Question

One lb of Refrigerant 134a contained within a piston–cylinder assembly undergoes a process from a state where the temperature is 60F and the refrigerant is saturated liquid to a state where the pressure is 140 lbf/in 2 and quality is 50%. Determine the change in specific entropy of the refrigerant, in Btu/lbR. Can this process be accomplished adiabatically?

Answer 1

Yes, it can can be completed adiabatically

Step-by-step explanation:

To solve the problem we will resort to the theory of thermodynamics,

It is necessary to develop this problem to resort to the A-11E tables in English Units for R134a (since the problem requires it, if it were SI just to change to that table)

State 1 indicates that the refrigerant is at 60 ° F,

In the first table (attached image of the value taken) the value of the entropy is

`S_f=S_1= 0.06675Btu/lbm.R`

For State 2 the refrigerant is at 50% quality and at a pressure of
`140lbf / in ^ 2`

In table 2 of the refrigerant (for the pressure values) we perform the reading and we have to

`Sf= 0.09214`

`Sg=0.21879`

We know that,

`S_2 = Sf +X_(quality)(S_g-S_f)`

`S_2 = 0.09214+0.5(0.21979-0.09214)`

`S_2 = 0.155965BTU/Lb.R`

The change in enthalpy would be given by

`\Delta S = S_2-S_1 = 0.155965BTU/Lb.R- 0.06675Btu/lbm.R`

`\Delta S = 0.089215Btu/lbm.R`

The change in enthalpy is positive, so the process can be completed adiabatically