Energy Efficiency in Electrical Utilities Full exam pattern: Section I 50 objective × 1 = 50 · Section II 8 short × 5 = 40 · Section III 6 long × 10 = 60 · Total 150, pass mark 75, 3 hours. Every question is a real BEE past-paper question with the official answer, chosen by chapter weightage and how often the examiner has repeated it (42 of the 64 have appeared in two or more exams). Sit it like the real thing: start the timer, answer Section I without notes, then write Sections II and III on paper before opening the model answers. ▶ Printable PDF (question paper + answer key)
Mock exam3:00:00
Section I — Objective type (50 × 1 = 50 marks)
Answer all 50. One mark each, no negative marking.
1.
If V1 is actual supply voltage and V2 is the rated voltage of a capacitor, the reactive kVAr produced would be in the ratio of
Answer: A — Confirmed vs Book-3 §1.4 Voltage effects — capacitor output varies with the square of the applied voltage, so the kVAr actually produced is in the ratio V₁²/V₂² of the rated value (V₁ = actual, V₂ = rated). Option (c) 1 − V₁²/V₂² gives the FRACTIONAL DROP in output, not the output itself — that form is used when a question asks 'by how much does the VAr output drop'.
Ch 1 · Electrical Systems · asked in 2013, 2022 · official key
2.
Which of the following is not a part of vapour compression refrigeration cycle ?
Answer: D — The vapour compression circuit has exactly four elements: compressor, condenser, expansion device and evaporator. The GENERATOR (along with the absorber, solution pump and solution heat exchanger) belongs to the vapour absorption machine, where heat boils refrigerant out of the strong solution - it is the heat-driven half of the 'thermal compressor' that replaces the mechanical one. Same principle as the 'absorber' version of this question: anything that needs heat rather than shaft work is an absorption component. [Another copy of this question in the bank was keyed C; the official answer D is used here.] [Also asked in 2015 with different choices: (a) compressor; (b) evaporator; (c) condenser; (d) absorber — correct there: (d) absorber.]
Ch 4 · HVAC & Refrigeration · asked in 2015, 2018 · official key
3.
In a transformer on load, if the secondary voltage is one-fourth the primary voltage, then the secondary current will be
Answer: A — Confirmed vs Book-3 §1.5 — 'Primary ampere-turns are equal to secondary ampere-turns', so V₁I₁ = V₂I₂. With V₂ = V₁/4, the secondary current must be four times the primary current. Option (c) is the trap: current and voltage move in OPPOSITE directions in a transformer, so the low-voltage side carries the larger current.
Ch 1 · Electrical Systems · asked in 2013, 2022 · official key
4.
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 ?
Answer: B — 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.
Ch 9 · DG Sets · asked in 2015, 2016, 2021 · official key
5.
If 30,000 kcal of heat is removed from a room every hour then the refrigeration tonnage will be nearly equal to
Answer: C — Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/hr (book value). TR = 30000/3024 ≈ 9.92 ≈ 10 TR. Option (a) 30 TR forgets to divide by 3024 and (b) 15 TR halves it; 30,000/3024 = 9.92, which rounds to 10 TR.
Ch 4 · HVAC & Refrigeration · asked in 2017, 2019, 2022 · official key
6.
During a leak test of a compressed air system, the compressor's average load time was 1.5 minute, average unload time was 10.5 minutes and flow rate was 35 m3/min. The leakage quantity is
Answer: A — Confirmed vs Book-3 §3.5 Avoiding Air Leaks — This is the book's worked leakage example: q = T/(T+t) x Q = 1.5/(1.5+10.5) x 35 = 0.125 x 35 = 4.375 m³/min, i.e. 6300 m³/day. The other options arise from using the unload time alone or the wrong cycle total; the denominator must be load time plus unload time.
Ch 3 · Compressed Air · asked in 2021 · official key
7.
L / G ratio in a cooling tower is the ratio of _________________.
Answer: C — L over G, in that order: L = water (liquid) mass flow, G = gas (air) mass flow. Option (d) reverses the ratio and is the trap — check which term is on top before ticking. Against design values, seasonal tuning of water box loading and fan blade angle is done to restore the design L/G and recover effectiveness.
Ch 7 · Cooling Towers · asked in 2015, 2018, 2025 · official key
8.
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?
Answer: B — 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.
Ch 9 · DG Sets · asked in 2009, 2022 · official key
9.
A process fluid at 40 m³/hr, with a density of 0.95, is flowing in a heat exchanger and is to be cooled from 35 °C to 29 °C. The fluid specific heat is 0.78 kCal/kg. The chilled water range across the heat exchanger is 4 °C, the chilled water flow rate is
Answer: A — Confirmed vs Book-3 §4.7 - Heat load = 40×0.95×1000×0.78×(35-29) = 177840 kcal/hr; chilled water flow = 177840/(1000×1×4) = 44.46 m³/hr. Option (b) copies the process flow rate; the chilled-water flow must carry the same duty (177,840 kcal/h) over its own 4 degC range, giving 44.46 m3/hr.
Ch 4 · HVAC & Refrigeration · asked in 2011, 2022 · official key
10.
Increasing the cycles of concentration of circulating water in a cooling tower will
Answer: B — Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation/(COC − 1); a higher COC lowers the blowdown quantity → (b). Drift loss is set by the eliminators and air velocity, and fan power by air flow – neither depends on COC. ⚠ Option d repaired: garbled 'decrease bar create power consumption' → 'decrease fan power consumption'.
Ch 7 · Cooling Towers · asked in 2009, 2018, 2021 · official key
11.
