Energy Efficiency in Electrical Utilities Available here with full solutions — 56 questions recovered from the 2015 exam:
Objective (1 mark)
43 of 50
Short (5 marks)
7 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
📖 §4.3 Types of refrigeration system (vapour compression vs vapour absorption)
1. Which of the following is not a part of vapour compression refrigeration cycle:
compressor
evaporator
condenser
absorber
Answer: D) absorber - The vapour compression cycle consists of compressor, condenser, expansion device and evaporator. An absorber (with generator, pump and solution heat exchanger) belongs to the vapour absorption cycle, where it replaces the compressor.
Learn the two component lists side by side. VCR: compressor -> condenser -> expansion device -> evaporator. VAR: generator + absorber + solution pump + solution heat exchanger -> condenser -> expansion device -> evaporator.
The absorber, together with the generator and pump, is the 'thermal compressor' that REPLACES the mechanical compressor - so it belongs only to the absorption cycle.
Memory hook: every part in the absorption machine that the compression machine lacks (generator, absorber, pump, solution heat exchanger) exists to move refrigerant vapour using heat instead of work.
📖 §1.8 Technical vs commercial losses in distribution (AT&C)
2. Which of the following can be attributed to commercial loss in electrical distribution system
lengthy low voltage lines
low load side power factor
faulty consumer service meters
undersize conductors
Answer: C) faulty consumer service meters - Technical losses arise from the physical network (long LT lines, undersized conductors, low PF). Commercial losses arise from metering, billing and collection deficiencies - defective/faulty consumer meters, meter tampering and theft.
Split the two families cleanly: TECHNICAL losses are physical I^2R and no-load losses - lengthy LT lines, undersized conductors, poor PF, overloaded transformers. COMMERCIAL losses are the metering-billing-collection failures - faulty/stopped/tampered meters, unmetered supply, theft and unrecovered arrears.
Options (a), (b) and (d) are all physical network causes; only the faulty consumer meter is a metering (commercial) defect.
Related formula: AT&C loss % = [1 - (billing efficiency x collection efficiency)] x 100.
📖 §1.5 Transformer losses (constant iron loss vs load-dependent copper loss)
3. Which loss in a distribution transformer is dominating; if the transformer is loaded at 68% of its rated capacity
core loss
copper loss
hysteresis loss
magnetic field loss
Answer: B) copper loss - Core (no-load) loss is constant, copper loss varies as (load fraction)^2 x full-load copper loss. At 68% loading copper loss = 0.68^2 = 0.4624 of full-load copper loss. Since full-load copper loss in a distribution transformer is typically 3-5 times the no-load loss, 0.46 x Cu(FL) still exceeds the iron loss, so copper loss dominates. (Maximum efficiency / equal losses occurs near 40-50% loading for such a transformer.)
Copper loss = (load fraction)^2 x full-load copper loss; iron loss is constant. At 0.68 loading the copper loss is 0.68^2 = 0.46 of its full-load value.
A distribution transformer's full-load copper loss is typically 3-5 times its no-load loss, so 0.46 x Cu(FL) still comfortably exceeds the iron loss - copper loss dominates.
The two decoys: hysteresis loss is a COMPONENT of core loss (so it cannot be a separate answer), and equal losses / maximum efficiency happen nearer 40-50% loading, well below 68%.
📖 §4.8 Factors affecting performance & energy efficiency of refrigeration plants
4. When evaporator temperature is reduced
refrigeration capacity increases
refrigeration capacity decreases
specific power consumption remains same
compressor will stop
Answer: B) refrigeration capacity decreases - Lowering the evaporator temperature lowers suction pressure, so the specific volume of the suction vapour rises and the mass flow handled by the compressor falls. Refrigeration capacity drops (roughly 3-4% per degC) while specific power consumption (kW/TR) rises.
Dropping the evaporator temperature drops the suction pressure; the suction vapour becomes less dense (higher specific volume), so the same compressor swept volume moves LESS refrigerant mass - capacity falls while the compressor work per unit of cooling rises.
So both the capacity and the COP fall, and kW/TR worsens; the book records roughly 3-4% capacity loss per degC of evaporator temperature reduction.
This is the operating rule behind the whole chapter: keep the chilled water/evaporator temperature as HIGH as the process will tolerate, and the condenser as cold as possible.
5. What is the function of drift eliminators in cooling towers
maximize water and air contact
capture water droplets escaping with air stream
enables entry of air to the cooling tower
eliminates uneven distribution of water into the cooling tower
Answer: B) capture water droplets escaping with air stream - Drift eliminators are baffles at the air outlet that change the air direction sharply so that entrained water droplets (drift) impinge and drain back into the tower, minimising drift loss. Maximising air-water contact is the job of the fill; distribution is the job of the nozzles/hot water basin.
Drift eliminators are closely spaced baffles at the air outlet that force the leaving air through sharp direction changes; the heavier water droplets cannot follow, impinge on the blades and drain back. Good cellular PVC eliminators cut drift to about 0.001–0.02% of circulation. Do not confuse the three internals: FILL maximises contact area, LOUVRES/nozzles handle entry and distribution, ELIMINATORS catch carryover.
6. Trivector meter measures three vectors representing
active, reactive and maximum demand
active, power factor and apparent power
active, harmonics and maximum demand
active, reactive and apparent power
Answer: D) active, reactive and apparent power - The trivector meter records the three components of the power triangle: active power (kW/kWh), reactive power (kVAr/kVArh) and apparent power (kVA/kVAh); from these it also derives power factor and maximum demand.
'Tri-vector' = the three sides of the power triangle: active (kW/kWh), reactive (kVAr/kVArh) and apparent (kVA/kVAh). PF and maximum demand are DERIVED from these, they are not among the three vectors.
That is exactly why (a) and (b) are wrong - they list a derived quantity in place of one of the three measured vectors.
The meter integrates demand over a fixed cycle (typically 30 minutes) to record maximum demand for billing.
📖 §1.2 Electricity billing (Time of Day tariff) / §1.9 Demand Side Management
7. Time of the Day metering (TOD) is a way to
reduce the peak demand of the distribution company
increase the revenue of the distribution company
increase the peak demand
increase the maximum demand in a industry
Answer: A) reduce the peak demand of the distribution company - TOD tariffs charge a higher rate during peak hours and a concessional rate during off-peak/night hours, encouraging consumers to shift load away from the system peak. The objective is peak clipping / load levelling for the utility, not revenue maximisation.
TOD tariff prices energy higher during the utility's peak hours and cheaper at night, so the consumer voluntarily shifts shiftable load off-peak. The utility's aim is peak clipping and load levelling - flattening the system load curve - not extra revenue.
For the consumer the benefit is a lower average energy rate; for the utility it defers new generation and transmission capacity.
Memory hook: TOD is a price signal, and the signal always points AWAY from the peak.
📖 §3.2 Compressor types (multi-staging and inter-cooling)
8. The purpose of inter-cooling in a multistage compressor is to
remove the moisture in the air
reduce the work of compression
separate moisture and oil vapour
none of the above
Answer: B) reduce the work of compression - Cooling the air between stages brings it back towards the isothermal line, reducing its specific volume before the next stage and hence the work input. BEE Book-3 notes about 15% power saving from perfect intercooling. Moisture condensation in the intercooler is a useful side-effect, not the purpose.
Compression heats the air; hot air occupies more volume; more volume means more work in the next stage. Cooling between stages brings the process back towards the ISOTHERMAL line, which is the minimum-work path - hence lower power for the same delivered air.
Book figure: perfect inter-cooling can save roughly 15% of the power over single-stage compression to the same pressure, and Table 3.7 shows a 5.5 degC rise in second-stage inlet temperature measurably raising specific power.
Moisture knocked out in the intercooler is a welcome side-effect, not the purpose - that distinction is what the question is testing.
📖 §1.4 Power factor improvement & benefits (I2R loss vs PF)
9. The percentage reduction in distribution loses when tail end power factor raised from 0.85 to 0.95 is
10.1%
19.9%
71%
84%
Answer: B) 19.9% - Distribution I^2R loss varies inversely as the square of power factor. Reduction = [1 - (PF1/PF2)^2] x 100 = [1 - (0.85/0.95)^2] x 100 = [1 - (0.8947)^2] x 100 = (1 - 0.8005) x 100 = 19.95% ~ 19.9%.
For a fixed kW, current I = kW/(sqrt(3) x V x cos(phi)), so I^2R loss varies as 1/PF^2. Percentage loss reduction = [1 - (PF_old/PF_new)^2] x 100.
