HSD, LDO and furnace oil are compression-ignition (diesel/heavy fuel oil) fuels, so (a), (b) and (d) are wrong.
Source: Book EOC
📖 §9.3 Operational factors — load pattern & DG set capacity; sequencing of loads (kW on engine, kVA on generator)
28. Two most important electrical parameters to be monitored for safe operation of a Diesel Generator set are:
voltage and ampere
kW and kVA
power factor and ampere
kVA and ampere
Answer: B) kW and kVA
Corrected (was a) — Book-3 §9.3: overload "should be carefully analysed" for a DG set and its transient limits apply "to both kW (as reflected on the engine) and kVA (as reflected on the generator)"; the diesel engine is designed for only 10% overload for 1 hr in 12 and the alternator for 50% overload for 15 s. kW therefore guards the engine and kVA the alternator, so together they are the two parameters to watch for safe operation (same answer given in the verified Set-A model solution and the 2010 official key on capacity utilisation). Voltage and ampere (a) alone do not show whether the engine (kW) is overloaded.
Source: 15th Exam
📖 §9.4 Energy performance assessment — specific fuel consumption & alternator loading
29. In a DG set, the generator capacity is 1000 kVA with a rated power factor 0.8. It is consuming 150 litre per hour diesel oil. If the specific fuel consumption is 0.25 litres/kWh at that load, then what is the kVA loading of the set at 0.88 PF?
682 kVA
800 kVA
750 kVA
none of the above
Answer: A) 682 kVA
Confirmed vs Book-3 §9.4 — kW generated = fuel rate ÷ SFC = 150 L/h ÷ 0.25 L/kWh = 600 kW. kVA loading = kW ÷ PF = 600/0.88 = 681.8 ≈ 682 kVA. Option (c) 750 kVA is the tempting value obtained by dividing 600 kW by the rated 0.8 PF instead of the actual 0.88 PF.
Source: 15th Exam
📖 §9.2 Unbalanced load effects
30. The maximum unbalanced load between phases should not exceed _______ % of the capacity of the DG set
10
5
1
none of the above
Answer: A) 10
Confirmed vs Book-3 §9.2 — "The maximum unbalanced load between phases should not exceed 10% of the capacity of the generating sets"; unbalance heats the alternator and produces unbalanced output voltages. 5% and 1% are not book limits.
Source: 15th Exam
📖 §9.2 Maximum single load on DG set
31. The capacity of largest motor that can be started in the given DG set is …… of kVA rating of DG set
25%
50%
75%
100%
Answer: B) 50%
Confirmed vs Book-3 §9.2 — With DOL starting (starting current ~6× rated, limited to 200% of alternator full-load capacity) the largest motor is about 50% of the genset kVA; it can go up to 75% (c) only with star-delta or auto-transformer starting, which is why (c) is the tempting wrong choice for the general case.
Source: 15th Exam
📖 §9.3 Waste heat recovery — jacket water heat balance; fuel input from calorific value
32. The jacket cooling water in a diesel engine flows at 12.9 m3/hr with a range of 10oC and accounts for 30% of the engine input energy. What will be the hourly diesel consumption in kg with a calorific value of 10,000 kcal/kg
43
12.9
17.3
none of the above
Answer: A) 43
Confirmed vs Book-3 §9.3 — Jacket-water heat = 12.9 m³/hr × 1000 kg/m³ × 1 kcal/kg °C × 10 °C = 1,29,000 kcal/hr, which is 30% of engine input → input = 1,29,000/0.30 = 4,30,000 kcal/hr → fuel = 4,30,000/10,000 = 43 kg/hr. Option (b) 12.9 merely repeats the water flow; (c) 17.3 has no basis.
33. Which of the following with respect to a turbocharger in a Diesel engine is true?
operates using energy of exhaust gases
decreases supply air pressure to engine
preheats the combustion air using energy from exhaust gases
all of the above
Answer: A) operates using energy of exhaust gases
Confirmed vs Book-3 §9.1 — A turbocharger is an "exhaust gas driven turbine" whose turbine wheel drives a compressor wheel that raises (not lowers) the pressure of the intake air, increasing rated output and lowering fuel consumption per kWh. It does not preheat the combustion air with exhaust energy (that is an air preheater/recuperator idea, and hot intake air actually derates the engine per Table 9.3), so (b), (c) and hence (d) are wrong.
Source: 14th Exam
📖 §9.2 Sizing — kVA = √3·V·I; §9.3 engine loading (kW) vs alternator loading; 1 HP = 0.746 kW
34. A DG set has a 300 HP engine drive and is connected to a 300 kVA alternator with 95% efficiency. When a plant load of 290 amps at 415 Volts and 0.76 power factor is connected, the engine loading works out to:
52%
74.51%
55.4%
None of the above
Answer: B) 74.51%
Confirmed vs Book-3 §9.2/§9.3 — Electrical output = √3 × 415 × 290 × 0.76 / 1000 = 158.4 kW. Engine shaft power needed = 158.4 / 0.95 (alternator efficiency) = 166.8 kW. Engine rating = 300 HP × 0.746 = 223.8 kW. Engine loading = 166.8/223.8 = 74.5%. Option (a) 52% is the tempting value obtained by comparing kW to the 300 kVA alternator rating, not the engine.
Source: 17th Sep-2016
📖 §9.4 Energy performance assessment — specific fuel consumption & alternator loading
35. In a DG set, the generator is consuming 70 litre per hour diesel oil. If the specific fuel consumption of this DG set is 0.33 litres/kWh at that load, what is the kVA loading of the set at 0.8 PF?
212 kVA
265 kVA
170 kVA
None of these
Answer: B) 265 kVA
Confirmed vs Book-3 §9.4 — kW generated = 70 L/h ÷ 0.33 L/kWh = 212.1 kW; kVA = 212.1/0.8 = 265 kVA (same question and key in the 16th-exam alternate set and March-2021 paper). Option (b) was transcribed as "262.5 kVA" in this record; it has been repaired to the computed 265 kVA so that exactly one option is correct. Option (a) 212 kVA is the tempting wrong choice — it is the kW value, not the kVA loading.
