Energy Efficiency in Electrical Utilities Available here with full solutions — 64 questions recovered from the 2025 exam:
Objective (1 mark)
50 of 50
Short (5 marks)
8 of 8
Long (10 marks)
6 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) — 50
📖 §7.2 Factors Affecting Performance – Fill Media Effects
1. The most influential component for cooling tower performance is:
Fill media
Drift eliminator
Casing
Fan motor
Answer: A) Fill media
Confirmed vs Book-3 §7.2 Fill Media Effects — Book: fill media is responsible for surface area, contact time and turbulence of heat exchange – i.e. the whole heat/mass transfer – and efficient fill directly cuts fan power and pumping head. Drift eliminator, casing and fan motor do not set the transfer area. Same as Book EOC Q8.
2. An equipment room measures 12 × 8 × 3.5 m. Ventilation required for 15 ACH is:
5060 m³/h
5020 m³/h
5040 m³/h
5080 m³/h
Answer: C) 5040 m³/h
Confirmed vs Book-3 §4.12 - Volume = 12×8×3.5 = 336 m³; ventilation = volume × ACH = 336 × 15 = 5040 m³/h. The trap is mis-multiplying the volume: 12 x 8 x 3.5 = 336 m3, and 336 x 15 ACH = 5040 m3/h exactly - the other options are arithmetic near-misses with no physical basis.
Confirmed vs Book-3 §8.2 — Luminous efficacy is 'the ratio of luminous flux emitted by a lamp to the power consumed by the lamp', unit lumens per Watt (lm/W). Candela is the unit of luminous intensity, luminance is brightness of a surface, and reflectance concerns reflected light — none is efficacy.
📖 §9.3 Waste heat recovery in DG sets — Table 9.5 energy balance for reciprocating engine
4. In a DG set, the component causing maximum energy loss is:
Coolant loss
Alternator loss
Radiation loss
Flue gas loss
Answer: A) Coolant loss
Corrected (was d) — Book-3 §9.3: Table 9.5 energy balance for a 500-kW diesel engine generator: electric power 35%, jacket (coolant) water 32%, exhaust heat 24%, radiated 9% (natural-gas engine: jacket 38%, exhaust 24%). So the jacket-cooling-water (coolant) stream is the single largest loss, ahead of flue gas. Flue gas (d) is tempting because the book calls the exhaust "more versatile" — it is at ~450 °C vs ~100 °C for cooling water — but it carries less energy (24% vs 32%). Alternator and radiation losses are small.
5. A 750 kVA transformer has 1200 W no-load loss and 7200 W full-load copper loss. At 60% load, total loss is:
4320 W
5520 W
7632 W
3792 W
Answer: D) 3792 W
Confirmed vs Book-3 §1.5 — P_TOTAL = P_NO-LOAD + (%Load/100)² × P_LOAD = 1200 + (0.6)² × 7200 = 1200 + 2592 = 3792 W.
Option (b) 5520 W scales the copper loss linearly (0.6 × 7200); the book is explicit that 'Copper loss varies with the square of the load current'.
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
6. Iron losses in an electric motor can be reduced by using
More copper and large conductors
Use of thinner gauge lower loss core steel
Use of low loss fan design
Optimised design and strict quality control
Answer: B) Use of thinner gauge lower loss core steel
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up — Table 2.2 attributes iron-loss reduction to thinner gauge, lower-loss core steel (less eddy current) plus a longer core to lower flux density. (a) more copper / larger conductors reduces stator I²R loss, not iron loss, and (c) low-loss fan design attacks friction & windage — both are the tempting confusions.
8. Energy savings by motor replacement can be worked out by:
KW output (ηold – ηnew)
KW output (ηnew – ηold)
KW output (1/ηold – 1/ηnew)
KW output (1/ηnew – 1/ηold)
Answer: C) KW output (1/ηold – 1/ηnew)
Confirmed vs Book-3 §2.5 Motor Selection — The book gives kW savings = kW output × [1/ηold − 1/ηnew]; savings arise from the difference in the INPUT power drawn for the same shaft output. (a)/(b), the plain difference of efficiencies, is the tempting wrong form — it has the wrong units and grossly under-states savings.
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
9. Stray losses in a motor are mainly caused by:
Leakage flux induced by load currents
Hysteresis and eddy currents
Frictional losses
Copper winding losses
Answer: A) Leakage flux induced by load currents
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up — Stray load losses are caused by leakage flux induced by the load currents in the laminations, vary as the square of load current and account for 4–5 % of total losses. (b) hysteresis and eddy currents are the core (iron) losses, a different, load-independent category — that is the usual confusion.
10. In a vapor compression refrigeration system, enthalpy changes occur across:
Compressor
Condenser
Evaporator
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §4.3 - Enthalpy changes in the evaporator (heat absorbed), compressor (work added) and condenser (heat rejected); only the expansion device is constant-enthalpy. Option (d) is right because the trap is picking a single component; only the expansion device is isenthalpic (Book §4.3: 'there is no heat loss or gain through the expansion device').
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
11. 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.
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme
12. Amorphous core transformers primarily reduce:
Load loss
No-load loss
Stray loss
None of the above
Answer: B) No-load loss
Confirmed vs Book-3 §1.5 Energy Efficient Transformers — an amorphous (metallic-glass) core gives 'expected reduction in core loss over conventional (Si Fe core) transformers…roughly around 70%'. Core loss = no-load loss.
Option (a) is wrong: load (copper) loss is fixed by winding resistance and is unaffected by the core material.
📖 §6.5 Efficient pumping system operation — Table 6.1 symptoms / best solutions
13. If a pump delivery valve is throttled to 30% of rated flow, best energy efficiency measure is:
Replacing the motor
Installing a larger impeller
Increasing pump speed
None of the above
Answer: D) None of the above
Confirmed vs Book-3 §6.5 — Table 6.1: a throttle-valve-controlled system signals an OVERSIZED pump; best solutions are trim impeller, smaller impeller, variable speed drive, two-speed drive or lower RPM — i.e. reduce output. A larger impeller or higher speed increases output and replacing the motor does nothing about the throttling loss, so none of the listed options is correct.
14. Calculate the FAD in CFM for an air compressor with a cylinder displacement of 150 CFM and volumetric efficiency of 90%:
165
135
150
None of the above
Answer: B) 135
Confirmed vs Book-3 §3.3 Compressor Efficiency — Volumetric efficiency = [FAD / compressor displacement] x 100, so FAD = 150 x 0.90 = 135 CFM.
