For an ideal refrigeration cycle COP is:
(A) $\dfrac{{{T_L}}}{{{T_H} + {T_L}}}$
(B) $\dfrac{{{T_L}}}{{{T_H} - {T_L}}}$
(C) $\dfrac{{{T_H}}}{{{T_H} - {T_L}}}$
(D) $\dfrac{{{T_L}}}{{{T_H} \times {T_L}}}$
Answer
618.3k+ views
Hint:The most commonly used refrigeration cycle is vapour compression cycle, which is used for refrigerator, air conditioner etc. By using the COP for an ideal refrigeration cycle formula and the first law of thermodynamics equation, the solution can be determined.
Useful formula:
The COP of an ideal refrigeration cycle is given by,
$\beta = \dfrac{{{Q_L}}}{W}$
Where, $\beta $ is the coefficient of performance, ${Q_L}$ is the heat removed per cycle and $W$ is the work done per cycle.
The equation of the first law of thermodynamics is given by,
$W = {Q_H} - {Q_L}$
Where, $W$ is the work done, ${Q_H}$ is the heat given to the source and ${Q_L}$ is the heat removed per cycle.
Complete step by step solution:
The COP (coefficient of performance) for an ideal refrigeration cycle is defined as the ratio of the heat removed per cycle from the substance to the work done per cycle to remove the heat.
Now,
The COP of an ideal refrigeration cycle is given by,
$\beta = \dfrac{{{Q_L}}}{W}\,...................\left( 1 \right)$
The equation of the first law of thermodynamics is given by,
$W = {Q_H} - {Q_L}\,..................\left( 2 \right)$
By substituting the equation (2) in the equation (1), then
$\beta = \dfrac{{{Q_L}}}{{{Q_H} - {Q_L}}}$
For easy simplification all the terms in the RHS is divided by the term ${Q_H}$ in the above equation, then the above equation is written as,
$\beta = \dfrac{{\left( {\dfrac{{{Q_L}}}{{{Q_H}}}} \right)}}{{\left( {\dfrac{{{Q_H}}}{{{Q_H}}}} \right) - \left( {\dfrac{{{Q_L}}}{{{Q_H}}}} \right)}}$
On further simplification, then the above equation is written as,
$\beta = \dfrac{{\left( {\dfrac{{{Q_L}}}{{{Q_H}}}} \right)}}{{1 - \left( {\dfrac{{{Q_L}}}{{{Q_H}}}} \right)}}$
From the Carnot cycle, the terms $\dfrac{{{Q_L}}}{{{Q_H}}}$ is written as $\dfrac{{{T_L}}}{{{T_H}}}$, then the above equation is written as,
$\beta = \dfrac{{\left( {\dfrac{{{T_L}}}{{{T_H}}}} \right)}}{{1 - \left( {\dfrac{{{T_L}}}{{{T_H}}}} \right)}}$
By cross multiplying the terms in the denominator in the RHS, then the above equation is written as,
$\beta = \dfrac{{\left( {\dfrac{{{T_L}}}{{{T_H}}}} \right)}}{{\dfrac{{{T_H} - {T_L}}}{{{T_H}}}}}$
By rearranging the terms, them the above equation is written as,
$\beta = \dfrac{{{T_L}}}{{{T_H}}} \times \dfrac{{{T_H}}}{{\left( {{T_H} - {T_L}} \right)}}$
By cancelling the same terms, then the above equation is written as,
$\beta = \dfrac{{{T_L}}}{{{T_H} - {T_L}}}$
Hence, the option (B) is the correct answer.
Note:The term ${Q_H}$ is divided with each of the terms because of the easy calculation, and also for converting the terms from heat $Q$ to the temperature $T$ with the help of the Carnot cycle equation. If the COP value is high, it represents the high efficiency.
Useful formula:
The COP of an ideal refrigeration cycle is given by,
$\beta = \dfrac{{{Q_L}}}{W}$
Where, $\beta $ is the coefficient of performance, ${Q_L}$ is the heat removed per cycle and $W$ is the work done per cycle.
The equation of the first law of thermodynamics is given by,
$W = {Q_H} - {Q_L}$
Where, $W$ is the work done, ${Q_H}$ is the heat given to the source and ${Q_L}$ is the heat removed per cycle.
Complete step by step solution:
The COP (coefficient of performance) for an ideal refrigeration cycle is defined as the ratio of the heat removed per cycle from the substance to the work done per cycle to remove the heat.
Now,
The COP of an ideal refrigeration cycle is given by,
$\beta = \dfrac{{{Q_L}}}{W}\,...................\left( 1 \right)$
The equation of the first law of thermodynamics is given by,
$W = {Q_H} - {Q_L}\,..................\left( 2 \right)$
By substituting the equation (2) in the equation (1), then
$\beta = \dfrac{{{Q_L}}}{{{Q_H} - {Q_L}}}$
For easy simplification all the terms in the RHS is divided by the term ${Q_H}$ in the above equation, then the above equation is written as,
$\beta = \dfrac{{\left( {\dfrac{{{Q_L}}}{{{Q_H}}}} \right)}}{{\left( {\dfrac{{{Q_H}}}{{{Q_H}}}} \right) - \left( {\dfrac{{{Q_L}}}{{{Q_H}}}} \right)}}$
On further simplification, then the above equation is written as,
$\beta = \dfrac{{\left( {\dfrac{{{Q_L}}}{{{Q_H}}}} \right)}}{{1 - \left( {\dfrac{{{Q_L}}}{{{Q_H}}}} \right)}}$
From the Carnot cycle, the terms $\dfrac{{{Q_L}}}{{{Q_H}}}$ is written as $\dfrac{{{T_L}}}{{{T_H}}}$, then the above equation is written as,
$\beta = \dfrac{{\left( {\dfrac{{{T_L}}}{{{T_H}}}} \right)}}{{1 - \left( {\dfrac{{{T_L}}}{{{T_H}}}} \right)}}$
By cross multiplying the terms in the denominator in the RHS, then the above equation is written as,
$\beta = \dfrac{{\left( {\dfrac{{{T_L}}}{{{T_H}}}} \right)}}{{\dfrac{{{T_H} - {T_L}}}{{{T_H}}}}}$
By rearranging the terms, them the above equation is written as,
$\beta = \dfrac{{{T_L}}}{{{T_H}}} \times \dfrac{{{T_H}}}{{\left( {{T_H} - {T_L}} \right)}}$
By cancelling the same terms, then the above equation is written as,
$\beta = \dfrac{{{T_L}}}{{{T_H} - {T_L}}}$
Hence, the option (B) is the correct answer.
Note:The term ${Q_H}$ is divided with each of the terms because of the easy calculation, and also for converting the terms from heat $Q$ to the temperature $T$ with the help of the Carnot cycle equation. If the COP value is high, it represents the high efficiency.
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