Question

An insulated piston-cylinder device contains 0.05 m3 of saturated R-134a vapor at 0.8 MPa. The
refrigerant is now allowed to expand in a reversible manner (i.e. isentropic) until the pressure drops to 0.4
MPa. Determine (a) the final temperature in the cylinder and (b) the work done by the refrigerant.

Answer 1

Here we assume that the process in our "well-insulated cylinder" is adiabatic. Since we are given that the expansion is reversible, we have reversible adiabatic process. We know that such a process is an isentropic (constant entropy) process, so that the final entropy equals the initial entropy. We can use this fact to find the answers to both parts of this problem.

First we find the data for the initial state. Since this is a saturated vapor at 0.8 MPa, we find the following saturated vapor properties from Table A-12 on page 930: v1 = 0.025621 m3/kg, u1 = 246.79 kJ/kg, and s1 = 0.91835 kJ/kg∙K.

We can find the (constant) mass from the initial volume, V1 = 0.05 m3 and the initial specific volume.

For this closed system, the first law is Q = DU + W. Since Q = 0, our first law reduces to W = -DU = m(u1 - u2). We can find u2, because we know the pressure P2 = 0.4 MPa and the entropy s2 = s1 = 0.91835 kJ/kg∙K at the final state. From the data on saturation properties at 0.4 MPa (400 kPa) in Table A-12 on page 930, we see that the value of s2 lies between the values of sf = 0.24761 kJ/kg∙K and sg = 0.92691 kJ/kg∙K. Thus we know that we are in the mixed region. This means that T2 = Tsat(P2 = 0.4 MPa) = 8.91oC.

We have to find the quality to determine the final internal energy.

We now find the work from the first law.

= 27.08 kJ