A package air conditioner of 5 TR capacity delivers a cooling effect of 4 TR. If Energy Efficiency Ratio (W/W) is 2.90, the power in kW drawn by compressor would be:
Answer: A — Confirmed vs Book-3 §4.9 - Cooling delivered = 4 TR = 4 x 3.51 = 14.04 kW. Power = cooling/EER = 14.04/2.90 = 4.84 kW. Option (b) 1.38 divides by 2.9 twice and (c) 1.724 uses the 5 TR nameplate; use the DELIVERED 4 TR = 14.07 kW and divide by the EER of 2.90.
Ch 4 · HVAC & Refrigeration · asked in 2011, 2018, 2021 · official key
12.
If the power consumed by an air conditioner compressor is 1.7 kW per ton of refrigeration, then its energy efficiency ratio (Watt/Watt) is ___________
Answer: B — Confirmed vs Book-3 §4.9 - EER = cooling/power = 3.517 kW per TR / 1.7 kW = 2.06 W/W. Option (a) 1.7 just repeats the input figure and (c) 0.59 inverts it; EER = 3.517 kW of cooling per TR divided by 1.7 kW input = 2.06 W/W (i.e. kW/TR = 3.516/COP rearranged).
Ch 4 · HVAC & Refrigeration · asked in 2018, 2021 · official key
13.
A fan is drawing 16 kW at 800 RPM. If the speed is reduced to 600 RPM then the power drawn by the fan would be
Answer: C — Power ∝ N³: 16 × (600/800)³ = 16 × 0.75³ = 16 × 0.4219 = 6.75 kW. The two wrong options are exactly the two classic errors — 12 kW is N¹ (treating power like flow) and 9 kW is N² (treating power like pressure). Chant it: 'flow one, pressure two, power three'.
Ch 5 · Fans & Blowers · asked in 2018, 2021, 2024 · official key
14.
Which of the following is not likely to create harmonics in an electrical system?
Answer: D — Confirmed vs Book-3 §1.10 — the book classes heaters as LINEAR loads ('Incandescent lamps, heaters and, to a great extent, motors are linear systems') because their impedance is constant, so they draw a sinusoidal current and create no harmonics. Options (a)–(c) are all power-electronic (non-linear) devices — soft starters, VFDs and UPS — which the book lists as harmonic sources. [Also asked in 2017 with different choices: (a) soft starters; (b) variable frequency drives; (c) uninterrupted power supply source (UPS); (d) induction motors — correct there: (d) induction motors.]
Ch 1 · Electrical Systems · asked in 2017, 2022 · official key
15.
A 5 kVAr, 415 V rated power factor capacitor was found to be having 5.5 kVAr operating capacity. The operating supply voltage at the same supply frequency would be approximately.
Answer: C — Confirmed vs Book-3 §1.4 — kVAr ∝ V², so V = 415 × √(5.5/5) = 415 × 1.0488 ≈ 435 V. Option (a) 400 V would REDUCE the output below 5 kVAr; a higher-than-rated output always means a higher-than-rated terminal voltage (which shortens capacitor life).
Ch 1 · Electrical Systems · asked in 2011, 2018 · official key
16.
Power factor is highest in case of
Answer: D — Confirmed vs Book-3 §8.3 — An incandescent lamp is a plain heated filament, a purely resistive load, so its power factor is ~1. Sodium vapour lamps and tube lights need inductive ballasts/ignitors (§8.2 control gear) and LED lamps need electronic drivers, all of which lower the power factor.
Ch 8 · Lighting · asked in 2015, 2019, 2021 · official key
17.
In which of the following fans the air does not change flow direction from suction to discharge?
Answer: D — Confirmed vs Book-3 §5.2 — Tube-axial, vane-axial and propeller are all axial-flow fans, in which 'air enters and leaves the fan with no change in direction' → all the above (d). Only centrifugal fans turn the airflow (twice).
Ch 5 · Fans & Blowers · asked in 2013, 2022 · official key
18.
A hotel building has four floors each of 1000 m² area. If the Lighting Power Density (LPD) is 10.8 W/m², the interior lighting power allowance for the hotel building is __________.
Answer: C — Confirmed vs Book-3 §8.2 (power density) — Allowance = area × LPD = (4 × 1000 m²) × 10.8 W/m² = 43,200 W. 21,600 W would be two floors; 1000 W ignores the LPD.
Ch 10 · Buildings & ECBC · asked in 2011, 2023 · official key
19.
The source of maximum harmonics among the following, in a plant power system is
Answer: D — Confirmed vs Book-3 §1.10 — VFDs head the book's list of non-linear loads, and the harmonic current injected scales with the load, so a 225 kW drive is by far the biggest source here. Option (b) is the trap: a 500 kW resistance furnace is large but LINEAR (constant impedance), so it injects no harmonics; the 5 kVA UPS is non-linear but negligible in size.
Ch 1 · Electrical Systems · asked in 2010, 2017 · official key
20.
Which of the following type of lamps is most suitable for color critical applications?
Answer: A — Confirmed vs Book-3 §8.3 Table 8.1 — Halogen CRI = Excellent (100), LED 80, CFL 85, LPSV Poor (10 — monochromatic, colours appear grey). Colour-critical work needs the highest CRI, so halogen.
Ch 8 · Lighting · asked in 2017, 2022, 2025 · official key
21.
Which of the following compressors do not use loading / un-loading method for capacity control?
Answer: B — Confirmed vs Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14) — Load/unload (two-step) and multi-step unloading are the control schemes the book describes for positive-displacement machines — reciprocating and screw compressors. Centrifugal capacity is matched instead by variable inlet guide vanes or by speed control (Table 3.14), so the centrifugal is the odd one out.
Ch 3 · Compressed Air · asked in 2021 · official key
22.
