Energy Efficiency in Electrical Utilities Available here with full solutions — 58 questions recovered from the 2010 exam:
Objective (1 mark)
44 of 50
Short (5 marks)
8 of 8
Long (10 marks)
6 of 6
This is not the complete paper. The questions below are the ones we could recover and verify; the rest of that year’s paper is not reproduced here. Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 44
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
1. For the same quantity of power handled by a distribution line, the lower the voltage
the higher the current drawn and higher the distribution loss
the lower the current drawn and lower the distribution loss
the lower the voltage drop and lower the distribution loss
the higher the voltage drop and lower the distribution loss
Answer: A) the higher the current drawn and higher the distribution loss
Confirmed vs Book-3 §1.1 — 'For the same quantity of power handled, lower the voltage, higher the current drawn and higher the voltage drop', and since P.Loss = I²R the distribution loss rises too.
Option (b) reverses this; the whole rationale for HV transmission is that a lower current means a lower I²R loss.
2. The ratio of overall maximum demand of the plant to the sum of individual maximum demand of various equipment is ______.
load factor
diversity factor
demand factor
maximum demand
Answer: B) diversity factor
Confirmed vs Book-3 §1.5 Rating of Transformer — 'Diversity factor is defined as the ratio of overall maximum demand of the plant to the sum of individual maximum demand of various equipment…Diversity factor will always be less than one.'
Option (a) load factor is average load ÷ maximum demand, and (c) demand factor is maximum demand ÷ connected load — different ratios.
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
3. Maximum demand charges for a billing cycle is calculated by the Utility based on:
the instantaneous demand drawn
the time integrated demand over the predefined recording cycle
the kVARh drawn per demand cycle
the kWh drawn during peak load period
Answer: B) the time integrated demand over the predefined recording cycle
Confirmed vs Book-3 §1.2 — 'while maximum demand is recorded, it is not the instantaneous demand drawn, as is often misunderstood, but the time integrated demand over the predefined recording cycle' (typically 30 minutes).
Option (a) is exactly the misconception the book warns against; a momentary current spike does not by itself set the billed demand.
4. The total amount of harmonics present in the system is expressed using ___.
Total Harmonic Factor
Total Harmonic Ratio
Total Harmonic Distortion
Crest Factor
Answer: C) Total Harmonic Distortion
Confirmed vs Book-3 §1.10 — 'Total Harmonic Distortion (THD) expresses the amount of harmonics', computed as the root-sum-square of the harmonic components as a percentage of the fundamental.
Option (d) Crest Factor is the peak-to-RMS ratio of a waveform — related to distortion but not a measure of total harmonic content.
📖 §1.1 Industrial End User — 'ONE Unit saved = TWO Units Generated'
6. Efficiency of power generation in a power plant is 30%, T&D losses are 23%, distribution loss within the factory is 6%, and equipment end use efficiency is 65%. The overall cascade system efficiency from fuel input to end-use will be
2.69%
14.11%
4.21
none of the above
Answer: B) 14.11%
Confirmed vs Book-3 §1.1 — multiply the stage efficiencies: 0.30 × (1 − 0.23) × (1 − 0.06) × 0.65 = 0.30 × 0.77 × 0.94 × 0.65 = 0.1411 = 14.11%.
Option (a) 2.69% comes from multiplying by the loss fractions (0.23, 0.06) instead of by the efficiencies (0.77, 0.94).
7. A 3 phase, 7.5 kW, 415 V, 15 A, 1480 RPM rated induction motor with full load efficiency of 90% draws 5 A at rated voltage and 0.5 power factor. The percentage loading of the motor is about
21.56%
23.96%
33.33%
none of the above
Answer: A) 21.56%
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — Input = √3 × 415 × 5 × 0.5 = 1.797 kW; rated input = 7.5/0.90 = 8.333 kW; % loading = 1.797 / 8.333 = 21.56 %. (b) 23.96 % is the trap of dividing the input by the 7.5 kW OUTPUT rating instead of by the equivalent rated input.
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
8. An induction motor installed with static PF correction capacitors across the motor terminals got damaged along with capacitors once it was disconnected from the supply. The possible reason among the following was
charging current of the capacitor was only 80% of the motor magnetising current
motor PF was over corrected or critically corrected (unity power factor)
motor was oversized
motor was undersized
Answer: B) motor PF was over corrected or critically corrected (unity power factor)
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — Over-correction (capacitor kVAr at or above the motor's no-load kVAr) makes the disconnected but still-spinning motor self-excite; the resulting over-voltage destroys both motor and capacitors — the book therefore caps the capacitor at 90 % of no-load kVAr. (a) under-correction at 80 % is exactly the safe condition the book recommends, so it cannot be the cause.
