Question

A gas reservoir has a bulk volume of 10000 ft\(^3\), a connate water saturation of 0.2, and a porosity of 0.2. No oil is present in the reservoir. The gas formation volume factor Bg is 0.005 RCF/SCF. The volume of gas in place (in SCF) is ________ x 10\(^4\). [RCF: Reservoir Cubic Feet; SCF: Standard Cubic Feet]

Hint:

Gas in place equals pore volume times (1 minus connate water saturation), divided by the gas formation volume factor Bg.

Answer 1

Correct Answer: 32

Step 1: Compute the pore volume of the reservoir:
The pore volume is the bulk volume multiplied by the porosity: \[ PV = V_b \times \phi = 10000 \times 0.2 = 2000 \text{ ft}^3 \]
Step 2: Compute the hydrocarbon, gas filled, pore volume:
Since connate water occupies a fraction Swc of the pore space and no oil is present, the remaining fraction, 1 minus Swc, is filled with gas at reservoir conditions: \[ HCPV = PV \times (1 - S_{wc}) = 2000 \times (1 - 0.2) = 2000 \times 0.8 = 1600 \text{ ft}^3 \] This 1600 ft\(^3\) is the volume of gas in the reservoir expressed in reservoir cubic feet, RCF.
Step 3: Convert the reservoir gas volume to standard conditions using Bg:
The gas formation volume factor Bg relates reservoir volume to standard volume as \[ B_g = \frac{V_{reservoir}}{V_{standard}} \] so the gas in place at standard conditions is \[ G = \frac{HCPV}{B_g} = \frac{1600}{0.005} = 320000 \text{ SCF} \]
Step 4: Express the answer in units of 10\(^4\) SCF:
Dividing 320000 SCF by 10\(^4\) gives \[ \frac{320000}{10000} = 32 \]
Final Answer:
\[ \boxed{32 \times 10^4 \text{ SCF}} \]