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All questions of Refrigeration & Air Conditioning for Mechanical Engineering Exam
A building in a cold climate is to be heated by a Carnot heat pump. The minimum outside temperature is –23°C. If the building is to be kept at 27°C and heat requirement is at the rate of 30 kW, what is the minimum power required for heat pump?- a)180 kW
- b)30 kW
- c)6 kW
- d)5 kW
Correct answer is option 'D'. Can you explain this answer?
A building in a cold climate is to be heated by a Carnot heat pump. The minimum outside temperature is –23°C. If the building is to be kept at 27°C and heat requirement is at the rate of 30 kW, what is the minimum power required for heat pump?
a)
180 kW
b)
30 kW
c)
6 kW
d)
5 kW
|
Sanvi Kapoor answered |
or
Sub-cooling with regenerative heat exchanger is used in a refrigeration cycle. The enthalpies at condenser outlet and evaporator outlet are 78 and 182 kJ/kg respectively. The enthalpy at outlet of isentropic compressor is 230 kJ/kg and enthalpy of subcooled liquid is 68 kJ/kg. The COP of the cycle is:- a)3.25
- b)2.16
- c)3.0
- d)3.5
Correct answer is option 'C'. Can you explain this answer?
Sub-cooling with regenerative heat exchanger is used in a refrigeration cycle. The enthalpies at condenser outlet and evaporator outlet are 78 and 182 kJ/kg respectively. The enthalpy at outlet of isentropic compressor is 230 kJ/kg and enthalpy of subcooled liquid is 68 kJ/kg. The COP of the cycle is:
a)
3.25
b)
2.16
c)
3.0
d)
3.5
|
Rahul Chatterjee answered |
The answer is b.
C.O.P = (182-68) / (230-182) = 2.375 << 2.16
C.O.P = (182-68) / (230-182) = 2.375 << 2.16
In a domestic refrigerator, a capillary tube controls the flow of refrigerant from the- a)expansion valve to the evaporator
- b)evaporator to the thermostat
- c)condenser to the expansion valve
- d)condenser to the evaporator
Correct answer is option 'D'. Can you explain this answer?
In a domestic refrigerator, a capillary tube controls the flow of refrigerant from the
a)
expansion valve to the evaporator
b)
evaporator to the thermostat
c)
condenser to the expansion valve
d)
condenser to the evaporator
|
Avinash Sharma answered |
Hence capillary tube is placed between condenser and evaporator.
The COP of a Carnot refrigeration cycle decreases on- a)Decreasing the difference in operating temperatures
- b)Keeping the upper temperature constant and increasing the lower temperature
- c)Increasing the upper temperature and keeping the lower temperature constant
- d)Increasing the upper temperature and decreasing the lower temperature
Correct answer is option 'C'. Can you explain this answer?
The COP of a Carnot refrigeration cycle decreases on
a)
Decreasing the difference in operating temperatures
b)
Keeping the upper temperature constant and increasing the lower temperature
c)
Increasing the upper temperature and keeping the lower temperature constant
d)
Increasing the upper temperature and decreasing the lower temperature
|
Anirban Khanna answered |
The correct answer is d.
where COP is the coefficient of performance, then N is equal to:
Can you explain the answer of this question below:A Carnot cycle refrigerator operates between 250K and 300 K. Its coefficient of performance is:- A:6.0
- B:5.0
- C:1.2
- D:0.8
The answer is b.
A Carnot cycle refrigerator operates between 250K and 300 K. Its coefficient of performance is:
A:
6.0
B:
5.0
C:
1.2
D:
0.8
|
Shivam Sharma answered |
COP = T2/ (T1 – T2)= 250/300-250= 5.The answer is b.
A heat pump works on a reversed Carnot cycle. The temperature in the condenser coils is 27°C and that in the evaporator coils is –23°C. For a work input of 1 kW, how much is the heat pumped?- a)1 kW
- b)5 kW
- c)6 kW
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
A heat pump works on a reversed Carnot cycle. The temperature in the condenser coils is 27°C and that in the evaporator coils is –23°C. For a work input of 1 kW, how much is the heat pumped?
a)
1 kW
b)
5 kW
c)
6 kW
d)
None of the above
|
|
Aditya Deshmukh answered |
For heat pump (COP) = Q1/W = T1/(T1 - T2) = 300/(300 - 250)
OR, Q1 = 6 X W = 6 X 1 = 6kW
Pure dehumidification can be achieved by passing air through the spray water maintained at- a)DBT
- b)WBT
- c)Dew point temperature
- d)Any of the above
Correct answer is option 'C'. Can you explain this answer?
Pure dehumidification can be achieved by passing air through the spray water maintained at
a)
DBT
b)
WBT
c)
Dew point temperature
d)
Any of the above
|
|
Avinash Sharma answered |
Pure dehumidification is achieved by passing air through the spray water which is maintained at dew point temperature. By doing this, excess moisture condenses and temperature remains constant.
The correct sequence of vapour compression (VC), vapour absorption (VA) and steam ejector (SE) refrigeration cycles in increasing order of the COP is:- a)VC, VA, SE
- b)VA, SE, VC
- c)SE, VC, VA
- d)SE, VA, VC
Correct answer is option 'B'. Can you explain this answer?
The correct sequence of vapour compression (VC), vapour absorption (VA) and steam ejector (SE) refrigeration cycles in increasing order of the COP is:
a)
VC, VA, SE
b)
VA, SE, VC
c)
SE, VC, VA
d)
SE, VA, VC
|
Sagarika Dey answered |
The correct sequence of VC, VA and SE in increasing order of COP is VA, SE and VC, the Value being of the order of 0.3 to 0.4 0.5 to 0.8 and 4 to 5 respectively.
A single-acting two-stage compressor with complete inter cooling delivers air at 16 bar. Assuming an intake state of 1 bar at 15°C, the pressure ratio per stage is:- a)16
- b)8
- c)4
- d)2
Correct answer is option 'C'. Can you explain this answer?
A single-acting two-stage compressor with complete inter cooling delivers air at 16 bar. Assuming an intake state of 1 bar at 15°C, the pressure ratio per stage is:
a)
16
b)
8
c)
4
d)
2
|
Raghav Saini answered |
Pressure ratio of each stage must be same
Which of the following is considered comfort conditions in air-conditioning?- a)10°C DBT and 50% RH
- b)15°C DBT and 40% RH
- c)20°C DBT and 60% RH
- d)25°C DBT and 70% RH
Correct answer is option 'C'. Can you explain this answer?
Which of the following is considered comfort conditions in air-conditioning?
a)
10°C DBT and 50% RH
b)
15°C DBT and 40% RH
c)
20°C DBT and 60% RH
d)
25°C DBT and 70% RH
|
Lavanya Menon answered |
A human body feels comfortable when the air is at 21°C DBT with 56% RH. In general, for summer air conditioning, the RH should not be less than 60% whereas for winter air conditioning it should not be more than 40%.
