Energy Efficiency in Electrical Utilities Available here with full solutions — 69 questions recovered from the 2011 exam:
Objective (1 mark)
40 of 50
Short (5 marks)
16 of 8
Long (10 marks)
13 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) — 40
📖 §2.4 Motor Efficiency
1. The kW rating indicated on the name plate of an induction motor indicates
rated input of the motor
rated output of the motor
maximum input power which the motor can draw
maximum instantaneous input power of the motor
Answer: B) rated output of the motor
Confirmed vs Book-3 §2.4 Motor Efficiency — The nameplate kW of an induction motor is the rated mechanical OUTPUT available at the shaft; the input is that output divided by the efficiency. (a) 'rated input' is the standard misconception — it would make a 75 kW, 90 % motor deliver only 67.5 kW.
📖 Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power)
2. In a textile mill, two 150 cfm belt-driven reciprocating compressors are 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 the 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 the 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 out of every 50 s means the pair is loaded only 40% of the time, so one 150 cfm machine can carry the whole demand and even that one is oversized.
The book advises that where a compressor runs unloaded for long periods the economical fix is to change the pulley size and reduce the RPM to de-rate it. Hence switching one off AND trimming the other's motor pulley (b) beats switching off alone (a); a VSD (c) is not justified on so small a machine.
3. The most energy-intensive dryer among the following is
refrigeration
desiccant (heat of compression)
desiccant (heatless purge)
desiccant (blower reactivated)
Answer: C) desiccant (heatless purge)
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Table 3.19: heatless purge = 20.7 kW per 1000 m³/hr and 'High' operating cost, against blower reactivated 18.0 kW, refrigeration 2.9 kW and heat of compression 0.8 kW.
The heatless purge type also throws away 12-15% of the dry compressed air as purge (versus 1-2% for the blower type and none for HOC), which is what makes it the most energy intensive.
4. For centrifugal fans, the relation between shaft input Power (kW) and Speed (N) is given by
kW2/kW1 = N2/N1
kW2/kW1 = (N2/N1)^2
kW2/kW1 = (N2/N1)^3
none of the above
Answer: C) kW2/kW1 = (N2/N1)^3
Confirmed vs Book-3 §5.3 — Fan laws: Q ∝ N, SP ∝ N², kW ∝ N³, so kW₂/kW₁ = (N₂/N₁)³. Option (b) (square law) is the PRESSURE relation; option (a) (linear) is the FLOW relation.
5. A fan is operating at 970 RPM developing a flow of 3000 Nm3/hr at a static pressure of 650 mmWC. If the speed is reduced to 700 RPM, the static pressure (mmWC) developed will be
338.5 (printed in book as 388.5 — typo)
244.3
469
none of the above
Answer: A) 338.5 mmWC (book prints option (a) as 388.5 — a typo for 338.5)
Confirmed vs Book-3 §5.3 — SP ∝ N²: SP₂ = 650 × (700/970)² = 650 × 0.5207 = 338.5 mmWC. The intended book key is option (a); the printed value 388.5 is a misprint of 338.5 (the Master Notes key also maps Q6 to (a)). Option (c) 469 = 650 × (700/970) is the LINEAR (flow-law) error; (b) 244.3 = 650 × (700/970)³ is the cube (power-law) error. Note: the 19th-exam re-use of this question omits 388.5 and its key is 'none of the above' (338.5).
6. The T2, T5, T8 and T12 fluorescent lamps are categorized based on
diameter of the tube
length of the tube
both diameter and length
none of the above
Answer: A) diameter of the tube
Confirmed vs Book-3 §8.3 — 'These four lamps vary in diameter': T12 = 38 mm (1.5" = 12/8"), T8 = 25 mm (1"), T5 = 16 mm (5/8"), T2 = 6 mm (1/4"). The T-number is the diameter in eighths of an inch; length is not part of the designation.
📖 §9.3 Operational factors — load pattern & DG set capacity; sequencing of loads (kW on engine, kVA on generator)
7. 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.
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme
8. In BEE Star labeled distribution transformers, which of the 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 lists max losses at 50% and 100% for 1–5 star).
