45827 Drilling Hydraulics

Chapter

3
Drilling Hydraulics
From rigging up to laying down the derrick, calculating hydraulics is animportant part of drilling a well economically. Chapter 3 demonstratescalculations for hydraulic and input horsepower required from the prime moverfor both gravity-feed and supercharged pumps. Equations can be used forcalculating fluid movement through the surface equipment, drillpipe, drillcollars, and bit, as well as back up the annulus to the flowline. These calculationswill help operators determine efficient and economical job procedures.

The tables in Chapter 3 include information about the effects of various jobparameters (the drilling fluid’s flow rate, weight, and viscosity as well as thesizes of the pipe, bit nozzle, and cuttings).

Calculating Hydraulic HorsepowerHydraulic horsepower (hhp) is a measure of the energy delivered to the fluid beingpumped. Hydraulic horsepower is different from the engine horsepower (hp)—hhp will always be less than the horsepower of the engine driving the pumpbecause of the pump efficiency. For example, a 1,000-hp motor driving a mudpump will normally produce 750 to 800 hhp.

Use Equation 3.1 to calculate hydraulic horsepower.

hhp = (P x Q) ÷ 1,714 ................................................................................................ (3.1)

where:

hhp = Hydraulic horsepower, hhpP = Pressure increase in pump, psiQ = Flow rate, gal/min1,714 = Constant

October 1996 Drilling Hydraulics 3-1

Halliburton Note

This file has been created for Halliburton use only. All information contained in this publication is confidential and proprietary property of Halliburton Energy Services, a division of Halliburton Company.

3-2 Genera

Example 1: How to calculate hydraulic horsepower

A mud pump delivers 420 gal/min at 1,000 psi and has a gravity-fed pumpintake. Therefore, the pump suction pressure is 0 psi. What is the hydrau-lic horsepower under these conditions?

Solution

hhp = (P x Q) ÷ 1,714 ............................................................................ (3.1)

= (1,000 psi x 420 gal/min) ÷ 1,714 = 245.041 hhp

Problem 1

What is the hydraulic horsepower of a pump delivering 1,260 gal/min at 333psi? This pump has a suction pressure of 0 psi.

Work Space

Answer ____________(The solution for Problem 1 is on Page 3-22.)

A comparison of the answers to Example 1 and Problem 1 shows that hydraulichorsepower output by high pressure and low rate is equal to the hydraulichorsepower output by low pressure and a high rate. Using high volume insteadof high pressure might seem to be more economical because high pressures aresometimes difficult to maintain. However, even though the hydraulic horse-power requirements are equal, the high-volume operation may not be as eco-nomical because reduced pressure may also decrease penetration rate.

When the mud pump is supercharged (pressurized suction) and is driven by aseparate prime mover, it is necessary to subtract the suction pressure from thedischarge pressure before calculating the hydraulic horsepower. Example 2demonstrates this calculation. When the same prime mover drives both thesupercharger and the mud pump, the total hydraulic horsepower is calculated asshown in Example 1.

l Hydraulics Manual October 1996

October 1996

Example 2: How to calculate hydraulic horsepower

What is the hydraulic horsepower of a pump delivering 420 gal/min at1,000 psi with a suction pressure of 100 psi?

Solution

P = Discharge pressure - Suction pressure = 1,000 psi - 100 psi = 900 psi

Q = 420 gal/min

hhp = (P x Q) ÷ 1,714 ............................................................................ (3.1)

= (900 psi x 420 gal/min) ÷ 1,714 = 220.538 hhp

A comparison of Examples 1 and 2 shows that the additional 100-psi suctionpressure decreased the hydraulic horsepower by about 24 hhp.

Problem 2

What is the hydraulic horsepower of a mud pump delivering 210 gal/min at1,000 psi with a suction pressure of 75 psi?

Work Space

Answer ____________(The solution for Problem 2 is on -22.)

Drilling Hydraulics 3-3

Calculating Input HorsepowerIf the efficiency of the pump is known, the horsepower required from the engineor the input horsepower (ihp) can be easily calculated.

The efficiency of the pump is always less than one, since the fluids beingpumped nearly always contain some air and are slightly compressible, and themechanical drive in the power end has some loss of energy. Overall efficiency(both volumetric and mechanical) of 70 to 80% is common, but some new pumpsapproach 100%.

