Energy Efficiency in Electrical Utilities Available here with full solutions — 64 questions recovered from the 2022 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
📖 §5.1 Introduction — Table 5.1 (ASME specific ratio)
1. The specific ratio as defined by ASME and used in differentiating fans, blowers and compressors, is given by
Confirmed vs Book-3 §5.1 — ASME specific ratio = discharge pressure / suction pressure (Table 5.1): fans up to 1.11, blowers 1.11–1.20, compressors > 1.20. Option (b) inverts the ratio and would give values below 1 — meaningless for classification.
Answer: A) inversely proportional to fan efficiency
Confirmed vs Book-3 §5.6 — Fan static efficiency = (Q × SP)/(102 × shaft kW), rearranged: shaft power = (Q × SP)/(102 × η). Power is therefore INVERSELY proportional to fan efficiency and DIRECTLY proportional to flow and static pressure. Options (c) and (d) reverse the direct proportionality to pressure and flow.
3. Increasing the suction pipe diameter in a pumping system will
Decrease NPSHA
Increase NPSHA
Decrease NPSHR
Increase NPSHR
Answer: B) Increase NPSHA
Confirmed vs Book-3 §6.5 — NPSHA = margin of eye pressure above vapour pressure; the book notes that as suction friction losses increase, NPSHA falls. A larger suction pipe lowers velocity and friction loss, so NPSHA increases. NPSHR is fixed by the pump design and does not change with pipe size. (Same as book end-of-chapter Q6.)
4. In a transformer on load, if the secondary voltage is one-fourth the primary voltage, then the secondary current will be
four times the primary current
sixteen times the primary current
one-fourth the primary current
two times the primary current
Answer: A) four times the primary current
Confirmed vs Book-3 §1.5 — 'Primary ampere-turns are equal to secondary ampere-turns', so V₁I₁ = V₂I₂. With V₂ = V₁/4, the secondary current must be four times the primary current.
Option (c) is the trap: current and voltage move in OPPOSITE directions in a transformer, so the low-voltage side carries the larger current.
📖 Book-3 Ch-6 Pumps and Pumping System (cross-chapter item filed under Ch-3)
5. The efficiency of a pump does not depend on
suction head
discharge head
motor efficiency
density of liquid
Answer: C) motor efficiency
Confirmed vs Book-3 Ch-6 Pumps and Pumping System (cross-chapter item filed under Ch-3) — Pump efficiency = hydraulic power (ρ g Q H) / shaft power input to the pump, so it depends on suction and discharge head through H and on the liquid through its density.
Motor efficiency belongs to the driving motor, not the pump; it enters only the combined pump-set (wire-to-water) efficiency. Note this is a pumps item filed within the compressed-air set.
6. Energy Star Label Rating scheme for Fluorescent lamp is based on
Lumens per Watt at 100, 2000 and 3500 hours of use
End of Lamp Life in terms of burning hours
Lumen depreciation at 2000 hours
Color Rendering Index
Answer: A) Lumens per Watt at 100, 2000 and 3500 hours of use
Confirmed vs Book-3 §8.8 Table 8.4 — The FTL star rating is set by lumens per Watt measured at 100, 2000 and 3500 hours of use (e.g. 5-star ≥ 92, ≥ 83 and ≥ 78 lm/W respectively). Lamp life, lumen depreciation alone or CRI are not the rating criteria.
📖 §6.2 System characteristics — static & friction head
7. Installing larger diameter pipe in pumping system results in reduction in ______
static head
frictional head
both a and b
neither a nor b
Answer: B) frictional head
Confirmed vs Book-3 §6.2 — Static head is 'simply the difference in height of the supply and destination reservoirs' and is independent of flow and pipe size. Friction head is the loss in pipes/valves/equipment, and the book notes that further reduction 'will require larger diameter pipe'. So only frictional head is reduced.
8. 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 %.
📖 §9.1 Table 9.1 — comparison of captive power plant types
9. Which of the following power plants has the highest efficiency?
Open cycle Gas Turbine
Diesel Engine
combined cycle gas turbine
Conventional coal plants
Answer: C) combined cycle gas turbine
Confirmed vs Book-3 §9.1 — Table 9.1: combined GT & ST 40–46%, diesel engine plant 43–45%, conventional steam plant 33–36%; an open-cycle gas turbine is lowest. The combined cycle (which recovers GT exhaust in an HRSG for a steam turbine) has the highest top-end efficiency; diesel (b) is the tempting distractor because the book says captive diesel plant "wins" in the small-capacity range and Table 9.1 puts its range close to combined cycle.
📖 §6.2 System characteristics — static & friction head
10. Friction losses in a pumping system is
inversely proportional to flow
inversely proportional to cube of flow
proportional to square of flow
inversely proportional square of flow
Answer: C) proportional to square of flow
Confirmed vs Book-3 §6.2 — Book: 'The friction losses are proportional to the square of the flow rate.' This is why the system curve is parabolic and why halving flow cuts friction head to a quarter. Inverse relations (a, b, d) are wrong.
11. Which of the following is not likely to create harmonics in an electrical system?
soft starters
variable frequency drives
uninterrupted power supply source (UPS)
electric heater
Answer: D) electric heater
Confirmed vs Book-3 §1.10 — the book classes heaters as LINEAR loads ('Incandescent lamps, heaters and, to a great extent, motors are linear systems') because their impedance is constant, so they draw a sinusoidal current and create no harmonics.
Options (a)–(c) are all power-electronic (non-linear) devices — soft starters, VFDs and UPS — which the book lists as harmonic sources.
12. In a vapor compression refrigeration system, the component across which the enthalpy remains constant
compressor
condenser
expansion valve
evaporator
Answer: C) expansion valve
Confirmed vs Book-3 §4.3 - Expansion (throttling) is an isenthalpic process; enthalpy remains constant across the expansion valve. Enthalpy rises in the evaporator (d) and compressor (a) and falls in the condenser (b); §4.3 states there is no heat loss or gain through the expansion device, so throttling is isenthalpic.