Working: [1 - (0.85/0.95)^2] x 100 = [1 - 0.8005] x 100 = 19.95% ~ 19.9%.
The mark is lost by using the ratio without squaring it (1 - 0.85/0.95 = 10.5%, which is why 10.1% appears as a distractor).
📖 §8.3 Light source and lamp types (fluorescent tube designations)
10. The nomenclature T2, T5, T8 and T12 for fluorescent lamps 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 - The 'T' number is the tube diameter in eighths of an inch: T12 = 12/8 in = 38 mm, T8 = 8/8 in = 26 mm, T5 = 5/8 in = 16 mm, T2 = 2/8 in = 7 mm. Slimmer tubes (T5) give higher efficacy with electronic ballasts.
T = tube diameter in eighths of an inch, so the whole family scales by diameter: T12 = 38 mm, T8 = 25 mm, T5 = 16 mm, T2 = 6 mm. Length varies independently with wattage, which is why option (c) is wrong. Slimmer tubes concentrate the arc and, on electronic ballasts, give higher efficacy — that is why T5/T8 displaced T12.
📖 §7.2 Cooling tower performance (approach as performance indicator)
11. The indicator of cooling tower performance is best assessed by
wet bulb temperature
dry bulb temperature
range
approach
Answer: D) approach - Approach = cold water temperature - ambient wet bulb temperature. It measures how close the tower brings the water to the theoretical limit (the WBT) and is therefore the true indicator of cooling tower performance/effectiveness. Range is set by the process heat load and circulation rate, not by the tower.
The book states it directly: although both should be monitored, 'Approach' is the better indicator of cooling tower performance. Range is fixed by the process heat load and circulation rate — the tower does not choose it — whereas approach measures how close the tower gets to the wet bulb, the thermodynamic floor. A 2.8 °C approach is about the closest any manufacturer will guarantee.
📖 §8.2 Basic parameters and terms (inverse square law)
12. The illuminance of a lamp at one meter distance is 10 Lm/m2. What will be the corresponding value at 0.7 meter distance
14.28
20.41
10
none of these
Answer: B) 20.41 - Illuminance follows the inverse square law: E2 = E1 x (d1/d2)^2 = 10 x (1/0.7)^2 = 10 x (1/0.49) = 20.408 ~ 20.41 lux.
Inverse square law: E = I/d², or in the handy form E1·d1² = E2·d2². So E2 = 10 × (1/0.7)² = 10/0.49 = 20.41 lux. Move CLOSER and illuminance goes UP, so any answer below 10 is wrong on inspection. The book's own worked example is at half the distance: 10 × (1/0.5)² = 40 lux. Note lm/m² and lux are the same unit.
📖 §5.3 Fan performance evaluation (system characteristics)
13. The fan system resistance is predominately due to
more bends used in the duct
more equipments in the system
volume of air handled
density of air
Answer: A) more bends used in the duct - System resistance is the sum of friction and dynamic (shock) losses. Bends, elbows, transitions and other fittings in the duct contribute the dominant dynamic pressure losses; adding bends shifts the system curve steeply upward.
System resistance = friction losses in straight duct + dynamic (shock) losses at bends, elbows, transitions, dampers and hoods; the shock losses at fittings dominate. Resistance varies as the square of flow, so the system curve is a parabola through the origin. Volume of air is what you push through the resistance, not the cause of it — that is why (c) is the tempting wrong answer.
14. The cooling tower size is _____________ with the entering WBT when heat load, range and approach are constant.
directly proportional
inversely proportional
constant
none of above
Answer: B) inversely proportional - For a fixed heat load, range and approach, a higher entering wet bulb temperature means the air can absorb more heat per unit mass (enthalpy driving potential rises steeply with WBT), so a SMALLER tower suffices. Tower size therefore varies inversely with entering WBT.
Book rule for tower size with the other three held constant: DIRECTLY with heat load, INVERSELY with range, INVERSELY with approach, INVERSELY with entering WBT. Warmer entering air holds far more moisture (enthalpy rises steeply with WBT), so each kg of air carries away more heat and a smaller tower will do. In order of importance for sizing: approach first, then flow rate, with range and wet bulb of lesser importance.
15. The components of two part tariff structure for HT & EHT category consumers are
one part for capacity (or demand) drawn and second part for actual energy drawn
one part for actual Power Factor and second part for actual energy drawn
one part for capacity (or demand) drawn and second part for actual reactive energy drawn
one part for actual apparent energy drawn and second part for actual reactive energy drawn
Answer: A) one part for capacity (or demand) drawn and second part for actual energy drawn - A two-part tariff has a fixed/demand charge based on the contracted or recorded maximum demand in kVA or kW, plus a variable energy charge for the kWh actually consumed. PF is handled separately as a penalty/incentive.
A two-part tariff = a FIXED part on capacity (contract/recorded maximum demand in kVA or kW, charged Rs./kVA/month) + a VARIABLE part on the energy actually consumed (Rs./kWh).
The demand part recovers the utility's capital cost of keeping capacity available; the energy part recovers fuel and running cost.
Power factor is handled OUTSIDE the two parts, as a separate penalty or incentive on the bill - which is why options mentioning PF or reactive energy as a 'part' are wrong.
📖 §3.4 Compressed air system components (air dryers)
16. The adsorption material used in an adsorption air dryer for compressed air is
calcium chloride
magnesium chloride
activated alumina
potassium chloride
Answer: C) activated alumina - Adsorption (desiccant) dryers use solid porous desiccants such as activated alumina, silica gel or molecular sieves, which adsorb moisture on their surface and are then regenerated. Calcium chloride and similar salts are deliquescent absorbents used in absorption dryers, not adsorption dryers.
Adsorption (desiccant) dryers use SOLID porous media that hold water on their surface and are then regenerated - the book names activated alumina and silica gel; molecular sieves are also used. They reach pressure dew points of about -20 degC to -40 degC.
Deliquescent salts such as calcium chloride dissolve as they take up water - that is ABSORPTION, a different dryer type, and the salts are consumed rather than regenerated.
Memory hook: adSORB = surface (solid stays solid); abSORB = soaks in (the medium is used up).
17. The actual measured load of 1000 kVA transformer is 400 kVA. Find out the total transformer loss corresponding to this load if no load loss is 1500 Watts and full load Copper Loss is 12,000 Watts
1920 watts
1500 watts
3420 watt
13500 watts
Answer: C) 3420 watt - Total loss = no-load (iron) loss + (load fraction)^2 x full-load copper loss. Load fraction = 400/1000 = 0.4. Copper loss = 0.4^2 x 12000 = 0.16 x 12000 = 1920 W. Total loss = 1500 + 1920 = 3420 W.
Formula: total loss = no-load (iron) loss + (load fraction)^2 x full-load copper loss.
Working: load fraction = 400/1000 = 0.4; copper loss = 0.4^2 x 12,000 = 0.16 x 12,000 = 1920 W; total = 1500 + 1920 = 3420 W.
Two traps sit in the options: 1920 W (forgetting to add the constant iron loss) and 13,500 W (using the full-load copper loss without scaling by the square of the load fraction). Never scale the iron loss - it is constant from no-load to full load.
reduces voltage by inserting resistance in rotor circuit
reduces voltage by inserting resistance in stator circuit
reduces voltage through a transformer
reduces the supply voltage due to change in connection configuration
Answer: D) reduces the supply voltage due to change in connection configuration - In star connection each winding receives V/sqrt(3) (58%) of line voltage, so starting current and torque fall to about one-third. No resistance or transformer is involved (those describe rotor-resistance, stator-resistance and auto-transformer starters respectively).
In star each winding sees V_line/sqrt(3) = 58% of the line voltage, so the starting current and starting torque both fall to about 1/3 of their direct-on-line values. Nothing is inserted in the circuit - the reduction comes purely from re-configuring the winding connection.
The three wrong options each describe a different starter: rotor-resistance (slip-ring motors), stator-resistance, and auto-transformer starting.
Remember why torque falls as the SQUARE of voltage: T is proportional to V^2, so 0.58^2 = 1/3.
📖 §2.9 Speed control of motors (slip power recovery)
19. Slip power recovery system is applicable in case of
squirrel cage induction motor
wound rotor motor
synchronous motor
DC shunt motor
Answer: B) wound rotor motor - Slip power recovery (static Scherbius / Kramer drive) taps the rotor slip power through slip rings and feeds it back to the supply. This requires access to the rotor winding, which only a wound rotor (slip ring) induction motor provides.
