fundamentals of reservoir fluid flow (ei function and pd) sample exercises 3

FUNDAMENTALS OFRESERVOIR FLUID FLOW

Unsteady-State

Solutions of Diffusivity Equation

FUNDAMENTALS OFRESERVOIR FLUID FLOW

Unsteady-State

Solutions of Diffusivity Equation

Ali F. M. Alta’ee

Reservoir Engineering II (PCB2053)

Example 1An oil well is producing at a constant flow rate of 300 STB/day underunsteady-state flow conditions. The reservoir has the following rock and fluid properties:

1. Calculate pressure at radii of 0.25, 5, 10, 50, 100, 500, 1000, 1500, 2000, and 2500 feet, for 1 hour.Plot the results as:

A. Pressure versus logarithm of radiusB. Pressure versus radius

2. Repeat part 1 for t = 12 hours and 24 hours. Plot the results as pressure versus logarithm of radius.

Solution• Step 1. From Equation (78) :

• Step 2. Perform the required calculations after one hour in the following tabulated form:

0.01 < x < 3.0, use Ei chart

x >10.9, Ei = zero

x < 0.01

• Step 3. Show results of the calculation graphically as illustrated in Figures:

Pressure profiles as a function of time

Pressure profiles as a function of time on a semi-log scale

• Step 4. Repeat the calculation for t = 12 and 24 hrs

6

An oil well is producing at a constant flow rate of 300 STB/day under unsteady-state flow conditions. Estimate the bottom-hole flowing pressure after 10 hours of production.The reservoir has the following rock and fluid properties:

Solution

Equation 6-83 can be used to calculate pwf only if the time exceeds the time limit

Since the specified time of 10 hr is greater than 0.000267 hrs, the pwf can then be estimated by Equation 6-83.

Example 2

E x a m p l e 3 . A w e l l a n d r e s e r v o i r a r e d e s c r i b e d b y th e f o l lo w i n g d a t a : D a t a p o r o s i t y , 1 9 % f o r m a t i o n v o l u m e f a c t o r f o r o i l , B

o 1 .4 r m 3 / s t m 3

n e t t h i c k n e s s o f f o r m a t i o n , h 1 0 0 m v i s c o s i t y o f r e s e r v o i r o i l , 1 .4 x 1 0 - 3 P a s c o m p r e s s i b i l i t y , c 2 .2 x 1 0 - 9 P a - 1 p e r m e a b i l i t y , k 1 0 0 m D w e l l b o r e r a d i u s , r

w 0 .1 5 m

e x t e rn a l r a d i u s , re 9 0 0 m

i n i t i a l r e s e rv o i r p r e s s u r e , Pi 4 0 0 b a r

w e l l f l o w r a t e ( c o n s t a n t ) 1 5 9 s t m 3 / d a y = 1 5 9

2 4 x 3 6 0 0s t m 3 / s e c o n d

s k i n f a c t o r 0 D e t e r m i n e t h e f o l l o w i n g : 1 ) t h e w e l l b o re f l o w i n g p r e s s u r e a f t e r 4 h o u r s p r o d u c t i o n 2 ) t h e p r e s s u r e i n t h e r e s e rv o i r a t a r a d i u s o f 9 m a f t e r 4 h o u r s p ro d u c t io n 3 ) t h e p r e s s u r e i n t h e r e s e rv o i r a t a r a d i u s o f 5 0 m a f t e r 4 h o u r s p r o d u c t i o n 4 ) t h e p r e s s u r e i n t h e r e s e rv o i r a t a r a d i u s o f 5 0 m a f t e r 5 0 h o u r s p ro d u c t i o n

Solution The line source solution is used to determine the pressures required at the specified radii and at the specified times (i.e. using the flowrate measured at the wellbore, the pressures at the other radii and times are calculated by the line source solution). SI units will be used so time will be converted to seconds. Checks are made to ensure that: i) there has been adequate time since the start of production to allow the line source solution to be accurate ii) the reservoir is infinite acting. Thereafter, the choice of Ei function or ln approximation to the Ei function has to be made.

A C h e c k E i a p p l i c a b i l i t y l i n e s o u r c e n o t a c c u r a t e u n t i l

t 1 0 0 c r w

2

k

t 1 0 0 x 0 . 1 9 x 1 . 4 x 1 0 - 3 x 2 . 2 x 1 0 9 x 0 . 1 5 2

1 0 0 x 1 0 - 1 5

t > 1 3 . 2 s t i m e i s 4 h o u r s , t h e r e f o r e l i n e s o u r c e i s a p p l i c a b l e .

