Energy Efficiency in Electrical Utilities Available here with full solutions — 65 questions recovered from the 2013 exam:
Objective (1 mark)
48 of 50
Short (5 marks)
9 of 8
Long (10 marks)
8 of 6
This is not the complete paper. The questions below are the ones we could recover and verify; the rest of that year’s paper is not reproduced here. Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 48
📖 §6.1 Pump types — centrifugal pump construction & working
1. The head generated by a centrifugal pump is
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.
Confirmed vs Book-3 §6.1 — Book: 'The pump generates the same head of liquid whatever the density of the liquid being pumped' — head (metres of liquid column) is set by impeller diameter, speed, eye size and number of impellers. Density affects the pressure produced and the power drawn (P = Q·H·ρ·g), not the head, so b–d are wrong.
2. Increasing the suction pipe diameter in a pumping system will
Decrease NPSHA
Increase NPSHA
Decrease NPSHR
Increase NPSHR
Answer: B) Increase NPSHA
Confirmed vs Book-3 §6.5 — NPSHA = margin of eye pressure above vapour pressure; the book notes that as suction friction losses increase, NPSHA falls. A larger suction pipe lowers velocity and friction loss, so NPSHA increases. NPSHR is fixed by the pump design and does not change with pipe size. (Same as book end-of-chapter Q6.)
3. If the speed of the pump is doubled, power goes up by
2 times
6 times
8 times
4 times
Answer: C) 8 times
Confirmed vs Book-3 §6.5 — Affinity law P∝N³: doubling speed gives 2³ = 8 times the power (flow doubles, head ×4). Book example in reverse: 3000→1500 rpm drops 40 kW to 5 kW (÷8). '4 times' is the head ratio, '2 times' the flow ratio.
4. Reduction in the delivery pressure of a compressor by 1 bar would reduce the power consumption by
1 to 5 %
6 to 10 %
11 to 15 %
none of the above
Answer: B) 6 to 10 %
Confirmed vs Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure (Table 3.9) — Book-3: 'A reduction in the delivery pressure by 1 bar in a compressor would reduce the power consumption by 6-10%'; the worked case of 8 to 7 kg/cm² gives 9% input power saving.
The question stem was repaired from the self-contradictory 'Increase in the delivery pressure... would reduce the power consumption' to the book's wording; the answer 6-10% is unchanged. Table 3.9's smaller 4% figure applies to a 0.7 bar cut, which is why 1-5% looks plausible but is too low for a full bar.
5. The gross efficiency of a coal based power generating unit with a gross heat rate of 2490 kcal/kWh is
40%
34.5 %
33.3%
45.2%
Answer: B) 34.5 %
Confirmed vs Book-3 §1.1 — 1 kWh ≡ 860 kCal, and heat rate is inversely proportional to efficiency, so η = 860/2490 = 34.5%.
Option (c) 33.3% would correspond to a heat rate of about 2580 kCal/kWh; always divide 860 by the given heat rate.
6. The efficiencies of a power plant and transmission system are 40% and 97% respectively. The distribution system loss is 23%. The cascade efficiency of generation, transmission and distribution is
8.92 %
29.87%
40 %
23%
Answer: B) 29.87%
Confirmed vs Book-3 §1.1 Cascade Efficiency — multiply the stage efficiencies: 0.40 × 0.97 × (1 − 0.23) = 0.40 × 0.97 × 0.77 = 0.2987 = 29.87%.
Option (a) 8.92% comes from multiplying by the 23% loss instead of by the 77% efficiency; cascade efficiency always uses efficiencies, never losses.
7. The rating of power factor correction capacitors at induction motor terminals should be
90% of no load magnetizing kVAr of induction motor
100 % of no load magnetizing kVAr of induction motor
80% of no load magnetizing kVAr of induction motor
none of the above
Answer: A) 90% of no load magnetizing kVAr of induction motor
Confirmed vs Book-3 §1.4 — the book's rule is that 'the rating of the capacitor should not be greater than the no-load magnetizing kVAr of the motor'; standard practice therefore sizes terminal capacitors at about 90% of that value to keep a safety margin against self-excitation.
