Energy Efficiency in Electrical Utilities Available here with full solutions — 74 questions recovered from the 2015 exam:
Objective (1 mark)
52 of 50
Short (5 marks)
10 of 8
Long (10 marks)
12 of 6
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) — 52
📖 Book-3 §3.4 Compressed Air System Components (Figure 3.6)
1. The purpose of inter-cooling in a multistage compressor is to
Increase the pressure of 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 must handle.
Raising pressure (a) is the compressor's job; moisture and oil separation (c) is an incidental benefit, not the purpose of inter-cooling.
📖 §2.7 Improving the Motor Loading by Operating in Star Mode
2. One low-investment measure to improve the efficiency of a squirrel cage induction motor which operates consistently below 40% of its rated capacity is by
operating it in star mode
replacing it with a correctly sized motor
operating in delta mode
none of the above
Answer: A) operating it in star mode
Confirmed vs Book-3 §2.7 Improving the Motor Loading by Operating in Star Mode — For motors consistently operating below 40 % of rated capacity the book recommends permanent star operation — an inexpensive measure needing only re-wiring at the terminal box and resetting the overload relay. (b) replacing with a correctly sized motor also works but is a HIGH-investment option, which is why it is not the answer to a 'low investment' question.
3. Which of the following is NOT a part of the vapour compression refrigeration system?
compressor
evaporator
condenser
absorber
Answer: D) absorber
Confirmed vs Book-3 §4.3 - A vapour compression system consists of compressor, condenser, expansion device and evaporator. The absorber is a component of the vapour absorption (not compression) system.
4. The T2, T5, T8 and T12 fluorescent lamps are categorized based on
diameter of the tube
length of the tube
both diameter and length
none of the above
Answer: A) diameter of the tube
Confirmed vs Book-3 §8.3 — 'These four lamps vary in diameter': T12 = 38 mm (1.5" = 12/8"), T8 = 25 mm (1"), T5 = 16 mm (5/8"), T2 = 6 mm (1/4"). The T-number is the diameter in eighths of an inch; length is not part of the designation.
📖 §9.1 Introduction — spark ignition vs compression ignition engines
5. A spark ignition engine is used for firing which of the following fuels?
high speed diesel
light diesel oil
natural gas
furnace oil
Answer: C) natural gas
Confirmed vs Book-3 §9.1 — "Spark ignition engines use a spark ... Typical fuels for such engines are gasoline, natural gas and sewage and landfill gas." HSD, LDO and furnace oil are compression-ignition (diesel/heavy fuel oil) fuels, so (a), (b) and (d) are wrong.
📖 §6.2 System characteristics — static & friction head
6. Installing larger diameter pipe in pumping system results in reduction in ______
static head
frictional head
both a and b
neither a nor b
Answer: B) frictional head
Confirmed vs Book-3 §6.2 — Static head is 'simply the difference in height of the supply and destination reservoirs' and is independent of flow and pipe size. Friction head is the loss in pipes/valves/equipment, and the book notes that further reduction 'will require larger diameter pipe'. So only frictional head is reduced.
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
7. In no load test of a poly-phase induction motor, the measured power by the wattmeter consists of:
core loss
copper loss
core loss, windage & friction loss
stator copper loss, iron loss, windage & friction loss
Answer: D) stator copper loss, iron loss, windage & friction loss
Confirmed vs Book-3 §2.4 Motor Efficiency — Field Tests for Determining Efficiency — The no-load wattmeter reading is the total no-load input, i.e. stator I²R at no-load current PLUS core (iron) loss PLUS friction & windage; the book subtracts (no-load current)² × stator resistance from it to leave core + F&W. (c) is the tempting answer because it names what is LEFT after that subtraction, not what the wattmeter actually reads.
8. A better indicator for cooling tower performance is:
Heat load in tower
Range
RH of air leaving cooling tower
Approach
Answer: D) Approach
Confirmed vs Book-3 §7.2 (ii) — Book: 'Although both range and approach should be monitored, the Approach is a better indicator of cooling tower performance' → (d). Range and heat load are fixed by the process, not by the tower (§7.2 Range).
📖 §8.3(1) Incandescent lamp (resistive filament); §8.2 Control gear
9. Power factor is highest in case of
sodium vapour lamps
LED lamps
tube lights
incandescent lamps
Answer: D) incandescent lamps
Confirmed vs Book-3 §8.3 — An incandescent lamp is a plain heated filament, a purely resistive load, so its power factor is ~1. Sodium vapour lamps and tube lights need inductive ballasts/ignitors (§8.2 control gear) and LED lamps need electronic drivers, all of which lower the power factor.
10. A 50 hp motor with a full load efficiency of 90 percent was found to be operating at 25 kW input. The percent Motor Load is:
75%
67%
60%
25%
Answer: C) 60%
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — 50 hp = 37.3 kW output; rated input = 37.3/0.90 = 41.4 kW; % load = 25 / 41.4 = 60 %. (a) 75 % comes from dividing 25 kW by the hp figure without converting to kW and without dividing by efficiency — the book's formula always compares measured input kW with rated kW ÷ rated efficiency.
11. Which Loss in a Distribution Transformer is predominant if the transformer is loaded to 75% of its rated capacity?
core loss
copper loss
hysteresis loss
magnetic field loss
Answer: B) copper loss
Confirmed vs Book-3 §1.5 — core loss is constant while copper loss varies as (%load)². At 75% load a typical 500 kVA unit (Table 1.3) has copper loss 0.75² × 6450 = 3630 W against a fixed core loss of only 900 W, so copper loss dominates.
Option (a) core loss dominates only at light loading (below roughly 35–40% load); (c) hysteresis is a COMPONENT of core loss, not a separate answer.
reduce the peak demand of the distribution company
increase the revenue of the distribution company
increase the peak demand
increase the maximum demand in an industry
Answer: A) reduce the peak demand of the distribution company
Confirmed vs Book-3 §1.2/§1.3 — 'Time of Day (TOD) rates like peak and non-peak hours…' are used by utilities to 'influence end user in better load management', so that consumers shift load away from the utility's peak.
Option (c)/(d) are the opposite of the intent — the higher peak-hour tariff is designed to discourage, not encourage, drawl at peak.
Confirmed vs Book-3 §4.2 - Specific (absolute) humidity = mass of water vapour per unit mass of dry air (g moisture/kg dry air). Option (c) 'per kg of air' is the trap - the reference is dry air; (b) and (d) describe relative humidity, which is a percentage rather than a mass ratio.
14. The percentage reduction in distribution losses when tail end power factor raised from 0.85 to 0.95 is:
10.1%
19.9%
71%
84%
Answer: B) 19.9%
Confirmed vs Book-3 §1.4 — reduction in distribution loss % = [1 − (PF₁/PF₂)²] × 100 = [1 − (0.85/0.95)²] × 100 = [1 − 0.8006] × 100 = 19.9%.
Option (c) 71% is a residual-loss figure from the 0.8→0.95 version of this question; always square the PF ratio and subtract from 1.
reduces voltage by inserting resistance in rotor circuit
reduces voltage by inserting resistance in stator circuit
reduces the voltage applied to motor windings at start
inserts capacitance in the stator
Answer: C) reduces the voltage applied to motor windings at start
Confirmed vs Book-3 §2.9 Soft Starter — A star-delta starter connects the windings in star at start, reducing the voltage across each winding by √3 and hence the starting current and torque, before switching to delta. (a)/(b) inserting resistance is how a slip-ring motor is started; the book notes star-delta gives only a partial solution because peaks can recur at transition.
📖 §1.4 Power Factor (see also Book-3 Ch-9 DG Sets)
16. Lower power factor of a DG set demands __________.
lower excitation currents
no change in excitation currents
higher excitation currents
none of the above
Answer: C) higher excitation currents
Confirmed vs Book-3 §1.4/Ch-9 — an alternator supplies the load's reactive kVAr from its field; the lower the power factor, the larger the kVAr for the same kW, so the field (excitation) current must be increased.
Option (a) is the reverse: only at high/unity power factor can the machine run with reduced excitation, which is why DG sets are de-rated at low PF.
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
17. In a no load test of a poly-phase induction motor, the measured power by the wattmeter consists of:
core loss only
copper loss only
core loss, windage & friction loss
full load copper loss
Answer: C) core loss, windage & friction loss
Confirmed vs Book-3 §2.4 Motor Efficiency — Field Tests for Determining Efficiency — With no shaft load the input covers the core (iron) loss plus friction & windage plus a small stator I²R at no-load current; of the choices offered, core loss with windage & friction is the correct grouping. (a) core loss only is incomplete — the book separates the two by repeating the test at variable voltage and taking the intercept as F&W.