In T-5 Fluorescent Lamp, '5' is indicative of:
Answer: D — Confirmed vs Book-3 §8.3 — T5 means a tube of 5/8 inch (16 mm) diameter; T8 = 1 inch (25 mm), T12 = 1.5 inch (38 mm). The number is neither wattage nor a saving percentage (the book's 5 % figure is the T5/T8 efficacy gain over T12, unrelated to the name).
Ch 8 · Lighting · asked in 2013, 2022 · official key
23.
The Solar Heat Gain Co-efficient (SHGC) of a window of a building is 0.30. This means
Answer: B — Confirmed vs Book-3 §10.5 — SHGC is the fraction of incident solar radiation that passes through the fenestration as heat; SHGC 0.30 means 30% of the sun's heat enters the interior (Figure 10.4: SHGC 0.39 → 39% transmitted). The remaining 70% is reflected or rejected, not 'incident', so (c)/(d) are wrong.
Ch 10 · Buildings & ECBC · asked in 2018, 2021, 2022 · official key
24.
__________ is used to capture water droplets in the air stream leaving the cooling tower.
Answer: C — Confirmed vs Book-3 §7.1 Components — Book: drift eliminators 'capture water droplets entrapped in the air stream' → (c). Splash and film fill are heat-transfer media, not droplet catchers.
Ch 7 · Cooling Towers · asked in 2023 · official key
25.
Which of the following equipment is having least compression ratio?
Answer: C — Confirmed vs Book-3 §5.1/§5.4 — Compression (specific) ratio: compressors > 1.20 > blowers 1.11–1.20 > fans ≤ 1.11; and among fans 'axial-flow fans produce lower pressure than centrifugal fans' (radial fans reach up to 1400 mmWC). So the least ratio is the axial fan (c).
Ch 5 · Fans & Blowers · asked in 2023 · official key
26.
The adsorption material used in an adsorption air dryer is
Answer: C — Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Book-3 names activated alumina and silica gel as the common adsorbents for compressed air drying; they bind moisture physically on a large porous inner surface and are then regenerated. Calcium, magnesium and potassium chloride are deliquescent absorption chemicals and are not used as the desiccant bed in these dryers.
Ch 3 · Compressed Air · asked in 2015, 2018 · official key
27.
In a vapor compression refrigeration system, the component across which the enthalpy remains constant
Answer: C — Confirmed vs Book-3 §4.3 - Expansion (throttling) is an isenthalpic process; enthalpy remains constant across the expansion valve. Enthalpy rises in the evaporator (d) and compressor (a) and falls in the condenser (b); §4.3 states there is no heat loss or gain through the expansion device, so throttling is isenthalpic.
Ch 4 · HVAC & Refrigeration · asked in 2017, 2022 · official key
28.
Lower power factor of a DG set demands __________.
Answer: C — Confirmed vs Book-3 §1.4/Ch-9 — an alternator supplies the load's reactive kVAr from its field; the lower the power factor, the larger the kVAr for the same kW, so the field (excitation) current must be increased. Option (a) is the reverse: only at high/unity power factor can the machine run with reduced excitation, which is why DG sets are de-rated at low PF.
Ch 1 · Electrical Systems · asked in 2015, 2023 · official key
29.
The unit of AAEPI is given by ___________
Answer: C — Confirmed vs Book-3 Ch-10 (ECBC / building energy index) - Air conditioning Annual Energy Performance Index is expressed as Wh per m2 per hour. Options (a), (b) and (d) mix up units used for building EPI and volumetric indices; the air-conditioning annual energy performance index is expressed per unit floor area per hour, i.e. (Wh/m2)/hr.
Ch 10 · Buildings & ECBC · asked in 2021 · official key
30.
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 __________.
Answer: C — 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.
Ch 9 · DG Sets · asked in 2023 · official key
31.
The percentage reduction in distribution losses when tail end power factor is raised from 0.8 to 0.95 is ________.
Answer: A — % reduction in distribution loss = [1 - (PF_old/PF_new)^2] x 100 = [1 - (0.8/0.95)^2] x 100 = [1 - 0.709] x 100 = 29.1% ~ 29%. The physics: for a fixed kW the line current is inversely proportional to PF, and loss goes as I^2, so loss goes as 1/PF^2. The 15.8% distractor is the un-squared version (1 - 0.8/0.95). Square the ratio, always.
Ch 1 · Electrical Systems · asked in 2016, 2017, 2025 · official key
32.
In which of the following fans air enters and leaves the fan with no change in direction ?
Answer: D — Forward-curved, backward-curved and radial are all CENTRIFUGAL impellers — air enters at the eye axially and is discharged radially, a 90° change of direction. Only the propeller fan (an axial machine) passes air straight through parallel to the shaft. Same idea as the tube-axial/vane-axial question, asked the other way round.
Ch 5 · Fans & Blowers · asked in 2009, 2018 · official key
33.
ECBC code is applicable to commercial buildings having connected load of ________.
Answer: A — Book-3 §10.2: after the EC (Amendment) Act 2010, ECBC applies to commercial buildings with connected load of 100 kW or more (or contract demand of 120 kVA or more). The exam question does not say 'minimum', but that is what it means — it asks for the load FROM WHICH the code applies, i.e. the threshold the Act names: 100 kW. A 250, 500 or 1000 kW building is of course also covered (they are above 100 kW), so those options are not 'wrong buildings' — they are simply not the threshold. Trap: 500 kW was the threshold in the original ECBC 2007; the 2010 amendment lowered it to 100 kW / 120 kVA, and the 2014 book and all later papers use the new figure.
Ch 10 · Buildings & ECBC · asked in 2021 · official key
34.
The purpose of inter-cooling in a multistage compressor is to
Answer: B — Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — Inter-stage coolers reduce the temperature of the air before it enters the next stage to reduce the work of compression and increase efficiency, by cutting the specific volume the next stage must handle. Raising pressure (a) is the compressor's job; moisture and oil separation (c) is an incidental benefit, not the purpose of inter-cooling.
