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3 years ago

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7 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?

1 answer:

julsineya [31]3 years ago

7 0

Answer:

$\Delta S = S_2-S_1 = 0.08835 \frac {BTU} {Lb ^ {\circ} R}$

Explanation:

A) In order to solve the table it is necessary to consult tables A11-E and A10E for refrigerant R134-a

In this way we obtain that:

$S_1 = S_f = 0.0648 \frac {BTU} {Lb. ^ {\circ} R}$

In this way,

$S_2 = S_f + x_2 (S_g-S_f) = 0.0902 + 0.5 (0.2161-0.0902)$

$S_2 = 0.15315 \frac {BTU} {Lb ^ {\circ} R}$

In this way the entropy change is,

$\Delta S = S_2-S_1 = 0.08835 \frac {BTU} {Lb. ^ {\°} R}$

B) Whenever entropy yields a positive result, the process can be carried out adiabatically.

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