# Reverse Carnot Cycle Efficiency

### Efficiency of the Reverse Carnot Cycle

An air conditioning device is working on a reverse Carnot cycle between the inside of a room at temperature T2 and the outside at temperature T1 > T2 with a monatomic ideal gas as the working medium. The air conditioner consumes the electrical power P. Heat leaks into the house according to the law $$\stackrel{.}{Q} = A(T_1 − T_2)$$.

1. Show that the efficiency of the air conditioner is $$η_{cool} = {T_2 \over T_1-T_2}$$.
2. Express the inside temperature T2 in terms of T1, A, and P.

### Solution

#### Efficiency

In the reverse Carnot cycle, work is done to extract heat from one system and expel it into another via four processes, two isothermal and two isentropic.

In process $$1 \rightarrow 2$$, the gas is isentropically compressed, and there is no heat flow into or out of the refrigerator.

In process $$2 \rightarrow 3$$, heat is expelled into the sink (e.g. outside air) isothermally (T2=T3). The amount of heat ejected per unit mass of gas is $$Q_C=T_2(S_2-S_3)$$.

In process $$3 \rightarrow 4$$, the gas is isentropically expanded. The pressure and temperature decrease to P4, T4. Heat transfer at this stage is zero.

In process $$4 \rightarrow 1$$, the gas expands isothermally (T4=T1), extracting heat from the source (e.g. room). This is where the cooling takes place. The heat extracted from the source per unit mass of gas is $$Q_H=T_1(S_1-S_4)=T_1(S_2-S_3)$$.

The work done during the process is simply $$W=Q_H-Q_C=(T_1-T_2)(S_2-S_3)$$.

The efficiency of the reverse Carnot cycle is the heat removed from the cold reservoir / the amount of work input: $$\eta_{cool}=\frac{Q_C}{W}$$, so $\eta_{cool}={T_2 \over T_1-T_2}\;\blacksquare$

#### Express T2 in terms of T1, A, and P

Re-arranging $$\stackrel{.}{Q} = A(T_1 − T_2)$$ we obtain $T_2=T_1-{\stackrel{.}{Q} \over A}$

In equilibrium, the heat flow in $$\stackrel{.}{Q}$$ must equal the power consumed $$P$$ times the efficiency $$\eta$$2:

\begin{eqnarray}
T_2&=&T_1 – {P \eta \over A} \nonumber \\
&=&T_1 – {P T_2 \over A(T_1-T_2)} \nonumber
\end{eqnarray}

This leads to a quadratic formula whose solution (taking the positive root) is:

$T_2 = 2T_1+{P \over 2A}\bigg[\sqrt{{4 T_1 A \over P} + 1} - 1 \bigg]\;\blacksquare$

1. http://www.mechlook.com/reversed-carnot-cycle/ []
2. Update (26/01/13): Sorry about the typo of the dropped η before; I’m sure you spotted it. []