Energy Efficiency in Electrical Utilities Available here with full solutions — 28 questions recovered from the 2017 exam:
Objective (1 mark)
16 of 50
Short (5 marks)
6 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) — 16
📖 §1.10 Harmonics (sources of harmonics)
1. Which of the following is not likely to create harmonics in an electrical system?
soft starters
variable frequency drives
uninterrupted power supply source (UPS)
induction motors
Answer: D) induction motors
Harmonics are generated by NON-LINEAR loads that chop or switch the current waveform: variable frequency drives, soft starters, UPS units, rectifiers, arc furnaces, electronic ballasts and SMPS.
A plain induction motor is a linear (inductive) load - it draws a lagging but essentially sinusoidal current, so it is a harmonics VICTIM (overheating, torque pulsation), not a harmonics source.
Memory hook: if it contains power electronics, it makes harmonics.
📖 §2.3 Motor characteristics (load torque characteristics) / §2.9 Speed control
2. Which of the following is an example of variable torque equipment ?
centrifugal pump
reciprocating compressor
screw compressor
roots blower
Answer: A) centrifugal pump
Variable-torque loads are the centrifugal machines - pumps, fans, blowers - where torque varies as speed^2 and power as speed^3. Constant-torque loads are the positive-displacement machines - reciprocating and screw compressors, roots blowers, conveyors - where torque is roughly independent of speed.
Only the centrifugal pump is centrifugal here, so it is the variable-torque equipment.
Why it matters: VFDs pay back fastest on variable-torque loads, because a 20% speed cut saves about 50% of the power (0.8^3 = 0.51).
📖 §1.4 Power factor improvement & benefits (I2R loss vs PF)
3. The percentage reduction in distribution losses when tail end power factor is raised from 0.8 to 0.95 is ________.
29%
15.8%
71%
none of the above
Answer: A) 29%
% reduction in distribution loss = [1 - (PF_old/PF_new)^2] x 100 = [1 - (0.8/0.95)^2] x 100 = [1 - 0.709] x 100 = 29.1% ~ 29%.
The physics: for a fixed kW the line current is inversely proportional to PF, and loss goes as I^2, so loss goes as 1/PF^2.
The 15.8% distractor is the un-squared version (1 - 0.8/0.95). Square the ratio, always.
4. The correction factor for actual free air discharge in a compressor capacity test will be --------- ---, when the compressed air discharge temperature is 15 °C higher than ambient air of 40 °C.
0.727
0.920
0.954
none of the above
Answer: C) 0.954
Correction factor = (273 + t_ambient)/(273 + t_discharge). Discharge = 40 + 15 = 55 degC, so factor = (273 + 40)/(273 + 55) = 313/328 = 0.954.
Working in degrees Celsius instead of Kelvin gives 40/55 = 0.727 - which is planted as option (a) precisely to catch that error.
Rule to memorise: any gas-law ratio in this paper - FAD correction, pipe sizing, Carnot COP - takes ABSOLUTE temperatures.
5. Which of the following happens to air when it is cooled through evaporation process in an air washer?
Humidity ratio of the air decreases.
Dry Bulb Temp of air decreases.
Dry Bulb Temp of air increases.
Enthalpy of outlet is air is less than enthalpy of inlet air.
Answer: B) Dry Bulb Temp of air decreases.
In an air washer the water evaporating into the air stream takes its latent heat from the air itself: dry bulb temperature falls, humidity ratio rises, enthalpy and wet-bulb temperature stay nearly constant.
So (a) is backwards (humidity rises), (c) is backwards (DBT falls) and (d) is wrong (the process is essentially adiabatic, so enthalpy is unchanged, not reduced).
The lowest temperature achievable is the air's wet-bulb temperature - the reason evaporative cooling works splendidly in dry climates and hardly at all in humid ones.
6. Which among the following is one of the parameters used to classify fans, blowers & compressors ?
air flow
speed RPM
specific ratio
none of the above
Answer: C) specific ratio
ASME classifies by SPECIFIC RATIO = discharge pressure ÷ suction pressure. Book figures to memorise: fan up to 1.11 (pressure rise up to 1136 mmWg), blower 1.11–1.20 (1136–2066 mmWg), compressor above 1.20. Flow and rpm say nothing about the class — a huge low-pressure fan and a small blower can move the same air.
7. The inner tube of a L-type Pitot tube facing the flow is measures ________ in the fan system
static pressure
velocity pressure
total pressure
all of the above
Answer: C) total pressure
An L-type (impact) pitot has two elements: the inner tube faces the flow and senses TOTAL pressure; the side/annular holes sense STATIC pressure. The manometer connected across the two reads the difference, which is velocity pressure. Remember Total = Static + Velocity, so the facing tube must be the total one.
