Energy Efficiency in Electrical Utilities Available here with full solutions — 63 questions recovered from the 2014 exam:
Objective (1 mark)
49 of 50
Short (5 marks)
8 of 8
Long (10 marks)
6 of 6
This is not the complete paper. The questions below are the ones we could recover and verify; the rest of that year’s paper is not reproduced here. Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 49
📖 §1.4 Selection and Location of Capacitors
1. The kVAr rating required for improving the power factor of a load operating at 500 kW and 0.85 power factor to 0.95 is
145 kVAr
500 kVAr
50 kVAr
100 kVAr
Answer: A) 145 kVAr
Confirmed vs Book-3 §1.4 — kVAr rating = kW[tanΦ₁ − tanΦ₂] = 500 × (tan cos⁻¹0.85 − tan cos⁻¹0.95) = 500 × (0.620 − 0.329) = 145 kVAr (Table 1.2 multiplier 0.291).
Option (b) 500 kVAr wrongly assumes the capacitor must equal the kW rating; only the difference in reactive components is compensated.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
2. In case of centrifugal pumps, impeller diameter changes are generally limited to reducing the diameter to about ____ of maximum size.
75%
50%
25%
none of the above
Answer: A) 75%
Confirmed vs Book-3 §6.5 — Book: 'Diameter changes are generally limited to reducing the diameter to about 75% of the maximum, i.e. a head reduction to about 50%. Beyond this, efficiency and NPSH are badly affected.' §6.6 repeats that impellers 'are rarely reduced below 75 percent of their original size'. 50% is the head reduction, not the diameter limit.
3. The capacity of the largest motor that can be started on a given DG set is
25%
50% kVA rating of DG set
75%
100%
Answer: B) 50% kVA rating of DG set
Confirmed vs Book-3 §9.2 — "the HP of the largest motor that can be started with direct on line starting is about 50% of the kVA rating of the generating set", because DOL starting current is ~6× rated and should not exceed 200% of alternator full-load capacity. 75% (c) applies only with star-delta or auto-transformer starting; 25%/100% are not book figures.
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
4. Which loss is considered the most unreliable or complicated to measure in electric motor efficiency testing?
stator Cu loss
rotor Cu loss
stator Iron loss
stray loss
Answer: D) stray loss
Confirmed vs Book-3 §2.4 Motor Efficiency — Field Tests for Determining Efficiency — Stray load losses are the hardest to determine — IEEE 112 gives a complicated method rarely used on the shop floor, and IS/IEC simply assume a fixed 0.5 % of input while the actual value is likely to be 1–3 %. (a)/(b) stator and rotor I²R losses are directly computable from measured resistance, current and slip, so they are not the unreliable ones.
5. A pure resistive load in an alternating current (AC) circuit draws
lagging reactive power
active power
leading reactive power
none of the above
Answer: B) active power
Confirmed vs Book-3 §1.4 Power Factor Basics — 'In case of pure resistive loads, the voltage (V), current (I), resistance (R) relations are linearly related' — such loads (incandescent lighting, resistance heating) draw only active power.
Options (a)/(c) require energy storage in a magnetic or electric field, which a pure resistance does not have.
6. Select the incorrect statement: The advantage of PF improvement by capacitor addition in an electric network is
active power component of the network is not affected
reactive power component of the network is not affected
I2R power losses are affected in the system
voltage level at the load end is affected
Answer: B) reactive power component of the network is not affected
Confirmed vs Book-3 §1.4 — the book's first listed advantage is that 'Reactive component of the network is reduced and so also the total current in the system'. Statement (b) claims the reactive component is NOT affected, which contradicts the book, so it is the incorrect statement asked for.
Options (c) and (d) restate the book's advantages (b) and (c), and (a) is true because capacitors do not change kW.
7. "Heat Rate" of a thermal power station is the heat input in kilo Calories or kilo Joules, for generating
one kW of electrical output
one kVAh of electrical output
one kWh of electrical output
one kVA of electrical output
Answer: C) one kWh of electrical output
Confirmed vs Book-3 §1.1 — ''HEAT RATE' is the heat input in kilo Calories or kilo Joules, for generating 'one' kilo Watt-hour of electrical output.'
Options (a)/(d) use kW and kVA, which are power (rate) units, not energy; heat rate is always expressed per unit of energy generated (kWh).
8. Improving power factor at motor terminals in a plant will
increase active power drawn by motor
reduce system distribution losses
reduce contract demand with utility
increase motor design power factor
Answer: B) reduce system distribution losses
Confirmed vs Book-3 §1.4 — 'Maximum benefit of capacitors is derived by locating them as close as possible to the load…This, in turn, will reduce power losses of the system substantially.' Correcting at the motor relieves the whole in-plant network of reactive current.
Option (c) is wrong: contract demand is a commercial agreement with the utility and is not automatically reduced by PF correction.