Option (a) 165 CFM adds 10% instead of taking 90%; FAD is always LESS than the swept displacement because of clearance volume re-expansion and valve/leakage losses.
15. If pump speed is reduced to 2/3rd of its original speed, power consumption will:
Decrease by half
Decrease to one-fourth
Decrease to approx. 30% of original
Remains same
Answer: C) Decrease to approx. 30% of original
Confirmed vs Book-3 §6.5 — Affinity law P∝N³, so P₂/P₁ = (2/3)³ = 8/27 = 0.296 ≈ 30% of original power. 'Half' would need N₂/N₁ = 0.79; 'one-fourth' would be the head ratio if speed were halved (H∝N²), not the power ratio here.
16. At higher altitudes, for same FAD, air compressors:
Consume less power
Consume more power
Show no difference
Work without lubrication
Answer: B) Consume more power
Confirmed vs Book-3 §3.5 Efficient Operation — Compressors located at higher altitudes consume more power to achieve a particular delivery pressure than those at sea level, because the lower barometric pressure raises the compression ratio.
Table 3.6 shows relative volumetric efficiency at 7 bar falling from 100% at sea level to 87.0% at 2500 m, so (a) and (c) are contradicted.
17. In a 4-stroke diesel engine, fuel is injected during:
Induction stroke
Compression stroke
Ignition and Power stroke
Exhaust stroke
Answer: C) Ignition and Power stroke
Confirmed vs Book-3 §9.1 — The book names the third stroke "ignition and power stroke: fuel is injected while the valves are closed (fuel injection actually starts at the end of the previous stroke), the fuel ignites spontaneously and the piston is forced downwards". Only air is drawn in on the induction stroke and only air is compressed on the compression stroke (CI engine), so (a)/(b) are wrong; nothing is injected during exhaust.
Answer: C) Causes excessive heating and reduces life
Confirmed vs Book-3 §2.7 Voltage Unbalance — Voltage unbalance produces a current unbalance 6–10 times as large and an additional temperature rise of 2 × (% unbalance)²; insulation life halves for every 10 °C rise, so it is a leading cause of premature motor failure. (b) is the trap — the harm is overheating and derating, not a useful change in slip; the book recommends unbalance at motor terminals not exceed 1 %.
Confirmed vs Book-3 §2.9 Soft Starter — Soft starters ramp the voltage to give smooth acceleration, cutting the DOL inrush of about +600 % of run current and the associated mechanical stress. (d) is the tempting answer — the listed advantages are less mechanical stress, improved power factor, lower maximum demand and less maintenance, not improved running efficiency.
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
20. An air compressor is driven by an IE3 premium efficiency motor. Compared to an IE2 motor of the same rating, which of the following statements is most accurate?
The IE3 motor will always consume less power under all load conditions.
The IE3 motor achieves higher efficiency mainly by reducing copper and iron losses.
The IE3 motor has lower inrush current during starting compared to IE2.
The IE3 motor achieves efficiency by increasing slip.
Answer: B) The IE3 motor achieves higher efficiency mainly by reducing copper and iron losses.
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up — Design improvements in energy-efficient motors target the dominant loss groups: stator + rotor I²R (55–60 % of losses, reduced by more/larger copper conductors) and core losses (20–25 %, reduced by thinner low-loss silicon steel and a longer core). (c) is wrong — EEMs do not have lower inrush; (d) is wrong because EEMs have LOWER slip (about 1 % faster than standard motors).
Confirmed vs Book-3 §4.8 - Condenser scaling insulates the tubes, reducing heat transfer, raising condensing pressure/discharge pressure and increasing compressor energy use — all of the above. Each of (a), (b) and (c) is a consequence of the same fouling, so 'all of the above' is right; §4.8 quantifies it - a 0.8 mm scale build-up on condenser tubes can raise energy consumption by as much as 35%.
Confirmed vs Book-3 §5.3 — The book states a fan 'in a system with narrow ducts and multiple short-radius elbows' must work harder than 'in a system with larger ducts'; larger ducts lower velocity and friction losses, so system resistance falls. Option (d) is wrong: lower resistance means the fan operates at LOWER static pressure (and higher flow) on its curve.
📖 §6.5 Efficient pumping system operation — Table 6.1 symptoms / best solutions
23. Which of the following is a common symptom indicating that a pump is oversized?
High discharge pressure
Throttle valve-controlled systems
Low suction pressure
High motor power consumption
Answer: B) Throttle valve-controlled systems
Confirmed vs Book-3 §6.5 — Table 6.1 lists 'throttle valve-controlled systems' and 'bypass line (partially or completely) open' as symptoms whose likely reason is an oversized pump. High discharge pressure or high motor power are consequences, not the diagnostic symptom the book names; low suction pressure relates to NPSH/cavitation.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
24. Reducing the diameter of an impeller in a centrifugal pump will:
Increase head
Decrease head
No effect on head
Increase flow
Answer: B) Decrease head
Confirmed vs Book-3 §6.5 — Impeller-diameter relations: Q∝D, H∝D², P∝D³. Trimming reduces tip speed and 'lowers both the flow and pressure generated by the pump' (§6.6). Head decreases (with the square of the diameter ratio); it cannot increase and flow also falls, so option d is wrong.
25. The main function of fill media in a cooling tower is to:
Reduce drift losses
Increase water–air contact
Reduce fan noise
Filter suspended solids
Answer: B) Increase water–air contact
Confirmed vs Book-3 §7.1 Components – Fill — Book: fills 'facilitate heat transfer by maximising water and air contact' (splash or film type). Reducing drift is the drift eliminator's job, and fill neither reduces fan noise nor filters solids (side-stream filters do that).
📖 §7.5 Energy Saving Opportunities in Cooling Towers
26. Energy-saving opportunities in cooling towers include:
Optimizing fan blade angle seasonally
Maintaining correct water chemistry
Cleaning fill media regularly
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §7.5 — All three appear in the book's §7.5 list / §7.3: 'optimise cooling tower fan blade angle on a seasonal and/or load basis', correct water chemistry (cooling water treatment prevents scale/algae that block nozzles) and cleaning fill/nozzles. Hence (d).
