Energy Efficiency in Electrical Utilities Available here with full solutions — 56 questions recovered from the 2016 exam:
Objective (1 mark)
42 of 50
Short (5 marks)
7 of 8
Long (10 marks)
7 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) — 42
📖 §4.2 Comfort Zone (Figure 4.3)
1. Which of the following is the most comfortable conditions for an office room? (DBT = Dry bulb temperature, RH = Relative humidity)
20°C DBT and 80% RH
26°C DBT and 100% RH
15°C DBT and 30% RH
25°C DBT and 55% RH
Answer: D) 25°C DBT and 55% RH
Confirmed vs Book-3 §4.2 - Human comfort zone is ~22-26°C DBT with ~50-60% RH; the other options have extreme temperature or humidity. The book's comfort zone is 22-27 degC DBT with 40-60% RH, so (a) is too humid, (b) is saturated air, and (c) is both too cold and too dry.
📖 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up
2. The effect of increasing the air gap in an induction motor will increase:
power factor
speed
capacity
magnetizing current
Answer: D) magnetizing current
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Table 2.2 Minimising Watts Loss / Loss Break-up — A larger air gap needs more magnetizing current to drive flux across it; that is why the book's loss-reduction advice is the opposite — 'lowering the operating flux density and possible shortening of air gap' reduces the magnetizing component of current. (a) power factor is the trap: more magnetizing current makes power factor WORSE, not better.
3. The formation of frost on cooling coils in a refrigerator:
improves C.O.P. of the system
increases heat transfer
reduces power consumption
increases power consumption
Answer: D) increases power consumption
Confirmed vs Book-3 §4.8 - Frost acts as an insulating layer, reducing heat transfer and forcing the compressor to work harder, increasing power consumption. Options (a)-(c) all assume frost helps; in fact the frost layer insulates the coil, depresses the evaporator temperature and, per Table 4.4, drives specific power consumption up.
4. In a refrigeration system, the expansion device is connected between the:
Compressor and condenser
Condenser and receiver
Condenser and evaporator
Evaporator and compressor
Answer: C) Condenser and evaporator
Confirmed vs Book-3 §4.3 - The expansion device drops high-pressure liquid from the condenser to low-pressure before the evaporator (constant enthalpy). Compressor-to-condenser (a) carries hot discharge gas and evaporator-to-compressor (d) is the suction line; stage 4-1 of Figure 4.4 places the expansion device between the condenser (liquid receiver) and the evaporator.
5. Which of the following is wrong with respect to Color Rendering Index (CRI)?
The CRI is expressed in a relative scale ranging from 0 - 100
CRI indicates how perceived colors match actual colors
CRI of Sodium Vapour lamp is much higher than that of a normal Incandescent Lamp
The higher the color rendering index, the less color shift or distortion occurs
Answer: C) CRI of Sodium Vapour lamp is much higher than that of a normal Incandescent Lamp
Confirmed vs Book-3 §8.2/Table 8.1 — CRI is a 0–100 scale showing how perceived colours match the reference source; higher CRI = less colour shift (statements a, b, d are true). Sodium vapour lamps have CRI 22 (HPSV) or 10 (LPSV) whereas incandescent is 100, so (c) is the wrong statement.
6. Which of the following is wrong with reference to heat rate of a coal fired thermal power plant?
Heat rate indicates the overall energy efficiency of a power plant
When calculating plant heat rate, the energy input to the system is GCV of the fuel
Lower the heat rate the better
860 kCal per kWh is practically achievable
Answer: D) 860 kCal per kWh is practically achievable
Confirmed vs Book-3 §1.1 — 1 kWh ≡ 860 kCal is the THEORETICAL heat equivalent (100% efficiency); real Indian coal plants run at 28–35% efficiency, i.e. about 2450–3070 kCal/kWh, so 860 kCal/kWh is not practically achievable.
Options (a)–(c) are all correct book statements, including 'lower the heat rate, higher is the generation efficiency'.
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
7. In electrical power system, transmission efficiency increases as:
both voltage and power factor increase
both voltage and power factor decrease
voltage increases but power factor decreases
voltage decreases but power factor increases
Answer: A) both voltage and power factor increase
Confirmed vs Book-3 §1.1 — for the same power, current falls as voltage rises AND as power factor rises (I = P/(√3·V·cosφ)); since loss = I²R, efficiency improves when both increase.
Option (c) is wrong because a falling power factor raises the current for the same kW and therefore raises the I²R loss.
