1. Laminar Flow in Pipes and Annuli
Drilling Engineering - PE 311
Laminar Flow in Pipes and Annuli
Newtonian Fluids

2. Frictional Pressure Drop in Pipes and Annuli
Under flowing conditions
In the annulus: Pwf = DPf(a) + rgTVD (1)
In the drillpipe: Pp – Pwf = DPf(dp) + DPb – rgTVD
Pwf = Pp - DPf(dp) – DPb + rgTVD (2)
From (1) and (2) give
Pp = DPf(dp) + DPf(a) + DPb

3. Frictional Pressure Drop in Pipes and Annuli
When attempting to quantify the pressure losses in side the drillstring and in the annulus it is worth considering the following matrix:

4. Momentum Equation Assumptions:
The drillstring is placed concentrically in the casing or open hole
The drillstring is not being rotated
Sections of open hole are circular in shape and of known diameter
Incompressible drilling fluid
Isothermal flow
Momentum equation:

5. Force Analysis

6. Force Analysis

7. Force Analysis

8. Pipe Flow – Newtonian Fluids
B.C. r = 0 --> t = 0: then C1 = 0
For Newtonian fluids t = mg = m du/dr
B.C. r = R --> u = 0: then

9. Pipe Flow – Newtonian Fluids

10. Pipe Flow – Newtonian Fluids
Maximum velocity:
Average fluid velocity in pipe u = umax / 2
From this equation, the pressure drop can be expressed as:
In field unit:

11. Pipe Flow – Newtonian Fluids
From equation:
Combining with the definition of Fanning friction factor:
Pressure drop can be calculated by using Fanning friction factor:
Field unit:
This equation can be used to calculate pressure under laminar or turbulent conditions

12. Pipe Flow – Newtonian Fluids
From this equation, the pressure drop can be expressed as:
Combining this equation and equation gives
where
This equation is used to calculate the Fanning friction factor when the flow is laminar. If Re < 2,100 then the flow is under laminar conditions

13. Pipe Flow – Newtonian Fluids

14. Pipe Flow – Newtonian Fluids
Combining these two equations and
The relationship between shear stress and shear rate can be express as
Where is called the nominal Newtonian shear rate. For Newtonian fluid, we can used this equation to calculate the shear rate of fluid as a function of velocity.

15. Pipe Flow – Newtonian Fluids
Determine whether a fluid with a viscosity of 20 cp and a density of 10 ppg flowing in a 5" 19.5 lb/ft (I.D. = 4.276") drillpipe at 400 gpm is in laminar or turbulent flow. What is the maximum flowrate to ensure that the fluid is in laminar flow ? Calculate the frictional pressure loss in the drillpipe in two cases:
Q = 400 GPM
Q = 40 GPM

16. Pipe Flow – Newtonian Fluids
Velocity:
Reynolds number:
NRe = : The fluid is under turbulent flow.

17. Pipe Flow – Newtonian Fluids

18. Pipe Flow – Newtonian Fluids
With turbulent flow and assuming smooth pipe, the Fanning friction factor:
Wrong calculation !!!!!!! Laminar flow !!!!!!!

19. Pipe Flow – Newtonian Fluids
If Q = 40 GPM, the fluid velocity is: u = 0.89 ft/s
The Reynolds number: Re = 1,772  Laminar flow
Frictional pressure loss:
f = 16 / Re = 0.009

20. Annular Flow – Newtonian Fluids
From the general equation for fluid flow:
For Newtonian fluids:

21. Annular Flow – Newtonian Fluids
We need two boundary conditions to find C1 and C2
B.C. 1: r = r1 --> u = 0
B.C. 2: r = r2 --> u = 0 r1 r2

22. Annular Flow – Newtonian Fluids
Flow rate can be calculated as:
Average velocity can be expressed as
Pressure drop:
SI Unit
Field Unit

23. Summary - Newtonian Fluids
Field unit:
Note: equivalent diameter for an annular section: de= (d2 – d1)

24. Annular Flow – Newtonian Fluids
Example: A 9-lbm/gal Newtonian fluid having a viscosity of 15 cp is being circulated in a 10,000-ft well containing a 7-in. ID casing and a 5-in OD drillsring at a rate of 80 gal/min. compute the static and circulating bottomhole pressure by assuming that a laminar flow pattern exists.
Solution:
Static pressure: P = r D = x 9 x 10,000 = 4,680 psig
Average velocity:

25. Annular Flow – Newtonian Fluids
Frictional pressure loss gradient
Or:
Circulating bottom hole pressure:
P = 4, x 10,000 = 4,731 Psig

26. Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Since the exact solution is so complicated, a narrow rectangular slot approximation is used to arrive at solutions still very useful for practical drilling engineering applications. We represent the annulus as a slot which has the same area and the same height with the annulus. This approximation is good if D1 / D2 > 0.3
An annular geometry can be represented by a rectangular slot with the height h and width w as given below

27. Annular Flow – Newtonian Fluids
Narrow Slot Approximation P1 P2 ty
ty + Dy
P1Wy - P2Wy + yWL - y+yW L = 0

28. Annular Flow – Newtonian Fluids
Narrow Slot Approximation
For Newtonian fluids:
Boundary conditions: y = 0 --> u = 0 and y = h --> v = 0

29. Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Flow rate:

30. Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Average velocity:
Frictional pressure losses gradient
Field unit

31. Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Determination of shear rate:
Field unit:

32. Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Field Unit:

33. Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Example: A 9-lbm/gal Newtonian fluid having a viscosity of 15 cp is being circulated in a 10,000-ft well containing a 7-in. ID casing and a 5-in OD drillsring (ID = 4.276’’) at a rate of 80 gal/min. Compute the frictional pressure loss and the shear rate at the wall in the drillpipe and in the annulus by using narrow slot approximation method. Assume that the flow is laminar. Also, calculate the pressure drop at the drill bit which has 3 nozzles: db = 13/32’’

34. Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Velocity of fluid in the drillpipe:
Velocity of fluid in the annulus:
Pressure drop in the drillpipe:
Pressure drop in the annulus:

35. Annular Flow – Newtonian Fluids
Narrow Slot Approximation
Drill bit area:
Pressure drop at the bit:
Total pressure drop:
Shear rate at the wall: