Energy Efficiency in Electrical Utilities Available here with full solutions — 70 questions recovered from the 2021 exam:
Objective (1 mark)
53 of 50
Short (5 marks)
9 of 8
Long (10 marks)
8 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) — 53
📖 §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.
3. 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.
📖 §8.3(1) Incandescent lamp (resistive filament); §8.2 Control gear
4. Power factor is highest in case of
sodium vapour lamps
LED lamps
tube lights
incandescent lamps
Answer: D) incandescent lamps
Confirmed vs Book-3 §8.3 — An incandescent lamp is a plain heated filament, a purely resistive load, so its power factor is ~1. Sodium vapour lamps and tube lights need inductive ballasts/ignitors (§8.2 control gear) and LED lamps need electronic drivers, all of which lower the power factor.
5. Which of the following devices do not produce any harmonics ?
UPS
incandescent bulb
arc furnace
electronic ballast
Answer: B) incandescent bulb
Confirmed vs Book-3 §8.3 — The incandescent filament is a linear resistive load, so current is sinusoidal and no harmonics are generated. UPS, arc furnaces and electronic ballasts (which convert supply to 20–30 kHz, §8.6f) are non-linear switching/arcing loads that inject harmonics.
6. A spark ignition engine is used for firing which type of fuels ________.
gasoline
bio-mass
natural gas
all of the above
Answer: D) all of the above
Confirmed vs Book-3 §9.1 — Spark-ignition engine fuels listed by the book are gasoline, natural gas and sewage/landfill (bio) gas; §9.5 also recommends partial use of biomass gas (with tar removal). So all three listed fuels are SI fuels → (d). Note the Book-EOC version of this question offers only one SI fuel (natural gas) against diesel/LDO/furnace oil.
📖 §1.3 Load management (see also Book-3 Ch-5 Fans and Blowers)
7. Flow control by damper operation in fan system will ___________
increase energy consumption
reduce energy consumption
increase system resistance
none of the above
Answer: B) reduce energy consumption
Confirmed vs Book-3 Ch-5 — closing a damper moves the operating point up the fan curve to a lower flow, and the fan's absorbed power falls, so energy consumption does reduce compared with full flow.
Option (c) increased system resistance is the MECHANISM, not the energy outcome the question asks about; note that damper control saves far less than speed (VFD) control, which follows the cube law.
📖 §10.8 ECBC Guidelines on Lighting — Lighting Power Density, Building Area Method (book example: hotel, 4 floors × 1000 m², LPD 10.8 W/m²)
8. A hotel building has four floors each of 1000m2 area. If the interior lighting power allowance for the hotel building is 43,000 W. The Lighting Power Density (LPD) is
10.75
0.09
43
data insufficient
Answer: A) 10.75
Confirmed vs Book-3 §8.2 (power density) — LPD = lighting power ÷ floor area = 43,000 W ÷ (4 × 1000 m²) = 10.75 W/m². Option (b) 0.09 is the inverted ratio; the data are sufficient.
9. 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).
10. 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.
11. 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.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
12. 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.
📖 Book-3 Ch-2 Electric Motors — speed control / Ch-1 power factor (cross-chapter item filed under Ch-3)
13. Which of following is not used for speed control ?
fluid coupling
eddy current
soft starter
variable frequency drive
Answer: C) soft starter
Confirmed vs Book-3 Ch-2 Electric Motors — Fluid couplings, eddy-current couplings and variable frequency drives all vary the output speed of a drive.
A soft starter only ramps the applied voltage during starting to limit inrush current and starting torque; once running, the motor returns to full speed, so it is not a speed-control device. Note this is a drives item filed within the compressed-air set.
14. Which of the following compressed air dryer requires the use of activated alumina ?
membrane dryer
heat of compression dryer
refrigerant dryers
all of the above
Answer: B) heat of compression dryer
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Activated alumina (with silica gel) is the adsorbent used in adsorption/desiccant dryers, and the heat of compression dryer is a desiccant dryer whose bed is regenerated by the compressor's own hot discharge air.
Membrane dryers work by selective permeation through hollow polymer fibres and refrigerant dryers by mechanical chilling — neither contains a desiccant, so 'all of the above' fails.
15. During a leak test of a compressed air system, the compressor's average load time was 1.5 minute, average unload time was 10.5 minutes and flow rate was 35 m3/min. The leakage quantity is
4.375
5.125
7.625
6.250
Answer: A) 4.375
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — This is the book's worked leakage example: q = T/(T+t) x Q = 1.5/(1.5+10.5) x 35 = 0.125 x 35 = 4.375 m³/min, i.e. 6300 m³/day.
The other options arise from using the unload time alone or the wrong cycle total; the denominator must be load time plus unload time.
16. Which of the following is not an adsorption type of air dryer for compressed air system ?
blower reactivated type
heat less purge type
heat of compression type
refrigerant type
Answer: D) refrigerant type
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — The book classifies adsorption (desiccant) dryers by their regeneration method: blower reactivated, heatless purge and heat of compression types.
The refrigerant dryer is not an adsorption dryer at all — it chills the air mechanically so water vapour condenses out, achieving about -20°C atmospheric dew point.
17. At which of the following dew points of the compressed air, the moisture content would be maximum ?
-10oC
-5oC
-40oC
-20oC
Answer: B) -5oC
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Table 3.18: moisture content rises with dew point — 2500 ppm at -5°C, 1600 at -10°C, 685 at -20°C and 80 ppm at -40°C.
Since 'lower the dew point, more dry is the air', the least negative dew point (-5°C) holds the maximum moisture; -40°C is the driest of the four.
18. Which of the following uses concept of evaporative cooling ?
cooling tower
domestic refrigerator
window air conditioner
deep freezer
Answer: A) cooling tower
Confirmed vs Book-3 §4.3 - A cooling tower cools water by evaporation of a small portion of the circulating water (evaporative cooling). A domestic refrigerator, window AC and deep freezer (b, c, d) are all vapour COMPRESSION machines; §4.3 explains that evaporative cooling - air brought into close contact with water and cooled towards the wet bulb temperature - is the cooling-tower principle.