9. For a 6 pole induction motor operating at 49.5 Hz, the percentage slip at a shaft speed of 950 RPM will be
4.0 %
5.0 %
0.04 %
none of the above
Answer: A) 4.0 %
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 49.5 / 6 = 990 rpm; slip = (990 − 950)/990 × 100 = 4.04 % ≈ 4 %. (b) 5 % is what you get by wrongly using 1000 rpm (i.e. assuming 50 Hz) — the book's slip formula must use the synchronous speed at the ACTUAL supply frequency.
📖 §1.4 Performance Assessment of Power Factor Capacitors
10. A plant had installed three phase shunt capacitors to improve power factor at MCC. Busbar three phase voltages at the main panel were balanced but at the MCC the line voltages were unbalanced. The main reason for this unbalanced voltage at MCC could be
PF capacitors were operating at higher supply frequency
PF improvement in all phase was not uniform due to blown fuse in one phase of the 3 phase PF capacitors
PF capacitors were operating at higher voltage than their rated values
PF capacitors were operating at lower voltage than their rated values
Answer: B) PF improvement in all phase was not uniform due to blown fuse in one phase of the 3 phase PF capacitors
Confirmed vs Book-3 §1.4 — a capacitor bank supplies kVAr per phase; if one phase's fuse blows, that phase gets no compensation while the other two do, so the phase currents and hence the voltage drops become unequal at the MCC.
Options (c)/(d) would change the kVAr in all three phases equally and so would not create an imbalance.
📖 §1.4 Performance Assessment of Power Factor Capacitors
11. A 50 kVAr, 415 V rated power factor capacitor was found to be having terminal supply voltage of 430 V. The capacity of the power factor capacitor at the operating supply voltage would be approximately
53.67 kVAr
50 kVAr
46.57 kVAr
none of the above
Answer: A) 53.67 kVAr
Confirmed vs Book-3 §1.4 Voltage effects — capacitor output varies with the square of the applied voltage: 50 × (430/415)² = 50 × 1.0735 = 53.67 kVAr.
Option (c) 46.57 kVAr is what would happen at a LOWER-than-rated voltage; note the book's caution that over-voltage shortens capacitor life.
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
12. A 75 kW squirrel cage induction motor with static PF correction capacitors across the motor terminals got damaged along with capacitors once supply was switched off due to power failure. The possible reason for the motor burn out could be
motor was oversized
motor was undersized
charging current of the capacitors was more than the magnetizing current of the motor
charging current of the capacitor was only 85% of the motor magnetizing current
Answer: C) charging current of the capacitors was more than the magnetizing current of the motor
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — The book limits terminal capacitors to not more than 90 % of the motor's no-load kVAr, warning that higher capacitors could result in over-voltages and motor burn-outs; on supply failure an over-sized capacitor bank self-excites the still-spinning motor and the resulting over-voltage destroys motor and capacitors. (a)/(b) sizing of the motor itself has nothing to do with a burn-out that occurs at the instant of switch-off.
13. An induction motor rated for 75 kW and 94 % efficiency, operating at full load, will
deliver 70.5 kW
deliver 75 kW
draw 75 kW
deliver 79.78 kW
Answer: B) deliver 75 kW
Confirmed vs Book-3 §2.4 Motor Efficiency — Nameplate kW is the rated shaft output, so a 75 kW motor at full load delivers 75 kW and draws 75/0.94 = 79.8 kW. (d) 79.78 kW is the deliberate trap — it is the INPUT power, not the delivered output.
14. Reduction in the delivery pressure of a compressor by 1 bar would reduce the power consumption by
1 to 5 %
6 to 10 %
11 to 15 %
none of the above
Answer: B) 6 to 10 %
Confirmed vs Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure (Table 3.9) — Book-3: 'A reduction in the delivery pressure by 1 bar in a compressor would reduce the power consumption by 6-10%'; the worked case of 8 to 7 kg/cm² gives 9% input power saving.
The question stem was repaired from the self-contradictory 'Increase in the delivery pressure... would reduce the power consumption' to the book's wording; the answer 6-10% is unchanged. Table 3.9's smaller 4% figure applies to a 0.7 bar cut, which is why 1-5% looks plausible but is too low for a full bar.
15. All other conditions remaining the same in a refrigeration system, at which of the following condenser temperatures will the power consumption be the least:
32.6 oC
35.9 oC
40.8 oC
43.4 oC
Answer: A) 32.6 oC
Confirmed vs Book-3 §4.8 - Lower condensing temperature reduces the compressor lift and hence power; the lowest listed condenser temperature (32.6 °C) gives least power. Every other option is a higher condensing temperature; Table 4.5 shows kW/TR rising from 1.17 at 26.7 degC to 1.41 at 40 degC, i.e. about 20% more power for the same duty.
16. The pressure of refrigerant in vapour compression system changes in
compressor
expansion valve
both (a) & (b)
evaporator
Answer: C) both (a) & (b)
Confirmed vs Book-3 §4.3 - Pressure rises in the compressor and drops across the expansion valve; evaporator and condenser are essentially constant-pressure processes. Evaporator (d) and condenser (b alone) are essentially constant-pressure heat exchangers, so neither alone is the answer; pressure is raised in the compressor and dropped across the expansion valve, hence 'both'.