36. A spark ignition engine is used for firing which type of fuels ________.
gasoline
bio-mass
natural gas
all of the above
Answer: D) all of the above
Confirmed vs Book-3 §9.1 — Spark-ignition engine fuels listed by the book are gasoline, natural gas and sewage/landfill (bio) gas; §9.5 also recommends partial use of biomass gas (with tar removal). So all three listed fuels are SI fuels → (d). Note the Book-EOC version of this question offers only one SI fuel (natural gas) against diesel/LDO/furnace oil.
Source: Mar 2021
📖 §9.1 Table 9.1 — comparison of captive power plant types
37. Which of the following power plants has the highest efficiency?
Open cycle Gas Turbine
Diesel Engine
combined cycle gas turbine
Conventional coal plants
Answer: C) combined cycle gas turbine
Confirmed vs Book-3 §9.1 — Table 9.1: combined GT & ST 40–46%, diesel engine plant 43–45%, conventional steam plant 33–36%; an open-cycle gas turbine is lowest. The combined cycle (which recovers GT exhaust in an HRSG for a steam turbine) has the highest top-end efficiency; diesel (b) is the tempting distractor because the book says captive diesel plant "wins" in the small-capacity range and Table 9.1 puts its range close to combined cycle.
Source: Jul 2022
📖 §9.3 Factors affecting waste heat recovery from flue gases — back pressure
38. A good DG set waste heat recovery device manufacturer will take precautions to prevent which of the following problem while DG set is in operation?
voltage unbalance on generator
Excessive back pressure on engine
excessive steam generation
turbulence in exhaust gases
Answer: B) Excessive back pressure on engine
Confirmed vs Book-3 §9.3 — "Back pressure in the gas path caused by additional pressure drop in the waste heat recovery unit is another key factor. Generally the maximum back pressure allowed is around 250–300 mm WC and the heat recovery unit should have a pressure drop lower than that." Excess back pressure derates the engine and raises fuel consumption; voltage unbalance (a) is a load issue, not a WHR-device issue.
Source: 17th Sep-2016
📖 §9.4 Energy performance assessment — specific fuel consumption & alternator loading
39. A DG set consumes 70 litres per hour of diesel oil. If the specific fuel consumption of this DG set is 0.33 litres/kWh at that load, then what is the kVA loading of the set at 0.8 PF?
212 kVA
265 kVA
170 kVA
none of the above
Answer: B) 265 kVA
Confirmed vs Book-3 §9.4 — kW = 70 L/h ÷ 0.33 L/kWh = 212.1 kW; kVA = 212.1 ÷ 0.8 = 265 kVA. Option (a) 212 is the kW figure mistaken for kVA; (c) 170 comes from wrongly multiplying 212 by 0.8.
Source: 16th Exam (alt set)
📖 §9.2 Air cooling vs water cooling — cross-ventilation of engine room
40. In an engine room 15 m long, 10 m wide and 4 m high, ventilation requirement in m3/hr for 20 air changes/hr is:
30
3000
12000
none of the above
Answer: C) 12000
Confirmed vs Book-3 §9.2 (engine-room ventilation to keep cooling air/radiator temperature within limits) — Room volume = 15 × 10 × 4 = 600 m³; ventilation = 600 m³ × 20 air changes/hr = 12,000 m³/hr. Option (b) 3000 would be 5 air changes; (a) 30 confuses m³/min-type units.
Source: 18th Exam
📖 §9.3 Factors affecting waste heat recovery from flue gases
41. Which of the following factors does not affect waste heat recovery in a DG Set?
DG Set loading in kW
DG Set reactive power loading
operation period of DG Set
back pressure of flue gas path
Answer: B) DG Set reactive power loading
Confirmed vs Book-3 §9.3 — The book lists exactly three factors: (a) DG set loading and exhaust gas temperature, (b) hours of operation, (c) back pressure on the DG set. Reactive power (kVAr) loading does not burn fuel or change exhaust quantity/temperature, so it does not affect WHR — every other option is a listed factor.
Source: 18th Exam
📖 §9.4 Energy performance assessment — specific fuel consumption (L/kWh)
42. In a DG set, the generator is generating 1000 kVA, at 0.7 PF. If the specific fuel consumption of this DG set is 0.25 lts/kWh at that load, then how much fuel is consumed while delivering generated power for one hour.
230 litre
250 litre
175 litre
none of the above
Answer: C) 175 litre
Confirmed vs Book-3 §9.4 — kW = 1000 kVA × 0.7 = 700 kW → 700 kWh in one hour × 0.25 L/kWh = 175 litres. Option (b) 250 litres is the tempting error of applying SFC to kVA instead of kW.
43. One of the thermal power plants operating with 2 nos. of 500 MW units has reported the operating heat rate of 11250 kJ/kWh. The Plant Load Factor (PLF) of the power plant is 73 %. The operating efficiency of the power plant will be
38 %
35 %
30 %
32 %
Answer: D) 32 %
Confirmed vs Book-3 §9.1 (efficiency–heat-rate relation) — Efficiency = 3600 kJ/kWh ÷ heat rate = 3600/11250 = 0.32 = 32%. The 73% PLF is a distractor: PLF measures capacity utilisation over the year, not conversion efficiency. 35% (b) or 38% (a) would need heat rates of 10,286 or 9,474 kJ/kWh.
44. A 1000 kW Gas engine is designed for 38 % efficiency. The operating load of the engine is 825 kW. If the GCV of gas is 8700 kcal/m3, the hourly gas consumption will be ____________ m3/hr.
214.6
260.13
188.89
272.74
Answer: A) 214.6
Confirmed vs Book-3 §9.1/§9.4 — Heat input = output ÷ efficiency = 825/0.38 = 2171 kW = 2171 × 860 = 18,67,105 kcal/hr; gas = 18,67,105/8700 = 214.6 m³/hr. Option (b) 260.1 is the tempting wrong value obtained by using the rated 1000 kW instead of the operating 825 kW load.
45. Thermal Power Plant efficiency is low due to ____________________.
Higher steam Pressure
Higher superheat temperature
Low GCV coal
Higher Heat loss in condenser
Answer: D) Higher Heat loss in condenser
Confirmed vs Book-3 §9.1 — A conventional steam plant reaches only 33–36% (Table 9.1) because the largest share of fuel energy is rejected as latent heat to cooling water in the condenser (Rankine-cycle limit). Higher steam pressure (a) and higher superheat (b) actually raise cycle efficiency; low-GCV coal changes fuel quantity, not cycle efficiency.