Ch 3 · Compressed Air · asked in 2015, 2018, 2019, 2025 · official key
35.
In an engine room 15 m long, 10 m wide and 4 m high, ventilation requirement for 20 air changes/hr is _____ m³/hr
Answer: C — Confirmed vs Book-3 (ventilation rule) — Room volume = 15 × 10 × 4 = 600 m³; at 20 air changes per hour the ventilation air required = 600 × 20 = 12,000 m³/hr. 30 (a) is the sum of dimensions and 3000 (b) is a slip.
Ch 4 · HVAC & Refrigeration · asked in 2018, 2022 · official key
36.
A pump with 230 mm diameter impeller is delivering a flow of 150 m3/hr. If the flow is to be reduced to 110 m3/hr by trimming the impeller, what should be the approximate impeller size?
Answer: C — Confirmed vs Book-3 §6.5 — Q∝D, so D₂ = 230×(110/150) = 168.7 ≈ 169 mm. 207 mm would still deliver ~135 m³/hr; 175/195 mm do not match the ratio. The trim (to 73%) sits at the book's ~75% practical limit.
Ch 6 · Pumps · asked in 2018, 2024 · official key
37.
What is window to wall ratio __________
Answer: A — Confirmed vs Book-3 §10.5 — WWR = vertical fenestration area ÷ gross exterior wall area (gross wall measured horizontally from the exterior surface and vertically from top of floor to bottom of roof). 'Net' wall area and the inverted ratios are wrong.
Ch 10 · Buildings & ECBC · asked in 2022 · official key
38.
If a 200 cfm compressor is pressurized (on load) in 10 seconds and unloads in 20 seconds during a leakage test, the air leakage would be __________.
Answer: A — Confirmed vs Book-3 §3.5 Avoiding Air Leaks — % leakage = T/(T+t) x 100 = 10/(10+20) x 100 = 33.3%, so leakage = 0.333 x 200 = 66.7 ≈ 67 cfm. The stem was missing the compressor capacity; the printed options fix it at 200 cfm (option (b) 100 cfm is what you get by wrongly using the ratio 10/20). The denominator must be the full load-plus-unload cycle.
Ch 3 · Compressed Air · asked in 2010, 2023 · official key
39.
The blow down loss in a cooling tower depends on
Answer: D — Confirmed vs Book-3 §7.2 (vi)–(vii) — Blow Down = Evaporation/(COC − 1) and COC = TDS in circulating water / TDS in make-up water, so blowdown depends on all three quantities → (d).
Ch 7 · Cooling Towers · asked in 2022 · official key
40.
Capacitors with automatic power factor controller when installed in a plant:
Answer: C — Confirmed vs Book-3 §1.4 — capacitors 'act as reactive power generators', supplying the magnetising kVAr locally so it need not be drawn from the grid; the APFC keeps that compensation matched to a fluctuating load. Option (d) is the opposite of what happens — the line current falls; note that the apparent power (a) also falls, but the direct, primary action of a capacitor is on the REACTIVE power.
Ch 1 · Electrical Systems · asked in 2022 · official key
41.
Friction losses in a pumping system is
Answer: C — Confirmed vs Book-3 §6.2 — Book: 'The friction losses are proportional to the square of the flow rate.' This is why the system curve is parabolic and why halving flow cuts friction head to a quarter. Inverse relations (a, b, d) are wrong.
Ch 6 · Pumps · asked in 2015, 2022 · official key
42.
Which of the following statements is not true regarding centrifugal pumps?
Answer: D — Confirmed vs Book-3 §6.3/§6.6 — For a centrifugal pump flow is zero at shut-off, head falls as flow rises, and efficiency peaks at the design (BEP) flow — a, b, c are all true. On throttling the duty point moves up the curve to lower flow and the book notes 'there is some reduction in pump power absorbed at the lower flow rate' — power decreases, so d is the false statement.
Ch 6 · Pumps · asked in 2017, 2019 · official key
43.
Increasing the suction pipe diameter in a pumping system will
Answer: B — Confirmed vs Book-3 §6.5 — NPSHA = margin of eye pressure above vapour pressure; the book notes that as suction friction losses increase, NPSHA falls. A larger suction pipe lowers velocity and friction loss, so NPSHA increases. NPSHR is fixed by the pump design and does not change with pipe size. (Same as book end-of-chapter Q6.)
Ch 6 · Pumps · asked in 2013, 2017, 2022, 2024 · official key
44.
The performance of winding of an induction motor can be assessed by which of the following factors?
Answer: D — Confirmed vs Book-3 §2.8 Rewinding Effects on Energy Efficiency — The book names two indicators of rewind quality: the no-load current and the stator resistance per phase, both compared with the original values at the same voltage. (a) load current is set by the driven machine and tells nothing about the winding work, so only the combination of (b) and (c) is correct.
Ch 2 · Electric Motors · asked in 2013, 2021, 2022 · official key
45.
Use of soft starters for induction motors results in
Answer: A — Confirmed vs Book-3 §2.9 Soft Starter — The book's listed advantages of soft start are less mechanical stress, improved power factor, lower maximum demand and less mechanical maintenance. (b) and (c) state the exact opposite of two of those advantages, so (d) 'all the above' cannot hold.
Ch 2 · Electric Motors · asked in 2018, 2022 · official key
46.
Which gas is used as refrigerant both in vapour compression and vapour absorption systems
Answer: D — Confirmed vs Book-3 §4.3 - Ammonia is used as the refrigerant in both vapour compression and vapour absorption (with water) systems. Lithium bromide (a) is the absorbent and water (b) the refrigerant only in LiBr machines, while HFC-134a (c) is compression-only; ammonia is used as the refrigerant in both types.
