198 questions — 138 objective (1 mark), 35 short (5 marks), 25 long (10 marks). Every answer is checked against the 2014 BEE guidebook and carries its book section reference plus an explanation. ▶ Practice this chapter interactively (timer, read-aloud, progress saving).
Objective questions (1 mark) — 138
📖 §4.7 Performance Assessment of Refrigeration Plants — Tonne of Refrigeration
1. 1 Tonne of Refrigeration (TR) is equal to:
860 kcal/hr = 1.00 kW = 3,412 BTU/hr
3024 kcal/hr = 3.51 kW = 12,000 BTU/hr
2575 kcal/hr = 3.00 kW = 10,200 BTU/hr
1000 kcal/hr = 1.16 kW = 3,968 BTU/hr
Answer: B) 3024 kcal/hr = 3.51 kW = 12,000 BTU/hr
Confirmed vs Book-3 §4.7 — the book defines 1 TR of refrigeration as 3024 kcal/hr of heat rejected, and its own objective question confirms 1 TR = 3.51 kW = 3024 kcal/hr = 12,000 BTU/hr ('all of the above'). Option (a) is the trap: 860 kcal/hr = 1 kW is the electrical-to-heat conversion constant, not a tonne of refrigeration. Memorise 3024 for kcal/hr working and 3.51 for kW working.
Source: AI practice
📖 §4.7 TR formula — coolant/chiller (evaporator) side
2. A chiller cools 30 m3/hr of water from 16 degC to 9 degC. What is the chiller tonnage (TR)?
46.3 TR
69.4 TR
60.0 TR
24.4 TR
Answer: B) 69.4 TR
Confirmed vs Book-3 §4.7 — TR = [Q × Cp × (Ti − To)] / 3024, with Q in kg/hr. Working: for water 30 m³/hr = 30,000 kg/hr, Cp = 1 kcal/kg·°C, ΔT = 16 − 9 = 7 °C, so TR = (30,000 × 1 × 7)/3024 = 210,000/3024 = 69.4 TR. Option (a) 46.3 TR is the book's own worked answer for a DIFFERENT case (20 m³/hr, 17 → 10 °C) — a tempting number to recall wrongly. Divide by 3024, never by 860.
Source: AI practice
📖 §4.7 Industry COP and TR conversion
3. A vapour compression chiller has a COP of 4.0 and a compressor input of 50 kW. What is the refrigeration capacity in TR?
57.0 TR
14.2 TR
200 TR
42.7 TR
Answer: A) 57.0 TR
Confirmed vs Book-3 §4.7 — industry COP = cooling effect (kW) ÷ power input to compressor (kW), so cooling effect = 4.0 × 50 = 200 kW; converting, TR = 200/3.51 = 57.0 TR. This mirrors the book's own worked objective (COP 3.0, input 100 kW → 300 kW → 85.3 TR). Option (c) 200 TR is the trap of reporting the cooling effect in kW as though it were already tonnes.
Source: AI practice
📖 §4.3 Types of Refrigeration System / §4.4 Table 4.1–4.2 Common Refrigerants
4. Which refrigerant is common to BOTH a vapour compression (VCR) chiller and a vapour absorption (VAR) system?
Ammonia
Lithium bromide
R-134a
Water
Answer: A) Ammonia
Confirmed vs Book-3 §4.3–§4.4 — ammonia (R-717) appears in the book's tables of commonly used vapour-compression refrigerants (boiling point −33.3 °C, COP-carnot 4.78), and §4.3 also describes ammonia-based ABSORPTION systems, which operate above atmospheric pressure and are capable of low-temperature operation below 0 °C. So ammonia is the fluid common to both cycles. Option (b) lithium bromide is wrong because in a LiBr machine LiBr is the ABSORBENT and pure water is the refrigerant.
Source: AI practice
📖 §4.7 COP-Carnot / §4.8 Design of Process Heat Exchangers
5. For higher COP in a refrigeration system, you should aim for:
Lower evaporator temp and higher condenser temp
Higher evaporator temp and lower condenser temp
Higher evaporator temp and higher condenser temp
Lower evaporator temp and lower condenser temp
Answer: B) Higher evaporator temp and lower condenser temp
Confirmed vs Book-3 §4.7 — COP-Carnot = Te / (Tc − Te) with both temperatures in KELVIN, and the book states directly that a higher COP-Carnot is obtained with a HIGHER evaporator temperature and a LOWER condenser temperature, because both shrink the temperature lift (Tc − Te). §4.8 quantifies it: a 1 °C rise in evaporator temperature saves almost 3% on power, and Table 4.5 shows kW/TR rising 20.5% when condensing temperature goes from 26.7 to 40 °C. Option (a) is the exact reverse and is what happens with fouled heat exchangers.
Source: AI practice
📖 §4.3 Vapour Compression Refrigeration — cycle energy balance
6. A system has a cooling load of 120 kW and a COP of 4. How much heat is rejected at the condenser/cooling tower?
120 kW
30 kW
150 kW
480 kW
Answer: C) 150 kW
Confirmed vs Book-3 §4.3 — the book states the condenser must reject the COMBINED heat inputs of the evaporator and the compressor. Working: compressor input = cooling load / COP = 120/4 = 30 kW; heat rejected = 120 + 30 = 150 kW. Option (a) 120 kW forgets the work of compression, which is exactly why condenser and cooling-tower duty must always be sized larger than the refrigeration duty.
7. Across the expansion (throttling) device in a refrigeration cycle, the enthalpy of the refrigerant:
Increases sharply
Decreases sharply
Remains constant (no heat gain or loss)
Doubles
Answer: C) Remains constant (no heat gain or loss)
Confirmed vs Book-3 §4.3 — the book describes the expansion device as both reducing the pressure of and controlling the flow of the high-pressure sub-cooled liquid into the evaporator, and states explicitly that there is NO heat loss or gain through the expansion device. With no heat and no work exchanged, enthalpy is unchanged (a throttling process); pressure and temperature drop instead. Option (b) confuses this with the evaporator, which is where heat is actually absorbed.
Source: AI practice
📖 §4.3 Absorption Refrigeration (VAR)
8. In a LiBr vapour absorption system, the refrigerant and absorbent are respectively:
Ammonia and water
Water and lithium bromide
R-22 and oil
Lithium bromide and water
Answer: B) Water and lithium bromide
Confirmed vs Book-3 §4.3 — in the absorption chiller the refrigerant is PURE WATER and the absorbent is LITHIUM BROMIDE solution; water evaporates at about 4 °C under a high vacuum of 754 mmHg, its vapour is absorbed into the LiBr solution in the absorber, and the diluted solution is re-concentrated in the generator using steam or waste heat. Such systems give a COP of 0.65–0.70 and need electricity only for the pumps. Option (d) simply reverses the two roles.
9. Which component is NOT part of a vapour compression refrigeration (VCR) system?
Compressor
Condenser
Evaporator
Absorber
Answer: D) Absorber
Confirmed vs Book-3 §4.3 — the VCR system consists of the evaporator, compressor, condenser and expansion device (plus a liquid receiver). The ABSORBER is one of the four vessels of the vapour ABSORPTION system, where refrigerant vapour is absorbed into the LiBr solution and the vacuum in the evaporator is maintained. This is the book's own objective question, and the condenser and evaporator in options (b) and (c) appear in BOTH cycles, so neither can be the odd one out.
Source: AI practice
📖 §4.2 Psychrometrics — dew-point temperature
10. The dew point temperature of air is the temperature at which:
Moisture in the air increases
Moisture in the air begins to condense
Moisture in the air decreases
Moisture in the air remains the same
Answer: B) Moisture in the air begins to condense
Confirmed vs Book-3 §4.2 — the dew-point temperature (Tdp) is one of the properties read from the psychrometric chart, and the book's own objective question states that at the dew-point temperature moisture in air BEGINS TO CONDENSE. Above the dew point the moisture stays as vapour (so option (d) describes the state before, not at, the dew point); the book's chart example reads Tdp = 23 °C for air at 32 °C DBT and 60% RH. This is the principle behind after-coolers and refrigerant dryers — cool the air to its dew point and the water drops out.
Source: AI practice
📖 §4.8 Maintenance of Heat Exchanger Surfaces
11. A 0.8 mm scale deposit on condenser tubes can increase energy consumption by up to:
5%
15%
35%
60%
Answer: C) 35%
Confirmed vs Book-3 §4.8 — the book states that a 0.8 mm scale build-up on condenser tubes can increase energy consumption by as much as 35%, because the scale insulates the tube, raises the condensing temperature and pressure, and so raises compressor work. Book Table 4.6 supports the magnitude: a dirty condenser alone pushes specific power up 20.4%, and a dirty condenser plus evaporator 38.7%. The book also notes that a 0.55 °C reduction in cooling-tower return water temperature cuts compressor power by 3%.
Source: AI practice
📖 §4.5 Compressor Types and Application
12. Hermetically sealed compressors are typically used in:
Large industrial chillers
Domestic refrigerators and room air conditioners (low capacity)
Centrifugal water chillers above 1000 TR
Ammonia absorption plants
Answer: B) Domestic refrigerators and room air conditioners (low capacity)
Confirmed vs Book-3 §4.5 — the book distinguishes the OPEN type (compressor and motor as separate units), normally used for industrial duty, from the HERMETIC type (motor and compressor in a single sealed unit), used in refrigerators, air conditioners and other low-capacity applications. Option (c) is wrong because large chillers above 1000 TR use centrifugal machines of open/semi-hermetic industrial construction, and option (d) is wrong because an absorption plant has no compressor at all.
Source: AI practice
📖 §4.3 Vapour Compression Refrigeration cycle
13. What is the primary function of the evaporator in a refrigeration cycle?
To compress the refrigerant
To absorb heat from the environment
To condense the refrigerant
To regulate the flow of refrigerant
Answer: B) To absorb heat from the environment
Confirmed vs Book-3 §4.3 - In the evaporator the low-pressure refrigerant boils, absorbing heat from the space/process being cooled (this Δh is the cooling effect). Compression is the compressor, condensing is the condenser, and flow regulation is the expansion valve. Compression (a) is the compressor's job, condensing (c) the condenser's, and flow regulation (d) the expansion device's; §4.3 stage 1-2 is the refrigerant boiling in the evaporator while absorbing heat.
Source: Sep 2024
📖 §4.3 VCR cycle stages (1-2-3-4)
14. In which component of an ideal refrigeration system, the refrigeration temperature will increase?
Compressor
Condenser
Evaporator
Expansion valve
Answer: A) Compressor
Confirmed vs Book-3 §4.3 - The compressor raises the refrigerant pressure and therefore its temperature, delivering superheated high-temperature gas to the condenser. The expansion valve drops temperature; the evaporator is the low-temperature side. The condenser (b) and expansion device (d) both reduce refrigerant temperature and the evaporator (c) runs at the lowest temperature in the cycle; §4.3 stage 2-3 notes a big temperature rise in the compressor as part of the input energy passes into the refrigerant.
Source: Sep 2024
📖 §4.3 Absorption Refrigeration
15. Which refrigerant is commonly used in vapor absorption refrigeration systems?
R-22
R-134a
H2O
LiBr
Answer: C) H2O
Confirmed vs Book-3 §4.3 - In a LiBr–water vapour absorption system the refrigerant is pure water (H2O) and lithium bromide is the absorbent. (Ammonia is the refrigerant common to both VCR and VAR; among the options here water is the refrigerant, LiBr is the absorbent, and R-22/R-134a are vapour-compression refrigerants.). Lithium bromide (d) is the tempting answer but §4.3 is explicit that it is the ABSORBENT; R-22 and R-134a (a, b) are vapour-compression refrigerants only.
Source: Sep 2024
📖 §4.7 COP & Figure 4.10 (evaporator temperature)
16. What is the effect of increasing the chilled water leaving temperature on the efficiency of a centrifugal chiller?
It increases the efficiency of the chiller
It decreases the efficiency of the chiller
It has no effect on efficiency
It increases the refrigerant flow rate
Answer: A) It increases the efficiency of the chiller
Confirmed vs Book-3 §4.7 - Raising the chilled-water (evaporator) leaving temperature raises the evaporator/suction pressure, reducing the compression ratio and compressor work, so COP/efficiency improves. Book rule: +1°C evaporator temperature ≈ 3% power saving. Option (b) is the intuitive but wrong choice; Figure 4.10 shows chiller COP rising as leaving chilled-water temperature rises, because the compressor works against a smaller temperature lift (about 3% power saving per 1 degC).
Source: Sep 2024
📖 §4.7 COP & kW/TR
17. An HVAC system operates with a COP (Coefficient of Performance) of 4. If the system provides 100 kW of cooling, what is the power input to chiller?
0.04 kW
25 kW
400 kW
None of the above
Answer: B) 25 kW
Confirmed vs Book-3 §4.7 - COP = cooling effect / power input → power input = cooling / COP = 100 / 4 = 25 kW. Option (c) 400 kW multiplies instead of divides, and (a) confuses the ratio with its reciprocal in MW. A COP of 4 means one unit of compressor power moves four units of heat.
Source: Sep 2024
📖 §4.16 Energy Saving Opportunities
18. What is the primary function of an economizer in an HVAC system?
To increase indoor air pollution
To reduce the cost of heating equipment
To use outdoor air for cooling when conditions are favorable, saving energy
To increase the use of mechanical cooling systems
Answer: C) To use outdoor air for cooling when conditions are favorable, saving energy
Confirmed vs Book-3 §4.16 - An air-side economizer brings in cool/dry outdoor air for free cooling when outdoor conditions are favourable, reducing mechanical-cooling (chiller) energy. ECBC requires it on systems with fan >1200 L/s and cooling >22 kW; it can save ~10% energy. Options (a), (b) and (d) all increase energy or pollution; an economizer uses cool outdoor air directly for free cooling, which belongs with the §4.16 building heat-load minimisation measures.
Source: Sep 2024
📖 §4.12 Ventilation Systems (ACH)
19. An equipment room measures 12 × 8 × 3.5 m. Ventilation required for 15 ACH is:
5060 m³/h
5020 m³/h
5040 m³/h
5080 m³/h
Answer: C) 5040 m³/h
Confirmed vs Book-3 §4.12 - Volume = 12×8×3.5 = 336 m³; ventilation = volume × ACH = 336 × 15 = 5040 m³/h. The trap is mis-multiplying the volume: 12 x 8 x 3.5 = 336 m3, and 336 x 15 ACH = 5040 m3/h exactly - the other options are arithmetic near-misses with no physical basis.
Source: Sep 2025
📖 §4.2 Psychrometrics (dew point)
20. If dew point temperature equals air temperature, relative humidity is:
0%
45%
50%
100%
Answer: D) 100%
Confirmed vs Book-3 §4.2 - When air is cooled to its dew point (DBT = dew point), the air is fully saturated, so relative humidity = 100%. Options (a)-(c) would require the air to be unsaturated, but if the dew point has risen to the dry bulb temperature no further cooling is possible without condensation, which is the definition of saturation.
Source: Sep 2025
📖 §4.3 / Table 4.3 Refrigerants
21. In vapour compression and vapour absorption systems, the common refrigerant is:
Lithium bromide
R-134a
Ammonia
None of the above
Answer: C) Ammonia
Confirmed vs Book-3 §4.3 - Ammonia (R-717) is used both as a refrigerant in vapour-compression chillers and in ammonia-water absorption systems, making it common to both. Lithium bromide (a) is the ABSORBENT, not a refrigerant, and R-134a (b) is a vapour-compression-only HFC; only ammonia serves as refrigerant in both VCR chillers and NH3-water absorption machines.
Source: Sep 2025
📖 §4.3 VCR cycle stages (1-2-3-4)
22. In a vapor compression refrigeration system, enthalpy changes occur across:
Compressor
Condenser
Evaporator
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §4.3 - Enthalpy changes in the evaporator (heat absorbed), compressor (work added) and condenser (heat rejected); only the expansion device is constant-enthalpy. Option (d) is right because the trap is picking a single component; only the expansion device is isenthalpic (Book §4.3: 'there is no heat loss or gain through the expansion device').
23. When air is cooled by evaporation in an air washer:
Humidity ratio decreases
Dry bulb temperature decreases
Dry bulb temperature increases
Enthalpy increases
Answer: B) Dry bulb temperature decreases
Confirmed vs Book-3 §4.14 - Evaporative (adiabatic) cooling in an air washer lowers DBT toward the WBT while humidity ratio rises and enthalpy stays nearly constant. Options (a) and (c) reverse the physics - moisture is ADDED so the humidity ratio rises - and (d) is wrong because the process is adiabatic, so enthalpy stays essentially constant.
Source: Sep 2025
📖 §4.8 Maintenance of Heat Exchanger Surfaces
24. Scale in condenser tubes:
Increases energy use
Reduces heat transfer
Can lead to higher operating pressure
All of the above
Answer: D) All of the above
Confirmed vs Book-3 §4.8 - Condenser scaling insulates the tubes, reducing heat transfer, raising condensing pressure/discharge pressure and increasing compressor energy use — all of the above. Each of (a), (b) and (c) is a consequence of the same fouling, so 'all of the above' is right; §4.8 quantifies it - a 0.8 mm scale build-up on condenser tubes can raise energy consumption by as much as 35%.
Source: Sep 2025
📖 §4.7 COP & Figure 4.10 (evaporator temperature)
25. Increasing chilled water leaving temperature in a centrifugal chiller:
Increases efficiency
Decreases efficiency
No effect
Increases refrigerant flow
Answer: A) Increases efficiency
Confirmed vs Book-3 §4.7 - Raising the evaporator (chilled-water leaving) temperature lifts the COP/efficiency — about 3% power saving per 1°C rise. Option (b) is the common reflex answer, but Figure 4.10 shows COP rising with leaving chilled-water temperature; the refrigerant flow (d) is not what the question asks about.
Source: Sep 2025
📖 §4.3 VCR cycle stages (1-2-3-4)
26. In a vapour compression system, refrigerant changes from vapour to liquid in the:
Compressor
Evaporator
Condenser
Expansion valve
Answer: C) Condenser
Confirmed vs Book-3 §4.3 - The condenser rejects heat and condenses the high-pressure refrigerant vapour into liquid. The evaporator (b) does the reverse (liquid to vapour), the compressor (a) only raises pressure of the vapour, and the expansion valve (d) is an isenthalpic throttle - only the condenser de-superheats, condenses and sub-cools.
