Energy Efficiency in Electrical Utilities Available here with full solutions — 64 questions recovered from the 2024 exam:
Objective (1 mark)
50 of 50
Short (5 marks)
6 of 8
Long (10 marks)
8 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) — 50
📖 §1.1 Introduction to Electric Power Supply Systems
1. What is the primary function of substations in the electrical power supply system?
To generate electricity
To communicate over long distances
To facilitate voltage transformation
None of the above
Answer: C) To facilitate voltage transformation
Confirmed vs Book-3 §1.1 — the book lists substations as the elements that 'connect the pieces to each other', and 'Sub-stations, containing step-down transformers, reduce the voltage for distribution to industrial users'. So their primary job is voltage transformation.
Option (a) is wrong because electricity is generated only in the power generating plant; the substation neither generates nor communicates.
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
2. How does the transmission voltage level affect the efficiency of long-distance power transmission?
Higher voltage levels reduce transmission losses
Higher voltage levels increase transmission losses
Voltage levels do not affect transmission losses
None of the above
Answer: A) Higher voltage levels reduce transmission losses
Confirmed vs Book-3 §1.1 — 'The current drawn is inversely proportional to the voltage level for the same quantity of power handled' and 'P.Loss = I²R', so raising voltage cuts loss in the ratio of the square of the voltages.
Option (b) reverses the physics: higher voltage means lower current, hence lower I²R loss, which is why HV/EHV transmission is used.
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
3. What is the primary purpose of using high voltage direct current (HVDC) transmission over long distances?
To increase the frequency of electricity
To step down the voltage for distribution
To minimize transmission losses over long distances
All of the above
Answer: C) To minimize transmission losses over long distances
Confirmed vs Book-3 §1.1 — 'Where transmission is over 1000 kM, high voltage direct current transmission is also favored to minimize the losses.'
Options (a)/(b) are wrong: HVDC has no power frequency to raise, and voltage stepping down is done by substation transformers, not by the HVDC link itself.
📖 §1.11 Analysis of Electrical Power Systems (Table 1.11)
4. What is the impact of voltage imbalance among the three phases in an electrical system?
Improved motor efficiency
Increased motor losses and reduced equipment life
Reduced power consumption
Enhanced power factor
Answer: B) Increased motor losses and reduced equipment life
Confirmed vs Book-3 §1.11 Table 1.11 — voltage imbalance among phases causes 'motor vibration, premature motor failure' and the book states 'A 5% imbalance causes a 40% increase in motor losses'.
Option (a) is wrong because imbalance produces a negative-sequence current that heats the motor and lowers, not improves, efficiency.
5. How does adding capacitors to an electrical distribution system improve power factor?
By providing the reactive power
By increasing the active power consumption
By lowering the system voltage
By increasing the frequency of the system
Answer: A) By providing the reactive power
Confirmed vs Book-3 §1.4 — power-factor capacitors 'act as reactive power generators, and provide the needed reactive power to accomplish kW of work'.
Option (b) is wrong: capacitors do not change the active (kW) power drawn by the load — they only supply the magnetising kVAr locally, so the kVA and current from the source fall.
6. A transformer has a primary voltage of 220V and a secondary voltage of 110V. If the primary current is 5A, what is the secondary current assuming no losses?
2.5A
5A
10A
20A
Answer: C) 10A
Confirmed vs Book-3 §1.5 Normal Operation — 'Primary ampere-turns are equal to secondary ampere-turns', i.e. V₁I₁ = V₂I₂. So I₂ = 5 × 220/110 = 10 A.
Option (a) 2.5 A wrongly divides by the turns ratio; in a step-down transformer the secondary current must rise as the voltage falls.
7. A factory consumes 500,000 kWh of electricity per month with a power factor of 0.8. How much is the reactive power (kVAR)?
400,000 kVAR
375,000 kVAR
800,000 kVAR
500,000 kVAR
Answer: B) 375,000 kVAR
Confirmed vs Book-3 §1.4 — kVAr = kW × tan(cos⁻¹PF). At PF 0.8, tanφ = 0.75, so reactive energy = 500,000 × 0.75 = 375,000 kVArh.
Option (a) 400,000 comes from kWh × sinφ (0.6/0.8 confusion); the power triangle uses the tangent, not the sine, of the angle when converting kW to kVAr.
8. Which of the following best describes an induction motor's operation?
It uses direct current to create mechanical energy.
It generates a rotating magnetic flux that induces current in the rotor.
It operates synchronously with the AC supply frequency.
It requires external excitation to operate.
Answer: B) It generates a rotating magnetic flux that induces current in the rotor.
Confirmed vs Book-3 §2.2 Motor Types — A 3-phase stator winding sets up a flux of constant magnitude rotating at synchronous speed; as it sweeps the stationary shorted rotor bars it induces an e.m.f (Faraday) and hence rotor current (Lenz), which produces torque. Option (c) is the tempting wrong choice: the rotor never catches the stator field, it always runs below synchronous speed with slip > 0 — only the synchronous motor runs at supply speed, and only it needs external DC excitation (d).
📖 §2.7 Improving the Motor Loading by Operating in Star Mode
9. How does operating a motor in star mode affect its performance?
It reduces the voltage and derates the motor capacity.
It increases motor speed.
It improves the power factor at high loads.
It eliminates the need for external capacitors.
Answer: A) It reduces the voltage and derates the motor capacity.
Confirmed vs Book-3 §2.7 Improving the Motor Loading by Operating in Star Mode — Changing from delta to permanent star reduces the winding voltage by a factor of √3 and electrically downsizes the motor to about 1/3 of its delta rating (a 15 kW delta motor becomes 5 kW in star), so a chronically <40 % loaded motor then operates near full load with better efficiency and power factor. (b) is wrong — speed actually drops slightly in star mode, which is why the book warns against it where output depends on motor speed.
