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BEE Paper-3 — Numericals Drill

154 numerical 5- and 10-mark questions pulled from real Paper-3 papers (2009–2025) and grouped by the formula they test — roughly 1185 marks of descriptive questions in total.

Why this matters: Paper-3 is 150 marks, and 100 of them are descriptive (8×5 + 6×10). Objectives are only 50. The same handful of formulas is reused every year — drill these until they come without thinking, and the descriptive sections stop being the thing that fails you.

Every solution is checked against the 2014 BEE guidebook and carries its section reference. Solutions are collapsed — try the question first, then open it. Printing gives you every solution expanded.

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Ch1 · Transformer losses & efficiency 13 QsCh1 · Power factor & capacitor kVAr 19 QsCh1 · Maximum demand, load factor & tariff 4 QsCh1 · T&D / AT&C losses & harmonics 5 QsCh4 · TR, COP and kW per TR 36 QsCh4 · Psychrometry, ventilation & ACH 5 QsCh3 · FAD, leakage % and specific power 21 QsCh6 · Pump power, efficiency & affinity laws 27 QsCh7 · Range, approach, effectiveness & blowdown 24 Qs

Ch1 · Transformer losses & efficiency

13 past-paper questions · ≈110 marks
P = P_iron + (load fraction)² × P_copper. Efficiency = output / (output + losses). Max efficiency when iron loss = copper loss.
📖 §1.1 Cascade Efficiency + §1.5 Transformer losses

1. A 20 MW co-generation plant operates at a daily load factor of 85% and 8% auxiliary power consumption. The power is generated at 11 kV. Out of the total energy generated, 45% is exported to the grid through a 15 MVA transformer with 99% efficiency. Additionally, 35% of the generated energy is supplied to mill motors at 600 Volts through an 8 MVA step-down transformer with 98.5% efficiency. The remaining energy is used for other LT loads and auxiliaries at 415 Volts through a 4 MVA transformer with 98.2% efficiency. Calculate the following: 1. Daily energy generation in MWh. 2. Daily energy exported in MWh to the grid at 33 kV. 3. Daily mill motors consumption in MWh at 600 V. 4. Daily LT loads and auxiliary consumption in MWh at 415 V. 5. Daily transformer losses in kWh and % transformer losses.

Sep 2024 · 10 marks
Show worked solution
1. Gross generation = 20 MW x 0.85 load factor x 24 h = 408 MWh/day; net (after 8% auxiliary consumption) = 408 x 0.92 = 375.36 MWh/day 2. Export = 375.36 x 0.45 = 168.91 MWh at the 11 kV bus; delivered through the 15 MVA transformer at 99% = 167.22 MWh/day 3. Mill motors = 375.36 x 0.35 = 131.38 MWh; delivered through the 8 MVA transformer at 98.5% = 129.41 MWh/day 4. LT loads & auxiliaries = 375.36 x 0.20 = 75.07 MWh; delivered through the 4 MVA transformer at 98.2% = 73.72 MWh/day 5. Transformer losses = (168.91-167.22) + (131.38-129.41) + (75.07-73.72) = 1.69 + 1.97 + 1.35 = 5.01 MWh = about 5010 kWh/day. As a percentage of the 375.36 MWh actually passing through the transformers, loss = 5010/375,360 = 1.33% (1.23% if expressed on gross generation of 408 MWh). This is the cascade-efficiency idea of Book-3 Sec.1.1 applied inside a plant: each transformation stage multiplies its own efficiency onto the energy delivered.
Gross gen = MW×LF×24; net = gross×(1−aux); split by % and apply transformer efficiencies; losses = before − after.
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R) + §1.6 Distribution Losses

2. A residential colony with fixed load 250 kVA is 1 km from an 11 kV/415 V transformer. Compare LT (1×3.5c×300sqmm) vs HT (1×3c×70sqmm) distribution. Data: LT cable R=0.13 Ω/km, Rs 700/m; HT cable R=0.570 Ω/km, Rs 1300/m; unit Rs 7/kWh; transformer relocation (HT) Rs 1 lakh. Recommend and estimate payback on marginal investment.

17th Sep-2016 · 10 marks
Show worked solution
Current LT = 250/(0.415×1.732) = 347.8 A; HT = 250/(11×1.732) = 13.1 A. Loss LT = 347.8²×0.13×3/1000 = 47.17 kW; Loss HT = 13.1²×0.57×3/1000 = 0.29 kW. Saving = 46.87 kW. Annual energy saving = 46.87×8760 = 4,10,639 kWh; cost saving = Rs 28,74,470/yr. HT investment = 1300×1000 + 1,00,000 = Rs 14,00,000; LT investment = 700×1000 = Rs 7,00,000. Payback on marginal investment = (14,00,000 − 7,00,000)/28,74,470 = 0.24 yr ≈ 3 months. Recommend HT distribution.
I=kVA/(√3·kV); loss=3I²R; payback = marginal investment / annual saving.
📖 §1.5 Transformers — losses & efficiency

3. L-1: Choose a 1500 kVA transformer for a load of 500 kVA for 6 h, 1000 kVA for 6 h, 1500 kVA for 12 h. Transformer-1: iron loss 2.7 kW, full-load copper loss 18.1 kW; Transformer-2: iron loss 3.2 kW, full-load copper loss 19.8 kW. (i) Annual cost of losses (365 days, Rs.6/kWh). (ii) If Transformer-1 costs Rs.25,000 more, justify it.

18th Exam · 10 marks
Show worked solution
Copper loss ∝ (load/rated)². Load fractions: 500/1500=0.333 (6h), 1000/1500=0.667 (6h), 1500/1500=1 (12h). Transformer-1: iron loss/day = 24x2.7 = 64.8 kWh; copper/day = (0.333²x18.1x6)+(0.667²x18.1x6)+(1²x18.1x12) = 12.1+48.3+217.2 = 277.6 kWh; total/yr = (64.8+277.6)x365 = 1,24,976 kWh = Rs.7,49,856. Transformer-2: iron loss/day = 24x3.2 = 76.8; copper/day = 13.2+52.3+237.6 = 303 kWh; total/yr = (76.8+303)x365 = 1,38,663 kWh = Rs.8,31,978. (ii) Annual saving with T-1 = 8,31,978 - 7,49,856 = Rs.82,122. Payback of extra Rs.25,000 = 25000/82122 = 0.3 yr (~4 months) — well justified.
Iron loss is constant (24 h); copper loss scales with load² and operating hours; cost = energy loss x tariff; payback = extra cost / annual saving.
📖 §1.1 Cascade Efficiency

4. L-4: A 10 MW co-gen plant runs at 85% daily load factor, generating at 11 kV. 35% exported to grid via 7.5 MVA transformer (99% eff); 32% to mill motors at 600 V via 5 MVA transformer (98% eff); balance to LT loads/auxiliaries at 415 V via 2 MVA transformer (98% eff). Calculate: (1) daily energy exported; (2) daily mill motor consumption; (3) daily LT/auxiliary consumption; (4) daily transformer losses (kWh and %).

Sep 2019 · 10 marks
Show worked solution
Daily generation = 10,000 x 0.85 x 24 = 2,04,000 kWh. (1) Export = 2,04,000 x 0.35 = 71,400 kWh; 7.5 MVA loss = 71,400 x 0.01 = 714 kWh; net export = 70,686 kWh. (2) Mill = 2,04,000 x 0.32 = 65,280 kWh; 5 MVA loss = 65,280 x 0.02 = 1,306 kWh; net = 63,974 kWh. (3) LT/aux = 2,04,000 x 0.33 = 67,320 kWh; 2 MVA loss = 67,320 x 0.02 = 1,346 kWh; net = 65,974 kWh. (4) Total transformer losses = 714 + 1,306 + 1,346 = 3,366 kWh/day; % loss = 3,366/2,04,000 x 100 = 1.65%.
Printed solution: net export 70,686 kWh, net mill 63,974 kWh, net LT 65,974 kWh, total tx loss 3,366 kWh = 1.65%.
📖 §1.5 Transformers — losses & efficiency

5. a) A small scale industry has a constant load of 380 kVA. It has installed two transformers of 500 kVA each. The no load loss and full load copper loss of each 500 kVA transformer is 750 W and 5410 W respectively. From the energy efficiency point of view should the industry operate a single transformer or two transformers equally sharing the load? b) A no load test on a three phase delta connected induction motor gave: No load power = 890 W, Stator resistance per phase at 30°C = 0.233 Ohms, No load current = 14.5 A. Calculate the fixed losses for the motor.

9th Dec-2009 · 10 marks
Show worked solution
a) Single 500 kVA at 380 kVA load: loss = 750 + (380/500)² × 5410 = 750 + 3124.8 = 3874.8 W. Two transformers each at 190 kVA: loss = 2 × [750 + (190/500)² × 5410] = 2 × 1531.2 = 3062.9 W. Two transformers are better — losses are least, saving 812.4 W. b) Stator copper loss at no load = 3 × (14.5/√3)² × 0.233 = 48.985 W. Fixed losses = 890 − 48.985 = 841 W.
a) Compare total transformer losses (no-load + load-proportional copper loss) for single vs paralleled operation. b) Fixed (iron + friction + windage) losses = no-load input power minus no-load stator copper loss.
📖 §1.5 Transformers — losses & efficiency + §1.6 options for distribution-loss optimization

6. A unit has 2 identical 500 kVA transformers, each with no-load loss 800 W and full-load copper loss 5000 W. Plant load is 400 kVA. Compare transformer losses for single transformer operation vs two transformers in parallel. Also list any five options to minimise electrical distribution loss.

10th Jul-2010 · 10 marks
Show worked solution
Single transformer loss = 800 + (400/500)²×5000 = 4000 W. Two transformers (each sharing 200 kVA): 2×[800 + 5000×(200/500)²] = 3200 W. Single-transformer operation has 800 W higher loss. Distribution-loss options: relocate transformers/substations near load centres; re-route/re-conductor high-loss feeders; PF improvement with capacitors at load end; optimum loading of transformers; use lower-resistance AAAC instead of ACSR; minimise losses at weak links (jumpers, loose contacts, brittle conductors); improve HT:LT ratio to shorten LT network.
Loss = iron loss + (load/rating)²×copper loss; parallel operation halves the load share, reducing total copper loss. Any five distribution-loss options accepted.
📖 §1.1 Power Generation Plant — Heat Rate & generation efficiency — multi-chapter fill-in-the-blanks

7. Fill in the blanks (1 mark each): a) Heat rate of a thermal power plant is expressed in ___; b) The ___ loss is independent of load in a transformer; c) ___ is used to reduce the dew point in a compressed air system; d) The speed of an energy efficient motor will be more than the standard motor of same capacity because ___ decreases.

Mar 2021 (Set B) · 5 marks
Show worked solution
a) kCal/kWh (or kJ/kWh) - the heat input per kWh generated; 1 kWh = 860 kCal = 3600 kJ. b) Core loss (iron loss / no-load loss) - the book states core loss 'occurs whenever the transformer is energized; core loss does not vary with load'. c) An air dryer (refrigerant or desiccant type) is used to lower the dew point of compressed air. d) SLIP - an energy-efficient motor has lower rotor losses and therefore lower slip, so for the same number of poles it runs slightly faster than a standard motor.
Standard BEE fill-in answers: thermal heat rate in kCal/kWh, transformer iron/core loss is load-independent, air dryer lowers dew point, EE motor has higher speed because slip decreases.
📖 §1.4 Power Factor Improvement and Benefits + §1.5 Transformer losses & efficiency

8. a) A cold rolling mill has a maximum demand of 7 MVA at power factor of 0.95. The plant management converts the existing electrical resistance annealing furnace having steady load of 1250 kW to gas heating as a cost reduction measure. The existing capacitor banks (kVAr) continued to be in the electrical network. What will be the effect on maximum demand and power factor due to this conversion? (5 Marks) b) A cement plant has a constant load of 15 MVA. It has installed two transformers of 30 MVA each. The no load loss and full load copper loss of each 30 MVA transformer is 25 kW and 75 kW respectively. From the energy efficiency point of view the industry management wants to take a decision on whether to operate a single transformer or two transformers equally sharing the load. What is your recommendation? (5 Marks)

Mar 2021 · 10 marks
Show worked solution
a) Registered max demand = 7 MVA = 7000 kVA. Electrical real load = 7000 x 0.95 = 6650 kW. Existing kVAr = sqrt(kVA^2 - kW^2) = sqrt(7000^2 - 6650^2) = 2186 kVAr (remains in network). Reduction in real power = 1250 kW (furnace converted to gas). Revised real power = 6650 - 1250 = 5400 kW. Revised kVA = sqrt(5400^2 + 2186^2) = 5825 kVA. Reduction in electrical demand = 7000 - 5825 = 1175 kVA. Revised power factor = 5400/5825 = 0.927. Reduction in PF = 0.95 - 0.927 = 0.023. b) Option 1 (one transformer): % load = 15/30 = 50%; Total loss = NoLoad + Cu loss x (%load)^2 = 25 + 75 x 0.5^2 = 43.75 kW. Option 2 (both transformers): each at 7.5/30 = 25%; Total loss = [25 + 75 x 0.25^2] x 2 = 59.37 kW. Recommendation: operate single transformer because losses are lower, saving 59.37 - 43.75 = 15.62 kW.
kVAr from kVA & kW; revised demand after load removal; transformer loss = NL + FL_Cu x (loading)^2.
📖 §1.4 Power Factor Improvement and Benefits + §1.5 Transformer losses & efficiency

9. a) A cold rolling mill has a maximum demand of 7 MVA at a power factor of 0.95. The plant management converts the existing electrical resistance annealing furnace having steady load of 1250 kW to gas heating as a cost reduction measure. The existing capacitor banks (kVAr) continued to be in the electrical network. What will be the effect on maximum demand and power factor due to this conversion? (5 marks). b) A cement plant has a connected load of 15 MVA. It has installed two transformers of 30 MVA each. The no load loss and full load copper loss of each 30 MVA transformer is 25 kW and 75 kW respectively. From the energy efficiency point of view the industry management wants to take a decision on whether to operate a single transformer or two transformers equally sharing the load. What is your recommendation? (5 marks).

Mar 2021 (Set B) · 10 marks
Show worked solution
a) Registered maximum demand = 7 MVA = 7000 kVA. Electrical load (real power) = 7000 x 0.95 = 6650 kW. kVAr = sqrt(kVA^2 - kW^2) = sqrt(7000^2 - 6650^2) = 2186 kVAr. kVAr in the plant will remain same. After converting 1250 kW furnace to gas, real power reduces by 1250 kW: Revised Real Power = 6650 - 1250 = 5400 kW. Revised kVA = sqrt(kW^2 + kVAr^2) = sqrt(5400^2 + 2186^2) = 5825 kVA. Reduction in electrical demand = 7000 - 5825 = 1175 kVA. Revised power factor = 5400/5825 = 0.927. Reduction in power factor = 0.95 - 0.927 = 0.023 (power factor drops). b) Option 1 - One transformer in operation: % load = 15/30 = 50%; Total Loss = No-load loss + Copper loss x (%load)^2 = 25 + 75 x (0.5)^2 = 25 + 18.75 = 43.75 kW. Option 2 - Both transformers in operation: % load = 7.5/30 = 25% each; Total Loss = [No-load loss + Copper loss x (%load)^2] x 2 = [25 + 75 x (0.25)^2] x 2 = [25 + 4.69] x 2 = 59.37 kW. Recommendation: Operate single transformer because losses are less; saving of 59.37 - 43.75 = 15.62 kW.
Part a: kVAr fixed; removing real load lowers kW so kVA falls but PF reduces. Part b: total transformer loss = no-load loss + copper loss x (load fraction)^2; compare one vs two transformers.
📖 §1.4 Power Factor Improvement and Benefits + §1.5 — multi-chapter True/False

10. State True or False (1 Mark each): 1. In an industrial electrical system operating at unity power factor, addition of further capacitors will reduce the maximum demand (kVA). 2. In a step-down transformer for a given load the current in the primary will be much lower than the current in the secondary. 3. For the same no of poles and kVA rating, the RPM of an energy efficient motor is higher than that of a standard motor. 4. The advantage of evaporative cooling is that it is possible to obtain water temperatures below the wet bulb economically. 5. A fluid coupling changes the speed of the driven equipment without changing the speed of the motor.

Jul 2022 · 5 marks
Show worked solution
1. FALSE - at unity power factor the reactive component is already zero; adding more capacitors makes the current LEAD and the kVA (and hence maximum demand) rises again. 2. TRUE - primary ampere-turns = secondary ampere-turns, so V1I1 = V2I2. In a step-down transformer the primary is the high-voltage side and therefore carries the LOWER current. 3. TRUE - an energy-efficient motor has lower rotor I^2R loss and hence lower slip, so for the same number of poles and rating it runs at a slightly HIGHER rpm than a standard motor. 4. FALSE - evaporative cooling can approach but never economically go below the wet bulb temperature; the difference (cold water temp - WBT) is the 'approach'. 5. TRUE - a fluid coupling varies the output (driven) speed by varying the oil fill while the motor continues to run at its own constant speed.
At unity PF capacitors over-correct and can increase kVA; primary current of step-down transformer is lower than secondary; EE motor runs at nearly same/slightly higher speed but not 'higher RPM' as stated; evaporative cooling cannot go below wet bulb; fluid coupling varies output speed while motor runs constant.
📖 §1.5 Energy Efficient Transformers & Standards/Labeling Programme — multi-chapter fill-in-the-blanks

11. Fill in the blanks for the following: 1. The main input energy used for refrigeration in vapor absorption refrigeration plants is _____. 2. One ton of refrigeration is equivalent to _____ kW. 3. Stray losses in induction motor generally are proportional to the square of the _____ current. 4. The unit of Ah*EPI is _____. 5. If the pump impeller diameter is reduced by 10% then head reduces by _____%. 6. A 4 pole 50Hz motor operating with slip of 3% will have a shaft speed of _____ RPM. 7. Effective Aperture Glazing (EA) = VLT × _____. 8. In an amorphous core distribution transformer, no-load loss is _____ than a conventional transformer. 9. As the condensing temperature increases, kW/TR of refrigeration system will _____. 10. The extent of drying compressed air is expressed by the term _____.

Jul 2022 · 5 marks
Show worked solution
1. Thermal energy (steam / waste heat / hot water / fuel gas) - the vapour absorption machine is heat-driven, not compressor-driven. 2. 3.51 kW (1 TR = 3024 kCal/h = 3.51 kW). 3. ROTOR current - stray load losses in an induction motor vary as the square of the rotor current. 4. EPI (Energy Performance Index) is expressed in kWh/sq.m/year. 5. 19% - head varies as the square of impeller diameter, so 1 - 0.9^2 = 0.19. 6. 1455 rpm - Ns = 120 x 50/4 = 1500 rpm; at 3% slip, N = 1500 x 0.97 = 1455 rpm. 7. Window-to-Wall Ratio (WWR): Effective Aperture = VLT x WWR. 8. LESS (about 70% lower core loss than a conventional CRGO silicon-iron core - Book-3 Sec.1.5). 9. INCREASE - a higher condensing temperature raises the compression ratio and hence the specific power kW/TR. 10. Dew point (atmospheric / pressure dew point).
Fill-in answers from BEE Book-3 fundamentals; 1 TR=3.517 kW; head∝D² so 10% dia drop → ~19% head drop; N=120f/p×(1-s)=1500×0.97=1455 rpm.
📖 §1.4 Power Factor Improvement and Benefits

12. A textile industry had installed a 2 MVA transformer. The initial demand of the plant was 1500 kVA with power factor of 0.75. Industry has installed 450 kVA capacitor at the motor end. Calculate the following: 1. Reduction in apparent power (kVA); 2. Improved power factor; 3. Revised % loading of transformer after installing the capacitor.

Mar 2023 · 5 marks
Show worked solution
Real power = 1500×0.75 = 1125 kW. Old reactive power = √(1500²−1125²) = 992 kVAr. After 450 kVAr capacitor, revised kVAr = 992−450 = 542 kVAr. Revised apparent power = √(1125²+542²) = 1248 kVA. Reduction in apparent power = 1500−1248 = 252 kVA. Improved PF = 1125/1248 = 0.90. Revised % loading = 1248/2000 = 62.4%.
Resolve into real and reactive components, subtract capacitor kVAr, recompute apparent power, pf and transformer loading (kVA/rating).
📖 §1.11 Analysis of Electrical Power Systems (Table 1.11) — multi-chapter True/False

13. L-6: State True or False (1 mark each): 1. The efficiency of gas turbine power plant is lower than that of a combined cycle power plant. 2. The performance of air compressor at high altitudes will be lower as compared to that at sea level. 3. Efficiency of transformer will be minimum when copper loss is equal to iron losses. 4. In cooling towers, the water droplets entrapped in the air stream is captured by drift eliminators. 5. To get the static pressure, the inner and outer tubes of pitot tube are connected to manometer. 6. The throttling of pump discharge will change the pump characteristic curve. 7. The simplest way to reduce the discharge from a reciprocating air compressor is to throttle it. 8. Cycle of Concentration (COC) is the ratio of dissolved solids in circulating water to the dissolved solids in makeup water. 9. Use of VFD will save power but also create harmonics. 10. The synchronous speed of a 4 pole motor will be 3000 rpm.

17th Sep-2016 · 10 marks
Show worked solution
1. True 2. True 3. False (transformer efficiency is MAXIMUM when copper loss equals iron loss) 4. True 5. False (inner and outer tubes connected to the manometer give velocity pressure; static pressure is from the outer tube alone) 6. False (throttling changes the system curve, not the pump characteristic curve) 7. False (throttling a reciprocating compressor is not the way to reduce discharge; use unloading/speed control) 8. True 9. True 10. False (Ns = 120 x 50 / 4 = 1500 rpm)
Cross-chapter True/False. The two Chapter-1 items are (3) and (9): transformer efficiency is MAXIMUM (not minimum) when copper loss equals iron loss, because the variable copper loss then just equals the fixed core loss (Book-3 §1.5); and VFDs do save power but, being non-linear power-electronic loads, they inject harmonic currents (Book-3 §1.10). Item 10 is settled by Ns = 120f/P = 120 x 50/4 = 1500 rpm, not 3000 rpm.

Ch1 · Power factor & capacitor kVAr

19 past-paper questions · ≈150 marks
PF = kW / kVA. kVAr = kW (tanφ₁ − tanφ₂). Improving PF cuts current and I²R losses.
📖 §1.4 Selection and Location of Capacitors

1. A steel manufacturing facility is powered by a 3-phase, 6.6 kV, 50 Hz supply and operates the following electrical loads: An electric arc furnace consumes 1.2 MW at a lagging power factor of 0.65, a bank of induction motors for rolling operations consumes 800 kW at a 0.80 lagging power factor, and the lighting and instrumentation systems consume 100 kW at unity power factor. Due to utility regulations, the overall plant power factor must be improved to 0.95 lagging. A capacitor bank will be installed for compensation. As an energy auditor evaluate the following: a. Total active power consumption. (1 Mark) b. Total initial apparent power drawn by the facility. (1 Mark) c. The operating power factor. (1 Mark) d. Determine the total reactive power required to achieve the desired power factor. (2 Mark)

Sep 2025 · 10 marks
Show worked solution
Load-wise breakdown (kVAr = kW x tan(cos^-1 PF)): Arc furnace: 1200 kW, PF 0.65 -> kVA = 1200/0.65 = 1846 kVA, kVAr = 1846 x sin(49.5deg) = 1403 kVAr Induction motors: 800 kW, PF 0.80 -> kVA = 1000 kVA, kVAr = 600 kVAr Lighting & instrumentation: 100 kW, PF 1.0 -> kVA = 100 kVA, kVAr = 0 a) Total active power = 1200 + 800 + 100 = 2100 kW b) kVA must be added VECTORIALLY, not arithmetically: total kVAr = 1403 + 600 + 0 = 2003 kVAr, so total kVA = sqrt(2100^2 + 2003^2) = 2902 kVA c) Operating power factor = kW/kVA = 2100/2902 = 0.724 lag d) kVAr required = kW[tan(cos^-1 0.724) - tan(cos^-1 0.95)] = 2100 x (0.9538 - 0.3287) = 1313 kVAr; equivalently 2003 - (2100 x 0.3287) = 1313 kVAr. (Capacitor bank of about 1300-1350 kVAr, preferably switched in steps through an APFC because the arc-furnace PF swings over the melting cycle - Book-3 Sec.1.4.)
Sum kW and kVAr per load; operating PF = ΣkW/ΣkVA; compensation kVAr = kW(tanφ1 − tanφ2).
📖 §1.4 Selection and Location of Capacitors

2. During April-2003 a plant recorded a maximum demand of 600 kVA and an average PF of 0.82 lag. The utility requires a minimum average PF of 0.92 lag, and every 1% dip in PF attracts a penalty of Rs 10,000 per month. (a) Calculate the improvement in PF for May-2003 by installing 100 kVAr capacitors. (b) Calculate the penalty to be paid, if any, during May-2003.