9. A 4-pole squirrel cage induction motor operates with 1% slip at full load. What is the approximate full load RPM at a grid frequency of 49.5 Hz?
1485
1470
1500
none of the above
Answer: B) 1470
Confirmed vs Book-3 §2.3 Motor Characteristics — Ns = 120 × 49.5 / 4 = 1485 rpm; at 1 % slip the speed = 1485 × 0.99 = 1470 rpm. (a) 1485 rpm ignores the slip and (c) 1500 rpm ignores the reduced grid frequency — both are the standard traps in this question.
Answer: A) increases with increase in motor loading
Confirmed vs Book-3 §2.3 Motor Characteristics — As loading rises the active component of current grows while the magnetizing current stays essentially fixed, so power factor improves with load; Figure 2.2 shows power factor dropping sharply at part load. (d) states the reverse and is the misconception the book's section on under-loading is written to correct.
📖 Book-3 §3.5 Equivalent Lengths of Fittings (Table 3.12)
11. Which of the following pipe fittings used in compressed air pipe line offers maximum resistance
Gate Valve in open condition
Return bend
Elbow
Tee 90o long bend
Answer: B) Return bend
Confirmed vs Book-3 §3.5 Equivalent Lengths of Fittings (Table 3.12) — Table 3.12 equivalent lengths rise in the order Gate valve < Tee 90° long bend < Elbow < Return bend — at 50 mm NB, 0.40 / 0.61 / 1.07 / 1.68 m respectively.
The return bend therefore offers the maximum resistance of the four fittings listed; the elbow is second and an open gate valve the least.
12. An induction motor rated for 15 kW and 93% efficiency, operating at full load at the rated parameters, will
deliver 15 kW
deliver 16.12 kW
draw 15 kW
draw 13.95 kW
Answer: A) deliver 15 kW
Confirmed vs Book-3 §2.4 Motor Efficiency — Nameplate kW is the rated shaft output, so the motor delivers 15 kW and draws 15/0.93 = 16.12 kW. (b) 'deliver 16.12 kW' is the trap — 16.12 kW is the electrical INPUT, and no motor can deliver more than its rating at full load.
13. Modest flow variation, from 100% to 80%, in a centrifugal fan is achieved more efficiently with which of the following flow control methods
inlet damper
outlet damper
inlet guide vanes
none of the above
Answer: C) inlet guide vanes
Confirmed vs Book-3 §5.5 — Inlet guide vanes pre-swirl the inlet air and change the fan curve; they are 'energy efficient for modest flow reductions — from 100 percent flow to about 80 percent'. Inlet/outlet dampers (a, b) simply add resistance and are the least efficient control.
14. A 200 cfm compressor has a loading and unloading period of 10 seconds and 20 seconds respectively during a compressed air leakage test. The air leakage in the compressed air system would be
20.3 cfm
42.1 cfm
66.6 cfm
132.8 cfm
Answer: C) 66.6 cfm
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — % leakage = T/(T+t) x 100 = 10/(10+20) x 100 = 33.3%, so leakage = 0.333 x 200 = 66.6 cfm.
The wrong answers come from using 10/20 (giving 100 cfm) or from inverting the ratio; the denominator must be the complete load-plus-unload cycle.
15. Select the correct Statement: The advantage of PF improvement by capacitor addition in an electric network is
apparent power component of the network is reduced
active power component of the network is reduced
I2R power losses are reduced in the system from the point of installation to the load end
voltage level at the load end is not improved
Answer: A) apparent power component of the network is reduced
Confirmed vs Book-3 §1.4 — capacitors reduce the reactive component and hence the total current, so the APPARENT power (kVA) drawn from the source falls; the book's advantage (d) notes the resulting kVA/capacity relief.
Option (b) is false because kW is set by the work done; (c) is false because losses fall only UPSTREAM of the capacitor, and (d) is false because the load-end voltage does improve.
16. A 500 cfm reciprocating compressor earlier operating at load-unload pressure of 6.0 and 7.5 kg/cm2g was changed to 6.0 to 6.5 kg/cm2g for the same end use. This change will result in
increased unloading cycle time of the compressor
increased loading cycle time of the compressor
increased energy consumption of the compressor
decreased leakage loss in air distribution system
Answer: D) decreased leakage loss in air distribution system
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — Narrowing the band to 6.0-6.5 kg/cm²g drops the maximum system pressure from 7.5 to 6.5 kg/cm²g, and Table 3.16 shows leak flow through a given orifice rising steadily with gauge pressure — so leakage loss in the distribution system falls. Lower pressure also cuts power (6-10% per bar), so (c) is wrong.
