Energy Efficiency in Electrical Utilities Available here with full solutions — 49 questions recovered from the 2011 exam:
Objective (1 mark)
35 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) — 35
📖 §1.4 Selection and Location of Capacitors
1. If power factor is improved from PF1 to PF2 then the reduction in distribution losses in an electric network is proportional to:
ratio of PF1 to PF2
square root of (PF1/PF2)
square of (PF1/PF2)
none of the above
Answer: C) square of (PF1/PF2)
Confirmed vs Book-3 §1.4 — the book gives the reduction in distribution loss % as [1 − (PF₁/PF₂)²] × 100, so the loss after correction is the square of the power-factor ratio times the loss before: loss₂/loss₁ = (PF₁/PF₂)².
The governing term is therefore the SQUARE of (PF₁/PF₂); options (a) and (b) use the plain ratio and its square root, which under-state the benefit. Memorise the full form [1 − (PF₁/PF₂)²] × 100 for the numericals.
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme
2. In the BEE labeling programme for distribution transformers, the total transformer losses at
50% and 100% loading have been defined.
only 50% loading have been defined.
only 100% loading have been defined.
25%, 50% and 100% loading have been defined.
Answer: A) 50% and 100% loading have been defined.
Confirmed vs Book-3 §1.5 Standards & Labeling — 'For the BEE labeling programme total losses at 50% and 100% load have been defined', with 1 star the highest-loss and 5 star the lowest-loss segment (Table 1.4).
Option (d) adds 25% loading, which belongs to no BEE/IS 1180 labelling point.
📖 §1.4 Performance Assessment of Power Factor Capacitors
3. A 5 kVAr, 415 V rated power factor capacitor was found to be having 5.5 kVAr operating capacity. The operating supply voltage at the same supply frequency would approximately be
400 V
415 V
435 V
none of the above
Answer: C) 435 V
Confirmed vs Book-3 §1.4 — kVAr ∝ V², so V = 415 × √(5.5/5) = 415 × 1.049 ≈ 435 V.
Option (b) 415 V would give exactly the rated 5 kVAr; an output ABOVE rating can only come from an over-voltage, which the book warns shortens capacitor life.
📖 §1.4 Performance Assessment of Power Factor Capacitors
4. Busbar Voltages at the main electrical panel were balanced but at the following Motor Control Circuit (MCC), fitted with PF Correction capacitors, the voltages were unbalanced by about 3%. The possible reason for this could be
motors connected to MCC were operating at partial loads
motors connected to MCC were overloaded
excessive kVAr of Capacitors than required at MCC
blown fuse in one phase of the 3 phase capacitor bank connected to the MCC.
Answer: D) blown fuse in one phase of the 3 phase capacitor bank connected to the MCC.
Confirmed vs Book-3 §1.4 — a three-phase bank compensates each phase separately; a blown fuse in one phase removes that phase's kVAr only, so the three phase currents (and hence the volt-drops down the feeder) become unequal, producing the 3% imbalance at the MCC.
Options (a)/(b) alter all three phases symmetrically and (c) over-compensation would raise the voltage in all three phases equally — none of these creates an imbalance.
5. The iron losses in a transformer are proportional to:
kVA load
square of kVA load
cube of kVA load
none of the above
Answer: D) none of the above
Confirmed vs Book-3 §1.5 — 'Core loss occurs whenever the transformer is energized; core loss does not vary with load.' Iron loss is therefore independent of the kVA loading, so none of the stated proportionalities applies.
Option (b) 'square of kVA load' describes the COPPER loss (P = I²R), which is the classic confusion in this question.
6. The synchronous speed (rpm) of a 2 pole induction motor at 49.5 Hz supply frequency is:
3000
2970
1500
none of the above
Answer: B) 2970
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 49.5 / 2 = 2970 rpm. (a) 3000 rpm is the trap of using the nominal 50 Hz instead of the actual 49.5 Hz supply frequency — synchronous speed is directly proportional to frequency.
7. kW rating indicated on the name plate of an induction motor indicates
rated input of the motor
maximum input power which the motor can draw
rated output of the motor
maximum instantaneous input power of the motor
Answer: C) rated output of the motor
Confirmed vs Book-3 §2.4 Motor Efficiency — The kW on an induction motor nameplate is the rated mechanical OUTPUT at the shaft; the input is that value divided by efficiency. (a) 'rated input' is the standard misconception — it would make a 75 kW, 90 % efficient motor deliver only 67.5 kW.