17. A 1.5 ton air conditioner installed in a room and working continuously for one hour will remove heat of
3024 kcals
4536 kcals
3000 kcals
6048 kcals
Answer: B) 4536 kcals
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/h; 1.5 TR x 3024 = 4536 kcal in one hour. Option (a) 3024 kcal is the heat for 1 TR only and (d) 6048 kcal is for 2 TR; multiply 3024 kcal/h by the 1.5 TR capacity and the one-hour run time.
18. If the power consumed by a 1.5 TR refrigeration compressor is 2.5 kW, what is the energy efficiency ratio?
2.1
1.5
0.6
1.66
Answer: A) 2.1
Confirmed vs Book-3 §4.9 - Refrigeration effect = 1.5 x 3.516 = 5.274 kW. EER = 5.274/2.5 = 2.11. Option (d) 1.66 is the trap of using 3024 kcal/h with the wrong conversion; use cooling in watts (1.5 x 3.516 = 5.274 kW) divided by 2.5 kW input to get EER 2.11 W/W.
19. In the performance assessment of a refrigeration system, which performance ratio (energy efficiency) does not follow the trend "a higher ratio means a more efficient refrigeration system"?
Coefficient of performance (COP)
Energy Efficiency Ratio (EER)
kW per ton
none of the above
Answer: C) kW per ton
Confirmed vs Book-3 §4.7 - For kW/TR a LOWER value means a more efficient system, opposite to COP and EER where higher is better. COP and EER (a, b) are output/input ratios where higher is better; kW/TR is input/output, so a smaller number means a better chiller - the book quotes 0.65-0.9 kW/TR for good centrifugal machines.
20. 2 ton of refrigeration (TR) is equivalent to about
100.8 kcal/min
7032 W
400 BTU/min
all of the above
Answer: D) all of the above
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/hr = 50.4 kcal/min = 3516 W = 12,000 BTU/hr = 200 BTU/min, so 2 TR = 100.8 kcal/min = 7032 W = 400 BTU/min and all three statements describe the same duty. Option (c) was repaired from the garbled '428.7 BTU/min', which matches no standard conversion; with 400 BTU/min the key (d) 'all of the above' is unambiguous, and picking any single option is the trap of not recognising that they are the same quantity in different units.
21. A fan with 25 cm pulley diameter is driven by a 2940 rpm motor through a V-belt system. If the motor pulley is reduced from 20 cm to 15 cm keeping the motor rpm and fan pulley diameter the same, the fan speed will reduce by
1176 rpm
1764 rpm
588 rpm
none of the above
Answer: C) 588 rpm
Confirmed vs Book-3 §5.5 — Belt drive: fan rpm = motor rpm × (motor pulley ÷ fan pulley). Before: 2940 × 20/25 = 2352 rpm; after: 2940 × 15/25 = 1764 rpm; reduction = 588 rpm (c). Option (b) 1764 is the NEW speed, not the reduction; (a) 1176 = 2940 × 10/25 is a mis-subtraction.
22. In series operation of identical centrifugal fans, ideally
flow doubles
static pressure doubles
static pressure goes up by four times
flow goes up by four times
Answer: B) static pressure doubles
Confirmed vs Book-3 §5.5 — Series (push-pull) staging raises 'the static pressure capability at a given airflow' — ideally doubling it for two identical fans (though 'not double at every flow point'). Doubling FLOW (a) is the ideal for PARALLEL operation; four-times (c, d) has no basis in the fan laws.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
23. The hydraulic power of a motor pump set is 8 kW. If the power drawn by the motor is 16 kW at 90% efficiency, the pump efficiency will be
55.5%
50%
45%
none of the above
Answer: A) 55.5%
Confirmed vs Book-3 §6.1 — Pump shaft power = motor input × motor efficiency = 16×0.9 = 14.4 kW. Pump efficiency = hydraulic power / shaft power = 8/14.4 = 55.5%. 50% (8/16) ignores the motor efficiency; 45% wrongly multiplies 50% by 0.9.
📖 §6.6 Flow control strategies — pump control by varying speed / VSDs & VFDs
24. The energy saving with variable speed drives in a pumping system will be maximum for systems with
pure static head
pure friction head
high static head and low friction head
high static head with high friction head
Answer: B) pure friction head
Confirmed vs Book-3 §6.6 — In a friction-only system (Fig 6.15) reducing speed moves the duty point along an iso-efficiency line and 'the affinity laws are obeyed... making variable speed the ideal control method for systems with friction loss'. With high static head (Fig 6.16) flow is no longer proportional to speed, efficiency drops and the pump can reach shut-off, so savings are smallest.
25. A process fluid at 50 m3/hr, with a density of 0.96, is flowing in a heat exchanger and is to be cooled from 36oC to 29oC. The fluid specific heat is 0.78 kcal/kg. If the chilled water range across the heat exchanger is 5oC, the chilled water flow rate is
67.2 m3/hr
52.42 m3/hr
50 m3/hr
none of the above
Answer: B) 52.42 m3/hr
Confirmed vs Book-3 §4.7 - Heat load = 50 x 0.96 x 1000 x 0.78 x (36-29) = 262,080 kcal/h. Chilled water flow = 262,080/(1000 x 1 x 5) = 52.42 m3/hr. Option (a) 67.2 ignores the fluid density of 0.96 and (c) 50 simply copies the process flow; balance the duty 262,080 kcal/h against the chilled-water range of 5 degC.
