## [MCQ] Refrigeration& Air Conditioning

#### Module 1

1. At what K is the triple point of water?
a) 0 K
b) 4 K
c) 273.16 K
d) -273.16
Explanation: At 273.16 K, i.e. 0C water contains all the three forms, solid, liquid as well as gas partially.  3. What does 298 Kelvin mean on Celsius scale?
a) 27 C
b) 25 C
c) 15 C
d) 35 C
Explanation: As for conversion of Kelvin to Celsius 273 is to be deducted, hence 298 K = 298 – 273 = 25C.

4. What is the value of 110 C on the Fahrenheit scale?
a) 198 F
b) 383 F
c) – 73 F
d) 230 F 5. 1 Pascal = 1 _____
a) N/mm2
b) N/m3
c) N/m2
d) N/mm3
Explanation: Pascal is the unit generally used to express pressure, which means the amount of pressure in Newton, divided by the area i.e. square meters.

6. 1 Bar = ______
a) 1×105 N/m2
b) 1×109 N/m2
c) 1×105 N/mm2
d) 1×106 N/m2
Explanation: Bar is the unit used for expressing the higher level of pressure, which means 760 mm of Hg at the sea level, which equals to 100,000 Newton.

7. What are the pressure in the places at “a” and “b” respectively in the given graph? a) Gauge, Absolute
b) Absolute, Gauge
c) Atmospheric, Gauge
d) Gauge, Atmospheric
Explanation: Absolute pressure is the sum of Gauge pressure and Atmospheric pressure. Also, Atmospheric pressure is measured from the datum level and Gauge pressure starts above the Atmospheric pressure.

8. What does 1 Tonne (TR) in refrigeration mean?
a) Weight of gases
b) Weight of coolant
c) Capacity of 1 tonne air to be cooled to 0 C in 24 hours
d) Capacity of 1 tonne water to be cooled to 0 C in 24 hours
Explanation: 1 Tonne Rating (TR) means that the capacity is such that it can convert 1 tonne of water to ice i.e. 0C in 24 hours.

9. 1 Tonne = ______ KJ/s.
a) 2.67
b) 1.087
c) 3.5
d) 232.6 10. Which is the S.I. unit to measure pressure in refrigeration?
a) Newton
b) Joule
c) Pascal
d) Bar
Explanation: Generally, unit like N/mm2, N/m2, KN/mm2 and MN/mm2 etc. are used but the S.I. unit used is N/m2 i.e. Pascal.

11. 0 Kelvin = ____ Celsius.
a) -273 C
b) 273 C
c) -273 K
d) 0 C
Explanation: As for Temperature 273 K= 0 C, i.e. 0 K = -273 C from the temperature scale developed by Celsius.

12. What is the term C.O.P. referred in terms of refrigeration?
a) Capacity of Performance
b) Co-efficient of Plant
c) Co-efficient of Performance
d) Cooling for Performance
Explanation: Co-efficient of Performance is generally referred as C.O.P. for Refrigeration, which is used to measure the capacity or level up to which the refrigeration will occur.

13. C.O.P. can be expressed by which equation? Explanation: Co-efficient of Performance is the ratio of the Refrigeration effect produced to the work done or work supplied to produce the effect.
Whereas the ratio- Work done to Heat transfer is called the efficiency.

14. What is the term relative C.O.P. referred in terms of refrigeration? Explanation: Relative Co-efficient of Performance is generally referred as the ratio of Actual C.O.P. measured to the Theoretical C.O.P. assumed before calculations. It is in relation with the theoretical C.O.P. and generally expressed in “%” value.

15. Find the C.O.P. of a refrigeration system if the work input is 40 KJ/kg and refrigeration effect produced is 130 KJ/kg of refrigerant flowing.
a) 3.00
b) 2.25
c) 3.75
d) 3.25 16. Find the Relative C.O.P. of a refrigeration system if the work input is 60 KJ/kg and refrigeration effect produced is 130 KJ/kg of refrigerant flowing. Also Theoretical C.O.P. is 3.
a) 0.65
b) 0.79
c) 0.72
d) 0.89 17. Efficiency of the Refrigerator is _________ to the C.O.P of refrigerator.
a) inversely proportional
b) equal
c) independent
d) directly proportional
Explanation: Efficiency is the ratio of work done to heat supplied, whereas C.O.P is the ratio of Refrigeration effect to work done. Hence it is totally independent quantity.

18. What is the C.O.P. of a refrigeration system if the work input is 40 KJ/kg and work output is 80 KJ/kg and refrigeration effect produced is 130 KJ/kg of refrigerant flowing.
a) 3.00
b) 3.25
c) 2.25
d) 3.75 19. Find the Relative C.O.P. of a refrigeration system if the work input is 100 KJ/kg and refrigeration effect produced is 250 KJ/kg of refrigerant flowing. Also Theoretical C.O.P. is 3.
a) 0.65
b) 0.80
c) 0.83
d) 0.91 20. If a condenser and evaporator temperatures are 225 K and 100 K respectively, then reverse Carnot C.O.P is _________
a) 0.5
b) 1.5
c) 1.25
d) 1.75 21. If a condenser and evaporator temperatures are 312 K and X K respectively, and C.O.P. is given as 5 then find the value of X.
a) 52
b) 65
c) 78
d) 82 22. The C.O.P for reverse Carnot refrigerator is 6. The ratio of lowest temperature to highest temperature will be _____
a) twice
b) three times
c) four times
d) seven times 23. For a standard system with temperatures T1 and T2, where T1 < Ta < T2 (Ta – Atmospheric Temperature). Q1 is the heat extracted from a body at temperature T1, and Q2 is heat delivered to the body at temperature T2. What is the C.O.P. of the heat pump for given conditions?
a) Q2 / (Q2 − Q1)
b) (Q2 − Q1) / Q1
c) (Q2 − Q1) / Q2
d) Q1 / (Q2 − Q1)
Explanation: As, C.O.P. = Desired effect / Work done
Here, work-done = Q2 − Q1
The desired temperature is T2. So, the heat delivered to achieve the desired temperature is Q2.
C.O.P. of the heat pump = Q2 / (Q2 − Q1).

24. What is the difference between Heat Pump and Refrigerator?
a) Heat Pump Gives efficiency and refrigerator gives C.O.P.
b) Both are similar
c) Both are almost similar, just the desired effect is different
d) Work is output in refrigerator and work is input in heat pump
Explanation: Heat Pump and Refrigerator work on the same principle. Work needs to be given to get the desired effect. The characteristic which differentiates both of them is the temperature of the desired effect, heat pump desires for higher temperature whereas Refrigerator desires for lower temperature than atmospheric temperature.

25. What is the equation between efficiency of Heat engine and C.O.P. of heat pump?
a) ηE = (C.O.P.)P
b) ηE = 1 / (C.O.P.)P
c) ηE / (C.O.P.)P = 1
d) ηE x (C.O.P.)P = 0
Explanation: ηE = W / Q hence for Carnot engine it is equal to (T2 – T1) / T2.
(C.O.P.)P for Carnot cycle is equal to T2 / (T2 – T1) .
So, these terms are related reciprocally.

26. How is the Relative coefficient of performance represented?
a) Theoretical C.O.P. / Actual C.O.P.
b) Actual C.O.P. / Theoretical C.O.P.
c) Theoretical C.O.P. x Actual C.O.P.
d) 1 / Theoretical C.O.P. x Actual C.O.P.
Explanation: Relative C.O.P. is the ratio of an actual to the theoretical coefficient of performance. It is used to show the deviation of C.O.P. due to the ideal state and real state conditions.

27. C.O.P. of the heat pump is always _____
a) one
b) less than One
c) greater than One
d) zero
Explanation: The second law of Thermodynamics states that a 100% conversion of heat into work is not possible without ideal conditions. So, efficiency will be less than 1. As C.O.P. is the reciprocal of efficiency, it tends to be more than 1.

28. For the systems working on reversed Carnot cycle, what is the relation between C.O.P. of Refrigerator i.e. (C.O.P.)R and Heat Pump i.e. (C.O.P)P?
a) (C.O.P.)R + (C.O.P)P = 1
b) (C.O.P.)R = (C.O.P)P
c) (C.O.P.)R = (C.O.P)P – 1
d) (C.O.P.)R + (C.O.P)P + 1 = 0
Explanation: If we put the values of C.O.P. for standard system i.e. (C.O.P.)R = T1/ (T2 − T1) and
(C.O.P.)P = T2/ (T2 − T1),
(C.O.P.)P − (C.O.P.)R = 1.
{T2 / (T2 − T1)} − {T1 / (T2 − T1)} = 1.

29. If the reversed Carnot cycle operating as a heat pump between temperature limits of 364 K and 294 K, then what is the value of C.O.P?
a) 4.2
b) 0.19
c) 5.2
d) 0.23
Explanation: C.O.P. of reversed Carnot cycle is given by,
C.O.P. = T1 / (T2 – T1)
= 364 / (364 – 294)
= 5.2.

30. A reversed Carnot cycle is operating between temperature limits of (-) 33°C and (+) 27°C. If it acts as a heat engine gives an efficiency of 20%. What is the value of C.O.P. of a heat pump operating under the same conditions?
a) 6.5
b) 8
c) 5
d) 2.5
Explanation: Temperature limits are given in the question so, we can calculate C.O.P. using the formula
C.O.P. = T1 / (T2 – T1)
But as the efficiency of the heat engine is given so directly by the relation, we can find out the C.O.P.
C.O.P. = 1 / ηE
= 1 / (0.2) = 5.

31. For a refrigerating system that works on reversed Carnot cycle having vapor as refrigerant, which one of the following is not a process of the cycle in p-v diagram? a) Isentropic compression process
b) Isothermal compression process
d) Isothermal expansion process
Explanation: The reversed Carnot cycle has 4 processes namely, isothermal compression process, isentropic compression process, isentropic expansion process and isothermal expansion process.

32. What is the coefficient of performance of the refrigerant system working on reversed Carnot cycle, having Qa as heat absorbed by the air during isothermal expansion per kg of air and Qr as heat rejected during isothermal compression per kg of air?  33. What happens to the COP of a Carnot refrigerator in summer and in winter?
a) The COP is more in winter
b) The COP is more in summer
c) The COP remains unaffected
d) The COP fluctuates continuously during winter and summer
Explanation: As the higher temperature T2 will always be lower in winter when compared with that of summer i.e. the temperature of air available of heat rejection will be low, the COP in winter will be higher.

34. The COPr of a Carnot refrigerating machine is 7.89. What will be the COPp of heat pump?
a) 10.3
b) 7.89
c) 8.89
d) 6.89
Explanation: Relation between COP of heat pump and COP of the refrigerating machine is
COPr = COPp-1
7.89 = COPp-1
COPp=7.89+1=8.89.

35. Which one is not a limitation of the Carnot cycle with air or gas as refrigerant?
a) It has low efficiency when working between two fixed temperature limits
b) Machine has to run at high and low speeds during adiabatic and isothermal process. Such variation of speed is not possible
c) Extreme pressure and large volume are developed
d) It is not possible to carry out isothermal heat transfer process in practice
Explanation: Carnot cycle has the highest efficiency theoretically when working between two fixed temperature limits. Other limitations of Carnot cycle are machine has to run at high and low speeds during adiabatic and isothermal process. Such variation of speed is not possible, extreme pressure and large volume is developed, it is not possible to carry out isothermal heat transfer process in practice.

36. Out of the following reasons which one is responsible for Carnot cycle not being used in practice?
a) It gives low COP
b) It gives high refrigeration effect
c) It gives low refrigeration effect
d) It has low theoretical efficiency
Explanation: Carnot cycle is not used in everyday practice and one of the many reasons is that because it has low refrigeration effect, even though it has the highest theoretical efficiency and also it is not possible to carry out isothermal heat transfer in practice.