📖 §8.3(8) LED lamp advantages; §8.9 Case study; §8.7 Street lighting
8. Which of the following is wrong statement with reference to LED lamps?
LED lamps are as energy efficient as CFL bulbs or better
LED lamps are more durable than CFLs
LED lamps has no hazardous material like mercury
LED lamps are not suitable for Street Lighting purpose
Answer: D) LED lamps are not suitable for Street Lighting purpose
Confirmed vs Book-3 §8.3/§8.9 — Table 8.1 lists LED applications as 'office, industry, outdoor, retail...' and the §8.9 case study replaces street/security FTLs with 18 W LEDs (payback 3.4 years), so LEDs ARE suitable for street lighting — (d) is the wrong statement. LEDs are mercury-free, vibration-resistant (durable) and 50–130 lm/W vs CFL 40–70, so (a)–(c) are true.
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
9. In no load test of a poly-phase induction motor, the measured power by the wattmeter consists of:
core loss
copper loss
core loss, windage & friction loss
stator copper loss, iron loss, windage & friction loss
Answer: D) stator copper loss, iron loss, windage & friction loss
Confirmed vs Book-3 §2.4 Motor Efficiency — Field Tests for Determining Efficiency — The no-load wattmeter reading is the total no-load input, i.e. stator I²R at no-load current PLUS core (iron) loss PLUS friction & windage; the book subtracts (no-load current)² × stator resistance from it to leave core + F&W. (c) is the tempting answer because it names what is LEFT after that subtraction, not what the wattmeter actually reads.
10. A 10 MVA generator has power factor 0.86 lagging. The reactive power produced will be:
10 MVAr
8 MVAr
5 MVAr
1.34 MVAr
Answer: C) 5 MVAr
Confirmed vs Book-3 §1.4 power triangle — kVAr = kVA × sinφ = 10 × sin(cos⁻¹0.86) = 10 × 0.51 = 5.1 ≈ 5 MVAr.
Option (b) 8 MVAr is the ACTIVE power (10 × 0.86 = 8.6 MW); the reactive side of the triangle uses the sine, not the cosine.
11. The no-load loss and copper loss of a 500 kVA transformer is 900 watts and 6400 watts respectively. What is the total loss at 50% of transformer loading?
4100 watts
6850 watts
2500 watts
3650 watts
Answer: C) 2500 watts
Confirmed vs Book-3 §1.5 — total loss = no-load loss + (%load)² × full-load copper loss = 900 + (0.5)² × 6400 = 900 + 1600 = 2500 W.
Option (d) 3650 W would come from halving the copper loss (900 + 3200); the book's formula squares the load fraction.
Confirmed vs Book-3 §4.2 - Mass of moisture per kg of dry air is the specific/absolute humidity (humidity ratio). Relative humidity (b) is a percentage ratio of vapour pressures and dew point (d) is a temperature; only absolute/specific humidity is a mass of moisture per kg of DRY air.
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
13. The basic function of an air dryer in a compressor is to:
Prevent dust from entering the compressor
Remove moisture before the intercooler
Remove moisture in compressor suction
Remove moisture at the downstream of the after-cooler
Answer: D) Remove moisture at the downstream of the after-cooler
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — Air dryers remove the remaining traces of moisture AFTER the after-cooler — i.e. downstream of it — because instrument and pneumatic air must be essentially moisture free.
Keeping dust out (a) is the intake filter's job, and there is no moisture removal before the intercooler or at the suction (b, c) — atmospheric moisture is carried in with the intake air.
14. The term "cooling range" in a cooling tower refers to the difference in the temperature of:
dry bulb and wet bulb
hot water entering the tower and the wet bulb temperature of the surrounding air
cold water leaving the tower and the wet bulb temperature of the surrounding air
hot water entering the tower and the cooled water leaving the tower
Answer: D) hot water entering the tower and the cooled water leaving the tower
Confirmed vs Book-3 §7.2 (i) — Book: 'Range is the difference between the cooling tower water inlet and outlet temperature' → (d). Option (c) is the Approach; (b) is the ideal range used in the effectiveness formula; (a) is the wet-bulb depression of air.
📖 §5.1 Introduction — Table 5.1 (ASME specific ratio)
15. The distinction between fans and blowers is based on:
impeller diameter
specific ratio
speed
volume delivered
Answer: B) specific ratio
Confirmed vs Book-3 §5.1 — Fans, blowers and compressors are distinguished by the ASME specific ratio (discharge/suction pressure): fan up to 1.11, blower 1.11–1.20, compressor > 1.20. Impeller diameter, speed or volume (a, c, d) are not the classification basis.
16. A better indicator for cooling tower performance is:
Heat load in tower
Range
RH of air leaving cooling tower
Approach
Answer: D) Approach
Confirmed vs Book-3 §7.2 (ii) — Book: 'Although both range and approach should be monitored, the Approach is a better indicator of cooling tower performance' → (d). Range and heat load are fixed by the process, not by the tower (§7.2 Range).