Which of the following statements is not correct for the volumetric efficiency of a reciprocating air compressor? - a)It decreases with increase in delivery pressure
- b)It increases with decrease in pressure ratio
- c)It increases with decrease in clearance ratio
- d)It decreases with increase in ambient temperature
Correct answer is option 'A'. Can you explain this answer?
Which of the following statements is not correct for the volumetric efficiency of a reciprocating air compressor?
a)
It decreases with increase in delivery pressure
b)
It increases with decrease in pressure ratio
c)
It increases with decrease in clearance ratio
d)
It decreases with increase in ambient temperature
|
Nandita Datta answered |
One tonne refrigerator machine means that- a)one tonne is the total mass of machine.
- b)one tonne of refrigerator is used.
- c)one tonne of water can be converted onto ice.
- d)one tonne of ice when melts from and at 0°C in 24 hours, the refrigerating effect produced is equivalent to 210 kJ/min.
Correct answer is option 'D'. Can you explain this answer?
One tonne refrigerator machine means that
a)
one tonne is the total mass of machine.
b)
one tonne of refrigerator is used.
c)
one tonne of water can be converted onto ice.
d)
one tonne of ice when melts from and at 0°C in 24 hours, the refrigerating effect produced is equivalent to 210 kJ/min.
|
|
Lavanya Menon answered |
1 tonne of refrigeration is the rate of heat removal required to freeze a metric ton (1000 kg) of water at 0°C in 24 hours. Based on the heat of fusion being 333.55 kJ/kg, 1 tonne of refrigeration = 13,898 kJ/h = 3.861 kW
In refrigeration system refrigerant gains heat at- a)compressor
- b)evaporator
- c)condenser
- d)expansion valve
Correct answer is option 'B'. Can you explain this answer?
In refrigeration system refrigerant gains heat at
a)
compressor
b)
evaporator
c)
condenser
d)
expansion valve
|
Garima Kulkarni answered |
In refrigeration system, heat is absorbed at evaporator and rejected at condenser.
Which of the following is not true in respect of an ideal refrigerant?- a)High latent heat of vaporization and specific heat.
- b)Critical pressure and temperature well above the maximum operating pressure and temperature limits.
- c)Low value of specific volume.
- d)High value of thermal conductivity
Correct answer is option 'A'. Can you explain this answer?
Which of the following is not true in respect of an ideal refrigerant?
a)
High latent heat of vaporization and specific heat.
b)
Critical pressure and temperature well above the maximum operating pressure and temperature limits.
c)
Low value of specific volume.
d)
High value of thermal conductivity
|
Niharika Iyer answered |
Desirable properties of an ideal refrigerant:
-Low boiling point
- High critical temperature
- High latent heat of vaporization
- Low specific heat of liquid
- Low specific volume of vapour
- Non-corrosive to metal
- Non-flammable and non-explosive
- Non-toxic
- Low cost
- Easy to liquify at moderate pressure and temperature
- Easy of locating leaks by odour or suitable indicator
- Mixes well with oil
-Low boiling point
- High critical temperature
- High latent heat of vaporization
- Low specific heat of liquid
- Low specific volume of vapour
- Non-corrosive to metal
- Non-flammable and non-explosive
- Non-toxic
- Low cost
- Easy to liquify at moderate pressure and temperature
- Easy of locating leaks by odour or suitable indicator
- Mixes well with oil
A heat pump for domestic heating operates between a cold system at 0°C and the hot system at 60°C. What is the minimum electric power consumption if the heat rejected is 80000 kJ/hr?- a)2 kW
- b)3 kW
- c)4 kW
- d)5 kW
Correct answer is option 'C'. Can you explain this answer?
A heat pump for domestic heating operates between a cold system at 0°C and the hot system at 60°C. What is the minimum electric power consumption if the heat rejected is 80000 kJ/hr?
a)
2 kW
b)
3 kW
c)
4 kW
d)
5 kW
|
Sagnik Yadav answered |
For minimum power consumption,
A heat pump is used to heat a house in the winter and then reversed to cool the house in the summer. The inside temperature of the house is to be maintained at 20°C. The heat transfer through the house walls is 7·9 kJ/s and the outside temperature in winter is 5°C. What is the minimum power (approximate) required driving the heat pump?- a)40·5 W
- b)405 W
- c)42·5 W
- d)425 W
Correct answer is option 'B'. Can you explain this answer?
A heat pump is used to heat a house in the winter and then reversed to cool the house in the summer. The inside temperature of the house is to be maintained at 20°C. The heat transfer through the house walls is 7·9 kJ/s and the outside temperature in winter is 5°C. What is the minimum power (approximate) required driving the heat pump?
a)
40·5 W
b)
405 W
c)
42·5 W
d)
425 W
|
Akshat Mehta answered |
or
Efficiency of a Carnot engine is 75%. If the cycle direction is reversed, COP of the reversed Carnot cycle is - a)1.33
- b)0.75
- c)0.33
- d)1.75
Correct answer is option 'A'. Can you explain this answer?
Efficiency of a Carnot engine is 75%. If the cycle direction is reversed, COP of the reversed Carnot cycle is
a)
1.33
b)
0.75
c)
0.33
d)
1.75
|
Samarth Chaudhary answered |
If the cycle direction is reversed it will be a heat pump not refrigerator. Students make a common mistake here and calculated (COP)R. We know that the definition of refrigeration is producing a temperature below atmospheric temperature. In heat engine lower temperature is atmospheric temperature. When we reverse this cycle then lower temperature will be atmospheric and higher temperature will be more than atmospheric means it will be a Heat Pump not a refrigerator.
0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves it with 450 kJ of specific enthalpy. Velocities at inlet and exit are 6 m/s and 2 m/s respectively. Assuming adiabatic process, what is power input to the compressor?- a)120 kW
- b)118 kW
- c)115kW
- d)112 kW
Correct answer is option 'D'. Can you explain this answer?
0.70 kg/s of air enters with a specific enthalpy of 290 kJ and leaves it with 450 kJ of specific enthalpy. Velocities at inlet and exit are 6 m/s and 2 m/s respectively. Assuming adiabatic process, what is power input to the compressor?
a)
120 kW
b)
118 kW
c)
115kW
d)
112 kW
|
|
Dishani Desai answered |
Power input to compressor 
In on-off control refrigeration system, which one of the following expansion devices is used?- a)Capillary tube
- b)Thermostat
- c)Automatic expansion valve
- d)Float valve
Correct answer is option 'B'. Can you explain this answer?
In on-off control refrigeration system, which one of the following expansion devices is used?
a)
Capillary tube
b)
Thermostat
c)
Automatic expansion valve
d)
Float valve
|
Yash Patel answered |
On-off control is done by thermostat. In the On- off control of the air-handling unit, the fan motor is operated intermittently by the room thermostat the bulb of which is installed in the return air.
The coefficient of performance (COP) of a refrigerator working as a heat pump is given by:- a)(COP)heat pump = (COP)refrigerator+ 2
- b)(COP)heat pump = (COP)refrigerator+ 1
- c)(COP)heat pump = (COP)refrigerator – 1
- d)(COP)heat pump = (COP)refrigerator
Correct answer is option 'B'. Can you explain this answer?
