Question

A Reversible heat engine operates with efficencency 30 % find heat supplied and power output if heat rejected form engine is 70kW estimate COP of heat pump if the engine is Reserved to work as heat pump?​

Answer 1

7/3

0.3=1-Q2/Q1

0.3=1-70/Q1

Q1=100

W 2Q, -Q2

=30kw

WP=Q/w=70/30=7/3

Answer 2

A Reversible heat engine operates with efficiency 30 %. If heat rejected from heat engine is 70kW.

We have to find heat supplied and power output and also estimate COP of heat pump if the engine is reserved to work as heat pump?

heat rejected = 70 % of total heat supplied

⇒70 kW = 70 % of total heat supplied

⇒total heat supplied = 70/70 × 100 = 100kW

power output = efficiency × total hear supplied

= 30 % × 100 kW

= 30 × 100/100 kW

= 30 kW

hence power output is 30 kW.

now COP of engine if the engine is reversed to work as heat pump is given by,

COP = 1/η - 1

= 1/0.3 - 1

= 10/3 - 1

= (10 - 3)/3

= 7/3

Therefore the COP of heat pump if the engine is reversed to work as heat pump is 7/3.