The record's answerText had drifted to option (a) while the answer key said (d); the answerText is now aligned with (d). Option (a) has been repaired to 'increased' so exactly one option is correct — with a band one third as wide the receiver empties sooner, so the unload period actually shortens.
📖 §6.1 Pump types — centrifugal pump construction & working
18. With increase in the suction lift from open wells, the delivery flow rate
increases
decreases
remains same
none of the above
Answer: B) decreases
Confirmed vs Book-3 §6.1 — Book: 'The greater the depth of the water, the lesser is the flow from the pump. Also, when it pumps against increasing pressure, the less it will pump.' Higher suction lift raises total head, moving the duty point to lower flow on the H-Q curve (and reducing NPSHA).
20. The COP of a vapour compression refrigeration system is 3.5. If the motor delivers power of 10.8 kW at its shaft with a 90% motor efficiency, the cooling effect will be
34 kW
37.8 kW
0.36 kW
none of the above as cooling effect is always measured in TR
Answer: B) 37.8 kW
Confirmed vs Book-3 §4.7 - Cooling effect = COP × shaft power = 3.5 × 10.8 = 37.8 kW. Here the 10.8 kW is already the SHAFT output, so no efficiency correction applies and (a) 34 kW - the answer when 10.8 kW is a motor input - is the trap; cooling = 3.5 x 10.8 = 37.8 kW.
📖 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives
21. In a variable speed drive using hydraulic coupling
motor speed changes
driven equipment speed changes
both a & b
neither a nor b
Answer: B) driven equipment speed changes
Confirmed vs Book-3 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives — A hydraulic (fluid) coupling changes the speed of the driven equipment while the motor itself continues to run at its fixed supply-determined speed. (c) 'both' is the tempting answer, but only a VFD can change the MOTOR's speed, by changing supply frequency.
22. A fan with 30 cm pulley diameter is driven by a 1480 rpm motor through a v-belt. If the motor pulley is reduced from 20 cm to 18 cm at the same motor rpm and fan pulley diameter, the fan speed will reduce by
247 rpm
888 rpm
98 rpm
none of the above
Answer: C) 98 rpm
Confirmed vs Book-3 §5.5 — Fan rpm = motor rpm × motor pulley/fan pulley: 1480 × 20/30 = 986.7 rpm → 1480 × 18/30 = 888 rpm; reduction ≈ 98.7 ≈ 98 rpm (c). (b) 888 is the new speed, not the reduction; (a) 247 would need a much larger pulley change.
Pump raises an incompressible fluid to a higher level of pressure or head.
Compressor raises a compressible fluid to a higher level of pressure.
Blower moves gas volumes with moderate increase of pressure.
Pump raises relatively compressible fluid to a higher level of pressure or head.
Answer: D) Pump raises relatively compressible fluid to a higher level of pressure or head.
Confirmed vs Book-3 — A pump handles INCOMPRESSIBLE liquids (a is correct, d contradicts it); a compressor raises a compressible gas to a higher pressure (b); a blower moves gas volumes with a moderate pressure rise — specific ratio 1.11–1.20 (c). Hence the wrong statement is (d).
24. As per the fan laws, by reducing the fan RPM by 10%, the fan power requirement:
decreases by 27%
decreases by 19%
does not change
decreases by 73%
Answer: A) decreases by 27%
Confirmed vs Book-3 §5.3 — 'Reducing the RPM by 10% decreases the power requirement by 27%' (0.9³ = 0.729). (b) 19% is the static-pressure reduction; (d) 73% is the REMAINING power fraction, not the reduction.
25. A centrifugal fan operating at 800 RPM develops a flow of 3000 Nm3/hr at a static pressure of 600 mmWC. If the fan speed is reduced to 600 RPM, the static pressure will become:
450 mmWC
519.6 mmWC
337.5 mmWC
none of the above
Answer: C) 337.5 mmWC
Confirmed vs Book-3 §5.3 — SP ∝ N²: 600 × (600/800)² = 600 × 0.5625 = 337.5 mmWC (c). (a) 450 = 600 × 0.75 is the linear (flow-law) error; (b) 519.6 = 600 × √0.75 is a square-root error.
📖 §6.6 Flow control strategies — pump control by varying speed / VSDs & VFDs
26. There are four pumps working against the same friction heads. For which static head will the variable speed drive be most economical?