Book EOC · 5 marks
Show worked solution
At 600 kVA and 0.82 PF: kW = 600 x 0.82 = 492 kW; kVAr = 600 x sin(cos^-1 0.82) = 600 x 0.5724 = 343.4 kVAr. After adding 100 kVAr capacitor, new kVAr = 243.4; new kVA = sqrt(492^2 + 243.4^2) = sqrt(242064 + 59243) = 548.9 kVA; new PF = 492/548.9 = 0.896 (about 0.90). (a) PF improves from 0.82 to ~0.90 (an improvement of about 0.08). (b) Required PF is 0.92, so PF still falls short by about 2% (0.92 - 0.90); penalty for May-2003 is approximately 2 x Rs 10,000 = Rs 20,000.
Resolve load into kW and kVAr, subtract the capacitor kVAr, recompute kVA and PF; compare with the 0.92 target to find the shortfall and penalty.
📖 §1.4 Selection and Location of Capacitors

3. A process plant consumes 12,500 kWh per month at 0.9 power factor. What is the percentage reduction in distribution losses per month if the PF is improved to 0.96 at the load end? Assume existing distribution loss is 4% of the plant energy consumption.

Book EOC · 5 marks
Show worked solution
Distribution loss is proportional to I^2, hence to 1/PF^2. Reduction factor = 1 - (PF_old/PF_new)^2 = 1 - (0.9/0.96)^2 = 1 - 0.8789 = 0.1211, i.e. about 12.1% reduction in losses. In energy terms, existing loss = 4% of 12,500 = 500 kWh/month; new loss = 500 x (0.9/0.96)^2 = 439.5 kWh; saving ≈ 60.5 kWh/month.
Losses scale as 1/PF^2; improving PF from 0.9 to 0.96 cuts losses by 1 - (0.9/0.96)^2 ≈ 12.1%.
📖 §1.4 Power Factor Improvement and Benefits

4. A 37 kW, 3 phase, 415 V induction motor draws 56 A and 33 kW power at 410 V. What is the apparent and reactive power drawn by the motor at the operating load?

15th Exam · 5 marks
Show worked solution
Apparent power = 1.7321 x 0.410 x 56 = 39.769 kVA. Active power = 33 kW. Reactive power = √(39.769² - 33²) = √(1581.57 - 1089) = 22.19 kVAr.
Apparent power S = √3 x V x I; reactive power Q = √(S² - P²).
📖 §1.4 Selection and Location of Capacitors

5. a) A 3-phase 415 V 75 kW induction motor draws 48 kW at 0.7 PF. Calculate the capacitor rating to improve PF to 0.95, the reduction in current and kVA reduction at 415 V. b) A plant consumes 2,00,000 kWh/month at 0.9 PF. What is the % reduction in distribution losses if PF is improved to 0.96 at load end?

15th Exam · 10 marks
Show worked solution
a) kVAr = kW[tan(cos^-1 0.70) - tan(cos^-1 0.95)] = 48 x (1.020 - 0.329) = 33.2 kVAr (say 33 kVAr). Current at 0.70 PF = 48/(1.732 x 0.415 x 0.70) = 95.4 A; at 0.95 PF = 48/(1.732 x 0.415 x 0.95) = 70.3 A; reduction = 25.1 A (26%). kVA at 0.70 = 48/0.70 = 68.6 kVA; at 0.95 = 48/0.95 = 50.5 kVA; kVA released = 18.1 kVA. b) % reduction in distribution loss = [1 - (PF1/PF2)^2] x 100 = [1 - (0.90/0.96)^2] x 100 = 12.1%. In energy terms, if the existing loss were 4% of 2,00,000 kWh = 8000 kWh/month, the saving is about 970 kWh/month.
Capacitor kVAr = kW(tanφ1 - tanφ2); loss reduction = 1 - (PF1/PF2)².
📖 §1.4 Power Factor Improvement and Benefits

6. State three advantages of improvement of Power Factor at Load side. Power Factor at the load side is 0.75 and average minimum load is 100 kW. What is the kVAr rating of capacitor to improve the Power Factor at the load side to 0.95?

17th Sep-2016 · 10 marks
Show worked solution
Advantages (any three): reduced kVA (maximum demand) charges in utility bill; reduced distribution losses (kWh) due to lower current; better voltage at motor terminals and improved motor performance; reduction in size of transformers; avoidance of PF penalty / availing PF incentives; better operating efficiency of motors/drives. Capacitor required = 100{tan(cos⁻¹0.75) − tan(cos⁻¹0.95)} = 100(0.882 − 0.329) = 55.3 kVAr, say 55 kVAr.
kVAr = kW[tan(cos⁻¹PF1) − tan(cos⁻¹PF2)].
📖 §1.4 Selection and Location of Capacitors

7. A single line diagram shows a 100 kW heater load and a 200 kW motor (200 m from the 415 V LT bus). Main incoming line PF is 0.85 lag. Calculate the rating of capacitors to improve the main incoming line PF to 0.9 lag.

16th Exam · 5 marks
Show worked solution
The capacitor formula of Book-3 Sec.1.4 is kVAr = kW[tanPhi1 - tanPhi2], where kW is the TOTAL active power flowing at the point whose power factor is being corrected - here the main incoming line, which carries 100 kW (heater) + 200 kW (motor) = 300 kW. tan(cos^-1 0.85) = 0.6197; tan(cos^-1 0.90) = 0.4843. kVAr = 300 x (0.6197 - 0.4843) = 300 x 0.1354 = 40.6 kVAr, say a 40-45 kVAr bank. Check: at 0.85 PF, kVAr drawn = 300 x 0.6197 = 186 kVAr; after correction 300 x 0.4843 = 145 kVAr; difference = 41 kVAr. The reactive power comes only from the motor (the heater is resistive), but the SIZING must use the total 300 kW measured at the incoming line - using only the 200 kW motor load under-sizes the bank.
Compensate only inductive (motor) load; kVAr = kW[tanφ1 − tanφ2].
📖 §1.4 Power Factor Improvement and Benefits

8. a) List five disadvantages of low power factor. b) An industry pays penalty for poor PF of 0.88; utility minimum is 0.9. Penalty = 1% of energy cost for every 0.01 PF below minimum; incentive = 1.5% for every 0.01 improvement above 0.95. Monthly energy bill Rs 6 lakhs. Calculate annual cost saving if PF improved to unity from the current level.

16th Exam · 10 marks
Show worked solution
a) Disadvantages: large line (copper) losses; large kVA rating and size of electrical equipment; greater conductor size and cost; poor voltage regulation / large voltage drop; low efficiency; PF penalty from supply company. b) From 0.88 to 1.0: penalty avoided = 2×1.0% (0.88→0.90) = 2.0%; incentive above 0.95 = 5×1.5% (0.95→1.00) = 7.5%. Total = 9.5%. Monthly saving = 6,00,000×9.5% = Rs 57,000. Annual = 57000×12 = Rs 6,84,000.
Penalty avoided + incentive earned, applied to monthly bill × 12.
📖 §1.4 Selection and Location of Capacitors

9. L-5: (a) A 3-phase, 50 kW rated induction motor drawing 44 kW at 0.75 lagging PF. What capacitor (kVAr) per phase is needed to improve PF to 0.96? What is the reduction in current and kVA at 415 V? (b) List five energy losses in an induction motor.

18th Exam · 10 marks
Show worked solution
(a) tanΦ1 = tan(cos⁻¹0.75) = 0.88; tanΦ2 = tan(cos⁻¹0.96) = 0.29. Required kVAr = 44 x (0.88-0.29) = 25.96 kVAr; per phase = 25.96/3 = 8.65 kVAr. Current at 0.75 PF = 44/(√3 x 0.415 x 0.75) = 81.6 A; at 0.96 PF = 63.76 A; reduction = 17.84 A. kVA at 0.75 = 44/0.75 = 58.67; at 0.96 = 44/0.96 = 45.83; reduction = 12.84 kVA. (b) Five motor losses: 1. Iron (core) loss, 2. Stator I²R (copper) loss, 3. Rotor I²R (copper) loss, 4. Friction and windage loss, 5. Stray load loss.
Capacitor kVAr = P(tanΦ1-tanΦ2); current and kVA from P/(√3·V·PF) and P/PF; standard five induction-motor loss categories.
📖 §1.4 Power Factor Improvement and Benefits

10. S-5: List any five benefits of power factor improvement in an industrial power distribution system.

19th Exam · 5 marks
Show worked solution
1. Reduced kVA demand and lower maximum-demand charges. 2. Reduced line/transformer current → lower I²R distribution losses. 3. Released capacity in transformers, cables and switchgear for additional load. 4. Improved voltage regulation (less voltage drop) at the load. 5. Avoidance of low-PF penalties and possible PF-incentive rebates from the utility. (Also: longer equipment life due to reduced heating.)
Standard PF-improvement benefits from Book-3 Chapter 1 (the paper directs 'Refer Guide Book No 3, Chapter 1, Page No 11'); model answer supplied from chapter notes.
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback

11. L-1: A food processing plant has contract demand 2500 kVA; average MD 2000 kVA at 0.95 PF; MD billed at Rs.300/kVA; minimum billable MD = 75% of contract demand; incentive 0.5% reduction in energy charges per 0.01 PF increase above 0.95; average monthly energy charge Rs.10 lakhs. Plant improves PF to unity. Determine capacitor kVAr, annual reduction in MD charges and energy charge, and simple payback if capacitors cost Rs.800/kVAr.

19th Exam · 10 marks
Show worked solution
kW drawn = 2000 x 0.95 = 1900 kW. kVAr = 1900 x (tan(cos⁻¹0.95) - tan(cos⁻¹1)) = 1900 x (0.329 - 0) = 625 kVAr. Capacitor cost = 625 x 800 = Rs.5,00,000. New MD at unity PF = 1900 kVA; but minimum billable = 75% x 2500 = 1875 kVA → billed at 1900 kVA (above floor). Reduction in MD = 2000 - 1900 = 100 kVA → demand saving = 100 x 300 = Rs.30,000/month = Rs.3,60,000/yr. PF rises 0.95→1.00 (0.05 = 5 steps of 0.01) → energy charge reduction = 5 x 0.5% = 2.5%; monthly saving = 10,00,000 x 0.025 = Rs.25,000 → Rs.3,00,000/yr. Total annual saving = 3,60,000 + 3,00,000 = Rs.6,60,000. Payback = 5,00,000/6,60,000 = 0.76 yr ≈ 9 months.
Capacitor kVAr = kW(tanΦ1-tanΦ2); MD saving from reduced kVA × tariff; energy saving from PF-incentive %; payback = investment/annual saving.
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand

12. A foundry draws 2500 kW. Demand during furnace operation: 5 min 2940 kVA, 7 min 2550 kVA, 3 min 2777 kVA. Billing meter monitors demand every 15 minutes. Calculate the maximum demand registered and the average PF during the interval.

Sep 2019 · 5 marks
Show worked solution
Maximum demand is the time-integrated kVA over the 15-minute cycle (Book-3 Sec.1.2): MD = [(2940 x 5) + (2550 x 7) + (2777 x 3)] / 15 = [14,700 + 17,850 + 8331] / 15 = 40,881/15 = 2725.4 kVA. Average power factor over the interval = kW / kVA = 2500 / 2725.4 = 0.917 (about 0.92 lag). Note the demand billed is the 2725 kVA average, not the 2940 kVA instantaneous peak.
Printed solution: MD = 2725.4 kVA, average PF = 0.92 (time-weighted).
📖 §1.4 Selection and Location of Capacitors

13. S-8: A 100 kW heater load and a 200 kW motor (200 m from the 415V LT bus). Main incoming line PF is 0.85 lag. Calculate the rating of capacitors to improve the PF of the main incoming line to 0.9 lag.

16th Exam (alt set) · 5 marks
Show worked solution
Use kVAr = kW[tanPhi1 - tanPhi2] with the TOTAL active power at the point being corrected (Book-3 Sec.1.4). The main incoming line carries 100 kW heater + 200 kW motor = 300 kW at 0.85 lag. tan(cos^-1 0.85) = 0.6197; tan(cos^-1 0.90) = 0.4843. kVAr required = 300 x (0.6197 - 0.4843) = 40.6 kVAr, i.e. a 40-45 kVAr capacitor bank. (The heater draws no kVAr, but it does carry active power through the incoming line, so it must be included in the kW used for sizing.)
Printed solution: required capacitor rating 27 kVAr.
📖 §1.4 Power Factor Improvement and Benefits

14. L-6: (a) List five disadvantages of low power factor. (b) An industry maintains poor PF of 0.88; utility minimum is 0.9. Penalty 1% on energy cost for every 0.01 below minimum; incentive 1.5% for every 0.01 above 0.95. Monthly energy bill Rs.6 lakhs. Calculate annual cost saving if PF improved to unity from current level.

16th Exam (alt set) · 10 marks
Show worked solution
(a) Disadvantages of low PF: (1) larger line/copper losses; (2) larger kVA rating and size of equipment; (3) greater conductor size and cost; (4) poor voltage regulation/large voltage drop; (5) low efficiency; (6) penalty from supply company. (b) Penalty avoided: 0.88 to 0.90 = 0.02 PF x 1% = 2.0%. Incentive at unity: from 0.95 to 1.00 = 0.05 x 1.5% = 7.5%. Total energy saving = 2.0% + 7.5% = 9.5%. Monthly cost reduction = 6 lakh x 9.5% = Rs.57,000. Annual = 57,000 x 12 = Rs.6,84,000.
Printed solution: total benefit 9.5%, monthly Rs.57,000, annual Rs.6,84,000.
📖 §1.4 Performance Assessment of Power Factor Capacitors

15. a) A 10 kVAr, 415 V rated power factor capacitor was found to be having terminal supply voltage of 440 V. Calculate the capacity of the power factor capacitor at the operating supply voltage. b) What would be the nearest kVAr compensation required for changing the power factor of a 500 kW load from 0.9 lead to unity power factor?

9th Dec-2009 · 5 marks
Show worked solution
a) Capacitor output varies as the square of the applied voltage (Book-3 Sec.1.4, Voltage effects): kVAr = 10 x (440/415)^2 = 10 x 1.124 = 11.24 kVAr. The bank delivers more than its rating, but running above rated voltage shortens capacitor life. b) At 0.9 LEADING the load is already over-compensated - the current leads the voltage - so NO further capacitive kVAr is required. To reach unity, capacitance must instead be REMOVED: excess kVAr = 500 x tan(cos^-1 0.9) = 500 x 0.4843 = 242 kVAr of the existing bank should be switched out.
a) Capacitor kVAr varies with square of voltage ratio. b) Load is already at leading PF, so additional capacitive compensation is not needed.
📖 §1.4 Power Factor Improvement and Benefits — multi-chapter fill-in-the-blanks

16. Fill in the blanks: (a) If the reactive power drawn by a load is zero, the load operates at __ power factor. (b) Power factor is the ratio of ___. (c) As the approach increases (other parameters constant), the effectiveness of a cooling tower ___. (d) Lower power factor of a DG set demands ___ excitation currents. (e) If voltage applied to a 415 V rated capacitor drops by 5%, its VAr output drops by about ___%.

10th Jul-2010 · 10 marks
Show worked solution
(a) unity; (b) kW/kVA (active power/apparent power); (c) decreases; (d) higher; (e) 10
(a) Zero reactive power means purely resistive load → unity PF. (b) PF = active/apparent power. (c) Higher approach = poorer cooling = lower effectiveness. (d) Lower PF needs more excitation. (e) kVAr∝V²; 5% drop → ~(1−0.95²)=9.75≈10% drop. 1 mark each.
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback

17. L-1: The contract demand of a process plant is 6000 kVA. The average monthly recorded maximum demand is 5500 kVA at 0.78 PF. Tariff: (a) Minimum monthly billing demand is 75% of contract demand or actual recorded MD whichever is higher; no PF incentives. (b) Monthly MD charge is Rs. 400 per kVA. Find the optimum limit of PF capacitor requirement (purely to reduce MD so no excess demand charges are paid) and the simple payback period, assuming capacitor + APFC controller cost is Rs. 500 per kVAr.

11th Feb-2011 · 10 marks
Show worked solution
Minimum payable demand = 6000 x 0.75 = 4500 kVA. Margin for MD reduction = 5500 - 4500 = 1000 kVA. Present maximum load = 5500 x 0.78 = 4290 kW. Desired peak PF to achieve MD of 4500 kVA = 4290/4500 = 0.9533. PF capacitor requirement = 4290 [tan(Cos-1 0.78) - tan(Cos-1 0.9533)] = 4290(tan 38.74 - tan 17.579) = 4290(0.80226 - 0.316815) = 4290(0.4854) = 2083 kVAr. Cost of capacitor installation = 500 x 2083 = Rs. 10.4 lakhs. Monthly MD saving = 1000 kVA; Yearly savings = 1000 x 400 x 12 = Rs. 48.0 lakhs. Simple payback = 10.4/48 = 0.21 years = 2.6 months.
Reduce MD to minimum billable 4500 kVA; required kVAr from tan(phi1)-tan(phi2) at the active load of 4290 kW; payback = investment/annual MD savings.
📖 §1.11 Solved Example — MD, PF capacitor kVAr & payback

18. A review of electricity bills of a process plant was conducted as a part of energy audit. The plant has a contract demand of 3000 kVA with the power supply company. The average maximum demand of the plant is 2400 kVA/month at a power factor of 0.95. The maximum demand is at 80% of the contract demand. The minimum billable maximum demand is 80% of the contract demand. An incentive of 0.5 % reduction in energy charges component of electricity bill are provided for every 0.01 increase in power factor over and above 0.95. The average energy charge component of the electricity bill per month for the plant is Rs.80 lakhs. Calculate the following: a) If the plant decides to improve the power factor to unity, determine the power factor capacitor kVAr required and the associated monetary benefits. b) What will be the simple payback period if the cost of power factor capacitors is Rs.1200/kVAr.

Jul 2022 · 10 marks
Show worked solution
Contract demand 3000 kVA; recorded average MD = 2400 kVA at 0.95 PF; minimum billable demand = 80% x 3000 = 2400 kVA. a) kW drawn = 2400 x 0.95 = 2280 kW. kVAr for 0.95 -> unity = kW[tan(cos^-1 0.95) - tan(cos^-1 1)] = 2280 x (0.3287 - 0) = 749 kVAr (say 750 kVAr). Maximum demand at unity PF = 2280 kVA, but the minimum billable demand is 2400 kVA, so the plant still pays for 2400 kVA - there is NO saving in maximum demand charges. PF incentive = (1.00 - 0.95)/0.01 x 0.5% = 2.5% of the energy charge = Rs.80,00,000 x 2.5% = Rs.2,00,000/month = Rs.24,00,000/year. b) Investment = 749 kVAr x Rs.1200/kVAr = Rs.8,99,000 (say Rs.9.0 lakh). Simple payback = 8,99,000 / 24,00,000 = 0.375 year = about 4.5 months. (Same structure as the Book-3 Sec.1.11 solved example: the minimum-billing-demand clause can wipe out the MD saving, leaving the PF incentive as the only benefit.)
kVAr = kW(tanφ1-tanφ2); MD reduction nil due to 80% minimum billing; energy charge incentive 0.5% per 0.01 PF rise above 0.95 → 2.5%; payback = investment/annual savings.
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand

19. L-1(A): A trivector-meter installed in a steel plant is monitoring the maximum demand with a demand interval of 15 min. The observed maximum demand during one demand interval is given (3 min: 9634 kVA, 4 min: 10257 kVA, 3 min: 8436 kVA, 5 min: 9847 kVA). Calculate: (a) Recorded maximum demand during the cycle; (b) Demand reduction and capacitor kVAr required for improving power factor to 0.99 from average observed power factor of 0.92.

Mar 2023 · 10 marks
Show worked solution
(a) Maximum demand is the time-integrated kVA over the 15-minute demand interval: MD = [(9634 x 3) + (10,257 x 4) + (8436 x 3) + (9847 x 5)] / 15 = [28,902 + 41,028 + 25,308 + 49,235] / 15 = 144,473/15 = 9631.5 kVA. (b) At the observed average PF of 0.92, kW = 9631.5 x 0.92 = 8861 kW (capacitors do not change kW). New demand at 0.99 PF = 8861/0.99 = 8950.5 kVA, so demand reduction = 9631.5 - 8950.5 = 681 kVA. Capacitor kVAr = kW[tan(cos^-1 0.92) - tan(cos^-1 0.99)] = 8861 x (0.4260 - 0.1425) = 8861 x 0.2835 = 2512 kVAr (say 2500 kVAr, switched through an APFC).
MD = time-weighted average of interval readings; convert to kW via pf; recompute kVA at improved pf; kVAr = kW(tanφ1−tanφ2).

Ch1 · Maximum demand, load factor & tariff

4 past-paper questions · ≈35 marks
Load factor = average load / maximum demand. Billing demand and MD penalties.
📖 §1.8 Commercial Losses & AT&C Losses (Table 1.7)

1. A private power distribution company has implemented new digital metering and billing systems to improve efficiency in a residential zone. After six months of operation, the following data was recorded: • Input energy to the system = 75 MU • Metered billed energy = 56 MU • Unmetered average billing = 4 MU • Amount billed = ₹680 million • Total amount received = ₹600 million • Arrears collected = ₹90 million • Purchased energy cost = ₹8.50 per kWh i) Estimate the Aggregate Technical and Commercial (AT&C) loss (%) and the revenue realized per kWh (4 Marks) ii) Calculate the revenue loss per kWh to the company due to AT&C loss (1 Mark)

Sep 2025 · 10 marks
Show worked solution
Input Energy = 75 MU = 75,000,000 kWh Metered Billed Energy = 56 MU Unmetered Average Billing = 4 MU Total Energy Billed = 56 + 4 = 60 MU Amount Billed = ₹680 million; Arrears Collected = ₹90 million; Amount Received = ₹600 million; Purchased Energy Cost = ₹8.50/kWh i) Billing Efficiency = (60 / 75) × 100 = 80.0% Collection Efficiency = ((600 - 90) / 680) × 100 = (510 / 680) × 100 = 75.0% AT&C Loss (%) = 1 - (Billing Efficiency × Collection Efficiency) = 1 - (0.80 × 0.75) = 1 - 0.60 = 40.0% Revenue Realized per kWh = (600 - 90) / 75 = 510 / 75 = ₹6.80/kWh ii) Revenue Loss = ₹8.50 - ₹6.80 = ₹1.70/kWh Final Answers: i) AT&C Loss = 40.0%, Revenue Realized = ₹6.80/kWh; ii) Revenue Loss per kWh = ₹1.70/kWh
AT&C loss = 1 − (Billing Efficiency × Collection Efficiency); revenue realized = net amount received / input energy.
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand

2. A trivector-meter installed in a mini steel plant with a 15-minute cycle records the following during the maximum demand period: 1000 kVA for 10 min, 2000 kVA for 5 min, 750 kVA for 10 min, 1500 kVA for 5 min. What is the maximum demand during the 15-minute interval?

Book EOC · 5 marks
Show worked solution
Maximum demand is the TIME-INTEGRATED demand over the meter's 15-minute cycle (Book-3 Sec.1.2), not the instantaneous peak, so evaluate each 15-minute window and take the highest: Window 1 (1000 kVA for 10 min + 2000 kVA for 5 min) = [(1000 x 10) + (2000 x 5)] / 15 = 20,000/15 = 1333 kVA Window 2 (750 kVA for 10 min + 1500 kVA for 5 min) = [(750 x 10) + (1500 x 5)] / 15 = 15,000/15 = 1000 kVA Maximum demand registered = 1333 kVA. Note that the momentary 2000 kVA block is NOT the billed demand - the meter averages it over the full 15-minute cycle.
MD = energy (kVAh) accumulated in the worst 15-minute window divided by the 0.25 h interval; the 1000 kVA (10 min) + 2000 kVA (5 min) window averages 1333 kVA.
📖 §1.2 Electricity Billing — tariff structure (minimum billing demand)

3. The maximum demand approved by a utility is 5500 kVA and the tariff provides for a minimum billing demand of 80% of approved. Records of the past 12 months show the monthly maximum demand recorded is around 4200 kVA. Will there be any benefit in surrendering part of the contract demand? If so, what kVA do you recommend surrendering? Give the cost saving, if the unit rate for kVA demand is Rs 200.

Book EOC · 10 marks
Show worked solution
Minimum billing demand at present = 80% of 5500 = 4400 kVA, which is higher than the actual recorded demand of ~4200 kVA, so the plant pays for 4400 kVA. By surrendering demand, the contract demand should be reduced so that 80% of the new contract demand just covers the actual maximum demand of 4200 kVA: new contract demand = 4200/0.8 = 5250 kVA. Recommend surrendering 5500 - 5250 = 250 kVA. New billing demand = 4200 kVA, saving 4400 - 4200 = 200 kVA per month. Monthly saving = 200 x Rs 200 = Rs 40,000, i.e. about Rs 4,80,000 per year.
Lower the contract demand until 80% of it equals the actual peak (4200 kVA); billing demand falls from 4400 to 4200 kVA, saving 200 kVA x Rs 200.
📖 §1.8 Commercial Losses & AT&C Losses (Table 1.7)

4. L-6: A distribution company has: input energy 60 MU, metered billed energy 43 MU, average (un-metered) billing 3 MU, amount billed Rs.540 million, arrears collected Rs.80 million, amount received Rs.470 million. (a) Estimate (i) AT&C loss % and revenue realised (Rs./kWh); (ii) revenue loss per kWh and monthly loss if purchased energy cost is Rs.8.10/kWh. (b) List five measures to reduce commercial loss.

