Energy Efficiency in Electrical Utilities Available here with full solutions — 76 questions recovered from the 2018 exam:
Objective (1 mark)
52 of 50
Short (5 marks)
13 of 8
Long (10 marks)
11 of 6
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) — 52
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
1. The purpose of inter-cooling in a multistage compressor is to
Increase the pressure of air
Reduce the work of compression
Separate moisture and oil vapour
None of the above
Answer: B) Reduce the work of compression
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.
Transformers operating near saturation level create harmonics
Devices that draw sinusoidal currents when a sinusoidal voltage is applied create harmonics
Harmonics are multiples of the supply frequency
Harmonics occur as spikes at intervals which are multiples of the supply frequency
Answer: B) Devices that draw sinusoidal currents when a sinusoidal voltage is applied create harmonics
Confirmed vs Book-3 §1.10 — 'As the value of impedance in above devices is constant, they are called linear'; linear devices drawing sinusoidal current from sinusoidal voltage do NOT create harmonics, so statement (b) is the incorrect one.
Options (c)/(d) are true — harmonics are integer multiples of the fundamental (5th = 250 Hz on a 50 Hz system) — and (a) is true because a saturated transformer is non-linear.
3. The Energy Performance Index (EPI) of a building as per ECBC and the Energy Conservation Act, 2001 is:
kWh per square meter per year
kWh per square meter
kW per square meter
kWh per year
Answer: A) kWh per square meter per year
Confirmed vs Book-3 §10.14 — EPI is the specific energy usage of a building in kWh per square metre per year (annual energy ÷ built-up area). kWh/m² alone lacks the time base and kW/m² is a power density, not energy.
4. The Solar Heat Gain Co-efficient (SHGC) of a window of a building is 0.30. This means
The window allows 70% of the sun's heat to pass through into interior of the building
The window allows 30% of the sun's heat to pass through into the building interior
70% of the sun's heat is incident on the window
The window reflects back to exterior a minimum of 30 % of the sun's heat
Answer: B) The window allows 30% of the sun's heat to pass through into the building interior
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.
📖 §5.7 Energy savings (item 4: hollow FRP impeller)
5. FRP fans consume less energy than aluminium fans because
they are lighter
they have better efficiencies
they encounter less system resistance
they deliver less air flow
Answer: A) they are lighter
Confirmed vs Book-3 §5.7 — ECO item 4: replace metallic/GRP impellers with 'the more energy efficient hollow FRP impeller with aerofoil design' (axial fans, e.g. cooling towers). The hollow FRP blade is much lighter, so less driving power is needed — the standard BEE key is 'they are lighter'. FRP does not change the system resistance (c) or reduce the air delivered (d).
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
6. The hydraulic power in a pumping system depends on
Pump efficiency
Motor efficiency
Both motor and pump efficiency
None of the above
Answer: D) None of the above
Confirmed vs Book-3 §6.1 — Hydraulic power Pₕ = Q×(h_d − h_s)×ρ×g/1000 depends only on flow, total head and liquid density. Pump efficiency converts it to shaft power (Pₛ = Pₕ/η_pump) and motor efficiency to input power — they are downstream of, not inputs to, hydraulic power. Hence 'none of the above'.
7. The most energy intensive heat transfer loop of a vapour compression refrigeration system is:
Indoor air loop
Chilled water loop
Refrigerant loop
Condenser water loop
Answer: D) Condenser water loop
Confirmed vs Book-3 §4.1 - The refrigerant loop contains the compressor — the largest energy-consuming component — making it the most energy-intensive heat-transfer loop of the system. The indoor air, chilled water and condenser water loops (a, b, d) only move fans and pumps; the refrigerant loop contains the compressor, which is by far the largest power consumer in the chain of Figure 4.1.
Confirmed vs Book-3 §8.2 — 'Lux (lx) is the metric unit of measure for illuminance of a surface' (1 lm/m²). Lumens measure luminous flux emitted by the source, LPD is lighting power density in W/m², and radians are angles.
9. In a rolling mill, the loading on the transformer was 1200 kVA at power factor 0.86. The plant improved the power factor to 0.98 by adding capacitors. What is the reduction in kVA?
144
147
171
163.3
Answer: B) 147
Corrected (was d) — Book-3 §1.4: capacitors do not change kW, so kW = 1200 × 0.86 = 1032 kW. At 0.98 PF the loading becomes 1032/0.98 = 1053 kVA, so the reduction is 1200 − 1053 = 147 kVA.
The printed option 163.3 does not follow from the data; option (a) 144 comes from rounding kW to 1030. Always work kW first, then divide by the new PF.
📖 §6.7 Boiler feed water pumps (multistage pump = single-stage pumps in series)
10. If two identical pumps operate in series, their shut-off head is
Not affected
More than double
Doubled
Less than double
Answer: C) Doubled
Confirmed vs Book-3 §6.7 — Series operation adds heads at the same flow (the book equates a multistage pump with single-stage pumps in series). At zero flow each identical pump gives its shut-off head, so the combination gives exactly double. Parallel pumps would leave shut-off head unchanged and add flow.
11. 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 ___________
1.7
2.06
0.59
none of the above
Answer: B) 2.06
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).
12. The adsorption material used in an adsorption air dryer is
Calcium chloride
Magnesium chloride
Activated alumina
Potassium chloride
Answer: C) Activated alumina
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.
13. The cooling tower size is _____ to the entering Wet Bulb Temperature (WBT), when the heat load, range and approach are constant.
Directly proportional
Inversely proportional
Constant
None of above
Answer: B) Inversely proportional
Confirmed vs Book-3 §7.2 Factors that affect cooling tower size — Book sidebar: when heat load, range and approach are constant, tower size varies 'inversely with the entering WBT' → (b). Book example: a tower for 4.45°C approach to 21.11°C WBT is larger than the same duty at 26.67°C WBT.
14. If the wet bulb temperature of air is 38 °C, then its relative humidity is __________%.
38 %
90 %
100 %
Insufficient data
Answer: D) Insufficient data
Confirmed vs Book-3 §4.2 - RH cannot be found from WBT alone; the dry-bulb temperature (or another property) is needed to fix the psychrometric state. Hence insufficient data. Options (a)-(c) assume one property fixes the state; the psychrometric chart needs two independent properties, so wet bulb alone cannot give RH.
15. Small diameter by-pass lines are installed in pumps sometimes to ________________.
Save energy
Control pump delivery head
Prevent pump running at zero flow
Reduce pump power consumption
Answer: C) Prevent pump running at zero flow
Confirmed vs Book-3 §6.6 — Book: the small by-pass line 'sometimes installed to prevent a pump running at zero flow is not a means of flow control, but required for the safe operation of the pump' — at shut-off the pump overheats ('becomes a water heater'). It does not save energy or control head.
📖 §5.7 Energy savings (item 4: hollow FRP impeller)
16. Fiberglass Reinforced Plastic (FRP) fans consume less energy than aluminum fans because
They are lighter
They have better efficiencies
They encounter less system resistance
They deliver less air flow
Answer: B) They have better efficiencies
Confirmed vs Book-3 §5.7 — The book lists replacing metallic/GRP impellers with 'hollow FRP impeller with aerofoil design' as an ECO for axial fans 'where significant savings have been reported'. The hollow FRP blade is lighter, so less shaft power is needed (standard BEE key: 'they are lighter'). System resistance (c) is a duct-system property, unaffected by blade material.
17. Ratio of luminous flux (lumen) emitted by a lamp to the power consumed (watt) by the lamp is called
Luminous intensity
Luminous efficacy
Reflectance
Luminance
Answer: B) Luminous efficacy
Confirmed vs Book-3 §8.2 — Verbatim: luminous efficacy 'is the ratio of luminous flux emitted by a lamp to the power consumed by the lamp' (lm/W). Luminous intensity is flux per solid angle (candela), luminance is surface brightness, reflectance is the reflected fraction.
