Question 4 of 7: Actual and Standard Gas Flow Rates

Question 4: Actual and Standard Gas Flow Rates (10 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.

Approach. The mass (and molar) flow rate is fixed by continuity regardless of which pressure/temperature it is evaluated at — divide by the plant-condition density from Question 3(b) for part (a), and re-express the same molar flow rate through a field-units standard molar volume for part (b).

1. (a) Actual volumetric flow rate at plant conditions. $$Q_{actual}=\frac{\dot m}{\rho_b}=\frac{33.34\ \text{kg/s}}{187.5\ \text{kg/m}^3}=\boxed{0.1778\ \text{m}^3/\text{s}}\ \ (=15{,}358\ \text{m}^3/\text{d})$$ Cross-check via the molar route: $Q_{actual}=\dot n\,V_{m,b}=(118{,}230\ \text{kmol/d})(0.1299\ \text{m}^3/\text{kmol})/86{,}400\ \text{s/d}=0.1778\ \text{m}^3/\text{s}$ — matches.
2. (b) Standard flow rate at 60°F/14.69 psia. Convert the same molar flow rate to field units and apply the ideal-gas standard molar volume ($Z_{sc}\approx1$): $$\dot n=118{,}230\ \text{kmol/d}\times2.2046=260{,}653\ \text{lbmol/d}$$ $$V_{m,sc}=\frac{RT_{sc}}{p_{sc}}=\frac{10.732(519.67)}{14.69}=379.7\ \text{ft}^3/\text{lbmol}$$ $$Q_{sc}=\dot n\,V_{m,sc}=260{,}653(379.7)=9.90\times10^{7}\ \text{scf/d}=\boxed{98.96\ \text{MMSCFD}}$$