Energy Efficiency in Electrical Utilities Available here with full solutions — 43 questions recovered from the 2013 exam:
Objective (1 mark)
29 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) — 29
📖 §1.1 Introduction to electric power supply systems (cascade efficiency)
1. The efficiencies of a power plant and transmission systems are 40%, and 97% respectively. The percentage loss of the distribution system of the same network is 23%. The cascade efficiency of generation, transmission and distribution system is given by
8.92 %
29.87%
40 %
23%
Answer: B) 29.87% (0.40 x 0.97 x 0.77 = 0.2987)
Cascade efficiency = product of the individual stage efficiencies, never their sum or average: 0.40 x 0.97 x 0.77 = 0.2987 = 29.87%.
The trap is the 23%: it is a LOSS, so the distribution efficiency to use is (100 - 23) = 77%, i.e. 0.77. Feeding 0.23 into the product gives 8.92%, which is why that distractor is there.
Memory hook: efficiencies multiply, losses subtract first.
📖 §1.4 Power factor improvement & benefits (kVA reduction)
2. If the maximum demand is 3500 kVA at 0.88 p.f., the maximum demand will reduce by ______ kVA if PF is improved to 0.98 :
3143
357
3897
maximum demand will not reduce
Answer: B) 357 kVA (kW = 3500 x 0.88 = 3080; new kVA = 3080/0.98 = 3143; reduction = 3500 - 3143 = 357)
Improving PF does not change the kW the load needs; it only shrinks the kVA. Hold kW constant and re-divide: kW = 3500 x 0.88 = 3080 kW; new kVA = 3080 / 0.98 = 3143 kVA; saving = 3500 - 3143 = 357 kVA.
Common mark-losers: multiplying by 0.98 instead of dividing, or answering 3143 (the new demand) when the question asks for the REDUCTION.
Shortcut: new kVA = old kVA x (PF_old / PF_new) = 3500 x 0.88/0.98 = 3143.
3. The performance of rewinding of an induction motor can be assessed by which of the following factors?
no load current
stator resistance per phase
load current
both no load current and stator resistance per phase
Answer: D) both no load current and stator resistance per phase
The book's test for rewind quality is a BEFORE-and-AFTER comparison of two measurable quantities: no-load current / no-load loss (which rises if the core laminations were heat-damaged during burn-out, i.e. higher iron loss) and stator resistance per phase (which rises if thinner conductor was used, i.e. higher copper loss).
Because the two indicators point at two different loss mechanisms - core loss and copper loss - you need both, so 'both' is the only complete answer.
Book figure: a careless rewind typically costs 1-5% efficiency, and the damage compounds with each successive rewind.
📖 §2.3 Motor characteristics (synchronous speed and slip)
4. For a synchronous speed of 1500 rpm, at a given mains frequency of 50 Hz, the induction motor will have _________ number of poles.
8
6
4
2
Answer: C) 4 poles (P = 120 x 50 / 1500 = 4)
Ns = 120f/P, so P = 120f/Ns = (120 x 50)/1500 = 4 poles.
Learn the 50 Hz ladder by heart - 2 poles = 3000, 4 = 1500, 6 = 1000, 8 = 750 rpm - and remember these are SYNCHRONOUS speeds; the nameplate full-load speed is always a little lower (e.g. 1440-1480 rpm on a 4-pole).
The common slip is writing Ns = 120P/f, which inverts the formula.
📖 §2.4 Motor efficiency (percentage loading by input-power method)
5. A 7.5 kW, 415 V, 14.5 A, 1460 RPM rated 3 phase induction motor with full load efficiency of 90%, draws 9.1 A and 4.6 kW of input power. The percentage loading of the motor is about
Loading by the kW method compares measured input with RATED INPUT, not with rated output: rated input = rated output/full-load efficiency = 7.5/0.90 = 8.33 kW.
% load = 4.6/8.33 x 100 = 55.2%.
Dividing 4.6 by the 7.5 kW output gives 61.3% - that is exactly the distractor at option (b). Always convert the shaft rating to an input rating first by dividing by the efficiency. The 9.1 A reading is not needed here.
📖 §2.4 Motor efficiency (slip, rotor input and rotor copper loss)
6. The power input to a rotor of three phase induction motor is 42.3 kW. If the induction motor is operating at a slip of 1.30 % the total mechanical power developed will be :
42.3 kW
41.75 kW
5.48 kW
47.79 kW
Answer: B) 41.75 kW (mechanical power = (1 - s) x rotor input = 0.987 x 42.3 = 41.75 kW)
The rotor power split: rotor copper loss = s x rotor input, and mechanical power developed = (1 - s) x rotor input.
Working: (1 - 0.013) x 42.3 = 0.987 x 42.3 = 41.75 kW; the 0.55 kW difference is the rotor I^2R loss.
Marks are lost by using s as 1.30 instead of 0.0130 - convert the percentage to a fraction first. Memory hook: slip is the fraction of rotor input burnt in the rotor bars, so low slip = efficient rotor.
📖 §2.7 PF capacitors at motor terminals (Table 2.5 capacitor ratings)
7. The rating of the p.f correction capacitors at motor terminals for a 37 kW, 2 poles induction motor will be ______ in comparison to the same sized induction motor of 6 poles
more
less
same
sometime less or more
Answer: B) less (a 2-pole motor has a lower no-load magnetizing kVAr than a 6-pole motor of the same rating)
Book-3 Table 2.5 lists capacitor kVAr against motor SPEED: for the same HP the required kVAr rises steadily as speed falls (3000 -> 1500 -> 1000 -> 750 rpm). A 2-pole motor is 3000 rpm, a 6-pole motor is 1000 rpm, so the 2-pole machine needs the smaller capacitor.