Source: Sep 2025
📖 §4.7 Ton of Refrigeration (TR)
27. If 27,216 kcal of heat is removed per hour, refrigeration tonnage is:
10 TR
7 TR
11 TR
None of the above
Answer: D) None of the above
Confirmed vs Book-3 §4.7 - TR = 27,216 / 3024 = 9.0 TR exactly (1 TR = 3024 kcal/h). Since 9 TR is not listed, the answer is 'None of the above'. 27,216/3024 = 9.0 TR exactly, and 9 TR is not offered, so (d) is correct; option (a) 10 TR is the trap for anyone who rounds carelessly or uses 3000 kcal/h per TR.
Source: Sep 2025
📖 §4.7 COP-Carnot
28. Higher COP can be achieved with
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
Confirmed vs Book-3 §4.7 - COP improves as the temperature lift (condenser minus evaporator) reduces. A higher evaporator temperature and lower condenser temperature minimize the lift and maximize COP.
Source: Book EOC
📖 §4.3 Absorption Refrigeration (COP 0.65-0.70)
29. Li-Br water absorption refrigeration systems have a COP in the range of
0.4 - 0.5
0.65 - 0.70
0.75 - 0.80
none of the above
Answer: B) 0.65 - 0.70
Confirmed vs Book-3 §4.3 - Single-effect lithium bromide-water vapour absorption machines typically have a COP of about 0.65-0.70. Option (c) 0.75-0.80 is the standard trap - that band belongs to double-effect machines; the book quotes 0.65-0.70 for LiBr-water systems delivering 6.7 degC chilled water with 30 degC cooling water.
Source: Book EOC
📖 §4.5 Compressor Types (hermetic / open)
30. Hermetic systems are used in
domestic refrigerators
centrifugal chillers
screw chillers
large reciprocating chillers
Answer: A) domestic refrigerators
Confirmed vs Book-3 §4.5 - In a hermetically sealed system the motor and compressor are enclosed in a common welded housing; this is used in small units such as domestic refrigerators and small air conditioners.
Source: Book EOC
📖 §4.3 & Table 4.3 Refrigerant / absorbent
31. Which of the following can be used as a refrigerant in both vapour compression and vapour absorption systems?
ammonia
R-11
R-12
lithium Bromide
Answer: A) ammonia
Confirmed vs Book-3 §4.3 - Ammonia (R-717) is used as the refrigerant in vapour compression systems and also as the refrigerant in ammonia-water vapour absorption systems. (In Li-Br systems, water is the refrigerant and lithium bromide is the absorbent.)
Source: Book EOC
📖 §4.7 Ton of Refrigeration (TR)
32. One ton of refrigeration (TR) is equal to
3.51 kW
3024 kcal/hr
12,000 BTU/hr
all of the above
Answer: D) all of the above
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/hr = 12,000 BTU/hr = 3.51 kW; all three statements describe the same quantity. All three figures are the same quantity expressed in different units, so any single choice (a), (b) or (c) is incomplete - the book itself gives 3024 kcal/hr and 3.51 kW.
Source: Book EOC
📖 §4.7 TR formula (coolant side)
33. The refrigeration load in TR when 20 m3/hr of water is cooled from 17 degC to 10 degC is about
46.3
41.6
116.6
none of the above
Answer: A) 46.3
Confirmed vs Book-3 §4.7 - Heat removed = mass x sp. heat x dT = 20,000 kg/hr x 1 kcal/kg degC x (17-10) = 140,000 kcal/hr. TR = 140,000/3024 = 46.3 TR. Option (b) 41.6 comes from using 3.5 kW/TR incorrectly and (c) 116.6 from forgetting to divide by 3024 properly; the correct route is mass x Cp x deltaT / 3024.
Source: Book EOC
📖 §4.2 Psychrometrics (dew point)
34. At the dew point temperature, moisture in air
increases
decreases
remains same
begins to condense
Answer: D) begins to condense
Confirmed vs Book-3 §4.2 - The dew point is the temperature at which air becomes saturated (100% RH); any further cooling causes the moisture to begin condensing out. Options (a) and (b) confuse absolute moisture with saturation: at the dew point the moisture content is unchanged but the air is saturated, so any further cooling makes it condense out.
Source: Book EOC
📖 §4.3 VCR components vs VAR
35. Which of the following is NOT a part of the vapour compression refrigeration system?
compressor
evaporator
condenser
absorber
Answer: D) absorber
Confirmed vs Book-3 §4.3 - A vapour compression system consists of compressor, condenser, expansion device and evaporator. The absorber is a component of the vapour absorption (not compression) system.
Source: Book EOC
📖 §4.7 COP & kW/TR
36. The COP of a vapour compression refrigeration system is 3.0. If the power input to the compressor is 100 kW, the tonnage of the refrigeration system is given by
85.3
9.48
300
none of the above
Answer: A) 85.3
Confirmed vs Book-3 §4.7 - Refrigeration effect = COP x power input = 3.0 x 100 = 300 kW. TR = 300/3.517 = 85.3 TR (1 TR = 3.517 kW). Option (c) 300 is the cooling effect in kW, not TR - the trap is forgetting to divide by 3.517 kW/TR; (b) 9.48 comes from dividing rather than multiplying by the COP.
Source: Book EOC
📖 §4.1 Introduction (definition of refrigeration)
37. The term refrigeration means
addition of cooling
removal of heat
removal and relocation of heat
replacement of heat
Answer: C) removal and relocation of heat
Confirmed vs Book-3 §4.1 - Refrigeration is the process of removing heat from a space/substance and relocating (rejecting) it to another medium, thereby maintaining a lower temperature. Option (b) 'removal of heat' is incomplete because the heat must also be rejected (relocated) to a higher-temperature sink; 'addition of cooling' (a) is not a physical process.
Source: Book EOC
📖 §4.7 COP-Carnot
38. Higher chiller COP can be achieved with
higher evaporator temperature and higher condensing temperature
higher evaporator temperature and lower condensing temperature
lower evaporator temperature and higher condensing temperature
lower evaporator temperature and lower condensing temperature
Answer: B) higher evaporator temperature and lower condensing temperature
Confirmed vs Book-3 §4.7 - COP rises as the temperature lift (condensing minus evaporating) reduces: higher evaporator and lower condenser temperature. Options (a), (c) and (d) all keep or widen the temperature lift; COP-Carnot = Te/(Tc - Te) improves only when Te rises AND Tc falls, which shrinks the denominator.
Source: 15th Exam
📖 §4.8 Table 4.5 (condenser temperature)
39. All other conditions remaining the same in a refrigeration system, at which of the following condenser temperatures will the power consumption be the least:
32.6 oC
35.9 oC
40.8 oC
43.4 oC
Answer: A) 32.6 oC
Confirmed vs Book-3 §4.8 - Lower condensing temperature reduces the compressor lift and hence power; the lowest listed condenser temperature (32.6 °C) gives least power. Every other option is a higher condensing temperature; Table 4.5 shows kW/TR rising from 1.17 at 26.7 degC to 1.41 at 40 degC, i.e. about 20% more power for the same duty.
Source: 15th Exam
📖 §4.3 VCR cycle stages (1-2-3-4)
40. The pressure of refrigerant in vapour compression system changes in
compressor
expansion valve
both (a) & (b)
evaporator
Answer: C) both (a) & (b)
Confirmed vs Book-3 §4.3 - Pressure rises in the compressor and drops across the expansion valve; evaporator and condenser are essentially constant-pressure processes. Evaporator (d) and condenser (b alone) are essentially constant-pressure heat exchangers, so neither alone is the answer; pressure is raised in the compressor and dropped across the expansion valve, hence 'both'.
Source: 15th Exam
📖 §4.7 Ton of Refrigeration (TR)
41. A 1.5 ton air conditioner installed in a room and working continuously for one hour will remove heat of
3024 kcals
4536 kcals
3000 kcals
6048 kcals
Answer: B) 4536 kcals
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/h; 1.5 TR x 3024 = 4536 kcal in one hour. Option (a) 3024 kcal is the heat for 1 TR only and (d) 6048 kcal is for 2 TR; multiply 3024 kcal/h by the 1.5 TR capacity and the one-hour run time.
Source: 15th Exam
📖 §4.9 Energy Efficiency Ratio (EER)
42. If the power consumed by a 1.5 TR refrigeration compressor is 2.5 kW, what is the energy efficiency ratio?
2.1
1.5
0.6
1.66
Answer: A) 2.1
Confirmed vs Book-3 §4.9 - Refrigeration effect = 1.5 x 3.516 = 5.274 kW. EER = 5.274/2.5 = 2.11. Option (d) 1.66 is the trap of using 3024 kcal/h with the wrong conversion; use cooling in watts (1.5 x 3.516 = 5.274 kW) divided by 2.5 kW input to get EER 2.11 W/W.
Source: 15th Exam
📖 §4.7 kW/TR vs COP/EER
43. In the performance assessment of a refrigeration system, which performance ratio (energy efficiency) does not follow the trend "a higher ratio means a more efficient refrigeration system"?
Coefficient of performance (COP)
Energy Efficiency Ratio (EER)
kW per ton
none of the above
Answer: C) kW per ton
Confirmed vs Book-3 §4.7 - For kW/TR a LOWER value means a more efficient system, opposite to COP and EER where higher is better. COP and EER (a, b) are output/input ratios where higher is better; kW/TR is input/output, so a smaller number means a better chiller - the book quotes 0.65-0.9 kW/TR for good centrifugal machines.
Source: 15th Exam
📖 §4.7 Ton of Refrigeration (TR)
44. 2 ton of refrigeration (TR) is equivalent to about
100.8 kcal/min
7032 W
400 BTU/min
all of the above
Answer: D) all of the above
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/hr = 50.4 kcal/min = 3516 W = 12,000 BTU/hr = 200 BTU/min, so 2 TR = 100.8 kcal/min = 7032 W = 400 BTU/min and all three statements describe the same duty. Option (c) was repaired from the garbled '428.7 BTU/min', which matches no standard conversion; with 400 BTU/min the key (d) 'all of the above' is unambiguous, and picking any single option is the trap of not recognising that they are the same quantity in different units.
Source: 15th Exam
📖 §4.7 TR formula (coolant side)
45. A process fluid at 50 m3/hr, with a density of 0.96, is flowing in a heat exchanger and is to be cooled from 36oC to 29oC. The fluid specific heat is 0.78 kcal/kg. If the chilled water range across the heat exchanger is 5oC, the chilled water flow rate is
67.2 m3/hr
52.42 m3/hr
50 m3/hr
none of the above
Answer: B) 52.42 m3/hr
Confirmed vs Book-3 §4.7 - Heat load = 50 x 0.96 x 1000 x 0.78 x (36-29) = 262,080 kcal/h. Chilled water flow = 262,080/(1000 x 1 x 5) = 52.42 m3/hr. Option (a) 67.2 ignores the fluid density of 0.96 and (c) 50 simply copies the process flow; balance the duty 262,080 kcal/h against the chilled-water range of 5 degC.
Source: 15th Exam
📖 §4.1 Figure 4.1 Heat Transfer Loops
46. The order of movement of thermal energy in an HVAC system is:
Indoor air - Condenser water - Chilled water - Cooling tower - Refrigerant
Chilled water - Indoor air - Refrigerant - Cooling tower - Condenser water
Indoor air - Chilled water - Refrigerant - Condenser water - Cooling tower
Indoor air - Chilled water - Refrigerant - Cooling tower - Condenser water
Answer: C) Indoor air - Chilled water - Refrigerant - Condenser water - Cooling tower
Confirmed vs Book-3 §4.1 - Heat flows: indoor air → chilled water → refrigerant (chiller) → condenser water → cooling tower (to atmosphere). Options (a), (b) and (d) scramble the sequence; Figure 4.1 fixes the order as indoor air, chilled water, refrigerant, condenser water, cooling tower, with heat moving left to right to the outdoors.
Source: 15th Exam
📖 §4.2 Psychrometric Chart (specific humidity)
47. The unit of specific humidity in a psychrometric chart is
grams moisture/kg of dry air
moisture percentage in air
grams moisture/kg of air
all of the above
Answer: A) grams moisture/kg of dry air
Confirmed vs Book-3 §4.2 - Specific humidity (humidity ratio) is expressed as grams (or kg) of moisture per kg of dry air. Option (c) 'per kg of air' is the classic trap - the reference mass is DRY air, not moist air; (b) percentage describes relative humidity, not specific humidity.
Source: 14th Exam
📖 §4.7 COP & kW/TR
48. The COP of a vapour compression refrigeration system is 3.1. If the motor draws 9.3 kW at an operating efficiency of 88%, the tonnage of the refrigeration system is about
8.2
9.3
7.2
none of the above
Answer: C) 7.2
Confirmed vs Book-3 §4.7 - Shaft power = 9.3 x 0.88 = 8.184 kW. Refrigeration effect = COP x shaft power = 3.1 x 8.184 = 25.37 kW. TR = 25.37/3.516 = 7.2 TR. Option (b) 9.3 is the motor input in kW misread as TR; apply the 88% operating efficiency first (9.3 x 0.88 = 8.184 kW shaft), then COP and 3.516 kW/TR.
Source: 14th Exam
📖 §4.7 TR formula (coolant side)
49. Chilled water enters an evaporator at 10°C and leaves at 6°C. The flow rate of chilled water is 200 m3/hr. The tons of refrigeration capacity is
265
200
661
2.65
Answer: A) 265
Confirmed vs Book-3 §4.7 - TR = (200 x 1000 x 1 x (10-6))/3024 = 800,000/3024 = 264.6 ≈ 265 TR. Option (c) 661 uses the 10 degC inlet instead of the 4 degC range, and (d) 2.65 misplaces the decimal by 100; TR = 200,000 x 4/3024 = 265.
Source: 14th Exam
📖 §4.2 Psychrometric Chart (enthalpy vs WBT lines)
50. In an air conditioning system analysis, which one temperature is sufficient to determine the enthalpy of air?
dry bulb temperature
wet bulb temperature
ambient temperature
none of the above
Answer: B) wet bulb temperature
Confirmed vs Book-3 §4.2 - Lines of constant wet-bulb temperature nearly coincide with constant-enthalpy lines on the psychrometric chart, so WBT essentially fixes the enthalpy of air. Dry bulb temperature alone (a) fixes only the sensible state and says nothing about moisture; on the psychrometric chart the constant-WBT lines run almost parallel to constant-enthalpy lines, so WBT effectively fixes enthalpy.
Source: 14th Exam
📖 §4.3 VAR - thermal energy as driving force
51. The driving force for refrigeration in a vapour absorption refrigeration system is
mechanical energy
electrical energy
thermal energy
chemical energy
Answer: C) thermal energy
Confirmed vs Book-3 §4.3 - VAR systems are heat-driven; the driving input is low-grade thermal energy (heat). Mechanical energy (a) is what drives the vapour COMPRESSION system; VAR needs electricity (b) only for its solution pump, and the actual driving input is heat - steam, hot water, gas or oil.
52. Which of the following happens to air when it is cooled through an evaporation process?
humidity ratio of the air decreases
dry bulb temperature of air decreases
dry bulb temperature of air increases
enthalpy of outlet air is less than enthalpy of inlet air
Answer: B) dry bulb temperature of air decreases
Confirmed vs Book-3 §4.14 - Evaporative cooling lowers the dry-bulb temperature while raising humidity ratio at nearly constant enthalpy. Options (a) and (c) invert the process, and (d) is wrong because evaporative cooling is nearly adiabatic - sensible heat is simply converted into latent heat at constant enthalpy.
Source: 14th Exam
📖 §4.7 COP & kW/TR
53. The relation between COP and kW/TR for a refrigeration system is given by
kW/TR = 3.516/COP
kW/TR = COP/3.516
kW/TR = 860/COP
none of the above
Answer: A) kW/TR = 3.516/COP
Confirmed vs Book-3 §4.7 - one TR = 3.516 kW of cooling, and COP = cooling effect (kW) / compressor power (kW). Hence kW/TR = 3.516/COP; a plant with COP 3.516 needs exactly 1 kW/TR. Option (b) inverts the relation and (c) uses 860 kcal/kWh, which converts electrical energy to heat, not TR to kW.
Source: 14th Exam
📖 §4.2 Psychrometric Chart
54. If the wet bulb temperature of air is 38°C, then its relative humidity in % is
38 %
90%
100%
insufficient data
Answer: D) insufficient data
Confirmed vs Book-3 §4.2 - RH cannot be found from wet-bulb temperature alone; dry-bulb temperature (or another property) is also required. Options (a)-(c) assume WBT alone fixes the state; the psychrometric chart needs two independent properties, so without the dry bulb (or dew point) the RH cannot be read.
Source: 14th Exam
📖 §4.11 Heat Pumps and Their Applications
55. Identify the statement that is not applicable to heat pumps
transfers heat by refrigerant through a cycle of evaporation and condensation
an air conditioner can work as a heat pump
no external energy is required
a vapour absorption refrigeration system can also work as a heat pump
Answer: C) no external energy is required
Confirmed vs Book-3 §4.11 - A heat pump still needs external (compressor/heat) energy to move heat; statement (c) is not applicable. Statements (a), (b) and (d) are all in §4.11 - the book says a heat pump takes 3 units from the surroundings and needs 1 unit of compressor work to deliver 4, so external drive energy is essential.
Source: 14th Exam
📖 §4.5 Compressor Types and Application
56. A refrigeration system using which of the following compressors is likely to be most efficient?
reciprocating
screw
scroll
all the above
Answer: B) screw
Corrected (was c) - Book-3 §4.5: the guidebook ranks compressor efficiency explicitly - "the maximum efficiency of reciprocating compressors is lower than that of centrifugal and screw compressors", and screw part-load efficiency "is generally higher than either centrifugal or reciprocating", in some cases giving higher full-load efficiency too. Scroll (c) is tempting because it is the newest type, but §4.5 makes no efficiency claim for it - it is described only as compact, with two moving parts and close machining tolerances.
