Energy Efficiency in Electrical Utilities Available here with full solutions — 69 questions recovered from the 2017 exam:
Objective (1 mark)
50 of 50
Short (5 marks)
8 of 8
Long (10 marks)
11 of 6
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
1. A 500 cfm reciprocating compressor has a loading and unloading period of 5 seconds and 20 seconds respectively during a compressor leakage test. The air leakage in the compressor air system will be ________
125 cfm
100 cfm
200 cfm
none of the above
Answer: B) 100 cfm
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — % leakage = T/(T+t) x 100 = 5/(5+20) x 100 = 20%; leakage quantity = 0.20 x 500 = 100 cfm.
The tempting 125 cfm (a) comes from using the load:unload ratio 5/20 instead of load/(load+unload), i.e. dividing by the unload time alone rather than by the full cycle.
2. Increasing the suction pipe diameter in a pumping system will
Decrease NPSHA
Increase NPSHA
Decrease NPSHR
Increase NPSHR
Answer: B) Increase NPSHA
Confirmed vs Book-3 §6.5 — NPSHA = margin of eye pressure above vapour pressure; the book notes that as suction friction losses increase, NPSHA falls. A larger suction pipe lowers velocity and friction loss, so NPSHA increases. NPSHR is fixed by the pump design and does not change with pipe size. (Same as book end-of-chapter Q6.)
3. Which of the following is not a climate zone as per ECBC classification?
Hot - dry
Warm - humid
Cold
Cold humid
Answer: D) Cold humid
Confirmed vs Book-3 §10.3 — For ECBC, India is divided into five climatic zones: Composite, Hot-Dry, Warm-Humid, Moderate (Temperate) and Cold. 'Cold-humid' does not exist in the ECBC classification.
4. The inner tube of an L-type pitot tube is used to measure …… in the air duct
total pressure
static pressure
velocity pressure
dynamic pressure
Answer: A) total pressure
Confirmed vs Book-3 §5.6 — 'Total pressure is measured using the inner tube of pitot tube and static pressure is measured using the outer tube.' Connecting both to one manometer gives velocity pressure (TP − SP). Options (c)/(d) (velocity = dynamic pressure) are the manometer DIFFERENCE, not what the inner tube alone senses.
📖 §6.6 Flow control strategies — pumps in parallel switched to meet demand
5. In pumping systems where static head is a high proportion of the total, the appropriate solution is
install two or more pumps to operate in parallel
install two or more pumps to operate in series
install two or more pumps to operate independently
install variable frequency drive for the pump
Answer: B) install two or more pumps to operate in series
Confirmed vs Book-3 §6.6 — Book: 'Another energy efficient method of flow control, particularly for systems where static head is a high proportion of the total, is to install two or more pumps to operate in parallel', switching them on/off to meet demand. A VFD is risky here — slowing a pump against high static head can push it to shut-off. Series adds head, not flow.
📖 §1.8 Estimation of Technical Losses in Distribution System
6. What is the reduction in distribution loss if the current flowing through the distribution line is reduced by 10%?
10%
81%
19%
None of the above
Answer: C) 19%
Confirmed vs Book-3 §1.8 — line loss ∝ I², so at 0.9 I the loss becomes 0.81 of the original, a reduction of 19%.
Option (a) 10% assumes loss varies directly with current; option (b) 81% is the remaining loss, not the reduction.
📖 §2.4 Motor Efficiency — Field Tests for Determining Efficiency
7. In no load test of a poly-phase induction motor, the measured power by the wattmeter consists of:
core loss
copper loss
core loss, windage & friction loss
stator copper loss, iron loss, windage & friction loss
Answer: D) stator copper loss, iron loss, windage & friction loss
Confirmed vs Book-3 §2.4 Motor Efficiency — Field Tests for Determining Efficiency — The no-load wattmeter reading is the total no-load input, i.e. stator I²R at no-load current PLUS core (iron) loss PLUS friction & windage; the book subtracts (no-load current)² × stator resistance from it to leave core + F&W. (c) is the tempting answer because it names what is LEFT after that subtraction, not what the wattmeter actually reads.
📖 §6.1 Pump types — centrifugal pump construction & working (displacement vs dynamic pumps); §6.5 Effect of speed variation — Affinity laws (Q∝N, H∝N², P∝N³) (affinity laws apply to rotodynamic pumps only)
8. If the speed of a reciprocating pump is reduced by 50 %, the head
is reduced by 25%
is reduced by 50%
is reduced by 75%
remains same
Answer: D) remains same
Confirmed vs Book-3 §6.1/§6.5 — The affinity laws (H∝N²) are stated by the book for 'rotodynamic pump performance parameters'. A reciprocating pump is a positive-displacement pump: its head is set by the system pressure it discharges into, while speed reduction only cuts the volume delivered (flow ∝ speed). So the head remains the same.
