Energy Efficiency in Electrical Utilities Available here with full solutions — 51 questions recovered from the 2019 exam:
Objective (1 mark)
37 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) — 37
📖 §7.1 Components of Cooling Tower – Drift eliminators
1. What is the function of drift eliminators in cooling towers?
maximize water and air contact
capture water droplets escaping with air stream
enables entry of air to the cooling tower
eliminates uneven distribution of water into the cooling tower
Answer: B) capture water droplets escaping with air stream
Confirmed vs Book-3 §7.1 Components – Drift eliminators — Book: drift eliminators 'capture water droplets entrapped in the air stream that otherwise would be lost to the atmosphere' → (b). Maximising contact is the fill (a); air entry/equalising is the air inlet/louvers (c); uniform water distribution is the nozzles (d).
📖 §6.3 Pump curves — pump operating point; §6.6 Flow control strategies — flow control valve (throttling)
2. Which of the following statements is not true regarding centrifugal pumps?
Flow is zero at shut off head
Maximum efficiency will be at design rated flow of the pump
Head decreases with increase in flow
Power increases with throttling
Answer: D) Power increases with throttling
Confirmed vs Book-3 §6.3/§6.6 — For a centrifugal pump flow is zero at shut-off, head falls as flow rises, and efficiency peaks at the design (BEP) flow — a, b, c are all true. On throttling the duty point moves up the curve to lower flow and the book notes 'there is some reduction in pump power absorbed at the lower flow rate' — power decreases, so d is the false statement.
3. A medium voltage end consumer receives 83 million units with a transmission and distribution cascade efficiency of 82%. The million units generated will be ____________.
101.2
68.1
83
None of the above
Answer: A) 101.2
Confirmed vs Book-3 §1.1 Cascade Efficiency — units generated = units received ÷ cascade efficiency = 83/0.82 = 101.2 MU.
Option (b) 68.1 MU multiplies by 0.82 instead of dividing; generation must always exceed the units delivered to the consumer.
4. A 1000 kW Gas engine is designed for 38 % efficiency. The operating load of the engine is 825 kW. If the GCV of gas is 8700 kcal/m3, the hourly gas consumption will be ____________ m3/hr.
214.6
260.13
188.89
272.74
Answer: A) 214.6
Confirmed vs Book-3 §9.1/§9.4 — Heat input = output ÷ efficiency = 825/0.38 = 2171 kW = 2171 × 860 = 18,67,105 kcal/hr; gas = 18,67,105/8700 = 214.6 m³/hr. Option (b) 260.1 is the tempting wrong value obtained by using the rated 1000 kW instead of the operating 825 kW load.
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
5. In an electrical power system, transmission efficiency increases as _______________.
both voltage and power factor increases
both voltage and power factor decreases
voltage increases but power factor decreases
Voltage decreases but power factor increases
Answer: A) both voltage and power factor increases
Confirmed vs Book-3 §1.1 — I = P/(√3·V·cosφ), and loss = I²R; raising either the voltage or the power factor lowers the current and hence the I²R loss, so transmission efficiency improves when both increase.
Option (d) is wrong because a lower voltage raises the current, wiping out any gain from a better power factor.
6. Which of the following is expressed in terms of percentage?
Absolute humidity
Relative humidity
Specific Gravity
All of the above
Answer: B) Relative humidity
Confirmed vs Book-3 §4.2 - Relative humidity is the ratio (expressed as %) of actual vapour pressure to saturation vapour pressure; absolute/specific humidity are mass ratios. Absolute humidity (a) is in g/kg dry air and specific gravity (c) is a dimensionless ratio to water; only relative humidity is quoted as a percentage on the psychrometric chart.
7. Which of the following is not true 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 with actual colors.
LED lamps are having comparatively higher CRI than Incandescent Lamps.
The higher the color rendering index, the less color shift or distortion occurs
Answer: C) LED lamps are having comparatively higher CRI than Incandescent Lamps.
Confirmed vs Book-3 §8.2/Table 8.1 — CRI is a 0–100 relative scale, indicates how perceived colours match actual colours, and a higher CRI means less colour shift (a, b, d true). Incandescent CRI = 100 vs LED 80, so 'LEDs have higher CRI than incandescent' is the untrue statement.