📖 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD
10. In a VFD-controlled motor, what happens when the supply frequency is reduced while maintaining the same voltage?
The motor speed increases.
The motor efficiency improves.
The motor draws higher current and may overheat.
The motor torque decreases.
Answer: C) The motor draws higher current and may overheat.
Confirmed vs Book-3 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD — The book states that when supply frequency is reduced the equivalent circuit impedance falls, so the motor draws higher current and the flux rises towards saturation; that is why V and f must be varied together at a constant ratio (V/f control). (d) is the trap — with correct V/f the torque stays almost constant up to base speed; torque only falls above base speed due to field weakening.
11. Which of the following describes the function of a soft starter in a motor system?
It increases the motor's full-load speed.
It converts AC power to DC power.
It reduces the inrush current during motor start-up.
It improves the motor's efficiency at low speeds.
Answer: C) It reduces the inrush current during motor start-up.
Confirmed vs Book-3 §2.9 Soft Starter — A soft starter delivers a controlled release of power giving smooth, stepless acceleration, so it limits the direct-on-line inrush that can reach about +600 % of normal run current. (d) is wrong — it is a starting device, not a speed controller, and gives no running-efficiency gain at low speed.
12. A motor operates at 75% load with an efficiency of 88%. If the motor's rated power is 20 kW, what is the actual output power?
15 kW
13.2 kW
17.6 kW
14.4 kW
Answer: A) 15 kW
Confirmed vs Book-3 §2.4 Motor Efficiency — Nameplate kW is the rated shaft output, so at 75 % loading the output is 0.75 × 20 = 15 kW. (b) 13.2 kW is the tempting error — it multiplies the output again by the 88 % efficiency, but efficiency relates output to electrical input, it does not reduce the shaft output.
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
13. What is the main purpose of an after-cooler in a compressed air system?
To remove moisture from the air by cooling it
To increase the pressure of the compressed air
To filter out dust and particles
To lubricate the compressed air
Answer: A) To remove moisture from the air by cooling it
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — The after-cooler is a water-cooled heat exchanger whose stated objective is to remove the moisture in the air by reducing its temperature; roughly 60-75% of the moisture in compressed air drops out here.
Pressure is raised only by the compressor itself, so (b) is wrong, and dust is stopped upstream by the intake air filter, so (c) is wrong.
14. How does a desiccant air dryer remove moisture from compressed air?
By cooling the air
By using adsorbents like silica gel or activated carbon
By increasing the pressure
By reducing the air flow rate
Answer: B) By using adsorbents like silica gel or activated carbon
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Air dryers remove the traces of moisture left after the after-cooler using adsorbents such as silica gel, activated alumina or activated carbon; adsorption binds water physically on a large porous inner surface without any chemical reaction.
Cooling (a) describes the refrigerant dryer, and neither pressure (c) nor flow rate (d) is the drying mechanism in a desiccant unit.
📖 Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14)
15. Which of the following is an efficient method to control the capacity of a centrifugal compressor?
Automatic on/off control
Variable inlet guide vanes
Load and unload control
Multi-step control
Answer: B) Variable inlet guide vanes
Confirmed vs Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14) — The book states that the capacity of centrifugal compressors can be controlled using variable inlet guide vanes, with speed control as the other efficient route (Table 3.14).
Automatic on/off, load-unload and multi-step unloading are the schemes listed for positive-displacement machines, so (a), (c) and (d) do not apply to a centrifugal.
16. What is the effect of increasing the intake air temperature on the efficiency of an air compressor?
Increases efficiency by reducing power consumption
Decreases efficiency by increasing power consumption
No significant effect on efficiency
Increases the volumetric capacity of the compressor
Answer: B) Decreases efficiency by increasing power consumption
Confirmed vs Book-3 §3.5 Efficient Operation — Thumb rule: every 4°C drop in inlet air temperature lowers energy consumption by 1%, so a rise raises power. Table 3.3 shows relative air delivery falling from 102.0% at 10°C to 91.2% at 43.3°C, with 5.8% more power.
Hot, less dense intake air also reduces the mass delivered, so (d) — an increase in capacity — is exactly backwards.
📖 Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
17. What is the main benefit of using a variable speed drive (VSD) with a screw compressor?
Increases the maximum pressure capacity
Reduces the size of the compressor
Eliminates unloaded running condition
Simplifies maintenance
Answer: C) Eliminates unloaded running condition
Confirmed vs Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System — The checklist says to retrofit variable speed drives in big compressors, say over 100 kW, to eliminate the 'unloaded' running condition altogether; unloading can consume up to 30% of full-load power.
A VSD does not change the machine's maximum pressure (a) or its physical size (b) — it removes the idle-running loss.
18. Which of the following describes the primary function of an air receiver in a compressed air system?
To increase the air pressure
To act as a reservoir and dampen pulsations
To filter out impurities
To cool the compressed air
Answer: B) To act as a reservoir and dampen pulsations
Confirmed vs Book-3 §3.5 Air Receivers (IS 7938-1976 sizing) — The receiver dampens the pulsations leaving the compressor, serves as a reservoir for sudden heavy demand, prevents short cycling and lets moisture and oil vapour precipitate.
It cannot raise pressure (a) — it only stores air at the delivered pressure; filtering is done by line filters (c) and cooling by the after-cooler (d).
19. What is the primary function of the evaporator in a refrigeration cycle?
To compress the refrigerant
To absorb heat from the environment
To condense the refrigerant
To regulate the flow of refrigerant
Answer: B) To absorb heat from the environment
Confirmed vs Book-3 §4.3 - In the evaporator the low-pressure refrigerant boils, absorbing heat from the space/process being cooled (this Δh is the cooling effect). Compression is the compressor, condensing is the condenser, and flow regulation is the expansion valve. Compression (a) is the compressor's job, condensing (c) the condenser's, and flow regulation (d) the expansion device's; §4.3 stage 1-2 is the refrigerant boiling in the evaporator while absorbing heat.