8. A 7.5 kW, 415 V, 14.0 A, 1480 RPM, three phase rated squirrel cage induction motor, after decoupling from the driven equipment, was found to be drawing 3.5 A at no load. The current drawn by the motor at no load is high because of
very high supply frequency at the time of no load test
faulty ammeter reading
very poor power factor as the load is almost inductive
loose motor terminal connections
Answer: C) very poor power factor as the load is almost inductive
Confirmed vs Book-3 §2.3 Motor Characteristics — At no load the current is almost purely magnetizing (inductive), giving a very poor no-load power factor, so 3.5 A on a 14 A motor is normal — the book recommends recording no-load current for exactly this reason. (b)/(d) an instrument fault or loose connection would not produce a consistent, expected no-load reading.
9. A six pole induction motor operating at 49.6 Hz, with 980 RPM actual speed, will have operating % slip of
1.21%
2%
0%
none of the above
Answer: A) 1.21%
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 49.6 / 6 = 992 rpm; slip = (992 − 980)/992 × 100 = 1.21 %. (b) 2 % results from assuming Ns = 1000 rpm at 50 Hz — the slip must always be referred to the synchronous speed at the measured frequency.
10. The total loss for a transformer loading at 60% and with no load and full load losses of 3 kW and 25 kW respectively, is
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 comes from scaling the copper loss linearly (0.6 × 25 = 15); the load fraction must be squared.
📖 §2.10 Star Labeling of Energy Efficient Induction Motors
11. Eff1 (as per IS 12615:2004) induction motor is
endorsed by BEE as high efficiency label
having same efficiency as of Eff2
having less efficiency than Eff 2 motor
not covered in BEE labeling scheme for motors
Answer: A) endorsed by BEE as high efficiency label
Confirmed vs Book-3 §2.10 Star Labeling of Energy Efficient Induction Motors — Eff1 was the high-efficiency class under IS 12615:2004 and is the class endorsed by BEE under its labelling scheme for energy-efficient motors (now carried forward as IE2/IE3 classes under IS 12615:2011). (c) reverses the classes — Eff1 is more efficient than Eff2, not less.
📖 Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power)
12. In a textile mill, two 150 cfm belt driven reciprocating compressors are seen to be working constantly with a loading time of 20 seconds and unloading time of 30 seconds. The best economic option for energy savings would be:
switch off one compressor
switch off one compressor and reduce motor pulley size of the other compressor appropriately
adopt variable speed drive for one of the compressors
none of the above
Answer: B) switch off one compressor and reduce motor pulley size of the other compressor appropriately
Confirmed vs Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power) — Loading 20 s in every 50 s means the two machines together are loaded only 40% of the time, so a single 150 cfm compressor covers the demand and is still oversized.
The book prescribes changing the pulley size to reduce RPM and de-rate an oversized compressor, so switching one off AND trimming the other's motor pulley (b) saves more than simply switching one off (a); a VSD (c) is not economic at this size.
14. In an air washer of textile humidification system airflow of 3000 m3/h at 25 oC and 10% relative humidity is humidified to 60% relative humidity by adding water through spray nozzles. The specific humidity of air at inlet and outlet are 0.002 kg/kg and 0.0062 kg/kg respectively. The amount of water required in kg/hr is
14.9
6
10
none of the above
Answer: A) 14.9
Confirmed vs Book-3 §4.14 - Mass of dry air ~3000*1.18 = 3540 kg/hr; water = 3540*(0.0062-0.002) = 14.9 kg/hr. Options (b) and (c) are round-number distractors; the book's worked example gives mass of air x change in humidity ratio = 3000 x 1.184 x (0.0062 - 0.002) = 14.9 kg/h.
15. In a vapour compression refrigeration system, the component where the refrigerant fluid experiences no heat loss or gain is
compressor
condenser
expansion valve
evaporator
Answer: C) expansion valve
Confirmed vs Book-3 §4.3 - Expansion through the throttling valve is adiabatic (isenthalpic) - no heat exchange. Heat is absorbed in the evaporator (d), rejected in the condenser (b) and work is added in the compressor (a); only the expansion device has no heat gain or loss, per §4.3.
16. The refrigeration load in TR when 20 m3/hr of water is cooled from 13 oC to 8 oC is about
33
80.3
39.6
none of the above
Answer: A) 33
Confirmed vs Book-3 §4.7 - Load = 20000 kg/hr * 1 * (13-8) = 100000 kCal/hr; /3024 = 33 TR. Options (b) and (c) come from using the wrong temperature difference; load = 20,000 x (13 - 8)/3024 = 33 TR.
