SKM DRILLING ENGINEERING. Chapter 3 - Drilling Hydraulics

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1 1 SKM DRILLING ENGINEERING Chapter 3 - Drilling Hydraulics Assoc. Prof. Abdul Razak Ismail Petroleum Engineering Dept. Faculty of Petroleum & Renewable Energy Eng. Universiti Teknologi Malaysia

2 Contents Review of flow in pipes (Fluid Mechanics) Drilling mud flow (circulating) system Newtonian fluid flow calculations Bingham plastic fluid flow calculations p across bit nozzles p calculation for typical system

3 3 Review of flow In Pipes Real fluid flow is much complex compare to perfect fluid flow. Between fluid particles Shear stress fluid Between fluid particles and pipe s wall Energy equilibrium principles are used to solve the problems. Partial differential equation (Euler s equation) has no general solution to solve problems. Results from experiment (analytical) and semi-empirical method needs to be used to solve flow problems. There are 2 types of steady flow of real fluid exists: Laminar flow Turbulent flow

4 4 Dye This one? or Laminar (viscous) flow this one? or Transition flow this one? Turbulent flow All three types of flow actually do occurred in real fluid flow. Laminar flow v Turbulent flow v The problem is: what is v and v.why we need to know?

5 5 This phenomenon was first investigated in 1883 by Osborne Reynolds in an experiment which has a classic in fluid mechanic. Laminar flow Transition flow Turbulent flow After a few experiments, he found out a mathematical relationship: vd

6 6 This mathematical relationship can be used to determine the types of flow. vd 2000 laminar flow vd transition flow vd 4000 turbulent flow Subsequently until now, this mathematical relationship is known as Reynolds number, Re (or NRe ). vd Re dimensionless laminar flow : Re 2000 transition flow : 2000 Re 4000 turbulent flow : Re 4000

7 7 vd where: v d fluid density fluid average velocity pipe inside diameter fluid absolute viscosity If kinematic viscosity,, is inserted in the equation: vd Re

8 8 Fluid velocity profile in a pipe: Laminar flow Turbulent flow Q Q vmaks v vavg 12 vmaks v vavg Q A v vavg Q A

9 9 Mechanical Energy of a Flowing Fluid Consider the situation below: Qin element ( mg Station 1 HL v1 Turbine HE z1 element ( mg Station 2 Pump v2 HA Control volume 1 Qout z2 2 Datum The energy possessed by a flowing fluid consists of internal energy and energies due to pressure, velocity, and position energy at energy energy energy energy at section 1 added lost extracted section 2 This equation, for steady flow of incompressible fluids in which the change in internal energy is negligible, simplifies to p1 v12 p2 v22 z H H H z 1 A L E 2 2g 2g

10 10 Energy Losses In Pipe Def.: Any energy losses in closed conduits due to friction, HL. This types of losses can be divided into 2 main categories: Major losses, HL-major, and Minor losses, HL-minor. From Bernoulli s equation: p1 v12 p2 v22 2 g z1 H A H L H E 2 g z2 Energy added to the system, HA, is frequently due to pump fluid head, HP, energy extracted, HE, is frequently due to turbine fluid head, HT, Bernoulli s equation can be simplify as: v12 p2 v22 z1 H P z2 H T H L major H L minor 2g 2g p1

11 11 Major Losses In Pipe Def.: The head loss due to friction in long, straight sections of pipe. The losses do happen in pipe, either in laminar or turbulent flow. a. Laminar flow Problem solved analytically derived purely from mathematical relationship Hagen-Porseuille equation 32 vl p f d2 in the forms of head loss, HL 32 vl HL d2 Darcy-Weisbach equation vd by replacing Re into Hagen-Porseuille equation 2 64 v L HL Re d 2 g

12 12 b. Turbulent flow From Darcy-Weisbach equation for laminar flow 2 64 v L HL Re d 2 g 2 v L HL f d 2g 64 a simple mathematical relationship. Re For turbulent flow, f has to be solved empirically experiment need to be done. In laminar and turbulent flow, f is known as friction coefficient or friction factor. Where, for laminar flow, f