Source: 14th Exam
📖 §4.3 & Table 4.3 Refrigerant / absorbent
57. The refrigerant which can be used both in vapour compression chillers and vapour absorption chillers is
R22
R21
ammonia
pure water
Answer: C) ammonia
Confirmed vs Book-3 §4.3 - Ammonia (R717) is used as a refrigerant in vapour compression systems and as the refrigerant in ammonia-water vapour absorption systems. R22 (a) and R21 (b) are vapour-compression-only halocarbons, and pure water (d) is the refrigerant only in LiBr absorption machines; ammonia works in both, as Table 4.3 shows.
Source: 14th Exam
📖 §4.9 Energy Efficiency Ratio (EER)
58. A 1.5 TR room air conditioner having EER (W/W) of 3.0 will draw input power of ______ kW
1.75
3.00
1.50
2.00
Answer: A) 1.75
Confirmed vs Book-3 §4.9 - Cooling = 1.5 x 3.516 = 5.274 kW. Input power = cooling/EER = 5.274/3.0 = 1.758 ≈ 1.75 kW. Option (c) 1.50 mistakes the TR rating for kW and (b) 3.00 quotes the EER itself; convert 1.5 TR to 5.274 kW of cooling and divide by the EER of 3.0.
Source: Set-A
📖 §4.3 VCR cycle stages (1-2-3-4)
59. Identify the wrong statement regarding the Vapour Compression Refrigeration system:
condenser rejects heat to atmosphere
evaporator removes heat from process or space
compressor sends superheated vapor to condenser
high pressure sub-cooled liquid refrigerant returns back to evaporator
Answer: D) high pressure sub-cooled liquid refrigerant returns back to evaporator
Confirmed vs Book-3 §4.3 - After the expansion valve the refrigerant enters the evaporator as a LOW-pressure liquid-vapour mixture, not as high-pressure sub-cooled liquid; (d) is wrong. Statements (a), (b) and (c) match the four cycle stages of Figure 4.4; (d) is wrong because the sub-cooled liquid must first pass through the expansion device, entering the evaporator as a LOW-pressure wet mixture.
Source: Set-A
📖 §4.8 Tables 4.4, 4.5 & 4.6
60. Statement A: Reducing condensing temperature by 5.5°C results in a 20-25% decrease in compressor power. Statement B: 5.5°C increase in evaporator temperature reduces compressor power by 20-25%. What is your judgement?
statements A & B are TRUE
statements A & B are FALSE
statement A is TRUE & B is FALSE
statement A is FALSE & B is TRUE
Answer: A) statements A & B are TRUE
Confirmed vs Book-3 §4.8 - Both reducing condensing temperature and raising evaporator temperature by ~5.5°C each cut compressor power by roughly 20-25%; both statements are true. Both statements match the book's direction of effect - Table 4.5 shows condenser temperature raising kW/TR by 20.5%, and §4.8 gives about 3% power saving per 1 degC rise in evaporator temperature - so (b), (c) and (d) are all inconsistent with §4.8.
Source: Set-A
📖 §4.7 TR formula (coolant side)
61. The refrigeration load in TR when 10 m3/hr of water is cooled from 15°C to 7°C is about
10
8
26.5
none of the above
Answer: C) 26.5
Confirmed vs Book-3 §4.7 - TR = (10 x 1000 x 1 x (15-7))/3024 = 80,000/3024 = 26.46 ≈ 26.5 TR. Options (a) and (b) simply echo the flow and temperature figures; the load is mass x Cp x deltaT / 3024 = 10,000 x 8/3024 = 26.5 TR.
Source: Set-A
📖 §4.1 Figure 4.1 Heat Transfer Loops
62. The most energy intensive heat transfer loop of a vapour compression refrigeration system is:
Indoor air loop
Chilled water loop
Refrigerant loop
Condenser water loop
Answer: C) Refrigerant loop
Confirmed vs Book-3 §4.1 - The refrigerant loop contains the compressor — the largest energy-consuming component — making it the most energy-intensive heat-transfer loop of the system. The indoor air, chilled water and condenser water loops (a, b, d) only move fans and pumps; the refrigerant loop contains the compressor, which is by far the largest power consumer in the chain of Figure 4.1.
Source: 19th Exam
📖 §4.2 Comfort Zone (Figure 4.3)
63. Which of the following is the most comfortable conditions for an office room? (DBT = Dry bulb temperature, RH = Relative humidity)
20°C DBT and 80% RH
26°C DBT and 100% RH
15°C DBT and 30% RH
25°C DBT and 55% RH
Answer: D) 25°C DBT and 55% RH
Confirmed vs Book-3 §4.2 - Human comfort zone is ~22-26°C DBT with ~50-60% RH; the other options have extreme temperature or humidity. The book's comfort zone is 22-27 degC DBT with 40-60% RH, so (a) is too humid, (b) is saturated air, and (c) is both too cold and too dry.
Source: 17th Sep-2016
📖 §4.8 Maintenance of Heat Exchanger Surfaces
64. The formation of frost on cooling coils in a refrigerator:
improves C.O.P. of the system
increases heat transfer
reduces power consumption
increases power consumption
Answer: D) increases power consumption
Confirmed vs Book-3 §4.8 - Frost acts as an insulating layer, reducing heat transfer and forcing the compressor to work harder, increasing power consumption. Options (a)-(c) all assume frost helps; in fact the frost layer insulates the coil, depresses the evaporator temperature and, per Table 4.4, drives specific power consumption up.
Source: 17th Sep-2016
📖 §4.3 VCR cycle stages (4-1 expansion device)
65. In a refrigeration system, the expansion device is connected between the:
Compressor and condenser
Condenser and receiver
Condenser and evaporator
Evaporator and compressor
Answer: C) Condenser and evaporator
Confirmed vs Book-3 §4.3 - The expansion device drops high-pressure liquid from the condenser to low-pressure before the evaporator (constant enthalpy). Compressor-to-condenser (a) carries hot discharge gas and evaporator-to-compressor (d) is the suction line; stage 4-1 of Figure 4.4 places the expansion device between the condenser (liquid receiver) and the evaporator.
Confirmed vs Book-3 §4.2 - Mass of moisture per kg of dry air is the specific/absolute humidity (humidity ratio). Relative humidity (b) is a percentage ratio of vapour pressures and dew point (d) is a temperature; only absolute/specific humidity is a mass of moisture per kg of DRY air.
Source: 17th Sep-2016
📖 §4.9 Energy Efficiency Ratio (EER)
67. If EER of One Ton Split AC unit is 3.51, what is its power rating?
1.0 kW
1.5 kW
0.8 kW
2.0 kW
Answer: A) 1.0 kW
Confirmed vs Book-3 §4.9 - 1 TR = 3.51 kW of cooling (book value). Power = cooling/EER = 3.51/3.51 = 1.0 kW. Options (b)-(d) ignore the definition EER = refrigeration effect (W)/power input (W); with 1 TR = 3.51 kW and EER 3.51 the input is exactly 1.0 kW, the familiar '1 kW per TR' benchmark.
Source: 17th Sep-2016
📖 §4.15 Standards and Labeling of Room ACs
68. The parameter used in Star labeling of air conditioner is:
COP
EER
KW/TR
EPI
Answer: B) EER
Confirmed vs Book-3 §4.15 - BEE star rating of air conditioners is based on EER (Energy Efficiency Ratio, W/W). COP (a) and kW/TR (c) are audit indicators but are not on the label, and EPI (d) belongs to building star rating; §4.15 fixes the AC star bands in EER (W/W), e.g. 5-star split >= 3.50.
Source: 17th Sep-2016
📖 §4.7 TR formula (coolant side)
69. The refrigeration load in TR when 30 m3/hr of water is cooled from 14°C to 6.5°C is about:
74.4
64.5
261.6
none of the above
Answer: A) 74.4
Confirmed vs Book-3 §4.7 - TR = (m × Cp × ΔT)/3024 = (30×1000 × 1 × (14−6.5))/3024 = 225000/3024 = 74.4 TR. Option (b) 64.5 uses the wrong temperature range and (c) 261.6 omits the division by 3024; TR = 30,000 x 7.5/3024 = 74.4.
Source: 17th Sep-2016
📖 §4.3 Absorption Refrigeration (water / LiBr)
70. In a lithium bromide absorption refrigeration system:
lithium bromide is used as a refrigerant and water as an absorbent
water is used as a refrigerant and lithium bromide as an absorbent
ammonia is used as a refrigerant and lithium bromide as an absorbent
none of these
Answer: B) water is used as a refrigerant and lithium bromide as an absorbent
Confirmed vs Book-3 §4.3 - In a LiBr VAR system, water is the refrigerant and lithium bromide solution is the absorbent. Option (a) swaps the two roles; §4.3 is explicit that pure water is the refrigerant and lithium bromide solution the absorbent, and (c) describes an ammonia machine, which uses water as absorbent.
Source: 17th Sep-2016
📖 §4.3 VCR components vs VAR
71. Which of the following is not a part of vapour compression refrigeration cycle:
Compressor
Condenser
Generator
Evaporator
Answer: C) Generator
Confirmed vs Book-3 §4.3 - the vapour compression cycle has only four components: evaporator, compressor, condenser and expansion device. The GENERATOR belongs to the vapour absorption machine, where heat re-concentrates the dilute LiBr solution. Compressor, condenser and evaporator are all named in the Figure 4.4 VCR schematic, so only (c) is outside the cycle.
Source: 16th Exam
📖 §4.8 Table 4.4 (evaporator temperature)
72. When evaporator temperature is reduced:
refrigeration capacity increases
refrigeration capacity decreases
specific power consumption remains same
compressor will stop
Answer: B) refrigeration capacity decreases
Confirmed vs Book-3 §4.8 - Lower evaporator temperature lowers suction pressure and refrigerant density, reducing refrigeration capacity and raising specific power consumption. Options (a) and (c) contradict Table 4.4, where capacity falls from 67.58 to 23.12 tons and kW/TR rises 106% as evaporator temperature drops from +5 to -20 degC; the compressor does not stop (d), it just runs less efficiently.
Source: 16th Exam
📖 §4.2 Psychrometric Chart (specific humidity)
73. The unit of specific humidity of air is:
grams moisture/kg of dry air
moisture percentage in air
grams moisture/kg of air
percentage
Answer: A) grams moisture/kg of dry air
Confirmed vs Book-3 §4.2 - Specific (absolute) humidity = mass of water vapour per unit mass of dry air (g moisture/kg dry air). Option (c) 'per kg of air' is the trap - the reference is dry air; (b) and (d) describe relative humidity, which is a percentage rather than a mass ratio.
Source: 16th Exam
📖 §4.7 COP & kW/TR
74. 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:
34.0 kW
37.8 kW
30.6 kW
9.72 kW
Answer: A) 34.0 kW
Confirmed vs Book-3 §4.7 - COP = cooling effect (kW) / power input to compressor (kW), so cooling effect = COP x shaft power. Shaft power = 10.8 x 0.90 = 9.72 kW, giving cooling effect = 3.5 x 9.72 = 34.0 kW. Option (b) 37.8 kW is the trap of using the motor input 10.8 kW without applying the 90% motor efficiency.
Source: 16th Exam
📖 §4.9 Energy Efficiency Ratio (EER)
75. If EER of One Ton Split AC is 3.5, what is its power rating?
1.0 kW
1.5 kW
0.8 kW
None of the above
Answer: A) 1.0 kW
Confirmed vs Book-3 §4.9 - 1 TR = 3.517 kW cooling; power = cooling/EER = 3.517/3.5 = 1.0 kW. Options (b)-(d) miss the definition EER = cooling watts / input watts; 3.517 kW of cooling divided by an EER of 3.5 gives almost exactly 1.0 kW, the benchmark for a 5-star 1 TR unit.
Source: 16th Exam (alt set)
📖 §4.14 Humidification Systems
76. 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
Confirmed vs Book-3 §4.14 - Adiabatic (evaporative) humidification adds moisture (raising specific humidity) while cooling the air (lowering dry-bulb temperature). Options (a)-(c) all reduce or hold the moisture content; §4.14 defines humidification as a reduction in dry bulb temperature together with an INCREASE in specific humidity, exactly what an air washer does.
Source: 16th Exam (alt set)
📖 §4.2 Psychrometrics (dew point)
77. If we increase the temperature of air without changing specific humidity, dew point temperature of air will
increase
decrease
remain constant
can't say
Answer: C) remain constant
Confirmed vs Book-3 §4.2 - Dew point depends only on the moisture content (specific humidity / partial pressure of water vapour). If specific humidity is unchanged, dew point stays constant even though DBT rises (RH falls).
Source: 18th Exam
📖 §4.14 Air washer / evaporative cooling
78. Which of the following happens to air when it is cooled through evaporation process in an air washer?
Humidity ratio of the air decreases.
Dry Bulb Temp of air decreases.
Dry Bulb Temp of air increases.
Enthalpy of outlet air is less than enthalpy of inlet air.
Answer: B) Dry Bulb Temp of air decreases.
Confirmed vs Book-3 §4.14 - Evaporative (adiabatic) cooling in an air washer adds moisture and lowers the dry-bulb temperature at nearly constant enthalpy; humidity ratio increases, so DBT decreases.
Source: 18th Exam
📖 §4.3 VCR cycle stages (3-4 condenser)
79. In a vapor compression refrigeration system, the component where the refrigerant changes its phase from vapor to liquid is
compressor
condenser
expansion valve
evaporator
Answer: B) condenser
Confirmed vs Book-3 §4.3 - In the condenser the high-pressure refrigerant vapour rejects heat and condenses to liquid (vapour → liquid phase change). The compressor (a) only superheats the vapour, the expansion valve (c) throttles liquid, and the evaporator (d) does the opposite phase change (liquid to vapour); condensation happens in stage 3-3b of Figure 4.4.
Source: 18th Exam
📖 §4.3 VCR cycle stages (4-1 expansion device)
80. In a vapor compression refrigeration system, the component across which the enthalpy remains constant
compressor
condenser
expansion valve
evaporator
Answer: C) expansion valve
Confirmed vs Book-3 §4.3 - Expansion (throttling) is an isenthalpic process; enthalpy remains constant across the expansion valve. Enthalpy rises in the evaporator (d) and compressor (a) and falls in the condenser (b); §4.3 states there is no heat loss or gain through the expansion device, so throttling is isenthalpic.
Source: Jul 2022
📖 §4.7 Ton of Refrigeration (TR)
81. If 30,000 kcal of heat is removed from a room every hour then the refrigeration tonnage will be nearly equal to
30 TR
15 TR
10 TR
100 TR
Answer: C) 10 TR
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/hr (book value). TR = 30000/3024 ≈ 9.92 ≈ 10 TR. Option (a) 30 TR forgets to divide by 3024 and (b) 15 TR halves it; 30,000/3024 = 9.92, which rounds to 10 TR.
82. If temperature of air increases, the amount of water vapor required for complete saturation will
Increase
Decrease
not change
Can't say
Answer: A) Increase
Confirmed vs Book-3 §4.2 - Warmer air can hold more moisture — the saturation vapour pressure (and saturation humidity ratio) rises with temperature, so more water vapour is needed to fully saturate it.
Source: 18th Exam
📖 §4.3 Absorption Refrigeration (COP 0.65-0.70)
83. COP of a single effect absorption refrigeration system is likely to be in the range of
0.6 to 0.7
1 to 1.2
1.5 to 2
3.0 to 4.0
Answer: A) 0.6 to 0.7
Confirmed vs Book-3 §4.3 - Li-Br-water single-effect vapour absorption systems have a COP of about 0.65–0.70 (book §4.3). Options (b)-(d) quote COP values typical of vapour COMPRESSION chillers; a single-effect absorption machine is heat-driven and the book fixes its COP at 0.65-0.70.
Source: 18th Exam
📖 §4.7 Ton of Refrigeration (TR)
84. One tonne of refrigeration has the ability to remove ______ kcal of heat in a 24-hour period.
50 kcal
3024 kcal
72576 kcal
12000 kcal
Answer: C) 72576 kcal
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/hr (book). Over 24 hours = 3024 x 24 = 72,576 kcal. Option (b) 3024 kcal is the hourly figure and (d) 12,000 is the BTU/hr value; over a full day 3024 x 24 = 72,576 kcal.
Source: 18th Exam
📖 §4.8 Table 4.4 (evaporator temperature)
85. When evaporator temperature is increased
refrigeration capacity decreases
refrigeration capacity increases
specific power consumption remains same
power consumption increases
Answer: B) refrigeration capacity increases
Confirmed vs Book-3 §4.8 - Raising the evaporator temperature increases the refrigerant suction density/mass flow and the COP, so refrigeration capacity increases and specific power (kW/TR) falls.
Source: 18th Exam
📖 §4.7 COP & kW/TR
86. If the COP of a vapour compression system is 3.5 and the motor draws a power of 10.8 kW at 90% motor efficiency, the cooling effect of vapour compression system will be
34 kW
37.8 kW
0.36 kW
none of the above
Answer: A) 34 kW
Confirmed vs Book-3 §4.7 - Compressor shaft power = 10.8 x 0.9 = 9.72 kW. Cooling effect = COP x compressor power = 3.5 x 9.72 = 34.0 kW. Option (b) 37.8 kW is the trap of using the 10.8 kW motor input without applying the 90% efficiency; use shaft power 9.72 kW x COP 3.5 = 34 kW.
Source: 18th Exam
📖 §4.2 Psychrometrics (dew point)
87. When the dew point temperature is equal to the air temperature then the relative humidity is
0%
50%
100%
Unpredictable
Answer: C) 100%
Confirmed vs Book-3 §4.2 - When DBT equals the dew point (and wet-bulb) temperature, the air is saturated, so relative humidity = 100%. Options (a) and (b) describe unsaturated air; when the dry bulb falls to the dew point the air holds all the moisture it can at that temperature, which is 100% RH by definition.
Source: 18th Exam
📖 §4.7 Ton of Refrigeration (TR)
88. One ton of refrigeration is not equal to __________.
3024 kCal/hr
3.51 kW
12000 Btu/hr
860 kCal/hr
Answer: D) 860 kCal/hr
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/hr = 3.51 kW = 12000 Btu/hr. 860 kcal/hr equals 1 kW (a power-to-heat conversion), not 1 TR. So 1 TR ≠ 860 kcal/hr. Options (a), (b) and (c) are the three book-quoted equivalents of 1 TR; 860 kcal/hr is the heat equivalent of 1 kW of electricity, a different conversion altogether, so (d) is the odd one out.
