Question

Refrigerant-134a at 400 psia has a specific volume of 0.1384 ft3/lbm. Determine the temperature of the refrigerant based on (a) the ideal-gas equation, (b) the generalized compressibility chart, and (c) the refrigerant tables.

Answer 1

Answer:

a

The temperature based on ideal gas is T = 526 R

b

The temperature  based on compressibility chart  is T = 694 R

c

The temperature based on refrigerant tables is T = 700 R

Explanation:

From the mathematical relation of an ideal gas we can define the temperature of the Refrigerant-134a as follows

                $T = \frac{Pv}{R}$

Where  $P$ is the pressure of the refrigerant  and given as  = 400 psia

            v is the volume of the refrigerant and it is given as $= 0.1384 ft^3/lbm$

           R is the gas constant and has a value of $0.1052\ psia \ \cdot ft^3/lbm \ \cdot R$ for Refrigerant-134a

            Substituting values

               $T = \frac{(400psia)(0.1384\ ft^3/lbm)}{0.1052 pais \cdot ft^3/lbm \cdot R}$

                 $= 526 R$

Considering the generalized compressibility chart

  looking at table of molar mass, gas constants and critical-point properties.The temperature of Refrigerant-134a denoted by $T_{cr}$ is 673.6 R and the critical pressure of Refrigerant-134a is 588.7 psia

    Mathematically  the reduced pressure is

                     $P_R = \frac{P}{P_{cr}}$

    substituting given and value obtained from table

               $P_R = \frac{400psia}{588.7psia} =0.679$

The pseudo-reduced specific volume  is Mathematically given as

                  $V_R = \frac{v_{actual}}{RT_{cr}/P_{cr}}$

 $Substituting \ 588.7\ psia \ for \ P_{cr} \ 673.6\ R \ for \ T_{cr} \ 0.1052\ psia \ \cdot \ ft^3 /lbm \cdot R\\\\ \ for \ R \ 0.1384ft^3/lbm \ for \ v_{actual}$

               $V_R = \frac{0.1384}{(0.1052)(673.6)/588.7} = \frac{(0.1384)(588.7)}{(0.1052)(673.6)} =1.15$

Now the value of pseudo-reduced specific volume and reduced pressure would gives us a pointer to select the correct value of reduced temperature($T_R$) from the generalized compressibility chart

and this is 1.03 for $T_R$

   Hence the temperature of the Refrigerant-134a

                 $T = T_R T_{cr}$

         Substituting values

           $T = (1.03)(673.6 R)$

              $= 694\ R$

Considering the refrigerant tables

           looking at the table of super-heated Refrigerant-134a the temperature for a pressure of 400 psia and

specific volume of $0.1384ft^3/lbm$ is 240°F

   Converting to Rankine scales we add 460

              $T = 240 +460 = 700\ R$