Source: Sep 2019
📖 §9.2 Sizing — kVA = √3·V·I; §9.4 specific fuel consumption
46. In a DG set, a 3 phase alternator is supplying on an average 100 A at 420 V and 0.9 pf to a load. If the specific fuel consumption of this DG set is 0.30 lts/kWh at that load, then how much fuel is consumed while delivering generated power for one hour?
11.34 litre
19.64 litre
21.82 litre
218.23 litre
Answer: B) 19.64 litre
Confirmed vs Book-3 §9.2/§9.4 — Power = √3 × 420 × 100 × 0.9 / 1000 = 65.47 kW; fuel = 65.47 kWh × 0.30 L/kWh = 19.64 litres in one hour. Option (c) 21.82 L is the tempting error of ignoring PF (√3 × 420 × 100 = 72.7 kVA × 0.3); (a) 11.34 L omits √3.
47. The gross efficiency of a coal based power plant with an operating gross heat rate of 2450 kCal/kWh is
28.48%
35.10%
30%
none of the above
Answer: B) 35.10%
Confirmed vs Book-3 §9.1 — Efficiency = 860 kcal/kWh ÷ gross heat rate = 860/2450 = 0.351 = 35.10%, consistent with the 33–36% conventional steam plant range in Table 9.1. Option (a) 28.48% would correspond to a 3020 kcal/kWh heat rate.
Source: 10th Jul-2010
📖 §9.2 Sizing — kVA = √3·V·I; §9.4 specific fuel consumption
48. In a DG set, a 3 phase alternator is loaded at 450 A, at 415 volts and 0.85 PF. If the specific fuel consumption is 0.25 lts/kWh, how much fuel is consumed delivering generated power for one hour.
39.68 litre
80.86 litre
68.74 litre
none of the above
Answer: C) 68.74 litre
Confirmed vs Book-3 §9.2/§9.4 — kW = √3 × 415 × 450 × 0.85 / 1000 = 274.9 kW; fuel = 274.9 × 0.25 = 68.7 litres/hr. Option (b) 80.86 L is the tempting error of using kVA (323.4) instead of kW; (a) 39.68 L omits √3.
49. Which combination of readings as indicated by the panel mounted instruments of a DG set would give the indications of proper capacity utilisation of diesel engine and generator?
kW & Voltage
kVA & kVAr
kW & kVA
none of the above
Answer: C) kW & kVA
Confirmed vs Book-3 §9.3 — "Alongside alternator loading, the engine loading in terms of kW or BHP needs to be maintained above 50%"; transient limits apply "to both kW (as reflected on the engine) and kVA (as reflected on the generator)". Hence kW shows engine (prime mover) utilisation and kVA shows alternator utilisation. kVA & kVAr (b) both describe the alternator only; kW & voltage (a) says nothing about alternator loading.
Source: 10th Jul-2010
📖 §9.4 Energy performance assessment — specific power generation kWh/litre (1 kWh = 860 kcal)
50. How many units per liter will be available from a DG set if the operating efficiency is 40%? The calorific value of diesel is 10,000 kCal per liter
3.50
6.98
cannot be worked out as DG set loading is not indicated
4.65
Answer: D) 4.65
Confirmed vs Book-3 §9.4 — Units per litre = (CV × efficiency) / 860 = (10,000 × 0.40)/860 = 4.65 kWh/litre, in line with the 3–4 units/litre implied by the 0.29–0.36 L/kWh SFCs of Table 9.7. Loading is not needed because efficiency is already given, so (c) is wrong; (b) 6.98 comes from dividing by 573 (kcal/kWh error); (a) 3.5 assumes ~30% efficiency.
Source: 10th Jul-2010
📖 §9.3 Factors affecting waste heat recovery from flue gases — back pressure
51. The main precaution to be taken care by the waste heat recovery device manufacturer to prevent the problem in a DG set during operation is:
temperature rise
back pressure
over loading of waste heat recovery tubes
turbulence of exhaust gases
Answer: B) back pressure
Confirmed vs Book-3 §9.3 — The WHR unit sits in the exhaust path; its pressure drop adds back pressure on the engine, and the book states the maximum allowed is around 250–300 mm WC, so the recovery unit must be designed for a lower pressure drop. Tube overloading (c) and gas turbulence (d) are boiler-side issues, not the engine-protection concern.
Source: 11th Feb-2011
📖 §9.4 Energy performance assessment — specific fuel consumption (L/kWh)
52. A DG set is generating 900 kVA at 0.8 PF. If the specific fuel consumption of this DG set is 0.3 lts/kWh at that load, then how much fuel is consumed while delivering generated power for one hour.
270 litres
300 litres
216 litres
none of the above
Answer: C) 216 litres
Confirmed vs Book-3 §9.4 — kW = 900 kVA × 0.8 = 720 kW; fuel in one hour = 720 × 0.3 = 216 litres. Option (a) 270 litres is the tempting error of applying SFC to the kVA (900 × 0.3).
Source: 11th Feb-2011
📖 §9.3 Operational factors — load pattern & DG set capacity; sequencing of loads (kW on engine, kVA on generator)
53. Two most important electrical parameters, which are to be monitored on generator panel, among the following, for safe operation of a Diesel generator set are:
voltage and ampere
kVA and ampere
power factor and voltage
kW and kVA
Answer: D) kW and kVA
Corrected (was a) — Book-3 §9.3: DG-set overload limits are tight (engine 10% for 1 hr in 12; alternator 50% for 15 s) and transient/overload limits apply "to both kW (as reflected on the engine) and kVA (as reflected on the generator)". Monitoring kW protects the engine and kVA protects the alternator, so kW & kVA are the two parameters for safe operation — consistent with the verified Set-A model solution (kVA and kW) and the 2010 official key (kW & kVA for capacity utilisation). Voltage and ampere (a) do not reveal engine (kW) overload.
Source: 11th Feb-2011
📖 §9.4 Energy performance assessment — specific fuel consumption & alternator loading
54. A DG set is consuming 70 litres per hour diesel oil. If the specific fuel consumption is 0.33 litres/kWh, what is the kVA loading at 0.8 power factor ?
