Question

An insulated piston–cylinder device contains 0.05 m3 of saturated refrigerant- 134a vapor at 0.8-MPa pressure. The refrigerant is now allowed to expand in a reversible manner 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

Answer:

a) $T = 8.91\,^{\textdegree}C$, b) $W_{out} = 27.744\,kJ$

Explanation:

The piston-cylinder device is modelled after the First Law of Thermodynamics:

$-W_{out} + P_{1}\cdot V_{1} - P_{2}\cdot V_{2} + m\cdot (u_{1}-u_{2}) = 0$

$-W_{out} = m \cdot (h_{2}-h_{1})$

$W_{out} = m\cdot (h_{1}-h_{2})$

The properties of the refrigerant 134a are, respectively:

Initial State (Saturated Vapor)

$P = 800\,MPa$

$T = 31.31\,^{\textdegree}C$

$\nu = 0.025645\,\frac{m^{3}}{kg}$

$h = 267.34\,\frac{kJ}{kg}$

$s = 0.91853\,\frac{kJ}{kg\cdot K}$

Final State (Liquid-Vapor Mixture)

$P = 400\,kPa$

$T = 8.91\,^{\textdegree}C$

$h = 253.11\,\frac{kJ}{kg}$

$s = 0.91853\,\frac{kJ}{kg\cdot K}$

$x = 0.987$

a) The final temperature in the cylinder is $T = 8.91\,^{\textdegree}C$.

b) The work done by the refrigerant is:

$W_{out} = \left(\frac{0.05\,m^{3}}{0.025645\,\frac{m^{3}}{kg} }\right)\cdot \left(267.34\,\frac{kJ}{kg} - 253.11\,\frac{kJ}{kg}\right)$

$W_{out} = 27.744\,kJ$