📖 §6.2 System characteristics — static & friction head
17. Friction loss in a piping system carrying fluid is proportional to
fluid flow
(fluid flow)2
1/fluid flow
1/(fluid flow)2
Answer: B) (fluid flow)2
Confirmed vs Book-3 §6.2 — Book: 'The friction losses are proportional to the square of the flow rate' — hence the parabolic system curve and the valve loss 'proportional to flow squared'. Linear (a) and inverse (c, d) relations are wrong.
📖 §6.2 System characteristics — static & friction head; §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³)
18. In a pumping system the static head is 10 m and the dynamic head is 15 m. If the pump speed is doubled, then the total head will be
50 m
70 m
40 m
none of the above
Answer: B) 70 m
Confirmed vs Book-3 §6.2/§6.5 — Static head is independent of flow, so it stays 10 m. Dynamic (friction) head ∝ flow² and flow ∝ speed, so doubling speed quadruples the dynamic head: 15×4 = 60 m. Total = 10 + 60 = 70 m. 50 m wrongly doubles the dynamic head; 40 m quadruples only the static part.
📖 §8.6(e) Reduction of lighting feeder voltage + Table 8.3
19. The advantage of installing a dedicated servo transformer for lighting feeders is;
'Voltage' fluctuations in lighting circuit can be minimized by isolating from the power feeders.
reduction of voltage related problems, which in turn increases the efficiency of the lighting system.
with proper control device 'over voltage' that might occur during lean load or off-peak can be avoided, in turn less energy consumption and improved lamp life can be achieved
all the above
Answer: D) all the above
Confirmed vs Book-3 §8.6(e) — The book notes that 'higher night-time voltage reduces lamp life' and that reactors/transformers on the lighting feeder save 5–15 %. A dedicated servo/lighting transformer isolates lighting from power-feeder fluctuations, avoids lean-load over-voltage, improves efficiency and lamp life — all three statements hold, so 'all the above'.
20. The COP of a vapour compression refrigeration system is 3.0. If the compressor motor output is 9.555 kW, the tonnage (TR) of the refrigeration system is
8.15
28.665
3
none of the above
Answer: A) 8.15
Confirmed vs Book-3 §4.7 - Refrigeration effect = COP*power = 3*9.555 = 28.665 kW; /3.516 = 8.15 TR. Option (b) 28.665 is the cooling effect in kW, not TR, and (c) simply repeats the COP; divide 28.665 kW by 3.516 kW/TR to get 8.15 TR.
21. 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 oC to 29 oC. The fluid specific heat is 0.78 kCal/kg. If the chilled water range across the heat exchanger is 4 oC, the chilled water flow rate is
44.46 m3/hr
40 m3/hr
35 m3/hr
none of the above
Answer: A) 44.46 m3/hr
Confirmed vs Book-3 §4.7 - Heat load = 40*1000*0.95*0.78*(35-29) = 177840 kCal/hr; chilled water flow = 177840/(1000*1*4) = 44.46 m3/hr. Option (b) 40 copies the process-fluid flow and ignores the density and specific heat; balance the duty 177,840 kcal/h against the 4 degC chilled-water range.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
22. Input power to the motor driving a pump is 30 kW. The motor efficiency is 0.9. The power transmitted to the water is 16.2 kW. The pump efficiency is
54%
60%
90%
none of the above
Answer: B) 60%
Confirmed vs Book-3 §6.1 — Pump shaft power = 30×0.9 = 27 kW; pump efficiency = hydraulic/shaft = 16.2/27 = 60%. 54% (16.2/30) ignores motor efficiency; 90% is the motor efficiency itself.
23. The illuminance is 10 lm/m2 from a lamp at 1 meter distance. The illuminance at half the distance will be
40 lm/m2
10 lm/m2
5 lm/m2
none of the above
Answer: A) 40 lm/m2
Confirmed vs Book-3 §8.2 — Book's worked example: E = (1.0/0.5)² × 10 = 40 lm/m². Halving the distance multiplies illuminance by 4 (inverse square law), not by 2.
📖 §9.3 Factors affecting waste heat recovery from flue gases — back pressure
24. The main precaution to be taken care by the waste heat recovery device manufacturer to prevent the problem in a DG set during operation is:
temperature rise
back pressure
over loading of waste heat recovery tubes
turbulence of exhaust gases
Answer: B) back pressure
Confirmed vs Book-3 §9.3 — The WHR unit sits in the exhaust path; its pressure drop adds back pressure on the engine, and the book states the maximum allowed is around 250–300 mm WC, so the recovery unit must be designed for a lower pressure drop. Tube overloading (c) and gas turbulence (d) are boiler-side issues, not the engine-protection concern.