Input horsepower can be calculated with Equation 3.2.

ihp = hhp ÷ E ............................................................................................................. (3.2)

where

ihp = Input horsepower, hp

E = Efficiencyhhp = Hydraulic horsepower, hhp

Example 3 and Problem 3 are applications of Equation 3.2. Problem 4 is a prac-tice problem that requires Equation 3.1 and Equation 3.2.

Example 3: How to calculate input horsepower

What input horsepower is necessary for a pump that is delivering 300hhp if the pump efficiency is 75%?

Solution

hhp = 300 hhp

E = 0.75

ih = hhp ÷÷÷÷÷ E ........................................................................................ (3.2)

= 300 hhp ÷ 0.75 = 400 hp

3-4 General Hydraulics Manual October 1996

October 1996

Problem 3

What input horsepower is necessary for a pump delivering 250 hhp when thepump efficiency is 80%?

Work Space


Problem 4

What input horsepower is required for a mud pump delivering 840 gal/min at1,100 psi if the pump efficiency is 75% and the suction pressure is 100 psi?

Work Space



Friction and Pressure Losses in DrillpipeSince the hydraulic power from the mud pump must be transmitted to thebottom of the hole (to aid the bit in removing chips) and back up the annulus toremove the cuttings, controlling the amount of hydraulic power reaching thesepoints is important. Pressure losses occur throughout the system, from the mudpump to the mud return line, which may be several miles in each direction. Forexample, a 33,000-ft well drilled in western Oklahoma had a mud round trip of12.5 miles.

Table 3.1 shows a typical listing of the pressure losses occurring from the pumpto the bit on a 15,000-ft well. The diameter of the hole was 7 7/8 in., requiring asurface pressure of 2,900 psi to circulate the 16.5-lb/gal mud at250 gal/min. The well contains the following equipment:

• 12,000 ft of 9 5/8-in. casing

• 14,300 ft of 4 1/2-in., 16.6-lb/ft drillpipe

• 700 ft of 6 1/4-in. OD, 2 3/4-in. ID drill collars

• two 3/8-in. bit jets

Table 3.2 shows typical pressure losses occurring from the bit to the surface forthe same well conditions described for Table 3.1.

Tables 3.1 and 3.2 show significant pressure losses. Drillpipe is one of the pri-mary sources of pressure losses because of friction. Losses of hydraulic energycaused by friction can never be recovered, and they must occur so that theappropriate amount of hydraulic energy can be supplied to the required point inthe system.

This situation is similar to buying a loaf of bread at the store where the totalprice includes freight cost. The freight cost cannot be eaten but the loaf of breadmust be at the store; therefore, the freight must be paid. On a drilling rig, acertain amount of hydraulic power is needed at the bit; therefore, the freightmust be paid by supplying additional hydraulic power at the surface to pay theway to the bottom and to lift the cuttings to the surface.

Several factors influence the amount of hydraulic energy expended to overcomefriction losses, including fluid velocity. Figure 3.1 (Page 3-8) illustrates two typesof flow: laminar flow (slow) and turbulent flow (fast). Normally, laminar flow inthe drillpipe occurs at extremely low pump rates with fluid velocities of 1 to2 ft/sec. Even heavy, thick muds flow smoothly. However, the drag of the fluidagainst the wall of the pipe and the internal friction of the fluid itself (viscosity)combine to create friction losses.

In normal drilling operations, turbulent flow is much more common. At thesehigher fluid velocities, drag from the pipe walls and changes in ID caused bytool joints cause eddies to form in the flow patterns. These eddies are smallwhirlpools that move within the main flowstream causing counterflow andcrossflow, which creates turbulence. Turbulent flow requires more energy thansmooth or laminar flow. The faster the flow, the greater the amount of energythat is absorbed by the turbulence and the greater the hydraulic power loss.