The coefficient of performance (COP) of a refrigerator working as a heat pump is given by:
a)
(COP)heat pump = (COP)refrigerator+ 2
b)
(COP)heat pump = (COP)refrigerator+ 1
c)
(COP)heat pump = (COP)refrigerator – 1
d)
(COP)heat pump = (COP)refrigerator
|
|
Samarth Chaudhary answered |
The COP of refrigerator is one less than COP of heat pump, if same refrigerator starts working as heat pump i.e. (COP)heat pump = (COP)refrigerator + 1
One tonnes of refrigeration is equal to- a)210 kJ/min
- b)3.5 kJ/min
- c)105 kJ/min
- d)250 kJ/min
Correct answer is option 'A'. Can you explain this answer?
One tonnes of refrigeration is equal to
a)
210 kJ/min
b)
3.5 kJ/min
c)
105 kJ/min
d)
250 kJ/min
|
Ashutosh Sharma answered |
One Tonnes of Refrigeration (TR) is equal to 210 kJ/min.
Explanation:
- One Tonnes of Refrigeration (TR) is a unit of power used in refrigeration and air conditioning industries.
- It is defined as the amount of cooling required to freeze one metric tonne (1000 kg) of water from 0°C to -1°C in 24 hours.
- TR is commonly used to rate the capacity of refrigeration and air conditioning systems.
- The conversion factor from TR to kJ/min is 210.
- Therefore, one TR is equal to 210 kJ/min.
Formula:
- The formula to calculate the cooling capacity in TR is given as: Q = m × Cp × ΔT / 24
- Where Q is the cooling capacity in TR, m is the mass of water to be frozen (in kg), Cp is the specific heat of water (4.18 kJ/kg°C), and ΔT is the temperature difference (in °C) between the initial and final temperatures of water.
- For example, if we want to freeze 1000 kg of water from 0°C to -1°C in 24 hours, the cooling capacity required will be:
Q = 1000 × 4.18 × (0-(-1)) / 24 = 34.9 kJ/min
- To convert this cooling capacity in TR, we need to divide it by 211. That is:
Q(TR) = 34.9 / 210 = 0.166 TR
Conclusion:
- One Tonnes of Refrigeration (TR) is equal to 210 kJ/min.
- It is used to rate the capacity of refrigeration and air conditioning systems.
- The formula to calculate the cooling capacity in TR is Q = m × Cp × ΔT / 24.
- To convert the cooling capacity in TR to kJ/min, we need to multiply it by 210.
Explanation:
- One Tonnes of Refrigeration (TR) is a unit of power used in refrigeration and air conditioning industries.
- It is defined as the amount of cooling required to freeze one metric tonne (1000 kg) of water from 0°C to -1°C in 24 hours.
- TR is commonly used to rate the capacity of refrigeration and air conditioning systems.
- The conversion factor from TR to kJ/min is 210.
- Therefore, one TR is equal to 210 kJ/min.
Formula:
- The formula to calculate the cooling capacity in TR is given as: Q = m × Cp × ΔT / 24
- Where Q is the cooling capacity in TR, m is the mass of water to be frozen (in kg), Cp is the specific heat of water (4.18 kJ/kg°C), and ΔT is the temperature difference (in °C) between the initial and final temperatures of water.
- For example, if we want to freeze 1000 kg of water from 0°C to -1°C in 24 hours, the cooling capacity required will be:
Q = 1000 × 4.18 × (0-(-1)) / 24 = 34.9 kJ/min
- To convert this cooling capacity in TR, we need to divide it by 211. That is:
Q(TR) = 34.9 / 210 = 0.166 TR
Conclusion:
- One Tonnes of Refrigeration (TR) is equal to 210 kJ/min.
- It is used to rate the capacity of refrigeration and air conditioning systems.
- The formula to calculate the cooling capacity in TR is Q = m × Cp × ΔT / 24.
- To convert the cooling capacity in TR to kJ/min, we need to multiply it by 210.
The by-pass factor of a cooling coil decreases with - a)decrease in fin spacing and increase in number of rows.
- b)increase in fin spacing and increase in number of rows. .
- c)increase in fin spacing and decrease in number of rows.
- d)decrease in fin spacing and decrease in number of rows.
Correct answer is option 'A'. Can you explain this answer?
The by-pass factor of a cooling coil decreases with
a)
decrease in fin spacing and increase in number of rows.
b)
increase in fin spacing and increase in number of rows. .
c)
increase in fin spacing and decrease in number of rows.
d)
decrease in fin spacing and decrease in number of rows.
|
Harshad Iyer answered |
The BPF depends upon the following factors:
1. The number of fins provided in a unit length i.e. the pitch of the cooling coil fins.
2. The number of rows in a coil in the direction of flow.
3. The velocity of flow of air.
1. The number of fins provided in a unit length i.e. the pitch of the cooling coil fins.
2. The number of rows in a coil in the direction of flow.
3. The velocity of flow of air.
When the lower temperature is fixed, COP of a refrigerating machine can be improved by:- a)Operating the machine at higher speeds
- b)Operating the machine at lower speeds
- c)Raising the higher temperature
- d)Lowering the higher temperature
Correct answer is option 'D'. Can you explain this answer?
When the lower temperature is fixed, COP of a refrigerating machine can be improved by:
a)
Operating the machine at higher speeds
b)
Operating the machine at lower speeds
c)
Raising the higher temperature
d)
Lowering the higher temperature
|
|
Hrishikesh Chakraborty answered |
The correct answer is option 'D' - Lowering the higher temperature. Let's break down why this is the correct choice:
- Understanding COP:
Coefficient of Performance (COP) is a key parameter in refrigeration systems that represents the efficiency of the system in transferring heat. It is defined as the ratio of cooling effect (Qc) to the work input (W) required to achieve that cooling effect, COP = Qc/W.
- Relationship between temperature and COP:
COP is influenced by the temperatures at which the refrigeration cycle operates. Lowering the higher temperature in the cycle can lead to an improvement in COP.
- Explanation:
When the higher temperature in the refrigeration cycle is lowered, the temperature difference between the hot and cold reservoirs decreases. This results in a more efficient heat transfer process, as the refrigeration machine has to do less work to achieve the same cooling effect.
- Impact on efficiency:
By lowering the higher temperature, the COP of the refrigeration machine increases, indicating that the system is operating more efficiently. This means that for the same amount of work input, the machine can provide a greater cooling effect.
- Conclusion:
In conclusion, to improve the COP of a refrigerating machine when the lower temperature is fixed, lowering the higher temperature in the cycle is the most effective method. This adjustment can lead to a more efficient operation of the refrigeration system, resulting in energy savings and better performance.
- Understanding COP:
Coefficient of Performance (COP) is a key parameter in refrigeration systems that represents the efficiency of the system in transferring heat. It is defined as the ratio of cooling effect (Qc) to the work input (W) required to achieve that cooling effect, COP = Qc/W.
