What is a heat engine?
A heat engine is a just 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: two distinct processes, a forward process and a reverse process together with some logistics that overall gives the illusion of moving heat from one place to another.
What we did not talk about was the role of temperature in all of this.
Let’s take a 1L piston/cylinder with normal air at standard room temperature and pressure (25°C, 1atm) and compress it into 0.5L: it gets hot. This is thermal energy that wasn’t there before, originating from the mechanical energy of the piston. If we can trust AI, it says that the extra thermal energy is 70.23J.
Let’s assume that one side of the cylinder is thermally bonded to a large thermal block (Reservoir A) that keeps the gas inside at 25°C, sucking all that extra heat (the 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 forcibly pull the piston so the volume goes back to 1L. This reverse action reduces thermal energy by exactly 70.23J, which gets sucked from the second reservoir.
Sounds like a whole lot of nothing, but in fact we just pumped 70.23J from Reservoir B to Reservoir A. Pretty cool, huh². 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 from 25°C (298K) to 50°C (323K) the amount pumped from B to A would increase to 76.12J.
Even so, we are just pumping the same amount of heat from A to B.
We need to get a little bit devious, and use different temperatures at A and B. 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 thermal energy in and out: 5.89J disappeared. Something doesn’t add up.
Well, to balance things out, we must have exactly the same mismatch in the mechanical energy in the compression and expansion phases, except this time we have an excess of 5.89J. Whew. Energy is conserved after all.
Not much, but we can use that energy 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. 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.
Let that sink in: a heat engine is just a heat pump where we arbitrage the conversion rate between mechanical energy and thermal energy, due to temperature differences at each conversion.
If we complete the cycle at a fixed rate, say 10 cycles per second (600rpm) we have a heat engine that requires 761W input (to keep reservoir A at 50°C) and gets about 59W mechanical energy out.
Hmm. The 59W is OK, we can do something with that, but 761W in? That doesn’t sound very efficient. And that doesn’t include real world losses from mechanical friction, heat leakage, finite thermal resistances (between the reservoir and the gas, reducing the effective temperatures). In reality, we might be lucky to get 30W out, around 4%.
Since the arbitrage depends on differences in temperature, it quickly becomes obvious that we need much larger temperature difference to get a better efficiency.
Mathematically, if we note that the heat energy (Q) in and out of the system out is proportional to temperature (T), it means that there must be a constant (k) such that:
Q = kT
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)
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 arrived at the famous Carnot theory for heat engines: the efficiency (excluding friction, leakage) is just a function of the temperature in Kelvin. And nobody had to mention entropy! OK, entropy is in the mix because that is what fundamentally causes Q = kT, but that is another story.
But this explains why heat engines need a large temperature difference. Small temperature differences of 10° or 20°C are small relative to the temperature in Kelvin (around 300K), which in turn mean the arbitrage is tiny: Even if Qc and Qh are large, the difference is small. We need large differences of 100°C or more that create huge differences in Qc and Qh, making the arbitrage attractive. It also explains why gas fired power stations (running at around 1750°C) are so much more efficient than coal (850°C).
So a heat engine is really just a special case of a heat pump, where Th > Tc, and in practice Th >> Tc.
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.