Question

state carnots theorem and find expression for coefficient of performance of a carnot engine.

Answer 1

Carnot theorem states that no heat engine working in a cycle between two constant temperature reservoirs can be more efficient than a reversible engine working between the same reservoirs.

In other words it means that all the engines operating between a given constant temperature source and a given constant temperature sink, none, has a higher efficiency than a reversible engine.

Proof:



Suppose there are two engines EAEA and EBEBoperating between the given source at temperature T1 and the given sink at temperature T2.

Let EAEA be any irreversible heat engine and EBEB be any reversible heat engine. We have to prove that efficiency of heat engine EBEBis more than that of heat engine EAEA.

Suppose both the heat engines receive same quantity of heat Q from the source at temperature T1.

Let WAWA and WBWB be the work output from the engines and their corresponding heat rejections be (Q–WA)(Q–WA) and (Q–WB)(Q–WB)respectively.

Assume that the efficiency of the irreversible engine be more than the reversible engine i.e. ηA>ηBηA>ηB

Hence, WAQ>WBQWAQ>WBQ

i.e WA>WBWA>WB



Now let us couple both the engines and EBEBis reversed which will act as a heat pump. It receives (Q–WB)(Q–WB) from sink and WAWA from irreversible engine EAEA and pumps heat Q to the source at temperature T1.

The net result is that heat WA–WBWA–WB is taken from sink and equal amount of work is produce. This violates second law of thermodynamics.

Hence the assumption we made that irreversible engine having higher efficiency than the reversible engine is wrong.

Hence it is concluded that reversible engine working between same temperature limits is more efficient than irreversible engine thereby proving Carnot’s theorem.

A general expression for the efficiency of a heat engine can be written as:



We know that all the energy that is put into the engine has to come out either as work or waste heat. So work is equal to Heat at High temperature minus Heat rejected at Low temperature. Therefore, this expression becomes:

Efficiency=QHot-QColdQHot

Answer 2

According to Carnot's theorem, if a heat engine is working between two heat reservoirs, its efficiency cannot be greater than Carnot's engine working between the same heat reservoirs.

The coefficient of performance is used in reference for the efficiency of Carnot's refrigerator. It is calculated as follows:

If $T_H$ is the temperature of hot reservoir in which $Q_H$ amount of heat is given and $T_C$ is the temperature of the cool reservoir from which $Q_C$ amount of heat is with drawn, then

$COP = \frac{ Q_H}{W} = \frac{Q_H}{Q_H-Q_C}=\frac{T_H}{T_H-T_C}$

Learn more about: Carnot's engine

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