📖 §5.1 Introduction — Table 5.1 (ASME specific ratio)
19. Specific Ratio is maximum for
backward curved fan
forward curved fan
blowers
Compressors
Answer: D) Compressors
Confirmed vs Book-3 §5.1 Table 5.1 — Specific ratio: fans ≤ 1.11, blowers 1.11–1.20, compressors > 1.20 → maximum for compressors (d). Backward- and forward-curved fans (a, b) both fall in the fan band ≤ 1.11.
📖 Book-3 Ch5 Fans & Blowers — flow control (pulley change, damper, inlet guide vanes); not a Ch6 pumps topic
20. Which of the following can be used to regulate the flow of fans ?
pulley change
damper control
inlet guide vane regulation
all of the above
Answer: D) all of the above
Confirmed vs Book-3 Ch5 — Fan flow can be regulated by changing speed (pulley change on belt drives), by damper (throttle) control and by inlet guide vane regulation — all listed fan flow-control methods in Book-3 Ch5, so 'all of the above'. (This fan question sits in the pumps set as it appeared in that exam block.)
📖 §10.2 Building definition — Energy Conservation (Amendment) Act 2010
22. ECBC code is applicable to commercial buildings having connected load of ________.
100 kW
500 kW
250 kW
1000 kW
Answer: A) 100 kW
Book-3 §10.2: after the EC (Amendment) Act 2010, ECBC applies to commercial buildings with connected load of 100 kW or more (or contract demand of 120 kVA or more). The exam question does not say 'minimum', but that is what it means — it asks for the load FROM WHICH the code applies, i.e. the threshold the Act names: 100 kW. A 250, 500 or 1000 kW building is of course also covered (they are above 100 kW), so those options are not 'wrong buildings' — they are simply not the threshold. Trap: 500 kW was the threshold in the original ECBC 2007; the 2010 amendment lowered it to 100 kW / 120 kVA, and the 2014 book and all later papers use the new figure.
📖 §6.1 Pump types — centrifugal pump construction & working
23. Which of the following pump is not a positive displacement pump ?
piston pump
rotary vane
diaphragm pump
centrifugal pump
Answer: D) centrifugal pump
Confirmed vs Book-3 §6.1 — Book classification: dynamic pumps (centrifugal, special effect) vs displacement pumps (rotary, reciprocating). Piston and diaphragm pumps are reciprocating and a rotary vane pump is rotary — all positive displacement. The centrifugal pump is a dynamic (rotodynamic) pump and 'is not positive acting'.
24. The power factor of an electrical system having an active power of 100 kW and reactive power of 80 kVAr will be ___________
0.81
0.88
0.78
cannot be determined
Answer: C) 0.78
Confirmed vs Book-3 §1.4 power triangle — kVA = √(kW² + kVAr²) = √(100² + 80²) = 128.06 kVA, so PF = 100/128.06 = 0.78.
Option (b) 0.88 would come from 100/(100+80)×… ; always build the vector sum first, never add kW and kVAr arithmetically.
25. A 22 kW motor rated for 415 V, 42 A and 0.8 power factor will have an efficiency of ___________
91 %
92 %
89.9 %
none of the above
Answer: A) 91 %
Confirmed vs Book-3 Ch-2 (Electric Motors) - Input = 1.732 x 415 x 42 x 0.8 = 24159 W = 24.16 kW; efficiency = 22/24.16 = 91%. Options (b) and (c) come from dropping the root-3 factor or the power factor; input = 1.732 x 415 x 42 x 0.8 = 24.16 kW, so efficiency = 22/24.16 = 91%.
📖 §9.4 Energy performance assessment — specific fuel consumption & alternator loading
26. A DG set is consuming 70 litres per hour diesel oil. If the specific fuel consumption is 0.33 litres/kWh, what is the kVA loading at 0.8 power factor ?
212 kVA
265 kVA
170 kVA
none of the above
Answer: B) 265 kVA
Confirmed vs Book-3 §9.4 — kW = 70/0.33 = 212.1 kW; kVA = 212.1/0.8 = 265 kVA. Option (a) 212 kVA is the kW figure; (c) 170 comes from multiplying by PF instead of dividing.
27. Which one of the following lamps has maximum CRI ?
Incandescent lamp
LED lamp
CFL lamp
HPSV lamp
Answer: A) Incandescent lamp
Confirmed vs Book-3 §8.3 Table 8.1 — Incandescent CRI = Excellent (100): it renders the standard colour chips identically to the reference source. LED = 80, CFL = 85, HPSV = Fair (22).
28. In a pumping system, if the temperature of the liquid handled decreases, then
NPSHa increases
NPSHa decreases
NPSHa remains constant
NPSHa and NPSHr are independent of temperature
Answer: A) NPSHa increases
Confirmed vs Book-3 §6.5 — NPSHA is the margin of eye pressure above the liquid's vapour pressure. A cooler liquid has a lower vapour pressure, so the margin — NPSHA — increases and cavitation risk falls. NPSHA is therefore temperature-dependent; NPSHR (pump property) is unchanged.
📖 Book-3 Ch-4 HVAC and Refrigeration — COP (cross-chapter item filed under Ch-3)
29. If the COP of a vapour compression system is 3.5 and the motor draws a power of 10.7 kW at 80% motor efficiency, the cooling effect of vapour compression system will be ___________
30 kW
42 kW
27 kW
none of the above
Answer: A) 30 kW
Confirmed vs Book-3 Ch-4 HVAC and Refrigeration — Shaft power delivered to the refrigeration compressor = 10.7 x 0.80 = 8.56 kW; cooling effect = COP x shaft power = 3.5 x 8.56 = 29.96 ≈ 30 kW.
The distractors come from applying the COP to the 10.7 kW motor input directly, or from dividing by the motor efficiency instead of multiplying. Note this is a refrigeration item filed within the compressed-air set.