📖 §3.4 Compressed air system components (air receiver)
8. Which of the following is false ?. Air receivers _____
reduce frequent on/off operation of compressors.
knock out some oil and moisture
increase compressor efficiency
act as reservoir to- take care of sudden demands
Answer: C) increase compressor efficiency
An air receiver does four things: damps discharge pulsations, stores air to meet sudden peak demands, precipitates some oil and moisture as the air cools, and cuts frequent load/unload cycling of the compressor.
What it does NOT do is change the compressor's own efficiency - the compressor's specific power (kW per m3/min) is set by its design and operating pressure, not by the storage downstream. That is why (c) is the false statement.
Sizing guide from the book: about one minute's free air delivery of the compressor.
9. Which among the following types of fans is predominantly used in cooling towers ?
centrifugal fan
axial fan
radial fan
all the above
Answer: B) axial fan
Cooling towers move enormous air volumes against very low pressure, which is exactly the axial (propeller) fan's duty — the book notes propeller fans are used in induced draft towers, with propeller or centrifugal in forced draft. Larger towers use variable-pitch propeller blades, and pitch adjustment is a recognised energy saving measure. Centrifugal fans suit high-pressure ducted systems, not open towers.
10. Which of the following type of lamps is most suitable for color critical applications ?
halogen lamps
LED lamps
CFLs
metal halide lamps
Answer: A) halogen lamps
Colour-critical work needs a high colour rendering index, and the book's Table 8.1 rates halogen CRI as 'Excellent (100)' — it is a continuous-spectrum incandescent source, so colours look as they do in daylight. Metal halide is good but not 100; CFLs and 2014-era LEDs are lower. The trade-off is efficacy: halogen manages only about 18–24 lm/W and lasts 2000–4000 hours, so it is used for display, flood and stadium lighting where colour beats efficiency.
📖 §9.3 Operational factors (waste heat recovery from flue gases)
11. Which of the following factors does not affect waste heat recovery in a DG Set ?
DG Set loading in kW
DG Set reactive power loading
operation period of DG Set
back pressure of flue gas path
Answer: B) DG Set reactive power loading
The book lists exactly three factors: DG set loading and exhaust gas temperature, hours of operation, and back pressure in the gas path. Reactive (kVAr) loading changes the alternator's excitation and losses but not the exhaust gas quantity or temperature, so it does not affect recovery. Practical rule from the same section: keep loading above about 60% for a steady, worthwhile flue gas profile.
📖 §7.2 Cooling tower performance (blowdown and COC)
12. The blow down requirement in m3/hr of a cooling tower for site Cycle of Concentration of 2.5 and approach of 4oC is:
10
0.63
1.6
Data not sufficient to calculate
Answer: D) Data not sufficient to calculate
Blowdown = Evaporation/(COC − 1), and evaporation itself needs 0.00085 × 1.8 × circulation × range. The question gives COC and APPROACH but neither circulation rate nor range, so the chain cannot be started — hence 'data not sufficient'. Approach never appears in any water-balance formula; it is offered here purely to tempt a calculation. Always list what the formula demands before deciding a question is answerable.
13. Which of the following is not true regarding system characteristic curve in a pumping system with large dynamic head ?
System curve represents a relationship between discharge and head loss in a system of pipes
System curve is dependent on the pump characteristic curve
The basic shape of system curve is parabolic
System curve will start at zero flow and zero head if there is no static lift
Answer: B) System curve is dependent on the pump characteristic curve
The system curve is a property of the PIPEWORK alone — H = static head + k·Q² — so it is parabolic, starts at the origin when there is no static lift, and would exist even with no pump connected. The pump curve is a property of the machine. Their INTERSECTION is the operating point; that is the only place the two meet, which is why 'system curve depends on the pump curve' is the false statement.
📖 §9.4 Energy performance assessment of DG sets (specific fuel consumption)
14. In a DG set, the generator is generating 1000 kVA, at 0.7 PF. If the specific fuel consumption of this DG set is 0.25 lts/ kWh at that load, then how much fuel is consumed while delivering generated power for one hour.
230 litre
250 litre
175 litre
none of the above
Answer: C) 175 litre
Convert kVA to kW with the power factor first: 1000 × 0.7 = 700 kW, then fuel = 700 × 0.25 = 175 litres in one hour. Option (b) 250 litres is what you get by multiplying the kVA directly — SFC is always quoted per kWh of REAL energy, so the PF step is compulsory. Remember the inverse form too: specific power generation in kWh/litre = 1/0.25 = 4 kWh/litre.