27. Increasing chilled water leaving temperature in a centrifugal chiller:
Increases efficiency
Decreases efficiency
No effect
Increases refrigerant flow
Answer: A) Increases efficiency
Confirmed vs Book-3 §4.7 - Raising the evaporator (chilled-water leaving) temperature lifts the COP/efficiency — about 3% power saving per 1°C rise. Option (b) is the common reflex answer, but Figure 4.10 shows COP rising with leaving chilled-water temperature; the refrigerant flow (d) is not what the question asks about.
Answer: B) Distribute and control light from lamps
Confirmed vs Book-3 §8.2 — The luminaire 'distributes, filters or transforms the light emitted from one or more lamps' (i.e. controls it). It does not produce light (the lamp does), store energy, or change lamp wattage.
29. The Envelope Performance Factor (EPF) in ECBC is used to:
Compare energy efficiency of proposed and baseline building designs
Determine cooling tower sizing
Calculate lighting power density
Measure indoor air quality
Answer: A) Compare energy efficiency of proposed and baseline building designs
Confirmed vs Book-3 §10.5 — Under the Envelope Trade-off approach, the Envelope Performance Factor is calculated for the proposed design and the baseline design; the proposed building's EPF must be equal to or better than the baseline. It has nothing to do with cooling-tower sizing, LPD or indoor air quality.
30. In a cooling tower, if any three of the four parameters: heat load, range, approach, and wet-bulb temperature, are kept constant, the required tower size will vary
Directly with the heat load
Inversely with the range
Inversely with the approach
All the above
Answer: D) All the above
Confirmed vs Book-3 §7.2 Factors that affect cooling tower size — Book sidebar: when three of heat load, range, approach and WBT are held constant, tower size varies directly with heat load, inversely with range, inversely with approach and inversely with entering WBT. All three listed relations are book statements → (d).
📖 §6.6 Flow control strategies — pump control by varying speed / VSDs & VFDs
31. A centrifugal pump has BEP efficiency of 65%. At shut-off head, efficiency is:
0%
65%
50%
30%
Answer: A) 0%
Confirmed vs Book-3 §6.6 — At shut-off head flow is zero, so hydraulic power Q·H·ρ·g is zero and pump efficiency = hydraulic/shaft power = 0%. The book describes this condition (Fig 6.16): 'pump efficiency and flow rate are zero and with energy still being input to the liquid, the pump becomes a water heater'. The 65% BEP figure is a distractor.
Answer: A) No reactive power is drawn from the supply
Confirmed vs Book-3 §1.4 — PF = kW/kVA = cosφ. At unity PF the angle φ = 0, so the kVAr side of the power triangle is zero: no reactive power is exchanged with the supply.
Options (b)/(c) describe leading and lagging conditions, both of which imply a non-zero reactive component.
📖 §8.6(g) Lighting controllers / §8.7 Energy efficient lighting controls
33. Examples of lighting controls include:
Dimmer switches
Timers
Photo-sensors
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §8.6(g) — 'Lighting controllers ... includes dimmers, motion & occupancy sensors, photosensors and timers.' Dimmer switches, timers and photo-sensors are all named, so 'all of the above'.
34. Which type of compressed air dryer consumes the least power for capacities higher than 250 CFM?
Refrigeration type
Blower reactivated type
Heat of compression type
Heatless purge type
Answer: C) Heat of compression type
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 the desiccant with the compressor's own 135°C discharge heat, so operating cost is zero to very minimal.
It is offered from 400 to 5000 cfm, which covers the >250 CFM duty asked about; the refrigerant dryer uses less power than the desiccant types but cannot be compared here as it only reaches -20°C dew point.
📖 §9.4 Energy performance assessment — specific fuel consumption (L/kWh)
35. A DG set operates at 1250 kVA, 0.8 PF, with specific fuel consumption of 0.23 L/kWh. Quantity of fuel used is_____________.
175 L/h
230 L/h
250 L/h
300 L/h
Answer: B) 230 L/h
Confirmed vs Book-3 §9.4 — Real power = 1250 kVA × 0.8 = 1000 kW; fuel = 1000 kW × 0.23 L/kWh = 230 L/h. Option (c) 250 L/h would need SFC 0.25; (a) 175 L/h is the 1000 kVA × 0.7 PF × 0.25 case from another exam. Always convert kVA to kW with PF before applying SFC.
36. In a UPS, DC to AC conversion is carried out by:
Converter
Charger
Battery
Inverter
Answer: D) Inverter
Confirmed vs Book-3 §1.10 — the UPS is listed as a non-linear load precisely because its output stage is an inverter, the device that converts DC (battery/DC link) into AC.
Option (a) 'converter' (rectifier) does the opposite — AC to DC — and the battery only stores energy; it cannot change the waveform.
37. If a plant receives 96 Million Units (MU) with a T&D efficiency of 80%, generation required is:
101.2 MU
76.9 MU
120 MU
68.1 MU
Answer: C) 120 MU
Confirmed vs Book-3 §1.1 Cascade Efficiency — energy delivered = energy generated × η. So generation = 96/0.80 = 120 MU.
Option (b) 76.9 MU multiplies instead of divides (96 × 0.80); losses must be added on top of the delivered units, so generation is always larger than the units received.
📖 §1.1 (a.c. fundamentals; see also Book-3 Ch-2 §2.2 — Ns = 120f/P)
38. Synchronous speed of a motor is inversely proportional to:
Number of poles
Frequency
Voltage
Temperature
Answer: A) Number of poles
Confirmed vs Book-3 §1.1/Ch-2 — synchronous speed Ns = 120f/P, so for a fixed supply frequency the speed is inversely proportional to the number of poles (4-pole → 1500 rpm, 2-pole → 3000 rpm at 50 Hz).
Option (b) is wrong because speed is directly (not inversely) proportional to frequency.
39. If 27,216 kcal of heat is removed per hour, refrigeration tonnage is:
10 TR
7 TR
11 TR
None of the above
Answer: D) None of the above
Confirmed vs Book-3 §4.7 - TR = 27,216 / 3024 = 9.0 TR exactly (1 TR = 3024 kcal/h). Since 9 TR is not listed, the answer is 'None of the above'. 27,216/3024 = 9.0 TR exactly, and 9 TR is not offered, so (d) is correct; option (a) 10 TR is the trap for anyone who rounds carelessly or uses 3000 kcal/h per TR.