37. If a refrigeration system having T1 and T2 as lower and higher temperatures respectively then, what is the value of C.O.P of the refrigeration system working on the reversed Carnot cycle?
a) T2 / (T2 − T1)
b) (T2 − T1) / T1
c) (T2 − T1) / T2
d) T1 / (T2 − T1)
Explanation: As, C.O.P. = Desired effect / Work done
Here, work-done = Q1 − Q1
The desired temperature is T1. So, the heat delivered to achieve the desired temperature is Q1.
C.O.P. of the heat pump = Q1 / (Q2 − Q1).
According to Carnot’s theorem,
C.O.P. = T1 / (T2 − T1).

38. In a refrigerating machine working on the reversed Carnot cycle, if the lower temperature is fixed, then what can be done to increase the C.O.P.?
a) Increasing higher temperature
b) Operating machine at higher speed
c) Decreasing higher temperature
d) Operating machine at a lower speed
Explanation: As C.O.P. of the refrigerator working on the Carnot cycle is given by,
C.O.P. = T1 / (T2 − T1)
So, If T1 is fixed, so by decreasing the denominator C.O.P. can be increased. If the higher temperature is reduced then by keeping numerator constant, the denominator decreases and leads to an increase in C.O.P.

39. If the condenser and evaporator temperatures are 320 K and 240 K respectively, then what is the value of the C.O.P.?
a) 0.25
b) 4
c) 0.33
d) 3
Explanation: Given: T1 = 240 K
T2 = 320 K
C.O.P. = T1 / (T2 − T1)
= 240 / (320 − 240)
= 240 / (80)
= 3.

40. The efficiency of Carnot heat engine is 40%. What is the value of C.O.P. of a refrigerator operating on reversed Carnot cycle?
a) 2.5
b) 1.5
c) 4
d) 10
Explanation: ηE = 40% = 0.4
C.O.P. of heat pump = 1 / ηE = 1 / 0.4 = 2.5
As we know, (C.O.P.)R = (C.O.P.)P − 1
C.O.P. of refrigerator = 2.5 − 1
= 1.5.

41. The C.O.P. of a reversed Carnot refrigerator is 5. What is the ratio of highest temperature to lower temperature?
a) −1.2
b) 0.8
c) 1.2
d) −0.8
Explanation: As C.O.P. of the refrigerator working on the Carnot cycle is given by,
C.O.P. = T1 / (T2 − T1)
5 = T1 / (T2 − T1)
(T2 − T1) / T1 = 1 / 5
(T2 / T1) − (T1 / T1) = 0.2
T2 / T1 − 1 = 0.2
T2 / T1 = 1.2.

42. For the systems working on reversed Carnot cycle, what is the relation between C.O.P. of Refrigerator i.e. (C.O.P.)R and Heat Pump i.e. (C.O.P.)P?
a) (C.O.P.)R + (C.O.P.)P = 1
b) (C.O.P.)R = (C.O.P.)P
c) (C.O.P.)R = (C.O.P.)P − 1
d) (C.O.P.)R + (C.O.P.)P + 1 = 0
Explanation: If we put the values of C.O.P. for standard system i.e. (C.O.P.)R = T1 / (T2 − T1) and
(C.O.P.)P = T2 / (T2 − T1),
(C.O.P.)P − (C.O.P.)R = 1.
{T2 / (T2 − T1)} − {T1 / (T2 − T1)} = 1.

43. Bell-Coleman cycle is also known as _____________
a) Carnot cycle
b) Reversed Brayton or Joule’s cycle
c) Rankine cycle
d) Otto cycle
Explanation: Bell-Coleman cycle was developed by Bell-Coleman and Light Foot by reversing the Joule’s air cycle and hence is also known as reversed Brayton or Joule’s cycle. It was one of the easiest types of refrigerators used in ships for carrying frozen meat.

44. For a refrigerating system that works on Bell-Coleman cycle, which one of the following is not a process of the cycle in p-v diagram? a) Isentropic compression process
b) Constant pressure cooling process
c) Isothermal expansion process
d) Constant pressure expansion process
Explanation: The Bell-Coleman cycle has 4 processes namely isentropic compression process, constant pressure cooling process, isentropic expansion process and constant pressure expansion process.

45. In transport aviation, the air conditioning systems are based on ______ cycle.
a) Reversed Carnot cycle
b) Reversed Brayton cycle
c) Reversed Joule’s cycle
d) Otto cycle
Explanation: It is because in vapor-cycle the disadvantage was that due to the leakage loss of fluid it would cause the aircraft to be completely without cooling.

46. Dense air Bell-Coleman refrigerator is preferred than open cycle air refrigerator.
a) True
b) False
Explanation: In open cycle air refrigerator the main drawback is the freezing of the moisture in the air during expansion stroke which is liable to choke up the valves. Hence dense air refrigerator is preferred.

47. Which one of the following is not a true disadvantage of the Bell-Coleman cycle?
a) High running cost
b) Low COP
c) The danger of frosting at the expander valve is more
d) The size of the system is small
Explanation: Mass of air required per ton of refrigeration is large as compared to other systems of refrigeration. Hence the size of the system is large.

48. Which one of the following is not a true advantage of the Bell-Coleman cycle?
a) Air is used a refrigerant which is easily available
b) It is safe as air is non-inflammable
c) Air is nontoxic, non-corrosive and stable
d) Weight of air refrigeration equipment per ton of refrigeration is much more in aircraft than other refrigeration system
Explanation: Weight of air refrigeration equipment per ton of refrigeration is much less in aircraft than other refrigeration systems. Hence it is light. It is because of air compressor is already available in the air-craft.

49. Which one of the following is not a component of a simple air cooling system?
a) Main compressor
b) Cooling fan
c) Heat exchanger
d) Generator

Explanation: The main components of simple air cooling system are the main compressor driven by the gas turbine, a cooling fan, heat exchanger and a cooling turbine.

50. For a simple air cooling system which one of the following is not a process of the cycle in T-S diagram? a) Ramming process
b) Compression process
c) Heating process
d) Refrigeration process

Explanation: A simple air cooling system has 5 processes in T-S diagram namely ramming process, compression process, expansion process, refrigeration processes and cooling process.

51. The COP of simple air cooling system is given by?
T6 = Inside temperature of cabin
T5′ = Exit temperature of cooling turbine
T3′ = Temperature at the exit of compressor
T2′ = Stagnation temperature  52. The simple air cooling system is good for _____ flight speed.
a) low
b) high
c) moderate
d) any
Explanation: The simple air cooling system is good for low flight speed so as fan can maintain airflow over the air cooler, which is difficult for it while at high speeds.

53. What is the main difference between simple air cooling system and simple air evaporative cooling system?
a) Simple air evaporative cooling system has an evaporator
b) Simple air evaporative cooling system has two evaporators
c) Simple air evaporative cooling system has an extra compressor
d) Simple air evaporative cooling system has three evaporator
Explanation: The difference between the simple air cooling system and simple air evaporative cooling system is that it has an evaporator between the heat exchanger and the cooling turbine.

54. If cooling of 45 minutes or less is required, it may be advantageous to use evaporative cooling system.
a) True
b) False
Explanation: The evaporative cooling system provides an additional cooling effect through evaporation of refrigerants such as water etc. Hence if cooling of 45 minutes or less is required, it may be advantageous to use evaporative cooling system.

55. Regeneration cooling system is a modification of _________
a) simple air cooling system
b) simple evaporative cooling system
c) boot-strap cooling system
d) boot-strap evaporative cooling system

Explanation: Regenerative air cooling system is a modification of simple air cooling system by the addition of a regenerative heat exchanger.

55. Cooling system used for supersonic aircrafts and rockets is?
a) Simple air cooling system
b) Simple evaporative cooling system
c) Boot-strap cooling system
d) Regeneration cooling system

Explanation: Cooling system used for supersonic aircrafts and rockets is of regeneration cooling system type, due to the regenerative heat exchanger and they give lower turbine discharge temperatures than simple air cooling system.

56. Which of the following is not a process in the T-s diagram of the regeneration cooling system? a) Isentropic ramming
b) Isentropic compression
c) Cooling of air by ram air in the heat exchanger and then cooling of air in regenerative heat exchanger
d) Isothermal expansio

Explanation: T-s diagram of regenerative air cooling system consists of Isentropic ramming, Isentropic compression, Cooling of air by ram air in the heat exchanger and then cooling of air in regenerative heat exchanger, Isentropic compression, isentropic expansion and then heating of air.

57. Simple air cooling system gives maximum cooling effect on ground surface whereas regeneration cooling system has more effect during high speeds.
a) True
b) False

Explanation: Simple air cooling system gives maximum cooling effect on ground surface whereas regeneration cooling system has more effect during high speeds because it gives low turbine discharge temperature than the simple air cooling system.

58. Calculate the power required (P) for maintaining the cabin temperature, when the COP of the regenerative cooling system is given to be 0.25 and the refrigeration load (Q) is 25.
a) 305
b) 350
c) 530
d) 503 59. In Boot-strap air cooling system how many heat exchangers are there?
a) 1
b) 2
c) 3
d) 0
Explanation: In Boot-strap air cooling system there are two heat exchangers and a cooling turbine driving the secondary compressor instead of the cooling fan.

60. Which of the following is not a process in the T-s diagram of the Boot-strap air cooling system? a) Isentropic ramming
b) Isentropic compression
c) Cooling of ram air
d) Isothermal expansion
Explanation: T-s diagram of Boot-strap air cooling system consists of Isentropic ramming, Isentropic compression, Cooling of ram air in first heat exchanger, Isentropic compression, cooling of ram air in the second heat exchanger, isentropic expansion and then heating of air.

61. What is the difference between Boot-strap air cooling system and Boot-strap evaporative cooling system?
a) Boot strap evaporative system has an evaporator
b) Boot strap evaporative system has two evaporators
c) Boot-strap evaporator eliminates the need for evaporator
d) Boot-strap evaporator system has three evaporators
Explanation: Boot strap evaporative system has an evaporator between the second heat exchanger and the cooling turbine.

62. Mass of air per tonne of refrigeration will be _____ in Boot-strap air cooling system than Boot-strap evaporative system.
a) less
b) more
c) equal to
d) can’t say
Explanation: Since the temperature of air leaving the cooling turbine in boot-strap system is lower than the simple boot-strap system, therefore the mass of air per tonne of refrigeration will be more in boot-strap air cooling system.

63. The air cooling system that is used mostly in transport type aircraft is?
a) Boot-strap air cooling system
b) Simple air cooling system
c) Simple evaporative cooling system
d) Regenerative air cooling system
Explanation: Boot-strap air cooling system uses a secondary air compressor along with an after cooler for achieving higher pressures of compression and more cooling effect. The cooling turbine replaces the use of cooling fan as well. Hence the air cooling system that is used mostly in transport type aircraft is Boot-strap air cooling system.

64. A Boot-strap air cooling system is used for a transport aircraft to take 10TR of refrigeration load (Q). The power required for the refrigerating system P is 800 KW. What is its COP?
a) 0.06426
b) 0.04375
c) 0.05435
d) 0.04367 #### Module 2

1. A vapor compression refrigeration system is an improved type of air refrigeration system in which a suitable working substance, termed as refrigerant is used.
a) True
b) False
Explanation: A vapor compression refrigeration system is an improved type of air refrigeration system in which a suitable working substance, termed as refrigerant is used. Refrigerants such as ammonia, CO2 and Sulphur dioxide etc. are used as refrigerants in a vapor compression refrigeration system.

