A heat pump needs a source of energy and most of the modern ones run on electricity. You just plug them into your house's electrical system and they'll start chugging away. Another interesting way to power a heat pump is with a heat engine. A thermodynamic diagram of this process is shown in the following figure.

There are three heat reservoirs, the outside environment at temperature \(T_0\), the inside of the house at temperature \(T_1\), and a high temperature heat source at \(T_2\). The high temperature source could come from the combustion of some fuel such as natural gas. There is a heat engine operating between \(T_2\gg T_1\) that produces work \(W\). This work is used to drive the heat pump that operates between \(T_1\gt T_0\) pumping heat into the house.

Does such a setup make sense? The answer is yes, at least from a theoretical perspective. It greatly improves on the efficiency of the heat pump alone, providing a much greater quantity of heat to the house for the same expenditure of energy. The method was first proposed by Lord Kelvin, one of the founders of thermodynamics, way back in 1853. Only a few thermodynamics books mention it, such as the one by Franzo H. Crawford published in 1963. A more modern reference is a paper by E. T. Jaynes published in 2003. Links to these references are given below.

So let's analyze this system to see how good it really is. The energy we have to pay for is \(Q_2\) which comes from the combustion of fuel. The energy that we get going into the house is \(Q_1 + Q_1^{\prime}\) where \(Q_1\) is the heat engine waste and \(Q_1^{\prime}\) is the heat coming from the pump. So the efficiency, or heat gain, is the ratio of \(Q_1 + Q_1^{\prime}\) to \(Q_2\). This is what we will calculate. The goal is to express the ratio in terms of the temperatures of the three reservoirs.

First we'll assume that both the pump and engine are reversible. This means they are as efficient as possible and produce no net entropy. Starting with the engine, we see from the diagram that the work it produces is given by (apply the first law of thermodynamics for a cyclic process)

\[W = Q_2 - Q_1\]

Since the engine produces no net entropy we must have (apply the second law of thermodynamics)

\[\frac{Q_2}{T_2} = \frac{Q_1}{T_1}\]

Using this equation, the work produced by the engine can be expressed as

\[W = Q_2\left(1 - \frac{T_1}{T_2}\right)\]

This work is used by the pump to deliver heat \(Q_1^{\prime}\) to the house so we have (again, apply the first law of thermodynamics for a cyclic process)

\[Q_1^{\prime} = Q_0 + W\]

where \(Q_0\) is the heat extracted from the outside environment. Since the pump produces no net entropy we have (apply the second law of thermodynamics)

\[\frac{Q_0}{T_0} = \frac{Q_1^{\prime}}{T_1}\]

With this equation we can write the heat delivered to the house as

\[Q_1^{\prime} = Q_1^{\prime}\frac{T_0}{T_1} + W\]

Using the above expression for \(W\) and solving for \(Q_1^{\prime}\) we get

\[Q_1^{\prime} = Q_2\frac{T_1}{T_2}\frac{T_2-T_1}{T_1-T_0}\]

The total heat delivered to the house is then

\[Q_H=Q_1^{\prime}+Q_1=Q_2\frac{T_1}{T_2}\frac{T_2-T_0}{T_1-T_0}\]

so the heat gain is

\[G(T_0,T_1,T_2)=\frac{Q_H}{Q_2}=\frac{T_1}{T_2}\frac{T_2-T_0}{T_1-T_0}\]

which can also be written as

\[G(T_0,T_1,T_2)=\frac{1-T_0/T_2}{1-T_0/T_1}\]

As an example, assume the fuel we are using is methane, which has a combustion temperature in air of 1950 ℃ = 2223 K. Assume the outside temperature is 0 ℃ = 273 K, and assume the inside temperature is 20 ℃ = 293 K. Plugging these values into the heat gain formula we get and incredible \(G(273,293,2223) = 12.85\). So for each unit of heat purchased this setup produces almost 13 units of heat in the house. It is important to remember however that we are assuming an ideal heat engine and pump that produce no entropy. With a real engine and pump we would probably only get about half this value.

