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24-Pet-A2 Petroleum Reservoir Fluids · Undated paper

Question 6 of 6: Gas Stream z-Factor, Density and Formation Volume Factor

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Notes on this paper

EGBC National Exam — Petroleum Engineering, 17-Pet-A2 Petroleum Reservoir Fluids, National Examinations May 2019. 3 hours duration, closed book, formula sheet supplied, personal scientific calculator permitted. SIX questions are printed on the paper; per the exam's own instructions, any FIVE constitute a complete answer paper. Every question is solved in full below (all six, not just the five a candidate would normally submit) so this set also serves as complete study material.

Reference texts: McCain, W.D., The Properties of Petroleum Fluids, 3rd ed. (PennWell); Ahmed, T., Reservoir Engineering Handbook, 5th ed.; Standing, M.B., Volumetric and Phase Behavior of Oil Field Hydrocarbon Systems; Craft, B.C. & Hawkins, M.F., Applied Petroleum Reservoir Engineering, 3rd ed.

Check: two graphical-correlation questions (4B's Carr et al. gas-viscosity chart, and 6a's Standing-Katz z-factor chart) are answered using the standard published Lee-Gonzalez-Eakin (1966) gas-viscosity correlation and the Dranchuk & Abou-Kassem (1975) z-factor equation of state respectively – both are the accepted numerical proxies for reading these two charts precisely, and are flagged again at each point of use.

Question 6: Gas Stream z-Factor, Density and Formation Volume Factor (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.

Gas composition and critical properties
ComponentyiTc (°F)Pc (psia)M
Methane0.82−116.5673.116.04
Ethane0.0590.09708.330.07
Propane0.04206.26617.444.10
n-Butane0.03305.62550.758.12
CO20.0388.0107344.01
N20.03−232.849228.01

Pres = 2400 psia, Tres = 180°F (639.67°R), gas rate = 1.5 MMSCFD, Vsc = 379.4 SCF/lb-mol.

Find. (a) z; (b) Mav; (c) ρg; (d) lb-mol/day; (e) lb/day; (f) specific volume; (g) Bg in reservoir bbl/SCF.

Approach. Build pseudo-critical properties with Kay's mixing rule from the given component data, get Tr, Pr, read z (here, via the Dranchuk & Abou-Kassem correlation as the verifiable proxy for the Standing-Katz chart), then work through the real-gas-law quantities in sequence.

Check: part (a) explicitly calls for the Standing-Katz graphical z-factor correlation. The numeric value below uses the Dranchuk & Abou-Kassem (1975) equation of state, the standard analytical fit to the same Standing-Katz data (accurate to within about 1% over this Tr, Pr range) – the accepted numerical substitute for this chart.
  1. (b) Apparent molecular weight via Kay's rule, $M_{av}=\sum y_iM_i$: $$M_{av}=0.82(16.04)+0.05(30.07)+0.04(44.10)+0.03(58.12)+0.03(44.01)+0.03(28.01)=\boxed{20.32\text{ lb/lb-mol}}$$
  2. Pseudo-critical properties (Kay's mixing rule, Tc converted °F→°R by $+459.67$): $$T_{pc}=\sum y_iT_{ci}=381.7^\circ\text{R},\qquad P_{pc}=\sum y_iP_{ci}=675.5\text{ psia}$$
  3. Reduced conditions. $$T_r=\frac{639.67}{381.7}=1.676,\qquad P_r=\frac{2400}{675.5}=3.553$$
  4. (a) z-factor from the Dranchuk & Abou-Kassem correlation at this Tr, Pr: $$z\approx\boxed{0.853}$$
  5. (c) Gas density via $\rho=pM/(zRT)$: $$\rho_g=\frac{(2400)(20.32)}{(0.853)(10.732)(639.67)}=\boxed{8.33\text{ lb/ft}^3}$$
  6. (d) Daily moles produced via $V_{sc}=n\times379.4$: $$n=\frac{1{,}500{,}000}{379.4}=\boxed{3954\text{ lb-mol/day}}$$
  7. (e) Daily mass produced. $$\dot m=n\times M_{av}=3954\times20.32=\boxed{80{,}355\text{ lb/day}}$$
  8. (f) Specific volume (reciprocal of density): $$v=\frac{1}{\rho_g}=\frac{1}{8.33}=\boxed{0.1201\text{ ft}^3/\text{lb}}$$
  9. (g) Gas formation volume factor. From the real gas law, 1 SCF of gas at standard conditions ($p_{sc}\approx14.7$ psia, $T_{sc}\approx520^\circ$R, $z_{sc}\approx1$) occupies, at reservoir conditions: $$B_g=\frac{V_{res}}{V_{sc}}=\frac{z\,T/p}{z_{sc}T_{sc}/p_{sc}}=0.02827\frac{zT}{p}\ \left[\frac{\text{ft}^3}{\text{SCF}}\right]$$ $$B_g=0.02827\times\frac{(0.853)(639.67)}{2400}=0.006428\ \text{ft}^3/\text{SCF}$$ Converting to barrels ($1\text{ bbl}=5.615\text{ ft}^3$): $$B_g=\frac{0.006428}{5.615}=\boxed{0.001145\text{ reservoir bbl/SCF}}$$
Question 6 – results
QuantityValue
(a) z-factor0.853
(b) Apparent MW20.32 lb/lb-mol
(c) Gas density8.33 lb/ft³
(d) Moles produced3954 lb-mol/day
(e) Mass produced80,355 lb/day
(f) Specific volume0.1201 ft³/lb
(g) Bg0.001145 rb/SCF (0.006428 ft³/SCF)
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