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

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

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:

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:

Force Analysis

Force Analysis

Force Analysis

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

Pipe Flow – Newtonian Fluids

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:

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

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

Pipe Flow – Newtonian Fluids

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.

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

Pipe Flow – Newtonian Fluids Velocity: Reynolds number: NRe = 17725 : The fluid is under turbulent flow.

Pipe Flow – Newtonian Fluids

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

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

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

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

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

Summary - Newtonian Fluids Field unit: Note: equivalent diameter for an annular section: de= 0.816 (d2 – d1)

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 = 0.052 r D = 0.052 x 9 x 10,000 = 4,680 psig Average velocity:

Annular Flow – Newtonian Fluids Frictional pressure loss gradient Or: Circulating bottom hole pressure: P = 4,680 + 0.0051 x 10,000 = 4,731 Psig

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

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

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

Annular Flow – Newtonian Fluids Narrow Slot Approximation Flow rate:

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

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

Annular Flow – Newtonian Fluids Narrow Slot Approximation Field Unit:

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’’

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:

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