212 kVA
265 kVA
170 kVA
none of the above
Answer: B) 265 kVA
Confirmed vs Book-3 §9.4 — kW = 70/0.33 = 212.1 kW; kVA = 212.1/0.8 = 265 kVA. Option (a) 212 kVA is the kW figure; (c) 170 comes from multiplying by PF instead of dividing.
55. The gross efficiency of a coal based power generating unit with a gross heat rate of 2600 kcal/kWh is
41.4%
38.7%
33.1%
30.8%
Answer: C) 33.1%
Confirmed vs Book-3 §9.1 — Gross efficiency = 860/2600 = 0.3308 = 33.1%, within the 33–36% conventional steam plant band of Table 9.1. 38.7% (b) would need a 2222 kcal/kWh heat rate.
Source: Jul 2022
📖 §9.2 Sizing — kVA = √3·V·I; §9.4 specific fuel consumption
56. In a DG set, a 3-phase alternator is supplying on an average 100 A at 420 V and 0.9 pf to a load. If the specific fuel consumption of this DG set is 0.30 lit/kWh at that load, then how much fuel is consumed while delivering generated power for one hour?
11.34 litre
19.64 litre
21.82 litre
65.50 litre
Answer: B) 19.64 litre
Confirmed vs Book-3 §9.2/§9.4 — Power = √3 × 420 × 100 × 0.9 = 65,466 W = 65.47 kW; fuel = 65.47 × 0.30 = 19.64 litres per hour. (c) 21.82 L ignores the 0.9 PF; (d) 65.50 is the kW value mis-read as litres.
Source: Jul 2022
📖 §9.1 Diesel generator captive power plants — Table 9.1
57. The maximum thermal efficiency of a diesel engine power plant is in the range of _____
43-45 %
53-55%
63-65 %
73-75%
Answer: A) 43-45 %
Confirmed vs Book-3 §9.1 — "Higher efficiency (as high as 43–45%)" is listed among the advantages of diesel power plants and Table 9.1 gives 43–45% for diesel engine plants. 53–55% and above are beyond reciprocating engines (the book's best lean-burn gas engine reaches close to 45%).
Source: Jul 2022
📖 (Misfiled in Ch9 — cooling-tower topic, Book-3 cooling tower chapter) Cycles of concentration = TDS circulating ÷ TDS make-up
58. A higher operative C.O.C of a cooling tower will depend on __________.
TDS in circulating water
TDS in make-up water
both a & b
none of the above
Answer: C) both a & b
Confirmed (topic belongs to Book-3 cooling-tower chapter, not Ch9) — Cycles of concentration (COC) = TDS of circulating water ÷ TDS of make-up water, so the attainable COC depends on both: a lower make-up TDS or a higher permissible circulating-water TDS allows more cycles. Hence (c) both a & b.
59. A super thermal power station of 2500 MW installed capacity generated 14,000 million units in a year. It's annual Plant Load Factor (PLF) is __________.
60%
79%
64%
none of the above
Answer: C) 64%
Confirmed vs Book-3 §9.1 (PLF concept, Table 9.1) — Maximum possible generation = 2500 MW × 8760 h = 21,900 GWh = 21,900 million units; PLF = 14,000/21,900 = 0.639 ≈ 64%. Option (b) 79% would correspond to 17,300 MU; (a) 60% to 13,140 MU.
Source: Mar 2023
Short questions (5 marks) — 14
📖 §9.5 Solved example (b) – WHR and steam generation
1. AI PRACTICE: A 600 kW DG set has an exhaust temperature of 450 degC dropping to 230 degC in the WHR boiler. Using 8 kg gas/kWh and Cp 0.25 kcal/kg degC, with steam enthalpy 650.57 kcal/kg and feedwater at 80 degC, calculate (a) the waste heat recovered and (b) the steam generated per hour.
Model answer: (a) WHR = 600 x 8 x 0.25 x (450 - 230) = 600 x 8 x 0.25 x 220 = 2,64,000 kcal/hr. (b) Steam = WHR / (enthalpy - feedwater temp) = 2,64,000 / (650.57 - 80) = 2,64,000 / 570.57 = 462.7 kg/hr.
Confirmed vs Book-3 §9.5 solved example (b), identical figures — Two-step method: (1) WHR (kcal/hr) = kWh/hr x 8 kg gas/kWh x 0.25 kcal/kg degC x (T_in - T_out), using the ACTUAL boiler outlet temperature stated (230 degC), not the 180 degC floor. (2) Steam (kg/hr) = WHR / (steam enthalpy - feed water temperature). Always show both steps and carry the units.
Source: AI practice
📖 §9.5 Long question L-1 – cost of power generation
2. AI PRACTICE: A 6 MW DG plant runs at a load of 5 MW. Furnace oil costs Rs.28/litre and the set consumes 1230 L/hr. Calculate the cost of generation in Rs/kWh. If a gas option costs Rs.20/Sm3 and yields 3.7 kWh/Sm3, which is cheaper?
Model answer: Diesel cost = (fuel L/hr x Rs/L) / kWh generated = (1230 x 28) / 5000 = 34,440 / 5000 = Rs.6.89/kWh. Gas cost = 20 / 3.7 = Rs.5.41/kWh. The gas option is cheaper (Rs.5.41 vs Rs.6.89 per kWh).
Confirmed vs Book-3 §9.5 (L-1), identical figures — Cost of generation = (fuel litres/hr x Rs/litre) / units generated per hour, where 5 MW load = 5000 kWh/hr. For the gas option, divide the gas price per Sm3 by the kWh obtained per Sm3. Then compare the two Rs/kWh figures and state the conclusion explicitly.
Source: AI practice
📖 §9.3 Table 9.5 Energy balance for reciprocating engine
3. AI PRACTICE: List the four streams of a DG set heat balance with their approximate percentages for a 500 kW diesel engine generator, and state which stream offers the most attractive waste heat recovery and why.
