Question

A Carnot heat pump moves heat to a warm house at 24°C at a rate of 300kW. The heat is pumped from the outside air at 7°C. What is the power required by the heat pump?

What is the rate of change of entropy for each of the thermal reservoirs (the house, and the outside air)?

Does this heat pump satisfy the second law of thermodynamics increase of entropy principle?

Answer 1

Given information

Heat pump: Working on Carnot cycle

Desired Efeect i.e. Qh=300 kW

Source resservior tempratur, Th=24 C

=24+273

=297 K

Sink resservior tempratur, Tl =7 C

=7+273

=280 K

As we know that,

COP for heat pump is given as

COP=Desired Effect/Work input

For carnot cycle,

COP= Th/(Th-Tl)

COP=(297)/(297-280)

COP=17.47

As we know that,

COP=Desired Effect/Work input

COP=Qh/Win

Win=300/17.47

Win=17.17 kW

where Qh=Win+Ql...............from 1st law of thermodynamic

Ql=Qh-Win

Ql=300-17.17

Ql=282.83 kW

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Entropy Analysis of above system

We know that entropy of a reserviour can be given as,s=Q/T

for sink reservior(i.e. outside air) s1= -282.83/280

=-1.01 kW/k

for source reservior (i.e house),s2=300/297

=+1.01

thus  s=s1+s2

s=-1.01+1.01

=0

where s =s universe

and suniverse >=0

thus this process satisfy the increas of entropy principle.

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