8. Flow control with ___________ in a fan system will not change the fan characteristic curve.
Inlet guide vane
speed change with variable frequency drive
speed change with hydraulic coupling
discharge damper
Answer: D) discharge damper
Confirmed vs Book-3 §5.5 — A discharge damper only adds system resistance; the fan stays on the same fan curve and moves along it ('forces the fan to move up or down along its characteristic curve … without changing fan speed'). Inlet guide vanes change the fan-curve characteristics, and speed change (VFD or hydraulic coupling) puts the fan on a different fan curve (a, b, c).
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
9. The primary purpose of inter-cooling in a multistage compressor is to _____________.
remove the moisture in the air
reduce the work of compression
separate moisture and oil vapour
none of the above
Answer: B) reduce the work of compression
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — Inter-stage coolers 'reduce the temperature of the air before it enters the next stage to reduce the work of compression and increase efficiency' by cutting the specific volume the next stage handles.
Removing moisture (a) is the stated objective of the AFTER-cooler, and moisture/oil separation (c) is an incidental benefit rather than the primary purpose of inter-cooling.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
10. A pump discharge has to be reduced from 120 m3/hr to 110 m3/hr by trimming the impeller. What should be the percentage reduction in impeller size?
10.52 %
8.34%
9.71%
17.1%
Answer: B) 8.34%
Confirmed vs Book-3 §6.5 — Q∝D: D₂/D₁ = 110/120 = 0.9167, so reduction = 8.33% ≈ 8.34%. 10.52% and 9.71% do not follow from the ratio; 17.1% would be the corresponding head reduction (1 − 0.9167²) — the tempting wrong option.
11. If 30240 kcal of heat is removed from a room every hour then the refrigeration tonnage will be nearly equal to ____________.
30.24 TR
3.024 TR
1 TR
10 TR
Answer: D) 10 TR
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/hr; 30240/3024 = 10 TR. Options (a) and (b) omit the division by 3024; 30,240/3024 = exactly 10 TR, which is why the examiner chose that number.
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
12. The basic function of an air dryer in an air compressor is to
Prevent dust from entering the compressor
Remove moisture before the intercooler
Remove moisture in compressor suction
Remove moisture in air supplied to the plants
Answer: D) Remove moisture in air supplied to the plants
Confirmed vs Book-3 §3.4 Compressed Air System Components (Figure 3.6) — The dryer's function is to remove the remaining water vapour so that the air supplied to plant instruments and pneumatic equipment is relatively free of moisture — water otherwise erodes valves, corrodes pipework and spoils paint spraying.
Dust exclusion (a) is the intake filter's job, and no moisture removal happens at the suction or before the intercooler (b, c).
📖 §8.3(1) Incandescent lamp (resistive filament); §8.2 Control gear
13. Power factor is highest in the case of _________________.
Sodium vapour lamps
Induction lamps
LED Lamps
Incandescent lamps
Answer: D) Incandescent lamps
Confirmed vs Book-3 §8.3 — The incandescent filament is a purely resistive load, giving power factor ≈ 1. Sodium vapour and induction lamps need ballasts/ignitors and LEDs need electronic drivers, all of which draw reactive/distorted current and lower the power factor.
📖 §10.14 Star rating of buildings — Energy Performance Index (EPI)
14. Energy performance index (EPI) kWh/m2/yr is the ratio of total building annual energy consumption to __________.
Built up area
Carpet area
Roof Area
Window and Wall Area
Answer: A) Built up area
Confirmed vs Book-3 §10.14 — EPI (kWh/m²/yr) = total building annual energy consumption ÷ built-up area. Built-up area = carpet area + wall thickness + balconies (parking basements excluded); carpet area (b) is the classic wrong pick.
15. In a pumping system, if the temperature of the liquid handled increases, then ___________________.
NPSHa increases
NPSHa decreases
NPSHa remains constant
NPSHa and NPSHr are independent of temperature
Answer: B) NPSHa decreases
Confirmed vs Book-3 §6.5 — NPSHA is the margin of eye pressure above the liquid's vapour pressure. Raising the liquid temperature raises its vapour pressure, so the margin — NPSHA — decreases and cavitation risk rises. NPSHA therefore is not independent of temperature; NPSHR (pump property) is unchanged.
📖 §7.2 Factors Affecting Performance – Fill Media Effects
16. Which of the following component has maximum effect on cooling tower performance?
Fill media
drift
louvers
casing
Answer: A) Fill media
Confirmed vs Book-3 §7.2 Fill Media Effects — Book: fill media provides the surface area, contact time and turbulence that govern heat exchange, and efficient fill reduces both fan power and pumping head → (a). Drift eliminators, louvers and casing are containment components. Same as Book EOC Q8.