Model answer: From Table 9.5, for a 500 kW DIESEL engine generator with a conventional cooling system the heat balance is: Electric power 35%; Jacket water 32%; Exhaust heat 24% (of which 16% is recoverable and 8% is lost); Radiated heat lost to atmosphere 9% - total 100%. (For a 500 kW natural gas engine generator the split is 30 / 38 / 24 / 8.) Waste heat is tapped mainly from the exhaust gases and the jacket cooling water, with additional potential from the lube oil and turbo coolers. The exhaust gas is the more attractive source: although the jacket water is the LARGER stream by quantity (32% against 24%), the exhaust is at around 450 degC against only about 100 degC for the cooling water, so it is far more versatile - it can raise steam in a waste heat recovery boiler, while jacket water is fit mainly for hot water or vapour absorption machines. Recovery is limited to a 180 degC exit gas temperature to avoid acid dew point corrosion.
Corrected — Book-3 §9.3 / Table 9.5: memorise the diesel split 35 / 32 / 24 / 9. Do NOT write that exhaust is the largest loss: jacket water (32%) exceeds exhaust (24%) in the book's table. Exhaust wins on TEMPERATURE (450 degC vs 100 degC) and hence versatility, which is exactly the reasoning the book uses.
4. A process plant continuously operates a furnace oil operated DG set of capacity 3.0 MW to avoid any process safety incident in case of tripping of critical equipment on power failure. Total critical load on DG set is 2.5 MW and exhaust flue gas at 430 deg.C is vented as original design intent was to operate DG set intermittently only during power failure. Since it is being operated continuously, the process team developed a scheme to generate saturated steam at 5 bar(g) using the waste heat boiler. Other operating parameters:
Specific heat of flue gas: 0.24 kcal/kg-Deg.C
Final stack temperature to avoid Sulphur dewing: 210.0 Deg.C
Flue gas flow: 17.5 TPH
Sat. temp. of steam at 5 barg: 159.0 Deg.C
Latent heat at 5 barg: 498.0 kcal/kg
Feed water temperature: 130.0 Deg.C
Calculate the quantity of steam generated from waste heat boiler in TPH.
5. A DG set operates at 750 kW loading with 440 C exhaust gas temperature, generating 8 kg gas/kWh with sp. heat 0.25 kcal/kg C. A heat recovery boiler reduces the exhaust gas temperature to 190 C. How much steam will be generated at 3 kg/cm2 with enthalpy 650 kcal/kg, with boiler feed water at 70 C?
Model answer: Exhaust gas mass = 750 kW x 8 kg/kWh = 6000 kg/hr. Heat recovered = mass x sp.heat x temp drop = 6000 x 0.25 x (440 - 190) = 6000 x 0.25 x 250 = 375000 kcal/hr. Heat needed per kg steam = enthalpy of steam - enthalpy of feed water = 650 - 70 = 580 kcal/kg. Steam generated = 375000 / 580 = 646.6 kg/hr, i.e. about 647 kg/hr of steam.
Gas = 6000 kg/hr; heat recovered = 6000 x 0.25 x 250 = 375000 kcal/hr; steam = 375000/(650-70) = about 647 kg/hr.
Source: Book EOC
📖 §9.2 Selection and installation factors — sizing of a genset
6. The connected load of a plant is 1200 kW and the diversity factor is 1.8. What is the desirable set rating with respect to 0.8 PF and a set load factor of 75%?
Model answer: Maximum demand = connected load / diversity factor = 1200/1.8 = 666.7 kW. Required engine kW capacity = demand / load factor = 666.7/0.75 = 888.9 kW. Required set rating in kVA = kW / PF = 888.9/0.8 = 1111.1 kVA. So the desirable DG set rating is about 1111 kVA (select the next standard size, e.g. 1125 or 1250 kVA).
7. A cement industry has a 6 MW furnace-oil DG set; furnace oil costs Rs 28/litre; average loading 5 MW with hourly furnace oil consumption 1230 litres. Estimate the cost of power generation in Rs/kWh. Management plans to convert to a gas-operated DG set; estimated gas cost Rs 20/Sm3 and power generation 3.7 kWh/Sm3. Calculate the generation cost per kWh with gas.
Model answer: Furnace oil case: fuel cost per hour = 1230 litres x Rs 28 = Rs 34440/hr. Energy generated per hour = 5 MW = 5000 kWh. Generation cost = 34440/5000 = Rs 6.89/kWh (fuel cost only). Gas case: 1 Sm3 gives 3.7 kWh, and 1 Sm3 costs Rs 20, so cost per kWh = 20/3.7 = Rs 5.41/kWh. Therefore generating with furnace oil costs about Rs 6.89/kWh while gas costs about Rs 5.41/kWh, i.e. gas is cheaper by about Rs 1.48/kWh, supporting the conversion (subject to conversion capital cost and gas availability).
Furnace oil: (1230 x 28)/5000 = Rs 6.89/kWh. Gas: 20/3.7 = Rs 5.41/kWh. Gas is about Rs 1.48/kWh cheaper.
Source: Book EOC
📖 §9.5 Solved example (a) — maximum power factor at full kVA load
8. A 180 kVA, 0.80 PF rated DG set has a diesel engine rating of 210 BHP. What is the maximum power factor that can be maintained at full load on the alternator without overloading the DG set? (Alternator losses and exciter power = 5.66 kW, no derating.)
Model answer: Engine rated power = 210 x 0.746 = 156.66 kW. Power available for alternator = 156.66 - 5.66 = 151 kW. Maximum PF = 151/180 = 0.84.
9. A DG set operates at 700 kW load with 450°C exhaust gas temperature, generating 7.8 kg exhaust gas/kWh. Specific heat of gas = 0.25 kCal/kg°C. A heat recovery boiler drops exhaust to 220°C. How much steam is generated at 3 kg/cm² with enthalpy 650.57 kcal/kg? Boiler feed water at 65°C.
Model answer: Heat recovered = 700 x 7.8 x 0.25 x (450-220) = 3,13,950 kCal/hr. Steam generation = 3,13,950/(650.57-65) = 536.14 kg/hr.
Heat available = mass x sp.heat x Δtemp; steam = heat/(steam enthalpy - feed water enthalpy).
Source: Set-A
📖 §9.3 Waste heat recovery — flue gas mass & temperature drop across WHR unit
10. A DG set rated 1000 kVA, 415 V, 1390 A, 0.8 PF, 1500 RPM. Full-load SFC = 4.0 kWh/litre of fuel; air drawn = 14 kg/kg fuel. WHR potential = 2.6×10^5 kcal/hr at exhaust gas temperature 583°C. Estimate exhaust temperature to chimney after WHR. Specific gravity 0.86, specific heat of flue gas 0.25 kcal/kg°C.