📖 §10.8 ECBC guidelines on lighting — Building Area Method
17. As per the building area method given in ECBC, compute the lighting power allowance; given the allowed LPD is 12 watt per square meter and enclosed office area is 500 square meter.
6 kW
4.16 kW
6 W
4.16 W
Answer: A) 6 kW
Confirmed vs Book-3 §10.8 — Interior lighting power allowance = gross lighted floor area × allowed LPD = 500 m² × 12 W/m² = 6000 W = 6 kW (same method as the book's hotel example: 1000 × 10.8 × 4 = 43,200 W). Watch the unit: 6 W (c) is a 1000× slip.
Improves with increase in excitation and may even become leading at high excitations
Decreases with increase in excitation
Is independent of its excitation
None of the above
Answer: A) Improves with increase in excitation and may even become leading at high excitations
Confirmed vs Book-3 §2.2 Motor Types — A synchronous motor's power factor is set by its DC field excitation: raising excitation improves the power factor and at over-excitation the machine draws leading current, which is why synchronous motors are used for plant PF correction. (c) 'independent of excitation' describes the induction motor, whose PF cannot be controlled this way.
19. As per ECBC compute the Effective Aperture (EA); given Window Wall Ratio (WWR) is 0.40 and Visible Light Transmittance (VLT) is 0.25.
0.10
0.65
0.33
0.15
Answer: A) 0.10
Confirmed vs Book-3 §10.5 — Effective Aperture = VLT × WWR = 0.25 × 0.40 = 0.10 (light-admitting potential of the glazing). The book's own example uses 0.4 × 0.26 = 0.104 (> 0.1, complies). Adding (0.65) or dividing (1.6/0.625) are the tempting errors.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
20. Increasing the impeller diameter in a pump:
Increases the flow
decreases the head
decreases the power
all of the above
Answer: A) Increases the flow
Confirmed vs Book-3 §6.5 — Impeller-diameter relations: Q∝D, H∝D², P∝D³. A larger impeller raises tip speed, so flow, head AND power all increase. Options b and c state decreases, so 'all of the above' is wrong; only a is true.
21. The percentage reduction in distribution losses when tail end power factor is raised from 0.8 to 0.95 is:
29%
15.8%
71%
84%
Answer: A) 29%
Confirmed vs Book-3 §1.4 — reduction in distribution loss % = [1 − (PF₁/PF₂)²] × 100 = [1 − (0.8/0.95)²] × 100 = [1 − 0.709] × 100 = 29%.
Option (c) 71% is the RESIDUAL loss fraction (0.709), not the reduction — a very common slip.
22. In a Three Phase Transformer, the secondary side line current is 139.1A, and secondary voltage is 415V. The rating of the transformer would be:
50 kVA
150 kVA
100 kVA
63 kVA
Answer: C) 100 kVA
Confirmed vs Book-3 §1.5 — for a 3-phase transformer, kVA = √3 × V × I = 1.732 × 415 × 139.1 = 99,980 VA ≈ 100 kVA.
Option (d) 63 kVA would result from omitting the √3 factor and using single-phase VA (415 × 139.1 ≈ 57.7 kVA).
📖 §1.4 Performance Assessment of Power Factor Capacitors
23. Shunt capacitors connection is normally adopted for:
Distribution Voltage improvement
Power factor improvement
Both a and b
None of these
Answer: B) Power factor improvement
Corrected (was c) — Book-3 §1.4 Performance Assessment of PF Capacitors: 'Shunt capacitor connections are adopted for almost all industry/end user applications, while series capacitors are adopted for voltage boosting in distribution networks.'
So the purpose for which shunt capacitors are normally adopted is power factor improvement; voltage boosting is the role of SERIES capacitors, which is why (a) and hence (c) 'both' are ruled out. (The load-end voltage does rise slightly as a by-product — book advantage (c) — but that is not what shunt banks are installed for.)
📖 §1.4 Performance Assessment of Power Factor Capacitors
24. A company installed a new 100 kVAr, 415Volt capacitor but the power analyzer indicates that it is operating at 93 kVAr. The reason could be:
Operation is at low load
Higher Voltage at terminals
Lower voltage at terminals
None of the above
Answer: C) Lower voltage at terminals
Confirmed vs Book-3 §1.4 — capacitor output ∝ V², so 93/100 = (V/415)² gives V = 415 × √0.93 ≈ 400 V — the terminal voltage is BELOW the rated 415 V.
Option (b) higher voltage would give MORE than 100 kVAr (and shorten capacitor life), so it cannot explain a 93 kVAr reading.
25. The kVA reduction by improving the power factor of a plant operating at 400 kW load from 0.85 to 0.95 is:
40
49
72
None of the above
Answer: B) 49
Confirmed vs Book-3 §1.4 — kVA₁ = 400/0.85 = 470.6 and kVA₂ = 400/0.95 = 421.1, so the kVA reduction is 470.6 − 421.1 ≈ 49 kVA.