13 13 Friction Factor a. Laminar flow Darcy-Weisbach equation 2 HL f L v d 2g b. where f 64 Re Turbulent flow In the literature (from 1900 s current date), there are many studies that have been conceded by various researchers. Blasius s equation (1913) von Karman s equation modified by Prandtl Nikuradse s equation (for smooth and rough pipes) Colebrook-White equation (1940 s) Moody Barr s equation (1975)

14 14 Laminar Friction Factor. (cont. 2) Transition f 64 Re Turbulent Complete Turbulent vd Re vd Moody Chart

15 15 Normal practice in determination of f 1. Calculate Re to determine the types of flow. 2. HL calculation: used Darcy-Weisbach equation. 2 L v HL f d 2g 3. For laminar flow: f 64 Re 4. For turbulent flow: a. Determine pipe relative roughness, d Where: pipes absolute roughness d pipe internal diameter d e is depend on pipe s material, normally is given in tabular forms. Material (new) Riverted steel Concrete Wood stave Cast iron Galvanized iron Asphalted cast iron Commercial steel or wrought iron Drawn tubing Glass Absolute roughness, ft mm (smooth) (smooth) b. Obtain f from Moody Re, d

16 16 Attention 1. In this subject, SKM1043, the f that we are using, is the American friction factor, famerican. 2. The value of famerican is different to the one that used by the British f American 4 f British 64 Re 16 Re needs to refer different Moody Chart 3. Sometimes: f American 4 f British

17 17 Since the mud enters the drill string and leaves the annulus at essentially the same elevation, the only pressure required is to overcome the frictional losses in the system. Hence, the discharge pressure at the pump is defined by: pt p s p p p c pb p ac p ap... (3.1) where: pt = ps = pp = pc = pb = pac = pap = pump discharge pressure pressure loss in surface piping, standpipe, and mud hose pressure loss inside drill pipe pressure loss inside drill collars pressure loss across bit nozzles pressure loss in annulus around drill collars pressure loss in annulus around drill pipe

18 18 The solution of Eq. (3.1) is rather tedious; separate calculations are needed for each section There are 4 different types of model used to calculate frictional pressure losses in mud circulating system: Newtonian Bingham plastic Power-law API Power-law Due to the limitation of the syllabus, Power-Law and API Power-Law models will not be discussed in this subject. All calculations will be focused on Newtonian and plastic fluid models.

19 19 Newtonian Fluid Flow Calculations Similar to generalized flow system approach, calculation of p for pipe flow requires a knowledge of which flow pattern pertains to the specific case, since different equations apply for each situation. Definition of the existing flow pattern is given by a dimensionless quantity known as the Reynolds number (NRe ): N Re where: 928 vd N Re = Reynolds s number v = average velocity of flow, ft/sec = fluid density, ppg d = pipe inside diameter, in. = fluid viscosity, cp q = circulating volume, gal/min... (3.2)

20 20 Similar to generalized flow system approach, that if laminar flow : N Re 2000 transition flow : 2000 N Re 4000 turbulent flow : N Re 4000 The p in laminar flow is given by the Hagan-Poiseuille law; this, in practical units, is Lv... (3.3) p 2 1,500d where: p = laminar flow p, lb/in2 L = length of pipe, ft turbulent flow For turbulent flow, Fanning s equation applies: f Lv 2... (3.4) p 25.8 d where: p = turbulent flow p, lb/in2 f = Fanning friction factor

21 21 The friction factor f is a function of and pipe roughness, and has been evaluated experimentally for numerous materials (see Fig. 7.1) p calculation for Newtonian fluid flow systems in the following manner: a. Calculate NRe from Equation (3.2). b. If NRe < 2000, use Equation (3.3) to calculate the pressure drop. c. If NRe > 2000, use Equation (3.4). In this case the friction factor f is obtained from Figure 7.1 or its equivalent.