We can analyze the system without the assumption of a ideal engine and pump by focusing on the overall change in entropy. Note, first of all, that the net result of a cycle is that \(Q_2\) leaves the \(T_2\) reservoir, and \(Q_0\) leaves the \(T_0\) reservoir, so the heat entering the house is \(Q_H=Q_2+Q_0\). The total entropy change of the system is

\[\Delta S = \frac{Q_H}{T_1} - \frac{Q_0}{T_0} - \frac{Q_2}{T_2}\]

In general, this must be greater than or equal to zero (it is only equal when the engine and pump are reversible). Therefore, we have

\[\frac{Q_H}{T_1} \ge \frac{Q_0}{T_0} + \frac{Q_2}{T_2} = \frac{Q_H - Q_2}{T_0} + \frac{Q_2}{T_2}\]

Rearranging this expression gives us

\[Q_H \le Q_2\frac{T_1}{T_2}\frac{T_2-T_0}{T_1-T_0}\]

When engine and pump are reversible, this becomes an equality and we get the previous result. The general expression for the heat gain is then

\[G(T_0,T_1,T_2) \le \frac{1-T_0/T_2}{1-T_0/T_1}\]

This shows that the result we calculated earlier is the maximum possible heat gain, for real engines and pumps it will be smaller.

How could you actually realize such a system? One way is to use a natural gas powered electric generator to power an electric heat pump. Electric generators are generally very efficient and you can also get very efficient electric heat pumps. You won't get anywhere near the ideal heat gain we calculated above, but it should be better than running just a heat pump alone off the electric grid. An additional benefit is that you have an electric backup generator in case the power ever goes out.

### References

Sir William Thomson, Baron Kelvin, "Power required for the thermodynamic heating of buildings", Cambridge and Dublin Mathematical Journal, November, 1853. A copy of this paper is found in Kelvin's Volume V of Mathematical and Physical Papers, p124-133.

Franzo Hazlett Crawford, "Heat, Thermodynamics and Statistical Physics", 1963, p217-219.

Edwin Thompson Jaynes, "Note on thermal heating efficiency", American Journal of Physics, Vol 71, No 2, February 2003, p180-182.

Out of all the conventional ways to heat a house using a heat pump is by far the best. The reason is that it takes less energy to move a quantity of heat than it does to generate that heat in the first place. A heat pump can take heat from the cold outside air, yes there is still heat in that air, and move or pump it into the warmer house. Heating a house with a heat pump is like having an inside out refrigerator where you're heating the inside and cooling the outside.

The basic thermodynamic operation of a heat pump is shown in the following figure.

We have two heat reservoirs, the outside of the house at temperature \(T_0\) and the inside at temperature \(T_1\gt T_0\). With an input of energy \(W\) (this is the part you have to pay for) the pump removes a quantity of heat \(Q_0\) from the outside air and deposits a quantity of heat \(Q_1=Q_0+W\) into the inside air. The efficiency of this process for a given heat pump is called its coefficient of performance \(\mathbf{COP}\). It is defined as follows

\[\mathbf{COP}=\frac{Q_1}{W}\]

In other words, it's the ratio of what you get to what you pay for. If \(\mathbf{COP}=2\) for example, then one unit of input energy will provide two units of heat energy to the inside of the house.

There is no single value for \(\mathbf{COP}\) since it is generally a function of the temperatures \(T_0\) and \(T_1\). Ideally there should be a plot of \(\mathbf{COP}\) for different values of \(T_0\) and \(T_1\) but we have not seen a single heat pump manufacturer that does this. Instead they use (for the case of heating performance) something called a heating seasonal performance factor \(\mathbf{HSPF}\). I should mention here that there are different types of heat pumps. Some use the air as a heat source while others use ground or water as a heat source. The HSPF mainly applies to air source heat pumps.

\(\mathbf{HSPF}\) is expressed as the heat output over an entire heating season measured in BTUs (British thermal units) divided by the total electricity used measured in watt-hours. This gives it the somewhat bizarre units of BTU/watt-hour. You can convert \(\mathbf{HSPF}\) to the average \(\mathbf{COP}\) over a heating season, which is unitless, by multiplying \(\mathbf{HSPF}\) by \(0.2931\), the number of watt-hours in a BTU.

High efficiency heat pumps have an \(\mathbf{HSPF}\) of at least 8 or an average \(\mathbf{COP}\) of \(8\cdot 0.2931=2.3448\). By contrast, an electrical resistance heater, which can do no more than turn all of its input energy into heat, would have \(\mathbf{COP}=1\) or \(\mathbf{HSPF}=1/0.2931=3.4118\).

It is interesting to compare a real heat pump with an ideal heat pump which produces no entropy so that

\[\frac{Q_1}{T_1}=\frac{Q_2}{T_2}\]

with the temperatures measured in Kelvin. The \(\mathbf{COP}\) can then be expressed in terms of the temperatures as

\[\mathbf{COP}=\frac{T_1}{T_1-T_0}\]

No heat pump operating between the temperatures \(T_1\gt T_0\) can have a greater \(\mathbf{COP}\) than this. As an example, if the outside temperature is 0°C and the inside temperature is 20°C then \(T_0=273.15 K\), \(T_0=293.15 K\) and the maximum \(\mathbf{COP}\) that any heat pump can have is 14.6575.