Confirmed vs Book-3 §8.2 — 'Lux (lx) is the metric unit of measure for illuminance of a surface' (1 lm/m²). Lumens measure luminous flux emitted by the source, LPD is lighting power density in W/m², and radians are angles.
📖 §6.1 Pump types — centrifugal pump construction & working (displacement vs dynamic pumps); §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³) (affinity laws apply to rotodynamic pumps only)
19. If the speed of a reciprocating pump is reduced by 50 %, the head
is reduced by 25%
is reduced by 50%
is reduced by 75%
remains same
Answer: D) remains same
Confirmed vs Book-3 §6.1/§6.5 — The affinity laws (H∝N²) are stated by the book for 'rotodynamic pump performance parameters'. A reciprocating pump is a positive-displacement pump: its head is set by the system pressure it discharges into, while speed reduction only cuts the volume delivered (flow ∝ speed). So the head remains the same.
20. If the observed temperature in air receiver is higher than ambient air temperature the correction factor for free air delivery will be:
(273+t2)/(273+t1)
greater than 1
less than 1
equal to 1
Answer: C) less than 1
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method) — Book-3 §3.6: when the discharge/receiver temperature t₂ exceeds the ambient t₁, the measured FAD must be corrected by the factor (273+t₁)/(273+t₂), which is necessarily LESS than 1 — hot air occupies more volume, so the uncorrected figure overstates FAD.
Option (a) was printed as the correct ratio itself, giving two right answers; it has been repaired to the inverted form (273+t₂)/(273+t₁), which is greater than 1 and is the classic trap, leaving (c) as the single correct choice.
21. If EER of One Ton Split AC is 3.5, what is its power rating?
1.0 kW
1.5 kW
0.8 kW
None of the above
Answer: A) 1.0 kW
Confirmed vs Book-3 §4.9 - 1 TR = 3.517 kW cooling; power = cooling/EER = 3.517/3.5 = 1.0 kW. Options (b)-(d) miss the definition EER = cooling watts / input watts; 3.517 kW of cooling divided by an EER of 3.5 gives almost exactly 1.0 kW, the benchmark for a 5-star 1 TR unit.
📖 §6.2 System characteristics — static & friction head
22. Friction losses in a pumping system is
inversely proportional to flow
inversely proportional to cube of flow
proportional to square of flow
inversely proportional square of flow
Answer: C) proportional to square of flow
Confirmed vs Book-3 §6.2 — Book: 'The friction losses are proportional to the square of the flow rate.' This is why the system curve is parabolic and why halving flow cuts friction head to a quarter. Inverse relations (a, b, d) are wrong.
23. Find the Total Harmonic Distortion (THD) for current for the following readings: fundamental (50 Hz) = 250 A, third harmonic = 50 A, fifth harmonic = 35 A.
58 %
48 %
24%
34 %
Answer: C) 24%
Confirmed vs Book-3 §1.10 — THD_current = √(ΣIₙ²)/I₁ × 100 = √(50² + 35²)/250 × 100 = 61.03/250 × 100 = 24.4% ≈ 24%.
Option (d) 34% comes from adding the harmonics arithmetically ((50+35)/250 = 34%); harmonic components must be combined as a root-sum-square.
📖 §10.8 ECBC Guidelines on Lighting — Lighting Power Density, Building Area Method (book example: hotel, 4 floors × 1000 m², LPD 10.8 W/m²)
24. A hotel building has four floors each of 1000m2 area. If the interior lighting power allowance for the hotel building is 43,000 W. The Lighting Power Density (LPD) is
10.75
0.09
43
data insufficient
Answer: A) 10.75
Confirmed vs Book-3 §8.2 (power density) — LPD = lighting power ÷ floor area = 43,000 W ÷ (4 × 1000 m²) = 10.75 W/m². Option (b) 0.09 is the inverted ratio; the data are sufficient.
25. If the COP of a vapour compression system is 3.5 and the motor draws a power of 10.8 kW at 90% motor efficiency, the cooling effect of vapour compression system will be
34 kW
37.8 kW
0.36 kW
none of the above
Answer: A) 34 kW
Confirmed vs Book-3 §4.7 - Compressor shaft power = 10.8 x 0.9 = 9.72 kW. Cooling effect = COP x compressor power = 3.5 x 9.72 = 34.0 kW. Option (b) 37.8 kW is the trap of using the 10.8 kW motor input without applying the 90% efficiency; use shaft power 9.72 kW x COP 3.5 = 34 kW.
📖 §6.7 Boiler feed water pumps (multistage pump = single-stage pumps in series)
26. If two identical pumps operate in series, their shut-off head is
Not affected
More than double
Doubled
Less than double
Answer: C) Doubled
Confirmed vs Book-3 §6.7 — Series operation adds heads at the same flow (the book equates a multistage pump with single-stage pumps in series). At zero flow each identical pump gives its shut-off head, so the combination gives exactly double. Parallel pumps would leave shut-off head unchanged and add flow.
27. The adsorption material used in an adsorption air dryer is
Calcium chloride
Magnesium chloride
Activated alumina
Potassium chloride
Answer: C) Activated alumina
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — Book-3 names activated alumina and silica gel as the common adsorbents for compressed air drying; they bind moisture physically on a large porous inner surface and are then regenerated.
Calcium, magnesium and potassium chloride are deliquescent absorption chemicals and are not used as the desiccant bed in these dryers.
28. The cooling tower size is _____ to the entering Wet Bulb Temperature (WBT), when the heat load, range and approach are constant.
Directly proportional
Inversely proportional
Constant
None of above
Answer: B) Inversely proportional
Confirmed vs Book-3 §7.2 Factors that affect cooling tower size — Book sidebar: when heat load, range and approach are constant, tower size varies 'inversely with the entering WBT' → (b). Book example: a tower for 4.45°C approach to 21.11°C WBT is larger than the same duty at 26.67°C WBT.
📖 §9.4 Energy performance assessment — specific fuel consumption & alternator loading
29. A DG set is consuming 70 litres per hour diesel oil. If the specific fuel consumption is 0.33 litres/kWh, what is the kVA loading at 0.8 power factor ?
212 kVA
265 kVA
170 kVA
none of the above
Answer: B) 265 kVA
Confirmed vs Book-3 §9.4 — kW = 70/0.33 = 212.1 kW; kVA = 212.1/0.8 = 265 kVA. Option (a) 212 kVA is the kW figure; (c) 170 comes from multiplying by PF instead of dividing.
30. The two-part tariff structure for HT category consumers are
one part for capacity drawn and second part for actual energy drawn
one part for actual power and second part for actual reactive power drawn
one part for capacity drawn and second part for actual reactive power drawn
one part for actual apparent energy drawn and second part for actual reactive energy drawn
Answer: A) one part for capacity drawn and second part for actual energy drawn
Confirmed vs Book-3 §1.2 — 'the electricity billing…in High Tension (HT) category, is often done on two-part tariff structure, i.e. one part for capacity (or demand) drawn and the second part for actual energy drawn during the billing cycle.'
Option (c) is wrong: reactive energy is billed separately as a PF penalty/bonus, not as the second part of the two-part tariff.
📖 §2.7 Improving the Motor Loading by Operating in Star Mode
31. The inexpensive way to improve energy efficiency of a motor which operates consistently at below 40% of rated capacity is by ___
Operating in Star mode
Replacing with correct sized motor
Operating in delta mode
Operating in VFD mode
Answer: A) Operating in Star mode
Confirmed vs Book-3 §2.7 Improving the Motor Loading by Operating in Star Mode — For motors consistently loaded below 40 % of rating, permanent star operation is the inexpensive fix — re-wiring the terminal box and resetting the overload relay, with no new equipment. (d) VFD operation would also cut losses but requires substantial capital, failing the 'inexpensive' criterion.
📖 §1.8 Technical vs commercial losses in distribution (AT&C)
32. Which of the following can be attributed to commercial loss in electrical distribution system
lengthy low voltage lines
low load side power factor
faulty consumer service meters
undersize conductors
Answer: C) faulty consumer service meters - Technical losses arise from the physical network (long LT lines, undersized conductors, low PF). Commercial losses arise from metering, billing and collection deficiencies - defective/faulty consumer meters, meter tampering and theft.
Split the two families cleanly: TECHNICAL losses are physical I^2R and no-load losses - lengthy LT lines, undersized conductors, poor PF, overloaded transformers. COMMERCIAL losses are the metering-billing-collection failures - faulty/stopped/tampered meters, unmetered supply, theft and unrecovered arrears.
Options (a), (b) and (d) are all physical network causes; only the faulty consumer meter is a metering (commercial) defect.
Related formula: AT&C loss % = [1 - (billing efficiency x collection efficiency)] x 100.
