Energy Efficiency in Electrical Utilities Available here with full solutions — 37 questions recovered from the 2018 exam:
Objective (1 mark)
23 of 50
Short (5 marks)
8 of 8
Long (10 marks)
6 of 6
This is not the complete paper. The questions below are the ones we could recover and verify; the rest of that year’s paper is not reproduced here. Every answer shown is checked against the 2014 BEE guidebook and carries its book section reference and an explanation.
Full paper pattern: Section-I 50×1 = 50 marks · Section-II 8×5 = 40 · Section-III 6×10 = 60 · Total 150, pass mark 75, 3 hours. ▶ Practice these interactively
Other years
Objective questions (1 mark) — 23
📖 §8.3 Light source and lamp types (incandescent lamps)
1. Which of the following incandescent bulbs will have the least resistance ?
220 V, 60 W
220 V, 100 W
115 V, 60 W
115 V, 100 W
Answer: D) 115 V, 100 W
Resistance from the rating plate: R = V²/P. So 220 V/60 W = 807 Ω, 220 V/100 W = 484 Ω, 115 V/60 W = 220 Ω and 115 V/100 W = 132 Ω — lowest voltage with highest wattage gives the least resistance. Rule of thumb: R rises with the SQUARE of voltage and falls inversely with wattage, so scan for low V and high W before calculating.
📖 §1.4 Power factor improvement & benefits (kVA reduction)
2. In a rolling mill, the loading on the transformer was 1200 kVA with the power factor of 0.86. The plant improved the power factor to 0.98 by adding capacitors. What is the reduction in kVA ?
144
147
171
163.3
Answer: B) 147
kW is unchanged by PF correction: kW = 1200 x 0.86 = 1032 kW. New kVA = 1032/0.98 = 1053 kVA. Reduction = 1200 - 1053 = 147 kVA.
One-step shortcut: reduction = kVA_old x [1 - PF_old/PF_new] = 1200 x (1 - 0.86/0.98) = 147 kVA.
Do not multiply 1200 by 0.98, and do not quote the new kVA when the question asks for the reduction.
📖 §2.4 Motor efficiency (efficiency from nameplate data)
3. A 22 kW, 415 V, 45 A, 0.8 pf, 1475 rpm, 4 pole 3 phase induction motor operating at 420 V, 40 A and 0.8 pf. What will be the motor efficiency?
85.0 %
94.5 %
89.9 %
None of the above
Answer: A) 85.0 %
Efficiency = output/input, and the input must be computed from the NAMEPLATE ratings: sqrt(3) x 415 x 45 x 0.8 = 25.88 kW against 22 kW output = 85.0%.
The 420 V and 40 A are the present operating point, deliberately supplied to tempt you into 1.732 x 420 x 40 x 0.8 = 23.28 kW, which would give 94.5% - option (b), the planted wrong answer.
Sanity rule: a standard 22 kW motor sits around 88-91% efficiency, so any answer above 94% on nameplate data should make you re-read the question.
📖 §4.7 Performance assessment of refrigeration plants (the TR constant)
4. One ton of refrigeration is not equal to__________.
3024 kCal/hr
3.51 kW
12000 Btu/hr
860 kCal/hr
Answer: D) 860 kCal/hr
Memorise the identity: 1 TR = 3024 kcal/hr = 3.5 kW = 12,000 Btu/hr. It is the heat needed to freeze one short ton of water in 24 hours.
860 kcal/hr is a different constant altogether - it is the heat equivalent of 1 kW (1 kWh = 860 kcal), which is why it is planted here.
Cross-check the set: 3024 kcal/hr / 860 kcal/kWh = 3.517 kW, and 3024 x 3.968 = 12,000 Btu/hr. Everything hangs together except 860.
📖 §4.3 Types of refrigeration system (VCR components)
5. Which of the following is not a part of vapour compression refrigeration cycle ?
Compressor
Evaporator
Condenser
Generator
Answer: D) Generator
The vapour compression circuit has exactly four elements: compressor, condenser, expansion device and evaporator.
The GENERATOR (along with the absorber, solution pump and solution heat exchanger) belongs to the vapour absorption machine, where heat boils refrigerant out of the strong solution - it is the heat-driven half of the 'thermal compressor' that replaces the mechanical one.
Same principle as the 'absorber' version of this question: anything that needs heat rather than shaft work is an absorption component.
📖 §4.9 Performance assessment of window, split and package AC units (EER)
6. If the power consumed by an air conditioner compressor is 1.7 kW per ton of refrigeration, then its energy efficiency ratio (Watt/Watt) is _____
1.7
2.1
0.59
None of the above
Answer: B) 2.1
EER = refrigeration effect in watts / power input in watts (the BEE star-labelling definition), so it is the exact reciprocal of kW/TR once the TR is converted to kW.
Working: 1 TR = 3.517 kW of cooling delivered for 1.7 kW of input, so EER = 3.517/1.7 = 2.07 ~ 2.1 W/W.
Answering 0.59 means you inverted it (1.7/3.517 x ...), and answering 1.7 means you quoted the input rather than the ratio. Direction check: higher EER = better; higher kW/TR = worse.
7. The cooling tower size is _____________ to the entering Wet Bulb Temprature (WBT), when the heat load, range and approach are constant.
Directly proportional
Inversely proportional
Constant
None of above
Answer: B) Inversely proportional
Repeat of the Sep-2015 question. With heat load, range and approach fixed, tower size varies INVERSELY with entering wet bulb temperature — higher WBT air carries more enthalpy per kg, so less air and a smaller tower are needed. The full book set: directly with heat load, inversely with range, inversely with approach, inversely with entering WBT.