26. The inner tube of an L-type pitot tube is used to measure …… in the air duct
total pressure
static pressure
velocity pressure
dynamic pressure
Answer: A) total pressure
Confirmed vs Book-3 §5.6 — 'Total pressure is measured using the inner tube of pitot tube and static pressure is measured using the outer tube.' Connecting both to one manometer gives velocity pressure (TP − SP). Options (c)/(d) (velocity = dynamic pressure) are the manometer DIFFERENCE, not what the inner tube alone senses.
📖 §6.4 Matching pump and system head-flow characteristics (Fig 6.11); §6.3 Pump curves — pump operating point
27. The intersection point of the centrifugal pump characteristic curve and the design system curve is the
pump efficiency point
best efficiency point
system efficiency point
none of the above
Answer: B) best efficiency point
Confirmed vs Book-3 §6.4 — Book: for the estimated duty 'we will chose a pump curve which intersects the system curve (Point A) at the pump's best efficiency point (BEP)'. So the intersection of the pump curve with the DESIGN system curve is the BEP — pump efficiency is highest at that one flow. 'Pump efficiency point' and 'system efficiency point' are not book terms.
28. A plant wants to replace the existing 100 TR water cooled vapour compression refrigeration system with a waste heat driven vapour absorption chiller. The capacity of the existing cooling tower
needs no change
is to be doubled
is to be raised to 1.2 times
none of the above
Answer: B) is to be doubled
Confirmed vs Book-3 §7.2 Heat Load — Book heat-rejection table: Refrigeration, Compression = 63 kcal/min/TR; Refrigeration, Absorption = 127 kcal/min/TR – almost exactly double. So replacing a 100 TR VCR with a VAM of the same capacity roughly doubles the heat rejected to the cooling tower → (b). '1.2 times' (c) understates it badly.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
29. Shaft power of the motor driving a pump is 30 kW. The motor efficiency is 0.92 and pump efficiency is 0.5. The power drawn by the motor will be
65.2 kW
15 kW
30 kW
32.6 kW
Answer: D) 32.6 kW
Confirmed vs Book-3 §6.1 — Motor input power = pump shaft power / motor efficiency = 30/0.92 = 32.6 kW. Pump efficiency (0.5) is a distractor — it gives hydraulic power (15 kW, option b), not motor input. 65.2 kW wrongly divides by both efficiencies.
31. In a plant, the loading on a transformer was 1000 kVA with a power factor of 0.88. The plant improved the power factor to 0.99 by adding capacitors on the load side. The release in transformer loading (kVA) will be
111
889
999
none of the above
Answer: A) 111
Confirmed vs Book-3 §1.4 — kW is unchanged by capacitors: kW = 1000 × 0.88 = 880 kW. At 0.99 PF the kVA becomes 880/0.99 = 889 kVA, so the transformer is relieved of 1000 − 889 = 111 kVA.
Option (b) 889 kVA is the NEW loading, not the capacity released — the question asks for the reduction.
📖 §7.2 Factors Affecting Performance – Wet Bulb Temperature
32. The wet bulb temperature normally chosen for designing of cooling tower is
average maximum wet bulb for rainy months
average maximum wet bulb for summer months
average minimum wet bulb for summer months
average maximum wet bulb for winter months
Answer: B) average maximum wet bulb for summer months
Confirmed vs Book-3 §7.2 Wet Bulb Temperature — Book: 'The temperature selected is generally close to the average maximum wet bulb for the summer months' (a value not exceeded more than 5% of the time) → (b). Designing for winter or minimum WBT would undersize the tower for the worst case.
33. Which one of the following types of cooling towers consumes least power for the same operating conditions?
counter flow film fill cooling tower
cross-flow splash fill cooling tower
counter flow splash fill cooling tower
none of the above
Answer: A) counter flow film fill cooling tower
Confirmed vs Book-3 §7.2 Choosing a Cooling Tower — Book Table 7.4 (16,000 m³/hr, 41.5→32.5°C, WBT 27.6°C): power at motor terminal per tower = 253 kW (counter-flow film fill), 310 kW (counter-flow splash) and 330 kW (cross-flow splash); 'the power consumption is least in Counter Flow Film Fill' → (a). Film fill needs less air and lower pumping head (Table 7.3).
34. Which among the following is the most energy efficient lamp for the same wattage rating?
HPMV
GLS
CFL
Metal halide
Answer: D) Metal halide
Corrected (was c) — Book-3 §8.3 Table 8.1: 'most energy efficient for the same wattage' means highest lumens per Watt. Metal halide = 75–125 lm/W (avg 100) beats CFL 40–70 (avg 60), HPMV 44–57 (avg 50) and GLS 8–18 (avg 14). CFL tempts because it is the efficient replacement for GLS, but it is not the most efficacious lamp in this list.