20. In which component of an ideal refrigeration system, the refrigeration temperature will increase?
Compressor
Condenser
Evaporator
Expansion valve
Answer: A) Compressor
Confirmed vs Book-3 §4.3 - The compressor raises the refrigerant pressure and therefore its temperature, delivering superheated high-temperature gas to the condenser. The expansion valve drops temperature; the evaporator is the low-temperature side. The condenser (b) and expansion device (d) both reduce refrigerant temperature and the evaporator (c) runs at the lowest temperature in the cycle; §4.3 stage 2-3 notes a big temperature rise in the compressor as part of the input energy passes into the refrigerant.
21. Which refrigerant is commonly used in vapor absorption refrigeration systems?
R-22
R-134a
H2O
LiBr
Answer: C) H2O
Confirmed vs Book-3 §4.3 - In a LiBr–water vapour absorption system the refrigerant is pure water (H2O) and lithium bromide is the absorbent. (Ammonia is the refrigerant common to both VCR and VAR; among the options here water is the refrigerant, LiBr is the absorbent, and R-22/R-134a are vapour-compression refrigerants.). Lithium bromide (d) is the tempting answer but §4.3 is explicit that it is the ABSORBENT; R-22 and R-134a (a, b) are vapour-compression refrigerants only.
22. What is the effect of increasing the chilled water leaving temperature on the efficiency of a centrifugal chiller?
It increases the efficiency of the chiller
It decreases the efficiency of the chiller
It has no effect on efficiency
It increases the refrigerant flow rate
Answer: A) It increases the efficiency of the chiller
Confirmed vs Book-3 §4.7 - Raising the chilled-water (evaporator) leaving temperature raises the evaporator/suction pressure, reducing the compression ratio and compressor work, so COP/efficiency improves. Book rule: +1°C evaporator temperature ≈ 3% power saving. Option (b) is the intuitive but wrong choice; Figure 4.10 shows chiller COP rising as leaving chilled-water temperature rises, because the compressor works against a smaller temperature lift (about 3% power saving per 1 degC).
23. An HVAC system operates with a COP (Coefficient of Performance) of 4. If the system provides 100 kW of cooling, what is the power input to chiller?
0.04 kW
25 kW
400 kW
None of the above
Answer: B) 25 kW
Confirmed vs Book-3 §4.7 - COP = cooling effect / power input → power input = cooling / COP = 100 / 4 = 25 kW. Option (c) 400 kW multiplies instead of divides, and (a) confuses the ratio with its reciprocal in MW. A COP of 4 means one unit of compressor power moves four units of heat.
24. What is the effect of decreasing the RPM of a fan by 10% on its power requirement?
Decreases the power requirement by 27%
Decreases the power requirement by 19%
Increases the power requirement by 10%
No significant effect
Answer: A) Decreases the power requirement by 27%
Confirmed vs Book-3 §5.3 — Fan law: Power ∝ (Speed)³, so at 90% speed power = 0.9³ = 0.729, a 27% reduction; the book states this figure verbatim. Option (b) 19% is the STATIC-PRESSURE reduction (0.9² = 0.81), not power — the classic distractor.
25. How does an increase in system resistance affect the operation of a centrifugal fan?
Increases the airflow
Reduces the airflow
Reduces the static pressure
No effect on fan performance
Answer: B) Reduces the airflow
Confirmed vs Book-3 §5.3 — Raising system resistance (e.g. partly closing a damper) creates a steeper system curve SC₂; the operating point moves from A to B on the same fan curve: LOWER flow Q₂ against HIGHER pressure P₂. Option (c) is wrong because static pressure rises, not falls; system resistance varies as (flow)².
26. What is the primary purpose of trimming the impeller in a centrifugal pump?
To increase the pump speed
To adjust the pump capacity to match system requirements
To reduce the pump speed
To increase the NPSH required
Answer: B) To adjust the pump capacity to match system requirements
Confirmed vs Book-3 §6.6 — Impeller trimming machines the impeller diameter to reduce the energy added to the liquid — a permanent correction for a pump that is oversized for its system. It lowers both flow and head (Q∝D, H∝D²), not speed; NPSHR is not the target. Trimming is rarely taken below 75% of original diameter.
27. How does increasing the diameter of the suction pipe affect the NPSHA in a pumping system?
Reduces NPSHA
Increases NPSHA
Decreases NPSHR
Increases NPSHR
Answer: B) Increases NPSHA
Confirmed vs Book-3 §6.5 — NPSHA is the margin by which suction pressure at the impeller eye exceeds vapour pressure and is a characteristic of the system. A larger suction pipe cuts velocity and friction loss in the suction line, so NPSHA rises. NPSHR is a pump-design property and is unaffected by the piping.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
28. A pump has a flow rate of 200 cubic meters per hour and operates against a head of 30 meters. If the pump efficiency is 70%, what is the input power required?
60.5 kW
23.36 kW
95.2 kW
100 kW
Answer: B) 23.36 kW
Confirmed vs Book-3 §6.1 — Hydraulic power = Q(m³/s)·H·ρ·g/1000 = (200/3600)×30×1000×9.81/1000 = 16.35 kW. Pump shaft (input) power = hydraulic power / pump efficiency = 16.35/0.70 = 23.36 kW. Option a (60.5) and d (100) do not follow from the formula; 95.2 would need a much lower efficiency.
Confirmed vs Book-3 §6.5 — The book lists three undesirable effects of cavitation: (1) collapsing bubbles erode the vane surface, (2) noise and vibration increase (shorter seal/bearing life), (3) cavities choke impeller passages and reduce head — in extreme cases total loss of head. It never improves efficiency or reduces noise; NPSHR is a pump-design property, not an effect.