2. The vapor compression refrigeration system is similar to a ________
a) latent heat pump
b) latent heat engine
c) generator
d) evaporator
Explanation: The vapor compression refrigeration system is a latent heat pump, as it pumps its latent heat from the brine and delivers it to the cooler.

3. The first vapor compression refrigeration system was developed by ________
a) L. G. Hamilton
b) Jacob Perkins
c) Stephen Rossi
d) Lenard Sandwik
Explanation: The first vapor compression refrigeration system was developed in 1834 by Jacob Perkins using hand operations.

4. The COP of vapor compression refrigeration compared to simple air refrigeration system is ________
a) high
b) low
c) same
d) can’t say
Explanation: COP of vapor compression refrigeration compared to simple air refrigeration system is high due to the use of refrigerant and as it can be employed over large range of temperatures.

5. Which of the following is not an advantage of vapor compression refrigeration system over air refrigeration system?
a) Smaller size for a given capacity of refrigeration
b) It has less running cost
c) It can be employed over wide range of temperature
d) There is no problem of leakage in vapor compression refrigeration system
Explanation: In vapor compression refrigeration system the prevention of leakage of the refrigerant is a major problem. The advantages of it are smaller size for a given capacity of refrigeration, it has less running cost and it can be employed over wide range of temperature.

6. The low pressure and temperature vapor refrigerant enters the ______ of the vapor compression system.
a) compressor
b) condenser
d) evaporator
Explanation: In the compressor the low pressure and temperature vapor refrigerant from the evaporator is drawn through the suction valve, where it is compressed to high pressure and temperature.

7. The high pressure and temperature vapor refrigerant enters the ______ of the vapor compression system.
a) compressor
b) condenser
d) evaporator
Explanation: In the condenser the high pressure and temperature vapor refrigerant from the compressor is condensed, during which the refrigerant gives up its latent heat.

8. In any compression refrigeration system there are how many pressure conditions?
a) 1
b) 2
c) 3
d) 4
Explanation: In any compression refrigeration system there are 2 pressure conditions, namely high pressure and low pressure side.

9. Which of the following is not an advantage of vapor compression cycle over reversed Carnot cycle?
a) It’ COP and refrigeration effect can’t be increased
b) Use of expansion valve reduces the size and cost of plant
c) It is a practical cycle on which plant can be built
d) Wet compression of Carnot cycle is avoided
Explanation: It’ COP and refrigeration effect can be increased by use of superheated vapor at entry of compressor. The advantages are use of expansion valve reduces the size and cost of plant, it is a practical cycle on which plant can be built and wet compression of Carnot cycle is avoided.

10. The pipe line emanating from compressor up to the condenser is called ________
a) suction line
b) pipe line
c) liquid line
d) delivery line
Explanation: The pipe line emanating from compressor up to the condenser is called delivery line or discharge line or hot gas line. Whereas suction line is the low pressure vapor line from the exit end evaporator leading to the compressor suction side and liquid line carries the liquid from receiver to the control valve

11. In vapor compression cycle having dry saturated vapor at the end of compression the work done is given by? H2 – Enthalpy of vapor after compression
H1 – Enthalpy of vapor before compression
a) H2 – H1
b) H2 + H1
c) H2 × H1
d) H2 ÷ H1
Explanation: Work is done on the compressor in the compression process (1 – 2), and the external work-done in the remainder process of the cycle is zero. Hence the net work-done equals to the compressor work.

12. In vapor compression cycle having dry saturated vapor at the end of compression the refrigeration effect is given by? H1 – Enthalpy of vapor before compression
H4 – Enthalpy of vapor before entering evaporator
H2 – Enthalpy of vapor after compression
a) H1 – H4
b) H2 – H4
c) H1 × H4
d) H2 × H4
Explanation: Refrigeration effect is the heat absorbed by the vaporizing refrigerant in the evaporator during the process (4 – 1), therefore refrigeration effect = H1 – H4.

13. Entropy of dry-saturated vapor when Sf is 0.2513 and Sfg is 4.7878, is _______
Sf – Entropy of saturated liquid at pressure p
Sfg – Entropy change during vaporization
a) 6.987
b) 5.987
c) 5.0391
d) 4.776
Explanation: Entropy of dry-saturated vapor when Sf is 0.2513 and Sfg is 4.5819, is given by
Sg = Sf + Sfg = 0.2513+4.7878 = 5.0391.

14. One of the assumptions of the vapor compression cycle is that all of the processes are reversible.
a) True
b) False

Explanation: Vapor compression cycle with vapor being dry saturated after compression is a theoretical cycle having various assumptions. One of the assumptions is that all the processes are reversible.

15. In which case does the compression process remains in superheated state?
a) Wet state
b) Dry state
c) Semi-dry state
d) Always
Explanation: In case of dry state the compression process remains in superheated state i.e. for compression of dry-saturated vapor or superheated vapor.

16. Dry compression is preferred over wet compression.
a) True
b) False
Explanation: Dry compression is preferred over wet compression since it gives high volumetric efficiency and mechanical efficiency of the compressor with less chances of damage to it. However, in some cases wet compression is preferred when the compression is done using a screw or rotary compressor.

17. The ratio of COP of vapor compression cycle and COP of Carnot cycle is known as?
a) Relative COP
b) Ideal COP
c) Performance index
d) Theoretical COP
Explanation: The ratio of COP of vapor compression cycle and COP of Carnot cycle is known as the Refrigeration efficiency or performance index (P.I). Whereas relative COP is the ratio of Actual COP divided by Ideal COP.

18. Which case is when the refrigerant compression process is carried while it’s wet?
a) Wet state
b) Dry state
c) Semi-dry state
d) Always
Explanation: In case of wet state the vapor compression process remains in wet state i.e. for compression of refrigerant is carried out while it’s still wet and for dry state compression is done in dry state.

19. In case of reciprocating compressors, wet compression is avoided.
a) True
b) False
Explanation: In case of reciprocating compressors, wet compression is avoided as the liquid droplets in the refrigerant would enter the compressor and damage the valves and other moving parts.

20. Which one of the following is not an assumption in theoretical vapour compression cycle?
a) There are no pressure losses in the condenser, evaporator, compressor, valves and the connecting part lines
b) All processes are reversible
c) There are mechanical and fluid friction losses
d) There is no heat transfer between the system and surroundings except in the evaporator and condenser
Explanation: There are no mechanical or fluid friction losses, is one of the assumptions made in theoretical vapor compression cycle. The other assumptions are all processes are reversible, there is no heat transfer between the system and surroundings except in the evaporator and condenser and there are no pressure losses in the condenser, evaporator, compressor, valves and the connecting part lines.

21. Power consumption in dry compression is less than wet compression.
a) True
b) False

Explanation: It is found that the power consumption is about 8%-10% less with wet compression compared to dry-compression. Hence power consumption in dry compression is less than wet compression.

22. Due to what in wet-compression affects the heat transfer rates from the vapor refrigerant to the cooling medium?
a) Liquid droplets
b) Work consumed
c) Power required
d) In wet-compression the heat transfer remains unaffected
Explanation: In wet – compression the liquid droplets may carry the lubricating oil from the cylinder walls of the compressor. This oil would then pass into the condenser and adversely affect the heat transfer rates from the vapor refrigerant to the cooling medium.

23. The following p – h diagram indicates? a) Vapor compression when vapor is dry at the end of the compression
b) Vapor compression when vapor is wet at the end of the compression
c) Vapor compression when vapor is dry – saturated at the end of the compression
d) Vapor compression when vapor is dry before of compression
Explanation: As from the diagram we can see that the compression line does not touch the saturated line it means that the vapor after compression is wet. For dry compression, it should’ve touched the saturation line.

24. Which of these is not an effect of superheat in suction vapour?
a) increase in refrigeration effect
b) increase in compression work
c) decrease the capacity of compressor per ton of refrigeration
d) condenser size decreases
Explanation: The effects of superheat in suction vapour are increase in refrigeration effect, increase in compression work, decreasing the capacity of compressor per ton of refrigeration and increasing the condenser size as the heat rejected in the condenser increases.

25. The COP of the cycle may increase or decrease or remain unchanged depending upon the range of pressure and the degree of superheating.
a) True
b) False
Explanation: The COP of the cycle may increase or decrease or remain unchanged depending upon the range of pressure and the degree of superheating as it can affect the refrigeration effect, compression work, heat rejected in condenser etc. and thus the COP.

26. In which case will the refrigeration effect will not increase and the COP of cycle with superheated vapour in suction phase will be less than the COP of ideal cycle?
a) Superheating the vapour from state 1 to 1′
b) Superheating the vapour from state 1 to 2
c) Superheating the vapour from state 1′ to 2′
d) COP always increases when superheating vapour in suction vapour
Explanation: As we can see from the below picture of a cycle with superheated vapour in suction vapour, the refrigeration effect is not increased and thus the COP will be less than that of ideal COP. 27. Which one of the following is not effect of liquid sub cooling?
a) increase the refrigeration effect
b) reduce the piston displacement
c) increase the COP
d) increase the compressor power per ton of refrigeration
Explanation: Sub cooling the liquid reduces the compressor power per ton of refrigeration due to reduced mass flow rate of refrigerant. It also increases the refrigeration effect, reduce the piston displacement and increase the COP.

28. Which one of the following is not effect of change in suction pressure?
a) reduces the net refrigeration effect
b) reduces the mass flow rate of refrigerant per ton of refrigeration
c) reduces the COP
d) increases the compressor power per ton of refrigeration
Explanation: Change in suction pressure reduces the net refrigeration effect thus increasing the mass flow rate of refrigerant per ton of refrigeration.

29. Effect of high side pressure or discharge pressure or condenser pressure is to reduce the COP of the cycle.
a) True
b) False
Explanation: Effect of high side pressure or discharge pressure or condenser pressure is to reduce the COP of the cycle as the refrigeration effect is reduced and the compression work increases as well as the compressor power increases per ton of refrigeration due to increased mass flow and additional energy supplied in compression.

30. What is the effect of using a flash chamber in VCR system?
a) C.O.P. increases
b) C.O.P. decreases
c) C.O.P. remains the same
d) Mass of refrigerant flowing through evaporator increases
Explanation: As C.O.P. of the VCR system containing flash chamber is given by,
C.O.P. = h1 − hf3 / h2 − h1
Which is same as the simple VCR cycle. The purest forms of refrigerant enter the respective system but effectively do not give enormous impact on C.O.P., So C.O.P. remains the same.
The mass of refrigerant required to flow through the evaporator decreases as even the smaller amount can produce the same refrigeration effect from its pure liquid form, resulting in a reduction in the size of evaporator.

31. From the following line diagram and p-h chart, what is the refrigeration effect in terms of the mass of mixture i.e., m2?  a) m2 (h1 − hf3)
b) m2 (h2 − h1)
c) m2 (h1 − h2)
d) m2 (h4 − h1)
Explanation: Refrigerating effect is considered in the evaporator i.e. heat absorbed by the evaporator is called as refrigerating effect.
R.E. = m1 (h1 − hf4’)
And also, m1 / m2 = h1 − h4 / h1 − hf4’
Hence, R.E. = m2 (h1 − hf3)

32. What is the value of C.O.P. in the VCR with flash chamber?  a) h1 – h2 / h1 – h4
b) h1 – hf3 / h2 – h1
c) h2 – h1 / h4 – h1
d) h1 – hf4’ / h1 – h4
Explanation: C.O.P. = Refrigerating effect / Work done
Refrigerating effect in terms of m2 = m2 (h1 – hf3)
Work = m2 (h2 – h1)
C.O.P. = m2 (h1 – hf3) / m2 (h2 – h1)
C.O.P. = (h1 – hf3) / (h2 – h1)

33. What is the effect of using accumulator in the VCR system?
a) C.O.P. increases
b) C.O.P. decreases
c) Total dry compression of refrigerant occurs
d) Sub-cooling happens
Explanation: As C.O.P. of the VCR system containing accumulator is given by,
C.O.P. = h1 – hf3 / h2 – h1
Which is same as the simple VCR cycle. The purest forms of refrigerant enter the respective system but effectively do not give enormous impact on C.O.P. So C.O.P. remains the same.
The C.O.P., R.E., and power required are the same, but it protects the liquid refrigerant from entering the compressor. Dry compression is always ensured by using accumulator or pre-cooler.

