Page 217 - Fundamentals of Reservoir Engineering

OILWELL TESTING 155

2) What will be the pressure drop at the well after flowing at the steady rate of 400 stb/d
for 3 hours, assuming transient conditions still prevail.

EXERCISE 7.1 SOLUTION

Converting the reservoir and fluid properties to Darcy units.
k = 0.05 D

h = 30 ft × 30.48 = 914.4 cm

2
r w 2 = 0.25 sq.ft. = 0.25 × (30.48) cm 2
= 232.3 cm 2
-6
-6
c = 10 × 10 /psi = 10 × 10 × 14.7/atm
-5
= 14.7 × 10 /atm
q = 400stb/d = 400× 1.25× 1.84 r.cc/sec = 920 r.cc/sec.

The approximation ei(x) ≈ − In(γx) applies for x < 0.01 i.e.

φµ cr w 2 < 0.01
4k t
or

φµ cr 2
t > w
0.04k
5
−
××
.3 314.710 × 232.3
×
t >
.04 .05
×
for t > 15.4 seconds
Now in a practical sense, nobody is concerned with what happens in the well during the
first 15 seconds, after which the pressure decline can be calculated using the
logarithmic approximation for ei(x) i.e.

2
qµ φµ cr
p − p wf = ei w
i
4kh 4k t
π
qµ 4k t
= ln 2
4kh γ φ µ cr w
π
After producing for 3 hours at a steady rate of 400 stb/d the pressure drop at the
wellbore is

9203 4 .0533600 10 5
×
×
×
×
×
p − p wf = ln
i
×
×
4π × .05 914.4 1.781 .3 3 14.7 232.3
×
×
×
= 50.8atm, or 747psi