18. Use of soft starters for induction motors results in
lower mechanical stress
lower power factor
higher maximum demand
All the above
Answer: A) lower mechanical stress
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.
📖 §10.2 Building definition — Energy Conservation (Amendment) Act 2010
19. Energy Conservation Act covers buildings having a connected load of
100 kW and above
100 kVA and above
500 kW and above
All buildings with HT connection
Answer: A) 100 kW and above
Confirmed vs Book-3 §10.2 — The EC Act covers commercial buildings with a connected load of 100 kW or contract demand of 120 kVA and above. 500 kW (c) and 100 kVA (b) are distractors; HT connection is not the criterion.
20. In a solar PV system the conversion from DC to AC is carried out by
Converter
Charger
Battery
Inverter
Answer: D) Inverter
Confirmed vs Book-3 §1.10 — the inverter is the stage that converts DC into AC (solar array/DC link → AC grid), and the book lists inverter-fed drives and UPS among non-linear, harmonic-producing loads for exactly this reason.
Option (a) 'converter' (rectifier) performs the reverse AC→DC conversion; a battery only stores DC.
📖 Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7)
21. The inlet air temperature to a two stage reciprocating air compressor is 35 °C. At which of the following 2nd stage inlet temperatures will the compressor consume least power?
75 °C
65 °C
60 °C
50 °C
Answer: D) 50 °C
Confirmed vs Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7) — Table 3.7 shows that the nearer the second-stage inlet temperature is to the first-stage inlet ('perfect cooling'), the lower the specific power: a 5.5°C rise at the second-stage inlet costs about 2% more specific energy.
With a 35°C first-stage inlet, 50°C is the closest of the four, so it gives the least power consumption; 75°C would be the worst.
22. __________ can be achieved using infrared, acoustic, ultrasonic or microwave sensors for energy efficient lighting control.
Time-based control
Daylight-linked control
Occupancy-linked control
Localized switching
Answer: C) Occupancy-linked control
Confirmed vs Book-3 §8.7 — 'Occupancy-linked control can be achieved using infra-red, acoustic, ultrasonic or microwave sensors, which detect either movement or noise.' Time-based control uses timed-turnoff switches; daylight-linked control uses photoelectric cells; localized switching uses local switches in large spaces.
23. A 7.5 kW, 415 V, 15 A, 970 RPM, 3 phase rated induction motor with full load efficiency of 86 % draws 7.5 A and 3.23 kW of input power. The percentage loading of the motor is about
37 %
43 %
50 %
None of the above
Answer: A) 37 %
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — % loading = 3.23 / (7.5/0.86) = 3.23 / 8.721 = 37 %. (c) 50 % would follow from the current ratio 7.5/15, which the book forbids ('loading should not be estimated as the ratio of currents') because power factor collapses at part load.
24. A two pole induction motor operating at 50 Hz, with 1 % slip will run at an actual speed of
3000 RPM
3030 RPM
2970 RPM
None of the above
Answer: C) 2970 RPM
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 50 / 2 = 3000 rpm; at 1 % slip the actual speed = 3000 × 0.99 = 2970 rpm. (b) 3030 rpm adds the slip — an induction motor can never exceed synchronous speed while motoring.
25. 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:
4.84
1.38
1.724
None of the above
Answer: A) 4.84
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.
📖 §1.4 Performance Assessment of Power Factor Capacitors
26. 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.
400 V
415 V
435 V
None of the above
Answer: C) 435 V
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).
27. 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.
28. Aggregate Technical & Commercial loss in distribution system covers
I2R losses of all transformers
Transmission & distribution loss
Only transmission losses
Energy and monetary loss
Answer: D) Energy and monetary loss
Confirmed vs Book-3 §1.8 — AT&C loss = {1 − (Billing Efficiency × Collection Efficiency)} × 100; it combines the energy not billed (technical + theft) with the money billed but not collected, i.e. both an energy and a monetary loss.
Option (b) is only the technical part; AT&C deliberately goes beyond T&D loss by adding the commercial/collection dimension.
29. The isothermal power of a 500 CFM air compressor is 72 kW and the efficiency is 78 %. The actual power drawn by the compressor will be
56 kW
92 kW
72 kW
None of the above
Answer: B) 92 kW
Confirmed vs Book-3 §3.3 Compressor Efficiency — Isothermal efficiency = isothermal power / actual measured input power, so actual power = 72 / 0.78 = 92.3 ≈ 92 kW.
Option (a) 56 kW comes from multiplying 72 x 0.78 instead of dividing; the actual input must always exceed the isothermal power, since isothermal power ignores friction.
30. A heat pump used in a heat recovery application extracts 66220 kcal/hr and the power consumed by the heat pump is 23 kW. The estimated heat supplied by the heat pump is
2916 kcal/hr
47300 kcal/hr
86860 kcal/hr
86000 kcal/hr
Answer: D) 86000 kcal/hr
Corrected (was c) - Book-3 §4.11: heat delivered by a heat pump = heat extracted + compressor work (Qe + W, Figure 4.13). W = 23 kW x 860 = 19,780 kcal/hr, so heat supplied = 66,220 + 19,780 = 86,000 kcal/hr exactly (77 kW + 23 kW = 100 kW). Option (c) 86,860 does not follow from any standard kWh-to-kcal conversion; option (b) wrongly subtracts the work input instead of adding it.
31. A coal fired boiler primary air fan maintains a velocity pressure of 70 mmWC, air temperature 38 °C, air density 1.135 kg/m3 and pitot tube constant 0.85. The velocity of air in m/sec will be
25.6
29.56
28.67
None of the above
Answer: B) 29.56
Confirmed vs Book-3 §5.6 — V = Cp × √(2 × 9.81 × ΔP/γ) with ΔP in mmWC: 0.85 × √(2 × 9.81 × 70/1.135) = 0.85 × √1210 = 0.85 × 34.78 = 29.56 m/s (b). (a) 25.6 m/s is the book's worked example (47 mmWC, Cp 0.9) — a memory trap; (c) 28.67 results from Cp 0.9 with 47 mmWC and other mixes.
32. A two stage air compressor drawing 75 kW has heat rejection of 862 kCal/kWh. The required capacity of the cooling tower when the operating temperature difference is 5 °C will be ________ TR.
21.55
107.5
22.93
57.4
Answer: A) 21.55
Confirmed vs Book-3 §7.2 Heat Load — Heat rejected = 75 kW × 862 kcal/kWh = 64,650 kcal/hr (book table: two-stage compressor with intercooler and after cooler = 862 kcal/kW/hr). TR = 64,650/3000 = 21.55 TR (exam key uses 3000 kcal/hr per TR; with 3024 it is 21.4 TR – option (a) either way). The 5°C ΔT is only needed for the cooling-water flow, not the TR load.
33. The star rating scheme of Fluorescent Tube light as per BEE Standards & Labelling Scheme is based on
Lumen Output
Lux per Watt
Lux per Watt per m2
Lumen per Watt at different operating hours
Answer: D) Lumen per Watt at different operating hours
Confirmed vs Book-3 §8.8 Table 8.4 — The FTL star rating is defined by lumens per Watt at 100, 2000 and 3500 hours of use (5-star ≥ 92 / ≥ 83 / ≥ 78 lm/W). Lumen output alone or lux-based ratios are not the criterion.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
34. 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?
195 mm
175 mm
169 mm
207 mm
Answer: C) 169 mm
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.