22 22

23 23 Plastic Fluid Flow Calculations Drilling fluids is non-newtonian fluid Newtonian fluid equations must be altered for application to typical drilling mud systems

24 24 Surface Equipment Losses ( ps) The surface equipment consist of standpipe, hose, swivel, kelly joint, and the piping between the pump and standpipe. In practice, there are only four types of surface equipment; each type is characterized by the dimensions of standpipe, kelly, rotary hose and swivel. Table 3.1 summarizes the four types of surface equipment. Table 3.1: Types of surface equipment & value of constant E Standpipe Type Hose Swivel, etc. Kelly Eq. length, ID E ID Length ID Length ID Length ID Length ft ft ft ft. 2,600 ft. 2.5 x ft ft ft ft. 946 ft. 9.6 x ft ft ft ft. 610 ft. 5.3 x ft ft ft ft. 424 ft. 4.2 x 10-5

25 25 To determine surface equipment losses ( ps): Use the following formula: ps E m0.8 q p... (3.5) where: ps = q = m = E = p surface pressure losses, psi flow rate, gpm mud density, ppg a constant depending on type of surface equipment used mud plastic viscosity, cp

26 Fluid Flow Inside the Pipe Shearing stress or pressure A. Laminar Flow Region p Yt mv 3 (True laminar flow) Yb, Bingham yield Transition from plug to laminar flow Plug flow Yt, True yield Rate of shear or velocity Fig. 3.1: Flow behavior of plastic and Newtonian fluids p Yt mv 3 where: 144 p = pressure drop, lb/ft2 4 3 Yt = Yb, lb/ft2 m = L/(1500d2), slope of linear portion (from Eq. (3.3)) 26

27 27 For practical values of v, the behavior of plastic fluids may be expressed as: p vl LYb p 300d 1500d 2 L p 300d where: pv Yb 5 d laminar flow... (3.6) p = plastic viscosity, cp. Yb = yield point, lb/100ft2. Eq. (3.6) may be used in cases where laminar flow exists Determination of flow characteristic (laminar or turbulent) is made by comparing the actual velocity with a calculated critical velocity

28 28 Average Velocity Calculation q ft 3 /sec v A ft 2 Avg. velocity inside the pipe Avg. velocity in the annulus 1 ft 3 1 min q gal/min 7.48 gal 60 sec ( /4)(d /12)2 q... (3.7a) v d OD ID Hole q... (3.7b) v (d h - d p ) where : v average velocity, ft/sec. q = flow rate, gpm d = diameter, in. Drill pipe Annulus Annulus Area Ah Ap 4 (d h2 d p2(od ) )

29 29 Critical Velocity Calculation If Eqs. (3.3) and (3.6) are equated, an equivalent Newtonian viscosity in terms of d, v, p and Yb is obtained: 5dYb p v Substituting the above Eq. for in the Reynolds s number of Eq. (3.2), equating the resulting equation to 2000, and solving for v gives: 1.08 p 1.08 p2 9.3 d 2Yb... (3.8) vc d where: vc = critical velocity, ft/sec, above which turbulent flow exists and below which the flow is laminar. Eq. (3.8) assumes that turbulence occurs at NRe = Therefore, if: v vc, flow is laminar v vc, flow is turbulent

30 30 B. Turbulent Flow Region Before Fanning Eq. can be used, alteration to NRe expression have to be done (after Beck, Nuss & Dunn) p... (3.9) t 3.2 where: t = turbulent viscosity of plastic fluids, cp Substitution of µ t, for µ in the general NRe expression (Eq. (3.2)) gives: NRe NRe 928 vd t 2,970 vd p... (3.10) By using Fig. 7.1, determine f This f may then be used in Eq. (3.4) for calculation of pressure

31 31 In summary, p calculation for plastic fluid flow systems can be done as follows: (1) Calculate the average velocity, v, from Eq. (3.7a) or (3.7b) (2) Calculate vc from Eq. (3.8) (3) If v vc flow is laminar, Eq. (3.6) applies (4) If v vc flow is turbulent, requiring: a. Calculation of NRe from Eq. (3.10) b. Determination of f from Fig. 7.1 at the calculated for the conduit in question c. Calculation of pressure drop from Eq. (3.4)

32 32 Example 3.1 Mud is flowing through 4 1/2 inch OD, internal flush drill pipe. Calculate the frictional pressure drop per 1000 ft of pipe. Mud properties Mud density, m Pipe ID Bingham yield, Yb Circulating rate, q Plastic viscosity, p = = = = = 10 lb/gal in. 10 lb/100 ft2 400 gal/min 30 cp

