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Refrigerator or Heat Pump is considered one the most difficult concept.
2 Questions around this concept.
For a refrigerator which of the following statements is true
1) A refrigerator is basically a heat engine running in the reverse direction
2) Coefficient of performance = Heat extracted / Work done
3) The relation b/w coefficient of performance and efficiency of the refrigerator is
A refrigerator or heat pump is basically a heat engine run in the reverse direction.
It consists of three parts
1. Source: At higher temperature T1
2. Working substance: It is called refrigerant. I.e liquid ammonia and freon works as a working substance.
3. Sink: At lower temperature T2.
As shown in the above figure, The working substance takes heat Q2 from a sink (contents of refrigerator) at lower temperature T2, has a net amount of work done W on it by an external agent (usually compressor of refrigerator) and gives out a larger amount of heat Q1 to a hot body at temperature T1 (usually atmosphere).
The cold body is cooled more and more with the help of a refrigerator. Because the refrigerator transfers heat from a cold to a hot body at the expense of mechanical energy supplied to it by an external agent.
- Coefficient of performance $\left(\beta^\beta\right.$ -
The coefficient of performance is defined as the ratio of the heat extracted from the cold body to the work needed to transfer it to the hot body.
$
\beta=\frac{\text { Heat extracted }}{\text { work done }}=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}
$
A perfect refrigerator is one which transfers heat from cold to a hot body without doing work.
i.e. $W=0$ so that $Q_1=Q_2$ and hence $\beta=\infty$
- Carnot refrigerator-
For Carnot refrigerator $\frac{Q_1}{Q_2}=\frac{T_1}{T_2}$
$
\therefore \frac{Q_1-Q_2}{Q_2}=\frac{T_1-T_2}{T_2} \text { or } \frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2}
$
So using
$
\beta=\frac{Q_2}{Q_1-Q_2}
$
we get
$
\beta=\frac{T_2}{T_1-T_2}
$
where $\mathrm{T}_1=$ temperature of surrounding, $\mathrm{T}_2=$ temperature of cold body and $T_1>T_2$
when $\mathrm{T}_2=0$ then $\beta=0$
I.e if the cold body is at the temperature equal to absolute zero, then the coefficient of performance will be zero
- The relation between $\beta$ and $\eta$ of the refrigerator
$
\beta=\frac{1-\eta}{\eta}
$
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