30. What is the relationship between pump speed and flow rate in a centrifugal pump according to the Affinity Laws?
Flow rate is proportional to the pump speed
Flow rate is proportional to the square of the pump speed
Flow rate is proportional to the cube of the pump speed
Flow rate is independent of pump speed
Answer: A) Flow rate is proportional to the pump speed
Confirmed vs Book-3 §6.5 — Affinity laws for a rotodynamic pump: Q∝N (flow directly proportional to speed), H∝N² (head ∝ speed²), P∝N³ (power ∝ speed³). Book example: halving speed 3000→1500 rpm halves flow 100→50 m³/hr. The square and cube relations belong to head and power, not flow.
📖 §6.2 System characteristics — static & friction head
31. What is the impact of using larger diameter pipes on the system resistance in a pumping system?
Reduces system resistance by lowering friction head losses
Increases system resistance
Increases power
Increases static head
Answer: A) Reduces system resistance by lowering friction head losses
Confirmed vs Book-3 §6.2 — Friction (dynamic) head is the loss in pipes, valves and equipment; after removing unnecessary fittings and length, further reduction in friction head needs larger diameter pipe (lower velocity; friction loss ∝ 1/D⁵). Static head is set by elevation difference and is unaffected by pipe size, so options b–d are wrong.
32. A cooling tower reduces the temperature of water from 40°C to 30°C. If the mass flow rate of water is 5 kg/s, what is the heat removed by the cooling tower?
500 kW
209 kW
2 kW
5 kW
Answer: B) 209 kW
Confirmed vs Book-3 §7.2 (iv) — Cooling capacity = mass flow × specific heat × temperature difference (Range). Q = 5 kg/s × 4.186 kJ/kg°C × (40−30)°C = 209.3 kW → option (b). Option (a) 500 kW ignores Cp; (c)/(d) are dimensionally meaningless. In book units this is 18,000 kg/h × 1 kcal/kg°C × 10°C = 1,80,000 kcal/h ≈ 209 kW.
33. The approach temperature of a cooling tower is 5°C, and the range is 10°C. If the inlet water temperature is 40°C, what is the outlet water temperature?
25°C
30°C
35°C
45°C
Answer: B) 30°C
Confirmed vs Book-3 §7.2 (i) — Range = hot water inlet − cold water outlet, so outlet = 40 − 10 = 30°C → (b). The 5°C approach (cold water − WBT) is a distractor: it would only be needed to find the WBT (30 − 5 = 25°C, which is option (a), a common trap).
📖 §7.5 Energy Saving Opportunities in Cooling Towers
34. How can the performance of a cooling tower be improved?
Proper water treatment
Regular maintenance
Optimizing air and water flow
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §7.5 — The book's energy-saving list covers all three: water treatment (algae/scale block nozzles and cut ΔT – §7.3), periodic cleaning of nozzles/fill and fan balance, and matching water and air flow (monitor L/G, blade angle, VFD). Hence (d) all of the above.
35. What is the primary function of a luminaire in a lighting system?
To generate light
To store electrical energy
To distribute light emitted from lamps
To control the voltage supply
Answer: C) To distribute light emitted from lamps
Confirmed vs Book-3 §8.2 — A luminaire is 'a device that distributes, filters or transforms the light emitted from one or more lamps' and includes everything for fixing/protecting the lamps except the lamps themselves. The lamp (not the luminaire) generates the light, so option (a) is wrong; storage/voltage control are not lighting functions.
36. Which type of lamp has the highest luminous efficacy among the following?
Low pressure sodium vapour lamp
Halogen lamp
LED lamp
Compact fluorescent lamp (CFL)
Answer: A) Low pressure sodium vapour lamp
Confirmed vs Book-3 §8.3 — Table 8.1 gives LPSV (SOX) 101–175 lm/W (avg 150), the highest of every lamp listed; the text calls LPSV 'the most efficacious light sources'. LED is 50–130 (avg 90), CFL 40–70 (avg 60), halogen 18–24 (avg 20). LPSV's penalty is CRI 10 (monochromatic light), not efficacy.
📖 §8.6(d) Selection of high efficiency lamps & luminaires
37. How does the use of high-efficiency luminaries contribute to energy conservation?
By decreasing the power consumption
By increasing the luminous efficacy
By improving light distribution characteristics
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §8.6(d) — Efficient lamps/luminaires save energy by (i) drawing less power for the same lumens, (ii) higher luminous efficacy (lm/W, Table 8.1) and (iii) better light distribution so more of the flux reaches the working plane (higher utilisation factor, §8.5). All three mechanisms apply, hence 'all of the above'.
📖 §9.2 Site condition effects on performance derating — Table 9.3 altitude & intake-temperature corrections
38. How does altitude affect the performance of a DG set?
Increases power output
Reduces fuel consumption
Reduces power output
No change in fuel consumption
Answer: C) Reduces power output
Confirmed vs Book-3 §9.2 — Table 9.3 gives altitude correction factors on engine output that fall steadily with height (e.g. 0.980 at 610 m, 0.855 at 1525 m, 0.494 at 4880 m for a non-supercharged engine; supercharged engines derate less). Thinner air means less oxygen mass per stroke, so output is derated. Options (a)/(b)/(d) are wrong: altitude neither raises output nor lowers fuel needed per kWh.
📖 §10.5 Building envelope — Solar Heat Gain Coefficient (SHGC)
39. What is the Solar Heat Gain Coefficient (SHGC) used for in building energy analysis?
To measure light transmittance
To measure the heat gain through fenestration due to solar radiation
To measure air leakage through windows
To measure the thermal emittance of roofing materials
Answer: B) To measure the heat gain through fenestration due to solar radiation
Confirmed vs Book-3 §10.5 — SHGC is the ratio of solar heat gain that passes through fenestration to the total incident solar radiation on it (directly transmitted + absorbed-and-re-radiated/conducted/convected inward). Option (a) is VLT (visible portion only), (c) is infiltration/weather-stripping and (d) is thermal emittance of roofs, not SHGC.
