Page 252 - Well Logging and Formation Evaluation

242 Well Logging and Formation Evaluation

CHAPTER 11: RESERVOIR ENGINEERING ISSUES

Exercise 11.1: Density of Air

In order to solve this we will use Equation 11.3, and calculate the mass
and volume of 1000 moles of gas. The mass is simply the molecular
weight, i.e., 29kg. At standard conditions, Z = 1. The volume is given by:

V = 1000*8.31*288.5 (1.01325*10 5 ) = 23.66 m 3

Hence the density = 29/23.66, or 1.226kg/m 3

Exercise 11.2: Material Balance of Undersaturated Oil Reservoir

From equation 11.10 the composite compressibility is given by:

-5
.
C = (44 10 * 02 +10 10 -5 ) 08 136 10 1. = . * -4 bar
*
.
*
From Equation 11.11:
)]
7
7
-4
.
(1 3
.
*
.
.
*
N p = [10 *. -1 25 )+10 * 1 3 1 36 10 * (300 - 250 1 25
= 471 10. * 5 m 3
Exercise 11.3: Radial Flow
Just apply Equation 11.19, making sure the units are correct. Hence:
2
0
k = f m**Ck = . * (10 *10 -3 ) ( * 2 *10 -4 *10 -5 ) ( . *0 3 10 -12 ) = 13 .3
p
(
k
h
t = r 2 ( * 14 )* gk *exp * * * * DPQ m))
( *
4
*
2
= ( .0 15 ) * (1 4 ) ( * .1 781 ) ( * 13 .3 )*exp 4 ([ * .14 * . *10 -12 * *30) *
3
0
3
( 30 *10 ) ( 200 24 *60 60) *10 *10 )]
5
3
-
(
*
= 306,934secs (3.56 days)