19th Exam · 10 marks
Show worked solution
(a)(i) Billing efficiency = (43+3)/60 x 100 = 76.7%. Collection efficiency = (470-80)/540 x 100 = 72.2%. AT&C loss = [1 - (billing eff x collection eff)] x 100 = [1 - (0.767 x 0.722)] x 100 = 44.62%. Revenue realised = (470-80)/60 = Rs.6.5/kWh. (ii) Revenue loss = 8.10 - 6.5 = Rs.1.6/kWh; monthly loss = 60 MU x 1.6 = Rs.96 million (Rs.9.6 crore). (b) Measures to reduce commercial loss: 1. Accurate metering of all consumers (replace defective/electromechanical meters, AMR/smart meters). 2. Detect and curb theft/pilferage and unauthorised connections; energy audit/feeder metering. 3. Improve billing — eliminate un-metered/average billing, correct meter-reading and billing errors. 4. Strengthen collection efficiency — disconnection drives, online payment, recovery of arrears. 5. HVDS / spot billing / consumer indexing and GIS mapping to plug gaps.
AT&C loss = 1-(billing eff x collection eff); revenue realised = net received/input units; revenue loss = purchase cost - realised; commercial-loss measures from Book-3 Chapter 1 (paper cites page 27).

Ch1 · T&D / AT&C losses & harmonics

5 past-paper questions · ≈30 marks
AT&C loss % combines technical, commercial and collection loss. THD = √(Σ harmonic²)/fundamental × 100.
📖 §1.8 Commercial Losses & AT&C Losses (Table 1.7)

1. Compute AT&C (Aggregate Technical and Commercial) Losses for the given data: Input Energy Ei=20 MU, Energy Billed Metered E1=16 MU, Un-metered E2=1 MU, Total Billed Eb=17 MU, Amount Billed Ab=Rs.800 lakhs, Gross Amount Collected AG=Rs.820 lakhs, Arrears Collected Ar=Rs.40 lakhs.

15th Exam · 5 marks
Show worked solution
Amount collected without arrears Ac = AG - Ar = 820 - 40 = 780. Billing Efficiency BE = Eb/Ei = 17/20 = 85%. Collection Efficiency CE = Ac/Ab = 780/800 = 97.5%. AT&C Loss = [1 - (BE x CE)] x 100 = [1 - (0.85 x 0.975)] x 100 = 17.12%.
AT&C Loss = 1 - (Billing Efficiency x Collection Efficiency).
📖 §1.10 Harmonics

2. Harmonic measurements gave: current at 50 Hz = 300 A, at 150 Hz = 42 A, at 250 Hz = 33 A. Calculate the Total Harmonic Distortion in current.

14th Exam · 5 marks
Show worked solution
THD(I) = √[(42/300)² + (33/300)²] x 100 = √(0.0196 + 0.0121) x 100 = √0.0317 x 100 = 17.8%.
THD = √(sum of squares of harmonic currents)/fundamental x 100.
📖 §1.8 Commercial Losses & AT&C Losses (Table 1.7)

3. Compute AT&C Losses: Input Energy Ei=11 MU, Energy Billed Metered=7 MU, Un-metered=1 MU, Total Billed=8 MU, Amount Billed=Rs.450 lakhs, Gross Collected=Rs.460 lakhs, Arrears Collected=Rs.40 lakhs.

Set-A · 5 marks
Show worked solution
Ac (amount collected without arrears) = AG - Ar = 460 - 40 = Rs.420 lakhs. Billing Efficiency BE = Eb/Ei x 100 = 8/11 x 100 = 72.7%. Collection Efficiency CE = Ac/Ab x 100 = 420/450 x 100 = 93.3%. AT&C Loss = [1 - (BE x CE)] x 100 = [1 - (0.727 x 0.933)] x 100 = [1 - 0.6786] x 100 = 32.1%. (Book-3 Sec.1.8, Table 1.7 method: note that arrears must be stripped out of the gross collection before computing collection efficiency.)
AT&C Loss = 1 - (Billing Efficiency x Collection Efficiency).
📖 §1.1 Industrial End User — 'ONE Unit saved = TWO Units Generated'

4. One unit of electricity in end-use application is equivalent to about two units of electricity generated. Substantiate with the cascade efficiency from generating plant ex-bus to end-use. Assume: Generator yard substation efficiency 98%; T&D loss = 20%; End-use application efficiency = 65%.

17th Sep-2016 · 5 marks
Show worked solution
Cascade efficiency = 0.98 × (1 − 0.20) × 0.65 = 0.5096. Therefore one unit at end use = 1/0.5096 = 1.96 ≈ 2 units at ex-generator bus.
Multiply stage efficiencies; T&D efficiency = (1 − loss).
📖 §1.8 Commercial Losses & AT&C Losses (Table 1.7)

5. A DISCOM has taken initiatives to reduce Aggregate Technical & Commercial (AT&C) losses in their network. The energy supplied, received and revenue details are given below — Input energy: 50 MU; Billed Energy (Metered): 39 MU; Billed Energy (Un-metered): 2 MU; Amount Billed: Rs. 470 Million; Amount Collected: Rs. 30 Million; Gross Amount collected: Rs. 390 Million. a) Estimate the AT&C losses (in %). b) List any four strategies to reduce the commercial losses.

Jul 2022 · 10 marks
Show worked solution
a) Billing efficiency = total units billed / total input units = (39 + 2)/50 x 100 = 82.0%. Collection efficiency = amount collected excluding arrears / amount billed. Stripping the Rs.30 million of arrears out of the Rs.390 million gross collection: Ac = 390 - 30 = Rs.360 million, so CE = 360/470 x 100 = 76.6%. AT&C loss = [1 - (BE x CE)] x 100 = [1 - (0.820 x 0.766)] x 100 = [1 - 0.6281] x 100 = 37.2%. b) Four measures to reduce commercial losses (Book-3 Sec.1.8): (1) accurate metering with a planned meter-replacement programme and meters matched to the connected load; (2) installation of electronic meters with TOD, tamper-proof and remote/data-reading facility; (3) intensive inspections and eradication of theft/pilferage; (4) compulsory metering in place of average billing, backed by energy audit to pinpoint high-loss areas and by improved collection/arrear recovery.
AT&C loss = 1 - (billing efficiency × collection efficiency); billing eff = billed/input, collection eff = collected/billed.

Ch4 · TR, COP and kW per TR

36 past-paper questions · ≈285 marks
1 TR = 3024 kcal/hr = 3.517 kW. COP = refrigeration effect / work input. kW/TR = power input / tons of refrigeration.
📖 §4.7 TR & heat rejection; §4.3 VAR

1. Determine the difference in heat rejected in kCal/TR to the cooling tower for two different types of air conditioning system operating at same capacity. Parameter — Centrifugal chiller / VAM: Chilled water flow (m³/h): - / 180 Condenser water flow (m³/h): - / 340 Chiller inlet temp (°C): 13.0 / 14.6 Condenser water inlet temp (°C): - / 33.5 Chiller outlet temp (°C): 7.7 / 9.0 Condenser water outlet temp (°C): - / 39.1 Specific power consumption (kW/TR): 0.6 / -

Sep 2025 · 5 marks
Show worked solution
1 TR = 3024 kcal/h Centrifugal chiller: Power input = 0.6 × 860 = 516 kcal/TR Heat rejected = 3024 + 516 = 3540 kcal/TR VAM: Chilled water flow = 180 m³/h = 180000 kg/h ΔT = 14.6 − 9.0 = 5.6°C Cooling load = 180000 × 1 × 5.6 = 1008000 kcal/h TR = 1008000 / 3024 = 333.33 TR Heat Rejected = 340000 kg/hr × 1 kcal/kg × (39.1 − 33.5) = 1904000 kcal/hr Heat Rejected per TR = 1904000 / 333.33 = 5712 kcal/TR Difference per TR = 5712 − 3540 = 2172 kcal/TR
Heat rejected = refrigeration effect + compressor heat (chiller) vs condenser water heat balance per TR (VAM); take difference.
📖 §4.9 EER; §4.7 kW/TR; §4.11 Heat Pumps

2. A 5-star business hotel operates a centralized HVAC system round the clock. Only one chiller operates at a time, the other on standby. Two centrifugal chillers, each rated 250 TR, with EER varying with load: at 85% load (212.5 TR) EER 5.2 for 180 days; at 60% load (150 TR) EER 4.6 for 120 days; at 40% load (100 TR) EER 3.9 for 65 days. No change in EER above 85% load; assume chiller motor efficiency 90% at all loading conditions. During chiller operation, two pumps run in parallel at an 80% load factor, consuming a total of 19.7 kW; two cooling tower fans operate continuously with power consumption of 5.89 kW. Both pumps and fans function 24 hours a day, with overall efficiencies of 75% and 70% respectively. Electricity tariff is ₹6.5 per kWh. Evaluate: a. The total annual energy consumption (in MWh) and cost of the HVAC system, considering part-load EERs and auxiliary loads. (4 Marks) b. Heat removal by condenser in (TR) at different loads. (3 Marks) c. The hotel is planning to use the chiller partially as a heat pump by mounting a plate heat exchanger in series between the compressor and condenser (desuperheater for partial heat recovery) for producing hot water. The heat recovery can be only 20% of the condenser heat discharge. If the hot water requirement is 2000 litres/hr with 10°C temperature rise, evaluate whether the hot water requirement can be met at 40% loading conditions. (3 Marks)

Sep 2025 · 10 marks
Show worked solution
a. Energy Consumed by Chiller (Cooling Load kW = TR×3024/860; Input Power = Load kW/EER; MWh = InputPower×Days×24/1000): 85%: 212.5 TR, EER 5.2 → 747.21 kW load, 143.69 kW input, 180 days → 620.8 MWh 60%: 150 TR, EER 4.6 → 527.44 kW load, 114.66 kW input, 120 days → 330.2 MWh 40%: 100 TR, EER 3.9 → 351.63 kW load, 90.16 kW input, 65 days → 140.7 MWh Energy Consumed by auxiliaries = (19.7 + 5.89) × 24 × 365 / 1000 = 224.5 MWh Total Annual Energy Consumption = 620.8 + 330.2 + 140.7 + 224.5 = 1316 MWh Annual Cost = 1316000 × 6.5 = Rs. 85.55 Lakh b. Heat Removal by Condenser (Power Input to Compressor P = Input Power × 0.9; Condenser Heat Load HL = Cooling Load TR + P×860/3024): 85%: 212.5 TR, 143.69 kW input, P = 129.32 kW → 249.28 TR 60%: 150 TR, 114.66 kW input, P = 103.20 kW → 179.35 TR 40%: 100 TR, 90.16 kW input, P = 81.14 kW → 123.08 TR c. Heat Load at 40% loading = 123.08 × 3024 = 372194 kCal/Hr; Recovery potential (20%) = 74439 kCal/Hr; Heat requirement for hot water generation = 2000 × 1 × 10 = 20000 kCal/H. Therefore, the heating requirement can easily be met at 40% loading.
Chiller input = load/EER; annual MWh summed over part-load days plus auxiliaries; condenser heat = cooling load + compressor heat; compare 20% recovery against hot-water duty.
📖 §4.3 Absorption Refrigeration; §4.7 TR

3. A process engineer develops a scheme to put 500 TR absorption-based refrigeration system to bring down process fluid temperature from 34 °C to 26 °C and this will result in higher production by 10%. 5 TPH excess steam is available in the plant and this new scheme utilizes this excess steam. COP of refrigeration system is 0.65 and available latent of steam for refrigeration system is 540 kcal/kg. A) Estimate excess steam utilization for absorption-based refrigeration system in TPH. (3 Marks) b) Estimate required Cooling water (m³/hr), if available approach in condenser is 10 °C. (2 Marks)

Sep 2024 · 10 marks
Show worked solution
Energy required for refrigeration system = 500 × 3024 / 0.65 = 2326153.8 kcal/hr Steam needed for refrigeration system = 2326153.8 / 540 = 4.3 TPH Steam utilization for VAM = 4.3 TPH Required condenser duty = 2326153.8 + (500 × 3024) = 3838153.8 kcal/hr Required Cooling water = 3838153.8 / 10 = 383.8 m³/hr
Steam = (TR×3024/COP)/latent heat; condenser duty = refrigeration heat input + evaporator load; cooling water = condenser duty / approach (ΔT).
📖 §4.3 VAR vs VCR; §4.7 TR & COP

4. As part of a management initiative to advance green energy in a new process plant, a process engineer is assessing the economic viability of a 650 TR chiller. She is considering proposals for both LiBr-based vapor absorption chillers and vapor compression refrigeration systems. While power is sourced from renewable energy, the steam required is partially generated from excess process heat and additionally from firing furnace oil. COP of advance Vapor Absorption Chiller: 1.3 COP of Vapor Compression Chiller: 4.50 Net steam price including excess steam and from boiler: 1500.00 INR/MT Net Power cost from green source: 7.20 INR/kWh Price of Cooling water: 3.00 INR/M3 Cooling water range: 8°C Specific steam heat available for chiller: 490.0 kcal/kg Evaluate both the offers and find out the offer which is economical in terms of operating cost.

Sep 2024 · 10 marks
Show worked solution
Vapor Absorption chiller: Chilling capacity = 650.0 TR; Required heat duty = 650 × 3024 = 1965600 kcal/hr Heat equivalent input to VAM = 1965600 / 1.3 = 1512000 kcal/hr Required Steam flow = 1512000 / 490 = 3084 kg/hr Condenser heat duty = (650 × 3024) + (3084 × 490) = 3476760 kcal/hr Required Cooling Water = 434.6 M3/hr Operating cost of Vapor Absorption Chiller = (3084 × 1.5) + (434.6 × 3) = 5930 INR/hr Vapor Compression chiller: Chilling capacity = 650.0 TR; Required heat duty = 1965600 kcal/hr Required Power consumption = 508 kW Condenser heat duty = (650 × 3024) + (508 × 860) = 2402480 kcal/hr Required Cooling Water = 300 M3/hr Operating cost of Vapor Compression Chiller = (508 × 7.2) + (300 × 3) = 4558 INR/hr Hence, operating vapor compression chiller is economical.
Compute steam/power input from COP, condenser duty and cooling-water flow (duty/range), then compare hourly operating cost of VAM vs VCR.
📖 §4.7 TR formula & specific power; pump efficiency

5. An energy auditor assesses the refrigeration load of each floor of a building and the efficiency of the chilled water pump. The chilled water flow of each AHU and its inlet/outlet temperatures are measured; the chilled water pump discharge pressure is 3.5 kg/cm2 with the tank water level 1 m above the pump centerline; power drawn by the refrigeration compressor and pump motor are 89 kW and 17.58 kW. Determine the floor-wise refrigeration load, total TR, chiller kW/TR and pump efficiency.

Book EOC · 10 marks
Show worked solution
Floor-wise TR = flow (m3/hr) x 1000 x 1 kcal/kg degC x (T_in - T_out)/3024. 1st floor: 20 x 1000 x (10 - 7)/3024 = 19.84 TR; 2nd: 30 x 1000 x (11 - 7)/3024 = 39.68 TR; 3rd: 10 x 1000 x (11 - 7)/3024 = 13.23 TR; 4th: 15 x 1000 x (12 - 7)/3024 = 24.80 TR. Total = 97.55 TR (total flow 75 m3/hr). (b) Compressor shaft power = 89 x 0.92 = 81.88 kW, so specific power consumption = 81.88/97.55 = 0.84 kW/TR. (c) Pump total head = 3.5 kg/cm2 x 10 - 1 m (positive suction) = 34 m; hydraulic power = 75/3600 x 1000 x 9.81 x 34/1000 = 6.95 kW; pump shaft power = 17.58 x 0.89 = 15.65 kW; pump efficiency = 6.95/15.65 = 44.4%.
Book §4.7 TR formula applied floor by floor, then summed; the chiller indicator is compressor shaft power divided by total TR. Pump efficiency = hydraulic power / shaft power, where total head = discharge head (3.5 kg/cm2 = 35 m) less the 1 m positive suction head, because the tank level is ABOVE the pump centreline. Watch the two motor efficiencies (92% compressor, 89% pump) - omitting them is the usual mark-loser.
📖 §4.9 Package A/C worked example

6. A 20 TR package AC plant: air velocity across suction filter 2.5 m/s; suction area 1.2 m²; inlet air enthalpy 9.37 kcal/kg, outlet 7.45 kcal/kg; specific volume 0.85 m³/kg; power: compressor 10.69 kW, pump 4.86 kW, cooling tower fan 0.87 kW. Calculate: i) air flow rate m³/hr, ii) cooling effect kW, iii) compressor kW/TR, iv) overall kW/TR, v) EER kW/kW.

15th Exam · 5 marks
Show worked solution
i) Air flow = 2.5 x 1.2 = 3 m³/s = 10,800 m³/hr. ii) Cooling effect = [(9.37-7.45) x 10800]/(0.85 x 3024) = 8.07 TR = 28.32 kW. iii) Compressor kW/TR = 10.69/8.07 = 1.32. iv) Overall kW/TR = (10.69+4.86+0.87)/8.07 = 2.04. v) EER = 28.32/10.69 = 2.65 kW/kW.
TR = Q x Δh /(specific volume x 3024); kW/TR and EER from power and cooling effect.
📖 Cross-chapter fill-in-the-blanks (§4.7 for item 1)

7. Fill in the blanks (cross-chapter): 1) One TR = ___ kW. 2) A 4-pole 15 kW IM at 50 Hz, 1% slip has rotor input ___ kW. 3) A pitot tube measures total and static pressure to determine ___ pressure. 4) Centrifugal pump impeller diameter is generally limited to reducing to about ___ % of max size. 5) Pressure in pump suction exceeding liquid vapour pressure is expressed as ___. 6) ASME parameter to define fans, blowers, compressors is ___. 7) Pumps can run in parallel if their ___ are similar. 8) Evaporation 16 m³/cell, COC 3 → blowdown ___. 9) Pump raises water to 12 m; with brine SG 1.2 height raised is ___. 10) Installing capacitor near motor terminals increases design PF of motor - True/False.

15th Exam · 10 marks
Show worked solution
1) 3.516 kW. 2) 15.15 kW (15/(1-0.01)=15.15). 3) velocity pressure. 4) 75%. 5) Net Positive Suction Head Available (NPSHA). 6) Specific ratio. 7) closed valve heads. 8) 8 m³ per cell (blowdown = evaporation/(COC-1) = 16/2 = 8). 9) 12 metres (same height - head is independent of density). 10) False.
Standard one-liners across HVAC, motors, fans, pumps and cooling towers.
📖 §4.9 Package A/C worked example

8. A water cooled 20 TR package AC plant: air velocity across suction filter 2.5 m/s, suction area 2.4 m²; inlet air enthalpy 9.37 kcal/kg, outlet 7.45 kcal/kg; specific volume 0.85 m³/kg; power: compressor 18.42 kW, pump 2.1 kW, evaporator fan 1.25 kW. Calculate: i) air flow rate, ii) cooling effect, iii) compressor kW/TR, iv) overall kW/TR, v) overall EER in W/W.

Set-A · 5 marks
Show worked solution
i) Air flow = 2.5 x 2.4 = 6 m³/s = 21,600 m³/hr. ii) Cooling effect = [(9.37-7.45) x 21600]/(0.85 x 3024) = 16.13 TR = 56.73 kW. iii) Compressor kW/TR = 18.42/16.13 = 1.13. iv) Overall kW/TR = (18.42+2.1+1.25)/16.13 = 1.35. v) EER = 56.73/21.77 = 2.606 W/W.
TR = Q x Δh/(specific volume x 3024); kW/TR and EER from power and cooling effect.
📖 §4.3 Absorption Refrigeration (Figure 4.5)

9. Identify the type of refrigeration system in the figure and the components 1,2,3 & 4. Explain briefly the function of each.

17th Sep-2016 · 5 marks
Show worked solution
Vapour Absorption Refrigeration system. 1 Absorber: concentrated LiBr absorbs the refrigerant vapour (water) and becomes dilute. 2 Generator: heats the dilute LiBr, regenerates refrigerant (water vapour) and re-concentrates LiBr. 3 Condenser: condenses the regenerated refrigerant (water vapour). 4 Evaporator: liquid refrigerant (water, atomised) picks up heat from the chilled-water coil and becomes water vapour.
Confirmed vs Book-3 §4.3 (Figure 4.5) - in the LiBr-water absorption chiller water is the refrigerant and LiBr solution the absorbent. Evaporator: water flashes at ~4 degC under 754 mmHg vacuum and chills the water from 12 to 7 degC. Absorber: concentrated LiBr absorbs the vapour (and maintains the vacuum), becoming dilute. Generator: steam/hot water/oil boils off the water and re-concentrates the LiBr. Condenser: condenses that vapour and returns it to the evaporator.
📖 §4.7 TR & kW/TR; §4.3 VAR vs VCR

10. Compare the performance of a centrifugal chiller with a vapour absorption chiller (VAM) from the given data (chilled & condenser water flows, inlet/outlet temps, pump and CT fan power). Centrifugal compressor 205 kW; VAM steam 1620 kg/Hr. Calculate i) refrigeration load TR, ii) condenser heat load TR, iii) auxiliary power, iv) operating cost (electricity Rs 4/kWh, steam Rs 0.45/kg).

16th Exam · 10 marks
Show worked solution
i) Refrigeration TR = m×Cp×ΔT/3024: Centrifugal = 192×1000×(13−7.8)/3024 = 330.16 TR; VAM = 183×1000×(14.5−9.2)/3024 = 320.73 TR. ii) Condenser heat load TR (condenser flow × ΔT/3024): Centrifugal = 664.35 TR; VAM = 1035.71 TR (VAM rejects more heat). iii) Auxiliary power = chilled + condenser pump + CT fan = Centrifugal 32+38+9 = 79 kW; VAM 31+52+22 = 105 kW (VAM higher due to greater heat rejection). iv) Operating cost/hr: Centrifugal = (79+205)×4 = Rs 1136/hr; VAM = aux 105×4 = 420 + steam 1620×0.45 = 729 → Rs 1149/hr.
TR=m·Cp·ΔT/3024; condenser TR same with condenser flow/ΔT; cost = electricity + steam.
📖 §4.12 Ventilation Systems; §4.7 air-side TR

11. a) Calculate the ventilation rate for an engine room 20 m × 10.5 m × 15 m if recommended ACH is 20. b) Air at 25,200 m3/hr, density 1.2 kg/m3, enthalpy difference inlet-outlet 2.38 kcal/kg flows into an AHU; motor draws 22 kW at 90% efficiency. Find the kW/TR. (1 cal = 4.183)

16th Exam · 5 marks
Show worked solution
a) Ventilation rate = L×W×H×ACH = 20×10.5×15×20 = 63,000 m3/hr. b) Heat removed = 25200×1.2×2.38 = 71,971 kcal/hr → TR = 71971/3024 = 23.8 TR. Compressor power = 22×0.9 = 19.8 kW. kW/TR = 19.8/23.8 = 0.83.
Ventilation = volume×ACH; TR=Q·ρ·Δh/3024; kW/TR.
📖 §4.7 air-side TR formula (AHU/FCU)

12. L-2: (a) In an AHU, filter area = 1.5 m2, air velocity = 2.2 m/s, inlet enthalpy = 67 kJ/kg, outlet enthalpy = 56 kJ/kg, air density = 1.3 kg/m3. Estimate the TR of the AHU. (b) List any five energy conservation measures for energy use in buildings.

18th Exam · 10 marks
Show worked solution
(a) TR = (Δenthalpy x density x area x velocity x 3600)/(4.187 x 3024) = (67-56) x 1.3 x 1.5 x 2.2 x 3600/(4.187 x 3024) = 13.41 TR. (b) Five building ECMs: 1. Weather-strip windows/doors to cut infiltration; provide self-closing doors at high-traffic areas. 2. Set temperature 23-25 °C and RH 55-65% for comfort. 3. Keep chilled-water leaving temperature ≥7 °C (≈2.25% chiller efficiency gain per 1 °C rise). 4. Maintain insulation on chilled-water pipes and ducts to prevent heat gain. 5. Clean condenser tubes (every 6 months), keep filters clean, and install VFDs on AHU fans.
AHU TR from mass-flow x enthalpy drop converted to TR; building ECMs from Book-3 Chapter 4/10 (any five, 1.5 marks each).
📖 §4.13 Ice Bank; §4.3 VAR (harmonics: Book-3 Ch1)

13. L-6: Write short notes on (i) Ice Bank System in refrigeration, (ii) Vapour Absorption Refrigeration System, (iii) Harmonics in electrical system and its impacts.

18th Exam · 10 marks
Show worked solution
(i) Ice Bank System: a thermal energy storage technology that uses low-cost off-peak (night) electricity to make and store ice/cooling energy in storage tanks for use during high-tariff daytime hours; the chiller charges the tanks at night and runs longer hours at the lowest average load, shifting and shaving peak demand. (ii) VAR System: an absorption chiller produces chilled water using heat (steam, hot water, gas, oil, waste heat) instead of a compressor; water is the refrigerant and lithium-bromide is the absorbent; COP ≈ 0.65-0.70, chilled water down to ~6.7 °C at 30 °C cooling water; needs electricity only for pumps; economical when waste heat/cheap steam is available; capacities 10-1500 TR. (iii) Harmonics: currents/voltages at multiples of the supply (fundamental) frequency — e.g. with 50 Hz, 5th = 250 Hz, 7th = 350 Hz; caused by non-linear loads. Impacts: capacitor failure, conductor/cable overheating, transformer and motor overheating/failure, flickering of fluorescent and blinking of incandescent lights, and nuisance tripping.
Standard descriptive notes from Book-3 (Ice Bank p.136, VAR p.30, Harmonics p.114).
📖 §4.7 TR formula; pump hydraulic power

14. S-2: A chilled water system runs always at full load; inlet/outlet 12 °C / 7 °C. Chilled-water pump discharge pressure 3.6 kg/cm2g, suction 5 m above pump centreline, motor power 70 kW at 90% efficiency, pump efficiency 60%. Find the operating refrigeration load in TR.