Pump A - 0 m
Pump B - 10 m
Pump C - 20 m
Pump D - 25 m
Answer: A) Pump A - 0 m
Confirmed vs Book-3 §6.6 — VSD savings are greatest in a friction-only system (zero static head): the duty point moves along an iso-efficiency line and the affinity laws (P∝N³) are fully obeyed (Fig 6.15). As static head rises the efficiency drop with speed 'reduces the economic benefits of variable speed control' (Fig 6.16). So Pump A (0 m) is the best VSD candidate.
📖 §7.2 Cooling Tower Performance (ii) Approach; §7.3 IV Performance Assessment
28. When you do a walk through energy audit of a cooling tower which salient parameter will you quickly spot check for its water cooling performance?
makeup water tap is on or off
hot water entry temperature to the cooling tower
cooling tower fan is on or off
cold well and ambient wet bulb temperature
Answer: D) cold well and ambient wet bulb temperature
Confirmed vs Book-3 §7.2 (ii) — Approach (cold-well water temperature − ambient WBT) is the book's best indicator of performance, so those two readings are the quickest spot check → (d). Hot water inlet (b) only reflects process load; fan/make-up status (a, c) says nothing about cooling performance.
29. Which of the following is correct statement in the case of cooling towers:
Range is the difference between the cooling tower water inlet and ambient wet bulb temperature.
Approach is the difference between the cooling tower outlet cold water temperature and hot inlet water temperature.
Range is the only indicator of cooling tower performance.
Cooling tower capacity is expressed as heat rejected in Ton of Refrigeration (TR)
Answer: D) Cooling tower capacity is expressed as heat rejected in Ton of Refrigeration (TR)
Confirmed vs Book-3 §7.2 (i)–(iv) — Book: Range = inlet − outlet water temp (so (a) is wrong); Approach = outlet cold water − WBT (so (b) is wrong); Approach, not Range, is the better indicator (so (c) is wrong); 'Cooling capacity is the heat rejected in kcal/hr or TR' → (d).
📖 Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7)
30. Which of the following is incorrect statement?
Inadequate cooling in after-coolers causes more condensation in air receivers and distribution lines
Performance of inter-coolers have no effect on work of compression
In a battery of air compressors, the compressor with lower part load power consumption should be modulated.
For the same capacity, a compressor consumes more power at higher delivery pressure
Answer: B) Performance of inter-coolers have no effect on work of compression
Confirmed vs Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7) — Inter-coolers 'reduce the work of compression (power requirements) by reducing the specific volume through cooling the air', and Table 3.7 shows a 5.5°C rise in second-stage inlet costs about 2% more specific energy — so (b) is plainly false.
Statements (a), (c) and (d) are all made in the book: poor after-cooling causes condensation and corrosion downstream, the compressor with lower part-load power should be modulated, and higher delivery pressure means more power for the same capacity.
31. Name the fan which is more suitable for high pressure application
propeller type fan
tube-axial fan
backward curved centrifugal fan
forward curved centrifugal fan
Answer: C) backward curved centrifugal fan
Confirmed vs Book-3 §5.2 Table 5.3 — Backward-curved centrifugal: 'High pressure, high flow, high efficiency' (FD fans etc.); forward-curved: medium pressure, 'best suited for moving large volumes of air against relatively low pressures'. Propeller and tube-axial (a, b) are low/medium-pressure axial fans; 'centrifugal fans are suitable for low to moderate flow at high pressures'.
32. The basic functions of an electronic ballast fitted to a fluorescent tube light exclude one of the following
to stabilize the gas discharge
to supply power to the lamp at supply frequency
to ignite the tube light
to supply power to the lamp at very high frequency
Answer: B) to supply power to the lamp at supply frequency
Confirmed vs Book-3 §8.6(f) — Basic functions of an electronic ballast: 'to ignite the lamp, to stabilize the gas discharge, and to supply the power to the lamp' — and it does so after converting the supply frequency to about 20,000–30,000 Hz. Supplying the lamp at supply (50 Hz) frequency is what a conventional electromagnetic choke does, so (b) is excluded.
📖 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD
33. Energy savings potential of variable torque applications compared to constant torque application is:
higher
lower
equal
none of the above
Answer: A) higher
Confirmed vs Book-3 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD — For variable torque loads (centrifugal fans and pumps) power varies as the cube of speed, so a small speed reduction gives a large power saving; the book calls this the largest potential for VSD savings. Constant torque loads are also suitable but their power falls only in proportion to speed, so their savings potential is lower, making (b)/(c) wrong.
📖 §9.2 Sizing — kVA = √3·V·I; §9.4 specific fuel consumption
34. In a DG set, a 3 phase alternator is loaded at 450 A, at 415 volts and 0.85 PF. If the specific fuel consumption is 0.25 lts/kWh, how much fuel is consumed delivering generated power for one hour.