17. In a vapour compression refrigeration system, the quantum of energy transferred at condenser is more than the energy transferred at ___________.
Compressor
Expansion Valve
Evaporator
All of the above
Answer: C) Evaporator
Confirmed vs Book-3 §4.3 - Condenser heat = evaporator heat + compressor work, so condenser rejects more energy than the evaporator absorbs. The book states the condenser must reject the combined evaporator heat plus the compressor work, so it handles more than the evaporator (c); the expansion valve (b) transfers no heat at all.
18. Which one of the following is true to estimate the range of cooling tower?
Range = Cooling water inlet temperature - Wet bulb temperature
Range = Cooling water outlet temperature - Wet bulb temperature
Range = Wet Bulb Temperature - Cooling Water Inlet Temperature
None of the above
Answer: D) None of the above
Confirmed vs Book-3 §7.2 (i) — Range = cooling water inlet (hot) temperature − outlet (cold) temperature; WBT does not appear. (b) outlet − WBT is the Approach; (a) inlet − WBT is the ideal range; (c) is negative. So none of the listed expressions gives the range → (d).
19. Modest flow variation between 80% to 100%, in a centrifugal fan is achieved more efficiently with ___________________.
Inlet damper
Outlet damper
Inlet guide vanes
Impeller Change
Answer: C) Inlet guide vanes
Confirmed vs Book-3 §5.5 — 'Guide vanes are energy efficient for modest flow reductions — from 100 percent flow to about 80 percent. Below 80 percent flow, energy efficiency drops sharply.' Dampers (a, b) are 'not particularly energy efficient'; impeller change (d) is a permanent derating, not a control method for variation.
20. ____________ is used as refrigerant both in vapour compression and vapour absorption systems.
Lithium Bromide
Water
HFC 134A
Ammonia
Answer: D) Ammonia
Confirmed vs Book-3 §4.3 - Ammonia (R-717) is used as a refrigerant in vapour compression systems and as the refrigerant in ammonia-water vapour absorption systems. Lithium bromide (a) is the absorbent, water (b) is the refrigerant only in LiBr machines, and HFC-134a (c) is compression-only; ammonia is the one fluid used as refrigerant in both systems.
21. In electrical distribution system, commercial loss covers discrepancies due to ________________.
Meter Reading
Metering
Collection Efficiency
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §1.8 Commercial Losses — the book attributes commercial losses to discrepancies in 'Meter Reading', 'Metering' and 'Collection efficiency'.
Each option names one of the three listed heads, so only 'all of the above' covers the book's definition.
Confirmed vs Book-3 §2.9 Soft Starter — A soft starter only ramps the voltage during starting and stopping; once at full voltage it gives no speed control. VFDs, hydraulic (fluid) couplings and eddy-current drives are all listed by the book as speed-control methods, so (b) is the odd one out.
23. When compared to standard motors, energy efficient motors will have ____________.
Higher slip
Higher starting torque
Lower No load current
All the above
Answer: C) Lower No load current
Confirmed vs Book-3 §2.6 Energy Efficient Motors — Energy-efficient designs lower the operating flux density and shorten the air gap to cut the magnetizing component of current, so the no-load current is lower than in a standard motor. (b) is the classic trap — the book warns that 'starting torque for efficient motors may be LOWER than for standard motors'; (a) is also wrong because EEMs have lower slip.
24. For a given air requirement, providing higher volume air receiver will ________________.
Increase energy consumption
Reduce energy consumption
Reduce Unload Power
Reduce Pressure fluctuations
Answer: D) Reduce Pressure fluctuations
Confirmed vs Book-3 §3.5 Air Receivers (IS 7938-1976 sizing) — The receiver's book-stated role is to dampen pulsations and even out the pressure variations from the compressor, buffering sudden short-duration demands; IS 7938-1976 sizes it at 1/10 to 1/6 of output in m³/min.
A bigger receiver by itself does not change compressor energy consumption (a, b) or unload power (c) — those follow from the pressure settings and the capacity control scheme.
25. Harmonics generation will be more in _________________.
Inverter drives
LED Lamps
Transformers
Resistance heaters
Answer: A) Inverter drives
Confirmed vs Book-3 §1.10 Causes of Harmonics — 'Common non-linear loads include variable speed drives (AC as well as DC), induction furnaces, LED based and CFL lamps, certain types of UPS & computer power supplies.' Inverter drives are the largest such source in industry.
Transformers and resistance heaters are linear (constant impedance) and generate negligible harmonics; LED lamps do generate harmonics but at a far smaller magnitude than a drive.