Slip power recovery (static Scherbius / Kramer drive) extracts the slip power from the rotor circuit through slip rings and returns it to the supply instead of wasting it as heat in a rotor resistor.
That requires physical ACCESS to the rotor winding, which only a wound-rotor (slip-ring) induction motor provides - a squirrel cage rotor is short-circuited internally with no terminals to tap.
Efficiency point: with a plain rotor rheostat the slip power (s x rotor input) is burnt as heat; slip power recovery makes sub-synchronous speed control efficient.
20. Rotating magnetic field is produced in a ___________
single-phase induction motor
three-phase induction motor
DC series motor
all of the above
Answer: B) three-phase induction motor - Three balanced currents displaced 120 electrical degrees in three space-displaced windings produce a constant-magnitude rotating field at synchronous speed Ns = 120f/P. A single-phase winding alone produces only a pulsating field (needing an auxiliary winding to start), and a DC motor has a stationary field.
Three balanced phase currents, displaced 120 electrical degrees in time and fed to three windings displaced 120 degrees in space, produce a constant-magnitude field that rotates at Ns = 120f/P. This is the whole basis of the induction motor.
A single-phase winding alone produces only a PULSATING field - it cannot self-start and needs an auxiliary/capacitor winding to create the second phase. A DC machine has a stationary field with a rotating armature.
Memory hook: three phases in space + three phases in time = rotation.
📖 §8.2 Control gear (ballast) & §8.3 Light source and lamp types
21. Power factor is highest in case of
sodium vapour lamps
mercury vapour lamps
fluorescent lamps
incandescent lamps
Answer: D) incandescent lamps - An incandescent lamp is a purely resistive filament load, so its power factor is unity. All discharge lamps (sodium, mercury, fluorescent) need an inductive ballast/choke and operate at a low lagging PF (typically 0.4-0.5 uncompensated).
An incandescent filament is a pure resistance, so voltage and current stay in phase and PF = 1.0. Every discharge lamp — fluorescent, mercury vapour, sodium — needs a current-limiting ballast/choke to counter the arc's negative resistance, and that inductance drags the PF down to roughly 0.4–0.5 unless a capacitor is fitted. Efficacy and power factor are opposite here: the worst lamp for efficacy has the best power factor.
Answer: C) reduction in reactive power - Capacitors supply the magnetising kVAr locally, so the reactive power (and hence the kVA and the line current) drawn from the supply falls. The active power kW required by the load and its active (in-phase) current component are unchanged, so (a), (b) and therefore (d) are wrong.
A shunt capacitor supplies the magnetising kVAr locally, so the REACTIVE power (and hence kVA and line current) drawn from the supply falls. The load's kW requirement and its in-phase (active) current component are untouched.
Hence 'reduction in active current' is wrong: total current falls, but its active component does not.
Benefits chain to remember: less kVAr -> less kVA -> less current -> lower I^2R loss, better voltage regulation, released transformer/cable capacity, no PF penalty.
📖 §2.4 Motor efficiency (slip and rotor losses) / §2.6 Energy efficient motors
23. Motor efficiency will be improved by
reducing the slip
increasing the slip
reducing the diameter of the motor
decreasing the length of the motor
Answer: A) reducing the slip - Rotor copper loss = slip x air-gap power, and motor efficiency is approximately (1 - s) neglecting other losses. Lower slip means lower rotor I^2R loss and higher efficiency; that is why energy-efficient motors run at slightly higher speed (lower slip) than standard motors.
Rotor copper loss = s x air-gap (rotor input) power, so the rotor's own efficiency is (1 - s). Cutting slip directly cuts the rotor I^2R loss and lifts overall efficiency.
That is why an energy-efficient motor runs slightly FASTER (lower slip) than the standard motor it replaces - a useful clue in the exam. Caution for centrifugal loads: that small speed rise can increase the load's power draw as speed^3, partly eating the saving.
Motor diameter and length are design parameters, not operating variables, so (c) and (d) are irrelevant.
Answer: C) higher excitation currents - A lagging load draws magnetising kVAr from the alternator, so the field (excitation) current must be increased to maintain terminal voltage. This raises field copper loss and rotor heating, and derates the alternator. [Note: the printed paper labels the last two options 'c)' and 'c)'; the fourth has been corrected to 'd)'.]
A lagging load draws magnetising kVAr from the alternator, so the field current must be raised to hold terminal voltage. The book notes AC generators are designed for 0.8 lag; lower PF demands higher excitation currents, increases losses and rotor heating, and forces oversizing at lower operating efficiency. The economical fix is capacitors at the load, not a bigger alternator.
Answer: C) water mass flow rate and air mass flow rate - In cooling tower thermodynamics L/G is the liquid-to-gas MASS flow ratio, i.e. kg of water circulated per kg of air. The tower characteristic KaV/L is plotted against L/G; it is a mass ratio, not a volumetric one.
L/G is the liquid-to-gas MASS flow ratio — kg of water circulated per kg of air, typically 0.5 to 2.5. The tower characteristic KaV/L is plotted against L/G, so both terms must be masses for the group to be dimensionless. Option (d) is the deliberate trap: volumetric flows would not work, because air density changes through the tower while mass does not.
📖 §6.2 System characteristics (static and friction head)
26. Installing larger diameter pipe in pumping system results in
increase in static head
decrease in static head
increase in frictional head
decrease in frictional head
Answer: D) decrease in frictional head - Friction head from the Darcy equation h = f L v^2 / (2 g D). For a given flow, a larger diameter reduces velocity as 1/D^2, so friction head falls roughly as 1/D^5. Static head depends only on the elevation/pressure difference and is unaffected by pipe size.
Friction head follows Darcy: hf = f L v²/(2gD). For a fixed flow, velocity falls as 1/D², so friction head falls roughly as 1/D⁵ — doubling the pipe diameter cuts friction loss by a factor of about 32. Static head is pure geometry (the difference in levels plus any vessel pressure) and is completely unaffected by pipe size; that independence is what makes options (a) and (b) impossible.
📖 §2.9 Speed control of motors (variable frequency drives)
27. Installation of Variable frequency drives (VFD) allows the motor to be operated with
constant current
lower start-up current
higher voltage
none of the above
Answer: B) lower start-up current - A VFD starts the motor at low frequency and low voltage (constant V/f), so the motor develops rated torque while drawing near-rated (typically 100-150%) current instead of the 6-7 times DOL inrush. It also gives soft start/stop and speed control.
A VFD starts the motor at low frequency and low voltage on a constant V/f ratio, so full torque is available while the current stays near rated - typically 100-150% - instead of the 6-7 times rated DOL inrush.
Extra benefits: soft start/stop reducing mechanical shock, and speed control which on centrifugal fans and pumps saves power as speed^3 (the affinity laws).
A VFD never raises the supply voltage, and the current it draws varies with load, so (a) and (c) are wrong.
28. In a 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 - At no load the motor still draws the magnetising current, so the wattmeter reading = stator I^2R (copper) loss + core/iron loss + friction and windage loss. Rotor copper loss is negligible because slip is almost zero. Option (c) is incomplete since it omits the stator copper loss that the wattmeter actually measures.
At no load the motor still draws its magnetising current, so the wattmeter reads stator I^2R loss + iron (core) loss + friction & windage loss. Rotor copper loss is negligible because slip is almost zero.
The standard exam step follows straight from this: iron + F&W = no-load input MINUS the no-load stator copper loss (3 x I_phase^2 x R_phase), computed at the measured winding temperature.
Option (c) is the trap - it is right in spirit but incomplete, because it drops the stator copper loss the wattmeter genuinely measures.
📖 §3.4 Compressed air system components (after coolers) / §3.5 moisture removal
29. In a large compressed air system, about 70% to 80% of moisture in the compressed air is removed at the
air dryer
after cooler
air receiver
inter cooler
Answer: B) after cooler - The aftercooler downstream of the last stage cools the discharge air close to ambient, condensing the bulk (about 70-80%) of the moisture, which is drained through a moisture trap. The dryer then removes the remaining moisture to the required pressure dew point.
The after-cooler sits immediately downstream of the last compression stage and drops the discharge air back near ambient; the sharp temperature fall condenses the bulk of the moisture, which is drained through a moisture trap. The dryer then handles only the remainder, down to the required dew point.
Note the exact book figure is 'about 60 to 75% of moisture is removed at the after cooler' - the question's 70-80% is a loose paraphrase, but the after-cooler is unambiguously the right component.
The intercooler cools BETWEEN stages (to cut compression work); the receiver only precipitates a little residual moisture as the air cools further.