- Relationship between temperature and COP:
COP is influenced by the temperatures at which the refrigeration cycle operates. Lowering the higher temperature in the cycle can lead to an improvement in COP.
- Explanation:
When the higher temperature in the refrigeration cycle is lowered, the temperature difference between the hot and cold reservoirs decreases. This results in a more efficient heat transfer process, as the refrigeration machine has to do less work to achieve the same cooling effect.
- Impact on efficiency:
By lowering the higher temperature, the COP of the refrigeration machine increases, indicating that the system is operating more efficiently. This means that for the same amount of work input, the machine can provide a greater cooling effect.
- Conclusion:
In conclusion, to improve the COP of a refrigerating machine when the lower temperature is fixed, lowering the higher temperature in the cycle is the most effective method. This adjustment can lead to a more efficient operation of the refrigeration system, resulting in energy savings and better performance.
The correct nomenclature for ethylene is- a)R 150
- b)R 740
- c)R 1150
- d)R 12
Correct answer is option 'A'. Can you explain this answer?
The correct nomenclature for ethylene is
a)
R 150
b)
R 740
c)
R 1150
d)
R 12
|
|
Tanishq Rane answered |
Correct nomenclature for ethylene is R 150
Explanation:
The nomenclature system used for naming refrigerants is called the "Refrigerant Numbering System" or the "R-number system". This system was developed by the American Society of Heating, Refrigerating and Air-Conditioning Engineers (ASHRAE) and is widely used in the industry.
Refrigerant R 150:
R 150 is the correct nomenclature for ethylene. Ethylene is a hydrocarbon compound with the chemical formula C2H4. It is a colorless gas with a sweet odor and is commonly used as a refrigerant in low-temperature applications.
Significance of R-number:
The R-number assigned to a refrigerant provides important information about its chemical composition, thermodynamic properties, and environmental impact. The R-number system helps in identifying and categorizing different refrigerants based on their characteristics.
Other options:
- R 740: This nomenclature is not correct for ethylene. R 740 refers to carbon dioxide (CO2), which is another commonly used refrigerant.
- R 1150: This nomenclature is not correct for ethylene. R 1150 refers to ethane (C2H6), which is also a hydrocarbon compound but has different properties compared to ethylene.
- R 12: This nomenclature is not correct for ethylene. R 12 refers to dichlorodifluoromethane (CCl2F2), which is an ozone-depleting refrigerant commonly known as Freon-12.
Conclusion:
The correct nomenclature for ethylene is R 150. It is important to use the correct nomenclature when working with refrigerants to ensure proper identification and handling.
Explanation:
The nomenclature system used for naming refrigerants is called the "Refrigerant Numbering System" or the "R-number system". This system was developed by the American Society of Heating, Refrigerating and Air-Conditioning Engineers (ASHRAE) and is widely used in the industry.
Refrigerant R 150:
R 150 is the correct nomenclature for ethylene. Ethylene is a hydrocarbon compound with the chemical formula C2H4. It is a colorless gas with a sweet odor and is commonly used as a refrigerant in low-temperature applications.
Significance of R-number:
The R-number assigned to a refrigerant provides important information about its chemical composition, thermodynamic properties, and environmental impact. The R-number system helps in identifying and categorizing different refrigerants based on their characteristics.
Other options:
- R 740: This nomenclature is not correct for ethylene. R 740 refers to carbon dioxide (CO2), which is another commonly used refrigerant.
- R 1150: This nomenclature is not correct for ethylene. R 1150 refers to ethane (C2H6), which is also a hydrocarbon compound but has different properties compared to ethylene.
- R 12: This nomenclature is not correct for ethylene. R 12 refers to dichlorodifluoromethane (CCl2F2), which is an ozone-depleting refrigerant commonly known as Freon-12.
Conclusion:
The correct nomenclature for ethylene is R 150. It is important to use the correct nomenclature when working with refrigerants to ensure proper identification and handling.
Operating temperature of a cold storage is –2°C From the surrounding at ambient temperature of 40°C heat leaked into the cold storage is 30 kW. If the actual COP of the plant is 1/10th of the maximum possible COP, then what will be the power required to pump out the heat to maintain the cold storage temperature at –2°C?- a)1.90 kW
- b)3.70 k W
- c)20 .28 kW
- d)46.50 kW
Correct answer is option 'D'. Can you explain this answer?
Operating temperature of a cold storage is –2°C From the surrounding at ambient temperature of 40°C heat leaked into the cold storage is 30 kW. If the actual COP of the plant is 1/10th of the maximum possible COP, then what will be the power required to pump out the heat to maintain the cold storage temperature at –2°C?
a)
1.90 kW
b)
3.70 k W
c)
20 .28 kW
d)
46.50 kW
|
|
Samarth Chaudhary answered |
Which of the following evaporators use surge tank as a accumulator?- a)Flooded evaporators
- b)Shell and coil evaporator
- c)Double tube evaporators
- d)Frosting evaporator
Correct answer is option 'A'. Can you explain this answer?
Which of the following evaporators use surge tank as a accumulator?
a)
Flooded evaporators
b)
Shell and coil evaporator
c)
Double tube evaporators
d)
Frosting evaporator
|
|
Nitin Joshi answered |
Flooded evaporators use surge tanks as accumulators.
Explanation:
Evaporators are heat exchangers that are used in refrigeration and air conditioning systems to remove heat from a substance, typically a liquid, and convert it into a vapor. This process is achieved by transferring heat from the substance to a refrigerant, which evaporates and carries the heat away.
Accumulators are devices used in refrigeration systems to store excess refrigerant and prevent it from entering the compressor. They help to separate the liquid and vapor phases of the refrigerant, ensuring that only vapor enters the compressor, which is more efficient and prevents damage to the compressor.
Flooded evaporators are a type of evaporator where the entire heat exchanger is completely filled with liquid refrigerant. The refrigerant enters the evaporator as a liquid and is completely evaporated before leaving the evaporator as a vapor. This type of evaporator is often used in large-scale industrial refrigeration systems.
Surge tanks, also known as surge drums or flash drums, are vessels that are connected to the evaporator and act as accumulators. They are used to store excess refrigerant that is not immediately evaporated and prevent it from entering the compressor. The surge tank helps to maintain a steady flow of refrigerant to the evaporator and prevent any liquid refrigerant from entering the compressor, which could cause damage.
In the case of flooded evaporators, the surge tank serves as an accumulator by collecting any excess liquid refrigerant and separating it from the vapor before it is sent to the compressor. The surge tank allows the liquid refrigerant to settle and be returned to the evaporator as needed, while only allowing vapor to enter the compressor. This helps to ensure the efficiency and longevity of the refrigeration system.
In summary, flooded evaporators use surge tanks as accumulators to store excess liquid refrigerant and prevent it from entering the compressor. This helps to maintain the efficiency and reliability of the refrigeration system.