40. A 4 pole 50 Hz induction motor is running at 1470 rpm. What is the slip value?
20%
2%
30%
40%
Answer: B) 2%
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 50 / 4 = 1500 rpm; slip = (1500 − 1470)/1500 × 100 = 2 %. (a) 20 % is a decimal slip mis-read; normal full-load slip of a squirrel cage motor is only 1–3 %.
41. In a vapor compression refrigeration system, the component where the refrigerant changes its phase from vapor to liquid is
compressor
condenser
expansion valve
evaporator
Answer: B) condenser
Confirmed vs Book-3 §4.3 - In the condenser the high-pressure refrigerant vapour rejects heat and condenses to liquid (vapour → liquid phase change). The compressor (a) only superheats the vapour, the expansion valve (c) throttles liquid, and the evaporator (d) does the opposite phase change (liquid to vapour); condensation happens in stage 3-3b of Figure 4.4.
43. Which of the following flow controls in a fan system will change the system resistance curve:
Inlet guide vane
speed change with variable frequency drive
speed change with hydraulic coupling
discharge damper
Answer: D) discharge damper
Confirmed vs Book-3 §5.5 — Dampers change volume 'by adding or removing system resistance', i.e. they shift the SYSTEM resistance curve (SC₁→SC₂ in Fig 5.7). Inlet guide vanes change the FAN curve characteristics; speed changes by VFD or hydraulic coupling (b, c) move the fan to a new fan curve while the system curve is unchanged.
44. When the dew point temperature is equal to the air temperature then the relative humidity is
0%
50%
100%
Unpredictable
Answer: C) 100%
Confirmed vs Book-3 §4.2 - When DBT equals the dew point (and wet-bulb) temperature, the air is saturated, so relative humidity = 100%. Options (a) and (b) describe unsaturated air; when the dry bulb falls to the dew point the air holds all the moisture it can at that temperature, which is 100% RH by definition.
📖 §10.2 Building definition — Energy Conservation (Amendment) Act 2010
45. 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.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
46. A pump with 200 mm impeller is delivering a flow of 120 m3/hr. If the flow is to be reduced to 100 m3/hr by trimming the impeller, what should be the approximate impeller size ?
60 mm
240 mm
167 mm
145 mm
Answer: C) 167 mm
Confirmed vs Book-3 §6.5 — Q∝D, so D₂ = 200×(100/120) = 166.7 ≈ 167 mm. 60 mm applies the ratio inversely; 240 mm is an enlargement; 145 mm would give only 87 m³/hr.
47. Which gas is used as refrigerant both in vapour compression and vapour absorption systems
Lithium Bromide
Water
HFC 134A
Ammonia
Answer: D) Ammonia
Confirmed vs Book-3 §4.3 - Ammonia is used as the refrigerant in both vapour compression and vapour absorption (with water) systems. Lithium bromide (a) is the absorbent and water (b) the refrigerant only in LiBr machines, while HFC-134a (c) is compression-only; ammonia is used as the refrigerant in both types.
📖 §1.4 Power factor improvement & benefits (I2R loss vs PF)
48. The percentage reduction in distribution losses when tail end power factor is raised from 0.8 to 0.95 is ________.
29%
15.8%
71%
none of the above
Answer: A) 29%
% reduction in distribution loss = [1 - (PF_old/PF_new)^2] x 100 = [1 - (0.8/0.95)^2] x 100 = [1 - 0.709] x 100 = 29.1% ~ 29%.
The physics: for a fixed kW the line current is inversely proportional to PF, and loss goes as I^2, so loss goes as 1/PF^2.
The 15.8% distractor is the un-squared version (1 - 0.8/0.95). Square the ratio, always.
49. Which of the following happens to air when it is cooled through evaporation process in an air washer?
Humidity ratio of the air decreases.
Dry Bulb Temp of air decreases.
Dry Bulb Temp of air increases.
Enthalpy of outlet is air is less than enthalpy of inlet air.
Answer: B) Dry Bulb Temp of air decreases.
In an air washer the water evaporating into the air stream takes its latent heat from the air itself: dry bulb temperature falls, humidity ratio rises, enthalpy and wet-bulb temperature stay nearly constant.
So (a) is backwards (humidity rises), (c) is backwards (DBT falls) and (d) is wrong (the process is essentially adiabatic, so enthalpy is unchanged, not reduced).
The lowest temperature achievable is the air's wet-bulb temperature - the reason evaporative cooling works splendidly in dry climates and hardly at all in humid ones.
50. 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.
1. A commercial training hall with dimensions 18 m × 12 m is being planned. Calculate the number of 18 W LED lamps, each providing 1800 lumens, required to achieve an illuminance level of 300 Lux. The lamps will be installed at a height of 3 meters from the working plane. The utilisation factor (UF) of the system is 0.70, and the light loss factor (LLF) is 0.80.
Model answer: Area of room (A): 18 × 12 = 216 m²
Total lumens required (Φ_total): Φ_total = E × A = 300 × 216 = 64800 lumens
Effective lumens per lamp (Φ_lamp_effective): Φ_lamp_effective = Lumen Output × UF × LLF = 1800 × 0.70 × 0.80 = 1008 lumens
Number of lamps required (N): N = Φ_total / Φ_lamp_effective = 64800 / 1008 = 64.3 ≈ 65 lamps
Book-3 §8.5 lumen method: N = (E × A)/(F × UF × LLF) = (300 × 216)/(1800 × 0.70 × 0.80) = 64,800/1008 = 64.3 → 65 lamps. Height (3 m) is only needed for the room index/UF, which is already given as 0.70.
2. A private power distribution company has implemented new digital metering and billing systems to improve efficiency in a residential zone. After six months of operation, the following data was recorded:
• Input energy to the system = 75 MU
• Metered billed energy = 56 MU
• Unmetered average billing = 4 MU
• Amount billed = ₹680 million
• Total amount received = ₹600 million
• Arrears collected = ₹90 million
• Purchased energy cost = ₹8.50 per kWh
i) Estimate the Aggregate Technical and Commercial (AT&C) loss (%) and the revenue realized per kWh (4 Marks)
ii) Calculate the revenue loss per kWh to the company due to AT&C loss (1 Mark)
Model answer: Input Energy = 75 MU = 75,000,000 kWh
Metered Billed Energy = 56 MU
Unmetered Average Billing = 4 MU
Total Energy Billed = 56 + 4 = 60 MU
Amount Billed = ₹680 million; Arrears Collected = ₹90 million; Amount Received = ₹600 million; Purchased Energy Cost = ₹8.50/kWh
i) Billing Efficiency = (60 / 75) × 100 = 80.0%
Collection Efficiency = ((600 - 90) / 680) × 100 = (510 / 680) × 100 = 75.0%
AT&C Loss (%) = 1 - (Billing Efficiency × Collection Efficiency) = 1 - (0.80 × 0.75) = 1 - 0.60 = 40.0%
Revenue Realized per kWh = (600 - 90) / 75 = 510 / 75 = ₹6.80/kWh
ii) Revenue Loss = ₹8.50 - ₹6.80 = ₹1.70/kWh
Final Answers: i) AT&C Loss = 40.0%, Revenue Realized = ₹6.80/kWh; ii) Revenue Loss per kWh = ₹1.70/kWh
AT&C loss = 1 − (Billing Efficiency × Collection Efficiency); revenue realized = net amount received / input energy.