Reason: a slow, many-pole motor has a bigger air gap and a larger magnetising (no-load) kVAr for the same kW output.
Sizing rule to memorise: capacitor kVAr must not exceed 90% of the motor's no-load kVAr, otherwise self-excitation over-voltage can burn the motor out.
📖 §2.4 Motor efficiency (slip method of load estimation)
8. Which parameters need to be measured to assess the percentage loading of a motor by slip method neglecting voltage correction?
motor speed
synchronous speed
operating motor speed and frequency
operating current
Answer: C) operating motor speed and frequency
Slip method: % load = (Ns - N)/(Ns - N_full-load) x 100, where Ns = 120f/P. To evaluate it you need the OPERATING speed (a tachometer or stroboscope reading) and the supply FREQUENCY, because Ns itself depends on f.
Supply frequency matters: a fall of even 1 Hz shifts Ns by 30 rpm on a 4-pole motor and can badly distort the estimated load.
Synchronous speed alone (option b) is not measured, it is calculated - which is precisely why frequency has to be measured.
📖 §3.5 Efficient operation of compressed air systems (cool air intake)
9. Which of the following is correct for air compressors?
for every 5.5 °C drop in the inlet air temperature, the increase in energy consumption is by 2%
for every 4 °C rise in the inlet air temperature, the increase in energy consumption is by 1%
for every 4 °C rise in the inlet air temperature, the decrease in energy consumption is by 1%
the energy consumption remains same irrespective of inlet air temperature
Answer: B) for every 4 °C rise in the inlet air temperature, the increase in energy consumption is by 1%
The book's thumb rule is printed as 'every 4 degC DROP in inlet air temperature results in 1% LOWER energy consumption' - which is the same statement read the other way round: every 4 degC RISE costs 1% more power.
Physics behind it: hotter air is less dense, so the compressor must handle a larger volume to deliver the same mass of air.
Practical action: draw suction air from a cool, shaded, well-ventilated outside point rather than from inside the hot compressor room. Do not confuse this rule with the separate 5.5 degC inter-stage figure in Table 3.7.
📖 §4.7 Performance assessment of refrigeration plants (COP and TR)
10. The COP of a vapour compression refrigeration system is 3.1. If the motor draws power of 9.3 kW at an operating efficiency of 88%, the tonnage of refrigeration system is about
8.2
9.3
7.2
none of the above
Answer: C) 7.2 TR (shaft power = 9.3 x 0.88 = 8.184 kW; refrigeration effect = 3.1 x 8.184 = 25.37 kW; TR = 25.37/3.516 = 7.2)
COP is defined on the power delivered to the COMPRESSOR SHAFT, so strip the motor loss first: shaft kW = 9.3 x 0.88 = 8.184 kW.
Refrigeration effect = COP x shaft kW = 3.1 x 8.184 = 25.37 kW; TR = 25.37/3.517 = 7.2 TR (1 TR = 3.517 kW = 3024 kcal/hr).
Skipping the 88% motor efficiency gives 3.1 x 9.3/3.517 = 8.2 TR, which is exactly option (a). Constant to memorise: 1 TR = 3.5 kW.
📖 §4.7 Performance assessment of refrigeration plants (TR from chilled water)
11. Chilled water enters an evaporator at 10 °C and leaves at 6 °C. The flow rate of chilled water was measured as 200 m3/hr. The tons of refrigeration capacity is
265
200
661
2.65
Answer: A) 265 TR (heat load = 200 x 1000 x 1 x (10 - 6) = 8,00,000 kCal/hr; TR = 8,00,000/3024 = 264.5 ≈ 265)
TR = flow (m3/hr) x 1000 (kg/m3) x Cp (1 kcal/kg degC) x deltaT (degC) / 3024.
Working: 200 x 1000 x 1 x (10 - 6)/3024 = 8,00,000/3024 = 264.5 ~ 265 TR.
Two habits that protect the mark: 1 TR = 3024 kcal/hr (use 3024 whenever you are in kcal, and 3.517 whenever you are in kW), and deltaT is the chilled-water temperature DROP across the evaporator, so the 200 m3/hr flow alone tells you nothing.
📖 §4.2 Psychrometrics and air-conditioning processes
12. In an air conditioning system analysis which one temperature is sufficient to determine the enthalpy of air?
dry bulb temperature
wet bulb temperature
ambient temperature
none of the above
Answer: B) wet bulb temperature (lines of constant WBT very nearly coincide with lines of constant enthalpy)
On the psychrometric chart the constant wet-bulb lines very nearly coincide with the constant-enthalpy lines, so WBT alone effectively fixes the enthalpy of moist air.
DBT alone cannot: at one dry-bulb temperature the air can hold anything from zero moisture to saturation, so its enthalpy is undefined until a second property is known.
Memory hook: WBT carries the LATENT information (it responds to moisture), DBT carries only the sensible. That is also why evaporative cooling follows a constant-WBT line.
independent of the density of the liquid being pumped
directly proportional to the density of the liquid being pumped
inversely proportional to the density of the liquid being pumped
proportional to the square of the density of the liquid being pumped
Answer: A) independent of the density of the liquid being pumped
A centrifugal pump imparts velocity, so it generates the same HEAD in metres of liquid column whatever the liquid. Density only enters when head is converted to pressure (P = ρgh) or to power (Ph = Q × H × ρ × g/1000). So pumping brine to the same height needs the same head but more kW. Hook: 'same metres, more kilowatts'.
📖 §4.2 Psychrometrics and air-conditioning processes (evaporative cooling)
14. Which of the following happens to air when it is cooled through evaporation process?
humidity ratio of the air decreases
dry bulb temperature of air decreases
dry bulb temperature of air increases
enthalpy of outlet air is less than enthalpy of inlet air
Answer: B) dry bulb temperature of air decreases (evaporative cooling: DBT falls, humidity ratio rises, enthalpy is nearly constant)
In evaporative (adiabatic) cooling the latent heat needed to evaporate the water is drawn from the AIR ITSELF, so the dry bulb temperature falls while the moisture content rises. The process runs along a constant wet-bulb / constant-enthalpy line towards saturation.