Ch 4 · HVAC & Refrigeration · asked in 2013, 2022, 2025 · official key
47.
Adsorption drying of air is achieved using __________.
Answer: A — Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Book-3: 'The most common adsorption materials used for compressed air drying are activated alumina and silica gel.' Carbon and zirconium molecular sieves are not the materials the book names for compressed-air adsorption drying, so (a) is the correct choice and 'none of the above' is wrong.
Ch 3 · Compressed Air · asked in 2023 · official key
48.
The inexpensive way to improve energy efficiency of a motor which operates consistently at below 40% of rated capacity is by ___
Answer: A — Confirmed vs Book-3 §2.7 Improving the Motor Loading by Operating in Star Mode — For motors consistently loaded below 40 % of rating, permanent star operation is the inexpensive fix — re-wiring the terminal box and resetting the overload relay, with no new equipment. (d) VFD operation would also cut losses but requires substantial capital, failing the 'inexpensive' criterion.
Ch 2 · Electric Motors · asked in 2015, 2022 · official key
49.
A pump discharge has to be reduced from 120 m3/hr to 100 m3/hr by trimming the impeller. What should be the percentage reduction in impeller size?
Answer: B — Confirmed vs Book-3 §6.5 — Q∝D, so D₂/D₁ = Q₂/Q₁ = 100/120 = 0.833; percentage reduction = (1 − 0.833)×100 = 16.7%. 83.3% is the remaining diameter ratio, not the reduction; 33.3% would be the head reduction (1 − 0.833²).
Ch 6 · Pumps · asked in 2017, 2022 · official key
50.
Which of the following is not true regarding system characteristic curve in a pumping system with large dynamic head?
Answer: B — Confirmed vs Book-3 §6.2 — The system curve is the head-loss vs flow relationship of the piping (a); friction ∝ Q² makes it parabolic (c); with no static lift it starts at zero head and zero flow (d, Fig 6.5). It depends on elevation, pipe size/length, fittings and equipment — not on pump speed, which shifts the PUMP curve. So b is false. [Also asked in 2017 with different choices: (a) System curve represents a relationship between discharge and head loss in a system of pipes; (b) System curve is dependent on the pump characteristic curve; (c) The basic shape of system curve is parabolic; (d) System curve will start at zero flow and zero head if there is no static lift — correct there: (b) System curve is dependent on the pump characteristic curve.]
Ch 6 · Pumps · asked in 2017, 2022 · official key
Section II — Short answer type (8 × 5 = 40 marks)
Answer all questions. Write the formula line with units first — step marks are given even if the arithmetic slips.
S1
Compute AT&C (Aggregate Technical and Commercial) Losses for the given data: Input Energy Ei=20 MU, Energy Billed Metered E1=16 MU, Un-metered E2=1 MU, Total Billed Eb=17 MU, Amount Billed Ab=Rs.800 lakhs, Gross Amount Collected AG=Rs.820 lakhs, Arrears Collected Ar=Rs.40 lakhs.
5 marks · Ch 1 · Electrical Systems · asked in 2011, 2014
Model answer Amount collected without arrears Ac = AG - Ar = 820 - 40 = 780. Billing Efficiency BE = Eb/Ei = 17/20 = 85%. Collection Efficiency CE = Ac/Ab = 780/800 = 97.5%. AT&C Loss = [1 - (BE x CE)] x 100 = [1 - (0.85 x 0.975)] x 100 = 17.12%.
AT&C Loss = 1 - (Billing Efficiency x Collection Efficiency). Book-3 §1.8 Commercial Losses & AT&C Losses (Table 1.7)
S2
A multi storied office has centralized air conditioning system by using the chilled water. The chilled water inlet and outlet temperatures are 13°C and 9°C respectively. The chilled water pump discharge pressure is 4.2 kg/cm²g and the suction is 10 meters above the pump centerline. The power drawn by the chilled water pump's motor is 75 kW and an efficiency of 92%. The chilled water pump efficiency at the operating point from pump characteristic curve is 65%. Find out the operating refrigeration load in TR.
5 marks · Ch 4 · HVAC & Refrigeration · asked in 2023
Model answer Total head of the Chilled Water Pump = (4.2×10) − 10 = 32 Meter. Shaft Power of the Pump = 75×0.92 = 69 kW. Flow rate = (69×1000×0.65)/(32×1000×9.81) = 0.14287 m³/s = 514.33 m³/hr. Refrigeration load = 514330×4/3024 = 680 TR.
Head from discharge pressure and static lift; hydraulic power gives flow; cooling load = flow × ΔT × density/specific heat converted to TR (3024 kcal/h per TR). Book-3 §4.7 TR formula; pump hydraulic power
S3
List five energy saving measures in compressed air system.
5 marks · Ch 3 · Compressed Air · asked in 2024, 2025
Model answer Any five from the Book-3 §3.7 checklist, with the book's own figures: 1. Carry out periodic leak tests and arrest leaks — leakage of 40-50% is not uncommon; leakage % = load time/(load+unload) x 100. 2. Reduce compressor delivery pressure to the minimum the plant needs — a 1 bar reduction saves 6-10% of input power (Table 3.9). 3. Ensure cool, clean, dry intake air, drawn from outside if the compressor room is hot — every 4°C rise in inlet temperature costs about 1% more power, and every 250 mmWC pressure drop across a choked filter costs about 2%. 4. Minimise unloaded running: if demand is below 50% of capacity, change over to a smaller compressor, reduce speed by trimming the motor pulley, or retrofit a VSD on machines above ~100 kW (unloading draws up to 30% of full-load power). 5. Avoid misuse of compressed air (body/floor cleaning, agitation, drying) and replace pneumatic conveying and pneumatic tools with blowers or electric tools where safe — pneumatic conveying uses about 8 times and pneumatic tools about 20 times the energy. Other valid answers: size piping generously and use a ring main (0.3 bar drop in the header, 0.5 bar in distribution); clean inter-coolers and after-coolers; recover heat from hot compressed air; keep the load-unload pressure band narrow; sequence multiple compressors so only one small machine modulates; do periodic FAD tests.