Option (c) 75% and 100% is the common distractor; 75% is not a defined labelling point in IS 1180/BEE star rating.
9. A 10 HP/7.5 kW, 415 V, 14.5 A, 1460 RPM, 3 phase rated induction motor, after decoupling from the driven equipment, was found to be drawing 3 A at no load. The current drawn by the motor at no load is high because of
faulty ammeter reading
very high supply frequency
loose motor terminal connections
poor power factor as the load is almost reactive
Answer: D) poor power factor as the load is almost reactive
Confirmed vs Book-3 §2.3 Motor Characteristics — At no load the current is almost entirely magnetizing (reactive), so a 3 A no-load draw on a 14.5 A motor is normal and simply reflects a very poor no-load power factor. (b)/(c) a frequency error or loose connection would show up as speed or heating faults, not as a routine high no-load current — the book advises keeping a record of no-load input power and current precisely because it is expected to be high.
10. A package air conditioner of 5 TR capacity delivers a cooling effect of 4 TR. If Energy Efficiency Ratio (W/W) is 2.90, the power in kW drawn by compressor would be:
4.84
1.38
1.724
None of the above
Answer: A) 4.84
Confirmed vs Book-3 §4.9 - Cooling delivered = 4 TR = 4 x 3.51 = 14.04 kW. Power = cooling/EER = 14.04/2.90 = 4.84 kW. Option (b) 1.38 divides by 2.9 twice and (c) 1.724 uses the 5 TR nameplate; use the DELIVERED 4 TR = 14.07 kW and divide by the EER of 2.90.
📖 §1.4 Performance Assessment of Power Factor Capacitors
11. 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 be approximately.
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 × 1.0488 ≈ 435 V.
Option (a) 400 V would REDUCE the output below 5 kVAr; a higher-than-rated output always means a higher-than-rated terminal voltage (which shortens capacitor life).
12. 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
13. 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
14. 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.
15. 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.
16. 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.
17. 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.
18. 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
19. 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.
20. 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.
21. 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.
22. 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
23. 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)
24. 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
25. 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'.
26. 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.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
27. 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.
28. 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
29. 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
30. 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.
31. 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)
32. 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.
33. 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.
📖 §9.4 Energy performance assessment — specific fuel consumption (L/kWh)
34. 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).
35. 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.
36. 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.
📖 §10.8 ECBC Guidelines on Lighting — Lighting Power Density, Building Area Method (book example: hotel, 4 floors × 1000 m², LPD 10.8 W/m²)
37. 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 __________.
1000 W
21600 W
43200 W
none of the above
Answer: C) 43200 W
Confirmed vs Book-3 §8.2 (power density) — Allowance = area × LPD = (4 × 1000 m²) × 10.8 W/m² = 43,200 W. 21,600 W would be two floors; 1000 W ignores the LPD.
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
39. 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.
1. Discuss in brief any three methods by which energy can be saved in an air conditioning system.
Model answer: 1) Raise the chilled water / cold air set-point temperature (and reduce condenser water/refrigerant condensing temperature) to lower the temperature lift and improve COP. 2) Reduce the cooling load - improve building insulation, use solar films/shading, control fresh-air infiltration and ventilation, and switch off unnecessary loads. 3) Operate chillers at optimum loading and use efficient part-load control (VFDs on pumps/AHU fans, optimum sequencing of multiple chillers), keep heat-exchanger surfaces (condenser/cooling tower) clean, and maintain correct refrigerant charge. These reduce kW/TR and overall energy use.
Reduce temperature lift, cut the cooling load, and optimize chiller/auxiliary operation (clean surfaces, VFDs, sequencing) to lower kW/TR.
2. 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).
3. 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).
4. Compute AT&C Losses: Input Energy Ei=11 MU, Energy Billed Metered=7 MU, Un-metered=1 MU, Total Billed=8 MU, Amount Billed=Rs.450 lakhs, Gross Collected=Rs.460 lakhs, Arrears Collected=Rs.40 lakhs.
Model answer: Ac (amount collected without arrears) = AG - Ar = 460 - 40 = Rs.420 lakhs.
