RAC
1. At what K is the triple point of water?
a) 0 K
b) 4 K
c) 273.16 K
d) -273.16
Answer: c
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
Answer: b
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
Answer: c
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
Answer: a
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
Answer: d
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
Answer: d
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
Answer: c
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
Answer: a
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
Answer: c
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?
Answer: b
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
Answer: c
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)
Answer: a
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
Answer: c
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
Answer: b
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.
Answer: b
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
Answer: c
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
Answer: c
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
Answer: c
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
Answer: c
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
c) Adiabatic expansion process
d) Isothermal expansion process
Answer: c
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
Answer: a
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
Answer: c
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
Answer: a
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
Answer: c
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)
Answer: d
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
Answer: c
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
Answer: d
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
Answer: b
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
Answer: c
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
Answer: c
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
Answer: b
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
Answer: c
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
Answer: b
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
Answer: a
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
Answer: d
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
Answer: d
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
Answer: a
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
Answer: a
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
Answer: a
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
Answer: b
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
Answer: d
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
Answer: a
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
Answer: b
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
Answer: a
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
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