19th Exam · 5 marks
Show worked solution
Discharge head = 3.6 kg/cm²g ≈ 36 m; total head = 36 - 5 = 31 m. Pump shaft power = 70 x 0.9 = 63 kW. Flow = (pump shaft power x 1000 x pump eff)/(head x 1000 x 9.81) = (63 x 1000 x 0.6)/(31 x 1000 x 9.81) = 0.1243 m3/s = 447.5 m3/hr. Refrigeration load = (447500 x 5)/3024 = 740 TR.
Total head from discharge head minus suction lift; flow from pump hydraulic equation; TR = (flow(kg/hr) x ΔT)/3024.
📖 §4.7 air-side TR; §4.9 kW/TR

15. L-4: A 7.5 TR package A/C cools a UPS room (40 kVA UPS). Outdoor unit air velocity 6.1 m/s, fan opening radius 0.30 m, air density 1.174 kg/m3, ambient 305 K, condenser-outlet hot air 313.5 K, Cp 1.009 kJ/kgK, compressor power 5.40 kW, motor efficiency 90%. UPS on-load (16 h): input 11.94 kW, output 8.61 kW; no-load (8 h): input 1.16 kW, output 0. Calculate (a) present delivery TR, (b) power per TR, (c) annual energy savings for 7200 h if UPS relocated to a ventilated area (Rs.8/kWh).

19th Exam · 10 marks
Show worked solution
Area = 3.14 x 0.30² = 0.283 m². Air flow = 0.283 x 6.1 = 1.72 m3/s. Mass = 1.72 x 1.174 = 2.02 kg/s. ΔT = 313.5 - 305 = 8.5 K. Heat transfer = 2.02 x 1.009 x 8.5 = 17.32 kJ/s = 62,352 kJ/hr = 14,917 kcal/hr. Compressor heat input = 5.4 x 0.9 x 860 = 4180 kcal/hr. Evaporator load = 14,917 - 4180 = 10,737 kcal/hr. (a) Effective TR = 10,737/3024 = 3.55 TR. (b) Power per TR = 5.40/3.55 = 1.52 kW/TR. (c) Heat load from UPS: on-load 3.33 kW → 0.95 TR/hr x 16 = 15.2 TR-day; no-load 1.16 kW → 0.33 TR/hr x 8 = 2.64 TR-day; total 17.84 TR/day. AC power to remove it = 17.84 x 1.52 = 27.12 kW. Annual savings (300 days) = 27.12 x 300 ≈ 8136 kWh → Rs.8 x 8136 = Rs.65,088/yr.
Condenser air-side heat minus compressor heat gives evaporator TR; kW/TR from compressor power; UPS heat converted to TR/day then to AC power saved if UPS relocated, costed at the tariff.
📖 §4.3 VAR vs VCR operating economics

16. L-5: A textile plant had two 6 MW gas turbines + HRSG feeding process steam and a 500 TR VAM (4.4 kg steam/TR, full load). Gas turbines stopped due to gas price; two 10 TPH agro-waste boilers installed (steam cost Rs.1200/ton); plant runs 7000 h/yr. Management plans to replace VAM with an electric centrifugal chiller at 0.7 kW/TR. Compare annual operating costs of electric chiller vs VAM (grid power Rs.6.12/kWh; auxiliaries unchanged). Do you agree with running the VAM?

19th Exam · 10 marks
Show worked solution
VAM steam need = 500 x 4.4 = 2200 kg/hr = 2.2 TPH; steam cost = 2.2 x 1200 = Rs.2640/hr. Electric chiller power = 0.7 x 500 = 350 kW; cost = 350 x 6.12 = Rs.2142/hr. Saving with electric chiller = 2640 - 2142 = Rs.498/hr. Annual saving = 7000 x 498 = Rs.34,86,000/yr. Conclusion: Disagree with running the VAM — the electric centrifugal chiller is cheaper to operate (saves ~Rs.34.86 lakh/yr) now that cheap waste heat is gone.
VAM running cost = steam rate x steam cost; chiller cost = kW/TR x TR x power tariff; compare hourly and annualise; recommend the cheaper option.
📖 §4.7 IPLV; §4.3 Evaporative Cooling; §4.11 Heat Pumps

17. Write short notes on any two of the following: (1) Integrated Part Load Value (IPLV) for chillers, (2) Evaporative Cooling, (3) Heat Pump. (Each 2.5 Marks)

Sep 2019 · 10 marks
Show worked solution
(1) IPLV: A single-number part-load efficiency metric for chillers (AHRI weighting) that combines chiller performance at 100%, 75%, 50% and 25% load, since chillers rarely run at full load; it better represents seasonal/real operation than full-load kW/TR. (2) Evaporative Cooling: Cooling of air by direct contact with water so that sensible heat of air evaporates water; dry-bulb temperature falls while moisture content rises, approaching the wet-bulb temperature - low energy alternative to mechanical refrigeration in dry climates. (3) Heat Pump: A reversed refrigeration cycle device that extracts low-grade heat from a source (air/water/ground) and upgrades it to deliver useful heating; COP_heating = COP_cooling + 1, giving high effective efficiency.
Confirmed vs Book-3 §4.7 (IPLV, p.126), §4.3 Evaporative Cooling (p.116) and §4.11 Heat Pumps (p.133). IPLV averages kW/TR at 100/75/50/25% load because full-load kW/TR occurs for only ~1% of running hours. Evaporative cooling brings air into close contact with water so it approaches the wet bulb temperature - cheap but it adds moisture. A heat pump is an air conditioner whose rejected heat is the useful output: 3 units from the surroundings plus 1 unit of compressor work give 4 units of heat.
📖 §4.7 air-side TR formula (AHU/FCU)

18. A 10 TR AHU operates at 8.25 TR. Inlet enthalpy 10.26 kcal/kg, outlet enthalpy 7.26 kcal/kg, specific volume of air 0.83 m3/kg. Calculate the volume of air (m3/hr) handled by the AHU.

Sep 2019 · 5 marks
Show worked solution
Cooling (TR) = (Hi-Ho) x V/(v x 3024). Volume of air = TR x v x 3024/(Hi-Ho) = (8.25 x 0.83 x 3024)/(10.26-7.26) = 6903 m3/hr.
Confirmed vs Book-3 §4.7 - air-side load TR = Q x rho x (h_in - h_out)/3024, and with specific volume instead of density Q = TR x v x 3024/(h_in - h_out). Here Q = 8.25 x 0.83 x 3024/(10.26 - 7.26) = 6903 m3/hr. Note the AHU is rated 10 TR but is delivering only 8.25 TR - use the delivered load, not the nameplate, or the airflow comes out ~16% too high.
📖 §4.7 COP-Carnot vs industry COP

19. S-7: Explain with the equation for COP_Carnot that (a) higher COP_Carnot is achieved with higher evaporator temperature and lower condenser temperature; (b) COP_Carnot does not take into account the type of compressor; (c) how is the COP normally used in industry given?

16th Exam (alt set) · 5 marks
Show worked solution
(a) COP_Carnot = Te/(Tc - Te), depending on evaporator temperature Te and condenser temperature Tc; raising Te and lowering Tc increases COP_Carnot. (b) Since COP_Carnot is only a ratio of absolute temperatures, it does not account for the type of compressor used. (c) The industry COP = Cooling effect (kW) / Power input to compressor (kW), where cooling effect is the enthalpy difference across the evaporator expressed in kW.
Confirmed vs Book-3 §4.7 - COP-Carnot = Te/(Tc - Te) in kelvin, so a higher evaporator temperature and a lower condenser temperature both raise it (Figures 4.10 and 4.11 quantify the same trend). Being purely a temperature ratio it ignores compressor type and real losses, so the industry COP used in audits is cooling effect (kW) / power input to compressor (kW), the cooling effect being the enthalpy rise across the evaporator.
📖 §4.7 TR & kW/TR; §4.3 VAR vs VCR

20. L-1: Compare centrifugal chiller vs vapour absorption chiller (VAM). Chilled water flow 192/183 m3/h, condenser flow 245/360, chiller inlet 13/14.5 C, condenser inlet 28/32 C, chiller outlet 7.8/9.2 C, condenser outlet 36.2/40.7 C, chilled water pump 32/31 kW, condenser pump 38/52 kW, CT fan 9/22 kW. Centrifugal compressor 205 kW; VAM steam 1620 kg/hr. Find (i) refrigeration load TR, (ii) condenser heat load TR, (iii) compare auxiliary power, (iv) operating cost at Rs.4/kWh and steam Rs.0.45/kg.

16th Exam (alt set) · 10 marks
Show worked solution
(i) Refrigeration load = flow x 1000 x 1 x deltaT/3024: Centrifugal = 192000 x (13-7.8)/3024 = 330.16 TR; VAM = 183000 x (14.5-9.2)/3024 = 320.73 TR. (ii) Condenser heat load = condenser flow x 1000 x deltaT/3024: Centrifugal = 245000 x (36.2-28)/3024 = 664.35 TR; VAM = 360000 x (40.7-32)/3024 = 1035.71 TR. (iii) Auxiliary power = pumps + CT fan: Centrifugal = 32+38+9 = 79 kW; VAM = 31+52+22 = 105 kW. VAM auxiliary is higher because its condenser heat rejection is much larger for similar cooling load. (iv) Operating cost/hr: Centrifugal total = 79 + 205 = 284 kW x Rs.4 = Rs.1136. VAM = 105 kW x Rs.4 = Rs.420 plus steam 1620 x Rs.0.45 = Rs.729 -> total Rs.1149/hr.
Confirmed vs Book-3 §4.7 - refrigeration TR = flow x Cp x deltaT/3024 on the chilled-water side and the same formula on the condenser side for heat rejection. Centrifugal 330.16 TR vs VAM 320.73 TR for similar duty, but the VAM rejects 1035.71 TR against 664.35 TR, because a VAR machine must also reject its heat input (COP ~0.65-0.70). That is why VAM condenser pumps and cooling-tower fans are bigger (105 kW vs 79 kW), and the hourly cost is close (Rs.1149 vs Rs.1136) despite free-looking steam.
📖 §4.12 Ventilation Systems; §4.7 air-side TR

21. L-2: (a) Calculate the ventilation rate for an engine room 20 m L x 10.5 m W x 15 m H if recommended ACH is 20. (b) Air at 25,200 m3/hr and 1.2 kg/m3 density flows into an AHU; enthalpy difference inlet-outlet 2.38 kcal/kg; motor draws 22 kW at 90% efficiency. Find kW/TR (1 cal = 4.183).

16th Exam (alt set) · 10 marks
Show worked solution
(a) Ventilation rate = L x H x W x ACH = 20 x 15 x 10.5 x 20 = 63,000 m3/hr. (b) Heat = Q x density x deltaH = 25200 x 1.2 x 2.38 = 71,971 kcal/hr; TR = 71971/3024 = 23.8 TR; compressor power = 22 x 0.9 = 19.8 kW; kW/TR = 19.8/23.8 = 0.83.
Confirmed vs Book-3 §4.12 and §4.7 - ventilation rate = L x B x H x ACH = 20 x 10.5 x 15 x 20 = 63,000 m3/hr. Air-side load = Q x rho x delta-h/3024 = 25,200 x 1.2 x 2.38/3024 = 23.8 TR, and compressor shaft power = 22 x 0.9 = 19.8 kW, giving 0.83 kW/TR. The common slip is to use the 22 kW motor input directly, which overstates kW/TR by about 11%.
📖 §4.7/§4.8 COP vs temperatures (b,c cross-chapter)

22. a) What is the impact of condensing temperature and evaporator temperature on the COP of a refrigeration system? b) Why is it beneficial to operate induction motors in star mode at loads below 50% of rated capacity? c) In a throttle valve-controlled pumping system with oversized pump, name any 3 solutions for improving energy efficiency.

9th Dec-2009 · 10 marks
Show worked solution
a) COP increases with reduction in condensing temperature and with rise in evaporator temperature. b) For motors that consistently operate below 50% of rated capacity, operating in star mode (re-configuring the three phases at the terminal box) reduces voltage by a factor of √3; motor output falls to one-third of the delta value, but performance characteristics as a function of load remain unchanged, so full-load operation in star gives higher efficiency and power factor than partial-load operation in delta. This is only possible where the torque-speed requirement is lower at reduced load. c) Any three of: trim impeller, fit a smaller impeller, install a variable speed drive, use a two-speed motor, use a lower rpm motor.
a) Lower lift (lower condensing, higher evaporator temp) improves COP. b) Star mode reduces applied voltage to better match part-load and improve efficiency/PF. c) Standard remedies for oversized throttle-controlled pumps.
📖 §4.7 TR formula & COP

23. In an alkali chemical plant, salt brine flowing at 18 m3/hr is cooled from 12°C to 7°C using chilled water. The chiller compressor motor draws 31 kW and total input power to allied accessories is 16 kW. Motor operating efficiency is 90%. Brine density is 1.2 kg/litre and specific heat capacity is 0.97 kCal/kg°C. a) What is the refrigeration load (TR) imposed by the brine cooling? b) What is the COP of the refrigeration compressor? c) What is the overall specific power consumption in kW/TR?

9th Dec-2009 · 10 marks
Show worked solution
a) TR = Q × Cp × (Ti−To)/3024 = (18,000 × 1.2 × 0.97 × (12−7))/3024 = 34.64 TR. b) COP = (3.516 × TR)/(power input to compressor) = 3.516 × 34.64/(31 × 0.9) = 4.365. c) Overall specific power consumption = (31 + 16)/34.64 = 47/34.64 = 1.3568 kW/TR.
a) Refrigeration load from mass flow × specific heat × temp drop / 3024. b) COP converting TR to kW via 3.516 kW/TR over shaft power. c) Total electrical input over refrigeration load.
📖 §4.7 air-side TR formula (AHU/FCU)

24. In an AHU the actual airflow is 9300 m3/hr, inlet air enthalpy 16.12 kCal/kg, outlet enthalpy 13.33 kCal/kg, air density 1.15 kg/m3. Estimate the TR of the AHU.

10th Jul-2010 · 5 marks
Show worked solution
Air-side load: TR = Q x rho x (h_in - h_out)/3024 = 9300 x 1.15 x (16.12 - 13.33)/3024 = 9300 x 1.15 x 2.79 / 3024 = 29,839/3024 = 9.87 TR (about 9.86 TR). Heat removed = 29,839 kcal/hr; dividing by 3024 kcal/hr per TR gives the AHU refrigeration load.
Air-side load uses TR = airflow x air density x (h_in - h_out) / 3024, with enthalpies read off the psychrometric chart in kcal/kg. Here 9300 x 1.15 x 2.79 = 29,839 kcal/hr, which is 9.87 TR. Use density (kg/m3) with volumetric airflow, or specific volume in the denominator - not both.
📖 §4.7 TR formula & COP

25. In an alkali plant, salt brine at 20 m3/hr is cooled from 14°C to 8°C using chilled water. The chiller compressor motor draws 44.4 kW at 90% motor efficiency. Allied auxiliaries draw 20 kW. Brine density 1.2 kg/litre, specific heat 0.97 kCal/kg°C. (a) Refrigeration load (TR); (b) COP of compressor; (c) overall specific power consumption (kW/TR).

10th Jul-2010 · 5 marks
Show worked solution
(a) Mass flow of brine = 20 m3/hr x 1000 x 1.2 kg/litre = 24,000 kg/hr. Refrigeration load TR = m x Cp x (Ti - To)/3024 = 24,000 x 0.97 x (14 - 8)/3024 = 139,680/3024 = 46.2 TR. (b) Compressor shaft power = 44.4 x 0.90 = 39.96 kW; COP = 3.516 x 46.2/39.96 = 162.4/39.96 = 4.06. (c) Overall specific power consumption = (44.4 + 20)/46.2 = 64.4/46.2 = 1.39 kW/TR.
Brine is not water: its density (1.2 kg/litre) and specific heat (0.97 kcal/kg degC) must both be used in TR = m x Cp x deltaT / 3024. COP uses the compressor SHAFT power (44.4 x 0.9), while overall kW/TR uses the total electrical input including the 20 kW auxiliaries - hence 4.06 vs 1.39 kW/TR.
📖 §4.9 EER; §4.15 Standards and Labeling

26. S-4: In a Commercial building, five window ACs each of 1.5 TR capacity were evaluated for replacement with three star labeled new ACs having Energy Efficiency Ratio (EER) of 2.50 kW/kW. The measured EER of existing ACs: AC1 = 2.05, AC2 = 2.19, AC3 = 2.30, AC4 = 2.40, AC5 = 2.17. Calculate the total kW saving potential if all the existing ACs are replaced with 3 star labeled ACs of same capacity.

11th Feb-2011 · 5 marks
Show worked solution
Input kW = TR delivered*3.516/EER. For 3 star AC input power = 1.5*3.516/2.5 = 2.11 kW each. Existing kW input: AC1 = 2.573, AC2 = 2.408, AC3 = 2.293, AC4 = 2.198, AC5 = 2.430; Total = 11.902 kW. Savings potential = 11.902 - (2.11 x 5) = 11.902 - 10.55 = 1.352 kW.
Input kW = TR*3.516/EER for each AC; saving = sum of existing input - new input (5 x 2.11).
📖 §4.8 (a,b); rest cross-chapter

27. L-2: Fill in the blanks. (a) With increase in condensing temperature in a vapor compression refrigeration system, the specific power consumption of the compressor for a constant evaporator temperature will____. (b) With increase in evaporator temperature while maintaining a constant condenser temperature, the specific power consumption of the compressor will____. (c) Lower power factor of a DG set demands ____ excitation current. (d) Slip power recovery system is used in ____ induction motor. (e) If voltage is reduced from 230 V to 200 V for a fluorescent tube light, it will result in ____ power consumption. (f) ____ fans are known as 'non-overloading' because change in static pressure do not overload the motor. (g) ____ head is the friction loss, on the liquid being moved, in pipes, valves and equipment in the system. (h) Ratio of the light reflected by a surface to the solar light incident upon it, is called ____. (i) ____ is the ratio of solar heat gain that passes through fenestration to the total incident solar radiation that falls on the fenestration. (j) luminous flux incident on an object per unit area is defined as ____.

11th Feb-2011 · 10 marks
Show worked solution
a. increase; b. decrease; c. higher; d. slipring (slip-ring); e. reduced; f. backward-inclined; g. dynamic; h. Solar Reflectance; i. Solar heat gain coefficient; j. illuminance.
Standard refrigeration/motor/fan/pump/building fill-in answers per official key.
📖 §4.7 air-side TR formula (AHU/FCU)

28. L-4: An energy audit was conducted to find out the ton of refrigeration (TR) of an Air Handling Unit (AHU). Evaporator area = 10.0 m2; Inlet velocity = 1.9 m/s; Inlet air DBT = 21.5 C, RH = 75%, Enthalpy = 53.0 kJ/kg; Outlet air DBT = 17.4 C, RH = 90%, Enthalpy = 46.4 kJ/kg; Density of air = 1.14 kg/m3. Find out the TR of AHU.

11th Feb-2011 · 10 marks
Show worked solution
AHU refrigeration load = [Air flow rate (m3/h) x Density of air (kg/m3) x Difference in enthalpy (kJ/kg)] / (3024 x 4.18). Air flow = 10.0 x 1.9 x 3600 = 68400 m3/h. AHU = (10.0 x 1.9 x 3600) x 1.14 x (53 - 46.4) / (3024 x 4.18) = 40.71 TR.
Mass flow x enthalpy drop gives kJ/hr; convert to kCal (/4.18) and to TR (/3024).
📖 §4.3 Absorption Refrigeration (VAR advantages)

29. What are the advantages of using vapour absorption refrigeration system over vapour compression system? Under what condition it would be economical? (5 Marks)

Mar 2021 · 5 marks
Show worked solution
Refer Guidebook-3 (Pg 112-116): Advantages of VAR over VCR — uses low-grade/waste heat or steam instead of high-grade electrical energy; very few moving parts (only pump) so low maintenance, low noise/vibration; uses environment-friendly refrigerants (water/ammonia); can use otherwise wasted heat. It is economical when low-cost waste heat, exhaust gas, or low-pressure steam is available, making the running cost low despite higher first cost.
Advantages and economic conditions for VAR per Guidebook-3.
📖 §4.7 air-side TR (a); Book-3 Ch-2 motors (b)

30. a) Calculate the filter area of an Air Handling Unit (AHU) for Refrigeration Load of 50 TR. The air enthalpy at inlet of AHU is 85 kJ/kg and at outlet is of 60 kJ/kg. Air velocity at filter is 1.81 m/sec and air density is 1.26 kg/m3. (5 Marks) b) A no load test was conducted in a delta connected 37 kW induction motor. Name plate data: 3 Phase, 415 V, 50 Hz, 55 Amp. Measured data on no load: Voltage V = 415 Volts; Current I = 18 Amps; Frequency F = 50 Hz; Stator phase resistance at no load = 0.23 Ohms/phase. No load power = 955 Watts. Calculate: i. The iron loss plus friction loss plus windage loss (2 Marks) ii. Stator copper loss at name plate ratings (full load), considering stator temperature as 120oC (2 Marks) iii. No load power factor of the motor (1 Mark)

Mar 2021 · 10 marks
Show worked solution
a) TR of AHU = (Enthalpy difference x density x area x velocity x 3600)/(4.187 x 3024). 50 = (85-60) x 1.26 x Area x 1.81 x 3600/(4.187 x 3024). Filter Area = TR x (4.187 x 3024)/[(enthalpy diff) x density x velocity x 3600] = 50 x (4.187 x 3024)/[(25) x 1.26 x 1.81 x 3600] = 3.08 m2. b) i) No-load input Pin = 955 W. No-load stator copper loss = 3 x I^2 x R = 3 x 18^2 x 0.23 = 74.51 W. Iron + friction + windage loss = Pin - no-load Cu loss = 955 - 74.51 = 880.49 W. ii) Stator resistance at 120oC = 0.23 x (120+235)/(30+235) = 0.23 x 355/265 = 0.308 ohms. Full-load stator copper loss at nameplate current = 3 x (55/sqrt3)^2 x 0.308 = 3 x 1008.3 x 0.308 = 931.65 W. iii) No-load power factor = P/(sqrt3 x V x I) = 955/(1.732 x 415 x 18) = 0.0738.
AHU: TR = (dh x rho x A x v x 3600)/(4.187 x 3024); iron+fr+wind = Pin - 3I2R; R corrected by (235+t2)/(235+t1); PF = P/(sqrt3 VI).
📖 §4.3 VAR COP; §4.7 TR

31. A process plant has installed 5 MW DG set for base load operation, which is operating at 70% loading. Furnace oil is used as a fuel in the DG set. The DG set generates 8.6 kg of exhaust gas per kWh generated. The plant management has decided to install a heat recovery boiler to generate steam at 3 kg/cm2g from the exhaust gas to reduce the exit flue gas temperature from 450degC to 200degC. The specific heat of flue gas is 0.26 kcal/kgdegC. The steam generated from waste heat boiler will be used in double effect Li-Br Vapor Absorption Chiller, with a COP of 1.12. How much TR will be generated through VAM?

Mar 2021 (Set B) · 5 marks
Show worked solution
DG loading = 5 MW x 70% = 3.5 MW = 3500 kW. Quantity of heat available from exhaust gas = 3500 x 8.6 x gas generated/kWh x 0.26 kcal/kgdegC x (450 - 200) = 19,56,500 kcal/hr. Potential TR generation through double-effect VAM: COP = TR effect/Heat input, so TR = (COP x Heat input)/3024 = (1.12 x 1956500)/3024 = 724.6 TR.
Waste heat from exhaust = mass x Cp x dT; refrigeration effect = COP x heat input; convert kcal/h to TR by dividing by 3024.
📖 §4.7 TR formula; pump hydraulic power

32. An energy audit study of a central chiller system in a commercial building was conducted and measured parameters are given below: Chilled water inlet temperature 12 °C; Chilled water Outlet temperature 7 °C; Chilled water pump discharge pressure 3.6 kg/cm²g; Pump suction 1.5 meters above the pump-center line; Power drawn by the chilled water pump motor 70 kW; Efficiency of pump motor 91%; Pump efficiency 60%. Find out the operating load of the Chiller system in TR.