Option (c) 72 is roughly the capacitor kVAr needed (400 × (0.620 − 0.329) = 116) confusion; the question asks for the drop in apparent power, not the capacitor size.
📖 §1.8 Estimation of Technical Losses in Distribution System
26. For a supply end Voltage of 10.6 kV and receiving end Voltage of 9.8 kV, the percentage regulation works out to:
0.80
8.16
7.55
None of these
Answer: B) 8.16
Confirmed vs Book-3 §1.8 Voltage Regulation — 'Percentage regulation = 100 (Es − Er)/Er' = 100 × (10.6 − 9.8)/9.8 = 8.16%.
Option (c) 7.55% divides by the SENDING end voltage (10.6); the book's formula uses the receiving-end voltage as the reference.
27. An Induction motor rated 15 kW and 90% efficiency, at full load will:
Draw 15 kW
Draw 13.5 kW
Deliver 16.66 kW
Deliver 15 kW
Answer: D) Deliver 15 kW
Confirmed vs Book-3 §2.4 Motor Efficiency — Nameplate kW is the rated output, so a 15 kW motor at full load DELIVERS 15 kW and draws 15/0.9 = 16.67 kW. (c) 'deliver 16.66 kW' is the standard inversion trap — 16.66 kW is the input drawn, not the shaft output.
28. A 50 hp motor with a full load efficiency of 90 percent was found to be operating at 25 kW input. The percent Motor Load is:
75%
67%
60%
25%
Answer: C) 60%
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — 50 hp = 37.3 kW output; rated input = 37.3/0.90 = 41.4 kW; % load = 25 / 41.4 = 60 %. (a) 75 % comes from dividing 25 kW by the hp figure without converting to kW and without dividing by efficiency — the book's formula always compares measured input kW with rated kW ÷ rated efficiency.
📖 §9.2 Sizing — kVA = √3·V·I; §9.3 engine loading (kW) vs alternator loading; 1 HP = 0.746 kW
29. A DG set has a 300 HP engine drive and is connected to a 300 kVA alternator with 95% efficiency. When a plant load of 290 amps at 415 Volts and 0.76 power factor is connected, the engine loading works out to:
52%
74.51%
55.4%
None of the above
Answer: B) 74.51%
Confirmed vs Book-3 §9.2/§9.3 — Electrical output = √3 × 415 × 290 × 0.76 / 1000 = 158.4 kW. Engine shaft power needed = 158.4 / 0.95 (alternator efficiency) = 166.8 kW. Engine rating = 300 HP × 0.746 = 223.8 kW. Engine loading = 166.8/223.8 = 74.5%. Option (a) 52% is the tempting value obtained by comparing kW to the 300 kVA alternator rating, not the engine.
30. At which of the following discharge pressures, the same reciprocating air compressor will consume maximum power?
3 bar
5 kgf/cm2
90 psi
500 kPa
Answer: C) 90 psi
Confirmed vs Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure (Table 3.9) — Convert to a single unit: 3 bar ≈ 3.06 kg/cm², 5 kgf/cm² ≈ 4.9 bar, 90 psi ≈ 6.2 bar, 500 kPa = 5.0 bar. The highest is 90 psi.
Book-3: 'For the same capacity, a compressor consumes more power at higher pressures', so the 90 psi setting draws the maximum power. The trap is comparing the raw numbers without unit conversion.
📖 §9.4 Energy performance assessment — specific fuel consumption & alternator loading
31. In a DG set, the generator is consuming 70 litre per hour diesel oil. If the specific fuel consumption of this DG set is 0.33 litres/kWh at that load, what is the kVA loading of the set at 0.8 PF?
212 kVA
265 kVA
170 kVA
None of these
Answer: B) 265 kVA
Confirmed vs Book-3 §9.4 — kW generated = 70 L/h ÷ 0.33 L/kWh = 212.1 kW; kVA = 212.1/0.8 = 265 kVA (same question and key in the 16th-exam alternate set and March-2021 paper). Option (b) was transcribed as "262.5 kVA" in this record; it has been repaired to the computed 265 kVA so that exactly one option is correct. Option (a) 212 kVA is the tempting wrong choice — it is the kW value, not the kVA loading.
33. As per the Inverse Square Law of illumination what will be the illuminance at half the distance?
50%
4 times
double
No change
Answer: B) 4 times
Confirmed vs Book-3 §8.2 — Illuminance is inversely proportional to the square of distance (E = I/d²). At half the distance E ∝ 1/(0.5)² = 4 times; the book's own example: 10 lm/m² at 1 m becomes 40 lm/m² at 0.5 m. 'Double' is the common trap (that would be a linear law).