Billing Efficiency BE = Eb/Ei x 100 = 8/11 x 100 = 72.7%.
Collection Efficiency CE = Ac/Ab x 100 = 420/450 x 100 = 93.3%.
AT&C Loss = [1 - (BE x CE)] x 100 = [1 - (0.727 x 0.933)] x 100 = [1 - 0.6786] x 100 = 32.1%.
(Book-3 Sec.1.8, Table 1.7 method: note that arrears must be stripped out of the gross collection before computing collection efficiency.)
AT&C Loss = 1 - (Billing Efficiency x Collection Efficiency).
5. 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.
6. 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.
7. 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)
8. 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
9. 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.
10. 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.
11. 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.
12. A DG set is operating at 600 kW load with 450 °C exhaust gas temperature. The DG set generates 8 kg of exhaust gas per kWh generated. The specific heat of gas at 450 °C is 0.25 kcal/kg °C. A heat recovery boiler is installed after which the exhaust temperature drops to 230 °C. How much steam will be generated at 3 kg/cm² with enthalpy of 650.57 kcal/kg? Assume boiler feed water temperature as 80 °C.
Model answer: Waste heat recovery = 600 kWh × 8 kg gas/kWh × 0.25 kcal/kg °C × (450 − 230) °C = 2,64,000 kcal/hr. Steam generation = 2,64,000 / (650.57 − 80) = 462.7 kg/hr.
Confirmed vs Book-3 §9.5 solved example (b) — two-step WHR method: heat recovered = kWh × 8 × 0.25 × ΔT = 2,64,000 kcal/hr; steam = heat ÷ (steam enthalpy − feed-water enthalpy) = 2,64,000/570.57 = 462.7 kg/hr. The exit temperature (230 °C) stays above the 180 °C acid-dew-point floor.
📖 §1.5 Transformers (no-load loss, load loss and loss at any load)
13. Briefly explain transformer losses and how the total transformer losses at any load level can be computed.
Model answer: Transformer losses consist of two parts: No-load loss and Load loss.
No-load loss (also called core loss) is the power consumed to sustain the magnetic field in the transformer's steel core. Core loss occurs whenever the transformer is energized; core loss does not vary with load. Core losses are caused by two factors: hysteresis and eddy current losses. Hysteresis loss is that energy lost by reversing the magnetic field in the core as the magnetizing AC rises and falls and reverses direction. Eddy current loss is a result of induced currents circulating in the core.
Load loss (also called copper loss) is associated with full-load current flow in the transformer windings. Copper loss is power lost in the primary and secondary windings of a transformer due to the ohmic resistance of the windings. Copper loss varies with the square of the load current (P = I²R).
For a given transformer, the manufacturer can supply values for no-load loss, P(NO-LOAD), and load loss, P(LOAD). The total transformer loss, P(TOTAL), at any load level can then be calculated from:
P(TOTAL) = P(NO-LOAD) + (% Load/100)² x P(LOAD)
The one formula to carry into the hall: P_total = P_no-load + (%Load/100)^2 x P_load(full-load copper loss).
No-load / core / iron loss (hysteresis + eddy current) is CONSTANT the moment the transformer is energised and does not care about load; load / copper loss is I^2R and therefore varies with the SQUARE of the load fraction.
Forgetting to square the load fraction is the single most common error in this family of questions. Maximum efficiency occurs at the load where copper loss = iron loss.
📖 §6.5 Efficient pumping system operation (NPSH and cavitation)
14. Distinguish between NPSH available and NPSH required in case of a centrifugal pump?
Model answer: NPSH Required (NPSHR): The minimum pressure required at the suction port of the pump to keep the pump from cavitating.
NPSHA is a function of the pumping system and must be calculated, whereas NPSHR is a function of the pump and must be provided by the pump manufacturer. NPSHA must be greater than NPSHR for the pump system to operate without cavitating. Put another way, you must have more suction side pressure available than the pump requires.