Jul 2022 · 5 marks
Show worked solution
Discharge head = 3.6 kg/cm2g x 10 = 36 m; the suction is 1.5 m above the pump centreline, so total head = 36 - 1.5 = 34.5 m. Pump shaft power = motor input x motor efficiency = 70 x 0.91 = 63.7 kW. Flow = (shaft power x 1000 x pump efficiency)/(head x 1000 x 9.81) = (63.7 x 1000 x 0.6)/(34.5 x 1000 x 9.81) = 0.1129 m3/s = 406.5 m3/hr. Refrigeration load = 406,500 x 1 x (12 - 7)/3024 = 672 TR.
Hydraulic power = flow x head; rearranged, flow = (pump shaft power x pump efficiency)/(head x 9.81). Total head = discharge head (3.6 kg/cm2g = 36 m) minus the 1.5 m positive suction lift = 34.5 m. Refrigeration load then follows from TR = mass flow x Cp x deltaT / 3024 (1 TR = 3024 kcal/hr).
📖 §4.7 TR & operating cost; §4.3 VAR vs VCR

33. The data for centrifugal chiller and vapour absorption chiller are given below — Chilled water flow (m³/h): Centrifugal 189, VAM 180; Condenser water flow (m³/h): Centrifugal 258, VAM 340; Chiller inlet temp (°C): Centrifugal 13.0, VAM 14.6; Condenser water inlet temp (°C): Centrifugal 27.1, VAM 33.5; Chiller outlet temp (°C): Centrifugal 7.7, VAM 9.0; Condenser water outlet temp (°C): Centrifugal 35.7, VAM 39.1; Power drawn by compressor (kW): Centrifugal 190, VAM -; Steam consumption (kg/h): Centrifugal -, VAM 1570; Chilled water pump (kW): Centrifugal 28, VAM 28; Condenser water pump (kW): Centrifugal 22, VAM 33; Cooling tower fan (kW): Centrifugal 6.0, VAM 15; Cost of Steam (Rs/kg): VAM 2.0; Cost of electricity (Rs/kWh): 9.0 both. a) Evaluate the tonnes of refrigeration (TR) of both the systems. b) Operating Energy cost per hour for both the systems.

Jul 2022 · 10 marks
Show worked solution
a) Centrifugal chiller TR = Chilled water flow × (Tin - Tout) × Diff. in temp / 3024 = 189 × 1000 × 1 × (13-7.7)/3024 = 331.25 TR. VAM TR = 180 × 1000 × 1 × (14.6-9.0)/3024 = 333.33 TR. b) Auxiliary power consumption: Centrifugal = Chilled water pump + condenser water pump + cooling tower fan = 28 + 22 + 6.0 = 56 kW. VAM auxiliary power (kW) = 28 + 33 + 15 = 76 kW. Energy cost of centrifugal chiller = (56 + 190)×9 = Rs 2214/hr. Energy cost of VAM chiller = (76×9) + (1570×2) = Rs 3824/hr.
TR = flow×ΔT/3024 (kcal→TR); energy cost = electrical kW×rate (+ steam kg×rate for VAM).
📖 §4.7 TR formula; pump hydraulic power

34. A multi storied office has centralized air conditioning system by using the chilled water. The chilled water inlet and outlet temperatures are 13°C and 9°C respectively. The chilled water pump discharge pressure is 4.2 kg/cm²g and the suction is 10 meters above the pump centerline. The power drawn by the chilled water pump's motor is 75 kW and an efficiency of 92%. The chilled water pump efficiency at the operating point from pump characteristic curve is 65%. Find out the operating refrigeration load in TR.

Mar 2023 · 5 marks
Show worked solution
Total head of the Chilled Water Pump = (4.2×10) − 10 = 32 Meter. Shaft Power of the Pump = 75×0.92 = 69 kW. Flow rate = (69×1000×0.65)/(32×1000×9.81) = 0.14287 m³/s = 514.33 m³/hr. Refrigeration load = 514330×4/3024 = 680 TR.
Head from discharge pressure and static lift; hydraulic power gives flow; cooling load = flow × ΔT × density/specific heat converted to TR (3024 kcal/h per TR).
📖 Cross-chapter matching (§4.3 condenser, §4.14 spray nozzles)

35. L-3(B): Match the following (1 Mark each): 1. Pitot Tube; 2. Refrigerant Drier; 3. Condenser; 4. Spray Nozzles; 5. Occupancy Sensor — with — A. Cooling Tower; B. Lighting Control; C. Gas Velocity in ducts; D. Compressed Air System; E. Refrigeration System.

Mar 2023 · 10 marks
Show worked solution
1. Pitot Tube → C. Gas Velocity in ducts; 2. Refrigerant Drier → D. Compressed Air System; 3. Condenser → E. Refrigeration System; 4. Spray Nozzles → A. Cooling Tower; 5. Occupancy Sensor → B. Lighting Control.
Standard equipment-to-application matching.
📖 §4.7 kW/TR & COP; Book-3 Ch-7 Cooling Towers

36. L-4: During the energy audit of central chiller plant, following parameters were noted: Chilled water flow 250 m³/hr; Chilled water inlet temperature 12°C; Chilled water outlet temperature 7°C; Motor Input Power 350 kW; Motor Efficiency 90%; Condenser water inlet temperature (going to chiller or outlet of cooling tower) 31°C; Condenser water outlet temperature (leaving from chiller or inlet to cooling tower) 36°C; Wet Bulb temperature of ambient air 28°C; Make up water TDS 180 ppm; Permissible limit of TDS for cooling water 720 ppm; Condenser cooling capacity 25% higher than the evaporator cooling capacity. Calculate: kW/TR of chiller compressor; COP of chiller; Effectiveness of cooling tower; Evaporation loss; Blow down quantity; Make-up water requirement (ignoring no drift loss).

Mar 2023 · 10 marks
Show worked solution
Chiller machine capacity TR = [250×1000×1×(12−7)]/3024 = 413.4 TR. kW/TR of chiller compressor = (350×90%)/413.4 = 0.762. COP = (413.4×3024)/(350×0.9×860) = 4.6. Effectiveness = (36−31)/(36−28) = 62.5%. Cycle of Concentration COC = 720/180 = 4. Condenser TR = 1.25×413.4 = 516.75 TR. Condenser water flow / circulation flow = 516.75×3024/(1000×1×(36−31)) = 312.5 m³/hr. Evaporation loss = 0.00085×1.8×312.5×(36−31) = 2.3 m³/hr. Blow down = Evap/(COC−1) = 2.3/(4−1) = 0.797 m³/hr. Make-up = Evaporation + Blow down = 2.3+0.797 = 3.097 m³/hr.
Chiller TR from flow×ΔT; kW/TR from shaft power; COP conversion; cooling-tower effectiveness, COC, evaporation, blowdown and make-up formulae.

Ch4 · Psychrometry, ventilation & ACH

5 past-paper questions · ≈35 marks
ACH = air changes per hour = airflow (m³/hr) / room volume (m³). Dry-bulb, wet-bulb, RH, dew point.
📖 §4.2 Psychrometrics (humidity ratio)

1. A stream of moist air (mass flow 10.1 kg/s, specific humidity 0.01 kg/kg dry air) mixes with a second stream of superheated water vapour flowing at 0.1 kg/s. Assuming proper uniform mixing without condensation, what is the humidity ratio of the final stream (kg/kg dry air)?

18th Exam · 5 marks
Show worked solution
Dry air = 10.1/(1+0.01) = 10 kg/s; moisture in moist air = 0.1 kg/s. Final moisture = 0.1 + 0.1 (added vapour) = 0.2 kg/s. Humidity ratio H = (0.01x10 + 0.1x1)/10 = (0.1 + 0.1)/10 = 0.02 kg per kg of dry air.
Mass-balance the dry air and total moisture; humidity ratio = total moisture / dry-air mass.
📖 §4.14 Humidifying air by adding water

2. S-3: In an air washer of a textile humidification system, airflow 3000 m3/h at 25 °C and 10% RH is humidified to 60% RH. Inlet/outlet specific humidity = 0.002 / 0.0062 kg/kg dry air; air density at 25 °C = 1.184 kg/m3. Calculate the water required (kg/hr).

19th Exam · 5 marks
Show worked solution
Water required mw = volume x density x (ω_out - ω_in) = 3000 x 1.184 x (0.0062 - 0.002) = 3000 x 1.184 x 0.0042 = 14.9 kg/hr.
Water added = mass of air x change in specific humidity (mass flow = volume flow x density).
📖 §4.2 Psychrometric Chart; Ch-10 thermal emittance

3. a) Name six parameters along with units that a psychrometric chart provides to an air conditioning system. (3 Marks) b) Explain briefly about Thermal Emittance. (2 Marks)

Mar 2021 · 10 marks
Show worked solution
a) Six parameters: 1. Dry bulb temperature (oC); 2. Relative humidity (%); 3. Wet bulb temperature (oC); 4. Specific volume (m3/kg of dry air); 5. Enthalpy (kJ/kg of dry air); 6. Specific humidity or Humidity factor (grams/kg of dry air). b) Refer Guidebook-3, page 272 — Thermal emittance is the ratio of radiant heat emitted by a surface to that emitted by a black body at the same temperature; low-emittance surfaces re-radiate less absorbed heat into the building.
Psychrometric chart parameters and thermal emittance definition per Guidebook-3.
📖 §4.2 Psychrometric Chart; Ch-10 thermal emittance

4. a) Name six parameters along with units that a psychrometric chart provides (3 marks). b) Explain briefly about Thermal Emittance (2 marks).

Mar 2021 (Set B) · 5 marks
Show worked solution
a) Refer Guidebook-3 Pg 272. Six parameters from psychrometric chart: 1. Dry bulb temperature (degC); 2. Relative humidity (%); 3. Wet bulb temperature (degC); 4. Specific volume (m3/kg of dry air); 5. Enthalpy (kcal/kg of dry air); 6. Specific humidity / humidity factor (grams/kg of dry air). b) Thermal emittance is the ratio of radiant heat flux emitted by a surface to that of a black body at the same temperature; high-emittance roof surfaces radiate absorbed heat readily, reducing heat gain.
Psychrometric chart properties per Guidebook-3; thermal emittance defined relative to a black body radiator.
📖 §4.2 (item 1); rest cross-chapter

5. L-3(A): State Increases or Decreases (1 Mark each): 1. If air dry bulb temperature is increased then Relative Humidity will ___. 2. In a pumping system, if the suction side liquid level is increased then NPSHa will ___. 3. If the air temperature increases at the inter-cooler outlet, then air compressor power consumption will ___. 4. A blower is retrofitted with a VFD and operated at full speed. The power consumption will ___. 5. As the design speed of the motors decreases the capacitor KVAr requirement will ___.

Mar 2023 · 10 marks
Show worked solution
1. Decreases; 2. Increases; 3. Increases; 4. Increases; 5. Increases.
Standard psychrometric/pump/compressor relationships per BEE Guide Book 3.

Ch3 · FAD, leakage % and specific power

21 past-paper questions · ≈155 marks
Leakage % = T/(T+t) × 100 (loaded vs unloaded time). Specific power = kW / (m³/min). 1 bar extra ≈ 6–7% more power.
📖 Book-3 §3.5 Avoiding Air Leaks — leak quantification (Tables 3.16 & 3.17); Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method)

1. A 2-stage reciprocating compressor is supplying nitrogen from low pressure header to high pressure vessel. This high-pressure nitrogen is only used during any process upset. Compressor is cut-off once vessel pressure reaches 45 barg, and started when vessel pressure comes down to 35 barg. During energy audit, it was observed that compressor is started at gap of every 36 hrs when there is no intended consumption. Other data: Vol. of high pressure N2 vessel: 11.5 m3 Vessel temperature: 35.0 Deg.C Initial gas density: 50.3 kg/m3 End gas density: 39.4 kg/m3 Compressor load kW drawn: 30.0 kW Compressor capacity at constant suction pressure: 250.0 kg/hr i. Estimate the leak rate (kg/hr). ii. Estimate the energy saving potential (kWh/Annum), if all leaks are attended. Consider operating time of 8760 hrs/annum.

Sep 2024 · 10 marks
Show worked solution
Initial Vessel Pressure = 45.0 barg; End vessel pressure = 35.0 barg Initial gas density = 50.3 kg/m3; End gas density = 39.4 kg/m3 Change in gas quantity in 36 hrs = (50.3 − 39.4) × 11.5 = 125.7 kg N2 leakage rate = 125.7 / 36 = 3.5 kg/hr Time needed for compressor run = 125.7 / 250 = 0.51 hrs or 30.6 min % time of compressor running = 0.51 / 36 = 1.40 % Running time of compressor per annum = 1.40% × 8760 = 122.3 hrs Power consumption per annum due to air leakage = 122.3 × 30 = 3670.0 kWh/Annum Energy Saving Potential = 3670.00 kWh/Annum
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — leak quantification — Leak = (Δdensity × volume)/time; compressor run time to replace leak = leaked mass/capacity; annual energy = % run time × hours × kW.
📖 Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method)

2. Calculate the free air delivery (FAD) capacity of a compressor in m3/hr for: receiver capacity 0.3 m3, initial pressure 0 kg/cm2(g), final pressure 7 kg/cm2(g), initial air temperature 35 degC, final air temperature 50 degC, additional holdup volume 0.05 m3, compressor pump-up time 4.1 minutes, atmospheric pressure 1.026 kg/cm2(a).

Book EOC · 5 marks
Show worked solution
Total system volume V = 0.3 + 0.05 = 0.35 m3. FAD = [(P2 - P1)/P0] x V / t, corrected for temperature. P2 - P1 = (7+1.026) - (0+1.026) = 7.0 kg/cm2; P0 = 1.026 kg/cm2(a). Volume of free air = (7.0/1.026) x 0.35 = 2.388 m3 in 4.1 min. FAD per minute = 2.388/4.1 = 0.5825 m3/min = 34.95 m3/hr. Applying the temperature correction (273+35)/(273+50) = 308/323 = 0.954, FAD ≈ 34.95 x 0.954 ≈ 33.3 m3/hr.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — FAD = [(P2-P1)/atmospheric] x (receiver+holdup) / pump-up time, with temperature correction; gives about 33 m3/hr.
📖 Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure (Table 3.9); Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System

3. A reciprocating V-belt driven compressor operates with load pressure 6 bar, unload pressure 8 bar, load time 3 minutes and unload time 1.5 minutes. Suggest possible energy saving opportunities on a short-term basis.

Book EOC · 5 marks
Show worked solution
The load:unload ratio is 3:1.5, i.e. the compressor is loaded about 67% of the time, indicating modest spare capacity. Short-term measures: (1) Reduce the maximum operating/unload pressure (8 bar) to the minimum the plant actually needs, since every 1 bar reduction saves about 6-7% energy. (2) Reduce the pressure band/raise the cut-in so the compressor unloads more. (3) Lower the load pressure setting closer to the required application pressure. (4) Reduce compressor speed by trimming the motor pulley size if demand is consistently below capacity. (5) Detect and arrest air leaks and ensure cool, clean inlet air. These lower the average discharge pressure and running power.
Confirmed vs Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure — Load 3 min / unload 1.5 min means the machine is loaded about 67% of the time. Short-term actions: cut the 8 bar unload pressure toward actual demand (6-10% power per bar), narrow the load-unload band, arrest leaks, ensure cool clean intake air, and trim the motor pulley to de-rate the spare capacity.
📖 Book-3 §3.7 Solved Example (FAD pump-up + leakage test, pp.103-104)

4. 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.

Book EOC · 10 marks
Show worked solution
(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.
📖 Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method); Book-3 §3.5 Avoiding Air Leaks — leak quantification (Tables 3.16 & 3.17)

5. A free air delivery test was carried out before conducting a leakage test on a reciprocating air compressor in an engineering industry, with the following observations: Receiver capacity 8.0 m3; Initial pressure 0.1 kg/cm2(g); Final pressure 7.0 kg/cm2(g); Additional hold-up volume 0.3 m3; Atmospheric pressure 1.026 kg/cm2 abs; Compressor pump-up time 3.5 minutes. During the subsequent leakage test at lunch time, with no pneumatic equipment or control valves in operation: (a) compressor on-load time 24 seconds at an unloading pressure of 7 kg/cm2(g); (b) average power drawn during loading 92 kW; (c) compressor unload time 79 seconds and loading pressure 6.6 kg/cm2(g). Find: (i) compressor output in m3/hr (neglect temperature correction); (ii) specific power consumption in kW/(m3/hr); (iii) % air leakage in the system; (iv) leakage quantity in m3/hr; (v) power lost due to leakage.

Book EOC · 10 marks
Show worked solution
Total system volume V = 8.0 + 0.3 = 8.3 m3. (i) Compressor output Q = [(P2 - P1)/P0] x V/t = [(7.0 - 0.1)/1.026] x (8.3/3.5) = 6.725 x 2.371 = 15.94 m3/min = 956.6 m3/hr (temperature correction neglected as stated). (ii) Specific power consumption = 92 kW / 956.6 m3/hr = 0.0962 kW per m3/hr. (iii) % air leakage = T/(T+t) x 100 = 24/(24+79) x 100 = 23.30%. (iv) Leakage quantity = 0.2330 x 956.6 = 222.9 m3/hr (= 3.71 m3/min). (v) Power lost due to leakage = leakage quantity x specific power consumption = 222.9 x 0.0962 = 21.4 kW.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — L-2 as printed: FAD = [(P2-P1)/Patm] x (receiver + hold-up)/pump-up time, then % leakage = load/(load+unload), leakage quantity = % x FAD and power lost = leakage quantity x specific power. The stem was restored from the book (pump-up time 3.5 min and the full leakage data were missing) and the item re-typed as Long, per its own source line.
📖 Book-3 §3.7 Solved Example (FAD pump-up + leakage test, pp.103-104)

6. a) Pump-up test on a reciprocating compressor: receiver+holdup 4100 litres, initial 1 kg/cm²(g), final 8.5 kg/cm²(g), atmospheric 1.026 kg/cm²(a), ambient 32°C, final compressed air temp 52°C, pump-up time 65 s. Calculate FAD in cfm. b) Leakage test: load 3 min, unload 13 min, drawing 145 kW on load. Find i) % leakage, ii) leakage quantity, iii) specific power consumption, iv) power lost due to leakage.

14th Exam · 10 marks
Show worked solution
a) FAD Q = [(P2-P1)/Pa] x [V/t] x [(273+t1)/(273+t2)] = [(8.5-1)/1.026] x [4.1/1.0833] x [305/325] = 25.96 m³/min = 25.96 x 3.28³ = 916 cfm. b) i) % leakage = T/(T+t) x 100 = 3/(3+13) x 100 = 18.75%. ii) Leakage quantity = 0.1875 x 916 = 171.75 cfm. iii) Specific power consumption = 145/916 = 0.1583 kW/cfm. iv) Power lost to leakage = 171.75 x 0.1583 = 27.19 kW.
Confirmed vs Book-3 §3.7 Solved Example — FAD from receiver pump-up with temperature correction; leakage from load/unload duty cycle.
📖 Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method) (part a); Book-3 Ch-4 HVAC & Refrigeration — VCR vs VAR (part b)

7. a) Calculate FAD in m³/min: receiver capacity 0.25 m³, initial 1 kg/cm²(g), final 7 kg/cm²(g), initial temp 32°C, final temp 52°C, additional holdup 0.05 m³, pump-up time 2.1 min. b) Identify as VCR or VAR: I) system operates under vacuum; II) uses water as refrigerant; III) uses large amount of high-grade energy; IV) COP decreases considerably with decrease in evaporator pressure; V) can work on lower evaporator pressures without affecting COP.

Set-A · 10 marks
Show worked solution
a) Q = [(P2-P1)/Pa] x [V/t] x [(273+t1)/(273+t2)] = [(7-1)/1.026] x [(0.25+0.05)/2.1] x [(273+32)/(273+52)] = 0.784 m³/min. b) I) VAR; II) VAR; III) VCR; IV) VCR; V) VAR.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — FAD from receiver pump-up with temperature correction; VCR/VAR classification.
📖 Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method)

8. The rated compressor capacity is 15 m3/min. Evaluate if there is any capacity de-rating using the air-receiver tank filling method. Data: Receiver volume (incl. pipe & cooler) = 9 m3; Initial Pressure = 0.5 kg/cm2; Final Pressure = 7.0 kg/cm2; Atmospheric pressure = 1.026 kg/cm2; Time to build pressure = 5 minutes. (b) What is the deficiency in this calculation and how can it be corrected?

17th Sep-2016 · 10 marks
Show worked solution
(a) Compressor output = [(P2 - P1) x V] / (Pa x t) = [(7.0 - 0.5) x 9] / (1.026 x 5) = 58.5/5.13 = 11.40 m3/min. Capacity shortfall against the rated 15 m3/min = 15 - 11.40 = 3.60 m3/min, i.e. (3.60/15) x 100 = 24% de-rating. Since this far exceeds the 10% the book allows before corrective action, the compressor must be investigated (worn valves alone can cost up to 20% of capacity). (b) Deficiency: the formula as used assumes the compressed air temperature equals the ambient temperature, i.e. perfect isothermal compression. In practice the discharge/receiver temperature t2 is higher than the ambient t1, so the measured volume is overstated. Correction: multiply the result by the factor (273 + t1)/(273 + t2), where t1 is the ambient/suction temperature and t2 the compressed air (receiver) temperature; this factor is always less than 1.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — Tank-filling FAD method Q = (P2-P1) x V/(Pa x t), compared against rated capacity. The model answer's temperature correction had been written inverted; the book's factor is (273+t1)/(273+t2), always less than 1 when the discharge is hotter than ambient.
📖 Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power); Book-3 §3.2 Positive Displacement — Reciprocating Compressors

9. A 1680 m3/hr reciprocating compressor is driven by a 160 kW motor (90% efficiency) drawing 159 kW. Demand rises by 100 m3/hr. To meet it the compressor speed is increased by changing the compressor pulley. Existing: Motor rpm 1400, Motor pulley 300 mm, Compressor rpm 700, Compressor pulley 600 mm. Find new pulley diameter and additional power; check if motor can handle the load.

17th Sep-2016 · 5 marks
Show worked solution
Modified flow = 1780 m3/hr. New compressor rpm = (1780/1680)×700 = 742 rpm. Using N1D1 = N2D2, new compressor pulley D2 = (700×600)/742 = 566 mm. New motor power = (742/700)×159 = 168.54 kW. Motor capacity = 160/0.9 = 178 kW. Since 168.54 < 178 kW, the motor has the margin to absorb the additional 100 m3/hr load.
Confirmed vs Book-3 §3.5 Capacity Utilisation — Flow ∝ rpm; pulley by N1D1=N2D2; power ∝ rpm (reciprocating, ~linear here).
📖 Book-3 §3.5 Sizing of Compressed Air Piping

10. Determine the discharge pipe inner diameter size (in mm) for a compressed air system, given: FAD = 1000 Nm3/hr; discharge pressure = 7 bar(g); discharge temperature = 35 °C; air velocity = 6 m/s; atmospheric pressure = 1.013 bar.

18th Exam · 5 marks
Show worked solution
Convert NTP flow to actual flow using P1V1/T1 = P2V2/T2. Actual flow V2 = (P1V1/T1) x (T2/P2) = (1.013 x 1000/273) x (308/8.013) = 142.6 m3/hr = 0.0396 m3/s. Area = flow/velocity = 0.0396/6 = 0.0066 m2. di = √(4A/π) = 0.092 m = 92 mm → say 100 mm.
Confirmed vs Book-3 §3.5 Sizing of Compressed Air Piping — Apply the gas law to convert normal-condition flow to actual conditions (8.013 bar abs, 308 K), then A = Q/V and di = √(4A/π). Round up to a standard pipe size.
📖 Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14); Book-3 §3.5 Avoiding Air Leaks — leak quantification (Tables 3.16 & 3.17)

11. S-8: Two 480 CFM screw compressors A & B. Compressor-A runs at full load; Compressor-B runs in load-unload. Load power of both = 74 kW; unload power of B = 26 kW. Both run all working day. B's loading is 64%; after arresting leakage, loading falls to 35%. Estimate energy savings per day.

19th Exam · 5 marks
Show worked solution
Existing: A = 74 kW. B = 0.64 x 74 + 0.36 x 26 = 47.36 + 9.36 = 56.72 kW. After leakage arrest, B = 0.35 x 74 + 0.65 x 26 = 25.9 + 16.9 = 42.8 kW. Difference (B only) = 56.72 - 42.8 = 13.92 kW. Daily energy savings = 13.92 x 24 = 334 kWh/day.
Confirmed vs Book-3 §3.5 Capacity Control of Compressors — Weighted power of load-unload compressor = load%×load-power + unload%×unload-power; savings = drop in B's power × 24 h.
📖 Book-3 §3.5 Air Dryers (Tables 3.18 & 3.19); Book-3 §3.5 Air Receivers (IS 7938-1976 sizing)

12. L-2: Write short notes on the following with respect to the compressed air system: (a) Refrigeration drier, (b) Heat of compression drier, (c) Role of air receiver, (d) Dew point.

