CO 2 -Saturated Formation Water — Mass Fraction and Water FVF: 24-Pet-A2 Petroleum Reservoir Fluids, December 2018, Q2

Question 2: CO₂-Saturated Formation Water — Mass Fraction and Water FVF (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. Basis: 1 Sm$^3$ formation water with 20 Sm$^3$ CO$_2$ dissolved in it (20 Sm$^3$/Sm$^3$).

Quantity	Value
Dissolved CO$_2$-to-water ratio	20 Sm$^3$/Sm$^3$
Water density, standard conditions, $\rho_{w,sc}$	1000 kg/m$^3$
Water density, reservoir conditions, $\rho_{w,res}$	900 kg/m$^3$
Molecular weight, CO$_2$ / water	44 / 18 g/mol
Molar volume at standard conditions, $V_m$	22.4 L/mol

Find. (a) Mass fraction of CO$_2$ in the water, $w_{CO_2}$; (b) water formation volume factor $B_w$ at reservoir conditions.

Approach. Convert the standard-condition CO$_2$ volume to moles, then mass, using the ideal molar volume; divide by the total mass to get the mass fraction. For $B_w$, recognize that the mass of water plus its dissolved CO$_2$ is conserved between standard and reservoir conditions, so $V=m/\rho$ gives the reservoir volume directly, and $B_w$ is that reservoir volume per unit standard water volume.

Convert the dissolved CO₂ volume to mass. On the 1 Sm$^3$ water basis, the dissolved gas is $20\ \text{Sm}^3=20{,}000\ \text{L}$ at standard conditions. Moles of CO$_2$: $n_{CO_2}=\dfrac{20{,}000\ \text{L}}{22.4\ \text{L/mol}}=892.86\ \text{mol}$. Mass: $m_{CO_2}=n_{CO_2}\times M_{CO_2}=892.86\times44=39{,}286\ \text{g}=\boxed{39.29\ \text{kg}}$.
Part (a) — mass fraction of CO₂. The water mass on the same 1 Sm$^3$ basis is $m_w=\rho_{w,sc}\times V=1000\ \text{kg/m}^3\times1\ \text{m}^3=1000\ \text{kg}$. The mass fraction is $w_{CO_2}=\dfrac{m_{CO_2}}{m_{CO_2}+m_w}=\dfrac{39.29}{39.29+1000}=\dfrac{39.29}{1039.29}$, so $\boxed{w_{CO_2}\approx0.0378}$ (3.78 wt%).
Part (b) — formation volume factor of water. No CO$_2$ leaves solution between standard and reservoir conditions (the water stays fully saturated), so the total mass carried into the reservoir volume is the same $m_{tot}=m_{CO_2}+m_w=1039.29\ \text{kg}$. Using $V=m/\rho$ at the reservoir density: $V_{w,res}=\dfrac{1039.29\ \text{kg}}{900\ \text{kg/m}^3}=1.1548\ \text{m}^3$. The formation volume factor is this reservoir volume per unit standard water volume (the 1 Sm$^3$ basis): $B_w=\dfrac{V_{w,res}}{V_{w,sc}}=\dfrac{1.1548}{1}$, so $\boxed{B_w\approx1.1548\ \text{rm}^3/\text{Sm}^3}$.

Quantity	Value
(a) Mass fraction of CO$_2$, $w_{CO_2}$	0.0378 (3.78 wt%)
(b) Water formation volume factor, $B_w$	1.1548 rm$^3$/Sm$^3$

Topic: Gas solubility in formation water and the water formation volume factor.

Key relations: $n=V/V_m$ (ideal molar volume); $w_i=m_i/\sum m$; $V=m/\rho$; $B_w=V_{w,res}/V_{w,sc}$.

Why this works: Because no gas is lost or added between the standard-condition reference state and the reservoir state, the SAME total mass simply occupies a different volume at the reservoir density — $B_w$ is nothing more than that mass-conservation statement written as a volume ratio, exactly as the hint's $V=m/\rho$ suggests.

Common pitfall: Computing $B_w$ from the water density ratio alone ($\rho_{sc}/\rho_{res}=1.111$) while forgetting that the dissolved CO$_2$ mass must be carried along with the water mass into the reservoir-condition volume; omitting it understates $B_w$.

Source: McCain, The Properties of Petroleum Fluids, 3rd ed., Ch. 8 (formation water properties); Ahmed, Reservoir Engineering Handbook, 5th ed., Ch. 2.

Question 2 of 7: CO 2 -Saturated Formation Water — Mass Fraction and Water FVF

Question 2: CO2-Saturated Formation Water — Mass Fraction and Water FVF (20 marks)

Question 2: CO₂-Saturated Formation Water — Mass Fraction and Water FVF (20 marks)