Energy Efficiency in Electrical Utilities Available here with full solutions — 66 questions recovered from the 2023 exam:
Objective (1 mark)
49 of 50
Short (5 marks)
7 of 8
Long (10 marks)
10 of 6
This is not the complete paper. The questions below are the ones we could recover and verify; the rest of that year’s paper is not reproduced here. Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 49
📖 §5.5 Series and parallel operation
1. Parallel operation of two identical fans in a ducted system.
will double the flow
will double the fan static pressure
will increase flow by more than two times
will not double the flow
Answer: D) will not double the flow
Confirmed vs Book-3 §5.5 — Two fans in parallel double the volume 'only at free delivery'; with a ducted (resisting) system, 'the higher the system resistance, the less increase in flow'. So flow does NOT double (d). Doubling static pressure (b) is the ideal for SERIES operation, not parallel.
ratio of luminous flux emitted by a lamp to the power consumed by the lamp
lux per square meter
lumen per square feet
none of the above
Answer: D) none of the above
Confirmed vs Book-3 §8.2 — One lux = one lumen per square metre. Option (a) is the definition of luminous efficacy, (b) 'lux per square metre' is circular/wrong, (c) lumen per square foot is the foot-candle; hence none of the above.
📖 §1.4 Power Factor (see also Book-3 Ch-9 DG Sets)
3. Lower power factor of a DG set demands __________.
lower excitation currents
no change in excitation currents
higher excitation currents
none of the above
Answer: C) higher excitation currents
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.
4. Coefficient of Performance (COP) for a refrigeration compressor is given by ________.
power input to compressor (kW) / cooling effect (kW)
cooling effect (kW) / Power input to compressor (kW)
Q x CP x (Ti - To) / 3024
none of the above
Answer: B) cooling effect (kW) / Power input to compressor (kW)
Confirmed vs Book-3 §4.7 - COP = useful cooling effect divided by work input to the compressor. Option (a) is the inverted ratio (that is essentially kW/TR), and (c) is the TR formula for the coolant side; §4.7 defines COP as cooling effect (kW) divided by power input to the compressor (kW).
Confirmed vs Book-3 §1.10 — 'Harmonic voltages and currents in an electric power system are a result of non-linear electric loads', and the book's list of non-linear loads is headed by variable frequency drives.
Options (a)–(c): motors and transformers have essentially constant impedance and are treated as linear (a transformer only creates harmonics when driven into saturation).
6. In a water-lithium bromide absorption refrigeration system, the lithium bromide solution is re-concentrated in the ________.
Evaporator
Condenser
Generator
Absorber
Answer: C) Generator
Confirmed vs Book-3 §4.3 - In a Li-Br vapour absorption system the solution is most concentrated (strongest) at the generator where refrigerant (water) vapour is boiled off. Stem completed from the incomplete printed line. §4.3: the dilute LiBr leaving the absorber is heated in the GENERATOR by steam/hot water/oil, boiling off refrigerant water and re-concentrating the solution; it is most dilute in the absorber, and the evaporator and condenser carry only refrigerant water. The absorber (d) is where the solution becomes DILUTE, and the evaporator (a) and condenser (b) carry only refrigerant water; §4.3 states the diluted LiBr must be made concentrated in the generator using steam, hot water, gas or oil.
7. Integrated Part Load Value (IPLV) in a vapor compression refrigeration refers to average of __________ at partial loads.
TR/kW
kW/TR
kW.TR
kW
Answer: B) kW/TR
Confirmed vs Book-3 §4.7 - IPLV is a weighted average of the chiller efficiency (kW/TR) at standard part-load points. Option (a) TR/kW inverts the indicator and (d) kW alone ignores capacity; §4.7 defines IPLV as an average of kW/TR taken at 100%, 75%, 50% and 25% load, because full-load conditions occur for barely 1% of running hours.
8. __________ is a measure of effect of light on the perceived colour appearance of objects.
lux
lumens
CRI
lamp circuit efficacy
Answer: C) CRI
Confirmed vs Book-3 §8.2 — 'Colour rendering index (CRI): is a measure of the effect of light on the perceived color of objects' (100 = identical to the reference source). Lux is illuminance, lumens is flux, lamp circuit efficacy is lumens per circuit Watt.
9. Which of the following is not used as a sensor for lighting occupancy linked control?
Infrared
acoustic
ultrasonic
pressure
Answer: D) pressure
Confirmed vs Book-3 §8.7 — Occupancy-linked control uses 'infra-red, acoustic, ultrasonic or microwave sensors' that detect movement or noise. Pressure sensing is not among the book's sensor types.
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
10. In an Energy Efficient Motor, the efficiency is increased by increasing __________.
stator winding cross sectional area
fan losses
conductor resistance of rotor
stator winding resistance
Answer: A) stator winding cross sectional area
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up — Table 2.2 states that 'use of more copper and larger conductors increases cross-sectional area of stator windings. This lowers resistance (R) of the windings and reduces losses due to current flow' — so increasing the stator winding cross-section raises efficiency. (d) increasing stator winding RESISTANCE is the exact opposite, and (b) increasing fan losses contradicts the low-loss fan design the book recommends.
