4 turbulent flow
- 1. Drilling Engineering Prepared by: Tan Nguyen Drilling Engineering – PE 311 Turbulent Flow in Pipes and Annuli
- 2. Drilling Engineering Prepared by: Tan Nguyen When attempting to quantify the pressure losses in side the drillstring and in the annulus it is worth considering the following matrix: Frictional Pressure Drop in Pipes and Annuli
- 3. Drilling Engineering Prepared by: Tan Nguyen Laminar Flow: In this type of flow, layers of fluid move in streamlines. There is no microscopic or macroscopic intermixing of the layers. Laminar flow systems are generally represented graphically by streamlines. Turbulent Flow: In turbulent flow, there is an irregular random movement of fluid in transverse direction to the main flow. This irregular, fluctuating motion can be regarded as superimposed on the mean motion of the fluid. Turbulent Flow in Pipes – Newtonian Fluids Introduction
- 4. Drilling Engineering Prepared by: Tan Nguyen Reynolds number, Re, is a dimensionless number that gives a measure of the ratio of inertial forces to viscous forces. Reynolds number is used to characterize different flow regimes, such as laminar or turbulent flow. Laminar occurs at low Reynolds number, where viscous forces are dominant, and is characterized by smooth, constant fluid motion; turbulent flow occurs at high Reynolds number and is dominated by inertial forces, which tend to produce chaotic eddies, vortices and other flow instabilities. For pipe In field unit: Definition of Reynolds Number µ ρ du _ Re = μ duρ _ 928 Re = Turbulent Flow in Pipes – Newtonian Fluids
- 5. Drilling Engineering Prepared by: Tan Nguyen If Re < 2,100 Laminar flow Re = 2,100 – 4,000 Transition flow Re > 4,000 Turbulent Note that this critical Reynolds number is correct only for Newtonian fluids Determination of Laminar/Turbulent Flow cp.fluid,ofviscosity inI.D.,pipe ft/svelocity,fluidavg. lbm/galdensity,fluidwhere 928 Re = = = = μ d u ρ μ duρ _ Turbulent Flow in Pipes – Newtonian Fluids
- 6. Drilling Engineering Prepared by: Tan Nguyen Relationship between shear stress and friction factor: Pipe flow under laminar conditions: Therefore, Newtonian fluids flow in pipe under laminar flow conditions: Hence, This equation will be used to calculate the friction factor of Newtonian fluids flow in pipe under laminar flow conditions. Determination of Friction Factor - Laminar Flow Turbulent Flow in Pipes – Newtonian Fluids
- 7. Drilling Engineering Prepared by: Tan Nguyen For turbulent flow, the friction factor can be calculated by using Colebrook correlation. Where ε is the absolute roughness. ε/d is the relative roughness. For smooth pipe, the relative roughness ε/d < 0.0004, the following equations can be used to calculate the friction factor in turbulent flow Re = 2,100 – 100,000: Blasius approximation: Re = 2,100 – 100,000: +−= fN d f Re 255.1 /269.0log4 1 ε Determination of Friction Factor - Turbulent Flow ( ) 395.0log4 1 Re −−= fN f 25.0 Re 0791.0 N f = Turbulent Flow in Pipes – Newtonian Fluids
- 8. Drilling Engineering Prepared by: Tan Nguyen Determination of Friction Factor Turbulent Flow in Pipes – Newtonian Fluids
- 9. Drilling Engineering Prepared by: Tan Nguyen Determination of Friction Factor Turbulent Flow in Pipes – Newtonian Fluids
- 10. Drilling Engineering Prepared by: Tan Nguyen From equation: , an equation of dp/dL can be expressed as In field unit: . This equation can be used to calculate the frictional pressure drop gradient for Newtonian and non-Newtonian fluids. Combining this equation and the Blasius approximation gives Note that the Moody friction factor is four times higher than the Fanning friction factor. Determination of the Frictional Pressure Loss 25.0 2 928 8.25 0791.0 = µ ρ ρ dv d u dL dp 75.4 25.075.175.0 25.0 25.075.175.0 86241800 d q d u dL dp µρµρ == Turbulent Flow in Pipes – Newtonian Fluids
