Volumetric Reserves: 2021 #14

A true vertical depth well log (all depths ss) shows a structure top at 7,980 ft and an OWC about 8,140 ft. The log also shows three porosity intervals 35 ft, 30 ft, and 30 ft thick, with porosity interval tops at 8,020 ft, 8,070 ft, and 8,100 ft, each separated by shale breaks. A true statement regarding calculating volumetric reserves in the above situation is (select any that apply):

This type of problem tests your knowledge of P2 and P3 reserves, logging data, and general oilfield knowledge. In the end, there is no shortcut to understanding the basics of logging data. So if it's not something you work with often, get familar with the applicable SPE Handbook and SPE Textbook material. The Guidebook has a good summary as well.

Diagenetic Porosity: 2017 #12

Porosity calculations are shown on a single GB page: 15 LOG 4. It starts:

Total Porosity: (measured with nuclear tools).
…equals primary porosity + secondary porosity.

Primary Porosity (apparent, intergranular).
Found from Wylie’s (acoustic) porosity equation below.

Secondary Porosity (isolated pores, vugs, and fractures).
also called diagenetic porosity.
…may be overlooked by acoustic-logs.
…equals total porosity – primary porosity.

At a glance, it's easy to see diagenetic porosity is found by nuclear tool porosity minus sonic porosity.

In this problem, nuclear tool porosity is given: 20 pu, and a sonic tool slowness of 79 microseconds/ft over the zone. The known slowness in sandstone & oil (shown in the GB variable box) are 55.5 & 232 microseconds/ft.

Wylie's Equation is next in the GB:
Por sonic = (dt - dtma)/(dtf - dtma) = (79 - 55.5)/(232 - 55.5) = 13 pu.

Since diagenetic = secondary = total - primary: 20 - 13 = 7 pu.

Diagenetic Porosity: 2016 #12

An oil reservoir has average diagenetic porosity of 10% and a CNL log measures a porosity of 15%. The oil reservoir is 200 acres, 10 ft thick, with residual oil saturation & initial water saturation of 25% each. The initial oil formation volume factor is 1.2 RB/STB. The maximum oil production from primary porosity would be closest to: (A) 320 MSTB (B) 340 MSTB (C) 620 MSTB (D) 640 MSTB

Definitions from 15 LOG 4:
Total Porosity: (measured with nuclear tools) …equals primary porosity + secondary porosity.
Primary Porosity (apparent, intergranular) used for reserves or maximum producible oil…equals total porosity – secondary porosity.
Secondary Porosity (isolated pores, vugs, and fractures) also called diagenetic porosity …may be overlooked by acoustic-logs …equals total porosity – primary porosity

Therefore, for this problem, primary porosity is: 0.15 – 0.10 = 0.05 pu. And maximum oi production from primary porosity is: 7758(200)10(1-.25-.25)0.05(1/1.2) = 323MSTB or (A).