📖 Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
3. List five energy saving measures in compressed air system.
Model answer: Any five from the Book-3 §3.7 checklist, with the book's own figures:
1. Carry out periodic leak tests and arrest leaks — leakage of 40-50% is not uncommon; leakage % = load time/(load+unload) x 100.
2. Reduce compressor delivery pressure to the minimum the plant needs — a 1 bar reduction saves 6-10% of input power (Table 3.9).
3. Ensure cool, clean, dry intake air, drawn from outside if the compressor room is hot — every 4°C rise in inlet temperature costs about 1% more power, and every 250 mmWC pressure drop across a choked filter costs about 2%.
4. Minimise unloaded running: if demand is below 50% of capacity, change over to a smaller compressor, reduce speed by trimming the motor pulley, or retrofit a VSD on machines above ~100 kW (unloading draws up to 30% of full-load power).
5. Avoid misuse of compressed air (body/floor cleaning, agitation, drying) and replace pneumatic conveying and pneumatic tools with blowers or electric tools where safe — pneumatic conveying uses about 8 times and pneumatic tools about 20 times the energy.
Other valid answers: size piping generously and use a ring main (0.3 bar drop in the header, 0.5 bar in distribution); clean inter-coolers and after-coolers; recover heat from hot compressed air; keep the load-unload pressure band narrow; sequence multiple compressors so only one small machine modulates; do periodic FAD tests.
Confirmed vs Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System — Any five of the printed §3.7 checklist items earn full marks; the strongest answers quote the book's numbers (4 degC/1%, 250 mmWC/2%, 1 bar/6-10%, up to 30% unload power).
4. Match the Following:
1. Solar Heat Gain Coefficient (SHGC)
2. U-value
3. HVAC System Efficiency
4. Cool Roof
5. Thermal Bridging
a. Coefficient of Performance (COP)
b. Solar Reflectance
c. Fenestration Heat Gain
d. Building Insulation
e. Conductive path for unwanted heat transfer
Model answer: Answer Key:
1 → c. Fenestration Heat Gain
2 → d. Building Insulation
3 → a. Coefficient of Performance (COP)
4 → b. Solar Reflectance
5 → e. Conductive path for unwanted heat transfer
Matching per Book-3 §10.5: SHGC → fenestration (solar) heat gain; U-value → insulation/heat flow of the envelope; cool roof → high solar reflectance & emittance; thermal bridging → conductive path of unwanted heat; HVAC efficiency → COP (§10.6).
5. A steel manufacturing facility is powered by a 3-phase, 6.6 kV, 50 Hz supply and operates the following electrical loads: An electric arc furnace consumes 1.2 MW at a lagging power factor of 0.65, a bank of induction motors for rolling operations consumes 800 kW at a 0.80 lagging power factor, and the lighting and instrumentation systems consume 100 kW at unity power factor. Due to utility regulations, the overall plant power factor must be improved to 0.95 lagging. A capacitor bank will be installed for compensation. As an energy auditor evaluate the following:
a. Total active power consumption. (1 Mark)
b. Total initial apparent power drawn by the facility. (1 Mark)
c. The operating power factor. (1 Mark)
d. Determine the total reactive power required to achieve the desired power factor. (2 Mark)
Model answer: Load-wise breakdown (kVAr = kW x tan(cos^-1 PF)):
Arc furnace: 1200 kW, PF 0.65 -> kVA = 1200/0.65 = 1846 kVA, kVAr = 1846 x sin(49.5deg) = 1403 kVAr
Induction motors: 800 kW, PF 0.80 -> kVA = 1000 kVA, kVAr = 600 kVAr
Lighting & instrumentation: 100 kW, PF 1.0 -> kVA = 100 kVA, kVAr = 0
a) Total active power = 1200 + 800 + 100 = 2100 kW
b) kVA must be added VECTORIALLY, not arithmetically: total kVAr = 1403 + 600 + 0 = 2003 kVAr, so total kVA = sqrt(2100^2 + 2003^2) = 2902 kVA
c) Operating power factor = kW/kVA = 2100/2902 = 0.724 lag
d) kVAr required = kW[tan(cos^-1 0.724) - tan(cos^-1 0.95)] = 2100 x (0.9538 - 0.3287) = 1313 kVAr; equivalently 2003 - (2100 x 0.3287) = 1313 kVAr.
(Capacitor bank of about 1300-1350 kVAr, preferably switched in steps through an APFC because the arc-furnace PF swings over the melting cycle - Book-3 Sec.1.4.)
Sum kW and kVAr per load; operating PF = ΣkW/ΣkVA; compensation kVAr = kW(tanφ1 − tanφ2).
6. Determine the difference in heat rejected in kCal/TR to the cooling tower for two different types of air conditioning system operating at same capacity.
Parameter — Centrifugal chiller / VAM:
Chilled water flow (m³/h): - / 180
Condenser water flow (m³/h): - / 340
Chiller inlet temp (°C): 13.0 / 14.6
Condenser water inlet temp (°C): - / 33.5
Chiller outlet temp (°C): 7.7 / 9.0
Condenser water outlet temp (°C): - / 39.1
Specific power consumption (kW/TR): 0.6 / -
📖 §5.3 Fan laws; §5.6 Volume calculation (Q = V × A)
7. An energy audit in an industrial unit revealed a fan directly coupled with motor was operating at 37 Hz through VFD for 500 hours/month and supplying air through a 150 mm diameter duct. The fan is designed to deliver an air flow of 1300 m³/h with a rated input power of 3 kW at 50 Hz. Calculate the air velocity and annual energy savings ignoring the motor losses.
