Energy Efficiency in Electrical Utilities Available here with full solutions — 57 questions recovered from the 2009 exam:
Objective (1 mark)
43 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) — 43
📖 §1.1 Transmission & Distribution Lines — voltage vs. line loss (P=I²R)
1. In the city electrical distribution scheme, a proposal is being prepared to upgrade 33 kV network to 66 kV. The distribution loss, corresponding to the same quantum of load in the proposed upgraded system will be
less by 25%
less by 33%
less by 75%
none of the above
Answer: C) less by 75%
Confirmed vs Book-3 §1.1 — line loss varies as the inverse square of the voltage: (33/66)² = 1/4, so the loss falls to 25% of its old value, i.e. it is LESS by 75%.
Option (a) 'less by 25%' quotes the remaining loss instead of the reduction — read the wording carefully.
📖 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings
2. Rating of power factor correction capacitors at induction motor terminals should be
100% of no load magnetizing kVAr of induction motor
90% of no load magnetizing kVAr of induction motor
120% of no load magnetizing kVAr of induction motor
none of the above
Answer: B) 90% of no load magnetizing kVAr of induction motor
Confirmed vs Book-3 §2.7 Power Factor Correction — Table 2.5 Capacitor Ratings — Same rule: the capacitor is selected not to exceed 90 % of the motor's no-load (magnetizing) kVAr, which can only be established by a no-load test. (c) 120 % would leave the machine critically or over-corrected, risking self-excitation over-voltage on supply interruption.
the advantage of PF improvement by capacitor addition in an electric network is that active power component of the network is reduced
the power factor indicated in the monthly electricity bill is the lowest power factor recorded at any time during the billing month
PF capacitors operating at lower voltage than their rated values have higher operating kVArs than their rated values
the power factor of an induction motor decreases with decrease in percentage motor loading
Answer: D) the power factor of an induction motor decreases with decrease in percentage motor loading
Confirmed vs Book-3 §1.4 — an induction motor's magnetising (reactive) current is nearly constant while its kW falls with load, so kW/kVA — the power factor — falls as loading falls; this is why lightly loaded motors are a major cause of poor plant PF.
Option (a) is wrong (capacitors do not reduce kW), (b) is wrong (the bill shows the AVERAGE PF, kWh/kVAh) and (c) is wrong (below rated voltage a capacitor gives LESS kVAr, since kVAr ∝ V²).
4. If the efficiencies of a power plant, transmission and distribution systems are 30%, 95% & 85% respectively, the cascade efficiency of power generation, and transmission system is given by
24.23%
28.5%
80.75%
95%
Answer: B) 28.5%
Confirmed vs Book-3 §1.1 Cascade Efficiency — the question asks only for generation AND transmission: 0.30 × 0.95 = 0.285 = 28.5%.
Option (a) 24.23% is the full cascade INCLUDING the 85% distribution stage (0.30 × 0.95 × 0.85) — that stage was not asked for.
5. What is the % slip of a 4 pole induction motor if the shaft speed at 49.5 Hz supply frequency is 1460 rpm?
1.68
2.66
1.71
none of the above
Answer: A) 1.68
Corrected (was c) — Book-3 §2.3 Motor Characteristics: the slip formula uses the SYNCHRONOUS speed in the denominator. Ns = 120 × 49.5 / 4 = 1485 rpm, so slip = (1485 − 1460)/1485 × 100 = 1.68 % — option (a). The recorded 1.71 % comes from dividing by the rotor speed 1460 instead of by Ns; (b) 2.66 % comes from wrongly assuming Ns = 1500 rpm at 50 Hz.
6. During induction motor operation, magnetic field is established in
stator winding only
rotor winding only
stator and rotor windings
at carbon brushes
Answer: C) stator and rotor windings
Confirmed vs Book-3 §2.2 Motor Types — The stator carries the rotating flux and the induced rotor current creates its own alternating field, so a magnetic field exists in both stator and rotor windings; their interaction produces torque. (d) 'at carbon brushes' applies to a slip-ring or DC machine and has nothing to do with field production.
📖 §1.2 Electricity Billing — trivector meter & Maximum Demand
8. If the apparent power drawn over a recording cycle of 30 minutes is 3000 kVA for 10 minutes, 2400 kVA for 15 minutes and 2900 kVA for 5 minutes, the MD recorder will compute MD as
3000 kVA
2400 kVA
2683 kVA
none of the above
Answer: C) 2683 kVA
Confirmed vs Book-3 §1.2 — MD = time-integrated demand over the cycle = [(3000×10) + (2400×15) + (2900×5)]/30 = 80,500/30 = 2683 kVA.