26. Thermal Power Plant efficiency is low due to ____________________.
Higher steam Pressure
Higher superheat temperature
Low GCV coal
Higher Heat loss in condenser
Answer: D) Higher Heat loss in condenser
Confirmed vs Book-3 §9.1 — A conventional steam plant reaches only 33–36% (Table 9.1) because the largest share of fuel energy is rejected as latent heat to cooling water in the condenser (Rankine-cycle limit). Higher steam pressure (a) and higher superheat (b) actually raise cycle efficiency; low-GCV coal changes fuel quantity, not cycle efficiency.
27. Among the following, _____________ has highest design efficiency.
High tension motors
Power transformers
Alternators
Electric melting furnaces
Answer: B) Power transformers
Confirmed vs Book-3 §1.5 — 'The efficiency [of transformers] varies anywhere between 96 to 99 percent' — the highest of any electrical machine, because a transformer is static (no friction or windage loss).
Options (a)/(c): rotating machines lose energy to friction, windage and slip, so their design efficiency is lower; furnaces are far lower still.
28. The difference between wet bulb temperature and cooling water inlet temperature in a cooling tower is called _______________.
Approach
Range
Effectiveness
None of the above
Answer: D) None of the above
Corrected (was a) — Book-3 §7.2: Approach is defined as the difference between the cooling tower OUTLET (cold) water temperature and the ambient WBT, and Range is inlet − outlet water temperature. The difference between the INLET (hot) water temperature and the WBT is what the book calls the 'ideal range' (denominator of effectiveness = Range + Approach) – it is neither approach, range nor effectiveness → (d). Tempting option (a) is right only if 'inlet' is misread as the cold water leaving the tower. ⚠ Answer changed a→d on book definitions (inlet − WBT = ideal range, not approach). If the examiner's key treats 'cooling water inlet' as the cold water entering the plant, (a) would be intended – learn both readings.
📖 §1.8 Estimation of Technical Losses in Distribution System
29. Technical loss in a distribution system can be reduced by _________________.
Maintaining low HT/LT ratio
Accurate meter reading
High voltage supply to consumers
Improving Collection Efficiency
Answer: C) High voltage supply to consumers
Confirmed vs Book-3 §1.8 Measures to reduce technical losses — the book recommends the High Voltage Distribution System (HVDS), replacing long LT lines with 11 kV lines and pole-mounted transformers at the load centres, because loss = I²R and higher voltage means lower current.
Options (b)/(d) — accurate meter reading and collection efficiency — reduce COMMERCIAL losses, not technical losses.
📖 Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11)
30. Pressure drop can be reduced in a compressed air distribution line by providing ______________.
After Coolers
Small diameter distribution pipes
High pressure air flow
Large Diameter Distribution pipes.
Answer: D) Large Diameter Distribution pipes.
Confirmed vs Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11) — Table 3.11 (170 m³/h): pressure drop per 100 m falls from 1.80 bar at 40 mm bore to 0.02 bar at 100 mm, with equivalent power loss dropping from 9.5 kW to 0.1 kW. Larger diameter distribution pipes are therefore the remedy.
Small-bore pipes (b) and high velocity (c) increase the drop, and an after-cooler (a) has nothing to do with distribution pressure loss.
31. Power consumption is very high for ___________ type of compressed air dryers.
Refrigeration type
Blower reactivated type
Heat of compression type
Heatless purge type
Answer: D) Heatless purge type
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Table 3.19 gives 20.7 kW per 1000 m³/hr for the heatless purge type — the highest of all four — plus a 12-15% loss of dry compressed air used for purging.
Blower reactivated is 18.0 kW, refrigeration 2.9 kW and heat of compression only 0.8 kW, so the heatless purge dryer is the very high consumer.
📖 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives
32. A DC excitation is used to vary the speed of _____________.
Eddy Current Coupling
fluid coupling
variable frequency drive
None of the above
Answer: A) Eddy Current Coupling
Confirmed vs Book-3 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives — In an eddy-current drive the freely revolving secondary member is separately excited by a DC field winding, and varying that DC excitation varies the output speed; its drawback is poor efficiency at low speeds. (b) a fluid coupling is purely hydrodynamic — speed is varied by the oil fill, not by excitation.
33. The isothermal power of 500 CFM air compressor is 72 kW and the efficiency is 76 %. The actual power drawn by the compressor will be ________.
