 Back on the PV diagram, in the segment from state 2 to state 3, we dump heat from the gas to the cold reservoir. This cools the gas and reduces the pressure, which allows us to compress the gas using less work than we obtain from the expansion segment. That's the key to the engine's operation. We have to transfer that same amount of heat from the hot reservoir to the gas in the heating segment from state 4 to state 1. This heat is wasted in the sense that we don't get any work out of it. This motivates us to look for a more efficient way to implement the heating and cooling of the gas. The wasted heat is the gas heat capacity, 3 halves Nk, times the temperature change delta T. This gets dumped into the cold reservoir. We can never get this back because heat only flows from hot to cold, and the cold reservoir is the coldest object the gas comes into contact with. The reheating of the gas requires the same amount of heat to be transferred from the hot reservoir. And this will cost us, typically in burn fuel. We can cut this waste in half by cooling the gas in two steps, each with temperature change delta T equal to one half the total temperature change T hot minus T cold. For the first step, we connect the gas to a reservoir at temperature T medium, midway between T hot and T cold. The temperature change is half what it was before, so the transferred heat, QE, is also half what it was before. Then we connect the gas to the cold reservoir to reduce its temperature by delta T again. This transfers another QE of heat. So now only half the total heat extracted from the gas is wasted. The other half is stored in the T medium reservoir. To warm the gas from T cold, we reconnect it to the medium temperature reservoir and extract the heat we stored during the cooling process. Then we connect the hot reservoir and transfer the second half of the heat. We have to make up for this by burning fuel. But we've recycled half of the heat required for heating the gas and so cut our fuel costs for this process in half. An obvious extension is to add more intermediate heat reservoirs. Suppose we transfer heat in four steps, each with temperature change delta T equal to one fourth T hot minus T cold. In the first step, we attach the gas to a medium hot reservoir and transfer one fourth of the total heat to it. Then we transfer the same amount of heat to the medium reservoir, then the same amount to a medium cold reservoir. And finally, the same amount to the cold reservoir. We've now wasted only one fourth of the total heat transferred. For heating, we can extract three fourths of the required heat from what we stored in the intermediate reservoirs and then generate only the last one fourth. A device that implements intermediate heat storage is called a regenerator and it's essential to making a sterling cycle engine as efficient as possible. Generalizing what we've done, we could heat and cool the gas in M steps, each with one over M of the total temperature change and hence one over M of the total heat transfer. Only one of these heat transfers would cost us, the rest would be recycled. A practical regenerator implementation might use a tube through which gas is transported between hot and cold chambers. The intermediate heat reservoirs are placed one after another along the tube, gradually heating or cooling the gas depending on the direction of flow. In principle, using an arbitrarily large number of intermediate reservoirs, we can reduce the amount of generated heat needed for the heating and cooling process to an arbitrarily small amount. If so, we end up getting the heating and cooling parts of the cycle for free, so to speak. There's another way to get free heating and cooling of the gas. We can make use of adiabatic expansion and compression. In an adiabatic process, the gas is thermally insulated. There is no transfer of heat. Suppose the piston moves from position X1 to X2. From the ideal gas law, the pressure equals NKT over V. We multiply by the piston cross-sectional area to get the force, which replaces V with X. And the internal energy of a monatomic gas is three-halves NKT. The change in internal energy, DU, equals three-halves NK times the change in temperature, DT. By the first law, this is the negative of the work done by the gas, minus F DX. Substituting for F, this becomes minus NKT DX over X. Consider the second and fourth expressions. Cancelling the common N and K factors, dividing by T, and multiplying by two-thirds, we get DT over T equals minus two-thirds DX over X. The corresponding relation between temperature and piston position is T2 equals T1 times X1 over X2 to the two-thirds power. When the gas temperature changes from T1 to T2, the internal energy change is three-halves NK times the temperature change. And this is converted into work done on the environment. For compression, we simply reverse the direction of the force and move the piston from X2 to X1. The gas returns to its original temperature, T1, and the work done on the environment during expansion is now done on the gas. An expansion compression cycle returns everything to its original state. We say that adiabatic expansion and compression are reversible. Replacing the heating and cooling segments of the sterling cycle with adiabatic processes, we obtain the Carnot cycle. We start with a gas attached to a hot reservoir in the piston and position X1. The gas undergoes isothermal expansion to piston position X2. This is identical to the sterling cycle and produces the same work on the environment. An equal amount of energy flows as heat from the hot reservoir to the gas. Now, instead of attaching the cold reservoir, we insulate the gas and expand the gas adiabatically to position X3, causing the temperature to drop from T-hot to T-cold. This produces additional work, three-halves NK T-hot minus T-cold. Now, we attach the cold reservoir. Since the gas is already cooled to this temperature, there is no heat transfer. The gas is isothermally compressed from piston position X3 to X4. This requires work NK T-cold log X3 over X4 to be done on the system. An equal amount of heat is transferred to the cold reservoir. The gas is insulated again and adiabatically compressed back to the initial piston position X1, with the gas temperature increasing back to T-hot. This increases the internal energy by three-halves NK T-hot minus T-cold, which equals the work done on the system. On a PV diagram, the first segment of this cycle is identical to the thrilling cycle. Isothermal expansion at T-hot takes the system from state one to state two. Then, adiabatic expansion takes the system from state two to state three and reduces the temperature from T-hot to T-cold. Isothermal compression at T-cold takes the system from state three to state four. Finally, adiabatic compression heats the gas from T-cold to T-hot and takes the system from state four back to state one. Isothermal expansion converts heat Qs to work We equal to NK T-hot log X2 over X1, just as in the first segment of the thrilling cycle. Adiabatic expansion converts internal energy to work We equals to three-halves NK T-hot minus T-cold, as the gas cools to T-cold. Isothermal compression converts work Ws to heat Qe equal to NK T-cold log X3 over X4 and adiabatic compression converts work Ws equal to three-halves NK T-hot minus T-cold into internal energy as the gas warms back up to T-hot. This is equal to the work we got out of the system during adiabatic expansion, so we get this reheating for free. From the relation between volume and temperature for the adiabatic segments, we find that X3 over X4 equals X2 over X1. So the heat flow and work for the isothermal compression segment are identical to this thrilling cycle. As for the thrilling cycle, the area enclosed by these four curves equals the net work produced by the cycle. If the temperatures and the piston positions X1 and X2 are the same, then the thrilling and Carnot cycles generate the same net work. Let's summarize our findings for the Carnot cycle. The net work generated is NK T-hot minus T-cold times log X2 over X1. To get this, our hot reservoir has to provide heat energy NK T-hot log X2 over X1 during isothermal expansion. The cycle efficiency is the ratio of the output work to the input heat energies. These expressions have factors of NK log X2 over X1 in common, so the ratio depends only on the temperature factors, which we can write as one minus T-cold over T-hot. The efficiency is less than one, or 100%, by the ratio of the cold-hot temperatures. For this thrilling cycle, the net work generated is the same as for the Carnot cycle. But the heat supplied by the hot reservoir is greater because in addition to the isothermal expansion term, we also have a term for the heating of the gas. It follows that the efficiency is less than that of the Carnot cycle. Now, as we saw, in principle, the heating term can be made arbitrarily small by using a regenerator. So we found two heat engine cycles having different efficiencies, and we've seen how the efficiency of one can be increased. This raises the question, is there a cycle with greater efficiency than the Carnot cycle? Or can we find a trick, like the regenerator, to improve the Carnot cycle? The answer to this question leads us to the second law of thermodynamics, which is the topic of the next video in this series.