📖 §7.2 Factors Affecting Performance – Wet Bulb Temperature
35. Which of the following ambient conditions will not evaporate maximum amount of water in a cooling tower
41°C DBT and 38°C WBT
38°C DBT and 37°C WBT
36°C DBT and 30°C WBT
36°C DBT and 31°C WBT
Answer: B) 38°C DBT and 37°C WBT
Confirmed vs Book-3 §7.2 Wet Bulb Temperature — Evaporation is driven by the wet-bulb depression (DBT − WBT). Options (c) and (d) have 6 and 5°C depression and evaporate a lot; (b) 38/37°C has only 1°C – near-saturated air – so it is the condition that will NOT evaporate the maximum → (b).
36. The 5th and 7th harmonic in a 50 Hz power environment will have:
voltage and current distortions with 55 Hz & 57 Hz
voltage and current distortions with 500 Hz & 700 Hz
voltage and current distortions with 250 Hz & 350 Hz
no voltage and current distortion at all
Answer: C) voltage and current distortions with 250 Hz & 350 Hz
Confirmed vs Book-3 §1.10 — the harmonic order multiplies the fundamental: 5 × 50 = 250 Hz and 7 × 50 = 350 Hz.
Option (b) 500/700 Hz would be the 10th and 14th harmonics; option (a) adds instead of multiplying.
37. The illuminance is 10 lm/m2 from a lamp at 1 meter distance. The illuminance at half the distance will be
40 lm/m2
10 lm/m2
5 lm/m2
none of the above
Answer: A) 40 lm/m2
Confirmed vs Book-3 §8.2 — Book's worked example: E = (1.0/0.5)² × 10 = 40 lm/m². Halving the distance multiplies illuminance by 4 (inverse square law), not by 2.
38. Increasing the cycles of concentration of circulating water in a cooling tower will
increase blow down quantity
decrease blow down quantity
increase drift losses
decrease fan power consumption
Answer: B) decrease blow down quantity
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'.
📖 Ventilation by air changes (Book-3 Ch9 DG set room / Ch4 HVAC): flow = room volume × ACH
39. In an engine room 15 m long, 10 m wide and 4 m high, ventilation requirement for 20 air changes/hr is _____ m³/hr
30
3000
12000
none of the above
Answer: C) 12000
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.
📖 §8.3 Light source and lamp types (fluorescent tube designations)
40. The T2,T5,T8 and T12 fluorescent tube light are categorized based on
diameter of the tube
length of the tube
both diameter and length of the tube
power consumption
Answer: A) diameter of the tube
Diameter in eighths of an inch again: T12 = 38 mm, T8 = 25 mm, T5 = 16 mm, T2 = 6 mm. This exact question has now appeared in three papers (2013, 2015, 2017, 2018) — the free mark is worth locking in. Length and wattage vary separately, so 'both diameter and length' is always wrong.
📖 §8.3 Light source and lamp types (incandescent lamps)
41. Which of the following incandescent bulbs will have the least resistance ?
220 V, 60 W
220 V, 100 W
115 V, 60 W
115 V, 100 W
Answer: D) 115 V, 100 W
Resistance from the rating plate: R = V²/P. So 220 V/60 W = 807 Ω, 220 V/100 W = 484 Ω, 115 V/60 W = 220 Ω and 115 V/100 W = 132 Ω — lowest voltage with highest wattage gives the least resistance. Rule of thumb: R rises with the SQUARE of voltage and falls inversely with wattage, so scan for low V and high W before calculating.
📖 §2.4 Motor efficiency (efficiency from nameplate data)
42. A 22 kW, 415 V, 45 A, 0.8 pf, 1475 rpm, 4 pole 3 phase induction motor operating at 420 V, 40 A and 0.8 pf. What will be the motor efficiency?
85.0 %
94.5 %
89.9 %
None of the above
Answer: A) 85.0 %
Efficiency = output/input, and the input must be computed from the NAMEPLATE ratings: sqrt(3) x 415 x 45 x 0.8 = 25.88 kW against 22 kW output = 85.0%.
The 420 V and 40 A are the present operating point, deliberately supplied to tempt you into 1.732 x 420 x 40 x 0.8 = 23.28 kW, which would give 94.5% - option (b), the planted wrong answer.
Sanity rule: a standard 22 kW motor sits around 88-91% efficiency, so any answer above 94% on nameplate data should make you re-read the question.
📖 §4.7 Performance assessment of refrigeration plants (the TR constant)
43. One ton of refrigeration is not equal to__________.
3024 kCal/hr
3.51 kW
12000 Btu/hr
860 kCal/hr
Answer: D) 860 kCal/hr
Memorise the identity: 1 TR = 3024 kcal/hr = 3.5 kW = 12,000 Btu/hr. It is the heat needed to freeze one short ton of water in 24 hours.
860 kcal/hr is a different constant altogether - it is the heat equivalent of 1 kW (1 kWh = 860 kcal), which is why it is planted here.
Cross-check the set: 3024 kcal/hr / 860 kcal/kWh = 3.517 kW, and 3024 x 3.968 = 12,000 Btu/hr. Everything hangs together except 860.
📖 §4.3 Types of refrigeration system (VCR components)
44. Which of the following is not a part of vapour compression refrigeration cycle ?
Compressor
Evaporator
Condenser
Generator
Answer: D) Generator
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.
📖 §6.6 Flow control strategies (pumps in parallel)
45. It is acceptable to run pumps in parallel provided their_________ are similar
Suction heads
Discharge heads
Closed valve heads
Total head at full flow
Answer: C) Closed valve heads
Closed-valve (shutoff) head is the head at zero flow — the top of the pump curve. Match those and both pumps keep contributing; mismatch them and the weaker pump is pushed back to zero flow, churning and overheating while the stronger one carries the load. Pumps of different sizes may be paralleled provided this one figure matches; equal duty-point heads are not enough.
46. L / G ratio in a cooling tower is the ratio of _________________.
Length and girth
Length and Temperature gradient
Water flow rate and air mass flow rate
Air mass flow rate and water flow rate
Answer: C) Water flow rate and air mass flow rate
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.
47. 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
12 kW
9 kW
6.75 kW
None of the above
Answer: C) 6.75 kW
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'.
48. In which of the following fans air enters and leaves the fan with no change in direction ?
Forward curved
Backward curved
Radial
Propeller
Answer: D) Propeller
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.
📖 §6.5 Efficient pumping system operation (NPSH and cavitation)
49. The value, by which the pressure in the pump suction exceeds the liquid vapour pressure, is expressed as
Net positive suction head available
Static head
Dynamic head
Suction head
Answer: A) Net positive suction head available
Book wording: NPSHa is the value by which the liquid pressure at the pump suction/impeller eye exceeds the liquid's vapour pressure at the pumping temperature, expressed as a head in metres. Static head and dynamic head are system heads with no vapour-pressure reference at all. Keep NPSHa > NPSHr or the liquid flashes to vapour, the bubbles collapse in the impeller and you get the gravel-in-the-casing noise of cavitation.
50. A fan is operating at 970 RPM developing a flow of 3000 Nm3/hour at a static pressure of 650 mmWC. If the speed is reduced to 700 RPM, the static pressure (mmWC) developed will be
244.3
650
469
None of the above
Answer: D) None of the above
Pressure ∝ N²: SP2 = 650 × (700/970)² = 650 × 0.5207 = 338.5 mmWC, which is not offered, so 'none of the above' is correct. Check where the distractors come from: 469 mmWC is N¹ (650 × 700/970) and 244 mmWC is N³ — both are the wrong exponent. Always compute the number before assuming the nearest option is intended.
📖 §1.1 Introduction to electric power supply systems (heat rate and generation efficiency)
51. One of the thermal power plants operating with 2 nos. of 500 MW units has reported the operating heat rate of 11250 kJ/kW. 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 %
Definition to memorise: 1 kWh = 860 kcal = 3600 kJ of thermal energy. So generation efficiency = 3600 / heat rate (kJ/kWh) = 3600/11250 = 0.32 = 32%.