39.68 litre
80.86 litre
68.74 litre
none of the above
Answer: C) 68.74 litre
Confirmed vs Book-3 §9.2/§9.4 — kW = √3 × 415 × 450 × 0.85 / 1000 = 274.9 kW; fuel = 274.9 × 0.25 = 68.7 litres/hr. Option (b) 80.86 L is the tempting error of using kVA (323.4) instead of kW; (a) 39.68 L omits √3.
35. Slip ring induction motors, in general, have a …… design efficiency in comparison with the squirrel cage induction motors for similar ratings
lower
higher
same
none of the above
Answer: A) lower
Confirmed vs Book-3 §2.4 Motor Efficiency — The book states plainly that squirrel cage motors are normally more efficient than slip-ring motors, so a slip-ring machine has a lower design efficiency for the same rating — the slip rings, brushes and rotor windings add loss. (b) 'higher' is the reversed reading of the same sentence.
📖 §1.4 Performance Assessment of Power Factor Capacitors
36. A power factor capacitor designed for 10 kVAr at 415 V was found to be operating at 405 V. The effective capacity of the capacitor would be
9.75 kVAr
10 kVAr
9.52 kVAr
none of the above
Answer: C) 9.52 kVAr
Confirmed vs Book-3 §1.4 Voltage effects — output kVAr ∝ V²: 10 × (405/415)² = 10 × 0.9524 = 9.52 kVAr.
Option (a) 9.75 kVAr assumes a linear voltage relation (405/415 × 10); the squared law is why even a small under-voltage noticeably de-rates a bank.
37. Which combination of readings as indicated by the panel mounted instruments of a DG set would give the indications of proper capacity utilisation of diesel engine and generator?
kW & Voltage
kVA & kVAr
kW & kVA
none of the above
Answer: C) kW & kVA
Confirmed vs Book-3 §9.3 — "Alongside alternator loading, the engine loading in terms of kW or BHP needs to be maintained above 50%"; transient limits apply "to both kW (as reflected on the engine) and kVA (as reflected on the generator)". Hence kW shows engine (prime mover) utilisation and kVA shows alternator utilisation. kVA & kVAr (b) both describe the alternator only; kW & voltage (a) says nothing about alternator loading.
38. Among the following electrical equipment _______ has the highest design efficiency
synchronous motor
DC shunt motor
induction motor
transformer
Answer: D) transformer
Confirmed vs Book-3 §1.5 — the transformer is a static machine with no friction or windage loss, and the book gives its efficiency as 96–99%, higher than any rotating machine.
Options (a)–(c) are all rotating machines whose efficiency is reduced by friction, windage and (for induction motors) slip losses.
📖 §7.1 Components of Cooling Tower – Fill; §7.2 Fill Media Effects
39. The surface area of heat exchange in a cooling tower is enhanced by
fill media
louvers
drift eliminator
cold water basin
Answer: A) fill media
Confirmed vs Book-3 §7.1/§7.2 — Fill maximises water–air contact; splash fill creates droplet surface and film fill provides sheet surface (30–45 vs 150 m²/m³, Table 7.3) → (a). Louvers equalise air flow, drift eliminators trap droplets, the basin only collects water.
📖 §9.4 Energy performance assessment — specific power generation kWh/litre (1 kWh = 860 kcal)
40. How many units per liter will be available from a DG set if the operating efficiency is 40%? The calorific value of diesel is 10,000 kCal per liter
3.50
6.98
cannot be worked out as DG set loading is not indicated
4.65
Answer: D) 4.65
Confirmed vs Book-3 §9.4 — Units per litre = (CV × efficiency) / 860 = (10,000 × 0.40)/860 = 4.65 kWh/litre, in line with the 3–4 units/litre implied by the 0.29–0.36 L/kWh SFCs of Table 9.7. Loading is not needed because efficiency is already given, so (c) is wrong; (b) 6.98 comes from dividing by 573 (kcal/kWh error); (a) 3.5 assumes ~30% efficiency.
41. The 5th and 7th harmonic in a 50 Hz power environment will have:
voltage and current distortions with 55 Hz & 57 Hz
voltage and current distortions with 500 Hz & 700 Hz
voltage and current distortions with 250 Hz & 350 Hz
no voltage and current distortion at all
Answer: C) voltage and current distortions with 250 Hz & 350 Hz
Confirmed vs Book-3 §1.10 — the harmonic order multiplies the fundamental: 5 × 50 = 250 Hz and 7 × 50 = 350 Hz.
Option (b) 500/700 Hz would be the 10th and 14th harmonics; option (a) adds instead of multiplying.