📖 §8.3 Light source and lamp types (fluorescent tube designations)
15. The T2,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
Diameter in eighths of an inch again: T12 = 38 mm, T8 = 25 mm, T5 = 16 mm, T2 = 6 mm. This exact question has now appeared in three papers (2013, 2015, 2017, 2018) — the free mark is worth locking in. Length and wattage vary separately, so 'both diameter and length' is always wrong.
📖 §1.4 Power factor basics (combining loads on the power triangle)
16. The combined power factor of a set of incandescent bulbs totaling 20 kW and two motors, each of 20 kW with power factor of 0.80 is
0.88
0.90
0.80
none of the above
Answer: A) 0.88
⚠ BEE's official key marks (a) 0.88, but the arithmetic gives 0.894 — which is nearer 0.90. Learn the METHOD, and if this exact question appears, answer 0.88 because that is what the examiner marks. Method — never average power factors, add kW and kVAr separately: bulbs 20 kW at unity PF contribute 0 kVAr; each motor 20 kW at 0.8 PF has tan(cos⁻¹0.8) = 0.75, so 20 × 0.75 = 15 kVAr each. Totals: 60 kW and 30 kVAr → kVA = √(60² + 30²) = 67.08 → PF = 60/67.08 = 0.894. The mark-losing trap is averaging (1.0 + 0.8 + 0.8)/3 = 0.87.
1. Determine the discharge pipe inner diameter size (in mm) for compressed air system, having following parameters. Compressed Air Flow at NTP (FAD) = 1000 Nm3/hr; Discharge Air Pressure = 7 bar(g); Discharge Air Temperature = 35 °C; Air Velocity = 6 m/s; Atmospheric Pressure = 1.013 bar
Model answer: Actual Condition vs NTP Condition:
P2 x V2 / T2 = P1 x V1 / T1
(1.013 + 7) x V2 / (273 + 35) = 1.013 x 1000 / 273
V2, actual flow rate = 142.6 m3/hr = 0.0396 m3/s (3 Marks)
Flow rate (m3/s) = Area (m2) x Velocity (m/s)
Area = Flow rate / Velocity = 0.0396 / 6 = 0.0066 m2
A = pi x (di^2 / 4) = 0.0066 m2
di = 0.092 m = 92 mm, say 100 mm (2 Marks)
Two steps. First convert the normal (NTP) flow to ACTUAL conditions with the gas law P1V1/T1 = P2V2/T2: V2 = 1.013 x 1000/273 x 308/(1.013 + 7) = 142.6 m3/hr = 0.0396 m3/s. Note the discharge pressure must be made ABSOLUTE (7 bar g + 1.013 = 8.013 bar a) and both temperatures Kelvin.
Then A = Q/v = 0.0396/6 = 0.0066 m2, and d = sqrt(4A/pi) = sqrt(4 x 0.0066/3.1416) = 0.092 m = 92 mm, rounded up to the next standard size, 100 mm.
The error that costs the marks is sizing the pipe on the 1000 Nm3/hr figure directly - compressed air occupies about 1/8 of that volume at 7 bar, so the pipe would be grossly oversized. Always round UP to a standard bore, never down.
📖 §4.2 Psychrometrics (humidity ratio and mixing of air streams)
2. A stream of moist air with a mass flow rate of 10.1 kg/s and with a specific humidity of 0.01 kg per kg dry air, mixes with a second stream of superheated water vapor, flowing at 0.1 kg/s. If we assume proper and uniform mixing without condensation, then what will be humidity ratio of the final stream, in kg per kg dry air?
Model answer: Humidity ratio of final stream:
H = (M1H1 + M2H2) / Dry air = [(0.01 x 10.1) + (0.1 x 1)] / [10.1 x (1 - 0.01)] = 0.02 kg per kg of dry air
Dry air can also be calculated as [10.1 kg/s - (moisture i.e. 10.1 x 0.01)] ......5 marks
OR
Mass of moist air = 10.1 kg/s; specific humidity = 0.01 kg/kg dry air.
Let X = amount of dry air; X + X x 0.01 = 10.1 kg/s, so X = 10 kg/s.
Moisture in moist air = 0.1 kg/s; superheated steam = 0.1 kg/s.
H = [(0.01 x 10) + (0.1 x 1)] / 10 = 0.02 kg per kg of dry air ......5 marks
Humidity ratio is per kg of DRY air, so first separate the dry air from the moisture: with W = 0.01, 10.1 kg/s of moist air = 10 kg/s dry air + 0.1 kg/s moisture (solve X + 0.01X = 10.1).
Mixing: W_final = (existing moisture + added moisture)/dry air = (0.1 + 0.1)/10 = 0.02 kg per kg dry air.
The mark-loser is dividing by the 10.1 kg/s of MOIST air instead of the 10 kg/s of dry air. Dry air is conserved through the mixing; moist air mass is not.
