Rotary Drilling Hydraulics

7.1 Introduction

Proper utilization of mud pump horsepower is of considerable importance to rotary drilling operations. Analytical appraisal of the rig's circulating system requires an understanding of the components which consume power, so that the available energy may be used as advantageously as possible. The standard hydraulics approach to such analyses is hindered by numerous factors, among which are:

  • 1) Mud flow property peculiarities, as discussed in Chapter 6
  • 2) Irregularities of the circulating system.

Drilling mud leaves the pump discharge, passes through the surface lines, standpipe, and mud hose, and finally enters the drill string at the top of the kelly joint. Here it begins the long downward travel through the drill pipe and drill collars, is expelled through the water courses or nozzles of the bit, and returns up the annulus?: The annular area is relatively small around the drill collars and becomes larger in the portion containing drill pipe. Since the mud enters the drill string and leaves the annulus at essentially the same elevation, the only pressure required is that necessary to overcome the frictional losses in the system. Hence the discharge pressure at the pump is defined by:

(7.1) AVt

= Aps + App + Apc + Apb + Apac -1- Apap

where Apf

= pump discharge pressure

Aps

= pressure loss in surface piping, standpipe,

and mud hose

A Pp

= pressure loss inside drill pipe

A pc

= pressure loss inside drill collars

A pb

= pressure loss across bit water courses or

nozzles

A Pac

= pressure loss in annulus around drill

collars

A pap

= pressure loss in annulus around drill pipe

The solution to Equation (7.1) is rather tedious, in that separate calculations for each section are required. However, with a little practice and understanding the task is not particularly formidable.

Before discussing plastic fluid flow calculations let us first review the fundamental equations of Newtonian fluid flow.

7.2 Newtonian Fluid Flow Calculations

Fluid flow through pipes is considered as either laminar or turbulent. In laminar (viscous) flow the fluid moves in parallel layers or laminae which are at all times parallel to the direction of flow. In turbulent flow, secondary irregularities and eddys are imposed on the main or average flow pattern. Calculation of pressure drop 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:

where Re = Reynold's number v = average velocity of flow, ft/sec, = q/2A5d2 p = fluid density, lb/gal d = pipe inside diameter, in. p = fluid viscosity, cp q = circulating volume, gal/min

It is commonly considered that if:

Re < 2000, flow is laminar Re > 4000, flow is turbulent 2000 < Re < 4000, flow is in transition, and is neither laminar nor turbulent.

The onset of turbulence is accelerated by any irregularity or entrance condition which will distort the flow pattern.

The pressure drop in laminar flow is given by the Hagan-Poiseuille law; this, in practical units, is pLv

1500 d2

where Ap = laminar flow pressure drop, lb/in.2 L = length of pipe, ft

For turbulent flow, Fanning's equation applies:

fpLv2

25.8 d where Ap = turbulent flow pressure drop, lb/in.2 / = Fanning friction factor

The friction factor / is a function of Re and pipe roughness, and has been evaluated experimentally for numerous materials (see Figure 7.1). Care must be exercised, because in some texts 4/ is plotted, instead of the / used here. This may be readily checked by noting that for all pipes, / ^ 0.013 at Re = 2000.

In summary, one can calculate pressure drop for Newtonian fluid flow systems in the following manner:

  • 1) Calculate Re from Eq. 7.2.
  • 2) If Re < 2000, use Eq. (7.3) to calculate the pressure drop.
  • 3) IfRe > 2000, useEq. (7.4). In this case the friction factor/is obtained from Figure 7.1 or its equivalent.

Detailed treatments of Newtonian fluid flow calculations may be found in numerous hydraulics texts and handbooks.

7.3 Plastic Fluid Flow Calculations

As was pointed out in Chapter 6, the flow behavior of plastic drilling fluids is complicated by the variation of apparent viscosity with rate of shear or flow. Consequently, the Newtonian fluid equations must be altered for application to typical drilling mud systems. An early worker on this subject was Pigott,1 whose data have been widely used for the construction of hydraulic tables and curves. Let us first consider the development of Beck, Nuss, and Dunn2 which incorporates the classic work of Bingham3 with the commonly applied Newtonian fluid flow equations just presented.

Laminar Flow Region

The typical pressure-velocity behavior of a plastic fluid flowing through a pipe was shown in Figure 6.2. A definite pressure (Yt) is required to initiate flow. True laminar flow is represented by the linear portion of the curve, the equation of which is:

where 144Ap = pressure drop, lb/ft2 4

Ö Yt = Yb, the Bingham yield value, lb/ft2 (commonly called yield point)

m, = slope of linear portion which is proportional to the plastic viscosity, ßP. Note: m = pL/1500 d2, from Eq. (7.3).

For practical values of g, the behavior of Bingham fluids may be expressed as:

1500d2 300d1

300ri where Yh = yield point, lb/100 ft2 pP = plastic viscosity, cp

Equation (7.6) 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.

