Help with thermodynamics -- work of Carnot engine expansion

AI Thread Summary
The discussion revolves around clarifying the calculation of work during the isentropic expansion of a Carnot engine. The first law of thermodynamics is referenced, emphasizing that for adiabatic processes, work equals mass times the change in internal energy. The confusion arises from calculating work from step 1-3 instead of the intended 2-3, leading to questions about whether the internal energy changes are equivalent. It is confirmed that the work for the isentropic expansion must be calculated independently, and the heat input is zero, making the work equal to the negative change in internal energy. Finally, it is noted that when calculating work, the internal energy must be consistent in units, particularly when factoring in the mass of the working fluid.
nate9519
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Homework Statement

I don't need help solving I just need some clarification. Since the carnot cycle is adiabatic, the first law would reduce to work= mass*(change in internal energy). On my test I was asked to calculate the work of the isentropic expansion of a carnot engine with air as the working fluid. I got confused and calculated from step 1-3( isothermal expansion and isentropic expansion) instead of 2-3( just isentropic expansion.) But since step 1-2 is isothermal would that mean the change in internal energy from 1-3 is the same as 2-3 meaning I would still have gotten the right answer

Homework Equations

The Attempt at a Solution

 
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nate9519 said:

Homework Statement

I don't need help solving I just need some clarification. Since the carnot cycle is adiabatic, the first law would reduce to work= mass*(change in internal energy). On my test I was asked to calculate the work of the isentropic expansion of a carnot engine with air as the working fluid. I got confused and calculated from step 1-3( isothermal expansion and isentropic expansion) instead of 2-3( just isentropic expansion.) But since step 1-2 is isothermal would that mean the change in internal energy from 1-3 is the same as 2-3 meaning I would still have gotten the right answer

Homework Equations

The Attempt at a Solution

Sure.
 
nate9519 said:

Homework Statement

I don't need help solving I just need some clarification. Since the carnot cycle is adiabatic, the first law would reduce to work= mass*(change in internal energy). On my test I was asked to calculate the work of the isentropic expansion of a carnot engine with air as the working fluid. I got confused and calculated from step 1-3( isothermal expansion and isentropic expansion) instead of 2-3( just isentropic expansion.) But since step 1-2 is isothermal would that mean the change in internal energy from 1-3 is the same as 2-3 meaning I would still have gotten the right answer

Homework Equations

The Attempt at a Solution

I need some clarification of the steps involved here. You were asked to calculate the Work during the isentropic expansion (2 - 3). You are saying the change in internal energy from 1 - 3 is the same as the change in internal energy from 2 - 3. I agree with your statement, but how is it related to the work? Your statement: "Since the carnot cycle is adiabatic, the first law would reduce to work= mass*(change in internal energy)." is very confusing. Are you talking about the entire Carnot cycle (1 - 2 - 3 - 4 - 1)? The entire cycle is not adiabatic, and the total change in internal energy for the entire cycle is zero, and that is not related to the work done.
 
sorry for confusion. No I didn't mean the entire cycle is adiabatic. I meant from 2-3. My book uses a piston cylinder to demonstrate the carnot cycle and since that is a closed system I thought the first law reduces to mass*(change in internal energy) for the work from 2-3
 
OK. So the work from 2-3 (adiabatic expansion) needs to be calculated independently. It is still equal to the negative of the change in internal energy, because the heat input is zero.
 
yes and that would give you an answer in kilojoules per unit mass. But the problem stated that the working fluid was 15kg of air so if I multiplied delta u by 15kg to get the answer in kJ that would be correct as well, right?
 
Make sure that the internal energy is calculated in the same units. Both the quantities, change in internal energy and work must come out in kJ.
 
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