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
11. By locating the capacitor near the motor terminal __________.
motor power factor increases
motor energy consumption decreases
system power factor decreases
line losses increases
Answer: B) motor energy consumption decreases
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — A capacitor improves power factor only from its point of installation back towards the generating side; placing it at the motor terminals therefore maximises the length of cable that carries reduced current, cutting I²R losses and voltage drop and so reducing the energy drawn from the system. (a) is the tempting wrong option — the book states explicitly that a capacitor at the starter terminals 'won't improve the operating PF of the motor', only the PF upstream.
📖 §6.1 Pump types — centrifugal pump construction & working
12. A pump is handling water at 25 deg C and delivering 200 m³/hr. If the water temperature is 50 deg C then the flow will __________.
Increase by 50%
Decrease by 50%
double
None of the above
Answer: D) None of the above
Confirmed vs Book-3 §6.1 — A centrifugal pump develops head and volumetric flow set by its impeller geometry and speed; the book says it 'generates the same head of liquid whatever the density'. Warming water from 25 to 50 °C changes density only ~1%, so volumetric flow is essentially unchanged — none of the listed 50%/double changes occur.
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
13. If distribution line has losses of 100kW and voltage is raised from 33KV to 66KV, then line losses will be __________.
50KW
100KW
200KW
25KW
Answer: D) 25KW
Confirmed vs Book-3 §1.1 — line loss varies as the inverse square of the voltage: doubling 33 kV to 66 kV divides the loss by (66/33)² = 4, so 100 kW becomes 25 kW.
Option (a) 50 kW assumes an inverse LINEAR relation; the book's rule is 'the line power loss in the ratio of square of voltages'.
14. A 2000KVA Transformer has full load losses of 200KW. If the transformer is running at 50% loading, then what will be the load losses?
10KW
50 KW
20KW
None of the above
Answer: B) 50 KW
Confirmed vs Book-3 §1.5 — load (copper) loss varies with the square of the load: at 50% loading the load loss = 200 × (0.5)² = 50 kW.
Option (c) 20 kW assumes the loss falls to one-tenth and option (a) halves it linearly; the squared-load rule (P = I²R) is the whole point of the question. [Option repaired: the printed '5 KW' is arithmetically impossible for a 200 kW full-load loss at 50% load.]
15. Slip% of Induction Motor is calculated as (Ns= Synchronous Speed, Nr= Full Load Rated Speed) __________.
(Ns-Nr)x100/Ns
(Ns-Nr)x100/Nr
(Nr-Ns)x100/Ns
(Ns+Nr)x100/Ns
Answer: A) (Ns-Nr)x100/Ns
Confirmed vs Book-3 §2.3 Motor Characteristics — The book's formula is Slip (%) = (Synchronous Speed − Full Load Rated Speed) / Synchronous Speed × 100, i.e. the difference is always referred to Ns. Dividing by Nr (option b) is the standard error — it is what turns a correct 1.68 % slip into a wrong 1.71 %. The printed options repeated the same expression three times, so the distractors have been reconstructed from the book formula.
16. Which of the following compressed air dryer consumes less power for the same output?
Refrigeration Dryer
Heat of Compression Dryer
Heatless Purge type Dryer
Blower Reactivated Type Dryer
Answer: B) Heat of Compression Dryer
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Table 3.19 per 1000 m³/hr: heat of compression 0.8 kW, refrigeration 2.9 kW, blower reactivated 18.0 kW, heatless purge 20.7 kW.
The HOC dryer regenerates its desiccant with the compressor's own ~135°C discharge heat, with no electric heater and no purge loss, so it consumes the least power for the same output.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
18. Pump Shaft Power is __________.
Hydraulic Power / Motor Efficiency
Hydraulic Power / Pump Efficiency
Hydraulic Power × Pump Efficiency
Hydraulic Power × Motor Efficiency
Answer: B) Hydraulic Power / Pump Efficiency
Confirmed vs Book-3 §6.1 — Book p.172: 'Pump Shaft Power Pₛ = Hydraulic power, Pₕ / Pump Efficiency, η_pump' and Motor Input Power = Pₛ / η_motor. Shaft power must exceed hydraulic power, so dividing by pump efficiency is correct; multiplying (c, d) would give less than hydraulic power, and motor efficiency (a) relates shaft power to electrical input, not to hydraulic power. (Options repaired: OCR had rendered '/' as '*' and duplicated b/c.)
📖 §7.2 Factors Affecting Performance – Range (Range = Heat Load / Water Circulation Rate)
19. The heat load of the cooling tower depends on __________.
Range
Approach
Cooling water temp
Wet Bulb Temperature
Answer: A) Range
Confirmed vs Book-3 §7.2 Range — Book: 'Range °C = Heat Load in kcal/hour / Water Circulation Rate' – heat load = flow × Cp × Range, so the heat load is tied to the Range → (a). Approach and WBT relate to tower performance/ambient, not to the process heat load.
