Answer:
a) 24.01%
b) 0.35%
c) 3.43%
d) 0%
e) 37.31%
Explanation:
A) water vapor=> H2O
From table @4000 lbf/in^2, 1000°F v = 0.1752 ft^3/ib
m = 18.02 ib/ibmol
P = 4000 lbf/in^2
From the ideal gas model
PV/m = mRT/m
V = RT/Pm
Where R = 1945 (ft.ibf)/(ibmol.°R)
T = 1000°F = 1459.67°R
Therefore
V = (1945 × 1459.67) ÷ (4000 × 18.02) = 2255190.19 ÷ 72080 = 0.21727 ft^3/ib
Error =( V - v)/v × 100% = (0.21727 - 0.1752) ÷ 0.1752 = 0.2401 ×100% = 24.01%
B) water vapor=> H2O
From table @ 5 lbf/in^2, 250°F v = 84.21 ft^3/ib
m = 18.02 ib/ibmol
P = 5 lbf/in^2
From the ideal gas model
PV/m = mRT/m
V = RT/Pm
Where R = 1945 (ft.ibf)/(ibmol.°R)
T = 250°F = 709.67°R
Therefore
V = (1945 × 709.67) ÷ (5 × 18.02) =84.51 ft^3/ib
Error =( V - v)/v × 100% = (84.51 - 84.21) ÷ 84.21 = 0.003538 ×100% = 0.35%
C) ammonia =>
From table @40 lbf/in^2, 60°F v = 7.9134 ft^3/ib
m = 17.03 ib/ibmol
P = 40 lbf/in^2
From the ideal gas model
PV/m = mRT/m
V = RT/Pm
Where R = 1945 (ft.ibf)/(ibmol.°R)
T = 60°F = 519.67°R
Therefore
V = (1945 × 519.67) ÷ (40 × 17.03) = 8.189 ft^3/ib
Error =( V - v)/v × 100% = (8.189 - 7.9134) ÷ 7.9134 = 0.03432 ×100% = 3.43%
D) Air at 1 atm =>
No table so we assume air is ideal
m = 28.97 ib/ibmol
P = 14.71 lbf/in^2
From the ideal gas model
PV/m = mRT/m
V = RT/Pm
Where R = 1945 (ft.ibf)/(ibmol.°R)
T = 560°R
Therefore
V = (1945 × 560) ÷ (14.71 × 28.97) = 14.099 ft^3/ib
Since Air is ideal error is zero =0%
E) refrigerant =>
From table @300 lbf/in^2, 180°F v = 0.1633 ft^3/ib
m = 102.03 ib/ibmol
P = 300 lbf/in^2
From the ideal gas model
PV/m = mRT/m
V = RT/Pm
Where R = 1945 (ft.ibf)/(ibmol.°R)
T = 180°F = 639.67°R
Therefore
V = (1945 × 639.67) ÷ (300 × 102.03) = 0.22422 ft^3/ib
Error =( V - v)/v × 100% = (0.22422 - 0.1633) ÷ 0.1633 = 0.37306 ×100% = 37.31%