📖 §8.3(1) Incandescent lamp — Figure 8.3 energy flow
25. The lamp which gives 10% visible radiation is
CFL
flourescent tube light
HPSV
incandescent lamp
Answer: D) incandescent lamp
Confirmed vs Book-3 §8.3 Figure 8.3 — Incandescent lamp energy flow: ≈10 % visible radiation, ~20 % conduction/convection loss, ~70 % infrared. Fluorescent/CFL are 3–5 times as efficient and HPSV is 67–121 lm/W, so only the incandescent lamp gives just 10 % visible output.
26. The electronic ballast in lighting application does not have one of the following characteristics
lower operational losses than conventional ballasts
tuned circuit to deliver power at 28-32 kHz
requiring a starter
low temperature rise
Answer: C) requiring a starter
Confirmed vs Book-3 §8.6(f) — Electronic ballasts: losses ~1 W vs 10–15 W (lower losses, low temperature rise), operate the lamp at high frequency (book 20–30 kHz), and 'the starter is eliminated' — so 'requiring a starter' is the characteristic it does not have.
📖 §8.3 Table 8.1 Luminous performance of lamps (LED row)
27. The lumens output varies from _______ Lumens/Watt in case of White LED lamps.
30-50
75-125
101-175
67-121
Answer: A) 30-50
Confirmed vs Book-3 §8.3 Table 8.1 — By elimination: 75–125 lm/W is metal halide, 101–175 is LPSV and 67–121 is HPSV, so 30–50 (the older-edition figure for white LEDs used by this 2011 paper) is the intended key. Note the 2014 Table 8.1 now lists LED at 50–130 lm/W (avg 90; up to 200 in the laboratory) — quote 50–130 if asked directly about LED efficacy.
28. The blowdown quantity required in cooling towers is given by
evaporation loss/ (cycle of concentration -1)
(cycle of concentration -1)/ evaporation loss
evaporation loss/ (1 - cycle of concentration)
evaporation loss/ (cycle of concentration +1)
Answer: A) evaporation loss/ (cycle of concentration -1)
Confirmed vs Book-3 §7.2 (vii) — Book: 'Blow Down = Evaporation Loss / (C.O.C. − 1)' → (a). (b) inverts the ratio, (c) gives a negative value for COC > 1, and (d) uses +1, all contradicting the book relation.
📖 §10.8 ECBC guidelines on lighting — Building Area Method (worked example)
29. A hotel building has four floors each of 1000 m2 area. If the Lighting Power Density (LPD) is 10.8 W/m2, the interior lighting power allowance for the hotel building is
10800 W
21600 W
43200 W
none of the above
Answer: C) 43200 W
Confirmed vs Book-3 §10.8 — This is the book's worked example: interior lighting power allowance = (1000 m² × 10.8 W/m²) × 4 floors = 43,200 W. 10,800 W (a) is one floor only; 21,600 W (b) is two floors.
📖 §9.4 Energy performance assessment — specific fuel consumption (L/kWh)
30. A DG set is generating 900 kVA at 0.8 PF. If the specific fuel consumption of this DG set is 0.3 lts/kWh at that load, then how much fuel is consumed while delivering generated power for one hour.
270 litres
300 litres
216 litres
none of the above
Answer: C) 216 litres
Confirmed vs Book-3 §9.4 — kW = 900 kVA × 0.8 = 720 kW; fuel in one hour = 720 × 0.3 = 216 litres. Option (a) 270 litres is the tempting error of applying SFC to the kVA (900 × 0.3).
📖 §9.3 Operational factors — load pattern & DG set capacity; sequencing of loads (kW on engine, kVA on generator)
31. Two most important electrical parameters, which are to be monitored on generator panel, among the following, for safe operation of a Diesel generator set are:
voltage and ampere
kVA and ampere
power factor and voltage
kW and kVA
Answer: D) kW and kVA
Corrected (was a) — Book-3 §9.3: DG-set overload limits are tight (engine 10% for 1 hr in 12; alternator 50% for 15 s) and transient/overload limits apply "to both kW (as reflected on the engine) and kVA (as reflected on the generator)". Monitoring kW protects the engine and kVA protects the alternator, so kW & kVA are the two parameters for safe operation — consistent with the verified Set-A model solution (kVA and kW) and the 2010 official key (kW & kVA for capacity utilisation). Voltage and ampere (a) do not reveal engine (kW) overload.