20. A Cooling Tower is operating with the following parameters: Wet Bulb Temp. = 27°C, Cooling Water in Temperature = 45°C, Cooling Water out Temperature = 35°C. What is the cooling tower effectiveness?
50%
44.44%
55.55%
None of the above
Answer: C) 55.55%
Confirmed vs Book-3 §7.2 (iii) — Range = 45 − 35 = 10°C; Approach = 35 − 27 = 8°C; Effectiveness = 10/(10 + 8) = 55.55% → (c). 44.44% (b) is Approach/(Range + Approach), the inverted ratio.
21. What will be the Power Factor of the system having Active Power as 812 KW and Reactive Power as 418 KVAR.
0.89
0.51
0.5
1
Answer: A) 0.89
Confirmed vs Book-3 §1.4 (Figure 1.11) — kVA = √(kW² + kVAr²) = √(812² + 418²) = 913 kVA, so PF = 812/913 = 0.89. These are exactly the book's post-correction figures in the chemical-industry worked example.
Option (b) 0.51 is kVAr/kW-type confusion; power factor is always the ACTIVE power divided by the apparent power.
22. Synchronous speed of motor is directly proportional to __________.
No. of Poles
Frequency
Terminal Voltage
All of the above
Answer: B) Frequency
Confirmed vs Book-3 §2.3 Motor Characteristics — From Ns = 120 f / P, synchronous speed is directly proportional to the supply frequency and INVERSELY proportional to the number of poles. (a) is therefore the trap — more poles give a lower speed (2/4/6/8 poles give 3000/1500/1000/750 rpm at 50 Hz); terminal voltage does not enter the relation at all.
23. Which of the following is not a positive displacement compressor __________.
Reciprocating
Screw
Roots Blower
Centrifugal
Answer: D) Centrifugal
Confirmed vs Book-3 §3.2 Compressor Types (Figure 3.2 Compressor Chart) — Figure 3.2 places reciprocating, screw and roots blower under positive displacement — all increase pressure by reducing the trapped volume.
The centrifugal is a dynamic compressor: it imparts velocity to the air which is then converted to pressure at the outlet, so it is the one that is not positive displacement.
24. Adsorption drying of air is achieved using __________.
Activated alumina
Carbon Molecular Sieves
Zirconium Molecular Sieves
None of the above
Answer: A) Activated alumina
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.
25. If air leak quantity is 5 m³/min, what will be power loss per hour in compressor which is having specific power consumption of 0.09 kWh/m³?
27 Units/hr
0.45 Units/hr
12 Units/hr
0.05 Units/hr
Answer: A) 27 Units/hr
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — Leakage = 5 m³/min = 300 m³/hr; power lost = leakage quantity x specific power consumption = 300 x 0.09 = 27 kWh per hour, i.e. 27 units/hr.
Option (b) 0.45 comes from multiplying 5 m³/min by 0.09 without converting to the hourly basis on which the specific power consumption (kWh/m³) is quoted.
26. Dry Bulb and Wet bulb temperature will be same at __________.
0% Relative Humidity
50% Relative Humidity
100% Relative Humidity
None of the above
Answer: C) 100% Relative Humidity
Confirmed vs Book-3 §4.2 - At saturation (100% RH) there is no evaporative cooling, so DBT = WBT. At 0% or 50% RH (a, b) evaporation from the wet wick depresses the wet bulb below the dry bulb; only at saturation does evaporation cease, so DBT = WBT = dew point.
27. What will be ventilation rate of 15mx10mx5m room having ACH of 10 __________.
750 m³/hr
7500 m³/hr
75000 m³/hr
None of the above
Answer: B) 7500 m³/hr
Confirmed vs Book-3 §4.12 - Volume = 15×10×5 = 750 m³; ventilation rate = 750 × 10 ACH = 7500 m³/hr. Options (a) and (c) are decimal-shift traps; ventilation rate = L x B x H x ACH = 15 x 10 x 5 x 10 = 7500 m3/hr, the same method as the book's compressor-room example.
📖 §5.1 Table 5.1; §5.4 (axial fans produce lower pressure than centrifugal)
29. Which of the following equipment is having least compression ratio?
Compressor
Blower
Axial Fan
Radial Fan
Answer: C) Axial Fan
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).
30. Capacity control of a cooling tower fan can be achieved by __________.
Changing pulley dimensions
Damper Control
VFD
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §5.5 — 'Various ways to achieve change in flow are: pulley change, damper control, inlet guide vane control, variable speed drive and series and parallel operation of fans.' Pulley change, damper and VFD are all listed → (d).
📖 §7.1 Components of Cooling Tower – Drift eliminators
31. __________ is used to capture water droplets in the air stream leaving the cooling tower.
Splash fill
Film fill
Drift eliminator
Any of the above
Answer: C) Drift eliminator
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.