Source: 19th Exam
📖 §4.9 EER & §4.7 kW/TR
89. If the power consumed by an air conditioner compressor is 1.7 kW per ton of refrigeration, then its energy efficiency ratio (Watt/Watt) is ___________
1.7
2.06
0.59
none of the above
Answer: B) 2.06
Confirmed vs Book-3 §4.9 - EER = cooling/power = 3.517 kW per TR / 1.7 kW = 2.06 W/W. Option (a) 1.7 just repeats the input figure and (c) 0.59 inverts it; EER = 3.517 kW of cooling per TR divided by 1.7 kW input = 2.06 W/W (i.e. kW/TR = 3.516/COP rearranged).
Source: Mar 2021 (Set B)
📖 §4.2 Psychrometric Chart
90. If the wet bulb temperature of air is 38 °C, then its relative humidity is __________%.
38 %
90 %
100 %
Insufficient data
Answer: D) Insufficient data
Confirmed vs Book-3 §4.2 - RH cannot be found from WBT alone; the dry-bulb temperature (or another property) is needed to fix the psychrometric state. Hence insufficient data. Options (a)-(c) assume one property fixes the state; the psychrometric chart needs two independent properties, so wet bulb alone cannot give RH.
Source: 19th Exam
📖 §4.9 Energy Efficiency Ratio (EER)
91. A package air conditioner of 5 TR capacity delivers a cooling effect of 4 TR. If Energy Efficiency Ratio (W/W) is 2.90, the power in kW drawn by compressor would be:
4.84
1.38
1.724
None of the above
Answer: A) 4.84
Confirmed vs Book-3 §4.9 - Cooling delivered = 4 TR = 4 x 3.51 = 14.04 kW. Power = cooling/EER = 14.04/2.90 = 4.84 kW. Option (b) 1.38 divides by 2.9 twice and (c) 1.724 uses the 5 TR nameplate; use the DELIVERED 4 TR = 14.07 kW and divide by the EER of 2.90.
Source: 19th Exam
📖 §4.11 Heat Pumps (Figure 4.13 energy balance)
92. A heat pump used in a heat recovery application extracts 66220 kcal/hr and the power consumed by the heat pump is 23 kW. The estimated heat supplied by the heat pump is
2916 kcal/hr
47300 kcal/hr
86860 kcal/hr
86000 kcal/hr
Answer: D) 86000 kcal/hr
Corrected (was c) - Book-3 §4.11: heat delivered by a heat pump = heat extracted + compressor work (Qe + W, Figure 4.13). W = 23 kW x 860 = 19,780 kcal/hr, so heat supplied = 66,220 + 19,780 = 86,000 kcal/hr exactly (77 kW + 23 kW = 100 kW). Option (c) 86,860 does not follow from any standard kWh-to-kcal conversion; option (b) wrongly subtracts the work input instead of adding it.
Source: 19th Exam
📖 §4.2 Psychrometrics (relative humidity)
93. Which of the following is expressed in terms of percentage?
Absolute humidity
Relative humidity
Specific Gravity
All of the above
Answer: B) Relative humidity
Confirmed vs Book-3 §4.2 - Relative humidity is the ratio (expressed as %) of actual vapour pressure to saturation vapour pressure; absolute/specific humidity are mass ratios. Absolute humidity (a) is in g/kg dry air and specific gravity (c) is a dimensionless ratio to water; only relative humidity is quoted as a percentage on the psychrometric chart.
Source: Sep 2019
📖 §4.7 Ton of Refrigeration (TR)
94. If 30240 kcal of heat is removed from a room every hour then the refrigeration tonnage will be nearly equal to ____________.
30.24 TR
3.024 TR
1 TR
10 TR
Answer: D) 10 TR
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/hr; 30240/3024 = 10 TR. Options (a) and (b) omit the division by 3024; 30,240/3024 = exactly 10 TR, which is why the examiner chose that number.
Source: Sep 2019
📖 §4.3 VCR heat balance (condenser duty)
95. In a vapour compression refrigeration system, the quantum of energy transferred at condenser is more than the energy transferred at ___________.
Compressor
Expansion Valve
Evaporator
All of the above
Answer: C) Evaporator
Confirmed vs Book-3 §4.3 - Condenser heat = evaporator heat + compressor work, so condenser rejects more energy than the evaporator absorbs. The book states the condenser must reject the combined evaporator heat plus the compressor work, so it handles more than the evaporator (c); the expansion valve (b) transfers no heat at all.
Source: Sep 2019
📖 §4.3 & Table 4.3 Refrigerant / absorbent
96. ____________ is used as refrigerant both in vapour compression and vapour absorption systems.
Lithium Bromide
Water
HFC 134A
Ammonia
Answer: D) Ammonia
Confirmed vs Book-3 §4.3 - Ammonia (R-717) is used as a refrigerant in vapour compression systems and as the refrigerant in ammonia-water vapour absorption systems. Lithium bromide (a) is the absorbent, water (b) is the refrigerant only in LiBr machines, and HFC-134a (c) is compression-only; ammonia is the one fluid used as refrigerant in both systems.
Source: Sep 2019
📖 §4.7 Ton of Refrigeration (TR)
97. 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.
Source: 9th Dec-2009
📖 §4.3 & Table 4.3 Refrigerant / absorbent
98. 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.
Source: 9th Dec-2009
📖 §4.7 TR formula (coolant side)
99. 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.
Source: 9th Dec-2009
📖 §4.5 Centrifugal Compressors
100. 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.
Source: 9th Dec-2009
📖 §4.3 VAR COP vs VCR COP
101. 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.
Source: 9th Dec-2009
📖 §4.8 Table 4.5 (condenser temperature)
102. 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.
Source: 9th Dec-2009
📖 §4.7 COP & kW/TR
103. 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.
Source: 9th Dec-2009
📖 §4.11 Heat Pumps and Their Applications
104. 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.
Source: 9th Dec-2009
📖 §4.5 Screw Compressors; §4.17 Case Study
105. 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%).
Source: 9th Dec-2009
📖 §4.7 COP & kW/TR
106. Coefficient of Performance (COP) for a refrigeration compressor is given by ________.
power input to compressor (kW) / cooling effect (kW)
cooling effect (kW) / Power input to compressor (kW)
Q x CP x (Ti - To) / 3024
none of the above
Answer: B) cooling effect (kW) / Power input to compressor (kW)
Confirmed vs Book-3 §4.7 - COP = useful cooling effect divided by work input to the compressor. Option (a) is the inverted ratio (that is essentially kW/TR), and (c) is the TR formula for the coolant side; §4.7 defines COP as cooling effect (kW) divided by power input to the compressor (kW).
Source: Mar 2023
📖 §4.9 EER & §4.7 kW/TR
107. If the energy efficiency ratio (Watt/Watt) of a split air conditioner is 2.3, then power consumed by it per ton of refrigeration will be
1.53 kW
0.66 kW
2.3 kW
none of the above
Answer: A) 1.53 kW
Confirmed vs Book-3 §4.9 - 1 TR = 3.517 kW cooling. Power = 3.517/2.3 = 1.53 kW per TR. Option (c) 2.3 repeats the EER and (b) 0.66 inverts the calculation; kW/TR = 3.516/EER = 3.516/2.3 = 1.53 kW.
Source: 10th Jul-2010
📖 §4.7 TR formula (coolant side)
108. The refrigeration load in TR when 84 litre/minute of water is cooled from 21°C to 15°C is about
0.166
1.66
16.66
10
Answer: D) 10
Confirmed vs Book-3 §4.7 - Heat = 84×60 kg/hr × 1 × (21−15) = 5040×6 = 30240 kCal/hr. TR = 30240/3024 = 10 TR. Options (a)-(c) misplace the decimal; 84 litre/min = 5040 kg/hr, and 5040 x 6/3024 = exactly 10 TR.
Source: 10th Jul-2010
📖 §4.7 COP & kW/TR
109. The COP of a vapour compression refrigeration system is 3.5. If the motor delivers power of 10.8 kW at its shaft with a 90% motor efficiency, the cooling effect will be
34 kW
37.8 kW
0.36 kW
none of the above as cooling effect is always measured in TR
Answer: B) 37.8 kW
Confirmed vs Book-3 §4.7 - Cooling effect = COP × shaft power = 3.5 × 10.8 = 37.8 kW. Here the 10.8 kW is already the SHAFT output, so no efficiency correction applies and (a) 34 kW - the answer when 10.8 kW is a motor input - is the trap; cooling = 3.5 x 10.8 = 37.8 kW.
Source: 10th Jul-2010
📖 §4.9 Energy Efficiency Ratio (EER)
110. A package air conditioner of 5 TR capacity delivers a cooling effect of 4 TR. If EER (W/W) is 2.90, the power in kW drawn by compressor would be:
4.85
1.38
1.724
none of the above
Answer: A) 4.85
Confirmed vs Book-3 §4.9 - Cooling = 4 TR = 4*3.516 = 14.064 kW; Power = 14.064/2.90 = 4.85 kW. Option (c) 1.724 uses the 5 TR nameplate rating instead of the 4 TR actually delivered; power = 4 x 3.516/2.90 = 4.85 kW.
Source: 11th Feb-2011
📖 §4.14 Humidifying air by adding water
111. In an air washer of textile humidification system airflow of 3000 m3/h at 25 oC and 10% relative humidity is humidified to 60% relative humidity by adding water through spray nozzles. The specific humidity of air at inlet and outlet are 0.002 kg/kg and 0.0062 kg/kg respectively. The amount of water required in kg/hr is
14.9
6
10
none of the above
Answer: A) 14.9
Confirmed vs Book-3 §4.14 - Mass of dry air ~3000*1.18 = 3540 kg/hr; water = 3540*(0.0062-0.002) = 14.9 kg/hr. Options (b) and (c) are round-number distractors; the book's worked example gives mass of air x change in humidity ratio = 3000 x 1.184 x (0.0062 - 0.002) = 14.9 kg/h.
Source: 11th Feb-2011
📖 §4.3 VCR cycle stages (4-1 expansion device)
112. In a vapour compression refrigeration system, the component where the refrigerant fluid experiences no heat loss or gain is
compressor
condenser
expansion valve
evaporator
Answer: C) expansion valve
Confirmed vs Book-3 §4.3 - Expansion through the throttling valve is adiabatic (isenthalpic) - no heat exchange. Heat is absorbed in the evaporator (d), rejected in the condenser (b) and work is added in the compressor (a); only the expansion device has no heat gain or loss, per §4.3.
Source: 11th Feb-2011
📖 §4.7 TR formula (coolant side)
113. The refrigeration load in TR when 20 m3/hr of water is cooled from 13 oC to 8 oC is about
33
80.3
39.6
none of the above
Answer: A) 33
Confirmed vs Book-3 §4.7 - Load = 20000 kg/hr * 1 * (13-8) = 100000 kCal/hr; /3024 = 33 TR. Options (b) and (c) come from using the wrong temperature difference; load = 20,000 x (13 - 8)/3024 = 33 TR.
Source: 11th Feb-2011
📖 §4.7 COP & kW/TR
114. The COP of a vapour compression refrigeration system is 3.0. If the compressor motor output is 9.555 kW, the tonnage (TR) of the refrigeration system is
8.15
28.665
3
none of the above
Answer: A) 8.15
Confirmed vs Book-3 §4.7 - Refrigeration effect = COP*power = 3*9.555 = 28.665 kW; /3.516 = 8.15 TR. Option (b) 28.665 is the cooling effect in kW, not TR, and (c) simply repeats the COP; divide 28.665 kW by 3.516 kW/TR to get 8.15 TR.
Source: 11th Feb-2011
📖 §4.7 TR formula (coolant side)
115. A process fluid at 40 m3/hr, with a density of 0.95, is flowing in a heat exchanger and is to be cooled from 35 oC to 29 oC. The fluid specific heat is 0.78 kCal/kg. If the chilled water range across the heat exchanger is 4 oC, the chilled water flow rate is
44.46 m3/hr
40 m3/hr
35 m3/hr
none of the above
Answer: A) 44.46 m3/hr
Confirmed vs Book-3 §4.7 - Heat load = 40*1000*0.95*0.78*(35-29) = 177840 kCal/hr; chilled water flow = 177840/(1000*1*4) = 44.46 m3/hr. Option (b) 40 copies the process-fluid flow and ignores the density and specific heat; balance the duty 177,840 kcal/h against the 4 degC chilled-water range.
Source: 11th Feb-2011
📖 Book-3 Ch-3 Compressed Air (dryers) - cross-chapter
116. The most energy intensive dryer among the following
refrigeration
desiccant (heat of compression)
desiccant (heatless purge)
desiccant (blower reactivated)
Answer: C) desiccant (heatless purge)
Confirmed vs Book-3 Ch-3 (Compressed Air, dryers) - Heatless purge desiccant dryers consume the most energy because a large fraction of compressed air is used for purging. Refrigeration and heat-of-compression dryers (a, b) use little or no purge air, and blower-reactivated units (d) recover part of it; the heatless purge type diverts the largest share of dried compressed air, making it the most energy intensive.
Source: Jul 2022
📖 §4.3 Evaporative Cooling
117. Which of the following uses concept of evaporative cooling ?
cooling tower
domestic refrigerator
window air conditioner
deep freezer
Answer: A) cooling tower
Confirmed vs Book-3 §4.3 - A cooling tower cools water by evaporation of a small portion of the circulating water (evaporative cooling). A domestic refrigerator, window AC and deep freezer (b, c, d) are all vapour COMPRESSION machines; §4.3 explains that evaporative cooling - air brought into close contact with water and cooled towards the wet bulb temperature - is the cooling-tower principle.
Source: Mar 2021
📖 Book-3 Ch-2 Electric Motors - cross-chapter
118. A 22 kW motor rated for 415 V, 42 A and 0.8 power factor will have an efficiency of ___________
91 %
92 %
89.9 %
none of the above
Answer: A) 91 %
Confirmed vs Book-3 Ch-2 (Electric Motors) - Input = 1.732 x 415 x 42 x 0.8 = 24159 W = 24.16 kW; efficiency = 22/24.16 = 91%. Options (b) and (c) come from dropping the root-3 factor or the power factor; input = 1.732 x 415 x 42 x 0.8 = 24.16 kW, so efficiency = 22/24.16 = 91%.
Source: Mar 2021 (Set B)
📖 §4.3 Absorption Refrigeration (Absorber)
119. In a water Lithium bromide refrigeration system, the concentration of the lithium bromide gets diluted in ___________
evaporator
condenser
generator
absorber
Answer: D) absorber
Confirmed vs Book-3 §4.3 - In the absorber the refrigerant vapour (water) is absorbed into LiBr solution, diluting its concentration. The generator (c) does the opposite - it boils off water and CONCENTRATES the solution - while the evaporator (a) and condenser (b) handle only refrigerant water; the absorber is where refrigerant vapour is absorbed and the LiBr becomes dilute.
Source: Mar 2021 (Set B)
📖 §4.3 Absorption Refrigeration (COP 0.65-0.70)
120. Single stage Li Br water absorption refrigeration systems have a COP in the range of
0.40 - 0.5
0.65 - 0.70
0.75 - 0.80
0.2 - 0.3
Answer: B) 0.65 - 0.70
Confirmed vs Book-3 §4.3 - Single-effect (single-stage) LiBr-water absorption chillers have a COP of about 0.65 to 0.70. Option (c) 0.75-0.80 belongs to double-effect machines and (d) 0.2-0.3 to half-effect; §4.3 gives 0.65-0.70 for single-stage LiBr-water absorption chillers.
Source: Mar 2021
📖 §4.5 Compressor Types (hermetic systems)
121. Hermetic system is used in ________.
domestic refrigerator
centrifugal chillers
screw chillers
large reciprocating chillers
Answer: A) domestic refrigerator
Confirmed vs Book-3 §4.5 - the printed stem 'Cooling system' is a transcription slip for 'Hermetic system': a hermetic unit has motor and compressor in one sealed casing and the book confines it to refrigerators, air conditioners and other low-capacity applications. Centrifugal, screw and large reciprocating chillers (b, c, d) all use open-type machines with the motor separate from the compressor, as industrial service needs field-serviceable drives.
Source: Mar 2021
📖 §4.9 Energy Efficiency Ratio (EER)
122. A package air conditioner of 5 TR capacity delivers a cooling effect of 4 TR. If the Energy Efficiency Ratio (W/W) is 2.90, the power in kW drawn by the compressor would be ________.
4.84
1.38
1.724
none of the above
Answer: A) 4.84
Confirmed vs Book-3 §4.9 - Cooling delivered = 4 TR x 3.517 = 14.07 kW; Power = 14.07/2.90 = 4.84 kW. Option (c) 1.724 uses the 5 TR nameplate instead of the 4 TR delivered, and (b) inverts the arithmetic; power = 4 x 3.517/2.90 = 4.84 kW.
Source: Mar 2021
📖 §4.7 COP-Carnot
123. COP of an air conditioner will be least with ________.
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: A) lower evaporator temperature and higher condenser temperature
Confirmed vs Book-3 §4.7 - COP falls as the temperature lift increases; lowest COP occurs with lowest evaporator and highest condenser temperature. COP-Carnot = Te/(Tc - Te), so the worst case is the largest temperature lift; (b) gives the highest COP and (c) and (d) give intermediate values.
Source: Mar 2021
📖 §4.2 Psychrometrics (saturation / WBT)
124. When the evaporation of water from a wet surface at atmospheric condition is zero, ___________
RH is 0%
RH is 100%
wet bulb temperature is greater than dry bulb temperature
none of the above
Answer: B) RH is 100%
Confirmed vs Book-3 §4.2 - At 100% relative humidity the air is saturated, so no further evaporation occurs. Option (a) 0% RH would give the fastest evaporation, and (c) is impossible because the wet bulb can never exceed the dry bulb; evaporation stops only when the air is saturated at 100% RH.
Source: Mar 2021 (Set B)
📖 Book-3 Ch-10 ECBC & building EPI - cross-chapter
125. The unit of AAEPI is given by ___________
kWh/m2/yr
x kWh/hr
(Wh/m2)/hr
m3/Wh/yr
Answer: C) (Wh/m2)/hr
Confirmed vs Book-3 Ch-10 (ECBC / building energy index) - Air conditioning Annual Energy Performance Index is expressed as Wh per m2 per hour. Options (a), (b) and (d) mix up units used for building EPI and volumetric indices; the air-conditioning annual energy performance index is expressed per unit floor area per hour, i.e. (Wh/m2)/hr.