📖 §1.3 Electrical Load Management and Maximum Demand Control
9. An Industrial Consumer has a load pattern of 2000 kW, 0.8 lag for 12 hrs and 1000 kW unity power factor for 12 hrs. The load factor is:
0.5
0.75
0.6
0.2
Answer: B) 0.75
Confirmed vs Book-3 §1.3 Load Curve — load factor = average load ÷ maximum demand = [(2000×12)+(1000×12)]/24 ÷ 2000 = 1500/2000 = 0.75.
Option (a) 0.5 is the ratio of the two load levels (1000/2000); the load factor must use the time-weighted AVERAGE load, and power factor plays no part in it.
10. Which of the following is not likely to create harmonics in an electrical system?
soft starters
variable frequency drives
uninterrupted power supply source (UPS)
electric heater
Answer: D) electric heater
Confirmed vs Book-3 §1.10 — the book classes heaters as LINEAR loads ('Incandescent lamps, heaters and, to a great extent, motors are linear systems') because their impedance is constant, so they draw a sinusoidal current and create no harmonics.
Options (a)–(c) are all power-electronic (non-linear) devices — soft starters, VFDs and UPS — which the book lists as harmonic sources.
11. A 10 HP/7.5 kW, 415 V, 14.5 A, 1460 RPM, 3 phase rated induction motor, after decoupling from the driven equipment, was found to be drawing 3 A at no load. The current drawn by the motor at no load is high because of
faulty ammeter reading
very high supply frequency
loose motor terminal connections
poor power factor as the load is almost reactive
Answer: D) poor power factor as the load is almost reactive
Confirmed vs Book-3 §2.3 Motor Characteristics — At no load the current is almost entirely magnetizing (reactive), so a 3 A no-load draw on a 14.5 A motor is normal and simply reflects a very poor no-load power factor. (b)/(c) a frequency error or loose connection would show up as speed or heating faults, not as a routine high no-load current — the book advises keeping a record of no-load input power and current precisely because it is expected to be high.
12. If we increase the temperature of air without changing specific humidity, dew point temperature of air will
increase
decrease
remain constant
can't say
Answer: C) remain constant
Confirmed vs Book-3 §4.2 - Dew point depends only on the moisture content (specific humidity / partial pressure of water vapour). If specific humidity is unchanged, dew point stays constant even though DBT rises (RH falls).
13. In a vapor compression refrigeration system, the component where the refrigerant changes its phase from vapor to liquid is
compressor
condenser
expansion valve
evaporator
Answer: B) condenser
Confirmed vs Book-3 §4.3 - In the condenser the high-pressure refrigerant vapour rejects heat and condenses to liquid (vapour → liquid phase change). The compressor (a) only superheats the vapour, the expansion valve (c) throttles liquid, and the evaporator (d) does the opposite phase change (liquid to vapour); condensation happens in stage 3-3b of Figure 4.4.
14. In a vapor compression refrigeration system, the component across which the enthalpy remains constant
compressor
condenser
expansion valve
evaporator
Answer: C) expansion valve
Confirmed vs Book-3 §4.3 - Expansion (throttling) is an isenthalpic process; enthalpy remains constant across the expansion valve. Enthalpy rises in the evaporator (d) and compressor (a) and falls in the condenser (b); §4.3 states there is no heat loss or gain through the expansion device, so throttling is isenthalpic.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
17. A pump discharge has to be reduced from 120 m3/hr to 100 m3/hr by trimming the impeller. What should be the percentage reduction in impeller size?
83.3%
16.7%
50.0%
33.3%
Answer: B) 16.7%
Confirmed vs Book-3 §6.5 — Q∝D, so D₂/D₁ = Q₂/Q₁ = 100/120 = 0.833; percentage reduction = (1 − 0.833)×100 = 16.7%. 83.3% is the remaining diameter ratio, not the reduction; 33.3% would be the head reduction (1 − 0.833²).
18. If temperature of air increases, the amount of water vapor required for complete saturation will
Increase
Decrease
not change
Can't say
Answer: A) Increase
Confirmed vs Book-3 §4.2 - Warmer air can hold more moisture — the saturation vapour pressure (and saturation humidity ratio) rises with temperature, so more water vapour is needed to fully saturate it.