📖 §8.3 Light source and lamp types (fluorescent tube designations)
8. The T5, T8 and T12 fluorescent tube light are categorized based on
Diameter of the tube
Length of the tube
Both diameter and length of the tube
Power consumption
Answer: A) Diameter of the tube
Tube diameter in eighths of an inch: T5 = 5/8 in = 16 mm, T8 = 1 in = 25 mm, T12 = 1.5 in = 38 mm. The book also lists U-bent (T12, T8) and circular (T9 = 38 mm, T5 = 16 mm) versions using the same convention, so the number never refers to length or wattage.
9. If the wet bulb temperature of air is 38 °C, then it's relative humidity is __________%.
38 %
90 %
100 %
Insufficient data
Answer: D) Insufficient data
Two independent properties are needed to fix the state of moist air; wet-bulb temperature is only one of them. RH compares the actual moisture with the saturation moisture AT THE DRY-BULB TEMPERATURE, which has not been given.
Supply the DBT as well and the psychrometric chart delivers RH, humidity ratio, enthalpy, dew point and specific volume together.
The single exception: if DBT = WBT the air is saturated at 100% RH - so option (c) would be right only if the question had told you the dry bulb was also 38 degC.
📖 §6.6 Flow control strategies (pumps in parallel)
10. It is acceptable to run pumps in parallel provided their_________ are similar
Suction heads
Discharge heads
Closed valve heads
Total head at full flow
Answer: C) Closed valve heads
Closed-valve (shutoff) head is the head at zero flow — the top of the pump curve. Match those and both pumps keep contributing; mismatch them and the weaker pump is pushed back to zero flow, churning and overheating while the stronger one carries the load. Pumps of different sizes may be paralleled provided this one figure matches; equal duty-point heads are not enough.
11. L / G ratio in a cooling tower is the ratio of _________________.
Length and girth
Length and Temperature gradient
Water flow rate and air mass flow rate
Air mass flow rate and water flow rate
Answer: C) Water flow rate and air mass flow rate
L over G, in that order: L = water (liquid) mass flow, G = gas (air) mass flow. Option (d) reverses the ratio and is the trap — check which term is on top before ticking. Against design values, seasonal tuning of water box loading and fan blade angle is done to restore the design L/G and recover effectiveness.
📖 §3.2/§3.5 Multi-staging and inter-stage cooling (Table 3.7)
12. The inlet air temperature to a two stage reciprocating air compressor is 35 °C. At which of the following 2nd stage inlet temperature's the compressor will consume least power ?
75 °C
65 °C
60 °C
50 °C
Answer: D) 50 °C
The nearer the second-stage inlet temperature gets back to the first-stage inlet (35 degC here), the closer the machine works to isothermal compression and the lower the specific power - so the LOWEST offered second-stage inlet, 50 degC, means least power.
Book data point: Table 3.7 shows that a 5.5 degC increase in the second-stage inlet temperature raises the specific power consumption measurably - the effect is not negligible.
Practical meaning: a fouled or under-cooled intercooler shows up directly as extra kW, so intercooler efficacy and cooling-water flow are prime audit checks.
13. A fan is drawing 16 kW at 800 RPM. If the speed is reduced to 600 RPM then the power drawn by the fan would be
12 kW
9 kW
6.75 kW
None of the above
Answer: C) 6.75 kW
Power ∝ N³: 16 × (600/800)³ = 16 × 0.75³ = 16 × 0.4219 = 6.75 kW. The two wrong options are exactly the two classic errors — 12 kW is N¹ (treating power like flow) and 9 kW is N² (treating power like pressure). Chant it: 'flow one, pressure two, power three'.
14. In which of the following fans air enters and leaves the fan with no change in direction ?
Forward curved
Backward curved
Radial
Propeller
Answer: D) Propeller
Forward-curved, backward-curved and radial are all CENTRIFUGAL impellers — air enters at the eye axially and is discharged radially, a 90° change of direction. Only the propeller fan (an axial machine) passes air straight through parallel to the shaft. Same idea as the tube-axial/vane-axial question, asked the other way round.
📖 §6.5 Efficient pumping system operation (NPSH and cavitation)
15. The value, by which the pressure in the pump suction exceeds the liquid vapour pressure, is expressed as
Net positive suction head available
Static head
Dynamic head
Suction head
Answer: A) Net positive suction head available
Book wording: NPSHa is the value by which the liquid pressure at the pump suction/impeller eye exceeds the liquid's vapour pressure at the pumping temperature, expressed as a head in metres. Static head and dynamic head are system heads with no vapour-pressure reference at all. Keep NPSHa > NPSHr or the liquid flashes to vapour, the bubbles collapse in the impeller and you get the gravel-in-the-casing noise of cavitation.
📖 §7.1 Introduction (evaporative cooling) & §7.2 wet bulb temperature
16. Which of the following ambient conditions will evaporate minimum amount of water in a cooling tower ?
35 °C DBT and 30 °C WBT
38 °C DBT and 31 °C WBT
38 °C DBT and 37 °C WBT
35 °C DBT and 29 °C WBT
Answer: C) 38 °C DBT and 37 °C WBT
Evaporation is driven by the wet-bulb DEPRESSION (DBT − WBT), which measures how far the air is from saturation. The depressions are 5, 7, 1 and 6 °C, so 38 °C DBT with 37 °C WBT is nearly saturated air and can absorb almost no more moisture. Note this is about the RATE of evaporative cooling, not the empirical make-up formula 0.00085 × 1.8 × flow × range, which contains no ambient term at all.
17. A fan is operating at 970 RPM developing a flow of 3000 Nm3/hour at a static pressure of 650 mmWC. If the speed is reduced to 700 RPM, the static pressure (mmWC) developed will be
244.3
650
469
None of the above
Answer: D) None of the above
Pressure ∝ N²: SP2 = 650 × (700/970)² = 650 × 0.5207 = 338.5 mmWC, which is not offered, so 'none of the above' is correct. Check where the distractors come from: 469 mmWC is N¹ (650 × 700/970) and 244 mmWC is N³ — both are the wrong exponent. Always compute the number before assuming the nearest option is intended.