📖 §4.8 Factors affecting performance & energy efficiency of refrigeration plants
33. When evaporator temperature is reduced
refrigeration capacity increases
refrigeration capacity decreases
specific power consumption remains same
compressor will stop
Answer: B) refrigeration capacity decreases - Lowering the evaporator temperature lowers suction pressure, so the specific volume of the suction vapour rises and the mass flow handled by the compressor falls. Refrigeration capacity drops (roughly 3-4% per degC) while specific power consumption (kW/TR) rises.
Dropping the evaporator temperature drops the suction pressure; the suction vapour becomes less dense (higher specific volume), so the same compressor swept volume moves LESS refrigerant mass - capacity falls while the compressor work per unit of cooling rises.
So both the capacity and the COP fall, and kW/TR worsens; the book records roughly 3-4% capacity loss per degC of evaporator temperature reduction.
This is the operating rule behind the whole chapter: keep the chilled water/evaporator temperature as HIGH as the process will tolerate, and the condenser as cold as possible.
34. 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 - Drift eliminators are baffles at the air outlet that change the air direction sharply so that entrained water droplets (drift) impinge and drain back into the tower, minimising drift loss. Maximising air-water contact is the job of the fill; distribution is the job of the nozzles/hot water basin.
Drift eliminators are closely spaced baffles at the air outlet that force the leaving air through sharp direction changes; the heavier water droplets cannot follow, impinge on the blades and drain back. Good cellular PVC eliminators cut drift to about 0.001–0.02% of circulation. Do not confuse the three internals: FILL maximises contact area, LOUVRES/nozzles handle entry and distribution, ELIMINATORS catch carryover.
35. Trivector meter measures three vectors representing
active, reactive and maximum demand
active, power factor and apparent power
active, harmonics and maximum demand
active, reactive and apparent power
Answer: D) active, reactive and apparent power - The trivector meter records the three components of the power triangle: active power (kW/kWh), reactive power (kVAr/kVArh) and apparent power (kVA/kVAh); from these it also derives power factor and maximum demand.
'Tri-vector' = the three sides of the power triangle: active (kW/kWh), reactive (kVAr/kVArh) and apparent (kVA/kVAh). PF and maximum demand are DERIVED from these, they are not among the three vectors.
That is exactly why (a) and (b) are wrong - they list a derived quantity in place of one of the three measured vectors.
The meter integrates demand over a fixed cycle (typically 30 minutes) to record maximum demand for billing.
📖 §8.2 Basic parameters and terms (inverse square law)
36. The illuminance of a lamp at one meter distance is 10 Lm/m2. What will be the corresponding value at 0.7 meter distance
14.28
20.41
10
none of these
Answer: B) 20.41 - Illuminance follows the inverse square law: E2 = E1 x (d1/d2)^2 = 10 x (1/0.7)^2 = 10 x (1/0.49) = 20.408 ~ 20.41 lux.
Inverse square law: E = I/d², or in the handy form E1·d1² = E2·d2². So E2 = 10 × (1/0.7)² = 10/0.49 = 20.41 lux. Move CLOSER and illuminance goes UP, so any answer below 10 is wrong on inspection. The book's own worked example is at half the distance: 10 × (1/0.5)² = 40 lux. Note lm/m² and lux are the same unit.
📖 §5.3 Fan performance evaluation (system characteristics)
37. The fan system resistance is predominately due to
more bends used in the duct
more equipments in the system
volume of air handled
density of air
Answer: A) more bends used in the duct - System resistance is the sum of friction and dynamic (shock) losses. Bends, elbows, transitions and other fittings in the duct contribute the dominant dynamic pressure losses; adding bends shifts the system curve steeply upward.
System resistance = friction losses in straight duct + dynamic (shock) losses at bends, elbows, transitions, dampers and hoods; the shock losses at fittings dominate. Resistance varies as the square of flow, so the system curve is a parabola through the origin. Volume of air is what you push through the resistance, not the cause of it — that is why (c) is the tempting wrong answer.
38. The actual measured load of 1000 kVA transformer is 400 kVA. Find out the total transformer loss corresponding to this load if no load loss is 1500 Watts and full load Copper Loss is 12,000 Watts
1920 watts
1500 watts
3420 watt
13500 watts
Answer: C) 3420 watt - Total loss = no-load (iron) loss + (load fraction)^2 x full-load copper loss. Load fraction = 400/1000 = 0.4. Copper loss = 0.4^2 x 12000 = 0.16 x 12000 = 1920 W. Total loss = 1500 + 1920 = 3420 W.
Formula: total loss = no-load (iron) loss + (load fraction)^2 x full-load copper loss.
Working: load fraction = 400/1000 = 0.4; copper loss = 0.4^2 x 12,000 = 0.16 x 12,000 = 1920 W; total = 1500 + 1920 = 3420 W.
Two traps sit in the options: 1920 W (forgetting to add the constant iron loss) and 13,500 W (using the full-load copper loss without scaling by the square of the load fraction). Never scale the iron loss - it is constant from no-load to full load.
📖 §2.9 Speed control of motors (slip power recovery)
39. Slip power recovery system is applicable in case of
squirrel cage induction motor
wound rotor motor
synchronous motor
DC shunt motor
Answer: B) wound rotor motor - Slip power recovery (static Scherbius / Kramer drive) taps the rotor slip power through slip rings and feeds it back to the supply. This requires access to the rotor winding, which only a wound rotor (slip ring) induction motor provides.
Slip power recovery (static Scherbius / Kramer drive) extracts the slip power from the rotor circuit through slip rings and returns it to the supply instead of wasting it as heat in a rotor resistor.
That requires physical ACCESS to the rotor winding, which only a wound-rotor (slip-ring) induction motor provides - a squirrel cage rotor is short-circuited internally with no terminals to tap.
Efficiency point: with a plain rotor rheostat the slip power (s x rotor input) is burnt as heat; slip power recovery makes sub-synchronous speed control efficient.
40. Rotating magnetic field is produced in a ___________
single-phase induction motor
three-phase induction motor
DC series motor
all of the above
Answer: B) three-phase induction motor - Three balanced currents displaced 120 electrical degrees in three space-displaced windings produce a constant-magnitude rotating field at synchronous speed Ns = 120f/P. A single-phase winding alone produces only a pulsating field (needing an auxiliary winding to start), and a DC motor has a stationary field.
Three balanced phase currents, displaced 120 electrical degrees in time and fed to three windings displaced 120 degrees in space, produce a constant-magnitude field that rotates at Ns = 120f/P. This is the whole basis of the induction motor.
A single-phase winding alone produces only a PULSATING field - it cannot self-start and needs an auxiliary/capacitor winding to create the second phase. A DC machine has a stationary field with a rotating armature.
Memory hook: three phases in space + three phases in time = rotation.
Answer: C) reduction in reactive power - Capacitors supply the magnetising kVAr locally, so the reactive power (and hence the kVA and the line current) drawn from the supply falls. The active power kW required by the load and its active (in-phase) current component are unchanged, so (a), (b) and therefore (d) are wrong.
A shunt capacitor supplies the magnetising kVAr locally, so the REACTIVE power (and hence kVA and line current) drawn from the supply falls. The load's kW requirement and its in-phase (active) current component are untouched.
Hence 'reduction in active current' is wrong: total current falls, but its active component does not.
Benefits chain to remember: less kVAr -> less kVA -> less current -> lower I^2R loss, better voltage regulation, released transformer/cable capacity, no PF penalty.
📖 §2.4 Motor efficiency (slip and rotor losses) / §2.6 Energy efficient motors
42. Motor efficiency will be improved by
reducing the slip
increasing the slip
reducing the diameter of the motor
decreasing the length of the motor
Answer: A) reducing the slip - Rotor copper loss = slip x air-gap power, and motor efficiency is approximately (1 - s) neglecting other losses. Lower slip means lower rotor I^2R loss and higher efficiency; that is why energy-efficient motors run at slightly higher speed (lower slip) than standard motors.
Rotor copper loss = s x air-gap (rotor input) power, so the rotor's own efficiency is (1 - s). Cutting slip directly cuts the rotor I^2R loss and lifts overall efficiency.
That is why an energy-efficient motor runs slightly FASTER (lower slip) than the standard motor it replaces - a useful clue in the exam. Caution for centrifugal loads: that small speed rise can increase the load's power draw as speed^3, partly eating the saving.
Motor diameter and length are design parameters, not operating variables, so (c) and (d) are irrelevant.
📖 §2.9 Speed control of motors (variable frequency drives)
43. Installation of Variable frequency drives (VFD) allows the motor to be operated with
constant current
lower start-up current
higher voltage
none of the above
Answer: B) lower start-up current - A VFD starts the motor at low frequency and low voltage (constant V/f), so the motor develops rated torque while drawing near-rated (typically 100-150%) current instead of the 6-7 times DOL inrush. It also gives soft start/stop and speed control.