Confirmed vs Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System — Any five of the printed §3.7 checklist items earn full marks; the strongest answers quote the book's numbers (4 degC/1%, 250 mmWC/2%, 1 bar/6-10%, up to 30% unload power). Book-3 Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
S4
A process plant is situated 100 m up the side of the hill. The plant requires 100 kL of water per hour. The management decides to install a pump at the ground level, with suction 3 metre below the ground level. The friction head is 12 metre. Evaluate the rating of the motor required considering 10% extra margin with respect to actual input pump power. The design pump efficiency is 65% and motor efficiency 93%.
5 marks · Ch 6 · Pumps · asked in 2021
Model answer Q = 100/3600 x 1000 = 100/3.6 m3/s = 0.0277 m3/s. Density = 1000 kg/m3. Static head = 100 + 3 = 103 m; Friction head = 12 m; Total Head = 103 + 12 = 115 m. Hydraulic power = Q x density x g x H / 1000 = 0.0277 x 1000 x 115 x 9.81/1000 = 31.25 kW. Pump input (shaft) power = 31.25/0.65 = 48.07 kW. Motor rating (shaft power) = 48.07 kW. Motor rating with 10% margin = 48.07 x 1.10 = 52.87 kW. Motor input = design rated power/motor efficiency = 48.07/0.93 = 51.69 kW.
Total head = static (lift + suction) + friction; hydraulic power via rho*g*Q*H; divide by pump efficiency for shaft power; add 10% margin and divide by motor efficiency. Book-3 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
S5
Estimate the cooling tower capacity (TR) and approach: water flow 2 m³/min, specific heat 1 kcal/kg°C, inlet water 43°C, outlet water 35°C, ambient WBT 30°C.
5 marks · Ch 7 · Cooling Towers · asked in 2013
Model answer Capacity (TR) = (flow x density x sp.heat x ΔT)/3024 = (2x60) x 1000 x 1.0 x (43-35)/3024 = 317.5 TR. Approach = 35 - 30 = 5°C.
TR from heat-load formula; approach = cold water out - ambient WBT. Book-3 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity
S6
Find out the Effective Aperture (EA) of the following two glazing and comment about compliance with ECBC. Case #1: Window to Wall Ratio (WWR) 0.2, Visible Light Transmittance (VLT) Transparent (perfectly clear glass, i.e. VLT = 1.0). Case #2: Window to Wall Ratio (WWR) 0.45, Visible Light Transmittance (VLT) 0.2.
5 marks · Ch 10 · Buildings & ECBC · asked in 2023
Model answer EA = VLT × WWR. Case #1: EA = 1.0 × 0.2 = 0.2; as EA > 0.1, glazing complies with ECBC. Case #2: EA = 0.2 × 0.45 = 0.09; as EA < 0.1, glazing does not comply with ECBC.
Book-3 §10.5: EA = VLT × WWR; transparent glazing has VLT = 1 so the effective aperture equals the opening. Book examples: 0.4 × 0.26 = 0.104 > 0.1 complies; 0.6 × 0.15 = 0.09 < 0.1 does not comply. Book-3 §10.5 Building envelope — Effective Aperture (EA = VLT × WWR), ECBC threshold 0.1
S7
A V-belt driven centrifugal fan is supplying air to a chemical process. Calculate the fan static efficiency for the following operating parameters: Ambient temperature 40degC; Density of air 1.127 kg/m3; Diameter of discharge air duct 1 meter; Velocity pressure measured by Pitot tube in discharge duct 47 mm WC; Pitot tube coefficient 0.9; Static pressure at fan inlet -22 mm WC; Static pressure at fan outlet 188 mm WC; Power drawn by motor 72 kW; Belt transmission efficiency 95%; Motor efficiency at operating load 90%.
5 marks · Ch 5 · Fans & Blowers · asked in 2021
Model answer Air velocity = Cp x (2 x 9.81 x dP/density)^0.5 = 0.9 x (2 x 9.81 x 47/1.127)^0.5 = 25.7 m/s. Area of discharge duct = 3.14 x 1 x 1/4 = 0.785 m2. Volume = 25.7 x 0.785 = 20.17 m3/s. Power input to fan shaft = 72 x 0.95 x 0.9 = 61.6 kW. Total static pressure = 188 - (-22) = 210 mm WC. Fan static efficiency = (Volume x total static pressure in mm WC) / (102 x shaft power in kW) = (20.17 x 210)/(102 x 61.6) = 67.4%.
Same method as the Book-3 §5.9 solved example: velocity from pitot VP, Q = V × A, shaft kW = motor kW × belt × motor efficiency, total static pressure = outlet − inlet (mind the negative inlet), η_static = (Q × ΔP)/(102 × shaft kW) = 67.4%. Book-3 §5.6 Pitot velocity, volume & fan static efficiency
S8
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)
5 marks · Ch 9 · DG Sets · asked in 2021
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). Book-3 §9.3 Waste heat recovery — WHR formula, Table 9.6 flue gas at part load; trigeneration (VAM)
Section III — Long answer type (6 × 10 = 60 marks)
Answer all questions. Write the formula line with units first — step marks are given even if the arithmetic slips.