Option (a) 3000 kVA is the highest instantaneous block; the book stresses the meter registers the AVERAGE over the whole 30-minute recording cycle.
📖 Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11)
9. The pressure drop in mains header at the farthest point of an industrial compressed air network shall not exceed
2 bar
0.3 bar
0.5 bar
1.0 bar
Answer: B) 0.3 bar
Confirmed vs Book-3 §3.5 Minimum Pressure Drop in Air Lines (Table 3.11) — Book-3: 'Typical acceptable pressure drop in industrial practice is 0.3 bar in mains header at the farthest point and 0.5 bar in distribution system.'
The 0.5 bar figure offered in (c) is the distribution allowance, not the mains-header limit asked for.
10. The Free Air Delivery capacity of a reciprocating compressor is directly proportional to
pressure
volume
speed
all of the above
Answer: C) speed
Confirmed vs Book-3 §3.2 Positive Displacement — Book-3 §3.2: 'the compressor capacity is directly proportional to the speed', while flow output 'remains nearly constant over a range of discharge pressures'.
Because FAD does not scale with delivery pressure, neither (a) nor 'all of the above' can hold; speed is the correct answer.
📖 Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7)
11. The inlet air temperature to a two stage reciprocating air compressor is 35°C. At which of the following 2nd stage inlet temperatures the compressor will consume least power?
75°C
65°C
60°C
50°C
Answer: D) 50°C
Confirmed vs Book-3 §3.5 Efficacy of Inter and After Coolers (Table 3.7) — Table 3.7 shows specific power falling as the second-stage inlet temperature approaches the first-stage inlet ('perfect cooling'); a 5.5°C rise at the second-stage inlet costs about 2% more specific energy.
With the first stage at 35°C, the lowest offered second-stage inlet — 50°C — gives the least power consumption.
12. At which of the following discharge pressures, the reciprocating air compressor will consume maximum power
3 bar
3.5 kg/cm2
150 psi
6 kg/cm2
Answer: C) 150 psi
Confirmed vs Book-3 §3.5 Pressure Settings / Reducing Delivery Pressure (Table 3.9) — Convert to common units: 3 bar ≈ 3.06 kg/cm², 3.5 kg/cm² ≈ 3.43 bar, 150 psi ≈ 10.34 bar (10.5 kg/cm²), 6 kg/cm² ≈ 5.88 bar. The highest by far is 150 psi.
Since 'for the same capacity, a compressor consumes more power at higher pressures', 150 psi draws the maximum power; the trap is comparing the bare numbers across mixed units.
13. Which of the following is not true of air receivers?
smoothens pulsating air output
stores large volumes of air
a source for draining moisture
increases the pressure of air
Answer: D) increases the pressure of air
Confirmed vs Book-3 §3.5 Air Receivers (IS 7938-1976 sizing) — Air receivers smooth pulsating output, store large volumes for sudden demand and provide a point for draining precipitated moisture — (a), (b) and (c) are all true of them.
A receiver holds air at the pressure the compressor delivers and cannot itself increase pressure, so (d) is the statement that is not true.
14. A 1.5 ton air conditioner installed in a room and working continuously for two hours will remove heat of
3024 kCals
6048 kCals
9072 kCals
none of the above
Answer: C) 9072 kCals
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kCal/h; 1.5 TR × 2 h = 1.5 × 3024 × 2 = 9072 kCal. Option (a) is for 1 TR in one hour and (b) for 1.5 TR in one hour; multiply 3024 kcal/h by 1.5 TR and by 2 hours to get 9072 kcal.
15. Which of the following can be used as refrigerant both in vapour compressor and vapour absorption systems
Ammonia
R-11
R-12
Lithium Bromide
Answer: A) Ammonia
Confirmed vs Book-3 §4.3 - Ammonia is used as the refrigerant in both vapour compression and vapour absorption (NH3-water) systems. R-11 and R-12 (b, c) are CFCs used only in vapour compression, and lithium bromide (d) is the absorbent, not a refrigerant; ammonia (R-717, Table 4.1) serves in both.