📖 §6.6 Flow control strategies (pumps in series and parallel)
30. If two identical pumps operate in series, then the combined shutoff head is
it does not affect head
more than double
doubled
less than double
Answer: C) doubled - In series operation the same flow passes through both pumps and the heads add at every flow point. At shutoff (zero flow) the combined head is therefore exactly twice the shutoff head of one pump.
Series adds HEAD at the same flow; parallel adds FLOW at the same head. At shutoff (zero flow) two identical pumps in series therefore give exactly twice one pump's shutoff head. Do not import the fan rule 'not quite double at a given airflow' — that applies at a working flow point where system resistance intervenes, not at the closed-valve point.
31. If the speed of a reciprocating pump is reduced by 50 %, the head
is reduced by 50%
is reduced by 12.5%
remains same
none of the above
Answer: C) remains same - A reciprocating pump is a positive displacement machine: its discharge is proportional to speed, but the head/pressure it develops is set by the system resistance, not by speed. Halving the speed halves the flow while the head remains essentially unchanged. (The affinity law H proportional to N^2 applies only to centrifugal pumps.)
A reciprocating pump is positive displacement: it delivers a fixed volume per stroke, so flow ∝ speed, but the pressure it develops is whatever the SYSTEM demands (limited only by the relief valve). Halve the speed and you halve the flow while the head sits still. The affinity law H ∝ N² is a centrifugal-machine law only — 12.5% is offered precisely to catch anyone who applies it here.
📖 §3.6 Compressor capacity assessment (temperature correction in the FAD test)
32. If the observed temperature in air receiver is higher than ambient air temperature the correction factor for free air delivery will be:
less than one
greater than one
equal to one
equal to zero
Answer: A) less than one - FAD is referred to ambient (inlet) conditions: FAD = [(P2 - P1)/P0] x (V/t) x (T0/T2), where T0 is the ambient absolute temperature and T2 the receiver air absolute temperature. If T2 > T0 the temperature correction factor T0/T2 is less than one, i.e. the measured volume must be corrected downward.
FAD is defined at the compressor INLET (ambient) conditions, so the pumped-up receiver volume must be corrected by (273 + t_ambient)/(273 + t_receiver).
If the receiver air is hotter than ambient, that ratio is less than 1 - the hot air in the receiver is less dense, so it represents less free air than its volume suggests, and the measured figure must be corrected DOWNWARD.
Two rules for the exam: always use ABSOLUTE temperature (add 273; using degC directly is a guaranteed lost mark), and remember an uncorrected FAD overstates the compressor's true capacity.
📖 §4.7 Performance assessment of refrigeration plants (COP)
33. If the COP of a vapour compression system is 3.5 and the motor draws 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 - Power delivered to the compressor shaft = motor input x motor efficiency = 10.8 x 0.90 = 9.72 kW. Cooling effect = COP x compressor power input = 3.5 x 9.72 = 34.02 kW ~ 34 kW. (Option b, 37.8 kW, is the trap value 3.5 x 10.8 obtained by ignoring motor efficiency.)
COP = cooling effect / power input AT THE COMPRESSOR SHAFT, so convert the motor input first: 10.8 x 0.90 = 9.72 kW.
Cooling effect = 3.5 x 9.72 = 34.02 kW ~ 34 kW (which is 34/3.517 = 9.7 TR, if the question had asked for tonnage).
Option (b) 37.8 kW is exactly 3.5 x 10.8 - the answer you get by ignoring motor efficiency. Whenever a question quotes 'power drawn by the motor' plus a motor efficiency, multiply before using COP.
📖 §4.2 Psychrometrics and air-conditioning processes (humidification) / §4.14 Humidification systems
34. Humidification involves
reducing wet bulb temperature and specific humidity
reducing dry bulb temperature and specific humidity
increasing wet bulb temperature and decreasing specific humidity
reducing dry bulb temperature and increasing specific humidity
Answer: D) reducing dry bulb temperature and increasing specific humidity - In evaporative (adiabatic) humidification water evaporates into the air stream; the latent heat is drawn from the air itself, so the dry bulb temperature falls while the moisture content (specific humidity) rises. The process follows the constant wet-bulb / constant-enthalpy line on the psychrometric chart.
Adiabatic (evaporative) humidification takes the latent heat of vaporisation out of the air stream itself: the moisture content (specific humidity) RISES and the dry bulb temperature FALLS, while enthalpy and wet-bulb temperature stay essentially constant.
On the chart it is a move up-and-left along a constant wet-bulb line towards the saturation curve.
Do not confuse it with steam humidification, which adds moisture at nearly constant dry-bulb temperature and raises the enthalpy - the exam wording 'humidification' in a textile air washer always means the adiabatic case.
lower evaporator temperature and higher condenser temperature
higher evaporator temperature and lower condenser temperature
higher evaporator temperature and higher condenser temperature
lower evaporator temperature and lower condenser temperature
Answer: B) higher evaporator temperature and lower condenser temperature - COP(Carnot) = Te/(Tc - Te). Raising Te and lowering Tc both increase the numerator and shrink the temperature lift (Tc - Te), so the COP rises. This is why chilled water temperature should be kept as high as the process allows and condenser/cooling water as cold as possible.
COP_Carnot = T_evap/(T_cond - T_evap), with both temperatures in KELVIN. Raising T_evap increases the numerator AND shrinks the lift; lowering T_cond shrinks the lift again - both push COP up.
Worked feel for the numbers: at 5 degC evaporator and 40 degC condenser, COP = 278/35 = 7.9; drop the condenser to 35 degC and it becomes 278/30 = 9.3, an 18% gain for 5 degC.
Using degC instead of Kelvin here is the classic destroyed answer (5/35 = 0.14 is meaningless). Operationally: run chilled water as warm as the process allows and keep condenser/cooling water as cold as possible - clean tubes, clean cooling tower.
36. Flow control by damper operation in fan system will
increase energy consumption
reduce energy consumption
reduce system resistance
none of the above
Answer: A) increase energy consumption - A damper throttles flow by ADDING artificial resistance, moving the operating point up the fan curve. The energy consumed per unit of air delivered (specific energy consumption, kW per m3/hr) therefore rises, and the throttling energy is wasted as pressure drop. Option (c) is plainly wrong, since a damper increases, not reduces, system resistance; compared with speed control (VFD), damper control is the energy-wasting method.
A damper does not slow the fan; it adds artificial resistance so the operating point rides UP the fan curve to a higher pressure and lower flow. Power falls a little but kW per m³/hr rises, and the throttled pressure drop is pure waste. Compare with a VFD, where power falls as N³. Hook: 'a damper is a brake you pay electricity to hold on'.
37. Find the Total Harmonic Distortion (THD) for current for the following current readings. Current at 50 Hz fundamental frequency = 250 A, Third harmonic current = 50 A, fifth harmonic current = 35 A
58 %
48 %
24%
34 %
Answer: C) 24% - THD(i) = sqrt(I3^2 + I5^2 + ...) / I1 x 100 = sqrt(50^2 + 35^2)/250 x 100 = sqrt(2500 + 1225)/250 x 100 = sqrt(3725)/250 x 100 = 61.03/250 x 100 = 24.4% ~ 24%.
THD_i = sqrt(I3^2 + I5^2 + ...) / I1 x 100, always referred to the 50 Hz fundamental.
Working: sqrt(50^2 + 35^2)/250 x 100 = sqrt(2500 + 1225)/250 x 100 = 61.03/250 x 100 = 24.4% ~ 24%.
The 34% distractor is what you get by adding the harmonic currents (50 + 35 = 85) instead of root-sum-squaring them - square first, add, then take the root.
📖 §5.6 Fan performance assessment (gas density calculation)
38. Calculate the density of air at 11400 mmWC absolute pressure and 65 degC. (Molecular weight of air: 28.92 kg/kg mole and Gas constant: 847.84 mmWC m3/kg mole K)
1.2 kg/m3
1.5 kg/m3
1.15 kg/m3
none of the above
Answer: C) 1.15 kg/m3 - Density = (P x M)/(R x T) where P = 11400 mmWC, M = 28.92 kg/kg mole, R = 847.84 mmWC m3/kg mole K and T = 65 + 273 = 338 K. Density = (11400 x 28.92)/(847.84 x 338) = 329,688/286,569 = 1.1504 ~ 1.15 kg/m3.
Density formula to memorise: ρ = (P × M)/(R × T), with P in mmWC absolute, M = 28.92 kg/kg-mole for air, R = 847.84 mmWC·m³/kg-mole·K and T in KELVIN. Here ρ = (11400 × 28.92)/(847.84 × 338) = 329,688/286,569 = 1.15 kg/m³. The mark is lost by leaving T at 65 instead of 338 K, or by using gauge instead of absolute pressure.
