# Homework Help: Thermodynamics - Specific Entropy

1. Nov 3, 2014

### ConnorM

1. The problem statement, all variables and given/known data
1)
A quantity of propane is contained in a piston-cylinder assembly. This working
fluid undergoes a process starting from an initial state of 2.0 MPa and 60oC. At
the end of the process the pressure of the propane is 1.0 MPa and the specific
entropy is 0.089 kJ/kgK higher than at the start of the process.

Determine the temperature of the propane at the end of the process.

2. Relevant equations
For the first question I am using the propane tables found here,
( http://www.engr.mun.ca/muzychka/ThermodynamicPropertiesTables.PDF )

I think this equation may be useful,
s2 - s1 = so(T2) - so(T1) - (R/M)(ln(p2/p1))

3. The attempt at a solution
For the first question I am pretty sure all I have to do is just interpolate to find what so(T1) is, then would I just rearrange the equation to find so(T2) and then interpolate to determine T2? I just don't quite understand interpolating but I think what I have said above is correct.

2. Nov 3, 2014

### ConnorM

I looked at the propane table and found that it was initially superheated, so then I went to the superheated propane table and looked under the chart where pressure was 2 MPa, from here I went to 60oC and found my initial entropy to be 1.722 kJ/kgK. Then I subbed my known values in to the equation s2 - s1 = so(T2) - so(T1) - (R/M)(ln(p2/p1))

0.089 kJ/kgK = so(T2) - 1.722 kJ/kgK - (8.314 kJ/kgK / 44.1 kg/kmol ) ln (2/1)

From this I got 1.94167 kJ/kgK
Then I used this value with the table for 1 MPa and found that the T2 would be between 60 and 70 degrees C. I interpolated by,

( 70 C - 60 C )/( T - 60 C ) = ( 1.997 kJ/kgK - 1.936 kJ/kgK )/( 1.94167 kJ/kgK - 1.936 kJ/kgK )

Finally I found the T = 60.93 C
But when I submitted this it was wrong?

3. Nov 5, 2014

### Staff: Mentor

Hi Conner. It is not necessary to use the equation you wrote to solve this problem. If the specific entropy increases by 0.089 kj/kgK, and the initial specific entropy was 1.722, what is the final specific entropy? Now, go to your table and find the temperature that corresponds to this final specific entropy and the final pressure. What do you get?

Chet

4. Nov 5, 2014

### ConnorM

Hey I ended up figuring it out after ignoring my equation and just using the change in entropy and initial entropy to determine the finaly entropy. After looking at the table I got an answer of 40 C.

What I am confused about is that the equation I was trying to use was used in some other calculations for different species like carbon dioxide as an ideal gas. I wasn't sure why I was supposed to omit it in this calculation.

5. Nov 5, 2014

### Staff: Mentor

I think you are using the equation incorrectly. s0 represents the specific entropy at temperature T and 1 atm pressure, not at the pressure of the system in either state. If you replace the s0 terms with $C_pln(T_2/T_1)$, you should get close to the right answer, where the heat capacity is the specific heat capacity.

Chet