Model answer: Given: Design air flow = 1300 m³/h at 50 Hz; Operating frequency = 37 Hz; Power at 50 Hz = 3 kW; Operating hours = 500 hours/month; Duct diameter = 150 mm = 0.15 m
Flow rate at 37 Hz: Q2 = 1300 × (37/50) = 962 m³/h = 962/3600 = 0.2672 m³/s
Duct area: A = π/4 × (0.15)² = 0.01767 m²
Air velocity: V = Q/A = 0.2672 / 0.01767 = 15.12 m/s
Power at 37 Hz: P2 = 3 × (37/50)³ = 3 × 0.405 = 1.215 kW
Monthly energy savings: (3 − 1.215) × 500 = 892.5 kWh/month
Annual energy savings: 892.5 × 12 = 10710 kWh/year
Fan laws with speed ∝ frequency: Q₂ = Q₁ × (f₂/f₁), P₂ = P₁ × (f₂/f₁)³; velocity = Q/A with A = (π/4)D²; savings = (P₁ − P₂) × hours. Note the 6000 h version of this question (Sep 2019) gives 10,680 kWh/yr.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); total head = static lift + suction lift (h_d − h_s)
8. A pump is used to fill a rectangular overhead tank measuring 5 m × 3.5 m with a height of 10 m. The inlet pipe to the tank is positioned at a height of 25 m above ground level. The following additional data is available: The pump draws water from an underground sump situated 4 meters below the pump level and delivers it to a tank whose overflow line is positioned 8 meters above the tank bottom. The motor driving the pump draws 7.5 kW of power. The operating efficiencies of the motor and the pump are 90% and 70% respectively. Calculate the time taken by the pump to fill the tank up to the overflow level.
Model answer: Step 1: Volume of water filled = 5 × 3.5 × 8 = 140 m³; Mass of water = 140 × 1000 = 140000 kg
Step 2: Total head (H) = 25 + 4 = 29 m
Step 3: Shaft Power = 7.5 × 0.90 = 6.75 kW
Step 4: Water Power = 6.75 × 0.70 = 4.725 kW = 4725 W
Step 5: Time taken:
Pump efficiency = Mass flow × g × Head / Shaft Power
0.7 = mass flow × 9.81 × 29 / (6.75 × 1000)
Mass flow = 16.6 kg/sec = 59791 kg/hr = 59.79 m³/hr
Time = 140 / 59.79 = 2.34 hrs = 140.5 minutes
Hydraulic power = ṁ·g·H; derive mass flow from pump efficiency, then time = volume/flow.
1. a) In a large-scale steel manufacturing facility, a cooling tower is used to reject heat from continuous casting operations. The circulating water flow rate is 2000 m³/hr. The cooling tower is currently operating at a Cycles of Concentration (COC) of 3. The evaporation loss is estimated at 1.0% of the circulating flow, and the drift loss is 0.1% of the circulating flow. The facility is planning to improve the COC from 3 to 6 through advanced water treatment. Evaluate the following:
i. Calculate the make-up water requirement at the current COC of 3.
ii. Calculate the revised make-up water requirement if the COC is increased to 6.
iii. Estimate the total water savings per day.
iv. Discuss one limitation or risk associated with increasing the COC.
b) True or False:
i. Cooling towers primarily reject heat through evaporative cooling.
ii. The approach temperature in a cooling tower is the difference between the hot water temperature and the ambient dry bulb temperature.
iii. Blowdown in a cooling tower is required to prevent the build-up of dissolved solids.
iv. Drift losses in a cooling tower refer to water carried away with the exhaust air.
v. Cooling tower effectiveness improves with higher approach temperatures.
vi. Cycles of concentration in a cooling tower relate to how many times the water is reused before discharge.
Model answer: a) Given: Circulating Water Flow (CWF) = 2000 m³/hr; Initial COC = 3; Final COC = 6; Evaporation Loss (E) = 1% of 2000 = 20 m³/hr; Drift Loss (D) = 0.1% of 2000 = 2 m³/hr
i) Current Make-up at COC = 3: B = 20 / (3 − 1) = 10 m³/hr; Make-up = E + D + B = 20 + 2 + 10 = 32 m³/hr
ii) Revised Make-up at COC = 6: B = 20 / (6 − 1) = 4 m³/hr; Make-up = 20 + 2 + 4 = 26 m³/hr
iii) Hourly Savings = 32 − 26 = 6 m³/hr; Daily Savings = 6 × 24 = 144 m³/day
iv) Increasing COC can lead to higher concentrations of dissolved solids in the water, which may cause scaling, corrosion, and microbiological fouling in the system. Effective water treatment and frequent monitoring are necessary to avoid operational issues.
b) i) True; ii) False (It is the difference between the cold-water temperature and the wet bulb temperature.); iii) True; iv) True; v) False (Lower approach means better effectiveness.); vi) True
Blowdown B = E/(COC−1); make-up = E + D + B; higher COC reduces blowdown and make-up.
📖 §9.5 Solved example (a) — maximum power factor; §9.2 unbalanced load ≤10%; §9.1 turbocharger; §9.3 power factor & losses
2. a) A manufacturing plant operates a 180 kVA diesel generator set rated at 0.8 lagging PF. The prime mover is a diesel engine rated 240 BHP. The alternator has total losses (including exciter power) of 5.44 kW. Assume no derating for site conditions. The generator is required to supply a mixed industrial load at its full kVA rating. The plant manager wishes to improve system efficiency by operating at a higher power factor. The diesel engine operates at a brake thermal efficiency of 32% when loaded near its rated capacity. The calorific value of the diesel fuel is 10,500 kCal/kg, and the specific gravity of the fuel is 0.85. Calculate the following:
i) Maximum power factor that can be maintained at full kVA load without exceeding the engine capacity. (3 Marks)
ii) Corresponding diesel fuel consumption (litres per hour) at this maximum power factor. (2 Marks)
b) True or False:
i) Improving power factor of the load on a DG set reduces the apparent power drawn and increases the system's overall fuel efficiency.
ii) Alternator losses are independent of the load power factor.
iii) Turbocharger in a diesel engine helps to reduce engine noise.
iv) A diesel generator set must always be operated at unity power factor for maximum efficiency.
v) DG sets are designed to handle unbalanced load between phases to 25% of their capacity.