📖 §7.2 Factors Affecting Performance – Wet Bulb Temperature
19. Which among the following inlet air conditions would result in the best cooling tower performance?
air with lowest wet bulb temperature and high relative humidity
air with lowest wet bulb temperature and low relative humidity
air with same dry bulb and wet bulb temperature
air with high dry bulb temperature and high moisture.
Answer: B) air with lowest wet bulb temperature and low relative humidity
Confirmed vs Book-3 §7.2 Wet Bulb Temperature — Book: WBT of entering air 'is a controlling factor from the aspect of minimum cold water temperature'; the lower the WBT (and the drier the air), the greater the evaporative driving force and the colder the water → (b). Air with DBT = WBT (c) is saturated and can evaporate almost nothing.
20. As the 'approach' increases while other parameters remain constant, the effectiveness of a cooling tower:
increases
remains unchanged
decreases
none of the above
Answer: C) decreases
Confirmed vs Book-3 §7.2 (iii) — Effectiveness = Range/(Range + Approach). With range constant, a larger approach increases the denominator, so effectiveness falls → (c). A low approach is what makes a tower 'effective'.
22. COP of a single effect absorption refrigeration system is likely to be in the range of
0.6 to 0.7
1 to 1.2
1.5 to 2
3.0 to 4.0
Answer: A) 0.6 to 0.7
Confirmed vs Book-3 §4.3 - Li-Br-water single-effect vapour absorption systems have a COP of about 0.65–0.70 (book §4.3). Options (b)-(d) quote COP values typical of vapour COMPRESSION chillers; a single-effect absorption machine is heat-driven and the book fixes its COP at 0.65-0.70.
📖 §6.3 Pump curves — pump operating point; §6.6 Flow control strategies — flow control valve (throttling)
23. Which of the following statements is not true regarding centrifugal pumps?
Flow is zero at shut off head
Maximum efficiency will be at design rated flow of the pump
Head decreases with increase in flow
Power increases with throttling
Answer: D) Power increases with throttling
Confirmed vs Book-3 §6.3/§6.6 — For a centrifugal pump flow is zero at shut-off, head falls as flow rises, and efficiency peaks at the design (BEP) flow — a, b, c are all true. On throttling the duty point moves up the curve to lower flow and the book notes 'there is some reduction in pump power absorbed at the lower flow rate' — power decreases, so d is the false statement.
📖 §6.2 System characteristics — static & friction head
24. 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 speed
The basic shape of a 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 speed
Confirmed vs Book-3 §6.2 — The system curve is the head-loss vs flow relationship of the piping (a); friction ∝ Q² makes it parabolic (c); with no static lift it starts at zero head and zero flow (d, Fig 6.5). It depends on elevation, pipe size/length, fittings and equipment — not on pump speed, which shifts the PUMP curve. So b is false.
25. The source of maximum harmonics among the following, in a plant power system is
100 CFL lamps of 11 W to 25 W
500 kW, 3 Phase, 415 V, 50 Hz resistance furnace
5 kVA UPS for computer system
variable frequency drive for 225 kW motive load
Answer: D) variable frequency drive for 225 kW motive load
Confirmed vs Book-3 §1.10 — VFDs head the book's list of non-linear loads, and the harmonic current injected scales with the load, so a 225 kW drive is by far the biggest source here.
Option (b) is the trap: a 500 kW resistance furnace is large but LINEAR (constant impedance), so it injects no harmonics; the 5 kVA UPS is non-linear but negligible in size.
26. The lamp based on high frequency electromagnetic field from outside, exciting the mercury gas sealed in the bulb, to produce UV radiation and light is
Induction lamp
Fluorescent lamp
Mercury vapour lamp
Metal halide lamp
Answer: A) Induction lamp
Confirmed vs Book-3 §8.3 — Induction lamp: 'high frequency electromagnetic fields are induced from outside the sealed chamber', electrons collide with mercury atoms to produce UV, which the phosphor converts to visible light — no electrodes inside the bulb. A fluorescent lamp uses internal filaments/electrodes; HPMV and metal halide are HID arc lamps.