34. What is the value of C.O.P. in the VCR with a flash chamber?  a) h1 – h2 / h1 – h4
b) h1 – hf3 / h2 – h1
c) h2 – h1 / h4 – h1
d) h1 – hf4’ / h1 – h4
Explanation: C.O.P. = Refrigerating effect / Work done
Refrigerating effect in terms of m2 = m2 (h1 – hf3)
Work = m2 (h2 – h1)
C.O.P. = m2 (h1 – hf3) / m2 (h2 – h1)
C.O.P. = (h1 – hf3) / (h2 – h1)

35. Which of the following cooling towers possess maximum heat transfer from air to water?
a) Natural Draft
b) Mechanical Draft
c) Natural and Mechanical Draft
d) Atmospheric Draft
Explanation: Natural Draft is where air circulates naturally without any external force, and the atmospheric draft is a type of natural draft. So, the mechanical draft is the type of cooling tower which gives the maximum heat transfer due to usage of fans for the circulation. More the force more heat can be transferred.

36. How is the performance of the cooling tower indicated?
a) Wet-bulb temperature
b) Dry bulb temperature
c) Approach
d) Range
Explanation: Approach can be defined as the temperature difference between water leaving the cooling tower and ambient wet-bulb temperature. If the approach is low, then the performance is good and vice-versa. This approach value denotes how close the cooling tower gets the water to the wet-bulb temperature of the ambient or surrounding air.

37. What is the difference between the temperature of entering and leaving water in the cooling tower?
a) Wet-bulb temperature
b) Dry bulb temperature
c) Approach
d) Range
Explanation: Range can be defined as the difference between entering water temperature and leaving water temperature. This is also an indicator of the performance of cooling tower but for the effective performance, a better indicator, i.e., the approach is used.

38. What is the formula for the effectiveness of cooling tower from the following?
a) Range / (Range + Approach)
b) Approach / (Range + Approach)
c) Range / Approach
d) Approach / Range
Explanation: Approach can be defined as the temperature difference between water leaving the cooling tower and ambient wet-bulb temperature. The range can be defined as the difference between entering water temperature and leaving water temperature.
Hence, the effectiveness can be given as,
Effectiveness = Range / (Range + Approach).

39. A cooling tower brings water temperature to _________
a) WBT
b) DBT
c) DPT
d) Ambient WBT
Explanation: Approach can be defined as the temperature difference between water leaving the cooling tower and ambient wet-bulb temperature. If the approach is low, then the performance is good and vice-versa. This approach value denotes how close the cooling tower gets the water to the wet-bulb temperature of the ambient or surrounding air. So, cooling tower brings the water temperature to ambient wet bulb temperature for the operation.

40. What is the correct representation of range from the following?
a) Range = Cooling tower water outlet temperature – Wet bulb temperature
b) Range = Cooling tower water inlet temperature – Wet bulb temperature
c) Range = Cooling tower water outlet temperature – Dry bulb temperature
d) Range = Heat load in kcal per hour / Water circulation in liters per hour
Explanation: Range can be defined as the difference between entering water temperature and leaving water temperature. It is also defined as the heat load per water circulation. As in the given options, the first three are incorrect as one of them represents approach so correct representation is,
Range = Heat load in kcal per hour / Water circulation in liters per hour
Range = Cooling tower water inlet temperature – Cooling tower water outlet temperature.

41. What is L / G in terms of the cooling tower?
a) Length / Girth
c) Air mass flow rate / Water flow rate
d) Water flow rate / Air mass flow rate
Explanation: The L / G is called a liquid-gas ratio. It is defined as the ratio of water flow rate to the air mass flow rate in the cooling tower. This L / G ratio affects the effectiveness of the cooling tower.

42. Mechanical draft cooling tower size is ________ the size of the natural draft cooling tower.
a) smaller than
b) larger than
c) equal to the
d) very much larger than
Explanation: The mechanical draft cooling towers are smaller in size than that of the natural draft cooling towers with the same capacity due to the fact that a large volume of forced air increases the cooling capacity.

43. The cooling capacity of mechanical draft cooling towers cannot be varied.
a) False
b) True
Explanation: The main advantage of mechanical draft cooling towers is cooling capacity can be controlled. This can be achieved by controlling the amount of forced air entering the cooling tower.

44. Mechanical Draft cooling towers do not depend on atmospheric air.
a) True
b) False
Explanation: One of the advantages of mechanical draft cooling towers is that they do not depend upon the atmospheric air. Fans do the operation. Due to the fact of independence, this type of cooling towers can be even installed inside the building.

45. Among the following refrigerants, which is having the lowest freezing point?
a) R – 21
b) R – 11
c) R – 12
d) R – 22
Explanation: Freezing temperature is a crucial property to carry out refrigeration. Among the given refrigerants, R – 22 is having the lowest freezing point, valued at -160°C. R – 11, R – 12 and R – 21 having freezing points as -111°C, -157.5°C and -135°C respectively.

46. What is the boiling point of Ammonia at atmospheric pressure?
a) -10.5°C
b) -30°C
c) -33.3°C
d) -77.7°C
Explanation: Ammonia has a boiling point at -33.3°C. This low boiling point makes Ammonia possible to get refrigeration below 0°C even at atmospheric pressure.

47. R – 717 is the refrigerant number of which of the following refrigerant?
a) Ammonia
b) Water
c) Carbon dioxide
d) Sulphur dioxide
Explanation R – 717 is the refrigerant number of Ammonia. Whereas, refrigerant number of Water, Carbon dioxide, and Sulphur dioxide is R – 118, R – 744 and R – 764 respectively.

48. What is the boiling point of Carbon dioxide at atmospheric pressure?
a) -20.5°C
b) -73.6°C
c) -50°C
d) -68°C
Explanation: Carbon dioxide has a boiling point at -73.6°C at given conditions. Carbon dioxide is having the lowest boiling point among all the refrigerants which are commonly used due to high capacity o refrigeration.

49. What is the freezing point of R – 12 at atmospheric pressure?
a) -86.6°C
b) -73.6°C
c) -157.5°C
d) -160°C
Explanation: R – 12 is having a boiling point at -157.5°C at given conditions. R – 12 is having the second-lowest freezing point after R – 22, which is valued at -160°C.

50. What is the refrigerant number of Water?
a) R – 717
b) R – 744
c) R – 118
d) R – 100
Explanation: Water is having refrigerant number as R – 118. Whereas, R – 717, 744, 100 corresponds to Ammonia, Carbon dioxide, and ethyl chloride.

51. Among the following refrigerants, which is having the lowest boiling point?
a) Ammonia
b) R – 12
c) Carbon dioxide
d) Sulphur dioxide
Explanation: Boiling point temperature is a crucial property to carry out refrigeration. Among the given refrigerants, Carbon dioxide is having the lowest boiling point, valued at –73.6°C. Ammonia, R -12 and Sulphur dioxide are having boiling points as -33.3°C, -29°C and -10°C respectively.

52. Among the following refrigerants, which is having the lowest C.O.P for refrigeration system working under the temperature limits of -15°C and 30°C as evaporator and condenser temperature respectively?
a) Ammonia
b) R – 12
c) R – 30
d) Carbon dioxide
Explanation: Coefficient of performance is the ratio of refrigeration produced and work done to get that refrigeration effect. C.O.P. is a crucial entity. The more the C.O.P., more the refrigeration effect is produced, and lower temperatures can be achieved. Among the given options, Carbon dioxide is having least C.O.P. for given temperatures conditions, which is valued at 2.56. Whereas, Ammonia, R – 12 and R – 30 is having C.O.P. as 4.76, 4.70 and 4.90 respectively.

53. R – 11 is having the highest C.O.P for refrigeration system working under the temperature limits of -15°C and 30°C as evaporator and condenser temperature respectively.
a) False
b) True
Explanation: Coefficient of performance is the ratio of refrigeration produced and work done to get that refrigeration effect. C.O.P. is a crucial entity. The more the C.O.P., more the refrigeration effect is produced, and lower temperatures can be achieved. For a Carnot cycle, C.O.P. under given conditions is 5.74 and R – 11 posses 5.09 C.O.P. very close to the ideal cycle.

54. A refrigerant should have a low latent heat of vaporization at the evaporator temperature.
a) True
b) False
Explanation: A refrigerant should have a high latent heat of vaporization at the evaporator temperature. More the latent heat of vaporization more is the capacity of refrigerant to absorb heat in the evaporator. So, Higher the value of latent heat of vaporization, higher is the refrigeration effect produced.

#### Module 3

1. What is the formula for COP of an ideal vapour absorption refrigeration system?
Qg = the heat given to the refrigerant in the generator
Qe= the heat absorbed by the refrigerant in the evaporator  2. An ideal vapour absorption refrigeration system may be regarded as a combination of _________
a) Carnot engine and Carnot pump
b) Carnot refrigerant and Carnot pump
c) Carnot engine and Carnot refrigerant
d) Carnot engine alone 3. Find the COP of a vapour absorption refrigeration system using the following data.
Qg = 6600
Qe = 2100
a) 0.45
b) 0.65
c) 0.31
d) 0.35 4. Find the COP of a vapour absorption refrigeration system using the following data.
Te = 268
Tg = 393.2
Tc = 303
a) 1.53
b) 1.76
c) 1.66
d) 1.98 5. In case the heat is discharged at different temperatures in condenser and absorber then the formula for COP would be?
Ta = Temperature at which heat is discharged in the absorber
Te and Tc are the temperature limits for Carnot refrigerator
Tg and Tc are the temperature limits for Carnot engine  6. Find the COP of an ideal vapour absorption refrigeration system using the following data.
Qg = 1880
Qe = 1045
a) 0.47
b) 0.65
c) 0.69
d) 0.55 7. Which of the following is correct about VARS and VCRS?
a) VARS use mechanical energy, and VCRS use heat energy
b) VARS use heat energy, and VCRS use mechanical energy
c) Both use mechanical energy
d) Both use heat energy
Explanation: VCRS use mechanical energy i.e., VCRS uses compressor which withdraws energy from the evaporator and drawn into condenser after compression. VARS use heat energy i.e., heat exchanger or heat generators are used and by which desired effect is achieved.

8. The compressor from VCRS is replaced by which of the following in the VARS?
a) Absorber, Pump
b) Generator, Pressure reducing valve
c) Absorber, Pump, Generator, and Pressure reducing valve
d) Absorber, Rectifier, Generator, and Pressure reducing valve
Explanation: Compressor from the VCRS is replaced by an absorber, a pump, a generator, and a pressure reducing valve. These all components together do the same work as a compressor but by using the heat energy.

9. What is the purpose of using an absorber? a) Heat absorption
b) Heat rejection
c) Pressure reduction
d) Work done
Explanation: Vapour refrigerant from the evaporator enters the absorber. Cooling water is circulated to transfer the heat, and refrigerant’s strong solution is discharged to the generator.