Table 3.1—Pressure Losses Inside Dr illstring

ComponentDepth at

Bottom of Component (ft)

Pressure Drop Across

Component (psi)

Total Pressure Drop to Bottom of Component

(psi)

Hydrostatic Pressure at Bottom of

Component (psi)

Total Pressure at Bottom of Component

(psi)

Pressure Gauge

0 0 0 0 2,900

Surface Connections

0 25 25 0 2,875

Drillpipe 14,300 673 698 12,247 14,449

Drill Collars 15,000 150 848 12,846 14,898

Bit 15,000 1,852 2,700 12,846 13,046

Table 3.2—Pressure Losses Outside Dr illstring

ComponentDepth at Top of Component (ft)

Pressure Drop Across

Component (psi)

Total Pressure Drop to Top of

Component (psi)

Hydrostatic Pressure at Top of Component

(psi)

Total Pressure at Top of

Component (psi)

Bit 15,000 1,852 2,700 12,846 13,046

Drill Collar in Open Hole

14,300 54 2,754 12,247 12,393

Drillpipe in Open Hole

12,000 34 2,788 10,227 10,389

Drillpipe in Casing

0 112 2,900 0 0

October 1996 Drilling Hydraulics 3-7

3-8 Genera

Laminar FlowÑSlow

LessEnergy

Required

Turbulent FlowÑFast

MoreEnergy

Required

Calculating Pressure Losses Inside Drillpipe DuringTurbulent FlowUse Equation 3.3 to calculate the pressure losses inside the drillpipe duringturbulent flow.

P = 7.7 x 10-5 x MW0.8 x Q1.8 x PV0.2 x L ÷ ÷ ÷ ÷ ÷ D4.8 ......................................................... (3.3)

where

P = Pressure losses in the drillpipe, psi7.7 x 10-5 = Constant

MW = Mud weight, lb/galQ = Flow rate, gal/minPV = Plastic viscosity, cpL = Length of pipe, ftD = Drillpipe ID, in.

Figure 3.1—Pressure losses in drillpipe


October 1996

Examples 4 and 5 show applications of Equation 3.3 and Equation 3.1 forhydraulic horsepower. Problem 5 requires Equation 3.1 and Equation 3.3.

Example 4: How to calculate pressure losses inside the drillpipeduring turbulent flow

10.0-lb/gal mud that has a plastic viscosity of 25 cp is being pumped at 210gal/min in 10,000 ft of 4 1/2-in., 16.60-lb/ft drillpipe. What is the pressuredrop?

Solution

MW= 10.0 lb/gal

Q = 210 gal/min

PV = 25 cp

L = 10,000 ft

D = 3.826 in.

P = (7.7 x 10-5) x (MW0.8 x Q1.8) x (PV0.2 x L) ÷ D4.8 ............................. (3.3)= (7.7 x 10-5) x (100.8 x 2101.8) x (250.2 x 10,000) ÷ 3.8264.8

= 0.000077 x (6.301 x 15,135.429) x (1.904 x 10,000) ÷ 626.828= 223.056 psi

Example 5: How to calculate hydraulic horsepower to overcomefriction loss

For the conditions described in Example 4, how much hydraulichorsepower is required to overcome the pressure drop caused byfriction?

Solution

P = 223.056 psi

Q = 210 gal/min

hhp = (P x Q) ÷ 1,714 ............................................................................ (3.1)

= 223.056 psi x 210 gal/min ÷ 1,714 = 27.330 hhp


3-10 Genera

Problem 5

What is the hydraulic power loss (hhp) in 5,000 ft of 3 1/2-in., 15.50-lb/ft drillpipewhen operators are circulating 15.0-lb/gal mud with a plastic viscosity of 50 cpat 150 gal/min?

Work Space


Factors Influencing Friction and Pressure Losses inDrillpipeTable 3.3 shows the pressure drop and the hydraulic horsepower loss for 10,000ft of 4 1/2-in., 16.6-lb/ft drillpipe at three different rates when 10.0 lb/gal mudwith a plastic viscosity of 25 cp is pumped. Doubling the flow rate increases thepressure drop by approximately three and one-half times, while tripling the flowrate increases the pressure drop by approximately seven and one-quarter times.Table 3.3 also shows that power losses can be even more significant: doublingthe rate increases the hydraulic horsepower loss by approximately seven times,and tripling the rate increases the loss by nearly 22 times.

Another important factor in friction losses or pressure losses is the fluid density.The heavier the mud, the greater the pressure and power losses. Table 3.4 com-pares the pressure losses for three mud weights and three flow rates. Pressurelosses were measured in 10,000 ft of drillpipe when 10.0-lb/gal mud with a plasticviscosity of 25 cp was pumped. Some of these pressure drops are excessive and areshown for comparison only. Doubling the mud weight (from 10 lb/gal to 20 lb/gal) almost doubles the pressure loss (1,613 psi compared to 2,809 psi).


Octobe

The mud’s viscosity also influences the pressure and power losses in the system,although not as significantly as the other factors. For example, increasing theplastic viscosity from 25 to 40 cp only increases the pressure losses by approxi-mately 10%.