Option (b) 100% sits exactly on the limit and risks the 'damaging over voltage or transient torques' the book warns about.
Answer: C) is the ratio of active and apparent power
Confirmed vs Book-3 §1.4 — 'The ratio of kW to kVA is called the power factor', i.e. active power ÷ apparent power (PF = cosΦ), always ≤ 1.
Option (d) is wrong: a purely inductive or capacitive load draws only reactive power, so its power factor is zero, not unity.
9. The Energy Performance Index (EPI) of a building as per ECBC and the Energy Conservation Act, 2001 is:
kWh per square meter per year
kWh per square meter
kW per square meter
kWh per year
Answer: A) kWh per square meter per year
Confirmed vs Book-3 §10.14 — EPI is the specific energy usage of a building in kWh per square metre per year (annual energy ÷ built-up area). kWh/m² alone lacks the time base and kW/m² is a power density, not energy.
10. For a synchronous speed of 1500 rpm at a mains frequency of 50 Hz, the induction motor will have _________ number of poles.
8
6
4
2
Answer: C) 4
Confirmed vs Book-3 §2.3 Motor Characteristics — From Ns = 120 f / P, P = 120 × 50 / 1500 = 4 poles — consistent with the book's list of Indian synchronous speeds 3000/1500/1000/750 rpm for 2/4/6/8 poles at 50 Hz. (b) 6 poles would give 1000 rpm, which is the common mis-pick.
11. A 7.5 kW, 415 V, 14.5 A, 1460 RPM rated 3-phase induction motor with full-load efficiency 90% draws 9.1 A and 4.6 kW of input power. The percentage loading of the motor is about
55.2 %
61.3 %
67.5 %
none of the above
Answer: A) 55.2 %
Corrected (was b) — Book-3 §2.7 Motor Loading — Measuring Load: % loading = input kW / (rated kW ÷ full-load efficiency) = 4.6 / (7.5/0.90) = 4.6 / 8.333 = 55.2 %, which is option (a). The recorded 61.3 % does not follow from the book formula; 9.1/14.5 = 62.8 % is the current-ratio method that the book expressly forbids ('loading should not be estimated as the ratio of currents').
12. The power input to the rotor of a three phase induction motor is 42.3 kW. If the motor operates 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
Confirmed vs Book-3 §2.4 Motor Efficiency — Mechanical power developed = (1 − s) × rotor input = (1 − 0.013) × 42.3 = 41.75 kW; the rotor I²R loss is slip × rotor input = 0.55 kW. (c) 5.48 kW is the trap of computing something other than (1−s)×Pr; (a) 42.3 kW ignores the rotor copper loss altogether.
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
13. If the apparent power drawn over a recording cycle of 30 minutes is 5000 kVA for 0.5 minutes, 3400 kVA for 20 minutes and 1800 for 9.5 minutes, the MD recorder will compute MD as
5000 kVA
3400 kVA
2920 kVA
1800 kVA
Answer: C) 2920 kVA
Confirmed vs Book-3 §1.2 — maximum demand is the time-integrated demand over the recording cycle: [(5000×0.5)+(3400×20)+(1800×9.5)]/30 = 87,600/30 = 2920 kVA.
Option (a) 5000 kVA is the instantaneous peak; the book stresses MD 'is not the instantaneous demand drawn, as is often misunderstood'.
14. Which parameters need to be measured to assess the percentage loading of a motor by the slip method neglecting voltage correction?
motor speed
synchronous speed
operating motor speed and frequency
operating current
Answer: C) operating motor speed and frequency
Confirmed vs Book-3 §2.7 Motor Loading — Measuring Load — The slip method needs the operating (actual) motor speed AND the supply frequency, because the synchronous speed used in slip % = (Ns − Nr)/Ns must be computed as 120 f / P at the ACTUAL frequency. (a) speed alone is insufficient — at 49.5 Hz a 4-pole motor's Ns is 1485 rpm, not 1500 rpm, and the slip changes materially.
Confirmed vs Book-3 §3.3 Compressor Efficiency — Isothermal power (kW) = P₁ x Q₁ x logₑ(r) / 36.7, where P₁ is the absolute intake pressure, Q₁ the free air delivered in m³/hr and r = P₂/P₁ the pressure ratio.
