Question

A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this refrigeration cycle b)-If the cycle is absorbing 15 kW at the -12C temperature, how much power is required? c)-If a Carnot heat pump operates between the same temperatures as the above refrigeration cycle, determine the COP of heat pump? d)-What is the rate of heat rejection at the 40°C temperature if the heat pump absorbs 15 kw at the -12 C temperature?

Answer 1

a)COP=5.01

b)
`W_(in)=2.998` KW

c)COP=6.01

d)
`Q_R=17.99 KW`

Step-by-step explanation:

Given

`T_L`= -12°C,
`T_H`=40°C

For refrigeration

We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

`COP=(T_L)/(T_H-T_L)` ,T in Kelvin.

`COP=(261)/(313-261)`

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that
`COP=(RE)/(W_(in))`

RE is the refrigeration effect

So

5.01=
`(15)/(W_(in))`

b)
`W_(in)=2.998` KW

For heat pump

So COP of heat pump is given as follows

`COP=(T_h)/(T_H-T_L)` ,T in Kelvin.

`COP=(313)/(313-261)`

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at low temperature+work in put

`Q_R=Q_A+W_(in)`

Given that
`Q_A=15`KW

We know that
`COP=(Q_R)/(W_(in))`

`COP=(Q_R)/(Q_R-Q_A)`

`6.01=(Q_R)/(Q_R-15)`

d)
`Q_R=17.99 KW`