Another major factor influencing the pressure and power losses is the drillpipeID. With a small ID, the fluid velocity must be greater for a given rate, creatinghigher turbulence and greater pressure and power loss. Table 3.5 (Page 3-12)compares the pressure losses for 10,000 ft of 3 1/2-in., 15.50-lb/ft to 4 1/2-in.,16.60-lb/ft drillpipe at three different circulating rates when 10.0-lb/gal mudwith a 25-cp plastic viscosity is pumped. Some of these pressure drops areexcessive and are shown for comparison only. Decreasing the drillpipe size from4 1/2-in., 16.60-lb/ft to 3 1/2-in., 15.50-lb/ft increases the pressure losses by ap-proximately six and one-third times. Even at reduced circulating rates com-monly used with smaller drillpipe, pressure losses are still high.

Because of the influence of the pipe ID, drillpipe size for a given hole sizeshould be reasonably large, yet not so large that it causes an excessive back-pressure on the formation from high annular-return velocity.

Table 3.3—Friction Losses in Drill pipe10.0-lb/gal Mud with 25-cp Plastic Viscosity

Flow Rate (gal/min)

Pressure Drop (psi)

Power Loss (hhp)

210 223 27

420 778 191

630 1,613 593

Table 3.4—Pressure Losses in Dr illpipe10.0-lb/gal Mud with 25-cp Plastic Viscosity

Flow Rate (gal/min)

Pressure Drop (psi) for 10.0 lb/gal mud



210 223 308 338

420 778 1,076 1,354

630 1,613 2,231 2,809

r 1996 Drilling Hydraulics 3-11

The factors influencing pressure losses in drillpipe include the following:

• flow rate

• fluid density

• mud viscosity

• drillpipe ID

Pressure and power losses in the drill collars are similar to those for drillpipeand are influenced by the same factors as the drillpipe. The reduced ID in thedrill collars increases the velocity and causes a higher friction loss per foot thanin drillpipe. Fortunately, drill collars are much shorter than drillpipe.

Bit HydraulicsAs drilling fluid enters the water courses of the bit and exits the bit nozzle, itsflow rate is greatly increased. Fluid travels through the drillpipe at about10 ft/sec. As it exits the nozzles of the bit, the fluid may be traveling in excess of400 ft/sec (more than 270 mile/hr). The increase in kinetic energy for eachgallon of 10.0-lb/gal fluid is even greater—from about 15 ft/lb to approximately25,000 ft/lb of energy.

Just as it is with the drillpipe, the system must pay for transportation of the fluidthrough jets with a pressure drop in the entire system. With two-thirds of thetotal pressure drop as an average, the drop across the bit is about twice thepressure drop across the rest of the system. In a well-balanced hydraulic system,the pressure drop across the nozzles ranges from one-half to three-quarters ofthe entire system’s pressure drop.

Table 3.5—Pressure Losses in Drillpipe10.0-lb/gal Mud with 25-cp Plastic Viscosity

Flow Rate (gal/min)

Pressure Drop (psi) in 4 1/2-in., 16.60-lb/ft Drillpipe

Pressure Drop (psi) in 3 1/2-in., 15.50-lb/ft Drillpipe

210 223 1,421

420 778 4,948

630 1,613 10,266


October 1996

Calculating Nozzle VelocityTo find the pressure drop through the bit, first calculate the nozzle velocity at thejets. Use Equation 3.4 to find the nozzle velocity.

NV = 0.32 x Q ÷ AN .................................................................................................. (3.4)

where

NV = Nozzle velocity, ft/secQ = Flow rate, gal/minAN = Total area of all nozzles, in.2

0.32 = Constant

Example 6 is an application of Equation 3.4.

Example 6: How to calculate nozzle velocity

What is the nozzle velocity if three Number 8 jets are used with a flowrate of 210 gal/min?

Solution

Diameter of Number 8 jet = 8/32 in. = 0.25 in.

Area of one jet = 0.7854 x 0.25 in. x 0.25 in. = 0.049 in.2

AN = 3 jets x 0.049 in.2 = 0.147 in.2

Q = 210 gal/min

NV = 0.32 x Q ÷ AN .............................................................................. (3.4)= 0.32 x 210 gal/min ÷ 0.147 in.2 = 457.143 ft/sec

Calculating Pressure Losses Through the BitOnce the velocity through the nozzle(s) has been calculated, use Equation 3.5 tocalculate the pressure losses through the bit based on the change in kinetic energy.