📖 §6.2 System characteristics — static & friction head
14. Which of the following is not true regarding system characteristic curve in a pumping system with large dynamic head?
System curve represents a relationship between discharge and head loss in a system of pipes
System curve is dependent on the pump speed
The basic shape of a system curve is parabolic
System curve will start at zero flow and zero head if there is no static lift
Answer: B) System curve is dependent on the pump speed
Confirmed vs Book-3 §6.2 — The system curve is the head-loss vs flow relationship of the piping (a); friction ∝ Q² makes it parabolic (c); with no static lift it starts at zero head and zero flow (d, Fig 6.5). It depends on elevation, pipe size/length, fittings and equipment — not on pump speed, which shifts the PUMP curve. So b is false.
15. Use of soft starters for induction motors results in
lower mechanical stress
lower power factor
higher maximum demand
All the above
Answer: A) lower mechanical stress
Confirmed vs Book-3 §2.9 Soft Starter — The book's listed advantages of soft start are less mechanical stress, improved power factor, lower maximum demand and less mechanical maintenance. (b) and (c) state the exact opposite of two of those advantages, so (d) 'all the above' cannot hold.
16. Which of the following parameters is not required for evaluating volumetric efficiency of reciprocating air compressor?
Power input
FAD
Cylinder Stroke
Cylinder bore
Answer: A) Power input
Confirmed vs Book-3 §3.3 Compressor Efficiency — Volumetric efficiency = FAD / compressor displacement, and displacement = (π/4) x D² x L x S x X x n, so bore, stroke, speed and cylinder count plus the measured FAD are all needed.
Power input is used for isothermal/adiabatic/mechanical efficiency only, so it is the parameter not required here.
17. Capacitors with automatic power factor controller when installed in a plant:
reduces apparent power drawn from grid
reduces the voltage of the plant
reduces the reactive power drawn from grid
increases the load current of the plant
Answer: C) reduces the reactive power drawn from grid
Confirmed vs Book-3 §1.4 — capacitors 'act as reactive power generators', supplying the magnetising kVAr locally so it need not be drawn from the grid; the APFC keeps that compensation matched to a fluctuating load.
Option (d) is the opposite of what happens — the line current falls; note that the apparent power (a) also falls, but the direct, primary action of a capacitor is on the REACTIVE power.
18. A cooling tower is said to be performing well when:
approach is closer to zero
range is closer to zero
approach is larger than design
range is larger than design
Answer: A) approach is closer to zero
Confirmed vs Book-3 Ch7 — Approach = cold-water outlet temperature − ambient wet-bulb; the closer the approach to zero, the better the tower is performing (a). A larger-than-design approach (c) means poor performance; range depends on heat load, so 'range closer to zero' (b) is not a performance criterion.
📖 Book-3 Ch-3 Compressed Air (dryers) - cross-chapter
19. The most energy intensive dryer among the following
refrigeration
desiccant (heat of compression)
desiccant (heatless purge)
desiccant (blower reactivated)
Answer: C) desiccant (heatless purge)
Confirmed vs Book-3 Ch-3 (Compressed Air, dryers) - Heatless purge desiccant dryers consume the most energy because a large fraction of compressed air is used for purging. Refrigeration and heat-of-compression dryers (a, b) use little or no purge air, and blower-reactivated units (d) recover part of it; the heatless purge type diverts the largest share of dried compressed air, making it the most energy intensive.
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme
20. In BEE Star labelled distribution transformers, which of following losses are defined?
total loss at 50% and 100% loading
total loss at 75 % loading
total loss at 75% and 100% loading
total loss at 100% loading
Answer: A) total loss at 50% and 100% loading
Confirmed vs Book-3 §1.5 Standards & Labeling — 'For the BEE labeling programme total losses at 50% and 100% load have been defined' (Table 1.4).
Option (c) 75% and 100% is the standard distractor; 75% loading is not a BEE star-rating reference point.
📖 §1.4 Performance Assessment of Power Factor Capacitors
21. If V1 is actual supply voltage and V2 is the rated voltage of a capacitor, the reactive kVAr produced would be in the ratio of
V1²/V2²
V1²/V2
1 - V1²/V2²
1 + V1²/V2²
Answer: A) V1²/V2²
Confirmed vs Book-3 §1.4 Voltage effects — capacitor output varies with the square of the applied voltage, so the kVAr actually produced is in the ratio V₁²/V₂² of the rated value (V₁ = actual, V₂ = rated).
Option (c) 1 − V₁²/V₂² gives the FRACTIONAL DROP in output, not the output itself — that form is used when a question asks 'by how much does the VAr output drop'.
22. The blow down loss in a cooling tower depends on
TDS in circulating water
TDS in make-up water
evaporation loss
all the above
Answer: D) all the above
Confirmed vs Book-3 §7.2 (vi)–(vii) — Blow Down = Evaporation/(COC − 1) and COC = TDS in circulating water / TDS in make-up water, so blowdown depends on all three quantities → (d).
23. Identify the wrong statement from the following regarding Vapor Compression Refrigeration system
condenser rejects heat to atmosphere
evaporator removes heat from process or space
compressor sends superheated vapor to condenser
high pressure sub-cooled liquid refrigerant returns back to evaporator
Answer: D) high pressure sub-cooled liquid refrigerant returns back to evaporator
Confirmed vs Book-3 §4.3 - Liquid passes through an expansion device which drops it to low pressure before entering the evaporator; it does not return as high-pressure sub-cooled liquid. Statements (a), (b) and (c) all match Figure 4.4; (d) is the wrong one because the high-pressure sub-cooled liquid must pass through the expansion device first, arriving at the evaporator as a low-pressure wet vapour.
24. Which of the following is a positive displacement compressor?
Screw compressor
Reciprocating compressor
Centrifugal compressor
Both a & b
Answer: D) Both a & b
Confirmed vs Book-3 §3.2 Compressor Types (Figure 3.2 Compressor Chart) — Figure 3.2: positive-displacement compressors comprise the reciprocating and rotary (screw, vane, roots, scroll, liquid ring) families — they raise pressure by reducing the volume of the trapped gas.