56 kW
94.7 kW
89 kW
72 kW
Answer: B) 94.7 kW
Confirmed vs Book-3 §3.3 Compressor Efficiency — Isothermal efficiency = isothermal power / actual input power, so actual power = 72 / 0.76 = 94.7 kW.
Option (a) 56 kW results from multiplying rather than dividing by the efficiency; the actual input is always larger than the isothermal power because the latter excludes friction.
📖 Book-3 Ch-2 Electric Motors — speed control / Ch-1 power factor (cross-chapter item filed under Ch-3)
34. Power factor improvement of a 75-kW compressor motor will ___________.
Reduce input power to the motor
Increase input power to the motor
Reduce the compressor motor shaft power
None of the above
Answer: A) Reduce input power to the motor
Confirmed vs Book-3 Ch-2 Electric Motors — Improving power factor cuts the reactive (magnetising) current drawn through the supply system, so line current and the associated I²R distribution losses fall and the power drawn from the network reduces.
The motor's shaft power is fixed by the compressor load and is unchanged, so (c) is wrong. Note this is a power-factor item filed within the compressed-air set.
35. A 500-kVA transformer is designed for No load loss of 750 watts and load loss of 5700 Watts. The calculated total transformer loss is 1662 watts. What will be the percentage loading of the transformer?
54.8 %
29 %
40 %
25.7 %
Answer: C) 40 %
Confirmed vs Book-3 §1.5 — total loss = no-load + (%load)² × load loss ⇒ 1662 = 750 + x² × 5700 ⇒ x² = 912/5700 = 0.16 ⇒ x = 0.40, i.e. 40% loading.
Option (b) 29% comes from taking 912/5700 = 16% without the square root; the square root step is essential because copper loss varies as the square of the load.
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
36. Rating of PF correction capacitors for Induction Motors terminal should be
100 % kVAr of the induction motor
20 % of Motor Rating
25 % of Motor rating
90 % of the no-load kVAr induction motor
Answer: D) 90 % of the no-load kVAr induction motor
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — The book fixes the size of a terminal-connected capacitor from the motor's no-load reactive kVAr, selecting it 'to not exceed 90 % of the no-load kVAR of the motor' because higher capacitors could cause over-voltages and burn-outs. (a) 100 % is exactly the over-correction the book warns against; percentages of motor kW rating (b)/(c) are not the book's basis.
Confirmed vs Book-3 §8.5 — 'LLF = Light Loss Factor. This factor takes account of the depreciation over time of lamp output and dirt accumulation on the fitting and walls'; LLF = lamp-lumen MF × luminaire MF × room-surface MF (typical 0.8 AC office, 0.7 clean industrial, 0.6 dirty industrial).
📖 Book-3 §3.5 Avoiding Misuse of Compressed Air — air amplifiers (Figure 3.7); Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14)
1. Two 270 cfm compressors operate at 7 kg/cm2(g), on-load 80% of the time; load power 40 kW and unload power 15 kW each. Cleaning air requirement is 60% of air generated. Calculate the daily energy consumption for cleaning air alone (continuous operation).
Model answer: Air delivered by 2 compressors = 270 x 0.80 x 2 = 432 cfm. Loading power = 40+40 = 80 kW; unloading power = 15+15 = 30 kW. Average kW = [80x(0.8x24) + 30x(0.2x24)]/24 = 70 kW. SEC = 70/432 = 0.162 kW/cfm. Cleaning air = 0.60 x 432 = 259 cfm. Energy for cleaning air/day = 259 x 0.162 x 24 = 1007 kWh/day (alternate method gives ~1008 kWh/day).
Confirmed vs Book-3 §3.5 Avoiding Misuse of Compressed Air — air amplifiers — Printed worked solution: average compressor power 70 kW, SEC 0.162 kW/cfm, ~1007-1008 kWh/day for cleaning air.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
2. A centrifugal pump pumps 80 m3/hr of water into a container at 3 kg/cm2(g). Discharge head 5 kg/cm2(g); water level 5 m below pump centre line. Motor draws 22 kW, motor efficiency 90%, water density 1000 kg/m3. Find pump efficiency.
Model answer: Discharge head = 5 kg/cm2 = 50 m; suction head = -5 m. Power input to pump = 22 x 0.9 = 19.8 kW. Liquid kW = (80/3600) x (50-(-5)) x 9.81 = 11.98 kW. Pump efficiency = 11.98/19.8 = 60.56%.