📖 §4.12 Ventilation systems (air changes per hour)
18. In an engine room 15 m long, 10 m wide and 4 m high, ventilation requirement in m3/hr for 20 air changes/hr is
6000
9000
12000
None of the above
Answer: C) 12000
Ventilation rate = room volume x ACH = (15 x 10 x 4) x 20 = 600 x 20 = 12,000 m3/hr. ACH is already per hour, so no extra time conversion.
Always multiply all THREE dimensions - leaving out the height (or using floor area) is what produces the 6000 and 9000 distractors.
Book reference values from Table 4.7: boiler room 15-30 ACH, compressor room 10-12 ACH - quote these if a question asks you to select the ACH yourself.
📖 §10.5 ECBC guidelines on building envelope (Solar Heat Gain Coefficient)
19. The Solar Heat Gain Coefficient (SHGC) of window of a building is 0.30. This means that
The window allows 70 % of the sun's heat to pass through into interior of the buildings
The window allows 30 % of the sun's heat to pass through into the building interior
70 % of the sun's heat is incident on the window
The window reflects back to exterior a minimum of 30 % of the sun's heat
Answer: B) The window allows 30 % of the sun's heat to pass through into the building interior
SHGC is the fraction of incident solar heat that gets through the fenestration (directly transmitted plus absorbed-and-re-radiated inward), so 0.30 means 30% enters and 70% is reflected or rejected. Lower SHGC = less cooling load; ECBC sets maximum SHGC values by climate zone alongside U-factor limits. Do not confuse it with VLT (visible light transmittance, the daylight fraction) — a good glazing wants low SHGC with high VLT, and Effective Aperture = WWR × VLT.
📖 §1.1 Introduction to electric power supply systems (heat rate and generation efficiency)
20. One of the thermal power plants operating with 2 nos. of 500 MW units has reported the operating heat rate of 11250 kJ/kW. The Plant Load Factor (PLF) of the power plant is 73 %. The operating efficiency of the power plant will be
38 %
35 %
30 %
32 %
Answer: D) 32 %
Definition to memorise: 1 kWh = 860 kcal = 3600 kJ of thermal energy. So generation efficiency = 3600 / heat rate (kJ/kWh) = 3600/11250 = 0.32 = 32%.
In kcal units the same relation is efficiency = 860 / heat rate (kcal/kWh). Heat rate is INVERSELY proportional to efficiency - the lower the heat rate, the better the plant.
The 500 MW units and the 73% PLF are deliberate distractors: PLF measures capacity utilisation, not thermal efficiency, and plays no part in this calculation.
21. The power measured in a boiler ID fan is 52 kW operating at 49 Hz. As an energy conservation measure the Variable Frequency Drive (VFD) was installed and the fan was operated at 34 Hz. The estimated power savings will be
36 kW
17.2 kW
34.7 kW
35 .7 kW
Answer: C) 34.7 kW
Frequency is a proxy for speed, so kW ∝ f³: power at 34 Hz = 52 × (34/49)³ = 52 × 0.334 = 17.3 kW, hence SAVING = 52 − 17.3 = 34.7 kW. Read the question wording carefully — 17.2 kW is the new power drawn, not the saving, and it is deliberately offered as option (b).
📖 §5.6 Fan performance assessment (pitot tube velocity)
22. A coal fired boiler primary air fan is maintaining a velocity pressure of 70 mmWC and the air temperature is 38 °C. The density of the air is 1.135 kg/m3 and the pitot tube constant is 0.85. The velocity of air in m/sec will be
25.6
29.56
28.67
None of the above
Answer: B) 29.56
v = Cp × √(2 × 9.81 × Δp/ρ) = 0.85 × √(2 × 9.81 × 70/1.135) = 0.85 × √1210 = 0.85 × 34.8 = 29.56 m/s. Here the density is already corrected for 38 °C, so do NOT apply 273/(273+t) a second time. Distractor 25.6 m/s is the standard book example (Δp = 47 mmWC, Cp = 0.9) — different data, so it cannot be the answer.
📖 §7.2 Cooling tower performance (cooling capacity in TR)
23. A two stage air compressor drawing 75 kW has heat rejection of 862 kCal/kWh. The required capacity of the cooling tower when the operating temperature difference of 5 °C will be ________TR.
21.55
107.5
22.93
57.4
Answer: A) 21.55
Heat rejected = 75 kW × 862 kcal/kWh = 64,650 kcal/hr, then TR = 64,650/3024 = 21.4 TR (the official key rounds via 3000 to 21.55, so option (a) stands either way). The 5 °C temperature difference is a distractor — it would only be needed to find the cooling WATER FLOW, not the TR. Multiplying by the 5 °C gives 107.5, which is exactly why option (b) is offered.
1. The operating data of an induced draft-cooling tower is as follows: Observed range: 8 °C; Cooling water flow rate: 12,500 m3/hr; Drift loss: 0.1 % of circulation rate; Wet Bulb Temperature: 27 °C; Ambient Dry Bulb Temperature: 35 °C; Effectiveness: 67 %; Cycle of Concentration: 3. Estimate the evaporation loss; make up water requirement and TR load of cooling tower.
Model answer: Evaporation loss = 0.00085 x 1.8 x 12500 x 8 = 153 m3/hr
Blow Down = 153 / (3 - 1) = 76.5 m3/hr
Make up = 153 + 76.5 + (12500 x 0.001) = 242 m3/hr
Heat load = 12500 x 1000 x 8 / 3024 = 33069 TR
Evaporation = 0.00085 × 1.8 × 12,500 × 8 = 153 m³/hr; Blowdown = 153/(3 − 1) = 76.5 m³/hr; Drift = 0.1% of 12,500 = 12.5 m³/hr; Make-up = E + B + D = 242 m³/hr. Heat load = 12,500 × 1000 × 8/3024 = 33,069 TR. The effectiveness of 67%, the WBT and the DBT are all distractors — none of them enters the water balance. Drift is quoted as a percentage of CIRCULATION rate, not of evaporation.
