Compressibility is a relative volume change of a fluid or solid in a response to a pressure change. We can relate this into a reservoir engineering aspect. Overburden pressure is rock weight and it typically has a gradient of 1 psi/ft. Rock metric and formation fluid in pore spaces supports the weight of rock above. When petroleum is produced from reservoir rocks, pressure of fluid in pore space decreases, but overburden is still the same. This will result in the reduction of bulk volume of rock and pore spaces. The reduction on volume in relation to pressure is called “pore volume compressibility (cf)” or “formation compressibility” and it can be mathematically expressed like this.

Where

Vp = pore volume

dVp = change in volume

dp = change in pressure, psi

cf = rock compressibility, 1/psi

Note: The actual measurement of rock compressibility is expensive and it is required to have a formation sample. In practical, utilizing Hall correlation to determine rock compressibility is acceptable.

Hall’s rock compressibility correlation is a function only of porosity. The correlation is based on laboratory data and is considered reasonable for normally pressured sandstones.

http://infohost.nmt.edu/~petro/faculty/Engler524/PET524-1c-porosity.pdf

Rock compressibility factor is very important for reservoir modelling.

cf is typically in the range from 3 x 10^{-6} to 6 x 10^{-6 }1/psi.

**Example**

Use the following data to determine volume change in reservoir rock per 100 psi of pressure drop.

Reservoir area = 2,000,000 square feet

Porosity = 15%

Rock compressibility = 3 x 10^{-6 }1/psi

Formation thickness = 150 ft

**Solution**

Reservoir rock volume = 2,000,000 x 150 = 300 x 10^{6 }square feet

Vp = reservoir rock volume x porosity

Vp = 300 x 10^{6 }x 0.15 = 45 x 10^{6 }ft^{3}

d Vp /dp = cf × Vp

d Vp /dp = 3 x 10^{-6 }x 45 x 10^{6} = 135 ft^{3}/psi

dp = 100 psi

d Vp = 13,500 ft^{3}

% change in reservoir pore volume @ 100 psi decline = dVp ÷ Vp =13,500 ÷ 45 x 10^{6 }= 0.03 %

**References**