Model answer: Rated kW = 800. Oil consumption = (800×0.86)/4 = 172 kg/hr. Mass of flue gas = (14+1)×172 = 2580 kg/hr. ΔT across WHR = 260000/(2580×0.25) = 403°C. Exit flue gas temp after WHR = 583 − 403 = 180°C.
Flue gas mass=(air+1)×fuel; ΔT=Q/(m·Cp); exit = inlet − ΔT.
Source: 16th Exam
📖 §9.4 Energy performance assessment — trial measurements (fuel by dip level, kWh, PF) & analysis (% loading, kWh/litre)
11. S-6: During a DG-set performance test: 1500 kVA set, test duration 36 min, units generated 442 kWh, average PF 0.92, diesel tank 90 x 90 x 90 cm, initial dip 63 cm and final dip 79 cm (from top). Calculate (1) diesel consumption (litres), (2) average load (kW), (3) % loading, (4) specific power generation (kWh/litre).
Model answer: 1. Drop in level = 79 - 63 = 16 cm = 0.16 m. Diesel consumed = 0.9 x 0.9 x 0.16 m³ = 0.1296 m³ = 129.6 litres. 2. Average load = (442 kWh/36 min) x 60 = 736.7 kW. 3. % Loading = (kVA load/rated kVA) = (736.7/0.92)/1500 = 800.8/1500 = 53%. 4. Specific power generation = 442/129.6 = 3.41 kWh/litre.
Diesel volume from tank cross-section x level drop; average kW from kWh over test time; loading from kVA(=kW/PF) vs rated; specific generation = units/litres.
Source: 19th Exam
📖 §9.3 Waste heat recovery — flue gas mass & temperature drop across WHR unit
12. S-3: DG generator 1000 kVA, 415V, 1390A, 0.8 PF, 1500 RPM. Full load SEC = 4.0 kWh/litre, air drawn = 14 kg/kg fuel. WHR potential = 2.6x10^5 kCal/hr at exhaust gas temp 583 C. Estimate exhaust temperature to chimney after WHR. Specific gravity of fuel oil 0.86, specific heat of flue gas 0.25 kCal/kg.C.
Model answer: Rated kW = 1000 x 0.8 = 800 kW. Oil consumption = (800/4) = 200 litre/hr; in kg/hr = 200 x 0.86 = 172 kg/hr. Mass of flue gas = (14+1) x 172 = 15 x 172 = 2580 kg/hr. Delta T across WHR = 260000/(2580 x 0.25) = 403 C. Exit flue gas temp = 583 - 403 = 180 C.
Printed solution: flue gas mass 2580 kg/hr, delta T 403 C, exit temperature 180 C.
Source: 16th Exam (alt set)
📖 §9.3 Waste heat recovery — WHR formula, Table 9.6 flue gas at part load; trigeneration (VAM)
13. A 5 MW DG set running at 70% load for base load operation generates 8.6 kg of exhaust gas per kWh. The exhaust gas is reduced to 200oC. The specific heat of flue gas is 0.26 kcal/kg-oC. The steam generated from waste heat boiler will be used in double effect Li-Br Vapour Absorption Chiller with a COP of 1.12. How much TR will be generated through VAM? (5 Marks)
Model answer: Loading of DG set = 70% x 5 MW = 3.5 MW = 3500 kW. Quantity of heat from exhaust gas = 3500 kW x 8.6 kg/kWh x 0.26 kcal/kg-oC x (450 - 200) oC = 19,56,500 kcal/hr. Potential TR via double-effect VAM: COP = (TR x Heat input)/3024 -> TR = COP x Heat input/3024 = (1.12 x 1956500)/3024 = 724.6 TR.
Heat in exhaust = mass x Cp x dT; TR = COP x heat/3024 (1 TR = 3024 kcal/h).
Source: Mar 2021
📖 §9.5 Solved example (b) — WHR steam generation
14. A DG set is operating at 600 kW load with 450 °C exhaust gas temperature. The DG set generates 8 kg of exhaust gas per kWh generated. The specific heat of gas at 450 °C is 0.25 kcal/kg °C. A heat recovery boiler is installed after which the exhaust temperature drops to 230 °C. How much steam will be generated at 3 kg/cm² with enthalpy of 650.57 kcal/kg? Assume boiler feed water temperature as 80 °C.
Model answer: Waste heat recovery = 600 kWh × 8 kg gas/kWh × 0.25 kcal/kg °C × (450 − 230) °C = 2,64,000 kcal/hr. Steam generation = 2,64,000 / (650.57 − 80) = 462.7 kg/hr.
Confirmed vs Book-3 §9.5 solved example (b) — two-step WHR method: heat recovered = kWh × 8 × 0.25 × ΔT = 2,64,000 kcal/hr; steam = heat ÷ (steam enthalpy − feed-water enthalpy) = 2,64,000/570.57 = 462.7 kg/hr. The exit temperature (230 °C) stays above the 180 °C acid-dew-point floor.
Source: Book EOC
Long questions (10 marks) — 10
📖 §9.5 Solved example (a) — maximum power factor; §9.2 unbalanced load ≤10%; §9.1 turbocharger; §9.3 power factor & losses
1. a) A manufacturing plant operates a 180 kVA diesel generator set rated at 0.8 lagging PF. The prime mover is a diesel engine rated 240 BHP. The alternator has total losses (including exciter power) of 5.44 kW. Assume no derating for site conditions. The generator is required to supply a mixed industrial load at its full kVA rating. The plant manager wishes to improve system efficiency by operating at a higher power factor. The diesel engine operates at a brake thermal efficiency of 32% when loaded near its rated capacity. The calorific value of the diesel fuel is 10,500 kCal/kg, and the specific gravity of the fuel is 0.85. Calculate the following:
i) Maximum power factor that can be maintained at full kVA load without exceeding the engine capacity. (3 Marks)
ii) Corresponding diesel fuel consumption (litres per hour) at this maximum power factor. (2 Marks)
b) True or False:
i) Improving power factor of the load on a DG set reduces the apparent power drawn and increases the system's overall fuel efficiency.
ii) Alternator losses are independent of the load power factor.
iii) Turbocharger in a diesel engine helps to reduce engine noise.
iv) A diesel generator set must always be operated at unity power factor for maximum efficiency.
v) DG sets are designed to handle unbalanced load between phases to 25% of their capacity.
