Science, asked by sanjaydastse06, 2 months ago

A heat pump working on the Carnot cycle takes in heat from a reservoir at 5 °C and delivers heat to a
reservoir at 60 °C. The heat pump is driven by a reversible heat engine which takes in heat from a reservoir at
840 °C and rejects heat to a reservoir at 60 °C. The reversible heat engine also drives a machine that absorbs
30KW. If the heat pump extracts 17 KJ/s from the 5 °C reservoir, determine (a) the rate of heat supply from the
840 °C source, and (b) the rate of heat rejection to the 60 °C sink

Answers

Answered by dipanshuduhan7
0

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Answered by studay07
1

Answer:

a. the rate of heat supply from the

840 °C source =Q1=??

b. the rate of heat rejection to the 60 °C sink =Q2 +Q3 =??

A.

Efficiency of the heat engine

η=1 - T2/T1

=1-333/1113

=0.7

Also, η=(Q1 - Q2) /Q1

So ( Q1 - Q2) =ηQ1

The C.O.P of heat pump

( COP) =T2 /(T2 - T3)

=333/(333-278)

=6.05

Also the C.O.P of the heat pump is

COP = (Heat supplied) / (work supplied )

=Q4/Wp

6.05 = 17 /Wp

Wp=2.81 kW

Total work

W=We - Wp

30kW =We - 2.81

We =32.81kW

But We = Q1 - Q2

=ηQ1

32.81= 0.7 Q1

Q1 =46.87 kW

B. We = Q1 - Q2

Q2 =46.87 -32. 81

=14.06 kW

Q3 =Q4 + Wp

=17+2.81

=19.81

Q2 +Q3 =33.87 kW

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