35. _____ is a measure of effect of light on the perceived colour of objects
lux
lumens
CRI
lamp circuit efficacy
Answer: C) CRI
Confirmed vs Book-3 §8.2 — 'Colour rendering index (CRI): is a measure of the effect of light on the perceived color of objects.' Lux is illuminance, lumens is luminous flux, and lamp circuit efficacy is lumens per circuit Watt including control-gear losses.
📖 §9.3 Operational factors — load pattern & DG set capacity; sequencing of loads (kW on engine, kVA on generator)
36. Two most important electrical parameters to be monitored for safe operation of a Diesel Generator set are:
voltage and ampere
kW and kVA
power factor and ampere
kVA and ampere
Answer: B) kW and kVA
Corrected (was a) — Book-3 §9.3: overload "should be carefully analysed" for a DG set and its transient limits apply "to both kW (as reflected on the engine) and kVA (as reflected on the generator)"; the diesel engine is designed for only 10% overload for 1 hr in 12 and the alternator for 50% overload for 15 s. kW therefore guards the engine and kVA the alternator, so together they are the two parameters to watch for safe operation (same answer given in the verified Set-A model solution and the 2010 official key on capacity utilisation). Voltage and ampere (a) alone do not show whether the engine (kW) is overloaded.
📖 §9.4 Energy performance assessment — specific fuel consumption & alternator loading
37. In a DG set, the generator capacity is 1000 kVA with a rated power factor 0.8. It is consuming 150 litre per hour diesel oil. If the specific fuel consumption is 0.25 litres/kWh at that load, then what is the kVA loading of the set at 0.88 PF?
682 kVA
800 kVA
750 kVA
none of the above
Answer: A) 682 kVA
Confirmed vs Book-3 §9.4 — kW generated = fuel rate ÷ SFC = 150 L/h ÷ 0.25 L/kWh = 600 kW. kVA loading = kW ÷ PF = 600/0.88 = 681.8 ≈ 682 kVA. Option (c) 750 kVA is the tempting value obtained by dividing 600 kW by the rated 0.8 PF instead of the actual 0.88 PF.
38. The maximum unbalanced load between phases should not exceed _______ % of the capacity of the DG set
10
5
1
none of the above
Answer: A) 10
Confirmed vs Book-3 §9.2 — "The maximum unbalanced load between phases should not exceed 10% of the capacity of the generating sets"; unbalance heats the alternator and produces unbalanced output voltages. 5% and 1% are not book limits.
📖 §9.3 Waste heat recovery — jacket water heat balance; fuel input from calorific value
39. The jacket cooling water in a diesel engine flows at 12.9 m3/hr with a range of 10oC and accounts for 30% of the engine input energy. What will be the hourly diesel consumption in kg with a calorific value of 10,000 kcal/kg
43
12.9
17.3
none of the above
Answer: A) 43
Confirmed vs Book-3 §9.3 — Jacket-water heat = 12.9 m³/hr × 1000 kg/m³ × 1 kcal/kg °C × 10 °C = 1,29,000 kcal/hr, which is 30% of engine input → input = 1,29,000/0.30 = 4,30,000 kcal/hr → fuel = 4,30,000/10,000 = 43 kg/hr. Option (b) 12.9 merely repeats the water flow; (c) 17.3 has no basis.
40. If the speed of a centrifugal fan is reduced to 80% of its rated speed then the power drawn will be _______% of its rated power:
80%
51.2 %
40 %
64 %
Answer: B) 51.2 %
Confirmed vs Book-3 §5.3 — Power ∝ N³: (0.8)³ = 0.512 → 51.2% of rated power. Option (d) 64% = 0.8² is the static-pressure fraction; (a) 80% is the flow fraction.
41. The order of movement of thermal energy in an HVAC system is:
Indoor air - Condenser water - Chilled water - Cooling tower - Refrigerant
Chilled water - Indoor air - Refrigerant - Cooling tower - Condenser water
Indoor air - Chilled water - Refrigerant - Condenser water - Cooling tower
Indoor air - Chilled water - Refrigerant - Cooling tower - Condenser water
Answer: C) Indoor air - Chilled water - Refrigerant - Condenser water - Cooling tower
Confirmed vs Book-3 §4.1 - Heat flows: indoor air → chilled water → refrigerant (chiller) → condenser water → cooling tower (to atmosphere). Options (a), (b) and (d) scramble the sequence; Figure 4.1 fixes the order as indoor air, chilled water, refrigerant, condenser water, cooling tower, with heat moving left to right to the outdoors.
42. Which one of the following devices will help to eliminate the hunting problems normally associated with capacitor switching?
Maximum Demand Controller
Intelligent Power Factor Controller (IPFC)
Soft Starter
Eddy Current Drives
Answer: B) Intelligent Power Factor Controller (IPFC)
Confirmed vs Book-3 §1.4 Automatic Power Factor Controllers — when the load fluctuates, fixed banks make the PF swing between lagging and leading; the APFC/IPFC has programmable delay-ON and delay-OFF times and 'keeps the numbers of operations equal across all banks', which stops the hunting.
