Respuesta :
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
[tex]T = \frac{Pv}{R}[/tex]
Where [tex]P[/tex] is the pressure of the refrigerant and given as = 400 psia
v is the volume of the refrigerant and it is given as [tex]= 0.1384 ft^3/lbm[/tex]
R is the gas constant and has a value of [tex]0.1052\ psia \ \cdot ft^3/lbm \ \cdot R[/tex] for Refrigerant-134a
Substituting values
[tex]T = \frac{(400psia)(0.1384\ ft^3/lbm)}{0.1052 pais \cdot ft^3/lbm \cdot R}[/tex]
[tex]= 526 R[/tex]
Considering the generalized compressibility chart
looking at table of molar mass, gas constants and critical-point properties.The temperature of Refrigerant-134a denoted by [tex]T_{cr}[/tex] is 673.6 R and the critical pressure of Refrigerant-134a is 588.7 psia
Mathematically the reduced pressure is
[tex]P_R = \frac{P}{P_{cr}}[/tex]
substituting given and value obtained from table
[tex]P_R = \frac{400psia}{588.7psia} =0.679[/tex]
The pseudo-reduced specific volume is Mathematically given as
[tex]V_R = \frac{v_{actual}}{RT_{cr}/P_{cr}}[/tex]
[tex]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}[/tex]
[tex]V_R = \frac{0.1384}{(0.1052)(673.6)/588.7} = \frac{(0.1384)(588.7)}{(0.1052)(673.6)} =1.15[/tex]
Now the value of pseudo-reduced specific volume and reduced pressure would gives us a pointer to select the correct value of reduced temperature([tex]T_R[/tex]) from the generalized compressibility chart
and this is 1.03 for [tex]T_R[/tex]
Hence the temperature of the Refrigerant-134a
[tex]T = T_R T_{cr}[/tex]
Substituting values
[tex]T = (1.03)(673.6 R)[/tex]
[tex]= 694\ R[/tex]
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 [tex]0.1384ft^3/lbm[/tex] is 240°F
Converting to Rankine scales we add 460
[tex]T = 240 +460 = 700\ R[/tex]
Given,
P=400 psia
V=0.1384 [tex]ft^3/lbm[/tex]
Part(a):
As we know,
[tex]T=\frac{PV}{R} \\=\frac{400\times 0.1384}{0.1052}\\T=526 R[/tex]
Part(b):
By using generalized compressibility chart:
[tex]P_{cr}=673.6R[/tex]
Reduced Pressure [tex](P_R)[/tex] then,
[tex]P_R=\frac{P}{P_{cr}}\\=\frac{400}{673.6}\\P_{cr}=0.679[/tex]
Part(c):
To calculate Pseudo-reduced volume[tex](V_R)[/tex] .
[tex]V_R=\frac{0.1384\times 588.7}{0.1252\times 675.6}\\=1.15[/tex]
Now, calculaing the temprature,
[tex]T=T_RT_{cr}\\=1.03\times 673.6\\=694 R[/tex]
Part(c):
By using refrigerant tables.
[tex]T=240F\\=460+240\\T=700R[/tex]
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