All three quantities appear explicitly in the formula, so no single one of (a), (b) or (c) can be singled out — the answer is all of the above.
16. Which of these desiccant compressed air dryers uses dry compressed air for regenerating the desiccant?
blower reactivated type
heatless purge type
heat of compression type
all of the above
Answer: B) heatless purge type
Confirmed vs Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19) — In the heatless purge dryer, 'pure dry compressed air is used for purging through the saturated dessicant', which is why its purge loss is 12-15% and its operating cost is very high.
The blower reactivated type uses a blower plus external heater (1-2% purge) and the heat of compression type uses the compressor's own hot discharge with no purge at all, so (a) and (c) are wrong.
17. The COP of a vapour compression refrigeration system is 3.1. If the motor draws 9.3 kW at an operating efficiency of 88%, the tonnage of the refrigeration system is about
8.2
9.3
7.2
none of the above
Answer: C) 7.2
Confirmed vs Book-3 §4.7 - Shaft power = 9.3 x 0.88 = 8.184 kW. Refrigeration effect = COP x shaft power = 3.1 x 8.184 = 25.37 kW. TR = 25.37/3.516 = 7.2 TR. Option (b) 9.3 is the motor input in kW misread as TR; apply the 88% operating efficiency first (9.3 x 0.88 = 8.184 kW shaft), then COP and 3.516 kW/TR.
18. Chilled water enters an evaporator at 10°C and leaves at 6°C. The flow rate of chilled water is 200 m3/hr. The tons of refrigeration capacity is
265
200
661
2.65
Answer: A) 265
Confirmed vs Book-3 §4.7 - TR = (200 x 1000 x 1 x (10-6))/3024 = 800,000/3024 = 264.6 ≈ 265 TR. Option (c) 661 uses the 10 degC inlet instead of the 4 degC range, and (d) 2.65 misplaces the decimal by 100; TR = 200,000 x 4/3024 = 265.
19. The driving force for refrigeration in a vapour absorption refrigeration system is
mechanical energy
electrical energy
thermal energy
chemical energy
Answer: C) thermal energy
Confirmed vs Book-3 §4.3 - VAR systems are heat-driven; the driving input is low-grade thermal energy (heat). Mechanical energy (a) is what drives the vapour COMPRESSION system; VAR needs electricity (b) only for its solution pump, and the actual driving input is heat - steam, hot water, gas or oil.
20. The relation between COP and kW/TR for a refrigeration system is given by
kW/TR = 3.516/COP
kW/TR = COP/3.516
kW/TR = 860/COP
none of the above
Answer: A) kW/TR = 3.516/COP
Confirmed vs Book-3 §4.7 - one TR = 3.516 kW of cooling, and COP = cooling effect (kW) / compressor power (kW). Hence kW/TR = 3.516/COP; a plant with COP 3.516 needs exactly 1 kW/TR. Option (b) inverts the relation and (c) uses 860 kcal/kWh, which converts electrical energy to heat, not TR to kW.
21. The inclined manometer connected to a pitot tube is used for measuring which pressure in a gas stream?
velocity
static
total
all of the above
Answer: A) velocity
Confirmed vs Book-3 §5.6 — 'When the inner and outer tube ends are connected to a manometer, we get the velocity pressure. For measuring low velocities, it is preferable to use an inclined tube manometer.' The manometer reads the DIFFERENCE (TP − SP) = velocity pressure, not static or total alone (b, c).