P = MW x NV2 ÷÷÷÷÷ 1,120 .............................................................................................. (3.5)

where

P = Pressure loss through the bit, psiMW = Mud weight, lb/galNV = Nozzle velocity, ft/sec1,120 = Constant

Example 7 shows a calculation using Equation 3.5.


3-14 Gener

Example 7: How to calculate pressure drop across the bit

Use the conditions for Example 6. What is the pressure drop acrossthe bit if the mud weight is 10.0 lb/gal?

Solution

MW= 10.0 lb/gal

NV = 457.143 ft/sec (from Example 6)

P = MW x NV2 ÷ 1,120 ....................................................................... (3.5)= 10.0 lb/gal x (457.143 ft/sec)2 ÷ 1,120 = 1,865.889 psi

Problem 6

Two Number 16 jets are used with 15.0-lb/gal mud at a 210-gal/min flow rate.

A. What is the nozzle velocity?

Work Space

Answer ____________

B. What is the pressure loss through the bit?

Work Space


al Hydraulics Manual October 1996

October 1996

Factors Influencing the Pressure Drop Across the BitA review of Equation 3.4 for nozzle velocity and Equation 3.5 for pressure lossesthrough the bit shows that three things affect the pressure drop across the bit:

• mud weight

• flow rate

• nozzle size

Increasing the nozzle area reduces both the velocity and the pressure drop acrossthe bit. When the mud weight is increased, the pressure drop increases propor-tionately. Larger flow rates also increase the pressure drop across the bit.

Table 3.6 shows nozzle exit velocities for various flow rates and nozzle sizes.Some of the high velocities shown in the table are unattainable; they are shownto illustrate the importance of nozzle selection. Doubling the flow rate doublesthe exit velocity at the bit nozzles (from 210 gal/min to 420 gal/min) and in-creases the velocity (from 457 ft/sec to 914 ft/sec for three Number 8 nozzles).Doubling the nozzle diameter (from three Number 8 jets to three Number 16 jets)quadruples the nozzle area (from 0.147 in.2 to 0.589 in.2) and reduces the velocityby a factor of four (457 ft/sec ÷ 4 ≅ 114 ft/sec).

Table 3.7, Page 3-16, shows the pressure drops with the same nozzles and ratesas Table 3.6. Several of the pressure drops shown are not attainable, but can beused for comparison. Notice that fluid density (mud weight) is an importantfactor affecting the pressure drop across the nozzles. Doubling the mud weight(from 10 lb/gal to 20 lb/gal) also doubles the pressure drop across the nozzles(1,866 psi x 2 = 3,732 psi at 210 gal/min using three Number 8 jets).

Nozzle size also affects pressure. At 210 gal/min with 10-lb/gal mud, doublingthe nozzle diameters (three Number 8 jets to three Number 16 jets) reduces thepressure drop by a factor of 16 (1,866 psi ÷ 16 = 116.6 psi). Flow rate also affectsthe pressure drop. With 10.0-lb/gal mud and three Number 8 jets, doubling theflow rate (from 210 gal/min to 420 gal/min) increases the pressure drop by afactor of four (1,866 psi x 4 = 7,464 psi).

Table 3.6—Nozzle Exit Velocities— Exit Velocity (ft/sec), Nozzle Area (in. 2)

Flow Rate (gal/min)

Three No. 8 Jets 0.147



210 457 203 114

420 914 406 228

840 1,828 812 456


3-16 Gen

Power is necessary to remove cuttings from the bottom of the hole. Since it isnormally necessary to maintain the hydrostatic pressure at the bottom of thehole at a higher value than the pressure in the pores of the formation, the cut-tings are securely held against the face of the rock. The high-velocity jettingaction causes the drilling fluid to penetrate among the cuttings and even into theminute fractures on the bottom of the hole, partially equalizing the differentialpressure existing at the rock face that otherwise tends to hold the cuttings down.Reducing this differential pressure then allows the high-velocity fluid to removethe cuttings quickly and easily. Figure 3.2 illustrates this effect.

