Mechanical Engineering - Power Engineering - Refrigeration

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Question - 1

TR machine means that

  • A one ton of the refrigerant is used
  • B one ton of water at DOC is converted into ice at DOC
  • C one ton of water at DOC is converted into ice at DOC in 24 h
  • D All of the above

Question - 2

COP of a refrigerator working between two fixed temperature limits T1 and T2 (T1 > T2) will have

  • A COP increased when the working substance has high specific heat
  • B COP increases with the increase in latent heat of vaporization during heat absorption process at T2.
  • C Cannot be predicted
  • D Fixed COP

Question - 3

A Carnot refrigerator requires 1.5 kW of work for 6 kW of refrigerating effect. If it absorbs heat at 240 K then the temperature of heat reservoir at which the heat is rejected equal to

  • A 320 K
  • B 300 K
  • C 400 K
  • D 280 K

Question - 4

A Carnot refrigerator has COP = 5. The ratio of higher temperature T1 to lower temperature T2 is

  • A 6/5
  • B 5/4
  • C 5/6
  • D 3/2

Question - 5

Bell Coleman air refrigeration cycle consists of

  • A two adiabatics, one constant volume and one constant pressure process
  • B two adiabatics and two constant volume processes
  • C two adiabatics and two constant pressure processes
  • D None of the above

Question - 6

Heat is absorbed from cold chamber to produce refrigerating effect from

  • A evaporator
  • B compressor
  • C expansion valve
  • D condenser

Question - 7

The power required to run the vapour compression cycle is 5 kW and heat rejected to condenser is 15 kW. It's COP is

  • A 2
  • B 3
  • C 1/2
  • D 1/3

Question - 8

A simple saturated cycle is one in which the refrigerant vapour is

  • A wet at entry to compressor
  • B dry-saturated at exit of compressor
  • C dry-saturated at entry to compressor
  • D All of the above

Question - 9

20 TR means

  • A 20 kW
  • B 70 kW
  • C 35 kW
  • D None of these

Question - 10

In a simple vapour compression cycle the enthalpy of refrigerant are at entry to compressor = 350 kJ/kg at exit to compressor = 450 kJ/kg, at exit of condenser = 150 kJ/kg. The COP is

  • A 2
  • B \(\frac{7}{3}\)
  • C 3
  • D \(\frac{9}{7}\)
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