Currently, in the United States, the minimum HSPF set by the government is 8.2, but will rise to 8.8 in 2023. You will pay more for higher HSPF. For example, a Senville 12,000 BTU mini split heat pump with HSPF of 8.5 costs 799.99 USD, while a Fujitsu 12,000 BTU mini split heat pump with HSPF of 12.5 costs 1539.00 USD. For this price difference of 739.01 USD the \(\mathbf{COP}\) rises from 2.49 to 3.66.

Of course, reliability is another factor to consider when buying a heat pump. Consumer Reports has a reliability rating for numerous brands.

A blog post we found worthwhile reading on the topic of installing a heat pump yourself is "Our DIY Heat Pump Install – Free Heating and Cooling for Life?".

It's winter in Colorado. Time to crank up the furnace and blow some hot air around, but is this really the best way to heat a house? It's called forced air heating, and most houses in Colorado seem to use it.

What's wrong with it? To begin with, air is a terrible way to try to transfer heat. It's not easy to heat up air. It has a low thermal conductivity and specific heat. But the worst thing is that it's a gas that can easily pass through cracks around the doors and windows. And yes, you do have to have cracks.

The house has to be able to exchange air with the outside or you will die. If it's completely sealed the pressure will increase as the house is heated, the walls will start to bulge, and may eventually explode, but you would probably be dead way before that happened due to the increased pressure and the build up of CO_{2} and other gases.

So what happens thermodynamically when you heat a house with forced air heating? For simplicity, let's assume we have only a one room cabin. This is essentially a box with rigid walls that have some holes and cracks around the windows and doors. The holes and cracks will keep the pressure inside and outside the room equal. This simply means that inside air can escape, and outside air can come in. So our thermodynamic system is a container that holds air at constant pressure and volume, but with variable mass or number of moles.

Air, at the terrestrial temperatures and pressures that human beings find comfortable, is very nearly an ideal gas. So we can use the ideal gas law, to relate the temperature and number of moles to the pressure and volume as follows:

\(nT=\frac{pV}{R}\)

where: \(n=\) number of moles, \(T=\) temperature in degrees Kelvin, \(p=\) pressure, \(V=\) volume, \(R=\) gas constant.

The right hand side of the equation is a constant, so it is telling us that if the temperature is to go up, the amount of gas in the room must go down. What effect does this have on the internal energy of the air in the room? The internal energy of an ideal gas is proportional to the temperature and the amount of gas. We can write this as the following equation

\(u=cnT\)

where: \(u=\) internal energy of the air in the room, \(c=\) molar specific heat at constant volume.

Substituting for \(nT\) from the ideal gas law, we get

\(u=\frac{cpV}{R}\)

The right side of this equation is a constant, so the internal energy remains constant. This means that all the heat energy used to raise the air temperature will escape to the outside through the holes and cracks in the walls (assuming for simplicity that the walls are perfect insulators so no heat conduction through them is possible). We do end up with air at a higher temperature, but there is less of it so the total energy remains constant.

This suggests that a possibly more efficient way to heat the room may be with radiant heat. A radiant heater emits infrared radiation. Air is composed mostly of nitrogen and oxygen, and these gases do not absorb the infrared radiation put out by these heaters, but the objects and the people in the room do. The air will of course heat up by thermal contact with the heater and the objects and people in the room, but it will never get as hot as the air blowing out of a furnace vent.

If you're interested in reading more about this subject, here are a couple of references:

Why do we have Winter Heating?, R. Emden, Nature 141, 908–909 (1938).

Thermodynamics of heating a room, H. J. Kreuzera and S. H. Payne, American Journal of Physics 79, 74 (2011).

On account of today being Albert Einstein's birthday, here is a list of books he recommended.

The recommendations can be found in Einstein's Ideas and Opinions which is a collection of his most important non-technical writings in one volume. This book was put together in 1954, one year before Einstein's death. Here are 6 of the recommendations. Included with each recommendation in the list below is the page number where the reference is found (Crown Publisher's 1982 edition):

- An Inquiry Into Meaning and Truth by Bertrand Russell. p.20
- Patterns of Culture by Ruth Benedict. p.52
- The Outline of History by H. G. Wells. p.58
- Lehrbuch der Physik by Arnold Berliner. p.70
- Logic and Scientific Method by Morris Raphael Cohen and Ernest Nagel. p.80
- The Anatomy of Peace by Emery Reves. p.122

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