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24-Pet-A2 Petroleum Reservoir Fluids · December 2015

Question 3 of 7: Separator Test — Well-Stream Specific Gravity

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

98-Pet-A2 — Petroleum Reservoir Fluids · National Exams, December 2015 · 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. (black-oil PVT laboratory data, well-stream recombination); Ahmed, T., Reservoir Engineering Handbook, 5th ed. (p/Z material balance, well-stream gravity); Danesh, A., PVT and Phase Behaviour of Petroleum Reservoir Fluids (equilibrium K-value flash calculations).

Question 3: Separator Test — Well-Stream Specific Gravity (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. $\text{GOR}=40{,}000$ SCF/STB; oil API $=50$; separator gas $\gamma_g=0.6$; separator oil $M_o=144\ \text{lb}_m/\text{lb-mol}$; 1 STB water $\approx350\ \text{lb}_m$ (i.e. water specific gravity basis $=1.0$); 1 lb-mol gas $=379.4$ SCF at standard conditions.

Find. The specific gravity $\gamma_{ws}$ of the produced well-stream fluid (gas plus stock-tank oil recombined, air $=1$).

Approach. Work on the basis of 1 STB of stock-tank oil: convert the API gravity to oil specific gravity, use the water-mass basis to get the oil's mass and hence its moles; convert the GOR to moles of gas via the standard molar volume; the well-stream apparent molecular weight is total mass over total moles, and its specific gravity follows by dividing by air's molecular weight.

  1. Oil specific gravity and mass. $\gamma_o=\dfrac{141.5}{131.5+API}=\dfrac{141.5}{131.5+50}=\dfrac{141.5}{181.5}=0.7796$. Mass of 1 STB of oil $=\gamma_o\times(\text{mass of 1 STB water})=0.7796\times350$, so $\boxed{m_o=272.9\ \text{lb}_m}$ per STB.
  2. Moles of oil. $n_o=\dfrac{m_o}{M_o}=\dfrac{272.9}{144}$, giving $\boxed{n_o=1.895\ \text{lb-mol}}$ per STB.
  3. Moles and mass of gas. $n_g=\dfrac{\text{GOR}}{379.4}=\dfrac{40{,}000}{379.4}$, giving $\boxed{n_g=105.4\ \text{lb-mol}}$ per STB. Gas molecular weight $M_g=28.97\gamma_g=28.97(0.6)=17.38\ \text{lb}_m/\text{lb-mol}$, so $m_g=n_gM_g=105.4\times17.38=1832.6\ \text{lb}_m$.
  4. Well-stream apparent molecular weight and gravity. Total mass $=m_o+m_g=272.9+1832.6=2105.5\ \text{lb}_m$; total moles $=n_o+n_g=1.895+105.4=107.3\ \text{lb-mol}$. $M_{ws}=\dfrac{2105.5}{107.3}=19.62\ \text{lb}_m/\text{lb-mol}$. So $\gamma_{ws}=\dfrac{M_{ws}}{28.97}=\dfrac{19.62}{28.97}$, giving $\boxed{\gamma_{ws}=0.677}$.
QuantityValue
Oil specific gravity $\gamma_o$0.7796
Moles oil / gas per STB1.895 / 105.4 lb-mol
Well-stream apparent MW $M_{ws}$19.62 lb$_m$/lb-mol
Well-stream specific gravity $\gamma_{ws}$0.677