Centrifugal machines are dynamic: they add velocity which is then converted to pressure at the outlet. So both the screw and the reciprocating machine qualify.
25. The ratings of the PF correction capacitors at motor terminals for a 37 kW induction motor at 3000 rpm synchronous speed will be __________ in comparison to the same sized induction motor at 1500 rpm synchronous speed
more
less
same
dependent on the connected load
Answer: B) less
Confirmed vs Book-3 §1.4 — 3000 rpm synchronous speed is a 2-pole machine and 1500 rpm is a 4-pole machine; the no-load magnetising kVAr (on which the terminal capacitor is sized) is smaller for the higher-speed, lower-pole machine.
So the 3000 rpm motor needs LESS kVAr; option (a) reverses the relationship between pole number and magnetising current.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
26. A pump discharge has to be reduced from 120 m³/hr to 100 m³/hr by trimming the impeller. What should be the percentage reduction in impeller size?
83.3%
16.7%
50.0%
33.3%
Answer: B) 16.7%
Confirmed vs Book-3 §6.5 — Q∝D: D₂/D₁ = 100/120 = 0.833, so the diameter reduction = 16.7%. 83.3% is the remaining ratio; 33.3% ≈ head reduction (1 − 0.833²), a tempting confusion of the H∝D² law.
📖 Ventilation by air changes (Book-3 Ch9 DG set room / Ch4 HVAC): flow = room volume × ACH
27. In an engine room 15 m long, 10 m wide and 4 m high, ventilation requirement for 20 air changes/hr is _____ m³/hr
30
3000
12000
none of the above
Answer: C) 12000
Confirmed vs Book-3 (ventilation rule) — Room volume = 15 × 10 × 4 = 600 m³; at 20 air changes per hour the ventilation air required = 600 × 20 = 12,000 m³/hr. 30 (a) is the sum of dimensions and 3000 (b) is a slip.
📖 §2.7 Improving the Motor Loading by Operating in Star Mode
28. The inexpensive way to improve energy efficiency of a motor which operates consistently at below 40% of rated capacity is by ___
Operating in Star mode
Replacing with correct sized motor
Operating in delta mode
Operating in VFD mode
Answer: A) Operating in Star mode
Confirmed vs Book-3 §2.7 Improving the Motor Loading by Operating in Star Mode — For motors consistently loaded below 40 % of rating, permanent star operation is the inexpensive fix — re-wiring the terminal box and resetting the overload relay, with no new equipment. (d) VFD operation would also cut losses but requires substantial capital, failing the 'inexpensive' criterion.
📖 §6.5 Efficient pumping system operation — Table 6.1 symptoms / best solutions; §6.11 Energy conservation opportunities in pumping systems (over-designed pump: VSD, downsize impeller, or replace with correct-sized pump)
29. If the delivery valve of the pump is throttled such that it delivers 30% of the rated flow, one of the best options for improved energy efficiency would be
Trimming of the impeller
Replacing the motor
Replacing with a smaller pump operating with VFD
none
Answer: C) Replacing with a smaller pump operating with VFD
Confirmed vs Book-3 §6.5/§6.11 — A pump throttled to 30% of rated flow is grossly oversized and most of its head is dissipated across the valve. Trimming is limited to ~75% diameter (Q∝D → ≥75% flow), so it cannot reach 30%; replacing the motor does nothing about the hydraulic waste. The book's remedy for an over-designed pump — 'replace with correct sized pump' and VSD for variable flow — is option c.
📖 Book-3 §3.3 Compressor Performance — Free Air Delivery (FAD)
30. For an air compressor of rated capacity of 100 CFM and system leakage of 10%, free air delivery is ____
111.11 CFM
90 CFM
100 CFM
None of the above
Answer: C) 100 CFM
Confirmed vs Book-3 §3.3 Compressor Performance — FAD is the volume of air drawn from the atmosphere, compressed and delivered by the machine — it is measured at the compressor, so leaks downstream in the distribution network do not change it.
The rated 100 CFM therefore stands; 90 CFM would be the air actually reaching end-use points, which is a distribution loss and not the FAD.
31. If 30,000 kcal of heat is removed from a room every hour then the refrigeration tonnage will be exactly equal to
30 TR
15 TR
10 TR
100 TR
Answer: C) 10 TR
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/h (book value); 30000/3024 ≈ 9.9 ≈ 10 TR. Option (a) omits the division by 3024 and (b) halves it; 30,000/3024 = 9.92 TR, which the paper rounds to 10 TR (30,240 kcal/h would be exactly 10 TR).
32. A 500 cfm reciprocating compressor has a loading and unloading period of 5 seconds and 20 seconds respectively during a measured period of leakage test. The air leakage in the compressed air system would be ____
125 cfm
100 cfm
200 cfm
none of the above
Answer: B) 100 cfm
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — % leakage = T/(T+t) x 100 = 5/(5+20) x 100 = 20%, giving leakage = 0.20 x 500 = 100 cfm.
Option (a) 125 cfm comes from the load:unload ratio 5/20; the denominator must be the total cycle of 25 seconds.
33. The Solar Heat Gain Coefficient (SHGC) of window of a building is 0.30. This means that
The window reflects back to exterior a minimum of 30 % of the sun's heat
The window allows 30 % of the sun's heat to pass through into the building interior
70 % of the sun's heat is reflected on the window
The window allows 70 % of the sun's heat to pass through into interior of the building
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 = solar heat gain through fenestration ÷ total incident solar radiation; SHGC 0.30 means 30% of the sun's heat passes into the interior. The statement says nothing about the reflected share, so (a)/(c) are wrong and (d) inverts the ratio.
📖 §10.14 Star rating of buildings — Energy Performance Index (EPI)
35. Energy performance index is calculated based on
total building annual energy consumption /built up area
total building annual energy consumption /carpet area
total building annual energy consumption for HVAC and lighting /carpet area
none of the above
Answer: A) total building annual energy consumption /built up area
Confirmed vs Book-3 §10.14 — EPI = total building annual energy consumption ÷ built-up area (kWh/m²/yr); all energy, not just HVAC and lighting, and built-up (not carpet) area.