42. The source of maximum harmonics among the following in a plant power system could be:
100 CFL lamps of 11 W to 25 W
500 kW, 3 Phase, 415 V, 50 Hz resistance furnace
5 kVA UPS for computer system
Variable Frequency Drive for 225 kW motive load
Answer: D) Variable Frequency Drive for 225 kW motive load
Confirmed vs Book-3 §1.10 — variable frequency drives head the book's list of non-linear loads, and the harmonic current injected scales with the load size, so a 225 kW VFD dwarfs the other options.
Option (b) a resistance furnace is a LINEAR load (constant impedance) and generates no harmonics despite its 500 kW rating; the 5 kVA UPS is non-linear but tiny.
43. Which of the following are emerging technological solutions in electric power distribution system?
Intelligent meters for improved system operation and customer relationship management
Intelligent or smart meters to replace older systems to allow customers a clear picture of their energy use profile
SCADA system to control and data acquisition for complete T&D system
all the above
Answer: D) all the above
Confirmed vs Book-3 §1.8 Measures to Reduce Commercial Losses — the book calls for 'Installation of Electronic meters with (TOD, tamper proof, data and remote reading facility)', accurate metering plans and energy audit as a tool, all of which smart meters and SCADA deliver.
Each option describes one of these emerging solutions, so only 'all the above' is complete.
📖 §8.6(e) Reduction of lighting feeder voltage + Table 8.3; Book EOC Q6
44. Which of the following options reduces the electricity consumption in lighting system in a wide spread plant?
installing separate lighting transformer and maintaining optimum voltage
maintaining 260 V for the lighting circuit with 220 V rated lamps
replacing 150 W HPSV lamps with 250 W HPMV lamps
none of the above
Answer: A) installing separate lighting transformer and maintaining optimum voltage
Confirmed vs Book-3 §8.6(e) — A separate lighting transformer maintaining optimum voltage saves 5–15 % and extends lamp life. Running 220 V lamps at 260 V increases power input (Table 8.3) and shortens life; swapping 150 W HPSV for 250 W HPMV increases watts and lowers efficacy (Table 8.1: HPSV 90 vs HPMV 50 lm/W).
1. Calculate the free air delivery (FAD) of a compressor in m3/hr from a pump-up test: Receiver 0.5 m3; initial 0.0 kg/cm2(g); final 7.0 kg/cm2(g); atmospheric 1.026 kg/cm2(a); ambient air 40°C; final compressed air 60°C; additional holdup volume 0.005 m3; pump-up time 5 min 30 s.
Model answer: Q = [(P2 - P1)/P0] x [V/t] x [(273+t1)/(273+t2)]
V = 0.5 + 0.005 = 0.505 m3; t = 5 min 30 s = 5.5 min; P2 - P1 = 7.0 - 0.0 = 7.0 kg/cm2; P0 = 1.026 kg/cm2(a); t1 = 40 degC (ambient), t2 = 60 degC (compressed air).
Q = (7.0/1.026) x (0.505/5.5) x (313/333) = 6.8226 x 0.09182 x 0.9399 = 0.5888 m3/min
FAD = 0.5888 x 60 = 35.33 m3/hr. (2 marks for the formula, 3 marks for the calculation.)
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — Q = [(P2-P1)/P0] x [V/t] x [(273+t1)/(273+t2)]; the answer now carries the full working rather than the bare result.
2. A 7.5 kW, 415 V, 2 pole, 50 Hz, 3 phase squirrel cage induction motor has full load efficiency 90% and power factor 0.88. At full load rated values find: (a) input power in kW; (b) current drawn; (c) RPM at full load slip of 1%.
Model answer: (a) Pin = 8.333 kW; (b) I = 13.17 A; (c) N = 2970 RPM
(a) Pin = 7.5/0.90 = 8.333 kW. (b) I = 8333/(√3×415×0.88) = 13.17 A. (c) Ns = 120×50/2 = 3000 RPM; N = 3000×(1−0.01) = 2970 RPM.
3. In an AHU the actual airflow is 9300 m3/hr, inlet air enthalpy 16.12 kCal/kg, outlet enthalpy 13.33 kCal/kg, air density 1.15 kg/m3. Estimate the TR of the AHU.
Model answer: Air-side load: TR = Q x rho x (h_in - h_out)/3024 = 9300 x 1.15 x (16.12 - 13.33)/3024 = 9300 x 1.15 x 2.79 / 3024 = 29,839/3024 = 9.87 TR (about 9.86 TR). Heat removed = 29,839 kcal/hr; dividing by 3024 kcal/hr per TR gives the AHU refrigeration load.