Option (a) a Maximum Demand Controller only sheds loads; it does not switch capacitor steps.
43. The occupancy sensors in a lighting installation are best suited for
conference halls
large production shops/hangars
entrances of offices/buildings
street lighting
Answer: A) conference halls
Corrected (was c) — Book-3 §8.7: occupancy sensors detect movement or noise 'in room spaces', switch lights ON when occupied and OFF after a set delay, with a built-in time delay because 'occupants often remain still or quiet for short periods' — i.e. intermittently used rooms such as conference halls. Large production shops/hangars call for localized switching (§8.7: 'applications which contain large spaces'); street lighting uses timer/daylight/lux-based controls (§8.7); building entrances need continuous lighting for security and see only transient movement, so a sensor there gives little saving.
📖 Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11)
44. The pressure drop in mains header at the farthest point of an industrial compressed air network shall not exceed
2 bar
0.3 bar
0.5 bar
1.0 bar
Answer: B) 0.3 bar
Confirmed vs Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11) — Book-3: 'Typical acceptable pressure drop in industrial practice is 0.3 bar in mains header at the farthest point and 0.5 bar in distribution system.'
The 0.5 bar figure offered in (c) is the distribution allowance, not the mains-header limit asked for.
45. Which of the following is not true of air receivers?
smoothens pulsating air output
stores large volumes of air
a source for draining moisture
increases the pressure of air
Answer: D) increases the pressure of air
Confirmed vs Book-3 §3.5 Air Receivers (IS 7938-1976 sizing) — Air receivers smooth pulsating output, store large volumes for sudden demand and provide a point for draining precipitated moisture — (a), (b) and (c) are all true of them.
A receiver holds air at the pressure the compressor delivers and cannot itself increase pressure, so (d) is the statement that is not true.
📖 §6.1 Pump types — centrifugal pump construction & working
46. With increase in the suction lift from open wells, the delivery flow rate
increases
decreases
remains same
none of the above
Answer: B) decreases
Confirmed vs Book-3 §6.1 — Book: 'The greater the depth of the water, the lesser is the flow from the pump. Also, when it pumps against increasing pressure, the less it will pump.' Higher suction lift raises total head, moving the duty point to lower flow on the H-Q curve (and reducing NPSHA).
47. The compressor capacity of a reciprocating compressor is directly proportional to __________.
Speed
Pressure
Volume
All
Answer: A) Speed
Confirmed vs Book-3 §3.2 Positive Displacement — Book-3 §3.2: 'the compressor capacity is directly proportional to the speed' — halving the RPM (e.g. by reducing the motor pulley) halves the delivered air, which is the basis of the pulley-change de-rating measure.
Output stays nearly constant over a range of discharge pressures, so capacity is not proportional to pressure, and 'All' is therefore wrong.
harmonics occur as spikes at intervals which are multiples of the supply frequency
harmonics are not multiples of the fundamental frequency
induction motors are not the major sources of harmonics
transformers operating near saturation level create harmonics
Answer: B) harmonics are not multiples of the fundamental frequency
Confirmed vs Book-3 §1.10 — 'A harmonic is a component frequency of the signal that is an integer multiple of the fundamental frequency' (5th harmonic on 50 Hz = 250 Hz). Statement (b) says harmonics are NOT multiples of the fundamental, which contradicts the definition, so it is the incorrect statement.
Statement (c) is true because the book classes motors as largely linear, and (d) is true because a saturated transformer behaves non-linearly. [Options re-lettered a)–d); they were printed without labels.]
📖 §4.7 Performance assessment (COP Carnot)
49. Higher COP can be achieved 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: B) higher evaporator temperature and lower condenser temperature - COP(Carnot) = Te/(Tc - Te). Raising Te and lowering Tc both increase the numerator and shrink the temperature lift (Tc - Te), so the COP rises. This is why chilled water temperature should be kept as high as the process allows and condenser/cooling water as cold as possible.
COP_Carnot = T_evap/(T_cond - T_evap), with both temperatures in KELVIN. Raising T_evap increases the numerator AND shrinks the lift; lowering T_cond shrinks the lift again - both push COP up.
Worked feel for the numbers: at 5 degC evaporator and 40 degC condenser, COP = 278/35 = 7.9; drop the condenser to 35 degC and it becomes 278/30 = 9.3, an 18% gain for 5 degC.
Using degC instead of Kelvin here is the classic destroyed answer (5/35 = 0.14 is meaningless). Operationally: run chilled water as warm as the process allows and keep condenser/cooling water as cold as possible - clean tubes, clean cooling tower.
1. A 37 kW, 3 phase, 415 V induction motor draws 56 A and 33 kW power at 410 V. What is the apparent and reactive power drawn by the motor at the operating load?
Model answer: Apparent power = 1.7321 x 0.410 x 56 = 39.769 kVA. Active power = 33 kW. Reactive power = √(39.769² - 33²) = √(1581.57 - 1089) = 22.19 kVAr.
Apparent power S = √3 x V x I; reactive power Q = √(S² - P²).