34. Find the air density at 35°C temperature at one atmospheric pressure. It is given that at one atmospheric pressure the air density at 20°C is 1.2041 kg/m3.
1.1455
1.2657
1.2024
none of the above
Answer: A) 1.1455
Confirmed vs Book-3 §5.6 — From γ = PM/RT at constant pressure, density ∝ 1/T(K): ρ₃₅ = 1.2041 × (273+20)/(273+35) = 1.2041 × 293/308 = 1.1455 kg/m³. Option (b) 1.2657 inverts the temperature ratio (density cannot rise with heating at constant pressure).
36. Which Loss in a Distribution Transformer is predominant if the transformer is loaded to 75% of its rated capacity?
core loss
copper loss
hysteresis loss
magnetic field loss
Answer: B) copper loss
Confirmed vs Book-3 §1.5 — core loss is constant while copper loss varies as (%load)². At 75% load a typical 500 kVA unit (Table 1.3) has copper loss 0.75² × 6450 = 3630 W against a fixed core loss of only 900 W, so copper loss dominates.
Option (a) core loss dominates only at light loading (below roughly 35–40% load); (c) hysteresis is a COMPONENT of core loss, not a separate answer.
37. The voltage unbalance in three phase supply is 1.5%. If the motor is operating at 100°C, the additional temperature rise in °C due to voltage unbalance is:
4.5
9
0
none of the above
Answer: A) 4.5
Confirmed vs Book-3 §2.7 Voltage Unbalance — Additional temperature rise = 2 × (% voltage unbalance)² = 2 × 1.5² = 4.5 °C. (b) 9 °C is the trap of multiplying by the unbalance instead of squaring it (the book's own example gives 8 °C for 2 % unbalance, confirming the square law).
38. Which of the following cannot be controlled by automatic power factor controllers?
KW
voltage
Power factor
KiloVAr
Answer: A) KW
Confirmed vs Book-3 §1.4 — an APFC 'monitors the displacement power factor' and switches capacitor banks in or out via relay outputs, so it controls kVAr, the resulting power factor, and indirectly the voltage.
Active power (kW) is set by the process load; capacitors supply no kW, so the controller cannot control it.
39. The parameter used in Star labeling of air conditioner is:
COP
EER
KW/TR
EPI
Answer: B) EER
Confirmed vs Book-3 §4.15 - BEE star rating of air conditioners is based on EER (Energy Efficiency Ratio, W/W). COP (a) and kW/TR (c) are audit indicators but are not on the label, and EPI (d) belongs to building star rating; §4.15 fixes the AC star bands in EER (W/W), e.g. 5-star split >= 3.50.
41. In a lithium bromide absorption refrigeration system:
lithium bromide is used as a refrigerant and water as an absorbent
water is used as a refrigerant and lithium bromide as an absorbent
ammonia is used as a refrigerant and lithium bromide as an absorbent
none of these
Answer: B) water is used as a refrigerant and lithium bromide as an absorbent
Confirmed vs Book-3 §4.3 - In a LiBr VAR system, water is the refrigerant and lithium bromide solution is the absorbent. Option (a) swaps the two roles; §4.3 is explicit that pure water is the refrigerant and lithium bromide solution the absorbent, and (c) describes an ammonia machine, which uses water as absorbent.
📖 §9.3 Factors affecting waste heat recovery from flue gases — back pressure
42. A good DG set waste heat recovery device manufacturer will take precautions to prevent which of the following problem while DG set is in operation?
voltage unbalance on generator
Excessive back pressure on engine
excessive steam generation
turbulence in exhaust gases
Answer: B) Excessive back pressure on engine
Confirmed vs Book-3 §9.3 — "Back pressure in the gas path caused by additional pressure drop in the waste heat recovery unit is another key factor. Generally the maximum back pressure allowed is around 250–300 mm WC and the heat recovery unit should have a pressure drop lower than that." Excess back pressure derates the engine and raises fuel consumption; voltage unbalance (a) is a load issue, not a WHR-device issue.
📖 §1.1 Industrial End User — 'ONE Unit saved = TWO Units Generated'
1. One unit of electricity in end-use application is equivalent to about two units of electricity generated. Substantiate with the cascade efficiency from generating plant ex-bus to end-use. Assume: Generator yard substation efficiency 98%; T&D loss = 20%; End-use application efficiency = 65%.
Model answer: Cascade efficiency = 0.98 × (1 − 0.20) × 0.65 = 0.5096. Therefore one unit at end use = 1/0.5096 = 1.96 ≈ 2 units at ex-generator bus.