Source: Mar 2021 (Set B)
📖 §4.5 Compressor Types (hermetic systems)
126. Hermetic system is used in ___________
domestic refrigerator
centrifugal chillers
screw chillers
large reciprocating chillers
Answer: A) domestic refrigerator
Confirmed vs Book-3 §4.5 - Hermetically sealed compressors are used in small systems such as domestic refrigerators. Centrifugal, screw and large reciprocating chillers (b, c, d) use open-type machines with a separate motor; §4.5 confines hermetic (sealed motor-compressor) units to refrigerators, air conditioners and other low-capacity applications.
Source: Mar 2021 (Set B)
📖 Book-3 Ch-3 Compressed Air (dew point) - cross-chapter
127. Which of the following dew points of the compressed air, the moisture content would be maximum?
-10degC
-5degC
-40degC
none of the above
Answer: B) -5degC
Confirmed vs Book-3 Ch-3 (Compressed Air, dew point) - A higher (less negative) dew point corresponds to more residual moisture in the air. The lower (more negative) the pressure dew point the drier the air, so -40 degC (c) holds the least moisture; -5 degC is the highest dew point offered and therefore carries the most residual moisture.
Source: Mar 2021 (Set B)
📖 §4.3 VCR cycle stages (1-2-3-4)
128. Identify the wrong statement from the following regarding Vapor Compression Refrigeration system
condenser rejects heat to atmosphere
evaporator removes heat from process or space
compressor sends superheated vapor to condenser
high pressure sub-cooled liquid refrigerant returns back to evaporator
Answer: D) high pressure sub-cooled liquid refrigerant returns back to evaporator
Confirmed vs Book-3 §4.3 - Liquid passes through an expansion device which drops it to low pressure before entering the evaporator; it does not return as high-pressure sub-cooled liquid. Statements (a), (b) and (c) all match Figure 4.4; (d) is the wrong one because the high-pressure sub-cooled liquid must pass through the expansion device first, arriving at the evaporator as a low-pressure wet vapour.
Source: Jul 2022
📖 §4.7 Ton of Refrigeration (TR)
129. If 30,000 kcal of heat is removed from a room every hour then the refrigeration tonnage will be exactly equal to
30 TR
15 TR
10 TR
100 TR
Answer: C) 10 TR
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/h (book value); 30000/3024 ≈ 9.9 ≈ 10 TR. Option (a) omits the division by 3024 and (b) halves it; 30,000/3024 = 9.92 TR, which the paper rounds to 10 TR (30,240 kcal/h would be exactly 10 TR).
Source: Jul 2022
📖 §4.3 & Table 4.3 Refrigerant / absorbent
130. Which gas is used as refrigerant both in vapour compression and vapour absorption systems
Lithium Bromide
Water
HFC 134A
Ammonia
Answer: D) Ammonia
Confirmed vs Book-3 §4.3 - Ammonia is used as the refrigerant in both vapour compression and vapour absorption (with water) systems. Lithium bromide (a) is the absorbent and water (b) the refrigerant only in LiBr machines, while HFC-134a (c) is compression-only; ammonia is used as the refrigerant in both types.
Source: Jul 2022
📖 §4.7 COP & kW/TR
131. The COP of a vapour compression refrigeration system is 3.3. If the motor draws power of 10 kW at an operating efficiency of 90%, the tonnage of refrigeration system is about:
0.8
8.5
7.2
9.6
Answer: B) 8.5
Confirmed vs Book-3 §4.7 - Refrigeration effect = COP×shaft power = 3.3×(10×0.9) = 29.7 kW; TR = 29.7/3.517 ≈ 8.45 ≈ 8.5 TR. Option (c) 7.2 forgets the COP of 3.3 and (d) 9.6 skips the motor efficiency; shaft power = 10 x 0.9 = 9 kW, cooling = 29.7 kW, and TR = 29.7/3.517 = 8.5.
Source: Jul 2022
📖 §4.7 TR formula (coolant side)
132. A process fluid at 40 m³/hr, with a density of 0.95, is flowing in a heat exchanger and is to be cooled from 35 °C to 29 °C. The fluid specific heat is 0.78 kCal/kg. The chilled water range across the heat exchanger is 4 °C, the chilled water flow rate is
44.46 m³/hr
40.41 m³/hr
35.37 m³/hr
none of the above
Answer: A) 44.46 m³/hr
Confirmed vs Book-3 §4.7 - Heat load = 40×0.95×1000×0.78×(35-29) = 177840 kcal/hr; chilled water flow = 177840/(1000×1×4) = 44.46 m³/hr. Option (b) copies the process flow rate; the chilled-water flow must carry the same duty (177,840 kcal/h) over its own 4 degC range, giving 44.46 m3/hr.
Source: Jul 2022
📖 §4.3 Absorption Refrigeration (Generator)
133. In a water-lithium bromide absorption refrigeration system, the lithium bromide solution is re-concentrated in the ________.
Evaporator
Condenser
Generator
Absorber
Answer: C) Generator
Confirmed vs Book-3 §4.3 - In a Li-Br vapour absorption system the solution is most concentrated (strongest) at the generator where refrigerant (water) vapour is boiled off. Stem completed from the incomplete printed line. §4.3: the dilute LiBr leaving the absorber is heated in the GENERATOR by steam/hot water/oil, boiling off refrigerant water and re-concentrating the solution; it is most dilute in the absorber, and the evaporator and condenser carry only refrigerant water. The absorber (d) is where the solution becomes DILUTE, and the evaporator (a) and condenser (b) carry only refrigerant water; §4.3 states the diluted LiBr must be made concentrated in the generator using steam, hot water, gas or oil.
Source: Mar 2023
📖 §4.7 Integrated Part Load Value (IPLV)
134. Integrated Part Load Value (IPLV) in a vapor compression refrigeration refers to average of __________ at partial loads.
TR/kW
kW/TR
kW.TR
kW
Answer: B) kW/TR
Confirmed vs Book-3 §4.7 - IPLV is a weighted average of the chiller efficiency (kW/TR) at standard part-load points. Option (a) TR/kW inverts the indicator and (d) kW alone ignores capacity; §4.7 defines IPLV as an average of kW/TR taken at 100%, 75%, 50% and 25% load, because full-load conditions occur for barely 1% of running hours.
Source: Mar 2023
📖 §4.2 Psychrometrics (DBT = WBT at saturation)
135. Dry Bulb and Wet bulb temperature will be same at __________.
0% Relative Humidity
50% Relative Humidity
100% Relative Humidity
None of the above
Answer: C) 100% Relative Humidity
Confirmed vs Book-3 §4.2 - At saturation (100% RH) there is no evaporative cooling, so DBT = WBT. At 0% or 50% RH (a, b) evaporation from the wet wick depresses the wet bulb below the dry bulb; only at saturation does evaporation cease, so DBT = WBT = dew point.
Source: Mar 2023
📖 §4.12 Ventilation Systems (ACH)
136. What will be ventilation rate of 15mx10mx5m room having ACH of 10 __________.
750 m³/hr
7500 m³/hr
75000 m³/hr
None of the above
Answer: B) 7500 m³/hr
Confirmed vs Book-3 §4.12 - Volume = 15×10×5 = 750 m³; ventilation rate = 750 × 10 ACH = 7500 m³/hr. Options (a) and (c) are decimal-shift traps; ventilation rate = L x B x H x ACH = 15 x 10 x 5 x 10 = 7500 m3/hr, the same method as the book's compressor-room example.
Source: Mar 2023
📖 §4.7 Ton of Refrigeration (TR)
137. 100 kCal/min heat transfer rate is equivalent to __________.
1.98 TR
0.98 TR
19.8 TR
None of the above
Answer: A) 1.98 TR
Confirmed vs Book-3 §4.7 - 1 TR = 3024 kcal/h = 50.4 kcal/min. 100 kcal/min ÷ 50.4 = 1.98 TR. Option (b) 0.98 halves the result and (c) 19.8 shifts the decimal; 1 TR = 3024 kcal/h = 50.4 kcal/min, so 100/50.4 = 1.98 TR.
Source: Mar 2023
📖 §4.14 Humidification vs dehumidification; §4.2
138. Dehumidification 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: B) reducing dry bulb temperature and specific humidity
Corrected (was d) - Book-3 §4.14 defines HUMIDIFICATION as "reduction in dry bulb temperature and increase in specific humidity", so option (d) is the definition of humidification, not dehumidification. Dehumidification removes moisture: the air is cooled below its dew point on the coil, so both dry bulb temperature and specific humidity fall (b). Note: if the paper's stem actually read 'Humidification', the key (d) would be right - read the stem carefully in the exam.
Source: Mar 2023
Short questions (5 marks) — 35
📖 §4.9 Energy Efficiency Ratio (EER) / §4.15 Standards and Labeling of Room Air Conditioners, Tables 4.8 & 4.9
1. Define EER as used in BEE star labeling of room air conditioners, and state the 5-star EER thresholds (2014-15 band) for split and unitary ACs.
Model answer: Definition (Book-3 §4.9): EER = Refrigeration effect (Watts) ÷ Power input (Watts), expressed in W/W, measured at FULL-LOAD conditions. It is the ratio adopted in the BEE star labeling programme; being a cooling-out/power-in ratio in consistent units, it is numerically the same as the full-load COP. Higher EER = more efficient.
Scope (§4.15): the labeling standard covers single-phase SPLIT and UNITARY room air conditioners of the vapour compression type for household use, up to a rated cooling capacity of 11 kW, within the scope of IS 1391 Parts 1 and 2. Ratings run from a minimum of 1 star to a maximum of 5 stars in one-star intervals.
5-star thresholds for the band valid 01 Jan 2014 – 31 Dec 2015:
• SPLIT type (Table 4.8): 5 star = EER 3.50 W/W and above (4 star = 3.30–3.49).
• UNITARY type (Table 4.9): 5 star = EER 3.30 W/W and above (4 star = 3.10–3.29).
Note the split bands sit one step (0.20 W/W) higher than the unitary bands at every star level.
Confirmed vs Book-3 §4.9 and §4.15 — quote the formula in W/W, the 11 kW / single-phase / vapour-compression scope, and the two 5-star numbers 3.50 (split) and 3.30 (unitary). Do not quote later BEE bands or ISEER — the 2014 guidebook's Tables 4.8/4.9 are the examinable values.
Source: AI practice
📖 §4.7 TR formula — air side (AHU / FCU / package A/C); §4.9 worked example
2. In a package air conditioner, air flows at 2.5 m/s through a 1.2 m2 duct. The specific volume is 0.85 m3/kg, inlet enthalpy 9.37 kcal/kg and outlet enthalpy 7.45 kcal/kg. Calculate the air flow (m3/hr) and the cooling effect in TR.
Model answer: Step 1 — Air flow: Q = velocity × area = 2.5 × 1.2 = 3.0 m³/s. Converting, Q = 3.0 × 3600 = 10,800 m³/hr.
Step 2 — Cooling effect, air-side formula (Book-3 §4.7, as applied in the book's package-A/C example):
TR = [Q × (h_in − h_out)] / (specific volume × 3024)
= [10,800 × (9.37 − 7.45)] / (0.85 × 3024)
= (10,800 × 1.92) / 2570.4
= 20,736 / 2570.4
= 8.07 TR.
Step 3 — In kW: 8.07 × 3.51 = 28.3 kW of cooling.
(The specific volume converts the volumetric flow into a mass flow of dry air: 10,800/0.85 = 12,706 kg/hr, and 12,706 × 1.92 kcal/kg = 24,396... more directly 20,736/0.85 = 24,395 kcal/hr, ÷ 3024 = 8.07 TR.)
Confirmed vs Book-3 §4.7/§4.9 — same method as the book's 10 TR package A/C example, which computes [(9.37 − 7.45) × 4751]/(0.8405 × 3024) = 3.6 TR. Enthalpies come from the psychrometric chart via measured DBT and WBT; keep enthalpy in kcal/kg so that the 3024 divisor is valid, and remember to convert m³/s to m³/hr first.
Source: AI practice
📖 §4.7 Integrated Part Load Value (IPLV)
3. What is IPLV and why is it a better performance reference than full-load kW/TR for a chiller?
Model answer: IPLV (Integrated Part Load Value) is the average of the chiller's kW/TR measured at partial loads, capturing FOUR points — 100%, 75%, 50% and 25% of capacity — with equal weight given to each.
Why it is a better reference than full-load kW/TR (Book-3 §4.7): the plain kW/TR figure is derived at 100% capacity under design (most critical) conditions, and those conditions may occur for only about 1% of the total operating time in a year. A chiller therefore spends almost all its life at part load, where its kW/TR is quite different from the design value, so full-load kW/TR alone will mis-state annual energy use. IPLV averages performance across the load range and so reflects real operation far more closely. A LOWER IPLV in kW/TR is better.
Limitation the book adds: most equipment actually operates between 50% and 75% load, so IPLV — which weights all four points equally — is 'the most appropriate but not the best' reference; a specific analysis for each case, plus a yearly operating profile of the heat exchangers, should be prepared.
Confirmed vs Book-3 §4.7 — for full marks state the four load points (100/75/50/25%), the equal weighting, the '1% of operating time at design condition' justification, and the 50–75% real-operation caveat the book itself raises.
Source: AI practice
📖 §4.7 TR & heat rejection; §4.3 VAR
4. Determine the difference in heat rejected in kCal/TR to the cooling tower for two different types of air conditioning system operating at same capacity.
Parameter — Centrifugal chiller / VAM:
Chilled water flow (m³/h): - / 180
Condenser water flow (m³/h): - / 340
Chiller inlet temp (°C): 13.0 / 14.6
Condenser water inlet temp (°C): - / 33.5
Chiller outlet temp (°C): 7.7 / 9.0
Condenser water outlet temp (°C): - / 39.1
Specific power consumption (kW/TR): 0.6 / -
Heat rejected = refrigeration effect + compressor heat (chiller) vs condenser water heat balance per TR (VAM); take difference.
Source: Sep 2025
📖 §4.16 Energy Saving Opportunities
5. Discuss in brief any three methods by which energy can be saved in an air conditioning system.
Model answer: 1) Raise the chilled water / cold air set-point temperature (and reduce condenser water/refrigerant condensing temperature) to lower the temperature lift and improve COP. 2) Reduce the cooling load - improve building insulation, use solar films/shading, control fresh-air infiltration and ventilation, and switch off unnecessary loads. 3) Operate chillers at optimum loading and use efficient part-load control (VFDs on pumps/AHU fans, optimum sequencing of multiple chillers), keep heat-exchanger surfaces (condenser/cooling tower) clean, and maintain correct refrigerant charge. These reduce kW/TR and overall energy use.
Reduce temperature lift, cut the cooling load, and optimize chiller/auxiliary operation (clean surfaces, VFDs, sequencing) to lower kW/TR.
Source: Book EOC
📖 §4.7 Performance Assessment of Refrigeration Plants
6. What are the parameters required to be measured while estimating the chiller performance in kW/TR?
Model answer: To estimate chiller efficiency (kW/TR) measure: (1) chilled water flow rate, (2) chilled water inlet (return) and outlet (supply) temperatures across the evaporator (to get refrigeration load TR = flow x sp.heat x dT), and (3) the electrical power input to the chiller compressor (kW) using a power meter. kW/TR = measured kW / measured TR. (Condenser water flow and inlet/outlet temperatures are also measured for a heat balance.)
Need chilled water flow, evaporator in/out temperatures (for TR) and compressor input power (kW); kW/TR = kW divided by TR.
Source: Book EOC
📖 §4.7 Integrated Part Load Value (IPLV)
7. Explain the term Integrated Part Load Value (IPLV).
Model answer: IPLV is a single number that expresses a chiller's efficiency (kW/TR or COP) averaged over a range of operating capacities, because chillers rarely run at 100% load. It is a weighted average of performance at standard part-load points (typically 100%, 75%, 50% and 25% capacity) using weighting factors that represent the proportion of time the chiller is expected to spend at each load. IPLV thus better represents real seasonal/annual operating efficiency than the single full-load rating.
IPLV is a weighted-average part-load efficiency (e.g. at 100/75/50/25% load) reflecting real operation rather than only full-load performance.
Source: Book EOC
📖 §4.2 Psychrometric Chart (Figure 4.2)
8. Name five parameters that a psychrometric chart provides for an air conditioning engineer.
Model answer: A psychrometric chart gives, for moist air: (1) dry bulb temperature, (2) wet bulb temperature, (3) relative humidity, (4) specific/absolute humidity (humidity ratio / moisture content), (5) dew point temperature, plus (6) enthalpy (total heat) and (7) specific volume of the air. Any one state point fixes all these properties.
Dry bulb, wet bulb, relative humidity, humidity ratio (moisture content), dew point, enthalpy and specific volume.
Source: Book EOC
📖 §4.7 Ton of Refrigeration (TR)
9. Define one 'Ton of Refrigeration (TR)'.
Model answer: One ton of refrigeration is the amount of cooling produced by melting one short ton (2000 lb) of ice at 0 degC in 24 hours. It equals a heat removal rate of 3024 kcal/hr (= 12,000 BTU/hr = 3.517 kW).
1 TR = heat to melt 1 ton of ice in 24 h = 3024 kcal/hr = 12,000 BTU/hr = 3.517 kW.
Source: Book EOC
📖 §4.3 VCR vs VAR
10. State any three major differences between vapour compression refrigeration (VCR) and vapour absorption refrigeration (VAR) systems.
Model answer: 1. VCR uses electric power for the compressor; VAR uses a heat source. 2. VCR uses hydrogen-fluorine-carbon compounds as refrigerant; VAR uses water (or ammonia). 3. VCR works under pressure; VAR works under vacuum. 4. VCR has high COP; VAR has low COP. 5. VAR requires about double the cooling tower capacity of VCR.
Energy input, refrigerant, operating pressure, COP and cooling-tower sizing differences.