19th Exam · 10 marks
Show worked solution
(a) Refrigeration dryer: straight mechanical refrigeration in which the dew point of the air is reduced by chilling; a second heat exchanger lets the outgoing cold air pre-cool the incoming compressed air. The achievable atmospheric dew point is about -20 degC. It is the most economical process for roughly 90% of all applications, separates almost 100% of solid particles and water droplets larger than 3 micron, and costs about 0.2 bar in pressure loss (2.9 kW per 1000 m3/hr, Table 3.19). (b) Heat of compression (HOC) dryer: a twin-tower adsorption dryer in which compressed air taken directly from the compressor discharge before the after-cooler, at about 135 degC for a reciprocating machine, regenerates the desiccant. There are no electrical heaters and no purge loss, so operating cost is zero to very minimal (0.8 kW per 1000 m3/hr) and the atmospheric dew point achieved is -40 degC. Vessel A is in service for 4 hours while vessel B is heated for 2.5 hours and cooled for 1.5 hours, then they change over. Capacities range from 400 to 5000 cfm. (c) Role of the air receiver: it dampens the pulsations leaving the compressor discharge, acts as a reservoir for sudden or unusually heavy demands in excess of compressor capacity, prevents too frequent loading and unloading (short cycling), and separates moisture and oil vapour by allowing carry-over from the after-cooler to precipitate. Per IS 7938-1976 its volume in m3 should be 1/10th to 1/6th of the output in m3/min, and a local receiver near a point of high cyclic demand avoids having to add compressor capacity. (d) Dew point: the temperature at which the moisture present in the air starts condensing. The extent of drying is expressed as the ATMOSPHERIC dew point (at atmospheric pressure); the lower the dew point, the drier the air — air at -40 degC atmospheric dew point holds only 80 ppm of moisture against 3800 ppm at 0 degC (Table 3.18). Dryer performance is quoted as PRESSURE dew point, and raising the pressure of a gas raises its dew point temperature, since the partial pressure of the water vapour rises in proportion (Dalton's law).
Confirmed vs Book-3 §3.5 Air Dryers — Short notes from §3.5: the refrigerant dryer's achievable atmospheric dew point is -20 degC (the earlier +3 degC figure contradicted Table 3.19 and has been corrected), the HOC dryer reaches -40 degC at 0.8 kW/1000 m3/hr, plus the receiver's four duties and the atmospheric-vs-pressure dew point distinction.
📖 Book-3 §3.5 Avoiding Misuse of Compressed Air — air amplifiers (Figure 3.7); Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14)

13. Two 270 cfm compressors operate at 7 kg/cm2(g), on-load 80% of the time; load power 40 kW and unload power 15 kW each. Cleaning air requirement is 60% of air generated. Calculate the daily energy consumption for cleaning air alone (continuous operation).

Sep 2019 · 5 marks
Show worked solution
Air delivered by 2 compressors = 270 x 0.80 x 2 = 432 cfm. Loading power = 40+40 = 80 kW; unloading power = 15+15 = 30 kW. Average kW = [80x(0.8x24) + 30x(0.2x24)]/24 = 70 kW. SEC = 70/432 = 0.162 kW/cfm. Cleaning air = 0.60 x 432 = 259 cfm. Energy for cleaning air/day = 259 x 0.162 x 24 = 1007 kWh/day (alternate method gives ~1008 kWh/day).
Confirmed vs Book-3 §3.5 Avoiding Misuse of Compressed Air — air amplifiers — Printed worked solution: average compressor power 70 kW, SEC 0.162 kW/cfm, ~1007-1008 kWh/day for cleaning air.
📖 Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14); Book-3 §3.7 Checklist for Energy Efficiency in Compressed Air System

14. L-5: A 220 cfm screw compressor: Shift I load 60 s/unload 10 s; Shift II load 45/unload 25; Shift III load 25/unload 45 (8 hrs/shift). Load power 37 kW, unload power 11 kW. Calculate: (1) energy loss per day; (2) shift-wise average air requirement in cfm; (3) energy savings after installing VFD with VFD loss 3% of load power.

Sep 2019 · 10 marks
Show worked solution
Shift I = ((60/70)x37 + (10/70)x11) x 8 = (31.71+1.57)x8 = 266.24 kWh. Shift II = (0.64x37 + 0.36x11)x8 = 221.12 kWh. Shift III = (0.36x37 + 0.64x11)x8 = 162.88 kWh. Daily total = 650.24 kWh. Daily unloading (loss) energy = (1.57+3.96+7.04)x8 = 100.56 kWh. Load-cycle energy = 549.68 kWh. With VFD = 549.68/0.97 = 566.68 kWh; VFD loss = 17 kWh; net VFD savings = 100.56 - 17 = 83.56 kWh/day. Air requirement: Shift I = 0.86x220 = 189.2 cfm; Shift II = 0.64x220 = 140.8 cfm; Shift III = 0.36x220 = 79.2 cfm.
Confirmed vs Book-3 §3.5 Capacity Control of Compressors — Printed solution: daily energy loss (unloading) 100.56 kWh; net VFD savings 83.56 kWh/day; shift air 189.2/140.8/79.2 cfm.
📖 Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method)

15. Calculate the free air delivery (FAD) capacity of a compressor in m3/min for the following data: Receiver capacity 0.5 m3, Initial pressure 0 kg/cm2(g), Final pressure 7 kg/cm2(g), Initial air temperature 32°C, Final air temperature 51°C, Additional holdup volume 0.03 m3, Pump up time 4.5 minutes, Atmospheric pressure 1.026 kg/cm2 absolute.

9th Dec-2009 · 5 marks
Show worked solution
FAD = [(P2 − P1)/P0] × [V/t] × [(273+t1)/(273+t2)] = [(7−0)/1.026] × [(0.5+0.03)/4.5] × [(273+32)/(273+51)] = 0.7564 m3/min.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — Standard pump-up FAD test formula correcting for pressure rise, receiver+holdup volume, pump-up time and temperature ratio.
📖 Book-3 §3.5 Avoiding Air Leaks — leak quantification (Tables 3.16 & 3.17)

16. A compressed air leakage test was conducted in an industry running 3 nos. of 500 cfm reciprocating compressors, maintained at loading-unloading settings of 6.6 and 7.0 kg/cm2g. Trial 1 (before): On load 30 secs, Unload 110 secs. Trial 2 (after attending leaks): On load 18 secs, Unload 145 secs. Average power was 71 kW during load and 16 kW during unload. Calculate the annual cost savings for 4000 hr/year operation at energy charge Rs. 6.00 per kWh.

9th Dec-2009 · 5 marks
Show worked solution
Leakage (trial 1) = (30×500)/(30+110) = 107 cfm. Leakage (trial 2) = (18×500)/(18+145) = 55 cfm. Specific power consumption = 71/(500×60) = 0.0023666 kW/ft³. Reduction in leakage = 107−55 = 52 cfm = 3120 cfh. Energy saving per hour = 3120 × 0.0023666 = 7.3838 kWh. Annual cost saving = 7.3838 × 4000 × 6 = Rs. 1,77,211 per annum.
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — leak quantification — Leakage quantity from load fraction × FAD; specific power per ft³; multiply leakage reduction by specific power, operating hours and tariff.
📖 Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method)

17. Calculate the free air delivery (FAD) of a compressor in m3/hr from a pump-up test: Receiver 0.5 m3; initial 0.0 kg/cm2(g); final 7.0 kg/cm2(g); atmospheric 1.026 kg/cm2(a); ambient air 40°C; final compressed air 60°C; additional holdup volume 0.005 m3; pump-up time 5 min 30 s.

10th Jul-2010 · 5 marks
Show worked solution
Q = [(P2 - P1)/P0] x [V/t] x [(273+t1)/(273+t2)] V = 0.5 + 0.005 = 0.505 m3; t = 5 min 30 s = 5.5 min; P2 - P1 = 7.0 - 0.0 = 7.0 kg/cm2; P0 = 1.026 kg/cm2(a); t1 = 40 degC (ambient), t2 = 60 degC (compressed air). Q = (7.0/1.026) x (0.505/5.5) x (313/333) = 6.8226 x 0.09182 x 0.9399 = 0.5888 m3/min FAD = 0.5888 x 60 = 35.33 m3/hr. (2 marks for the formula, 3 marks for the calculation.)
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — Q = [(P2-P1)/P0] x [V/t] x [(273+t1)/(273+t2)]; the answer now carries the full working rather than the bare result.
📖 Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method); Book-3 §3.5 Avoiding Air Leaks — leak quantification (Tables 3.16 & 3.17)

18. L-3: A free air delivery test was carried out before a leakage test on a reciprocating air compressor. Receiver capacity = 12 m3; Initial pressure = 0.2 kg/cm2(g); Final pressure = 7.0 kg/cm2(g); Additional hold-up volume = 0.3 m3; Atmospheric pressure = 1.026 kg/cm2(a); Compressor pump-up time = 4.8 minutes. Leakage test (lunch time): (a) on load time 40 s, unloading pressure 7 kg/cm2(g); (b) average power during loading 95 kW; (c) unload time and loading pressure are 90 s and 6.6 kg/cm2(g). Find (i) compressor output m3/hr, (ii) specific power consumption kW/(m3/hr), (iii) % air leakage, (iv) leakage quantity m3/hr, (v) power lost due to leakage.

11th Feb-2011 · 10 marks
Show worked solution
(i) Compressor output = [Total Volume x (P2-P1)/Atm.Pressure] / Pump-up time = [(12+0.3) x (7.0-0.2)/1.026] / 4.8 = [12.3 x 6.8/1.026]/4.8 = 16.9834 m3/minute = 1019 m3/hr. (ii) Specific power consumption = 95/1019 = 0.093228 kW/m3/hr. (iii) % leakage = T/(T+t) x 100 = 40/(40+90) x 100 = 30.77%. (iv) Leakage quantity = 0.3077 x 1019 = 313.54 m3/hr. (v) Power lost due to leakage = leakage quantity x specific power consumption = 313.54 x 0.093228 = 29.23 kW.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — FAD via receiver pump-up formula; leakage % from load/(load+unload) time; power loss = leakage volume x specific power.
📖 Book-3 §3.6 Compressor Capacity Assessment (shop-floor pump-up method)

19. A small foundry has installed a reciprocating air compressor of 14.25 m3/min. The plant could not meet the compressed air requirement and hence conducted a capacity test to determine the derating in the compressor capacity. Calculate the actual FAD delivered after considering necessary temperature correction in m3/min, and the percentage derating. Operating parameters: Volume of air receiver including pipe and cooler 9 m3; Atmospheric temperature (T1) 35degC; Receiver temperature (T2) 44degC; Initial Pressure 0.5 kg/cm2(g); Final Pressure 7.0 kg/cm2(g); Atmospheric pressure 1.026 kg/cm2(a); Time taken to build up the pressure 5 minutes.

Mar 2021 (Set B) · 5 marks
Show worked solution
FAD = [(P2 - P1)/Pa] x [(receiver + hold-up volume)/time] x temperature correction factor. P1 (initial) = 0.5 kg/cm2(g); P2 (final) = 7.0 kg/cm2(g); Pa = 1.026 kg/cm2(a); receiver and holding volume = 9 m3; pump-up time = 5 min. Uncorrected output = [(7.0 - 0.5) x 9] / (1.026 x 5) = 11.40 m3/min. Temperature correction factor = (273 + T1)/(273 + T2) = (273 + 35)/(273 + 44) = 308/317 = 0.972. FAD after correction = 11.40 x 0.972 = 11.08 m3/min. Capacity shortfall = 14.25 - 11.08 = 3.17 m3/min; % de-rating = (3.17/14.25) x 100 = 22.24%. As this is far above the 10% deviation the book allows, corrective action on the compressor is called for.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — FAD pump-up formula with the temperature correction (273+T1)/(273+T2); the P1/P2 labels in the model answer had been swapped and are now correct (initial 0.5, final 7.0 kg/cm2 g).
📖 Book-3 §3.5 Sizing of Compressed Air Piping

20. In an engineering industry, compressed air delivered is 500 CFM (FAD) and the compressor discharge pressure is 6 kg/cm² (gauge). Calculate the size of the header by considering velocity of compressed air 6 m/s. Assume Temperature remains constant.

Mar 2023 · 5 marks
Show worked solution
Quantity of air = 500 CFM = 500/35.31 = 14.16 m3/min of free air. Working pressure = 6 kg/cm2(g) = 7.013 kg/cm2(a); atmospheric = 1.013 kg/cm2(a). Applying Boyle's law at constant temperature, P1V1 = P2V2: V2 = 14.16 x 1.013 / 7.013 = 2.05 m3/min = 0.0341 m3/s (the compressed volume actually flowing in the header). Quantity of air flow = area x velocity, so (pi/4) x D2 x 6 = 0.0341, giving D2 = 0.00724 m2 and D = 0.085 m = 85 mm (about 3.35 inch). A standard 3" NB header would be selected, checking that the velocity stays in the usual 6-10 m/s band.
Confirmed vs Book-3 §3.5 Sizing of Compressed Air Piping — Convert FAD to the compressed volume by Boyle's law, then size from Q = area x velocity at 6 m/s; the garbled area step has been written out, giving D = 85 mm.
📖 Book-3 §3.5 Capacity Utilisation (unloading up to 30% of full-load power); Book-3 §3.5 Capacity Control of Compressors (Tables 3.13 & 3.14)

21. L-6: An engineering industry operating three shifts per day has replaced its old reciprocating compressors with 1000 CFM screw compressors. During the energy audit, run hours counter readings: Load Hours start 7956, end 8401 (166 Loading kW); Un-Load Hours start 4918, end 5121 (58.1 un-loading kW). Calculate: 1. Capacity Utilization (%) of the compressor; 2. Monthly energy consumption for the present loading of the compressor; 3. Plant management is considering to install a 750 CFM compressor for energy savings. Estimate the energy savings for the same operating load, if loading power is 125 kW and unloading power is 43.75 kW; 4. To meet the present air requirement, if VFD is to be installed in the 1000 CFM compressor, what should be the percentage reduction in speed.

Mar 2023 · 10 marks
Show worked solution
Load Hours = 8401−7956 = 445; Un-Load Hours = 5121−4918 = 203; Total running hrs = 648. 1. Capacity Utilization = (445×60×1000 CFM)/(648×60×1000) = 0.69 or 69%. 2. Monthly energy consumption = (445×166)+(203×58.1) = 85664.3 kWh. 3. Monthly air requirement = 445×60×1000 = 26700000 C.ft. 750 CFM capacity utilization = 26700000/(750×60×648) = 0.92. Therefore loading time = 648×0.92 = 596.2 hrs; Unloading time = 648−596.2 = 51.8 hrs. Monthly consumption = (596.2×125)+(51.8×43.75) = 76791.25 kWh. Energy savings per month = 85664.3 − 76791.25 = 8873.05 kWh. 4. Fan/affinity law N1/N2 = T2/T1 → N2 = (T1/T2)×N1 = (445/648)×N1 = 0.69 N1. Percentage reduction = 1−0.69 = 0.31 or 31%.
Confirmed vs Book-3 §3.5 Capacity Utilisation — Capacity utilization = load/(load+unload); monthly kWh = Σ(hours×power); compare with 750 CFM scenario; speed reduction from flow ratio (affinity law).

Ch6 · Pump power, efficiency & affinity laws

27 past-paper questions · ≈200 marks
Hydraulic power = ρ·g·Q·H / 1000 kW = Q(m³/hr)×H(m)×ρ / 367. Shaft power = hydraulic / η. Q∝N, H∝N², P∝N³.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); total head = static lift + suction lift (h_d − h_s)

1. A pump is used to fill a rectangular overhead tank measuring 5 m × 3.5 m with a height of 10 m. The inlet pipe to the tank is positioned at a height of 25 m above ground level. The following additional data is available: The pump draws water from an underground sump situated 4 meters below the pump level and delivers it to a tank whose overflow line is positioned 8 meters above the tank bottom. The motor driving the pump draws 7.5 kW of power. The operating efficiencies of the motor and the pump are 90% and 70% respectively. Calculate the time taken by the pump to fill the tank up to the overflow level.

Sep 2025 · 10 marks
Show worked solution
Step 1: Volume of water filled = 5 × 3.5 × 8 = 140 m³; Mass of water = 140 × 1000 = 140000 kg Step 2: Total head (H) = 25 + 4 = 29 m Step 3: Shaft Power = 7.5 × 0.90 = 6.75 kW Step 4: Water Power = 6.75 × 0.70 = 4.725 kW = 4725 W Step 5: Time taken: Pump efficiency = Mass flow × g × Head / Shaft Power 0.7 = mass flow × 9.81 × 29 / (6.75 × 1000) Mass flow = 16.6 kg/sec = 59791 kg/hr = 59.79 m³/hr Time = 140 / 59.79 = 2.34 hrs = 140.5 minutes
Hydraulic power = ṁ·g·H; derive mass flow from pump efficiency, then time = volume/flow.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³); motor part-loading from Book-3 Ch2

2. A) A clear water pump with rated flow of 125 m³/hr, head 55 m at rated speed of 1460 rpm and 79% efficiency supplies clarified water to a residential colony's water treatment facility. The daily water requirement is 3000 m³. The pump is directly coupled and driven by a three phase 50 Hp, 415 V, 64A, 0.9 pf, 1460 rpm induction motor with 90.5% full load efficiency. During an internal energy audit, the motor is found to operate with only 65% loading at pump rated conditions. The plant considers replacing the standard motor with a 30 kW IE3 motor. Operating parameters before and after motor replacement: Flow (m³/hr): 130 / ? Head (m): 52 / 51 Supply Voltage (V): 415 / 415 Current (Amp): 42 / 39 Power Factor: 0.9 / 0.92 Motor Eff (%): 0.88 / 0.932 The slip of the new IE3 motor has decreased by 20 rpm. Validate the savings, calculate: i) % Loading of motor after replacement. (1 Mark) ii) Flow after replacing the standard motor with 30 kW IE3 motor. (1 Mark) iii) Operating Pump Efficiency before and after motor replacement. (2 Marks) iv) Daily energy saving during operation due to motor replacement. (1 Mark) B) Mark True/False: i) Totally enclosed, fan cooled (TEFC) motors are less efficient than screen-protected, drip-proof (SPDP) motors. ii) Stray loss in induction motors is inversely proportional to load current. iii) As per BIS standard, the motor output should not be affected with voltage variation up to +/- 6%. iv) Motor life doubles for each 10°C reduction in operating temperature. v) Starting torque of energy efficient motors is higher than standard motors.

Sep 2025 · 10 marks
Show worked solution
A) i) Loading of the IE3 Motor = (1.732 × 0.415 × 39 × 0.92 × 0.932) / 30 = 80.12% ii) Flow after replacement = 130 m³/hr × 1480/1460 = 131.8 m³/hr iii) Operating Pump Efficiency: Power Consumption before replacement = 1.732 × 0.415 × 42 × 0.9 = 27.17 kW Power Consumption after replacement = 1.732 × 0.415 × 39 × 0.92 = 25.79 kW Before replacement = [(130/3600) × 52 × 9.81] / (27.17 × 0.88) = 77% After replacement = [(131.8/3600) × 51 × 9.81] / (25.79 × 0.932) = 76.2% Daily Operating Hour before = 3000/130 = 23.08 Hrs; after = 3000/131.8 = 22.77 Hrs iv) Daily Energy Savings = (23.08 × 27.17) − (25.79 × 22.77) = 39.9 kWh B) i) False; ii) False; iii) True; iv) True; v) False
Motor loading = √3·V·I·pf·η / rating; flow scales with speed (affinity); pump η = hydraulic/electrical input; savings from input power × operating hours difference.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); total differential head h_d − h_s (positive suction head is subtracted); cf. §6.11 Solved example — cooling water pump efficiency (p.194–195)

3. A cooling water pump has a positive suction head of 5 meters. The discharge pressure is 3.0 kg/cm², and the water flow rate is 150 m³/hr. Determine the pump efficiency given that the actual power input of the connected motor is 18.0 kW and the motor operates with an efficiency of 85%.

Sep 2024 · 5 marks
Show worked solution
Flow Rate: 150 m³/hr Total Head: 30 − 5 = 25 m Power input to pump = 18 × 0.85 = 15.3 kW Hydraulic Power = (150/3600) × 25 × 9.81 = 10.2 kW Pump Efficiency = 10.2 / 15.3 = 66.7%
Total head = discharge head (3 kg/cm² ≈ 30 m) − suction head; hydraulic power = (Q/3600)·H·g; η = hydraulic/shaft power.
📖 §6.3 Pump curves — pump operating point; §6.4 pump efficiency highest at one flow (BEP)

4. Analyse the following data collected for a water pump. If the operating head is 16m explain what will happen to other parameters. Design Parameters / Values: Flow (Q): 40 lps Head (H): 20 m Power (P): 15 kW Efficiency: 51%

Sep 2024 · 5 marks
Show worked solution
1. If the operating head is 16 m instead of 20 m, the operating flow will be higher than the rated flow. 2. Since the operating point has deviated from the BEP, the operating efficiency will be less than design efficiency. 3. Since the flow has increased and pump efficiency decreased than rated, the operating power demand will be more than the rated power.
On a pump H-Q curve, lower head pushes the operating point right (higher flow), away from BEP, lowering efficiency and raising power.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); 1 kg/cm² ≈ 10 m water; suction lift (level below centreline) adds to head

5. A pump is delivering 50 m3/hr of water with a discharge pressure of 3.5 kg/cm2. The water is drawn from a sump where water level is 5 meter below the pump centerline. The power drawn by the motor is 9.5 kW at 90% motor efficiency. Find out the pump efficiency.

Book EOC · 5 marks
Show worked solution
Total head = discharge head + suction lift = (3.5 kg/cm2 x 10 m/(kg/cm2)) + 5 m = 35 + 5 = 40 m. Flow = 50 m3/hr = 50/3600 = 0.01389 m3/s. Hydraulic power = (rho x g x Q x H)/1000 = (1000 x 9.81 x 0.01389 x 40)/1000 = 5.45 kW. Pump shaft power = motor input x motor efficiency = 9.5 x 0.90 = 8.55 kW. Pump efficiency = hydraulic power / shaft power = 5.45 / 8.55 = 0.637 = about 64%.
Total head 40 m, hydraulic power about 5.45 kW, shaft power 8.55 kW, so pump efficiency is roughly 64%.
📖 §6.4 Factors affecting pump performance — effect of oversizing & energy loss in throttling; §6.11 Energy conservation opportunities in pumping systems (over-designed pump: VSD, downsize/replace impeller, or correct-sized pump; optimise stages)

6. A cooling water pump connected to a pillar furnace has specifications Q = 12.5 lps, H = 60 m, P = 13.4 kW. The furnace manufacturer requires only 12.5 lps at 3.0 kg/cm2. What energy conservation measure can be proposed and estimate the reduction in power consumption.

Book EOC · 10 marks
Show worked solution
Required duty: 12.5 lps at 3.0 kg/cm² ≈ 30 m head, but the pump develops 60 m — it is oversized in head by 100% and the excess head is being throttled/wasted. Measure: reduce the developed head to ~30 m — install a variable speed drive, trim/downsize the impeller, reduce the number of stages if multistage, or replace with a correctly sized pump (book §6.11; note that trimming or speed reduction alone also reduces flow, so the pump must be re-matched to give 12.5 lps at 30 m). Estimate: hydraulic power at 60 m = 1000×9.81×0.0125×60/1000 = 7.36 kW, so present pump efficiency ≈ 7.36/13.4 = 0.55. Hydraulic power needed at 30 m = 1000×9.81×0.0125×30/1000 = 3.68 kW; at the same efficiency input ≈ 3.68/0.55 = 6.7 kW. Reduction in power ≈ 13.4 − 6.7 = 6.7 kW (about 50%), since power falls roughly in proportion to head at constant flow.
Required head is only 30 m versus 60 m supplied; trim the impeller or use a VFD; power drops roughly in proportion to head, giving about a 6.7 kW (about 50%) reduction.
📖 §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³); §6.1 Hydraulic power, pump shaft power & motor input power (p.172)

7. A centrifugal pump pumping water operates at 35 m3/hr and 1440 RPM. Pump operating efficiency is 68% and motor efficiency is 90%. The discharge pressure gauge shows 4.4 kg/cm2 and the suction is 2 m below the pump centerline. If the speed of the pump is reduced by 50%, estimate the new flow, head and power.

Book EOC · 10 marks
Show worked solution
Original total head = discharge + suction lift = (4.4 x 10) + 2 = 44 + 2 = 46 m. Original flow Q1 = 35 m3/hr at speed N1 = 1440 RPM. New speed N2 = 0.5 x 1440 = 720 RPM. Using the affinity laws: New flow Q2 = Q1 x (N2/N1) = 35 x 0.5 = 17.5 m3/hr. New head H2 = H1 x (N2/N1)^2 = 46 x 0.25 = 11.5 m. Original hydraulic power = rho x g x Q1 x H1 = 1000 x 9.81 x (35/3600) x 46 / 1000 = 4.39 kW. Original input (motor) power = 4.39 / (0.68 x 0.90) = 7.17 kW. New power varies as cube of speed: P2 = P1 x (N2/N1)^3 = 7.17 x 0.125 = 0.90 kW (input). So new flow about 17.5 m3/hr, new head about 11.5 m and new input power about 0.9 kW.
Affinity laws: flow halves to 17.5 m3/hr, head falls to one-quarter (11.5 m), power falls to one-eighth (about 0.9 kW input).
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); part (b) Book-3 Ch3 Compressed air — pressure optimisation

8. a) Pump suction head 3 m below centreline, discharge pressure 2.8 kg/cm², flow 120 m³/hr. Find pump efficiency if actual motor input is 15.0 kW at 0.90 motor efficiency. b) A V-belt reciprocating instrument air compressor maintains 7 kg/cm²g. 20% of air goes to boiler-house control valves needing 6.5 kg/cm²g; balance 80% needs 2 kg/cm²g. What do you advise?

15th Exam · 10 marks
Show worked solution
a) Discharge head = 2.8 kg/cm² = 28 m; suction head = -3 m; total head = 28 - (-3) = 31 m. Hydraulic power = (120/3600) x 1000 x 9.81 x 31 /1000 = 10.137 kW. Pump shaft power = 15 x 0.9 = 13.5 kW. Pump efficiency = 10.137/13.5 = 75%. b) Advise: 1) Provide a separate small compressor at 7 kg/cm²g near the control valves and reduce the main distribution pressure from 7 to 2 kg/cm²g for the bulk pneumatic instruments. 2) Reduced pressure lowers leakage loss; the compressor will begin to unload, so reduce the motor pulley size to match the lower demand.
Hydraulic power = Q x ρ x g x H; efficiency = hydraulic/shaft. Part b is a pressure-optimisation advisory.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); 1 kg/cm² ≈ 10 m; suction lift adds to head

9. Water level is 4 m below the pump centreline. Discharge pressure 2.60 kg/cm². Flow 1.5 m³/min. Find pump efficiency if motor draws 14 kW at 0.88 motor efficiency.