Table 3.7—Pressure Drop Across Bit Jets

Flow Rate Mud Weight

Pressure Drop (psi)

Nozzle Area (in. 2)

(gal/min) (lb/gal) Three No. 8 Jets 0.147


Three No.16 Jets 0.589

10 1,866 368 116

210 15 2,799 552 174

20 3,732 736 232

10 7,464 1,472 465

420 15 11,195 2,208 697

20 14,927 2,944 929

10 29,854 5,888 1,860

840 15 44,781 8,832 2,789

20 59,708 11,776 3,719

eral Hydraulics Manual October 1996

October 1

Figure 3.2—Jets help relieve chip holddown.

High-velocityfluid-jet actionreleases chips

fractured bybit teeth.

Solids filtered fromthe mud seal across

cracks in fracturedrocks.

Mud filtrate invasion into fractures lessens chiphold-down.

Hydrostatic pressurecauses chiphold-down.

Pore pressure is lessthan hydrostaticpressure.

996 Drilling Hydraulics 3-17

Annular HydraulicsAnnular hydraulic principles differ from drillpipe or bit hydraulic principles inseveral ways. First, annular flow is much slower; in fact, flow in the annulus isnormally considered to be laminar as opposed to the turbulent flow inside thepipe. Second, the shape of the flow passage is not just round but generally isconsidered more doughnut-shaped, as shown in Figure 3.3. Even this assump-tion does not generally hold true. Since the pipe is not always centered in thehole, the annular flow passage is actually more crescent-shaped (Figure 3.4).However, for calculation purposes (since the cross-sectional areas are the same)the doughnut approach is usually used. Third, the irregularities in the hole(washed-out sections or sloughing formations) cause the openhole section todisturb the laminar flow pattern. Fourth, the rotation of the pipe (as crescentrevolves around the inside of the hole) also tends to break up the laminar flow inthe annulus.

Figure 3.4

Figure 3.3

Crescent-ShapedAnnulus

Drillpipe

Doughnut-ShapedAnnulus

Drillpipe


October 1996

During slow, hard rock drilling, an annular velocity of 1 1/2 to 2 ft/sec (about 100to 125 ft/min) is adequate to bring cuttings to the surface. For fast top-holedrilling, experience shows that higher annular velocities will be required, espe-cially when laminar flow exists.

The principles that apply to flow in the drillpipe also apply to the annular flow.The higher the flow rate, the greater the pressure drop caused by friction. Simi-larly, high-viscosity mud also increases the annular pressure drop slightly; thebackpressure at the bottom increases, which tends to hold down the chips.

Calculating Slip VelocityAnnular velocities must be sufficient to carry the cuttings to the surface. Cuttingsdo not travel at the same speed as the fluid. The difference in the fluid velocityand the cutting velocity is known as the slip velocity. A higher slip velocity meansthat a cutting “slips” more in the fluid carrying it.

To calculate slip velocity, use Equation 3.6.

(( ))SV =

175 PD PW -MWMW cp0.333 0.333

×× ××××

0 667.

where

SV = Slip velocity, ft/min

175 = ConstantPD = Particle (cutting) diameter, in.PW = Particle weight, lb/gal (normally about 21.0 lb/gal)MW = Mud weight, lb/gal

cp = Mud viscosity, cp

Example 8 (Page 3-20) is an application of Equation 3.6.

........................................................................... (3.6)


3-20 Gener

Example 8: How to calculate slip velocity

What is the slip velocity of 0.20-in. diameter cuttings in 10 lb/gal of 40-cp mud?

Solution

PD = 0.20 in.

PW = 21.0 lb/gal

MW = 10.0 lb/gal

cp = 40 cp

(( ))

(( ))

SV =175 PD PW -MW

MW CP

=175 0.20 2.10-10.0

10.0

=175 0.20 4.9500

2.153 3.146=

0.667

0.333 0.333

0.667

0.333

×× ××××

×× ××××

×× ××××

400 333.

23.561 ft / min

Problem 7

What is the slip velocity of 0.30-in. diameter cuttings when 9.5-lb/gal mud withan 89-cp viscosity is circulated?

Work Space


........................................................... (3.6)

al Hydraulics Manual October 1996

October 1996

Factors that Influence Slip VelocityTable 3.8 compares slip velocities for various mud weights, viscosities, andcutting sizes. Some of the viscosities shown in Table 3.8 are unattainable with themud weights listed, but are included for comparative purposes.

Viscosity causes a small change in slip velocity. Changing the viscosity of10.0-lb/gal mud from 40 to 80 cp only decreases slip velocity from 23.6 ft/min to18.7 ft/min, a reduction of approximately 21%; the cuttings are traveling closerto the mud velocity in 80-cp mud than in 40-cp mud.