B C h e c k r e s e r v o i r i s i n f i n i t e a c t i n g

t h e r e s e r v o i r i s i n f i n i t e a c t i n g i f t h e t i m e , t c r e

2

4 k

i . e . t 0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 x 9 0 0 2

4 x 1 0 0 x 1 0 - 1 5

t < 1 1 8 5 0 3 0 s t < 3 2 9 h o u r s t h e r e f o r e l i n e s o u r c e s o l u t i o n i s a p p l i c a b l e .

1 ) t h e b o t t o m h o l e f l o w i n g p r e s s u r e a f t e r 4 h o u r s p r o d u c t i o n , P w f a t 4 h o u r s i ) c h e c k l n a p p r o x i m a t i o n t o E i f u n c t i o n

t h e l n a p p r o x i m a t i o n i s v a l i d i f t h e t i m e , t 2 5 c r w

2

k

t 2 5 x 0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 x 0 . 1 5 2

1 0 0 x 1 0 - 1 5

t > 3 . 3 s t h e r e f o r e l n a p p r o x i m a t i o n i s v a l i d .

i i ) P w f P i q B o

4 khl n

c r w2

4k t

( t a k i n g a c c o u n t o f t h e c o n v e r s i o n f r o m s t o c k t a n k t o

r e s e r v o i r c o n d i t i o n s v i a t h e f o r m a t i o n v o l u m e f a c t o r f o r o i l , B o , f l o w r a t e s i n r e s e r v o i r m 3 / s a n d p r e s s u r e s i n P a s c a l ) .

q B o

4 k h

1 5 9 x 1 . 4 x 1 0 3 x 1 . 4

2 4 x 3 6 0 0 x 4 x 1 0 0 x 1 0 1 5 x 1 0 0 = 2 8 7 0 3

c r 2

4 k t

0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 r 2

4 x 1 0 0 x 1 0 - 1 5 x 4 x 3 6 0 0 = 1 0 1 5 9 7 x 1 0 - 9 r 2

P w f = 4 0 0 x 1 0 5 + 2 8 7 0 3 x l n ( 1 . 7 8 1 x 1 0 1 5 9 7 x 1 0 - 9 x 0 . 1 5 2 ) = 4 0 0 x 1 0 5 - 3 5 6 2 4 9 = 3 9 6 4 3 7 5 1 P a = 3 9 6 . 4 b a r

2 ) t h e p r e s s u r e a f t e r 4 h o u r s p r o d u c t i o n a t a r a d i u s o f 9 m f r o m t h e w e l l b o r e i ) c h e c k l n a p p r o x i m a t i o n t o E i f u n c t i o n

t h e l n a p p r o x i m a t i o n i s v a l i d i f t h e t i m e , t 2 5 c r 2

k

t 2 5 x 0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 x 9 2

1 0 0 x 1 0 - 1 5

t > 1 1 8 5 0 s t > 3 . 3 h o u r s t h e r e f o r e l n a p p r o x i m a t i o n i s v a l i d .

i i ) P P i q B o

4 k hl n

c r 2

4 k t

( t a k i n g a c c o u n t o f t h e c o n v e r s i o n f r o m s t o c k t a n k t o

r e s e r v o i r c o n d i t i o n s v i a t h e f o r m a t i o n v o l u m e f a c t o r f o r o i l , B o

a n d a l s o t h e f a c t t h a t t h e r a d i u s , r , i s n o w a t 9 m f r o m t h e w e l l b o r e ) .

q B o

4 k h

1 5 9 x 1 . 4 x 1 0 3 x 1 . 4

2 4 x 3 6 0 0 x 4 x 1 0 0 x 1 0 1 5 x 1 0 0 = 2 8 7 0 3

c r 2

4 k t

0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 r 2

4 x 1 0 0 x 1 0 - 1 5 x 4 x 3 6 0 0 = 1 0 1 5 9 7 x 1 0 - 9 r 2

P = 4 0 0 x 1 0 5 + 2 8 7 0 3 x l n ( 1 . 7 8 1 x 1 0 1 5 9 7 x 1 0 - 9 x 9 2 ) = 4 0 0 x 1 0 5 - 1 2 1 2 0 9 = 3 9 8 7 8 7 9 1 P a = 3 9 8 . 8 b a r