3. A pump is filling water in to a rectangular overhead tank of 5 m x 4 m with a height of 8 m. The inlet pipe to the tank is located at height of 20 m above ground. The following additional data is collected: Pump suction: 3 m below pump level; Overhead tank overflow line: 7.5 m from the bottom of the tank; Power drawn by motor: 5.5 kW; Motor efficiency: 92%; Time taken by the pump to fill the overhead tank upto overflow level: 180 minutes. Assess the pump efficiency.
Model answer: Volume of the tank = 5 x 4 x 7.5 = 150 m3
Flow = 150 / 3 = 50 m3/hr ......1.5 marks
Hydraulic power = Q (m3/s) x total head (m) x 1000 x 9.81 / 1000
= (50/3600) x (20 - (-3)) x 1000 x 9.81 / 1000
Hydraulic power = 3.13 kW ......2.5 marks
Power input to pump = 5.5 x 0.92 = 5.06 kW
Pump efficiency = 3.13 / 5.06 = 61.9% ......1 mark
Volume actually pumped is to the OVERFLOW line, not the full tank height: 5 × 4 × 7.5 = 150 m³ in 3 hours = 50 m³/hr (using 8 m gives 160 m³ and a wrong answer). Suction is 3 m below the pump so total head = 20 − (−3) = 23 m. Ph = (50/3600) × 23 × 9.81 = 3.13 kW; shaft = 5.5 × 0.92 = 5.06 kW; η = 61.9%. The 20 m is the inlet pipe height above ground — the tank's own height is not added.
📖 §5.5 Flow control strategies (damper vs VFD, cube law)
4. The operating boiler load and associated Induced-draft fan power consumption of a boiler is given below. Boiling loading 80% - Damper position #1 - Operating hours a day 4 - Fan motor power (with damper operation) 31 kW; Boiler loading 70% - Damper position #2 - Operating hours a day 12 - Fan motor power 29 kW; Boiler loading 60% - Damper position #3 - Operating hours a day 8 - Fan motor power 26 kW. The fan consumes 35 kW at 100% boiler loading with damper in full open condition. Estimate the daily energy savings that can be achieved if the damper is replaced by a VFD for induced draft fan to meet the desired requirements. Assume that the air requirement is proportional to boiler loading.
Model answer: Fan motor power with VFD, D = A^3 x 35 (A = fan flow, same as boiler loading, as a fraction).
Power savings E = C - D; Energy savings F = B x E.
A=80%, B=4 hrs, C=31 kW, D=17.9 kW, E=13.1 kW, F=52.32 kWh
A=70%, B=12 hrs, C=29 kW, D=12 kW, E=17 kW, F=203.94 kWh
A=60%, B=8 hrs, C=26 kW, D=7.6 kW, E=18.4 kW, F=147.52 kWh
Total Daily Savings = 403.78 kWh ......5 marks
With a VFD the fan power at part flow = (flow fraction)³ × full-load power, i.e. 35 kW × A³, and the saving is the measured damper power minus that. 80%: 35 × 0.512 = 17.9 kW → save 13.1 × 4 h = 52.3 kWh; 70%: 35 × 0.343 = 12.0 kW → save 17.0 × 12 h = 204 kWh; 60%: 35 × 0.216 = 7.6 kW → save 18.4 × 8 h = 147.5 kWh; total ≈ 404 kWh/day. Base the cube on the 35 kW full-open figure, never on the throttled reading at that load.
5. Fill in the blanks for the following: a) Voltage levels can be varied without isolating the connected load to the transformer using ______________ b) Use of ________ starter is appropriate in case of high number of motor starts and stops per hour. c) Operating a highly under loaded motor in star mode reduces voltage by a factor of ________. d) ____________ is the ratio of dissolved solids in circulating water to the dissolved solids in makeup water. e) In SI units ____________ is the measure of light output of a lamp.
Model answer: a) On load tap changer (OLTC)
b) Soft starter
c) sqrt(3) (i.e. square root of three)
d) Cycles of Concentration (COC)
e) Lumens
......5 marks (each one carries one mark)
Blank (a): an ON-LOAD TAP CHANGER (OLTC) shifts taps without de-energising - an off-load tap changer requires isolating the load, which is exactly the distinction being tested.
Blank (b): soft starters suit frequent starts/stops because they limit both inrush current and mechanical shock. Blank (c): re-connecting an under-loaded motor from delta to star drops the winding voltage by a factor of sqrt(3) (to 58%), cutting the iron loss - only safe below about 30-40% load.
Blanks (d) and (e): Cycles of Concentration (COC) = TDS in circulating water / TDS in make-up water; the SI measure of a lamp's light output is the lumen (luminous flux), while lux is illuminance on a surface - do not swap the two.