3. A 10 TR AHU operates at 8.25 TR. Inlet enthalpy 10.26 kcal/kg, outlet enthalpy 7.26 kcal/kg, specific volume of air 0.83 m3/kg. Calculate the volume of air (m3/hr) handled by the AHU.
Model answer: Cooling (TR) = (Hi-Ho) x V/(v x 3024). Volume of air = TR x v x 3024/(Hi-Ho) = (8.25 x 0.83 x 3024)/(10.26-7.26) = 6903 m3/hr.
Confirmed vs Book-3 §4.7 - air-side load TR = Q x rho x (h_in - h_out)/3024, and with specific volume instead of density Q = TR x v x 3024/(h_in - h_out). Here Q = 8.25 x 0.83 x 3024/(10.26 - 7.26) = 6903 m3/hr. Note the AHU is rated 10 TR but is delivering only 8.25 TR - use the delivered load, not the nameplate, or the airflow comes out ~16% too high.
4. A fan is designed for 1300 m3/hr, 50 Hz, drawing 3 kW. Operated with VFD at 37 Hz for 6000 hours. Calculate the air velocity in a 150 mm diameter duct and the annual energy savings.
Model answer: Flow at 37 Hz = 1300 x (37/50) = 962 m3/hr. Duct area = 0.0177 m2. Velocity = (962/3600)/0.0177 = 15.09 m/s. Power at 37 Hz = (37/50)^3 x 3 = 1.22 kW. Annual savings = 6000 x (3-1.22) = 10,680 kWh.
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
5. A foundry draws 2500 kW. Demand during furnace operation: 5 min 2940 kVA, 7 min 2550 kVA, 3 min 2777 kVA. Billing meter monitors demand every 15 minutes. Calculate the maximum demand registered and the average PF during the interval.
Model answer: Maximum demand is the time-integrated kVA over the 15-minute cycle (Book-3 Sec.1.2):
MD = [(2940 x 5) + (2550 x 7) + (2777 x 3)] / 15 = [14,700 + 17,850 + 8331] / 15 = 40,881/15 = 2725.4 kVA.
Average power factor over the interval = kW / kVA = 2500 / 2725.4 = 0.917 (about 0.92 lag).
Note the demand billed is the 2725 kVA average, not the 2940 kVA instantaneous peak.
📖 §7.6 Case Study – VFD for CT fan (cube law); §7.4 Flow Control Strategies (two-speed fans)
6. A 4-cell cooling tower has 45 kW CT fans per cell operating at 40 kW at 1450 rpm. Fans are replaced with two-speed motors at 1450 rpm and 740 rpm. High-speed mode 5300 hours, low-speed mode 1800 hours/year. Estimate annual energy savings vs continuous fixed-speed 1450 rpm operation.
Model answer: Present (all at 1450 rpm) = 4 x 40 x (5300+1800) = 11,36,000 kWh. After two-speed: high speed = 4 x 40 x 5300 = 8,48,000 kWh; low speed = (740/1450)^3 x 40 x 4 x 1800 = 38,281 kWh. Annual savings = 11,36,000 - (8,48,000 + 38,281) = 2,49,719 kWh.
Printed solution using fan affinity law (P proportional to N^3): savings 2,49,719 kWh/year.
1. Write short notes on any two of the following: (1) Integrated Part Load Value (IPLV) for chillers, (2) Evaporative Cooling, (3) Heat Pump. (Each 2.5 Marks)
Model answer: (1) IPLV: A single-number part-load efficiency metric for chillers (AHRI weighting) that combines chiller performance at 100%, 75%, 50% and 25% load, since chillers rarely run at full load; it better represents seasonal/real operation than full-load kW/TR. (2) Evaporative Cooling: Cooling of air by direct contact with water so that sensible heat of air evaporates water; dry-bulb temperature falls while moisture content rises, approaching the wet-bulb temperature - low energy alternative to mechanical refrigeration in dry climates. (3) Heat Pump: A reversed refrigeration cycle device that extracts low-grade heat from a source (air/water/ground) and upgrades it to deliver useful heating; COP_heating = COP_cooling + 1, giving high effective efficiency.
Confirmed vs Book-3 §4.7 (IPLV, p.126), §4.3 Evaporative Cooling (p.116) and §4.11 Heat Pumps (p.133). IPLV averages kW/TR at 100/75/50/25% load because full-load kW/TR occurs for only ~1% of running hours. Evaporative cooling brings air into close contact with water so it approaches the wet bulb temperature - cheap but it adds moisture. A heat pump is an air conditioner whose rejected heat is the useful output: 3 units from the surroundings plus 1 unit of compressor work give 4 units of heat.