40. What is the primary function of an economizer in an HVAC system?
To increase indoor air pollution
To reduce the cost of heating equipment
To use outdoor air for cooling when conditions are favorable, saving energy
To increase the use of mechanical cooling systems
Answer: C) To use outdoor air for cooling when conditions are favorable, saving energy
Confirmed vs Book-3 §4.16 - An air-side economizer brings in cool/dry outdoor air for free cooling when outdoor conditions are favourable, reducing mechanical-cooling (chiller) energy. ECBC requires it on systems with fan >1200 L/s and cooling >22 kW; it can save ~10% energy. Options (a), (b) and (d) all increase energy or pollution; an economizer uses cool outdoor air directly for free cooling, which belongs with the §4.16 building heat-load minimisation measures.
📖 §10.14 Star rating of buildings — Energy Performance Index (EPI)
41. Energy Performance Index is the ratio of total building annual energy consumption to ------
Carpet area
Built up area
roof area
Windows and Walls area
Answer: B) Built up area
Confirmed vs Book-3 §10.14 — EPI (kWh/sq m/year) = total building annual energy consumption ÷ built-up area. Built-up area = carpet area + wall-thickness area + balconies (excluding parking basements), so carpet area (a) is the tempting wrong choice; roof or window/wall areas are never the denominator.
📖 §9.4 Energy performance assessment — specific fuel consumption (L/kWh)
42. In a D G set, the generator is generating 1000kVA at 0.7PF. If the specific fuel consumption of this D G set is 0.25 lits per kWh, then how much fuel in litres will be consumed while delivering generated power for one hour?
230
250
175
225
Answer: C) 175
Confirmed vs Book-3 §9.4 — Fuel is consumed against real power (kW = kVA × PF), not kVA. 1000 kVA × 0.7 = 700 kW → 700 kWh in one hour × 0.25 L/kWh = 175 litres. Option (b) 250 L wrongly multiplies the kVA (1000 × 0.25); (a) and (d) do not follow from the data.
43. The power measured in an Induced Draft (I D) fan operating at 49 Hz is 52 kW. A Variable Frequency Drive (VFD) is installed and the fan was operated at 34 Hz, The estimated Power saving will be_________
35.7 kW
17.3 kW
34.6 kW
36 kW
Answer: C) 34.6 kW
Confirmed vs Book-3 §5.3 — Speed ∝ frequency and Power ∝ (Speed)³: P₂ = 52 × (34/49)³ = 52 × 0.334 = 17.4 kW. Saving = 52 − 17.4 = 34.6 kW. Option (b) 17.3 kW is the NEW power, not the saving — a common trap.
Confirmed vs Book-3 §1.10 — the book lists 'Variable frequency drives (VFDs), electronic ballasts, UPS and Computers, induction and arc furnaces' as non-linear (harmonic-producing) devices.
Options (c)/(d): transformers and resistance heaters are essentially linear — impedance is constant — so they draw a sinusoidal current and generate negligible harmonics.
46. A 500 cfm reciprocating compressor has a loading and unloading period of 5 seconds and 20 seconds respectively during a compressor leakage test. The air leakage in the compressor air system will be ________
125 cfm
100 cfm
200 cfm
none of the above
Answer: B) 100 cfm
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — % leakage = T/(T+t) x 100 = 5/(5+20) x 100 = 20%; leakage quantity = 0.20 x 500 = 100 cfm.
The tempting 125 cfm (a) comes from using the load:unload ratio 5/20 instead of load/(load+unload), i.e. dividing by the unload time alone rather than by the full cycle.
47. What is the efficiency of motor with the following nameplate details 22 kW, 415V, 42 A, 0.8 p.f, 1475 rpm?
94.5%
91%
89.9%
None of the above
Answer: B) 91%
Confirmed vs Book-3 §2.4 Motor Efficiency — Rated input = √3 × 415 × 42 × 0.8 = 24.15 kW, so rated efficiency = 22 / 24.15 = 91 %. (c) 89.9 % and (a) 94.5 % do not follow from the nameplate data; the common slip is to forget the √3 or the power factor in the input calculation.
📖 §8.2 Installed power density (W/m²) — LPD arithmetic (ECBC context)
48. A hotel building has 14 floors, each of 1000m2 area, If the Lighting Power Density is 10.8 per m2 the interior lighting power allowance for the hotel building is_________
110800 W
129600 W
151200 W
186600 W
Answer: C) 151200 W
Confirmed vs Book-3 §8.2 (power density) — Lighting power allowance = total floor area × LPD = (14 floors × 1000 m²) × 10.8 W/m² = 14,000 × 10.8 = 151,200 W. Option (b) 129,600 W corresponds to 12 floors and (a)/(d) do not match any multiple of 10.8.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
49. A pump with 230mm diameter impeller is delivering a flow of 150 m3/hr. If the flow is to be reduced to 110m3/hr by trimming the impeller, what should be the approximate size of the impeller?
207mm
175 mm
169 mm
195 mm
Answer: C) 169 mm
Confirmed vs Book-3 §6.5 — Impeller-diameter affinity law Q∝D, so D₂ = D₁×(Q₂/Q₁) = 230×(110/150) = 168.7 ≈ 169 mm. 207 mm would give 150×207/230 = 135 m³/hr; 175 and 195 mm do not satisfy the ratio. Note 169/230 = 73%, at the edge of the book's ~75% trim limit — an exam-level approximation.
50. What is the main advantage of using a rotary screw air compressor over a reciprocating compressor?
Lower initial cost
Continuous, pulsation-free air delivery
Higher maximum pressure
None of the above
Answer: B) Continuous, pulsation-free air delivery
Confirmed vs Book-3 §3.2 Positive Displacement — Rotary compressors have rotors in place of pistons and give a continuous, pulsation-free discharge, need lower starting torque, smaller foundations and have fewer wearing parts; reciprocating output is pulsating.
