Engineering, 12.03.2020 06:18 coopera1744

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.

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Answer from: rosyjesus297

The temperature based on ideal gas is T = 526 R

The temperature based on compressibility chart is T = 694 R

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 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

$T = 240 +460 = 700\ R$

Answer from: Quest

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explanation:

Answer from: Quest

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Refrigerant-134a at 400 psia has a specific volume of 0.1384 ft3/lbm. Determine the temperature of t...