📖 §2.7 Factors affecting motor efficiency (voltage unbalance)
6. A 75 kW, 415 V, 140 Amp, 4 pole, 50 Hz, 3-phase squirrel cage induction motor has a full load efficiency of 87.6%. The measured operating motor terminal voltages in a 3-phase supply are 415 V, 418 V & 420 V. The current drawn in 3-phase supply are 137 Amp, 132 Amp & 137 Amp. Estimate the additional temperature rise of motor, due to unbalanced voltage supply.
Model answer: Additional temperature rise:
Phase R: V = 415, deviation from mean = -2.67
Phase Y: V = 418, deviation from mean = 0.33
Phase B: V = 420, deviation from mean = 2.33
Mean = 417.67 V
Voltage unbalance = Maximum deviation from mean / mean voltage
= 2.67 x 100 / 417.67 = 0.639% ......3 Marks
Additional temperature rise = 2 x (% voltage unbalance)^2
= 2 x (0.639)^2 = 0.8166% ......2 Marks
Two formulas, both examinable: % voltage unbalance = (maximum deviation from the mean voltage / mean voltage) x 100, and additional temperature rise (%) = 2 x (% voltage unbalance)^2.
Working: mean = (415 + 418 + 420)/3 = 417.67 V; deviations -2.67, +0.33, +2.33; take the LARGEST magnitude, 2.67; unbalance = 2.67/417.67 x 100 = 0.639%; extra temperature rise = 2 x 0.639^2 = 0.82%.
Mistakes that cost marks: averaging the deviations instead of taking the maximum, using the current readings (they are a distractor here), and forgetting to square the unbalance. Book limit: keep unbalance below 1% at the motor terminals, else derate.
📖 §1.5 Transformer losses & efficiency (loss evaluation over a load cycle)
1. It is required to choose a transformer to cater to a load which varies over a 24 hour period in the following manner: 500 kVA for 6 hours, 1000 kVA for 6 hours and 1500 kVA for 12 hours. Quotations have been received for two transformers, each rated at 1,500 kVA. Transformer-1 has an iron loss of 2.7 kW and a full load copper loss of 18.1 kW, while Transformer-2 has an iron loss of 3.2 kW and a full-load copper loss of 19.8 kW. (i) Calculate the annual cost of losses for each transformer at 365 days of operation if electrical energy cost is Rs. 6 per kWh. (ii) If the transformer-1 is to be purchased at an additional cost of Rs.25,000 over transformer-2, how would you justify it to the finance department?
Model answer: (i) Cost of Losses
Transformer 1:
Energy loss per day due to iron loss = 24 x 2.7 = 64.8 kWh
Energy loss per day due to copper loss = [(500/1500)^2 x 18.1 x 6] + [(1000/1500)^2 x 18.1 x 6] + [(1500/1500)^2 x 18.1 x 12]
= 12.1 + 48.3 + 217.2 = 277.6 kWh
Total energy loss per annum = (64.8 + 277.6) x 365 = 1,24,976 kWh
Annual cost of energy losses = Rs. 6 x 124976 = Rs. 7,49,856 ......(3 Marks)
Transformer 2:
Energy loss per day due to iron loss = 24 x 3.2 = 76.8 kWh
Energy loss per day due to copper loss = [(500/1500)^2 x 19.8 x 6] + [(1000/1500)^2 x 19.8 x 6] + [(1500/1500)^2 x 19.8 x 12]
= 13.2 + 52.3 + 237.6 = 303 kWh
Total energy loss per annum = (76.8 + 303) x 365 = 1,38,663 kWh
Annual cost of energy losses = Rs. 6 x 1,38,663 = Rs. 8,31,978 ......(3 Marks)
(ii) The capital cost of transformer-1 is Rs.25,000 more than that of transformer-2.
Annual saving in energy cost due to losses = (Rs. 8,31,978 - Rs. 7,49,856) = Rs. 82,122
Payback of additional investment = 25000 / 82,122 = around 4 months = 0.3 years ......(4 Marks)
Iron loss runs all 24 hours regardless of load; copper loss must be computed load-block by load-block with the SQUARE of each load fraction and then multiplied by that block's hours.
Transformer-1 per day: iron 24 x 2.7 = 64.8 kWh; copper [(1/3)^2 x 18.1 x 6] + [(2/3)^2 x 18.1 x 6] + [1^2 x 18.1 x 12] = 12.1 + 48.3 + 217.2 = 277.6 kWh; annual cost = (64.8 + 277.6) x 365 x 6 = Rs.7,49,856.
Transformer-2 works out to Rs.8,31,978, so the Rs.25,000 premium for Transformer-1 is repaid by Rs.82,122/year - about 4 months. Forgetting to square the 500/1500 and 1000/1500 fractions is the mark-killer.