In kcal units the same relation is efficiency = 860 / heat rate (kcal/kWh). Heat rate is INVERSELY proportional to efficiency - the lower the heat rate, the better the plant.
The 500 MW units and the 73% PLF are deliberate distractors: PLF measures capacity utilisation, not thermal efficiency, and plays no part in this calculation.
52. The power measured in a boiler ID fan is 52 kW operating at 49 Hz. As an energy conservation measure the Variable Frequency Drive (VFD) was installed and the fan was operated at 34 Hz. The estimated power savings will be
36 kW
17.2 kW
34.7 kW
35 .7 kW
Answer: C) 34.7 kW
Frequency is a proxy for speed, so kW ∝ f³: power at 34 Hz = 52 × (34/49)³ = 52 × 0.334 = 17.3 kW, hence SAVING = 52 − 17.3 = 34.7 kW. Read the question wording carefully — 17.2 kW is the new power drawn, not the saving, and it is deliberately offered as option (b).
1. S-1: Operating data of an induced-draft cooling tower: range 8 °C, cooling water flow 12,500 m3/hr, drift loss 0.1% of circulation, WBT 27 °C, ambient DBT 35 °C, effectiveness 67%, COC 3. Estimate evaporation loss, make-up water requirement and TR load.
Model answer: Evaporation loss = 0.00085 x 1.8 x 12500 x 8 = 153 m3/hr. Blowdown = 153/(3-1) = 76.5 m3/hr. Drift = 12500 x 0.001 = 12.5 m3/hr. Make-up = evaporation + blowdown + drift = 153 + 76.5 + 12.5 = 242 m3/hr. TR load = (12500 x 1000 x 8)/3024 = 33,069 TR.
Use book evaporation formula (0.00085 x 1.8 x circulation x range), blowdown = evap/(COC-1), make-up = evap+blowdown+drift, heat-load TR = m·Cp·ΔT/3024.
2. S-2: A chilled water system runs always at full load; inlet/outlet 12 °C / 7 °C. Chilled-water pump discharge pressure 3.6 kg/cm2g, suction 5 m above pump centreline, motor power 70 kW at 90% efficiency, pump efficiency 60%. Find the operating refrigeration load in TR.
Model answer: Discharge head = 3.6 kg/cm²g ≈ 36 m; total head = 36 - 5 = 31 m. Pump shaft power = 70 x 0.9 = 63 kW. Flow = (pump shaft power x 1000 x pump eff)/(head x 1000 x 9.81) = (63 x 1000 x 0.6)/(31 x 1000 x 9.81) = 0.1243 m3/s = 447.5 m3/hr. Refrigeration load = (447500 x 5)/3024 = 740 TR.
Total head from discharge head minus suction lift; flow from pump hydraulic equation; TR = (flow(kg/hr) x ΔT)/3024.
3. S-3: In an air washer of a textile humidification system, airflow 3000 m3/h at 25 °C and 10% RH is humidified to 60% RH. Inlet/outlet specific humidity = 0.002 / 0.0062 kg/kg dry air; air density at 25 °C = 1.184 kg/m3. Calculate the water required (kg/hr).
Model answer: Water required mw = volume x density x (ω_out - ω_in) = 3000 x 1.184 x (0.0062 - 0.002) = 3000 x 1.184 x 0.0042 = 14.9 kg/hr.
Water added = mass of air x change in specific humidity (mass flow = volume flow x density).
4. S-4: In a thermal power station, steam input to a turbine on fully condensing mode is 100 TPH. Condenser heat rejection = 555 kcal/kg of steam condensed; cooling water inlet/outlet = 27 °C / 37 °C. Find the circulating cooling water flow.
Model answer: Heat rejected = 100,000 kg/hr x 555 kcal/kg = 55.5 million kcal/hr. Cooling water flow = heat rejected/(ΔT x Cp) = 55,500,000/((37-27) x 1) = 5,550,000 kg/hr = 5550 m3/hr.
Equate condenser heat rejection to heat gained by cooling water (m·Cp·ΔT) and solve for flow.
5. S-5: List any five benefits of power factor improvement in an industrial power distribution system.
Model answer: 1. Reduced kVA demand and lower maximum-demand charges. 2. Reduced line/transformer current → lower I²R distribution losses. 3. Released capacity in transformers, cables and switchgear for additional load. 4. Improved voltage regulation (less voltage drop) at the load. 5. Avoidance of low-PF penalties and possible PF-incentive rebates from the utility. (Also: longer equipment life due to reduced heating.)
Standard PF-improvement benefits from Book-3 Chapter 1 (the paper directs 'Refer Guide Book No 3, Chapter 1, Page No 11'); model answer supplied from chapter notes.
📖 §9.4 Energy performance assessment — trial measurements (fuel by dip level, kWh, PF) & analysis (% loading, kWh/litre)
6. 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.
7. S-7: How does a motor lose its efficiency upon rewinding? What two parameters will indicate the efficacy of the rewinding?
Model answer: Efficiency loss on rewinding: the burnout/stripping of the old winding (especially using high temperature or mechanical force) can damage the stator core lamination insulation, increasing iron (core) losses; changes in winding wire gauge, number of turns, or winding tightness alter the copper resistance and increase I²R losses; poor slot fill and handling raise stray and friction losses. A typical rewound motor can lose 1-2% (or more) efficiency. The two parameters that indicate the efficacy/quality of the rewind are: (1) the no-load (core) loss / no-load current and (2) the stator winding resistance (copper loss / I²R) — both measured before and after to confirm losses have not increased.
Book-3 Chapter 2 content (paper says 'Refer Guide Book No 3, Chapter 2, Page No 61'); model answer supplied from chapter notes — core-loss test and winding-resistance test indicate rewind quality.
📖 Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14); Book-3 §3.5 Avoiding Air Leaks — leak quantification (Tables 3.16 & 3.17)
8. S-8: Two 480 CFM screw compressors A & B. Compressor-A runs at full load; Compressor-B runs in load-unload. Load power of both = 74 kW; unload power of B = 26 kW. Both run all working day. B's loading is 64%; after arresting leakage, loading falls to 35%. Estimate energy savings per day.
Model answer: Existing: A = 74 kW. B = 0.64 x 74 + 0.36 x 26 = 47.36 + 9.36 = 56.72 kW. After leakage arrest, B = 0.35 x 74 + 0.65 x 26 = 25.9 + 16.9 = 42.8 kW. Difference (B only) = 56.72 - 42.8 = 13.92 kW. Daily energy savings = 13.92 x 24 = 334 kWh/day.
Confirmed vs Book-3 §3.5 Capacity Control of Compressors — Weighted power of load-unload compressor = load%×load-power + unload%×unload-power; savings = drop in B's power × 24 h.
9. The operating data of an induced draft-cooling tower is as follows: Observed range: 8 °C; Cooling water flow rate: 12,500 m3/hr; Drift loss: 0.1 % of circulation rate; Wet Bulb Temperature: 27 °C; Ambient Dry Bulb Temperature: 35 °C; Effectiveness: 67 %; Cycle of Concentration: 3. Estimate the evaporation loss; make up water requirement and TR load of cooling tower.
Model answer: Evaporation loss = 0.00085 x 1.8 x 12500 x 8 = 153 m3/hr
Blow Down = 153 / (3 - 1) = 76.5 m3/hr
Make up = 153 + 76.5 + (12500 x 0.001) = 242 m3/hr
Heat load = 12500 x 1000 x 8 / 3024 = 33069 TR
Evaporation = 0.00085 × 1.8 × 12,500 × 8 = 153 m³/hr; Blowdown = 153/(3 − 1) = 76.5 m³/hr; Drift = 0.1% of 12,500 = 12.5 m³/hr; Make-up = E + B + D = 242 m³/hr. Heat load = 12,500 × 1000 × 8/3024 = 33,069 TR. The effectiveness of 67%, the WBT and the DBT are all distractors — none of them enters the water balance. Drift is quoted as a percentage of CIRCULATION rate, not of evaporation.