That kills the other three options: the humidity ratio increases (not decreases), the DBT falls (not rises), and the enthalpy is essentially unchanged (not lower).
Book note: the air can be cooled to a temperature approaching its wet-bulb temperature - which is the theoretical limit of any evaporative device, including a cooling tower.
15. In which of the following fans the air does not change flow direction from suction to discharge?
tube axial fan
vane axial fan
propeller fan
all of the above
Answer: D) all of the above (all three are axial flow fans, in which air moves straight through along the shaft axis)
Axial family = propeller, tube-axial, vane-axial: air enters and leaves parallel to the shaft, so there is no 90° turn. Centrifugal fans (forward-curved, backward-curved, radial) take air in axially and throw it out radially — that is the turn. Memory hook: 'axial = straight through, centrifugal = right-angle bend'.
📖 §5.6 Fan performance assessment (fan static efficiency)
16. The pressure to be considered for calculating the power required for centrifugal fans is ___
vapour pressure
dynamic pressure
total static pressure
velocity pressure
Answer: C) total static pressure
Field fan power always uses the STATIC pressure rise: kW(shaft) = Q(m³/s) × ΔPst(mmWC) / (102 × η). Total pressure = static + velocity pressure; mechanical (total) efficiency uses total pressure, static efficiency uses static only. The mark is lost by picking velocity pressure — that is only the pitot reading used to get velocity, not the pressure the fan has to work against.
📖 §6.1 Pump types (hydraulic, shaft and motor input power)
17. If the power drawn by the motor driving a pump is 20 kW at a 91% efficiency, and the hydraulic power of a motor pump set is 12.5 kW, the pump efficiency will be ___
68.7%
62.5%
56.8%
none of the above
Answer: A) 68.7% (shaft power = 20 x 0.91 = 18.2 kW; pump efficiency = 12.5/18.2 = 68.7 %)
Chain: motor input kW × η(motor) = shaft kW; hydraulic kW ÷ shaft kW = pump efficiency. So shaft = 20 × 0.91 = 18.2 kW and η(pump) = 12.5/18.2 = 68.7%. The trap is dividing 12.5 by 20 (= 62.5%, offered as option b) — that is the combined MOTOR-PUMP set efficiency, not the pump efficiency the question asks for.
📖 §6.6 Flow control strategies (impeller trimming)
18. The preferred method of flow control for reducing pump flow permanently in a pumping system is -------
throttling
speed control
impeller trimming
none of the above
Answer: C) impeller trimming
Trimming machines the impeller diameter down permanently, so the pump stops adding energy it never needed — the cheapest fix for a chronically oversized pump. Throttling wastes the excess as pressure drop across a valve and speed control (VFD) costs more and is aimed at VARYING demand. The book limits trimming to about 75% of the maximum impeller diameter; below that efficiency collapses.
📖 §6.6 Flow control strategies (impeller trimming laws)
19. A water pump is delivering 20 m3/hr at ambient conditions. The impeller diameter is trimmed by 10%. This will reduce the pump discharge by
18 m3/hr
2 m3/hr
0.2 m3/hr
none of the above
Answer: B) 2 m3/hr (flow ∝ diameter; new flow = 0.9 x 20 = 18 m3/hr, i.e. a reduction of 2 m3/hr)
Trimming laws: Q ∝ D, H ∝ D², P ∝ D³. New flow = 0.9 × 20 = 18 m³/hr, so the REDUCTION is 20 − 18 = 2 m³/hr. Option (a) 18 m³/hr is the new flow, deliberately placed to catch anyone who answers the wrong question — read 'reduce by' versus 'reduce to' every time.
📖 §6.5 Efficient pumping system operation (NPSH and cavitation)
20. Increasing the suction pipe diameter in a pumping system will
reduce NPSHA
increase NPSHA
decrease NPSHR
increase NPSHR
Answer: B) increase NPSHa
NPSHa = atmospheric head + static suction head − vapour pressure head − suction friction losses. Fatten the suction pipe and velocity falls as 1/D², friction loss falls roughly as 1/D⁵, so NPSHa RISES and cavitation margin improves. NPSHr belongs to the pump (set by the manufacturer's impeller design) and nothing you do to the pipework changes it — that is why (c) and (d) are wrong.
📖 §7.2 Cooling tower performance (range and approach)
21. The range of a cooling tower with inlet and outlet temperature as 41 °C and 32 °C respectively and wet bulb temperature as 29 °C is
9 °C
3 °C
29 °C
12 °C
Answer: A) 9 °C (Range = CW inlet - CW outlet = 41 - 32 = 9 °C)
Range = hot water in − cold water out = 41 − 32 = 9 °C, and it uses only the two WATER temperatures. Approach = cold water out − WBT = 32 − 29 = 3 °C, which is why option (b) is sitting there. Memory hook: 'Range is water-to-water, Approach is water-to-air'. Swapping the two is the single most common cooling tower mistake in this paper.
22. Find the correct equation, if M = makeup water (from the mains water supply), E = losses due to evaporation, B = losses due to blow-down and D = drift losses of a cooling tower:
M = E + B + D
M = E + B - D
M = E - B + D
M = E - B - D
Answer: A) M = E + B + D
Make-up must replace everything that leaves the circuit, so M = E + B + D: evaporation (the useful loss that does the cooling), blowdown (deliberate, to hold cycles of concentration) and drift (droplets carried out past the eliminators). Only evaporation removes pure water and concentrates dissolved solids; blowdown and drift carry solids away with them, which is why COC is defined on blowdown plus drift, not evaporation.