A VFD starts the motor at low frequency and low voltage on a constant V/f ratio, so full torque is available while the current stays near rated - typically 100-150% - instead of the 6-7 times rated DOL inrush.
Extra benefits: soft start/stop reducing mechanical shock, and speed control which on centrifugal fans and pumps saves power as speed^3 (the affinity laws).
A VFD never raises the supply voltage, and the current it draws varies with load, so (a) and (c) are wrong.
📖 §3.4 Compressed air system components (after coolers) / §3.5 moisture removal
44. In a large compressed air system, about 70% to 80% of moisture in the compressed air is removed at the
air dryer
after cooler
air receiver
inter cooler
Answer: B) after cooler - The aftercooler downstream of the last stage cools the discharge air close to ambient, condensing the bulk (about 70-80%) of the moisture, which is drained through a moisture trap. The dryer then removes the remaining moisture to the required pressure dew point.
The after-cooler sits immediately downstream of the last compression stage and drops the discharge air back near ambient; the sharp temperature fall condenses the bulk of the moisture, which is drained through a moisture trap. The dryer then handles only the remainder, down to the required dew point.
Note the exact book figure is 'about 60 to 75% of moisture is removed at the after cooler' - the question's 70-80% is a loose paraphrase, but the after-cooler is unambiguously the right component.
The intercooler cools BETWEEN stages (to cut compression work); the receiver only precipitates a little residual moisture as the air cools further.
📖 §4.2 Psychrometrics and air-conditioning processes (humidification) / §4.14 Humidification systems
45. Humidification involves
reducing wet bulb temperature and specific humidity
reducing dry bulb temperature and specific humidity
increasing wet bulb temperature and decreasing specific humidity
reducing dry bulb temperature and increasing specific humidity
Answer: D) reducing dry bulb temperature and increasing specific humidity - In evaporative (adiabatic) humidification water evaporates into the air stream; the latent heat is drawn from the air itself, so the dry bulb temperature falls while the moisture content (specific humidity) rises. The process follows the constant wet-bulb / constant-enthalpy line on the psychrometric chart.
Adiabatic (evaporative) humidification takes the latent heat of vaporisation out of the air stream itself: the moisture content (specific humidity) RISES and the dry bulb temperature FALLS, while enthalpy and wet-bulb temperature stay essentially constant.
On the chart it is a move up-and-left along a constant wet-bulb line towards the saturation curve.
Do not confuse it with steam humidification, which adds moisture at nearly constant dry-bulb temperature and raises the enthalpy - the exam wording 'humidification' in a textile air washer always means the adiabatic case.
lower evaporator temperature and higher condenser temperature
higher evaporator temperature and lower condenser temperature
higher evaporator temperature and higher condenser temperature
lower evaporator temperature and lower condenser temperature
Answer: B) higher evaporator temperature and lower condenser temperature - COP(Carnot) = Te/(Tc - Te). Raising Te and lowering Tc both increase the numerator and shrink the temperature lift (Tc - Te), so the COP rises. This is why chilled water temperature should be kept as high as the process allows and condenser/cooling water as cold as possible.
COP_Carnot = T_evap/(T_cond - T_evap), with both temperatures in KELVIN. Raising T_evap increases the numerator AND shrinks the lift; lowering T_cond shrinks the lift again - both push COP up.
Worked feel for the numbers: at 5 degC evaporator and 40 degC condenser, COP = 278/35 = 7.9; drop the condenser to 35 degC and it becomes 278/30 = 9.3, an 18% gain for 5 degC.
Using degC instead of Kelvin here is the classic destroyed answer (5/35 = 0.14 is meaningless). Operationally: run chilled water as warm as the process allows and keep condenser/cooling water as cold as possible - clean tubes, clean cooling tower.
47. Flow control by damper operation in fan system will
increase energy consumption
reduce energy consumption
reduce system resistance
none of the above
Answer: A) increase energy consumption - A damper throttles flow by ADDING artificial resistance, moving the operating point up the fan curve. The energy consumed per unit of air delivered (specific energy consumption, kW per m3/hr) therefore rises, and the throttling energy is wasted as pressure drop. Option (c) is plainly wrong, since a damper increases, not reduces, system resistance; compared with speed control (VFD), damper control is the energy-wasting method.
A damper does not slow the fan; it adds artificial resistance so the operating point rides UP the fan curve to a higher pressure and lower flow. Power falls a little but kW per m³/hr rises, and the throttled pressure drop is pure waste. Compare with a VFD, where power falls as N³. Hook: 'a damper is a brake you pay electricity to hold on'.
📖 §5.6 Fan performance assessment (gas density calculation)
48. Calculate the density of air at 11400 mmWC absolute pressure and 65 degC. (Molecular weight of air: 28.92 kg/kg mole and Gas constant: 847.84 mmWC m3/kg mole K)
1.2 kg/m3
1.5 kg/m3
1.15 kg/m3
none of the above
Answer: C) 1.15 kg/m3 - Density = (P x M)/(R x T) where P = 11400 mmWC, M = 28.92 kg/kg mole, R = 847.84 mmWC m3/kg mole K and T = 65 + 273 = 338 K. Density = (11400 x 28.92)/(847.84 x 338) = 329,688/286,569 = 1.1504 ~ 1.15 kg/m3.
Density formula to memorise: ρ = (P × M)/(R × T), with P in mmWC absolute, M = 28.92 kg/kg-mole for air, R = 847.84 mmWC·m³/kg-mole·K and T in KELVIN. Here ρ = (11400 × 28.92)/(847.84 × 338) = 329,688/286,569 = 1.15 kg/m³. The mark is lost by leaving T at 65 instead of 338 K, or by using gauge instead of absolute pressure.
📖 §1.4 Power factor improvement (capacitor output vs applied voltage)
49. A company installed a 130 kVAr, 600 Volt capacitor but the power meter indicates that it is only operating at 119 kVAr. The reason out of the following could be
operating at low load
high voltage
low voltage
low current
Answer: C) low voltage - Capacitor output kVAr = 2 pi f C V^2, i.e. proportional to the square of applied voltage. Delivered/rated = 119/130 = 0.9154, so V/Vrated = sqrt(0.9154) = 0.957, i.e. about 574 V against the rated 600 V. The capacitor is under-delivering because it is operating at a lower than rated voltage; capacitor kVAr does not depend on the load.
Capacitor kVAr = 2(pi)fCV^2, so output falls with the square of the voltage - the capacitance itself has not changed.
Check: 119/130 = 0.9154, so V/V_rated = sqrt(0.9154) = 0.957, i.e. about 574 V on a 600 V unit. A 4% voltage shortfall costs ~8% of the kVAr.
Capacitor output is independent of the connected load, so 'operating at low load' and 'low current' cannot explain the shortfall.
📖 §2.4 Motor efficiency (efficiency from nameplate data)
50. A 22 kW, 415 V, 45 A, 0.8 pf, 1475 rpm, 4 pole 3 phase induction motor operating at 420 V, 40 A and 0.8 pf. What will be the motor efficiency?
85.0 %
94.5 %
89.9 %
None of the above
Answer: A) 85.0 %
Efficiency = output/input, and the input must be computed from the NAMEPLATE ratings: sqrt(3) x 415 x 45 x 0.8 = 25.88 kW against 22 kW output = 85.0%.
The 420 V and 40 A are the present operating point, deliberately supplied to tempt you into 1.732 x 420 x 40 x 0.8 = 23.28 kW, which would give 94.5% - option (b), the planted wrong answer.
Sanity rule: a standard 22 kW motor sits around 88-91% efficiency, so any answer above 94% on nameplate data should make you re-read the question.
📖 §4.3 Types of refrigeration system (VCR components)
51. Which of the following is not a part of vapour compression refrigeration cycle ?
Compressor
Evaporator
Condenser
Generator
Answer: D) Generator
The vapour compression circuit has exactly four elements: compressor, condenser, expansion device and evaporator.
The GENERATOR (along with the absorber, solution pump and solution heat exchanger) belongs to the vapour absorption machine, where heat boils refrigerant out of the strong solution - it is the heat-driven half of the 'thermal compressor' that replaces the mechanical one.
Same principle as the 'absorber' version of this question: anything that needs heat rather than shaft work is an absorption component.
52. L / G ratio in a cooling tower is the ratio of _________________.
Length and girth
Length and Temperature gradient
Water flow rate and air mass flow rate
Air mass flow rate and water flow rate
Answer: C) Water flow rate and air mass flow rate
L over G, in that order: L = water (liquid) mass flow, G = gas (air) mass flow. Option (d) reverses the ratio and is the trap — check which term is on top before ticking. Against design values, seasonal tuning of water box loading and fan blade angle is done to restore the design L/G and recover effectiveness.
