XIV-B

FLUID FLOW DESIGN PRACTICES

SINGLE-PHASE LIQUID FLOW Section

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PROPRIETARY INFORMATION - For Authorized Company Use OnlyDate

December, 1998

EXXON RESEARCH AND ENGINEERING COMPANY - FLORHAM PARK, N.J.

EXXONENGINEERING

CONTENTSSection Page

SCOPE .................................................................................................................................................... 3

REFERENCES ......................................................................................................................................... 3

DESIGN PRACTICES (BESIDES OTHER SECTIONS OF THIS SECTION)....................................... 3

OTHER LITERATURE...................................................................................................................... 3

DEFINITIONS........................................................................................................................................... 3

BASIC DESIGN CONSIDERATIONS........................................................................................................ 3

GENERAL CONSIDERATIONS ........................................................................................................ 3

PRINCIPLES OF PRESSURE DROP CALCULATION....................................................................... 4

NON-NEWTONIAN LIQUIDS............................................................................................................ 5

HORIZONTAL STRAIGHT PIPE ....................................................................................................... 5

EFFECT OF FITTINGS..................................................................................................................... 6

EXPANSIONS AND CONTRACTIONS.............................................................................................. 6

NON-HORIZONTAL PIPES .............................................................................................................. 6

COMBINING AND DIVIDING OF STREAMS ..................................................................................... 6

ORIFICES, FLOW NOZZLES AND VENTURIS ................................................................................. 7

PERFORATED PIPE DISTRIBUTORS.............................................................................................. 7

DESIGN PROCEDURES .......................................................................................................................... 8

PRESSURE DROP ACROSS SINGLE PIPING COMPONENTS........................................................ 8Straight Pipe.................................................................................................................................. 8Perforated Pipe Distributors ..........................................................................................................13

INTEGRATED PRESSURE DROP CALCULATION FOR PIPING SYSTEMS....................................15

SAMPLE PROBLEMS.............................................................................................................................17

PROBLEM 1 - INTEGRATED PRESSURE DROP CALCULATION ...................................................17

PROBLEM 2 - ORIFICE PRESSURE DROP CALCULATION ...........................................................21

PROBLEM 3 - PERFORATED - PIPE DISTRIBUTOR ......................................................................21

NOMENCLATURE ..................................................................................................................................24

COMPUTER PROGRAMS.......................................................................................................................25

GUIDANCE AND CONSULTING......................................................................................................25

LITERATURE..................................................................................................................................25

AVAILABLE PROGRAMS................................................................................................................25

Changes shown by ç

DESIGN PRACTICES FLUID FLOW

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SINGLE-PHASE LIQUID FLOW

DateDecember, 1998 PROPRIETARY INFORMATION - For Authorized Company Use Only


EXXONENGINEERING

CONTENTS (Cont)Section Page

TABLESTable 1A Design Basis for “Average” Carbon Steel Lines (Customary)...............................................26Table 1B Design Basis for “Average” Carbon Steel Lines (Metric) ......................................................26Table 2 Typical Process Line Equivalent Lengths............................................................................27Table 3 Representative Equivalent Length of Various Valves, in Pipe Diameters (L/D) .....................28

FIGURESFigure 1 Friction Factors for Any Type of Commercial Pipe (Customary) (Metric) ..............................29Figure 2A Friction Factors for Clean Commercial Steel and Wrought Iron Pipe (Customary) ................30Figure 2B Friction Factors for Clean Commercial Steel and Wrought Iron Pipe (Metric) .......................31Figure 3A Liquid Pressure Drop in Commercial Steel Pipe (Customary) ..............................................32Figure 3B Liquid Pressure Drop in Commercial Steel Pipe (Metric) .....................................................39Figure 4A Approximate Liquid Pressure Drop in Commercial Pipe (Customary)...................................46Figure 4B Approximate Liquid Pressure Drop in Commercial Pipe (Metric) ..........................................47Figure 4C Pressure Drop in Commercial Pipe for Water at 75°F (Customary) ......................................48Figure 4D Pressure Drop in Commercial Pipe for Water at 24°C (Metric).............................................49Figure 5A Equivalent Lengths L and L/D and Resistance Coefficient K for Valves (Customary)............50Figure 5B Equivalent Lengths L and L/D and Resistance Coefficient K for Valves (Metric) ...................51Figure 5C Resistance Coefficient for Bends, Ells and Tees (Customary) .............................................52Figure 5D Resistance Coefficient for Bends, Ells and Tees (Metric).....................................................53Figure 5E Resistance Coefficients for Return Bends and Miter Bends .................................................54Figure 6 Resistance Coefficient for Cross-Section Changes .............................................................55Figure 7A Flow Coefficient for Orifices with Flange Taps.....................................................................56Figure 7B Flow Coefficient for Orifices with Flange Taps.....................................................................56Figure 8 Flow Coefficient for Flow Nozzles with Flange Taps............................................................57Figure 9A Equivalents of Resistance Coefficient K and Flow Coefficient Cv for Valves (Customary) .....58Figure 9B Equivalents of Resistance Coefficient K and Flow Coefficient Cv for Valves (Metric).............59Figure 10 Pressure Recovery Factor for Orifices, Nozzles and Venturis ..............................................60Figure 11 J Factor for Calculating Distributor Head Loss ....................................................................61

Revision Memo

12/98 Highlights of this revision are:

1. Drafting and other errors in figures have been corrected: Figures 2A, 2B,3A, 3B, 4A, 4B, 4C, 4D, 5C.



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SCOPE

This section presents methods for determining pressure drop through piping and related equipment for Newtonian liquids. Forgeneral design considerations other than pressure drop, see Section XIV-A.