📖 §4.7 Performance assessment of refrigeration plants (TR via pump hydraulics)
10. A plant is operating a chilled water system always at full load. The chilled water inlet and outlet temperatures are 12 °C and 7 °C respectively. The chilled water pump discharge pressure is 3.6 kg/cm2g and the suction is 5 meters above the pump centerline. The power drawn by the chilled water pump's motor is 70 kW and an efficiency of 90 %. The chilled water pump efficiency at the operating point from pump characteristic curve is 60 %. Find out the operating refrigeration load in TR.
Model answer: Total head = 36 - 5 = 31 m
Pump shaft power = 70 x 0.9 = 63 kW
Flow rate = (63 x 1000) x 0.6 / (31 x 1000 x 9.81) = 0.124297 m3/s = 447.5 m3/hr
Refrigeration load = (447500 x 5) / 3024 = 740 TR
Route: pump power -> flow -> TR. Head = discharge head - static suction lift correction = 36 - 5 = 31 m (3.6 kg/cm2g ~ 36 m water column). Pump shaft power = motor input x motor efficiency = 70 x 0.9 = 63 kW.
Flow from hydraulic power: Q = (shaft kW x 1000 x pump efficiency)/(H x rho x g) = (63,000 x 0.6)/(31 x 1000 x 9.81) = 0.1243 m3/s = 447.5 m3/hr.
Then TR = 447,500 kg/hr x 1 kcal/kg degC x (12 - 7)/3024 = 740 TR. The two mark-losers are omitting the pump efficiency (it multiplies the useful hydraulic power) and forgetting that 1 kg/cm2 ~ 10 m of water head.
📖 §4.14 Humidification systems (air washer water requirement)
11. In an air washer of a textile humidification system with an airflow of 3000 m3/h at 25 °C and 10 % relative humidity is humidified to 60 % relative humidity by adding water through spray nozzles. The specific humidity of air at inlet and outlet are 0.002 kg/kg of dry air and 0.0062 kg/kg of dry air respectively. The density of air at 25 °C is 1.184 kg/m3. Calculate the amount of water required in kg/hr.
Model answer: The amount of water required:
mw = V x rho x (W_out - W_in)
= 3000 x 1.184 x (0.0062 - 0.002)
= 14.9 kg/h
Water added = volumetric air flow x air density x (W_out - W_in), where W is the specific humidity in kg per kg of dry air.
Working: 3000 x 1.184 x (0.0062 - 0.0020) = 3552 x 0.0042 = 14.9 kg/hr.
The relative humidity figures (10% and 60%) are context, not calculation inputs - the specific humidities have already been read off the chart for you. Convert to kg/kg before subtracting; using g/kg leaves the answer 1000 times too large.
📖 §9.4 Energy performance assessment of DG sets (dip-level fuel measurement)
12. During the performance evaluation of a DG set, the following parameters were noted: Capacity of DG set 1500 kVA; Test duration 36 minutes; Units generated 442 kWh; Average Power factor 0.92 pf; Length of diesel tank 90 cm; Width of diesel tank 90 cm; Height of the diesel tank 90 cm; Initial tank dip level (from top) 63 cm; Final tank dip level (from top) 79 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)
Model answer: 1. Diesel Consumption = 0.9 x 0.9 x 0.16 = 0.1296 m3 = 129.6 Litres
(level drop = 79 - 63 = 16 cm = 0.16 m)
2. Average load (kW) = (442 / 36) x 60 = 736.7 kW
3. Percentage Loading (%) = (736.7 / 0.92) / 1500 = 53 %
4. Specific power generation = 442 / 129.6 = 3.41 kWh/Litre
Fuel from the dip: the level DROP is 79 − 63 = 16 cm, so volume = 0.9 × 0.9 × 0.16 = 0.1296 m³ = 129.6 litres (using 63 or 79 cm as a depth is the classic error). Load = 442 kWh over 36 min = 442 × 60/36 = 736.7 kW. % loading compares like with like — convert kW back to kVA first: (736.7/0.92)/1500 = 53%. Specific generation = 442/129.6 = 3.41 kWh/litre. Dividing 736.7 by 1500 directly gives 49% and is marked wrong.
📖 §3.5 Efficient operation (load/unload operation, cost of leakage)
13. 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 64 %. 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: Existing Case:
Energy consumed per hour by Compressor-A = 74 kW
Energy consumed per hour by Compressor-B = 0.64 x 74 + 0.36 x 26 = 56.72 kW
Total energy consumed (Compressor A & B) = 74 + 56.72 = 130.72 kW/hr
Energy consumed per day = 130.72 x 24 hrs = 3137.3 kWh/day
Leakage Calculation:
Energy consumed per hour by Compressor-B (before) = 0.64 x 74 + 0.36 x 26 = 56.72 kW
Energy consumed per hour by Compressor-B (after) = 0.35 x 74 + 0.65 x 26 = 42.8 kW
Difference in power consumption = 56.72 - 42.8 = 13.92 kW/hr
Savings by arresting leakage per day = 13.92 x 24 = 334 kWh/day
Model a load/unload compressor as a weighted average: kW = (load fraction x load power) + (unload fraction x unload power). Compressor-B before: 0.64 x 74 + 0.36 x 26 = 56.72 kW; after leak arrest: 0.35 x 74 + 0.65 x 26 = 42.80 kW.
Saving = 56.72 - 42.80 = 13.92 kW, and over 24 hours = 334 kWh/day. Compressor-A is fully loaded throughout, so it cancels out and can be ignored for the SAVING (it matters only for the total consumption of 3137 kWh/day).
The mark-loser is forgetting the unload power altogether: an unloaded compressor still eats about 25-35% of its full-load power, which is exactly why cutting leaks pays.
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback
1. L-1: A food processing plant has contract demand 2500 kVA; average MD 2000 kVA at 0.95 PF; MD billed at Rs.300/kVA; minimum billable MD = 75% of contract demand; incentive 0.5% reduction in energy charges per 0.01 PF increase above 0.95; average monthly energy charge Rs.10 lakhs. Plant improves PF to unity. Determine capacitor kVAr, annual reduction in MD charges and energy charge, and simple payback if capacitors cost Rs.800/kVAr.
Model answer: kW drawn = 2000 x 0.95 = 1900 kW. kVAr = 1900 x (tan(cos⁻¹0.95) - tan(cos⁻¹1)) = 1900 x (0.329 - 0) = 625 kVAr. Capacitor cost = 625 x 800 = Rs.5,00,000. New MD at unity PF = 1900 kVA; but minimum billable = 75% x 2500 = 1875 kVA → billed at 1900 kVA (above floor). Reduction in MD = 2000 - 1900 = 100 kVA → demand saving = 100 x 300 = Rs.30,000/month = Rs.3,60,000/yr. PF rises 0.95→1.00 (0.05 = 5 steps of 0.01) → energy charge reduction = 5 x 0.5% = 2.5%; monthly saving = 10,00,000 x 0.025 = Rs.25,000 → Rs.3,00,000/yr. Total annual saving = 3,60,000 + 3,00,000 = Rs.6,60,000. Payback = 5,00,000/6,60,000 = 0.76 yr ≈ 9 months.
Capacitor kVAr = kW(tanΦ1-tanΦ2); MD saving from reduced kVA × tariff; energy saving from PF-incentive %; payback = investment/annual saving.