Table 3.8 shows that doubling the cutting size approximately doubles the slipvelocity. A 0.20-in. diameter cutting in 20-lb/gal, 40-cp mud has a slip velocity of3.8 ft/min compared to 7.6 ft/min for a cutting twice its size. The larger thecutting, the greater the difference in its velocity and the fluid velocity.

Table 3.8 also shows that the higher the mud weight, the lower the slip velocity.With 40-cp mud carrying 0.20-in. cuttings, increasing the mud weight from 10lb/gal to 20 lb/gal decreases the slip velocity from 23.6 ft/min to 3.8 ft/min.Heavier mud, therefore, has a much greater carrying capacity because the slipvelocity is much less.

Table 3.8—Slip Velocity

Slip Velocity (ft/min)

Viscosity (cp)

Mud Weight (lb/gal)

Cutting Size (in.)

0.20 0.30 0.40

40 10 23.6 35.3 47.1

20 3.8 5.7 7.6

80 10 18.7 28.1 37.4

20 3.0 4.5 6.0


Solutions to Problems

Pages 3-22 and 3-23 list the solutions to the practice problems in Chapter 3. Mostof the problems can be solved more than one way. Different methods may giveslightly different answers, depending on how the numbers are rounded off.

NOTE If you use and understand a particular method, always use it, even if it is notused in this text.

Solution for Problem 1P = 333 psiQ = 1,260 gal/minhhp = (P x Q) ÷ 1,714 ................................................................................................ (3.1)

= (333 psi x 1,260 gal/min) ÷ 1,714 = 244.796 hhp ................................ Answer

Solution for Problem 2P = Discharge pressure - Suction pressure

= 1,000 psi - 75 psi = 925 psiQ = 210 gal/minhhp = (P x Q) ÷ 1,714 ................................................................................................ (3.1)

= (925 psi x 210 gal/min) ÷ 1,714 = 113.331 hhp ................................... Answer

Solution for Problem 3hhp = 250 hhpE = 0.80

ihp = hhp ÷ E ............................................................................................................ (3.2)= 250 hhp ÷ 0.80 = 312.5 hp ...................................................................... Answer

Solution for Problem 4P = Discharge pressure - Suction pressure = 1,100 psi - 100 psi = 1,000 psiQ = 840 gal/minhhp = (P x Q) ÷ 1,714 ................................................................................................ (3.1)

= (1,000 psi x 840 gal/min) ÷ 1,714 = 490.082 hhp

E = 0.75ihp = hhp ÷ E ........................................................................................................... (3.2)

= 490.082 hhp ÷ 0.75 = 653.442 hp ........................................................... Answer


October 1996

Solution for Problem 5MW = 15.0 lb/gal

Q = 150 gal/minPV = 50 cpL = 5,000 ftD = 2.602 in.P = (7.7 x 10-5) x (MW0.8 x Q1.8) x (PV0.2 x L) ÷ D4.8 .............................................................................. (3.3)

= (0.000077 x 8.727) x (8,259.700 x 2.187) x (5,000) ÷ 98.509 = 616.051 psihhp = (P x Q) ÷ 1,714 ............................................................................................... (3.1)

= (616.051 psi x 150 gal/min ) ÷ 1,714 = 53.913 hhp ............................ Answer

Solution for Problem 6A. Q = 210 gal/min

AN = 2 x 0.196 in.² = 0.392 in.²NV = 0.32 x Q ÷ AN ........................................................................................... (3.4)

= 0.32 x 210 gal/min ÷ 0.392 in.² = 171.429 ...................................... Answer

B. MW= 1.50 lb/galNV = 171.429 ft/secP = MW x NV² ÷ 1,120 ................................................................................... (3.5)

= 15.0 x 171.429² ÷ 1,120 = 393.586 ..................................................... Answer

Solution for Problem 7PD = 0.30 in.PW = 21.0 lb/gal

MW = 9.5 lb/galcp = 89 cp

175 x PD x (PW - MW)0.667

MW0.333 x cp0.333

175 x 0.30 in. x (21.0 - 9.5)0.667

9.50.333 x 890.333

SV = ............................................................................ (3.6)

= = 28.373 ft/sec ......................................... Answer



Notes

45827 Drilling Hydraulics

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