REFERENCES

DESIGN PRACTICES (BESIDES OTHER SECTIONS OF THIS SECTION)

Section III Fractionating Towers

Section XI Compressors

Section XII Instrumentation

OTHER LITERATURE

1. Perry, R. H. and Chilton, C. H., Chemical Engineers’ Handbook, 5th ed., Section 5, Fluid and Particle Dynamics, McGraw-Hill, New York (1973).

2. Crane Co., Technical Paper No. 410, Flow of Fluids through Valves, Fittings and Pipe, 1991.

3. Simpson, L. L., Process Piping: Functional Design, Chem. Eng., 76 No. 8, (Deskbook Issue) 167-181, (April 14, 1969).

4. Patel, R. D., Non-Newtonian Flows, in Handbook of Fluids in Motion, Ann Arbor Science Publishers (1983).

5. Westaway, C. R. and Loomis, A. W., eds., Cameron Hydraulic Data, Ingersoll-Rand, 15th ed (1997).

6. Fluid Meters, Their Theory and Application, ASME Report, 5th Ed., (1959).

7. Greskovich, E. J. and O’Bara, J. T., Perforated-Pipe Distributors, I. & E.C. Process Design and Dev. 7 (4) 593-595 (1968).

8. Zenz, F. A., Minimize Manifold Pressure Drop, Hydrocarbon Proc. & Petr. Ref. 41 (12) 125-130 (1962).

9. Golan, L. P. and Hawkins, L. E., Single Phase Flow Distribution in Manifolds, ER&E Report EE.74E.75 (August, 1975).

DEFINITIONSSee Section XIV-A.

BASIC DESIGN CONSIDERATIONS

The considerations discussed below provide the basis for calculation procedures given later in this section.

GENERAL CONSIDERATIONS

In most piping designs, the primary requirement is to find an inside diameter that will permit a certain required throughput at agiven pressure drop. This usually involves a trial and error procedure. A diameter is chosen and the pressure drop iscalculated for the required throughput. If the calculated pressure drop is too great, a larger diameter is taken for the next trial.If the pressure drop is smaller than necessary, a smaller diameter is chosen.

Typical pressure drops that may be used for pipe sizing are shown in Table 1. In case of expensive construction materials, aneconomic analysis would be desirable to find the optimum line size. In cases of very high pressure and steam traced lines, itmay also be desirable to find the optimum line size.


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EXXONENGINEERING

BASIC DESIGN CONSIDERATIONS (Cont)

PRINCIPLES OF PRESSURE DROP CALCULATION

The basic equation for calculating pressure drop for liquid flow in pipes and fittings is the generalized Bernoulli equation, whichassumes constant density:

{

( )44 344 21

443442143421 loss head or Friction

FK +

change Elevation

g

z g +

changeenergyKinetic

g 2

V

change Pressure

P K12

cc

211 ∆∆α

=ρ∆

Eq. (1)

where: K11 = 144 (Customary)

K11 = 10-3 (Metric)

F = Friction or head loss, ft lbf/lbm (kPa⋅m3/kg)

g = Acceleration of gravity, ft/s2 (m/s2)

gc = Dimensional constant, 32.174 ft lbm /lbf s2 (1.0 kg m/N s2)

K12 = 1.0 (Customary)

K12 = 10-3 (Metric)

∆P = Pressure change, lbf/in.2 (kPa): inlet pressure - outlet pressure

V = Velocity of the fluid, ft/s (m/s)

∆ V2

2

= Change in fluid kinetic energy (outlet-inlet) ft2/s2 (m2/s2)

z = Elevation, ft (m)

∆z = Change in elevation (outlet elev - inlet elev), ft (m)

ρ = Density, lbm/ft3 (kg/m3)

α = Constant depending on velocity profile (α = 1.1 for turbulent flow, α = 2.0 for laminar flow)

All design equations presented in this section are derived from this equation. The significance of the terms are as follows. The“Pressure change” term is the pressure drop (inlet minus outlet pressure), and for most cases this drop is positive. The“Kinetic energy change” is the outlet fluid kinetic energy minus the inlet fluid kinetic energy per unit mass of fluid flowing, andmay be positive or negative. The “Elevation change” is the outlet elevation minus the inlet elevation (really the change inpotential energy per unit mass flowing) and may also be positive or negative. The “Friction or head loss” term is alwayspositive, and represents the irreversible conversion of mechanical energy to internal energy. Inspection of Eq. (1) shows thatthe kinetic and elevation terms may result in a positive or negative pressure drop, but the friction term always results in apressure decrease, or positive pressure drop. The relative importance of the terms in the equation varies from application toapplication. For constant-diameter horizontal pipes, only the friction term on the right-hand side of Equation (1) is non-zero.For vertical or inclined pipes, one must include the elevation term; and for cross-section changes, the kinetic energy term.