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19); Book-3 §3.5 Air Receivers (IS 7938-1976 sizing)
2. L-2: Write short notes on the following with respect to the compressed air system: (a) Refrigeration drier, (b) Heat of compression drier, (c) Role of air receiver, (d) Dew point.
Model answer: (a) Refrigeration dryer: straight mechanical refrigeration in which the dew point of the air is reduced by chilling; a second heat exchanger lets the outgoing cold air pre-cool the incoming compressed air. The achievable atmospheric dew point is about -20 degC. It is the most economical process for roughly 90% of all applications, separates almost 100% of solid particles and water droplets larger than 3 micron, and costs about 0.2 bar in pressure loss (2.9 kW per 1000 m3/hr, Table 3.19).
(b) Heat of compression (HOC) dryer: a twin-tower adsorption dryer in which compressed air taken directly from the compressor discharge before the after-cooler, at about 135 degC for a reciprocating machine, regenerates the desiccant. There are no electrical heaters and no purge loss, so operating cost is zero to very minimal (0.8 kW per 1000 m3/hr) and the atmospheric dew point achieved is -40 degC. Vessel A is in service for 4 hours while vessel B is heated for 2.5 hours and cooled for 1.5 hours, then they change over. Capacities range from 400 to 5000 cfm.
(c) Role of the air receiver: it dampens the pulsations leaving the compressor discharge, acts as a reservoir for sudden or unusually heavy demands in excess of compressor capacity, prevents too frequent loading and unloading (short cycling), and separates moisture and oil vapour by allowing carry-over from the after-cooler to precipitate. Per IS 7938-1976 its volume in m3 should be 1/10th to 1/6th of the output in m3/min, and a local receiver near a point of high cyclic demand avoids having to add compressor capacity.
(d) Dew point: the temperature at which the moisture present in the air starts condensing. The extent of drying is expressed as the ATMOSPHERIC dew point (at atmospheric pressure); the lower the dew point, the drier the air — air at -40 degC atmospheric dew point holds only 80 ppm of moisture against 3800 ppm at 0 degC (Table 3.18). Dryer performance is quoted as PRESSURE dew point, and raising the pressure of a gas raises its dew point temperature, since the partial pressure of the water vapour rises in proportion (Dalton's law).
Confirmed vs Book-3 §3.5 Air Dryers — Short notes from §3.5: the refrigerant dryer's achievable atmospheric dew point is -20 degC (the earlier +3 degC figure contradicted Table 3.19 and has been corrected), the HOC dryer reaches -40 degC at 0.8 kW/1000 m3/hr, plus the receiver's four duties and the atmospheric-vs-pressure dew point distinction.
📖 §5.6 Gas density (γ = PM/RT) and fan power formula
3. L-3: In a boiler, the forced-draught fan develops a total static pressure of 300 mmWC. Determine the shaft power (kW) to drive the fan if 10,000 kg/hr coal is burnt with 13 kg air/kg coal. Boiler-house temperature 20 °C, static efficiency 80%. Use R = 847.84 mmWC·m3/kg·mole·K and M = 28.92 kg/kg·mole.
Model answer: Mass of air = 10000 x 13/3600 = 36.11 kg/s. Atmospheric pressure P = 1 kg/cm² = 10,000 mmWC; T = 293 K. Density = P·M/(R·T) = (10000 x 28.92)/(847.84 x 293) = 1.164 kg/m3. Volume = 36.11/1.164 = 31.02 m3/s. Shaft power = (volume x total pressure)/(102 x fan efficiency) = (31.02 x 300)/(102 x 0.8) = 114 kW.
4. L-4: A 7.5 TR package A/C cools a UPS room (40 kVA UPS). Outdoor unit air velocity 6.1 m/s, fan opening radius 0.30 m, air density 1.174 kg/m3, ambient 305 K, condenser-outlet hot air 313.5 K, Cp 1.009 kJ/kgK, compressor power 5.40 kW, motor efficiency 90%. UPS on-load (16 h): input 11.94 kW, output 8.61 kW; no-load (8 h): input 1.16 kW, output 0. Calculate (a) present delivery TR, (b) power per TR, (c) annual energy savings for 7200 h if UPS relocated to a ventilated area (Rs.8/kWh).
Model answer: Area = 3.14 x 0.30² = 0.283 m². Air flow = 0.283 x 6.1 = 1.72 m3/s. Mass = 1.72 x 1.174 = 2.02 kg/s. ΔT = 313.5 - 305 = 8.5 K. Heat transfer = 2.02 x 1.009 x 8.5 = 17.32 kJ/s = 62,352 kJ/hr = 14,917 kcal/hr. Compressor heat input = 5.4 x 0.9 x 860 = 4180 kcal/hr. Evaporator load = 14,917 - 4180 = 10,737 kcal/hr. (a) Effective TR = 10,737/3024 = 3.55 TR. (b) Power per TR = 5.40/3.55 = 1.52 kW/TR. (c) Heat load from UPS: on-load 3.33 kW → 0.95 TR/hr x 16 = 15.2 TR-day; no-load 1.16 kW → 0.33 TR/hr x 8 = 2.64 TR-day; total 17.84 TR/day. AC power to remove it = 17.84 x 1.52 = 27.12 kW. Annual savings (300 days) = 27.12 x 300 ≈ 8136 kWh → Rs.8 x 8136 = Rs.65,088/yr.
Condenser air-side heat minus compressor heat gives evaporator TR; kW/TR from compressor power; UPS heat converted to TR/day then to AC power saved if UPS relocated, costed at the tariff.
5. L-5: A textile plant had two 6 MW gas turbines + HRSG feeding process steam and a 500 TR VAM (4.4 kg steam/TR, full load). Gas turbines stopped due to gas price; two 10 TPH agro-waste boilers installed (steam cost Rs.1200/ton); plant runs 7000 h/yr. Management plans to replace VAM with an electric centrifugal chiller at 0.7 kW/TR. Compare annual operating costs of electric chiller vs VAM (grid power Rs.6.12/kWh; auxiliaries unchanged). Do you agree with running the VAM?
Model answer: VAM steam need = 500 x 4.4 = 2200 kg/hr = 2.2 TPH; steam cost = 2.2 x 1200 = Rs.2640/hr. Electric chiller power = 0.7 x 500 = 350 kW; cost = 350 x 6.12 = Rs.2142/hr. Saving with electric chiller = 2640 - 2142 = Rs.498/hr. Annual saving = 7000 x 498 = Rs.34,86,000/yr. Conclusion: Disagree with running the VAM — the electric centrifugal chiller is cheaper to operate (saves ~Rs.34.86 lakh/yr) now that cheap waste heat is gone.
VAM running cost = steam rate x steam cost; chiller cost = kW/TR x TR x power tariff; compare hourly and annualise; recommend the cheaper option.
6. L-6: A distribution company has: input energy 60 MU, metered billed energy 43 MU, average (un-metered) billing 3 MU, amount billed Rs.540 million, arrears collected Rs.80 million, amount received Rs.470 million. (a) Estimate (i) AT&C loss % and revenue realised (Rs./kWh); (ii) revenue loss per kWh and monthly loss if purchased energy cost is Rs.8.10/kWh. (b) List five measures to reduce commercial loss.
Model answer: (a)(i) Billing efficiency = (43+3)/60 x 100 = 76.7%. Collection efficiency = (470-80)/540 x 100 = 72.2%. AT&C loss = [1 - (billing eff x collection eff)] x 100 = [1 - (0.767 x 0.722)] x 100 = 44.62%. Revenue realised = (470-80)/60 = Rs.6.5/kWh. (ii) Revenue loss = 8.10 - 6.5 = Rs.1.6/kWh; monthly loss = 60 MU x 1.6 = Rs.96 million (Rs.9.6 crore). (b) Measures to reduce commercial loss: 1. Accurate metering of all consumers (replace defective/electromechanical meters, AMR/smart meters). 2. Detect and curb theft/pilferage and unauthorised connections; energy audit/feeder metering. 3. Improve billing — eliminate un-metered/average billing, correct meter-reading and billing errors. 4. Strengthen collection efficiency — disconnection drives, online payment, recovery of arrears. 5. HVDS / spot billing / consumer indexing and GIS mapping to plug gaps.