L1
A review of electricity bills of a process plant was conducted as a part of energy audit. The plant has a contract demand of 3000 kVA with the power supply company. The average maximum demand of the plant is 2400 kVA/month at a power factor of 0.95. The maximum demand is at 80% of the contract demand. The minimum billable maximum demand is 80% of the contract demand. An incentive of 0.5 % reduction in energy charges component of electricity bill are provided for every 0.01 increase in power factor over and above 0.95. The average energy charge component of the electricity bill per month for the plant is Rs.80 lakhs. Calculate the following: a) If the plant decides to improve the power factor to unity, determine the power factor capacitor kVAr required and the associated monetary benefits. b) What will be the simple payback period if the cost of power factor capacitors is Rs.1200/kVAr.
10 marks · Ch 1 · Electrical Systems · asked in 2011, 2022
Model answer Contract demand 3000 kVA; recorded average MD = 2400 kVA at 0.95 PF; minimum billable demand = 80% x 3000 = 2400 kVA. a) kW drawn = 2400 x 0.95 = 2280 kW. kVAr for 0.95 -> unity = kW[tan(cos^-1 0.95) - tan(cos^-1 1)] = 2280 x (0.3287 - 0) = 749 kVAr (say 750 kVAr). Maximum demand at unity PF = 2280 kVA, but the minimum billable demand is 2400 kVA, so the plant still pays for 2400 kVA - there is NO saving in maximum demand charges. PF incentive = (1.00 - 0.95)/0.01 x 0.5% = 2.5% of the energy charge = Rs.80,00,000 x 2.5% = Rs.2,00,000/month = Rs.24,00,000/year. b) Investment = 749 kVAr x Rs.1200/kVAr = Rs.8,99,000 (say Rs.9.0 lakh). Simple payback = 8,99,000 / 24,00,000 = 0.375 year = about 4.5 months. (Same structure as the Book-3 Sec.1.11 solved example: the minimum-billing-demand clause can wipe out the MD saving, leaving the PF incentive as the only benefit.)
kVAr = kW(tanφ1-tanφ2); MD reduction nil due to 80% minimum billing; energy charge incentive 0.5% per 0.01 PF rise above 0.95 → 2.5%; payback = investment/annual savings. Book-3 §1.11 Solved Example — MD, PF capacitor kVAr & payback
L2
a) Calculate the filter area of an Air Handling Unit (AHU) for Refrigeration Load of 50 TR. The air enthalpy at inlet of AHU is 85 kJ/kg and at outlet is of 60 kJ/kg. Air velocity at filter is 1.81 m/sec and air density is 1.26 kg/m3. (5 Marks) b) A no load test was conducted in a delta connected 37 kW induction motor. Name plate data: 3 Phase, 415 V, 50 Hz, 55 Amp. Measured data on no load: Voltage V = 415 Volts; Current I = 18 Amps; Frequency F = 50 Hz; Stator phase resistance at no load = 0.23 Ohms/phase. No load power = 955 Watts. Calculate: i. The iron loss plus friction loss plus windage loss (2 Marks) ii. Stator copper loss at name plate ratings (full load), considering stator temperature as 120oC (2 Marks) iii. No load power factor of the motor (1 Mark)
10 marks · Ch 4 · HVAC & Refrigeration · asked in 2021
Model answer a) TR of AHU = (Enthalpy difference x density x area x velocity x 3600)/(4.187 x 3024). 50 = (85-60) x 1.26 x Area x 1.81 x 3600/(4.187 x 3024). Filter Area = TR x (4.187 x 3024)/[(enthalpy diff) x density x velocity x 3600] = 50 x (4.187 x 3024)/[(25) x 1.26 x 1.81 x 3600] = 3.08 m2. b) i) No-load input Pin = 955 W. No-load stator copper loss = 3 x I^2 x R = 3 x 18^2 x 0.23 = 74.51 W. Iron + friction + windage loss = Pin - no-load Cu loss = 955 - 74.51 = 880.49 W. ii) Stator resistance at 120oC = 0.23 x (120+235)/(30+235) = 0.23 x 355/265 = 0.308 ohms. Full-load stator copper loss at nameplate current = 3 x (55/sqrt3)^2 x 0.308 = 3 x 1008.3 x 0.308 = 931.65 W. iii) No-load power factor = P/(sqrt3 x V x I) = 955/(1.732 x 415 x 18) = 0.0738.
AHU: TR = (dh x rho x A x v x 3600)/(4.187 x 3024); iron+fr+wind = Pin - 3I2R; R corrected by (235+t2)/(235+t1); PF = P/(sqrt3 VI). Book-3 §4.7 air-side TR (a); Book-3 Ch-2 motors (b)
L3
In a steel industry, cooling water of 7500 m³/hr and 4200 m³/hr from two different sections with temperatures of 38 °C and 55 °C respectively, are fed to cooling tower after proper mixing. If the measured heat rejection by the cooling tower is 38,000 TR, calculate the effectiveness and evaporation loss of the cooling tower at 28 °C WBT.
10 marks · Ch 7 · Cooling Towers · asked in 2022
Model answer Mixed Hot Water Temp, °C = [(Flow1 × Temp1) + (Flow2 × Temp2)] / (Total Flow) = [(7500 × 38) + (4200 × 55)] / 11700 = 44.1 °C. Range of Cooling Tower, °C = Heat Rejection / (Flow × Density × Sp. Heat) = (38000 × 3024) / (11700 × 1000 × 1) ≈ 9.82 °C [book: 9.82]. Cold Water Temp, °C = Hot Water Temp - Range = 44.1 - 9.82 = 34.28 °C. Approach, °C = Cold Water Temp - WBT of Air = 34.28 - 28 = 6.28 °C. Effectiveness = Range / (Range + Approach) = 9.82 / (9.82 + 6.28) = 60.99% or 61%. Evaporation Loss (m³/hr) = 0.00085 × 1.8 × circulation rate (m³/hr) × Range = 0.00085 × 1.8 × 11700 × 9.82 = 175.8 m³/hr.