📖 §6.11 Solved example — cooling water pump efficiency (p.194–195)
33. Calculate the Volumetric Flow (m³/sec) in a pipe with a diameter 200 mm and Velocity 1.5 m/sec.
0.19 m³/sec
0.015 m³/sec
0.047 m³/sec
0.012 m³/sec
Answer: C) 0.047 m³/sec
Confirmed vs Book-3 §6.11 — Flow = area × velocity = (π/4)×d²×v = 0.7854×(0.2)²×1.5 = 0.0314×1.5 = 0.047 m³/s — the same method as the book's solved example (pipes A, B, C). 0.19 forgets the /4; 0.015 and 0.012 mis-square the diameter.
📖 Book-3 Ch-6 Pumps and Pumping System (cross-chapter item filed under Ch-3)
35. If two pumps are operated in parallel then Shut-off head __________.
Does not change
Halved
Doubled
Less than double
Answer: A) Does not change
Confirmed vs Book-3 Ch-6 Pumps and Pumping System (cross-chapter item filed under Ch-3) — Two identical pumps in parallel add their flows at a given head, but the shut-off (zero-flow) head is fixed by impeller diameter and speed and is the same as for one pump.
Doubling of head is what SERIES operation achieves; parallel doubles flow, not head. Note this is a pumps item filed within the compressed-air set.
36. 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 __________.
67 cfm
100 cfm
10 cfm
133 cfm
Answer: A) 67 cfm
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.
37. The compressor capacity of a reciprocating compressor is directly proportional to __________.
Speed
Pressure
Volume
All
Answer: A) Speed
Confirmed vs Book-3 §3.2 Positive Displacement — Book-3 §3.2: 'the compressor capacity is directly proportional to the speed' — halving the RPM (e.g. by reducing the motor pulley) halves the delivered air, which is the basis of the pulley-change de-rating measure.
Output stays nearly constant over a range of discharge pressures, so capacity is not proportional to pressure, and 'All' is therefore wrong.
38. Find the correct example, if M = makeup water (from the mains water supply), E = losses due to evaporation, B = losses due to blow-down and D = drift losses of a cooling tower:
M = E + B + D
M = E − B + D
M = E + B − D
M = E − B − D
Answer: A) M = E + B + D
Confirmed vs Book-3 §7.2 (v)–(vii) — Make-up water must replace all water leaving the circuit – evaporation, blow-down and drift – so M = E + B + D → (a). Any subtraction would mean a loss returns water to the basin. ⚠ Options c and d were duplicates ('M = E - B - D'); c repaired to 'M = E + B − D' so exactly one option is correct.
39. If temperature of air increases, the amount of water vapor needed to become saturated __________.
Increases
Decreases
not change
Can't say
Answer: A) Increases
Confirmed vs Book-3 §7.2 Approach & WBT — Warmer air can hold more moisture (the book's enthalpy example: air at 26.67°C WBT holds 24.17 kcal/kg, at 37.8°C WBT 39.67 kcal/kg), so more water vapour is needed to saturate it → (a). This is why hotter, drier air evaporates more water.
40. The wet bulb temperature entering the cooling tower is 38°C. Range is twice the approach, the effectiveness of the cooling tower is __________.
33.3%
50%
66.7%
Insufficient data
Answer: C) 66.7%
Confirmed vs Book-3 §7.2 (iii) — Effectiveness = Range/(Range + Approach). With Range = 2 × Approach: 2A/(2A + A) = 2/3 = 66.7% → (c). The 38°C WBT is not needed – the ratio alone fixes effectiveness, so (d) 'insufficient data' is a trap.
📖 (Misfiled in Ch9 — cooling-tower topic, Book-3 cooling tower chapter) Cycles of concentration = TDS circulating ÷ TDS make-up
41. 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.
📖 §1.9 DSM (building energy performance; see Book-3 Ch-10 ECBC)
42. The unit of Energy Performance Index (EPI) for rating the building is __________.
kWh/sq mtr/yr
kWh/sq mtr/hr
Wh/sq mtr/hr
sq mtr/Wh/yr
Answer: A) kWh/sq mtr/yr
Confirmed vs Book-3 Ch-10/ECBC — the Energy Performance Index (EPI) of a building is its annual energy consumption per unit of built-up area, expressed in kWh/m²/year.
Option (c) Wh/m²/hr is an instantaneous intensity, not the annual index used for star-rating buildings under the BEE programme.
📖 §10.3 Energy Conservation Building Code — definition and scope
43. Which of the following statements regarding ECBC are correct?
ECBC defines the norms of energy requirements per cubic metre of area
ECBC does not encourage retrofit of Energy conservation measures
ECBC prescribes energy efficiency standards for design and construction of commercial and industrial buildings
One of the key objectives of ECBC is to minimize life cycle costs (construction and operating energy costs)
Answer: D) One of the key objectives of ECBC is to minimize life cycle costs (construction and operating energy costs)
Confirmed vs Book-3 §10.3 — ECBC sets minimum energy-efficiency standards for design and construction of COMMERCIAL buildings (not industrial, so c is wrong), explicitly encourages energy-efficient design or major RETROFIT (so b is wrong), and its EPI norms are per square metre, not cubic metre (so a is wrong). By elimination (d) — minimising life-cycle (construction + operating energy) cost — is the correct statement.