📖 §9.1 Introduction (spark ignition vs compression ignition)
39. A spark ignition engine is used for firing which type of fuels:
high speed diesel
light diesel oil
natural gas
furnace oil
Answer: C) natural gas - Spark ignition (Otto cycle) engines need a volatile, high-octane fuel that will not pre-ignite under compression - petrol, natural gas, LPG, biogas. Diesel, LDO and furnace oil are compression-ignition (Diesel cycle) fuels.
Spark ignition (Otto cycle) needs a volatile, high-octane fuel that will not self-ignite under compression — petrol, natural gas, LPG, biogas. Compression ignition (Diesel cycle) relies on the fuel igniting spontaneously in hot compressed air, which suits HSD, LDO and furnace oil. Hook: 'spark = gas, compression = diesel'; CI engines run higher compression ratios and so achieve higher thermal efficiency.
📖 §10.8 ECBC guidelines on lighting (Lighting Power Density)
40. A hotel building has four floors each of 1000 m2 area. If the interior lighting power allowance for the hotel building is 43000 W, the Lighting Power Density (LPD) is:
10.75
0.09
43
data insufficient
Answer: A) 10.75 (W/m2) - LPD = total interior lighting power / total lit floor area = 43,000 W / (4 floors x 1000 m2) = 43,000/4,000 = 10.75 W/m2. ECBC prescribes LPD limits by building type using exactly this calculation.
LPD = interior lighting power (W) ÷ total lit floor area (m²) = 43,000/(4 × 1000) = 10.75 W/m². The trap is forgetting to multiply by the number of floors — using 1000 m² gives 43 W/m², which is option (c). ECBC works the other way round in practice: multiply the code's LPD limit for the building type by the area to get the allowed lighting power, and the design must not exceed it.
📖 §1.4 Power factor improvement (capacitor output vs applied voltage)
41. A company installed a 130 kVAr, 600 Volt capacitor but the power meter indicates that it is only operating at 119 kVAr. The reason out of the following could be
operating at low load
high voltage
low voltage
low current
Answer: C) low voltage - Capacitor output kVAr = 2 pi f C V^2, i.e. proportional to the square of applied voltage. Delivered/rated = 119/130 = 0.9154, so V/Vrated = sqrt(0.9154) = 0.957, i.e. about 574 V against the rated 600 V. The capacitor is under-delivering because it is operating at a lower than rated voltage; capacitor kVAr does not depend on the load.
Capacitor kVAr = 2(pi)fCV^2, so output falls with the square of the voltage - the capacitance itself has not changed.
Check: 119/130 = 0.9154, so V/V_rated = sqrt(0.9154) = 0.957, i.e. about 574 V on a 600 V unit. A 4% voltage shortfall costs ~8% of the kVAr.
Capacitor output is independent of the connected load, so 'operating at low load' and 'low current' cannot explain the shortfall.
📖 §2.4 Motor efficiency (efficiency from nameplate data)
43. A 22 kW, 415 kV, 45 A, 0.8 PF, 1475 RPM, 4 pole 3 phase induction motor operating at 420 V, 40 A and 0.8 PF. What will be the rated efficiency
85.0%
94.5%
89.9%
88.2%
Answer: A) 85.0% - Rated efficiency uses the NAMEPLATE values (the 420 V/40 A figures describe the present operating point and are a distractor). Rated input = 1.732 x 415 x 45 x 0.8 = 25,875.7 W = 25.88 kW. Rated output = 22 kW. Efficiency = 22/25.88 x 100 = 85.02% ~ 85.0%. (The nameplate '415 kV' is an obvious misprint for 415 V.)
RATED efficiency uses nameplate values only - the 420 V / 40 A figures describe the present operating point and are pure distraction.
Rated input = sqrt(3) x V x I x PF = 1.732 x 415 x 45 x 0.8 = 25,876 W = 25.88 kW; efficiency = 22/25.88 x 100 = 85.0%.
Traps: dropping the sqrt(3) (gives 147%, obviously absurd - a sanity check worth doing), and reading '415 kV' literally, which is a misprint for 415 V.
📖 §6.2 System characteristics (static and friction head)
1. The total system resistance of a piping loop is 50 meters and the static head is 15 meters at designed water flow. Calculate the system resistance offered at 75%, 50% and 25% of water flow
Model answer: Solution:
Total system resistance of piping loop = 50 m; Static head = 15 m.
So dynamic (friction) head at designed water flow = 50 - 15 = 35 m. ..... (2 marks)
Static head is independent of flow; dynamic head varies as the square of flow: Dynamic head = 35 x (% flow)^2.
At 75% flow: dynamic head = 35 x 0.75^2 = 19.68 m; total resistance = 19.68 + 15 = 34.68 m
At 50% flow: dynamic head = 35 x 0.50^2 = 8.75 m; total resistance = 8.75 + 15 = 23.75 m
At 25% flow: dynamic head = 35 x 0.25^2 = 2.19 m; total resistance = 2.19 + 15 = 17.19 m
..... (3 marks - 1 mark each)
Split the head first: friction (dynamic) = 50 − 15 = 35 m, static = 15 m and static NEVER changes with flow. Only the friction part scales with flow squared: 75% → 35 × 0.5625 = 19.7, total 34.7 m; 50% → 35 × 0.25 = 8.75, total 23.75 m; 25% → 35 × 0.0625 = 2.19, total 17.19 m. Squaring the whole 50 m instead of just the 35 m is the standard error and gives 28.1 m at 75% flow.
📖 §9.3 Waste heat recovery & §9.4 Energy performance assessment of DG sets
2. In a DG set, the generator is rated at 1000 kVA, 415V, 1390 A, 0.8 PF, 1500 RPM. The full load specific energy consumption of this DG set as measured by the energy auditor is 4.0 kWh per liter of fuel and air drawn by the DG set is 14 kg/kg of fuel. The energy auditor has recommended a waste heat recovery (WHR) system. Also the auditor indicated that the waste heat recovery potential is 2.6x10^5 kCal/hr at the existing engine exhaust gas temperature of 583 degC. Estimate the exhaust temperature to chimney after installation of proposed WHR system. The specific gravity of fuel oil is 0.86 and specific heat of flue gas is 0.25 kCal/kg degC.
Model answer: Solution:
1. Rated kVA of diesel generator (given) = 1000
2. Rated kW at 0.8 PF = 1000 x 0.8 = 800 kW ..... (0.5 mark)
3. Specific fuel consumption (given) = 4 kWh/litre
4. Specific gravity of fuel oil (given) = 0.86
5. Oil consumption at full load = (800 x 0.86)/4 = 172 kg/hr ..... (1 mark)
6. Air supplied per kg of fuel (given) = 14 kg
7. Mass of flue gas = 14 + 1 = 15 kg per kg of fuel
8. Mass of flue gas per hour = 15 x 172 = 2580 kg/hr ..... (1 mark)
9. Waste heat recovery potential (given) = 2,60,000 kCal/hr
10. Delta T across the WHR system = Heat (kCal/hr)/(mass of flue gas kg/hr x specific heat kCal/kg degC) = 260000/(2580 x 0.25) = 403 degC ..... (1.5 marks)
11. Flue gas temperature before WHR system (given) = 583 degC
12. Exit flue gas temperature to chimney after WHR system = 583 - 403 = 180 degC ..... (1 mark)
Chain the mass balance before touching the heat. kW = 1000 kVA × 0.8 = 800 kW; fuel = 800/4 = 200 litres/hr × 0.86 = 172 kg/hr; flue gas = (14 + 1) × 172 = 2580 kg/hr — the +1 is the fuel itself and is the step most candidates forget. Then ΔT = 260,000/(2580 × 0.25) = 403 °C, so the chimney temperature is 583 − 403 = 180 °C. Sanity-check the result: the book warns against dropping below roughly 180 °C because of cold-end corrosion.
📖 §5.3 Fan laws & §5.5 Flow control strategies (pulley change)
3. The input power to a fan is 30 kW for a 2500 Nm3/hr fluid flow. The fan pulley diameter is 300 mm. If the flow to be reduced by 15% by changing the fan pulley, what should be the diameter of fan pulley and power input to fan.