36. 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.
37. The gross efficiency of a coal based power generating unit with a gross heat rate of 2600 kcal/kWh is
41.4%
38.7%
33.1%
30.8%
Answer: C) 33.1%
Confirmed vs Book-3 §9.1 — Gross efficiency = 860/2600 = 0.3308 = 33.1%, within the 33–36% conventional steam plant band of Table 9.1. 38.7% (b) would need a 2222 kcal/kWh heat rate.
38. The COP of a vapour compression refrigeration system is 3.3. If the motor draws power of 10 kW at an operating efficiency of 90%, the tonnage of refrigeration system is about:
0.8
8.5
7.2
9.6
Answer: B) 8.5
Confirmed vs Book-3 §4.7 - Refrigeration effect = COP×shaft power = 3.3×(10×0.9) = 29.7 kW; TR = 29.7/3.517 ≈ 8.45 ≈ 8.5 TR. Option (c) 7.2 forgets the COP of 3.3 and (d) 9.6 skips the motor efficiency; shaft power = 10 x 0.9 = 9 kW, cooling = 29.7 kW, and TR = 29.7/3.517 = 8.5.
40. In T-5 Fluorescent Lamp, '5' is indicative of:
5 watt power rating
5% energy saving with respect to T8
5/8 generation lamp
Tube diameter
Answer: D) Tube diameter
Confirmed vs Book-3 §8.3 — T5 means a tube of 5/8 inch (16 mm) diameter; T8 = 1 inch (25 mm), T12 = 1.5 inch (38 mm). The number is neither wattage nor a saving percentage (the book's 5 % figure is the T5/T8 efficacy gain over T12, unrelated to the name).
41. 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.
📖 §9.2 Sizing — kVA = √3·V·I; §9.4 specific fuel consumption
42. In a DG set, a 3-phase alternator is supplying on an average 100 A at 420 V and 0.9 pf to a load. If the specific fuel consumption of this DG set is 0.30 lit/kWh at that load, then how much fuel is consumed while delivering generated power for one hour?
11.34 litre
19.64 litre
21.82 litre
65.50 litre
Answer: B) 19.64 litre
Confirmed vs Book-3 §9.2/§9.4 — Power = √3 × 420 × 100 × 0.9 = 65,466 W = 65.47 kW; fuel = 65.47 × 0.30 = 19.64 litres per hour. (c) 21.82 L ignores the 0.9 PF; (d) 65.50 is the kW value mis-read as litres.
43. The total loss for a transformer loading at 60% with no load and full load losses of 3 kW and 25 kW respectively, would be
3 kW
12 kW
18 kW
25 kW
Answer: B) 12 kW
Confirmed vs Book-3 §1.5 — total loss = no-load loss + (%load)² × full-load copper loss = 3 + (0.6)² × 25 = 3 + 9 = 12 kW.
Option (c) 18 kW scales the copper loss linearly with load; the book's formula squares the load fraction because P = I²R.
44. A process fluid at 40 m³/hr, with a density of 0.95, is flowing in a heat exchanger and is to be cooled from 35 °C to 29 °C. The fluid specific heat is 0.78 kCal/kg. The chilled water range across the heat exchanger is 4 °C, the chilled water flow rate is
44.46 m³/hr
40.41 m³/hr
35.37 m³/hr
none of the above
Answer: A) 44.46 m³/hr
Confirmed vs Book-3 §4.7 - Heat load = 40×0.95×1000×0.78×(35-29) = 177840 kcal/hr; chilled water flow = 177840/(1000×1×4) = 44.46 m³/hr. Option (b) copies the process flow rate; the chilled-water flow must carry the same duty (177,840 kcal/h) over its own 4 degC range, giving 44.46 m3/hr.
45. In which of the following fans the air does not change flow direction from suction to discharge?
tube axial fan
vane axial fan
propeller fan
all the above
Answer: D) all the above
Confirmed vs Book-3 §5.2 — Tube-axial, vane-axial and propeller are all axial-flow fans, in which 'air enters and leaves the fan with no change in direction' → all the above (d). Only centrifugal fans turn the airflow (twice).
📖 §10.5 Building envelope — Window-Wall Ratio (WWR)
46. What is window to wall ratio __________
Vertical fenestration area / gross exterior wall area
Vertical fenestration area / Net exterior wall area
gross exterior wall area/ Vertical fenestration area
Net exterior wall area/ Vertical fenestration area
Answer: A) Vertical fenestration area / gross exterior wall area
Confirmed vs Book-3 §10.5 — WWR = vertical fenestration area ÷ gross exterior wall area (gross wall measured horizontally from the exterior surface and vertically from top of floor to bottom of roof). 'Net' wall area and the inverted ratios are wrong.
📖 §9.1 Diesel generator captive power plants — Table 9.1
47. The maximum thermal efficiency of a diesel engine power plant is in the range of _____
43-45 %
53-55%
63-65 %
73-75%
Answer: A) 43-45 %
Confirmed vs Book-3 §9.1 — "Higher efficiency (as high as 43–45%)" is listed among the advantages of diesel power plants and Table 9.1 gives 43–45% for diesel engine plants. 53–55% and above are beyond reciprocating engines (the book's best lean-burn gas engine reaches close to 45%).
48. The advantage of multi-staging compression over single stage compression is
Lower power consumption per unit of air delivered
High volumetric efficiency
Decreased temperature of discharge
All of above
Answer: D) All of above
Confirmed vs Book-3 §3.2 Positive Displacement — Two-stage machines discharge at 140-160°C against 205-240°C for single stage, and the checklist states that a two-stage or multistage compressor 'consumes less power for the same air output than a single stage compressor'.
Inter-cooling and the lower pressure ratio per stage also raise volumetric efficiency, so all three listed advantages hold.