23. If the wet bulb temperature of air is 38 °C, then its relative humidity in % is
38 %
90%
100%
insufficient data
Answer: D) insufficient data (RH cannot be found from WBT alone; the dry bulb temperature is also needed)
Moist air needs TWO independent properties to be fixed. RH is the ratio of the actual vapour pressure to the saturation vapour pressure at the DRY-bulb temperature, so a wet-bulb reading alone cannot give it.
Give the pair - DBT and WBT - and the chart yields RH, humidity ratio, enthalpy, dew point and specific volume all at once.
The only case where one reading suffices is DBT = WBT, which means saturated air at 100% RH; since the question gives no dry-bulb value, the data really is insufficient.
📖 §7.2 Cooling tower performance (blowdown and COC)
24. For a cooling tower if blowdown is 10 m3/hour and Cycles of Concentration (CoC) is 2.5 the evaporation loss is equal to:
25 m3/hour
15 m3/ hour
0.25 m3/hour
6.67 m3/hour
Answer: B) 15 m3/hour (Blowdown = Evaporation loss / (CoC - 1), so E = 10 x (2.5 - 1) = 15 m3/hour)
Rearrange the book relation Blowdown = Evaporation/(COC − 1) into Evaporation = Blowdown × (COC − 1) = 10 × (2.5 − 1) = 15 m³/hr. The denominator is (COC − 1), never COC — using 2.5 gives 25 m³/hr, which is option (a) and the trap. Physically COC − 1 is the excess concentration the blowdown has to carry away.
📖 §8.3 Light source and lamp types (fluorescent tube designations)
25. In T-5 Fluorescent Lamp, “5” is indicative of:
Tube diameter
5 watt loss
5% Energy Saving with respect to T8
5th generation lamp
Answer: A) Tube diameter (5 eighths of an inch, i.e. 16 mm)
The T-number is tube diameter in EIGHTHS OF AN INCH: T12 = 12/8 in = 38 mm, T8 = 8/8 in = 25 mm, T5 = 5/8 in = 16 mm, T2 = 2/8 in = 6 mm. The book credits T5 and T8 with about a 5% efficacy gain over the 40 W T12 — note that the 5% figure has nothing to do with the '5' in T5, which is the trap built into option (c).
26. Which of the following with respect to turbocharger in a Diesel engine is true?
operates using energy of exhaust gases
decreases supply air pressure to engine
preheats the combustion air using energy from exhaust gases
all of the above
Answer: A) operates using energy of exhaust gases
A turbocharger is an exhaust-gas turbine on a common shaft with an inlet-air compressor: waste exhaust energy drives the turbine, which pressurises the intake air so more fuel can be burnt per stroke. It therefore INCREASES supply air pressure, ruling out (b). Option (c) is the subtle trap — the air does get hot from compression, but that is unwanted, and an intercooler/aftercooler is fitted to COOL it and raise the charge density.
27. The refrigerant which can be used both in vapour compression chillers and vapour absorption chiller is
R22
R21
ammonia
pure water
Answer: C) ammonia
Ammonia (R-717) is the one refrigerant on the list used in BOTH cycles: as the working refrigerant in vapour compression plants (large industrial and cold-storage systems) and as the refrigerant in ammonia-water vapour absorption machines, where water is the absorbent.
Contrast with the other common absorption pair, lithium bromide-water, where WATER is the refrigerant and LiBr is the absorbent - that pair cannot go below 0 degC, whereas ammonia systems work well below 0 degC and above atmospheric pressure.
R22 and R21 are halocarbons used only in vapour compression, and pure water works as a refrigerant only inside a LiBr absorption machine, not on its own.
📖 §1.5 Transformers (turns ratio and current ratio)
28. In a transformer on load, if the secondary voltage is one-fourth the primary voltage, then the secondary current will be
four times the primary current
equal to the primary current
one-fourth the primary current
two times the primary current
Answer: A) four times the primary current (V1I1 = V2I2)
A transformer conserves VA: V1 x I1 = V2 x I2, so current goes UP exactly as voltage comes down. V2 = V1/4 therefore I2 = 4 x I1.
Remember the ratio chain: V1/V2 = N1/N2 = I2/I1 - the current ratio is the INVERSE of the voltage ratio.
Practical consequence: this is why LT-side currents (and hence I2R losses and cable sizes) are so much larger than HT-side currents.
📖 §1.4 Power factor improvement & benefits (capacitor output vs voltage)
29. If V1 is actual supply voltage and V2 is the rated voltage of a capacitor, the reactive KVAr produced would be in the ratio of
V2²/V1²
V1²/V2²
1 - V2²/V1²
1 + V2²/V1²
Answer: B) V1²/V2²
Capacitor output kVAr = 2(pi)fCV^2, so it varies with the SQUARE of the applied voltage. Delivered kVAr / rated kVAr = V1^2 / V2^2.
The mistake that loses the mark is treating the relationship as linear in V. A capacitor run 10% under its rated voltage gives only 0.9^2 = 81% of its nameplate kVAr.
Practical note: this is why capacitors installed at a low-voltage tail end under-deliver, and why the rated voltage must always match the actual bus voltage.
