How Does Running Two Heat Engines Affect the Second Law of Thermodynamics?

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SUMMARY

This discussion centers on the implications of running two heat engines, specifically a Carnot engine and another engine with a higher efficiency, on the Second Law of Thermodynamics. The Carnot engine operates between a firebox at 750K and an ambient temperature of 300K, achieving 60% efficiency. The second engine, referred to as engine S, has an efficiency of 70%. The analysis reveals that when both engines operate together, the total energy output from the firebox and the energy transferred to the environment violate the Clausius statement of the Second Law, demonstrating the spontaneous transfer of heat from a cold reservoir to a hot reservoir without the necessary input work.

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  • Basic thermodynamic principles and energy transfer concepts
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Erik Horwath
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a firebox is at 750K and the ambient temp is 300K. The efficiency of a Carnot engine doing 150J of work as it transports energy between these constant temperature baths is 60%. The Carnot engine must take in energy 150/.60=250J from the hot reservoir and must put out 100J of energy by heat into the environment. To follow Carnot's reasoning, suppose that xome other heat engine S could have efficiency 70%. a)Find the energy input and wasted energy output of engine S as it does 150J of work. b)Let engine S operate as in part a) and run the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, and the total energy transferred to the environment. Show that the Clausius statemetn of the second law of thermodynamics is violated.

Part a) is obvious. They show exactly how to do it within the problem. But the wording in part b) is confusing to me. If the engine is operating in reverse, then it is a heat pump, not a heat engine. Therefore the work under consideration must be input work, not output work. So when it says "let engine S operate as in part a)" does that mean 150J of work in done on the system rather than by the systm? Further, the expression for the coefficient of performance of a heat pump is different from that for the effiecency of a heat engine. The former is e=Wnet/Qin and is always less than unity and the latter is cop=Qout/Win and is generally >1. I don't understand what the problem is asking exactly. Thanks for your help! :smile:
 
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You should show that the whole heat is turned into work.
 
Maxos said:
You should show that the whole heat is turned into work.

I'm still confused as to how to proceed. I think you are confusing the Clausius statement of the second law (which essentially says that the spontaneous transfer of heat does not occur from a cold reservoir to a hot reservoir) with the Kelvin-Planck statement (which says that it is not possible to design a perfectly efficient heat engine - i.e. one that converts all input heat to work). The problem asks me to show that running two heat engines simultaneously forward and in reverse (between the same two energy reservoirs) results in the transfer of heat from a cold reservoir to a hot reservoir without the positive input work.

The question is an odd - the answer in the back of the text is: "-35.7J, -35.7J. The net effect is the transport of energy by heat from the cold to the hot reservoir without the expenditure of work."
 

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