10. What is the purpose of using a pump? a) Heat absorption
b) Heat rejection
c) Pressure reduction
d) Pressure increment
Explanation: The strong solution discharging from the absorber is pumped to the generator using a liquid pump. The liquid pump increases the pressure up to 10 bar.

11. What is the purpose of using a generator? a) Heat supplied
b) Heat rejection
c) Pressure reduction
d) Pressure increment
Explanation: Generator generates heat. From the generator, heat is supplied to the refrigerant, and the strong refrigerant is heated further by using some external source like gas or steam.

12. What is the fundamental equation of the C.O.P. of Ideal VAR system?
a) Heat absorbed in the evaporator / Work done by a pump
b) Heat absorbed in the evaporator / Heat supplied in the generator
c) Heat absorbed in the evaporator / Work done by compressor + Heat supplied in the generator
d) Heat absorbed in the evaporator / Work done by pump + Heat supplied in the generator
Explanation: VAR uses the generator to do work as well as pump to pressurize the liquid to the generator. So, total work done is the summation of work done by a pump and generator. Refrigeration effect is the heat absorbed in the evaporator.
So, C.O.P. = Refrigeration effect / Total work done
= Heat absorbed in the evaporator / Work done by pump + Heat supplied in the generator.

13. What is the relation between QC, QG, and QE if they represent heat given to the refrigerant in the generator, heat discharging to atmosphere from condenser to absorber and heat absorbed by the refrigerant in the evaporator?
a) QC + QG = QE
b) QC = QG + QE
c) QC + QE = QG
d) QC = QG – QE
Explanation: According to the first law of thermodynamics, heat discharged to the atmosphere or cooling water from the condenser to absorber is equal to the heat given to the refrigerant in a generator, heat absorbed by the refrigerant in the evaporator and heat added to the refrigerant due to pump work. Heat added due to pump work is neglected due to minimal effect.
So, QC = QG + QE

14. What is the equation of VAR system in terms of entropy?
a) QC + QE / TC = QG / TG + QE / TE
b) QG + QC / TC = QG / TG + QE / TE
c) QG + QE / TC = QG / TG + QE / TE
d) QG + QE / TC = QC / TC + QE / TE
Explanation: As vapour absorption system is considered as a perfectly reversible system,
Hence the initial entropy of the system should be equal to the entropy of the system after some change in its conditions.
As entropy is heat absorbed divided by the temperature.
From the first law of thermodynamics, QC = QG + QE
So, the equation in terms of entropy QC / TC = QG / TG + QE / TE
Modified as, QG + QE / TC = QG / TG + QE / TE

15. What is the ratio of QG / QE in terms of temperature?
If TC, TG and TE if they represent temperature at which heat is given to the refrigerant in the generator, temperature at which heat is discharging to atmosphere from condenser to absorber and temperature at which heat is absorbed by refrigerant in the evaporator
a) [TC – TE / TE] [TG / TG – TC]
b) [TC – TE / TE] [TG / TG – TE]
c) [TC – TE / TE] [TC / TG – TC]
d) [TC – TE / TC] [TG / TG – TC]
Explanation: By using the phenomenon of perfectly reversible system,
QG + QE / TC = QG / TG + QE / TE
QG / TG – QG / TC = QE / TC – QE / TE
QG [TC – TG / TC x TG] = QE [TE – TC / TC x TE]
By rearranging, QG / QE = [TC – TE / TE] [TG / TG – TC]

16. What is the value of maximum C.O.P. in terms of heat?
a) QG / QE
b) QE / QG
c) QG / QC
d) QC / QE
Explanation: As VAR uses the generator to do work as well as pump to pressurize the liquid to the generator. So, total work done is the summation of work done by a pump and generator. Refrigeration effect is the heat absorbed in the evaporator.
So, C.O.P. = Refrigeration effect / Total work done
= Heat absorbed in the evaporator / Work done by pump + Heat supplied in the generator
For maximum C.O.P., due to the minimal effect of work done by a pump, QP is neglected.
QE = Heat absorbed in the evaporator and QG = Heat supplied in the generator
C.O.P. maximum = QE / QG

17. What is the value of maximum C.O.P. in terms of temperature?
a) [TC – TE / TE] [TG / TG – TC]
b) [TC – TE / TE] [TG / TG – TC]
c) [TC – TE / TE] [TG / TG – TC]
d) [TG – TC / TG] [TE / TC – TE]
Explanation: As VAR uses the generator to do work as well as pump to pressurize the liquid to the generator. So, total work done is the summation of work done by a pump and generator. Refrigeration effect is the heat absorbed in the evaporator.
So, C.O.P. = Refrigeration effect / Total work done
= Heat absorbed in the evaporator / Work done by pump + Heat supplied in the generator
For maximum C.O.P., due to the minimal effect of work done by a pump, QP is neglected.
QE = Heat absorbed in the evaporator and QG = Heat supplied in generator
C.O.P. maximum = QE / QG
By using the phenomenon of perfectly reversible system,
QG + QE / TC = QG / TG + QE / TE
QG / TG – QG / TC = QE / TC – QE / TE
QG [TC – TG / TC x TG] = QE [TE – TC / TC x TE]
By rearranging, QE / QG = [TG – TC / TG] [TE / TC – TE]
(C.O.P.) max = QE / QG = [TG– TC / TG] [TE / TC – TE]

18. Which of the following is true?
a) (C.O.P.)max = (C.O.P.)Carnot / ηCarnot
b) (C.O.P.)max = (C.O.P.)Carnot + ηCarnot
c) (C.O.P.)max = (C.O.P.)Carnot x ηCarnot
d) (C.O.P.)max = (C.O.P.)Carnot – ηCarnot
Explanation: As, (C.O.P.) max = QE / QG = [TG – TC / TG] [TE / TC – TE]
Where, it may be noted that, [TE / TC – TE] is the coefficient of performance of a Carnot refrigerator working between TE and TC temperatures.
And, [TG – TC / TG] is the efficiency of a Carnot engine working between TG and TC temperatures.
Hence, (C.O.P.)max = (C.O.P.)Carnot x ηCarnot

19. C.O.P. of the VAR system is lower than the C.O.P. of VCR system.
a) True
b) False
Explanation: Though the VAR system has many advantages over VCR, VAR gives lower C.O.P. as compared to VCR. Compressor used in the VCR system given higher pressure difference, which increases the temperature more and thus increasing the heat rejection. Overall impacting the C.O.P. to increase but in case of VAR system, heat supplied is limited so overall C.O.P. is not as high as in the VCR.

20. In a VAR system, heating, cooling, and refrigeration take place at temperatures 200°C, 30°C, and -10°C respectively. What is the value of maximum C.O.P. of the system?
a) 2.202
b) 2.808
c) 3.404
d) 2.404
Explanation: Given: TG = 200°C = 200 + 273 = 473 K
TC = 30°C = 273 + 30 = 303 K
TE = – 10°C = 273 – 10 = 263 K
As, (C.O.P.)max = QE / QG = [TG – TC / TG] [TE / TC – TE]
= [263 / 303 – 263] [473 – 303 / 473]
= [263/ 40] [173 / 473]
= [45499 / 18920]
= 2.404.

21. What is the value of maximum C.O.P. if the VAR system has a capacity of 12 TR and heat given to the refrigerant in the generator is 40 kW?
a) 0.9523
b) 0.4878
c) 1.05
d) 2.05
Explanation: Given: QE = 12 TR = 12 x 3.5 = 42 kW
QG = 40 kW
As, C.O.P. maximum = QE / QG
= 42 / 40
= 1.05.

22. The device which divides the high pressure side and the low pressure side of a refrigerating system is known as _____________
a) condenser device
b) evaporator device
d) expansion device
Explanation: The expansion device is an important device that divides the high pressure side and the low pressure side of a refrigerating system. It is also known as metering device or throttling device.

23. Which of the following function is not performed by the expansion device?
a) It reduces the high pressure liquid refrigerant to low pressure liquid refrigerant
b) It maintains the desired pressure difference between the high and low pressure sides of the system
c) It controls the flow of refrigerant according to the load on the evaporator
d) It compresses the vapor
Explanation: The expansion device performs the following functions, it reduces the high pressure liquid refrigerant to low pressure liquid refrigerant, it maintains the desired pressure difference between the high and low pressure sides of the system and It controls the flow of refrigerant according to the load on the evaporator.

24. The expansion device used with flooded evaporators is known as expansion valves.
a) True
b) False
Explanation: The expansion device used with flooded evaporators is known as float valves and the expansion device used with dry expansion evaporators are called expansion valves.

25. The expansion device is placed between which two components?
a) Condenser and evaporator
b) Compressor and condenser
c) Evaporator and compressor

Explanation: The expansion device is placed between receiver (containing liquid refrigerant at high pressure) and evaporator (containing liquid refrigerant at low pressure).

26. Which of the following is not an advantage of capillary tube?
a) The cost of capillary tube is less
b) A high starting motor is not required
d) A capillary tube designed for a specific condition will also work efficiently for other conditions
Explanation: A capillary tube designed for a specific condition will not work efficiently for other conditions as the length is directly and inner diameter is indirectly proportional to the frictional resistance. The longer the tube and smaller the diameter, greater is the pressure drop created in the refrigerant flow.

27. Capillary tube, as an expansion device is used in?
a) Water coolers
b) Domestic refrigerators
c) Room air conditioners
d) All of the mentioned
Explanation: The capillary tube as an expansion device is used in small capacity hermetic sealed refrigeration units such as in water coolers, domestic refrigerators and room air conditioners, etc.

28. Which one of the following is also known as a constant superheat valve?
a) Capillary tube
b) Hand-operated expansion valve
c) Thermostatic Expansion valve
d) Low side float valve
Explanation: Thermostatic expansion valve is also called a constant superheat valve because it maintains a constant superheat of the vapor refrigerant at the end of the evaporator coil, by controlling the flow of liquid refrigerant through the evaporator.

29. Thermostatic expansion valves are usually set for a superheat of?
a) 10°C
b) 5°C
c) 8°C
d) 15°C
Explanation: Thermostatic expansion valves are usually set for a superheat of 5°C for better efficiency of the refrigeration cycle.

30. The low-side float valve is located between the condenser and evaporator.
a) True
b) False
Explanation: The low-side float valve is located in the low pressure side i.e. between the evaporator and the compressor suction line. Whereas the high side float valve is located in the high pressure side i.e. between the condenser and evaporator.

31. The thermostatic expansion valve operates on the changes in the ___________
a) degree of superheat at exit from the evaporator
b) temperature of the evaporator
c) pressure in the condenser
d) pressure in the evaporator
Explanation: The thermostatic expansion valve operates on the changes in the degree of superheat at exit from the evaporator. Thermostatic expansion valves are usually set for a superheat of 5° C for better efficiency of the refrigeration cycle.

#### Module 4 & 5

1. What is the mixture of a number of gases?
a) Moist air
b) Dry air
c) Fresh air
d) Saturated air
Explanation: Dry air is the mixture of a number of gases like nitrogen, oxygen, carbon dioxide, hydrogen, etc. These gases have different proportion in this mixture by which the properties may alter.

2. Which of the following has the maximum contribution in the dry air?
a) Nitrogen
b) Argon
c) Carbon dioxide
d) Hydrogen
Explanation: Dry air is the mixture of a number of gases like nitrogen, oxygen, carbon dioxide, hydrogen, etc. In this mixture, nitrogen has the maximum contribution. Nitrogen is approximately 78.03 % by volume and 75.47 % by mass in this mixture.