14th Exam · 5 marks
Show worked solution
Discharge head = 2.60 kg/cm² = 26 m; suction head = -4 m; total head = 26-(-4) = 30 m. 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%.
Hydraulic power = Q x ρ x g x H; efficiency = hydraulic/shaft power.
📖 Item 1: §6.5 Pump suction performance — cavitation & NPSH; item 6: §6.2 System characteristics — static & friction head (key treats 3 m as positive suction head: 21 − 3 = 18 m); item 9: §6.6 Flow control strategies — pumps in parallel switched to meet demand; others Ch4/5/2/7/8

10. Fill in the blanks: 1) Cavitation occurs when local static pressure falls below the ___ pressure of the liquid. 2) In an ammonia VAR system the absorbent is ___. 3) System resistance of a fan is proportional to the ___ of flow rate. 4) DBT=30°C and WBT=30°C → RH = ___. 5) Slip ring induction motors are ___ efficient than squirrel cage of same rating. 6) Suction static head 3 m, friction head 21 m → total head ___. 7) Lowest theoretical temperature water can be cooled to in a cooling tower is the ___ of atmospheric air. 8) Measure of illuminance in metric units is ___. 9) Pumps run in parallel if their ___ heads are similar. 10) When heat load, range and WBT held constant, cooling tower size is ___ proportional to approach.

14th Exam · 10 marks
Show worked solution
1) Vapour. 2) Water. 3) Square. 4) 100%. 5) Less. 6) 18 m (note: keyed answer 18 m). 7) Wet bulb temperature. 8) Lux. 9) Closed valve heads. 10) Inversely.
Standard fill-in answers across pumps, refrigeration, fans, motors, cooling towers and lighting.
📖 §6.11 Solved example — cooling water pump efficiency (p.194–195)

11. 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.

14th Exam · 5 marks
Show worked solution
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 Hydraulic power, pump shaft power & motor input power (p.172)

12. A pump delivers 64 m³/hr of water with a discharge head of 26 m. Water is drawn from a sump 3 m below the pump centreline. Motor draws 8.89 kW at 88% motor efficiency. Find the pump efficiency.

Set-A · 5 marks
Show worked solution
Q = 64/3600 m³/s; total head = 26-(-3) = 29 m. Hydraulic power = (64/3600) x 29 x 1000 x 9.81/1000 = 5.0576 kW. Pump shaft power = 8.89 x 0.88 = 7.8232 kW. Pump efficiency = 5.0576/7.8232 = 64.65%.
Hydraulic power = Q x ρ x g x H; efficiency = hydraulic/shaft power.
📖 §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³); §6.1 Hydraulic power, pump shaft power & motor input power (p.172)

13. A centrifugal pump runs at 60 m3/hr, 1470 RPM, pump efficiency 65%, motor efficiency 89%. Discharge gauge 3.4 kg/cm2, suction 3 m below pump centerline. Auditor recommends replacing the motor with a 4-pole 91% efficient motor at 1% slip. Determine new flow and motor power; throttle fully open and system head purely frictional. Comment.

17th Sep-2016 · 10 marks
Show worked solution
Existing: Head = 34 − (−3) = 37 m; pump power = (60/3600)×37×1000×9.81/(1000×0.65) = 9.3 kW. New motor speed = 1500 − 0.01×1500 = 1485 rpm. New flow = 60×(1485/1470) = 60.61 m3/hr. New pump power = 9.3×(1485/1470)³ = 9.59 kW. Existing motor input = 9.3/0.89 = 10.46 kW; new motor input = 9.59/0.91 = 10.54 kW. Comment: power consumption is slightly more, so not recommended (though flow is also marginally higher).
Affinity laws Q∝N, P∝N³; head from gauge + suction lift.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); §6.6 Flow control strategies — pumps in parallel switched to meet demand

14. Three identical parallel cooling-water pumps (two running, one standby) all develop 3.4 kg/cm2(a). Flow meter at common header reads: Pumps 1&2 = 545 m3/hr; 2&3 = 535; 3&1 = 550. Motor power: P1 33 kW, P2 31.5 kW, P3 32.5 kW. Motor efficiency 92% (P1,P2) and 91.5% (P3). Suction 3 m below pump centerline. Find i) individual pump efficiencies, ii) specific energy consumption (kWh/m3), iii) best operating combination.

16th Exam · 10 marks
Show worked solution
Solving X+Y=545, Y+Z=535, X+Z=550 gives X=280, Y=265, Z=270 m3/hr. Total head = 3.4 kg/cm2(a) → 2.4 kg/cm2(g) = 24 m discharge − (−3) suction = 27 m. Liquid kW = flow(m3/s)×27×1000×9.81/1000: P1 20.60, P2 20.22, P3 19.87. Pump input = motor kW × motor eff: P1 30.36, P2 28.98, P3 29.74. Pump eff = liquid/input: P1 67.9%, P2 69.8%, P3 66.8%. Specific energy (motor kW/flow): P1 0.118, P2 0.119, P3 0.120 kWh/m3. Best combination: Pumps 1 & 2.
Solve simultaneous flows; head from gauge+suction; pump eff=liquid kW/input kW.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)

15. A pump fills a rectangular overhead tank 5 m x 4 m, height 8 m. Inlet pipe is 20 m above ground; pump suction is 3 m below pump level; overflow line is 7.5 m from tank bottom; motor power 5.5 kW at 92% efficiency; fills to overflow in 180 minutes. Assess the pump efficiency.

18th Exam · 5 marks
Show worked solution
Volume = 5 x 4 x 7.5 = 150 m3. Flow = 150/3 h = 50 m3/hr. Total head = 20 - (-3) = 23 m. Hydraulic power = (50/3600) x 23 x 1000 x 9.81/1000 = 3.13 kW. Pump input = 5.5 x 0.92 = 5.06 kW. Pump efficiency = 3.13/5.06 = 61.9%.
Flow from tank volume/time; total head = lift + suction below pump; hydraulic kW = Q·H·ρ·g; efficiency = hydraulic power / shaft (pump input) power.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)

16. A centrifugal pump pumps 80 m3/hr of water into a container at 3 kg/cm2(g). Discharge head 5 kg/cm2(g); water level 5 m below pump centre line. Motor draws 22 kW, motor efficiency 90%, water density 1000 kg/m3. Find pump efficiency.

Sep 2019 · 5 marks
Show worked solution
Discharge head = 5 kg/cm2 = 50 m; suction head = -5 m. Power input to pump = 22 x 0.9 = 19.8 kW. Liquid kW = (80/3600) x (50-(-5)) x 9.81 = 11.98 kW. Pump efficiency = 11.98/19.8 = 60.56%.
Printed solution table: liquid power 11.98 kW, pump input 19.8 kW, pump efficiency 60.56%.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172); §6.6 Flow control strategies — pumps in parallel switched to meet demand

17. L-3: Three identical cooling water pumps in parallel (two running, one standby), all combinations give 3.4 kg/cm2(a) discharge. Flow: pumps 1&2 = 545, 2&3 = 535, 3&1 = 550 m3/hr. Motor power 33/31.5/32.5 kW; motor eff 92%/92%/91.5%; suction 3 m below pump centre line. Find (i) individual pump efficiencies, (ii) specific energy consumption kWh/m3, (iii) best operating combination.

16th Exam (alt set) · 10 marks
Show worked solution
Solving X+Y=545, Y+Z=535, X+Z=550 gives X=280, Y=265, Z=270 m3/hr. Discharge head = 3.4 kg/cm2(a) = 2.4 kg/cm2(g) = 24 m; suction head = -3 m; total head = 27 m. Liquid kW = flow(m3/s) x 27 x 1000 x 9.81/1000: Pump1 = 20.60, Pump2 = 20.22, Pump3 = 19.87 kW. Pump input power = motor power x motor eff: 30.36, 28.98, 29.74 kW. Pump efficiency = liquid/input: 67.9%, 69.8%, 66.8%. SEC = motor power/flow: 0.118, 0.119, 0.120 kWh/m3. Best operating combination = pumps 1 & 2.
Printed solution: pump effs 67.9/69.8/66.8%, SEC 0.118/0.119/0.120, best = pumps 1&2.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)

18. A pump is delivering 40 m3/hr of water with a discharge pressure of 29 metre. The water is drawn from a sump where water level is 6 metre below the pump centerline. The power drawn by the motor is 7.5 kW at 89% motor efficiency. Find out the pump efficiency.

9th Dec-2009 · 5 marks
Show worked solution
Hydraulic power Ph = (40/3600) × [29−(−6)] × 1000 × 9.81 / 1000 = 3.815 kW. Pump shaft power = 7.5 × 0.89 = 6.675 kW. Pump efficiency = 3.815 / 6.675 = 57.15%.
Total head = discharge head − suction head = 29 − (−6) = 35 m. Hydraulic power = Q×H×ρ×g/1000; pump shaft power = motor power × motor efficiency; pump efficiency = hydraulic/shaft power.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)

19. A centrifugal pump pumps 90 m3/hr of water with discharge pressure 3 kg/cm2(g) and a negative suction head of 3 m. Motor power drawn is 13 kW. Find pump efficiency. Motor efficiency 91%, water density 1000 kg/m3.

10th Jul-2010 · 5 marks
Show worked solution
Discharge head = 3 kg/cm²(g) ≈ 30 m; suction head = −3 m (negative/lift). Total head = 30 − (−3) = 33 m. Hydraulic power = Q(m³/s)×H×ρ×g/1000 = (90/3600)×33×1000×9.81/1000 = 8.09 kW. Pump shaft power = motor input × motor efficiency = 13×0.91 = 11.83 kW. Pump efficiency = 8.09/11.83 = 68.4%.
Total head = h_d − h_s with negative suction head added; hydraulic power = Q·H·ρ·g/1000; pump efficiency = hydraulic power / (motor power × motor efficiency).
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%); §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³)

20. A water pump is delivering 400 m3/hr. The impeller diameter is trimmed by 8% and its speed reduced by 10%. Find the water flow at the changed conditions.

10th Jul-2010 · 5 marks
Show worked solution
Flow varies directly with impeller diameter (Q∝D) and with speed (Q∝N), so the two effects multiply. 8% trim → D₂/D₁ = 0.92; 10% speed reduction → N₂/N₁ = 0.90. Q₂ = 400 × 0.92 × 0.90 = 331.2 ≈ 331 m³/hr. (Note: the printed key gives 288 m³/hr = 400 × 0.8 × 0.9, which corresponds to a 20% trim; with the question data as stated the affinity laws give 331 m³/hr — show the working and state the assumption.)
Apply Q∝D then Q∝N: 400×0.92×0.90 = 331 m³/hr. The official key's 288 m³/hr uses a 0.8 diameter factor (20% trim) — an inconsistency with the '8%' in the question.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)

21. S-5: The following data of a water pump of a process plant have been collected. Flow: 70 m3/hr, Total head: 24 meters, Power drawn by motor 7.2 kW, Motor efficiency 89%. Determine the pump efficiency.

11th Feb-2011 · 5 marks
Show worked solution
Hydraulic power = Q(m³/s)×H×ρ×g/1000 = (70/3600)×24×1000×9.81/1000 = 4.578 kW. Pump shaft (input) power = motor power × motor efficiency = 7.2×0.89 = 6.41 kW. Pump efficiency = 4.578/6.41 = 71.4%. (The printed key uses 7.2×0.90 = 6.48 kW, giving 70.65% — with the stated 89% motor efficiency the answer is ≈71.4%; either is accepted if the method is shown.)
Pump efficiency = hydraulic power / shaft power, where shaft power = motor input × motor efficiency. Note the official key applied 0.90 instead of the stated 0.89.
📖 §6.10 Agricultural pumping system — demonstrated ECMs

22. S-6: List any 5 energy conservation opportunities in agriculture pump sets.

11th Feb-2011 · 5 marks
Show worked solution
1. Installation of low friction foot valves; 2. Installation of low friction HDPE suction and delivery pipes; 3. Installation of long bends; 4. Installation of high efficiency pumps and motors; 5. Lower discharge head.
Standard demonstrated ECMs for agricultural pumping per the official key.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)

23. A process plant is situated 100 m above the ground level on the top of the hill. The plant requires 100 kL of water per hour. The management decides to install a pump at the ground level, with suction 3 meter below the ground level. The friction head is 12 meter. Evaluate the rating of the motor required considering 10% extra margin with respect to actual input pump power. The design pump efficiency is 65%. Also calculate the motor input power if the motor efficiency is 93%. (5 Marks)

Mar 2021 · 10 marks
Show worked solution
Q = 100 kL/hr = 100 m³/hr = 100/3600 = 0.0278 m³/s; ρ = 1000 kg/m³. Static head = delivery height + suction depth = 100 + 3 = 103 m; friction head = 12 m; Total head = 103 + 12 = 115 m. Hydraulic power = Q×ρ×g×H/1000 = 0.0278×1000×9.81×115/1000 = 31.3 kW. Pump shaft (input) power = 31.3/0.65 = 48.1 kW. Motor rating with 10% margin = 48.1×1.10 = 52.9 kW (select next standard size, e.g. 55 kW). Motor input power at the actual load = 48.1/0.93 = 51.7 kW.
Total head = static (lift + suction) + friction; hydraulic power = Q·ρ·g·H/1000; ÷ pump efficiency for shaft power; +10% for motor rating; ÷ motor efficiency for input power.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%); §6.1 Hydraulic power, pump shaft power & motor input power (p.172)

24. Rated capacity of a fresh water shut-discharge pump flow rate is 485 m3/hr and discharge pressure is 13.5 kg/cm2 at rated speed of 2950 rpm. It has been observed that 450 m3/hr water is sufficient to dispose the ash. The suction pressure of the pump is 0.5 kg/cm2. The management has decided to trim the impeller to satisfy the reduced flow requirement. Calculate: a) % reduction of impeller diameter (5 Marks) b) Annual energy savings after modification, if the pump is operating for 6 hours/day and 330 days in a year. (5 Marks)

Mar 2021 · 10 marks
Show worked solution
a) Q∝D, so D₂/D₁ = 450/485 = 0.928; % reduction in impeller diameter = (1 − 0.928)×100 = 7.2%. b) Rated differential head = (13.5 − 0.5) kg/cm² = 13 kg/cm² ≈ 130 m. Hydraulic power (rated) = (485/3600)×1000×130×9.81/1000 = 171.8 kW. After trimming, discharge head ∝ D²: H₂ = 0.928²×135 = 116.3 m, so new differential head = 116.3 − 5 = 111.3 m. Hydraulic power (new) = (450/3600)×1000×111.3×9.81/1000 = 136.4 kW. Power saving = 171.8 − 136.4 = 35.4 kW. Annual energy saving = 35.4×6×330 ≈ 70,000 kWh/year (official key prints 69,894 kWh using 35.3 kW). Alternative by P∝D³: P₂ = 0.928³×171.8 = 137.3 kW, saving 34.5 kW ≈ 68,300 kWh/year — both accepted.
Trim: Q∝D gives the diameter ratio; H∝D² gives new head; hydraulic power = Q·ρ·g·H/1000 before/after; annual saving = kW saved × 6 h × 330 days.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)

25. A process plant is situated 100 m up the side of the hill. The plant requires 100 kL of water per hour. The management decides to install a pump at the ground level, with suction 3 metre below the ground level. The friction head is 12 metre. Evaluate the rating of the motor required considering 10% extra margin with respect to actual input pump power. The design pump efficiency is 65% and motor efficiency 93%.

Mar 2021 (Set B) · 5 marks
Show worked solution
Q = 100/3600 x 1000 = 100/3.6 m3/s = 0.0277 m3/s. Density = 1000 kg/m3. Static head = 100 + 3 = 103 m; Friction head = 12 m; Total Head = 103 + 12 = 115 m. Hydraulic power = Q x density x g x H / 1000 = 0.0277 x 1000 x 115 x 9.81/1000 = 31.25 kW. Pump input (shaft) power = 31.25/0.65 = 48.07 kW. Motor rating (shaft power) = 48.07 kW. Motor rating with 10% margin = 48.07 x 1.10 = 52.87 kW. Motor input = design rated power/motor efficiency = 48.07/0.93 = 51.69 kW.
Total head = static (lift + suction) + friction; hydraulic power via rho*g*Q*H; divide by pump efficiency for shaft power; add 10% margin and divide by motor efficiency.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%); §6.1 Hydraulic power, pump shaft power & motor input power (p.172)

26. Rated capacity of bottom ash disposal pump flow rate is 485 m3/hr and discharge pressure is 13.5 kg/cm2 at rated speed of 2950 rpm. It has been observed that 450 m3/hr water is sufficient to dispose the ash. The suction pressure of the pump is 0.5 kg/cm2. The management has decided to trim the impeller to satisfy the reduced flow requirement. Calculate: a) % reduction of impeller diameter (5 marks). b) The annual energy savings after modification, if the pump is operating for 6 hours/day and 330 days in a year (5 marks).

Mar 2021 (Set B) · 10 marks
Show worked solution
a) Flow ∝ impeller diameter: D_new/D_old = 450/485 = 0.928; % impeller diameter reduction = (1 − 0.928)×100 = 7.2%. b) Rated hydraulic power = Q×ρ×(h_d − h_s)×g/1000 = (485/3600)×1000×(135 − 5)×9.81/1000 = 171.8 kW. New discharge head (H∝D²) = 0.928²×135 = 116.3 m; new differential head = 116.3 − 5 = 111.3 m. Hydraulic power after trimming = (450/3600)×1000×111.3×9.81/1000 = 136.4 kW. Power saving = 171.8 − 136.4 = 35.4 kW. Annual energy saving = 35.4×6×330 ≈ 70,000 kWh/year (key prints 69,884 kWh). Alternative P∝D³ method: P_new = 0.928³×171.8 = 137.3 kW, saving 34.5 kW → ≈ 68,300 kWh/year.
Affinity laws for trimming: Q∝D, H∝D², P∝D³; % diameter reduction from the flow ratio; annual saving = power saved × operating hours (6×330 = 1980 h).
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172) (ρ = slurry density); §6.7 Boiler feed water pumps (multistage pump = single-stage pumps in series) (series pumps add head)

27. L-5: A 30 MW coal fired thermal power plant uses conventional wet ash disposal system for ash evacuation. During an energy audit at ash slurry disposal pump house it was observed that only one ash slurry disposal series is continually operated out of installed three numbers of series. Data collected for one series: two pumps per series; rated parameters each pump flow 815 m³/hr, head 33 mWc; slurry flow rate measured at second series pump discharge 752 m³/hr; suction head to 1st pump +2.6 mtr; final slurry discharge pressure 5.2 kg/cm²g; differential pressure across 1st slurry pump 2.44 kg/cm²; differential pressure across 2nd slurry pump 2.5 kg/cm²; ash water ratio (by weight) of the slurry 1:15; ash slurry density 1032 kg/m³; electric power input to motor of 1st slurry pump 141 kW; electric power input to motor of 2nd slurry pump 139 kW; motor rating of each pump 200 kW; full load motor efficiency 93%; belt transmission efficiency 92%. Evaluate: (i) Individual pump efficiencies if the operating motor efficiency is 92% for all pumps; (ii) Specific energy consumption of each pump (kWh/m³); (iii) Specific energy consumption of the series (kWh/m³); (iv) As water conservation measure the energy auditor recommended to maintain ash water ratio at 1:7 and slurry density of 1067 kg/m³, calculate the incremental power consumption of each pump and series if all other parameters are unchanged.

Mar 2023 · 10 marks
Show worked solution
Common data: slurry flow Q = 752 m³/hr = 0.2089 m³/s; slurry density ρ = 1032 kg/m³; 1 kg/cm² ≈ 10 m. (i) Pump efficiencies. 1st pump: differential head = 2.44 kg/cm² ≈ 24.4 m; liquid (hydraulic) power = Q×H×ρ×g/1000 = 0.2089×24.4×1032×9.81/1000 = 51.6 kW; power at pump shaft = motor input × motor eff × belt eff = 141×0.92×0.92 = 119.3 kW; pump efficiency = 51.6/119.3 = 43.2%. 2nd pump: head = 2.5 kg/cm² ≈ 25 m; liquid power = 0.2089×25×1032×9.81/1000 = 52.9 kW; shaft power = 139×0.92×0.92 = 117.6 kW; pump efficiency = 52.9/117.6 = 45.0% (the printed key rounds both pumps to ≈43%). (ii) Specific energy consumption = motor input / flow: 1st = 141/752 = 0.188 kWh/m³; 2nd = 139/752 = 0.185 kWh/m³. (iii) Series SEC = (141+139)/752 = 0.372 kWh/m³. (iv) With ash:water 1:7 the slurry density rises to 1067 kg/m³; at unchanged flow, head and efficiencies the liquid power and hence input power scale with density (×1067/1032 = 1.034): 1st pump liquid power = 53.4 kW, input ≈ 141×1.034 = 145.8 kW (+4.8 kW); 2nd pump liquid power = 54.7 kW, input ≈ 143.7 kW (+4.7 kW). Incremental power for the series ≈ 9.5 kW (printed key: ≈10.4 kW with its rounding). Water conservation therefore slightly raises pumping power but saves far more water.
Liquid power = Q·H·ρ·g with slurry density; pump efficiency = liquid power / (motor input × motor η × belt η); SEC = kW / (m³/hr); higher slurry density raises liquid power and input power in proportion.

Ch7 · Range, approach, effectiveness & blowdown

24 past-paper questions · ≈185 marks
Range = hot − cold water. Approach = cold water − wet bulb. Effectiveness = Range/(Range+Approach)×100. Blowdown = evaporation/(COC−1).
📖 §7.2 Cooling Tower Performance (v)–(vii) Evaporation, COC & Blow down

1. a) In a large-scale steel manufacturing facility, a cooling tower is used to reject heat from continuous casting operations. The circulating water flow rate is 2000 m³/hr. The cooling tower is currently operating at a Cycles of Concentration (COC) of 3. The evaporation loss is estimated at 1.0% of the circulating flow, and the drift loss is 0.1% of the circulating flow. The facility is planning to improve the COC from 3 to 6 through advanced water treatment. Evaluate the following: i. Calculate the make-up water requirement at the current COC of 3. ii. Calculate the revised make-up water requirement if the COC is increased to 6. iii. Estimate the total water savings per day. iv. Discuss one limitation or risk associated with increasing the COC. b) True or False: i. Cooling towers primarily reject heat through evaporative cooling. ii. The approach temperature in a cooling tower is the difference between the hot water temperature and the ambient dry bulb temperature. iii. Blowdown in a cooling tower is required to prevent the build-up of dissolved solids. iv. Drift losses in a cooling tower refer to water carried away with the exhaust air. v. Cooling tower effectiveness improves with higher approach temperatures. vi. Cycles of concentration in a cooling tower relate to how many times the water is reused before discharge.

Sep 2025 · 10 marks
Show worked solution
a) Given: Circulating Water Flow (CWF) = 2000 m³/hr; Initial COC = 3; Final COC = 6; Evaporation Loss (E) = 1% of 2000 = 20 m³/hr; Drift Loss (D) = 0.1% of 2000 = 2 m³/hr i) Current Make-up at COC = 3: B = 20 / (3 − 1) = 10 m³/hr; Make-up = E + D + B = 20 + 2 + 10 = 32 m³/hr ii) Revised Make-up at COC = 6: B = 20 / (6 − 1) = 4 m³/hr; Make-up = 20 + 2 + 4 = 26 m³/hr iii) Hourly Savings = 32 − 26 = 6 m³/hr; Daily Savings = 6 × 24 = 144 m³/day iv) Increasing COC can lead to higher concentrations of dissolved solids in the water, which may cause scaling, corrosion, and microbiological fouling in the system. Effective water treatment and frequent monitoring are necessary to avoid operational issues. b) i) True; ii) False (It is the difference between the cold-water temperature and the wet bulb temperature.); iii) True; iv) True; v) False (Lower approach means better effectiveness.); vi) True
Blowdown B = E/(COC−1); make-up = E + D + B; higher COC reduces blowdown and make-up.
📖 §7.2 Cooling Tower Performance (v)–(vii) Evaporation, COC & Blow down; Heat Load – Steam Turbine Condenser

2. A steel industry has 100 MW of captive power plant with 2 nos. of identical extraction condensing steam turbine. Power demand is 80 MW. The turbine specific condensing load is 3.20 kg/kWh and heat rejection in condenser is 560 kCal/kg. Cold cooling water temperature: 32.0 °C Hot cooling water temperature: 39.2 °C Calculate the following: a) The Cooling Water circulation flow (m³/hr) through both condensers, if only one cooling tower supplies water to condenser of both the steam turbines. b) Make-up water flow rate (kg/hr) to basin, assuming blowdown loss is 1.0 % of circulation flow.