2. Compute AT&C (Aggregate Technical and Commercial) Losses for the given data: Input Energy Ei=20 MU, Energy Billed Metered E1=16 MU, Un-metered E2=1 MU, Total Billed Eb=17 MU, Amount Billed Ab=Rs.800 lakhs, Gross Amount Collected AG=Rs.820 lakhs, Arrears Collected Ar=Rs.40 lakhs.
Model answer: Amount collected without arrears Ac = AG - Ar = 820 - 40 = 780. Billing Efficiency BE = Eb/Ei = 17/20 = 85%. Collection Efficiency CE = Ac/Ab = 780/800 = 97.5%. AT&C Loss = [1 - (BE x CE)] x 100 = [1 - (0.85 x 0.975)] x 100 = 17.12%.
AT&C Loss = 1 - (Billing Efficiency x Collection Efficiency).
3. Define Range, approach and effectiveness in cooling tower operation.
Model answer: i) Range = difference between cooling tower water inlet and outlet temperature. ii) Approach = difference between the cooling tower outlet cold water temperature and the ambient wet bulb temperature (approach is a better indicator of cooling tower performance). iii) Effectiveness (%) = Range / ideal range = Range / (Range + Approach), where ideal range = inlet water temp - ambient WBT.
4. An induced draft cooling tower is designed for a range of 7°C. An energy manager finds the operating range as 4°C. What could be the reasons for this situation?
Model answer: 1. Excess cooling water flow rate. 2. Reduced heat load from the process. 3. Some cooling tower cell fans switched off. 4. Poor approach due to high humid conditions. 5. Nozzles may be blocked.
Lower-than-design range implies either reduced heat load, excess water flow or degraded heat-transfer.
5. State any three major differences between vapour compression refrigeration (VCR) and vapour absorption refrigeration (VAR) systems.
Model answer: 1. VCR uses electric power for the compressor; VAR uses a heat source. 2. VCR uses hydrogen-fluorine-carbon compounds as refrigerant; VAR uses water (or ammonia). 3. VCR works under pressure; VAR works under vacuum. 4. VCR has high COP; VAR has low COP. 5. VAR requires about double the cooling tower capacity of VCR.
Energy input, refrigerant, operating pressure, COP and cooling-tower sizing differences.
📖 §9.5 Solved example (a) — maximum power factor at full kVA load
6. A 180 kVA, 0.80 PF rated DG set has a diesel engine rating of 210 BHP. What is the maximum power factor that can be maintained at full load on the alternator without overloading the DG set? (Alternator losses and exciter power = 5.66 kW, no derating.)
Model answer: Engine rated power = 210 x 0.746 = 156.66 kW. Power available for alternator = 156.66 - 5.66 = 151 kW. Maximum PF = 151/180 = 0.84.
📖 §5.6 Fan static efficiency formula (Book EOC S-2)
8. A fan delivers 18,500 Nm³/hr at static pressure rise 45 mm WC. The 3-phase motor records 2.9 kW/phase; motor operating efficiency 88%. What is the fan static efficiency?
Model answer: Q = 18,500 Nm³/hr = 5.13888 m³/s; SP = 45 mmWC. Power input to motor = 2.9 x 3 = 8.7 kW. Power to fan shaft = 8.7 x 0.88 = 7.656 kW. Fan static efficiency = (5.13888 x 45)/(102 x 7.656) = 0.296 = 29.6%.
2. A 3-phase induction motor: rated 37 kW, 415 V, 66 A, 0.88 pf; operating 410 V, 49 A, 0.76 pf. Efficiency constant 50-100% load. Plant runs 7000 h/yr at Rs.6/unit. Proposed replacement: 30 kW EE motor at 92% efficiency costing Rs.75,000 (salvage of old Rs.10,000). a) Rated efficiency and loading of existing motor. b) Loading with EE motor. c) Payback period.
Model answer: Rated input = 1.732 x 0.415 x 66 x 0.88 = 41.746 kW; rated efficiency = 37/41.746 = 88.63%. Actual input = 1.732 x 0.410 x 49 x 0.76 = 26.44 kW; loading = 26.44/41.746 = 63.3%; shaft output = 37 x 0.633 = 23.44 kW. EE motor (30 kW): loading = 23.44/30 = 78%. Annual savings = 23.44 x (1/0.8863 - 1/0.92) x 7000 x 6 = Rs.40,740. Payback = (75,000 - 10,000)/40,740 = 1.59 years.
Rated input via √3VIcosφ; loading = actual/rated input; savings from efficiency difference applied to shaft load x hours x tariff.
3. a) A 3-phase 415 V 75 kW induction motor draws 48 kW at 0.7 PF. Calculate the capacitor rating to improve PF to 0.95, the reduction in current and kVA reduction at 415 V. b) A plant consumes 2,00,000 kWh/month at 0.9 PF. What is the % reduction in distribution losses if PF is improved to 0.96 at load end?
Model answer: a) kVAr = kW[tan(cos^-1 0.70) - tan(cos^-1 0.95)] = 48 x (1.020 - 0.329) = 33.2 kVAr (say 33 kVAr).