2. Match the following Terms in ECBC: 1 Lighting Power Density (LPD); 2 Energy Performance Index (EPI); 3 Effective Aperture (EA); 4 Visible Light Transmittance (VLT); 5 U-Factor — with A Rate of Heat Flow in Watt per m² per °C; B Light admitting potential of a Glazing System; C Watts per square meter; D kWh per square meter per year; E Ratio of light passing through glazing to light through perfectly transmissive glazing.
Model answer: 1-C; 2-D; 3-B; 4-E; 5-A.
Book-3: LPD = W/m² (§10.8); EPI = kWh/m²/year (§10.14); EA = light-admitting potential of a glazing system (§10.5); VLT = ratio of light through glazing to perfectly transmissive glazing; U-factor = heat flow per m² per °C (§10.5).
📖 §10.15 Energy efficiency measures in buildings · §10.5 envelope/glazing
3. List five energy saving measures in a commercial building.
Model answer: 1) Optimize air conditioning volumes (false ceiling, partition/segregation of critical areas). 2) Reduce solar heat gain through the envelope with efficient glazing. 3) Use energy efficient lighting systems. 4) Use occupancy/motion/sound sensors for lighting. 5) Use energy efficient pumping and air conditioning systems; provide barriers against hot-air leakage; optimize evaporator temperature to ~22°C; avoid heating appliances in cool spaces.
Book-3 §10.15 measures (temperature/humidity settings, efficient lighting, controls, AHU VFDs, weather stripping) plus §10.5 envelope measures (low-SHGC glazing, shading, insulation).
📖 §6.6 Flow control strategies — pump control by varying speed / VSDs & VFDs; §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³)
4. Explain how a Variable Frequency Drive saves power in a three phase electric motor driven pumping system? What will be the reduction in power drawn by a motor by reducing the speed by half?
Model answer: A VFD converts fixed-frequency, fixed-voltage line power into variable-frequency, variable-voltage output to control the speed of induction motors. By controlling pump speed instead of throttling flow with valves, energy savings are substantial. By the affinity law, power ∝ speed³, so halving speed reduces power by a factor of (1/2)³ = 1/8 (to one-eighth). A throttling device on a fixed-speed motor leaves the motor running at nearly full power.
5. A 415 V, 15 kW, 3-ph, 50 Hz induction motor at full load: 88% efficiency, 0.85 PF lagging. a) Find current drawn. b) If replaced by a 92.5% efficient motor with 0.92 PF, what are the power savings in kW and kVA?
Model answer: a) Input power = 15/0.88 = 17.05 kW. Line current = 17.05×1000/(√3×415×0.85) = 27.91 A. kVA = 17.05/0.85 = 20.06 kVA. b) New input = 15/0.925 = 16.216 kW; new kVA = 16.216/0.92 = 17.62 kVA. Savings = 17.05 − 16.216 = 0.834 kW and 20.06 − 17.62 = 2.44 kVA.
6. Identify the type of refrigeration system in the figure and the components 1,2,3 & 4. Explain briefly the function of each.
Model answer: Vapour Absorption Refrigeration system. 1 Absorber: concentrated LiBr absorbs the refrigerant vapour (water) and becomes dilute. 2 Generator: heats the dilute LiBr, regenerates refrigerant (water vapour) and re-concentrates LiBr. 3 Condenser: condenses the regenerated refrigerant (water vapour). 4 Evaporator: liquid refrigerant (water, atomised) picks up heat from the chilled-water coil and becomes water vapour.
Confirmed vs Book-3 §4.3 (Figure 4.5) - in the LiBr-water absorption chiller water is the refrigerant and LiBr solution the absorbent. Evaporator: water flashes at ~4 degC under 754 mmHg vacuum and chills the water from 12 to 7 degC. Absorber: concentrated LiBr absorbs the vapour (and maintains the vacuum), becoming dilute. Generator: steam/hot water/oil boils off the water and re-concentrates the LiBr. Condenser: condenses that vapour and returns it to the evaporator.
📖 Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power); Book-3 §3.2 Positive Displacement — Reciprocating Compressors
7. A 1680 m3/hr reciprocating compressor is driven by a 160 kW motor (90% efficiency) drawing 159 kW. Demand rises by 100 m3/hr. To meet it the compressor speed is increased by changing the compressor pulley. Existing: Motor rpm 1400, Motor pulley 300 mm, Compressor rpm 700, Compressor pulley 600 mm. Find new pulley diameter and additional power; check if motor can handle the load.
Model answer: Modified flow = 1780 m3/hr. New compressor rpm = (1780/1680)×700 = 742 rpm. Using N1D1 = N2D2, new compressor pulley D2 = (700×600)/742 = 566 mm. New motor power = (742/700)×159 = 168.54 kW. Motor capacity = 160/0.9 = 178 kW. Since 168.54 < 178 kW, the motor has the margin to absorb the additional 100 m3/hr load.