16. Chilled water enters evaporator at 12°C and leaves at 6°C. The flow rate of chilled water was measured as 300 m3/hr. The tons of refrigeration capacity is
0.595
595.24
35.7
none of the above
Answer: B) 595.24
Confirmed vs Book-3 §4.7 - Heat = 300×1000×1×(12−6) = 1,800,000 kCal/h; TR = 1,800,000/3024 = 595.24 TR. Option (c) 35.7 divides by 3024 twice and (a) 0.595 misplaces the decimal; TR = 300,000 x 6/3024 = 595.24.
17. Centrifugal compressors are most efficient when they are operating at_____.
50% load
full load
75% load
all load conditions
Answer: B) full load
Confirmed vs Book-3 §4.5 - Centrifugal (dynamic) compressors are most efficient at or near their design/full-load point. Options (a) and (c) describe part loads, where inlet-guide-vane throttling makes efficiency fall away sharply below 50%; the book says centrifugals are the most efficient type when operating NEAR FULL LOAD.
18. The Coefficient of Performance (COP) of Vapour Absorption Refrigeration System (VAR)
is higher than that of Vapour Compression Refrigeration (VCR) System
is lower than that of Vapour Compression Refrigeration (VCR) System
is same as that of Vapour Compression Refrigeration (VCR) System
is normally 4 to 4.5
Answer: B) is lower than that of Vapour Compression Refrigeration (VCR) System
Confirmed vs Book-3 §4.3 - VAR systems typically have COP well below 1, much lower than vapour compression systems (COP 3–5). Option (a) reverses the comparison and (d) quotes a VCR-range COP; LiBr-water absorption machines run at COP 0.65-0.70 against 3-5 for compression chillers, which is why VAR pays only with cheap waste heat.
19. Backward-inclined fans are known as _____ because change in static pressure does not overload the motor
overloading
non-overloading
radial
axial
Answer: B) non-overloading
Confirmed vs Book-3 §5.2 — 'Backward-inclined fans are known as "non-overloading" because changes in static pressure do not overload the motor' — their power peaks and then falls within the usable flow range. Forward-curved fans, whose power rises continuously with flow, are the overloading type.
Confirmed vs Book-3 §5.3 — Among the manufacturer's curves 'the curve static pressure (SP) vs. flow is especially important'; its intersection with the system curve defines the operating point. Total/dynamic/suction pressure vs flow (b, c, d) are not the standard fan characteristic.
📖 §6.2 System characteristics — static & friction head
22. For the same flow through which of the following diameter pipes, the pump will work with maximum pressure
80 mm
100 mm
120 mm
140 mm
Answer: A) 80 mm
Confirmed vs Book-3 §6.2 — For a fixed flow the smallest pipe (80 mm) gives the highest velocity and friction loss (∝ 1/D⁵), so the pump must work against the maximum pressure. Larger pipes reduce friction head — the book's rationale for larger diameter pipe. (Option d corrected from OCR '1400 mm' to '140 mm'.)
23. If inlet and outlet water temperatures of a cooling tower are 44°C and 38°C respectively and atmospheric DBT and WBT are 40°C and 35°C respectively, then the effectiveness of cooling tower is
54.5%
66.6%
75%
none of the above
Answer: B) 66.6%
Corrected (was a) — Book-3 §7.2 (iii): Effectiveness = Range/(Range + Approach) × 100. Range = 44 − 38 = 6°C; Approach = 38 − 35 = 3°C (cold water − WBT, DBT is a distractor); Effectiveness = 6/(6 + 3) = 66.6% → (b). 54.5% (a) = 6/11 would only result if the WBT were 33°C (approach 5°C) – it does not follow from the figures printed here. ⚠ Answer changed a→b: with the printed data (WBT 35°C) the book formula gives 66.6%. The old key value 54.5% corresponds to a 33°C WBT variant of this question.
24. 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
Confirmed vs Book-3 §5.2 — Axial fans (propeller, tubeaxial, vaneaxial) pass air straight through with no change in direction; propeller (d) is the only axial option. Forward-curved, backward-curved and radial (a, b, c) are centrifugal — airflow turns twice.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
25. The motor efficiency is 0.9 and the pump efficiency is 0.6. The input power to the motor driving the pump is 28 kW. The power transmitted to the water is
15.12 kW
28 kW
25.2 kW
none of the above
Answer: A) 15.12 kW
Confirmed vs Book-3 §6.1 — Power to water (hydraulic) = motor input × η_motor × η_pump = 28×0.9×0.6 = 15.12 kW. 25.2 kW is only the motor shaft output (28×0.9); 28 kW is the electrical input.
