Question

Using an engine of 30% thermal efficiency to derive a refrigerator having a COP of 5, what is the heat input to the engine for each MJ removed from the cold body by the refrigeartor

Answer 1

Concept
For a heat engine with the help of the efficiency of the engine, find the heat input to the engine.
Given
30% Efficiency of an engine
COP is 5.
To Find
Heat input to the engine.
Solution
We know,
The efficiency of the engine = $\frac {work done}{heat supplied} = 0.3 = \frac {W}{Q_2}$        - 1
∴ $W = Q_2 -Q_1$
where,
$Q_1$ is the heat rejected from the engine;
$Q_2$ is the heat supplied from the engine;
∴ COP OF REFRIGERATOR = $\frac {Q_1}{Q_2 - Q_1} = \frac {Q_1}{W} = 5$                    - 2
Multiplying the above two equations:
$\frac {W}{Q_2}* \frac {Q_1}{W} = \frac {Q_1}{W} = 0.3*5 = 1.5$
∴$Q_2 = \frac {Q_1}{1.5}$
we know,
for each megawatt removed $Q_1 = 1*10^6 watt$
$Q_2 = \frac {Q_1}{1.5} = \frac {1*10^6}{1.5} = 6.6 MW$
∴ The net heat input to the engine for each megawatt of heat removed from the refrigerator is 6.6 Megawatt.

Answer 2

Answer:

Heat input can calculated mathematically by using thermal efficiency of the heat engine and COP of the refrigerattor.

Explanation:

    Consider W= work Done, q₁ = heat output, q₂ = heat input

• Effieciency of heat engine =$\frac{Work Done}{Heat provided}$ =$\frac{W}{q_{2} }$  =30%=0.3                    where W=q₂-q₁
• COP (coedfficient of performnace of refrigeratoe)=$\frac{heat released}{work done}$=$\frac{q_{1} }{W}$ =5
• Multiple both the above equation                                                   $\frac{q_{1} }{q_{2}} =1.5$ $\frac{W}{q_{2} } *\frac{q_{1} }{W}$ = $0.3*5=1.5$
• ∴    $\frac{q_{1} }{q_{2} }= 1.5$

• We know that, 1MJ = 10⁶Joule

        so,  heat removed i.e. q₁ = 10⁶ joule

         Heat input , q₂ =$\frac{q_{1} }{1.5} =\frac{10^{6} }{1.5} = 0.66MJ$

    Therefore,  heat input to the engine is 0.66MJ for each MJ removed from the cold body by the refrigerator.