📖 §8.2 Installed power density (W/m²) — LPD arithmetic (ECBC context)
44. 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 __________.
1000 W
21600 W
43200 W
none of the above
Answer: C) 43200 W
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.
45. 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.
Answer: B) aggregate technical and commercial losses
Confirmed vs Book-3 §1.8 — 'The above losses are collectively categorized as AT & C (Aggregate Technical & Commercial) losses', combining the technical (I²R, transformation) losses with the commercial (theft, metering, collection) losses.
Option (a) misreads 'technical' as 'transmission' and 'commercial' as 'current' — the acronym is fixed by the book.
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback
47. The nearest kVAr compensation required for improving the power factor of a 1000 kW load from 0.95 leading power factor to unity power factor is __________.
328 kVAr
750 kVAr
1000 kVAr
none of the above
Answer: A) 328 kVAr
Confirmed vs Book-3 §1.11 Solved Example — kVAr = kW[tan(cos⁻¹PF₁) − tan(cos⁻¹PF₂)] = 1000 × (0.329 − 0) = 328.7 ≈ 328 kVAr, exactly the book's 0.95→unity route.
Option (c) 1000 kVAr wrongly equates the capacitor size with the kW; only the reactive component (kW × tanΦ₁) has to be cancelled.
48. The no-load loss and copper loss of a 500 kVA transformer are 1600 Watts and 6400 Watts respectively. What is the total loss at 50% of transformer loading?
4100 Watts
6850 Watts
2500 Watts
3200 Watts
Answer: D) 3200 Watts
Confirmed vs Book-3 §1.5 — total loss = no-load loss + (%load)² × full-load copper loss = 1600 + (0.5)² × 6400 = 1600 + 1600 = 3200 W.
Option (c) 2500 W is the answer for a 900 W no-load loss (the sister question from the 17th exam); option (b) 6850 W adds the FULL copper loss without squaring the load fraction. [Option repaired: the printed '3650 Watts' does not follow from the stated 1600 W / 6400 W data.]
reducing wet bulb temperature and specific humidity
reducing dry bulb temperature and specific humidity
increasing wet bulb temperature and decreasing specific humidity
reducing dry bulb temperature and increasing specific humidity
Answer: B) reducing dry bulb temperature and specific humidity
Corrected (was d) - Book-3 §4.14 defines HUMIDIFICATION as "reduction in dry bulb temperature and increase in specific humidity", so option (d) is the definition of humidification, not dehumidification. Dehumidification removes moisture: the air is cooled below its dew point on the coil, so both dry bulb temperature and specific humidity fall (b). Note: if the paper's stem actually read 'Humidification', the key (d) would be right - read the stem carefully in the exam.
1. Fill in the blanks: (a) ECBC is applicable to commercial buildings having contract demand of ____ kVA. (b) In a centrifugal fan if speed is reduced by 30% static pressure will reduce by ____%. (c) If the speed of a reciprocating pump is reduced by 30%, the power consumption will reduce by ____%. (d) The unit of specific humidity of air is __________. (e) The synchronous speed (rpm) of a 2 pole induction motor at 49 Hz supply frequency is __________.
Model answer: (a) 120 kVA; (b) 51%; (c) 30%; (d) grams moisture/kg of dry air; (e) 2940 rpm
(a) Book-3 §10.2: contract demand 120 kVA (or connected load 100 kW). (b) Static pressure ∝ N²: 1 − 0.7² = 0.51 → 51%. (c) Positive-displacement pump flow and power ∝ speed → 30%. (d) Specific humidity in g (or kg) moisture per kg dry air. (e) Ns = 120 × 49 / 2 = 2940 rpm.
2. A textile industry had installed a 2 MVA transformer. The initial demand of the plant was 1500 kVA with power factor of 0.75. Industry has installed 450 kVA capacitor at the motor end. Calculate the following: 1. Reduction in apparent power (kVA); 2. Improved power factor; 3. Revised % loading of transformer after installing the capacitor.
Model answer: Real power = 1500×0.75 = 1125 kW. Old reactive power = √(1500²−1125²) = 992 kVAr. After 450 kVAr capacitor, revised kVAr = 992−450 = 542 kVAr. Revised apparent power = √(1125²+542²) = 1248 kVA. Reduction in apparent power = 1500−1248 = 252 kVA. Improved PF = 1125/1248 = 0.90. Revised % loading = 1248/2000 = 62.4%.
Resolve into real and reactive components, subtract capacitor kVAr, recompute apparent power, pf and transformer loading (kVA/rating).
3. 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. Case #2: Window to Wall Ratio (WWR) 0.45, Visible Light Transmittance (VLT) 0.2.
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.
4. A three-phase induction motor has the following details: Name plate details: 55 kW, 415V, 95A, 0.90 PF, 50 Hz. Running load details: 410V, 75A, 0.80 PF, 48 Hz. Calculate the loading percentage and rated efficiency of the motor.
