Thermodynamics Energy transfer question

[Silverhobbiest]

Homework Statement

In a frictionless piston-cylinder system, there are 3 kg of R-134a initially at 280 kPa and 15 °C. Heat is transferred to the system in the amount 120 kJ. What will the final temperature of the refrigerant be (deg C)?

Homework Equations

Q - W = ΔU (internal energy)
Q - W_b (boundary work) = ΔH (enthalpy)
(if pressure is constant) W_b = P(v₂-v₁)

[where v is specific volume]

The Attempt at a Solution

I assumed that pressure was constant for this problem because the piston is implied to be free to move. Also, I know that there is boundary work because if the piston is free to move then the volume is changing.

So, I used the equation:
Q - W_b = ΔH

I identified the refrigerant to be in the superheated vapor phase and used the property tables to calculate the specific volume to the refrigerant at the initial state and got 0.078215 m³/kg and calculated the enthalpy to be 264.12kJ/kg.

I was then stuck on how to find boundary work because I have no idea how to find v2 with the given information. I planned on calculating boundary work so that I can use the equation to solve for h2 and then use the property tables to locate the temperature of the substance.

I became frustrated and attempted to ignore it entirely and got an incorrect answer of 59.4 degrees celsius.

How do I find boundary work? Is my approach correct? Please help me solve thi problem so that I can understand what I'm doing wrong.

 

[Chestermiller]

If the expansion takes place at constant pressure, the equation you used is incorrect. The correct equation to use for constant pressure expansion is $$Q=\Delta H$$Since you know Q, you also know $\Delta H$. That means that you know the final enthalpy per unit mass. You then use your tables to find the final temperature that gives that value for the final enthalpy.

 

[Silverhobbiest]

If the expansion takes place at constant pressure, the equation you used is incorrect. The correct equation to use for constant pressure expansion is $$Q=\Delta H$$Since you know Q, you also know $\Delta H$. That means that you know the final enthalpy per unit mass. You then use your tables to find the final temperature that gives that value for the final enthalpy.

I initially did it this way and got 59.4 °C, but that was incorrect. That is what led me to believe that there was boundary work involved because the process takes place inside of a piston cylinder system. So, why isn't boundary work supposed to be considered in this process?

 

[Chestermiller]

I initially did it this way and got 59.4 °C, but that was incorrect. That is what led me to believe that there was boundary work involved because the process takes place inside of a piston cylinder system. So, why isn't boundary work supposed to be considered in this process?

Maybe they expected you to assume constant volume, and maybe you made a mistake in the calculation. If you want, I'll redo the constant pressure calculation.

 

[Chestermiller]

I pretty much confirm your final temperature for the heating at constant pressure.

 

[Silverhobbiest]

I pretty much confirm your final temperature for the heating at constant pressure.

Okay, then I'll just email the instructor to see if it's another error in the system. Thank you very much for helping me!

 

[Chestermiller]

Okay, then I'll just email the instructor to see if it's another error in the system. Thank you very much for helping me!

For constant volume, I get something around 63 C.

 

[Silverhobbiest]

For constant volume, I get something around 63 C.

According to the system, the correct answer is 34.3 C. Do you know anyway this answer could have been obtained?

 

[Chestermiller]

According to the system, the correct answer is 34.3 C. Do you know anyway this answer could have been obtained?

No. Have you given the entire problem statement, word for word?

 

[Silverhobbiest]

No. Have you given the entire problem statement, word for word?

Yes I have.

 

[Chestermiller]

Yes I have.

Well, then I don't see how they get that answer.