Model answer: Solution:
1. Input power to fan = 30 kW
2. Fluid flow = 2500 Nm3/hr
3. Diameter of fan pulley = 300 mm
4. Governing equation is N1 D1 = N2 D2 ..... Eqn-1 ..... (1 mark)
5. Flow is proportional to speed, and flow is reduced by 15%, so N2 = 0.85 N1
6. From Eqn-1: D2 = D1 x (N1/N2) = 300 x (N1/0.85 N1) = 352 mm ..... (2 marks)
7. Power varies as the cube of speed: (kW1/kW2) = (N1^3/N2^3), hence kW2 = (N2/N1)^3 x kW1 = (0.85)^3 x 30 = 18.42 kW ..... (2 marks)
So the power requirement for the fan will be 18.4 kW and the fan pulley is to be changed to 352 mm diameter.
Fan laws: Q ∝ N, SP ∝ N², kW ∝ N³. Flow down 15% ⇒ N2 = 0.85 N1, so power = 0.85³ × 30 = 0.614 × 30 = 18.42 kW. For the pulley use N1 D1 = N2 D2 on the FAN (driven) pulley: D2 = 300 × (1/0.85) = 352 mm — the driven pulley gets BIGGER to run slower. Shrinking it to 255 mm is the standard trap; and using ×2 (N²) instead of ×3 (N³) for power gives 21.7 kW and loses the mark.
📖 §8.2 Basic parameters and terms (lux and luminous efficacy)
4. Define Lux and Luminous efficacy
Model answer: Answer:
Lux (lx) is the illuminance produced by a luminous flux of one lumen, uniformly distributed over a surface area of one square meter. It is also defined as the International System unit of illumination, equal to one lumen per square meter. ..... (2.5 marks)
Luminous efficacy is defined as the ratio of luminous flux emitted by a lamp to the power consumed by the lamp (lumens per watt). Efficacy is the energy efficiency of conversion from electricity to light form. ..... (2.5 marks)
Two definitions to write out exactly: LUX is illuminance, one lumen uniformly distributed over one square metre (lm/m²) — a property of the lit SURFACE. LUMINOUS EFFICACY is lumens emitted by the lamp per watt consumed by the lamp (lm/W) — a property of the SOURCE. The mark is lost by quoting lumens when the question asks for lux; note also lamp CIRCUIT efficacy includes ballast losses, so it is always the lower number.
5. During an energy audit of a power plant cooling tower, the following observations were made. Power plant generation = 785 MW; Circulation rate = 107000 m3/hr; Cooling tower range = 10.5 degC; Power plant design COC value = 3.8. As an auditor find out a) The total water consumption per hour, b) Specific water consumption in m3/MW generation. The plant is pursuing an up-gradation treatment plan to increase COC to 7.0. c) What would be the potential water savings in m3/hr and m3/MW generation?
Model answer: Answer:
1. Evaporation loss = 0.00085 x circulation rate (m3/hr) x (CT range in degC) x 1.8 = 0.00085 x 107000 x 10.5 x 1.8 = 1719 m3/hr ..... (0.5 mark)
2. Blow-down loss = Evaporation loss/(COC - 1) = 1719/(3.8 - 1) = 614 m3/hr ..... (0.5 mark)
3. Total as-run hourly consumption = 1719 + 614 = 2333 m3/hr ..... (0.5 mark)
4. Specific water consumption = 2333/785 = 2.97 m3/MW ..... (0.5 mark)
5. Blow-down at improved COC of 7.0 = 1719/(7 - 1) = 286.5 m3/hr ..... (0.5 mark)
6. Total water consumption at improved COC = 1719 + 286.5 = 2005.5 m3/hr ..... (0.5 mark)
7. Specific water consumption at improved COC = 2005.5/785 = 2.56 m3/MW ..... (0.5 mark)
8. Total water saving per hour = 2333 - 2005.5 = 327.5 m3/hr ..... (0.5 mark)
9. Water saving per MW generation = 327.5/785 = 0.417 m3/MW ..... (1 mark)
Evaporation = 0.00085 × 1.8 × circulation × range = 0.00085 × 1.8 × 107,000 × 10.5 = 1719 m³/hr and it does NOT change when COC changes — only blowdown does. Blowdown = 1719/(3.8 − 1) = 614 m³/hr, total 2333 m³/hr, specific 2333/785 = 2.97 m³/MW. At COC 7.0 blowdown falls to 1719/6 = 286.5, total 2005.5 m³/hr, so the saving is 327.5 m³/hr or 0.417 m³/MW. Raising COC saves blowdown water only — quoting a saving in evaporation is wrong.
📖 §4.7 Performance assessment of refrigeration plants (COP Carnot vs actual COP)
6. Explain with equation for COP(Carnot) that: (a) higher COP(Carnot) is achieved with higher evaporator temperature and lower condenser temperature. (b) COP(Carnot) does not take into account the type of compressor. (c) How is the COP normally used in the industry given?
Model answer: Answer:
a) The theoretical Coefficient of Performance (Carnot), COP(Carnot) - a standard measure of refrigeration efficiency of an ideal refrigeration system - depends on two key system temperatures, namely the evaporator temperature Te and the condenser temperature Tc, with COP being given as: COP(Carnot) = Te/(Tc - Te). ..... (2 marks)
b) This expression also indicates that a higher COP(Carnot) is achieved with a higher evaporator temperature and a lower condenser temperature (raising Te increases the numerator and reduces the lift Tc - Te; lowering Tc reduces the lift). But COP(Carnot) is only a ratio of temperatures, and hence does not take into account the type of compressor. ..... (2 marks)
c) Hence the COP normally used in the industry is given by COP = [Cooling effect (kW)/Power input to compressor (kW)], where the cooling effect is the difference in enthalpy across the evaporator, expressed as kW. ..... (1 mark)
Write the equation first: COP_Carnot = Te/(Tc - Te), both in KELVIN. Raising Te lifts the numerator and cuts the lift (Tc - Te); lowering Tc cuts the lift - so both moves raise the COP.
Part (b) argument: the expression contains only two temperatures, so it is blind to the machine's hardware - the type of compressor (reciprocating, screw, centrifugal), its isentropic efficiency and its part-load behaviour do not appear anywhere in it. COP_Carnot is therefore a ceiling, not a prediction.
Part (c): the industry figure is COP = cooling effect (kW) / power input to the compressor (kW), where the cooling effect is the enthalpy difference across the evaporator; its reciprocal-style cousin is kW/TR, the standard plant KPI.
📖 §1.4 Power factor improvement (capacitor sizing)
7. The following single line diagram depicts the location of a 100 kW heater load and a 200 kW motor (which is 200 metres away from the 415V, LT bus). The main incoming line power factor of the system is 0.85 lag. Calculate the rating of capacitors to improve PF of main incoming line to 0.9 lag. [refers to a figure in the original paper]
Model answer: Answer:
Total inductive load requiring PF compensation = 200 kW (since the other 100 kW heater is a resistive load and already operates at unity power factor). ..... (1 mark)
Operating PF cos(phi1) = 0.85 lag; Desired PF cos(phi2) = 0.90 lag
kVAr required = kW x [tan(cos^-1 phi1) - tan(cos^-1 phi2)] ..... (1 mark)
= 200 x [tan(cos^-1 0.85) - tan(cos^-1 0.90)]
= 200 x [tan(31.78 deg) - tan(25.84 deg)]
= 200 x (0.619 - 0.484)
= 200 x 0.135
= 27 kVAr ..... (3 marks)
Only the INDUCTIVE kW gets compensated: the 100 kW heater is resistive and already at unity PF, so it contributes no kVAr and must be excluded. Compensate the 200 kW motor alone.
kVAr = kW[tan(cos^-1 PF1) - tan(cos^-1 PF2)] = 200 x (tan 31.78deg - tan 25.84deg) = 200 x (0.619 - 0.484) = 27 kVAr.
Including the heater's 100 kW would give 40.5 kVAr and lose the marks. Memory hook: capacitors correct only what is magnetising, and a heater magnetises nothing.
📖 §4.7 Performance assessment of refrigeration plants (chiller vs VAM comparison)
1. Compare the performance of centrifugal chiller with vapour absorption chiller using the data given below: (1) Chilled water flow (m3/h): Centrifugal Chiller 192, VAM 183; (2) Condenser water flow (m3/h): 245, 360; (3) Chiller inlet water temperature (degC): 13, 14.5; (4) Condenser water inlet temperature (degC): 28, 32; (5) Chiller outlet water temperature (degC): 7.8, 9.2; (6) Condenser water outlet temperature (degC): 36.2, 40.7; (7) Chilled water pump consumption (kW): 32, 31; (8) Condenser water pump consumption (kW): 38, 52; (9) Cooling tower fan consumption (kW): 9, 22. If the compressor of centrifugal chiller consumes 205 kW, the steam consumption for VAM is 1620 kg/Hr. Calculate the following: i) Refrigeration load delivered (TR) for both systems? ii) Condenser Heat load (TR) for both systems? iii) Compare auxiliary power consumption for both systems, give reason? iv) If electricity cost is Rs.4.0/kWh and steam cost is Rs.0.45/kg compare the operating cost for both systems.