📖 §4.7 Performance assessment (TR of an AHU) + §4.16 Energy saving in buildings
2. a) In an air-handling unit (AHU), the filter area is 1.5 m2 while air velocity is 2.2 m/s. The inlet air has an enthalpy of 67 kJ/kg. At the outlet of AHU, air has an enthalpy of 56 kJ/kg. The density of air of 1.3 kg/m3. Estimate the TR of the air-handling unit? b) List out any five energy conservation measures for energy use in buildings
Model answer: a) TR of AHU = (Enthalpy difference 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 ......2.5 marks
b) Any five (each 1.5 marks, maximum five points):
1. Weather-stripping of Windows and Doors: Minimize exfiltration of cool air and infiltration of warm air through leaky windows and doors by incorporating effective means of weather stripping. Self-closing doors should also be provided where heavy traffic of people is anticipated.
2. Temperature and Humidity Setting: Ensure human comfort by setting the temperature to between 23 °C and 25 °C and the relative humidity between 55% to 65%.
3. Chilled Water Leaving Temperature: Maintain the chilled water leaving temperature at or above 7 °C. As a rule of thumb, the efficiency of a centrifugal chiller increases by about 2.25% for every 1 °C rise in the chilled water leaving temperature.
4. Chilled Water Pipes and Air Ducts: Ensure that the insulation of the chilled water pipes and ducting system is maintained in good condition. This helps to prevent heat gain from the surroundings.
5. Chiller Condenser Tubes: Ensure that mechanical cleaning of the tubes is carried out at least once every six months. Fouling in the condenser tubes in the form of slime and scales reduces the heat transfer and thereby reduces the energy efficiency of the chiller.
6. Cooling Towers: Ensure that the cooling towers are clean to allow for maximum heat transfer so that the temperature of the water returning to the condenser is less than or equal to the ambient temperature.
7. Air Handling Unit Fan Speed: Install devices such as frequency converters to vary the fan speed. This will reduce the energy consumption of the fan motor by as much as 15%.
8. Air Filter Condition: Maintain the filter in a clean condition. This will improve the heat transfer between air and chilled water and correspondingly reduce the energy consumption.
Note: Any other relevant point may also be considered. ......7.5 marks
Chain the AHU formula in one line: TR = delta-h (kJ/kg) x rho x area x velocity x 3600 / (4.187 x 3024). The 4.187 converts kJ to kcal and the 3024 converts kcal/hr to TR - miss either and the answer is out by a factor of 4 or 3000.
Working: (67 - 56) x 1.3 x 1.5 x 2.2 x 3600/(4.187 x 3024) = 1,69,884/12,661 = 13.41 TR.
For the second half, quote measures with numbers attached where you can: chilled water leaving temperature at or above 7 degC (about 2.25% chiller efficiency gain per 1 degC rise), 23-25 degC and 55-65% RH space conditions, condenser tube cleaning every six months, clean filters, VFDs on AHU fans (up to ~15% fan energy), weather-stripping and insulation.
3. Fill in the blanks for the following: 1. The dry bulb temperature is 30 °C and the wet bulb temperature is 30 °C. The relative humidity is _________%. 2. Cavitations may occur in a pump when the local static pressure in a fluid reaches a level below the _________ pressure of the liquid at the actual temperature. 3. As the "Approach" decreases, the other parameters remaining constant, the effectiveness of cooling tower will __________. 4. The ratio of luminous flux emitted by a lamp to the power consumed by the lamp is called_________________. 5. A centrifugal pump raises water to a height of 12 meter. If the same pump handles brine with specific gravity of 1.2, the height to which the brine will be raised is __________ m. 6. Harmonics in electricity supply are multiples of the ____________ frequency. 7. A motor which can conveniently be operated at lagging as well as leading power factors is the __________ motor. 8. As per Energy Conservation Building Code, the Effective Aperture (EA) is ________, given that Window Wall Ratio (WWR) is 0.40 and Visible Light Transmittance (VLT) is 0.25. 9. In an amorphous core distribution transformer, ______ loss is less than a conventional transformer. 10. In case of centrifugal pumps, impeller diameter changes are generally limited to reducing the diameter to about _______% of maximum size.
Model answer: 1. RH = 100%
2. Vapour (vapour pressure)
3. Increases
4. Luminous efficacy
5. 12 meter, or the same
6. Fundamental, or 50 Hz
7. Synchronous
8. 0.10 (EA = WWR x VLT = 0.40 x 0.25)
9. No load (other correct answers could be: fixed, iron, total)
10. 75% (or 80%)
......10 marks (each question carries one mark)
Book numbers worth memorising from this set: impeller trims are limited to about 75% of maximum diameter; Effective Aperture EA = WWR × VLT = 0.40 × 0.25 = 0.10; efficacy is lumens per watt. Concept traps: a centrifugal pump raises brine (SG 1.2) to the SAME 12 m because head is independent of density — only the kW rises; and as approach DECREASES the tower is working closer to the wet bulb, so effectiveness = Range/(Range+Approach) increases. Harmonics are multiples of the fundamental (50 Hz), and the synchronous motor is the one that runs lagging or leading.