33 33 Solution 3.1 Eq. (3.7a) : v q 2.45d 2 Eq. (3.8) : v c 1.08 p 1.08 p2 9.3 d 2Yb ft/sec 2.45(3.64)2 (1) v (2) (1.08)(30) (1.08) (30)2 (9.3)(10)(3.64)2 (10) vc 4.3 ft/sec (10)(3.64) (3) Since v vc, flow is turbulent. (a) N Re (2,970)(10)(12.3)(3.64) 44, (b) f from Curve II, Fig. 3.1 (0.0062)(10)(1000)(12.3)2 (c) p p 100 psi/1000 ft (25.8)(3.64) d

34 34 Hydraulically Equivalent Annulus Diameter For annular flow, it is necessary to use a hypothetical circular diameter, da, which is the hydraulic equivalent of the actual annular system The hydraulic radius is defined as: hydraulic radius, rh = cross-sectional area of flow system wetted perimeter of conduit Annulus (r12 r22 ) r1 r2 for an annulus rh = 2 (r1 r2 ) 2 r2 r for a circular pipe rh = 2 r 2 r2 r The frictional loss in an annulus is equal to the loss in a circular pipe having the same hydraulic radius; hence, in general terms: re = r1 r2 or de = d1 d2... (3.11) where re and de are the hydraulically equivalent radius and diameter r1

35 35 Pressure Drop Across Bit Nozzles Consider the diagram below for incompressible fluid: Fig. 3.2: Schematic sketch of incompressible fluid flowing through a converging tube or nozzle. Assuming steady state, adiabatic, and frictionless: v12 p2 v22 2g 2g p1 where: p1, p2 = turbulent flow pressure drop, lb/ft2 v1, v2 = density, lb/ft3 = velocities at points 1 and 2, ft/sec... ( a)

36 36 or p1 v12 p2 v22 w 2g w 2g... (a ) v22 v12 2g Practically, v22 v12 v22, therefore: p... (b) v22 2 g The ideal rate of flow, qi A2 v2. The actual flow rate q is:... (c) q Cqi where C is the flow or nozzle coefficient for particular design. p

37 37 By substituting Eq. (c) into Eq. (b), and rearranging it, the equation becomes: q2... (3.12) p gc A2 Altering Eq. (3.12) to practical units for mud flow, we: q2 pb 7, 430 C 2 d e 4... (3.13) where de = hydraulically equivalent nozzle diameter, in. The value of C is around

38 38 Multiple Nozzles The calculation of p across a multiple nozzle bit may be simplified by substituting the sum of the nozzle areas for A in Equation (3.12). For single nozzle: q2 p 2 gc 2 A2 For several nozzles, each of area A1: q12 pm 2 gc 2 A12

39 39 q For parallel flow, q1, where n = number of nozzles. n therefore: pm q12 A2 q12 A p q A1 n q1 A1 Cross sectional area of flow, A, is defined as A n A1 A2 n2 A12 or A na1... (3.14)

40 40 Similarly, for use in Eq. (3.13) d e nd 2... (3.15a) If the multiple nozzles vary in size, where: de ad12 bd 22 etc.... (3.15b) a = number of nozzles having diameter d1. b = number of nozzles having diameter d2. d e = hydraulically equivalent single nozzle diameter, in.

41 41 Example 3.2 A 10 lb/gal mud is being circulated at the rate of 500 gal/min. through a tri-cone bit having three 3/8 in. diameter jets. What is the pressure drop across the bit? Solution 3.2 Drill string p1 Hole Nozzle vn1 vn2 vn3 p2 de or d 3( 83 ) in. (equivalent single nozzle diameter) Using Eq. (3.13): (500) 2 (10) ( p1 p2 ) or p 2,100 psi 2 4 (7430)(0.95) (0.65)

42 42 Pressure Drop Calculations for a Typical Systems Example 3.3 Operating Data Depth = 6,000 ft (5,500 ft drill pipe, 500 ft drill collars) Drill pipe = 4 ½-in. internal flush, 16.6 lb/ft (ID = in.) Drill collars = 6 ¾ in. (ID = in.) Mud density, m = 10 lb/gal Plastic viscosity, p = 30 cp Bingham yield, Yb = 10 lb/100ft2 Bit = 7 7/8-in., 3 cone, jet rock bit Nozzle velocity required = at least 250 ft/sec through each nozzle (this value is obtained by a commonly applied rule of thumb). Assume C = 0.95 Surface equipment type = 2 What hydraulic (pump output) horsepower will be required for these conditions?