Model answer: Solution: Compression Chiller vs. VAM
1. Refrigeration load delivered (TR) = mass of chilled water flow x specific heat x delta T of chilled water = flow (m3/hr) x 1000 kg/m3 x 1 kcal/kg degC x (inlet temp - outlet temp)/3024
Centrifugal chiller = 192 x 1000 x (13 - 7.8)/3024 = 330.16 TR
VAM = 183 x 1000 x (14.5 - 9.2)/3024 = 320.73 TR ..... (2 marks)
2. Condenser heat load (TR) = mass of condenser water flow x specific heat x delta T of condenser water = flow (m3/hr) x 1000 kg/m3 x 1 kcal/kg degC x (outlet temp - inlet temp)/3024
Centrifugal chiller = 245 x 1000 x (36.2 - 28)/3024 = 664.35 TR
VAM = 360 x 1000 x (40.7 - 32)/3024 = 1035.71 TR ..... (2 marks)
3. Auxiliary power consumption (kW) = chilled water pump + condenser water pump + cooling tower fan
Centrifugal chiller = 32 + 38 + 9 = 79 kW
VAM = 31 + 52 + 22 = 105 kW
The auxiliary power consumption in case of the VAM system is higher because heat rejection in the VAM condenser is comparatively higher than the centrifugal chiller with approximately similar cooling load. ..... (2 marks)
4. Total energy consumption:
Centrifugal chiller = 284 kW (auxiliary power of 79 kW plus chiller consumption of 205 kW)
VAM = auxiliary power of 105 kW and steam consumption of 1620 kg/hr ..... (2 marks)
5. Operating energy cost per hour of operation:
Centrifugal chiller = 284 x 4 = Rs. 1136/-
VAM = (105 x 4 = Rs. 420/-) plus (1620 x 0.45 = Rs. 729/-) = Rs. 1149/- ..... (2 marks)
Both loads come from the same water-side heat balance: TR = flow (m3/hr) x 1000 x 1 x deltaT/3024. Chilled side: 192 x 1000 x 5.2/3024 = 330.2 TR (centrifugal) and 183 x 1000 x 5.3/3024 = 320.7 TR (VAM). Condenser side: 245 x 1000 x 8.2/3024 = 664.4 TR and 360 x 1000 x 8.7/3024 = 1035.7 TR.
The whole point of the question is in those condenser figures: a VAM must reject the evaporator load PLUS the driving heat input, so for a similar cooling duty it rejects far more heat - hence bigger cooling towers, bigger condenser pumps and higher auxiliary power (105 kW against 79 kW).
Cost comparison must put both on the same footing: centrifugal (205 + 79) x Rs.4 = Rs.1136/hr; VAM 105 x 4 + 1620 x 0.45 = 420 + 729 = Rs.1149/hr. Remember to convert the delta-T for the condenser as outlet MINUS inlet, and for the chiller as inlet MINUS outlet.
📖 §4.12 Ventilation systems (ACH) + §4.7 Performance assessment (kW/TR)
2. a) Calculate the ventilation rate for an engine room of 20 m length, 10.5 m width and 15 m height; if the recommended Air Changes per Hour (ACH) is 20. b) Air at 25,200 m3/hr and at 1.2 kg/m3 density is flowing into an air handling unit of an inspection room. The enthalpy difference between the inlet and outlet air is 2.38 kcal/kg. If the motor draws 22 kW with an efficiency of 90%, find out the kW/TR of the refrigeration system. (1 cal = 4.183)
Model answer: Solution:
a) Ventilation rate:
Room length = 20 m; Room height = 15 m; Room width = 10.5 m; Air changes per hour (ACH) = 20
Ventilation rate (m3/hr) = length x height x width x ACH = 20 x 15 x 10.5 x 20 = 63,000 m3/hr ..... (5 marks)
b) Refrigeration load:
Heat load = Q x density x (h2 - h1) = 25,200 x 1.2 x 2.38 = 71,971 kcal/hr ..... (2 marks)
TR = 71,971/3024 = 23.8 TR ..... (1 mark)
Power input to the compressor = 22 x 0.9 = 19.8 kW ..... (1 mark)
kW/TR = 19.8/23.8 = 0.83 ..... (1 mark)
Ventilation rate = room volume x air changes per hour = (20 x 10.5 x 15) x 20 = 3150 x 20 = 63,000 m3/hr. ACH is per HOUR, so no further time conversion is needed.
For part (b): heat load = Q x rho x delta-h = 25,200 x 1.2 x 2.38 = 71,971 kcal/hr, and TR = 71,971/3024 = 23.8 TR. Compressor shaft power = 22 x 0.9 = 19.8 kW, so kW/TR = 19.8/23.8 = 0.83.
Two guards: the enthalpy difference is already in kcal/kg here so the 4.187 conversion is not needed, and kW/TR uses the SHAFT power, not the motor input - a lower kW/TR is better, with 0.7-0.9 typical for a good centrifugal plant.
3. In a dairy plant 3 numbers of cooling water pumps, identical in characteristics are installed in parallel to supply cooling. During normal operation two of the pumps are operational while one pump is on standby. All pump combinations develop a discharge pressure of 3.4 kg/cm2 (a). The installed water flow meter at the common header during an energy audit reads the following: Pump No 1 & 2 = 545 m3/hr; Pump No 2 & 3 = 535 m3/hr; Pump No 3 & 1 = 550 m3/hr. The power drawn by motors of cooling water pump 1, 2 & 3 are 33 kW, 31.5 kW & 32.5 kW respectively. While the operating motor efficiency for pump no. 1 & 2 is 92% the motor efficiency for pump no. 3 is 91.5%. If the water level in suction of all pumps is 3 meter below pump central line. Calculate the following: i) Individual pump efficiencies ii) Specific energy consumption (kWh/m3) iii) Which is the best operating pump combination
Model answer: Solution:
Let the flows of pumps 1, 2 and 3 be X, Y and Z respectively.
X + Y = 545 ---(1); Y + Z = 535 ---(2); X + Z = 550 ---(3)
Subtracting (2) from (1): X - Z = 10 ---(4)
Adding (3) and (4): 2X = 560, so X = 280
Putting X in (1) and (2): Y = 265 and Z = 270
Therefore individual pump flow rates are 280 m3/hr, 265 m3/hr and 270 m3/hr respectively. ..... (3 marks)
Pump-wise calculation (Pump 1 / Pump 2 / Pump 3):
A) Flow rate (m3/hr) (calculated): 280 / 265 / 270
B) Discharge head = 3.4 kg/cm2 (a) = 2.4 kg/cm2 (g) = 24 m: 24 / 24 / 24
C) Suction head (m) (given): -3 / -3 / -3
D) Total head = discharge head - suction head = (B - C): 27 / 27 / 27 ..... (1 mark)
E) Liquid (hydraulic) kW = flow (m3/s) x total head (m) x density (1000 kg/m3) x 9.81/1000: 20.60 / 20.22 / 19.87 ..... (2 marks)
F) Power drawn by motor kW (given): 33 / 31.5 / 32.5
G) Motor efficiency % (given): 92.0% / 92.0% / 91.5%
H) Pump input (shaft) power kW = F x G: 30.36 / 28.98 / 29.74 ..... (1 mark)
I) Pump efficiency % = E/H: 67.9% / 69.8% / 66.8% ..... (1 mark)
J) Specific energy consumption (kWh/m3) = F/A: 0.118 / 0.119 / 0.120 ..... (1 mark)
Pump No. 1 & 2 are the best performing operating combination. ..... (1 mark)
Note: The total head has been calculated by subtracting the suction gauge pressure from the discharge gauge pressure. The candidates can also calculate total head as the difference of absolute pressures as follows: Discharge head = 3.4 kg/cm2 (a); Suction head = 1 - 0.3 = 0.7 kg/cm2; Total head developed = 3.4 - 0.7 = 2.7 kg/cm2 = 27 m.