📖 §5.5 Flow control strategies (damper loss recovery, pulley change)
4. A belt-driven centrifugal fan supplies air to a series of process stations as shown in the figure below. While doing an air balance check on the system, the damper on the main duct and all system dampers had to be partially closed to reduce air flow to the design values. Energy auditor has recommended that fan power can be saved by fully opening the main damper and reducing the fan speed by changing the fan pulley diameter. The following initial conditions were measured on the main air supply system: Air Volume Flow Rate: 68,400 m3/hr; Fan Differential Static Pressure: 112 mmWC; Pressure differential across main damper: 17 mmWC. The following initial conditions were measured on the air supply fan and motor: Motor input power: 26.8 kW; Supply Fan Speed: 600 rpm; Motor Speed: 1,460 rpm; Fan pulley Diameter: 560 mm; Motor pulley Diameter: 230 mm. Calculate: (a) The annual energy savings considering 6000 hours of operation per year. (b) The new fan pulley diameter. [refers to a figure in the original paper]
Model answer: Fan flow = 68400 / 3600 = 19 m3/s
Input fan motor power in case-1 (W1) = 26.8 kW
Theoretical air power with damper in original partially-closed position:
WTh1 = (m3/s) x (mmWC) / 102 = (19 x 112) / 102 = 20.86 kW ......2 marks
Reduction in differential static pressure across the fan with the main damper fully open = 112 - 17 = 95 mmWC
Theoretical air power with damper fully open:
WTh2 = (19 x 95) / 102 = 17.7 kW ......2 marks
The input fan motor power in case-2 (W2) is estimated by proportionality using theoretical fan powers:
(W1 / W2) = (WTh1 / WTh2)
W2 = W1 x (WTh2 / WTh1) = 26.8 x (17.7 / 20.86) = 22.7 kW ......2 marks
(a) Annual energy saving = Power reduction x operating hours
= (26.8 - 22.7) x 6000 = 24600 kWh ......2 marks
(b) Fan pulley diameter change for reduced speed:
(N1 / N2) = (p1 / p2)^0.5, therefore N2 = N1 x (p2 / p1)^0.5 = 600 x (95/112)^0.5 = 553 RPM
Pulley diameter relation: N1 D1 = N2 D2 (N = speed in rpm, D = pulley diameter)
D2 = (N1 / N2) x D1 = (600 / 553) x 560 = 608 mm ......2 marks
The damper pressure drop is the recoverable loss: with it fully open the fan only has to develop 112 − 17 = 95 mmWC. Scale the measured motor kW by the ratio of theoretical air powers: W2 = 26.8 × (19×95)/(19×112) = 26.8 × 95/112 = 22.7 kW, so saving = 4.1 kW × 6000 h = 24,600 kWh/yr. New speed from SP ∝ N²: N2 = 600 × √(95/112) = 553 rpm, then N1 D1 = N2 D2 gives D2 = 560 × 600/553 = 608 mm. Use the SQUARE-ROOT law for the speed here (pressure is the known quantity), not the cube law.
📖 §1.4 Power factor improvement (capacitor sizing, current & kVA relief) + §2.4 motor losses
5. a) A 3-Phase, 50 kW rated Induction motor drawing 44 kW in a manufacturing industry has a power factor of 0.75 lagging. What size of capacitor in kVAr in each phase is required to improve the operating power factor to 0.96? What is the reduction in current and kVA due to capacitor installation at operating voltage of 415 V? b) List five energy losses in an induction motor
Model answer: a) Motor input P = 44 kW
Original PF = cos(phi1) = 0.75; Final PF = cos(phi2) = 0.96
phi1 = cos^-1(0.75) = 41.41 deg; tan(phi1) = 0.88
phi2 = cos^-1(0.96) = 16.26 deg; tan(phi2) = 0.29
Required capacitor kVAr = P (tan phi1 - tan phi2) = 44 (0.88 - 0.29) = 25.96 kVAr ......2.5 marks
Rating of capacitors connected in each phase = 25.96 / 3 = 8.65 kVAr
Current drawn at 0.75 PF = 44 / (sqrt(3) x 0.415 x 0.75) = 81.6 A
Current drawn at 0.96 PF = 44 / (sqrt(3) x 0.415 x 0.96) = 63.76 A
Reduction in current drawn = 81.6 - 63.76 = 17.84 A
Initial kVA at 0.75 PF = 44 / 0.75 = 58.67 kVA
kVA at 0.96 PF = 44 / 0.96 = 45.83 kVA
Reduction in kVA = 58.67 - 45.83 = 12.84 kVA ......2.5 marks
b) Five energy losses in an induction motor:
1. Iron (core) loss
2. Stator I2R (copper) loss
3. Rotor I2R (copper) loss
4. Friction and windage loss
5. Stray load loss ......5 marks
Compensate the OPERATING kW (44 kW), never the 50 kW nameplate rating: kVAr = 44 x (tan(cos^-1 0.75) - tan(cos^-1 0.96)) = 44 x (0.88 - 0.29) = 25.96 kVAr, i.e. 8.65 kVAr per phase.