📖 §1.11 (lighting terminology; see Book-3 Ch-8 §8.2 Lighting System)
49. A device that distributes filters or transforms the light emitted from one or more lamps is
Control gear
Luminaire
Lamp
Starter
Answer: B) Luminaire
Confirmed vs Book-3 Ch-8 §8.2 — a luminaire is the complete lighting unit that 'distributes, filters or transforms the light emitted from one or more lamps', including the housing, reflector and the parts that hold and protect the lamp.
Option (c) the lamp is only the light SOURCE inside the luminaire; control gear (a) is the ballast/driver circuit.
📖 §8.6(e) Voltage & losses (general electrical: P = VI, I²R loss)
50. For the same quantity of power handled by a distribution line, lower the voltage
lower the current drawn and lower the distribution loss
lower the voltage drop and lower the distribution loss
higher the current drawn and higher the distribution loss
higher the voltage drop and lower the distribution loss
Answer: C) higher the current drawn and higher the distribution loss
Confirmed (general principle; not specific to Book-3 Ch.8) — For the same power P = V·I, a lower voltage means a proportionally higher current, and distribution loss I²R rises with the square of the current, so both current and losses are higher.
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19); Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
1. A plant has installed a refrigerant dryer for supplying dry air for their process applications and dryer coil is maintained at 5 °C & 100% RH. The average air flow through the dryer is 100 kg/min. The air properties are given below — Inlet at 35 °C & 50% RH: Enthalpy 81 kJ/kg of dry air, Absolute Humidity 18 grams/kg of dry air; Dryer coil 5 °C & 100% RH: Enthalpy 19 kJ/kg of dry air, Absolute Humidity 5.5 grams/kg of dry air. (i) Calculate the moisture removed per hour. (ii) Cooling capacity of coil in TR. (iii) List down any three energy saving measures in compressed air systems.
Model answer: (i) Moisture removed = mass flow x change in absolute humidity = 100 x (18 - 5.5) = 1250 grams/min = 1.25 kg/min = 75 kg/hr.
(ii) Cooling load = 100 x (81 - 19) = 6200 kJ/min = 6200/4.186 = 1481.1 kcal/min = 88,868 kcal/hr; TR = 88,868/3024 = 29.39 TR.
(iii) Three energy saving measures in compressed air systems (Book-3 §3.7 checklist): (1) carry out periodic leak tests and arrest leaks, since 40-50% leakage is not uncommon; (2) reduce the compressor delivery pressure to the minimum the plant needs, saving about 6-10% of power per bar; (3) draw cool, clean intake air from outside the compressor room and clean the inlet filters regularly — every 4°C rise in inlet temperature costs about 1% more power and every 250 mmWC filter pressure drop about 2%. (Also acceptable: eliminate misuse of compressed air, retrofit VSDs to eliminate unloaded running, or size piping generously as a ring main.)
Confirmed vs Book-3 §3.5 Air Dryers — Moisture removed = mass flow x change in absolute humidity; cooling load from the enthalpy difference (1 TR = 3024 kcal/h). Part (iii) now lists three actual measures instead of a page reference.
2. The size of an air-conditioned office is 12 m × 7 m. Desired illuminance level is 200 Lux. An architect has suggested to install 24 no's of 20 W LED lights at a height of 3 m from ground level. The working plane is 0.75 m above the floor. The other details of 20W LED lamps are: Output of LED Lamps 2000 lumens; Utilization factor 0.65; Light Loss Factor (LLF) 0.75. Calculate floor index & number of LED lights required to get the desired illuminance. As an energy manager do you agree with the architect decision = why?
Model answer: Mounting Height, Hm = 3 - 0.75 = 2.25 m. Room Index (RI) = (L×W)/[Hm×(L+W)] = (12×7)/[2.25×(12+7)] = 84/42.75 = 1.97. Number of LED lights = (E×A)/(F×UF×LLF) = (200×12×7)/(2000×0.65×0.75) = 16800/975 = 17.23 ≈ 18 lights. So total number of 20W LED lights required is 18. No, I don't agree with the architect because the number of LED lights required is only 18 against suggested 24 nos, which is an energy inefficient design.
Book-3 §8.5: Hm = 3 − 0.75 = 2.25 m; RI = (12×7)/[2.25×(12+7)] = 1.97; N = (200×84)/(2000×0.65×0.75) = 17.2 → 18 lamps, so 24 lamps would over-light the office (energy inefficient).
3. A centrifugal fan drawing 16 kW and operating at 1440 RPM is delivering air at 30000 m³/hr. The head developed by the fan is 400mmWC. If the speed is decreased by 200 rpm, calculate the following: a) Air flow in m³/hr; b) Static Pressure in mmWC; c) Power drawn in kW.
Model answer: New speed = 1440 − 200 = 1240 rpm.
a) Air flow = (1240/1440) × 30,000 = 25,833 m³/hr.
b) Static pressure = (1240/1440)² × 400 = 296.6 mmWC.
c) Power drawn = (1240/1440)³ × 16 = 0.6385 × 16 = 10.22 kW.
(If the fan power in the paper is 54 kW, as in the Sep-2024 repeat of this question, part (c) = 0.6385 × 54 = 34.48 kW.)
Book-3 §5.3 fan laws: Q ∝ N, SP ∝ N², kW ∝ N³ with N₂/N₁ = 1240/1440 = 0.861. The earlier answer text ('34.48 → 9.54 kW') was garbled — 34.48 kW belongs to the 54 kW version; with 16 kW the power is 10.22 kW.
📖 §1.4 Power Factor Improvement and Benefits + §1.5 — multi-chapter True/False
4. State True or False (1 Mark each): 1. In an industrial electrical system operating at unity power factor, addition of further capacitors will reduce the maximum demand (kVA). 2. In a step-down transformer for a given load the current in the primary will be much lower than the current in the secondary. 3. For the same no of poles and kVA rating, the RPM of an energy efficient motor is higher than that of a standard motor. 4. The advantage of evaporative cooling is that it is possible to obtain water temperatures below the wet bulb economically. 5. A fluid coupling changes the speed of the driven equipment without changing the speed of the motor.