📖 §4.7 Performance assessment of refrigeration plants (TR via pump hydraulics)
2. A plant is operating a chilled water system always at full load. The chilled water inlet and outlet temperatures are 12 °C and 7 °C respectively. The chilled water pump discharge pressure is 3.6 kg/cm2g and the suction is 5 meters above the pump centerline. The power drawn by the chilled water pump's motor is 70 kW and an efficiency of 90 %. The chilled water pump efficiency at the operating point from pump characteristic curve is 60 %. Find out the operating refrigeration load in TR.
Model answer: Total head = 36 - 5 = 31 m
Pump shaft power = 70 x 0.9 = 63 kW
Flow rate = (63 x 1000) x 0.6 / (31 x 1000 x 9.81) = 0.124297 m3/s = 447.5 m3/hr
Refrigeration load = (447500 x 5) / 3024 = 740 TR
Route: pump power -> flow -> TR. Head = discharge head - static suction lift correction = 36 - 5 = 31 m (3.6 kg/cm2g ~ 36 m water column). Pump shaft power = motor input x motor efficiency = 70 x 0.9 = 63 kW.
Flow from hydraulic power: Q = (shaft kW x 1000 x pump efficiency)/(H x rho x g) = (63,000 x 0.6)/(31 x 1000 x 9.81) = 0.1243 m3/s = 447.5 m3/hr.
Then TR = 447,500 kg/hr x 1 kcal/kg degC x (12 - 7)/3024 = 740 TR. The two mark-losers are omitting the pump efficiency (it multiplies the useful hydraulic power) and forgetting that 1 kg/cm2 ~ 10 m of water head.
📖 §4.14 Humidification systems (air washer water requirement)
3. In an air washer of a textile humidification system with an airflow of 3000 m3/h at 25 °C and 10 % relative humidity is humidified to 60 % relative humidity by adding water through spray nozzles. The specific humidity of air at inlet and outlet are 0.002 kg/kg of dry air and 0.0062 kg/kg of dry air respectively. The density of air at 25 °C is 1.184 kg/m3. Calculate the amount of water required in kg/hr.
Model answer: The amount of water required:
mw = V x rho x (W_out - W_in)
= 3000 x 1.184 x (0.0062 - 0.002)
= 14.9 kg/h
Water added = volumetric air flow x air density x (W_out - W_in), where W is the specific humidity in kg per kg of dry air.
Working: 3000 x 1.184 x (0.0062 - 0.0020) = 3552 x 0.0042 = 14.9 kg/hr.
The relative humidity figures (10% and 60%) are context, not calculation inputs - the specific humidities have already been read off the chart for you. Convert to kg/kg before subtracting; using g/kg leaves the answer 1000 times too large.
4. In a Thermal Power Station, the steam input to a turbine operating on a fully condensing mode is 100 TPH. The heat rejection requirement of the steam turbine condenser is 555 kcal/kg of steam condensed. The temperature of cooling water at the inlet and outlet of the turbine condenser is 27 °C and 37 °C respectively. Find out the circulating cooling water flow.
Model answer: The quantum of heat rejected in the turbine condenser
= Quantum of steam condensed (kg) x heat rejection (kcal/kg)
= 100,000 x 555 = 55.5 Million kcal/h
Heat gained by circulating cooling water = Heat rejected in the condenser
Circulating cooling water flow = 100,000 x 555 / [(37 - 27) x specific heat (1)]
= 5550 m3/hr
Heat rejected by the condenser = heat gained by the cooling water. Heat = 100 TPH × 1000 kg/T × 555 kcal/kg = 55.5 million kcal/hr, so water flow = 55,500,000/((37 − 27) × 1 kcal/kg°C) = 5,550,000 kg/hr = 5550 m³/hr. The 10 °C rise here IS the cooling tower range. Forgetting to convert TPH to kg/hr (the ×1000) is the usual factor-of-1000 disaster.
📖 §1.4 Power factor improvement and benefits (Book-3 Ch.1, benefits list)
5. List any five benefits of power factor improvement in an industrial power distribution system
Model answer: (The paper prints only "Refer Guide Book No 3, Chapter 1, Page No 11". Model answer from the 2014 BEE Book-3, Chapter 1:)
1. Reduction in kVA demand for the same kW load, hence lower maximum demand charges from the utility.
2. Reduction in the current drawn, therefore reduced I2R losses in cables, transformers and distribution lines.
3. Improved voltage regulation / reduced voltage drop at the load end.
4. Released capacity of the existing transformers, switchgear and cables, so additional load can be connected without augmentation.
5. Avoidance of low power factor penalty and earning of power factor incentives / rebates in the electricity bill.
6. Better utilisation and improved efficiency of the electrical distribution system as a whole.
Structure the answer as a chain and the five points write themselves: less kVAr -> less kVA for the same kW -> less current -> less I^2R loss -> less voltage drop -> released transformer/cable capacity -> no utility penalty (and often an incentive).
Quantify one point if you can: kVA = kW/PF, so lifting PF from 0.85 to 0.95 cuts kVA demand by about 10.5% and I^2R loss by about 20% for the same kW.
Capacitors improve PF only UPSTREAM of where they are installed - motor-terminal capacitors do not improve the motor's own PF, only the cable and transformer feeding it.