Critical Velocity Calculation

If Eqs. (7.3) and (7.6) are equated, an equivalent Newtonian viscosity in terms of d, v, pp, and Yb is obtained:

Substituting Eq. (7.7) for p in the Reynold's number Eq. (7.2), equating the resulting equation to 2000, and solving for v gives:

144Ap = -Yt + mv where vc = critical velocity, ft/sec, above which turbulent flow exists and below which the flow is laminar.

Equation (7.8) assumes that turbulence occurs at Re = 2000. This is reasonable, since drill pipe rotation and tool joint irregularities promote turbulence at this lower value. Therefore, if:

v < vc, flow is laminar v > vc, flow is turbulent

Turbulent Flow Calculations

Fanning's equation may be used for turbulent flow calculations providing the Reynold's number expression is altered by substitution of

(7.9) pt = = turbulent viscosity of plastic fluids

This is the experimental finding of Beck, Nuss, and Dunn, i.e., that the viscosity of plastic fluids in turbulent flow is constant, as defined by Eq. (7.9). This premise has been criticised by Havenaar,4 who proposed a different expression based on the volume percent

0.002

Plastic Fluids Friction Factor

2000 10,000

Reynolds number, R,

I Lowest values for drawn brass or glass tubing (Walker, Lewis, & McAdams)

I For clean internal-flush tubular goods (Walker, Lewis, & McAdams)

H For full-hole drill pipe or annuli in cased hole (Piggatt's data)

W For annuli in uncased hole (Piggott's data)

100,000

0.002

1,000,000

0.002

2000 10,000

Reynolds number, R,

I Lowest values for drawn brass or glass tubing (Walker, Lewis, & McAdams)

I For clean internal-flush tubular goods (Walker, Lewis, & McAdams)

H For full-hole drill pipe or annuli in cased hole (Piggatt's data)

W For annuli in uncased hole (Piggott's data)

100,000

0.002

1,000,000

Fig. 7.1. Friction factor vs. Reynolds number courtesy solids in the mud. Fortunately, reasonable errors in viscosity determination have little effect on turbulent flow calculations, and Eq. (7.9) will be used for our present purpose.

Substitution of pt for p in the general Reynold's number expression (Eq. 7.2) gives:

for mud flow calculations. After Ormsby,13 API.

Solution:

  • 1) »C =
  • 4.3 ft/sec
  • 1.08X30) + (1.08)V(30)2 + (9.3)(10)(3.64)2(10) (10) (3.64)

1 ft3

Pt Pp

Figure 7.1 allows determination of the friction factor/ for drill pipe and open hole combinations. This / may then be used in Eq. (7.4) for calculation of pressure drop.

In summary, one can calculate pressure drop for plastic "fluids as follows:

  • 1) Calculate vc from Eq. (7.8)
  • 2) If v < vc, flow is laminar, and Eq. (7.6) applies
  • 3) If v > vc, flow is turbulent, requiring:
  • a) Calculation of Re from Eq. (7.10)
  • b) Determination of / from Figure 7.1 at the calculated Re for the conduit in question
  • c) Calculation of pressure drop from Eq. (7.4a).

Example 7.1

Mud is flowing through 4| inch OD, internal flush drill pipe. Calculate the frictional pressure drop per 1000 ft of pipe.

Mud properties pm = 10 lb/gal

Q ft3/sec

A ft2 g

2.45d2 400

2.45(3.64)2 (3) v > vc, and flow is turbulent.

  • a) Re = (2970)(10)3((;2-3)(3-64) = 44,300
  • b) / = 0.0062 from Curve II, Figure 7.1

_ (0.0062) (10) (1000) (12.3)2

  • c) App - (25.8) (3.64)
  • 100 psi/1000 ft 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 concept is satisfactory for this purpose and is defined as: cross-sectional area of flow stream

Pipe i.d. = 3.640 in. Circulating rate q = 400 gal/min wetted perimeter of conduit = hydraulic radius

For an annulus,

where r1 and r2 are the large and small radii, respectively. For a circular pipe,

2rr r 2

The frictional loss in an annulus is equal to the loss in a circular pipe having the same hydraulic radius; hence, in general terms,

where ra and da are the hydraulically equivalent radius and diameter of a circular pipe which may be used in the previous pressure drop equations.

Example 7.2

Mud is flowing through the annulus between an 8-in. hole and 4j-inch drill pipe. What value of d should be used in Eqs. 7.2, 7.3, 7.4, 7.6, 7.7, 7.8? Ans. (8 - 4i) = 3| in.