Model answer: 1. FALSE - at unity power factor the reactive component is already zero; adding more capacitors makes the current LEAD and the kVA (and hence maximum demand) rises again.
2. TRUE - primary ampere-turns = secondary ampere-turns, so V1I1 = V2I2. In a step-down transformer the primary is the high-voltage side and therefore carries the LOWER current.
3. TRUE - an energy-efficient motor has lower rotor I^2R loss and hence lower slip, so for the same number of poles and rating it runs at a slightly HIGHER rpm than a standard motor.
4. FALSE - evaporative cooling can approach but never economically go below the wet bulb temperature; the difference (cold water temp - WBT) is the 'approach'.
5. TRUE - a fluid coupling varies the output (driven) speed by varying the oil fill while the motor continues to run at its own constant speed.
At unity PF capacitors over-correct and can increase kVA; primary current of step-down transformer is lower than secondary; EE motor runs at nearly same/slightly higher speed but not 'higher RPM' as stated; evaporative cooling cannot go below wet bulb; fluid coupling varies output speed while motor runs constant.
📖 §6.2 System characteristics — static & friction head
5. The total static resistance of a water supply piping system is 30 meters and the static head is 10 meters at designed water flow. Calculate the system resistance offered at 75%, 50% and 25% of design water flow.
Model answer: Static Head = 10 m (Static head will remain same irrespective of the flow). So, Dynamic Head at designed water flow: (30-10) = 20 m. At 75% flow: Static 10 + Dynamic (0.75²×20=11.25) = 21.25 m. At 50% flow: Static 10 + Dynamic (0.5²×20=5.0) = 15.0 m. At 25% flow: Static 10 + Dynamic (0.25²×20=1.25) = 11.25 m.
Static head constant; dynamic (friction) head varies with square of flow; total = static + dynamic.
6. An energy audit study of a central chiller system in a commercial building was conducted and measured parameters are given below: Chilled water inlet temperature 12 °C; Chilled water Outlet temperature 7 °C; Chilled water pump discharge pressure 3.6 kg/cm²g; Pump suction 1.5 meters above the pump-center line; Power drawn by the chilled water pump motor 70 kW; Efficiency of pump motor 91%; Pump efficiency 60%. Find out the operating load of the Chiller system in TR.
Model answer: Discharge head = 3.6 kg/cm2g x 10 = 36 m; the suction is 1.5 m above the pump centreline, so total head = 36 - 1.5 = 34.5 m. Pump shaft power = motor input x motor efficiency = 70 x 0.91 = 63.7 kW. Flow = (shaft power x 1000 x pump efficiency)/(head x 1000 x 9.81) = (63.7 x 1000 x 0.6)/(34.5 x 1000 x 9.81) = 0.1129 m3/s = 406.5 m3/hr. Refrigeration load = 406,500 x 1 x (12 - 7)/3024 = 672 TR.
Hydraulic power = flow x head; rearranged, flow = (pump shaft power x pump efficiency)/(head x 9.81). Total head = discharge head (3.6 kg/cm2g = 36 m) minus the 1.5 m positive suction lift = 34.5 m. Refrigeration load then follows from TR = mass flow x Cp x deltaT / 3024 (1 TR = 3024 kcal/hr).
7. Fill in the blanks for the following: 1. The main input energy used for refrigeration in vapor absorption refrigeration plants is _____. 2. One ton of refrigeration is equivalent to _____ kW. 3. Stray losses in induction motor generally are proportional to the square of the _____ current. 4. The unit of Ah*EPI is _____. 5. If the pump impeller diameter is reduced by 10% then head reduces by _____%. 6. A 4 pole 50Hz motor operating with slip of 3% will have a shaft speed of _____ RPM. 7. Effective Aperture Glazing (EA) = VLT × _____. 8. In an amorphous core distribution transformer, no-load loss is _____ than a conventional transformer. 9. As the condensing temperature increases, kW/TR of refrigeration system will _____. 10. The extent of drying compressed air is expressed by the term _____.
Model answer: 1. Thermal energy (steam / waste heat / hot water / fuel gas) - the vapour absorption machine is heat-driven, not compressor-driven.
2. 3.51 kW (1 TR = 3024 kCal/h = 3.51 kW).
3. ROTOR current - stray load losses in an induction motor vary as the square of the rotor current.
4. EPI (Energy Performance Index) is expressed in kWh/sq.m/year.
5. 19% - head varies as the square of impeller diameter, so 1 - 0.9^2 = 0.19.
6. 1455 rpm - Ns = 120 x 50/4 = 1500 rpm; at 3% slip, N = 1500 x 0.97 = 1455 rpm.
7. Window-to-Wall Ratio (WWR): Effective Aperture = VLT x WWR.
8. LESS (about 70% lower core loss than a conventional CRGO silicon-iron core - Book-3 Sec.1.5).
9. INCREASE - a higher condensing temperature raises the compression ratio and hence the specific power kW/TR.
10. Dew point (atmospheric / pressure dew point).
Fill-in answers from BEE Book-3 fundamentals; 1 TR=3.517 kW; head∝D² so 10% dia drop → ~19% head drop; N=120f/p×(1-s)=1500×0.97=1455 rpm.
📖 §10.14 EPI · §10.15 AC measures · §10.5 Building envelope & SHGC · §10.8 LPD methods
8. Write short notes on the following (each 2 Marks): a) Energy Performance Index (EPI); b) List any two Energy Efficiency measures in Building air conditioning system; c) Building Envelop from an energy efficiency point of view; d) Difference between building area method and space function method for deriving Lighting Power density (LPD); e) Solar Heat Gain Coefficient (SHGC).