- 11. Drilling Engineering Prepared by: Tan Nguyen Example: Determine the frictional pressure drop in 10000ft of 4.5-in commercial steel drillpipe having an internal diameter of 3.826in. If a 20 cp Newtonian fluid having a density of 9 lbm/gal is pumped through the drillpipe at a rate of 400 gal/min Solution: Mean velocity: Reynolds number: Since Re > 2,100, the flow is under turbulent flow conditions. Example sft d q u /16.11 826.3*488.2 400 488.2 22 === 17831 20 826.3*16.11*9*928928 Re === μ duρ _ Turbulent Flow in Pipes – Newtonian Fluids
- 12. Drilling Engineering Prepared by: Tan Nguyen From table 4.5, the absolute roughness for commercial steel pipe is ε = 0.000013 inches. The relative roughness ε/d = 0.000013/3.826 = 0.0000034 < 0.0004 --> smooth pipe Solve this equation for the Fanning friction factor: f = 0.00666 Thus the frictional pressure loss can be obtained by Example +−= fN d f re 255.1 /269.0log4 1 ε psiD d vf D dL dp pf 75610000* 826.3*8.25 16.11*9*00666.0 8.25 22 ====∆ ρ Turbulent Flow in Pipes – Newtonian Fluids
- 13. Drilling Engineering Prepared by: Tan Nguyen Using Blasius approximation ,the equation becomes Pressure drop: DP = dp/dL x D = (0.0777)(10,000) = 777 psi Example 25.0 Re 0791.0 N f = 25.0 2 928 8.25 0791.0 = µ ρ ρ dv d v dL dp ftpsi d v dL dp /0777.0 1800 25.0 25.075.175.0 == µρ Turbulent Flow in Pipes – Newtonian Fluids
- 14. Drilling Engineering Prepared by: Tan Nguyen Using Blasius approximation ,the equation becomes Pressure drop: DP = dp/dL x D = (0.0777)(10,000) = 777 psi Example 25.0 Re 0791.0 N f = 25.0 2 928 8.25 0791.0 = µ ρ ρ dv d v dL dp ftpsi d v dL dp /0777.0 1800 25.0 25.075.175.0 == µρ Turbulent Flow in Pipes – Newtonian Fluids
- 15. Drilling Engineering Prepared by: Tan Nguyen Hydraulic diameter is defined as: Equivalent diameter by using hydraulic diameter method: Equivalent Diameter for Annular Geometry – Hydraulic Diameter Method Turbulent Flow in Pipes/Annuli – NonNewtonian Fluids
- 16. Drilling Engineering Prepared by: Tan Nguyen From the momentum equation, frictional pressure drop for Newtonian fluid in the annulus is For pipe flow, d1 --> 0 then Comparing these two equations, the equivalent diameter an annulus can be obtained Equivalent Diameter for Annular Geometry – From Momentum Equation Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids − −+ = 1 2 2 1 2 22 1 2 2 ln 500,1 d d dd dd u dL dpf µ 2 500,1 d u dL dpf µ = ( )1 2 ln 2 1 2 22 1 2 2 d de dd ddd − −+=
- 17. Drilling Engineering Prepared by: Tan Nguyen From the narrow slot approximation, frictional pressure drop for Newtonian fluid in the annulus is For pipe flow Comparing these two equations, the equivalent diameter an annulus can be obtained Equivalent Diameter for Annular Geometry – Narrow Slot Approximation Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids 2 500,1 d u dL dpf µ = ( )12816.0 ddde −=
- 18. Drilling Engineering Prepared by: Tan Nguyen A fourth expression for the equivalent diameter of an annulus was developed by Crittendon. When using Crittendon correlation, a fictitious average velocity also must be used in describing the flow system. Equivalent Diameter for Annular Geometry – Crittendon Correlation Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
- 19. Drilling Engineering Prepared by: Tan Nguyen 1. Obtain apparent viscosity by combining the frictional pressure loss in pipe (or annulus) for both Newtonian and Bingham Plastic models 2. Use apparent viscosity to determine Reynolds Number Bingham Plastic Model Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids dd v d v ypa 22515001500 22 τµµ += )( 66.6 Pipe v dy pa τ µµ += )( )(5 12 Annulus v ddy pa − += τ µµ or
- 20. Drilling Engineering Prepared by: Tan Nguyen Another way to determine the flow regime (critical Reynolds number) is to use the Hedstrom number In field unit Bingham Plastic Model Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids 2 2 p y HE d N µ ρτ = 2 2 37100 p y HE d N µ ρτ =
- 21. Drilling Engineering Prepared by: Tan Nguyen Bingham Plastic Model Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids Turbulent