27. State which of the following statements is true?
for a given fan operating at a constant temperature, the power input to fan increases by 4 times when the fan speed becomes double
for a given fan operating at a constant temperature, the power input to fan increases by 8 times when the fan speed becomes double
for a given fan operating at a constant flow rate, the power input increases as the air temperature increases
for a given fan operating at a constant static pressure rise, the flow rate reduces as the air temperature increases
Answer: B) for a given fan operating at a constant temperature, the power input to fan increases by 8 times when the fan speed becomes double
Confirmed vs Book-3 §5.3 — Power ∝ N³, so doubling speed gives 2³ = 8 times the power (b); (a) 4× would be the PRESSURE rise. (c) is wrong: at constant flow, hotter (less dense) air needs LESS power; (d) is wrong: volume flow at a given speed is set by the fan laws, not by temperature.
28. One tonne of refrigeration has the ability to remove ______ kcal of heat in a 24-hour period.
50 kcal
3024 kcal
72576 kcal
12000 kcal
Answer: C) 72576 kcal
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/hr (book). Over 24 hours = 3024 x 24 = 72,576 kcal. Option (b) 3024 kcal is the hourly figure and (d) 12,000 is the BTU/hr value; over a full day 3024 x 24 = 72,576 kcal.
📖 §1.8 Estimation of Technical Losses in Distribution System
30. HVDS (High Voltage Distribution System) is preferred to
reduce technical loss in distribution system
improve voltage regulation
comply with regulatory mandate
reduce energy bill for the end consumer
Answer: A) reduce technical loss in distribution system
Confirmed vs Book-3 §1.8 Measures to reduce technical losses — HVDS replaces long 415 V LT lines with 11 kV lines feeding small (16/25 kVA) pole-mounted transformers at the load centres; at higher voltage the current, and hence the I²R loss, is far lower.
Option (b) voltage regulation does improve, but it is a by-product; the book's stated purpose of HVDS is reduction of technical distribution losses.
Confirmed vs Book-3 §4.8 - Raising the evaporator temperature increases the refrigerant suction density/mass flow and the COP, so refrigeration capacity increases and specific power (kW/TR) falls.
32. Improving power factor at motor terminals in a factory will
increase active power
release distribution transformer capacity
reduce contract demand
increase motor efficiency
Answer: B) release distribution transformer capacity
Confirmed vs Book-3 §1.4 — advantage (d): 'kVA loading on the source generators as also on the transformers and lines up to the capacitors reduces giving capacity relief. A high power factor can help in utilizing the full capacity of the electrical system.'
Option (c) is wrong — contract demand is a contractual figure with the utility, which must be renegotiated; PF correction only lowers the recorded kVA.
33. If the COP of a vapour compression system is 3.5 and the motor draws a power of 10.8 kW at 90% motor efficiency, the cooling effect of vapour compression system will be
34 kW
37.8 kW
0.36 kW
none of the above
Answer: A) 34 kW
Confirmed vs Book-3 §4.7 - Compressor shaft power = 10.8 x 0.9 = 9.72 kW. Cooling effect = COP x compressor power = 3.5 x 9.72 = 34.0 kW. Option (b) 37.8 kW is the trap of using the 10.8 kW motor input without applying the 90% efficiency; use shaft power 9.72 kW x COP 3.5 = 34 kW.
34. A parameter that indicates adequacy of lighting for a particular application is
installed load efficacy
installed power density
lux
lumens
Answer: C) lux
Confirmed vs Book-3 §8.2/§8.4 — Adequacy of lighting for a task is judged by the illuminance on the working plane in lux, compared with the recommended lux ranges (IS 3646, Table 8.2). Installed load efficacy (lux/W/m²) and installed power density (W/m²/100 lux) measure energy efficiency, and lumens is lamp output, not what reaches the task.
35. Which of the following is not an example of lighting controls?
dimmers
timers
photosensors
daylight harvesting
Answer: D) daylight harvesting
Confirmed vs Book-3 §8.6(g) — The book lists lighting controllers as 'dimmers, motion & occupancy sensors, photosensors and timers'. Daylight harvesting is the strategy of using natural daylight (§8.6a / §8.7 daylight-linked control); it is implemented BY photosensors and dimmers, so it is not itself a control device.
36. Which of the following flow controls in a fan system will change the system resistance curve:
Inlet guide vane
speed change with variable frequency drive
speed change with hydraulic coupling
discharge damper
Answer: D) discharge damper
Confirmed vs Book-3 §5.5 — Dampers change volume 'by adding or removing system resistance', i.e. they shift the SYSTEM resistance curve (SC₁→SC₂ in Fig 5.7). Inlet guide vanes change the FAN curve characteristics; speed changes by VFD or hydraulic coupling (b, c) move the fan to a new fan curve while the system curve is unchanged.