📖 Not a Ch-7 topic (thermal power plant heat rate); η = 3600 kJ/kWh ÷ heat rate
30. One of the thermal power plants operating with 2 numbers 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 by calculation (not Book-3 Ch-7 content) — Plant efficiency = 3600 kJ/kWh / 11,250 kJ/kWh = 0.32 = 32% → (d). PLF (73%) and unit count are distractors; they do not enter the heat-rate → efficiency conversion.
31. The two-part tariff structure for HT category consumers are
one part for capacity drawn and second part for actual energy drawn
one part for actual power and second part for actual reactive power drawn
one part for capacity drawn and second part for actual reactive power drawn
one part for actual apparent energy drawn and second part for actual reactive energy drawn
Answer: A) one part for capacity drawn and second part for actual energy drawn
Confirmed vs Book-3 §1.2 — 'the electricity billing…in High Tension (HT) category, is often done on two-part tariff structure, i.e. one part for capacity (or demand) drawn and the second part for actual energy drawn during the billing cycle.'
Option (c) is wrong: reactive energy is billed separately as a PF penalty/bonus, not as the second part of the two-part tariff.
32. The illuminance is 10 lm/m2 from a lamp at 1 meter distance. What will be the illuminance (in lm/m2) at 2 meter distance from lamp ?
2.75
2.5
40
20
Answer: B) 2.5
Confirmed vs Book-3 §8.2 — E1·d1² = E2·d2² → E2 = 10 × (1/2)² = 2.5 lm/m². Doubling the distance cuts illuminance to one quarter; 40 would be the answer for halving the distance and 20 assumes a (wrong) linear law.
33. In a water Lithium bromide refrigeration system, the concentration of the lithium bromide gets diluted in ___________
evaporator
condenser
generator
absorber
Answer: D) absorber
Confirmed vs Book-3 §4.3 - In the absorber the refrigerant vapour (water) is absorbed into LiBr solution, diluting its concentration. The generator (c) does the opposite - it boils off water and CONCENTRATES the solution - while the evaporator (a) and condenser (b) handle only refrigerant water; the absorber is where refrigerant vapour is absorbed and the LiBr becomes dilute.
34. 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'.
35. Which of the following is true for energy efficient motors ?
starting torque is higher than standard motors
slip is higher than standard motors
no-load current is higher than standard motors
speed is about 1 % higher than standard motors
Answer: D) speed is about 1 % higher than standard motors
Corrected (was a) — Book-3 §2.6 Energy Efficient Motors: the book says 'starting torque for efficient motors may be LOWER than for standard motors' and that 'less slippage in energy efficient motors results in speeds about 1 % faster than in standard counterparts'. So the true statement is the higher speed, not a higher starting torque. The printed options also repeated the same 'starting torque is higher' text twice, so they have been reconstructed from the book so that exactly one option is correct.
36. The theoretical synchronous speed of 4 pole motor operating at 50 Hz will be ___________
1500 rpm
3000 rpm
200 rpm
1450 rpm
Answer: A) 1500 rpm
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 50 / 4 = 1500 rpm — one of the standard Indian synchronous speeds listed in the book (3000/1500/1000/750 rpm for 2/4/6/8 poles). (d) 1450 rpm is a typical full-load RUNNING speed, not the synchronous speed; (b) 3000 rpm is the 2-pole value.
37. As per Energy Conservation Building Code, compute the Effective Aperture (EA) given that Window Wall Ratio (WWR) is 0.40 and Visible Light Transmittance (VLT) is 0.25.
0.1
1.6
0.65
0.625
Answer: A) 0.1
Confirmed vs Book-3 §10.5 — EA = VLT × WWR = 0.25 × 0.40 = 0.1, exactly the ECBC compliance threshold (book: EA > 0.1 complies, EA < 0.1 does not). 1.6 and 0.625 come from wrongly dividing; 0.65 from adding.
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
38. The purpose of after cooler in a multistage compressor is to ________.
remove the moisture in the air
reduce the work of compression
separate moisture and oil vapour
none of the above
Answer: A) remove the moisture in the air
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — The after-cooler's stated objective is 'to remove the moisture in the air by reducing the temperature in a water-cooled heat exchanger'; about 60-75% of the moisture is condensed out there.
Reducing the work of compression (b) is the INTER-cooler's purpose, and moisture/oil vapour separation by settling (c) is a role of the receiver.
39. The outer tube connection of the Pitot tube is used to measure ___________ in the fan system
static pressure
total pressure
velocity pressure
none of the above
Answer: A) static pressure
Confirmed vs Book-3 §5.6 — 'Static pressure is measured using the outer tube of pitot tube' (side holes at 90° to flow); the inner tube gives total pressure; both together give velocity pressure. So the outer connection reads static pressure (a), not total (b) or velocity (c).
📖 Book-3 Ch1 Electrical Systems — Technical Losses (T&D losses in the distribution system)
40. Which of the following contributes to increased technical losses ?
lower sized conductors
low power factor
loose connections
all of the above
Answer: D) all of the above
Confirmed vs Book-3 Ch1 — Technical (I²R) losses rise with undersized conductors (higher resistance), low power factor (higher current for the same kW) and loose connections (contact resistance/heating). All three contribute → (d).
41. Single stage Li Br water absorption refrigeration systems have a COP in the range of
0.40 - 0.5
0.65 - 0.70
0.75 - 0.80
0.2 - 0.3
Answer: B) 0.65 - 0.70
Confirmed vs Book-3 §4.3 - Single-effect (single-stage) LiBr-water absorption chillers have a COP of about 0.65 to 0.70. Option (c) 0.75-0.80 belongs to double-effect machines and (d) 0.2-0.3 to half-effect; §4.3 gives 0.65-0.70 for single-stage LiBr-water absorption chillers.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
42. Shaft power of the motor driving a pump is 20 kW. The motor efficiency is 0.9 and pump efficiency is 0.55 at shaft operating load. The power transmitted to the water is ________.