2. Write short notes on any two of the following: (1) Solar Heat Gain Coefficient (SHGC), (2) Visible Light Transmittance (VLT), (3) Cool Roof. (Each 2.5 Marks)
Model answer: (1) SHGC: Fraction of incident solar radiation admitted through a window (both directly transmitted and absorbed-then-reradiated inward); ranges 0-1, lower SHGC means less solar heat gain - ECBC specifies maximum SHGC for fenestration. (2) VLT: Visible Light Transmittance - the fraction of visible light (380-780 nm) transmitted through glazing; higher VLT gives more daylight. ECBC encourages high VLT with low SHGC for daylighting without heat gain. (3) Cool Roof: A roof surface with high solar reflectance and high thermal emittance that stays cooler under the sun, reducing roof heat gain, cooling load and the urban heat island effect.
Book-3 §10.5: SHGC = solar heat gain through fenestration ÷ incident solar radiation (transmitted + absorbed/re-radiated); VLT = visible light ratio vs perfectly transmissive glazing (daylighting); cool roof = high solar reflectance (≥ 0.7) and high emittance (≥ 0.75) to reflect and reject heat.
3. L-1(a): A fan delivers 24,000 Nm3/hr of air. Suction static pressure -15 mmWC, discharge static pressure 35 mmWC. Motor power 7 kW, motor efficiency 90%. What is the static efficiency of the fan? (b) Match the following: Heat Pump, Compressor, Pumping Pressure, Fan, Pump with NPSHR, Static Head, Static Pressure, Compressor, Free air delivery test.
Model answer: (a) Q = 24,000 Nm3/hr = 6.67 m3/s. Static pressure rise = 35-(-15) = 50 mmWC. Power input to fan shaft = 7 x 0.90 = 6.3 kW. Fan static efficiency = (Q x Pst)/(102 x shaft power) = (6.67 x 50)/(102 x 6.3) = 0.519 = 51.9%. (b) Matches: Heat Pump - Compressor; Compressor - Free air delivery test; Pumping Pressure - Static Head; Fan - Static Pressure; Pump - NPSHR.
📖 §10.4 Compliance approaches (Prescriptive vs Whole Building Performance) · §10.5 envelope terms
4. L-2(A): Classify each as Prescriptive Method or Whole Building Performance Method: (1) Compliance by meeting/exceeding specific levels for each individual element; (2) Allows trade-off option for building envelope; (3) Allows use of energy simulation software; (4) Computer model of proposed design compared with Standard Design; (5) Compliance if proposed design energy use is less than standard design. (B) Match building-envelope terms.
Model answer: A. (1) Prescriptive Method; (2) Prescriptive Method; (3) Whole Building Performance Method; (4) Whole Building Performance Method; (5) Whole Building Performance Method. B. (1) Building envelope - (c) Roof, walls, windows, skylights, doors and other openings; (2) Passive solar design strategy - (e) Cross ventilation; (3) Visual Light Transmittance - (a) Day lighting of building; (4) Weather stripping - (b) Exfiltration and Infiltration of air; (5) Cool roof - (d) Property of high solar reflectance and emittance.
Book-3 §10.4: Prescriptive = meet/exceed specified levels for each element and includes the envelope Trade-Off option; WBP = simulation software, computer model of Proposed Design vs Standard Design, complies if proposed energy use is not greater. Part B terms from §10.5 (envelope, cross-ventilation, VLT/daylighting, weather stripping, cool roof).
5. L-3: Replace 30 motors with energy efficient motors. Data: 7.5 kW, 75% load, old eff 86%, new eff 89%, 12 nos; 11.5 kW, 85% load, old 88%, new 91%, 7 nos; 15 kW, 70% load, old 89%, new 92%, 11 nos. Motor loading same in both cases; calculate annual energy savings for 4000 hours/year.
Model answer: Savings per motor = rated kW × % loading × (1/ηold − 1/ηnew), the book's motor-replacement relation.
7.5 kW, 75 % load, 12 nos: old input = 7.5 × 0.75 / 0.86 = 6.541 kW; new = 7.5 × 0.75 / 0.89 = 6.320 kW; saving = 0.221 kW × 4,000 h × 12 = 10,584 kWh.
11.5 kW, 85 % load, 7 nos: old = 11.5 × 0.85 / 0.88 = 11.108 kW; new = 11.5 × 0.85 / 0.91 = 10.742 kW; saving = 0.366 kW × 4,000 h × 7 = 10,254 kWh.
