What is a heat engine?
A heat engine is a special form of a heat pump.
OK, we need to take a step back here.
In the previous article, we noted that a heat pump makes use of a thermodynamic reversable process: a separable forward process (e.g. evaporating water) and a reverse process (e.g. condensing water). What we did not talk about was the role of temperature in all of this.
It turns out stuff involving phase changes, while fantastic for energy density, gets messy on the math, so it’s best to go with the text book stuff which usually starts with isothermal gas compression.
If we took a 1L cylinder of normal air at standard room temperature and pressure (25°C, 1atm) and compressed it into 0.5L, that would create an extra 70.23J of thermal energy. This is thermal energy that wasn’t there before, coming from the mechanical energy of the piston.
Let assume that one side of the cylinder is thermally shorted to a large thermal block (Reservoir A) that keeps the gas inside at 25°C, neatly extracting all that nice 70.23J into the reservoir without appreciably changing the temperature, due to the size of the reservoir¹.
Now, we detach and move the cylinder in good thermal contact with a different thermal block (Reservoir B) also at 25°C, and forceably pull the piston so the volume goes back to 1L. This reduces thermal energy by exactly 70.23J, which gets extracted from the reservoir.
Sounds like a whole lot of nothing, but in fact we just pumped 70.23J from Reservoir B to Reservoir A. Pretty cool². But hang on, aren’t we meant to be talking about heat engines, not heat pumps?
Well, it turns out that the amount of thermal energy in this push/pull action is proportional to the absolute temperature (in Kelvin). If we increased the temperature of the whole experiment to 50°C the amount pumped from B to A would increase to 76.12J, due to the increase in absolute temperature of 298K to 323K.
Still, it’s just pumping heat, though.
How about we get a little bit devious: let’s start with the cylinder at 0.5L, 50°C and expand back to 1L, which absorbs 76.12J from A, and then do the compression stroke at 25°C, which releases 70.23J into B.
Now we have a mismatch of mechanical energy in and out: 5.89J.
It turns out, we can make use of that energy in the form of mechanical work, to drive a motor, move a train (a very small train), create electricity or whatever. We now have a heat engine!
The process is really just arbitrage, taking advantage of the different conversion rates at different temperatures. It’s a bit like exchanging Aussie dollars to Japanese yen at ¥100/$A, and later exchanging back at ¥95/$1, you get a 5% boost.
Of course, in the example, the amount is tiny, especially relative to the amount of heat we put in: just 5.89J work out after putting 76.12J in: an efficiency of 7.7%. And that doesn’t include real world losses from friction, heat leakage, finite thermal resistances (between the reservoir and the gas, reducing the effective temperatures). So, with a starting point of 7.7%, we might expect 3-4% in reality.
It starts to get obvious that if we want a useful heat engine, we need a fairly big difference between Tc and Th. But how much is enough?
Since the raw energy (Q) in the forward and reverse processes is proportional to temperature (T), it means that there is a constant (we will call k) such that Q = kT, or k = Q/T.
If we consider the temperature of the forward process (expansion) as Tc, and the reverse process (compression) as Th, then:
Qh = k·Th (heat in)
Qc = k·Tc (heat out)
It’s important to note again that T is absolute temperature (Kelvin), so small changes like going from 26°C to 29°C (299K to 302K) don’t really mean much: there’s less than 1% difference between Qc and Qh and nothing much to gain from a little arbitrage.
If we define work W as being the difference between the Qh and Qc, we get:
W = Qh - Qc = k·Th - k·Tc = k(Th - Tc)
Now if we want to focus on efficiency (n), we could use the ratio of work to the heat in:
n = W / Qh = k(Th-Tc)/kTh = (Th - Tc)/Th
And thus, we have the famous Carnot theory for heat engines, which concludes that the efficiency (excluding friction, leakage) is just a function of the temperature in Kelvin. This also explains why heat engines need a large temperature difference, 100K or more, to get anything decent out. It also explains why gas fired power stations (running at around 1750°C) are so much more efficient than coal (850°C).
In text books, heat engines are often introduced first, with heat pumps second, and heat pumps are usually seen as just the reverse of a heat engine. But in reality, heat engines are still just heat pumps, moving heat from one place to another, it’s just that there is a difference in the heat in and heat out, that we can make use of as long as the temperature of the “hot side” '(the reverse process) is greater than the cold side (forward process).
¹ While it can seem unreasonable to make assumptions like this, real heat pumps (properly designed) actually do this fairly well, with negligible thermal resistance between the reversible process and the reservoirs, maintaining a fairly constant temperature due to the massive size of the reservoir, for example, your room that is being airconditioned as you read this.
² It’s actually not as silly as it seems. It is possible to have an object like a CPU that is generating 300W of heat, but you want to keep it at the same temperature as room ambient (say 25°C). Without a heat pump, the temperature would quickly increase, but a heat pump can move that 100W into the (relatively large) room ambient, and keep your gaming PC ticking over at 360 FPS just nicely.