Current relief: I = kW/(sqrt(3) x kV x PF), so 44/(1.732 x 0.415 x 0.75) = 81.6 A falls to 44/(1.732 x 0.415 x 0.96) = 63.76 A, a drop of 17.84 A; kVA falls 58.67 -> 45.83, i.e. 12.84 kVA released.
The five induction-motor losses: iron (core), stator I^2R, rotor I^2R, friction & windage, and stray load loss - the first and fourth are fixed, the rest vary with load.
6. Write short notes on: i) Ice Bank System in refrigeration ii) Vapour Absorption Refrigeration System iii) Harmonics in electrical system and its impacts
Model answer: (i) Ice Bank Systems (Book 3, page 136):
- Ice Bank System is a proven technology that has been utilized for decades. Thermal energy storage takes advantage of low cost, off-peak electricity, produced more efficiently throughout the night, to create and store cooling energy for use when electricity tariffs are higher, typically during the day.
- The essential element for either full- or partial-storage configurations are thermal energy storage tanks.
- How it works: during off-peak night time hours, the chiller charges the ICEBANK tanks for use during the next day's cooling. The lowest possible average load is obtained by extending the chiller hours of operation. ......3.33 marks
(ii) Vapour Absorption Refrigeration System (Book 3, page 30):
- The absorption chiller is a machine which produces chilled water by using heat such as steam, hot water, gas, oil etc.
- Chilled water is produced by the principle that liquid (refrigerant), which evaporates at low temperature, absorbs heat from surroundings when it evaporates.
- Pure water is used as refrigerant and lithium bromide solution is used as absorbent.
- Heat for the vapour absorption refrigeration system can be provided by waste heat extracted from process, diesel generator sets etc. Absorption systems require electricity to run pumps only.
- Depending on the temperature required and the power cost, it may even be economical to generate heat / steam to operate the absorption system.
Features: Li-Br-water absorption refrigeration systems have a COP in the range of 0.65 - 0.70 and can provide chilled water at 6.7 °C with a cooling water temperature of 30 °C. Systems capable of providing chilled water at 3 °C are also available. Ammonia based systems operate above atmospheric pressure and are capable of low temperature operation (below 0 °C). Absorption machines of capacities in the range of 10-1500 tons are available. Although the initial cost of absorption system is higher than compression system, operational cost is much lower if waste heat is used. ......3.33 marks
(iii) Harmonics in electrical system and its impacts (Book 3, page 114):
- Harmonics are multiples of the fundamental frequency of an electrical power system.
- If, for example, the fundamental frequency is 50 Hz, then the 5th harmonic is five times that frequency, or 250 Hz. Likewise, the 7th harmonic is seven times the fundamental or 350 Hz, and so on for higher order harmonics.
Some of the harmonic problems are:
1. Blinking of incandescent lights
2. Capacitor failure
3. Conductor failure
4. Flickering of fluorescent lights
5. Motor failures (overheating)
6. Transformer failures ......3.33 marks
Ice bank: thermal energy storage that makes ice at night on cheap off-peak power and melts it for daytime cooling - it shifts and levels load, lets the chiller run longer at a lower average load, and cuts maximum demand; the storage tank is the essential element of both full- and partial-storage schemes.
VAR: heat-driven chiller using LiBr-water (water is the refrigerant, LiBr the absorbent) with COP 0.65-0.70, chilled water at 6.7 degC with 30 degC cooling water, capacities 10-1500 TR, electricity needed only for the solution pump; ammonia-water versions go below 0 degC. Higher first cost, much lower running cost when waste heat is free.
Harmonics: integer multiples of the 50 Hz fundamental (5th = 250 Hz, 7th = 350 Hz), created by non-linear loads; effects to list are capacitor failure, transformer and motor overheating, conductor (especially neutral) overloading, flickering lamps and nuisance tripping. THD_i = sqrt(sum of harmonic currents squared)/I1 x 100.