📖 §6.2 System characteristics — static & friction head
1. The total system resistance of a piping loop is 50 meters and the static head is 15 meters at designed water flow. Calculate the system resistance at 75%, 50% and 25% of water flow.
Model answer: Static head = 15 m (constant); dynamic head at design = 50 − 15 = 35 m (∝ flow²). At 75%: dynamic = 35×0.75² = 19.68, total = 34.68 m. At 50%: dynamic = 35×0.5² = 8.75, total = 23.75 m. At 25%: dynamic = 35×0.25² = 2.19, total = 17.19 m.
2. The input power to a fan is 30 kW for a 2500 Nm3/hr fluid flow. The fan pulley diameter is 300 mm. If the flow is to be reduced by 15% by changing the fan pulley, what should be the diameter of the fan pulley and power input to fan?
Model answer: N2 = 0.85 N1. From N1D1 = N2D2: D2 = D1×(N1/N2) = 300/0.85 = 352 mm. Power: kW2 = (N2/N1)³×kW1 = 0.85³×30 = 18.42 kW ≈ 18.4 kW. So change fan pulley to 352 mm; fan power ≈ 18.4 kW.
Q ∝ N so N₂ = 0.85 N₁; belt ratio N₁D₁ = N₂D₂ → D₂ = 300/0.85 = 352 mm (larger fan pulley = slower fan); kW ∝ N³ → 30 × 0.85³ = 18.4 kW.
3. Explain with equation for COP_Carnot that: (a) higher COP_Carnot is achieved with higher evaporator and lower condenser temperature; (b) COP_Carnot does not account for compressor type; (c) how COP is normally used in industry.
Model answer: a) COP_Carnot = Te/(Tc − Te), where Te = evaporator temperature and Tc = condenser temperature (in Kelvin). Higher Te and lower Tc increase the ratio, hence higher COP. b) Since it is only a ratio of temperatures, it does not consider the type of compressor. c) The industry COP = Cooling effect (kW)/Power input to compressor (kW), where cooling effect is the enthalpy difference across the evaporator expressed in kW.
Confirmed vs Book-3 §4.7 - COP-Carnot = Te/(Tc - Te) with temperatures in kelvin, so raising Te or lowering Tc shrinks the denominator and raises COP; this is exactly why the book advises the highest practical chilled-water temperature and the lowest practical condensing temperature. Because it is only a temperature ratio it says nothing about the compressor, so industry uses COP = cooling effect (kW) / compressor power input (kW).
5. S-1: List five energy saving measures for air conditioning system.
Model answer: (1) Insulate all cold lines/vessels with economic insulation thickness to minimize heat gains. (2) Optimize air-conditioning volumes (false ceiling, air curtains, segregate critical areas). (3) Minimize AC loads (roof cooling/painting, efficient lighting, pre-cooling fresh air via air-to-air heat exchangers, sun film, variable volume air system, optimal thermostat setting). (4) Minimize part-load operation by matching loads to plant capacity on line and adopt variable speed drives. (5) Ensure regular maintenance, adequate chilled/cooling water flow, avoid bypass flow, frequent cleaning/descaling of heat exchangers; adopt VAR (non-CFC) where economical and continuously optimize condenser/evaporator parameters.
Confirmed vs Book-3 §4.16 - full marks for any five of the guidebook's listed measures: economic cold insulation, optimised air-conditioned volume (false ceilings, air curtains), minimised building heat load (roof cooling/painting, sun film, efficient lighting, pre-cooled fresh air), optimum thermostat setting and variable-volume air systems, and plant-side actions such as avoiding bypass flows, matching capacity to load with VSDs, cleaning condensers/evaporators and adopting VAR where economics permit.
Model answer: Lux (lx): the illuminance produced by a luminous flux of one lumen uniformly distributed over a surface area of one square metre; the SI unit of illumination, equal to one lumen per square metre. Luminous efficacy: the ratio of luminous flux emitted by a lamp to the power consumed by the lamp (lumen/Watt); it is the energy efficiency of conversion from electricity to light.
Book-3 §8.2 definitions (2.5 marks each): one lux = one lumen per square metre; luminous efficacy = lumens emitted per Watt consumed by the lamp.
7. S-6: Power plant cooling tower audit: generation 785 MW, circulation rate 107000 m3/hr, range 10.5 C, design COC 3.8. Find (a) total water consumption/hr, (b) specific water consumption (m3/MW). If COC raised to 7.0, (c) potential water savings in m3/hr and m3/MW.
Model answer: Evaporation loss = 0.00085 x 107000 x 10.5 x 1.8 = 1719 m3/hr. Blowdown = 1719/(3.8-1) = 614 m3/hr. (a) Total = 1719 + 614 = 2333 m3/hr. (b) Specific = 2333/785 = 2.97 m3/MW. At COC 7.0: blowdown = 1719/(7-1) = 286.5 m3/hr; total = 1719 + 286.5 = 2005.5 m3/hr; specific = 2005.5/785 = 2.56 m3/MW. (c) Water saving = 2333 - 2005.5 = 327.5 m3/hr; per MW = 327.5/785 = 0.417 m3/MW.
Printed solution table: total 2333 m3/hr, specific 2.97 m3/MW, savings 327.5 m3/hr and 0.417 m3/MW.
📖 §9.3 Waste heat recovery & §9.4 Energy performance assessment of DG sets
8. In a DG set, the generator is rated at 1000 kVA, 415V, 1390 A, 0.8 PF, 1500 RPM. The full load specific energy consumption of this DG set as measured by the energy auditor is 4.0 kWh per liter of fuel and air drawn by the DG set is 14 kg/kg of fuel. The energy auditor has recommended a waste heat recovery (WHR) system. Also the auditor indicated that the waste heat recovery potential is 2.6x10^5 kCal/hr at the existing engine exhaust gas temperature of 583 degC. Estimate the exhaust temperature to chimney after installation of proposed WHR system. The specific gravity of fuel oil is 0.86 and specific heat of flue gas is 0.25 kCal/kg degC.
Model answer: Solution:
1. Rated kVA of diesel generator (given) = 1000
2. Rated kW at 0.8 PF = 1000 x 0.8 = 800 kW ..... (0.5 mark)
3. Specific fuel consumption (given) = 4 kWh/litre
4. Specific gravity of fuel oil (given) = 0.86
5. Oil consumption at full load = (800 x 0.86)/4 = 172 kg/hr ..... (1 mark)
6. Air supplied per kg of fuel (given) = 14 kg
7. Mass of flue gas = 14 + 1 = 15 kg per kg of fuel
8. Mass of flue gas per hour = 15 x 172 = 2580 kg/hr ..... (1 mark)
9. Waste heat recovery potential (given) = 2,60,000 kCal/hr
10. Delta T across the WHR system = Heat (kCal/hr)/(mass of flue gas kg/hr x specific heat kCal/kg degC) = 260000/(2580 x 0.25) = 403 degC ..... (1.5 marks)
11. Flue gas temperature before WHR system (given) = 583 degC
12. Exit flue gas temperature to chimney after WHR system = 583 - 403 = 180 degC ..... (1 mark)
Chain the mass balance before touching the heat. kW = 1000 kVA × 0.8 = 800 kW; fuel = 800/4 = 200 litres/hr × 0.86 = 172 kg/hr; flue gas = (14 + 1) × 172 = 2580 kg/hr — the +1 is the fuel itself and is the step most candidates forget. Then ΔT = 260,000/(2580 × 0.25) = 403 °C, so the chimney temperature is 583 − 403 = 180 °C. Sanity-check the result: the book warns against dropping below roughly 180 °C because of cold-end corrosion.
9. During an energy audit of a power plant cooling tower, the following observations were made. Power plant generation = 785 MW; Circulation rate = 107000 m3/hr; Cooling tower range = 10.5 degC; Power plant design COC value = 3.8. As an auditor find out a) The total water consumption per hour, b) Specific water consumption in m3/MW generation. The plant is pursuing an up-gradation treatment plan to increase COC to 7.0. c) What would be the potential water savings in m3/hr and m3/MW generation?