📖 §2.3 Motor characteristics + §2.4 Motor efficiency (input power, current, speed)
1. A 15 kW, 415 V, 4 pole, 50 Hz, 3 Phase squirrel cage induction motor has a full load efficiency of 92% and power factor of 0.89. Find the following if the motor operates at full load rated values. a) input power in kW b) current drawn by the motor c) RPM at a full load slip of 0.8%
Model answer: a) Pin (Input power) = 15 / 0.92 = 16.304 kW
b) I (Input current) = 16.304 / (1.732 x 0.415 x 0.89) = 25.48 A
c) Ns = 120 x f / p = 120 x 50 / 4 = 1500 RPM
N = Ns (1 - S) = 1500 (1 - 0.008) = 1488 RPM
Three formulas, applied in order: input kW = output/efficiency = 15/0.92 = 16.304 kW; I = kW/(sqrt(3) x kV x PF) = 16.304/(1.732 x 0.415 x 0.89) = 25.48 A; Ns = 120f/P = 120 x 50/4 = 1500 rpm, so N = Ns(1 - s) = 1500 x (1 - 0.008) = 1488 rpm.
Note that the current must be computed from the INPUT kW, not the 15 kW shaft output - using 15 kW gives 23.4 A and loses the mark.
Other regulars: dropping the sqrt(3), and leaving voltage in volts instead of kV so the answer comes out 1000x wrong.
📖 §6.1 Pump types (hydraulic power) & §6.2 System characteristics
2. In a pumping system the water level is 4 m below the pump centerline. The discharge pressure is 2.60 kg/cm2. The flow rate of water is 1.5 m3/min. Find out the pump efficiency if the actual power drawn by the pump motor is 14 kW at a motor operating efficiency of 0.88.
Model answer: Discharge Head = 2.60 kg/cm2 = 26 metre head
Suction Head = - 4 metre
Total Head = 26 - (-4) = 30 metre
Hydraulic Power = (1.5/60) x 1000 x 9.81 x 30/1000 = 7.36 kW
Shaft input = 14 x 0.88 = 12.32 kW
Pump Efficiency = 100 x 7.36/12.32 = 59.74 %
Sign convention is the whole question: water 4 m BELOW the pump means suction head = −4 m, so total head = 26 − (−4) = 30 m, not 22 m. Convert pressure first: 2.60 kg/cm² = 26 m of water (1 kg/cm² ≈ 10 m). Ph = (1.5/60) × 30 × 1000 × 9.81/1000 = 7.36 kW; shaft = 14 × 0.88 = 12.32 kW; η = 7.36/12.32 = 59.7%. Also watch the flow unit — m³/min must be divided by 60, not 3600.
3. Harmonic measurements in an electrical system of an industry gave the following results. Current at 50 Hz : 300 A; Current at 150 Hz : 42 A; Current at 250 Hz : 33 A. Calculate the Total Harmonic Distortion in current for the system.
Model answer: I(THD) = √[(42/300)² + (33/300)²] x 100
= √(0.0196 + 0.0121) x 100
= 17.8 %
Formula to memorise: THD_current = sqrt(I3^2 + I5^2 + I7^2 + ...) / I1 x 100 - a root-sum-square of the harmonic currents, expressed as a % of the FUNDAMENTAL (50 Hz), not of the total current.
Working: sqrt(42^2 + 33^2)/300 x 100 = sqrt(1764 + 1089)/300 x 100 = 53.4/300 x 100 = 17.8%.
Marks are lost by adding the harmonics arithmetically (42 + 33 = 75 -> 25%) instead of squaring, summing and taking the root.
📖 §5.6 Fan performance assessment (pitot tube velocity & density correction)
4. Air flow measurements using the pitot tube, in the primary air fan of a coal fired boiler gave the following data: Air temperature = 38 °C; Velocity pressure = 47 mmWC; Pitot tube constant, Cp = 0.9; Air density at 0 °C (standard data) = 1.293 kg/m3. Find out the velocity of air in m/sec.
Model answer: Corrected air density = 273 x 1.293 / (273 + 38) = 1.135 kg/m3
Velocity, m/s = Cp x √(2 x 9.81 x Δp x γw / γair)
= 0.9 x √(2 x 9.81 x 47 x 1000 / (1000 x 1.135))
= 25.6 m/s
Two-step drill: correct density first, then apply velocity. Density at t °C = 1.293 × 273/(273+t) = 1.293 × 273/311 = 1.135 kg/m³. Velocity = Cp × √(2 × 9.81 × Δp/ρ) = 0.9 × √(2 × 9.81 × 47/1.135) = 0.9 × 28.5 = 25.6 m/s. The classic slip is using 1.293 kg/m³ straight from the data sheet without the 273/(273+t) correction — that inflates the velocity by about 6%.
📖 §8.6 General energy saving opportunities in lighting
5. List five measures to reduce energy consumption in lighting system for buildings, industry and street lighting
Model answer: Any five of the following:
- Reduce excessive illumination levels to standard levels using switching, delamping, etc. (know the electrical effects before delamping).
- Aggressively control lighting with clock timers, delay timers, photocells and/or occupancy sensors.
- Install efficient alternatives to incandescent lighting, mercury vapour lighting, etc. Efficiency (lumens/watt) of various technologies ranges from best to worst approximately as follows: low pressure sodium, high pressure sodium, metal halide, fluorescent, mercury vapour, incandescent.
- Select ballasts and lamps carefully with high power factor and long-term efficiency in mind.
- Upgrade obsolete fluorescent systems to compact fluorescents and electronic ballasts.
- Consider lowering the fixtures to enable using less of them.
- Consider daylighting, skylights, etc.
- Consider painting the walls a lighter colour and using fewer lighting fixtures or lower wattages.
- Use task lighting and reduce background illumination.
- Re-evaluate exterior lighting strategy, type and control. Control it aggressively.
- Change exit signs from incandescent to LED.
Group the answers so five come easily: (1) cut over-illumination to standard lux levels by de-lamping or switching; (2) control aggressively with timers, photocells and occupancy sensors; (3) replace with higher-efficacy sources — the book's efficacy order, best to worst, is low-pressure sodium, high-pressure sodium, metal halide, fluorescent, mercury vapour, incandescent; (4) improve the installation — electronic ballasts, lower mounting heights, lighter wall colours, task lighting, daylighting/skylights; (5) LED exit signs and a re-thought exterior lighting strategy. Quote the efficacy ranking — it earns marks by itself.