12.2 kW
9.9 kW
11 kW
12.7 kW
Answer: C) 11 kW
Confirmed vs Book-3 §6.1 — The 20 kW is the pump shaft power; hydraulic (water) power = shaft power × η_pump = 20×0.55 = 11 kW. Motor efficiency is a distractor (it relates shaft power to electrical input, 20/0.9 = 22.2 kW). 9.9 kW wrongly applies both efficiencies. (Option c repaired from garbled OCR '0.75 - 0.66 kW' to '11 kW', matching the Set-B print.)
📖 §10.14 Star rating — BPO buildings, Average Annual hourly EPI (AAhEPI)
43. The unit of AAhEPI is given by
kWh/m2/yr
m2 × kWh/hr
(Wh/m2)/hr
m2/Wh/yr
Answer: C) (Wh/m2)/hr
Confirmed vs Book-3 §10.14 — AAhEPI (Average Annual hourly Energy Performance Index, used for BPO/call-centre star rating) is in (Wh/sqm)/hr: [EPI ÷ (daily hours × days/week × 52 weeks)] × 1000. kWh/m²/yr (a) is the unit of ordinary EPI, the tempting wrong option. (Option b repaired from garbled 'm2/a kWh/hr'; question acronym corrected from 'AhhEPI'.)
Confirmed vs Book-3 §4.5 - the printed stem 'Cooling system' is a transcription slip for 'Hermetic system': a hermetic unit has motor and compressor in one sealed casing and the book confines it to refrigerators, air conditioners and other low-capacity applications. Centrifugal, screw and large reciprocating chillers (b, c, d) all use open-type machines with the motor separate from the compressor, as industrial service needs field-serviceable drives.
📖 Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14)
45. Which of the following compressors do not use loading / un-loading method for capacity control?
screw compressor
centrifugal compressor
reciprocating compressor
all of the above
Answer: B) centrifugal compressor
Confirmed vs Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14) — Load/unload (two-step) and multi-step unloading are the control schemes the book describes for positive-displacement machines — reciprocating and screw compressors.
Centrifugal capacity is matched instead by variable inlet guide vanes or by speed control (Table 3.14), so the centrifugal is the odd one out.
46. The output of a 900 kW rated motor operating with 90% efficiency is ________.
900 kW
1000 kW
810 kW
none of the above
Answer: A) 900 kW
Corrected (was b) — Book-3 §2.4 Motor Efficiency: the kW rating on the nameplate is the rated OUTPUT of the motor, so the output of a 900 kW rated motor at full load is 900 kW; 1000 kW (= 900/0.9) is the INPUT it draws. The book's own end-of-chapter question makes this explicit — a 75 kW, 90 % motor at full load 'delivers 75 kW'.
47. COP of an air conditioner will be least with ________.
lower evaporator temperature and higher condenser temperature
higher evaporator temperature and lower condenser temperature
higher evaporator temperature and higher condenser temperature
lower evaporator temperature and lower condenser temperature
Answer: A) lower evaporator temperature and higher condenser temperature
Confirmed vs Book-3 §4.7 - COP falls as the temperature lift increases; lowest COP occurs with lowest evaporator and highest condenser temperature. COP-Carnot = Te/(Tc - Te), so the worst case is the largest temperature lift; (b) gives the highest COP and (c) and (d) give intermediate values.
48. Which of the following is not true of energy efficient motors?
starting torque is higher than standard motors
starting torque is lower than standard motors
slip is lower than standard motors
speed is higher than standard motors
Answer: A) starting torque is higher than standard motors
Confirmed vs Book-3 §2.6 Energy Efficient Motors — The book warns that 'starting torque for efficient motors may be lower than for standard motors', so (a) — higher starting torque — is NOT true of energy-efficient motors. (c) lower slip and (d) about 1 % higher speed are both stated by the book as genuine EEM characteristics, so they cannot answer a 'not true' question.
49. When the evaporation of water from a wet surface at atmospheric condition is zero, ___________
RH is 0%
RH is 100%
wet bulb temperature is greater than dry bulb temperature
none of the above
Answer: B) RH is 100%
Confirmed vs Book-3 §4.2 - At 100% relative humidity the air is saturated, so no further evaporation occurs. Option (a) 0% RH would give the fastest evaporation, and (c) is impossible because the wet bulb can never exceed the dry bulb; evaporation stops only when the air is saturated at 100% RH.
📖 §10.14 Star Rating of Buildings — Energy Performance Index; AAhEPI (Book-3 Ch10 end-of-chapter Q9)
50. The unit of AAEPI is given by ___________
kWh/m2/yr
x kWh/hr
(Wh/m2)/hr
m3/Wh/yr
Answer: C) (Wh/m2)/hr
Confirmed vs Book-3 Ch-10 (ECBC / building energy index) - Air conditioning Annual Energy Performance Index is expressed as Wh per m2 per hour. Options (a), (b) and (d) mix up units used for building EPI and volumetric indices; the air-conditioning annual energy performance index is expressed per unit floor area per hour, i.e. (Wh/m2)/hr.
📖 Book-3 Ch-3 Compressed Air (dew point) - cross-chapter
51. Which of the following dew points of the compressed air, the moisture content would be maximum?
-10degC
-5degC
-40degC
none of the above
Answer: B) -5degC
Confirmed vs Book-3 Ch-3 (Compressed Air, dew point) - A higher (less negative) dew point corresponds to more residual moisture in the air. The lower (more negative) the pressure dew point the drier the air, so -40 degC (c) holds the least moisture; -5 degC is the highest dew point offered and therefore carries the most residual moisture.
52. The performance of winding of an induction motor can be assessed by which of the following factors?
load current
stator resistance
no load current
both b and c
Answer: D) both b and c
Confirmed vs Book-3 §2.8 Rewinding Effects on Energy Efficiency — The book names two indicators of rewind quality: the no-load current and the stator resistance per phase, both compared with the original values at the same voltage. (a) load current is set by the driven machine and tells nothing about the winding work, so only the combination of (b) and (c) is correct.