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
22. If the power drawn by the motor driving a pump is 20 kW at 91% efficiency, and the hydraulic power of the 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%
Confirmed vs Book-3 §6.1 — Pump shaft power = 20×0.91 = 18.2 kW; pump efficiency = hydraulic/shaft = 12.5/18.2 = 68.7%. 62.5% (12.5/20) ignores motor efficiency; 56.8% wrongly multiplies that by 0.91.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
23. Which of the following is not true for impeller trimming?
pressure ∝ diameter
head ∝ diameter²
power ∝ diameter³
flow ∝ diameter
Answer: A) pressure ∝ diameter
Confirmed vs Book-3 §6.5 — Impeller-diameter relations: Q∝D, H∝D², P∝D³ (book §6.5/§6.6). Pressure (head) varies with the SQUARE of diameter, so 'pressure ∝ diameter' is the false statement; b, c and d are all correct affinity relations.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
24. A water pump is delivering 20 m3/hr. 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
Confirmed vs Book-3 §6.5 — Q∝D: a 10% diameter trim gives Q₂ = 0.9×20 = 18 m³/hr, i.e. a REDUCTION of 2 m³/hr. Option a (18) is the new flow, not the reduction; 0.2 would be a 1% change.
25. The range of a cooling tower with inlet and outlet temperature 41°C and 32°C respectively and wet bulb temperature 29°C is
9 °C
3 °C
29 °C
12 °C
Answer: A) 9 °C
Confirmed vs Book-3 §7.2 (i) — Range = cooling tower water inlet − outlet = 41 − 32 = 9°C → (a). 3°C (b) is the Approach (32 − 29) and 12°C (d) is the ideal range (41 − 29) – both are distractors.
26. If the metered kWh is 95, kVAh is 100 and kVARh is 31, the power factor will be:
0.95
0.61
0.69
unity
Answer: A) 0.95
Confirmed vs Book-3 §1.2 — the trivector meter records kWh, kVArh and kVAh; average power factor = kWh/kVAh = 95/100 = 0.95.
Option (b)/(c) come from using kVArh in the ratio; the kVArh value (31) is only a cross-check (√(95²+31²) ≈ 100).
27. Identify the statement that is not applicable to heat pumps
transfers heat by refrigerant through a cycle of evaporation and condensation
an air conditioner can work as a heat pump
no external energy is required
a vapour absorption refrigeration system can also work as a heat pump
Answer: C) no external energy is required
Confirmed vs Book-3 §4.11 - A heat pump still needs external (compressor/heat) energy to move heat; statement (c) is not applicable. Statements (a), (b) and (d) are all in §4.11 - the book says a heat pump takes 3 units from the surroundings and needs 1 unit of compressor work to deliver 4, so external drive energy is essential.
28. Which of the following is not true for a fluorescent lamp with electronic ballast
presence of stroboscopic effect
energy savings
increased light output
no starter required
Answer: A) presence of stroboscopic effect
Confirmed vs Book-3 §8.6(f) — An electronic ballast runs the tube at 20–30 kHz, so the tube 'lights up instantly without flickering' — no stroboscopic effect. The other three ARE true: ~1 W loss vs 10–15 W (energy saving of 15–20 W/tube), efficacy improves at high frequency (more light), and the starter is eliminated.
29. Which of the following with respect to a 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
Confirmed vs Book-3 §9.1 — A turbocharger is an "exhaust gas driven turbine" whose turbine wheel drives a compressor wheel that raises (not lowers) the pressure of the intake air, increasing rated output and lowering fuel consumption per kWh. It does not preheat the combustion air with exhaust energy (that is an air preheater/recuperator idea, and hot intake air actually derates the engine per Table 9.3), so (b), (c) and hence (d) are wrong.
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme
30. Energy efficient distribution transformer core is made up of ______.
silicon alloyed iron (grain oriented)
copper
amorphous core - metallic glass alloy
none of the above
Answer: C) amorphous core - metallic glass alloy
Confirmed vs Book-3 §1.5 — 'the latest technology is to use amorphous material — a metallic glass alloy for the core', giving about 70% lower core loss and 98.5% efficiency even at 35% load.
Option (a) grain-oriented silicon iron is the CONVENTIONAL core the book contrasts with; copper is a winding material, not a core material.
31. 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
sixteen times the primary current
one-fourth the primary current
two times the primary current
Answer: A) four times the primary current
Confirmed vs Book-3 §1.5 — 'Primary ampere-turns are equal to secondary ampere-turns', so V₁I₁ = V₂I₂. With V₂ = V₁/4, the secondary current must be four times the primary current.
Option (c) is the trap: current and voltage move in OPPOSITE directions in a transformer, so the low-voltage side carries the larger current.