Model answer: Answer:
1. Evaporation loss = 0.00085 x circulation rate (m3/hr) x (CT range in degC) x 1.8 = 0.00085 x 107000 x 10.5 x 1.8 = 1719 m3/hr ..... (0.5 mark)
2. Blow-down loss = Evaporation loss/(COC - 1) = 1719/(3.8 - 1) = 614 m3/hr ..... (0.5 mark)
3. Total as-run hourly consumption = 1719 + 614 = 2333 m3/hr ..... (0.5 mark)
4. Specific water consumption = 2333/785 = 2.97 m3/MW ..... (0.5 mark)
5. Blow-down at improved COC of 7.0 = 1719/(7 - 1) = 286.5 m3/hr ..... (0.5 mark)
6. Total water consumption at improved COC = 1719 + 286.5 = 2005.5 m3/hr ..... (0.5 mark)
7. Specific water consumption at improved COC = 2005.5/785 = 2.56 m3/MW ..... (0.5 mark)
8. Total water saving per hour = 2333 - 2005.5 = 327.5 m3/hr ..... (0.5 mark)
9. Water saving per MW generation = 327.5/785 = 0.417 m3/MW ..... (1 mark)
Evaporation = 0.00085 × 1.8 × circulation × range = 0.00085 × 1.8 × 107,000 × 10.5 = 1719 m³/hr and it does NOT change when COC changes — only blowdown does. Blowdown = 1719/(3.8 − 1) = 614 m³/hr, total 2333 m³/hr, specific 2333/785 = 2.97 m³/MW. At COC 7.0 blowdown falls to 1719/6 = 286.5, total 2005.5 m³/hr, so the saving is 327.5 m³/hr or 0.417 m³/MW. Raising COC saves blowdown water only — quoting a saving in evaporation is wrong.
📖 §1.4 Power factor improvement (capacitor sizing)
10. The following single line diagram depicts the location of a 100 kW heater load and a 200 kW motor (which is 200 metres away from the 415V, LT bus). The main incoming line power factor of the system is 0.85 lag. Calculate the rating of capacitors to improve PF of main incoming line to 0.9 lag. [refers to a figure in the original paper]
Model answer: Answer:
Total inductive load requiring PF compensation = 200 kW (since the other 100 kW heater is a resistive load and already operates at unity power factor). ..... (1 mark)
Operating PF cos(phi1) = 0.85 lag; Desired PF cos(phi2) = 0.90 lag
kVAr required = kW x [tan(cos^-1 phi1) - tan(cos^-1 phi2)] ..... (1 mark)
= 200 x [tan(cos^-1 0.85) - tan(cos^-1 0.90)]
= 200 x [tan(31.78 deg) - tan(25.84 deg)]
= 200 x (0.619 - 0.484)
= 200 x 0.135
= 27 kVAr ..... (3 marks)
Only the INDUCTIVE kW gets compensated: the 100 kW heater is resistive and already at unity PF, so it contributes no kVAr and must be excluded. Compensate the 200 kW motor alone.
kVAr = kW[tan(cos^-1 PF1) - tan(cos^-1 PF2)] = 200 x (tan 31.78deg - tan 25.84deg) = 200 x (0.619 - 0.484) = 27 kVAr.
Including the heater's 100 kW would give 40.5 kVAr and lose the marks. Memory hook: capacitors correct only what is magnetising, and a heater magnetises nothing.
1. Compare the performance of a centrifugal chiller with a vapour absorption chiller (VAM) from the given data (chilled & condenser water flows, inlet/outlet temps, pump and CT fan power). Centrifugal compressor 205 kW; VAM steam 1620 kg/Hr. Calculate i) refrigeration load TR, ii) condenser heat load TR, iii) auxiliary power, iv) operating cost (electricity Rs 4/kWh, steam Rs 0.45/kg).
Model answer: (i) Refrigeration load = flow x 1000 x 1 x deltaT/3024: Centrifugal = 192000 x (13-7.8)/3024 = 330.16 TR; VAM = 183000 x (14.5-9.2)/3024 = 320.73 TR. (ii) Condenser heat load = condenser flow x 1000 x deltaT/3024: Centrifugal = 245000 x (36.2-28)/3024 = 664.35 TR; VAM = 360000 x (40.7-32)/3024 = 1035.71 TR. (iii) Auxiliary power = pumps + CT fan: Centrifugal = 32+38+9 = 79 kW; VAM = 31+52+22 = 105 kW. VAM auxiliary is higher because its condenser heat rejection is much larger for similar cooling load. (iv) Operating cost/hr: Centrifugal total = 79 + 205 = 284 kW x Rs.4 = Rs.1136. VAM = 105 kW x Rs.4 = Rs.420 plus steam 1620 x Rs.0.45 = Rs.729 -> total Rs.1149/hr.
Confirmed vs Book-3 §4.7 - refrigeration TR = flow x Cp x deltaT/3024 on the chilled-water side and the same formula on the condenser side for heat rejection. Centrifugal 330.16 TR vs VAM 320.73 TR for similar duty, but the VAM rejects 1035.71 TR against 664.35 TR, because a VAR machine must also reject its heat input (COP ~0.65-0.70). That is why VAM condenser pumps and cooling-tower fans are bigger (105 kW vs 79 kW), and the hourly cost is close (Rs.1149 vs Rs.1136) despite free-looking steam.
4. L-2: (a) Calculate the ventilation rate for an engine room 20 m L x 10.5 m W x 15 m H if recommended ACH is 20. (b) Air at 25,200 m3/hr and 1.2 kg/m3 density flows into an AHU; enthalpy difference inlet-outlet 2.38 kcal/kg; motor draws 22 kW at 90% efficiency. Find kW/TR (1 cal = 4.183).
Model answer: (a) Ventilation rate = L x H x W x ACH = 20 x 15 x 10.5 x 20 = 63,000 m3/hr. (b) Heat = Q x density x deltaH = 25200 x 1.2 x 2.38 = 71,971 kcal/hr; TR = 71971/3024 = 23.8 TR; compressor power = 22 x 0.9 = 19.8 kW; kW/TR = 19.8/23.8 = 0.83.
Confirmed vs Book-3 §4.12 and §4.7 - ventilation rate = L x B x H x ACH = 20 x 10.5 x 15 x 20 = 63,000 m3/hr. Air-side load = Q x rho x delta-h/3024 = 25,200 x 1.2 x 2.38/3024 = 23.8 TR, and compressor shaft power = 22 x 0.9 = 19.8 kW, giving 0.83 kW/TR. The common slip is to use the 22 kW motor input directly, which overstates kW/TR by about 11%.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); §6.6 Flow control strategies — pumps in parallel switched to meet demand
5. L-3: Three identical cooling water pumps in parallel (two running, one standby), all combinations give 3.4 kg/cm2(a) discharge. Flow: pumps 1&2 = 545, 2&3 = 535, 3&1 = 550 m3/hr. Motor power 33/31.5/32.5 kW; motor eff 92%/92%/91.5%; suction 3 m below pump centre line. Find (i) individual pump efficiencies, (ii) specific energy consumption kWh/m3, (iii) best operating combination.
Model answer: Solving X+Y=545, Y+Z=535, X+Z=550 gives X=280, Y=265, Z=270 m3/hr. Discharge head = 3.4 kg/cm2(a) = 2.4 kg/cm2(g) = 24 m; suction head = -3 m; total head = 27 m. Liquid kW = flow(m3/s) x 27 x 1000 x 9.81/1000: Pump1 = 20.60, Pump2 = 20.22, Pump3 = 19.87 kW. Pump input power = motor power x motor eff: 30.36, 28.98, 29.74 kW. Pump efficiency = liquid/input: 67.9%, 69.8%, 66.8%. SEC = motor power/flow: 0.118, 0.119, 0.120 kWh/m3. Best operating combination = pumps 1 & 2.
📖 §7.2 Factors Affecting Performance – Heat Load (heat-rejection table); Range = Heat load / Water circulation rate
6. L-5: An 18 MW cogeneration plant: max condenser load 7 MW, extraction steam 57 TPH for process and VAM. Condenser heat load 550 kcal/kg of steam, steam rate 5 kg/kW for condenser power. VAM heat load 127 kcal/min/TR, VAM capacity 1100 TR. Estimate cooling tower heat load (kcal/hr); for 6 C range and 5 C approach calculate the cooling water flow.
Model answer: Steam for condenser power = 7000 x 5 = 35,000 kg/hr. Condenser heat load = 35000 x 550 = 1,92,50,000 kcal/hr. VAM heat load = 1100 x 127 x 60 = 83,82,000 kcal/hr. Total CT heat load = 1,92,50,000 + 83,82,000 = 2,76,32,000 kcal/hr. Cooling water flow = total heat load/range = 27632000/6 = 46,05,333 litres/hr = 4605 m3/hr.
Printed solution: total heat load 2,76,32,000 kcal/hr; water flow 4605 m3/hr.
