Question 4 of 7: Two-Phase PVT Cell — n-Decane / Ethane Flash

Question 4: Two-Phase PVT Cell — n-Decane / Ethane Flash (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.

Find. (a) Moles of ethane charged, $n_{C_2}$; (b) equilibrium liquid ($x_i$) and vapour ($y_i$) mole fractions of both components.

Approach. Let $x=n_{C_2}$ (mol). The overall mole fractions are $z_{C_{10}}=1/(1+x)$ and $z_{C_2}=x/(1+x)$. Apply the formula-sheet flash relation $\sum_i z_i/[1+V(K_i-1)]=1$ at the given $V=0.10$ to solve for $x$, then recover $x_i=z_i/[1+V(K_i-1)]$ and $y_i=K_ix_i$.

1. Set up the flash equation. With $V=0.10$: for n-decane, $1+V(K_{C_{10}}-1)=1+0.10(0.005-1)=0.9005$; for ethane, $1+V(K_{C_2}-1)=1+0.10(8-1)=1.700$. The flash constraint is $\dfrac{z_{C_{10}}}{0.9005}+\dfrac{z_{C_2}}{1.700}=1$.
2. Part (a) — solve for moles of ethane. Substituting $z_{C_{10}}=1/(1+x)$, $z_{C_2}=x/(1+x)$ and multiplying through by $(1+x)$: $\dfrac{1}{0.9005}+\dfrac{x}{1.700}=1+x$. Rearranging, $\left(\dfrac{1}{0.9005}-1\right)=x\left(1-\dfrac{1}{1.700}\right)$, i.e. $0.11049=0.41176\,x$, so $\boxed{x=n_{C_2}\approx0.268\ \text{mol}}$ of ethane.
3. Part (b) — overall and phase compositions. Total moles $=1+0.268=1.268$, giving overall mole fractions $z_{C_{10}}=1/1.268=0.7884$ and $z_{C_2}=0.268/1.268=0.2116$. Liquid-phase compositions: $x_{C_{10}}=z_{C_{10}}/0.9005=0.7884/0.9005$, so $\boxed{x_{C_{10}}\approx0.8755}$; $x_{C_2}=z_{C_2}/1.700=0.2116/1.700$, so $\boxed{x_{C_2}\approx0.1245}$ (sum $=1.000$, checks). Vapour-phase compositions from $y_i=K_ix_i$: $y_{C_{10}}=0.005\times0.8755$, so $\boxed{y_{C_{10}}\approx0.0044}$; $y_{C_2}=8\times0.1245$, so $\boxed{y_{C_2}\approx0.9956}$ (sum $=1.000$, checks).