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.
33. If the wet bulb temperature of air is 38 °C, then its relative humidity is __________%.
38 %
90 %
100 %
Insufficient data
Answer: D) Insufficient data
Confirmed vs Book-3 §4.2 - RH cannot be found from WBT alone; the dry-bulb temperature (or another property) is needed to fix the psychrometric state. Hence insufficient data. Options (a)-(c) assume one property fixes the state; the psychrometric chart needs two independent properties, so wet bulb alone cannot give RH.
📖 §1.4 Performance Assessment of Power Factor Capacitors
34. 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
V1²/V2²
V1²/V2
1 - V1²/V2²
1 + V1²/V2²
Answer: A) V1²/V2²
Confirmed vs Book-3 §1.4 Voltage effects — capacitor output varies with the square of the applied voltage, so the kVAr actually produced is in the ratio V₁²/V₂² of the rated value (V₁ = actual, V₂ = rated).
Option (c) 1 − V₁²/V₂² gives the FRACTIONAL DROP in output, not the output itself — that form is used when a question asks 'by how much does the VAr output drop'.
35. The ratings of the PF correction capacitors at motor terminals for a 37 kW induction motor at 3000 rpm synchronous speed will be __________ in comparison to the same sized induction motor at 1500 rpm synchronous speed
more
less
same
dependent on the connected load
Answer: B) less
Confirmed vs Book-3 §1.4 — 3000 rpm synchronous speed is a 2-pole machine and 1500 rpm is a 4-pole machine; the no-load magnetising kVAr (on which the terminal capacitor is sized) is smaller for the higher-speed, lower-pole machine.
So the 3000 rpm motor needs LESS kVAr; option (a) reverses the relationship between pole number and magnetising current.
36. Which gas is used as refrigerant both in vapour compression and vapour absorption systems
Lithium Bromide
Water
HFC 134A
Ammonia
Answer: D) Ammonia
Confirmed vs Book-3 §4.3 - Ammonia is used as the refrigerant in both vapour compression and vapour absorption (with water) systems. Lithium bromide (a) is the absorbent and water (b) the refrigerant only in LiBr machines, while HFC-134a (c) is compression-only; ammonia is used as the refrigerant in both types.
37. In T-5 Fluorescent Lamp, '5' is indicative of:
5 watt power rating
5% energy saving with respect to T8
5/8 generation lamp
Tube diameter
Answer: D) Tube diameter
Confirmed vs Book-3 §8.3 — T5 means a tube of 5/8 inch (16 mm) diameter; T8 = 1 inch (25 mm), T12 = 1.5 inch (38 mm). The number is neither wattage nor a saving percentage (the book's 5 % figure is the T5/T8 efficacy gain over T12, unrelated to the name).
38. The performance of winding of an induction motor can be assessed by which of the following factors?
load current
stator resistance
no load current
both b and c
Answer: D) both b and c
Confirmed vs Book-3 §2.8 Rewinding Effects on Energy Efficiency — The book names two indicators of rewind quality: the no-load current and the stator resistance per phase, both compared with the original values at the same voltage. (a) load current is set by the driven machine and tells nothing about the winding work, so only the combination of (b) and (c) is correct.
39. 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 the above
Answer: D) all the above
Confirmed vs Book-3 §5.2 — Tube-axial, vane-axial and propeller are all axial-flow fans, in which 'air enters and leaves the fan with no change in direction' → all the above (d). Only centrifugal fans turn the airflow (twice).
40. Find the correct example, 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
Confirmed vs Book-3 §7.2 (v)–(vii) — Make-up water must replace all water leaving the circuit – evaporation, blow-down and drift – so M = E + B + D → (a). Any subtraction would mean a loss returns water to the basin. ⚠ Options c and d were duplicates ('M = E - B - D'); c repaired to 'M = E + B − D' so exactly one option is correct.
📖 §1.4 Power factor improvement & benefits (kVA reduction)
41. 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.
📖 §3.5 Efficient operation of compressed air systems (cool air intake)
42. 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.
📖 §4.2 Psychrometrics and air-conditioning processes
43. 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.
