Question

APPENDIX 6 PHYSICAL PROPERTIES OF FLUIDS Table A6-1 Properties of Hydrocarbons and Common Gases. Source: GPSA Engineering Data Book, 10th Ed. 1987 Critical Compound Temp. Normal Butane Normal Pentane ptane 652.0 Carbon Dioxide 1300 Hydrogen Sulfide H2S 546.9 -221.31 0.87476d 95.557 Mixture Heating Valuc 6 marks] Question 1.4: A pure n-butane exists in the two-phase region at 120°F and 70 psia. Calculate the density of the coexisting phase by using the following equations of state: a. Van der Waals b. Redlich-Kwong c.Soave-Redlich-Kwong Compare the results obtained

Answer 1

1) van der Waals equation

(P+음) (V-b) y2

for unit mass

Vー

$\left(P+{a\cdot \rho ^{2}}\right)(\frac{1}{\rho }-b)=R T$

R = 10.731573 ft³·psia/°R·lb.mol

T = 120 ^oF

P = 70 psia

for van der vaals

27 (RTc) Cl 64pc

RTc h_1

thus we get

a = 8242.10

b = 0.7445

(70+ 8242.1 . ρ2) (-_ 0.7445) 10.73 * 120

$\rho \approx 1.16498$

b) Redlich-Kwong

RT Ca

рґа

IPT 0.42748

$b \approx 0.08664 \frac{R T_{c}}{p_{c}}$

we get

a = 145959.86

b = 0.51601

рґа

we get

$\rho \approx 1.73366$

c) Soave Redlich Kwong

$p=\frac{R T}{V-b}-\frac{a \alpha}{V\left(V+b\right)}$

$p=\frac{\rho R T}{1-\rho \cdot b}-\frac{\rho ^{2}\cdot a\cdot \alpha}{\left(1+\rho\cdot b\right)}$

a = 0.42747/727 P.

0.08664RI

$\alpha=\left(1+\left(0.48508+1.55171 \omega-0.15613 \omega^{2}\right)\left(1-T_{r}^{0.5}\right)\right)^{2}$

$T_{r}=\frac{T}{T_{c}}$

$\omega = Acentric factor$

T_r = 0.392

w= 0.212

we get

$\alpha=1.6945$

a = 8348.96

b = 0.5160

ρ2 . 8348.96 . 1.6945 ρ * 10.73 * 120 70 = -f) . 0.5160 (1+f) . 0.5160)

$\rho \approx 1.74657$

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