32. Select the feature which does not apply to energy efficient motors by design:
energy efficient motors last longer
starting torque for efficient motors may be lower than for standard motors
energy efficient motors have high slips which results in speeds about 1% lower than standard motors
energy efficient motors have low slips which results in speeds about 1% higher than standard motors
Answer: C) energy efficient motors have high slips which results in speeds about 1% lower than standard motors
Confirmed vs Book-3 §2.6 Energy Efficient Motors — The book states that less slippage in energy-efficient motors results in speeds about 1 % FASTER than standard counterparts, so statement (c) — high slip and 1 % lower speed — is the one that does not apply. (b) is a genuine EEM feature the book warns about ('starting torque for efficient motors may be lower'), so it cannot be the answer to a 'does not apply' question.
lower the heat rate of a power generating unit, higher is the generation efficiency
one kilo Watt hour of electrical energy being equivalent to 3600 kilo Joules of thermal energy
'Heat Rate' is directly proportional to the efficiency of power generation
design 'Heat Rate' of a 210 MW thermal generating unit is lower than that of a 110 MW thermal generating unit
Answer: C) 'Heat Rate' is directly proportional to the efficiency of power generation
Confirmed vs Book-3 §1.1 — 'The 'HEAT RATE' is inversely proportional to efficiency of power generation i.e., lower the heat rate, higher is the generation efficiency.' Statement (c) says DIRECTLY proportional, which flatly contradicts the book, so it is the wrong statement.
Statements (a) and (b) are book facts (1 kWh ≡ 3600 kJ ≡ 860 kCal), and (d) is true as repaired: larger units are more efficient, so a 210 MW set has a LOWER design heat rate than a 110 MW set. [Option (d) repaired so that exactly one statement is wrong.]
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
34. Select the incorrect statement:
required PF capacitor kVAr at induction motor terminal increases with decrease in speed of the motor
PF capacitor improves power factor from the point of installation back to the load side
induction motor efficiency increases with increase in its rated capacity
the largest potential for electricity savings with variable speed drives is generally in variable torque applications
Answer: B) PF capacitor improves power factor from the point of installation back to the load side
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — The book states that a PF capacitor 'improves power factor from the point of installation back to the generating side' — i.e. upstream, not towards the load — so statement (b) is the incorrect one. (a), (c) and (d) all restate book facts: capacitor kVAr rises as motor speed falls (Table 2.5), efficiency rises with rated capacity, and variable torque applications offer the largest VSD savings.
Metal halide lamp can be considered as a variant of high pressure mercury vapour lamp (HPMV)
Efficacy of fluorescent tube light (FTL) remains constant throughout its operational life
HPSV lamps differ from mercury and metal-halide lamps in that they do not contain starting electrodes
LPSV lamps are the most efficacious light sources, but they produce the poorest quality light of all the lamp types
Answer: B) Efficacy of fluorescent tube light (FTL) remains constant throughout its operational life
Confirmed vs Book-3 §8.3 — Every lamp has 'the percent of output that a lamp loses over its life' and §8.6(h) says light output falls with ageing lamps, so FTL efficacy does NOT stay constant — (b) is the incorrect statement (Table 8.4 even rates FTLs at 100/2000/3500 h for this reason). (a), (c) and (d) are verbatim book statements: metal halide is a variant of HPMV, HPSV lamps contain no starting electrodes, and LPSV is the most efficacious but poorest-quality light.
2. S-2: Match the following load-shape objectives of any Demand Side Management (DSM) programme of a utility. (i) Peak Clipping, (ii) Valley filling, (iii) Load shifting, (iv) Conservation, (v) Load building - with the corresponding load-shape diagrams a-e.
Model answer: i - c; ii - d; iii - b; iv - e; v - a
Standard DSM load-shape matching as given in the official key.
3. S-3: The power input to a three phase induction motor is 52 kW. If the induction motor is operating at a slip of 1.9% and with total stator losses of 1.30 kW, find the total mechanical power developed.
Model answer: Stator input = 52 kW; Stator losses = 1.30 kW; Stator output = 52 - 1.30 = 50.7 kW = Rotor input; Slip = 1.9%; Mechanical Power Output = (1 - s) x Rotor Input = (1 - 0.019) x 50.7 = 0.981 x 50.7 = 49.737 kW.
