contestada

Refrigerant 22 is at an initial temperature of 60° F and an initial pressure of 80 lb/in^2. The Refrigerant 22 undergoes an isobaric process to a final state of 100° F. Determine the change in specific volume of the Refrigerant 22.

Respuesta :

Answer:

Change in specific volume will be [tex]0.076ft^3/lb[/tex]

Explanation:

We have given [tex]R_{22}[/tex] is at 60°F and an initial pressure of [tex]80lb/in^2[/tex]

Now according to [tex]R_{22}[/tex] property at 60°F and at a pressure of [tex]80lb/in^2[/tex] specific volume is 0.7291 [tex]ft^3/lb[/tex]

As it is an isobaric process so pressure will be constant that is [tex]P_1=P_2[/tex]

So at 100°F and at 80 [tex]80 lb/in^3[/tex] specific volume will be 0.8051 [tex]ft^3/lb[/tex]

So change in volume = 0.8051 -0.7291 = 0.076[tex]ft^3/lb[/tex]

Otras preguntas