Screw machines are not cheaper to buy (a), and very high pressures are the multi-stage reciprocating machine's territory (Table 3.1: up to 700 bar), so (c) is wrong.
1. A 3 phase Induction motor has the following details:
Name plate details: 55 kW, 415 V, 95 A, 0.9 p.f, 50 Hz
Running load details: 410 V, 75 A, 0.80 p.f, 48 Hz
Calculate the following:
a) loading percentage,
b) Rated efficiency,
Model answer: Actual power drawn by the motor = 1.732 × 410 × 75 × 0.80 / 1000 = 42.6 kW
Rated input power = 1.732 × 415 × 95 × 0.90 / 1000 = 61.5 kW
Percentage loading of motor = 42.6 / 61.5 = 69.3 %
Rated efficiency of motor = (55 / 61.5) × 100 = 89.4%
2. Match the following:
Column A:
1. Envelope Performance Factor (EPF)
2. Luminous Efficacy
3. Economizer
4. Thermal Mass
5. Energy Simulation Software
Column B:
a. Lighting System Efficiency
b. Heat Storage in Building Materials
c. ECBC Compliance
d. Building Energy Performance Modeling
e. Outdoor Air for Free Cooling
Model answer: 1. Envelope Performance Factor (EPF) → c. ECBC Compliance
2. Luminous Efficacy → a. Lighting System Efficiency
3. Economizer → e. Outdoor Air for Free Cooling
4. Thermal Mass → b. Heat Storage in Building Materials
5. Energy Simulation Software → d. Building Energy Performance Modeling
Per Book-3: EPF is the envelope trade-off compliance metric (§10.5); an economizer brings in outside air for free cooling in mild weather (§10.6); energy simulation software models whole-building performance (§10.4); luminous efficacy (lm/W) is lighting efficiency; thermal mass stores heat in building materials.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); total differential head h_d − h_s (positive suction head is subtracted); cf. §6.11 Solved example — cooling water pump efficiency (p.194–195)
3. A cooling water pump has a positive suction head of 5 meters. The discharge pressure is 3.0 kg/cm², and the water flow rate is 150 m³/hr. Determine the pump efficiency given that the actual power input of the connected motor is 18.0 kW and the motor operates with an efficiency of 85%.
Model answer: Flow Rate: 150 m³/hr
Total Head: 30 − 5 = 25 m
Power input to pump = 18 × 0.85 = 15.3 kW
Hydraulic Power = (150/3600) × 25 × 9.81 = 10.2 kW
Pump Efficiency = 10.2 / 15.3 = 66.7%
Total head = discharge head (3 kg/cm² ≈ 30 m) − suction head; hydraulic power = (Q/3600)·H·g; η = hydraulic/shaft power.
4. A process plant continuously operates a furnace oil operated DG set of capacity 3.0 MW to avoid any process safety incident in case of tripping of critical equipment on power failure. Total critical load on DG set is 2.5 MW and exhaust flue gas at 430 deg.C is vented as original design intent was to operate DG set intermittently only during power failure. Since it is being operated continuously, the process team developed a scheme to generate saturated steam at 5 bar(g) using the waste heat boiler. Other operating parameters:
Specific heat of flue gas: 0.24 kcal/kg-Deg.C
Final stack temperature to avoid Sulphur dewing: 210.0 Deg.C
Flue gas flow: 17.5 TPH
Sat. temp. of steam at 5 barg: 159.0 Deg.C
Latent heat at 5 barg: 498.0 kcal/kg
Feed water temperature: 130.0 Deg.C
Calculate the quantity of steam generated from waste heat boiler in TPH.
📖 Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
5. List five energy efficiency measures in Compressed air system.
Model answer: Any five energy efficiency measures from the Book-3 §3.7 checklist (p.101-102):
1. Locate the compressor so intake air is cool, clean and dry, drawing cold air from outside a hot compressor room — every 4°C rise in inlet temperature raises power by about 1%.
2. Clean inlet air filters regularly and fit manometers across them — efficiency drops about 2% for every 250 mmWC of filter pressure drop.
3. Reduce the delivery pressure wherever possible and keep the minimum possible range between load and unload settings — about 6-10% power saved per bar.
4. Carry out periodic leak tests and arrest leaks (40-50% leakage is not uncommon), and fit interlocked solenoid cut-off valves so idle machines get no air.
5. Minimise low-load running: below 50% demand change to a smaller compressor, reduce speed via the motor pulley, or retrofit a VSD above ~100 kW.
Also acceptable: periodic FAD tests; six-monthly valve inspection (worn valves cost up to 50% efficiency); periodic cleaning of inter-coolers; heat recovery from hot compressed air; generous pipe sizing with a ring main; discouraging misuse of compressed air.
Confirmed vs Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System — The expected answer is any five items from the printed §3.7 checklist; the model answer now spells them out with the book's quantified figures instead of pointing at page 101.
📖 §6.3 Pump curves — pump operating point; §6.4 pump efficiency highest at one flow (BEP)
6. Analyse the following data collected for a water pump. If the operating head is 16m explain what will happen to other parameters.
Design Parameters / Values:
Flow (Q): 40 lps
Head (H): 20 m
Power (P): 15 kW
Efficiency: 51%
Model answer: 1. If the operating head is 16 m instead of 20 m, the operating flow will be higher than the rated flow.
2. Since the operating point has deviated from the BEP, the operating efficiency will be less than design efficiency.
3. Since the flow has increased and pump efficiency decreased than rated, the operating power demand will be more than the rated power.
On a pump H-Q curve, lower head pushes the operating point right (higher flow), away from BEP, lowering efficiency and raising power.
1. A steel industry has 100 MW of captive power plant with 2 nos. of identical extraction condensing steam turbine. Power demand is 80 MW. The turbine specific condensing load is 3.20 kg/kWh and heat rejection in condenser is 560 kCal/kg.