3. Nitrogen and Argon have a major portion in the dry air.
a) False
b) True
Explanation: Dry air is the mixture of a number of gases like nitrogen, oxygen, carbon dioxide, hydrogen, etc. In this mixture, nitrogen has the maximum contribution. Nitrogen is approximately 78.03 % by volume and 75.47 % by mass in this mixture. Oxygen is approximately 20.99 % by volume and 23.19 % by mass in the mixture. But the argon has a very little portion in the mixture. So, Nitrogen and Oxygen have a major portion in the dry air.

4. Mixture of dry air and water vapor is __________
a) moist air
b) dry air
c) fresh air
d) saturated air
Explanation: If water vapor is present in a larger amount in the dry air, then the type of air is called as moist air. The amount of water vapor present in the air depends upon the absolute pressure and temperature of the mixture.

5. Specific humidity and absolute humidity have the same unit.
a) False
b) True
Explanation: Humidity ratio or Specific humidity is the mass of water vapor present in 1 kg of dry air. It is generally expressed in, g / kg of dry air and absolute humidity is the mass of water present in 1 m3 of dry air. It is generally expressed in, g / m3 of dry air.

6. The mixture of dry air and water vapor called when the air has diffused the maximum amount of water vapor into it is called ________
a) dry air
b) moist air
c) saturated air
d) specific humidity
Explanation: Saturated air is the mixture of dry air and water vapor when the air has diffused the maximum amount of water vapor into it. The water vapors occur in the form of superheated steam as an invisible gas.

7. What is the ratio of actual mass of water vapor in a unit mass of dry air to the mass of water vapor in the same mass of dry air when it is saturated at the same temperature?
a) Moist air
b) Dry air
c) Degree of saturation
d) Saturated air
Explanation: The ratio of actual mass of water vapor in a unit mass of dry air to the mass of water vapor in the same mass of dry air when it is saturated at the same temperature is called as the degree of saturation.

8. What is the mass of water vapor present in 1 kg of dry air called?
a) Specific Humidity
b) Relative humidity
c) Degree of saturation
d) Saturated air
Explanation: Humidity ratio or Specific humidity is the mass of water vapor present in 1 kg of dry air. It is generally expressed in, g / kg of dry air. Mathematically, W = mv / ma.

9. What is the ratio of actual mass of water vapor in a given volume of moist air to the mass of water vapor in the same volume of saturated air at the same pressure and temperature?
a) Specific Humidity
b) Relative humidity
c) Degree of saturation
d) Saturated air
Explanation: Relative humidity is the ratio of actual mass of water vapor in a given volume of moist air to the mass of water vapor in the same volume of saturated air. It can be represented as,
∅ = mv / ms = pv / ps.

10. What is the temperature of air recorded by a thermometer, when the moisture present in it starts condensing?
a) DBT
b) WBT
c) DPT
d) WBD
Explanation: DPT i.e., Dew Point Temperature is the temperature at which the moisture in it begins to condense. It can also be defined as the temperature corresponding to the partial pressure of water vapor.

11. What is the value of humidity ratio if pv = 0.387 bar and pa = 0.997 bar?
a) 0.482
b) 0.241
c) 0.122
d) 0.622
Explanation: Humidity ratio is the mass of water vapor present in 1 kg of dry air and has a unit g / kg of dry air.
W = 0.622 pv / pa
= 0.622 x 0.387 / 0.997
= 0.241 kg / kg of dry air.

12. What is the value of humidity ratio if mv = 1.423 and ma = 3.589?
a) 0.1
b) 0.2
c) 0.4
d) 0.5
Explanation: Humidity ratio is the mass of water vapor present in 1 kg of dry air and has a unit g / kg of dry air. It is also defined as the ratio of the mass of water vapor to the mass of dry air in a given volume of the air-vapor mixture. Hence, it is given as m/ ma.
W = 1.423 / 3.589 = 0.396 ≅ 0.4.

13. What is the value of relative humidity if mv = 2.901 and ms = 9.056?
a) 0.300
b) 0.330
c) 0.310
d) 0.320
Explanation: Relative humidity is the ratio of actual mass of water vapor in a given volume of moist air to the mass of water vapor in the same volume of saturated air. It can be represented as,
∅ = mv / ms = pv / ps = 2.901 / 9.056 = 0.320.

14. What is the value of relative humidity if pv = 0.468 bar and ps = 0.893 bar?
a) 52 %
b) 54 %
c) 56 %
d) 190 %
Explanation: Relative humidity is the ratio of actual mass of water vapor in a given volume of moist air to the mass of water vapor in the same volume of saturated air. It can be represented as,
∅ = mv / ms = pv / ps = 0.468 / 0.893 = 0.524 = 52.4 %.

15. What is the value of pb if pa = 1.48 bar and pv = 1.52 bar, according to Dalton’s law of partial pressures?
a) 1 bar
b) 2 bar
c) 3 bar
d) 4 bar
Explanation: The Dalton’s law states that the total pressure exerted by the mixture of air and water vapor is equal to the sum of the pressures, which each constituent would exert if it occupied the same space by itself. Mathematically, it can be represented as, pb = pa + pv = 1.48 + 1.52 = 3 bar.

16. If DBT = 30°C, WBT = 20°C and Barometer reading is 740 mm of Hg. Find out DPT using steam tables.
a) 5°C
b) 35°C
c) 15°C
d) 25°C
Explanation: For given conditions, pressure corresponding to WBT from steam table is pw = 0.02337 bar
pb = 740 mm of Hg = 740 x 133.3 / 100000 = 0.98642 bar
Pressure of water vapor = pv = pw – [(pb – pw) (td – tw) / 1544 – 1.44 tw]
= 0.02337 – [(0.98642 – 0.02337) (30 – 20) / 1544 – 1.44 x 20]
= 0.01701 bar
For 0.01701 bar water vapor pressure, dew point temperature is 15°C.

17. What do vertical and uniformly spaced lines indicate on the psychrometric chart?
a) DPT
b) WBT
c) DBT
d) Specific humidity
Explanation: The psychrometric chart has DBT on the x-axis. So, the vertical and uniformly spaced lines denote dry bulb temperature and spaced by 5°C.

18. Which of the following is represented by curved lines on the psychrometric chart?
a) Specific humidity
b) Relative humidity
c) WBT
d) DPT
Explanation: The psychrometric chart has DBT on the x-axis and specific humidity on the y-axis. The curved line on the grid shows the relative humidity. From 0 to 100 %, these lines are drawn at an interval of 10%.

19. What is represented by inclined straight lines but non-uniformly spaced on the psychrometric chart?
a) Specific humidity
b) Relative humidity
c) WBT
d) DPT
Explanation: The psychrometric chart has DBT on the x-axis and specific humidity on the y-axis. The inclined lines from the saturation curve and on the x-axis denote wet-bulb temperature.

20. What is represented by horizontal lines but non-uniformly spaced on the psychrometric chart?
a) Specific humidity
b) Relative humidity
c) WBT
d) DPT
Explanation: The psychrometric chart has DBT on the x-axis and specific humidity on the y-axis. The horizontal lines from the saturation curve and parallel to the abscissa denote dew point temperature.

21. What is represented by inclined straight lines but uniformly spaced on the psychrometric chart?
a) Enthalpy
b) Relative humidity
c) WBT
d) DPT
Explanation: The psychrometric chart has DBT on the x-axis and specific humidity on the y-axis. The inclined lines from the saturation curve and parallel to the wet-bulb temperature lines denote Enthalpy.

22. What is represented by obliquely inclined straight lines but uniformly spaced on the psychrometric chart?
a) Enthalpy
b) Relative humidity
c) Specific volume
d) DPT
Explanation: The psychrometric chart has DBT on the x-axis and specific humidity on the y-axis. The obliquely inclined lines from the saturation curve and on the dry-bulb temperature lines denote Specific volume.

23. Specific humidity lines are also known as_________
a) moisture content lines
b) relative humidity
c) specific volume
d) moist lines
Explanation: Specific humidity is the moisture present in the air mixture. So, specific humidity denotes moisture content. Hence, these lines are also known as moisture content lines.

24. Relative humidity on saturation curve has value of ________% at various dry bulb temperatures.
a) 0
b) 50
c) 10
d) 100
Explanation: Relative humidity is the ratio of actual mass of water vapor in the given volume of moist air to the mass of water vapor in the same amount of saturated air. So, on the saturation curve, for every dry bulb temperature, the value of relative humidity remains 100%.

25. Enthalpy lines and specific volume lines are the same.
a) False
b) True
Explanation: Though both of the lines representing given properties are inclined straight lines and uniformly spaced, but the slope of both the lines are different. The specific volume lines are obliquely inclined.

26. A psychrometric chart is a graphical representation of various physical properties of dry air.
a) True
b) False
Explanation: Psychrometric chart is a graphical representation of the various thermodynamic properties of moist air. This is used to find out properties of air in the field of air conditioning.

27. What is the alternate name for thermodynamic wet-bulb temperature?
a) Isobaric WBT
b) Isobaric Saturation Temperature
Explanation: The thermodynamic wet-bulb temperature is also known as adiabatic saturation temperature due to the process involved in it.

28. What is the temperature at which air can be brought to saturation state adiabatically?
a) Thermodynamic WBT
b) Thermodynamic DBT
c) Thermodynamic DPT
d) DPT
Explanation: The temperature at which air can be brought to saturation state adiabatically is called a thermodynamic wet-bulb temperature or adiabatic saturation temperature.

29. Which of the following is carried out to obtain saturation state at the thermodynamic WBT?
a) Condensation
b) Evaporation
c) Compression
d) Expansion
Explanation: The temperature at which air can be brought to saturation state adiabatically is called a thermodynamic wet-bulb temperature or adiabatic saturation temperature. This state is achieved by evaporation of water into the flowing air.

30. Which of the following is the major equipment in the simplest form is used to do saturation of air?
a) Chamber
b) Vessel
c) Insulated Chamber
d) Pipe
Explanation: The equipment used for saturation of the air in the simplest form is an insulated chamber containing an optimum amount of water.

31. The arrangement for extra-water to flow into the chamber from its top is called as ___________
a) chamber ingot
b) chamber gate
c) make-up gate
d) make-up water
Explanation: The equipment used for saturation of the air in the simplest form is an insulated chamber containing an optimum amount of water. In order to get more water into the chamber, an arrangement is provided, which is called make-up water.

32. What is the ratio of humidity ratio of entering and discharging air?
a) W2 / W1 = (hs2 – hfw) + ha2 – ha1 / (hs1 – hfw)
b) W1 / W2 = (hs1 – hfw) + ha2 – ha1 / (hs2 – hfw)
c) W1 / W2 = (hs2 – hfw) + ha2 – ha1 / (hs1 – hfw)
d) W1 / W2 = (hs2 – hfw) + ha1 – ha2 / (hs1 – hfw)
Explanation: Balancing enthalpies, h1 – W1 hfw = h2 – W2 hfw
As, h1 = ha1 + W1 hs1 and h2 = ha2 + W2 hs2
So, by putting values we get,
W1 / W2 = (hs2 – hfw) + ha2 – ha1 / (hs1 – hfw).

33. The term (h2 – W2 hfw) is known as_________
a) sigma term
b) sigma heat
c) heat factor
d) heat term
Explanation: The term (h2 – W2 hfw) is known as sigma heat, and this term remains constant during the adiabatic process.

34. If h1 = 32 kJ/kg, h2 = 41 kJ/kg, hfw = 49 kJ/kg, W1 = 0.0109 kg/kg of dry air and W2 = 0.0297 kg/kg of dry air then what is the value of sigma heat?
a) 41.0865
b) 39.5447
c) 38.9013
d) 45.8775
Explanation: Sigma heat = (h2 – W2 hfw)
So, putting values in the formula,
Sigma heat = [41 – (49 x 0.0297)] = 39.5447.