📖 §4.3 Types of refrigeration system (VCR vs VAR comparison)
6. Identify each of the following statement as applicable to Vapor Compression Refrigeration System (VCR) and to Vapor Absorption Refrigeration System (VAR). (Need not copy and write the following statements in the Answer book; only write against the statements A, B, C, D etc. whether it is applicable to VCR or VAR) A. No effect of reducing the load on performance. B. Uses low grade energy C. Liquid traces in suction line may damage the compressor. D. Moving parts are only in the pump and hence operation is smooth. E. The system can work on lower evaporator pressures also without affecting the COP. F. Performance is adversely affected at partial loads. G. Liquid traces of refrigerant present in piping at the exit of evaporator H. Using high-grade energy like mechanical work I. Moving parts are more; therefore, more equipment maintenance and noise J. The COP decreases considerably with decrease in evaporator pressure
Model answer: A. VAR
B. VAR
C. VCR
D. VAR
E. VAR
F. VCR
G. VAR
H. VCR
I. VCR
J. VCR
Sort every statement by one question: which cycle drives the refrigerant round? VCR uses HIGH-grade energy (mechanical work in a compressor) and has many moving parts, hence noise, maintenance and vulnerability to liquid slugging in the suction line. VAR uses LOW-grade energy (steam, hot water, waste heat) with moving parts only in the solution pump - so quiet, smooth and part-load tolerant.
Part-load behaviour is the discriminator worth memorising: VCR performance falls off badly at partial load, whereas VAR holds its COP right down to low load and at lower evaporator pressures.
Typical figures: VAR COP 0.65-0.70 (LiBr-water), chilled water at 6.7 degC with 30 degC cooling water; VCR COP is several times higher, but on purchased electricity rather than waste heat.
7. List any five factors that affect the rate of evaporation of water in cooling towers
Model answer: - Amount of water surface area exposed
- The time of exposure
- The relative velocity of air passing over the droplets
- The RH of air
- The direction of airflow relative to water
(Any other relevant point to be considered)
Evaporation rate is governed by how much air-water contact you create and how thirsty the air is: exposed water surface area (what the fill exists to maximise), contact TIME, relative velocity of air over the droplets, the relative humidity/wet bulb of the incoming air, and the direction of airflow relative to the water (counterflow versus crossflow). Note that RH is the air-side driver — saturated air cannot evaporate anything, which is the physics behind the 'minimum evaporation' MCQ.
📖 §7.2 Cooling tower performance (cooling capacity and approach)
8. Estimate the cooling tower capacity (TR) and approach with the following parameters: Water flow rate through CT = 2 m3/min; Specific heat of water = 1 kcal/kg °C; Inlet water temperature = 43 °C; Outlet water temperature = 35 °C; Ambient WBT = 30 °C
Model answer: Cooling tower capacity (TR) = (flow rate x density x sp. heat x diff. temp)/3024
= (2 x 60) x 1000 x 1.0 x (43 - 35)/3024
= 317.5 TR
Approach = 35 - 30 = 5 °C
TR = (flow m³/hr × 1000 × specific heat × range)/3024. Watch the units: 2 m³/min = 120 m³/hr, so TR = 120 × 1000 × 1 × (43 − 35)/3024 = 960,000/3024 = 317.5 TR. Approach = cold water out − WBT = 35 − 30 = 5 °C, using the OUTLET water, not the inlet. 3024 kcal/hr = 1 TR is the constant to memorise; forgetting the ×60 on m³/min is the other easy mark to drop.
📖 §2.4 Motor efficiency (no-load test and full-load efficiency assessment)
1. An efficiency assessment test was carried out for a standard 4 pole squirrel cage induction motor in a chemical plant. The motor specifications are as under: Motor rated specification: 3 phase delta connected, 37 kW, 415 Volt, 63 Amps, 1475 rpm. The following data was collected during the no-load test on the motor: Voltage = 415 Volts; Current = 17 Amps; Frequency = 50 Hz; Stator resistance per phase = 0.260 Ohms at 30 °C; No load power = 1152 Watts. Calculate the following: (i) Iron plus friction and windage losses. (ii) Stator resistance at 120 °C. (iii) Stator copper loss at full load at operating temperature of 120 °C. (iv) Full load slip and rotor input assuming rotor losses are slip times rotor input. (v) Motor input assuming that stray losses are 0.5% of the motor rated output power. (vi) Motor full load efficiency
Model answer: (i) Iron plus friction and windage loss, Pi+fw
No load power, Pnl = 1152 Watts
Stator copper loss at 30 °C, Pst.cu = 3 x (17/√3)² x 0.260 = 75.13 Watts
Pi+fw = Pnl - Pst.cu = 1152 - 75.13 = 1076.87 W
(ii) Stator resistance at 120 °C
R(120 °C) = 0.260 x (120 + 235)/(30 + 235) = 0.3483 ohms per phase
(iii) Stator copper losses at full load at 120 °C
Pst.cu(120 °C) = 3 x (63/√3)² x 0.3483 = 1382.3 Watts
(iv) Full load slip
S = (1500 - 1475)/1500 = 0.01666 or 1.66%
Rotor input, Pr = Poutput/(1 - S) = 37000/(1 - 0.01666) = 37000/0.98334 = 37626.86 Watts
(v) Motor full load input power
Pinput = Pr + Pst.cu(120 °C) + (Pi+fw) + Pstray
= 37626.86 + 1382.3 + 1076.87 + (0.005 x 37000)
= 40271.03 Watts
(where stray losses = 0.5% of rated output, assumed)
(vi) Motor efficiency at full load = (Poutput/Pinput) x 100
= (37000/40271.03) x 100 = 91.87 %
The five standing rules for this classic 10-marker: (1) for a DELTA-connected motor the phase current is I_line/sqrt(3), so stator copper loss = 3 x (I/sqrt(3))^2 x R_phase; (2) resistance rises with temperature as R2 = R1 x (235 + t2)/(235 + t1) for copper - 235 is the copper constant, and using 273 here is a standard mark-loser;
(3) iron + friction & windage = no-load input minus the no-load stator copper loss (1152 - 75.13 = 1076.87 W); (4) rotor input = output/(1 - s) with s = (1500 - 1475)/1500 = 0.0167; (5) stray loss is taken as 0.5% of rated output.