📖 §6.5 Effects of impeller diameter change (Q∝D, H∝D², P∝D³; trim limited to ~75%)
26. A water pump is delivering 200 m3/hr at ambient conditions. The impeller diameter is trimmed by 10%. The water flow at the changed conditions is
220 m3/hr
180 m3/hr
162 m3/hr
none of the above
Answer: B) 180 m3/hr
Confirmed vs Book-3 §6.5 — Q∝D: a 10% trim gives Q₂ = 0.9×200 = 180 m³/hr. 162 m³/hr (0.9²×200) wrongly uses the head relation; 220 would be an enlarged impeller.
27. Increasing the Cycles of Concentration (C.O.C) in circulating water in a cooling tower, the blow down quantity will
increase
decrease
not change
none of the above
Answer: B) decrease
Confirmed vs Book-3 §7.2 (vii) — Blow Down = Evaporation/(COC − 1); a higher COC means a larger denominator and therefore a smaller blowdown → (b). This is the basis of the book's 'COC improvement for water savings'.
28. At which of the following condenser temperatures, the power consumption of a vapour compression refrigeration system will be the least
26°C
28°C
29°C
25°C
Answer: D) 25°C
Confirmed vs Book-3 §4.8 - Lower condensing temperature reduces compressor lift and power; 25°C is the lowest, giving least power. Every other option is a higher condensing temperature, and Table 4.5 shows specific power consumption rising steadily with condenser temperature - lower condensing temperature means a smaller compression lift.
📖 §7.2 Factors Affecting Performance – Wet Bulb Temperature
29. Which of the following ambient conditions will not evaporate maximum amount of water in a cooling tower
41°C DBT and 38°C WBT
38°C DBT and 37°C WBT
36°C DBT and 30°C WBT
36°C DBT and 31°C WBT
Answer: B) 38°C DBT and 37°C WBT
Confirmed vs Book-3 §7.2 Wet Bulb Temperature — Evaporation is driven by the wet-bulb depression (DBT − WBT). Options (c) and (d) have 6 and 5°C depression and evaporate a lot; (b) 38/37°C has only 1°C – near-saturated air – so it is the condition that will NOT evaporate the maximum → (b).
30. If inlet and outlet water temperatures of a cooling tower are 39°C and 33°C respectively and atmospheric DBT and WBT are 35°C and 28°C respectively then the approach of cooling tower is
3°C
4°C
5°C
6°C
Answer: C) 5°C
Confirmed vs Book-3 §7.2 (ii) — Approach = outlet cold water temperature − ambient WBT = 33 − 28 = 5°C → (c). Using DBT (35 − 33 = 2°C) or range (39 − 33 = 6°C, option d) are the common traps.
📖 §8.6(e) Reduction of lighting feeder voltage + Table 8.3
32. If voltage is increased from 230 V to 250 V for a fluorescent tube light, it will result in
reduced power consumption
increased power consumption
decreased light levels
no change in power consumption and light levels
Answer: B) increased power consumption
Confirmed vs Book-3 §8.6(e) Table 8.3 — For fluorescent lamps a 10 % higher voltage increases power input by 8.1 % (and light output by only 8 %); higher voltage also reduces lamp life. So 230 → 250 V raises power consumption; it does not reduce light level.
33. Automatic power factor controller using kVAr control, requires sensing of
current
voltage
capacitance
both a and b
Answer: D) both a and b
Confirmed vs Book-3 §1.4 — an APFC computes the reactive power/displacement power factor, which requires both the current and the voltage (and the angle between them) to be sensed before it can switch capacitor steps.
Option (a) alone gives no angle information, so no kVAr or PF can be derived from current sensing by itself.
34. The COP of a vapour compression system is 3.0. If the motor draws power of 11 kW at 90% motor efficiency, the cooling effect of vapour compression system will be
29.7 kW
37.8 kW
0.36 kW
none of the above as cooling effect is always measured in TR
Answer: A) 29.7 kW
Confirmed vs Book-3 §4.7 - Shaft power = 11×0.9 = 9.9 kW; cooling effect = COP × power = 3.0 × 9.9 = 29.7 kW. Option (b) 37.8 kW ignores the 90% motor efficiency; shaft power = 11 x 0.9 = 9.9 kW, and cooling effect = COP x shaft power = 29.7 kW.
35. Which of the following can also act as a heat pump?
centrifugal pump
centrifugal compressor
air conditioner
none of the above
Answer: C) air conditioner
Confirmed vs Book-3 §4.11 - A reversible air conditioner can deliver heat (heat pump mode) as well as cooling. A centrifugal pump (a) moves liquid and a compressor alone (b) is only a component; §4.11 states a heat pump is the same as an air conditioner except that the rejected heat becomes the useful output.