Cold cooling water temperature: 32.0 °C
Hot cooling water temperature: 39.2 °C
Calculate the following:
a) The Cooling Water circulation flow (m³/hr) through both condensers, if only one cooling tower supplies water to condenser of both the steam turbines.
b) Make-up water flow rate (kg/hr) to basin, assuming blowdown loss is 1.0 % of circulation flow.
2. A process engineer develops a scheme to put 500 TR absorption-based refrigeration system to bring down process fluid temperature from 34 °C to 26 °C and this will result in higher production by 10%. 5 TPH excess steam is available in the plant and this new scheme utilizes this excess steam. COP of refrigeration system is 0.65 and available latent of steam for refrigeration system is 540 kcal/kg.
A) Estimate excess steam utilization for absorption-based refrigeration system in TPH. (3 Marks)
b) Estimate required Cooling water (m³/hr), if available approach in condenser is 10 °C. (2 Marks)
Model answer: Energy required for refrigeration system = 500 × 3024 / 0.65 = 2326153.8 kcal/hr
Steam needed for refrigeration system = 2326153.8 / 540 = 4.3 TPH
Steam utilization for VAM = 4.3 TPH
Required condenser duty = 2326153.8 + (500 × 3024) = 3838153.8 kcal/hr
Required Cooling water = 3838153.8 / 10 = 383.8 m³/hr
3. A 20 MW co-generation plant operates at a daily load factor of 85% and 8% auxiliary power consumption. The power is generated at 11 kV. Out of the total energy generated, 45% is exported to the grid through a 15 MVA transformer with 99% efficiency. Additionally, 35% of the generated energy is supplied to mill motors at 600 Volts through an 8 MVA step-down transformer with 98.5% efficiency. The remaining energy is used for other LT loads and auxiliaries at 415 Volts through a 4 MVA transformer with 98.2% efficiency. Calculate the following:
1. Daily energy generation in MWh.
2. Daily energy exported in MWh to the grid at 33 kV.
3. Daily mill motors consumption in MWh at 600 V.
4. Daily LT loads and auxiliary consumption in MWh at 415 V.
5. Daily transformer losses in kWh and % transformer losses.
Model answer: 1. Gross generation = 20 MW x 0.85 load factor x 24 h = 408 MWh/day; net (after 8% auxiliary consumption) = 408 x 0.92 = 375.36 MWh/day
2. Export = 375.36 x 0.45 = 168.91 MWh at the 11 kV bus; delivered through the 15 MVA transformer at 99% = 167.22 MWh/day
3. Mill motors = 375.36 x 0.35 = 131.38 MWh; delivered through the 8 MVA transformer at 98.5% = 129.41 MWh/day
4. LT loads & auxiliaries = 375.36 x 0.20 = 75.07 MWh; delivered through the 4 MVA transformer at 98.2% = 73.72 MWh/day
5. Transformer losses = (168.91-167.22) + (131.38-129.41) + (75.07-73.72) = 1.69 + 1.97 + 1.35 = 5.01 MWh = about 5010 kWh/day.
As a percentage of the 375.36 MWh actually passing through the transformers, loss = 5010/375,360 = 1.33% (1.23% if expressed on gross generation of 408 MWh).
This is the cascade-efficiency idea of Book-3 Sec.1.1 applied inside a plant: each transformation stage multiplies its own efficiency onto the energy delivered.
Gross gen = MW×LF×24; net = gross×(1−aux); split by % and apply transformer efficiencies; losses = before − after.
4. As part of a management initiative to advance green energy in a new process plant, a process engineer is assessing the economic viability of a 650 TR chiller. She is considering proposals for both LiBr-based vapor absorption chillers and vapor compression refrigeration systems. While power is sourced from renewable energy, the steam required is partially generated from excess process heat and additionally from firing furnace oil.
COP of advance Vapor Absorption Chiller: 1.3
COP of Vapor Compression Chiller: 4.50
Net steam price including excess steam and from boiler: 1500.00 INR/MT
Net Power cost from green source: 7.20 INR/kWh
Price of Cooling water: 3.00 INR/M3
Cooling water range: 8°C
Specific steam heat available for chiller: 490.0 kcal/kg
Evaluate both the offers and find out the offer which is economical in terms of operating cost.
5. L3 a) Match the following: (4 Marks)
1 Prescriptive Approach
2 Whole Building Performance Approach
3 Building envelope
4 Effective Aperture
Options:
1. Exterior façade
2. Trade-Off option
3. light admitting potential
4. Uses simulation to show compliance for the entire building
b. Fill in the following blank statements:
1. The Effective Aperture (EA) or light admitting potential of a glazing system is determined by multiplying the Visible Light Transmittance (VLT) of the glazing by the _____ of the building.
2. Thermal emittance is the relative ability of a material to _____ the absorbed heat.
3. If a window has a SHGC of 0.25 and the total incident solar radiation is 600 W/m², the solar heat gain through the window is _____
4. The emissivity of a material is the ratio of energy radiated by a particular material to energy radiated by a _____ at the same temperature.
5. As per ECBC the unit of Energy Performance Index (EPI) _____
6. Fenestration surface having a slope of less than 60 degrees from the horizontal plane is termed _____ (6 Marks)
Model answer: a) 1. Prescriptive Approach – b (Trade-Off option)
2. Whole Building Performance Approach – d (Uses simulation to show compliance for the entire building)
3. Building envelope – a (Exterior façade)
4. Effective Aperture – c (light admitting potential)
b) 1. ... multiplying the VLT of the glazing by the Window-Wall Ratio (WWR) of the building.
2. Thermal emittance is the relative ability of a material to radiate the absorbed heat.
3. Solar heat gain through the window = 0.25 × 600 = 150 W/m².
4. ... energy radiated by a black body at the same temperature.
