(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A closed, rigid tank filled with water, initially at 20 bar, a quality of 80%, and a volume of .5 m^3, is cooled until the pressure reaches 4 bar. Evaluate the heat transfer, in kJ, and draw the process in a T-v diagram

2. Relevant equations

ΔU=Q-W (but W goes to zero here because of the constant volume)

u= ugX + (1-x)uf

v= vgX + (1-x)vf

3. The attempt at a solution

Ok, so. I know that the volume is constant. My plan was to use the quality to calculate specific internal energies at 20 bar and 4 bar, thus finding the change in internal energy.

I ended up with u1(20 bar) = 2261.528 kJ/kg and u2(4 bar) = 2163.742 kJ/kg using the above equation using quality.

However, it wants the answer in kJ, so I had to find the specific volume to find the mass to cancel the kJ/kg. So, i used vg and vf at 20 bar (using the equation v=vgx + (1-x)vf) and I got .07993934 m^3/kg. I divided the given .5 m^3 by the specific volume I got, and received the answer of 6.255 kg. I then multiplied this by the change in internal energy, which I found to be -97.786 (cooling process, makes sense).

The final answer I calculated was -611.65 kJ. My professor gave the answers in class so we could check, but did not go through the problems specifically. This answer does not agree with his answer of -8282 kJ. Maybe -82.82? I have it written down as -8282.

Thanks for any help

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# Homework Help: Thermodynamics Rigid tank problem

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