Question 5 of 7: Two-Phase Flash of a Binary Mixture

Question 5: Two-Phase Flash of a Binary Mixture (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.

Given. Feed $z_A=0.40$, $z_B=0.60$; $K_A=10$, $K_B=0.04$ at cell conditions (500 psia, 100°F). Formula sheet: Rachford–Rice $\sum_i \dfrac{z_i(K_i-1)}{1+V(K_i-1)}=0$, $x_i=\dfrac{z_i}{1+V(K_i-1)}$, $y_i=K_ix_i$, $L+V=1$.

Find. Vapour fraction $V$, liquid fraction $L$, and phase compositions $x_A,x_B,y_A,y_B$.

Approach. Solve the two-component Rachford–Rice objective function for $V$ directly (it reduces to one linear equation for a binary system), then back out phase compositions from the flash relations.

1. Set up Rachford–Rice for two components. $\dfrac{z_A(K_A-1)}{1+V(K_A-1)}+\dfrac{z_B(K_B-1)}{1+V(K_B-1)}=0$, i.e. $\dfrac{0.40(9)}{1+9V}+\dfrac{0.60(-0.96)}{1-0.96V}=0$, or $\dfrac{3.6}{1+9V}=\dfrac{0.576}{1-0.96V}$.
2. Solve for V. Cross-multiplying: $3.6(1-0.96V)=0.576(1+9V) \Rightarrow 3.6-3.456V=0.576+5.184V \Rightarrow 3.024=8.64V$. So $\boxed{V=0.350}$ and $\boxed{L=1-V=0.650}$.
3. Liquid-phase composition. $x_A=\dfrac{z_A}{1+V(K_A-1)}=\dfrac{0.40}{1+0.35(9)}=\dfrac{0.40}{4.15}=0.0964$. $x_B=\dfrac{z_B}{1+V(K_B-1)}=\dfrac{0.60}{1+0.35(-0.96)}=\dfrac{0.60}{0.664}=0.9036$. Check: $x_A+x_B=1.000$. So $\boxed{x_A=0.0964,\ x_B=0.9036}$.
4. Vapour-phase composition. $y_A=K_Ax_A=10(0.0964)=0.9639$. $y_B=K_Bx_B=0.04(0.9036)=0.0361$. Check: $y_A+y_B=1.000$. So $\boxed{y_A=0.9639,\ y_B=0.0361}$.