📖 §9.4 Energy performance assessment of DG sets (dip-level fuel measurement)
6. During the performance evaluation of a DG set, the following parameters were noted: Capacity of DG set 1500 kVA; Test duration 36 minutes; Units generated 442 kWh; Average Power factor 0.92 pf; Length of diesel tank 90 cm; Width of diesel tank 90 cm; Height of the diesel tank 90 cm; Initial tank dip level (from top) 63 cm; Final tank dip level (from top) 79 cm. Calculate the following: 1. Diesel consumption (Litres) (1 Mark) 2. Average load (kW) (1 Mark) 3. Percentage Loading (%) (2 Marks) 4. Specific power generation (kWh/Litre) (1 Mark)
Model answer: 1. Diesel Consumption = 0.9 x 0.9 x 0.16 = 0.1296 m3 = 129.6 Litres
(level drop = 79 - 63 = 16 cm = 0.16 m)
2. Average load (kW) = (442 / 36) x 60 = 736.7 kW
3. Percentage Loading (%) = (736.7 / 0.92) / 1500 = 53 %
4. Specific power generation = 442 / 129.6 = 3.41 kWh/Litre
Fuel from the dip: the level DROP is 79 − 63 = 16 cm, so volume = 0.9 × 0.9 × 0.16 = 0.1296 m³ = 129.6 litres (using 63 or 79 cm as a depth is the classic error). Load = 442 kWh over 36 min = 442 × 60/36 = 736.7 kW. % loading compares like with like — convert kW back to kVA first: (736.7/0.92)/1500 = 53%. Specific generation = 442/129.6 = 3.41 kWh/litre. Dividing 736.7 by 1500 directly gives 49% and is marked wrong.
7. How does a motor lose its efficiency upon rewinding? (2.5 Marks) What two parameters will indicate the efficacy of the rewinding? (2.5 Marks)
Model answer: (The paper prints only "Refer Guide Book No 3, Chapter 2, Page No 61". Model answer from the 2014 BEE Book-3, Chapter 2:)
Loss of efficiency on rewinding: It is generally observed that rewound motors have a lower efficiency than the original, typically a drop of 1% to 5% (an average of about 1-2% for good practice, up to 5% for poor practice). The main causes are:
- Excessive heat applied while stripping the old winding (burn-out ovens / blow torch) damages the inter-laminar insulation of the stator core, increasing eddy current and hysteresis (iron) losses.
- Use of a smaller conductor cross-section / poorer winding-space utilisation than the original, which increases stator I2R (copper) loss.
- Change of winding configuration, number of turns, coil pitch or winding pattern from the original design.
- Mechanical damage to the stator slots and the air gap, and poor bearing/assembly practice, increasing friction and stray load losses.
Each subsequent rewind compounds the loss, so repeated rewinding of small motors is often uneconomical compared to replacement with an energy-efficient motor.
Two parameters that indicate the efficacy of the rewinding:
1. No-load current (and no-load loss / no-load input power) of the rewound motor, compared with the original value - a higher no-load current indicates increased core losses / air-gap damage.
2. Stator winding resistance per phase (and the resulting load or full-load current / I2R loss), compared with the original value - a higher resistance indicates a reduced conductor size and higher copper loss.
Three loss mechanisms explain the drop: burn-out heat damages the inter-laminar insulation of the stator core (higher eddy-current and hysteresis loss); a smaller conductor cross-section or poorer slot fill raises stator I^2R loss; and any change in turns, coil pitch or winding pattern alters the design flux and stray losses.
Book figure to quote: rewound motors typically lose 1-5% efficiency, and the loss compounds with each successive rewind - which is why replacing a small, repeatedly rewound motor with an energy-efficient motor usually wins.
The two efficacy indicators are the no-load current / no-load loss (flags core damage) and the stator resistance per phase (flags reduced conductor size) - each compared against the pre-rewind value, which is why keeping those records matters.
📖 §3.5 Efficient operation (load/unload operation, cost of leakage)
8. A medium sized engineering industry has installed two 480 CFM screw compressors, A & B. Compressor-A is operating at full load and Compressor-B is running in load-unload condition. The load power of both the compressor is 74 kW and the unload power of the Compressor-B is 26 kW. Both the compressors are operated during working day. The percentage loading of the Compressor-B during working day is 64 %. After arresting the leakage in the system the loading of the compressor was found to be 35 %. Estimate the energy savings per day.
Model answer: Existing Case:
Energy consumed per hour by Compressor-A = 74 kW
Energy consumed per hour by Compressor-B = 0.64 x 74 + 0.36 x 26 = 56.72 kW
Total energy consumed (Compressor A & B) = 74 + 56.72 = 130.72 kW/hr
Energy consumed per day = 130.72 x 24 hrs = 3137.3 kWh/day
Leakage Calculation:
Energy consumed per hour by Compressor-B (before) = 0.64 x 74 + 0.36 x 26 = 56.72 kW
Energy consumed per hour by Compressor-B (after) = 0.35 x 74 + 0.65 x 26 = 42.8 kW
Difference in power consumption = 56.72 - 42.8 = 13.92 kW/hr
Savings by arresting leakage per day = 13.92 x 24 = 334 kWh/day
Model a load/unload compressor as a weighted average: kW = (load fraction x load power) + (unload fraction x unload power). Compressor-B before: 0.64 x 74 + 0.36 x 26 = 56.72 kW; after leak arrest: 0.35 x 74 + 0.65 x 26 = 42.80 kW.
Saving = 56.72 - 42.80 = 13.92 kW, and over 24 hours = 334 kWh/day. Compressor-A is fully loaded throughout, so it cancels out and can be ignored for the SAVING (it matters only for the total consumption of 3137 kWh/day).
The mark-loser is forgetting the unload power altogether: an unloaded compressor still eats about 25-35% of its full-load power, which is exactly why cutting leaks pays.
📖 §1.4 PF improvement economics + §1.3 Maximum demand control
1. A food processing plant has a contract demand of 2500 kVA with the power supply company. The average maximum demand of the plant is 2000 kVA at a power factor of 0.95. The maximum demand is billed at the rate of Rs.300/kVA. The minimum billable maximum demand is 75 % 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 company is Rs.10 lakhs. The plant decides to improve the power factor to unity. Determine the power factor capacitor kVAr required, annual reduction in maximum demand charges and energy charge component. What will be the simple payback period if the cost of power factor capacitors is Rs.800/kVAr?