37. When the dew point temperature is equal to the air temperature then the relative humidity is
0%
50%
100%
Unpredictable
Answer: C) 100%
Confirmed vs Book-3 §4.2 - When DBT equals the dew point (and wet-bulb) temperature, the air is saturated, so relative humidity = 100%. Options (a) and (b) describe unsaturated air; when the dry bulb falls to the dew point the air holds all the moisture it can at that temperature, which is 100% RH by definition.
📖 Book-3 §3.3 Compressor Performance — Free Air Delivery (FAD)
38. For an air compressor of rated capacity of 100 CFM and system leakage of 10%, free air delivery is ____
111.11 CFM
90 CFM
100 CFM
None of the above
Answer: C) 100 CFM
Confirmed vs Book-3 §3.3 Compressor Performance — FAD is the volume of air drawn from the atmosphere, compressed and delivered by the machine — it is measured at the compressor, so leaks downstream in the distribution network do not change it.
The rated 100 CFM therefore stands; 90 CFM would be the air actually reaching end-use points, which is a distribution loss and not the FAD.
📖 §2.3 Motor characteristics (load torque characteristics) / §2.9 Speed control
39. 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)
40. 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.
41. 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.
42. 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.
43. 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.
📖 §3.4 Compressed air system components (air receiver)
44. 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.
45. 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.
📖 §9.3 Operational factors (waste heat recovery from flue gases)
46. 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)
47. 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.
📖 §9.4 Energy performance assessment of DG sets (specific fuel consumption)
48. 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)
49. 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)
50. 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. List five energy saving measures in a centralized chilled water based air conditioning system.
Model answer: 1. Insulate all cold lines/vessels using economic insulation thickness to minimise heat gains. 2. Optimise air-conditioned volumes (false ceilings, segregate critical areas, use air curtains). 3. Minimise AC load by roof cooling/roof painting, efficient lighting, pre-cooling fresh air via air-to-air heat exchangers. 4. Optimal thermostatic temperature setting of conditioned spaces. 5. Minimise part-load operation by matching load with online plant capacity; use variable speed drives for varying load.
Standard chilled-water-system ECMs from Book-3 Chapter 4 (each point one mark).
2. 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)?
Model answer: 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.
3. An ID fan consumes 35 kW at 100% boiler loading with damper fully open. Estimate daily energy savings if the damper is replaced by a VFD, for the schedule: 80% load, 4 h, 31 kW; 70% load, 12 h, 29 kW; 60% load, 8 h, 26 kW. Air requirement ∝ boiler loading.
Model answer: VFD power = (flow%)³ x 35 kW. At 80%: 0.8³x35 = 17.9 kW, saving 31-17.9 = 13.1 kW x 4 h = 52.32 kWh. At 70%: 0.7³x35 = 12 kW, saving 29-12 = 17 kW x 12 h = 203.94 kWh. At 60%: 0.6³x35 = 7.6 kW, saving 26-7.6 = 18.4 kW x 8 h = 147.52 kWh. Total daily savings ≈ 403.78 kWh.
With VFD, flow ∝ speed so power = 35 × (load fraction)³; damper power is the measured value. Savings: 13.1 kW × 4 h + 17 kW × 12 h + 18.4 kW × 8 h ≈ 403.8 kWh/day.
4. A 75 kW, 415 V, 140 A, 4 pole, 50 Hz, 3-phase squirrel-cage induction motor (full-load efficiency 87.6%) has measured terminal voltages 415, 418, 420 V and currents 137, 132, 137 A. Estimate the additional temperature rise of the motor due to unbalanced voltage supply.
Model answer: Mean voltage = (415 + 418 + 420)/3 = 417.67 V.
Maximum deviation from the mean = |415 − 417.67| = 2.67 V (larger than the 0.33 V and 2.33 V deviations).
% voltage unbalance (NEMA) = 2.67 / 417.67 × 100 = 0.64 %.
Additional temperature rise = 2 × (% voltage unbalance)² = 2 × (0.64)² = 0.82 °C.
This is within the book's recommended limit of 1 % unbalance at motor terminals, so no derating is called for; note that current unbalance can still be 6–10 times the voltage unbalance.
Voltage unbalance = max deviation from mean ÷ mean x 100; additional heating ≈ 2 x (% unbalance)².
5. 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.
6. 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)
7. 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.
8. 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.