15 kW, 70 % load, 11 nos: old = 15 × 0.70 / 0.89 = 11.798 kW; new = 15 × 0.70 / 0.92 = 11.413 kW; saving = 0.385 kW × 4,000 h × 11 = 16,931 kWh.
Total annual energy savings ≈ 37,769 kWh (≈ 37,800 kWh/year).
Printed solution: total 37,640 kWh/year (10,560 + 10,360 + 16,720).
6. L-4: A 10 MW co-gen plant runs at 85% daily load factor, generating at 11 kV. 35% exported to grid via 7.5 MVA transformer (99% eff); 32% to mill motors at 600 V via 5 MVA transformer (98% eff); balance to LT loads/auxiliaries at 415 V via 2 MVA transformer (98% eff). Calculate: (1) daily energy exported; (2) daily mill motor consumption; (3) daily LT/auxiliary consumption; (4) daily transformer losses (kWh and %).
Model answer: Daily generation = 10,000 x 0.85 x 24 = 2,04,000 kWh. (1) Export = 2,04,000 x 0.35 = 71,400 kWh; 7.5 MVA loss = 71,400 x 0.01 = 714 kWh; net export = 70,686 kWh. (2) Mill = 2,04,000 x 0.32 = 65,280 kWh; 5 MVA loss = 65,280 x 0.02 = 1,306 kWh; net = 63,974 kWh. (3) LT/aux = 2,04,000 x 0.33 = 67,320 kWh; 2 MVA loss = 67,320 x 0.02 = 1,346 kWh; net = 65,974 kWh. (4) Total transformer losses = 714 + 1,306 + 1,346 = 3,366 kWh/day; % loss = 3,366/2,04,000 x 100 = 1.65%.
Printed solution: net export 70,686 kWh, net mill 63,974 kWh, net LT 65,974 kWh, total tx loss 3,366 kWh = 1.65%.
📖 Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14); Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System
7. L-5: A 220 cfm screw compressor: Shift I load 60 s/unload 10 s; Shift II load 45/unload 25; Shift III load 25/unload 45 (8 hrs/shift). Load power 37 kW, unload power 11 kW. Calculate: (1) energy loss per day; (2) shift-wise average air requirement in cfm; (3) energy savings after installing VFD with VFD loss 3% of load power.
Model answer: Shift I = ((60/70)x37 + (10/70)x11) x 8 = (31.71+1.57)x8 = 266.24 kWh. Shift II = (0.64x37 + 0.36x11)x8 = 221.12 kWh. Shift III = (0.36x37 + 0.64x11)x8 = 162.88 kWh. Daily total = 650.24 kWh. Daily unloading (loss) energy = (1.57+3.96+7.04)x8 = 100.56 kWh. Load-cycle energy = 549.68 kWh. With VFD = 549.68/0.97 = 566.68 kWh; VFD loss = 17 kWh; net VFD savings = 100.56 - 17 = 83.56 kWh/day. Air requirement: Shift I = 0.86x220 = 189.2 cfm; Shift II = 0.64x220 = 140.8 cfm; Shift III = 0.36x220 = 79.2 cfm.
Confirmed vs Book-3 §3.5 Capacity Control of Compressors — Printed solution: daily energy loss (unloading) 100.56 kWh; net VFD savings 83.56 kWh/day; shift air 189.2/140.8/79.2 cfm.
📖 §7.2 Cooling Tower Performance (viii) L/G ratio; §7.2 Fill Media Effects (enthalpy values from Approach & WBT example)
8. L-6: (a) What is L/G ratio and how is it useful in cooling tower operation? (b) Functions of fill media. (c) Calculate L/G ratio: water flow 4540 m3/hr, approach 4.45 C, air entering enthalpy 24.17 kcal/kg at 26.67 C, air leaving enthalpy 39.67 kcal/kg at 37.8 C, hot water 47.77 C, cold water 31.11 C.
Model answer: (a) L/G ratio = ratio of water (liquid) to air (gas) mass flow rates. By energy balance L(T1-T2) = G(h2-h1), so L/G = (h2-h1)/(T1-T2). Against design, seasonal variation requires tuning of water/air flow (water box loading, blade angle) for best effectiveness. (b) Fill media increases the air-water contact surface and contact time, promoting heat and mass transfer (evaporative cooling). (c) L/G = (39.67-24.17)/(47.77-31.11) = 15.5/16.66 = 0.93.
Printed solution: L/G = 0.93; fill media function referenced to Page 209.