7.4 Pressure Drop Across Bit Nozzles and Watercourses

Figure 7.2 illustrates the flow of an incompressible fluid through a converging tube (nozzle, orifice, etc.). Assuming steady state, adiabatic, and frictionless conditions:

+ 2 g w '¿g w where Plt P2 = pressure, lb/ft2 w = density, lb/ft3 »„ v2 = velocities at points 1 and 2, ft/sec or

Practically, v2 (2)

w 2g

The ideal rate of flow, Q, = A2v2. The actual flow rate Q is:

(3) Q = CQi where C is the flow or nozzle coefficient for a particular design. With these substitutions, Eq. (2) becomes wQ2

2 gC2A2

Altering Eq. 7.12 to practical units for mud flow, we have:

An = qp P 7430C2d4

where d = nozzle or watercourse diameter, in.

Eckel and Bielstein6 have shown that C may be as high as 0.98 for properly designed jet bit nozzles; how-

Fig. 7.2. Schematic sketch of incompressible fluid flowing through a converging tube or nozzle.

ever, 0.95 is commonly used for field purposes. For ordinary watercourses, which are merely flat drilled holes, C = 0.80.

Multiple Nozzles

Normally a jet rock bit has the same number of nozzles as cones. The calculation of pressure drop across a multiple nozzle bit may be simplified by substituting the sum of the nozzle areas for A in Eq. 7.12. This fact follows from Eq. 7.12. For a single nozzle:

2 gC2A2

For several nozzles, each of area A i:

Therefore:

A Pm

Q2 A x2

AP Q2 Ax2 n2^!2^!2 It is desired to choose an A such that

A =nAu Q.E.D. Similarly, for use in Eq. 7.13, (7.15) d. = rfuT2

Where the multiple nozzles vary in size,

where a = number of nozzles having diameter dl b = number of nozzles having diameter d2, etc.

de — hydraulically equivalent single nozzle diameter, in.

Example 7.3

A 10 lb/gal mud is being circulated at the rate of 500 gal/ min through a tri-cone bit having three f-in. diameter jets. What is the pressure drop across the bit?

Solution:

d, = -^3(1)2 = 0.65 in. (equivalent single nozzle diameter) By Eq. (7.13),

7.5 Pressure Drop Calculations for a Typical System

To illustrate the pressure loss calculations for each system component, let us work a complete example. As stated in Eq. (7.1), the total pressure drop in the system is the sum of the losses in each section. Consider then the following operating conditions:

Operating Data

Depth = 6000 ft (5500 ft drill pipe, 500 ft drill collars)

Drill pipe = 4§ in. internal flush, 16.6 lb/ft (i.d. = 3.826 in.)

Mud density, pm = 10 lb/gal HP = 30 cp Yb = 10 lb/100 ft2

Nozzle velocity = at least 25 ft/sec per inch of bit diameter (this value is obtained by a commonly applied rule of thumb).

What hydraulic (pump output) horsepower will be required for these conditions?

Calculation Steps

  1. 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.
  2. The flow rate q is normally desired in gal/min:

q = Annulus area X velocity = 2.45 (dh2 - d/)v = 2.45 (62 - 20.2) (3) = 308 gal/min

3. Nozzle size: 3 nozzles (one for each cone) will be used, hence £ q will flow through each. For v = 250 ft/sec through each, d - aJÄ! ~ V(2.

The nearest stock nozzle is |f in.; this one is then chosen (see Figure 7.10). This nozzle allows an actual velocity of:

  • 225 ft/sec
  • 2.45)(13/32)2

4. Surface equipment losses: The surface equipment consists of the standpipe, swivel, kelly joint, and the piping between the pump and standpipe. Since this part of the system causes only a small fraction of the total losses, it will be satisfactory to choose the one of the precomputed cases shown in Figure 7.3 which most closely approximates the actual case. Assume Case 2 fits our example. Then,

5. Pressure drop inside drill pipe: The critical velocity is calculated from Eq. 7.8:

_ (1.08)(30) + 1.08V(30)2 + (9.3)(10)(3.826)2(10) (10)(3.826)

= 4.2 ft/sec The actual velocity inside drill pipe is:

" = 245d2 = (2.45^(^826)2 = 8'58 ft/seC Since 8.58 > 4.2, flow is turbulent and Eq. 7.4a applies: 2970pfJd

  • Eq. 7.10)
  • 2970) (10) (8.58) (3.826) 30
  • 32,500 From Figure 7.1, curve II
  • 0.0066 Applying equation 7.4a:

_ fpLv2 (0.0066) (10)(5500)(8.58)2 App 25.8 d (25.8) (3.826)

= 270 psi* 6. Pressure drop inside drill collars:

flow is turbulent, by inspection. ^ = (2970) (10) (15.9) (2.813) = ^^ 30

*It should be realized that the drill pipe may have two inside diameters; one for the pipe body and one for the tool joint. In such cases separate pressure drop calculations must be made for the total length of each i.d. This is commonly accomplished by computing an average i.d. for the pipe-tool joint combination. This complication was avoided in our examples by specifying the internally flush joint. The possible effect of tool joint design is, however, evident from the Hughes charts shown later.

Petroleum Engineering Charts

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