Source: 15th Exam
📖 §4.9 Package A/C worked example
11. A 20 TR package AC plant: air velocity across suction filter 2.5 m/s; suction area 1.2 m²; inlet air enthalpy 9.37 kcal/kg, outlet 7.45 kcal/kg; specific volume 0.85 m³/kg; power: compressor 10.69 kW, pump 4.86 kW, cooling tower fan 0.87 kW. Calculate: i) air flow rate m³/hr, ii) cooling effect kW, iii) compressor kW/TR, iv) overall kW/TR, v) EER kW/kW.
TR = Q x Δh /(specific volume x 3024); kW/TR and EER from power and cooling effect.
Source: 15th Exam
📖 §4.3 VCR vs VAR
12. Identify each statement as applicable to VCR or VAR: A) No effect of reducing load on performance. B) Uses low grade energy. C) Liquid traces in suction line may damage compressor. D) Moving parts only in pump, smooth operation. E) Can work on lower evaporator pressures without affecting COP. F) Performance adversely affected at partial loads. G) Liquid traces of refrigerant at exit of evaporator. H) Uses high-grade energy like mechanical work. I) Moving parts more; more maintenance and noise. J) COP decreases considerably with decrease in evaporator pressure.
Model answer: A) VAR, B) VAR, C) VCR, D) VAR, E) VAR, F) VCR, G) VAR, H) VCR, I) VCR, J) VCR.
VAR is heat-driven, smooth (pump only), load-tolerant; VCR is compressor-driven, sensitive to suction liquid and evaporator pressure.
Source: 14th Exam
📖 §4.11 Heat Pump Applications
13. List any five industrial applications of a heat pump.
Model answer: 1. Space heating. 2. Heating of process streams. 3. Water heating for washing, sanitation and cleaning. 4. Steam production. 5. Drying/dehumidification. (Also: evaporation, distillation, concentration.)
Standard industrial heat-pump applications from BEE guidebook.
Source: Set-A
📖 §4.9 Package A/C worked example
14. A water cooled 20 TR package AC plant: air velocity across suction filter 2.5 m/s, suction area 2.4 m²; inlet air enthalpy 9.37 kcal/kg, outlet 7.45 kcal/kg; specific volume 0.85 m³/kg; power: compressor 18.42 kW, pump 2.1 kW, evaporator fan 1.25 kW. Calculate: i) air flow rate, ii) cooling effect, iii) compressor kW/TR, iv) overall kW/TR, v) overall EER in W/W.
TR = Q x Δh/(specific volume x 3024); kW/TR and EER from power and cooling effect.
Source: Set-A
📖 §4.3 Absorption Refrigeration (Figure 4.5)
15. Identify the type of refrigeration system in the figure and the components 1,2,3 & 4. Explain briefly the function of each.
Model answer: Vapour Absorption Refrigeration system. 1 Absorber: concentrated LiBr absorbs the refrigerant vapour (water) and becomes dilute. 2 Generator: heats the dilute LiBr, regenerates refrigerant (water vapour) and re-concentrates LiBr. 3 Condenser: condenses the regenerated refrigerant (water vapour). 4 Evaporator: liquid refrigerant (water, atomised) picks up heat from the chilled-water coil and becomes water vapour.
Confirmed vs Book-3 §4.3 (Figure 4.5) - in the LiBr-water absorption chiller water is the refrigerant and LiBr solution the absorbent. Evaporator: water flashes at ~4 degC under 754 mmHg vacuum and chills the water from 12 to 7 degC. Absorber: concentrated LiBr absorbs the vapour (and maintains the vacuum), becoming dilute. Generator: steam/hot water/oil boils off the water and re-concentrates the LiBr. Condenser: condenses that vapour and returns it to the evaporator.
Source: 17th Sep-2016
📖 §4.7 COP-Carnot vs industry COP
16. Explain with equation for COP_Carnot that: (a) higher COP_Carnot is achieved with higher evaporator and lower condenser temperature; (b) COP_Carnot does not account for compressor type; (c) how COP is normally used in industry.
Model answer: a) COP_Carnot = Te/(Tc − Te), where Te = evaporator temperature and Tc = condenser temperature (in Kelvin). Higher Te and lower Tc increase the ratio, hence higher COP. b) Since it is only a ratio of temperatures, it does not consider the type of compressor. c) The industry COP = Cooling effect (kW)/Power input to compressor (kW), where cooling effect is the enthalpy difference across the evaporator expressed in kW.
Confirmed vs Book-3 §4.7 - COP-Carnot = Te/(Tc - Te) with temperatures in kelvin, so raising Te or lowering Tc shrinks the denominator and raises COP; this is exactly why the book advises the highest practical chilled-water temperature and the lowest practical condensing temperature. Because it is only a temperature ratio it says nothing about the compressor, so industry uses COP = cooling effect (kW) / compressor power input (kW).
Source: 16th Exam
📖 §4.12 Ventilation Systems; §4.7 air-side TR
17. a) Calculate the ventilation rate for an engine room 20 m × 10.5 m × 15 m if recommended ACH is 20. b) Air at 25,200 m3/hr, density 1.2 kg/m3, enthalpy difference inlet-outlet 2.38 kcal/kg flows into an AHU; motor draws 22 kW at 90% efficiency. Find the kW/TR. (1 cal = 4.183)
Model answer: a) Ventilation rate = L×W×H×ACH = 20×10.5×15×20 = 63,000 m3/hr. b) Heat removed = 25200×1.2×2.38 = 71,971 kcal/hr → TR = 71971/3024 = 23.8 TR. Compressor power = 22×0.9 = 19.8 kW. kW/TR = 19.8/23.8 = 0.83.
Ventilation = volume×ACH; TR=Q·ρ·Δh/3024; kW/TR.
Source: 16th Exam
📖 §4.16 Energy Saving Opportunities
18. List five energy saving measures in a centralized chilled water based air conditioning system.
Model answer: 1. Insulate all cold lines/vessels using economic insulation thickness to minimise heat gains. 2. Optimise air-conditioned volumes (false ceilings, segregate critical areas, use air curtains). 3. Minimise AC load by roof cooling/roof painting, efficient lighting, pre-cooling fresh air via air-to-air heat exchangers. 4. Optimal thermostatic temperature setting of conditioned spaces. 5. Minimise part-load operation by matching load with online plant capacity; use variable speed drives for varying load.
Standard chilled-water-system ECMs from Book-3 Chapter 4 (each point one mark).
Source: 18th Exam
📖 §4.2 Psychrometrics (humidity ratio)
19. A stream of moist air (mass flow 10.1 kg/s, specific humidity 0.01 kg/kg dry air) mixes with a second stream of superheated water vapour flowing at 0.1 kg/s. Assuming proper uniform mixing without condensation, what is the humidity ratio of the final stream (kg/kg dry air)?
Model answer: Dry air = 10.1/(1+0.01) = 10 kg/s; moisture in moist air = 0.1 kg/s. Final moisture = 0.1 + 0.1 (added vapour) = 0.2 kg/s. Humidity ratio H = (0.01x10 + 0.1x1)/10 = (0.1 + 0.1)/10 = 0.02 kg per kg of dry air.
Mass-balance the dry air and total moisture; humidity ratio = total moisture / dry-air mass.
Source: 18th Exam
📖 §4.7 TR formula; pump hydraulic power
20. S-2: A chilled water system runs always at full load; inlet/outlet 12 °C / 7 °C. Chilled-water pump discharge pressure 3.6 kg/cm2g, suction 5 m above pump centreline, motor power 70 kW at 90% efficiency, pump efficiency 60%. Find the operating refrigeration load in TR.
Model answer: Discharge head = 3.6 kg/cm²g ≈ 36 m; total head = 36 - 5 = 31 m. Pump shaft power = 70 x 0.9 = 63 kW. Flow = (pump shaft power x 1000 x pump eff)/(head x 1000 x 9.81) = (63 x 1000 x 0.6)/(31 x 1000 x 9.81) = 0.1243 m3/s = 447.5 m3/hr. Refrigeration load = (447500 x 5)/3024 = 740 TR.
Total head from discharge head minus suction lift; flow from pump hydraulic equation; TR = (flow(kg/hr) x ΔT)/3024.
Source: 19th Exam
📖 §4.14 Humidifying air by adding water
21. S-3: In an air washer of a textile humidification system, airflow 3000 m3/h at 25 °C and 10% RH is humidified to 60% RH. Inlet/outlet specific humidity = 0.002 / 0.0062 kg/kg dry air; air density at 25 °C = 1.184 kg/m3. Calculate the water required (kg/hr).
Model answer: Water required mw = volume x density x (ω_out - ω_in) = 3000 x 1.184 x (0.0062 - 0.002) = 3000 x 1.184 x 0.0042 = 14.9 kg/hr.
Water added = mass of air x change in specific humidity (mass flow = volume flow x density).
Source: 19th Exam
📖 §4.7 air-side TR formula (AHU/FCU)
22. A 10 TR AHU operates at 8.25 TR. Inlet enthalpy 10.26 kcal/kg, outlet enthalpy 7.26 kcal/kg, specific volume of air 0.83 m3/kg. Calculate the volume of air (m3/hr) handled by the AHU.
Model answer: Cooling (TR) = (Hi-Ho) x V/(v x 3024). Volume of air = TR x v x 3024/(Hi-Ho) = (8.25 x 0.83 x 3024)/(10.26-7.26) = 6903 m3/hr.
Confirmed vs Book-3 §4.7 - air-side load TR = Q x rho x (h_in - h_out)/3024, and with specific volume instead of density Q = TR x v x 3024/(h_in - h_out). Here Q = 8.25 x 0.83 x 3024/(10.26 - 7.26) = 6903 m3/hr. Note the AHU is rated 10 TR but is delivering only 8.25 TR - use the delivered load, not the nameplate, or the airflow comes out ~16% too high.
Source: Sep 2019
📖 §4.16 Energy Saving Opportunities
23. S-1: List five energy saving measures for air conditioning system.
Model answer: (1) Insulate all cold lines/vessels with economic insulation thickness to minimize heat gains. (2) Optimize air-conditioning volumes (false ceiling, air curtains, segregate critical areas). (3) Minimize AC loads (roof cooling/painting, efficient lighting, pre-cooling fresh air via air-to-air heat exchangers, sun film, variable volume air system, optimal thermostat setting). (4) Minimize part-load operation by matching loads to plant capacity on line and adopt variable speed drives. (5) Ensure regular maintenance, adequate chilled/cooling water flow, avoid bypass flow, frequent cleaning/descaling of heat exchangers; adopt VAR (non-CFC) where economical and continuously optimize condenser/evaporator parameters.
Confirmed vs Book-3 §4.16 - full marks for any five of the guidebook's listed measures: economic cold insulation, optimised air-conditioned volume (false ceilings, air curtains), minimised building heat load (roof cooling/painting, sun film, efficient lighting, pre-cooled fresh air), optimum thermostat setting and variable-volume air systems, and plant-side actions such as avoiding bypass flows, matching capacity to load with VSDs, cleaning condensers/evaporators and adopting VAR where economics permit.
Source: 16th Exam (alt set)
📖 §4.7 COP-Carnot vs industry COP
24. S-7: Explain with the 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 industry given?
Model answer: (a) COP_Carnot = Te/(Tc - Te), depending on evaporator temperature Te and condenser temperature Tc; raising Te and lowering Tc increases COP_Carnot. (b) Since COP_Carnot is only a ratio of absolute temperatures, it does not account for the type of compressor used. (c) The industry COP = Cooling effect (kW) / Power input to compressor (kW), where cooling effect is the enthalpy difference across the evaporator expressed in kW.
Confirmed vs Book-3 §4.7 - COP-Carnot = Te/(Tc - Te) in kelvin, so a higher evaporator temperature and a lower condenser temperature both raise it (Figures 4.10 and 4.11 quantify the same trend). Being purely a temperature ratio it ignores compressor type and real losses, so the industry COP used in audits is cooling effect (kW) / power input to compressor (kW), the cooling effect being the enthalpy rise across the evaporator.
Source: 16th Exam (alt set)
📖 §4.7 TR definition & air-side TR formula
25. 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.
Source: 9th Dec-2009
📖 §4.7 air-side TR formula (AHU/FCU)
26. In an AHU the actual airflow is 9300 m3/hr, inlet air enthalpy 16.12 kCal/kg, outlet enthalpy 13.33 kCal/kg, air density 1.15 kg/m3. Estimate the TR of the AHU.
Model answer: Air-side load: TR = Q x rho x (h_in - h_out)/3024 = 9300 x 1.15 x (16.12 - 13.33)/3024 = 9300 x 1.15 x 2.79 / 3024 = 29,839/3024 = 9.87 TR (about 9.86 TR). Heat removed = 29,839 kcal/hr; dividing by 3024 kcal/hr per TR gives the AHU refrigeration load.
Air-side load uses TR = airflow x air density x (h_in - h_out) / 3024, with enthalpies read off the psychrometric chart in kcal/kg. Here 9300 x 1.15 x 2.79 = 29,839 kcal/hr, which is 9.87 TR. Use density (kg/m3) with volumetric airflow, or specific volume in the denominator - not both.
Source: 10th Jul-2010
📖 §4.7 TR formula & COP
27. In an alkali plant, salt brine at 20 m3/hr is cooled from 14°C to 8°C using chilled water. The chiller compressor motor draws 44.4 kW at 90% motor efficiency. Allied auxiliaries draw 20 kW. Brine density 1.2 kg/litre, specific heat 0.97 kCal/kg°C. (a) Refrigeration load (TR); (b) COP of compressor; (c) overall specific power consumption (kW/TR).
Model answer: (a) Mass flow of brine = 20 m3/hr x 1000 x 1.2 kg/litre = 24,000 kg/hr. Refrigeration load TR = m x Cp x (Ti - To)/3024 = 24,000 x 0.97 x (14 - 8)/3024 = 139,680/3024 = 46.2 TR. (b) Compressor shaft power = 44.4 x 0.90 = 39.96 kW; COP = 3.516 x 46.2/39.96 = 162.4/39.96 = 4.06. (c) Overall specific power consumption = (44.4 + 20)/46.2 = 64.4/46.2 = 1.39 kW/TR.
Brine is not water: its density (1.2 kg/litre) and specific heat (0.97 kcal/kg degC) must both be used in TR = m x Cp x deltaT / 3024. COP uses the compressor SHAFT power (44.4 x 0.9), while overall kW/TR uses the total electrical input including the 20 kW auxiliaries - hence 4.06 vs 1.39 kW/TR.
Source: 10th Jul-2010
📖 §4.9 EER; §4.15 Standards and Labeling
28. S-4: In a Commercial building, five window ACs each of 1.5 TR capacity were evaluated for replacement with three star labeled new ACs having Energy Efficiency Ratio (EER) of 2.50 kW/kW. The measured EER of existing ACs: AC1 = 2.05, AC2 = 2.19, AC3 = 2.30, AC4 = 2.40, AC5 = 2.17. Calculate the total kW saving potential if all the existing ACs are replaced with 3 star labeled ACs of same capacity.
Model answer: Input kW = TR delivered*3.516/EER. For 3 star AC input power = 1.5*3.516/2.5 = 2.11 kW each. Existing kW input: AC1 = 2.573, AC2 = 2.408, AC3 = 2.293, AC4 = 2.198, AC5 = 2.430; Total = 11.902 kW. Savings potential = 11.902 - (2.11 x 5) = 11.902 - 10.55 = 1.352 kW.
Input kW = TR*3.516/EER for each AC; saving = sum of existing input - new input (5 x 2.11).
Source: 11th Feb-2011
📖 §4.11 Heat Pump Applications
29. S-7: Write any 5 industrial applications of a heat pump.
Model answer: Industrial heat pumps are mainly used for: Space heating; Heating of process streams; Water heating for washing, sanitation and cleaning; Steam production; Drying/dehumidification; Evaporation; Distillation; Concentration.
Any five of the listed industrial heat-pump applications per the official key.
Source: 11th Feb-2011
📖 §4.3 Absorption Refrigeration (VAR advantages)
30. What are the advantages of using vapour absorption refrigeration system over vapour compression system? Under what condition it would be economical? (5 Marks)
Model answer: Refer Guidebook-3 (Pg 112-116): Advantages of VAR over VCR — uses low-grade/waste heat or steam instead of high-grade electrical energy; very few moving parts (only pump) so low maintenance, low noise/vibration; uses environment-friendly refrigerants (water/ammonia); can use otherwise wasted heat. It is economical when low-cost waste heat, exhaust gas, or low-pressure steam is available, making the running cost low despite higher first cost.
Advantages and economic conditions for VAR per Guidebook-3.
31. a) Name six parameters along with units that a psychrometric chart provides (3 marks). b) Explain briefly about Thermal Emittance (2 marks).
Model answer: a) Refer Guidebook-3 Pg 272. Six parameters from psychrometric chart: 1. Dry bulb temperature (degC); 2. Relative humidity (%); 3. Wet bulb temperature (degC); 4. Specific volume (m3/kg of dry air); 5. Enthalpy (kcal/kg of dry air); 6. Specific humidity / humidity factor (grams/kg of dry air). b) Thermal emittance is the ratio of radiant heat flux emitted by a surface to that of a black body at the same temperature; high-emittance roof surfaces radiate absorbed heat readily, reducing heat gain.
Psychrometric chart properties per Guidebook-3; thermal emittance defined relative to a black body radiator.
Source: Mar 2021 (Set B)
📖 §4.3 VAR COP; §4.7 TR
32. A process plant has installed 5 MW DG set for base load operation, which is operating at 70% loading. Furnace oil is used as a fuel in the DG set. The DG set generates 8.6 kg of exhaust gas per kWh generated. The plant management has decided to install a heat recovery boiler to generate steam at 3 kg/cm2g from the exhaust gas to reduce the exit flue gas temperature from 450degC to 200degC. The specific heat of flue gas is 0.26 kcal/kgdegC. The steam generated from waste heat boiler will be used in double effect Li-Br Vapor Absorption Chiller, with a COP of 1.12. How much TR will be generated through VAM?
Model answer: DG loading = 5 MW x 70% = 3.5 MW = 3500 kW. Quantity of heat available from exhaust gas = 3500 x 8.6 x gas generated/kWh x 0.26 kcal/kgdegC x (450 - 200) = 19,56,500 kcal/hr. Potential TR generation through double-effect VAM: COP = TR effect/Heat input, so TR = (COP x Heat input)/3024 = (1.12 x 1956500)/3024 = 724.6 TR.