Confirmed vs Book-3 §3.5 Capacity Utilisation — Flow ∝ rpm; pulley by N1D1=N2D2; power ∝ rpm (reciprocating, ~linear here).
1. The rated compressor capacity is 15 m3/min. Evaluate if there is any capacity de-rating using the air-receiver tank filling method. Data: Receiver volume (incl. pipe & cooler) = 9 m3; Initial Pressure = 0.5 kg/cm2; Final Pressure = 7.0 kg/cm2; Atmospheric pressure = 1.026 kg/cm2; Time to build pressure = 5 minutes. (b) What is the deficiency in this calculation and how can it be corrected?
Model answer: (a) Compressor output = [(P2 - P1) x V] / (Pa x t) = [(7.0 - 0.5) x 9] / (1.026 x 5) = 58.5/5.13 = 11.40 m3/min.
Capacity shortfall against the rated 15 m3/min = 15 - 11.40 = 3.60 m3/min, i.e. (3.60/15) x 100 = 24% de-rating. Since this far exceeds the 10% the book allows before corrective action, the compressor must be investigated (worn valves alone can cost up to 20% of capacity).
(b) Deficiency: the formula as used assumes the compressed air temperature equals the ambient temperature, i.e. perfect isothermal compression. In practice the discharge/receiver temperature t2 is higher than the ambient t1, so the measured volume is overstated. Correction: multiply the result by the factor (273 + t1)/(273 + t2), where t1 is the ambient/suction temperature and t2 the compressed air (receiver) temperature; this factor is always less than 1.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — Tank-filling FAD method Q = (P2-P1) x V/(Pa x t), compared against rated capacity. The model answer's temperature correction had been written inverted; the book's factor is (273+t1)/(273+t2), always less than 1 when the discharge is hotter than ambient.
2. State three advantages of improvement of Power Factor at Load side. Power Factor at the load side is 0.75 and average minimum load is 100 kW. What is the kVAr rating of capacitor to improve the Power Factor at the load side to 0.95?
Model answer: Advantages (any three): reduced kVA (maximum demand) charges in utility bill; reduced distribution losses (kWh) due to lower current; better voltage at motor terminals and improved motor performance; reduction in size of transformers; avoidance of PF penalty / availing PF incentives; better operating efficiency of motors/drives. Capacitor required = 100{tan(cos⁻¹0.75) − tan(cos⁻¹0.95)} = 100(0.882 − 0.329) = 55.3 kVAr, say 55 kVAr.
📖 §9.4 Energy performance assessment — 2-hour trial, efficiency, SFC, turbocharger/WHR gas temperatures
3. DG set performance: Trial 2 hrs; Energy generated 1500 kWh; Level difference in day tank 51.6 cm; Day tank diameter 1 m; CV 10500 kcal/kg; Air drawn 30 kg/kg fuel; WHR potential 2.6×10^5 kcal/hr with flue gas after WHR at 180°C. a) Calculate average efficiency and specific fuel consumption. b) Calculate present flue gas exit temperature; specific gravity 0.86, specific heat of flue gas 0.25 kcal/kg°C.
Model answer: Fuel during 2 hr = (π/4×1²×0.516×1000) = 405 litres → 202.5 lit/hr; mass = 405×0.86/2 = 174.18 kg/hr. SFC = 1500/405 ≈ 3.7 kWh/lit (or ~4.3 kWh/kg). Efficiency = (750×860)/(174.18×10500) = 35.3%. Mass of flue gas = (30+1)×174.18 = 5399.5 kg/hr. ΔT across WHR = 260000/(5399.5×0.25) = 192.61°C. Present flue gas temp = 180 + 192.61 = 372.6°C.
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R) + §1.6 Distribution Losses
4. A residential colony with fixed load 250 kVA is 1 km from an 11 kV/415 V transformer. Compare LT (1×3.5c×300sqmm) vs HT (1×3c×70sqmm) distribution. Data: LT cable R=0.13 Ω/km, Rs 700/m; HT cable R=0.570 Ω/km, Rs 1300/m; unit Rs 7/kWh; transformer relocation (HT) Rs 1 lakh. Recommend and estimate payback on marginal investment.
Model answer: Current LT = 250/(0.415×1.732) = 347.8 A; HT = 250/(11×1.732) = 13.1 A. Loss LT = 347.8²×0.13×3/1000 = 47.17 kW; Loss HT = 13.1²×0.57×3/1000 = 0.29 kW. Saving = 46.87 kW. Annual energy saving = 46.87×8760 = 4,10,639 kWh; cost saving = Rs 28,74,470/yr. HT investment = 1300×1000 + 1,00,000 = Rs 14,00,000; LT investment = 700×1000 = Rs 7,00,000. Payback on marginal investment = (14,00,000 − 7,00,000)/28,74,470 = 0.24 yr ≈ 3 months. Recommend HT distribution.
