Answer to Question #112255 in Molecular Physics | Thermodynamics for Ali aziz

Question

A piston-cylinder device contains 6 kg of refrigerant-134a at 800 kPa
and 50oC. The refrigerant is now cooled at constant pressure until it
exist as a liquid at 24oC. Show the process on T-v diagram and
determine the heat loss from the system. State any assumption made.

Answer 1

Solution.

$P_2 = 800kPa;$

$T2 = 24^oC;$

$P_1 = 800kPa;$

$T_1 = 50^oC;$

$u_1 = 261.62 kJ/kg; v_1 = 0.02846 m^3/kg;$ Tabl A-12;

$u_2 = 82.37kj/kg; v_2 = 0.0008257m^3/kg;$

$m = 6 kg;$

$\Delta V = m(v_2 - v_1) ; \Delta V = 6\sdot(0.0008257 - 0.02846) =-0.1658m^3;$

$\Delta U = m( u_2 - u_1); \Delta U = 6\sdot(82.37 - 261.62) = -1075.5kJ;$

$\Delta Q = W + \Delta U;$

$W = P\sdot\Delta V; W = 8\sdot10^5\sdot(-0.1658) =- 132kJ;$

$\Delta Q = -132 -1075.5 = -1207.5kJ = -1.2075MJ;$

Answer: $\Delta Q = -1.2075MJ.$