The one-line discriminator: NPSHa belongs to the SYSTEM and you calculate it; NPSHr belongs to the PUMP and the manufacturer supplies it. Operating rule NPSHa > NPSHr, with a margin, or the liquid flashes at the impeller eye and cavitates. NPSHa falls with hotter liquid (higher vapour pressure), a longer or narrower suction line, and altitude; NPSHr rises with speed (NPSHr ∝ N²).
📖 §5.6 Fan performance assessment (fan static efficiency)
15. An energy audit of a fan was carried out. It was observed that the fan was delivering 16,000 Nm3/hr of air with static pressure rise of 55 mm WC. The power measurement of the 3-phase induction motor coupled with the fan recorded 2.1 kW/phase on an average. The motor operating efficiency was assessed as 86% from the motor performance curves. What would be the fan static efficiency?
Model answer: Q = 16,000 Nm3/hr = 4.444 m3/sec
SP = 55 mmWC
Power input to motor = 2.1 x 3 = 6.3 kW
Power input to fan shaft = 6.3 x 0.86 = 5.418 kW
Fan static η = (Volume in m3/sec x ΔPst in mmWC) / (102 x Power input to shaft)
= (4.444 x 55) / (102 x 5.418)
= 0.4423 = 44.23 %
Memorise: η(static) = Q(m³/s) × ΔPst(mmWC) / (102 × shaft kW). Here Q = 16,000/3600 = 4.444 m³/s; motor input = 2.1 × 3 = 6.3 kW (per-phase reading × 3); shaft = 6.3 × 0.86 = 5.418 kW; η = 4.444 × 55 / (102 × 5.418) = 44.2%. Two traps: forgetting to multiply the per-phase kW by 3, and MULTIPLYING by motor efficiency instead of dividing — remember power flows motor → shaft, so shaft kW is the smaller number.
📖 §9.3 Operational factors (waste heat recovery from flue gases)
16. A DG set is operating at 600 kW load with 450 °C exhaust gas temperature. The DG set generates 8 kg of exhaust gas/kWh generated. The specific heat of gas at 450 °C is 0.25 kCal/kg °C. A heat recovery boiler is installed after which the exhaust temperature drops to 230 °C. How much steam will be generated at 3 kg/cm2 with enthalpy of 650.57 kCal/kg. Assume boiler feed water temperature as 30 °C.
Model answer: Heat available = 600 kWh x 8 kg gas generated/kWh output x 0.25 kCal/kg °C x (450 °C - 230 °C) = 2,64,000 kCal/hr
Steam generation = 2,64,000 kCal/hr / (650.57 - 30) = 425.41 kg/hr
Identical heat input, 264,000 kcal/hr, but colder feedwater at 30 °C widens the enthalpy rise to 650.57 − 30 = 620.57 kcal/kg, so steam output falls to 425.4 kg/hr. Compare with the 80 °C version at 462.7 kg/hr: preheating the feedwater by 50 °C buys about 9% more steam from the same exhaust — a standard exam point about economisers and feedwater heating.
1. A free air delivery test was carried out before conducting a leakage test on a reciprocating air compressor in an engineering industry, with the following observations: Receiver capacity 8.0 m3; Initial pressure 0.1 kg/cm2(g); Final pressure 7.0 kg/cm2(g); Additional hold-up volume 0.3 m3; Atmospheric pressure 1.026 kg/cm2 abs; Compressor pump-up time 3.5 minutes. During the subsequent leakage test at lunch time, with no pneumatic equipment or control valves in operation: (a) compressor on-load time 24 seconds at an unloading pressure of 7 kg/cm2(g); (b) average power drawn during loading 92 kW; (c) compressor unload time 79 seconds and loading pressure 6.6 kg/cm2(g). Find: (i) compressor output in m3/hr (neglect temperature correction); (ii) specific power consumption in kW/(m3/hr); (iii) % air leakage in the system; (iv) leakage quantity in m3/hr; (v) power lost due to leakage.
Model answer: Total system volume V = 8.0 + 0.3 = 8.3 m3.
(i) Compressor output Q = [(P2 - P1)/P0] x V/t = [(7.0 - 0.1)/1.026] x (8.3/3.5) = 6.725 x 2.371 = 15.94 m3/min = 956.6 m3/hr (temperature correction neglected as stated).