Model answer: a) i) Convert BHP to kW (shaft power): 240 BHP × 0.746 = 179.04 kW; Net electrical power available = 179.04 − 5.44 = 173.6 kW; At full load (180 kVA), maximum PF = Real Power / Apparent Power = 173.6 / 180 = 0.964
ii) Thermal Input Required = Electrical Output / Efficiency = 179.04 / 0.32 = 559.5 kW; 1 kg diesel = 10,500 kcal = 10,500 × 4.1868 = 43,961.4 kJ/kg; Fuel consumption (kg/hr) = (559.5 × 3600) / 43961.4 = 45.8 kg/hr; In litres per hour = 45.8 / 0.85 = 53.88 L/hr
b) i) True; ii) False; iii) False; iv) False; v) False
Max PF = (shaft kW − alternator loss)/kVA; fuel = thermal input ÷ CV, converted to litres via density.
Source: Sep 2025
📖 §9.2 Parallel operation with grid; §9.3 Load characteristics — power factor & excitation
2. A 5 MW DG set with an average load of 3 MW running in parallel with the grid was found to be exporting 100 kVAr. Without calculating, explain the possible reasons for the export of reactive power to the grid and list the advantages/disadvantages.
Model answer: Possible reasons: the alternator is being over-excited (its AVR/excitation is set too high), so it generates more reactive power (kVAr) than the local load needs and pushes the surplus into the grid; or the local plant load is highly capacitive / over-compensated by capacitor banks, leaving excess leading/lagging reactive power that flows to the grid. Advantages: exporting kVAr helps support/raise the grid voltage and the set runs at a lagging power factor that the utility may welcome for voltage support. Disadvantages: the alternator and cables carry extra current for no real-power benefit, increasing I2R (copper) losses and heating, derating the set's useful kW output, possible over-excitation/penalty issues, and the utility may not pay for (or may penalise) exported reactive power. The remedy is to adjust the excitation/AVR to a power factor near unity for the exported power.
Cause: over-excitation of the alternator (AVR set high) or over-compensated/capacitive local load. Pro: grid voltage support. Con: extra current, higher copper losses, kW derating, no commercial benefit; fix by trimming excitation.
3. What are the sources of waste heat recovery in a DG set? Waste heat recovery from which source is more economically attractive and why?
Model answer: Sources of waste heat in a DG set (Book-3 Fig. 9.1 / §9.3): (1) exhaust gases (via HRSG or heat-recovery boiler); (2) jacket cooling water circulating around the cylinders; (3) lubricating-oil cooler; (4) charge-air (turbo) cooler. Engine exhaust and cooling water each provide about half of the useful thermal energy, but the exhaust is at a much higher temperature (around 450 °C versus about 100 °C for jacket water) and is therefore more versatile: it can raise steam in a waste heat boiler (or run a vapour absorption chiller) and offers a large usable temperature margin down to the 180 °C acid-dew-point limit. Jacket water and lube oil heat is low grade, useful mainly for hot water, pre-heating or single-effect VAM. Hence exhaust-gas heat recovery is the more economically attractive source.
Sources: exhaust gas, jacket cooling water, lube oil, charge-air cooler. Exhaust gas is most attractive because it is large in quantity and high temperature, giving high-grade recoverable heat (steam/VAM).
4. Explain the principle of a four stroke diesel engine.
Model answer: A four stroke diesel engine completes one power cycle in four piston strokes (two crankshaft revolutions): (1) Suction (intake) stroke - piston moves down, the inlet valve opens and only air is drawn into the cylinder. (2) Compression stroke - both valves closed, piston moves up and compresses the air to a high ratio (about 14-25:1), raising its temperature to about 700-900 C. (3) Power (expansion) stroke - near top dead centre fuel is injected as a fine spray into the hot compressed air and auto-ignites (compression ignition); the burning gases expand and push the piston down, producing the power stroke. (4) Exhaust stroke - the exhaust valve opens and the rising piston pushes out the burnt gases. The cycle then repeats. Ignition is by compression heat (no spark plug), and the fuel is admitted by injection, not premixed.
Four strokes: suction (air in), compression (high ratio, air heated), power (fuel injected, auto-ignites, gases expand), exhaust (burnt gas out); compression ignition, one cycle per two crank revolutions.
5. A process plant has 3.5 MW DG sets fired on furnace oil. The load factor is 80% and the set is operated 7000 hours/yr. Calculate the energy generated and the waste heat recovery potential from exhaust gas.
Model answer: Average load = rated × load factor = 3500 kW × 0.80 = 2800 kW. Energy generated per year = 2800 kW × 7000 hr = 1,96,00,000 kWh = 196 lakh kWh (19.6 million units)/yr. Waste heat recovery potential from exhaust (Book-3 §9.3 formula): WHR = (kWh output/hr) × 8 kg gas/kWh × 0.25 kcal/kg °C × (t_g − 180 °C), where 180 °C is the minimum exit temperature to avoid acid dew-point corrosion. The exhaust temperature is not given; taking the book's typical value of about 450 °C after turbocharger: WHR = 2800 × 8 × 0.25 × (450 − 180) = 15,12,000 kcal/hr (≈ 15.1 lakh kcal/hr ≈ 1758 kW). Annual recoverable exhaust heat = 15,12,000 × 7000 = 1.06 × 10^10 kcal/yr. (If the exam states an exhaust temperature, substitute it in the (t_g − 180) term.)
6. As Energy Manager, what are all the factors you look into for energy saving in operating DG sets?
Model answer: 1. Ensure steady load conditions and provide cold, dust-free intake air. 2. Improve air filtration. 3. Ensure fuel oil storage, handling and operation per manufacturer/oil-company guidelines. 4. Consider fuel oil additives. 5. Calibrate fuel injection pumps periodically. 6. Ensure compliance with maintenance checklists. 7. Ensure balanced electrical loading. 8. For base-load operation, consider a waste heat recovery system.
Standard DG-set energy conservation checklist.
Source: 15th Exam
📖 §9.3 Operational factors — kW (engine) & kVA (generator) monitoring; other parts from Book-3 motor/distribution/pump chapters
7. Answer any two: (i) Two most important electrical parameters to monitor for safe DG operation. (ii) Slip method of motor load assessment. (iii) Five options for electricity distribution loss optimization. (iv) Five energy conservation opportunities in pumping systems.