Explanation:
Evaporators are heat exchangers that are used in refrigeration and air conditioning systems to remove heat from a substance, typically a liquid, and convert it into a vapor. This process is achieved by transferring heat from the substance to a refrigerant, which evaporates and carries the heat away.
Accumulators are devices used in refrigeration systems to store excess refrigerant and prevent it from entering the compressor. They help to separate the liquid and vapor phases of the refrigerant, ensuring that only vapor enters the compressor, which is more efficient and prevents damage to the compressor.
Flooded evaporators are a type of evaporator where the entire heat exchanger is completely filled with liquid refrigerant. The refrigerant enters the evaporator as a liquid and is completely evaporated before leaving the evaporator as a vapor. This type of evaporator is often used in large-scale industrial refrigeration systems.
Surge tanks, also known as surge drums or flash drums, are vessels that are connected to the evaporator and act as accumulators. They are used to store excess refrigerant that is not immediately evaporated and prevent it from entering the compressor. The surge tank helps to maintain a steady flow of refrigerant to the evaporator and prevent any liquid refrigerant from entering the compressor, which could cause damage.
In the case of flooded evaporators, the surge tank serves as an accumulator by collecting any excess liquid refrigerant and separating it from the vapor before it is sent to the compressor. The surge tank allows the liquid refrigerant to settle and be returned to the evaporator as needed, while only allowing vapor to enter the compressor. This helps to ensure the efficiency and longevity of the refrigeration system.
In summary, flooded evaporators use surge tanks as accumulators to store excess liquid refrigerant and prevent it from entering the compressor. This helps to maintain the efficiency and reliability of the refrigeration system.
A standard vapour compression refrigeration cycle consists of the following 4 thermodynamic processes in sequence:- a)Isothermal expansion, isentropic compression, isothermal compression and isentropic expansion
- b)Constant pressure heat addition, isentropic compression, constant pressure heat rejection and isentropic expansion
- c)Isothermal expansion, constant pressure heat addition, isothermal compression and constant pressure heat rejection
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
A standard vapour compression refrigeration cycle consists of the following 4 thermodynamic processes in sequence:
a)
Isothermal expansion, isentropic compression, isothermal compression and isentropic expansion
b)
Constant pressure heat addition, isentropic compression, constant pressure heat rejection and isentropic expansion
c)
Isothermal expansion, constant pressure heat addition, isothermal compression and constant pressure heat rejection
d)
none of these
|
|
Lavanya Menon answered |
-Constant pressure heat addition, isentropic compression, constant pressure heat rejection and isentropic expansion is the correct sequence.
For a multistage compressor, the polytropic efficiency is:- a)The efficiency of all stages combined together
- b)The efficiency of one stage
- c)Constant throughout for all the stages
- d)A direct consequence of the pressure ratio
Correct answer is option 'A'. Can you explain this answer?
For a multistage compressor, the polytropic efficiency is:
a)
The efficiency of all stages combined together
b)
The efficiency of one stage
c)
Constant throughout for all the stages
d)
A direct consequence of the pressure ratio
|
Sanskriti Chakraborty answered |
For multistage compressor, the polytropic efficiency is the efficiency of all stages combined together.
The domestic refrigerator operating on vapour compression cycle uses the following device for expansion- a)thermostatic valve
- b)thermostat
- c)capillary tube
- d)throttling valve
Correct answer is option 'C'. Can you explain this answer?
The domestic refrigerator operating on vapour compression cycle uses the following device for expansion
a)
thermostatic valve
b)
thermostat
c)
capillary tube
d)
throttling valve
|
|
Pallabi Chavan answered |
Capillary tube is used as an expansion device in small capacity hermetic sealed refrigeration units such as in domestic refrigerators, water coolers, room air-conditioners and freezers.
The air with enthalpy of 100kJ/kg is compressed by an air compressor to a pressure and temperature at which its enthalpy becomes 200kJ/kg. The loss of heat is 40 kJ/kg from the compressor as· the air passes through it. Neglecting kinetic and potential energies, the power required for an air mass flow of 0.5kg/s is:- a)30kW
- b)50kW
- c)70 kW
- d)90 kW
Correct answer is option 'C'. Can you explain this answer?
The air with enthalpy of 100kJ/kg is compressed by an air compressor to a pressure and temperature at which its enthalpy becomes 200kJ/kg. The loss of heat is 40 kJ/kg from the compressor as· the air passes through it. Neglecting kinetic and potential energies, the power required for an air mass flow of 0.5kg/s is:
a)
30kW
b)
50kW
c)
70 kW
d)
90 kW
|
|
Sohail Shaikh answered |
All the data is give in specific which is in J/Kg but the answers are asked in Rate so you have to multiply by mass flow rate which is given in question and you will get -70 kW because work is done on the system.
A Carnot heat pump works between temperature limits of 277º C and 27º C. Its COP is- a)1.108
- b)1.2
- c)2.2
- d)9.26
Correct answer is option 'C'. Can you explain this answer?
A Carnot heat pump works between temperature limits of 277º C and 27º C. Its COP is
a)
1.108
b)
1.2
c)
2.2
d)
9.26
|
|
Suyash Patel answered |
[We may put T1 and T2 in ºC or in K but T1 – T2 will be same]
Assertion (A ):Power input per TR of a refrigeration system increases with decrease in evaporator temperature. Reason (R): COP of refrigeration system decreases with decrease in evaporator temperature. - a)Both A and R are individually true and R is the correct explanation of A
- b)Both A and R are individually true but R is not the correct explanation of A
- c)A is true but R is false
- d)A is false but R is true
Correct answer is option 'A'. Can you explain this answer?
Assertion (A ):Power input per TR of a refrigeration system increases with decrease in evaporator temperature.
Reason (R): COP of refrigeration system decreases with decrease in evaporator temperature.
a)
Both A and R are individually true and R is the correct explanation of A
b)
Both A and R are individually true but R is not the correct explanation of A
c)
A is true but R is false
d)
A is false but R is true
|
|
Mrinalini Sharma answered |
The correct answer is option 'A': Both assertion (A) and reason (R) are individually true and R is the correct explanation of A.
Explanation:
- The assertion (A) states that the power input per TR (ton of refrigeration) of a refrigeration system increases with a decrease in evaporator temperature. This means that as the evaporator temperature decreases, more power is required to achieve the desired cooling effect.
- The reason (R) provided is that the coefficient of performance (COP) of a refrigeration system decreases with a decrease in evaporator temperature. The COP is a measure of the efficiency of a refrigeration system and is defined as the ratio of cooling effect (refrigeration capacity) to the power input. As the evaporator temperature decreases, the cooling effect decreases while the power input remains constant or increases. Therefore, the COP decreases.
Explanation of the correct answer:
- The reason (R) provides a valid explanation for the assertion (A). The decrease in COP with a decrease in evaporator temperature directly leads to an increase in power input per TR of the refrigeration system. This is because a lower COP means that more power is required to achieve the same cooling effect.
- Option 'A' is the correct answer because both the assertion and the reason are individually true, and the reason explains why the assertion is true. The decrease in COP with a decrease in evaporator temperature is a well-known phenomenon in refrigeration systems and is a result of thermodynamic principles.
