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

Answer:

a)COP=5.01

b)$W_{in}=2.998$ KW

c)COP=6.01

d)$Q_R=17.99 KW$

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

$COP=\dfrac{T_L}{T_H-T_L}$  ,T in Kelvin.

$COP=\dfrac{261}{313-261}$

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that  $COP=\dfrac{RE}{W_{in}}$

RE is the refrigeration effect

So

5.01=$\dfrac{15}{W_{in}}$

b)$W_{in}=2.998$ KW

For heat pump

So COP of heat pump is given as follows

$COP=\dfrac{T_h}{T_H-T_L}$  ,T in Kelvin.

$COP=\dfrac{313}{313-261}$

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  $COP=\dfrac{Q_R}{W_{in}}$

$COP=\dfrac{Q_R}{Q_R-Q_A}$

$6.01=\dfrac{Q_R}{Q_R-15}$

d)$Q_R=17.99 KW$