1. Briefly explain any three different methods of flow control for fans.
Model answer: Pulley change: for a permanent flow change, alter fan speed by changing the v-belt pulley diameter (simplest permanent speed change). Damper control: dampers add/remove system resistance, forcing the fan up/down its curve to deliver more or less air without changing speed. Inlet guide vanes: curved vanes pre-swirl the inlet air, changing the fan curve; energy-efficient for modest reductions (100% to ~80% flow), efficiency drops sharply below 80%. Variable speed drive: reduce fan speed to match reduced flow; since power ∝ flow³, this is usually the most efficient capacity control.
Book-3 §5.5 lists pulley change, damper control, inlet guide vanes, variable speed drive (and series/parallel operation, variable-pitch axial blades). Any three with the key points — dampers add resistance (least efficient), IGVs efficient 100→80% flow, VSD most efficient (power ∝ N³).
2. 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.
Model answer: 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.
3. 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.
Model answer: (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. L-3 fill in the blanks: 1) DBT 30°C, WBT 30°C → RH = ___%. 2) Cavitation occurs when local static pressure falls below the ___ pressure of the liquid at actual temperature. 3) As 'Approach' decreases, cooling tower effectiveness will ___. 4) Ratio of luminous flux emitted to power consumed is called ___. 5) A centrifugal pump raises water to 12 m; with brine of SG 1.2, the height raised is ___ m. 6) Harmonics are multiples of the ___ frequency. 7) A motor that can run at lagging as well as leading PF is the ___ motor. 8) Per ECBC, Effective Aperture (EA) = ___ for WWR 0.40 and VLT 0.25. 9) In an amorphous-core transformer, ___ loss is less than conventional. 10) Centrifugal pump impeller is generally trimmed down to about ___% of maximum size.
Model answer: 1) 100%. 2) Vapour. 3) Increases. 4) Luminous efficacy. 5) 12 m (same — a centrifugal pump develops head independent of fluid density). 6) Fundamental (50 Hz). 7) Synchronous. 8) EA = VLT x WWR = 0.25 x 0.40 = 0.10. 9) No-load (iron/core/fixed) loss. 10) 75% (about 75-80%).
Definitional fill-ins across chapters: DBT = WBT → saturated (100% RH); cavitation when pressure < vapour pressure; lower approach → higher effectiveness; lumen/W = luminous efficacy; centrifugal pump head is independent of fluid density (12 m); harmonics = multiples of fundamental; synchronous motor PF control; EA = VLT × WWR; amorphous core → lower no-load loss; impeller trim ≈ 75%.
📖 §5.3 Fan laws; §5.5 Pulley change; §5.6 Fan power formula
5. L-4: A belt-driven centrifugal fan: air flow 68,400 m3/hr, fan differential static pressure 112 mmWC, pressure drop across main damper 17 mmWC, motor input power 26.8 kW, fan speed 600 rpm, motor speed 1460 rpm, fan pulley 560 mm, motor pulley 230 mm. The auditor recommends opening the main damper fully and reducing fan speed via pulley change. Calculate (a) annual energy savings for 6000 h/yr and (b) the new fan pulley diameter.
Model answer: Flow = 68400/3600 = 19 m3/s. Theoretical air power (partly-closed) WTh1 = 19x112/102 = 20.86 kW. With damper fully open, pressure = 112-17 = 95 mmWC; WTh2 = 19x95/102 = 17.7 kW. Case-2 motor power W2 = W1 x (WTh2/WTh1) = 26.8 x (17.7/20.86) = 22.7 kW. (a) Annual savings = (26.8-22.7) x 6000 = 24,600 kWh. (b) New speed N2 = N1 x (p2/p1)^0.5 = 600 x (95/112)^0.5 = 553 rpm. Using N1D1 = N2D2, new fan pulley D2 = (N1/N2) x D1 = (600/553) x 560 = 608 mm.
Opening the damper removes the 17 mmWC throttling loss, so air power falls from 19 × 112/102 = 20.86 kW to 19 × 95/102 = 17.7 kW; motor power scales in proportion (26.8 → 22.7 kW) → 24,600 kWh/yr. New speed from SP ∝ N²: 600 × √(95/112) = 553 rpm; fan pulley from N₁D₁ = N₂D₂ → 608 mm (larger driven pulley slows the fan).
6. 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.
Model answer: (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.
📖 §4.13 Ice Bank; §4.3 VAR (harmonics: Book-3 Ch1)
7. 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.
Model answer: (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).
📖 §1.5 Transformer losses & efficiency (loss evaluation over a load cycle)
8. 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.
9. 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)
10. 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
11. 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.