Input = 37,626.86 + 1382.3 + 1076.87 + 185 = 40,271 W, so efficiency = 37,000/40,271 = 91.87%. Note the copper loss is computed twice - once cold at 30 degC for the no-load split, once hot at 120 degC for the full-load loss.
📖 §6.5 & §6.6 Pumping (also §5.3 fan system curve, §7.2 cooling tower, §8.2 lux)
2. Fill in the blanks: 1. Cavitation may occur in a pump when the local static pressure in a fluid reaches a level below the _________ pressure of the liquid at the actual temperature. 2. In a vapour absorption system using ammonia as refrigerant, the absorbent is ______. 3. The system resistance of a fan system is proportional to the ______ of flow rate or velocity. 4. If the dry bulb temp. is 30 °C and the wet bulb temp. is 30 °C, then the % relative humidity will be _______. 5. Slip ring induction motors are comparatively ________ efficient than of the squirrel cage motors of same ratings. 6. In a pumping system with a horizontal discharge, the suction static head is 3 m and the friction head is 21 m. The total head developed by the pump will be ___________. 7. The lowest theoretical temperature to which water can be cooled in a cooling tower is the _________ of atmospheric air. 8. The measure of illuminance of a surface in metric units is ________. 9. It is acceptable to run pumps in parallel provided their _________ heads are similar. 10. When heat load, range and wet bulb temperature are held constant, the cooling tower size is ________ proportional to the approach.
Model answer: 1. Vapour
2. Water
3. Square
4. 100%
5. Less
6. 18 m
7. Wet bulb temperature
8. Lux
9. Closed valve heads
10. Inversely
The recurring traps in this mixed set: DBT = WBT means saturated air, so RH = 100%; system resistance of a fan varies as the SQUARE of flow; the lowest theoretical cold water temperature in a tower is the ambient WET BULB, never the dry bulb; illuminance is measured in LUX (lumen/m²), while the lamp's output is lumens. For blank 6, a horizontal discharge with 3 m suction static head and 21 m friction gives 21 − 3 = 18 m. Tower size is INVERSELY proportional to approach — a closer approach needs a bigger, costlier tower.
📖 §6.1 Pump types (hydraulic power) & §6.2 System characteristics
3. The cooling water circuit of a process industry is depicted in the figure below. Cooling water is pumped to three heat exchangers via pipes A, B and C where flow is throttled depending upon the requirement. The diameter of pipes and measured velocities with non-contact ultrasonic flow meter in each pipe are indicated in the figure. The following are the other data: Measured motor power : 50.7 kW; Motor efficiency at operating load: 90%; Pump discharge pressure : 3.4 kg/cm2; Suction head : 2 meters. Determine the efficiency of the pump [refers to a figure in the original paper]
Model answer: Flow in pipe A = (22/7) x (0.1)²/4 x 1.5 = 0.011786 m3/s
Flow in pipe B = (22/7) x (0.1)²/4 x 1.8 = 0.014143 m3/s
Flow in pipe C = (22/7) x (0.2)²/4 x 2.0 = 0.062857 m3/s
Total flow = 0.088786 m3/s
Total head = 34 m - 2 m = 32 m
Pump hydraulic power = 0.088786 x 32 x 9.81 = 27.9 kW
Pump efficiency = 27.9 x 100/(50.7 x 0.9) = 61 %
(Pipe A: 100 mm dia at 1.5 m/s; Pipe B: 100 mm dia at 1.8 m/s; Pipe C: 200 mm dia at 2.0 m/s, as read from the figure in the original paper.)
Total flow first: Q = (π/4)D²v summed over the three pipes = 0.011786 + 0.014143 + 0.062857 = 0.088786 m³/s. Convert 3.4 kg/cm² to 34 m and subtract a POSITIVE suction head of 2 m → total head 32 m. Ph = Q × H × 9.81 = 0.088786 × 32 × 9.81 = 27.9 kW, and shaft = 50.7 × 0.9 = 45.6 kW, so η = 61%. Two habits save marks here: work pipe diameters in metres before squaring, and use only the two 100 mm pipes' velocities separately — you cannot average velocities across different diameters.
📖 §5.6 Fan performance assessment & §5.5 Flow control strategies (fan laws)
4. a) The size of an engine room to be ventilated is 30 m x 20 m x 5 m. The number of air changes per hour is designed to be 20. If the static pressure rise across the ventilator fan is 15 mm WC and fan efficiency is 70 % find out the motor power drawn at a motor efficiency of 90%. b) A seal air fan for a coal mill is operating with suction damper in 25 % open condition. The power drawn at 50 Hz by fan motor is 120 kW. A VFD is to be installed eliminating the damper operation. It is found that the damper can be completely opened and the fan motor can be operated at 33 Hz. Calculate the power drawn by the fan motor at 33 Hz, assuming that motor and fan efficiency remains constant.
