Thermodynamic Work vs Mechanical Work

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Thermodynamic work and mechanical work are defined differently, leading to the conclusion that they are not necessarily equal. Thermodynamic work refers to the work done by a thermodynamic process, which can result in mechanical work, while mechanical work is defined as the work done by a force. The movement of a mass due to gas expansion or compression is influenced by these definitions, emphasizing the need to specify the system and its boundaries. Additionally, mechanical work represents useful energy output, which is always less than the input thermal energy due to inefficiencies. Understanding these distinctions is crucial for analyzing energy transfer in thermodynamic systems.
neil123
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Homework Statement
In a heat engine experiment, which showcases the Ericsson Cycle, the thermodynamic work and the mechanical work is computed. Should these values be equal?
Relevant Equations
Thermodynamic Work = Wisothermal + Wisobaric
Mechanical Work = mg(hf - hi)
At first I thought these values should be equal because I thought thermodynamic work is simply the work done as the gas compresses or expands. That is, the work done as the gas expands ( and the mass gets lifted up) should be equal to the work done against gravity as the mass goes up, and the work done as the gas compresses (and the mass also does down) should also be equal to the work done by gravity as the mass goes down.

However, some sources say that they shouldn't be equal because they are defined differently and that thermodynamic work is work done by a thermodynamic process (that can cause a mechanical work), while mechanical work is the work done by a force.

I don't really understand this. Isn't the reason the mass moves in the first place due to the expansion/compression of the gas? So why wouldn't the work done as the gas expands be equal to the work
 
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neil123 said:
At first I thought these values should be equal because I thought thermodynamic work is simply the work done as the gas compresses or expands.
You need to be more specific. The work done by what on what?
 
Work is work. It is the integral of the applied force over a displacement.

In thermodynamics, you have to very carefully define what you are calling your system, and where your system ends and its surrounding begins. This choice is entirely up to you.

You also have to specify what is doing work on what (e.g.,, surroundings on system or system on surrounding),

You can provide a specific example, and we can help you work it through.
 
As the economizer of that machine cannot return all the thermal energy that it contains into the intake air (heat transfer depends on temperature differential), some of the input heat from an external source will be wasted into the atmosphere.

The mechanical work is measured output useful mechanical energy, which magnitude will always be less than input thermal energy.

Please, see:
https://en.wikipedia.org/wiki/Erics...ith_Carnot,_Diesel,_Otto,_and_Stirling_cycles
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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