Model answer: a) i) Convert BHP to kW (shaft power): 240 BHP × 0.746 = 179.04 kW; Net electrical power available = 179.04 − 5.44 = 173.6 kW; At full load (180 kVA), maximum PF = Real Power / Apparent Power = 173.6 / 180 = 0.964
ii) Thermal Input Required = Electrical Output / Efficiency = 179.04 / 0.32 = 559.5 kW; 1 kg diesel = 10,500 kcal = 10,500 × 4.1868 = 43,961.4 kJ/kg; Fuel consumption (kg/hr) = (559.5 × 3600) / 43961.4 = 45.8 kg/hr; In litres per hour = 45.8 / 0.85 = 53.88 L/hr
b) i) True; ii) False; iii) False; iv) False; v) False
Max PF = (shaft kW − alternator loss)/kVA; fuel = thermal input ÷ CV, converted to litres via density.
3. A distribution company (DISCOM) plans to implement a comprehensive Demand Side Management (DSM) initiative to reduce its peak load and overall energy procurement cost. The program targets Residential consumers (LED replacement) and Industrial consumers (load shifting). The DISCOM supplies electricity to 10,000 households, each using 4 CFL bulbs (30 W each), used for 5 hours per day during evening peak hours. The DISCOM replaces each CFL bulb with a 9 W LED bulb. Procurement cost of each LED bulb is ₹100, provided to consumers at a subsidized rate of ₹70/bulb. Administrative cost per household is ₹10. DISCOM also serves 50 industrial consumers, each with a shiftable evening load of 100 kW, used from 6 PM to 10 PM. DISCOM incentivizes them to shift load from 10 PM to 2 AM by offering ₹2 per kWh shifted. Eighty percent of industrial consumers agree. Power purchase cost: Evening Peak (5–10 PM) ₹7/kWh; Late Night (10 PM–6 AM) ₹3/kWh. Calculate:
a) For residential consumers: i) total daily energy savings (kWh/day) from LED replacement (2 Marks); ii) daily cost savings for DISCOM (1 Mark); iii) total one-time cost to DISCOM including subsidies and administrative costs (2 Marks); iv) simple payback period in days (1 Mark).
b) For industrial consumers: i) total energy shifted in kWh/day (1 Mark); ii) net daily savings for DISCOM considering power cost reduction and incentive payout (1 Mark).
c) Estimate the carbon emission avoidance due to above two DSM activity, if emission factor of grid electricity is 0.716 tCO2/MWh (2 Marks).
Model answer: a) Residential (LED replacement)
i) Daily energy saving = 10,000 households x 4 lamps x 5 h x (30 - 9) W / 1000 = 4200 kWh/day
ii) Daily cost saving to DISCOM (all in the 5-10 PM peak block at Rs.7/kWh) = 4200 x 7 = Rs.29,400/day
iii) One-time cost = subsidy 10,000 x 4 x (100 - 70) = Rs.12,00,000 + administrative 10,000 x 10 = Rs.1,00,000 -> Rs.13,00,000
iv) Simple payback = 13,00,000 / 29,400 = 44.2 days (about 44 days)
b) Industrial (load shifting)
i) Energy shifted = 50 consumers x 100 kW x 4 h x 0.80 = 16,000 kWh/day
ii) Net daily saving = 16,000 x (7 - 3 - 2) = Rs.32,000/day (Rs.4/kWh avoided cost less Rs.2/kWh incentive)
c) Carbon avoidance: only the LED programme saves ENERGY (load shifting merely re-times it, so it avoids peak cost but not kWh). CO2 avoided = 4200 kWh/day x 0.716 tCO2/MWh / 1000 = 3.0 tCO2/day, i.e. about 1098 tCO2/year.
This is peak clipping + conservation in Book-3 Sec.1.9 terms - the classic DSM combination of an end-use efficiency measure with a load-shifting tariff incentive.
LED savings = households×bulbs×hrs×ΔW; payback = one-time cost / annual saving; load-shift saving = energy×(peak − offpeak − incentive); CO2 = energy saved × emission factor.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³); motor part-loading from Book-3 Ch2
4. A) A clear water pump with rated flow of 125 m³/hr, head 55 m at rated speed of 1460 rpm and 79% efficiency supplies clarified water to a residential colony's water treatment facility. The daily water requirement is 3000 m³. The pump is directly coupled and driven by a three phase 50 Hp, 415 V, 64A, 0.9 pf, 1460 rpm induction motor with 90.5% full load efficiency. During an internal energy audit, the motor is found to operate with only 65% loading at pump rated conditions. The plant considers replacing the standard motor with a 30 kW IE3 motor. Operating parameters before and after motor replacement:
Flow (m³/hr): 130 / ?
Head (m): 52 / 51
Supply Voltage (V): 415 / 415
Current (Amp): 42 / 39
Power Factor: 0.9 / 0.92
Motor Eff (%): 0.88 / 0.932
The slip of the new IE3 motor has decreased by 20 rpm. Validate the savings, calculate:
i) % Loading of motor after replacement. (1 Mark)
ii) Flow after replacing the standard motor with 30 kW IE3 motor. (1 Mark)
iii) Operating Pump Efficiency before and after motor replacement. (2 Marks)
iv) Daily energy saving during operation due to motor replacement. (1 Mark)
B) Mark True/False:
i) Totally enclosed, fan cooled (TEFC) motors are less efficient than screen-protected, drip-proof (SPDP) motors.
ii) Stray loss in induction motors is inversely proportional to load current.
iii) As per BIS standard, the motor output should not be affected with voltage variation up to +/- 6%.
iv) Motor life doubles for each 10°C reduction in operating temperature.
v) Starting torque of energy efficient motors is higher than standard motors.
Model answer: A) i) Loading of the IE3 Motor = (1.732 × 0.415 × 39 × 0.92 × 0.932) / 30 = 80.12%
ii) Flow after replacement = 130 m³/hr × 1480/1460 = 131.8 m³/hr
iii) Operating Pump Efficiency:
Power Consumption before replacement = 1.732 × 0.415 × 42 × 0.9 = 27.17 kW
Power Consumption after replacement = 1.732 × 0.415 × 39 × 0.92 = 25.79 kW
Before replacement = [(130/3600) × 52 × 9.81] / (27.17 × 0.88) = 77%
After replacement = [(131.8/3600) × 51 × 9.81] / (25.79 × 0.932) = 76.2%
Daily Operating Hour before = 3000/130 = 23.08 Hrs; after = 3000/131.8 = 22.77 Hrs
iv) Daily Energy Savings = (23.08 × 27.17) − (25.79 × 22.77) = 39.9 kWh
B) i) False; ii) False; iii) True; iv) True; v) False
Motor loading = √3·V·I·pf·η / rating; flow scales with speed (affinity); pump η = hydraulic/electrical input; savings from input power × operating hours difference.