36. A slide valve is used for capacity control in which of the following refrigeration compressors?
reciprocating
centrifugal
screw
scroll
Answer: C) screw
Confirmed vs Book-3 §4.5 - Screw compressors use a sliding slide valve to vary the effective rotor length and hence capacity. Reciprocating machines (a) unload cylinders, centrifugals (b) use inlet guide vanes and scrolls (d) have no capacity control of this kind; §4.5 and the §4.17 case study name the slide valve as the common screw-compressor control (10-100%).
📖 §9.2 Sizing — kVA = √3·V·I; §9.4 specific fuel consumption
37. In a DG set, a 3 phase alternator is supplying on an average 100 A at 420 V and 0.9 pf to a load. If the specific fuel consumption of this DG set is 0.30 lts/kWh at that load, then how much fuel is consumed while delivering generated power for one hour?
11.34 litre
19.64 litre
21.82 litre
218.23 litre
Answer: B) 19.64 litre
Confirmed vs Book-3 §9.2/§9.4 — Power = √3 × 420 × 100 × 0.9 / 1000 = 65.47 kW; fuel = 65.47 kWh × 0.30 L/kWh = 19.64 litres in one hour. Option (c) 21.82 L is the tempting error of ignoring PF (√3 × 420 × 100 = 72.7 kVA × 0.3); (a) 11.34 L omits √3.
📖 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD
38. The largest potential for electricity savings with variable speed drives is generally for:
variable torque applications
constant torque loads
constant power load
combination of above
Answer: A) variable torque applications
Confirmed vs Book-3 §2.9 Application of Variable Speed Drives — Concept & Principles of VFD — The book states the largest potential for electricity savings with VSDs is generally in variable torque applications such as centrifugal pumps and fans, where power changes as the cube of speed — a 20 % speed cut gives almost 50 % input power reduction. (b) constant torque loads are also suitable for VSDs but their power falls only linearly with speed, so the savings are smaller.
39. The electronic ballast fitted in a tube light fitting does not have one of the following characteristics
lower operational losses than conventional ballasts
tuned circuit to deliver power at 28-32 KHz
requiring a starter
low temperature rise
Answer: C) requiring a starter
Confirmed vs Book-3 §8.6(f) — Electronic ballasts have ~1 W loss (vs 10–15 W), run the lamp at high frequency (book: 20–30 kHz; exam quotes 28–32 kHz) and hence low temperature rise, and 'the starter is eliminated'. Requiring a starter is the characteristic they do NOT have.
📖 §1.3 Electrical Load Management and Maximum Demand Control
40. Maximum demand controller is used to
switch off non-essential loads in a logical sequence
switch off essential loads in a logical sequence
controls the reactive power of the plant
all the above.
Answer: A) switch off non-essential loads in a logical sequence
Confirmed vs Book-3 §1.3 Shedding of Non-Essential Loads — the MD controller provides 'Automatic load shedding in a predetermined sequence' and 'Automatic restoration of load' when the demand approaches a preset limit.
Option (b) is wrong: only NON-essential loads are shed; option (c) describes an APFC, which handles reactive power.
📖 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives
41. In a fluid coupling, connecting an induction motor and a fan
motor speed can be changed by the fluid coupling
fan speed can be changed by the fluid coupling
both motor and fan speed can be changed by the fluid coupling
none of the above is possible
Answer: B) fan speed can be changed by the fluid coupling
Confirmed vs Book-3 §2.9 Other Methods of Speed Control — Fluid Coupling / Eddy Current Drives — A fluid coupling varies the speed of the DRIVEN equipment without changing the speed of the motor — the impeller runs at motor speed and slip between impeller and runner sets the output speed. (a)/(c) are wrong precisely because the motor stays on a fixed-frequency supply and keeps its own speed.
📖 Not in Book-3 Ch5 text (belt drive slip; cf. §5.5 Pulley change)
42. In a "V" belt coupled fan drive, the measured speed at motor end 6" diameter pulley is 1480 rpm and that at fan end 10" diameter pulley is 820 RPM. What is the slippage loss in %?