53. 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'.
1. Fill in the blanks (1 mark each): a) Heat rate of a thermal power plant is expressed in ___; b) The ___ loss is independent of load in a transformer; c) ___ is used to reduce the dew point in a compressed air system; d) The speed of an energy efficient motor will be more than the standard motor of same capacity because ___ decreases.
Model answer: a) kCal/kWh (or kJ/kWh) - the heat input per kWh generated; 1 kWh = 860 kCal = 3600 kJ.
b) Core loss (iron loss / no-load loss) - the book states core loss 'occurs whenever the transformer is energized; core loss does not vary with load'.
c) An air dryer (refrigerant or desiccant type) is used to lower the dew point of compressed air.
d) SLIP - an energy-efficient motor has lower rotor losses and therefore lower slip, so for the same number of poles it runs slightly faster than a standard motor.
Standard BEE fill-in answers: thermal heat rate in kCal/kWh, transformer iron/core loss is load-independent, air dryer lowers dew point, EE motor has higher speed because slip decreases.
📖 §9.3 Waste heat recovery — WHR formula, Table 9.6 flue gas at part load; trigeneration (VAM)
2. A 5 MW DG set running at 70% load for base load operation generates 8.6 kg of exhaust gas per kWh. The exhaust gas is reduced to 200oC. The specific heat of flue gas is 0.26 kcal/kg-oC. The steam generated from waste heat boiler will be used in double effect Li-Br Vapour Absorption Chiller with a COP of 1.12. How much TR will be generated through VAM? (5 Marks)
Model answer: Loading of DG set = 70% x 5 MW = 3.5 MW = 3500 kW. Quantity of heat from exhaust gas = 3500 kW x 8.6 kg/kWh x 0.26 kcal/kg-oC x (450 - 200) oC = 19,56,500 kcal/hr. Potential TR via double-effect VAM: COP = (TR x Heat input)/3024 -> TR = COP x Heat input/3024 = (1.12 x 1956500)/3024 = 724.6 TR.
Heat in exhaust = mass x Cp x dT; TR = COP x heat/3024 (1 TR = 3024 kcal/h).
3. What are the advantages of using vapour absorption refrigeration system over vapour compression system? Under what condition it would be economical? (5 Marks)
Model answer: Refer Guidebook-3 (Pg 112-116): Advantages of VAR over VCR — uses low-grade/waste heat or steam instead of high-grade electrical energy; very few moving parts (only pump) so low maintenance, low noise/vibration; uses environment-friendly refrigerants (water/ammonia); can use otherwise wasted heat. It is economical when low-cost waste heat, exhaust gas, or low-pressure steam is available, making the running cost low despite higher first cost.
Advantages and economic conditions for VAR per Guidebook-3.
4. A small foundry has installed a reciprocating air compressor of 14.25 m3/min. The plant could not meet the compressed air requirement and hence conducted a capacity test to determine the derating in the compressor capacity. Calculate the actual FAD delivered after considering necessary temperature correction in m3/min, and the percentage derating. Operating parameters: Volume of air receiver including pipe and cooler 9 m3; Atmospheric temperature (T1) 35degC; Receiver temperature (T2) 44degC; Initial Pressure 0.5 kg/cm2(g); Final Pressure 7.0 kg/cm2(g); Atmospheric pressure 1.026 kg/cm2(a); Time taken to build up the pressure 5 minutes.
Model answer: FAD = [(P2 - P1)/Pa] x [(receiver + hold-up volume)/time] x temperature correction factor.
P1 (initial) = 0.5 kg/cm2(g); P2 (final) = 7.0 kg/cm2(g); Pa = 1.026 kg/cm2(a); receiver and holding volume = 9 m3; pump-up time = 5 min.
Uncorrected output = [(7.0 - 0.5) x 9] / (1.026 x 5) = 11.40 m3/min.
Temperature correction factor = (273 + T1)/(273 + T2) = (273 + 35)/(273 + 44) = 308/317 = 0.972.
FAD after correction = 11.40 x 0.972 = 11.08 m3/min.
Capacity shortfall = 14.25 - 11.08 = 3.17 m3/min; % de-rating = (3.17/14.25) x 100 = 22.24%. As this is far above the 10% deviation the book allows, corrective action on the compressor is called for.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — FAD pump-up formula with the temperature correction (273+T1)/(273+T2); the P1/P2 labels in the model answer had been swapped and are now correct (initial 0.5, final 7.0 kg/cm2 g).
📖 §5.6 Pitot velocity, volume & fan static efficiency
5. A V-belt driven centrifugal fan is supplying air to a chemical process. Calculate the fan static efficiency for the following operating parameters: Ambient temperature 40degC; Density of air 1.127 kg/m3; Diameter of discharge air duct 1 meter; Velocity pressure measured by Pitot tube in discharge duct 47 mm WC; Pitot tube coefficient 0.9; Static pressure at fan inlet -22 mm WC; Static pressure at fan outlet 188 mm WC; Power drawn by motor 72 kW; Belt transmission efficiency 95%; Motor efficiency at operating load 90%.
Model answer: Air velocity = Cp x (2 x 9.81 x dP/density)^0.5 = 0.9 x (2 x 9.81 x 47/1.127)^0.5 = 25.7 m/s. Area of discharge duct = 3.14 x 1 x 1/4 = 0.785 m2. Volume = 25.7 x 0.785 = 20.17 m3/s. Power input to fan shaft = 72 x 0.95 x 0.9 = 61.6 kW. Total static pressure = 188 - (-22) = 210 mm WC. Fan static efficiency = (Volume x total static pressure in mm WC) / (102 x shaft power in kW) = (20.17 x 210)/(102 x 61.6) = 67.4%.
Same method as the Book-3 §5.9 solved example: velocity from pitot VP, Q = V × A, shaft kW = motor kW × belt × motor efficiency, total static pressure = outlet − inlet (mind the negative inlet), η_static = (Q × ΔP)/(102 × shaft kW) = 67.4%.
