Page 412 - Petrophysics 2E

380 PETROPHYSICS: RESERVOIR ROCK PROPERTIES

Returning to Equation 6.24 and noting that water is essentially
incompressible leads to:

6F
-=V,orF2 - F1=V(pz - pl)=VPc (6.25)
6V

Therefore, for the transfer of a unit volume of water, the capillary
pressure, Pc, represents the change in the isothermal, reversible work
accompanying the process, or:

6F
pc = E (6.26)

If the element of volume of the porous system containing water and
oil is such that it contains a unit pore volume of water, the fractional
water saturation within the element multiplied by the pore volume is
numerically equal to the volume of water. Thus, dV = -Vp x dS,
where V represents the water transferred out of the porous medium.
Substituting this expression into Equation 6.26 yields:

dF= -Pc x Vp x dS, (6.27)

In Equation 6.27, dF, expressed in Nm or joules, represents the free
energy change of water per unit of pore space accompanying a change
of water saturation, dS,. The integral of Equation 6.27 is the area under
the capillary pressure curve (Figure 6.7). The capillary pressure curves
can be fit to hyperbolic equations by a least-squares fit of the capillary
pressure versus saturation data:

Pc=(l +Ax S,)/(B+C x S,) (6.28)

Area= ( BS, AB - C
A+BxS,
1+cxs, ) xdS,=-+ C C2 log (1 + C&) (6.29)

where constants A, B, and C are obtained from the least-squares fit of the
data.
The areas under the capillary pressure curves (Figure 6.7) can be
readily calculated by integration of Equation 6.30 below to yield the