7.66
8.29
6.67
insufficient data, cannot be worked out
Answer: B) 8.29
Confirmed vs official key (topic not covered in Book-3 Ch5) — Theoretical fan speed = 1480 × 6/10 = 888 rpm; actual 820 rpm, so speed lost to slip = 68 rpm. The official key (b) 8.29% expresses this relative to the ACTUAL fan speed: 68/820 = 8.29%. Relative to the theoretical speed it would be 68/888 = 7.66% (option a) — follow the official key in the exam. The data is sufficient, so (d) is wrong.
slip ring induction motors are normally less efficient than squirrel cage induction motors
high speed squirrel cage induction motors are normally less efficient than low speed squirrel cage induction motors
the capacitor requirement for PF improvement at induction motor terminal increases with decrease in rated speed of the induction motor
induction motor efficiency increases with increase in its rated capacity
Answer: B) high speed squirrel cage induction motors are normally less efficient than low speed squirrel cage induction motors
Confirmed vs Book-3 §2.4 Motor Efficiency — The book states that 'higher-speed motors are normally more efficient than lower-speed motors', so statement (b) reverses the book and is the incorrect one. (a), (c) and (d) are all book-supported — squirrel cage beats slip-ring on efficiency, capacitor kVAr rises as speed falls (Table 2.5), and motor efficiency increases with rated capacity.
Short questions (5 marks) — 8
📖 §1.4 Performance Assessment of Power Factor Capacitors
1. a) A 10 kVAr, 415 V rated power factor capacitor was found to be having terminal supply voltage of 440 V. Calculate the capacity of the power factor capacitor at the operating supply voltage. b) What would be the nearest kVAr compensation required for changing the power factor of a 500 kW load from 0.9 lead to unity power factor?
Model answer: a) Capacitor output varies as the square of the applied voltage (Book-3 Sec.1.4, Voltage effects): kVAr = 10 x (440/415)^2 = 10 x 1.124 = 11.24 kVAr. The bank delivers more than its rating, but running above rated voltage shortens capacitor life.
b) At 0.9 LEADING the load is already over-compensated - the current leads the voltage - so NO further capacitive kVAr is required. To reach unity, capacitance must instead be REMOVED: excess kVAr = 500 x tan(cos^-1 0.9) = 500 x 0.4843 = 242 kVAr of the existing bank should be switched out.
a) Capacitor kVAr varies with square of voltage ratio. b) Load is already at leading PF, so additional capacitive compensation is not needed.
📖 §6.1 Hydraulic power, pump shaft power & motor input power (p.172)
2. A pump is delivering 40 m3/hr of water with a discharge pressure of 29 metre. The water is drawn from a sump where water level is 6 metre below the pump centerline. The power drawn by the motor is 7.5 kW at 89% motor efficiency. Find out the pump efficiency.
Total head = discharge head − suction head = 29 − (−6) = 35 m. Hydraulic power = Q×H×ρ×g/1000; pump shaft power = motor power × motor efficiency; pump efficiency = hydraulic/shaft power.
3. Define one 'Ton of Refrigeration (TR)'. How do you calculate TR across the Air Handling Units?
Model answer: A ton of refrigeration is the quantity of heat to be removed to form one ton of ice in 24 hours when the initial water temperature is 0°C, equivalent to 50.4 kCal/min or 3024 kCal/h. Refrigeration load TR = Q × ρ × (h_in − h_out) / 3024, where Q is air flow in CMH, ρ is air density (kg/m³), h_in and h_out are enthalpies of inlet and outlet air (kCal/kg).
Definition of TR plus the AHU enthalpy-difference method for computing cooling load across the unit.
4. Estimate the cooling tower capacity (TR) and approach with the following parameters: Water flow rate = 120 m3/hr, Specific heat of water = 1 kCal/kg°C, Inlet water temperature = 42°C, Outlet water temperature = 36°C, Ambient WBT = 32°C.
5. A fan is delivering 20,000 Nm3/hr of air at static pressure difference of 70 mm WC. If the fan static efficiency is 55%, find out the shaft power of the fan.
Model answer: Q = 20,000/3600 = 5.56 m³/s. Fan static efficiency = (Q × Pst)/(102 × shaft power). 0.55 = (5.56 × 70)/(102 × P), giving shaft power P = 6.94 kW.
6. Calculate the free air delivery (FAD) capacity of a compressor in m3/min for the following data: Receiver capacity 0.5 m3, Initial pressure 0 kg/cm2(g), Final pressure 7 kg/cm2(g), Initial air temperature 32°C, Final air temperature 51°C, Additional holdup volume 0.03 m3, Pump up time 4.5 minutes, Atmospheric pressure 1.026 kg/cm2 absolute.