35. Evaporation from water to flowing air is carried out adiabatically.
a) False
b) True
Explanation: The temperature at which air can be brought to saturation state adiabatically is called a thermodynamic wet-bulb temperature or adiabatic saturation temperature. This state is achieved by evaporation of water into the flowing air.

36. Make-up water is used to keep the water level constant.
a) True
b) False
Explanation: In order to get more water into the chamber, an arrangement is provided, which is called make-up water. This water is added from the top to keep the water level constant and keep the energy transfer at an optimum pace.

37. What is the By-pass factor for heating coil, if td1 = temperature at entry, td2 = temperature at exit and td3 = coil temperature?
a) td3 – td1 / td3 – td1
b) td3 – td2 / td3 – td1
c) td3 – td2 / td2 – td1
d) td3 – td2 / td1 – td2
Explanation: By-pass factor is the amount of air by-passed in the process. So, for heating coil,
BPF = Temperature difference between coil and exit / Temperature difference between coil and entry
= td3 – td2 / td3 – td1.

38. What is the By-pass factor for cooling coil, if td1 = temperature at entry, td2 = temperature at exit and td3 = coil temperature?
a) td3 – td1 / td3 – td1
b) td3 – td2 / td3 – td1
c) td3 – td2 / td2 – td1
d) td2 – td3 / td1 – td3
Explanation: By-pass factor is the amount of air by-passed in the process. So, for cooling coil,
BPF = Temperature difference between exit and coil / Temperature difference between entry and coil
= td2 – td3 / td1 – td3.

39. If the value of BPF for one row of a coil is y then what is the value of BPF for n similar rows?
a) n / y
b) n + y
c) (y)n
d) n x y
Explanation: BPF for multiple similar rows of a coil is the power of one BPF to the number of rows. i.e. (y)n.

40. What is the value of sensible heat given out by the coil?
a) U Ac tm
b) U Ac
c) U tm
d) U Ac2 tm
Explanation: Sensible heat given out by the coil is the product of overall heat transfer coefficient, the surface area of the coil and logarithmic mean temperature difference.
So, Sensible heat = U Ac tm.

41. What is the formula of logarithmic mean temperature in terms of By-pass factor?
a) Tm = td2 – td1 / loge [1/BPF]
b) Tm = td2 – td1 / log10 [BPF]
c) Tm = td2 – td1 / loge [BPF]
d) Tm = td2 – td3 / loge [BPF]
Explanation: Logarithmic mean temperature difference for the given arrangement is,
Tm = td2 – td1 / loge [td3 – td1 / td3 – td2]
As, BPF for the coil = [td3 – td1 / td3 – td2]
Tm = td2 – td1 / loge [BPF].

42. What is the efficiency of the coil?
a) 1 + BPF
b) 1 / BPF
c) BPF
d) 1 – BPF
Explanation: As the by-pass factor is the inefficiency, so the contact factor or efficiency of the coil is given by 1 – BPF.

43. What is the contact factor for heating coil, if td1 = temperature at entry, t2 = temperature at exit and td3 = coil temperature?
a) td3 – td1 / td3 – td1
b) td2 – td1 / td3 – td1
c) td3 – td2 / td2 – td1
d) td3 – td2 / td1 – td2
Explanation: Contact factor or efficiency of the coil is 1 – By-pass factor. So, for heating coil,
ηH = 1 – BPF
= 1 – [td3 – td2 / td3 – td1]
= td2 – td1 / td3 – td1.

44. What is the contact factor for cooling coil, if td1 = temperature at entry, td2 = temperature at exit and td3 = coil temperature?
a) td1 – td2 / td1 – td3
b) td2 – td1 / td3 – td1
c) td3 – td2 / td2 – td1
d) td3 – td2 / td1 – td2
Explanation: Contact factor or efficiency of the coil is 1 – By-pass factor. So, for cooling coil,
ηC = 1 – BPF
= 1 – [td2 – td3 / td1 – td3]
= td1 – td2 / td1 – td3.

45. A coil with low BPF has better performance.
a) True
b) False
Explanation: As BPF is the amount of air bypassed in the process and is the inefficiency of the coil. So, lower the value of BPF then better is the performance and vice-versa.

46. The value of BPF for the heating and the cooling coil is different under the same temperature conditions.
a) True
b) False
Explanation: BPF for heating coil is, BPF = td3 – td2 / td3 – td1
BPF for cooling coil is, BPF = td2 – td3 / td1 – td3
As the numerator and denominator are reversed in the cooling coil than the heating coil. So, the value we get for either of the coils is the same

47. Which of the following denotes Humidification process on the psychrometric chart?
a) A horizontal line with an arrow towards right
b) Vertical line with arrow upwards
c) Horizontal line with an arrow towards right
d) Vertical line with arrow downwards
Explanation: Psychrometric chart has DBT on the x-axis and sp. Humidity on the y-axis. So, if a process is going up on the chart, then there is an increase in relative as well as specific humidity. So, the vertical line with an arrow upwards denotes humidification process.

48. Which of the following denotes dehumidification process on the psychrometric chart?
a) A horizontal line with an arrow towards the right
b) Vertical line with arrow upwards
c) A horizontal line with an arrow towards the right
d) Vertical line with arrow downwards
Explanation: Psychrometric chart has DBT on the x-axis and sp. Humidity on the y-axis. So, if a process is going down on the chart, then there is a decrease in relative as well as specific humidity. So, the vertical line with an arrow downwards denotes dehumidification process.

49. Which of the following is denoted by LH = (h2 – h1) = hfg (W2 – W1)?
a) Sensible cooling
b) Sensible heating
c) Humidification
d) Dehumidification
Explanation: Latent heat transfer i.e., LH = (h2 – h1) = hfg (W2 – W1) denotes Humidification. As in the humidification process, there is an increase in specific humidity, so W2 > W1 and also increase in enthalpy, h2 > h1.

50. Which of the following is denoted by LH = (h1 – h2) = hfg (W1 – W2)?
a) Sensible cooling
b) Sensible heating
c) Humidification
d) Dehumidification
Explanation: Latent heat transfer i.e., LH = (h1 – h2) = hfg (W1 – W2) denotes dehumidification. As in the dehumidification process, there is a decrease in specific humidity, so W1 > W2 and also decrease in enthalpy, h1 > h2.

51. What is the relation between DBT and WBT if the relative humidity is 100%?
a) DBT = WBT
b) DBT > WBT
c) DBT ≫ WBT
d) DBT < WBT
Explanation: When the dry bulb temperature is equal to the wet-bulb temperature, then the relative humidity tends to be 100%.

52. The minimum temperature to which moist air can be cooled under ideal conditions in a spray washer is ___________
a) DPT of inlet air
b) Water inlet temperature
c) WBT of inlet air
d) Water outlet temperature
Explanation: Wet-bulb temperature is the minimum temperature to which moist air can be cooled under ideal conditions in a spray washer. It is an integral part of the process.

53. Which of the following process is used in winter air conditioning?
a) Humidification
b) Dehumidification
c) Heating and Humidification
d) Cooling and Dehumidification
Explanation: In Winter, the weather is dry and cold, so the process of heating and humidification is done to get the desired comfort conditions.

54. Which of the following process is used in summer air conditioning?
a) Humidification
b) Dehumidification
c) Heating and Humidification
d) Cooling and Dehumidification
Explanation: In Summer, the weather is humid and hot, so the process cooling and dehumidification is done to get the desired comfort conditions.

55. What is the value of Latent heat transfer in the process of humidification having enthalpies as 20 and 40 kJ / kg of dry air respectively?
a) 60
b) 40
c) -20
d) 20

Explanation: Latent heat transfer i.e., LH = (h2 – h1) = hfg (W2 – W1) denotes Humidification. As in the humidification process, there is an increase in specific humidity, so W2 > W1 and also increase in enthalpy, h2 > h1. So, as per given data, h1 = 40 and h2 = 20 kJ / kg of dry air.
So, latent heat transfer = h2 – h1 = 40 – 20 = 20 kJ/ kg of dry air.

56. What is the value of Latent heat transfer in the process of dehumidification having enthalpies as 19.45 and 13.67 kJ/ kg of dry air respectively?
a) 5.78
b) -5.78
c) 33.12
d) -33.12
Explanation: Latent heat transfer i.e., LH = (h1 – h2) = hfg (W1 – W2) denotes dehumidification. As in the dehumidification process, there is a decrease in specific humidity, so W1 > W2 and also decrease in enthalpy, h1 > h2. So, as per given data, h1 = 19.45 and h2 = 13.67 kJ / kg of dry air.
So, latent heat transfer = h1 – h2 = 19.45 – 13.67 = 20 kJ / kg of dry air.

57. What is the value of optimum effective temperature in winter?
a) 17
b) 18
c) 19
d) 20
Explanation: From the table of design conditions for comfort, we get to know that in winter, the optimum effective temperature is 19°C.

58. What is the value of the optimum effective temperature in summer?
a) 23
b) 22
c) 21
d) 24
Explanation: From the table of summer design conditions for comfort, we get to know that in summer, the optimum effective temperature is 22°C.

59. Which of the following provides the relationship between the optimum indoor effective temperature and optimum outdoor effective temperature?
a) Clothing
b) Occupants
c) Duration of stay
d) Climatic and seasonal differences
Explanation: As season changes, then the relationship between the optimum indoor effective temperature and optimum outdoor effective temperature changes. This affects the optimum effective temperature for comfort.

60. A person with _________ clothing needs ___________ optimum temperature.
a) heavy, higher
b) heavy, lesser
c) light, higher
d) light, lesser
Explanation: Clothing is one of the factors affecting the effective temperature. So, to note, a person with light clothing needs lesser optimum temperature.

61. A person with heavy clothing needs ____________ optimum temperature as compared to a person with light clothing.
a) higher
b) lesser
c) zero
d) sam
Explanation: Clothing is one of the factors affecting the effective temperature. So to note, a person with heavy clothing needs a higher optimum temperature than a person with light clothing.

62. Men need a higher effective temperature than women.
a) True
b) False
Explanation: Women actually need higher effective temperature than men to be comfortable.

63. What is the value of an increase in effective temperature do women need than men?
Explanation: Women actually need higher effective temperature than men to be comfortable. This value differs approximately about 0.5⁰C.

64. Which of the following is true about effective temperature between young and older people?
a) Tyoung > Told
b) Tyoung < Told
c) Tyoung ≪ Told
d) Tyoung = Told
Explanation: As there is a difference between effective temperature as per gender then it also happens in the age sector. Young people need a higher effective temperature than older people.

65. What is the value of an increase in effective temperature do children need than older ones in case of maternity halls?
a) 1 to 2°C
b) 2 to 3°C
c) 3 to 4°C
d) 0 to 1°C
Explanation: As there is a difference between effective temperature as per gender then it also happens in the age sector. Young people need a higher effective temperature than older people, i.e., Tyoung > Told
Hence, maternity halls are maintained 2 to 3°C more than the effective temperature needed for adults.

66. People living in colder climate regions feel comfortable at lower effective temperature.
a) True
b) False
Explanation: Climatic and seasonal differences affect the effective temperature. People living in colder climate areas are comfortable with low effective temperatures than people living in warmer areas.

67. Which of the following is true about the optimum effective temperature for human comfort?
a) Same in winter and summer
b) Not dependent on season
c) Lower in winter than in summer
d) Higher in winter than in summer
Explanation: Optimum effective temperature is dependent on climatic and seasonal changes. So, for human comfort, the optimum effective temperature is lower in winter than in summer.