Sep 2024 · 10 marks
Show worked solution
Condenser heat load = (3.2 × 80000 × 560 / 1000000) = 143.36 MkCal/hr Circulated cooling water flow = (143.36 × 1000000 / 7.2 / 1000) = 19911 m³/hr Evaporation losses = 0.00085 × 1.8 × 19911 × 7.2 = 219.3 m³/hr Blowdown loss = 19911 × 0.01 = 199 m³/hr Water makeup to cooling tower = 219.3 + 199 = 418.4 m³/hr = 418400 kg/hr
Condenser heat = condensing load × power × heat rejection; circulation flow = heat / (ΔT range); make-up = evaporation (0.00085×1.8×flow×range) + blowdown.
📖 §7.2 Cooling Tower Performance (iv) Cooling capacity (Book EOC S-1)

3. Estimate the cooling tower capacity (TR) with: water flow rate 120 m3/h, sp. heat 1 kcal/kg C, inlet water temperature 37 C, outlet water temperature 32 C, ambient WBT 29 C.

Book EOC · 5 marks
Show worked solution
Heat rejected = mass flow × sp. heat × Range. Water flow = 120 m³/h = 120,000 kg/h. Range = 37 − 32 = 5°C. Heat load = 120,000 × 1 × 5 = 6,00,000 kcal/h. 1 TR = 3024 kcal/h, so capacity = 6,00,000 / 3024 = 198.4 TR (≈ 200 TR). (Approach = 32 − 29 = 3°C; the WBT is not needed for the TR figure.)
Heat load = 120000 kg/h x 1 x 5 C = 6.0 lakh kcal/h; dividing by 3024 kcal/h per TR gives about 198 TR.
📖 §7.2 Cooling Tower Performance (v)–(vii) Evaporation, COC & Blow down (Book EOC S-3)

4. In a cooling tower the cooling water circulation rate is 1250 m3/hr and the operating range is 7 C. If the blowdown rate is 1% of the circulation rate, calculate the evaporation loss and COC.

Book EOC · 5 marks
Show worked solution
Evaporation loss (m3/hr) = approx 0.00085 x 1.8 x circulation rate x range = 0.00085 x 1.8 x 1250 x 7 = 13.39 m3/hr (about 1.07% of circulation). Blowdown = 1% of 1250 = 12.5 m3/hr. COC = (Evaporation + Blowdown) / Blowdown = (13.39 + 12.5)/12.5 = 25.89/12.5 = 2.07. So evaporation loss is about 13.4 m3/hr and COC is about 2.0.
Evaporation = 0.00085 x 1.8 x 1250 x 7 = 13.4 m3/hr; COC = (Evap + Blowdown)/Blowdown = (13.4 + 12.5)/12.5 = approx 2.07.
📖 §7.3 Efficient System Operation – IV Performance Assessment of Cooling Towers (Book EOC L-1 – identical to book trial)

5. A cooling tower cools 1565 m3/hr of water from 44 C to 37.6 C at 29.3 C wet bulb temperature. The fan air flow rate is 989544 m3/hr (air density 1.08 kg/m3) and it operates at 2.7 cycles of concentration. Find: a) Range, b) Approach, c) % CT Effectiveness, d) L/G ratio in kg/kg, e) Cooling duty in TR, f) Evaporation losses in m3/hr, g) Blowdown in m3/hr, h) Make-up water in m3/hr.

Book EOC · 10 marks
Show worked solution
a) Range = 44 − 37.6 = 6.4°C. b) Approach = 37.6 − 29.3 = 8.3°C. c) Effectiveness = Range/(Range + Approach) × 100 = 6.4/(6.4 + 8.3) × 100 = 43.53%. d) Water mass L = 1565 × 1000 = 15,65,000 kg/hr; air mass G = 989544 × 1.08 = 10,68,708 kg/hr; L/G = 15,65,000/10,68,708 = 1.46 kg/kg. e) Cooling duty = 1565 × 1000 × 1 × 6.4 = 1,00,16,000 kcal/hr = 10016 × 10³ kcal/hr; in TR = 1,00,16,000/3024 ≈ 3312 TR. f) Evaporation loss = 0.00085 × 1.8 × 1565 × 6.4 = 15.32 m³/hr (≈ 0.97% of circulation). g) Blow down = Evaporation/(COC − 1) = 15.32/(2.7 − 1) = 9.01 m³/hr. h) Make-up water = Evaporation + Blow down = 15.32 + 9.01 = 24.33 m³/hr (book's printed '2433' is a typo for 24.33).
Book trial values (§7.3 IV): Range 6.4°C, Approach 8.3°C, Effectiveness 43.53%, L/G 1.46 kg/kg, Duty 10016 × 10³ kcal/hr ≈ 3312 TR, Evaporation 15.32 m³/hr, Blowdown 9.01 m³/hr, Make-up 24.33 m³/hr.
📖 §7.2 Cooling Tower Performance (i) Range; §7.3 Efficient System Operation (nozzle blockage, fans)

6. An induced draft cooling tower is designed for a range of 7°C. An energy manager finds the operating range as 4°C. What could be the reasons for this situation?

15th Exam · 5 marks
Show worked solution
1. Excess cooling water flow rate. 2. Reduced heat load from the process. 3. Some cooling tower cell fans switched off. 4. Poor approach due to high humid conditions. 5. Nozzles may be blocked.
Lower-than-design range implies either reduced heat load, excess water flow or degraded heat-transfer.
📖 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity

7. 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.

14th Exam · 5 marks
Show worked solution
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.
📖 §7.2 Cooling Tower Performance (v)–(vii) Evaporation, COC & Blow down (part b is transformer loss – not Ch-7)

8. a) Cooling tower circulation rate 1200 m³/hr, water enters at 38°C, ambient WBT 26°C, approach 4°C, blowdown 1% of circulation. Calculate evaporation loss and COC. b) An industry with 450 kVA load has two 500 kVA transformers (no-load loss 760 W, full-load copper loss 5400 W each). Should it run one transformer at full load or two sharing the load? Recommend with reasons.

Set-A · 10 marks
Show worked solution
a) Cold water out = 26+4 = 30°C; range = 38-30 = 8°C. Evaporation loss = 0.00085 x 1.8 x 1200 x 8 = 14.69 m³/hr. Blowdown = 1% x 1200 = 12 m³/hr. COC = Evap/Blowdown + 1: 12 = 14.69/(COC-1) → COC = 2.224. b) One 500 kVA at 450 load: loss = 760 + (450/500)² x 5400 = 760 + 4374 = 5134 W. Two 500 kVA at 50% (225 each): 2 x [760 + (225/500)² x 5400] = 2 x (760 + 1093.5) = 3707 W. Two transformers are better - losses are least (3707 W < 5134 W).
Evaporation loss formula; transformer loss = no-load + (load ratio)² x copper loss.
📖 §7.2 Cooling Tower Performance (v)–(vii) Evaporation, COC & Blow down

9. Power plant cooling tower audit: Generation 785 MW; Circulation 107000 m3/hr; Range 10.5°C; COC 3.8. Find a) total water consumption per hour, b) specific water consumption in m3/MW. If COC is increased to 7.0, c) potential water savings in m3/hr and m3/MW.

16th Exam · 10 marks
Show worked solution
Evaporation loss = 0.00085×107000×10.5×1.8 = 1719 m3/hr. Blowdown = 1719/(3.8−1) = 614 m3/hr. Total = 1719 + 614 = 2333 m3/hr. Specific = 2333/785 = 2.97 m3/MW. At COC 7.0: blowdown = 1719/6 = 286.5; total = 1719 + 286.5 = 2005.5 m3/hr; specific = 2.56 m3/MW. Water saving = 2333 − 2005.5 = 327.5 m3/hr and 327.5/785 = 0.417 m3/MW.
Evap=0.00085×circ×range×1.8; blowdown=evap/(COC−1).
📖 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity; (v) Evaporation loss (part b is fan efficiency – Fans chapter)

10. a) Two streams of cooling water 9000 m3/hr at 41°C and 6000 m3/hr at 52°C are mixed and fed to one cooling tower; measured heat rejection 45,000 TR at 31°C WBT. Calculate effectiveness and evaporation loss. b) In a 0.5 m × 0.5 m AC duct, average air velocity 28 m/s; suction static −20 mmWC, discharge 30 mmWC; 3-ph motor draws 10.8 A at 415 V, 0.9 PF; motor efficiency 88%. Find fan efficiency (neglect density correction).

16th Exam · 10 marks
Show worked solution
a) Mixed flow = 15000 m3/hr; mixed hot temp = (9000×41+6000×52)/15000 = 45.4°C. Range = 45000×3024/(15000×1000) = 9.072°C. Cold water temp = 45.4 − 9.072 = 36.33°C. Approach = 36.33 − 31 = 5.33°C. Effectiveness = Range/(Range+Approach) = 9.072/(9.072+5.33) = 63%. Evaporation loss = 0.00085×1.8×15000×9.072 = 208.2 m3/hr. b) Duct area = 0.25 m2; flow = 28×0.25 = 7 m3/s. Motor power = √3×415×10.8×0.9/1000 = 6.99 kW. Air power = 7×(30−(−20))/102 = 3.43 kW. Shaft power = 6.99×0.88 = 6.15 kW. Fan static efficiency = 3.43/6.15×100 = 55.76%.
Range from TR; effectiveness=Range/(Range+Approach); fan eff=air power/shaft power.
📖 §7.2 Factors Affecting Performance – Heat Load (heat-rejection table); Range = Heat load / Water circulation rate

11. An 18 MW cogeneration plant: max condenser load 7 MW; extraction steam 57 TPH for process and VAM. Condenser heat load 550 kcal/kg steam; steam rate 5 kg/kW for condenser power; VAM heat load 127 kcal/min/TR with VAM capacity 1100 TR. Estimate cooling tower heat load (kcal/hr). For 6°C range, calculate cooling water flow. Design approach 5°C.

16th Exam · 5 marks
Show worked solution
Steam for condenser = 7000×5 = 35,000 kg/hr. Condenser heat load = 35000×550 = 1,92,50,000 kcal/hr. VAM heat load = 1100×127×60 = 83,82,000 kcal/hr. Total CT heat load = 1,92,50,000 + 83,82,000 = 2,76,32,000 kcal/hr. For 6°C range: cooling water flow = 27632000/6 = 46,05,333 litres/hr ≈ 4605 m3/hr.
Heat load = condenser + VAM; water flow = heat load / range (Cp=1).
📖 §7.2 Cooling Tower Performance (v)–(vii) Evaporation, COC & Blow down; (iv) Cooling capacity

12. S-1: Operating data of an induced-draft cooling tower: range 8 °C, cooling water flow 12,500 m3/hr, drift loss 0.1% of circulation, WBT 27 °C, ambient DBT 35 °C, effectiveness 67%, COC 3. Estimate evaporation loss, make-up water requirement and TR load.

19th Exam · 5 marks
Show worked solution
Evaporation loss = 0.00085 x 1.8 x 12500 x 8 = 153 m3/hr. Blowdown = 153/(3-1) = 76.5 m3/hr. Drift = 12500 x 0.001 = 12.5 m3/hr. Make-up = evaporation + blowdown + drift = 153 + 76.5 + 12.5 = 242 m3/hr. TR load = (12500 x 1000 x 8)/3024 = 33,069 TR.
Use book evaporation formula (0.00085 x 1.8 x circulation x range), blowdown = evap/(COC-1), make-up = evap+blowdown+drift, heat-load TR = m·Cp·ΔT/3024.
📖 §7.6 Case Study – VFD for CT fan (cube law); §7.4 Flow Control Strategies (two-speed fans)

13. A 4-cell cooling tower has 45 kW CT fans per cell operating at 40 kW at 1450 rpm. Fans are replaced with two-speed motors at 1450 rpm and 740 rpm. High-speed mode 5300 hours, low-speed mode 1800 hours/year. Estimate annual energy savings vs continuous fixed-speed 1450 rpm operation.

Sep 2019 · 5 marks
Show worked solution
Present (all at 1450 rpm) = 4 x 40 x (5300+1800) = 11,36,000 kWh. After two-speed: high speed = 4 x 40 x 5300 = 8,48,000 kWh; low speed = (740/1450)^3 x 40 x 4 x 1800 = 38,281 kWh. Annual savings = 11,36,000 - (8,48,000 + 38,281) = 2,49,719 kWh.
Printed solution using fan affinity law (P proportional to N^3): savings 2,49,719 kWh/year.
📖 §7.2 Cooling Tower Performance (viii) L/G ratio; §7.2 Fill Media Effects (enthalpy values from Approach & WBT example)

14. L-6: (a) What is L/G ratio and how is it useful in cooling tower operation? (b) Functions of fill media. (c) Calculate L/G ratio: water flow 4540 m3/hr, approach 4.45 C, air entering enthalpy 24.17 kcal/kg at 26.67 C, air leaving enthalpy 39.67 kcal/kg at 37.8 C, hot water 47.77 C, cold water 31.11 C.

Sep 2019 · 10 marks
Show worked solution
(a) L/G ratio = ratio of water (liquid) to air (gas) mass flow rates. By energy balance L(T1-T2) = G(h2-h1), so L/G = (h2-h1)/(T1-T2). Against design, seasonal variation requires tuning of water/air flow (water box loading, blade angle) for best effectiveness. (b) Fill media increases the air-water contact surface and contact time, promoting heat and mass transfer (evaporative cooling). (c) L/G = (39.67-24.17)/(47.77-31.11) = 15.5/16.66 = 0.93.
Printed solution: L/G = 0.93; fill media function referenced to Page 209.
📖 §7.2 Cooling Tower Performance (v)–(vii) Evaporation, COC & Blow down

15. S-6: Power plant cooling tower audit: generation 785 MW, circulation rate 107000 m3/hr, range 10.5 C, design COC 3.8. Find (a) total water consumption/hr, (b) specific water consumption (m3/MW). If COC raised to 7.0, (c) potential water savings in m3/hr and m3/MW.

16th Exam (alt set) · 5 marks
Show worked solution
Evaporation loss = 0.00085 x 107000 x 10.5 x 1.8 = 1719 m3/hr. Blowdown = 1719/(3.8-1) = 614 m3/hr. (a) Total = 1719 + 614 = 2333 m3/hr. (b) Specific = 2333/785 = 2.97 m3/MW. At COC 7.0: blowdown = 1719/(7-1) = 286.5 m3/hr; total = 1719 + 286.5 = 2005.5 m3/hr; specific = 2005.5/785 = 2.56 m3/MW. (c) Water saving = 2333 - 2005.5 = 327.5 m3/hr; per MW = 327.5/785 = 0.417 m3/MW.
Printed solution table: total 2333 m3/hr, specific 2.97 m3/MW, savings 327.5 m3/hr and 0.417 m3/MW.
📖 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity; (v) Evaporation loss (part b is fan efficiency – Fans chapter)

16. L-4: (a) Cooling water 9000 m3/hr at 41 C and 6000 m3/hr at 52 C mixed and fed to one cooling tower. Measured heat rejection 45,000 TR; WBT 31 C. Calculate effectiveness and evaporation loss. (b) AC duct 0.5 x 0.5 m, air velocity 28 m/s, suction static -20 mmWC, discharge 30 mmWC, motor draws 10.8 A at 415 V, 0.9 PF; motor eff 88%. Find fan efficiency (neglect density correction).

16th Exam (alt set) · 10 marks
Show worked solution
(a) Mixed flow = 15000 m3/hr; mixed hot water temp = (9000x41 + 6000x52)/15000 = 45.4 C. Range = (45000 x 3024)/(15000 x 1000) = 9.072 C. Cold water temp = 45.4 - 9.072 = 36.328 C; approach = 36.328 - 31 = 5.328 C. Effectiveness = range/(range+approach) = 9.072/(9.072+5.328) = 63%. Evaporation loss = 0.00085 x 1.8 x 15000 x 9.072 = 208.2 m3/hr. (b) Area = 0.25 m2; airflow = 0.25 x 28 = 7 m3/s. Motor power = 1.732 x 415 x 10.8 x 0.9/1000 = 6.99 kW. Air power = 7 x (30-(-20))/102 = 3.43 kW. Shaft power = 6.99 x 0.88 = 6.15 kW. Fan static efficiency = 3.43/6.15 x 100 = 55.76%.
Printed solution: effectiveness 63%, evaporation 208.2 m3/hr; fan static efficiency 55.76%.
📖 §7.2 Factors Affecting Performance – Heat Load (heat-rejection table); Range = Heat load / Water circulation rate

17. L-5: An 18 MW cogeneration plant: max condenser load 7 MW, extraction steam 57 TPH for process and VAM. Condenser heat load 550 kcal/kg of steam, steam rate 5 kg/kW for condenser power. VAM heat load 127 kcal/min/TR, VAM capacity 1100 TR. Estimate cooling tower heat load (kcal/hr); for 6 C range and 5 C approach calculate the cooling water flow.

16th Exam (alt set) · 10 marks
Show worked solution
Steam for condenser power = 7000 x 5 = 35,000 kg/hr. Condenser heat load = 35000 x 550 = 1,92,50,000 kcal/hr. VAM heat load = 1100 x 127 x 60 = 83,82,000 kcal/hr. Total CT heat load = 1,92,50,000 + 83,82,000 = 2,76,32,000 kcal/hr. Cooling water flow = total heat load/range = 27632000/6 = 46,05,333 litres/hr = 4605 m3/hr.
Printed solution: total heat load 2,76,32,000 kcal/hr; water flow 4605 m3/hr.
📖 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity

18. Estimate the cooling tower capacity (TR) and approach with the following parameters: Water flow rate = 120 m3/hr, Specific heat of water = 1 kCal/kg°C, Inlet water temperature = 42°C, Outlet water temperature = 36°C, Ambient WBT = 32°C.

9th Dec-2009 · 5 marks
Show worked solution
Cooling tower capacity (TR) = (flow × density × sp.heat × temp diff)/3024 = 120 × 1000 × 1.0 × (42−36)/3024 = 238 TR. Approach = outlet temp − WBT = 36 − 32 = 4°C.
Capacity from heat removed divided by 3024 kCal/h per TR; approach is cold water outlet temperature minus ambient wet bulb temperature.
📖 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity; §7.3 Efficient System Operation

19. a) Define Range, approach and effectiveness in cooling tower operation. b) An induced draft cooling tower is designed for a range of 8°C. The energy auditor finds the operating range as 2°C. What could be the reasons for such a situation?

9th Dec-2009 · 10 marks
Show worked solution
a) Range = difference between cooling tower water inlet and outlet temperature. Approach = difference between cooling tower outlet cold water temperature and ambient wet bulb temperature (a better indicator of performance). Effectiveness (%) = ratio of range to the ideal range = Range/(Range + Approach). b) Possible reasons: excess cooling water flow rate; reduced heat load from the process; some cooling tower cell fans switched off; poor approach due to high humidity; nozzles blocked.
Standard cooling-tower definitions plus typical causes of a much lower-than-design operating range.
📖 §7.2 Cooling Tower Performance (v)–(vii) Evaporation, COC & Blow down

20. Find the blow down rate of a cooling tower: cooling water flow 600 m3/hr; operating range 8°C; TDS in circulating water 1500 ppm; TDS in make-up water 300 ppm.

10th Jul-2010 · 5 marks
Show worked solution
Evaporation loss = 0.00085 × 1.8 × circulation rate × range = 0.00085 × 1.8 × 600 × 8 = 7.344 m³/hr. COC = TDS in circulating water / TDS in make-up water = 1500/300 = 5. Blow down = Evaporation loss/(COC − 1) = 7.344/(5 − 1) = 1.836 m³/hr.
Evaporation loss = 0.00085×1.8×600×8 = 7.344 m3/hr. COC = 1500/300 = 5. Blowdown = Evap/(COC−1) = 7.344/(5−1) = 1.836 m3/hr. ⚠ answerText expanded from bare result to full book-method working.
📖 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity (part b is DG set – not Ch-7)

21. (a) Estimate the cooling tower approach and capacity (TR): water flow 150 m3/hr; sp.heat 1 kCal/kg°C; inlet water 42°C; outlet water 34°C; ambient WBT 30°C. (b) A 180 kVA, 0.80 PF rated DG set has a diesel engine rating of 240 BHP. What is the maximum power factor maintainable at full load on the alternator without overloading the diesel engine? (alternator losses + exciter power = 5.44 kW; no derating.)

10th Jul-2010 · 10 marks
Show worked solution
(a) Approach = outlet cold water − ambient WBT = 34 − 30 = 4°C. Range = 42 − 34 = 8°C. Capacity = flow × density × Cp × range / 3024 = 150 × 1000 × 1 × 8 / 3024 = 12,00,000/3024 = 396.8 TR. (b) Engine output = 240 BHP × 0.746 = 179.04 kW; power available to alternator output = 179.04 − 5.44 = 173.6 kW; maximum PF = kW/kVA = 173.6/180 = 0.964.
(a) Approach = outlet − WBT = 34−30 = 4°C; capacity = 150×1000×1×(42−34)/3024 = 396.8 TR. (b) Engine power = 240×0.746 = 179.04 kW; power available for alternator = 179.04−5.44 = 173.6 kW; max PF = 173.6/180 = 0.964. ⚠ answerText expanded from bare results to full working.
📖 §7.2 Cooling Tower Performance (i) Range; §7.3 Efficient System Operation

22. S-8: An induced draft-cooling tower is designed for a range of 8 C. The energy auditor finds the operating range as 2 C during the conduct of energy audit. In your opinion what could be the reasons for this situation?

11th Feb-2011 · 5 marks
Show worked solution
1. There may be excess cooling water flow rate; 2. There may be reduced heat load from the process; 3. Some of the cooling tower cells fan are switched off; 4. Approach may be poor because of high humid condition; 5. Cooling tower nozzles may be blocked.
Low range (cooling) caused by excess water flow, reduced heat load, or operational/maintenance issues.
📖 §7.3 Efficient System Operation – IV Performance Assessment of Cooling Towers (book trial method)

23. L-6: A cooling tower cools 1450 m3/hr of water from 43 C to 36.6 C at 30 C wet bulb temperature. The cooling tower fan air flow rate is 9,50,000 m3/hr (air density = 1.08 kg/m3) and operates at 2.7 cycles of concentration. Find (a) Range, (b) Approach, (c) % CT Effectiveness, (d) L/G Ratio in kg/kg, (e) Cooling Duty Handled in TR, (f) Evaporation Losses in m3/hr, (g) Blow down requirement in m3/hr, (h) Make up water requirement in m3/hr.

11th Feb-2011 · 10 marks
Show worked solution
CT water flow = 1450 m3/hr = 1450000 kg/hr; CT fan flow = 950000 m3/hr; fan flow mass @1.08 kg/m3 = 1026000 kg/hr. (d) L/G Ratio = 1450000/1026000 = 1.41325 kg/kg. (a) Range = 43 - 36.6 = 6.4 C. (b) Approach = 36.6 - 30 = 6.6 C. (c) % CT Effectiveness = 100 x Range/(Range + Approach) = 100 x 6.4/(6.4+6.6) = 49.23%. (e) Cooling duty = 1450 x 6.4 x 10^3 = 9280 x 10^3 kCal/hr; /3024 = 3068 TR. (f) Evaporation losses = 0.00085 x 1.8 x 1450 x 6.4 = 14.1984 m3/hr (% evaporation loss = 14.1984/1450 x 100 = 0.98%). (g) Blow down = Evaporation losses/(COC - 1) = 14.198/(2.7-1) = 8.352 m3/hr. (h) Make up water = Evaporation loss + Blow down loss = 14.198 + 8.352 = 22.55 m3/hr.
Standard CT formulae: range, approach, effectiveness, L/G, duty in TR, evaporation (0.00085 x 1.8 x flow x range), blowdown = evap/(COC-1), make-up = evap + blowdown.
📖 §7.2 Cooling Tower Performance (i)–(iv) Range, Approach, Effectiveness, Capacity; (v) Evaporation loss

24. In a steel industry, cooling water of 7500 m³/hr and 4200 m³/hr from two different sections with temperatures of 38 °C and 55 °C respectively, are fed to cooling tower after proper mixing. If the measured heat rejection by the cooling tower is 38,000 TR, calculate the effectiveness and evaporation loss of the cooling tower at 28 °C WBT.

Jul 2022 · 10 marks
Show worked solution
Mixed Hot Water Temp, °C = [(Flow1 × Temp1) + (Flow2 × Temp2)] / (Total Flow) = [(7500 × 38) + (4200 × 55)] / 11700 = 44.1 °C. Range of Cooling Tower, °C = Heat Rejection / (Flow × Density × Sp. Heat) = (38000 × 3024) / (11700 × 1000 × 1) ≈ 9.82 °C [book: 9.82]. Cold Water Temp, °C = Hot Water Temp - Range = 44.1 - 9.82 = 34.28 °C. Approach, °C = Cold Water Temp - WBT of Air = 34.28 - 28 = 6.28 °C. Effectiveness = Range / (Range + Approach) = 9.82 / (9.82 + 6.28) = 60.99% or 61%. Evaporation Loss (m³/hr) = 0.00085 × 1.8 × circulation rate (m³/hr) × Range = 0.00085 × 1.8 × 11700 × 9.82 = 175.8 m³/hr.
Mixed temp by flow-weighted average; range from heat rejection; effectiveness = range/(range+approach); evaporation = 0.00085×1.8×flow×range. ⚠ Question stem repaired: garbled 'heat rejection ... is at 1700 m³/hr' → '38,000 TR' (the value used in the printed solution: 38000 × 3024 / 11,700,000 = 9.82 °C range).