- 22. Drilling Engineering Prepared by: Tan Nguyen Apparent Viscosity for use in the Reynolds Number is obtained by comparing the laminar flow equations for Newtonian and Power Law fluids Pipe flow: Annular flow: Power Law Model Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids n n n a n d vK d v + = + 0416.0 /13 1440001500 )1(2 µ )( 0416.0 /13 96 )1( )1( Pipe n v Kd n n n a + = − − µ n n n a n dd vK dd v + − = − + 0208.0 /12 )(144000)(1000 )1( 12 2 12 µ )( 0208.0 /12 144 )( )1( )1( 12 Annulus n v ddK n n n a +− = − − µ
- 23. Drilling Engineering Prepared by: Tan Nguyen Reynolds number for power law fluids in pipe: Reynolds number for power law fluids In annulus: Friction factor for power law fluids under turbulent flow conditions Power Law Model Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids nn n d K v N + = − /13 0416.089100 )2( Re ρ nn n dd K v N + − = − /12 )(0208.0109000 12 )2( Re ρ 2.1 2/1 Re75.0 395.0 )log( 0.4 /1 n fN n f n −= −
- 24. Drilling Engineering Prepared by: Tan Nguyen Power Law Model Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
- 25. Drilling Engineering Prepared by: Tan Nguyen Summary
- 26. Drilling Engineering Prepared by: Tan Nguyen Summary Newtonian Model Bingham Plastic Model Power Law Model
- 27. Drilling Engineering Prepared by: Tan Nguyen Summary Newtonian Model Bingham Plastic Model Power Law Model
- 28. Drilling Engineering Prepared by: Tan Nguyen Summary Newtonian Model Bingham Plastic Model Power Law Model
- 29. Drilling Engineering Prepared by: Tan Nguyen Example – Newtonian Fluid in Annulus Example 1: A 9.0 lbm/gal brine having a viscosity of 1.0 cp is being circulated in a well at a rate of 200 gal/min. Apply the all the criteria for computing equivalent diameter. Determine the flow pattern and frictional pressure gradient. The drillpipe has an external diameter of 5.0 in. and the hole has a diameter of 10 in. Solution:
- 30. Drilling Engineering Prepared by: Tan Nguyen Example – Newtonian Fluid in Annulus
- 31. Drilling Engineering Prepared by: Tan Nguyen Example – Newtonian Fluid in Annulus
- 32. Drilling Engineering Prepared by: Tan Nguyen Example – Newtonian Fluid in Annulus Note that the Crittendon correlation is applied for the fourth method. In this method, we need to calculate the equivalent diameter based on Crittendon correlation and the fictitious average velocity. Hydraulic Method Momentum Mothod Narrow Slot Method Crittendon Method
- 33. Drilling Engineering Prepared by: Tan Nguyen Example – BHF - Annulus Example 2: A 10 lbm/gal mud having a plastic viscosity of 40 cp and a yield point of 15 lbf/100ft2 is circulated at a rate of 600 gal/min. Estimate the frictional pressure loss in the annulus opposite the drill collars if the drill collars are in a 6.5-in hole, have a length of 1,000 ft, and a 4.5 in. OD. Check for turbulence using both the apparent viscosity concept and the Hedstrom number approach. Use an narrow slot equivalent diameter to represent the annular geometry.
- 34. Drilling Engineering Prepared by: Tan Nguyen Example – BHF - Annulus Equivalent diameter using narrow slot approximation Reynolds number based on apparent viscosity
- 35. Drilling Engineering Prepared by: Tan Nguyen Example – BHF - Annulus Reynolds number for a plastic viscosity of 40 cp Using the graph for Hedstrom number, the critical Reynolds number is 3,300. The flow is turbulent
- 36. Drilling Engineering Prepared by: Tan Nguyen Example – BHF - Annulus Using Blasius approximation with Re = 3,154, the friction factor is f = 0.0098 Frictional pressure loss is given:
- 37. Drilling Engineering Prepared by: Tan Nguyen Example – BHF - Annulus
- 38. Drilling Engineering Prepared by: Tan Nguyen Example – PL - Annulus Example 3: A 15.6 lbm/gal cement slurry having a consistency index of 335 eq cp and a flow behavior index of 0.67 is being pumped at a rate of 672 gal/min between a 9.625-in. hole and a 7.0-in.casing. Determine the frictional pressure loss per 100 ft of slurry. Use the equivalent diameter based on the narrow slot approximation. Solution: The mean velocity: Reynolds number:
- 39. Drilling Engineering Prepared by: Tan Nguyen Power Law Model Turbulent Flow in Pipes/Annuli – nonNewtonian Fluids
- 40. Drilling Engineering Prepared by: Tan Nguyen Example– PL - Annulus