📖 §4.7 Performance assessment of refrigeration plants (chiller vs VAM comparison)
7. Compare the performance of centrifugal chiller with vapour absorption chiller using the data given below: (1) Chilled water flow (m3/h): Centrifugal Chiller 192, VAM 183; (2) Condenser water flow (m3/h): 245, 360; (3) Chiller inlet water temperature (degC): 13, 14.5; (4) Condenser water inlet temperature (degC): 28, 32; (5) Chiller outlet water temperature (degC): 7.8, 9.2; (6) Condenser water outlet temperature (degC): 36.2, 40.7; (7) Chilled water pump consumption (kW): 32, 31; (8) Condenser water pump consumption (kW): 38, 52; (9) Cooling tower fan consumption (kW): 9, 22. If the compressor of centrifugal chiller consumes 205 kW, the steam consumption for VAM is 1620 kg/Hr. Calculate the following: i) Refrigeration load delivered (TR) for both systems? ii) Condenser Heat load (TR) for both systems? iii) Compare auxiliary power consumption for both systems, give reason? iv) If electricity cost is Rs.4.0/kWh and steam cost is Rs.0.45/kg compare the operating cost for both systems.
Model answer: Solution: Compression Chiller vs. VAM
1. Refrigeration load delivered (TR) = mass of chilled water flow x specific heat x delta T of chilled water = flow (m3/hr) x 1000 kg/m3 x 1 kcal/kg degC x (inlet temp - outlet temp)/3024
Centrifugal chiller = 192 x 1000 x (13 - 7.8)/3024 = 330.16 TR
VAM = 183 x 1000 x (14.5 - 9.2)/3024 = 320.73 TR ..... (2 marks)
2. Condenser heat load (TR) = mass of condenser water flow x specific heat x delta T of condenser water = flow (m3/hr) x 1000 kg/m3 x 1 kcal/kg degC x (outlet temp - inlet temp)/3024
Centrifugal chiller = 245 x 1000 x (36.2 - 28)/3024 = 664.35 TR
VAM = 360 x 1000 x (40.7 - 32)/3024 = 1035.71 TR ..... (2 marks)
3. Auxiliary power consumption (kW) = chilled water pump + condenser water pump + cooling tower fan
Centrifugal chiller = 32 + 38 + 9 = 79 kW
VAM = 31 + 52 + 22 = 105 kW
The auxiliary power consumption in case of the VAM system is higher because heat rejection in the VAM condenser is comparatively higher than the centrifugal chiller with approximately similar cooling load. ..... (2 marks)
4. Total energy consumption:
Centrifugal chiller = 284 kW (auxiliary power of 79 kW plus chiller consumption of 205 kW)
VAM = auxiliary power of 105 kW and steam consumption of 1620 kg/hr ..... (2 marks)
5. Operating energy cost per hour of operation:
Centrifugal chiller = 284 x 4 = Rs. 1136/-
VAM = (105 x 4 = Rs. 420/-) plus (1620 x 0.45 = Rs. 729/-) = Rs. 1149/- ..... (2 marks)
Both loads come from the same water-side heat balance: TR = flow (m3/hr) x 1000 x 1 x deltaT/3024. Chilled side: 192 x 1000 x 5.2/3024 = 330.2 TR (centrifugal) and 183 x 1000 x 5.3/3024 = 320.7 TR (VAM). Condenser side: 245 x 1000 x 8.2/3024 = 664.4 TR and 360 x 1000 x 8.7/3024 = 1035.7 TR.
The whole point of the question is in those condenser figures: a VAM must reject the evaporator load PLUS the driving heat input, so for a similar cooling duty it rejects far more heat - hence bigger cooling towers, bigger condenser pumps and higher auxiliary power (105 kW against 79 kW).
Cost comparison must put both on the same footing: centrifugal (205 + 79) x Rs.4 = Rs.1136/hr; VAM 105 x 4 + 1620 x 0.45 = 420 + 729 = Rs.1149/hr. Remember to convert the delta-T for the condenser as outlet MINUS inlet, and for the chiller as inlet MINUS outlet.
📖 §4.12 Ventilation systems (ACH) + §4.7 Performance assessment (kW/TR)
8. a) Calculate the ventilation rate for an engine room of 20 m length, 10.5 m width and 15 m height; if the recommended Air Changes per Hour (ACH) is 20. b) Air at 25,200 m3/hr and at 1.2 kg/m3 density is flowing into an air handling unit of an inspection room. The enthalpy difference between the inlet and outlet air is 2.38 kcal/kg. If the motor draws 22 kW with an efficiency of 90%, find out the kW/TR of the refrigeration system. (1 cal = 4.183)
Model answer: Solution:
a) Ventilation rate:
Room length = 20 m; Room height = 15 m; Room width = 10.5 m; Air changes per hour (ACH) = 20
Ventilation rate (m3/hr) = length x height x width x ACH = 20 x 15 x 10.5 x 20 = 63,000 m3/hr ..... (5 marks)
b) Refrigeration load:
Heat load = Q x density x (h2 - h1) = 25,200 x 1.2 x 2.38 = 71,971 kcal/hr ..... (2 marks)
TR = 71,971/3024 = 23.8 TR ..... (1 mark)
Power input to the compressor = 22 x 0.9 = 19.8 kW ..... (1 mark)
kW/TR = 19.8/23.8 = 0.83 ..... (1 mark)
Ventilation rate = room volume x air changes per hour = (20 x 10.5 x 15) x 20 = 3150 x 20 = 63,000 m3/hr. ACH is per HOUR, so no further time conversion is needed.
For part (b): heat load = Q x rho x delta-h = 25,200 x 1.2 x 2.38 = 71,971 kcal/hr, and TR = 71,971/3024 = 23.8 TR. Compressor shaft power = 22 x 0.9 = 19.8 kW, so kW/TR = 19.8/23.8 = 0.83.
Two guards: the enthalpy difference is already in kcal/kg here so the 4.187 conversion is not needed, and kW/TR uses the SHAFT power, not the motor input - a lower kW/TR is better, with 0.7-0.9 typical for a good centrifugal plant.
9. In a dairy plant 3 numbers of cooling water pumps, identical in characteristics are installed in parallel to supply cooling. During normal operation two of the pumps are operational while one pump is on standby. All pump combinations develop a discharge pressure of 3.4 kg/cm2 (a). The installed water flow meter at the common header during an energy audit reads the following: Pump No 1 & 2 = 545 m3/hr; Pump No 2 & 3 = 535 m3/hr; Pump No 3 & 1 = 550 m3/hr. The power drawn by motors of cooling water pump 1, 2 & 3 are 33 kW, 31.5 kW & 32.5 kW respectively. While the operating motor efficiency for pump no. 1 & 2 is 92% the motor efficiency for pump no. 3 is 91.5%. If the water level in suction of all pumps is 3 meter below pump central line. Calculate the following: i) Individual pump efficiencies ii) Specific energy consumption (kWh/m3) iii) Which is the best operating pump combination
Model answer: Solution:
Let the flows of pumps 1, 2 and 3 be X, Y and Z respectively.
X + Y = 545 ---(1); Y + Z = 535 ---(2); X + Z = 550 ---(3)
Subtracting (2) from (1): X - Z = 10 ---(4)
Adding (3) and (4): 2X = 560, so X = 280
Putting X in (1) and (2): Y = 265 and Z = 270
Therefore individual pump flow rates are 280 m3/hr, 265 m3/hr and 270 m3/hr respectively. ..... (3 marks)
Pump-wise calculation (Pump 1 / Pump 2 / Pump 3):
A) Flow rate (m3/hr) (calculated): 280 / 265 / 270
B) Discharge head = 3.4 kg/cm2 (a) = 2.4 kg/cm2 (g) = 24 m: 24 / 24 / 24
C) Suction head (m) (given): -3 / -3 / -3
D) Total head = discharge head - suction head = (B - C): 27 / 27 / 27 ..... (1 mark)
E) Liquid (hydraulic) kW = flow (m3/s) x total head (m) x density (1000 kg/m3) x 9.81/1000: 20.60 / 20.22 / 19.87 ..... (2 marks)
F) Power drawn by motor kW (given): 33 / 31.5 / 32.5
G) Motor efficiency % (given): 92.0% / 92.0% / 91.5%
H) Pump input (shaft) power kW = F x G: 30.36 / 28.98 / 29.74 ..... (1 mark)
I) Pump efficiency % = E/H: 67.9% / 69.8% / 66.8% ..... (1 mark)
J) Specific energy consumption (kWh/m3) = F/A: 0.118 / 0.119 / 0.120 ..... (1 mark)
Pump No. 1 & 2 are the best performing operating combination. ..... (1 mark)
Note: The total head has been calculated by subtracting the suction gauge pressure from the discharge gauge pressure. The candidates can also calculate total head as the difference of absolute pressures as follows: Discharge head = 3.4 kg/cm2 (a); Suction head = 1 - 0.3 = 0.7 kg/cm2; Total head developed = 3.4 - 0.7 = 2.7 kg/cm2 = 27 m.