Model answer: a) Flow rate = 30 x 20 x 5 x 20 = 60,000 m3/hr
Motor power = (60,000/3600) x 15/(102 x 0.7 x 0.9) = 3.89 kW
b) Power at 50 Hz = 120 kW
Power at 33 Hz = 120 x (33/50)³ = 34.5 kW
(a) Air changes → flow: room volume × ACH = 30×20×5×20 = 60,000 m³/hr = 16.67 m³/s. Motor kW = Q × ΔPst / (102 × ηfan × ηmotor) = 16.67 × 15 / (102 × 0.7 × 0.9) = 3.89 kW. Divide by BOTH efficiencies when the question asks for motor power drawn; dividing by the fan efficiency only gives 5.0 kW and loses the mark. (b) VFD saving uses the cube law on FREQUENCY: 120 × (33/50)³ = 34.5 kW — squaring instead of cubing gives 52 kW and is wrong.
5. a) In an automobile industry a pump-up test was conducted to determine the free air delivery (FAD) of a reciprocating compressor and the following data were obtained: Receiver capacity and additional holdup volume in piping and after-cooler : 4100 litres; Initial pressure : 1 kg/cm2 (g); Final pressure : 8.5 kg/cm2 (g); Atmospheric Pressure : 1.026 kg/cm2 (a); Ambient air temperature : 32 °C; Final compressed air temperature : 52 °C; Compressor pump up time : 65 secs. Calculate the FAD of the compressor in cubic foot per minute. b) Further a leakage test was carried out in the same compressed air system and with the same compressor as in problem a) above and following were the observations: - Compressor was on load for 03 minutes - Compressor was unloaded for 13 minutes - Compressor was drawing 145 kW during load. Calculate the following: i. % leakage in compressed air system ii. Leakage quantity iii. Specific power consumption iv. Power lost due to leakage
Model answer: a) Q = [(P2 - P1)/P0] x (V/t) x [(273 + t1)/(273 + t2)]
Time = 65 sec = 1.0833 minutes
= [(8.5 - 1)/1.026] x (4.1/1.0833) x (305/325)
= 25.96 m3/min
= 25.96 x (3.28)³
= 916 cfm
b) i) % Leakage in the system
Load time (T) = 03 minutes; Unload time (t) = 13 minutes
% leakage = T/(T + t) x 100 = 3/(3 + 13) x 100 = 18.75 %
ii) Leakage quantity = 0.1875 x 916 = 171.75 cfm
iii) Operating capacity (FAD) = 916 cfm; Actual power consumption = 145 kW
Specific power consumption = 145/916 = 0.1583 kW/cfm
iv) Power lost due to leakage = leakage quantity x specific power consumption
= 171.75 x 0.1583 = 27.19 kW
FAD formula: Q = [(P2 - P1)/P0] x (V/t) x [(273 + t1)/(273 + t2)] - the last bracket is the temperature correction back to AMBIENT conditions, and omitting it is a standard mark-loser. Here Q = (7.5/1.026) x (4.1/1.0833) x (305/325) = 25.96 m3/min; x 3.28^3 = 916 cfm.
Leakage: % leakage = T/(T + t) x 100 where T = load time and t = unload time = 3/(3 + 13) x 100 = 18.75%, so leakage = 0.1875 x 916 = 171.75 cfm.
Specific power = 145/916 = 0.1583 kW/cfm, so power lost to leaks = 171.75 x 0.1583 = 27.19 kW. Watch the units: convert the pump-up time to minutes and the receiver volume to m3 before dividing.
📖 §6.6 Flow control strategies, §7.2 cooling capacity, §5.7 fan energy savings
6. Answer any two of the following: (i) In a throttle valve-controlled pumping system with oversized pump list any five options to improve energy efficiency? (Note: Name only options, no explanation required) (ii) Define one 'Ton of Refrigeration (TR)'. How do you calculate TR across the Air Handling Units? (iii) List five energy conservation opportunities in fan system.
Model answer: i) Trim impeller, replace with smaller impeller, install variable speed drive, change pulley if it is belt driven, change to two speed drive, and lower rpm drive.
ii) A ton of refrigeration is defined as the quantity of heat to be removed in order to form one ton of ice in 24 hours when the initial temperature of water is 0 °C. This is equivalent to 50.4 kCal/min or 3024 kCal/h in the metric system.
Refrigeration load in TR across an AHU is assessed as:
TR = Q x ρ x (h_in - h_out) / 3024
where Q is the air flow in CMH, ρ is density of air in kg/m3, h_in is enthalpy of inlet air in kCal/kg and h_out is enthalpy of outlet air in kCal/kg.
iii) Energy conservation opportunities in fan system:
- Use smooth, well-rounded air inlet cones for fan air intakes.
- Avoid poor flow distribution at the fan inlet.
- Minimize fan inlet and outlet obstructions.
- Clean screens, filters and fan blades regularly.
- Use aerofoil-shaped fan blades.
- Minimize fan speed.
- Use low-slip or flat belts.
- Check belt tension regularly.
- Eliminate variable pitch pulleys.
- Use variable speed drives for large variable fan loads.
- Use energy-efficient motors for continuous or near-continuous operation.
- Eliminate leaks in ductwork.
- Minimise bends in ductwork.
- Turn fans off when not needed.
(i) The oversized-pump list is always the same six: trim impeller, fit a smaller impeller, install a VFD, change the pulley on a belt drive, use a two-speed drive, use a lower-rpm motor. (ii) 1 TR = 3024 kcal/hr = 50.4 kcal/min — memorise 3024, it is the denominator for every TR calculation; across an AHU use ENTHALPY difference (kcal/kg) with air flow in CMH, not temperature difference. (iii) Fan savings are grouped as minimise resistance (bends, obstructions, clean filters), improve the drive (flat belts, no variable-pitch pulleys, VFD, EE motor) and cut demand (turn fans off, seal ducts).