AT&C loss = 1-(billing eff x collection eff); revenue realised = net received/input units; revenue loss = purchase cost - realised; commercial-loss measures from Book-3 Chapter 1 (paper cites page 27).
📖 §1.4 PF improvement economics + §1.3 Maximum demand control
7. A food processing plant has a contract demand of 2500 kVA with the power supply company. The average maximum demand of the plant is 2000 kVA at a power factor of 0.95. The maximum demand is billed at the rate of Rs.300/kVA. The minimum billable maximum demand is 75 % 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 company is Rs.10 lakhs. The plant decides to improve the power factor to unity. Determine the power factor capacitor kVAr required, annual reduction in maximum demand charges and energy charge component. What will be the simple payback period if the cost of power factor capacitors is Rs.800/kVAr?
Model answer: kW drawn = 2000 x 0.95 = 1900 kW
kVAr required to improve power factor from 0.95 to 1
= kW (tan phi1 - tan phi2)
= 1900 [tan(cos^-1 0.95) - tan(cos^-1 1)]
= 1900 (0.329 - 0) = 625 kVAr
Cost of capacitors @ Rs.800/kVAr = Rs. 5,00,000
Maximum demand at unity power factor = 1900 / 1 = 1900 kVA
75 % of contract demand = 1875 kVA
Reduction in demand charges = 100 kVA x Rs.300 = Rs.30,000/month x 12 = Rs. 3,60,000/year
Percentage reduction in energy charge from 0.95 to 1 @ 0.5 % for every 0.01 increase = 2.5 %
Monthly energy cost component of the bill = Rs. 10,00,000
Reduction in energy cost component = 10,00,000 x (2.5/100) = Rs. 25,000/month
Annual reduction = Rs. 25,000 x 12 = Rs. 3,00,000
Savings in electricity bill = Rs. 6,60,000
Investment = Rs. 5,00,000
Payback period = 5,00,000 / 6,60,000 = 0.76 years or about 9 months
Work the three parts in sequence: kW = 2000 x 0.95 = 1900 kW; kVAr = 1900 x (tan(cos^-1 0.95) - 0) = 1900 x 0.329 = 625 kVAr; investment = 625 x 800 = Rs.5,00,000.
Then test the new demand against the minimum billable floor: 75% x 2500 = 1875 kVA, and the improved demand 1900/1.0 = 1900 kVA is ABOVE that floor, so the full drop counts: 2000 - 1900 = 100 kVA x Rs.300 x 12 = Rs.3,60,000/year. PF incentive = 5 steps x 0.5% = 2.5% of Rs.10 lakh/month = Rs.3,00,000/year.
Payback = 5,00,000 / 6,60,000 = 0.76 years (~9 months). Always test the improved demand against the minimum billable floor before claiming the saving.
📖 §5.6 Fan performance assessment (gas density, fan shaft power)
8. In a boiler, the forced draught fan develops a total static pressure of 300 mmWC. Determine the shaft power (in kW) required to drive the fan if 10,000 kg of coal is burnt per hour with 13 kg of air per kg of coal burnt. The boiler house temperature is 20 °C and static efficiency of the fan is 80 %. The operating air density may be calculated from the following: R = 847.84 mmWC m3/kg mole K and Molecular weight of air, M = 28.92 kg/kg mole.
Model answer: Total pressure = 300 mm of WC
Mass of air handled, m = 10000 x 13 / 3600 = 36.11 kg/s
Atmospheric pressure, P = 1 kg/cm2 = 10 m of WC = 10,000 mm of WC
Temperature T = 20 + 273 = 293 K
Gas constant for air, R = 847.84 mmWC m3/kg mole K
Molecular weight of air, M = 28.92 kg/kg mole
Density = (P x M) / (R x T) = (10000 x 28.92) / (847.84 x 293) = 1.164 kg/m3
Volume = mass (kg/s) / density (kg/m3) = 36.11 / 1.164 = 31.02 m3/s
Power to fan shaft, kW = [Volume (m3/s) x Total pressure (mmWC)] / [102 x fan efficiency]
= (31.02 x 300) / (102 x 0.8)
= 114 kW
Three linked steps. Mass of air = 10,000 × 13/3600 = 36.11 kg/s. Density = (P × M)/(R × T) with P = atmospheric = 10,000 mmWC and T = 293 K: (10000 × 28.92)/(847.84 × 293) = 1.164 kg/m³. Volume = 36.11/1.164 = 31.02 m³/s, then shaft kW = Q × ΔP/(102 × η) = 31.02 × 300/(102 × 0.8) = 114 kW. The pressure inside the density formula is ATMOSPHERIC (10,000 mmWC), not the 300 mmWC fan pressure — mixing those two is the mark-losing mistake.
📖 §4.9 Performance assessment of package AC units + §4.7 kW/TR
9. A 7.5 TR package air conditioner is provided for a UPS room for removing the heat generated from the UPS of rated capacity 40 kVA. The following parameters were noticed while performing the assessment of the total system. UPS Parameters (40 kVA): On Load (16 hrs) - Input Power 11.94 kW, Output Power 8.61 kW; No Load (8 hrs) - Input Power 1.16 kW, Output Power 0.00 kW. Air conditioner parameters: Installed capacity of Air conditioner 7.5 TR; Outdoor unit (condenser) air velocity 6.1 m/s; Radius of the fan opening at the point of velocity measurement in outdoor unit 0.30 m; Air Density 1.174 kg/m3; Ambient temperature 305 K; Temperature of hot air (condenser outlet) 313.5 K; Specific heat of air 1.009 kJ/kg K; Power drawn by the compressor 5.40 kW; Efficiency of the compressor motor 90 %. Calculate a) Present delivery capacity of air conditioner (TR) (3 Marks) b) Power drawn per TR of refrigeration (3 Marks) c) Calculate the annual energy savings for 7200 hrs, if the UPS is relocated to a non-air-conditioned ventilated area. Assume energy cost Rs.8/kWh. (4 Marks)
Model answer: a) Present delivery capacity:
Capacity installed = 7.5 TR
Outdoor unit air velocity = 6.1 m/s; radius of the opening = 0.30 m
Area of cross section = 3.14 x 0.3^2 = 0.283 m2
Total air flow = 0.283 x 6.1 = 1.72 m3/s
Density of air = 1.174 kg/m3
Mass of air, m = 1.72 x 1.174 = 2.02 kg/s
T1 = 305 K; T2 = 313.5 K; dT = 8.5 K
Specific heat at constant pressure, cp = 1.009 kJ/kg K
Heat transfer = m x cp x dT = 17.32 kJ/s
Heat transfer per hour = 62352 kJ/hr = 14917 kcal/hr
Heat input from the compressor = 5.4 x 0.9 x 860 = 4180 kcal/hr
Evaporator heat load = 14917 - 4180 = 10737 kcal/hr
(the printed solution mis-types this line as "14949 - 4180", but the result 10737 corresponds to 14917 - 4180)
1 TR = 3024 kcal/hr
Effective TR = 3.55 TR
b) Power drawn by the compressor = 5.40 kW
Power taken per TR of refrigeration = 5.40 / 3.55 = 1.52 kW/TR
c) Heat load generated by UPS in the conditioned space:
On Load (16 hrs): input 11.94 kW, output 8.61 kW, heat load = 3.33 kW = 0.80 kcal/s = 2880 kcal/hr = 0.95 TR/hr = 15.2 TR/day
No Load (8 hrs): input 1.16 kW, output 0 kW, heat load = 1.16 kW = 0.28 kcal/s = 1008 kcal/hr = 0.33 TR/hr = 2.64 TR/day
Total = 17.84 TR/day
AC load generated by UPS per day = 17.84 TR
Power taken by AC to generate 17.84 TR at 1.52 kW/TR = 27.12 kW (kWh/day)
Annual energy savings at 300 days of operation = 8136 kWh
Cost of power = Rs. 8/kWh
Annual cost savings = Rs. 65,088/-
(Note: the question states 7200 hrs; the printed model answer works the annual saving on 300 days of operation, i.e. 27.12 kWh/day x 300 = 8136 kWh.)