Air-side load uses TR = airflow x air density x (h_in - h_out) / 3024, with enthalpies read off the psychrometric chart in kcal/kg. Here 9300 x 1.15 x 2.79 = 29,839 kcal/hr, which is 9.87 TR. Use density (kg/m3) with volumetric airflow, or specific volume in the denominator - not both.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
4. A centrifugal pump pumps 90 m3/hr of water with discharge pressure 3 kg/cm2(g) and a negative suction head of 3 m. Motor power drawn is 13 kW. Find pump efficiency. Motor efficiency 91%, water density 1000 kg/m3.
Model answer: Discharge head = 3 kg/cm²(g) ≈ 30 m; suction head = −3 m (negative/lift). Total head = 30 − (−3) = 33 m. Hydraulic power = Q(m³/s)×H×ρ×g/1000 = (90/3600)×33×1000×9.81/1000 = 8.09 kW. Pump shaft power = motor input × motor efficiency = 13×0.91 = 11.83 kW. Pump efficiency = 8.09/11.83 = 68.4%.
Total head = h_d − h_s with negative suction head added; hydraulic power = Q·H·ρ·g/1000; pump efficiency = hydraulic power / (motor power × motor efficiency).
5. Find the blow down rate of a cooling tower: cooling water flow 600 m3/hr; operating range 8°C; TDS in circulating water 1500 ppm; TDS in make-up water 300 ppm.
Model answer: Evaporation loss = 0.00085 × 1.8 × circulation rate × range = 0.00085 × 1.8 × 600 × 8 = 7.344 m³/hr. COC = TDS in circulating water / TDS in make-up water = 1500/300 = 5. Blow down = Evaporation loss/(COC − 1) = 7.344/(5 − 1) = 1.836 m³/hr.
Evaporation loss = 0.00085×1.8×600×8 = 7.344 m3/hr. COC = 1500/300 = 5. Blowdown = Evap/(COC−1) = 7.344/(5−1) = 1.836 m3/hr. ⚠ answerText expanded from bare result to full book-method working.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%); §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³)
7. A water pump is delivering 400 m3/hr. The impeller diameter is trimmed by 8% and its speed reduced by 10%. Find the water flow at the changed conditions.
Model answer: Flow varies directly with impeller diameter (Q∝D) and with speed (Q∝N), so the two effects multiply. 8% trim → D₂/D₁ = 0.92; 10% speed reduction → N₂/N₁ = 0.90. Q₂ = 400 × 0.92 × 0.90 = 331.2 ≈ 331 m³/hr. (Note: the printed key gives 288 m³/hr = 400 × 0.8 × 0.9, which corresponds to a 20% trim; with the question data as stated the affinity laws give 331 m³/hr — show the working and state the assumption.)
Apply Q∝D then Q∝N: 400×0.92×0.90 = 331 m³/hr. The official key's 288 m³/hr uses a 0.8 diameter factor (20% trim) — an inconsistency with the '8%' in the question.
8. In an alkali plant, salt brine at 20 m3/hr is cooled from 14°C to 8°C using chilled water. The chiller compressor motor draws 44.4 kW at 90% motor efficiency. Allied auxiliaries draw 20 kW. Brine density 1.2 kg/litre, specific heat 0.97 kCal/kg°C. (a) Refrigeration load (TR); (b) COP of compressor; (c) overall specific power consumption (kW/TR).
Model answer: (a) Mass flow of brine = 20 m3/hr x 1000 x 1.2 kg/litre = 24,000 kg/hr. Refrigeration load TR = m x Cp x (Ti - To)/3024 = 24,000 x 0.97 x (14 - 8)/3024 = 139,680/3024 = 46.2 TR. (b) Compressor shaft power = 44.4 x 0.90 = 39.96 kW; COP = 3.516 x 46.2/39.96 = 162.4/39.96 = 4.06. (c) Overall specific power consumption = (44.4 + 20)/46.2 = 64.4/46.2 = 1.39 kW/TR.
Brine is not water: its density (1.2 kg/litre) and specific heat (0.97 kcal/kg degC) must both be used in TR = m x Cp x deltaT / 3024. COP uses the compressor SHAFT power (44.4 x 0.9), while overall kW/TR uses the total electrical input including the 20 kW auxiliaries - hence 4.06 vs 1.39 kW/TR.
📖 §1.4 Power Factor Improvement and Benefits — multi-chapter fill-in-the-blanks
1. Fill in the blanks: (a) If the reactive power drawn by a load is zero, the load operates at __ power factor. (b) Power factor is the ratio of ___. (c) As the approach increases (other parameters constant), the effectiveness of a cooling tower ___. (d) Lower power factor of a DG set demands ___ excitation currents. (e) If voltage applied to a 415 V rated capacitor drops by 5%, its VAr output drops by about ___%.