5. As per ECBC the unit of EPI = kWh/Sq.mt/year.
6. Fenestration surface having a slope of less than 60 degrees from the horizontal plane is termed Skylight.
Book-3: Prescriptive method offers the envelope Trade-Off option; WBP uses simulation (§10.4); envelope = exterior façade; EA = VLT × WWR; thermal emittance = ability to radiate absorbed heat, emissivity relative to a black body; SHGC × incident radiation = 0.25 × 600 = 150 W/m²; EPI in kWh/sq m/year (§10.14); skylight = fenestration sloped < 60° from horizontal (§10.5).
6. L4 State whether True or False:
1. The lumen (lm) is the photometric equivalent of the Watt, weighted to match the eye response of the "standard observer," with blue light receiving the greatest weight.
2. The CRI of a lamp is 100 if it renders the color of the chips identical to the reference light source, indicating perfect color rendering.
3. A commercial building with a high window-to-wall ratio (WWR) and low SHGC glazing will experience higher cooling loads, as more solar heat will be transmitted through the windows.
4. Rotary screw compressors are preferable for fluctuating air demand.
5. Operating compressors at lower delivery pressures always results in higher energy efficiency.
6. Heat of compression dryers have higher operating costs compared to heatless purge dryers.
7. Using variable speed drives in compressors can eliminate unloaded running conditions and save energy.
8. Motor efficiency generally increases as the motor's rated capacity increases.
9. The power factor of an induction motor improves as the load on the motor decreases.
10. A decrease in supply voltage by 10% will decrease the torque of the motor by approximately 19%.
Lighting items: (1) False — Book-3 §8.1: yellowish-green (555 nm) light, not blue, receives the greatest weight (683 lm/W). (2) True — §8.2: if the lamp renders the standard colour chips identical to the reference source its CRI is 100. Item 3 (glazing) and items 4–10 (compressors, motors) belong to other chapters; keys as per the printed model solution.
7. A 2-stage reciprocating compressor is supplying nitrogen from low pressure header to high pressure vessel. This high-pressure nitrogen is only used during any process upset. Compressor is cut-off once vessel pressure reaches 45 barg, and started when vessel pressure comes down to 35 barg. During energy audit, it was observed that compressor is started at gap of every 36 hrs when there is no intended consumption. Other data:
Vol. of high pressure N2 vessel: 11.5 m3
Vessel temperature: 35.0 Deg.C
Initial gas density: 50.3 kg/m3
End gas density: 39.4 kg/m3
Compressor load kW drawn: 30.0 kW
Compressor capacity at constant suction pressure: 250.0 kg/hr
i. Estimate the leak rate (kg/hr).
ii. Estimate the energy saving potential (kWh/Annum), if all leaks are attended. Consider operating time of 8760 hrs/annum.
Model answer: Initial Vessel Pressure = 45.0 barg; End vessel pressure = 35.0 barg
Initial gas density = 50.3 kg/m3; End gas density = 39.4 kg/m3
Change in gas quantity in 36 hrs = (50.3 − 39.4) × 11.5 = 125.7 kg
N2 leakage rate = 125.7 / 36 = 3.5 kg/hr
Time needed for compressor run = 125.7 / 250 = 0.51 hrs or 30.6 min
% time of compressor running = 0.51 / 36 = 1.40 %
Running time of compressor per annum = 1.40% × 8760 = 122.3 hrs
Power consumption per annum due to air leakage = 122.3 × 30 = 3670.0 kWh/Annum
Energy Saving Potential = 3670.00 kWh/Annum
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — leak quantification — Leak = (Δdensity × volume)/time; compressor run time to replace leak = leaked mass/capacity; annual energy = % run time × hours × kW.
📖 §5.6 Pitot velocity, volume & fan static efficiency; §5.3 Fan laws
8. a) During performance guarantee test of an induced draft cooling tower, it was found that design approach of cooling tower is not achieved. As one of the probable causes, the team decided to check the efficiency of cooling tower fan. If design static efficiency is 70%, estimate the operating static efficiency using following parameters:
Pitot tube coefficient: 0.9
Velocity pressure: 49.0 mmWC
Air Density at operating condition: 1.129 kg/m3
Duct diameter: 2.1 m
Differential pressure across fan: 130.0 mmWC
Motor shaft Power: 190.0 kW
Motor Efficiency: 95.0 %
Gear Box Efficiency: 96.0 %
b) A centrifugal fan drawing 54 kW and operating at 1440 rpm is delivering air at 30,000 m³/hr. The head developed by the fan is 400 mm WC. If the speed is decreased by 200 rpm, calculate the following:
1. Air flow in m³/hr
2. Static pressure in mm WC
3. Power drawn in kW
Model answer: a) Air velocity = Cp × √(2 × 9.81 × ΔP/γ) = 0.9 × √(2 × 9.81 × 49/1.129) ≈ 26.3 m/s
Duct area = π/4 × (2.1)² = 3.46 ≈ 3.5 m²
Volume flow = 26.3 × 3.5 ≈ 92.0 m³/s
Air kW transferred = Q × ΔP/102 = 92 × 130/102 = 117.25 kW
Power input to fan shaft = motor shaft power × gearbox efficiency = 190 × 0.96 = 182.4 kW (190 kW is already the motor OUTPUT, so the 95% motor efficiency is not applied again)
Operating static efficiency = 117.25/182.4 × 100 = 64.28% (design 70% → fan is under-performing)
b) New speed = 1440 − 200 = 1240 rpm
1. Air flow = (1240/1440) × 30,000 = 25,833 m³/hr
2. Static pressure = (1240/1440)² × 400 = 296.6 mmWC
3. Power drawn = (1240/1440)³ × 54 = 34.48 kW
Static efficiency = (Q × ΔP_static)/(102 × shaft kW) per Book-3 §5.6; shaft kW = motor shaft output × gearbox efficiency (the model answer's printed expression '190 × 0.95 × 0.96 = 182.4' is a typo — 182.4 = 190 × 0.96). Part (b) applies Q ∝ N, SP ∝ N², kW ∝ N³ at 1240 rpm.