Model answer: kW drawn = 2000 x 0.95 = 1900 kW
kVAr required to improve power factor from 0.95 to 1
= kW (tan phi1 - tan phi2)
= 1900 [tan(cos^-1 0.95) - tan(cos^-1 1)]
= 1900 (0.329 - 0) = 625 kVAr
Cost of capacitors @ Rs.800/kVAr = Rs. 5,00,000
Maximum demand at unity power factor = 1900 / 1 = 1900 kVA
75 % of contract demand = 1875 kVA
Reduction in demand charges = 100 kVA x Rs.300 = Rs.30,000/month x 12 = Rs. 3,60,000/year
Percentage reduction in energy charge from 0.95 to 1 @ 0.5 % for every 0.01 increase = 2.5 %
Monthly energy cost component of the bill = Rs. 10,00,000
Reduction in energy cost component = 10,00,000 x (2.5/100) = Rs. 25,000/month
Annual reduction = Rs. 25,000 x 12 = Rs. 3,00,000
Savings in electricity bill = Rs. 6,60,000
Investment = Rs. 5,00,000
Payback period = 5,00,000 / 6,60,000 = 0.76 years or about 9 months
Work the three parts in sequence: kW = 2000 x 0.95 = 1900 kW; kVAr = 1900 x (tan(cos^-1 0.95) - 0) = 1900 x 0.329 = 625 kVAr; investment = 625 x 800 = Rs.5,00,000.
Then test the new demand against the minimum billable floor: 75% x 2500 = 1875 kVA, and the improved demand 1900/1.0 = 1900 kVA is ABOVE that floor, so the full drop counts: 2000 - 1900 = 100 kVA x Rs.300 x 12 = Rs.3,60,000/year. PF incentive = 5 steps x 0.5% = 2.5% of Rs.10 lakh/month = Rs.3,00,000/year.
Payback = 5,00,000 / 6,60,000 = 0.76 years (~9 months). Always test the improved demand against the minimum billable floor before claiming the saving.
📖 §3.4 Compressed air system components (dryers, receiver, dew point)
2. Write short notes on the following with respect to the compressed air system (each carries 2.5 Marks): a) Refrigeration drier b) Heat of compression drier c) Role of air receiver d) Dew point
Model answer: (The paper prints only Guide Book references: a) Book 3, Ch.3, Page 94; b) Book 3, Ch.3, Page 95; c) Book 3, Ch.3, Page 97; d) Book 3, Ch.3, Page 93. Model answer from the 2014 BEE Book-3, Chapter 3:)
a) Refrigeration drier: The compressed air is cooled in an air-to-air (and then air-to-refrigerant) heat exchanger down to about 2-3 °C, so that the moisture in the air condenses and is drained off through a moisture trap; the dried air is then reheated in the air-to-air exchanger by the incoming hot air. Refrigerant driers give a pressure dew point of about 2-3 °C at line pressure and are suitable for general plant air. They consume electrical power (roughly 3 % of the compressor power) and cause a pressure drop of about 0.2-0.5 bar.
b) Heat of compression drier: A regenerative desiccant drier that uses the heat available in the hot air leaving the compressor to regenerate the desiccant bed, instead of using separate external heaters or purge air. It is applicable only with oil-free (usually centrifugal or oil-free screw) compressors where the hot discharge air can be passed directly over the desiccant. Because no compressed purge air is lost and no external heating is used, it is the most energy-efficient type of desiccant drier and can achieve very low dew points (about -40 °C).
c) Role of air receiver: The air receiver (i) dampens pulsations from the compressor discharge, giving steady pressure in the distribution system; (ii) acts as a storage/reservoir to meet sudden or short-duration peak demands without starting an extra compressor; (iii) helps knock out and separate some of the oil and moisture carried over from the compressor, as the air cools in the receiver; (iv) reduces frequent loading/unloading (on-off) cycling of the compressor, which reduces unload power and improves compressor life; (v) allows the compressor to run at a lower pressure setting because the storage smooths demand fluctuations. Air receiver capacity is typically sized at about 1 minute (or about 6-10 times) of the free air delivery of the compressor.
d) Dew point: Dew point is the temperature at which the water vapour present in the compressed air starts to condense into liquid water, at a given pressure. It is a measure of the moisture content (dryness) of the compressed air - the lower the dew point, the drier the air. It must always be stated at a reference pressure: "pressure dew point" is measured at line pressure, whereas "atmospheric dew point" is at atmospheric pressure. Refrigerant driers typically give a pressure dew point of about 2-3 °C, while desiccant driers give about -20 °C to -40 °C. Specifying a dew point lower than the process actually needs wastes energy.
Refrigerant dryer: cools the air to about 2-3 degC so moisture condenses out, giving a pressure dew point of 2-3 degC; costs roughly 3% of the compressor power plus a 0.2-0.5 bar pressure drop. Heat-of-compression dryer: a regenerative desiccant dryer that reuses the hot discharge air to regenerate the bed, so no purge air and no external heater - the most energy-efficient type, but only usable with oil-free compressors.
Receiver: damps pulsations, stores air for peak demand, precipitates oil and moisture, and reduces load/unload cycling; size it at roughly one minute of FAD.
Dew point: the temperature at which water vapour in the air begins to condense AT A STATED PRESSURE - always distinguish pressure dew point (at line pressure) from atmospheric dew point. Specifying a dew point drier than the process needs is a direct waste of energy.
📖 §5.6 Fan performance assessment (gas density, fan shaft power)
3. In a boiler, the forced draught fan develops a total static pressure of 300 mmWC. Determine the shaft power (in kW) required to drive the fan if 10,000 kg of coal is burnt per hour with 13 kg of air per kg of coal burnt. The boiler house temperature is 20 °C and static efficiency of the fan is 80 %. The operating air density may be calculated from the following: R = 847.84 mmWC m3/kg mole K and Molecular weight of air, M = 28.92 kg/kg mole.