43 43 Solution 3.3 Gbr ni tak perlu ubah Circulation rate: This is obtained from the desired annular velocity necessary for proper hole cleaning (cutting removal). Assume that this is a fast drilling, soft rock area and that 180 ft/min (3 ft/sec) upward velocity based on a gauge hole is required (i.e. annular velocity around the drill pipe). 5,500 ft. The flow rate, q is: q (annulus area) velocity 2.45(d h2 - d p2 )v (3) 307 gpm 500 ft.

44 Eq. (3.16) ps E m0.8 q p (a) Surface equipment losses ( ps) Surface equipment type 2 Table 3.1 E ps ( )(10) 0.8 (307)1.8 (30) psi (b) Pressure losses inside drill pipe ( pp) The average velocity inside the drill pipe: v q ft/sec d 2.45(3.826) The critical velocity: vc 1.08 p 1.08 p2 9.3 m d 2Yb m d 1.08 (30) 1.08 (30) 2 (9.3)(10)(3.826)2 (10) (10)(3.826) 4.25 ft/sec

45 45 v vc turbulent flow (use Eq. 3.4) NRe 2,970 vd p N Re 32, 400 Curve II (2,970)(10)(8.58)(3.826) 32, , Fig. 7.1 f Applying Eq. (3.4): f Lv 2 (0.0066)(10)(5,500)(8.56)2 pp 269 psi 25.8 d (25.8)(3.826)

46 46 (c) Pressure losses inside drill collar ( pc) The average velocity inside the drill collar: q 307 v ft/sec d 2.45(2.813) The critical velocity: vc 1.08 p 1.08 p2 9.3 m d 2Yb m d 1.08 (30) 1.08 (30) 2 (9.3)(10)(2.813) 2 (10) (10)(2.813) 4.64 ft/sec

47 47 v vc turbulent flow (use Eq. 3.4) NRe 2,970 vd p N Re 44,100 Curve II (2,970)(10)(15.84)(2.813) 44,112 44, Fig. 7.1 f Applying Eqn. (3.4): f Lv 2 (0.0062)(10)(500)(15.84)2 pc 107 psi 25.8 d (25.8)(2.813)

48 48 (d) Pressure losses through bit ( pb) Three nozzles (one for each cone) will be used, hence 1/3 q will flow through each. For v = at least 250 ft/sec through each nozzle, q d 1 3 q 307 / in. 2.45v (2.45)(250) Nozzle sizes are sell in multiples of 1/32 in. Therefore, the nearest stock nozzle available is 13/32 in. (i.e in.): nozzle diameter of in. is chosen This nozzle allows an actual velocity of: 102 v 252 ft/sec ( 32 ) 1 3 q 1 3 q 1 3 q

49 49 Eq. (3.15a) d e nd 2 Eq. (3.15b) de ad12 bd 22 etc. Using Eq. (3.15) or (3.15a), the actual nozzle diameter: 2 d 3 ( 13 ) in. 32 q 2 m Eq. (3.13) pb 7, 430C 2 d 4 Pressure drop across the bit, pb : (307)2 (10) pb 573 psi 2 4 7, 430(0.95) (0.704)

50 50 (e) Pressure losses around drill collar ( pac) The average velocity around the drill collar: 307 v 7.62 ft/sec (2.45) (7 8 ) (6 4 ) The hydraulically equivalent diameter of the annulus: d a d1 d 2 d in. The critical velocity: vc 1.08 (30) 1.08 (30)2 (9.3)(10)(1 18 )2 (10) 1 8 (10)(1 ) 7.26 ft/sec

51 51 v vc turbulent flow (use Eq. 3.4) NRe 2,970 vd p (2,970)(7.62)(1 18 ) 8, 487 8, N Re 8, 400 Curve IV (for annuli in uncased hole) Fig. 7.1 f Applying Eqn. (3.4): f Lv 2 (0.0098)(10)(500)(7.62)2 pac 98 psi d (25.8)(1 8 )