3 ) t h e p r e s s u r e a f t e r 4 h o u r s p r o d u c t i o n a t a r a d i u s o f 5 0 m f r o m t h e w e l l b o r e i ) c h e c k l n a p p r o x i m a t i o n t o E i f u n c t i o n

t h e l n a p p r o x i m a t i o n i s v a l i d i f t h e t i m e , t 2 5 c r 2

k

t 2 5 x 0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 x 5 0 2

1 0 0 x 1 0 - 1 5

t > 3 6 5 7 5 0 s t > 1 0 1 . 6 h o u r s t h e r e f o r e l n a p p r o x i m a t i o n i s n o t v a l i d a n d t h e E i f u n c t i o n i s u s e d .

i i ) P P i q B o

4 k hE i

cr 2

4 k t

( t a k i n g a c c o u n t o f t h e c o n v e r s i o n f r o m s t o c k t a n k t o

r e s e r v o i r c o n d i t i o n s v i a t h e f o r m a t i o n v o l u m e f a c t o r f o r o i l , B o

a n d a l s o t h e f a c t t h a t t h e r a d i u s , r , i s n o w a t 5 0 m f r o m t h e w e l l b o r e ) .

q B o

4 k h

1 5 9 x 1 . 4 x 1 0 3 x 1 . 4

2 4 x 3 6 0 0 x 4 x 1 0 0 x 1 0 1 5 x 1 0 0 = 2 8 7 0 3

c r 2

4 k t

0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 5 0 2

4 x 1 0 0 x 1 0 - 1 5 x 4 x 3 6 0 0 = 0 . 2 5 4

P = 4 0 0 x 1 0 5 + 2 8 7 0 3 x E i ( - 0 . 2 5 4 ) E i ( - 0 . 2 5 4 ) = - 1 . 0 3 2 ( b y l i n e a r i n t e r p o l a t i o n o f t h e v a l u e s i n T a b l e 4 ) P = 4 0 0 x 1 0 5 + 2 8 7 0 3 x - 1 . 0 3 2 = 4 0 0 x 1 0 5 - 2 9 6 2 2 = 3 9 9 7 0 3 7 8 P a = 3 9 9 . 7 b a r

4 ) t h e p r e s s u r e a f t e r 5 0 h o u r s p r o d u c t i o n a t a r a d i u s o f 5 0 m f r o m t h e w e l l b o r e i ) c h e c k l n a p p r o x i m a t i o n t o E i f u n c t i o n

t h e l n a p p r o x i m a t i o n i s v a l i d i f t h e t i m e , t 2 5 c r 2

k

t 2 5 x 0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 x 5 0 2

1 0 0 x 1 0 - 1 5

t > 3 6 5 7 5 0 s t > 1 0 1 . 6 h o u r s t h e r e f o r e l n a p p r o x i m a t i o n i s n o t v a l i d a n d t h e E i f u n c t i o n i s u s e d .

i i ) P P i q B o

4 k hE i

cr 2

4 k t

( t a k i n g a c c o u n t o f t h e c o n v e r s i o n f r o m s t o c k t a n k t o

r e s e r v o i r c o n d i t i o n s v i a t h e f o r m a t i o n v o l u m e f a c t o r f o r o i l , B o

a n d a l s o t h e f a c t t h a t t h e r a d i u s , r , i s n o w a t 5 0 m f r o m t h e w e l l b o r e a n d t h e t i m e i s n o w 5 0 h o u r s a f t e r s t a r t o f p r o d u c t i o n ) .

q B o

4 k h

1 5 9 x 1 . 4 x 1 0 3 x 1 . 4

2 4 x 3 6 0 0 x 4 x 1 0 0 x 1 0 1 5 x 1 0 0 = 2 8 7 0 3

c r 2

4 k t

0 . 1 9 x 1 . 4 x 1 0 3 x 2 . 2 x 1 0 9 5 0 2

4 x 1 0 0 x 1 0 - 1 5 x 5 0 x 3 6 0 0 = 0 . 0 2 0

P = 4 0 0 x 1 0 5 + 2 8 7 0 3 x E i ( - 0 . 0 2 0 ) E i ( - 0 . 0 2 0 ) = - 3 . 3 5 5 P = 4 0 0 x 1 0 5 + 2 8 7 0 3 x - 3 . 3 5 5 = 4 0 0 x 1 0 5 - 9 6 3 0 0 = 3 9 9 0 3 7 0 0 P a = 3 9 9 . 0 b a r

time radius pressure (hours) (m) (bar)

0 all 400.0 4 0.15 396.4 4 9.00 398.8 4 50.00 399.7

50 50.00 399.0

Summary