6. List five Energy Efficiency measures in buildings.
Model answer: Five energy-efficiency measures in buildings (Book-3 §10.15): (1) Weather stripping of windows and doors plus self-closing doors to minimise exfiltration of conditioned air and infiltration of outside air. (2) Temperature and humidity setting of 23–25 °C and 55–65% RH; maintain chilled-water leaving temperature at or above 7 °C (centrifugal chiller efficiency rises ~2.5% per 1 °C). (3) Keep chilled-water pipe and duct insulation in good condition; clean chiller condenser tubes at least every six months; keep cooling towers and air filters clean. (4) Install frequency converters (VFDs) to vary AHU fan speed — saves up to 15% of fan-motor energy (also VFDs on chilled/cooling-water pumps, §10.10). (5) Lighting: daylighting with controls that reduce luminaire output by at least half, switch off lights when not in use, separate switches for peripheral lighting, replace incandescent lamps with CFL (75% saving), electronic ballasts (2 W vs 12 W loss), optical luminaires (up to 50%), clean fixtures four times a year, light colours on walls/ceilings and task lighting.
Book-3 §10.15 AC measures (weather stripping, 23–25 °C/55–65% RH, CHW ≥ 7 °C, insulation, condenser cleaning, AHU VFD 15%) and lighting measures (daylighting controls, CFL 75%, electronic ballast, optical luminaires, cleaning, light colours).
7. a) Name six parameters along with units that a psychrometric chart provides (3 marks). b) Explain briefly about Thermal Emittance (2 marks).
Model answer: a) Refer Guidebook-3 Pg 272. Six parameters from psychrometric chart: 1. Dry bulb temperature (degC); 2. Relative humidity (%); 3. Wet bulb temperature (degC); 4. Specific volume (m3/kg of dry air); 5. Enthalpy (kcal/kg of dry air); 6. Specific humidity / humidity factor (grams/kg of dry air). b) Thermal emittance is the ratio of radiant heat flux emitted by a surface to that of a black body at the same temperature; high-emittance roof surfaces radiate absorbed heat readily, reducing heat gain.
Psychrometric chart properties per Guidebook-3; thermal emittance defined relative to a black body radiator.
8. A process plant has installed 5 MW DG set for base load operation, which is operating at 70% loading. Furnace oil is used as a fuel in the DG set. The DG set generates 8.6 kg of exhaust gas per kWh generated. The plant management has decided to install a heat recovery boiler to generate steam at 3 kg/cm2g from the exhaust gas to reduce the exit flue gas temperature from 450degC to 200degC. The specific heat of flue gas is 0.26 kcal/kgdegC. The steam generated from waste heat boiler will be used in double effect Li-Br Vapor Absorption Chiller, with a COP of 1.12. How much TR will be generated through VAM?
Model answer: DG loading = 5 MW x 70% = 3.5 MW = 3500 kW. Quantity of heat available from exhaust gas = 3500 x 8.6 x gas generated/kWh x 0.26 kcal/kgdegC x (450 - 200) = 19,56,500 kcal/hr. Potential TR generation through double-effect VAM: COP = TR effect/Heat input, so TR = (COP x Heat input)/3024 = (1.12 x 1956500)/3024 = 724.6 TR.
Waste heat from exhaust = mass x Cp x dT; refrigeration effect = COP x heat input; convert kcal/h to TR by dividing by 3024.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
9. A process plant is situated 100 m up the side of the hill. The plant requires 100 kL of water per hour. The management decides to install a pump at the ground level, with suction 3 metre below the ground level. The friction head is 12 metre. Evaluate the rating of the motor required considering 10% extra margin with respect to actual input pump power. The design pump efficiency is 65% and motor efficiency 93%.
Model answer: Q = 100/3600 x 1000 = 100/3.6 m3/s = 0.0277 m3/s. Density = 1000 kg/m3. Static head = 100 + 3 = 103 m; Friction head = 12 m; Total Head = 103 + 12 = 115 m. Hydraulic power = Q x density x g x H / 1000 = 0.0277 x 1000 x 115 x 9.81/1000 = 31.25 kW. Pump input (shaft) power = 31.25/0.65 = 48.07 kW. Motor rating (shaft power) = 48.07 kW. Motor rating with 10% margin = 48.07 x 1.10 = 52.87 kW. Motor input = design rated power/motor efficiency = 48.07/0.93 = 51.69 kW.
Total head = static (lift + suction) + friction; hydraulic power via rho*g*Q*H; divide by pump efficiency for shaft power; add 10% margin and divide by motor efficiency.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
1. A process plant is situated 100 m above the ground level on the top of the hill. The plant requires 100 kL of water per hour. The management decides to install a pump at the ground level, with suction 3 meter below the ground level. The friction head is 12 meter. Evaluate the rating of the motor required considering 10% extra margin with respect to actual input pump power. The design pump efficiency is 65%. Also calculate the motor input power if the motor efficiency is 93%. (5 Marks)
Model answer: Q = 100 kL/hr = 100 m³/hr = 100/3600 = 0.0278 m³/s; ρ = 1000 kg/m³. Static head = delivery height + suction depth = 100 + 3 = 103 m; friction head = 12 m; Total head = 103 + 12 = 115 m. Hydraulic power = Q×ρ×g×H/1000 = 0.0278×1000×9.81×115/1000 = 31.3 kW. Pump shaft (input) power = 31.3/0.65 = 48.1 kW. Motor rating with 10% margin = 48.1×1.10 = 52.9 kW (select next standard size, e.g. 55 kW). Motor input power at the actual load = 48.1/0.93 = 51.7 kW.
Total head = static (lift + suction) + friction; hydraulic power = Q·ρ·g·H/1000; ÷ pump efficiency for shaft power; +10% for motor rating; ÷ motor efficiency for input power.