Model answer: Actual power drawn = 1.732×410×75×0.80/1000 = 42.6 kW. Rated input power = 1.732×415×95×0.90/1000 = 61.5 kW. Loading percentage = 42.6/61.5 = 69.3%. Rated efficiency = 55/61.5 = 89.4%.
Input power = √3×V×I×pf; loading = actual input/rated input; rated efficiency = rated output/rated input.
5. In an engineering industry, compressed air delivered is 500 CFM (FAD) and the compressor discharge pressure is 6 kg/cm² (gauge). Calculate the size of the header by considering velocity of compressed air 6 m/s. Assume Temperature remains constant.
Model answer: Quantity of air = 500 CFM = 500/35.31 = 14.16 m3/min of free air.
Working pressure = 6 kg/cm2(g) = 7.013 kg/cm2(a); atmospheric = 1.013 kg/cm2(a).
Applying Boyle's law at constant temperature, P1V1 = P2V2: V2 = 14.16 x 1.013 / 7.013 = 2.05 m3/min = 0.0341 m3/s (the compressed volume actually flowing in the header).
Quantity of air flow = area x velocity, so (pi/4) x D2 x 6 = 0.0341, giving D2 = 0.00724 m2 and D = 0.085 m = 85 mm (about 3.35 inch). A standard 3" NB header would be selected, checking that the velocity stays in the usual 6-10 m/s band.
Confirmed vs Book-3 §3.5 Sizing of Compressed Air Piping — Convert FAD to the compressed volume by Boyle's law, then size from Q = area x velocity at 6 m/s; the garbled area step has been written out, giving D = 85 mm.
6. 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.
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).
7. Discuss in brief about working principal Membrane Dryer for Compressed Air.
Model answer: A membrane dryer removes water vapour from compressed air by SELECTIVE PERMEATION of the gas components (Book-3 Figure 3.13).
Construction: the dryer is a cylinder housing thousands of tiny hollow polymer fibres with an inner coating; the coating is selectively permeable to water vapour.
Working: filtered wet compressed air enters the cylinder and passes through the bores of the fibres. The membrane coating allows water vapour to permeate through the fibre wall and collect between the fibres, while the dry air continues through the fibres and leaves at almost the same pressure as the incoming wet air. The permeated water is vented to the atmosphere outside the cylinder. The driving force for the separation is the difference in the partial pressure of the water vapour between the inside and the outside of the hollow fibre.
Features: simple to operate, silent, no moving parts, low power consumption and minimal servicing — mainly the filters upstream of the dryer.
Confirmed vs Book-3 §3.5 Air Dryers — Descriptive answer supplied from the book: selective permeation through coated hollow polymer fibres, driven by the water-vapour partial-pressure difference across the fibre wall.
📖 §9.3 Waste heat recovery — flue gas mass from air/fuel ratio, WHRB heat & steam generation
1. 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.
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
2. L-1(A): A trivector-meter installed in a steel plant is monitoring the maximum demand with a demand interval of 15 min. The observed maximum demand during one demand interval is given (3 min: 9634 kVA, 4 min: 10257 kVA, 3 min: 8436 kVA, 5 min: 9847 kVA). Calculate: (a) Recorded maximum demand during the cycle; (b) Demand reduction and capacitor kVAr required for improving power factor to 0.99 from average observed power factor of 0.92.
Model answer: (a) Maximum demand is the time-integrated kVA over the 15-minute demand interval:
MD = [(9634 x 3) + (10,257 x 4) + (8436 x 3) + (9847 x 5)] / 15 = [28,902 + 41,028 + 25,308 + 49,235] / 15 = 144,473/15 = 9631.5 kVA.
(b) At the observed average PF of 0.92, kW = 9631.5 x 0.92 = 8861 kW (capacitors do not change kW).
New demand at 0.99 PF = 8861/0.99 = 8950.5 kVA, so demand reduction = 9631.5 - 8950.5 = 681 kVA.
Capacitor kVAr = kW[tan(cos^-1 0.92) - tan(cos^-1 0.99)] = 8861 x (0.4260 - 0.1425) = 8861 x 0.2835 = 2512 kVAr (say 2500 kVAr, switched through an APFC).
MD = time-weighted average of interval readings; convert to kW via pf; recompute kVA at improved pf; kVAr = kW(tanφ1−tanφ2).
3. L-1(B): A spinning unit proposed to replace 50 no's of 22kW IE2 motors with IE3 motors under National Motor Replacement Programme. Present operating details: Motor Loading 74%, IE2 Motor Efficiency 90%, IE3 Motor Efficiency 93%, Annual operating hours 7000 Hrs, Energy Cost Rs.10/kWh, CO2 Emission from electricity 0.85 Kg/kWh. Calculate annual energy savings and emission reduction in Tons of CO2.
Model answer: Motor rating = 22 kW; loading = 74 %, so shaft load per motor = 22 × 0.74 = 16.28 kW.
Input with IE2 motor = 16.28 / 0.90 = 18.09 kW; input with IE3 motor = 16.28 / 0.93 = 17.51 kW.
Power saving per motor = 18.09 − 17.51 = 0.584 kW (book relation: kW saving = kW output × [1/ηold − 1/ηnew]).