Mixed temp by flow-weighted average; range from heat rejection; effectiveness = range/(range+approach); evaporation = 0.00085×1.8×flow×range. ⚠ Question stem repaired: garbled 'heat rejection ... is at 1700 m³/hr' → '38,000 TR' (the value used in the printed solution: 38000 × 3024 / 11,700,000 = 9.82 °C range). Book-3 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity; (v) Evaporation loss
L4
Rated capacity of a fresh water shut-discharge pump flow rate is 485 m3/hr and discharge pressure is 13.5 kg/cm2 at rated speed of 2950 rpm. It has been observed that 450 m3/hr water is sufficient to dispose the ash. The suction pressure of the pump is 0.5 kg/cm2. The management has decided to trim the impeller to satisfy the reduced flow requirement. Calculate: a) % reduction of impeller diameter (5 Marks) b) Annual energy savings after modification, if the pump is operating for 6 hours/day and 330 days in a year. (5 Marks)
10 marks · Ch 6 · Pumps · asked in 2021
Model answer a) Q∝D, so D₂/D₁ = 450/485 = 0.928; % reduction in impeller diameter = (1 − 0.928)×100 = 7.2%. b) Rated differential head = (13.5 − 0.5) kg/cm² = 13 kg/cm² ≈ 130 m. Hydraulic power (rated) = (485/3600)×1000×130×9.81/1000 = 171.8 kW. After trimming, discharge head ∝ D²: H₂ = 0.928²×135 = 116.3 m, so new differential head = 116.3 − 5 = 111.3 m. Hydraulic power (new) = (450/3600)×1000×111.3×9.81/1000 = 136.4 kW. Power saving = 171.8 − 136.4 = 35.4 kW. Annual energy saving = 35.4×6×330 ≈ 70,000 kWh/year (official key prints 69,894 kWh using 35.3 kW). Alternative by P∝D³: P₂ = 0.928³×171.8 = 137.3 kW, saving 34.5 kW ≈ 68,300 kWh/year — both accepted.
Trim: Q∝D gives the diameter ratio; H∝D² gives new head; hydraulic power = Q·ρ·g·H/1000 before/after; annual saving = kW saved × 6 h × 330 days. Book-3 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%); §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
L5
L-6: An engineering industry operating three shifts per day has replaced its old reciprocating compressors with 1000 CFM screw compressors. During the energy audit, run hours counter readings: Load Hours start 7956, end 8401 (166 Loading kW); Un-Load Hours start 4918, end 5121 (58.1 un-loading kW). Calculate: 1. Capacity Utilization (%) of the compressor; 2. Monthly energy consumption for the present loading of the compressor; 3. Plant management is considering to install a 750 CFM compressor for energy savings. Estimate the energy savings for the same operating load, if loading power is 125 kW and unloading power is 43.75 kW; 4. To meet the present air requirement, if VFD is to be installed in the 1000 CFM compressor, what should be the percentage reduction in speed.
Confirmed vs Book-3 §3.5 Capacity Utilisation — Capacity utilization = load/(load+unload); monthly kWh = Σ(hours×power); compare with 750 CFM scenario; speed reduction from flow ratio (affinity law). Book-3 Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power); Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14)
L6
L3 a) Match the following: (4 Marks) 1 Prescriptive Approach 2 Whole Building Performance Approach 3 Building envelope 4 Effective Aperture Options: 1. Exterior façade 2. Trade-Off option 3. light admitting potential 4. Uses simulation to show compliance for the entire building b. Fill in the following blank statements: 1. The Effective Aperture (EA) or light admitting potential of a glazing system is determined by multiplying the Visible Light Transmittance (VLT) of the glazing by the _____ of the building. 2. Thermal emittance is the relative ability of a material to _____ the absorbed heat. 3. If a window has a SHGC of 0.25 and the total incident solar radiation is 600 W/m², the solar heat gain through the window is _____ 4. The emissivity of a material is the ratio of energy radiated by a particular material to energy radiated by a _____ at the same temperature. 5. As per ECBC the unit of Energy Performance Index (EPI) _____ 6. Fenestration surface having a slope of less than 60 degrees from the horizontal plane is termed _____ (6 Marks)
10 marks · Ch 10 · Buildings & ECBC · asked in 2024
Model answer a) 1. Prescriptive Approach – b (Trade-Off option) 2. Whole Building Performance Approach – d (Uses simulation to show compliance for the entire building) 3. Building envelope – a (Exterior façade) 4. Effective Aperture – c (light admitting potential) b) 1. ... multiplying the VLT of the glazing by the Window-Wall Ratio (WWR) of the building. 2. Thermal emittance is the relative ability of a material to radiate the absorbed heat. 3. Solar heat gain through the window = 0.25 × 600 = 150 W/m². 4. ... energy radiated by a black body at the same temperature. 5. As per ECBC the unit of EPI = kWh/Sq.mt/year. 6. Fenestration surface having a slope of less than 60 degrees from the horizontal plane is termed Skylight.
Book-3: Prescriptive method offers the envelope Trade-Off option; WBP uses simulation (§10.4); envelope = exterior façade; EA = VLT × WWR; thermal emittance = ability to radiate absorbed heat, emissivity relative to a black body; SHGC × incident radiation = 0.25 × 600 = 150 W/m²; EPI in kWh/sq m/year (§10.14); skylight = fenestration sloped < 60° from horizontal (§10.5). Book-3 §10.4 Compliance approaches · §10.5 Envelope terms (EA, emittance, SHGC, skylight) · §10.14 EPI unit
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