Question 5 of 6: Relative Volumetric Feed Rates of Real Ethylene and Oxygen

Question 5: Relative Volumetric Feed Rates of Real Ethylene and Oxygen (Part B)

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. Stoichiometric molar feed $n_{\text{C}_2\text{H}_4}:n_{\text{O}_2}=1:\tfrac12$; both gases at $T=298.15\ \text{K}$, $P=100\ \text{bar}$; critical constants ethylene $T_c=282.3\ \text{K}$, $P_c=50.4\ \text{bar}$, $\omega=0.087$; oxygen $T_c=154.6\ \text{K}$, $P_c=50.43\ \text{bar}$, $\omega=0.022$.

Find. The volumetric feed ratio $\dot V_{\text{C}_2\text{H}_4}/\dot V_{\text{O}_2}$.

Approach. At common $T,P$ the volumetric ratio of two real gases reduces to the product of their mole ratio and their compressibility-factor ratio; obtain each $Z$ from the Peng–Robinson EOS (equivalent to the appended Lee/Kesler chart).

1. Reduce the ratio to compressibility factors. Since $\dot V_i = \dot n_i Z_i RT/P$, at common $T,P$: $$\frac{\dot V_{\text{C}_2\text{H}_4}}{\dot V_{\text{O}_2}} = \frac{\dot n_{\text{C}_2\text{H}_4}\,Z_{\text{C}_2\text{H}_4}}{\dot n_{\text{O}_2}\,Z_{\text{O}_2}} = \frac{2\,Z_{\text{C}_2\text{H}_4}}{Z_{\text{O}_2}}$$
2. Compressibility factors from Peng–Robinson. Solving the cubic at 298.15 K, 100 bar: $$Z_{\text{C}_2\text{H}_4} = 0.354\ (T_r=1.06,\ \text{near-critical, strongly non-ideal}), \qquad Z_{\text{O}_2} = 0.935\ (T_r=1.93,\ \text{nearly ideal})$$
3. Volumetric ratio. Combine: $$\frac{\dot V_{\text{C}_2\text{H}_4}}{\dot V_{\text{O}_2}} = \frac{2(0.354)}{0.935} = \boxed{0.76}$$ Although twice the moles of ethylene are fed, at 100 bar it sits just above its critical temperature and occupies only ~38% of the ideal molar volume, so it actually takes less volume than the oxygen — the required ratio is 0.76, not the ideal value of 2.