Waste heat from exhaust = mass x Cp x dT; refrigeration effect = COP x heat input; convert kcal/h to TR by dividing by 3024.
Source: Mar 2021 (Set B)
📖 §4.3 Absorption Refrigeration (VAR advantages)
33. What are the advantages of using vapour absorption refrigeration system over vapour compression system? Under what condition it would be economical?
Model answer: Refer Guidebook-3 (Pg 112-116). Advantages: uses low-grade/waste heat (steam, hot water, exhaust gas) instead of electricity, very low electrical power consumption, few moving parts so low noise/vibration and maintenance, uses environment-friendly refrigerant (water) with non-CFC, good part-load performance. It is economical where cheap waste heat or low-cost thermal energy is available and electricity is expensive.
VAR systems substitute thermal energy for shaft work; economical when waste/cheap heat is abundant relative to electricity cost.
Source: Mar 2021 (Set B)
📖 §4.7 TR formula; pump hydraulic power
34. An energy audit study of a central chiller system in a commercial building was conducted and measured parameters are given below: Chilled water inlet temperature 12 °C; Chilled water Outlet temperature 7 °C; Chilled water pump discharge pressure 3.6 kg/cm²g; Pump suction 1.5 meters above the pump-center line; Power drawn by the chilled water pump motor 70 kW; Efficiency of pump motor 91%; Pump efficiency 60%. Find out the operating load of the Chiller system in TR.
Model answer: Discharge head = 3.6 kg/cm2g x 10 = 36 m; the suction is 1.5 m above the pump centreline, so total head = 36 - 1.5 = 34.5 m. Pump shaft power = motor input x motor efficiency = 70 x 0.91 = 63.7 kW. Flow = (shaft power x 1000 x pump efficiency)/(head x 1000 x 9.81) = (63.7 x 1000 x 0.6)/(34.5 x 1000 x 9.81) = 0.1129 m3/s = 406.5 m3/hr. Refrigeration load = 406,500 x 1 x (12 - 7)/3024 = 672 TR.
Hydraulic power = flow x head; rearranged, flow = (pump shaft power x pump efficiency)/(head x 9.81). Total head = discharge head (3.6 kg/cm2g = 36 m) minus the 1.5 m positive suction lift = 34.5 m. Refrigeration load then follows from TR = mass flow x Cp x deltaT / 3024 (1 TR = 3024 kcal/hr).
Source: Jul 2022
📖 §4.7 TR formula; pump hydraulic power
35. A multi storied office has centralized air conditioning system by using the chilled water. The chilled water inlet and outlet temperatures are 13°C and 9°C respectively. The chilled water pump discharge pressure is 4.2 kg/cm²g and the suction is 10 meters above the pump centerline. The power drawn by the chilled water pump's motor is 75 kW and an efficiency of 92%. The chilled water pump efficiency at the operating point from pump characteristic curve is 65%. Find out the operating refrigeration load in TR.
Model answer: Total head of the Chilled Water Pump = (4.2×10) − 10 = 32 Meter. Shaft Power of the Pump = 75×0.92 = 69 kW. Flow rate = (69×1000×0.65)/(32×1000×9.81) = 0.14287 m³/s = 514.33 m³/hr. Refrigeration load = 514330×4/3024 = 680 TR.
Head from discharge pressure and static lift; hydraulic power gives flow; cooling load = flow × ΔT × density/specific heat converted to TR (3024 kcal/h per TR).
Source: Mar 2023
Long questions (10 marks) — 25
📖 §4.9 EER; §4.7 kW/TR; §4.11 Heat Pumps
1. A 5-star business hotel operates a centralized HVAC system round the clock. Only one chiller operates at a time, the other on standby. Two centrifugal chillers, each rated 250 TR, with EER varying with load: at 85% load (212.5 TR) EER 5.2 for 180 days; at 60% load (150 TR) EER 4.6 for 120 days; at 40% load (100 TR) EER 3.9 for 65 days. No change in EER above 85% load; assume chiller motor efficiency 90% at all loading conditions. During chiller operation, two pumps run in parallel at an 80% load factor, consuming a total of 19.7 kW; two cooling tower fans operate continuously with power consumption of 5.89 kW. Both pumps and fans function 24 hours a day, with overall efficiencies of 75% and 70% respectively. Electricity tariff is ₹6.5 per kWh. Evaluate:
a. The total annual energy consumption (in MWh) and cost of the HVAC system, considering part-load EERs and auxiliary loads. (4 Marks)
b. Heat removal by condenser in (TR) at different loads. (3 Marks)
c. The hotel is planning to use the chiller partially as a heat pump by mounting a plate heat exchanger in series between the compressor and condenser (desuperheater for partial heat recovery) for producing hot water. The heat recovery can be only 20% of the condenser heat discharge. If the hot water requirement is 2000 litres/hr with 10°C temperature rise, evaluate whether the hot water requirement can be met at 40% loading conditions. (3 Marks)
Model answer: a. Energy Consumed by Chiller (Cooling Load kW = TR×3024/860; Input Power = Load kW/EER; MWh = InputPower×Days×24/1000):
85%: 212.5 TR, EER 5.2 → 747.21 kW load, 143.69 kW input, 180 days → 620.8 MWh
60%: 150 TR, EER 4.6 → 527.44 kW load, 114.66 kW input, 120 days → 330.2 MWh
40%: 100 TR, EER 3.9 → 351.63 kW load, 90.16 kW input, 65 days → 140.7 MWh
Energy Consumed by auxiliaries = (19.7 + 5.89) × 24 × 365 / 1000 = 224.5 MWh
Total Annual Energy Consumption = 620.8 + 330.2 + 140.7 + 224.5 = 1316 MWh
Annual Cost = 1316000 × 6.5 = Rs. 85.55 Lakh
b. Heat Removal by Condenser (Power Input to Compressor P = Input Power × 0.9; Condenser Heat Load HL = Cooling Load TR + P×860/3024):
85%: 212.5 TR, 143.69 kW input, P = 129.32 kW → 249.28 TR
60%: 150 TR, 114.66 kW input, P = 103.20 kW → 179.35 TR
40%: 100 TR, 90.16 kW input, P = 81.14 kW → 123.08 TR
c. Heat Load at 40% loading = 123.08 × 3024 = 372194 kCal/Hr; Recovery potential (20%) = 74439 kCal/Hr; Heat requirement for hot water generation = 2000 × 1 × 10 = 20000 kCal/H. Therefore, the heating requirement can easily be met at 40% loading.
Chiller input = load/EER; annual MWh summed over part-load days plus auxiliaries; condenser heat = cooling load + compressor heat; compare 20% recovery against hot-water duty.
Source: Sep 2025
📖 §4.3 Absorption Refrigeration; §4.7 TR
2. A process engineer develops a scheme to put 500 TR absorption-based refrigeration system to bring down process fluid temperature from 34 °C to 26 °C and this will result in higher production by 10%. 5 TPH excess steam is available in the plant and this new scheme utilizes this excess steam. COP of refrigeration system is 0.65 and available latent of steam for refrigeration system is 540 kcal/kg.
A) Estimate excess steam utilization for absorption-based refrigeration system in TPH. (3 Marks)
b) Estimate required Cooling water (m³/hr), if available approach in condenser is 10 °C. (2 Marks)
Model answer: Energy required for refrigeration system = 500 × 3024 / 0.65 = 2326153.8 kcal/hr
Steam needed for refrigeration system = 2326153.8 / 540 = 4.3 TPH
Steam utilization for VAM = 4.3 TPH
Required condenser duty = 2326153.8 + (500 × 3024) = 3838153.8 kcal/hr
Required Cooling water = 3838153.8 / 10 = 383.8 m³/hr
3. As part of a management initiative to advance green energy in a new process plant, a process engineer is assessing the economic viability of a 650 TR chiller. She is considering proposals for both LiBr-based vapor absorption chillers and vapor compression refrigeration systems. While power is sourced from renewable energy, the steam required is partially generated from excess process heat and additionally from firing furnace oil.
COP of advance Vapor Absorption Chiller: 1.3
COP of Vapor Compression Chiller: 4.50
Net steam price including excess steam and from boiler: 1500.00 INR/MT
Net Power cost from green source: 7.20 INR/kWh
Price of Cooling water: 3.00 INR/M3
Cooling water range: 8°C
Specific steam heat available for chiller: 490.0 kcal/kg
Evaluate both the offers and find out the offer which is economical in terms of operating cost.
Compute steam/power input from COP, condenser duty and cooling-water flow (duty/range), then compare hourly operating cost of VAM vs VCR.
Source: Sep 2024
📖 §4.7 TR formula & specific power; pump efficiency
4. An energy auditor assesses the refrigeration load of each floor of a building and the efficiency of the chilled water pump. The chilled water flow of each AHU and its inlet/outlet temperatures are measured; the chilled water pump discharge pressure is 3.5 kg/cm2 with the tank water level 1 m above the pump centerline; power drawn by the refrigeration compressor and pump motor are 89 kW and 17.58 kW. Determine the floor-wise refrigeration load, total TR, chiller kW/TR and pump efficiency.
Model answer: Floor-wise TR = flow (m3/hr) x 1000 x 1 kcal/kg degC x (T_in - T_out)/3024. 1st floor: 20 x 1000 x (10 - 7)/3024 = 19.84 TR; 2nd: 30 x 1000 x (11 - 7)/3024 = 39.68 TR; 3rd: 10 x 1000 x (11 - 7)/3024 = 13.23 TR; 4th: 15 x 1000 x (12 - 7)/3024 = 24.80 TR. Total = 97.55 TR (total flow 75 m3/hr). (b) Compressor shaft power = 89 x 0.92 = 81.88 kW, so specific power consumption = 81.88/97.55 = 0.84 kW/TR. (c) Pump total head = 3.5 kg/cm2 x 10 - 1 m (positive suction) = 34 m; hydraulic power = 75/3600 x 1000 x 9.81 x 34/1000 = 6.95 kW; pump shaft power = 17.58 x 0.89 = 15.65 kW; pump efficiency = 6.95/15.65 = 44.4%.
Book §4.7 TR formula applied floor by floor, then summed; the chiller indicator is compressor shaft power divided by total TR. Pump efficiency = hydraulic power / shaft power, where total head = discharge head (3.5 kg/cm2 = 35 m) less the 1 m positive suction head, because the tank level is ABOVE the pump centreline. Watch the two motor efficiencies (92% compressor, 89% pump) - omitting them is the usual mark-loser.
Source: Book EOC
📖 Cross-chapter fill-in-the-blanks (§4.7 for item 1)
5. Fill in the blanks (cross-chapter): 1) One TR = ___ kW. 2) A 4-pole 15 kW IM at 50 Hz, 1% slip has rotor input ___ kW. 3) A pitot tube measures total and static pressure to determine ___ pressure. 4) Centrifugal pump impeller diameter is generally limited to reducing to about ___ % of max size. 5) Pressure in pump suction exceeding liquid vapour pressure is expressed as ___. 6) ASME parameter to define fans, blowers, compressors is ___. 7) Pumps can run in parallel if their ___ are similar. 8) Evaporation 16 m³/cell, COC 3 → blowdown ___. 9) Pump raises water to 12 m; with brine SG 1.2 height raised is ___. 10) Installing capacitor near motor terminals increases design PF of motor - True/False.
Model answer: 1) 3.516 kW. 2) 15.15 kW (15/(1-0.01)=15.15). 3) velocity pressure. 4) 75%. 5) Net Positive Suction Head Available (NPSHA). 6) Specific ratio. 7) closed valve heads. 8) 8 m³ per cell (blowdown = evaporation/(COC-1) = 16/2 = 8). 9) 12 metres (same height - head is independent of density). 10) False.
Standard one-liners across HVAC, motors, fans, pumps and cooling towers.
Source: 15th Exam
📖 §4.16 Energy Saving Opportunities
6. List five energy saving measures for air conditioning system.
Model answer: 1) Insulate all cold lines/vessels with economic insulation thickness to minimize heat gain. 2) Optimize air-conditioned volumes (false ceiling, segregation by air curtains). 3) Minimize AC loads (roof cooling/painting, efficient lighting, pre-cooling fresh air via air-to-air heat exchanger, VAV, optimal thermostat setting, sun film). 4) Minimize process heat loads / reduce TR (flow optimization, increased heat-transfer area, avoid chilled-water wastage, regular descaling). 5) Adopt variable speed drives, run plant at matched part-load, regular maintenance, and use VAR systems where economics permit.
Confirmed vs Book-3 §4.16 - the guidebook lists the measures under cold insulation, building envelope, building heat-load minimisation, process heat-load minimisation and plant-area actions. Any five of these carry full marks, e.g. economic insulation thickness, false ceilings/air curtains, roof cooling and sun film, pre-cooling fresh air with air-to-air heat exchangers, optimum thermostat setting, avoiding bypass flows, VSDs to avoid part-load running, and VAR where economics permit.
Source: 16th Exam
📖 §4.7 TR & kW/TR; §4.3 VAR vs VCR
7. Compare the performance of a centrifugal chiller with a vapour absorption chiller (VAM) from the given data (chilled & condenser water flows, inlet/outlet temps, pump and CT fan power). Centrifugal compressor 205 kW; VAM steam 1620 kg/Hr. Calculate i) refrigeration load TR, ii) condenser heat load TR, iii) auxiliary power, iv) operating cost (electricity Rs 4/kWh, steam Rs 0.45/kg).
TR=m·Cp·ΔT/3024; condenser TR same with condenser flow/ΔT; cost = electricity + steam.
Source: 16th Exam
📖 §4.7 air-side TR formula (AHU/FCU)
8. L-2: (a) In an AHU, filter area = 1.5 m2, air velocity = 2.2 m/s, inlet enthalpy = 67 kJ/kg, outlet enthalpy = 56 kJ/kg, air density = 1.3 kg/m3. Estimate the TR of the AHU. (b) List any five energy conservation measures for energy use in buildings.
Model answer: (a) TR = (Δenthalpy x density x area x velocity x 3600)/(4.187 x 3024) = (67-56) x 1.3 x 1.5 x 2.2 x 3600/(4.187 x 3024) = 13.41 TR. (b) Five building ECMs: 1. Weather-strip windows/doors to cut infiltration; provide self-closing doors at high-traffic areas. 2. Set temperature 23-25 °C and RH 55-65% for comfort. 3. Keep chilled-water leaving temperature ≥7 °C (≈2.25% chiller efficiency gain per 1 °C rise). 4. Maintain insulation on chilled-water pipes and ducts to prevent heat gain. 5. Clean condenser tubes (every 6 months), keep filters clean, and install VFDs on AHU fans.
AHU TR from mass-flow x enthalpy drop converted to TR; building ECMs from Book-3 Chapter 4/10 (any five, 1.5 marks each).
Source: 18th Exam
📖 §4.13 Ice Bank; §4.3 VAR (harmonics: Book-3 Ch1)
9. L-6: Write short notes on (i) Ice Bank System in refrigeration, (ii) Vapour Absorption Refrigeration System, (iii) Harmonics in electrical system and its impacts.
Model answer: (i) Ice Bank System: a thermal energy storage technology that uses low-cost off-peak (night) electricity to make and store ice/cooling energy in storage tanks for use during high-tariff daytime hours; the chiller charges the tanks at night and runs longer hours at the lowest average load, shifting and shaving peak demand. (ii) VAR System: an absorption chiller produces chilled water using heat (steam, hot water, gas, oil, waste heat) instead of a compressor; water is the refrigerant and lithium-bromide is the absorbent; COP ≈ 0.65-0.70, chilled water down to ~6.7 °C at 30 °C cooling water; needs electricity only for pumps; economical when waste heat/cheap steam is available; capacities 10-1500 TR. (iii) Harmonics: currents/voltages at multiples of the supply (fundamental) frequency — e.g. with 50 Hz, 5th = 250 Hz, 7th = 350 Hz; caused by non-linear loads. Impacts: capacitor failure, conductor/cable overheating, transformer and motor overheating/failure, flickering of fluorescent and blinking of incandescent lights, and nuisance tripping.
Standard descriptive notes from Book-3 (Ice Bank p.136, VAR p.30, Harmonics p.114).
Source: 18th Exam
📖 §4.7 air-side TR; §4.9 kW/TR
10. L-4: A 7.5 TR package A/C cools a UPS room (40 kVA UPS). Outdoor unit air velocity 6.1 m/s, fan opening radius 0.30 m, air density 1.174 kg/m3, ambient 305 K, condenser-outlet hot air 313.5 K, Cp 1.009 kJ/kgK, compressor power 5.40 kW, motor efficiency 90%. UPS on-load (16 h): input 11.94 kW, output 8.61 kW; no-load (8 h): input 1.16 kW, output 0. Calculate (a) present delivery TR, (b) power per TR, (c) annual energy savings for 7200 h if UPS relocated to a ventilated area (Rs.8/kWh).
Model answer: Area = 3.14 x 0.30² = 0.283 m². Air flow = 0.283 x 6.1 = 1.72 m3/s. Mass = 1.72 x 1.174 = 2.02 kg/s. ΔT = 313.5 - 305 = 8.5 K. Heat transfer = 2.02 x 1.009 x 8.5 = 17.32 kJ/s = 62,352 kJ/hr = 14,917 kcal/hr. Compressor heat input = 5.4 x 0.9 x 860 = 4180 kcal/hr. Evaporator load = 14,917 - 4180 = 10,737 kcal/hr. (a) Effective TR = 10,737/3024 = 3.55 TR. (b) Power per TR = 5.40/3.55 = 1.52 kW/TR. (c) Heat load from UPS: on-load 3.33 kW → 0.95 TR/hr x 16 = 15.2 TR-day; no-load 1.16 kW → 0.33 TR/hr x 8 = 2.64 TR-day; total 17.84 TR/day. AC power to remove it = 17.84 x 1.52 = 27.12 kW. Annual savings (300 days) = 27.12 x 300 ≈ 8136 kWh → Rs.8 x 8136 = Rs.65,088/yr.
Condenser air-side heat minus compressor heat gives evaporator TR; kW/TR from compressor power; UPS heat converted to TR/day then to AC power saved if UPS relocated, costed at the tariff.