Rotor input = stator input - stator losses; mechanical power = (1-slip) x rotor input.
4. S-4: In a Commercial building, five window ACs each of 1.5 TR capacity were evaluated for replacement with three star labeled new ACs having Energy Efficiency Ratio (EER) of 2.50 kW/kW. The measured EER of existing ACs: AC1 = 2.05, AC2 = 2.19, AC3 = 2.30, AC4 = 2.40, AC5 = 2.17. Calculate the total kW saving potential if all the existing ACs are replaced with 3 star labeled ACs of same capacity.
Model answer: Input kW = TR delivered*3.516/EER. For 3 star AC input power = 1.5*3.516/2.5 = 2.11 kW each. Existing kW input: AC1 = 2.573, AC2 = 2.408, AC3 = 2.293, AC4 = 2.198, AC5 = 2.430; Total = 11.902 kW. Savings potential = 11.902 - (2.11 x 5) = 11.902 - 10.55 = 1.352 kW.
Input kW = TR*3.516/EER for each AC; saving = sum of existing input - new input (5 x 2.11).
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
5. S-5: The following data of a water pump of a process plant have been collected. Flow: 70 m3/hr, Total head: 24 meters, Power drawn by motor 7.2 kW, Motor efficiency 89%. Determine the pump efficiency.
Model answer: Hydraulic power = Q(m³/s)×H×ρ×g/1000 = (70/3600)×24×1000×9.81/1000 = 4.578 kW. Pump shaft (input) power = motor power × motor efficiency = 7.2×0.89 = 6.41 kW. Pump efficiency = 4.578/6.41 = 71.4%. (The printed key uses 7.2×0.90 = 6.48 kW, giving 70.65% — with the stated 89% motor efficiency the answer is ≈71.4%; either is accepted if the method is shown.)
Pump efficiency = hydraulic power / shaft power, where shaft power = motor input × motor efficiency. Note the official key applied 0.90 instead of the stated 0.89.
📖 §6.10 Agricultural pumping system — demonstrated ECMs
6. S-6: List any 5 energy conservation opportunities in agriculture pump sets.
Model answer: 1. Installation of low friction foot valves; 2. Installation of low friction HDPE suction and delivery pipes; 3. Installation of long bends; 4. Installation of high efficiency pumps and motors; 5. Lower discharge head.
Standard demonstrated ECMs for agricultural pumping per the official key.
7. S-7: Write any 5 industrial applications of a heat pump.
Model answer: Industrial heat pumps are mainly used for: Space heating; Heating of process streams; Water heating for washing, sanitation and cleaning; Steam production; Drying/dehumidification; Evaporation; Distillation; Concentration.
Any five of the listed industrial heat-pump applications per the official key.
8. S-8: An induced draft-cooling tower is designed for a range of 8 C. The energy auditor finds the operating range as 2 C during the conduct of energy audit. In your opinion what could be the reasons for this situation?
Model answer: 1. There may be excess cooling water flow rate; 2. There may be reduced heat load from the process; 3. Some of the cooling tower cells fan are switched off; 4. Approach may be poor because of high humid condition; 5. Cooling tower nozzles may be blocked.
Low range (cooling) caused by excess water flow, reduced heat load, or operational/maintenance issues.
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback
1. L-1: The contract demand of a process plant is 6000 kVA. The average monthly recorded maximum demand is 5500 kVA at 0.78 PF. Tariff: (a) Minimum monthly billing demand is 75% of contract demand or actual recorded MD whichever is higher; no PF incentives. (b) Monthly MD charge is Rs. 400 per kVA. Find the optimum limit of PF capacitor requirement (purely to reduce MD so no excess demand charges are paid) and the simple payback period, assuming capacitor + APFC controller cost is Rs. 500 per kVAr.
Model answer: Minimum payable demand = 6000 x 0.75 = 4500 kVA. Margin for MD reduction = 5500 - 4500 = 1000 kVA. Present maximum load = 5500 x 0.78 = 4290 kW. Desired peak PF to achieve MD of 4500 kVA = 4290/4500 = 0.9533. PF capacitor requirement = 4290 [tan(Cos-1 0.78) - tan(Cos-1 0.9533)] = 4290(tan 38.74 - tan 17.579) = 4290(0.80226 - 0.316815) = 4290(0.4854) = 2083 kVAr. Cost of capacitor installation = 500 x 2083 = Rs. 10.4 lakhs. Monthly MD saving = 1000 kVA; Yearly savings = 1000 x 400 x 12 = Rs. 48.0 lakhs. Simple payback = 10.4/48 = 0.21 years = 2.6 months.
