Question 3 of 7: Separator–Laboratory Recombination

Question 3: Separator–Laboratory Recombination (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. GOR $=50{,}000$ SCF/STB; separator $B_o=1.3$ bbl/STB; separator oil density $\rho=40\ \text{lb}_m/\text{ft}^3$; separator oil molecular weight $M=90\ \text{lb}_m/\text{lb-mol}$; 1 lb-mol of gas $=379.4$ SCF at standard conditions; 1 bbl $=5.615\ \text{ft}^3$.

Find. Moles of gas per mole of separator liquid needed to recombine the collected gas and oil samples back to wellstream composition, per STB of stock-tank oil.

Approach. Work per 1 STB of stock-tank oil: convert the separator oil volume (given by $B_o$) to a mass via its density, convert mass to moles via $M$; convert the gas volume (given by GOR) to moles via the standard molar volume; the recombination ratio is the mole ratio of gas to liquid.

1. Separator-oil volume per STB. $V_{o} = B_o \times 5.615\ \text{ft}^3/\text{bbl} = 1.3\times5.615 = 7.300\ \text{ft}^3$ of separator liquid per STB.
2. Separator-oil mass and moles. $m_o=\rho V_o = 40\times7.300=292.0\ \text{lb}_m$. Moles of liquid: $n_{liq}=\dfrac{m_o}{M}=\dfrac{292.0}{90}$, so $\boxed{n_{liq}=3.244\ \text{lb-mol per STB}}$.
3. Separator-gas moles. The GOR gives the standard gas volume evolved per STB: $n_{gas}=\dfrac{\text{GOR}}{379.4}=\dfrac{50{,}000}{379.4}$, so $\boxed{n_{gas}=131.8\ \text{lb-mol per STB}}$.
4. Recombination ratio. $\dfrac{n_{gas}}{n_{liq}}=\dfrac{131.8}{3.244}$, giving $\boxed{40.6\ \text{mol gas per mol liquid}}$ — the ratio at which the separator gas and separator liquid samples must be mechanically recombined in the PVT cell to reconstitute the original wellstream.