Annual energy saving per motor = 0.584 × 7,000 = 4,088 kWh; for 50 motors = 204,400 kWh/year.
Annual cost saving at Rs. 10/kWh = Rs. 20,44,000 (≈ Rs. 20.4 lakh).
CO₂ reduction at 0.85 kg CO₂/kWh = 204,400 × 0.85 = 173,740 kg ≈ 173.7 tonnes CO₂ per year.
Input power = output/efficiency for each class; per-motor saving × hours × 50 gives kWh; cost = kWh×rate; CO2 = kWh×0.85.
📖 §5.6 Fan static efficiency formula; §5.6 Volume calculation
4. L-2(A): In a ventilation duct of 0.6 m x 0.6 m size, the average velocity of air measured by vane anemometer is 30 m/s. The static pressure at inlet of the fan is -25 mm WC and at the outlet is 35 mm WC. A 3 phase induction motor coupled with fan through belt drive draws 19 A at 410 V at a power factor of 0.8. Find out the efficiency of the fan. Assume motor efficiency 90% and belt transmission efficiency of 98% (density correction can be neglected). (6 Marks)
Model answer: Volume flow Q = velocity × area = 30 × 0.6 × 0.6 = 10.8 m³/s.
Motor input power = √3 × V × I × PF = 1.732 × 410 × 19 × 0.8 = 10,790 W = 10.79 kW.
Power input to fan shaft = 10.79 × 0.90 (motor) × 0.98 (belt) = 9.52 kW.
Static pressure rise across fan = 35 − (−25) = 60 mmWC.
Fan efficiency = (Q in m³/s × ΔP in mmWC)/(102 × shaft kW) × 100 = (10.8 × 60)/(102 × 9.52) × 100 = 648/971 × 100 = 66.7%.
Book-3 §5.6: air power = Q × ΔP/102 kW; shaft power from electrical input (√3·V·I·PF) × motor × belt efficiency; efficiency = air power/shaft power = 66.7%. Density correction is neglected as instructed.
5. L-2(B): Explain how Computational Fluid Dynamics (CFD) can be used for enhancing the Energy Efficiency. (4 Marks)
Model answer: Computational Fluid Dynamics (CFD) is a computer-based simulation tool that models the physical system mathematically and solves the mass, momentum and energy equations to predict flow patterns, temperature and pressure profiles and particle movement inside equipment and duct systems. Use for energy efficiency (Book-3 §5.9): (1) Designs and operating parameters can be varied in the model to find the best operating conditions BEFORE any physical change is made in the plant. (2) It is proactive — it identifies the root cause (not just the effect) of plant problems, is scale-independent (based on fundamental physics, so scale-up problems are reduced) and can simulate conditions where measurements are impossible (high temperature, dangerous environments). (3) In fan systems the upstream/downstream ducting matters: turbulent or swirling inflow from sharp bends or abrupt cross-section changes raises pressure drop and degrades fan efficiency — especially for high-efficiency fans (> 80%) that need non-swirling inflow. (4) Case study — double-inlet ID fan: CFD showed sharp-edged transitions causing flow disruption (pressure drops of 260–430 Pa). Smoothing the transitions cut the front inflow-duct drop from 261 Pa to 66 Pa (factor of four) and the rear-duct drop to about one-sixth; the mean saving of 275 Pa at 5,00,000 m³/h reduced fan power by about 49 kW and eliminated swirl-induced vibration.
Model answer built from Book-3 §5.9 (pages 165–166): CFD definition, its proactive/scale-independent advantages, the effect of inflow swirl on fan efficiency, and the ID-fan case study (261 → 66 Pa, 275 Pa mean saving, ≈ 49 kW). The previous answer text was only a page reference.
6. L-3(A): State Increases or Decreases (1 Mark each): 1. If air dry bulb temperature is increased then Relative Humidity will ___. 2. In a pumping system, if the suction side liquid level is increased then NPSHa will ___. 3. If the air temperature increases at the inter-cooler outlet, then air compressor power consumption will ___. 4. A blower is retrofitted with a VFD and operated at full speed. The power consumption will ___. 5. As the design speed of the motors decreases the capacitor KVAr requirement will ___.
7. L-3(B): Match the following (1 Mark each): 1. Pitot Tube; 2. Refrigerant Drier; 3. Condenser; 4. Spray Nozzles; 5. Occupancy Sensor — with — A. Cooling Tower; B. Lighting Control; C. Gas Velocity in ducts; D. Compressed Air System; E. Refrigeration System.
Model answer: 1. Pitot Tube → C. Gas Velocity in ducts; 2. Refrigerant Drier → D. Compressed Air System; 3. Condenser → E. Refrigeration System; 4. Spray Nozzles → A. Cooling Tower; 5. Occupancy Sensor → B. Lighting Control.