- Option 'B' is incorrect because the reason does explain the assertion correctly.
- Option 'C' is incorrect because the assertion is true, as explained above.
- Option 'D' is incorrect because the reason is true, as explained above.
Explanation:
- The assertion (A) states that the power input per TR (ton of refrigeration) of a refrigeration system increases with a decrease in evaporator temperature. This means that as the evaporator temperature decreases, more power is required to achieve the desired cooling effect.
- The reason (R) provided is that the coefficient of performance (COP) of a refrigeration system decreases with a decrease in evaporator temperature. The COP is a measure of the efficiency of a refrigeration system and is defined as the ratio of cooling effect (refrigeration capacity) to the power input. As the evaporator temperature decreases, the cooling effect decreases while the power input remains constant or increases. Therefore, the COP decreases.
Explanation of the correct answer:
- The reason (R) provides a valid explanation for the assertion (A). The decrease in COP with a decrease in evaporator temperature directly leads to an increase in power input per TR of the refrigeration system. This is because a lower COP means that more power is required to achieve the same cooling effect.
- Option 'A' is the correct answer because both the assertion and the reason are individually true, and the reason explains why the assertion is true. The decrease in COP with a decrease in evaporator temperature is a well-known phenomenon in refrigeration systems and is a result of thermodynamic principles.
- Option 'B' is incorrect because the reason does explain the assertion correctly.
- Option 'C' is incorrect because the assertion is true, as explained above.
- Option 'D' is incorrect because the reason is true, as explained above.
For cooling and dehumidification of unsaturated moist air, it must be passed over a coil at a temperature- a)of adiabatic saturation of incoming steam
- b)which is lower than the dew point of incoming stream
- c)which lies between dry bulb and wet bulb temperature
- d)which lies between wet bulb and dew point temperature of incoming stream
Correct answer is option 'B'. Can you explain this answer?
For cooling and dehumidification of unsaturated moist air, it must be passed over a coil at a temperature
a)
of adiabatic saturation of incoming steam
b)
which is lower than the dew point of incoming stream
c)
which lies between dry bulb and wet bulb temperature
d)
which lies between wet bulb and dew point temperature of incoming stream
|
|
Sinjini Nambiar answered |
Cooling and dehumidification : The dehumidification of air is only possible when the effective surface temperature of the cooling coil is less than the dew point temperature of the air entering the coil.
If an engine of 40% thermal efficiency drives a refrigerator having a coefficient of performance of 5, then the heat input to engine for each kJ of heat removed from the cold body of the refrigerator is- a)0.50 kJ
- b)0.75 kJ
- c)1 kJ
- d)1.25 k
Correct answer is option 'A'. Can you explain this answer?
If an engine of 40% thermal efficiency drives a refrigerator having a coefficient of performance of 5, then the heat input to engine for each kJ of heat removed from the cold body of the refrigerator is
a)
0.50 kJ
b)
0.75 kJ
c)
1 kJ
d)
1.25 k
|
|
Navya Saha answered |
Assertion (A): If a domestic refrigerator works inside an adiabatic room with its door open, the room temperature gradually decreases. Reason (R): Vapour compression refrigeration cycles have high COP compared to air refrigeration cycles.- a)Both A and R are individually true and R is the correct explanation of A.
- b)Both A and R are individually true but R is not the correct explanation of A.
- c)A is true but R is false.
- d)A is false but R is true.
Correct answer is option 'D'. Can you explain this answer?
Assertion (A): If a domestic refrigerator works inside an adiabatic room with its door open, the room temperature gradually decreases.
Reason (R): Vapour compression refrigeration cycles have high COP compared to air refrigeration cycles.
a)
Both A and R are individually true and R is the correct explanation of A.
b)
Both A and R are individually true but R is not the correct explanation of A.
c)
A is true but R is false.
d)
A is false but R is true.
|
|
Anjali Shah answered |
Answer:
Adiabatic Room and Domestic Refrigerator:
- An adiabatic room is a room that is thermally insulated, meaning it does not exchange heat with the outside environment.
- A domestic refrigerator is a refrigeration system used for storing food and beverages.
- If a domestic refrigerator works inside an adiabatic room with its door open, the room temperature gradually decreases.
Vapour Compression Refrigeration Cycles and COP:
- Vapour compression refrigeration cycles are the most common refrigeration cycles used in domestic refrigerators.
- COP (Coefficient of Performance) is a measure of a refrigeration system's efficiency.
- Compared to air refrigeration cycles, vapour compression refrigeration cycles have high COP.
Assertion and Reason:
- Assertion (A): If a domestic refrigerator works inside an adiabatic room with its door open, the room temperature gradually decreases.
- Reason (R): Vapour compression refrigeration cycles have high COP compared to air refrigeration cycles.
Explanation:
- The assertion is true because when a domestic refrigerator works inside an adiabatic room with its door open, the refrigeration system transfers heat from the room to the refrigerant fluid, thereby decreasing the room temperature.
- The reason is also true because vapour compression refrigeration cycles have high COP due to their ability to compress the refrigerant fluid to increase its temperature and then expand it to decrease its temperature, making the system more efficient than air refrigeration cycles.
- However, the reason is not the correct explanation of the assertion because the reason does not explain why the room temperature decreases when a domestic refrigerator works inside an adiabatic room with its door open. Instead, it explains why vapour compression refrigeration cycles have high COP compared to air refrigeration cycles.
Conclusion:
- Option 'D' is the correct answer because the assertion is false due to the incorrect explanation provided by the reason.
Adiabatic Room and Domestic Refrigerator:
- An adiabatic room is a room that is thermally insulated, meaning it does not exchange heat with the outside environment.
- A domestic refrigerator is a refrigeration system used for storing food and beverages.
- If a domestic refrigerator works inside an adiabatic room with its door open, the room temperature gradually decreases.
Vapour Compression Refrigeration Cycles and COP:
- Vapour compression refrigeration cycles are the most common refrigeration cycles used in domestic refrigerators.
- COP (Coefficient of Performance) is a measure of a refrigeration system's efficiency.
- Compared to air refrigeration cycles, vapour compression refrigeration cycles have high COP.
Assertion and Reason:
- Assertion (A): If a domestic refrigerator works inside an adiabatic room with its door open, the room temperature gradually decreases.
- Reason (R): Vapour compression refrigeration cycles have high COP compared to air refrigeration cycles.
Explanation:
- The assertion is true because when a domestic refrigerator works inside an adiabatic room with its door open, the refrigeration system transfers heat from the room to the refrigerant fluid, thereby decreasing the room temperature.
- The reason is also true because vapour compression refrigeration cycles have high COP due to their ability to compress the refrigerant fluid to increase its temperature and then expand it to decrease its temperature, making the system more efficient than air refrigeration cycles.
- However, the reason is not the correct explanation of the assertion because the reason does not explain why the room temperature decreases when a domestic refrigerator works inside an adiabatic room with its door open. Instead, it explains why vapour compression refrigeration cycles have high COP compared to air refrigeration cycles.
