fluid flow

1. 1

2. 2
Lecture8

3. Learning Objectives
3
Having worked through this the students will be able
to:
Applications the Solutions of the Diffusivity Equation
Examples

4. 4

5. 5
𝑃 𝑟 𝑡 = 𝑃𝑖 + 70.6
𝑞 × 𝜇 × 𝐵
𝑘 × ℎ
𝐸𝑖 −
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2
0.001055 × 𝑘 × 𝑡

6. 6
x =
𝜑×𝜇×𝑐𝑡×𝑟2
0.00105×𝑘×𝑡
x< 0.01
𝑃 𝑟 𝑡 = 𝑃𝑖 + 70.6
𝑞 × 𝜇 × 𝐵
𝑘 × ℎ
𝐸𝑖 −
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2
0.001055 × 𝑘 × 𝑡

7. 7

8. 8
x =
𝜑×𝜇×𝑐𝑡×𝑟2
0.00105×𝑘×𝑡
x> 0.01
𝑃 𝑟 𝑡 = 𝑃𝑖 + 70.6
𝑞 × 𝜇 × 𝐵
𝑘 × ℎ
𝐸𝑖 −
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2
0.001055 × 𝑘 × 𝑡

9. 9

10. 10
𝑃(𝑟, 𝑡) = 𝑃𝑖 − 162.6
𝑞 × 𝜇 × 𝐵
𝑘ℎ
log(
𝑘 × 𝑡
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2
) − 3.23
𝑃 𝑟 𝑡 = 𝑃𝑖 + 70.6
𝑞 × 𝜇 × 𝐵
𝑘 × ℎ
𝐸𝑖 −
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2
0.001055 × 𝑘 × 𝑡
x =
𝜑×𝜇×𝑐𝑡×𝑟2
0.00105×𝑘×𝑡
x< 0.01

11. 11

12. 12
x< 0.01
𝑃(𝑟, 𝑡) = 𝑃𝑖 − 162.6
𝑞 × 𝜇 × 𝐵
𝑘ℎ
log(
𝑘 × 𝑡
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2) − 3.23
x =
𝜑×𝜇×𝑐𝑡×𝑟2
0.00105×𝑘×𝑡

13. 13

14. 14

15. 15

16. Applications the Solutions of the Diffusivity Equation in
Case (a): The Infinite Acting System(Infinite Acting Reservoir)
𝑃 𝑟 𝑡 = 𝑃𝑖 + 70.6
𝑞 × 𝜇 × 𝐵
𝑘 × ℎ
𝐸𝑖 −
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2
0.00105 × 𝑘 × 𝑡
x =
𝜑×𝜇×𝑐𝑡×𝑟2
0.00105×𝑘×𝑡
for x< 0.01
𝑃(𝑟, 𝑡) = 𝑃𝑖 − 162.6
𝑞 × 𝜇 × 𝐵
𝑘ℎ
log(
𝑘 × 𝑡
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2
) − 3.23

17. 17
X =
𝜑𝜇𝑐𝑡𝑟2
0.00105𝑘𝑡
for x greater than 0.01
𝑃 𝑟 𝑡 = 𝑃𝑖 + 70.6
𝑞𝜇𝐵
𝑘ℎ
𝐸𝑖 −
𝜑𝜇𝑐𝑡𝑟2
0.00105𝑘𝑡

18. 18
Example 1.
An oil well has flowed for 30days at flow rate of 780 STB/day.The reservoir has the
following rock and fluid properties:
Flow rate Q 780 B/d
Reservoir pressure pi 2850 psi
Reservoir thickness h 58 ft
Reservoir permeability K 178 md
Oil viscosity µ 1.35 cp
formation porosity Ø 15%
Oil Formation Volume Factor, Bo 1.29 bbl/STB
Oil compressibility ct 22×10−6 psi−1
Drainage Radius, re 1600 ft
Well Radius, rw 3.5 in
(1) assuming infinite acting system, Calculate the bottom hole flowing pressure and
pressure at distance 100ft away from the wellbore .

19. 19
X =
𝜑𝜇𝑐𝑡𝑟2
0.00105𝑘𝑡
X =
0.15 ×1.35×22×10−6×
3.5
12
2
0.00105×178×30×24
X =2.82×10−9
x< 0.01
𝑃𝑤𝑓 = 2850 − 162.6
780 × 1.35 × 1.29
178 × 58
𝑙𝑜𝑔
178 × 30 × 24
0.15 × 1.35 × 22 × 10−6 ×
3.5
12
2 − 3.23
𝑃𝑤𝑓 =2672.5 psi
𝑃(𝑟, 𝑡) = 𝑃𝑖 − 162.6
𝑞 × 𝜇 × 𝐵
𝑘ℎ
log
𝑘 × 𝑡
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2 − 3.23

20. 20
Solution
𝑃 𝑟 = 1000𝑓𝑡, 𝑡 = 30 × 24 =?
X =
𝜑×𝜇×𝑐𝑡×𝑟2
0.00105𝑘×𝑡
X =
0.15 ×1.35×22×10−6× 1000 2
0.00105×178×30×24
= 0.033
Use the exponential logarithmic
chart and use the exponential
equation.
𝐸𝑖 (−0.33)
From chart reading @ −𝐸𝑖 −0.033 =2.88

21. 21
𝑃 𝑟 𝑡 = 𝑃𝑖 + 70.6
𝑞 × 𝜇 × 𝐵
𝑘 × ℎ
𝐸𝑖 −
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2
0.00105 × 𝑘 × 𝑡
𝑃 𝑟 = 1000𝑓𝑡, 𝑡 = 30 × 24 = 2850𝑖 − 70.6
780 × 1.35 × 1.29
178 × 58
2.88
𝑃 𝑟 = 1000𝑓𝑡, 𝑡 = 30 × 24 = 2823.2𝑝𝑠𝑖

22. 22
Example 2.
Flow rate Q 110 B/d
Reservoir pressure pi 2800 psi
Reservoir thickness h 80 ft
Reservoir permeability K 75 md
Oil viscosity µ 1.3 cp
formation porosity Ø 18%
Oil Formation Volume Factor, Bo 1.25 bbl/STB
Oil compressibility ct 1.62×10−5 psi−1
Drainage Radius, re 3500 ft
Well Radius, rw 0.333 in
(1) assuming infinite acting system, Calculate the bottom hole flowing pressure after
30days of production.

23. 23
X =
𝜑𝜇𝑐𝑡𝑟2
0.00105𝑘𝑡
X =
0.18 ×1.3×1.62×10−5× 0.333 2
0.00105×75×30×24
X =7.38×10−9
x< 0.01
𝑃𝑤𝑓 = 2850 − 162.6
110 × 1.3 × 1.25
75 × 80
𝑙𝑜𝑔
75 × 30 × 24
0.18 × 1.3 × 1.62 × 10−5 × 0.333 2
− 3.23
𝑃𝑤𝑓 =2746 psi
𝑃(𝑟, 𝑡) = 𝑃𝑖 − 162.6
𝑞 × 𝜇 × 𝐵
𝑘ℎ
log
𝑘 × 𝑡
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2 − 3.23

24. 24
X =
0.18 ×1.3×1.62×10−5×50002
0.00105×75×30×24
= 1.6638
𝑃 𝑟 𝑡 = 𝑃𝑖 + 70.6
𝑞 × 𝜇 × 𝐵
𝑘 × ℎ
𝐸𝑖 −
𝜑 × 𝜇 × 𝑐𝑡× 𝑟2
0.00105 × 𝑘 × 𝑡
𝑃 𝑟 𝑡 = 2799.8psi

25. 25

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