Reduce MD to minimum billable 4500 kVA; required kVAr from tan(phi1)-tan(phi2) at the active load of 4290 kW; payback = investment/annual MD savings.
2. L-2: Fill in the blanks. (a) With increase in condensing temperature in a vapor compression refrigeration system, the specific power consumption of the compressor for a constant evaporator temperature will____. (b) With increase in evaporator temperature while maintaining a constant condenser temperature, the specific power consumption of the compressor will____. (c) Lower power factor of a DG set demands ____ excitation current. (d) Slip power recovery system is used in ____ induction motor. (e) If voltage is reduced from 230 V to 200 V for a fluorescent tube light, it will result in ____ power consumption. (f) ____ fans are known as 'non-overloading' because change in static pressure do not overload the motor. (g) ____ head is the friction loss, on the liquid being moved, in pipes, valves and equipment in the system. (h) Ratio of the light reflected by a surface to the solar light incident upon it, is called ____. (i) ____ is the ratio of solar heat gain that passes through fenestration to the total incident solar radiation that falls on the fenestration. (j) luminous flux incident on an object per unit area is defined as ____.
Model answer: a. increase; b. decrease; c. higher; d. slipring (slip-ring); e. reduced; f. backward-inclined; g. dynamic; h. Solar Reflectance; i. Solar heat gain coefficient; j. illuminance.
Standard refrigeration/motor/fan/pump/building fill-in answers per official key.
3. L-3: A free air delivery test was carried out before a leakage test on a reciprocating air compressor. Receiver capacity = 12 m3; Initial pressure = 0.2 kg/cm2(g); Final pressure = 7.0 kg/cm2(g); Additional hold-up volume = 0.3 m3; Atmospheric pressure = 1.026 kg/cm2(a); Compressor pump-up time = 4.8 minutes. Leakage test (lunch time): (a) on load time 40 s, unloading pressure 7 kg/cm2(g); (b) average power during loading 95 kW; (c) unload time and loading pressure are 90 s and 6.6 kg/cm2(g). Find (i) compressor output m3/hr, (ii) specific power consumption kW/(m3/hr), (iii) % air leakage, (iv) leakage quantity m3/hr, (v) power lost due to leakage.
Model answer: (i) Compressor output = [Total Volume x (P2-P1)/Atm.Pressure] / Pump-up time = [(12+0.3) x (7.0-0.2)/1.026] / 4.8 = [12.3 x 6.8/1.026]/4.8 = 16.9834 m3/minute = 1019 m3/hr. (ii) Specific power consumption = 95/1019 = 0.093228 kW/m3/hr. (iii) % leakage = T/(T+t) x 100 = 40/(40+90) x 100 = 30.77%. (iv) Leakage quantity = 0.3077 x 1019 = 313.54 m3/hr. (v) Power lost due to leakage = leakage quantity x specific power consumption = 313.54 x 0.093228 = 29.23 kW.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — FAD via receiver pump-up formula; leakage % from load/(load+unload) time; power loss = leakage volume x specific power.
4. L-4: An energy audit was conducted to find out the ton of refrigeration (TR) of an Air Handling Unit (AHU). Evaporator area = 10.0 m2; Inlet velocity = 1.9 m/s; Inlet air DBT = 21.5 C, RH = 75%, Enthalpy = 53.0 kJ/kg; Outlet air DBT = 17.4 C, RH = 90%, Enthalpy = 46.4 kJ/kg; Density of air = 1.14 kg/m3. Find out the TR of AHU.
Model answer: AHU refrigeration load = [Air flow rate (m3/h) x Density of air (kg/m3) x Difference in enthalpy (kJ/kg)] / (3024 x 4.18). Air flow = 10.0 x 1.9 x 3600 = 68400 m3/h. AHU = (10.0 x 1.9 x 3600) x 1.14 x (53 - 46.4) / (3024 x 4.18) = 40.71 TR.
Mass flow x enthalpy drop gives kJ/hr; convert to kCal (/4.18) and to TR (/3024).
📖 §10.15 Energy efficiency measures · §10.13 BEMS · §10.3 ECBC scope & climatic zones
5. L-5: Write short notes on any two of the following: (a) Energy Efficiency Measures in Buildings; (b) Building Management System (BMS); (c) Energy Conservation Building Codes (ECBC).