📖 §5.6 Pitot velocity, volume & fan static efficiency
2. A V-belt driven centrifugal fan is supplying air to a chemical process. Calculate the fan static efficiency for: Ambient temperature 40oC; Density of air 1.127 kg/m3; Diameter of discharge air duct 1 m; Velocity pressure (Pitot) 47 mm WC; Pitot tube coefficient 0.9; Static pressure at fan inlet -22 mm WC; Static pressure at fan outlet 188 mm WC; Power drawn by motor 72 kW; Belt transmission efficiency 95%; Motor efficiency at operating load 90%. (5 Marks)
Model answer: Air velocity = Cp x (2 x 9.81 x dp/rho)^0.5 = 0.9 x (2 x 9.81 x 47/1.127)^0.5 = 25.7 m/s. Area of discharge duct = 3.14 x 1 x 1/4 = 0.785 m2. Volume = 25.7 x 0.785 = 20.17 m3/s. Power input to fan shaft = 72 x 0.95 x 0.9 = 61.6 kW. Fan static efficiency = (Volume m3/s x total static pressure mmWC)/(102 x power input to shaft kW) = [20.17 x (188-(-22))]/(102 x 61.6) = (20.17 x 210)/6283 = 67.4%.
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
3. a) List five losses in electrical motors and discuss about the measures taken by the motor manufacturers to make it energy efficient motor. (5 Marks) b) List four energy conservation opportunities in pumping system. (5 Marks)
Model answer: a) Book-3 §2.6 (Table 2.2 and the loss break-up). The five losses in an induction motor are: 1. Stator I²R (copper) loss and 2. Rotor I²R loss — together 55–60 % of total losses; 3. Core / iron loss (hysteresis + eddy current) — 20–25 %; 4. Friction & windage loss — 8–12 %; 5. Stray load loss — 4–5 %.
Manufacturers' measures in energy-efficient motors: more copper and larger-cross-section conductors and larger rotor bars (lower R, lower I²R); thinner, lower-loss silicon steel laminations and a longer core to lower flux density (lower eddy-current and hysteresis loss); smaller air gap and lower flux density to cut magnetizing current; low-loss fan design and superior bearings (lower friction & windage); optimised slot/tooth geometry and strict quality control (lower stray load loss). Result: full-load efficiency 3–7 percentage points higher, with mounting dimensions kept to IS 1231.
b) Pumping-system opportunities (Book-3 Ch. 6): avoid oversizing / re-size the pump to the actual duty; use a VFD instead of throttle-valve or bypass control; trim or replace the impeller to match the required head; and replace worn or low-efficiency pumps and eliminate unnecessary head losses (oversized valves, fittings, clogged strainers).
Motor losses and EE measures, and pumping ECOs from Guidebook-3.
📖 §8.7 Energy efficient lighting controls — Daylight linked control (part b: Ch.7 cooling towers)
4. a) List any three energy efficient lighting controls. Describe briefly about daylight linked control (5 marks). b) Explain briefly about various water losses in cooling towers and how they can be minimized (5 marks).
Model answer: a) Refer Guidebook-3 Page 243-244. Three energy efficient lighting controls: occupancy/motion sensors, timers/time-scheduled controls, and daylight linked (photoelectric) controls. Daylight linked control uses a photo sensor to measure available natural daylight and automatically dims or switches off artificial lighting near windows/skylights so total illuminance is maintained at the design level, saving energy when daylight is sufficient. b) Refer Guidebook-3 Page 205. Cooling tower water losses: Evaporation loss (water evaporated to provide cooling - unavoidable, inherent to process); Drift/windage loss (fine droplets carried out by air - minimized by efficient drift eliminators); Blowdown loss (water bled off to control dissolved solids - minimized by increasing cycles of concentration with proper water treatment); plus leakage/overflow losses minimized by good maintenance.
Book-3 §8.7: occupancy sensors, timed-turnoff switches and daylight-linked (photoelectric) control; photocells switch or dim so that daylight + electric light always reaches the design level, switching electric light off when daylight alone suffices. Part (b) is from the cooling-tower chapter.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%); §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
5. Rated capacity of a fresh water shut-discharge pump flow rate is 485 m3/hr and discharge pressure is 13.5 kg/cm2 at rated speed of 2950 rpm. It has been observed that 450 m3/hr water is sufficient to dispose the ash. The suction pressure of the pump is 0.5 kg/cm2. The management has decided to trim the impeller to satisfy the reduced flow requirement. Calculate: a) % reduction of impeller diameter (5 Marks) b) Annual energy savings after modification, if the pump is operating for 6 hours/day and 330 days in a year. (5 Marks)
Model answer: a) Q∝D, so D₂/D₁ = 450/485 = 0.928; % reduction in impeller diameter = (1 − 0.928)×100 = 7.2%. b) Rated differential head = (13.5 − 0.5) kg/cm² = 13 kg/cm² ≈ 130 m. Hydraulic power (rated) = (485/3600)×1000×130×9.81/1000 = 171.8 kW. After trimming, discharge head ∝ D²: H₂ = 0.928²×135 = 116.3 m, so new differential head = 116.3 − 5 = 111.3 m. Hydraulic power (new) = (450/3600)×1000×111.3×9.81/1000 = 136.4 kW. Power saving = 171.8 − 136.4 = 35.4 kW. Annual energy saving = 35.4×6×330 ≈ 70,000 kWh/year (official key prints 69,894 kWh using 35.3 kW). Alternative by P∝D³: P₂ = 0.928³×171.8 = 137.3 kW, saving 34.5 kW ≈ 68,300 kWh/year — both accepted.
Trim: Q∝D gives the diameter ratio; H∝D² gives new head; hydraulic power = Q·ρ·g·H/1000 before/after; annual saving = kW saved × 6 h × 330 days.