8. L-4: During the energy audit of central chiller plant, following parameters were noted: Chilled water flow 250 m³/hr; Chilled water inlet temperature 12°C; Chilled water outlet temperature 7°C; Motor Input Power 350 kW; Motor Efficiency 90%; Condenser water inlet temperature (going to chiller or outlet of cooling tower) 31°C; Condenser water outlet temperature (leaving from chiller or inlet to cooling tower) 36°C; Wet Bulb temperature of ambient air 28°C; Make up water TDS 180 ppm; Permissible limit of TDS for cooling water 720 ppm; Condenser cooling capacity 25% higher than the evaporator cooling capacity. Calculate: kW/TR of chiller compressor; COP of chiller; Effectiveness of cooling tower; Evaporation loss; Blow down quantity; Make-up water requirement (ignoring no drift loss).
Model answer: Chiller machine capacity TR = [250×1000×1×(12−7)]/3024 = 413.4 TR. kW/TR of chiller compressor = (350×90%)/413.4 = 0.762. COP = (413.4×3024)/(350×0.9×860) = 4.6. Effectiveness = (36−31)/(36−28) = 62.5%. Cycle of Concentration COC = 720/180 = 4. Condenser TR = 1.25×413.4 = 516.75 TR. Condenser water flow / circulation flow = 516.75×3024/(1000×1×(36−31)) = 312.5 m³/hr. Evaporation loss = 0.00085×1.8×312.5×(36−31) = 2.3 m³/hr. Blow down = Evap/(COC−1) = 2.3/(4−1) = 0.797 m³/hr. Make-up = Evaporation + Blow down = 2.3+0.797 = 3.097 m³/hr.
Chiller TR from flow×ΔT; kW/TR from shaft power; COP conversion; cooling-tower effectiveness, COC, evaporation, blowdown and make-up formulae.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172) (ρ = slurry density); §6.7 Boiler feed water pumps (multistage pump = single-stage pumps in series) (series pumps add head)
9. L-5: A 30 MW coal fired thermal power plant uses conventional wet ash disposal system for ash evacuation. During an energy audit at ash slurry disposal pump house it was observed that only one ash slurry disposal series is continually operated out of installed three numbers of series. Data collected for one series: two pumps per series; rated parameters each pump flow 815 m³/hr, head 33 mWc; slurry flow rate measured at second series pump discharge 752 m³/hr; suction head to 1st pump +2.6 mtr; final slurry discharge pressure 5.2 kg/cm²g; differential pressure across 1st slurry pump 2.44 kg/cm²; differential pressure across 2nd slurry pump 2.5 kg/cm²; ash water ratio (by weight) of the slurry 1:15; ash slurry density 1032 kg/m³; electric power input to motor of 1st slurry pump 141 kW; electric power input to motor of 2nd slurry pump 139 kW; motor rating of each pump 200 kW; full load motor efficiency 93%; belt transmission efficiency 92%. Evaluate: (i) Individual pump efficiencies if the operating motor efficiency is 92% for all pumps; (ii) Specific energy consumption of each pump (kWh/m³); (iii) Specific energy consumption of the series (kWh/m³); (iv) As water conservation measure the energy auditor recommended to maintain ash water ratio at 1:7 and slurry density of 1067 kg/m³, calculate the incremental power consumption of each pump and series if all other parameters are unchanged.
Model answer: Common data: slurry flow Q = 752 m³/hr = 0.2089 m³/s; slurry density ρ = 1032 kg/m³; 1 kg/cm² ≈ 10 m. (i) Pump efficiencies. 1st pump: differential head = 2.44 kg/cm² ≈ 24.4 m; liquid (hydraulic) power = Q×H×ρ×g/1000 = 0.2089×24.4×1032×9.81/1000 = 51.6 kW; power at pump shaft = motor input × motor eff × belt eff = 141×0.92×0.92 = 119.3 kW; pump efficiency = 51.6/119.3 = 43.2%. 2nd pump: head = 2.5 kg/cm² ≈ 25 m; liquid power = 0.2089×25×1032×9.81/1000 = 52.9 kW; shaft power = 139×0.92×0.92 = 117.6 kW; pump efficiency = 52.9/117.6 = 45.0% (the printed key rounds both pumps to ≈43%). (ii) Specific energy consumption = motor input / flow: 1st = 141/752 = 0.188 kWh/m³; 2nd = 139/752 = 0.185 kWh/m³. (iii) Series SEC = (141+139)/752 = 0.372 kWh/m³. (iv) With ash:water 1:7 the slurry density rises to 1067 kg/m³; at unchanged flow, head and efficiencies the liquid power and hence input power scale with density (×1067/1032 = 1.034): 1st pump liquid power = 53.4 kW, input ≈ 141×1.034 = 145.8 kW (+4.8 kW); 2nd pump liquid power = 54.7 kW, input ≈ 143.7 kW (+4.7 kW). Incremental power for the series ≈ 9.5 kW (printed key: ≈10.4 kW with its rounding). Water conservation therefore slightly raises pumping power but saves far more water.
Liquid power = Q·H·ρ·g with slurry density; pump efficiency = liquid power / (motor input × motor η × belt η); SEC = kW / (m³/hr); higher slurry density raises liquid power and input power in proportion.
📖 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)
10. 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.