calculation procedure for exponent of capillary pressure curve from laboratory data

[hojbarnaz]

formulaes above need to be adjusted for drainage conditions that start from Sw=1. use S*=(Sw-Swir)/(1-Swir). You may need to iterate a bit on Swir for each data set - but you can use the lowest Sw in most cases. For sample 2 I used 0.28

No simply plots pc vs S* on a log-log plot seeking straight line trends ( for the exponents). We note some curvature at very high Sw - which again we tend to discount if "spin-up" a time offset effects impact experiment. Whats left is surprisingly OK. I get a slope of about -0.8 which implies a Lambda of 1.25 ( which is also reasonable)

I ended up fitting J*ScosT= 9.3*(S*)^-1/1.25 to the valid data. 2 out of 3 match quite well and the third is acceptable.
Sample 183291 plots up OK using Swir of 0.08.

dear itag i have few question on your idea as you have said

simply plots pc vs S* on a log-log plot seeking straight line trends ( for the exponents).

now my question is who should be plot on x-axis and y-axis
secondly how you deduce exponent from log log plot as said above by you

second question is
I ended up fitting J*ScosT= 9.3*(S*)^-1/1.25 to the valid data.

what term are J*ScosT
and from where and how you deduce the exponent of your equation i.e -1/1.25
kindly share your idea in detail , i have tried a lot but unable to arrive at

[elvirus]

Can any body help me solve for Corey's exponents?
Keo=0.204;
Kew=0.128;
Sw=[14.3 63.3 66.0 68.1 68.7 69.1 72.9 74.8 76.4 76.9 77.8 79.4]';
Kro=[1.0 0.12 0.0513 0.0161 0.0113 0.0094 0.0025 0.0011 0.0004 0.0002 0.0000 0.0000]';
Krw=[0.000 0.342 0.376 0.398 0.400 0.402 0.436 0.464 0.503 0.530 0.529 0.627]';