2. List down any 5 energy conservation opportunities in fan systems.
Model answer: 1) Minimise excess air in combustion systems to reduce FD/ID fan load; 2) Minimise air in-leaks in hot flue gas path to reduce ID fan load; 3) Avoid cold air in-leaks that choke ID fan capacity; 4) Minimise system resistance/pressure drops via duct improvements; 5) Adopt inlet guide vanes in place of discharge damper control; (also: energy-efficient flat/cogged V-belts; two-speed motors or VSDs; fan speed reduction by pulley dia change; hollow FRP aerofoil impellers; impeller derating; higher-efficiency fan/impeller with cone).
Any five from the Book-3 §5.7 list: minimise excess air and in-leaks, high-efficiency impeller/fan, impeller derating, hollow FRP impellers, pulley speed reduction, VSD/two-speed motors, efficient belts, IGV instead of damper, lower system resistance.
3. A unit has 2 identical 500 kVA transformers, each with no-load loss 800 W and full-load copper loss 5000 W. Plant load is 400 kVA. Compare transformer losses for single transformer operation vs two transformers in parallel. Also list any five options to minimise electrical distribution loss.
Model answer: Single transformer loss = 800 + (400/500)²×5000 = 4000 W. Two transformers (each sharing 200 kVA): 2×[800 + 5000×(200/500)²] = 3200 W. Single-transformer operation has 800 W higher loss. Distribution-loss options: relocate transformers/substations near load centres; re-route/re-conductor high-loss feeders; PF improvement with capacitors at load end; optimum loading of transformers; use lower-resistance AAAC instead of ACSR; minimise losses at weak links (jumpers, loose contacts, brittle conductors); improve HT:LT ratio to shorten LT network.
Loss = iron loss + (load/rating)²×copper loss; parallel operation halves the load share, reducing total copper loss. Any five distribution-loss options accepted.
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
4. Efficiency assessment of a 37 kW, 415 V, 62 A, 1480 rpm, 3-phase delta SCIM. No-load test: 415 V, 18 A, 50 Hz, stator resistance 0.275 Ω/phase at 30°C, no-load power 1164 W. Calculate: (i) iron+friction&windage loss; (ii) stator resistance at 120°C; (iii) stator copper loss at full load at 120°C; (iv) full load slip and rotor input; (v) motor input (stray loss 0.5% of rated output); (vi) full load efficiency and power factor.
5. (a) How do you calculate the velocity of gas in a duct using the average differential pressure and density of the gas? (b) A V-belt centrifugal fan performance test: density of air at 0°C = 1.293 kg/m3; ambient 40°C; discharge duct dia 0.8 m; velocity pressure 45 mmWC; pitot coefficient 0.9; static pressure at fan inlet −20 mmWC, outlet 185 mmWC; motor power 75 kW; belt transmission efficiency 97%; motor efficiency 93%. Find the static fan efficiency.
Model answer: (a) V (m/s) = Cp × √(2×9.81×Δp×γ)/γ, with Cp = pitot constant (0.85 typical), Δp = average velocity pressure, γ = gas density at test condition. (b) Static fan efficiency ≈ 37.58%.
📖 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity (part b is DG set – not Ch-7)
6. (a) Estimate the cooling tower approach and capacity (TR): water flow 150 m3/hr; sp.heat 1 kCal/kg°C; inlet water 42°C; outlet water 34°C; ambient WBT 30°C. (b) A 180 kVA, 0.80 PF rated DG set has a diesel engine rating of 240 BHP. What is the maximum power factor maintainable at full load on the alternator without overloading the diesel engine? (alternator losses + exciter power = 5.44 kW; no derating.)
Model answer: (a) Approach = outlet cold water − ambient WBT = 34 − 30 = 4°C. Range = 42 − 34 = 8°C. Capacity = flow × density × Cp × range / 3024 = 150 × 1000 × 1 × 8 / 3024 = 12,00,000/3024 = 396.8 TR. (b) Engine output = 240 BHP × 0.746 = 179.04 kW; power available to alternator output = 179.04 − 5.44 = 173.6 kW; maximum PF = kW/kVA = 173.6/180 = 0.964.
(a) Approach = outlet − WBT = 34−30 = 4°C; capacity = 150×1000×1×(42−34)/3024 = 396.8 TR. (b) Engine power = 240×0.746 = 179.04 kW; power available for alternator = 179.04−5.44 = 173.6 kW; max PF = 173.6/180 = 0.964. ⚠ answerText expanded from bare results to full working.