Part (a) measures the capacity on the CONDENSER side and then subtracts the compressor heat: air flow = pi r^2 x v = 3.14 x 0.3^2 x 6.1 = 1.72 m3/s; mass = 1.72 x 1.174 = 2.02 kg/s; heat rejected = m x cp x dT = 2.02 x 1.009 x 8.5 = 17.32 kJ/s = 14,917 kcal/hr. The compressor adds 5.4 x 0.9 x 860 = 4180 kcal/hr of work into the refrigerant, so the true evaporator load = 14,917 - 4180 = 10,737 kcal/hr = 3.55 TR against 7.5 TR installed.
Part (b): kW/TR = 5.40/3.55 = 1.52 - poor, since a healthy package unit is nearer 1.0-1.2, confirming the unit is badly oversized and part-loaded.
Part (c): the UPS heat that must be removed is INPUT minus OUTPUT (11.94 - 8.61 = 3.33 kW on load, and the full 1.16 kW off load), never the input itself; convert with 860 kcal/kWh and 3024 kcal/TR, then multiply by the 1.52 kW/TR. Note the printed key annualises on 300 days, not the 7200 hrs stated in the question.
📖 §4.3 Types of refrigeration system (VAM vs electric chiller operating cost)
10. One of the textile processing plants has installed two numbers of 6 MW gas turbines and also Heat Recovery Steam Generator (HRSG) to generate steam from the hot gases. The steam generated from HRSG is utilized for process steam requirement and also for 500 TR Vapour Absorption Machine (VAM). The VAM consumes 4.4 kg steam per TR and is operated at full load. Due to increase in gas price the plant has stopped gas turbine operations and avails power supply from the grid. To meet the steam requirement the plant has installed two numbers of 10 TPH Agro Waste Boilers and steam is supplied to the process plant as well as to VAM machine. The average cost of steam is Rs.1200/- per ton from agro waste boiler. The plant operates for 7000 hours in a year. The management is planning to replace the VAM chillers by electrical centrifugal chiller which will operate at 0.7 kW/TR. Compare the annual operating costs of electrical chiller and VAM. The cost of grid power is Rs 6.12/kWh. Consider all the other auxiliary power remains same in both the cases. Do you agree with the management decision of operating VAM machine for chilling requirements?
Model answer: Capacity of VAM machine = 500 TR
Steam required per TR = 4.4 kg/TR
Total steam requirement = 500 x 4.4 = 2200 kg/hr = 2.2 TPH
Cost of steam from agro boiler = 2.2 x 1200 = Rs. 2640/hr
Power consumed by electric chiller = 0.7 x 500 = 350 kW
Cost of electricity = Rs. 6.12/kWh
Operating cost of electric chiller = 350 x 6.12 = Rs. 2142/hr
Savings by electric chiller = 2640 - 2142 = Rs. 498/hr
Annual operating savings = 7000 x 498 = Rs. 34,86,000/-
Conclusion: Disagree with the management decision of continuing to operate the VAM machine - the electrical centrifugal chiller is cheaper to operate by about Rs. 34.86 lakh per year.
Put both machines on a Rs./hour basis for the SAME 500 TR duty. VAM: 500 x 4.4 = 2200 kg/hr = 2.2 t/hr x Rs.1200/t = Rs.2640/hr. Electric centrifugal: 0.7 kW/TR x 500 = 350 kW x Rs.6.12 = Rs.2142/hr.
Difference = Rs.498/hr x 7000 hr = Rs.34.86 lakh a year in favour of the electric chiller, so the management's plan to replace the VAM is correct - i.e. DISAGREE with continuing to run the VAM.
The economics turned only because the free waste heat disappeared: a VAM wins when its heat is genuinely free (HRSG, waste heat), and loses the moment that steam has to be bought at Rs.1200/tonne. Auxiliary power is stated to be unchanged, so it cancels and can be left out.
📖 §1.8 AT&C losses in distribution (billing & collection efficiency)
11. A distribution company has taken initiatives to reduce Aggregate Technical & Commercial (AT&C) loss in their network. The energy supplied, received and revenue details are given below: Input energy = 60 MU; Metered Billed Energy = 43 MU; Average Billing = 3 MU; Amount Billed = Rs. 540 Million; Arrears collected = Rs. 80 Million; Amount received = Rs. 470 Million. a) Estimate the following (each carries 2.5 Marks): i) AT&C loss in % and revenue realized in Rs./kWh ii) Revenue loss per kWh and monthly loss, if the purchased energy cost is Rs. 8.10/kWh. b) List five measures to reduce commercial loss in the network (5 Marks)
Model answer: a)
Billing efficiency = (43 + 3) / 60 x 100 = 76.7 %
Collection efficiency = ((470 - 80) / 540) x 100 = 72.2 %
AT&C Loss = [1 - (Billing efficiency x Collection efficiency)] x 100
= [1 - (0.767 x 0.722)] x 100 = 44.62 %
Revenue realised per kWh = (470 - 80) / 60 = Rs. 6.5/kWh
Revenue loss per kWh = Rs. 8.10 - 6.5 = Rs. 1.6/kWh
Monthly revenue loss = 60 x 1.6 = Rs. 96 Million (Rs. 9,60,00,000/-)
b) (The paper prints only "Refer Guide Book No 3, Chapter 1, Page No 27". Measures from the 2014 BEE Book-3, Chapter 1:)
1. 100 % metering of all consumers and of all distribution transformers / feeders, with accurate, tamper-proof electronic meters.
2. Energy accounting and auditing at feeder and distribution-transformer level to locate high-loss pockets and pin-point theft.
3. Detection and prevention of theft / pilferage - removal of direct hooking on LT lines, use of aerial bunched conductors (ABC) and armoured service cables, regular raids and penalties.
4. Improvement of billing efficiency - elimination of unmetered supply and average / provisional billing, correction of faulty and stopped meters, spot billing and reduction of billing errors.
5. Improvement of collection efficiency - regular and timely bill distribution, computerised billing and collection, easy payment options, incentives for prompt payment, disconnection of chronic defaulters and recovery of arrears.
6. Consumer indexing and GIS mapping of consumers to feeders / distribution transformers so that energy input and energy billed can be reconciled.
Formula set to memorise: Billing efficiency = energy billed / energy input; Collection efficiency = amount collected (excluding arrears) / amount billed; AT&C loss % = [1 - (BE x CE)] x 100.
Working: BE = (43 + 3)/60 = 76.7%; CE = (470 - 80)/540 = 72.2%; AT&C = [1 - 0.767 x 0.722] x 100 = 44.6%. Revenue realised = (470 - 80)/60 MU = Rs.6.5/kWh, so the gap on a Rs.8.10/kWh purchase cost is Rs.1.6/kWh -> 60 MU x 1.6 = Rs.96 million.
The mark-loser is leaving the Rs.80 million of ARREARS in the collection figure - arrears belong to earlier billing periods and must be stripped out of both the numerator and the revenue-realised calculation.