Confirmed vs Book-3 §3.6 Compressor Capacity Assessment — Standard pump-up FAD test formula correcting for pressure rise, receiver+holdup volume, pump-up time and temperature ratio.
7. A 15 kW, 415 V, 26 A, 4 pole, 50 Hz, 3 phase squirrel cage induction motor has full load efficiency and power factor of 90% and 0.89. An energy auditor measures: Supply voltage 408 V, Current 15 A, PF 0.81, Supply frequency 49.9 Hz, RPM 1488. Find at the operating conditions: 1) Power input in kW, 2) % motor loading, 3) % slip.
Model answer: 1) Power input = √3 × V × I × PF = 1.732 × 408 × 15 × 0.81 = 8,586 W = 8.59 kW.
2) Rated input = rated output / full-load efficiency = 15 / 0.90 = 16.67 kW; % motor loading = 8.59 / 16.67 × 100 = 51.5 %. (Using the nameplate √3 × 415 × 26 × 0.89 = 16.63 kW gives the same 51.6 %. The current ratio 15/26 = 58 % must NOT be used — the book forbids estimating loading from currents.)
3) Synchronous speed at 49.9 Hz = 120 × 49.9 / 4 = 1,497 rpm; % slip = (1,497 − 1,488)/1,497 × 100 = 0.60 %.
Power input from √3×V×I×PF; loading vs rated input (rated kW/efficiency); slip from synchronous speed at measured frequency.
8. A compressed air leakage test was conducted in an industry running 3 nos. of 500 cfm reciprocating compressors, maintained at loading-unloading settings of 6.6 and 7.0 kg/cm2g. Trial 1 (before): On load 30 secs, Unload 110 secs. Trial 2 (after attending leaks): On load 18 secs, Unload 145 secs. Average power was 71 kW during load and 16 kW during unload. Calculate the annual cost savings for 4000 hr/year operation at energy charge Rs. 6.00 per kWh.
Model answer: Leakage (trial 1) = (30×500)/(30+110) = 107 cfm. Leakage (trial 2) = (18×500)/(18+145) = 55 cfm. Specific power consumption = 71/(500×60) = 0.0023666 kW/ft³. Reduction in leakage = 107−55 = 52 cfm = 3120 cfh. Energy saving per hour = 3120 × 0.0023666 = 7.3838 kWh. Annual cost saving = 7.3838 × 4000 × 6 = Rs. 1,77,211 per annum.
Confirmed vs Book-3 §3.5 Avoiding Air Leaks — leak quantification — Leakage quantity from load fraction × FAD; specific power per ft³; multiply leakage reduction by specific power, operating hours and tariff.
1. Fill in the blanks: a) Heat Rate of a thermal power plant is expressed in ____. b) With increase in design speed of induction motors, the required capacitive kVAr for reactive power compensation for the same capacity range will ____. c) An air dryer in a compressed air system reduces ____ point of air. d) A Pitot tube measures the difference between ____ and ____ pressures of the fluid. e) The friction loss in a pipe carrying a fluid is proportional to the ____ power of pipe diameter.
Model answer: a) kCal/kWh b) decrease c) dew (dew point) d) total and static e) fifth
Standard BEE definitions: heat rate in kCal/kWh; higher-speed motors need less magnetizing kVAr; air dryer lowers dew point; Pitot tube reads total minus static pressure; pipe friction loss ∝ 1/d⁵ (fifth power of diameter).
📖 §8.3(1) Incandescent lamp; §8.2 Control gear (discharge lamps); §8.6 Energy saving opportunities
2. a) Briefly explain the difference between a 'filament lamp' and a 'gas discharge lamp'. b) State any 3 best practices in a lighting system for energy savings.
Model answer: a) Filament lamps (e.g. incandescent) produce light by a filament heated to incandescence by the flow of electric current through it. A gas discharge lamp produces light not by heating a filament but by the excitation of gas contained in a tubular or elliptical outer bulb. b) Any three of: install energy efficient fluorescent lamps in place of conventional ones; CFLs in place of incandescent lamps; metal halide lamps in place of mercury/sodium vapour lamps; HPSV lamps where colour rendering is not critical; LED indicator lamps in place of filament lamps; optimum daylighting; grouping of lighting for control flexibility; microprocessor based controllers; exclusive lighting transformer; servo stabilizer on lighting feeder; high-frequency electronic ballasts in place of conventional ballasts.