📖 §7.2 Factors Affecting Performance – Heat Load (heat-rejection table) (condenser 555 kcal/kg); Range = Heat load / Water circulation rate (pump power is Pumps chapter)
6. In a Thermal Power Station, the steam input to a turbine operating on a fully condensing mode is 100 Tonnes/hr. The heat rejection requirement of the steam turbine condenser is 555 kcals/kg of steam condensed. The head developed by the cooling water pump is 2.5 kg/cm2. During 4500 hours of normal operation per year, the cooling water temperatures at the inlet and outlet of turbine condenser are measured to be 27degC and 35degC and during the winter period operation of 3000 hours per year the cooling water temperatures at the inlet and outlet of turbine condenser are measured to be 15degC and 25degC. Find out: i. The circulating cooling water flow for normal operation as well as for winter operation (5 marks). ii. Calculate the annual energy reduction during winter operation if the combined efficiency of the pump and motor is 70% (5 marks).
Model answer: i. Normal operation (27/35degC, 8degC rise, 4500 hrs): Heat rejected = 100000 x 555 = 55.5 million kcal/hr. Cooling water flow = 55.5x10^6/(35-27) x sp.heat 1 x (1/1000) = 6937.5 m3/hr. Head = 2.5 kg/cm2. Hydraulic power = (6937.5/3600)*25*9.81 = 472.62 kW. Input power = 472.62/0.7 = 675.16 kW. Winter operation (15/25degC, 10degC rise, 3000 hrs): Cooling water flow = 55.5x10^6/(25-15) x 1 x (1/1000) = 5550 m3/hr. Hydraulic power = (5550/3600)*25*9.81 = 378.09 kW. Input power = 378.09/0.7 = 540.13 kW. ii. Energy reduction during winter = (675.16 - 540.13) x 3000 = 405090 kWh.
Cooling water flow = heat rejected/(temperature rise x sp.heat x 1000); pump hydraulic power = (flow/3600) x head(m) x 9.81; input = hydraulic/combined efficiency; annual saving = power difference x winter hours.
📖 §1.4 Power Factor Improvement and Benefits + §1.5 Transformer losses & efficiency
7. a) A cold rolling mill has a maximum demand of 7 MVA at power factor of 0.95. The plant management converts the existing electrical resistance annealing furnace having steady load of 1250 kW to gas heating as a cost reduction measure. The existing capacitor banks (kVAr) continued to be in the electrical network. What will be the effect on maximum demand and power factor due to this conversion? (5 Marks) b) A cement plant has a constant load of 15 MVA. It has installed two transformers of 30 MVA each. The no load loss and full load copper loss of each 30 MVA transformer is 25 kW and 75 kW respectively. From the energy efficiency point of view the industry management wants to take a decision on whether to operate a single transformer or two transformers equally sharing the load. What is your recommendation? (5 Marks)
Model answer: a) Registered max demand = 7 MVA = 7000 kVA. Electrical real load = 7000 x 0.95 = 6650 kW. Existing kVAr = sqrt(kVA^2 - kW^2) = sqrt(7000^2 - 6650^2) = 2186 kVAr (remains in network). Reduction in real power = 1250 kW (furnace converted to gas). Revised real power = 6650 - 1250 = 5400 kW. Revised kVA = sqrt(5400^2 + 2186^2) = 5825 kVA. Reduction in electrical demand = 7000 - 5825 = 1175 kVA. Revised power factor = 5400/5825 = 0.927. Reduction in PF = 0.95 - 0.927 = 0.023. b) Option 1 (one transformer): % load = 15/30 = 50%; Total loss = NoLoad + Cu loss x (%load)^2 = 25 + 75 x 0.5^2 = 43.75 kW. Option 2 (both transformers): each at 7.5/30 = 25%; Total loss = [25 + 75 x 0.25^2] x 2 = 59.37 kW. Recommendation: operate single transformer because losses are lower, saving 59.37 - 43.75 = 15.62 kW.
kVAr from kVA & kW; revised demand after load removal; transformer loss = NL + FL_Cu x (loading)^2.
8. a) Calculate the filter area of an Air Handling Unit (AHU) for Refrigeration Load of 50 TR. The air enthalpy at inlet of AHU is 85 kJ/kg and at outlet is of 60 kJ/kg. Air velocity at filter is 1.81 m/sec and air density is 1.26 kg/m3. (5 Marks) b) A no load test was conducted in a delta connected 37 kW induction motor. Name plate data: 3 Phase, 415 V, 50 Hz, 55 Amp. Measured data on no load: Voltage V = 415 Volts; Current I = 18 Amps; Frequency F = 50 Hz; Stator phase resistance at no load = 0.23 Ohms/phase. No load power = 955 Watts. Calculate: i. The iron loss plus friction loss plus windage loss (2 Marks) ii. Stator copper loss at name plate ratings (full load), considering stator temperature as 120oC (2 Marks) iii. No load power factor of the motor (1 Mark)
Model answer: a) TR of AHU = (Enthalpy difference x density x area x velocity x 3600)/(4.187 x 3024). 50 = (85-60) x 1.26 x Area x 1.81 x 3600/(4.187 x 3024). Filter Area = TR x (4.187 x 3024)/[(enthalpy diff) x density x velocity x 3600] = 50 x (4.187 x 3024)/[(25) x 1.26 x 1.81 x 3600] = 3.08 m2. b) i) No-load input Pin = 955 W. No-load stator copper loss = 3 x I^2 x R = 3 x 18^2 x 0.23 = 74.51 W. Iron + friction + windage loss = Pin - no-load Cu loss = 955 - 74.51 = 880.49 W. ii) Stator resistance at 120oC = 0.23 x (120+235)/(30+235) = 0.23 x 355/265 = 0.308 ohms. Full-load stator copper loss at nameplate current = 3 x (55/sqrt3)^2 x 0.308 = 3 x 1008.3 x 0.308 = 931.65 W. iii) No-load power factor = P/(sqrt3 x V x I) = 955/(1.732 x 415 x 18) = 0.0738.
AHU: TR = (dh x rho x A x v x 3600)/(4.187 x 3024); iron+fr+wind = Pin - 3I2R; R corrected by (235+t2)/(235+t1); PF = P/(sqrt3 VI).