# Help with thermodynamics -- work of Carnot engine expansion

1. Nov 22, 2015

### nate9519

1. The problem statement, all variables and given/known data

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
2. Relevant equations

3. The attempt at a solution

2. Nov 22, 2015

### Staff: Mentor

Sure.

3. Nov 23, 2015

### Chandra Prayaga

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.

4. Nov 23, 2015

### nate9519

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

5. Nov 23, 2015

### Chandra Prayaga

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.

6. Nov 24, 2015

### nate9519

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?

7. Nov 24, 2015

### Chandra Prayaga

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.