(ii) Specific power consumption = 92 kW / 956.6 m3/hr = 0.0962 kW per m3/hr.
(iii) % air leakage = T/(T+t) x 100 = 24/(24+79) x 100 = 23.30%.
(iv) Leakage quantity = 0.2330 x 956.6 = 222.9 m3/hr (= 3.71 m3/min).
(v) Power lost due to leakage = leakage quantity x specific power consumption = 222.9 x 0.0962 = 21.4 kW.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — L-2 as printed: FAD = [(P2-P1)/Patm] x (receiver + hold-up)/pump-up time, then % leakage = load/(load+unload), leakage quantity = % x FAD and power lost = leakage quantity x specific power. The stem was restored from the book (pump-up time 3.5 min and the full leakage data were missing) and the item re-typed as Long, per its own source line.
📖 §7.3 Efficient System Operation – IV Performance Assessment of Cooling Towers (Book EOC L-1 – identical to book trial)
2. A cooling tower cools 1565 m3/hr of water from 44 C to 37.6 C at 29.3 C wet bulb temperature. The fan air flow rate is 989544 m3/hr (air density 1.08 kg/m3) and it 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 in TR, f) Evaporation losses in m3/hr, g) Blowdown in m3/hr, h) Make-up water in m3/hr.
Model answer: a) Range = 44 − 37.6 = 6.4°C. b) Approach = 37.6 − 29.3 = 8.3°C. c) Effectiveness = Range/(Range + Approach) × 100 = 6.4/(6.4 + 8.3) × 100 = 43.53%. d) Water mass L = 1565 × 1000 = 15,65,000 kg/hr; air mass G = 989544 × 1.08 = 10,68,708 kg/hr; L/G = 15,65,000/10,68,708 = 1.46 kg/kg. e) Cooling duty = 1565 × 1000 × 1 × 6.4 = 1,00,16,000 kcal/hr = 10016 × 10³ kcal/hr; in TR = 1,00,16,000/3024 ≈ 3312 TR. f) Evaporation loss = 0.00085 × 1.8 × 1565 × 6.4 = 15.32 m³/hr (≈ 0.97% of circulation). g) Blow down = Evaporation/(COC − 1) = 15.32/(2.7 − 1) = 9.01 m³/hr. h) Make-up water = Evaporation + Blow down = 15.32 + 9.01 = 24.33 m³/hr (book's printed '2433' is a typo for 24.33).
3. List down any 5 energy conservation opportunities in fan systems.
Model answer: 1) Minimise excess air in combustion systems to reduce FD/ID fan load; 2) Minimise air in-leaks in hot flue gas path to reduce ID fan load; 3) Avoid cold air in-leaks that choke ID fan capacity; 4) Minimise system resistance/pressure drops via duct improvements; 5) Adopt inlet guide vanes in place of discharge damper control; (also: energy-efficient flat/cogged V-belts; two-speed motors or VSDs; fan speed reduction by pulley dia change; hollow FRP aerofoil impellers; impeller derating; higher-efficiency fan/impeller with cone).
Any five from the Book-3 §5.7 list: minimise excess air and in-leaks, high-efficiency impeller/fan, impeller derating, hollow FRP impellers, pulley speed reduction, VSD/two-speed motors, efficient belts, IGV instead of damper, lower system resistance.
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback
4. 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.
5. 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.
6. 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.
7. 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
8. 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).
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback
9. 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.
10. Fill in the blanks for the following: 1. The ratio of solar heat gain that passes through fenestration to the total incident solar radiation that falls on the fenestration is called ________. 2. Presenting the load demand of a consumer against time of the day is known as ______ curve. 3. The vector sum of active power and reactive power is ____. 4. The ratio of isothermal power to actual measured input power of an air compressor is known as ------. 5. The type of main input energy used for refrigeration in vapor absorption refrigeration plants is ____. 6. One ton of refrigeration is equivalent to ______ kW. 7. Stray losses in an induction motor generally are proportional to the square of the ________ current. 8. The capacitor kVAR selected for PF Correction at the induction motor terminals should not exceed ____ % of the no-load kVAR of the motor. 9. The ratio of luminous flux emitted by a lamp to the power consumed by the lamp is called _________________. 10. In an amorphous core distribution transformer, ______ loss is less than a conventional transformer.