Get the individual flows by simultaneous equations: X+Y=545, Y+Z=535, X+Z=550 → X=280, Y=265, Z=270 m³/hr. Then head is common to all three: 3.4 kg/cm²(a) = 2.4 kg/cm²(g) = 24 m, plus 3 m of suction lift = 27 m. Hydraulic kW = Q(m³/s) × 27 × 9.81, shaft = measured kW × motor η, so η(pump) = 67.9 / 69.8 / 66.8%. Choose the best COMBINATION on specific energy consumption (kW ÷ m³/hr), not on single-pump efficiency — pumps 1 and 2 win at 0.118 kWh/m³. Subtracting a gauge from an absolute pressure is the classic slip; keep both in the same reference.
📖 §7.2 Cooling tower performance & §5.6 Fan performance assessment
10. a) In a chemical industry, cooling water of 9000 m3/hr and 6000 m3/hr from two independent heat exchangers with temperature of 41 degC and 52 degC respectively are fed to one cooling tower after proper mixing at top basin. If measured heat rejection by the cooling tower is 45,000 TR, calculate effectiveness and evaporation loss of the cooling tower at 31 degC WBT. b) In an air conditioning duct 0.5 m x 0.5 m, the average velocity of air measured by vane anemometer is 28 m/s. The static pressure at suction of the fan is -20 mmWC and at the discharge is 30 mmWC. A three phase induction motor draws 10.8 A at 415 V with a power factor of 0.9. Find out efficiency of the fan if motor efficiency = 88% (neglect density correction)
Model answer: Solution:
a) Cooling tower:
1. Flow rates (given): Stream 1 = 9000 m3/hr, Stream 2 = 6000 m3/hr
2. Temperatures (given): Stream 1 = 41 degC, Stream 2 = 52 degC
3. Mixed flow rate = 9000 + 6000 = 15,000 m3/hr
4. Mixed hot water temperature = [(Flow1 x Temp1) + (Flow2 x Temp2)]/(Flow1 + Flow2) = [(9000 x 41) + (6000 x 52)]/15000 = 45.4 degC ..... (1 mark)
5. Heat rejection (given) = 45,000 TR
6. Range of cooling tower = (Heat rejection TR x 3024)/(Flow m3/hr x 1000) = (45000 x 3024)/(15000 x 1000) = 9.072 degC ..... (1 mark)
7. WBT (given) = 31 degC
8. Cold water temperature = mixed hot water temperature - range = 45.4 - 9.072 = 36.328 degC ..... (0.5 mark)
9. Approach = cold water temperature - WBT = 36.328 - 31 = 5.328 degC ..... (0.5 mark)
10. Effectiveness = Range/(Range + Approach) = 9.072/(9.072 + 5.328) = 63% ..... (1 mark)
11. Evaporation loss (m3/hr) = 0.00085 x 1.8 x mixed flow m3/hr x range = 0.00085 x 1.8 x 15000 x 9.072 = 208.2 m3/hr ..... (1 mark)
b) Fan efficiency:
1. Area of the duct = 0.5 x 0.5 = 0.25 m2
2. Average velocity (given) = 28 m/s
3. Air flow = 0.25 x 28 = 7 m3/s ..... (1 mark)
4. Suction static pressure (given) = -20 mmWC
5. Discharge static pressure (given) = 30 mmWC
6. Power drawn by the motor = 1.732 x 415 x 10.8 x 0.9/1000 = 6.99 kW ..... (1 mark)
7. Air power kW = flow (m3/s) x (discharge pressure - suction pressure) mmWC/102 = 7 x (30 - (-20))/102 = 7 x 50/102 = 3.43 kW ..... (1 mark)
8. Power to fan shaft = motor drawn power x motor efficiency of 88% = 6.99 x 0.88 = 6.15 kW ..... (1 mark)
9. Fan static efficiency = air power x 100/shaft input = 3.43 x 100/6.15 = 55.76% ..... (1 mark)
(a) Mix the two streams on a flow-weighted basis first: (9000×41 + 6000×52)/15000 = 45.4 °C — a plain average of 46.5 °C is wrong. Range from the duty: 45,000 TR × 3024/(15,000 × 1000) = 9.07 °C, so cold water = 36.33 °C, approach = 5.33 °C and effectiveness = 9.07/(9.07+5.33) = 63%. Evaporation = 0.00085 × 1.8 × 15,000 × 9.07 = 208 m³/hr. (b) Fan total pressure = 30 − (−20) = 50 mmWC; SUBTRACT the negative suction pressure so it adds. Air power = 7 × 50/102 = 3.43 kW, shaft = 6.99 × 0.88 = 6.15 kW, η = 55.8%.
📖 §7.2 Cooling tower performance (cooling capacity and water flow)
11. One of the process industries has installed 18 MW cogeneration plant. The Cogeneration plant maximum condenser load is 7 MW and the extraction steam of 57 TPH is used for process and also for vapour absorption machine. The condenser heat load is 550 Kcal/kg of steam and the steam rate is 5 kg/kW for condenser power. The heat load of VAM is 127 Kcal/min/TR and the capacity of VAM is 1100 TR. Estimate cooling tower heat load in Kcal/hr. If the tower is designed for 6 degC range, calculate the water flow in cooling tower. The design approach temperature of the CT is 5 degC.
Model answer: Answer:
Condenser load = 7 MW
Steam rate for condenser = 5 kg/kW
Total steam required for condenser power = 7000 x 5 = 35,000 kg/hr ..... (2 marks)
Condenser heat load = 35,000 x 550 = 1,92,50,000 Kcal/hr ..... (2 marks)
Heat load of VAM = 1100 x 127 x 60 = 83,82,000 Kcal/hr ..... (2 marks)
Total cooling tower heat load = 1,92,50,000 + 83,82,000 = 2,76,32,000 Kcal/hr ..... (2 marks)
Range of tower = 6 degC
Cooling water flow required = 2,76,32,000/6 = 46,05,333 litres/hr or 4605 m3/hr ..... (2 marks)
Build the heat load from both sources: condenser = 7000 kW × 5 kg/kW × 550 kcal/kg = 19.25 million kcal/hr; VAM = 1100 TR × 127 kcal/min/TR × 60 = 8.382 million kcal/hr; total 27.63 million kcal/hr. Then flow = heat load /(range × 1000) = 27,632,000/6 = 4.605 million litres/hr = 4605 m³/hr. Two traps: the VAM figure is per MINUTE so multiply by 60, and the 18 MW plant rating and 5 °C approach are distractors — only the 7 MW condenser load and the range enter the arithmetic.
📖 §1.4 Power factor improvement & benefits (disadvantages of low PF, PF penalty/incentive)
12. a) List five disadvantages of low Power Factor? b) An industry is losing money as penalty on account of maintaining a poor power factor of 0.88. The power utility has specified a minimum power factor of 0.9 to avoid penalty. The penalty on energy cost is 1% for every 0.01 power factor less than the minimum prescribed. Also an incentive on energy cost is available @ 1.5% for every 0.01 improvement above 0.95. If the monthly energy bill of the industry is Rs 6 lakhs, calculate the annual cost saving potential if power factor is improved to unity from the current level.
Model answer: Answer:
a) Disadvantages of low power factor (any five - 1 mark each):
1. Large line losses (copper losses)
2. Large kVA rating and size of electrical equipments
3. Greater conductor size and cost
4. Poor voltage regulation and large voltage drop
5. Low efficiency
6. Penalty from electric power supply company on low power factor
b) Minimum PF to be maintained to avoid penalty = 0.9
Present penalty = 1.00% (on energy bill) for every 0.01 PF
For 0.02 PF (0.90 - 0.88) = 1.00 x 2 = 2.0% ..... (1 mark)
Incentives (from 0.95 to 1.00, i.e. 5 steps of 0.01) = 1.5 x 5 = 7.5% ..... (1 mark)
Energy saving potential = 2.0 + 7.5 = 9.5%
Cost reduction potential per month = Rs. 6 lakh x 9.5% = Rs. 57,000 ..... (2 marks)
Annual cost reduction = 57,000 x 12 = Rs. 6,84,000 ..... (1 mark)
Count the 0.01 steps carefully in two separate stretches: penalty 0.88 -> 0.90 is 2 steps at 1% = 2%; incentive 0.95 -> 1.00 is 5 steps at 1.5% = 7.5%. Total benefit = 9.5% of the energy bill.
Rs.6,00,000 x 9.5% = Rs.57,000/month -> Rs.6,84,000/year.
The band between 0.90 and 0.95 earns nothing (no penalty, no incentive) - counting those 5 steps as well is the classic mistake here. Low-PF penalties come on top of larger cables, bigger kVA equipment, poor voltage regulation and higher line losses.