Model answer: (a) Energy Efficiency Measures in Buildings - Air-Conditioning System: weather stripping of windows/doors (minimise infiltration; self-closing doors); temperature 23-25 C and RH 55-65%; maintain chilled water leaving temperature at or above 7 C (centrifugal chiller efficiency rises ~2.25% per 1 C rise in leaving temp); maintain insulation of chilled water pipes and ducts; clean chiller condenser tubes at least every six months; keep cooling towers clean; install frequency converters for AHU fan speed (saves up to 15%); keep air filters clean. Lighting System: switch off lights when not in use; separate switches for peripheral lighting (use daylight); install high-efficiency lighting (CFL for incandescent saves 75%); use electronic ballasts (losses 2W vs 12W conventional); optical luminaires (aluminium/silver/dielectric) save up to 50%; integrate lighting with AC (return air through luminaires); clean lights/fixtures (dust 4 times a year); use light colours for walls, floors, ceilings. (b) Building Management System (BMS) - Energy management systems range from simple ON/OFF timers up to a computerised central controller linked to numerous sensors (temperature, flow, pressure) and data sources (time, day, occupancy, meteorology, solar/internal gains). A microprocessor stores sensor data; performance equations (algorithms) compute deviations from desired conditions and control plant (e.g., adjusting chilled-water valve to AHU to hold set point). Trends can be stored, anticipation and self-correction built in. The function of a BMS/BEMS is economical and efficient monitoring and control of building services; one system can control a group of buildings. (c) Energy Conservation Building Codes (ECBC) - set minimum energy efficiency standards for design/construction of commercial and residential buildings without constraining function, comfort, health or productivity. India is grouped into five climatic zones: Composite (Delhi), Hot-Dry (Ahmedabad), Warm-Humid (Kolkata), Moderate/Temperate (Bangalore), Cold (Shillong). ECBC covers: building envelopes (except unconditioned storage/warehouses), mechanical systems and equipment (HVAC), service hot water heating, interior and exterior lighting, and electrical power and motors. It does not apply to buildings using neither electricity nor fossil fuel, equipment/systems using energy primarily for manufacturing processes, and multi-family buildings of three or fewer storeys plus single-family buildings.
Book-3 §10.15 (AC and lighting measures with the book's figures), §10.13 (BEMS: computerised central controller, sensors, microprocessor algorithms, self-correction, one system for a group of buildings), §10.3 (ECBC = minimum energy-efficiency standards for commercial buildings; five zones; covers envelope, HVAC, service hot water, lighting, electrical power & motors; excludes buildings using neither electricity nor fossil fuel and manufacturing-process equipment).
📖 §7.3 Efficient System Operation – IV Performance Assessment of Cooling Towers (book trial method)
6. L-6: A cooling tower cools 1450 m3/hr of water from 43 C to 36.6 C at 30 C wet bulb temperature. The cooling tower fan air flow rate is 9,50,000 m3/hr (air density = 1.08 kg/m3) and operates at 2.7 cycles of concentration. Find (a) Range, (b) Approach, (c) % CT Effectiveness, (d) L/G Ratio in kg/kg, (e) Cooling Duty Handled in TR, (f) Evaporation Losses in m3/hr, (g) Blow down requirement in m3/hr, (h) Make up water requirement in m3/hr.
Model answer: CT water flow = 1450 m3/hr = 1450000 kg/hr; CT fan flow = 950000 m3/hr; fan flow mass @1.08 kg/m3 = 1026000 kg/hr. (d) L/G Ratio = 1450000/1026000 = 1.41325 kg/kg. (a) Range = 43 - 36.6 = 6.4 C. (b) Approach = 36.6 - 30 = 6.6 C. (c) % CT Effectiveness = 100 x Range/(Range + Approach) = 100 x 6.4/(6.4+6.6) = 49.23%. (e) Cooling duty = 1450 x 6.4 x 10^3 = 9280 x 10^3 kCal/hr; /3024 = 3068 TR. (f) Evaporation losses = 0.00085 x 1.8 x 1450 x 6.4 = 14.1984 m3/hr (% evaporation loss = 14.1984/1450 x 100 = 0.98%). (g) Blow down = Evaporation losses/(COC - 1) = 14.198/(2.7-1) = 8.352 m3/hr. (h) Make up water = Evaporation loss + Blow down loss = 14.198 + 8.352 = 22.55 m3/hr.
Standard CT formulae: range, approach, effectiveness, L/G, duty in TR, evaporation (0.00085 x 1.8 x flow x range), blowdown = evap/(COC-1), make-up = evap + blowdown.