We have shown that the Carnot engine is the most efficient...

We have shown that the Carnot engine is the most efficient engine when operating between two heat reservoirs with temperatures $T_H$ and $T_L\left(T_H>\right.$ $T_L$ ), whereby its efficiency is given by $\eta_{\text {Carnot }}=1-\frac{T_L}{T_H}$. Now, suppose that you are given a series of reservoirs with temperatures ranging between $T_L$ and $T_H$ and are asked to construct a heat engine that interacts with any subset of these reservoirs. Show that the efficiency of an engine with internal irreversibilities (e.g. friction) but a well-defined temperature at every juncture (i.e. its process is quasistatic such that it is always in an equilibrium state) is smaller than that of an internally reversible engine following the same cycle of equilibrium states (hint: use Clausius' inequality). Next, prove that the efficiency of an internally reversible engine constructed with the given reservoirs is no larger than the Carnot efficiency $\eta_{\text {Carnot }}=1-\frac{T_L}{T_H}$. This shows that a Carnot engine operating between the highest and lowest temperature reservoirs is still the most efficient when a series of reservoirs with intermediate temperatures is available.

Key Concepts

Heat Reservoirs

Heat reservoirs are large thermal bodies whose temperatures remain effectively constant regardless of the heat exchanged with the engine. They provide the necessary thermal boundary conditions for the engine's operation. In the analysis of engine cycles, including scenarios with multiple intermediate temperatures, these reservoirs set the ultimate limits on efficiency according to the Carnot formula.

Irreversible Processes

Irreversible processes involve factors such as friction, turbulence, and other dissipative effects that lead to entropy production. Even if an engine operates in a quasistatic manner so that it is always near equilibrium, internal irreversibilities still reduce its efficiency compared to the idealized reversible case. The presence of such irreversibilities means that some energy is unavoidably lost to the environment in each cycle.

Clausius Inequality

Clausius' inequality is a mathematical formulation of the second law of thermodynamics, stating that for any cyclic process the sum of the ratio of heat exchanged to the temperature at which it is exchanged is less than or equal to zero, with equality only for reversible processes. This inequality is key to proving that any engine with irreversible processes must have a lower efficiency than its reversible counterpart.

Reversible Processes

Reversible processes are idealized thermodynamic processes that occur infinitely slowly, ensuring that the system remains in equilibrium at all stages. They do not produce any entropy, meaning there is no net loss of useful energy. In the context of heat engines, a reversible engine, such as a Carnot engine, achieves maximum possible efficiency because it avoids internal dissipation.

Carnot Efficiency

Carnot efficiency is the theoretical upper limit on the efficiency of any heat engine operating between two thermal reservoirs. It is defined as 1 - (T_low/T_high) where T_low and T_high are the absolute temperatures of the cold and hot reservoirs, respectively. This concept is central to understanding that no engine can be more efficient than a Carnot engine operating between the same two temperatures.

Transcript

00:01 And this problem, we want to look at a kind of a, well, a coronary cycle, but it's ideal, but not in the mutual sense that the high temperature is what's in the low temperature, what's actually in contact with the, or where we're getting the heat from.

00:23 So if we think about basically we have a high temperature reservoir, but then we have something in here to get heat transfer, frost it.

00:32 And so then we have a low temperature.

00:35 And then down here we have a, i mean, another high temperature.

00:38 And here we have a low temperature.

00:40 And so inside of here is the heat engine.

00:45 So what's happening is, is that, you know, we have a reservoir here and a reservoir here, but we have to get that, we have to get heat flux across there and then into the end of the heat engine to then get our workout.

00:59 So we have these, if you look at basically relationships from heat transfer, we have a heat transfer coefficient here.

01:10 And the rate of heat flow is this heat, heat, is the coefficient, is the rate of heat transfer coefficient, in times the change in temperature.

01:24 So the heat flow across here, there'll be a change in temperature, or we won't have any heat flow across here.

01:32 And likewise here, we won't have any heat flow unless there's a change in temperature.

01:36 So we can write these relationships.

01:43 And again, i guess i didn't write the q .l equals h -a -l -t -l minus t -l -sar minus t -l.

01:55 Since we have, again, basically it's just a sign convention here.

02:01 Tl star will be higher than tl.

02:04 So we'll have heat transfer in this direction.

02:09 So again, basically, this is just conductive heat transfer, and you probably haven't taken a class in heat transfer yet.

02:15 But when you do, you see that this is just the relationships.

02:20 I can't remember foia's law.

02:22 I can't remember exactly.

02:23 Basically says some law that the heat flux is proportion to the change in temperature between two surfaces if you have a body.

02:34 And then this depends on the material and all kinds of other things.

02:38 So we know that the power output here is the efficiency of the heat engine times the heat flux, the high from the hot reservoir.

02:56 The efficiency, again, we can write as 1 minus tl star over t high star, because that's basically this, you know, that's the definition of the efficiency of a kernel cycle.

03:12 Okay.

03:12 So now these are the operating, the high and the low operating temperatures.

03:22 And so we can then take this and plug that into here or no, plug that into here.

03:27 And pull a th out.

03:31 So then we can divide through by this.

03:35 And this is actually has a unit of power.

03:40 So this thing here has is dimensionless.

03:44 And so what we get is we get 1 minus tl star all over th star times 1 minus thh star over th.

03:54 And you notice that this is the dimensionless quantity here.

03:58 So when we're doing these calculations kind of by hand, it's good to get into dimensionless quantities.

04:06 And so then we can define r.

04:10 Sorry, when do i define r to be? oh, why did i? i wrote that wrong.

04:18 Let's see here.

04:19 This should be r should be just this, not one minus that.

04:25 Yeah, there we go.

04:27 So this is 1 minus r, and then we're defined this to be x.

04:32 And notice, r and x are both dimensionless quantities.

04:38 Now, let's see, we also know we can just get this relationship here, just for a future, for tl star over t h, we can multiply and divide by th star, and then use this relationship here, and we find that this is r times 1 minus x.

04:55 Now, for a reversible cycle, so we're assuming that we're assuming that this is r times 1 minus x.

05:00 That it's reversible in here, not necessarily this isn't all reversible, but internally it's reversible.

05:07 So for an internally reversible cycle, we have this relationship here between the temperature ratio and the heat flux, and the ratio of the rates of heat transfer.

05:19 Now, what we can see here is this term is 1 over r, right? and then this term, we can use these relationships up here.

05:32 And after we do a bunch of algebra and manipulating, we'll get down to this relationship here where we have now things in terms of heat transfer coefficients and x and r and then this ratio of the external, the reservoir high and low temperatures, which we're again, assuming all these are constant.

05:56 Now we can solve this thing for x, which is x is this value.

06:04 Here.

06:06 Solving for x, we get this relationship, just a bunch of algebra.

06:12 And then we can go back here to this relationship here and plug x into here...

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