Conclusion:
- Option 'D' is the correct answer because the assertion is false due to the incorrect explanation provided by the reason.
This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Vapor Compression Refrigeration Cycle-1”.- a)expansion engine
- b)throttling valve or capillary tube
- c)both of the mentioned
- d)None
Correct answer is option 'B'. Can you explain this answer?
This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Vapor Compression Refrigeration Cycle-1”.
a)
expansion engine
b)
throttling valve or capillary tube
c)
both of the mentioned
d)
None
|
|
Sinjini Bose answered |
Vapor Compression Refrigeration Cycle-1
Throttling Valve or Capillary Tube
- In a vapor compression refrigeration cycle, the throttling valve or capillary tube is a crucial component that is responsible for controlling the flow of refrigerant within the system.
- The main function of the throttling valve or capillary tube is to reduce the pressure of the high-pressure liquid refrigerant coming from the condenser to a low-pressure liquid-vapor mixture before it enters the evaporator.
- This pressure reduction causes the refrigerant to undergo a phase change from liquid to vapor, which is essential for the cooling process in the evaporator.
- The throttling valve or capillary tube also helps in regulating the flow rate of the refrigerant, ensuring that the system operates efficiently and maintains the desired temperature inside the refrigerated space.
- By controlling the pressure and flow of the refrigerant, the throttling valve or capillary tube plays a critical role in the overall performance of the vapor compression refrigeration cycle.
Throttling Valve or Capillary Tube
- In a vapor compression refrigeration cycle, the throttling valve or capillary tube is a crucial component that is responsible for controlling the flow of refrigerant within the system.
- The main function of the throttling valve or capillary tube is to reduce the pressure of the high-pressure liquid refrigerant coming from the condenser to a low-pressure liquid-vapor mixture before it enters the evaporator.
- This pressure reduction causes the refrigerant to undergo a phase change from liquid to vapor, which is essential for the cooling process in the evaporator.
- The throttling valve or capillary tube also helps in regulating the flow rate of the refrigerant, ensuring that the system operates efficiently and maintains the desired temperature inside the refrigerated space.
- By controlling the pressure and flow of the refrigerant, the throttling valve or capillary tube plays a critical role in the overall performance of the vapor compression refrigeration cycle.
Consider the following statements:Moisture should be removed from refrigerants to avoid 1. Compressor seal failure2. Freezing at the expansion valve 3. Restriction to refrigerant flow4. Corrosion of steel parts Of these statements: - a)1, 2, 3 and 4 are correct
- b)1 and 2 are correct
- c)2, 3 and 4 are correct
- d)1, 3 and 4 are correct.
Correct answer is option 'A'. Can you explain this answer?
Consider the following statements:
Moisture should be removed from refrigerants to avoid
1. Compressor seal failure
2. Freezing at the expansion valve
3. Restriction to refrigerant flow
4. Corrosion of steel parts
Of these statements:
a)
1, 2, 3 and 4 are correct
b)
1 and 2 are correct
c)
2, 3 and 4 are correct
d)
1, 3 and 4 are correct.
|
Debolina Menon answered |
Moisture in refrigerants can have several negative effects on the performance and lifespan of refrigeration systems. Let's examine each statement individually to understand why moisture should be removed from refrigerants.
1. Compressor seal failure:
Moisture in refrigerants can cause corrosion and damage to the compressor seals. Over time, this can lead to seal failure, resulting in refrigerant leakage and decreased system efficiency. Removing moisture from the refrigerant helps to prevent this issue.
2. Freezing at the expansion valve:
When moisture is present in the refrigerant, it can freeze at the expansion valve. This can block the flow of refrigerant and disrupt the normal operation of the system. By removing moisture, the risk of freezing at the expansion valve is minimized.
3. Restriction to refrigerant flow:
Moisture can mix with oil and form sludge or acidic compounds, which can restrict the flow of refrigerant through the system. This can lead to reduced cooling capacity and increased energy consumption. Removing moisture helps to maintain proper refrigerant flow and system performance.
4. Corrosion of steel parts:
Moisture in refrigerants can react with steel parts within the system, leading to corrosion and degradation of the components. This corrosion can weaken the structural integrity of the system, increasing the risk of leaks and failures. By removing moisture, the potential for corrosion is reduced, ensuring the longevity of the steel parts.
In summary, removing moisture from refrigerants is crucial to avoid compressor seal failure, freezing at the expansion valve, restriction to refrigerant flow, and corrosion of steel parts. These negative effects can significantly impact the performance and lifespan of refrigeration systems. Therefore, it is important to implement proper moisture removal techniques, such as using desiccant dryers or vacuum pumps, to ensure the optimal operation of refrigeration systems.
1. Compressor seal failure:
Moisture in refrigerants can cause corrosion and damage to the compressor seals. Over time, this can lead to seal failure, resulting in refrigerant leakage and decreased system efficiency. Removing moisture from the refrigerant helps to prevent this issue.
2. Freezing at the expansion valve:
When moisture is present in the refrigerant, it can freeze at the expansion valve. This can block the flow of refrigerant and disrupt the normal operation of the system. By removing moisture, the risk of freezing at the expansion valve is minimized.
3. Restriction to refrigerant flow:
Moisture can mix with oil and form sludge or acidic compounds, which can restrict the flow of refrigerant through the system. This can lead to reduced cooling capacity and increased energy consumption. Removing moisture helps to maintain proper refrigerant flow and system performance.
4. Corrosion of steel parts:
Moisture in refrigerants can react with steel parts within the system, leading to corrosion and degradation of the components. This corrosion can weaken the structural integrity of the system, increasing the risk of leaks and failures. By removing moisture, the potential for corrosion is reduced, ensuring the longevity of the steel parts.
In summary, removing moisture from refrigerants is crucial to avoid compressor seal failure, freezing at the expansion valve, restriction to refrigerant flow, and corrosion of steel parts. These negative effects can significantly impact the performance and lifespan of refrigeration systems. Therefore, it is important to implement proper moisture removal techniques, such as using desiccant dryers or vacuum pumps, to ensure the optimal operation of refrigeration systems.
The comfort chart is drawn between- a)DBT and RH
- b)DRT and WDT
- c)WBT and DBT
- d)WBT, RH and Air velocity
Correct answer is option 'B'. Can you explain this answer?
The comfort chart is drawn between
a)
DBT and RH
b)
DRT and WDT
c)
WBT and DBT
d)
WBT, RH and Air velocity
|
Rajat Basu answered |
Comfort chart is drawn between DBT-in x-axis and WBT-in y-axis.
Assertion (A ): T he isothermal efficiency of a reciprocating compressor becomes 100% if perfect cooling of the fluid during compression is attained.Reason (R): Wor k do ne in a reciprocating compressor is less if the process of compression is isothermal rather than polytropic. - a)Both A and R are individually true and R is the correct explanation of A
- b)Both A and R are individually true but R is not the correct explanation of A
- c)A is true but R is false
- d)A is false but R is true
Correct answer is option 'A'. Can you explain this answer?