Current at 0.70 PF = 48/(1.732 x 0.415 x 0.70) = 95.4 A; at 0.95 PF = 48/(1.732 x 0.415 x 0.95) = 70.3 A; reduction = 25.1 A (26%).
kVA at 0.70 = 48/0.70 = 68.6 kVA; at 0.95 = 48/0.95 = 50.5 kVA; kVA released = 18.1 kVA.
b) % reduction in distribution loss = [1 - (PF1/PF2)^2] x 100 = [1 - (0.90/0.96)^2] x 100 = 12.1%.
In energy terms, if the existing loss were 4% of 2,00,000 kWh = 8000 kWh/month, the saving is about 970 kWh/month.
Capacitor kVAr = kW(tanφ1 - tanφ2); loss reduction = 1 - (PF1/PF2)².
📖 §10.15 Energy efficiency measures in buildings (plus Ch3 compressed air, Ch6 pumps, Ch8 lighting)
4. List five energy conservation measures each for any two of: a) Energy use in buildings, b) Compressed air system, c) Pumps and pumping systems, d) Lighting systems.
Model answer: Buildings: weather-stripping of windows/doors; set temperature 23-25°C and RH 55-65%; maintain chilled water leaving temp ≥7°C; insulate chilled water pipes and ducts; clean condenser tubes every 6 months; keep cooling towers clean; vary AHU fan speed with VFD; keep air filters clean. Compressed air: keep intake air cool (every 4°C rise raises power 1%); clean inlet filters (2% loss per 250 mmWC drop); ring-main piping; carry out leak tests (40-50% leakage common); fit solenoid cut-off valves; reduce delivery pressure; use VFD on large compressors; separate HP/LP systems. Pumps: ensure adequate NPSH; operate near BEP; minimize throttling; use VSD for load variation; trim/replace oversized impellers; replace with energy-efficient pumps; reduce system resistance by pipe sizing. Lighting: switch off lights when not in use; separate switching for peripheral/daylit zones; use high-efficacy lamps (CFL saves ~75%); electronic ballasts (2 W vs 12 W); optical luminaires; clean fixtures; light-coloured surfaces; lighting controls (timers, daylight, occupancy sensors).
Buildings list is Book-3 §10.15 verbatim (weather stripping, 23–25 °C/55–65% RH, CHW ≥ 7 °C, insulation, condenser cleaning every 6 months, clean cooling towers, AHU VFD, clean filters); other systems from their own chapters.
📖 Cross-chapter fill-in-the-blanks (§4.7 for item 1)
5. Fill in the blanks (cross-chapter): 1) One TR = ___ kW. 2) A 4-pole 15 kW IM at 50 Hz, 1% slip has rotor input ___ kW. 3) A pitot tube measures total and static pressure to determine ___ pressure. 4) Centrifugal pump impeller diameter is generally limited to reducing to about ___ % of max size. 5) Pressure in pump suction exceeding liquid vapour pressure is expressed as ___. 6) ASME parameter to define fans, blowers, compressors is ___. 7) Pumps can run in parallel if their ___ are similar. 8) Evaporation 16 m³/cell, COC 3 → blowdown ___. 9) Pump raises water to 12 m; with brine SG 1.2 height raised is ___. 10) Installing capacitor near motor terminals increases design PF of motor - True/False.
Model answer: 1) 3.516 kW. 2) 15.15 kW (15/(1-0.01)=15.15). 3) velocity pressure. 4) 75%. 5) Net Positive Suction Head Available (NPSHA). 6) Specific ratio. 7) closed valve heads. 8) 8 m³ per cell (blowdown = evaporation/(COC-1) = 16/2 = 8). 9) 12 metres (same height - head is independent of density). 10) False.
Standard one-liners across HVAC, motors, fans, pumps and cooling towers.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); part (b) Book-3 Ch3 Compressed air — pressure optimisation
6. a) Pump suction head 3 m below centreline, discharge pressure 2.8 kg/cm², flow 120 m³/hr. Find pump efficiency if actual motor input is 15.0 kW at 0.90 motor efficiency. b) A V-belt reciprocating instrument air compressor maintains 7 kg/cm²g. 20% of air goes to boiler-house control valves needing 6.5 kg/cm²g; balance 80% needs 2 kg/cm²g. What do you advise?
Model answer: a) Discharge head = 2.8 kg/cm² = 28 m; suction head = -3 m; total head = 28 - (-3) = 31 m. Hydraulic power = (120/3600) x 1000 x 9.81 x 31 /1000 = 10.137 kW. Pump shaft power = 15 x 0.9 = 13.5 kW. Pump efficiency = 10.137/13.5 = 75%. b) Advise: 1) Provide a separate small compressor at 7 kg/cm²g near the control valves and reduce the main distribution pressure from 7 to 2 kg/cm²g for the bulk pneumatic instruments. 2) Reduced pressure lowers leakage loss; the compressor will begin to unload, so reduce the motor pulley size to match the lower demand.
Hydraulic power = Q x ρ x g x H; efficiency = hydraulic/shaft. Part b is a pressure-optimisation advisory.