Book-3 §8.3: filament lamp = wire heated to incandescence by current; discharge lamp = current through low-pressure mercury vapour/gas between electrodes (needs ballast/ignitor, §8.2). Best practices from §8.6/§8.7: efficient lamps per Table 8.1, electronic ballasts, feeder-voltage optimisation, daylighting, controls.
3. a) A small scale industry has a constant load of 380 kVA. It has installed two transformers of 500 kVA each. The no load loss and full load copper loss of each 500 kVA transformer is 750 W and 5410 W respectively. From the energy efficiency point of view should the industry operate a single transformer or two transformers equally sharing the load? b) A no load test on a three phase delta connected induction motor gave: No load power = 890 W, Stator resistance per phase at 30°C = 0.233 Ohms, No load current = 14.5 A. Calculate the fixed losses for the motor.
Model answer: a) Single 500 kVA at 380 kVA load: loss = 750 + (380/500)² × 5410 = 750 + 3124.8 = 3874.8 W. Two transformers each at 190 kVA: loss = 2 × [750 + (190/500)² × 5410] = 2 × 1531.2 = 3062.9 W. Two transformers are better — losses are least, saving 812.4 W. b) Stator copper loss at no load = 3 × (14.5/√3)² × 0.233 = 48.985 W. Fixed losses = 890 − 48.985 = 841 W.
a) Compare total transformer losses (no-load + load-proportional copper loss) for single vs paralleled operation. b) Fixed (iron + friction + windage) losses = no-load input power minus no-load stator copper loss.
📖 §4.7/§4.8 COP vs temperatures (b,c cross-chapter)
4. a) What is the impact of condensing temperature and evaporator temperature on the COP of a refrigeration system? b) Why is it beneficial to operate induction motors in star mode at loads below 50% of rated capacity? c) In a throttle valve-controlled pumping system with oversized pump, name any 3 solutions for improving energy efficiency.
Model answer: a) COP increases with reduction in condensing temperature and with rise in evaporator temperature. b) For motors that consistently operate below 50% of rated capacity, operating in star mode (re-configuring the three phases at the terminal box) reduces voltage by a factor of √3; motor output falls to one-third of the delta value, but performance characteristics as a function of load remain unchanged, so full-load operation in star gives higher efficiency and power factor than partial-load operation in delta. This is only possible where the torque-speed requirement is lower at reduced load. c) Any three of: trim impeller, fit a smaller impeller, install a variable speed drive, use a two-speed motor, use a lower rpm motor.
a) Lower lift (lower condensing, higher evaporator temp) improves COP. b) Star mode reduces applied voltage to better match part-load and improve efficiency/PF. c) Standard remedies for oversized throttle-controlled pumps.
5. a) Define Range, approach and effectiveness in cooling tower operation. b) An induced draft cooling tower is designed for a range of 8°C. The energy auditor finds the operating range as 2°C. What could be the reasons for such a situation?
Model answer: a) Range = difference between cooling tower water inlet and outlet temperature. Approach = difference between cooling tower outlet cold water temperature and ambient wet bulb temperature (a better indicator of performance). Effectiveness (%) = ratio of range to the ideal range = Range/(Range + Approach). b) Possible reasons: excess cooling water flow rate; reduced heat load from the process; some cooling tower cell fans switched off; poor approach due to high humidity; nozzles blocked.
Standard cooling-tower definitions plus typical causes of a much lower-than-design operating range.
6. In an alkali chemical plant, salt brine flowing at 18 m3/hr is cooled from 12°C to 7°C using chilled water. The chiller compressor motor draws 31 kW and total input power to allied accessories is 16 kW. Motor operating efficiency is 90%. Brine density is 1.2 kg/litre and specific heat capacity is 0.97 kCal/kg°C. a) What is the refrigeration load (TR) imposed by the brine cooling? b) What is the COP of the refrigeration compressor? c) What is the overall specific power consumption in kW/TR?
Model answer: a) TR = Q × Cp × (Ti−To)/3024 = (18,000 × 1.2 × 0.97 × (12−7))/3024 = 34.64 TR. b) COP = (3.516 × TR)/(power input to compressor) = 3.516 × 34.64/(31 × 0.9) = 4.365. c) Overall specific power consumption = (31 + 16)/34.64 = 47/34.64 = 1.3568 kW/TR.
a) Refrigeration load from mass flow × specific heat × temp drop / 3024. b) COP converting TR to kW via 3.516 kW/TR over shaft power. c) Total electrical input over refrigeration load.