Model answer: 1. Solar Heat Gain Coefficient (SHGC)
2. Load or hourly load
3. Apparent Power
4. Isothermal efficiency
5. Thermal energy (or steam or waste heat or gas or any energy related to thermal energy)
6. 3.51
7. rotor
8. 90
9. Luminous efficacy
10. No load or core
Numbers worth burning in: 1 TR = 3.51 kW = 3024 kcal/hr = 12,000 Btu/hr; capacitor kVAr at motor terminals <= 90% of the motor's no-load kVAr; isothermal efficiency = isothermal power / actual measured input power.
Definitions: apparent power (kVA) is the VECTOR sum of kW and kVAr; luminous efficacy = lumens / watt; SHGC = transmitted solar heat gain / incident solar radiation.
Trap in blank 7: stray losses vary as the square of the ROTOR current, not the stator current. In blank 10, amorphous-core transformers cut the NO-LOAD (core) loss to about 30% of a conventional unit - the copper loss is unchanged.
📖 §4.7 Performance assessment (heat balance and TR of an AHU)
11. a) In a Thermal Power Station, steam input to a turbine operating on a fully condensing mode is 110 Tonnes/Hr. The heat rejection requirement of the steam turbine condenser is 556 kCals/kg of steam condensed. The cooling water temperatures at the inlet to and outlet from the turbine condenser were measured to be 29 °C and 38 °C respectively. Find out the circulating cooling water flow. b) An energy audit was conducted to find out the ton of refrigeration (TR) of an Air Handling Unit (AHU). The audit observations are as under: Evaporator area (m2) = 9.5; Inlet velocity (m/s) = 1.9; Inlet air DBT (°C) = 21.5; RH (%) = 75.0; Enthalpy (kJ/kg) = 53.0; Outlet air DBT (°C) = 17.4; RH (%) = 90.0; Enthalpy (kJ/kg) = 46.4; Density of air (kg/m3) = 1.14. Find out the TR of AHU.
Model answer: a) Heat rejected = Heat pickup by cooling water
Steam flow rate x heat rejection = cooling water flow rate x Cp x ΔT
110 (TPH) x 1000 (kg/T) x 556 = Cooling water flow rate x 1 kCal/kg °C x (38 - 29) °C
Cooling water flow rate = 6795.55 m3/hr
b) TR = Q x ρ x (h_in - h_out)/3024
where Q is the air flow in m3/h, ρ is density of air in kg/m3, h_in is enthalpy of inlet air in kCal/kg, h_out is enthalpy of outlet air in kCal/kg
Q (m3/hr) = Area (m2) x Inlet velocity (m/s) x 3600 (s/hr) = 9.5 x 1.9 x 3600 = 64980 m3/hr
ρ = 1.14 kg/m3
h_in = 53.0 kJ/kg = 12.667 kCal/kg
h_out = 46.4 kJ/kg = 11.089 kCal/kg
TR = 64980 x 1.14 x (12.667 - 11.089)/3024 = 38.65 TR
Part (a) is a straight heat balance: heat rejected by steam = heat picked up by cooling water, i.e. 110 x 1000 x 556 = m x 1 x (38 - 29), giving 6795.6 m3/hr. Watch the tonnes-to-kg conversion.
Part (b): TR = Q x rho x (h_in - h_out)/3024, where Q = area x velocity x 3600 = 9.5 x 1.9 x 3600 = 64,980 m3/hr. Enthalpies MUST be converted from kJ/kg to kcal/kg (divide by 4.187): 53.0 -> 12.667 and 46.4 -> 11.089.
TR = 64,980 x 1.14 x 1.578/3024 = 38.65 TR. Mixing kJ with the 3024 kcal constant is the single biggest mark-loser in this question - keep the whole calculation in one unit system.