68. The heat production from a normal healthy man when asleep is about _________
a) 50 W
b) 40 W
c) 70 W
d) 60 W
Explanation: A normal healthy man when asleep produces heat nearly equal to 60 W. This might vary depending on various parameters.

69. When the heat stored in the body is ________ the human body feels comfortable.
a) zero
b) infinite
c) positive
d) negative
Explanation: The human body feels comfortable when the heat stored in the body is zero. This is the desired comfort condition.

70. Which of the following does not mainly a factor of dependency for the degree of warmth or cold?
a) Relative humidity
b) WBT
c) Air velocity
d) DBT
Explanation: The degree of warmth or cold felt by the human body depends mainly on dry bulb temperature, air velocity and relative humidity.

71. The effective temperature increases with a decrease in relative humidity at the same DBT.
a) True
b) False
Explanation: At the same value of dry bulb temperature, the effective temperature decreases with a decrease in relative humidity to achieve desired comfort conditions.

72. What is the inefficiency of the coil to attain the desired temperature called?
a) By-pass factor
b) Effectiveness
c) Efficiency
d) By-pressure factor
Explanation: The DBT of air leaving the apparatus should be equal to the coil temperature, but this is not possible due to the inefficiency of the coil. This type of phenomenon is called a by-pass factor.

73. What is the number of fins provided in a unit length called?
a) By-pass
b) Distance
c) Pitch
Explanation: By-pass factor depends upon the number of fins provided. So, the fins provided in a unit length are called as Pitch of the coil fins.

74. Which of the following is true about the effect of velocity on the by-pass factor?
a) decreases with an increase in velocity of air passing through it
b) remains unchanged with an increase in velocity of air passing through it
c) may increase or decrease with an increase in velocity of air passing through it depending upon the condition of air entering
d) increases with an increase in velocity of air passing through it
Explanation: The change in the velocity affects the temperature of leaving the air. Moreover, the inefficiency increases due to an increase in velocity as the by-pass factor increases.

75. The by-pass factor of the coil ___________ with _________ in fin spacing.
a) increases, increase
b) decreases, decrease
c) increases, decrease
d) decreases, increase
Explanation: One of the factors which affect the by-pass factor is fin spacing. If the fin spacing is not much, then the temperature leaving the apparatus becomes close to the coil temperature. So, the decrease in fin spacing results in a decrease in the by-pass factor by improving efficiency.

76. The by-pass factor of the coil ___________ with _________ in number of rows of fins.
a) increases, increase
b) decreases, decrease
c) increases, decrease
d) decreases, increase
Explanation: One of the factors which affect the by-pass factor is a number of rows of fins. More the number of rows of fins, better is the heat transfer rate and decreasing the inefficiency, i.e., by-pass factor. So, the increase in the number of rows results in a decrease in the by-pass factor by improving efficiency.

77. Which of the following does not affect the by-pass factor?
a) Pitch of the coil fins
b) Velocity of air
c) Weight of apparatus
d) Fin spacing
Explanation: The weight of the apparatus has nothing to do with the by-pass factor. Pitch of the coil does affect the by-pass factor. And the velocity of air as well as fin spacing affects the by-pass factor. These factors can reduce the efficiency of the process.

78. Number of rows of the fins in the opposite direction to the flow affects BPF.
a) False
b) True
Explanation: More the number of rows increases the heat transfer rate and decrease the by-pass factor. But these rows should be in the direction of flow to get desired results.

79. Air being passed by not being in contact with the coil known to be measured as BPF.
a) True
b) False
Explanation: When air flows, some of it gets in contact with the coil, and some passes directly unaffected. This by-pass process is measured as a by-pass factor and is affected by a lot of parameters.

80. What is the formula for m1 / m2 in terms of enthalpies for adiabatic mixing of two streams?
a) h1 – h2 / h1 – h3
b) h3 – h2 / h1 – h3
c) h3 – h2 / h2 – h3
d) h3 – h2 / h1 – h2
Explanation: For energy balance, m1 h1 + m2 h2 = m3 h3
m3 = m1 + m2
So, m1 h1 + m2 h2 = (m1 + m2) h3
By solving we get,
m1 / m2 = h3 – h2 / h1 – h3.

81. What is the formula for m1 / m2 in terms of specific humidity for adiabatic mixing of two streams?
a) W2 – W1 / W1 – W3
b) W3 – W1 / W1 – W3
c) W3 – W2 / W1 – W3
d) W3 – W2 / W2 – W3
Explanation: For energy balance, m1 W1 + m2 W2 = m3 W3
m3 = m1 + m2
So, m1 W1 + m2 W2 = (m1 + m2) W3
By solving we get,
m1 / m2 = W3 – W2 / W1 – W3.

82. Which of the following is true for the adiabatic mixing of two streams?
a) m3 = m3 – m1
b) m2 = m1 + m3
c) m1 = m3 + m2
d) m3 = m1 – m2
Explanation: As two streams are mixed, then the result of it is the summation of mixed masses, and using the energy balance ratio of masses can be obtained. So, m3 = m1 + m2.

83. When the adiabatic mixing is carried out, the air having ______ enthalpies and ____________ specific humidities are mixed.
a) similar, similar
b) different, similar
c) similar, different
d) different, different
Explanation: The two air streams getting mixed adiabatically have different enthalpies and different specific humidities to get the final condition of air.

85. What is the value of m1 / m2 if h1 = 81 kJ/ kg of dry air, h2 = 46 kJ/ kg of dry air and h3 = 58 kJ/ kg of dry air?
a) 1
b) 0
c) 0.51
d) 0.52
Explanation: As we know, m1 / m2 = h3 – h2 / h1 – h3
= 58 – 46 / 81 – 58
= 0.5217 = 0.52.

86. What is the value of m1 / m2 if W1 = 0.0157 kg/ kg of dry air, W2 = 0.0084 kg / kg of dry air and W3 = 0.0103 kJ/ kg of dry air?
a) 0.31
b) 0.28
c) 0.35
d) 0.52
Explanation: As we know, m1 / m2 = W3 – W2 / W1 – W3
= 0.0103 – 0.0084 / 0.0157 – 0.0103
= 0.0019 / 0.0054 = 0.3518 = 0.35.

87. What is the value of one of the mass before mixing if the other mass is 1.98 kg, and the final is 4.22 kg?
a) 2.24
b) 1.98
c) 4.22
d) 6.20
Explanation As two streams are mixed, then the result of it is the summation of mixed masses, and using the energy balance ratio of masses can be obtained. So, m3 = m1 + m2
If, m1 = 1.98 and m3 = 4.22 then, m2 = 4.22 – 1.98 = 2.24 kg.

88. What is the value of final enthalpy if the enthalpies before mixing were 50 and 28 kJ / kg of dry air respectively and m2 / m1 = 2?
a) 38.67
b) 48.67
c) 40.05
d) 52
Explanation: For energy balance, m1 h1 + m2 h2 = m3 h3
m3 = m1 + m2
So, m1 h1 + m2 h2 = (m1 + m2) h3
m1 / m2 = h3 – h2 / h1 – h3
0.5 = h3 – 28 / 50 – h3
50 – h3 = 2h3 – 56
3h3 = 50 + 56 = 116
h3 = 38.67.

89. Even a lesser quantity of steam may result in the formation of fog.
a) True
b) False
Explanation: Fog also results when steam or very fine water spray is injected into the air in a greater quantity than required to do the saturation of the air. So, even less quantity of steam without mixing properly can result in fog.

90. Fog can be cleared by cooling.
a) True
b) False
Explanation: Fog can be cleared by heating or mixing it with unsaturated air or mechanically separating water droplets from the air.

91. Which of the following has the highest DBT in the chart of outside summer design conditions?
a) Agra
b) Ambala
c) Banaras
d) Bangalore
Explanation: From the table of outside summer design conditions, the highest DBT among the given options is at Ambala with 43.3°C, whereas the temperatures of Agra, Banaras, Bangalore are 41.5°C, 40.8°C and 32.9°C respectively.

92. Which of the following has the lowest DBT in the chart of outside summer design conditions?
a) Cochin
b) Darjeeling
c) Jaipur
d) Srinagar
Explanation: From the table of outside summer design conditions, the lowest DBT among the given options is at Darjeeling with 17.2°C, whereas the temperatures of Cochin, Jaipur, Srinagar are 35°C, 40.8°C and 25°C respectively.

93. Which of the following has the highest WBT in the chart of outside summer design conditions?
a) Ahmednagar
b) Baroda
c) Cuttack
d) Trivandrum
Explanation: From the table of outside summer design conditions, the highest WBT among the given options is at Ahmednagar with 31.1°C, whereas the temperatures of Baroda, Cuttack, Trivandrum are 29.1°C, 30.6°C and 29.4°C respectively.

94. Which of the following has the lowest WBT in the chart of outside summer design conditions?
a) Cochin
b) Srinagar
c) Mahabaleshwar
d) Darjeeling
Explanation: From the table of outside summer design conditions, the lowest WBT among the given options is at Darjeeling with 14.5°C, whereas the temperatures of Cochin, Mahabaleshwar, Srinagar are 27.8°C, 19.2°C and 18.4°C respectively.

95. What is the value of optimum effective temperature in winter?
a) 17
b) 18
c) 19
d) 20
Explanation: From the table of design conditions for comfort, we get to know that in winter, the optimum effective temperature is 19°C.

96. Which of the following has the highest relative humidity in the chart of outside summer design conditions?
a) Aligarh
b) Bhopal
c) Chennai
d) Trivandrum
Explanation: From the table of outside summer design conditions, the highest RH among the given options is at Trivandrum with 90% whereas, Aligarh, Bhopal, Chennai have 35%, 24% and 47% respectively.

97. Which of the following has the lowest relative humidity in the chart of outside summer design conditions?
a) Jaipur
b) Chandigarh
c) Indore
d) Mysore
Explanation: From the table of outside summer design conditions, the lowest RH among the given options is at Jaipur with 18% whereas, Chandigarh, Indore, Mysore have 27%, 32% and 52% respectively.

98. What is the value of the optimum effective temperature in summer?
a) 23
b) 22
c) 21
d) 24
Explanation: From the table of summer design conditions for comfort, we get to know that in summer, the optimum effective temperature is 22°C.

99. Which of the following has the highest effective temperature in the chart of outside summer design conditions?
c) Banaras
d) Cuttack
Explanation: From the table of outside summer design conditions, the highest ET among the given options is at Ahmedabad with 32.5°C, whereas the temperatures of Allahabad, Banaras, Cuttack are 30.8°C, 31.9°C and 32°C respectively.

100. Which of the following has the lowest effective temperature in the chart of outside summer design conditions?
a) Srinagar
b) Shimla
c) Darjeeling
d) Surat
Explanation: From the table of outside summer design conditions, the lowest ET among the given options is at Darjeeling with 16.8°C, whereas the temperatures of Srinagar, Shimla, Surat are 22.2°C, 19.2°C and 30.5°C respectively.

101. Which of the following is the odd man out in terms of dry bulb temperature?
a) Ambala
b) Rajpur
c) Jodhpur
d) Pune
Explanation: Ambala, Rajpur and Jodhpur have a similar value of dry bulb temperature, which is 43.3°C, and Pune has 37.1°C. So, the odd man out of the given options is Pune.

102. Relative humidity at Agra is _____________ the relative humidity at Kanpur.
a) higher than
b) lower than
c) equal to
d) much higher than
Explanation: Relative humidity at Agra is equal to the relative humidity at Kanpur, having a value of 17%.

103. Which of the cities has the lowest WBT in the chart?
a) Surat
b) Shimla
c) Lucknow
d) Mumbai