Reversed Heat Engines

Applications of reverse heat engines

There are two main reverse heat engine applications: heat pumps and refrigerators, which are designed to remove heat from a cold region and transfer it to a hotter one.

Refrigerators

Refrigerators and air conditioners are used to cool down a space by removing heat. Work is done on the system, using a motor to pump warm air inside the fridge to the environment, which is a higher temperature region. The process involves a fluid that is circulated through a closed system:

1. The fluid passes through a nozzle and expands, which turns it into a gas that cools down. This is known as an adiabatic expansion, where energy or mass is not transferred to the environment.
2. The cooler gas is transferred into the inner space of the fridge, which has a higher temperature. Heat is transferred from the fridge to the gas, increasing its temperature. This is known as an isobaric expansion, which means that the pressure is constant.
3. The gas is then transferred to a compressor, which adds work to the system. The compressed gas is heated and becomes a liquid again.
4. The hot liquid goes through coils located on the outside of the fridge, and heat is transferred to the room. This is known as isobaric compression.

These steps can be used to construct a p-v diagram, as shown below in figure 2. The amount of heat removed from the fridge per work is given by the coefficient of performance (COP_ref). It is a measure of the amount of heat transfer from the cold region compared to the work input to the system.

Using the relation between work and heat transfer in a reverse heat engine, we get the following equation for the coefficient of performance:

\[COP_{ref} = \frac{Q_C}{W} = \frac{Q_C}{Q_H-Q_C}\]

For an ideal refrigerator, we assume that the amount of heat transfer in each region is equal to the temperature of the region, which gives us the following expression for COP:

\[COP_{ideal} = \frac{T_C}{T_H-T_C}\]

Heat pumps

Heat pumps are used to warm up a room. The system is usually comprised of compressed gas, and the sequential working process of a heat pump is as follows:

1. The electrically driven compressor inputs mechanical work into the system. This raises the temperature and pressure of the gas, which is thus forced to enter some condenser coils located in the higher temperature region.
2. As the temperature of the gas is higher than that of the surroundings, thermal energy is transferred to the room, and the gas condenses into a liquid.
3. The liquid goes through an expansion valve, which reduces its temperature before it returns to the evaporator coils to continue the cycle.
4. The last two steps are usually done in reverse order to operate a cooling cycle, and the refrigerant fluid passes through the expansion valve reducing its temperature to the evaporator transferring heat from indoors to outdoors which cools down the refrigerant.

The amount of heat transferred (Q_H) into a space per unit work input (W) is the coefficient of performance of a heat pump COP_hp.

\[COP_{hp} = \frac{Q_H}{W} = \frac{Q_H}{Q_H-Q_C}\]

\[COP_{ideal} = \frac{T_H}{T_H-T_C}\]

As seen from the equation above, heat pumps seem to have a greater performance when the temperature difference is small. The coefficient of performance is the ratio of heating to required work. Hence, a higher COP means that the heat pump provides the same work with less energy. Therefore, the higher the COP, the higher the efficiency.

Efficiency

The efficiency of a reverse heat engine is the amount of heat transfer that is actually converted into work. This is determined by dividing work by the heat transfer Q_H. Then, a relation can be written for the COP_hp, and the efficiency determined, as seen below.

\[\eta = \frac{W}{Q_H} \text{ or } \eta_{\%} = \frac{W}{Q_H} \cdot 100 \qquad COP_{hp} = \frac{1}{\eta}\]

Since the efficiency of a heat engine is always less than 1 (there will always be some heat lost), COP_hp is always greater than 1 (see the equations below). Therefore, a heat pump has more heat transfer Q_h than work put into it.

There is also a relation between the refrigerator coefficient of performance and the heat pump coefficient of performance. This can be derived using the equation of work and the heat pump coefficient formula, as seen below.

We begin using the equation that describes the heat transfer in a heat pump:

\[Q_H = Q_C + W\]

Then, we use the heat pump coefficient of performance and refrigerator coefficient of performance equations and re-arrange them in terms of Q_H and Q_C, respectively:

\[COP_{hp} = \frac{Q_H}{W} \Rightarrow Q_H = COP_{hp} \cdot W\]

\[COP_{ref} = \frac{Q_C}{W} \Rightarrow Q_C = COP_{ref} \cdot W\]

We now substitute them into the heat transfer equation mentioned earlier and divide by the work on both sides of the equation, which gives us:

\[COP_{hp} \cdot W = COP_{ref} \cdot W +W \qquad \frac{COP_{hp} \cdot W}{W} = \frac{COP_{ref} \cdot W}{W} + \frac{W}{W}COP_{hp} = COP_{ref} + 1 \text { or } COP_{ref} = COP_{hp}-1\]