52 52 (f) Pressure losses around drill pipe ( pap) The average velocity around the drill collar (as assume/given earlier): v 3 ft/sec The hydraulically equivalent diameter of the annulus: d a d1 d 2 d in. The critical velocity: vc 1.08(30) 1.08 (30)2 (9.3)(10)(3 83 )2 (10) v vc 3 8 (10)(3 ) 4.39 ft/sec laminar flow (use Eq. 3.6) p L 300d 5,500 pap 300 (3 83 ) 30 (3) 83 psi (3 8 ) pv Yb 5 d

53 53 (g) The total pressure drop in the system ( pt) pt ,166 psi (h) Horsepower output at the pump q p HP 1,714 v m... (3.17) where : q = flow rate, gpm v = volumetric efficiency m = mechanical efficiency Assuming volumetric and mechanical efficiencies of the pump are 90% and 85% respectively: HP 307 (1,166) 273 horsepower 1,714(0.90)(0.85)

54 54 Summary Bingham Plastic Model: Calculation Steps pt p s p p p c pb p ac p ap p vc s Eq. (3.8) E q. (3.1 6 ) o r F ig. 3.3 v Eq. (3.7a ) or (3.7b) de Eq. (3.15a) or (3.15b) pb No (lam in ar) p p, pc, pac, pap Eq. (3.6) if v vc (Eqn. 3.13) Yes (tu rb u len t) N Re Eq. (3.2) f (Fig. 7.1) p p, pc, pac, pap Eq. (3.4)

55 55 p p pu (3) m Example 3.4 Using a data as in Example 3.3, calculate the circulating pressure required. Solution 3.4 From Example 3.3: q = 307 gpm, bit = 3 13/32 in. nozzles (a) Surface equipment losses ( ps) q = 307 gpm Curve type 2 Fig. 7.3 pu 27 psi ps psi (3) (b) Pressure losses inside drill pipe ( pp) q = 307 gpm Fig. 7.5 (for 4.5 d/p) Curve 7 (assume ID = 3 ¾ ) p p (3) 32 pu 5, psi 1, psi 0.14

56 56 p p pu (3) m (c) Pressure losses inside drill collar ( pc) q = 307 gpm Curve 2 ¾ bore Fig. 7.7 (assume ID = 2 ¾ ) pc (3) 3 13 " psi Fig. 7.9 nozzle pb psi (d) Pressure losses through bit ( pb) q = 307 gpm pu (no viscosity effect) psi 9.5 p pu pu 550 psi m

57 57 p p pu (3) m (e) Pressure losses around drill collar ( pac) q = 307 gpm 6 ¾ drill collar Fig (bit size = 7 7/8 ) pac (3) psi 100 pu psi (f) Pressure losses around drill pipe ( pap) q = 307 gpm 4 ½ drill pipe Fig (bit size = 7 7/8 ) pap (3) pu 1.4 5, psi psi (g) The total pressure drop in the system ( pt) pt ,185 psi 0.14

58 58 Comparison of p Calculation Methods System component Plastic flow calculation (psi) Hughes Tools Co. charts (psi) Surface connections, ps Inside drill pipe, pp Inside drill collar, pc Bit nozzles, pb Outside drill collar, pac Outside drill pipe, pap ,166 1,185 Total circulating pressure, pt

59 59 Additional Information Besides Newtonian and Bingham Plastic Models, there are several other model used to predict pressure losses in mud circulating systems. Generally, each model is based on a set of assumptions which cannot be completely fulfilled in any drilling situation. Power law, Herschel-Bulkley (Yield Power API Power Law) models are the most widely used in the oil industry. Table 3.3 shows a summary of pressure loss equations

RIG HYDRAULICS. P 1 = Eρ 0.8 Q 1.8 (PV) 0.2. psi or bar. P 1 ρ. = pressure loss, psi or bar = Density, ppg or kg/l = flow rate, gpm, l/min

RIG HYDRAULICS. P 1 = Eρ 0.8 Q 1.8 (PV) 0.2. psi or bar. P 1 ρ. = pressure loss, psi or bar = Density, ppg or kg/l = flow rate, gpm, l/min RIG HYDRAULICS This chapter covers the following items: Pressure losses Surface connection losses Pressure drop across the bit Optimization of bit hydraulics Surface pressure drop Hydraulic criteria Comparison

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