Source: 19th Exam
📖 §4.3 VAR vs VCR operating economics
11. L-5: A textile plant had two 6 MW gas turbines + HRSG feeding process steam and a 500 TR VAM (4.4 kg steam/TR, full load). Gas turbines stopped due to gas price; two 10 TPH agro-waste boilers installed (steam cost Rs.1200/ton); plant runs 7000 h/yr. Management plans to replace VAM with an electric centrifugal chiller at 0.7 kW/TR. Compare annual operating costs of electric chiller vs VAM (grid power Rs.6.12/kWh; auxiliaries unchanged). Do you agree with running the VAM?
Model answer: VAM steam need = 500 x 4.4 = 2200 kg/hr = 2.2 TPH; steam cost = 2.2 x 1200 = Rs.2640/hr. Electric chiller power = 0.7 x 500 = 350 kW; cost = 350 x 6.12 = Rs.2142/hr. Saving with electric chiller = 2640 - 2142 = Rs.498/hr. Annual saving = 7000 x 498 = Rs.34,86,000/yr. Conclusion: Disagree with running the VAM — the electric centrifugal chiller is cheaper to operate (saves ~Rs.34.86 lakh/yr) now that cheap waste heat is gone.
VAM running cost = steam rate x steam cost; chiller cost = kW/TR x TR x power tariff; compare hourly and annualise; recommend the cheaper option.
12. Write short notes on any two of the following: (1) Integrated Part Load Value (IPLV) for chillers, (2) Evaporative Cooling, (3) Heat Pump. (Each 2.5 Marks)
Model answer: (1) IPLV: A single-number part-load efficiency metric for chillers (AHRI weighting) that combines chiller performance at 100%, 75%, 50% and 25% load, since chillers rarely run at full load; it better represents seasonal/real operation than full-load kW/TR. (2) Evaporative Cooling: Cooling of air by direct contact with water so that sensible heat of air evaporates water; dry-bulb temperature falls while moisture content rises, approaching the wet-bulb temperature - low energy alternative to mechanical refrigeration in dry climates. (3) Heat Pump: A reversed refrigeration cycle device that extracts low-grade heat from a source (air/water/ground) and upgrades it to deliver useful heating; COP_heating = COP_cooling + 1, giving high effective efficiency.
Confirmed vs Book-3 §4.7 (IPLV, p.126), §4.3 Evaporative Cooling (p.116) and §4.11 Heat Pumps (p.133). IPLV averages kW/TR at 100/75/50/25% load because full-load kW/TR occurs for only ~1% of running hours. Evaporative cooling brings air into close contact with water so it approaches the wet bulb temperature - cheap but it adds moisture. A heat pump is an air conditioner whose rejected heat is the useful output: 3 units from the surroundings plus 1 unit of compressor work give 4 units of heat.
Model answer: (i) Refrigeration load = flow x 1000 x 1 x deltaT/3024: Centrifugal = 192000 x (13-7.8)/3024 = 330.16 TR; VAM = 183000 x (14.5-9.2)/3024 = 320.73 TR. (ii) Condenser heat load = condenser flow x 1000 x deltaT/3024: Centrifugal = 245000 x (36.2-28)/3024 = 664.35 TR; VAM = 360000 x (40.7-32)/3024 = 1035.71 TR. (iii) Auxiliary power = pumps + CT fan: Centrifugal = 32+38+9 = 79 kW; VAM = 31+52+22 = 105 kW. VAM auxiliary is higher because its condenser heat rejection is much larger for similar cooling load. (iv) Operating cost/hr: Centrifugal total = 79 + 205 = 284 kW x Rs.4 = Rs.1136. VAM = 105 kW x Rs.4 = Rs.420 plus steam 1620 x Rs.0.45 = Rs.729 -> total Rs.1149/hr.
Confirmed vs Book-3 §4.7 - refrigeration TR = flow x Cp x deltaT/3024 on the chilled-water side and the same formula on the condenser side for heat rejection. Centrifugal 330.16 TR vs VAM 320.73 TR for similar duty, but the VAM rejects 1035.71 TR against 664.35 TR, because a VAR machine must also reject its heat input (COP ~0.65-0.70). That is why VAM condenser pumps and cooling-tower fans are bigger (105 kW vs 79 kW), and the hourly cost is close (Rs.1149 vs Rs.1136) despite free-looking steam.
Source: 16th Exam (alt set)
📖 §4.12 Ventilation Systems; §4.7 air-side TR
14. L-2: (a) Calculate the ventilation rate for an engine room 20 m L x 10.5 m W x 15 m H if recommended ACH is 20. (b) Air at 25,200 m3/hr and 1.2 kg/m3 density flows into an AHU; enthalpy difference inlet-outlet 2.38 kcal/kg; motor draws 22 kW at 90% efficiency. Find kW/TR (1 cal = 4.183).
Model answer: (a) Ventilation rate = L x H x W x ACH = 20 x 15 x 10.5 x 20 = 63,000 m3/hr. (b) Heat = Q x density x deltaH = 25200 x 1.2 x 2.38 = 71,971 kcal/hr; TR = 71971/3024 = 23.8 TR; compressor power = 22 x 0.9 = 19.8 kW; kW/TR = 19.8/23.8 = 0.83.
Confirmed vs Book-3 §4.12 and §4.7 - ventilation rate = L x B x H x ACH = 20 x 10.5 x 15 x 20 = 63,000 m3/hr. Air-side load = Q x rho x delta-h/3024 = 25,200 x 1.2 x 2.38/3024 = 23.8 TR, and compressor shaft power = 22 x 0.9 = 19.8 kW, giving 0.83 kW/TR. The common slip is to use the 22 kW motor input directly, which overstates kW/TR by about 11%.
Source: 16th Exam (alt set)
📖 §4.7/§4.8 COP vs temperatures (b,c cross-chapter)
15. 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.
Source: 9th Dec-2009
📖 §4.7 TR formula & COP
16. 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.
Source: 9th Dec-2009
📖 §4.8 (a,b); rest cross-chapter
17. L-2: Fill in the blanks. (a) With increase in condensing temperature in a vapor compression refrigeration system, the specific power consumption of the compressor for a constant evaporator temperature will____. (b) With increase in evaporator temperature while maintaining a constant condenser temperature, the specific power consumption of the compressor will____. (c) Lower power factor of a DG set demands ____ excitation current. (d) Slip power recovery system is used in ____ induction motor. (e) If voltage is reduced from 230 V to 200 V for a fluorescent tube light, it will result in ____ power consumption. (f) ____ fans are known as 'non-overloading' because change in static pressure do not overload the motor. (g) ____ head is the friction loss, on the liquid being moved, in pipes, valves and equipment in the system. (h) Ratio of the light reflected by a surface to the solar light incident upon it, is called ____. (i) ____ is the ratio of solar heat gain that passes through fenestration to the total incident solar radiation that falls on the fenestration. (j) luminous flux incident on an object per unit area is defined as ____.
Model answer: a. increase; b. decrease; c. higher; d. slipring (slip-ring); e. reduced; f. backward-inclined; g. dynamic; h. Solar Reflectance; i. Solar heat gain coefficient; j. illuminance.
Standard refrigeration/motor/fan/pump/building fill-in answers per official key.
Source: 11th Feb-2011
📖 §4.7 air-side TR formula (AHU/FCU)
18. L-4: An energy audit was conducted to find out the ton of refrigeration (TR) of an Air Handling Unit (AHU). Evaporator area = 10.0 m2; Inlet velocity = 1.9 m/s; Inlet air DBT = 21.5 C, RH = 75%, Enthalpy = 53.0 kJ/kg; Outlet air DBT = 17.4 C, RH = 90%, Enthalpy = 46.4 kJ/kg; Density of air = 1.14 kg/m3. Find out the TR of AHU.
Model answer: AHU refrigeration load = [Air flow rate (m3/h) x Density of air (kg/m3) x Difference in enthalpy (kJ/kg)] / (3024 x 4.18). Air flow = 10.0 x 1.9 x 3600 = 68400 m3/h. AHU = (10.0 x 1.9 x 3600) x 1.14 x (53 - 46.4) / (3024 x 4.18) = 40.71 TR.
Mass flow x enthalpy drop gives kJ/hr; convert to kCal (/4.18) and to TR (/3024).
19. a) Name six parameters along with units that a psychrometric chart provides to an air conditioning system. (3 Marks) b) Explain briefly about Thermal Emittance. (2 Marks)
Model answer: a) Six parameters: 1. Dry bulb temperature (oC); 2. Relative humidity (%); 3. Wet bulb temperature (oC); 4. Specific volume (m3/kg of dry air); 5. Enthalpy (kJ/kg of dry air); 6. Specific humidity or Humidity factor (grams/kg of dry air). b) Refer Guidebook-3, page 272 — Thermal emittance is the ratio of radiant heat emitted by a surface to that emitted by a black body at the same temperature; low-emittance surfaces re-radiate less absorbed heat into the building.
Psychrometric chart parameters and thermal emittance definition per Guidebook-3.
Source: Mar 2021
📖 §4.7 air-side TR (a); Book-3 Ch-2 motors (b)
20. a) Calculate the filter area of an Air Handling Unit (AHU) for Refrigeration Load of 50 TR. The air enthalpy at inlet of AHU is 85 kJ/kg and at outlet is of 60 kJ/kg. Air velocity at filter is 1.81 m/sec and air density is 1.26 kg/m3. (5 Marks) b) A no load test was conducted in a delta connected 37 kW induction motor. Name plate data: 3 Phase, 415 V, 50 Hz, 55 Amp. Measured data on no load: Voltage V = 415 Volts; Current I = 18 Amps; Frequency F = 50 Hz; Stator phase resistance at no load = 0.23 Ohms/phase. No load power = 955 Watts. Calculate: i. The iron loss plus friction loss plus windage loss (2 Marks) ii. Stator copper loss at name plate ratings (full load), considering stator temperature as 120oC (2 Marks) iii. No load power factor of the motor (1 Mark)
Model answer: a) TR of AHU = (Enthalpy difference x density x area x velocity x 3600)/(4.187 x 3024). 50 = (85-60) x 1.26 x Area x 1.81 x 3600/(4.187 x 3024). Filter Area = TR x (4.187 x 3024)/[(enthalpy diff) x density x velocity x 3600] = 50 x (4.187 x 3024)/[(25) x 1.26 x 1.81 x 3600] = 3.08 m2. b) i) No-load input Pin = 955 W. No-load stator copper loss = 3 x I^2 x R = 3 x 18^2 x 0.23 = 74.51 W. Iron + friction + windage loss = Pin - no-load Cu loss = 955 - 74.51 = 880.49 W. ii) Stator resistance at 120oC = 0.23 x (120+235)/(30+235) = 0.23 x 355/265 = 0.308 ohms. Full-load stator copper loss at nameplate current = 3 x (55/sqrt3)^2 x 0.308 = 3 x 1008.3 x 0.308 = 931.65 W. iii) No-load power factor = P/(sqrt3 x V x I) = 955/(1.732 x 415 x 18) = 0.0738.
AHU: TR = (dh x rho x A x v x 3600)/(4.187 x 3024); iron+fr+wind = Pin - 3I2R; R corrected by (235+t2)/(235+t1); PF = P/(sqrt3 VI).
Source: Mar 2021
📖 §4.7 TR & operating cost; §4.3 VAR vs VCR
21. The data for centrifugal chiller and vapour absorption chiller are given below — Chilled water flow (m³/h): Centrifugal 189, VAM 180; Condenser water flow (m³/h): Centrifugal 258, VAM 340; Chiller inlet temp (°C): Centrifugal 13.0, VAM 14.6; Condenser water inlet temp (°C): Centrifugal 27.1, VAM 33.5; Chiller outlet temp (°C): Centrifugal 7.7, VAM 9.0; Condenser water outlet temp (°C): Centrifugal 35.7, VAM 39.1; Power drawn by compressor (kW): Centrifugal 190, VAM -; Steam consumption (kg/h): Centrifugal -, VAM 1570; Chilled water pump (kW): Centrifugal 28, VAM 28; Condenser water pump (kW): Centrifugal 22, VAM 33; Cooling tower fan (kW): Centrifugal 6.0, VAM 15; Cost of Steam (Rs/kg): VAM 2.0; Cost of electricity (Rs/kWh): 9.0 both. a) Evaluate the tonnes of refrigeration (TR) of both the systems. b) Operating Energy cost per hour for both the systems.
Model answer: a) Centrifugal chiller TR = Chilled water flow × (Tin - Tout) × Diff. in temp / 3024 = 189 × 1000 × 1 × (13-7.7)/3024 = 331.25 TR. VAM TR = 180 × 1000 × 1 × (14.6-9.0)/3024 = 333.33 TR. b) Auxiliary power consumption: Centrifugal = Chilled water pump + condenser water pump + cooling tower fan = 28 + 22 + 6.0 = 56 kW. VAM auxiliary power (kW) = 28 + 33 + 15 = 76 kW. Energy cost of centrifugal chiller = (56 + 190)×9 = Rs 2214/hr. Energy cost of VAM chiller = (76×9) + (1570×2) = Rs 3824/hr.
TR = flow×ΔT/3024 (kcal→TR); energy cost = electrical kW×rate (+ steam kg×rate for VAM).
Source: Jul 2022
📖 §4.2 (item 1); rest cross-chapter
22. L-3(A): State Increases or Decreases (1 Mark each): 1. If air dry bulb temperature is increased then Relative Humidity will ___. 2. In a pumping system, if the suction side liquid level is increased then NPSHa will ___. 3. If the air temperature increases at the inter-cooler outlet, then air compressor power consumption will ___. 4. A blower is retrofitted with a VFD and operated at full speed. The power consumption will ___. 5. As the design speed of the motors decreases the capacitor KVAr requirement will ___.
23. L-3(B): Match the following (1 Mark each): 1. Pitot Tube; 2. Refrigerant Drier; 3. Condenser; 4. Spray Nozzles; 5. Occupancy Sensor — with — A. Cooling Tower; B. Lighting Control; C. Gas Velocity in ducts; D. Compressed Air System; E. Refrigeration System.
Model answer: 1. Pitot Tube → C. Gas Velocity in ducts; 2. Refrigerant Drier → D. Compressed Air System; 3. Condenser → E. Refrigeration System; 4. Spray Nozzles → A. Cooling Tower; 5. Occupancy Sensor → B. Lighting Control.
Standard equipment-to-application matching.
Source: Mar 2023
📖 §4.7 kW/TR & COP; Book-3 Ch-7 Cooling Towers
24. L-4: During the energy audit of central chiller plant, following parameters were noted: Chilled water flow 250 m³/hr; Chilled water inlet temperature 12°C; Chilled water outlet temperature 7°C; Motor Input Power 350 kW; Motor Efficiency 90%; Condenser water inlet temperature (going to chiller or outlet of cooling tower) 31°C; Condenser water outlet temperature (leaving from chiller or inlet to cooling tower) 36°C; Wet Bulb temperature of ambient air 28°C; Make up water TDS 180 ppm; Permissible limit of TDS for cooling water 720 ppm; Condenser cooling capacity 25% higher than the evaporator cooling capacity. Calculate: kW/TR of chiller compressor; COP of chiller; Effectiveness of cooling tower; Evaporation loss; Blow down quantity; Make-up water requirement (ignoring no drift loss).
Model answer: Chiller machine capacity TR = [250×1000×1×(12−7)]/3024 = 413.4 TR. kW/TR of chiller compressor = (350×90%)/413.4 = 0.762. COP = (413.4×3024)/(350×0.9×860) = 4.6. Effectiveness = (36−31)/(36−28) = 62.5%. Cycle of Concentration COC = 720/180 = 4. Condenser TR = 1.25×413.4 = 516.75 TR. Condenser water flow / circulation flow = 516.75×3024/(1000×1×(36−31)) = 312.5 m³/hr. Evaporation loss = 0.00085×1.8×312.5×(36−31) = 2.3 m³/hr. Blow down = Evap/(COC−1) = 2.3/(4−1) = 0.797 m³/hr. Make-up = Evaporation + Blow down = 2.3+0.797 = 3.097 m³/hr.
Chiller TR from flow×ΔT; kW/TR from shaft power; COP conversion; cooling-tower effectiveness, COC, evaporation, blowdown and make-up formulae.
Source: Mar 2023
📖 §4.3 Types of Refrigeration System (VCR & VAR)
25. Explain the operating principle of a vapour compression and a vapour absorption refrigeration system.
Model answer: Vapour Compression Refrigeration (VCR) - driving force is mechanical (electrical) energy. Four stages (Figure 4.4): (1-2) low-pressure liquid refrigerant boils in the evaporator, absorbing heat from air/water/process liquid and leaving slightly superheated; (2-3) the compressor raises its pressure and temperature; (3-4) in the condenser the gas is de-superheated, condensed and sub-cooled, rejecting heat to air or water; (4-1) the expansion device drops the pressure and meters flow back to the evaporator with no heat gain or loss. The condenser must reject the evaporator heat plus the compressor work. Typical COP 2.5-5 (0.65-0.9 kW/TR for centrifugal chillers).
Vapour Absorption Refrigeration (VAR) - driving force is thermal energy (steam, hot water, gas, oil or waste heat); the compressor is replaced by an absorber, solution pump and generator. Refrigerant water evaporates at about 4 degC under a vacuum of 754 mmHg in the evaporator, taking latent heat from the chilled water (12 degC to 7 degC). The vapour is absorbed by concentrated lithium-bromide solution in the absorber, which maintains the vacuum; the heat of absorption is carried away by cooling water. The diluted solution is pumped to the generator, where the heating medium boils off the water and re-concentrates the LiBr; the released vapour is condensed in the condenser and returned to the evaporator. LiBr-water machines give COP 0.65-0.70 and chilled water at 6.7 degC with 30 degC cooling water; ammonia-water machines work above atmospheric pressure and reach below 0 degC. Electricity is needed only for pumps, so VAR is economical where cheap waste heat or steam is available, although first cost is higher.
Book-3 §4.3 - the guidebook states the two principal industrial plants are VCR (mechanical energy as the driving force) and VAR (thermal energy as the driving force). Answer the VCR part with the four numbered stages of Figure 4.4 and the VAR part with the evaporator-absorber-generator-condenser sequence of Figure 4.5. Quote the book numbers - VAR COP 0.65-0.70, chilled water 6.7 degC at 30 degC cooling water, capacities 10-1500 TR - to secure the marks.