So I combine these two engines together. Essentially I just put a big box around them. They're both operating between these two heat sources. These two reservoirs. So I call this the, you know, your super engine plus my reverse Carnot engine. So what's happening now? What's the net heat that's being taken in or put out of here? So we have q we have-- let me see. We have q1 minus w 1 plus x, but in this direction, we have q so in this direction, we could rewrite this.
I want to make sure you're clear on the algebra. This could be rewritten as what? As q1 times 1 plus x times, or minus, w times 1 place x. Now, if you compare these terms, this is the same as this term. This term is bigger than this term. This term is clearly bigger, because we're multiplying it by something larger than 1. It's bigger than this storm. So if we combine these two, the upward movement, or the amount of heat I'm taking up from my reverse Carnot, is going to be greater than the amount of heat being put in by your friend's super engine.
And we can actually calculate the amount. We can just take this amount minus that amount, and that's the net upward movement. So the net upward movement from our cold reservoir is what? It's this value minus this value. So minus q1 minus w 1 plus x. If we take a minus, we're going to subtract it. So it's a minus and a plus. These cancel out. This minus cancels out with-- so this first term could be rewritten as q1 plus q1x. We could rewrite it that way.
So this cancels out with that. And so the net upward movement when we combine the two engines is q1 times x. Now what about the work transfer?
Well, whatever work this guy produces is exactly the amount of work that I need. So no outside work has to be done on the system. It just works. This guy produces work, this guy uses the work. Now what's in that heat transferred up to our hot reservoir? What's the amount of heat? Well, it's the difference between these two. And this is clearly a larger number than this one, so the upward movement dominates.
So what's this minus that? So this can be rewritten as q1 plus q1x, right? I just distributed the q1. We're going to subtract that out. Minus q1. You're left with q1x. So the net movement, when we combine the two engines, is q1x.
So what's happening here? I have no external energy or work has to be expended into this system. And it's just taking heat from a cold body, and it's moving it to a warm body. And it does this indefinitely. It'll do this as much as I want to. I can just build a bigger one. It'll do it on even a larger and larger scale. So if you think about it, I could heat my house with ice by just making the ice colder.
I could create steam from things that arbitrarily cold. This goes against the second law of thermodynamics. The net entropy in this world is going down. Because what's happening here? This is just a straight up transfer of q1x from a cold body to a hot body. Kelvin-Planck's law gives the relationship for. If 2 kg mass of water, with a specific heat of 4. A machine which is transferring heat from the lower body to the higher body without consuming any external work.
Such machine is an example of:. Suggested Test Series. Suggested Exams. More Thermodynamics Questions Q1. For specifying the state of superheated vapour, one needs.
When wet steam undergoes adiabatic expansion then. The difference between the actual temperature of superheated steam and saturation temperature corresponding to its pressure is known as. Change in enthalpy of a system is equal to heat transfer under the following conditions.
Increase in pressure. Expression for the specific entropy of wet steam is-. An adiabatic process in a thermodynamic system is one in which there is? All the answers here are wrong to an extent - or at least very misleading.
The Carnot cycle is not the highest efficiency cycle of all possible cycles, it is only one of an infinity of cycles all of which exhibit the highest possible efficiency. There is nothing particularly special about the Carnot cycle, except that it is a simple cycle, that it is relatively easy to conceptualize and thus it makes a good teaching example, and that it is the cycle Carnot chose to use to explain the concept so it has historical precedence.
Any cycle made up entirely of reversible processes will be a reversible cycle and all of these will have the same highest possible efficiency - the Carnot efficiency. As an example of one - look at the Stirling Cycle with an ideal regenerator. Also, the Carnot cycle has relatively little enclosed area on the P-V diagram so it does relatively little work per cycle making it a relatively poor cycle to implement in real machinery.
Thus, there are not a lot of intentionally designed Carnot Cycle engines lying about. We do everyone a disservice propagating the misconception that the Carnot Cycle is the one best cycle of all possible cycles efficiency wise.
It is only one of many. The Carnot cycle only defines the limit of efficiency for a heat engine operating between a high temperature heat source and a lower temperature heat sink. It defines the limit of efficiency, since it would use up all of the potential heat energy available to transfer from the high temperature heat source to the lower temperature heat sink. Sign up to join this community. The best answers are voted up and rise to the top. Stack Overflow for Teams — Collaborate and share knowledge with a private group.
Create a free Team What is Teams? Learn more. Why is the Carnot engine the most efficient? Ask Question. Asked 6 years, 11 months ago.
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