24-Pet-A2 Petroleum Reservoir Fluids · December 2016
Question 6 of 7: Natural Gas Properties — MW, Gravity, Density, and Gas FVF
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Pet-A2 — Petroleum Reservoir Fluids · National Exams, December 2016 · 3 hours, closed book, non-communicating calculator only · first five questions in the answer book are marked, all questions equal value, all parts of a multipart question equal weight.
Reference texts: Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed. (Ch. 1–2, PVT properties, reservoir/well-stream classification, material balance); Lyons, W.C. (ed.), Standard Handbook of Petroleum and Natural Gas Engineering, 3rd ed. (Standing–Katz Z-factor correlation, gas properties); McCain, W.D., The Properties of Petroleum Fluids, 3rd ed. (phase behaviour, black-oil PVT laboratory data); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (reservoir fluid classification, material balance, well-stream gravity); Danesh, A., PVT and Phase Behaviour of Petroleum Reservoir Fluids (equilibrium K-value flash calculations).
Question 6: Natural Gas Properties — MW, Gravity, Density, and Gas FVF (20 marks)
Check: the printed table (quoted exactly above) swaps two molecular weights. It lists carbon dioxide as 34.08 and hydrogen sulfide as 44.01, but the true values are CO$_2$ = 44.01 and H$_2$S = 34.08, the same values the Question 4 table on this paper prints. The correct molecular weights are used below. Using the table literally would give $M_{av}=19.99$ instead of 20.01 lb$_m$/lb-mol, a 0.08% difference that does not change $\gamma_g$, $Z$, $\rho$ or $B_g$ at the reported precision.
Check: the composition carries 0.84% CO$_2$ + 0.68% H$_2$S (1.52% total acid gas) — enough that a Wichert–Aziz sour-gas correction to the pseudo-criticals is sometimes applied in practice. The paper's own formula sheet supplies only the plain Standing pseudo-critical correlation (no acid-gas correction term), so that is what is used here, consistent with a closed-book exam limited to its own provided formulas.
Given. Composition table above (mole fractions sum to 1.0000; CO$_2$/H$_2$S molecular weights taken as 44.01/34.08, see callout); $p=2000$ psia; $T=100\,{}^{\circ}\text{F}=559.67\,{}^{\circ}\text{R}$. Formula sheet: $T_{pc}=168+325\gamma_g-12.5\gamma_g^2$, $p_{pc}=677+15.0\gamma_g-37.5\gamma_g^2$; $M_{av}=\sum y_iM_i$; $\rho=\dfrac{pM}{ZRT}$, $R=10.732$ psi-ft$^3$/(lb-mol-$^{\circ}$R); $B_g=0.02827\dfrac{ZT}{p}$ (ft$^3$/SCF).
Find. Apparent molecular weight $M_{av}$, gas specific gravity $\gamma_g$, gas density $\rho$, and gas formation volume factor $B_g$ at 2000 psia, 100°F.
Approach. Build the apparent molecular weight and specific gravity from composition, form the pseudo-critical properties (Standing) and hence $T_r,p_r$, solve the equivalent Standing–Katz $Z$-factor numerically (Dranchuk–Abou-Kassem form of the same chart), then evaluate density and $B_g$ from the real-gas relations.
Apparent molecular weight and gravity. $M_{av}=\sum y_iM_i=0.0062(28.01)+0.0084(44.01)+0.0068(34.08)+0.8667(16.04)+0.0391(30.07)+0.0280(44.10)+0.0224(58.12)+0.0224(72.15)$. Summing (mole fractions total 1.0000, confirming the basis): $\boxed{M_{av}=20.01\ \text{lb}_m/\text{lb-mol}}$. Then $\gamma_g=M_{av}/28.97=20.01/28.97$, giving $\boxed{\gamma_g=0.6906}$.
Pseudo-critical properties and reduced conditions. $T_{pc}=168+325(0.6906)-12.5(0.6906)^2=168+224.44-5.96=386.5\,{}^{\circ}\text{R}$; $p_{pc}=677+15.0(0.6906)-37.5(0.6906)^2=677+10.36-17.89=669.5$ psia. $T_r=\dfrac{559.67}{386.5}=1.448$; $p_r=\dfrac{2000}{669.5}=2.987$.
Gas deviation factor $Z$. Solving the Standing–Katz correlation numerically at $(T_r,p_r)=(1.448,2.987)$ — the pressure is well above the pseudo-critical while $T_r$ is only moderately above 1, placing this point in the correlation's steep low-$Z$ region — gives $\boxed{Z=0.744}$.
Gas density. $\rho=\dfrac{pM}{ZRT}=\dfrac{(2000)(20.01)}{(0.744)(10.732)(559.67)}$. Denominator $=(0.744)(10.732)(559.67)=4471.0$; numerator $=40{,}020$. So $\boxed{\rho=8.95\ \text{lb}_m/\text{ft}^3}$.
