Propane / n-Octane Flash Calculation: 24-Pet-A2 Petroleum Reservoir Fluids, December 2019, Q6

Question 6: Propane / n-Octane Flash Calculation (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Check: the source's P-T legend for this figure explicitly marks curves 3, 4 and 5's compositions "illegible," and the flash calculation below does not depend on reading the chart at all – it is solved analytically via the Wilson K-value correlation, the same method used in Q2, at the exact stated conditions (500 psia, 220°C).

Given. $z_{C3}=z_{C8}=0.5$ (mole basis, already stated as moles); $T=220\,{}^{\circ}\text{C}=428\,{}^{\circ}\text{F}=887.67\,{}^{\circ}\text{R}$; $P=500$ psia. Propane: $T_c=665.7\,{}^{\circ}\text{R}$, $P_c=616.0$ psia, $\omega=0.152$. n-Octane: $T_c=1023.7\,{}^{\circ}\text{R}$, $P_c=361.1$ psia, $\omega=0.398$.

Find. (a) equilibrium liquid/vapour compositions and $V/L$; (b) additional moles of C3 needed for $V/L=3$.

Approach. Compute Wilson K-values at the stated $(T,P)$, solve the Rachford–Rice equation $\sum z_i(K_i-1)/[1+V(K_i-1)]=0$ for the vapour mole fraction $V$, then back out phase compositions; for (b), re-solve Rachford–Rice for the feed composition that gives $V=0.75$ (since $V/L=3 \Rightarrow V=3(1-V)\Rightarrow V=0.75$) at the SAME $K$-values (composition-independent at fixed $T,P$).

(a) Flash at 500 psia, 220°C, equimolar feed

Wilson K-values. $$K_{C3}=\frac{P_{c,C3}}{P}\exp\!\left[5.373(1+\omega_{C3})\left(1-\frac{T_{c,C3}}{T}\right)\right] = 5.792$$ $$K_{C8}=\frac{P_{c,C8}}{P}\exp\!\left[5.373(1+\omega_{C8})\left(1-\frac{T_{c,C8}}{T}\right)\right] = 0.2285$$
Rachford–Rice for $V$. $$\frac{0.5(K_{C3}-1)}{1+V(K_{C3}-1)}+\frac{0.5(K_{C8}-1)}{1+V(K_{C8}-1)}=0$$ Solving numerically: $$\boxed{V/F = 0.544\ ,\quad L/F = 0.456\ ,\quad V/L = 1.19}$$
Phase compositions. $$x_i = \frac{z_i}{1+V(K_i-1)}\ ,\qquad y_i = K_ix_i$$ $$\boxed{x_{C3}=0.139,\ x_{C8}=0.861\ (\text{liquid})\qquad y_{C3}=0.803,\ y_{C8}=0.197\ (\text{vapour})}$$

Quantity	Value
$K_{C3}$, $K_{C8}$	5.79, 0.229
$V/F$	0.544
$V/L$	1.19
Liquid: $x_{C3}$, $x_{C8}$	0.139, 0.861
Vapour: $y_{C3}$, $y_{C8}$	0.803, 0.197

(b) Moles of C3 to add for $V/L=3$

Target vapour fraction. $$V/L=3 \;\Rightarrow\; \frac{V}{1-V}=3 \;\Rightarrow\; V=0.75$$
Solve Rachford–Rice for the required feed ratio (K-values unchanged, same $T,P$): at $V=0.75$, $$\frac{z_{C3}}{z_{C8}} = -\frac{(K_{C8}-1)/[1+0.75(K_{C8}-1)]}{(K_{C3}-1)/[1+0.75(K_{C3}-1)]} = 1.755$$
Total C3 with n-C8 fixed at 0.5 mol. $$n_{C3,total} = 1.755\times0.5 = 0.878\text{ mol}$$ $$\boxed{\Delta n_{C3} = 0.878-0.500 = 0.378\ \text{mol C3 to add}}$$

Quantity	Value
Total C3 required	0.878 mol
C3 to add	0.378 mol

Topic. Two-phase isothermal flash calculation for a binary system via Wilson K-values and Rachford–Rice.

Key relations. Wilson K-value correlation; Rachford–Rice objective function $\sum z_i(K_i-1)/[1+V(K_i-1)]=0$; $x_i=z_i/[1+V(K_i-1)]$, $y_i=K_ix_i$.

Why this works. At fixed temperature and pressure the K-values (equilibrium ratios) are, to the Wilson approximation, independent of feed composition – only the material-balance (Rachford–Rice) equation changes when the feed composition changes, which is exactly why part (b) can reuse the same $K_{C3}, K_{C8}$ computed in part (a) and simply re-solve for the feed ratio giving a different $V$.

Common pitfall. Forgetting that $V/L=3$ means $V=0.75$ (not $V=3$, which is impossible since $V\le1$) – always convert a vapour-to-liquid ratio to a vapour mole fraction via $V=(V/L)/(1+V/L)$ before using Rachford–Rice.

Source. Danesh, PVT and Phase Behaviour of Petroleum Reservoir Fluids, Ch.4 (flash calculations, K-value correlations).

Question 6 of 7: Propane / n-Octane Flash Calculation

Question 6: Propane / n-Octane Flash Calculation (20 marks)

(a) Flash at 500 psia, 220°C, equimolar feed

(b) Moles of C3 to add for $V/L=3$