Refrigerant 134a undergoes a process for which the pressure-volume relation is P

Question

Refrigerant 134a undergoes a process for which the pressure-volume relation is PV^n = Constant. The initial and final states of the refrigerant are fixed by P1= 200 kPa, T1= -10 degrees C and P2= 1000 kPa, T2 = 50 degrees C, respectively. (a) Draw a T-v diagram, label the saturated temperature, the specific volumes, and the temperatures of both the initial and final states. (b) Determine the work done during this expansion process, in KJ/kg. (c) Calculate the amount of heat that transferred into R-134a, in kJ/kg.

Explanation / Answer

Given PV^n = constant
==> PV= nRT gas law
==> V = nRT/P

So P(T/P)^n = constant

==> P^(1-n) T^n = constant

==> P1/P2 = (T2/T1)^(n/(1-n))

==> 1/5 = (323/263)^(n/(1-n))
==> n/(1-n) = -7.83
==> n = (n-1)*7.83

==> 7.83 = 6.83 n
==> n = 1.14

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