📖 §5.6 Fan performance assessment (fan static efficiency)
44. 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.
📖 §6.6 Flow control strategies (impeller trimming)
45. 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.
📖 §7.2 Cooling tower performance (blowdown and COC)
46. 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.
47. Which of the following happens to air when it is cooled through evaporation process in an air washer?
Humidity ratio of the air decreases.
Dry Bulb Temp of air decreases.
Dry Bulb Temp of air increases.
Enthalpy of outlet is air is less than enthalpy of inlet air.
Answer: B) Dry Bulb Temp of air decreases.
In an air washer the water evaporating into the air stream takes its latent heat from the air itself: dry bulb temperature falls, humidity ratio rises, enthalpy and wet-bulb temperature stay nearly constant.
So (a) is backwards (humidity rises), (c) is backwards (DBT falls) and (d) is wrong (the process is essentially adiabatic, so enthalpy is unchanged, not reduced).
The lowest temperature achievable is the air's wet-bulb temperature - the reason evaporative cooling works splendidly in dry climates and hardly at all in humid ones.
48. Which among the following is one of the parameters used to classify fans, blowers & compressors ?
air flow
speed RPM
specific ratio
none of the above
Answer: C) specific ratio
ASME classifies by SPECIFIC RATIO = discharge pressure ÷ suction pressure. Book figures to memorise: fan up to 1.11 (pressure rise up to 1136 mmWg), blower 1.11–1.20 (1136–2066 mmWg), compressor above 1.20. Flow and rpm say nothing about the class — a huge low-pressure fan and a small blower can move the same air.
1. A 15 kW, 415 V, 4 pole, 50 Hz, 3-phase squirrel cage induction motor has full-load efficiency 92% and PF 0.89. Find at full load: a) input power in kW, b) current drawn, c) RPM at full-load slip of 0.8%.
Model answer: a) Pin = 15/0.92 = 16.304 kW. b) I = 16,304/(√3 x 415 x 0.89) = 25.48 A. c) Ns = 120 x 50/4 = 1500 rpm; N = 1500(1-0.008) = 1488 rpm.
Input = output/efficiency; I = Pin/(√3 V cosφ); N = Ns(1-slip).
2. Pitot tube air-flow data in a boiler primary air fan: air temperature 38°C, velocity pressure 47 mmWC, pitot constant Cp = 0.9, air density at 0°C = 1.293 kg/m³. Find the air velocity in m/s.
Model answer: Corrected air density = 273 x 1.293/(273+38) = 1.135 kg/m³. Velocity = Cp x √(2 x 9.81 x Δp x γ)/γ = 0.9 x √(2 x 9.81 x 47 x 1.135)/1.135 ≈ 25.6 m/s.
3. Estimate the cooling tower capacity (TR) and approach: water flow 2 m³/min, specific heat 1 kcal/kg°C, inlet water 43°C, outlet water 35°C, ambient WBT 30°C.
Model answer: Capacity (TR) = (flow x density x sp.heat x ΔT)/3024 = (2x60) x 1000 x 1.0 x (43-35)/3024 = 317.5 TR. Approach = 35 - 30 = 5°C.
TR from heat-load formula; approach = cold water out - ambient WBT.
📖 §6.11 Solved example — cooling water pump efficiency (p.194–195)
4. Cooling water is pumped to three heat exchangers via pipes A, B, C. Pipe A: 0.1 m dia, 1.5 m/s; Pipe B: 0.1 m dia, 1.8 m/s; Pipe C: 0.2 m dia, 2.0 m/s. Measured motor power 50.7 kW, motor efficiency 90%, pump discharge pressure 3.4 kg/cm², suction head 2 m. Determine pump efficiency.
Model answer: Flow A = (π/4)(0.1)² x 1.5 = 0.011786 m³/s; Flow B = (π/4)(0.1)² x 1.8 = 0.014143 m³/s; Flow C = (π/4)(0.2)² x 2.0 = 0.062857 m³/s; total = 0.088786 m³/s. Total head = 34 - 2 = 32 m. Hydraulic power = 0.088786 x 32 x 9.81 = 27.9 kW. Pump efficiency = 27.9 x 100/(50.7 x 0.9) = 61%.
