%3Cdiv%20style%3D%22color%3A%26nbsp%3Brgb(34%2C%26nbsp%3B34%2C%26nbsp%3B34)%3B%2
ID: 2994032 • Letter: #
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KNOWN
T1 = 50 deg = 323 K
T2 = 400 deg = 673 K
P1 = 300 kPa
m = 0.75 kg
FIND:
P2
dU, change in internal energy
ASSUMPTIONS:
Rigid Tank
PROPERTIES:
Molecular weight of CO2 = 44.01 kg/kmol
Density at 50 deg = 1.668 kg/m^3
Density at 400 deg = 0.7949 kg/m^3
Cp at 50 deg = 0.8747 KJ/kg.K
Cp at 400 deg = 1.11385 kJ/kg.K
ANALYSIS
p is used as a for density
At Tx = 320 K
px = 1.683 kg/m^3 and Cpx = 0.872 KJ/kg.K
At Ty = 340 K
py = 1.583 and Cpy = 0.890
Now at T1 = 323 K
By linear Interpolating
(T1-Tx)/(Ty-Tx) = (p1-px)/(py-px)
p1 = 1.668
Similarly
Cp1 = 0.8747
At Tx = 660 px = 0.813 Cpx = 1.108
At Ty = 680 py = 0.789 Cpy = 1.117
At T2 = 673 K
p2 = 0.7949
Cp2 = 1.11385
Specific Volumes
v1 = 1/p1 = 0.5995 m^3/kg
v2 = 1/p2 = 1.2580 m^3/kg
P1*v1/T1 = P2*v2/T2
300*(0.5995)/323 = P2*1.2580/673
P2 = 297.88 kPa
dQ = m*(Cp2*T2
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