Assertion (A ): T he isothermal efficiency of a reciprocating compressor becomes 100% if perfect cooling of the fluid during compression is attained.
Reason (R): Wor k do ne in a reciprocating compressor is less if the process of compression is isothermal rather than polytropic.
a)
Both A and R are individually true and R is the correct explanation of A
b)
Both A and R are individually true but R is not the correct explanation of A
c)
A is true but R is false
d)
A is false but R is true
|
|
Aman Ghosh answered |
Assertion (A): The isothermal efficiency of a reciprocating compressor becomes 100% if perfect cooling of the fluid during compression is attained.
Reason (R): Work done in a reciprocating compressor is less if the process of compression is isothermal rather than polytropic.
To understand the given assertion and reason, let's discuss the concepts of isothermal efficiency and work done in a reciprocating compressor.
Isothermal Efficiency:
Isothermal efficiency is defined as the ratio of the work done on the fluid during compression to the ideal work done on the fluid during isothermal compression. It is represented by the symbol ηiso.
Isothermal efficiency can be calculated using the formula:
ηiso = (Ideal work done)/(Work done on the fluid during actual compression)
Work Done in a Reciprocating Compressor:
The work done in a reciprocating compressor is the energy required to compress the fluid from the initial state to the final state. It can be calculated using the formula:
Work = PdV
Where:
P is the pressure of the fluid
dV is the change in volume of the fluid
Now, let's analyze the given assertion and reason:
Assertion (A): The isothermal efficiency of a reciprocating compressor becomes 100% if perfect cooling of the fluid during compression is attained.
This assertion is true. When the fluid is perfectly cooled during compression, the process becomes isothermal. In an isothermal process, the temperature remains constant, resulting in a decrease in the work done on the fluid. As a result, the ideal work done during isothermal compression becomes equal to the work done on the fluid during actual compression. Therefore, the isothermal efficiency becomes 100%.
Reason (R): Work done in a reciprocating compressor is less if the process of compression is isothermal rather than polytropic.
This reason is also true. In a polytropic process, the work done on the fluid is more compared to an isothermal process. This is because in a polytropic process, the temperature and pressure change, resulting in a higher work requirement. In contrast, an isothermal process maintains a constant temperature, resulting in lower work requirements. Therefore, the work done in a reciprocating compressor is less if the process of compression is isothermal rather than polytropic.
Conclusion:
Both the assertion and reason are individually true, and the reason correctly explains the assertion. Therefore, option 'A' is the correct answer.
Reason (R): Work done in a reciprocating compressor is less if the process of compression is isothermal rather than polytropic.
To understand the given assertion and reason, let's discuss the concepts of isothermal efficiency and work done in a reciprocating compressor.
Isothermal Efficiency:
Isothermal efficiency is defined as the ratio of the work done on the fluid during compression to the ideal work done on the fluid during isothermal compression. It is represented by the symbol ηiso.
Isothermal efficiency can be calculated using the formula:
ηiso = (Ideal work done)/(Work done on the fluid during actual compression)
Work Done in a Reciprocating Compressor:
The work done in a reciprocating compressor is the energy required to compress the fluid from the initial state to the final state. It can be calculated using the formula:
Work = PdV
Where:
P is the pressure of the fluid
dV is the change in volume of the fluid
Now, let's analyze the given assertion and reason:
Assertion (A): The isothermal efficiency of a reciprocating compressor becomes 100% if perfect cooling of the fluid during compression is attained.
This assertion is true. When the fluid is perfectly cooled during compression, the process becomes isothermal. In an isothermal process, the temperature remains constant, resulting in a decrease in the work done on the fluid. As a result, the ideal work done during isothermal compression becomes equal to the work done on the fluid during actual compression. Therefore, the isothermal efficiency becomes 100%.
Reason (R): Work done in a reciprocating compressor is less if the process of compression is isothermal rather than polytropic.
This reason is also true. In a polytropic process, the work done on the fluid is more compared to an isothermal process. This is because in a polytropic process, the temperature and pressure change, resulting in a higher work requirement. In contrast, an isothermal process maintains a constant temperature, resulting in lower work requirements. Therefore, the work done in a reciprocating compressor is less if the process of compression is isothermal rather than polytropic.
Conclusion:
Both the assertion and reason are individually true, and the reason correctly explains the assertion. Therefore, option 'A' is the correct answer.
A heat pump working on a reversed Carnot cycle has a C.O.P. of 5. lf it works as a refrigerator taking 1 kW of work input, the refrigerating effect will be:- a)1 kW
- b)2 kW
- c)3 kW
- d)4 kW
Correct answer is option 'D'. Can you explain this answer?
A heat pump working on a reversed Carnot cycle has a C.O.P. of 5. lf it works as a refrigerator taking 1 kW of work input, the refrigerating effect will be:
a)
1 kW
b)
2 kW
c)
3 kW
d)
4 kW
|
|
Nandini Basak answered |
or heat rejected = 5 × work done
And heat rejected = refrigeration effect + work input
or, 5 × work input – work input = refrigeration effect
or, 4 × work input = refrigeration effect
or refrigeration effect = 4 × 1 kW = 4 kW
Consider the following statements:The volumetric efficiency of a compressor depends upon1. Clearance volume2. Pressure ratio 3. Index of expansion Of these correct statements are: - a)1 and 2
- b)1 and 3
- c)2 and 3
- d)1, 2, and 3
Correct answer is option 'D'. Can you explain this answer?
Consider the following statements:The volumetric efficiency of a compressor depends upon
1. Clearance volume
2. Pressure ratio
3. Index of expansion
Of these correct statements are:
a)
1 and 2
b)
1 and 3
c)
2 and 3
d)
1, 2, and 3
|
|
Sai Reddy answered |
The volumetric efficiency of a compressor depends upon
1. Clearance volume
2. Pressure ratio
3. Index of expansion
The usual assumption in elementary compressor cascade theory is that - a)Axial velocity through the cascade changes.
- b)For elementary compressor cascade theory, the pressure rise across the cascade is given by equation of state
- c)Axial velocity through the cascade does not change.
- d)With no change in axial velocity between inlet and outlet, the velocity diagram is formed.
Correct answer is option 'C'. Can you explain this answer?
The usual assumption in elementary compressor cascade theory is that
a)
Axial velocity through the cascade changes.
b)
For elementary compressor cascade theory, the pressure rise across the cascade is given by equation of state
c)
Axial velocity through the cascade does not change.
d)
With no change in axial velocity between inlet and outlet, the velocity diagram is formed.
|
|
Meera Bose answered |
The usual assumption in elementary compressor cascade theory is that axial velocity thr6ugh the cascade does not change.
Chapter doubts & questions for Refrigeration & Air Conditioning - GATE Mechanical (ME) Mock Test Series 2026 2025 is part of Mechanical Engineering exam preparation. The chapters have been prepared according to the Mechanical Engineering exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Mechanical Engineering 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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