Sum pipe flows; head = discharge head - suction head; efficiency = hydraulic/shaft.
📖 §6.1 Pump types (hydraulic power) & §6.2 System characteristics
5. 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.
6. 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.
📖 §8.6 General energy saving opportunities in lighting
7. 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)
8. 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.
9. 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.
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
1. A 50 HP/415 V, 60 A, 1475 rpm, 3-phase delta-connected squirrel cage motor gave no-load test data: 415 V, 18 A, 50 Hz, stator resistance/phase 0.27 ohm, no-load power 1080 W, ambient 35 degC. Calculate (i) iron + friction and windage losses, (ii) stator resistance at 120 degC, (iii) stator copper loss at 120 degC, (iv) full-load slip and rotor input (rotor losses = slip x rotor input), (v) motor input (stray losses = 0.5% of rated power), (vi) full-load efficiency and PF.
Model answer: (i) No-load stator copper loss = 3 × (I/√3)² × R = 3 × (18/√3)² × 0.27 = 87.5 W; iron + friction & windage loss = 1,080 − 87.5 = 992.5 W.
(ii) R₁₂₀ = R₃₅ × (235 + 120)/(235 + 35) = 0.27 × 355/270 = 0.355 Ω per phase.
(iii) Stator copper loss at full load at 120 °C = 3 × (60/√3)² × 0.355 = 1,278 W.
(iv) Ns = 1,500 rpm, so full-load slip = (1,500 − 1,475)/1,500 = 0.0167 (1.67 %); rated output = 50 HP = 37.3 kW; rotor input = output/(1 − s) = 37,300 / 0.9833 = 37,932 W.
(v) Motor input = rotor input + stator Cu loss + (iron + F&W) + stray = 37,932 + 1,278 + 992.5 + (0.005 × 37,300 = 186.5) = 40,389 W.
(vi) Full-load efficiency = 37,300 / 40,389 × 100 = 92.4 %; full-load PF = 40,389 / (√3 × 415 × 60) = 0.94.
Separate no-load losses into Cu and (iron+F&W); correct stator R to 120 degC; sum stator/rotor Cu, fixed and stray losses to get input, efficiency and PF.
2. A pump-up test on a reciprocating compressor gave: receiver + holdup volume 4100 litres, initial pressure 1 kg/cm2(g), final pressure 8.5 kg/cm2(g), atmospheric pressure 1.026 kg/cm2(a), ambient 32 degC, final compressed air temp 52 degC, pump-up time 65 sec. (a) Calculate the FAD in cfm. (b) A leakage test on the same system: on load 3 min, unloaded 13 min, drawing 145 kW on load. Calculate (i) % leakage, (ii) leakage quantity, (iii) specific power consumption, (iv) power lost due to leakage.
Model answer: (a) V = 4100 L = 4.1 m3. FAD = [(P2 - P1)/P_atm] x V / t = [(8.5 - 1.0)/1.026] x 4.1 / (65/60) min = (7.5/1.026) x 4.1 / 1.0833 = 7.310 x 4.1 / 1.0833 = 27.67 m3/min, with temperature correction (273+32)/(273+52)=305/325=0.938 gives ~25.96 m3/min = about 916 cfm. (b) % leakage = load time/(load + unload) x 100 = 3/(3+13) x 100 = 18.75%. (ii) Leakage quantity = 18.75% of FAD ≈ 0.1875 x 25.96 = 4.87 m3/min (≈172 cfm). (iii) Specific power consumption = 145 kW / FAD; using ~25.96 m3/min = 1557 m3/hr, SPC = 145/1557 = 0.093 kW per m3/hr. (iv) Power lost due to leakage = 18.75% of 145 kW = 27.2 kW.
Confirmed vs Book-3 §3.7 Solved Example — FAD from pump-up formula (temperature corrected); leakage % = load/(load+unload); leakage qty and power loss scale by that %; SPC = power/FAD.
📖 §2.4 Motor efficiency (no-load test and full-load efficiency assessment)
3. 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)
4. 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
5. 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)
6. 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.
7. 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
8. 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).