# Why is work done on the system considered when calculating thermodynamic efficiency?

## Main Question or Discussion Point

The thermodynamic efficiency $\eta$ is calculated by $\eta= \frac{W_{out}}{Q_{in}}$

Using the first law of thermodynamics we usually say that $W_{out}$ is $Q_c+Q_h$, where $Q_c$ is the heat dissipated into a cold reservoir, and $Q_h$ is the heat absorbed by the system because of a hot reservoir. Both are measured within the system, such that $Q_c<0$ and $Q_h>0$

However I object to that. $W_{out}$ is not $Q_c+Q_h$. That calculation is simply the magnitude of net energy in the process due to work. Namely, its considering the $W_{input}$, done by the surroundings on the gas, to calculate the $W_{output}$.

In case it's not clear yet, I'm saying $Q_c+Q_h$ is the same as $-(W{out}+W{in})$ where $W_{out}<0$ and $W_{in}>0$.

Consider the Carnot engine. I would say that the work output is the area underneath the expansion isotherm, and the expansion adiabat. All other works are done on the system and are not "outputs".

You might say that the other works are "negative outputs". But although this makes sense in a mathematical sense, it doesn't make practical sense that this should be a $W_{output}$.

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Bystander
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Q c Q_c is the heat dissipated into a cold reservoir
Q h Q_h is the heat absorbed by a hot reservoir
Based on these two definitions, you might want to "rethink" your understanding of "Q," and of the sign conventions you're applying to the statement of the first law.

I messed up there. What I mean by Qh is the well known thing. Qh is the heat that the hot reservoir deposits into the system and the system measures it as positive. I do understand what Qh and Qc are.

Bystander
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I messed up there.
You've got some other sign convention problems to deal with as well. Don't try to paraphrase textbook definitions for signs of "work on/by" the system, or of heat added/removed; learn them as they are written within the context they are written. It's thermo, not an exercise in creative writing.

Theres nothing wrong with my conventions. I'm not paraphrasing anything, I understand this, and this isn't my question… This is a nicely thought out question not an exercise in reading comprehension.

Qh>0. Qc<0. W>0 if done on the system W<0 if the system has done it to the surroundings. There are variants in other textbooks, but this is the usual conventions for signs.

Im specifically asking about the definition of $\eta$ as "work done by the system to the surroundings over the heat it has absorbed$. I don't see why the work done by the system to the surroundings should have the work done by the surroundings to the system as a consideration. Namely, imagine the Carnot engine diagram. The work done as output, according to my confused point of view, should be the area underneath the top expansion isotherm and the expansion adiabat. Not the area enclosed. If i'm some type of engineer and constructed some type of system which punches you each time it does expansion work, then the work output should only be the punches the system gives me not the punches I give to the system. Bystander Science Advisor Homework Helper Gold Member The work done as output, For the complete cycle (starting point all the way around the cycle to the same starting point); should be the area underneath the top expansion isotherm and the expansion adiabat. the positive work done by (or negative work done on, pick ONE convention and stick with throughout any analysis), Not the area enclosed. PLUS the negative work done by the system (or positive work done on the system), which IS the area enclosed. If you've invested any effort in reading my question you should see that I didn't regard that justification as valid. Namely when I said: "You might say that the other works are "negative outputs". But although this makes sense in a mathematical sense, it doesn't make practical sense that this should be a Woutput." My analogy with the "punching thermodynamic system" which you might have not read as well, shows what I think W_out should be. I want someone to criticise this POV, since its obviously wrong. Last edited: Bystander Science Advisor Homework Helper Gold Member "punching thermodynamic system" Deleted? this makes sense in a mathematical sense, it doesn't make practical sense that this should be a Woutput." You're asserting that the math doesn't satisfy your intuition? No. I'm saying that if$\eta$was invented for engineering purposes, and people wanted a measure of how much something could give as work compared to how much it used or absorbed or whatever satisfies you as the meaning of the denominator for$\eta$, then it makes more sense that the work is not the area enclosed, but the thing i previously described. . Obviously I'm wrong. But why? Bystander Science Advisor Homework Helper Gold Member Obviously I'm wrong. But why? :You CANNOT have your cake and eat it too. You HAVE to put the work in to RESET the engine to its starting point, just as you have to put heat into it to get the expansion. berkeman Mentor Thread closed temporarily for Moderation... Edit: Thread reopened. Last edited by a moderator: PeterDonis Mentor 2019 Award But although this makes sense in a mathematical sense, it doesn't make practical sense that this should be a$W_{output}$. It most certainly does. Instead of talking about abstractions, let's take a concrete example: the Carnot engine is a piston with working fluid inside, and the hot and cold heat reservoirs hooked up to it via heat exchangers, and driving a crankshaft. Put the whole setup inside a box, with the only interfaces being the two heat exchangers, so we can put in$Q_h$and take out$Q_c$--we assume those quantities are both known and fixed--and the crankshaft coming out, so we can measure the work done at the crankshaft where it exits the box. The key question then is: if we let the engine inside the box go through one full cycle and return to the same starting state, what work will we measure at the crankshaft where it exits the box? My answer is that we will measure$W_{out} = Q_h + Q_c$, i.e., we will measure a smaller amount of work than$Q_h$(because$Q_c$is negative). That is the practical sense in which that definition of$W_{out}$is the correct one. It seems from your posts that you think the answer to the question should be$Q_h$, i.e., that we should somehow be able to capture that amount of work at the crankshaft where it exits the box. If that is in fact your answer, please explain how you propose to capture that amount of work under the constraints given: remember, you cannot add any more interfaces to the box (so no more heat exchangers and no more ways of putting work in or getting work out), and the engine inside the box must return to the same state after one cycle. If I'm wrong and your answer is that we will in fact capture work$W_{out} = Q_h + Q_c < Q_h$at the crankshaft where it exits the box, then what's the problem? Doesn't that give a practical meaning to the standard definition of efficiency? Last edited: Okok. I have a quick question though: Consider theres a gas inside the piston cylinder thing. Work's will be calculated via pV work. Are you implying all expansion work will be translated into the kinetic energy of the crankshaft? What will happen during compression then? The slowing down of the crankshaft is interpreted as the work done to my gas? I think that clears things up now! If after one complete cycle the crankshaft has 0 kinetic energy then we say "no work was done" and$\eta=0$. If my crankshaft has a small amount of kinetic energy left, then "work was done" and$\eta>0##

Is this right?

Thanks!

Bystander
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Looks like you're getting the idea. These are the debits and credits in the bookkeeping.

PeterDonis
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Are you implying all expansion work will be translated into the kinetic energy of the crankshaft?
Temporarily, yes. See below.

What will happen during compression then? The slowing down of the crankshaft is interpreted as the work done to my gas?
Yes, exactly. So after one complete cycle, the net work captured at the crankshaft (which, as you say, we can measure as kinetic energy of the crankshaft) is the value I gave.

I have a question though. Is this definition of efficiency intended mainly for machinery that moves? (Like trucks, boats, whatever) . Or is there something fundamental about this way of defining the efficiency of an engine?

We could have invented a new definition if we were thinking about a machine like a stove or something right?

Thanks.

PeterDonis
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Is this definition of efficiency intended mainly for machinery that moves? (Like trucks, boats, whatever) . Or is there something fundamental about this way of defining the efficiency of an engine?
It's fundamental for anything that converts heat into work. (That is the usual definition of a "heat engine".) Since work involves motion, it would seem that some sort of motion would have to be involved.

We could have invented a new definition if we were thinking about a machine like a stove or something right?
How would you use a stove to do work?