A 20 ft...homogeneous reservoir has a porosity of 10%...core is taken...(see problem)...cementation factor? (A) 1.95; (B) 2.00; (C) 2.05 ; (D) 2.10
Solution: 15 LOG 1
1. F = Ro/Rw = 42/.42 = 100
2. F = a/por^m = 0.9/0.10^m
3. (m)log(por) = log(a/F)
4. m = log(a/F)/log(por)
5. m = log(0.9/100)/log(.1)
6. m = -2.05/-1 = 2.05 (C).
On page 15Log1 the equation for m = log(1/F)/log(phi), should it be m = log(a/F)/log(phi) based on this solution?
ReplyDeleteMost of the time a = 1 so that's the standard equation; in the rare case you are given an a you can do the derivation as you did here.
ReplyDeleteThere are a lot of non-precise stuff in the Guidebook trying to make things manageable. All I've got is 150 pages trying to cram in the whole TS & HS plus more. Gotta cut corners. But I recommend folk to take notes over their Guidebook to their own experience level. I've seen GB with more handwriting than text :-).