📖 §10.14 Star rating (EPI, built-up area) · §10.8 LPD · Ch9 DG set efficiency
5. A Commercial Office building accommodates two government departments. Total employees = 250, of which 70% average present at any time. The building is operational 6 days a week, 10 working Hrs a day. Supply through a 33 kV feeder, stepped down to 415 V. No separate parking, lawn, internal roads. Building information sheet (Annual Data April 24-March 25):
1. Contract Demand (kW): 130
2. Installed capacity DG Set(s) (kVA): 160
3a. Annual Electricity Consumption purchased from Utilities (kWh): 105753
3b. Annual Electricity Consumption through DG Set(s) (kWh): 2136
4a. Annual Cost of Electricity purchased from Utilities (Rs.): 1043557
4b. Annual Cost of Electricity generated through DG Set(s) (Rs.): 54405
5. Built Up Area (sq.m): 3591.96; Conditioned Area (sq.m): 2155.18
6. Installed capacity of Chiller (TR): 137.5
7. Installed lighting load (kW): 8.11
8. Office Appliances (kW): 11.0
9. Other Loads (kW): 12.5
10. HSD Consumption in DG (GCV 10800 Kcal/kg and density 0.85): 585 Litres
Calculate:
a. Total electricity consumed by the building and average electricity unit cost. (2 Mark)
b. EPI of the building for past one year. Recommend the appropriate BEE star rating if the bandwidth of the EPI ranges between 150-50 kWh/sq.m/year. (3 Marks)
c. Design diversity factor of the building, if design EER of the chiller is 3.5 and recorded maximum demand is 75% of the contract demand. (2 Marks)
d. Estimate the overall operating efficiency of the DG set. (2 Marks)
e. Calculate the lighting power density. (1 Mark)
Model answer: a. Total electricity consumed: 105753 + 2136 = 107889 kWh; Average unit cost = (1043557 + 54405) / 107889 = Rs. 10.18 per kWh
b. EPI = 107889 / 3591.96 = 30 kWh/m²/year; the EPI is below the bandwidth for star rating, hence 5 Star rated.
c. Diversity Factor = Maximum Demand / Connected Load; Maximum Demand = 130 × 0.75 = 97.5 kW; Connected Load = Lighting + appliances + others + AC = 8.11 + 11.0 + 12.5 + (137.5 × 3024/860/3.5) = 169.75 kW; Diversity Factor = 97.5/169.75 = 0.57 (or 169.75/97.5 = 1.74)
d. Overall DG Set efficiency = (2136 × 860) / (585 × 0.85 × 10800) = 34.2%
e. Lighting Power Density = 8.11 × 1000 / 3591.96 = 2.25 W/m²
EPI = annual energy ÷ built-up area (§10.14); DG η = kWh × 860 ÷ (litres × density × GCV); diversity factor = max demand ÷ connected load; LPD = installed lighting W ÷ area (§10.8). Chiller kW = TR × 3024 ÷ 860 ÷ EER.
6. A 5-star business hotel operates a centralized HVAC system round the clock. Only one chiller operates at a time, the other on standby. Two centrifugal chillers, each rated 250 TR, with EER varying with load: at 85% load (212.5 TR) EER 5.2 for 180 days; at 60% load (150 TR) EER 4.6 for 120 days; at 40% load (100 TR) EER 3.9 for 65 days. No change in EER above 85% load; assume chiller motor efficiency 90% at all loading conditions. During chiller operation, two pumps run in parallel at an 80% load factor, consuming a total of 19.7 kW; two cooling tower fans operate continuously with power consumption of 5.89 kW. Both pumps and fans function 24 hours a day, with overall efficiencies of 75% and 70% respectively. Electricity tariff is ₹6.5 per kWh. Evaluate:
a. The total annual energy consumption (in MWh) and cost of the HVAC system, considering part-load EERs and auxiliary loads. (4 Marks)
b. Heat removal by condenser in (TR) at different loads. (3 Marks)
c. The hotel is planning to use the chiller partially as a heat pump by mounting a plate heat exchanger in series between the compressor and condenser (desuperheater for partial heat recovery) for producing hot water. The heat recovery can be only 20% of the condenser heat discharge. If the hot water requirement is 2000 litres/hr with 10°C temperature rise, evaluate whether the hot water requirement can be met at 40% loading conditions. (3 Marks)
Model answer: a. Energy Consumed by Chiller (Cooling Load kW = TR×3024/860; Input Power = Load kW/EER; MWh = InputPower×Days×24/1000):
85%: 212.5 TR, EER 5.2 → 747.21 kW load, 143.69 kW input, 180 days → 620.8 MWh
60%: 150 TR, EER 4.6 → 527.44 kW load, 114.66 kW input, 120 days → 330.2 MWh
40%: 100 TR, EER 3.9 → 351.63 kW load, 90.16 kW input, 65 days → 140.7 MWh
Energy Consumed by auxiliaries = (19.7 + 5.89) × 24 × 365 / 1000 = 224.5 MWh
Total Annual Energy Consumption = 620.8 + 330.2 + 140.7 + 224.5 = 1316 MWh
Annual Cost = 1316000 × 6.5 = Rs. 85.55 Lakh
b. Heat Removal by Condenser (Power Input to Compressor P = Input Power × 0.9; Condenser Heat Load HL = Cooling Load TR + P×860/3024):
85%: 212.5 TR, 143.69 kW input, P = 129.32 kW → 249.28 TR
60%: 150 TR, 114.66 kW input, P = 103.20 kW → 179.35 TR
40%: 100 TR, 90.16 kW input, P = 81.14 kW → 123.08 TR
c. Heat Load at 40% loading = 123.08 × 3024 = 372194 kCal/Hr; Recovery potential (20%) = 74439 kCal/Hr; Heat requirement for hot water generation = 2000 × 1 × 10 = 20000 kCal/H. Therefore, the heating requirement can easily be met at 40% loading.
Chiller input = load/EER; annual MWh summed over part-load days plus auxiliaries; condenser heat = cooling load + compressor heat; compare 20% recovery against hot-water duty.