Model answer: a) Energy Performance Index (EPI), §10.14: the specific energy usage of a building = total annual energy consumption (purchased + generated electricity) ÷ built-up area, in kWh/sq m/year; it is the basis of BEE star rating (1–5 Star; label valid 5 years), with EPI bandwidths set per climatic zone and % air-conditioned area — e.g. composite zone, > 50% AC: band 190–90, 5 Star below 90, 1 Star at 165–190 kWh/m²/yr. Built-up area = carpet area + wall thickness + balconies, excluding parking basements. b) Two AC energy-efficiency measures, §10.15: (i) maintain chilled-water leaving temperature at or above 7 °C — centrifugal chiller efficiency improves ~2.5% per 1 °C rise; (ii) install frequency converters to vary AHU fan speed, cutting fan-motor energy by up to 15% (others: weather stripping, 23–25 °C/55–65% RH, 6-monthly condenser cleaning, clean filters). c) Building envelope, §10.5: the exterior façade — walls, windows, roof, skylights, doors and openings — separating conditioned space from the weather/unconditioned spaces. From an energy viewpoint it must handle external loads (solar gain through windows, heat loss/gain across surfaces, infiltration) and internal loads, admit daylight to cut electric lighting, and regulate heat transfer through insulation of roof/walls, proper glazing and framing, shading, cool roofs and sealing/weather stripping of all joints and openings. d) Building Area Method vs Space Function Method, §10.8: both give the interior lighting power allowance = Σ (gross lighted floor area × allowed LPD in W/m²). The Building Area Method applies one LPD by TYPE OF BUILDING to the whole gross lighted area (hotel: 4 × 1000 m² × 10.8 W/m² = 43,200 W); the Space Function Method applies a separate LPD by TYPE OF OPERATION/space and sums the allowances for all spaces (enclosed office 400 m² × 11.8 W/m² = 4,720 W). e) SHGC, §10.5: the ratio of solar heat gain that passes through fenestration to the total incident solar radiation, including directly transmitted and absorbed-then-re-radiated/conducted/convected heat; 0–1, lower means less solar gain (ECBC max e.g. 0.25 for WWR ≤ 40% in hot zones).
Book-3 §10.14 (EPI kWh/m²/yr, star bands), §10.15 (CHW ≥ 7 °C, AHU VFD 15%), §10.5 (envelope definition and design basics; SHGC ratio), §10.8 (building-area vs space-function LPD methods with the two worked examples).
1. A DISCOM has taken initiatives to reduce Aggregate Technical & Commercial (AT&C) losses in their network. The energy supplied, received and revenue details are given below — Input energy: 50 MU; Billed Energy (Metered): 39 MU; Billed Energy (Un-metered): 2 MU; Amount Billed: Rs. 470 Million; Amount Collected: Rs. 30 Million; Gross Amount collected: Rs. 390 Million. a) Estimate the AT&C losses (in %). b) List any four strategies to reduce the commercial losses.
Model answer: a) Billing efficiency = total units billed / total input units = (39 + 2)/50 x 100 = 82.0%.
Collection efficiency = amount collected excluding arrears / amount billed. Stripping the Rs.30 million of arrears out of the Rs.390 million gross collection: Ac = 390 - 30 = Rs.360 million, so CE = 360/470 x 100 = 76.6%.
AT&C loss = [1 - (BE x CE)] x 100 = [1 - (0.820 x 0.766)] x 100 = [1 - 0.6281] x 100 = 37.2%.
b) Four measures to reduce commercial losses (Book-3 Sec.1.8): (1) accurate metering with a planned meter-replacement programme and meters matched to the connected load; (2) installation of electronic meters with TOD, tamper-proof and remote/data-reading facility; (3) intensive inspections and eradication of theft/pilferage; (4) compulsory metering in place of average billing, backed by energy audit to pinpoint high-loss areas and by improved collection/arrear recovery.
2. The input parameter measured for a 15 kW, 3 phase, 415 V induction motor are 25 A and 12 kW at 410 V. Calculate the following: a) Apparent power drawn by the motor at the operating load (3 Marks); b) Reactive Power drawn by the motor at the operating load (1 Mark); c) Operating power factor (1 Mark).
Model answer: a) Apparent power = √3 × 0.410 × 25 = 17.75 kVA. b) Reactive power = sqrt(apparent power² - active power²) = sqrt(17.75² - 12²) = 13.07 kVAr. c) Operating power factor = Active power/Apparent power = 12/17.75 = 0.676.
3. In a steel industry, cooling water of 7500 m³/hr and 4200 m³/hr from two different sections with temperatures of 38 °C and 55 °C respectively, are fed to cooling tower after proper mixing. If the measured heat rejection by the cooling tower is 38,000 TR, calculate the effectiveness and evaporation loss of the cooling tower at 28 °C WBT.
Model answer: Mixed Hot Water Temp, °C = [(Flow1 × Temp1) + (Flow2 × Temp2)] / (Total Flow) = [(7500 × 38) + (4200 × 55)] / 11700 = 44.1 °C. Range of Cooling Tower, °C = Heat Rejection / (Flow × Density × Sp. Heat) = (38000 × 3024) / (11700 × 1000 × 1) ≈ 9.82 °C [book: 9.82]. Cold Water Temp, °C = Hot Water Temp - Range = 44.1 - 9.82 = 34.28 °C. Approach, °C = Cold Water Temp - WBT of Air = 34.28 - 28 = 6.28 °C. Effectiveness = Range / (Range + Approach) = 9.82 / (9.82 + 6.28) = 60.99% or 61%. Evaporation Loss (m³/hr) = 0.00085 × 1.8 × circulation rate (m³/hr) × Range = 0.00085 × 1.8 × 11700 × 9.82 = 175.8 m³/hr.
Mixed temp by flow-weighted average; range from heat rejection; effectiveness = range/(range+approach); evaporation = 0.00085×1.8×flow×range. ⚠ Question stem repaired: garbled 'heat rejection ... is at 1700 m³/hr' → '38,000 TR' (the value used in the printed solution: 38000 × 3024 / 11,700,000 = 9.82 °C range).