Model answer: Total pressure = 300 mm of WC
Mass of air handled, m = 10000 x 13 / 3600 = 36.11 kg/s
Atmospheric pressure, P = 1 kg/cm2 = 10 m of WC = 10,000 mm of WC
Temperature T = 20 + 273 = 293 K
Gas constant for air, R = 847.84 mmWC m3/kg mole K
Molecular weight of air, M = 28.92 kg/kg mole
Density = (P x M) / (R x T) = (10000 x 28.92) / (847.84 x 293) = 1.164 kg/m3
Volume = mass (kg/s) / density (kg/m3) = 36.11 / 1.164 = 31.02 m3/s
Power to fan shaft, kW = [Volume (m3/s) x Total pressure (mmWC)] / [102 x fan efficiency]
= (31.02 x 300) / (102 x 0.8)
= 114 kW
Three linked steps. Mass of air = 10,000 × 13/3600 = 36.11 kg/s. Density = (P × M)/(R × T) with P = atmospheric = 10,000 mmWC and T = 293 K: (10000 × 28.92)/(847.84 × 293) = 1.164 kg/m³. Volume = 36.11/1.164 = 31.02 m³/s, then shaft kW = Q × ΔP/(102 × η) = 31.02 × 300/(102 × 0.8) = 114 kW. The pressure inside the density formula is ATMOSPHERIC (10,000 mmWC), not the 300 mmWC fan pressure — mixing those two is the mark-losing mistake.
📖 §4.9 Performance assessment of package AC units + §4.7 kW/TR
4. A 7.5 TR package air conditioner is provided for a UPS room for removing the heat generated from the UPS of rated capacity 40 kVA. The following parameters were noticed while performing the assessment of the total system. UPS Parameters (40 kVA): On Load (16 hrs) - Input Power 11.94 kW, Output Power 8.61 kW; No Load (8 hrs) - Input Power 1.16 kW, Output Power 0.00 kW. Air conditioner parameters: Installed capacity of Air conditioner 7.5 TR; Outdoor unit (condenser) air velocity 6.1 m/s; Radius of the fan opening at the point of velocity measurement in outdoor unit 0.30 m; Air Density 1.174 kg/m3; Ambient temperature 305 K; Temperature of hot air (condenser outlet) 313.5 K; Specific heat of air 1.009 kJ/kg K; Power drawn by the compressor 5.40 kW; Efficiency of the compressor motor 90 %. Calculate a) Present delivery capacity of air conditioner (TR) (3 Marks) b) Power drawn per TR of refrigeration (3 Marks) c) Calculate the annual energy savings for 7200 hrs, if the UPS is relocated to a non-air-conditioned ventilated area. Assume energy cost Rs.8/kWh. (4 Marks)
Model answer: a) Present delivery capacity:
Capacity installed = 7.5 TR
Outdoor unit air velocity = 6.1 m/s; radius of the opening = 0.30 m
Area of cross section = 3.14 x 0.3^2 = 0.283 m2
Total air flow = 0.283 x 6.1 = 1.72 m3/s
Density of air = 1.174 kg/m3
Mass of air, m = 1.72 x 1.174 = 2.02 kg/s
T1 = 305 K; T2 = 313.5 K; dT = 8.5 K
Specific heat at constant pressure, cp = 1.009 kJ/kg K
Heat transfer = m x cp x dT = 17.32 kJ/s
Heat transfer per hour = 62352 kJ/hr = 14917 kcal/hr
Heat input from the compressor = 5.4 x 0.9 x 860 = 4180 kcal/hr
Evaporator heat load = 14917 - 4180 = 10737 kcal/hr
(the printed solution mis-types this line as "14949 - 4180", but the result 10737 corresponds to 14917 - 4180)
1 TR = 3024 kcal/hr
Effective TR = 3.55 TR
b) Power drawn by the compressor = 5.40 kW
Power taken per TR of refrigeration = 5.40 / 3.55 = 1.52 kW/TR
c) Heat load generated by UPS in the conditioned space:
On Load (16 hrs): input 11.94 kW, output 8.61 kW, heat load = 3.33 kW = 0.80 kcal/s = 2880 kcal/hr = 0.95 TR/hr = 15.2 TR/day
No Load (8 hrs): input 1.16 kW, output 0 kW, heat load = 1.16 kW = 0.28 kcal/s = 1008 kcal/hr = 0.33 TR/hr = 2.64 TR/day
Total = 17.84 TR/day
AC load generated by UPS per day = 17.84 TR
Power taken by AC to generate 17.84 TR at 1.52 kW/TR = 27.12 kW (kWh/day)
Annual energy savings at 300 days of operation = 8136 kWh
Cost of power = Rs. 8/kWh
Annual cost savings = Rs. 65,088/-
(Note: the question states 7200 hrs; the printed model answer works the annual saving on 300 days of operation, i.e. 27.12 kWh/day x 300 = 8136 kWh.)
Part (a) measures the capacity on the CONDENSER side and then subtracts the compressor heat: air flow = pi r^2 x v = 3.14 x 0.3^2 x 6.1 = 1.72 m3/s; mass = 1.72 x 1.174 = 2.02 kg/s; heat rejected = m x cp x dT = 2.02 x 1.009 x 8.5 = 17.32 kJ/s = 14,917 kcal/hr. The compressor adds 5.4 x 0.9 x 860 = 4180 kcal/hr of work into the refrigerant, so the true evaporator load = 14,917 - 4180 = 10,737 kcal/hr = 3.55 TR against 7.5 TR installed.
Part (b): kW/TR = 5.40/3.55 = 1.52 - poor, since a healthy package unit is nearer 1.0-1.2, confirming the unit is badly oversized and part-loaded.
Part (c): the UPS heat that must be removed is INPUT minus OUTPUT (11.94 - 8.61 = 3.33 kW on load, and the full 1.16 kW off load), never the input itself; convert with 860 kcal/kWh and 3024 kcal/TR, then multiply by the 1.52 kW/TR. Note the printed key annualises on 300 days, not the 7200 hrs stated in the question.