📖 §5.3 Fan laws; §5.5 Pulley change; §5.6 Fan power formula
5. A belt-driven centrifugal fan system delivers 12 m3/s. One branch (1.5 m3/s) needs 89 mmWC static; the rest needs only 66 mmWC but the fan runs at 89 mmWC. Auditor proposes reducing fan speed to give 66 mmWC and adding a booster fan (75% eff, motor 85% eff) for the branch. Main fan input = 16.2 kW; initial fan speed 1200 rpm, motor pulley 209 mm, fan pulley 305 mm; 6000 h/yr. Calculate annual energy and cost savings.
Model answer: Revised fan speed = 1200×(66/89)^0.5 = 1031 rpm; new fan pulley = 305×1200/1031 = 355 mm. Initial air kW = 12×89/102 = 10.5; revised air kW = 12×66/102 = 7.8. Revised motor input = 16.2×7.8/10.5 = 12 kW. Main fan saving = (16.2−12)×6000 = 25,200 kWh = Rs 1,76,400/yr. Booster: 1.5 m3/s, ΔP=23 mmWC, air kW=1.5×23/102=0.34; shaft=0.34/0.75=0.45; motor=0.45/0.85=0.53 kW → 0.53×6000=3180 kWh = Rs 22,260/yr. Net saving = 1,76,400 − 22,260 = Rs 1,54,140/yr.
SP ∝ N² gives the reduced main-fan speed (1031 rpm) and the pulley from N₁D₁ = N₂D₂ (355 mm); motor input scaled by the ratio of air powers (Q × ΔP/102); booster fan power = air kW/(η_fan × η_motor). Net saving ≈ Rs 1.54 lakh/yr at Rs 7/kWh.
📖 §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³); §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
6. A centrifugal pump runs at 60 m3/hr, 1470 RPM, pump efficiency 65%, motor efficiency 89%. Discharge gauge 3.4 kg/cm2, suction 3 m below pump centerline. Auditor recommends replacing the motor with a 4-pole 91% efficient motor at 1% slip. Determine new flow and motor power; throttle fully open and system head purely frictional. Comment.
Model answer: Existing: Head = 34 − (−3) = 37 m; pump power = (60/3600)×37×1000×9.81/(1000×0.65) = 9.3 kW. New motor speed = 1500 − 0.01×1500 = 1485 rpm. New flow = 60×(1485/1470) = 60.61 m3/hr. New pump power = 9.3×(1485/1470)³ = 9.59 kW. Existing motor input = 9.3/0.89 = 10.46 kW; new motor input = 9.59/0.91 = 10.54 kW. Comment: power consumption is slightly more, so not recommended (though flow is also marginally higher).
Affinity laws Q∝N, P∝N³; head from gauge + suction lift.
📖 §1.11 Analysis of Electrical Power Systems (Table 1.11) — multi-chapter True/False
7. L-6: State True or False (1 mark each):
1. The efficiency of gas turbine power plant is lower than that of a combined cycle power plant.
2. The performance of air compressor at high altitudes will be lower as compared to that at sea level.
3. Efficiency of transformer will be minimum when copper loss is equal to iron losses.
4. In cooling towers, the water droplets entrapped in the air stream is captured by drift eliminators.
5. To get the static pressure, the inner and outer tubes of pitot tube are connected to manometer.
6. The throttling of pump discharge will change the pump characteristic curve.
7. The simplest way to reduce the discharge from a reciprocating air compressor is to throttle it.
8. Cycle of Concentration (COC) is the ratio of dissolved solids in circulating water to the dissolved solids in makeup water.
9. Use of VFD will save power but also create harmonics.
10. The synchronous speed of a 4 pole motor will be 3000 rpm.
Model answer: 1. True
2. True
3. False (transformer efficiency is MAXIMUM when copper loss equals iron loss)
4. True
5. False (inner and outer tubes connected to the manometer give velocity pressure; static pressure is from the outer tube alone)
6. False (throttling changes the system curve, not the pump characteristic curve)
7. False (throttling a reciprocating compressor is not the way to reduce discharge; use unloading/speed control)
8. True
9. True
10. False (Ns = 120 x 50 / 4 = 1500 rpm)
Cross-chapter True/False. The two Chapter-1 items are (3) and (9): transformer efficiency is MAXIMUM (not minimum) when copper loss equals iron loss, because the variable copper loss then just equals the fixed core loss (Book-3 §1.5); and VFDs do save power but, being non-linear power-electronic loads, they inject harmonic currents (Book-3 §1.10). Item 10 is settled by Ns = 120f/P = 120 x 50/4 = 1500 rpm, not 3000 rpm.