4. The data for centrifugal chiller and vapour absorption chiller are given below — Chilled water flow (m³/h): Centrifugal 189, VAM 180; Condenser water flow (m³/h): Centrifugal 258, VAM 340; Chiller inlet temp (°C): Centrifugal 13.0, VAM 14.6; Condenser water inlet temp (°C): Centrifugal 27.1, VAM 33.5; Chiller outlet temp (°C): Centrifugal 7.7, VAM 9.0; Condenser water outlet temp (°C): Centrifugal 35.7, VAM 39.1; Power drawn by compressor (kW): Centrifugal 190, VAM -; Steam consumption (kg/h): Centrifugal -, VAM 1570; Chilled water pump (kW): Centrifugal 28, VAM 28; Condenser water pump (kW): Centrifugal 22, VAM 33; Cooling tower fan (kW): Centrifugal 6.0, VAM 15; Cost of Steam (Rs/kg): VAM 2.0; Cost of electricity (Rs/kWh): 9.0 both. a) Evaluate the tonnes of refrigeration (TR) of both the systems. b) Operating Energy cost per hour for both the systems.
Model answer: a) Centrifugal chiller TR = Chilled water flow × (Tin - Tout) × Diff. in temp / 3024 = 189 × 1000 × 1 × (13-7.7)/3024 = 331.25 TR. VAM TR = 180 × 1000 × 1 × (14.6-9.0)/3024 = 333.33 TR. b) Auxiliary power consumption: Centrifugal = Chilled water pump + condenser water pump + cooling tower fan = 28 + 22 + 6.0 = 56 kW. VAM auxiliary power (kW) = 28 + 33 + 15 = 76 kW. Energy cost of centrifugal chiller = (56 + 190)×9 = Rs 2214/hr. Energy cost of VAM chiller = (76×9) + (1570×2) = Rs 3824/hr.
TR = flow×ΔT/3024 (kcal→TR); energy cost = electrical kW×rate (+ steam kg×rate for VAM).
📖 §9.4 Energy performance assessment — trial data (fuel by dip level, kWh, PF), % loading, kWh/litre (part b: Book-3 compressed-air chapter)
5. During the performance evaluation of a DG set, the following parameters were noted — Capacity of DG set: 750 kVA; Test duration: 36 minutes; Units generated: 250 kWh; Average Power factor: 0.92 pf; Length of diesel tank: 100 cm; Width of diesel tank: 100 cm; Height of the diesel tank: 90 cm; Initial tank dip level (from top): 63 cm; Final tank dip level (from top): 53 cm. Calculate the following: 1. Diesel consumption (Litres) (1 Mark); 2. Average load (kW) (1 Mark); 3. Percentage Loading (%) (2 Marks); 4. Specific power generation (kWh/Litre) (1 Mark). b) A medium sized engineering industry has installed two 480 CFM screw compressors, A & B. Compressor-A is operating at full load and Compressor-B is running in load–unload condition. The load power of both the compressor is 74 kW and the unload power of the Compressor-B is 26 kW. Both the compressors are operated during working day. The percentage loading of the Compressor-B during working day is 70 %. After arresting the leakage in the system the loading of the compressor was found to be 35 %. Estimate the energy savings per day.
Model answer: a) 1. Diesel consumption = (1×1×0.1)×1000 = 100 Liters (level drop 63→53 = 10 cm = 0.1 m over 1m×1m tank). 2. Average load = (250/36)×60 = 416.67 kW. 3. Percentage Loading = (416.67×100)/(750×0.92) = 60.4 %. 4. Specific power generation = (250/100) = 2.5 kWh/Litre. b) Existing Case: Energy consumed per hour by Compressor-A = 74 kWh. Energy consumed per hour by Compressor-B = 0.70×74 + 0.30×26 = 59.6 kWh. Energy consumed per day = 133.6 × 24 hrs = 3206.4 kWh/day. Leakage Calculation (after arresting): Energy/hr Comp-A = 74 kWh; Energy/hr Comp-B = 0.35×74 + 0.65×26 = 42.8 kWh. Savings by arresting leakage per day = 16.8 × 24 = 403.2 kWh/day.
Diesel volume from tank dip×area; avg load = kWh/hours; %loading = load/(kVA×pf); compressor energy from load fraction × load power + unload fraction × unload power; savings = difference × 24h.
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback
6. A review of electricity bills of a process plant was conducted as a part of energy audit. The plant has a contract demand of 3000 kVA with the power supply company. The average maximum demand of the plant is 2400 kVA/month at a power factor of 0.95. The maximum demand is at 80% of the contract demand. The minimum billable maximum demand is 80% of the contract demand. An incentive of 0.5 % reduction in energy charges component of electricity bill are provided for every 0.01 increase in power factor over and above 0.95. The average energy charge component of the electricity bill per month for the plant is Rs.80 lakhs. Calculate the following: a) If the plant decides to improve the power factor to unity, determine the power factor capacitor kVAr required and the associated monetary benefits. b) What will be the simple payback period if the cost of power factor capacitors is Rs.1200/kVAr.
Model answer: Contract demand 3000 kVA; recorded average MD = 2400 kVA at 0.95 PF; minimum billable demand = 80% x 3000 = 2400 kVA.
a) kW drawn = 2400 x 0.95 = 2280 kW.
kVAr for 0.95 -> unity = kW[tan(cos^-1 0.95) - tan(cos^-1 1)] = 2280 x (0.3287 - 0) = 749 kVAr (say 750 kVAr).
Maximum demand at unity PF = 2280 kVA, but the minimum billable demand is 2400 kVA, so the plant still pays for 2400 kVA - there is NO saving in maximum demand charges.
PF incentive = (1.00 - 0.95)/0.01 x 0.5% = 2.5% of the energy charge = Rs.80,00,000 x 2.5% = Rs.2,00,000/month = Rs.24,00,000/year.
b) Investment = 749 kVAr x Rs.1200/kVAr = Rs.8,99,000 (say Rs.9.0 lakh).
Simple payback = 8,99,000 / 24,00,000 = 0.375 year = about 4.5 months.
(Same structure as the Book-3 Sec.1.11 solved example: the minimum-billing-demand clause can wipe out the MD saving, leaving the PF incentive as the only benefit.)
kVAr = kW(tanφ1-tanφ2); MD reduction nil due to 80% minimum billing; energy charge incentive 0.5% per 0.01 PF rise above 0.95 → 2.5%; payback = investment/annual savings.