12. a) The efficiency at various stages from power plant to end-use is given below. Efficiency of power generation in the power plant - 30 %; T & D losses - 23 %; Distribution loss of the plant - 6 %; Equipment end use efficiency - 65 %. What is the overall cascade system efficiency from generation to end-use? b) The energy audit observations at a cooling tower (CT) in a process industry are given below: Cooling Water (CW) Flow : 3000 m3/hr; CW in Temperature: 41 deg. C; CW Out Temperature: 31 deg C; Wet Bulb Temperature: 24 deg. C. Find out Range, Approach, Effectiveness and cooling tower capacity in kCal per hour of the CT?
Model answer: a) Overall cascade system efficiency from generation to end-use
= 0.30 x (1 - 0.23) x (1 - 0.06) x 0.65
= 0.1411 = 14.11 %
b) Range = (Inlet - Outlet) Cooling Water Temperature = (41 - 31) = 10 deg. C
Approach = (Outlet Cooling Water - Air Wet Bulb) Temperature = (31 - 24) = 7 deg C
% CT Effectiveness = 100 x [Range/(Range + Approach)] = 10/[10 + 7] x 100 = 58.8 %
Cooling capacity, kCal/hr = heat rejected = CW flow rate in kg per hour x (CW inlet hot water temp. to CT - CW outlet cold well temp.)
= 3000 x 1000 x (41 - 31) = 30,000,000 kCal per hour = 30 Million kCal per hour
Work in that order every time: Range = 41 − 31 = 10 °C; Approach = 31 − 24 = 7 °C; Effectiveness = Range/(Range + Approach) × 100 = 10/17 × 100 = 58.8%; Capacity = 3000 × 1000 × 10 = 30 million kcal/hr. The denominator is Range + Approach (= hot water − WBT, the ideal range) — dividing by Range alone or by the approach is the mark-losing slip. In part (a) the cascade efficiency multiplies SURVIVING fractions: 0.30 × (1−0.23) × (1−0.06) × 0.65 = 14.11%; using 0.23 and 0.06 directly instead of (1−loss) is the trap.
13. Write short notes on any three of the following: (i) Effect of supply voltage on capacitor kVAR rating (ii) Pump impeller trimming (iii) Affinity laws for centrifugal machines (iv) Trigeneration (v) Building fenestration
Model answer: i) Ideally capacitor voltage rating is to match the supply voltage. If the supply voltage is lower, the reactive kVAr produced will be in the ratio V1²/V2², where V1 is the actual supply voltage and V2 is the rated voltage.
ii) Impeller trimming refers to the process of machining the diameter of an impeller to reduce the energy added to the system fluid. Impeller trimming offers a useful correction to pumps that, through overly conservative design practices or changes in system loads, are oversized for their application. The laws with respect to impeller trimming are: Flow ∝ D, Head ∝ D², Power ∝ D³.
iii) The equations relating centrifugal machine performance parameters of flow, head and power absorbed, to speed are known as the Affinity Laws: Q ∝ N, H ∝ N², P ∝ N³. Where Q = flow rate, H = head or resistance, P = power absorbed, N = rotating speed.
iv) Trigeneration refers to simultaneous generation of steam (heat), power and refrigeration through integrated systems. For example, in a DG set besides power being generated, steam is produced with waste exhaust gases and chilled water is generated using jacket cooling water. Three different utilities are created using a single fuel as energy source.
v) Fenestration systems include windows, skylights, ventilators and doors that are more than one-half glazed. All openings (including the frames) in the building envelope that let in light. Total area of the fenestration is measured using the rough opening (including glazing, sash and frame). For glass doors where glazed vision area is less than 50% of the door area, the fenestration area is the glazed vision area; otherwise, it is the door area.
The two examinable formula sets: affinity laws for SPEED — Q ∝ N, H ∝ N², P ∝ N³; and trimming laws for DIAMETER — Q ∝ D, H ∝ D², P ∝ D³. Same exponents, different variable, and impeller cuts are limited to about 75% of maximum diameter. For capacitors, delivered kVAr scales as the square of the voltage ratio (V1/V2)², so a capacitor run below its rated voltage under-delivers badly. Fenestration = all openings more than half glazed, measured over the ROUGH opening including frame and sash.