Get the individual flows by simultaneous equations: X+Y=545, Y+Z=535, X+Z=550 → X=280, Y=265, Z=270 m³/hr. Then head is common to all three: 3.4 kg/cm²(a) = 2.4 kg/cm²(g) = 24 m, plus 3 m of suction lift = 27 m. Hydraulic kW = Q(m³/s) × 27 × 9.81, shaft = measured kW × motor η, so η(pump) = 67.9 / 69.8 / 66.8%. Choose the best COMBINATION on specific energy consumption (kW ÷ m³/hr), not on single-pump efficiency — pumps 1 and 2 win at 0.118 kWh/m³. Subtracting a gauge from an absolute pressure is the classic slip; keep both in the same reference.
📖 §7.2 Cooling tower performance & §5.6 Fan performance assessment
4. a) In a chemical industry, cooling water of 9000 m3/hr and 6000 m3/hr from two independent heat exchangers with temperature of 41 degC and 52 degC respectively are fed to one cooling tower after proper mixing at top basin. If measured heat rejection by the cooling tower is 45,000 TR, calculate effectiveness and evaporation loss of the cooling tower at 31 degC WBT. b) In an air conditioning duct 0.5 m x 0.5 m, the average velocity of air measured by vane anemometer is 28 m/s. The static pressure at suction of the fan is -20 mmWC and at the discharge is 30 mmWC. A three phase induction motor draws 10.8 A at 415 V with a power factor of 0.9. Find out efficiency of the fan if motor efficiency = 88% (neglect density correction)
Model answer: Solution:
a) Cooling tower:
1. Flow rates (given): Stream 1 = 9000 m3/hr, Stream 2 = 6000 m3/hr
2. Temperatures (given): Stream 1 = 41 degC, Stream 2 = 52 degC
3. Mixed flow rate = 9000 + 6000 = 15,000 m3/hr
4. Mixed hot water temperature = [(Flow1 x Temp1) + (Flow2 x Temp2)]/(Flow1 + Flow2) = [(9000 x 41) + (6000 x 52)]/15000 = 45.4 degC ..... (1 mark)
5. Heat rejection (given) = 45,000 TR
6. Range of cooling tower = (Heat rejection TR x 3024)/(Flow m3/hr x 1000) = (45000 x 3024)/(15000 x 1000) = 9.072 degC ..... (1 mark)
7. WBT (given) = 31 degC
8. Cold water temperature = mixed hot water temperature - range = 45.4 - 9.072 = 36.328 degC ..... (0.5 mark)
9. Approach = cold water temperature - WBT = 36.328 - 31 = 5.328 degC ..... (0.5 mark)
10. Effectiveness = Range/(Range + Approach) = 9.072/(9.072 + 5.328) = 63% ..... (1 mark)
11. Evaporation loss (m3/hr) = 0.00085 x 1.8 x mixed flow m3/hr x range = 0.00085 x 1.8 x 15000 x 9.072 = 208.2 m3/hr ..... (1 mark)
b) Fan efficiency:
1. Area of the duct = 0.5 x 0.5 = 0.25 m2
2. Average velocity (given) = 28 m/s
3. Air flow = 0.25 x 28 = 7 m3/s ..... (1 mark)
4. Suction static pressure (given) = -20 mmWC
5. Discharge static pressure (given) = 30 mmWC
6. Power drawn by the motor = 1.732 x 415 x 10.8 x 0.9/1000 = 6.99 kW ..... (1 mark)
7. Air power kW = flow (m3/s) x (discharge pressure - suction pressure) mmWC/102 = 7 x (30 - (-20))/102 = 7 x 50/102 = 3.43 kW ..... (1 mark)
8. Power to fan shaft = motor drawn power x motor efficiency of 88% = 6.99 x 0.88 = 6.15 kW ..... (1 mark)
9. Fan static efficiency = air power x 100/shaft input = 3.43 x 100/6.15 = 55.76% ..... (1 mark)
(a) Mix the two streams on a flow-weighted basis first: (9000×41 + 6000×52)/15000 = 45.4 °C — a plain average of 46.5 °C is wrong. Range from the duty: 45,000 TR × 3024/(15,000 × 1000) = 9.07 °C, so cold water = 36.33 °C, approach = 5.33 °C and effectiveness = 9.07/(9.07+5.33) = 63%. Evaporation = 0.00085 × 1.8 × 15,000 × 9.07 = 208 m³/hr. (b) Fan total pressure = 30 − (−20) = 50 mmWC; SUBTRACT the negative suction pressure so it adds. Air power = 7 × 50/102 = 3.43 kW, shaft = 6.99 × 0.88 = 6.15 kW, η = 55.8%.
📖 §7.2 Cooling tower performance (cooling capacity and water flow)
5. One of the process industries has installed 18 MW cogeneration plant. The Cogeneration plant maximum condenser load is 7 MW and the extraction steam of 57 TPH is used for process and also for vapour absorption machine. The condenser heat load is 550 Kcal/kg of steam and the steam rate is 5 kg/kW for condenser power. The heat load of VAM is 127 Kcal/min/TR and the capacity of VAM is 1100 TR. Estimate cooling tower heat load in Kcal/hr. If the tower is designed for 6 degC range, calculate the water flow in cooling tower. The design approach temperature of the CT is 5 degC.
Model answer: Answer:
Condenser load = 7 MW
Steam rate for condenser = 5 kg/kW
Total steam required for condenser power = 7000 x 5 = 35,000 kg/hr ..... (2 marks)
Condenser heat load = 35,000 x 550 = 1,92,50,000 Kcal/hr ..... (2 marks)
Heat load of VAM = 1100 x 127 x 60 = 83,82,000 Kcal/hr ..... (2 marks)
Total cooling tower heat load = 1,92,50,000 + 83,82,000 = 2,76,32,000 Kcal/hr ..... (2 marks)
Range of tower = 6 degC
Cooling water flow required = 2,76,32,000/6 = 46,05,333 litres/hr or 4605 m3/hr ..... (2 marks)
Build the heat load from both sources: condenser = 7000 kW × 5 kg/kW × 550 kcal/kg = 19.25 million kcal/hr; VAM = 1100 TR × 127 kcal/min/TR × 60 = 8.382 million kcal/hr; total 27.63 million kcal/hr. Then flow = heat load /(range × 1000) = 27,632,000/6 = 4.605 million litres/hr = 4605 m³/hr. Two traps: the VAM figure is per MINUTE so multiply by 60, and the 18 MW plant rating and 5 °C approach are distractors — only the 7 MW condenser load and the range enter the arithmetic.
📖 §1.4 Power factor improvement & benefits (disadvantages of low PF, PF penalty/incentive)
6. a) List five disadvantages of low Power Factor? b) An industry is losing money as penalty on account of maintaining a poor power factor of 0.88. The power utility has specified a minimum power factor of 0.9 to avoid penalty. The penalty on energy cost is 1% for every 0.01 power factor less than the minimum prescribed. Also an incentive on energy cost is available @ 1.5% for every 0.01 improvement above 0.95. If the monthly energy bill of the industry is Rs 6 lakhs, calculate the annual cost saving potential if power factor is improved to unity from the current level.
Model answer: Answer:
a) Disadvantages of low power factor (any five - 1 mark each):
1. Large line losses (copper losses)
2. Large kVA rating and size of electrical equipments
3. Greater conductor size and cost
4. Poor voltage regulation and large voltage drop
5. Low efficiency
6. Penalty from electric power supply company on low power factor
b) Minimum PF to be maintained to avoid penalty = 0.9
Present penalty = 1.00% (on energy bill) for every 0.01 PF
For 0.02 PF (0.90 - 0.88) = 1.00 x 2 = 2.0% ..... (1 mark)
Incentives (from 0.95 to 1.00, i.e. 5 steps of 0.01) = 1.5 x 5 = 7.5% ..... (1 mark)
Energy saving potential = 2.0 + 7.5 = 9.5%
Cost reduction potential per month = Rs. 6 lakh x 9.5% = Rs. 57,000 ..... (2 marks)
Annual cost reduction = 57,000 x 12 = Rs. 6,84,000 ..... (1 mark)
Count the 0.01 steps carefully in two separate stretches: penalty 0.88 -> 0.90 is 2 steps at 1% = 2%; incentive 0.95 -> 1.00 is 5 steps at 1.5% = 7.5%. Total benefit = 9.5% of the energy bill.
Rs.6,00,000 x 9.5% = Rs.57,000/month -> Rs.6,84,000/year.
The band between 0.90 and 0.95 earns nothing (no penalty, no incentive) - counting those 5 steps as well is the classic mistake here. Low-PF penalties come on top of larger cables, bigger kVA equipment, poor voltage regulation and higher line losses.