Model answer: (i) kVA and kW. (ii) Slip method (uses a tachometer when no power meter): % Load = Slip/(Ss - Sr) x 100, where Slip = synchronous speed - measured speed, Ss = synchronous speed at operating frequency, Sr = nameplate full-load speed. Slip also varies inversely with terminal voltage squared, so a voltage-correction factor (V/Vr)² can be applied. (iii) Distribution loss optimization: minimise length of distribution lines; adequate conductor size; install distribution transformers at load centres; maintain high power factor; high-voltage distribution system (HVDS); amorphous core transformers. (iv) Pumping ENCON: ensure adequate NPSH; operate near BEP; minimize throttling; use variable speed drives; stop running multiple pumps (auto-start spare/booster); conduct water balance; replace old pumps with energy-efficient ones.
DG monitoring, slip method, distribution loss and pumping ENCON lists from BEE guidebook.
Source: Set-A
📖 §9.4 Energy performance assessment — 2-hour trial, efficiency, SFC, turbocharger/WHR gas temperatures
8. DG set performance: Trial 2 hrs; Energy generated 1500 kWh; Level difference in day tank 51.6 cm; Day tank diameter 1 m; CV 10500 kcal/kg; Air drawn 30 kg/kg fuel; WHR potential 2.6×10^5 kcal/hr with flue gas after WHR at 180°C. a) Calculate average efficiency and specific fuel consumption. b) Calculate present flue gas exit temperature; specific gravity 0.86, specific heat of flue gas 0.25 kcal/kg°C.
Model answer: Fuel during 2 hr = (π/4×1²×0.516×1000) = 405 litres → 202.5 lit/hr; mass = 405×0.86/2 = 174.18 kg/hr. SFC = 1500/405 ≈ 3.7 kWh/lit (or ~4.3 kWh/kg). Efficiency = (750×860)/(174.18×10500) = 35.3%. Mass of flue gas = (30+1)×174.18 = 5399.5 kg/hr. ΔT across WHR = 260000/(5399.5×0.25) = 192.61°C. Present flue gas temp = 180 + 192.61 = 372.6°C.
📖 §9.4 Energy performance assessment — trial data (fuel by dip level, kWh, PF), % loading, kWh/litre (part b: Book-3 compressed-air chapter)
9. During the performance evaluation of a DG set, the following parameters were noted — Capacity of DG set: 750 kVA; Test duration: 36 minutes; Units generated: 250 kWh; Average Power factor: 0.92 pf; Length of diesel tank: 100 cm; Width of diesel tank: 100 cm; Height of the diesel tank: 90 cm; Initial tank dip level (from top): 63 cm; Final tank dip level (from top): 53 cm. Calculate the following: 1. Diesel consumption (Litres) (1 Mark); 2. Average load (kW) (1 Mark); 3. Percentage Loading (%) (2 Marks); 4. Specific power generation (kWh/Litre) (1 Mark). b) A medium sized engineering industry has installed two 480 CFM screw compressors, A & B. Compressor-A is operating at full load and Compressor-B is running in load–unload condition. The load power of both the compressor is 74 kW and the unload power of the Compressor-B is 26 kW. Both the compressors are operated during working day. The percentage loading of the Compressor-B during working day is 70 %. After arresting the leakage in the system the loading of the compressor was found to be 35 %. Estimate the energy savings per day.
Model answer: a) 1. Diesel consumption = (1×1×0.1)×1000 = 100 Liters (level drop 63→53 = 10 cm = 0.1 m over 1m×1m tank). 2. Average load = (250/36)×60 = 416.67 kW. 3. Percentage Loading = (416.67×100)/(750×0.92) = 60.4 %. 4. Specific power generation = (250/100) = 2.5 kWh/Litre. b) Existing Case: Energy consumed per hour by Compressor-A = 74 kWh. Energy consumed per hour by Compressor-B = 0.70×74 + 0.30×26 = 59.6 kWh. Energy consumed per day = 133.6 × 24 hrs = 3206.4 kWh/day. Leakage Calculation (after arresting): Energy/hr Comp-A = 74 kWh; Energy/hr Comp-B = 0.35×74 + 0.65×26 = 42.8 kWh. Savings by arresting leakage per day = 16.8 × 24 = 403.2 kWh/day.
Diesel volume from tank dip×area; avg load = kWh/hours; %loading = load/(kVA×pf); compressor energy from load fraction × load power + unload fraction × unload power; savings = difference × 24h.
Source: Jul 2022
📖 §9.3 Waste heat recovery — flue gas mass from air/fuel ratio, WHRB heat & steam generation
10. In a DG set, the generator is rated for 1000 kVA, 415V, 1390 A, 0.8 pf, 1500 rpm. The full load specific energy consumption of this DG set as measured by the energy auditor is 3.7 kWh per litre of fuel and air drawn by the DG set is 25 kg/kg of fuel. The energy auditor recommended for a waste heat recovery system. The exhaust gas temperature difference across the waste heat recovery boiler is 215°C. The flue gas temperature after waste heat recovery system is maintained at 180°C to avoid corrosion. Calculate the steam generation in kg/hr from waste heat recovery boiler if the heat gain by feed water is 580 kCal/kg, specific gravity of feed fuel oil 0.86 and specific heat of flue gas is 0.23 kCal/kg°C.
Model answer: 1) Rated kVA = 1000; rated kW at 0.8 PF = 800 kW. 2) Specific energy generation = 3.7 kWh/litre → fuel at full load = 800/3.7 = 216.2 litres/hr; with specific gravity 0.86 → 216.2 × 0.86 = 185.95 kg/hr. 3) Air supplied = 25 kg/kg fuel → mass of flue gas = fuel × (air + 1) = 185.95 × 26 = 4834.6 kg/hr. 4) Heat available for recovery in WHRB = m × Cp × ΔT = 4834.6 × 0.23 × 215 = 2,39,071 kcal/hr (exit gas held at 180 °C to avoid acid-dew-point corrosion). 5) Heat gain by feed water to steam = 580 kcal/kg → steam generated = 2,39,071 / 580 = 412 kg/hr.