A saturated mixture with a quality of 0.10 of R-134a is to expand in an insulated flexible container from a pressure of 1.00 MPa and a volume of 0.02 m3 until the pressure is 500 kPa and the quality if 0.5. Determine the final temperature and volume of the R-134a, and the work done in the process.
Given: R-134a; P1 = 1.0 MPa; x1 = 0.10; V1 = 0.02 m3; P2 = 500 kPa; x2 = 0.5.
Assume: Q = 0 (insulated); ?KE = ?PE = 0
What will be an ideal response?
Solution: The First Law for a closed system reduces to
W = m (u1 – u2)
The mass is found from the initial state:
v1 = 0.00280 m3/kg
m = V1 / v1 = 7.14 kg
v2 = 0.02085 m3/kg
V2 = 0.149 m3
As state 2 is saturated, T2 = Tsat(P2) = 15.74oC
While the work appears to be moving boundary work, we don’t know the functional relationship between P and V. Therefore, we will find the work from the First Law:
u1 = 118.76 kJ/kg
u2 = 153.29 kJ/kg
W = m (u1 – u2) = -247 kJ
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