📖 §4.3 Types of refrigeration system (VAM vs electric chiller operating cost)
5. One of the textile processing plants has installed two numbers of 6 MW gas turbines and also Heat Recovery Steam Generator (HRSG) to generate steam from the hot gases. The steam generated from HRSG is utilized for process steam requirement and also for 500 TR Vapour Absorption Machine (VAM). The VAM consumes 4.4 kg steam per TR and is operated at full load. Due to increase in gas price the plant has stopped gas turbine operations and avails power supply from the grid. To meet the steam requirement the plant has installed two numbers of 10 TPH Agro Waste Boilers and steam is supplied to the process plant as well as to VAM machine. The average cost of steam is Rs.1200/- per ton from agro waste boiler. The plant operates for 7000 hours in a year. The management is planning to replace the VAM chillers by electrical centrifugal chiller which will operate at 0.7 kW/TR. Compare the annual operating costs of electrical chiller and VAM. The cost of grid power is Rs 6.12/kWh. Consider all the other auxiliary power remains same in both the cases. Do you agree with the management decision of operating VAM machine for chilling requirements?
Model answer: Capacity of VAM machine = 500 TR
Steam required per TR = 4.4 kg/TR
Total steam requirement = 500 x 4.4 = 2200 kg/hr = 2.2 TPH
Cost of steam from agro boiler = 2.2 x 1200 = Rs. 2640/hr
Power consumed by electric chiller = 0.7 x 500 = 350 kW
Cost of electricity = Rs. 6.12/kWh
Operating cost of electric chiller = 350 x 6.12 = Rs. 2142/hr
Savings by electric chiller = 2640 - 2142 = Rs. 498/hr
Annual operating savings = 7000 x 498 = Rs. 34,86,000/-
Conclusion: Disagree with the management decision of continuing to operate the VAM machine - the electrical centrifugal chiller is cheaper to operate by about Rs. 34.86 lakh per year.
Put both machines on a Rs./hour basis for the SAME 500 TR duty. VAM: 500 x 4.4 = 2200 kg/hr = 2.2 t/hr x Rs.1200/t = Rs.2640/hr. Electric centrifugal: 0.7 kW/TR x 500 = 350 kW x Rs.6.12 = Rs.2142/hr.
Difference = Rs.498/hr x 7000 hr = Rs.34.86 lakh a year in favour of the electric chiller, so the management's plan to replace the VAM is correct - i.e. DISAGREE with continuing to run the VAM.
The economics turned only because the free waste heat disappeared: a VAM wins when its heat is genuinely free (HRSG, waste heat), and loses the moment that steam has to be bought at Rs.1200/tonne. Auxiliary power is stated to be unchanged, so it cancels and can be left out.
📖 §1.8 AT&C losses in distribution (billing & collection efficiency)
6. A distribution company has taken initiatives to reduce Aggregate Technical & Commercial (AT&C) loss in their network. The energy supplied, received and revenue details are given below: Input energy = 60 MU; Metered Billed Energy = 43 MU; Average Billing = 3 MU; Amount Billed = Rs. 540 Million; Arrears collected = Rs. 80 Million; Amount received = Rs. 470 Million. a) Estimate the following (each carries 2.5 Marks): i) AT&C loss in % and revenue realized in Rs./kWh ii) Revenue loss per kWh and monthly loss, if the purchased energy cost is Rs. 8.10/kWh. b) List five measures to reduce commercial loss in the network (5 Marks)
Model answer: a)
Billing efficiency = (43 + 3) / 60 x 100 = 76.7 %
Collection efficiency = ((470 - 80) / 540) x 100 = 72.2 %
AT&C Loss = [1 - (Billing efficiency x Collection efficiency)] x 100
= [1 - (0.767 x 0.722)] x 100 = 44.62 %
Revenue realised per kWh = (470 - 80) / 60 = Rs. 6.5/kWh
Revenue loss per kWh = Rs. 8.10 - 6.5 = Rs. 1.6/kWh
Monthly revenue loss = 60 x 1.6 = Rs. 96 Million (Rs. 9,60,00,000/-)
b) (The paper prints only "Refer Guide Book No 3, Chapter 1, Page No 27". Measures from the 2014 BEE Book-3, Chapter 1:)
1. 100 % metering of all consumers and of all distribution transformers / feeders, with accurate, tamper-proof electronic meters.
2. Energy accounting and auditing at feeder and distribution-transformer level to locate high-loss pockets and pin-point theft.
3. Detection and prevention of theft / pilferage - removal of direct hooking on LT lines, use of aerial bunched conductors (ABC) and armoured service cables, regular raids and penalties.
4. Improvement of billing efficiency - elimination of unmetered supply and average / provisional billing, correction of faulty and stopped meters, spot billing and reduction of billing errors.
5. Improvement of collection efficiency - regular and timely bill distribution, computerised billing and collection, easy payment options, incentives for prompt payment, disconnection of chronic defaulters and recovery of arrears.
6. Consumer indexing and GIS mapping of consumers to feeders / distribution transformers so that energy input and energy billed can be reconciled.
Formula set to memorise: Billing efficiency = energy billed / energy input; Collection efficiency = amount collected (excluding arrears) / amount billed; AT&C loss % = [1 - (BE x CE)] x 100.
Working: BE = (43 + 3)/60 = 76.7%; CE = (470 - 80)/540 = 72.2%; AT&C = [1 - 0.767 x 0.722] x 100 = 44.6%. Revenue realised = (470 - 80)/60 MU = Rs.6.5/kWh, so the gap on a Rs.8.10/kWh purchase cost is Rs.1.6/kWh -> 60 MU x 1.6 = Rs.96 million.
The mark-loser is leaving the Rs.80 million of ARREARS in the collection figure - arrears belong to earlier billing periods and must be stripped out of both the numerator and the revenue-realised calculation.