For liquids one may, in general, assume constant viscosity and density. Non-Newtonian liquids are an exception to this ruleand are discussed below. Another exception is non-isothermal flow, due either to heat exchange, or to heat production orconsumption in the liquid by chemical reaction or friction losses.Where the flow may be assumed to be isothermal across the pipe cross-section, but is not isothermal along the length of thepipe, the pressure drop can be determined by dividing the pipe into a number of lengths and calculating the pressure drop ineach section. When the flow cannot be assumed to be isothermal across the pipe cross-section and the viscosity dependsstrongly on temperature, special calculation methods must be used. When problems of this type arise, consult the Reactor andFluid Dynamics Section of Exxon Engineering.



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NON-NEWTONIAN LIQUIDS

Fluids in which viscosity is dependent on shear rate or time are called non-Newtonian. The following fluids may belong to thiscategory:

• Dense emulsions

• Dense slurries

• Polymer solutions

• Polymer melts

• Some heavy crude oils

For these fluids, the regular liquid flow equations are not applicable.

Most non-Newtonian fluids belong to one of the following classes:

• Bingham plastics - A minimum shear stress is required to obtain flow.

Examples: fine suspensions and pastes.

• Pseudoplastic fluids - Viscosity decreases with increasing shear rate (velocity gradient). These are commonly referred toas “shear thinning” fluids.

Examples: polymer solutions and melts, some slurries, emulsions.

• Dilatant fluids - Viscosity increases with increasing shear rate. These are commonly referred to as “shear thickening”fluids. These are less common than Bingham plastics and pseudoplastic fluids. Certain slurries exhibit dilatant behavior.

Viscosity may also be time-dependent:

• Thixotropic fluids - Viscosity decreases with time after application of a constant shear stress.

• Rheopectic fluids - Viscosity increases with time after application of a constant shear stress.

• Viscoelastic fluids - Liquids that partially return to their original form when the shear stress is released.

Because of the complicated rheological behavior of non-Newtonian fluids, it is not possible to give a single generally validdesign equation for pressure drop. In general, rheological properties will first have to be determined in the laboratory for asuitable shear stress range and time frame.For problems which arise in handling non-Newtonian liquids, consult the Reactor and Fluid Dynamics Section. An introductionto simple calculation methods for pressure drop is given in Reference (4).

HORIZONTAL STRAIGHT PIPE

Pressure drop in horizontal straight pipe of constant diameter is caused by friction and can be calculated from the Fanningfriction equation. The experimental factor in this equation, called the Fanning friction factor, f, is a function of Reynolds numberand relative pipe wall roughness (Figure 1). For a given class of pipe material, roughness is relatively independent of pipediameter; therefore the friction factor can be expressed as a function of Reynolds number and pipe diameter (Figures 2A and2B). For laminar flow (Re < 2100), the friction factor is independent of pipe wall roughness and can be expressed as a functionof Reynolds number alone [Eq. (4)].A transition region lies between Reynolds numbers of about 2,100 and 4,000. Here the flow may be intermittently laminar andturbulent or essentially fully turbulent, depending on such factors as change of cross-section or presence of valves, fittings orobstructions in the piping. In this regime, the friction factor is difficult to determine and lies somewhere between the limits forlaminar and turbulent flow. For most commercial applications, however, flow tends to be turbulent and the higher value of thefriction factor should be used.The accuracy of the Fanning friction equation is ± 15% for smooth tubing and ± 10% for commercial steel pipe. Fouling canreduce the cross-sectional area or increase pipe wall roughness with time. Therefore, when calculating pressure drop, oneshould allow for fouling.


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EXXONENGINEERING


Most studies of the effect of fouling on pressure drop have been for water piping. For such piping, instead of the Fanningcorrelation, an empirical correlation known as the Hazen-Williams correlation has been most widely used. The correlationcontains a coefficient known as the H-W “C” factor, which is used to account for surface condition and fouling. Reference 5compiles head loss tables versus pipe size and flow rate along with recommended “C” factors for various types of service. Forfurther information, consult the Reactor and Fluid Dynamics Section.Note: In some references [e.g., (2)] the Darcy friction factor, fD, is used instead of the Fanning friction factor. They are related

by the equation fD = 4f, and calculated results are identical.

EFFECT OF FITTINGS

Bends, tees, valves, orifices and other flow restrictions cause additional pressure drop in a pipe. Fittings that have the samenominal diameter as the pipe can be accounted for in terms of an equivalent length of straight pipe. This equivalent length canbe calculated from the resistance coefficients of the fittings [Figures 5A - 5E and 6 and Eq. (17)]. The equivalent length is thenadded to the actual length of the pipe and the sum is used in the Fanning equation for predicting the total friction pressure drop.It should be recognized during design that the actual resistance coefficient of bends, tees, and valves may deviate from thevalues presented in Figures 5 and 6 by as much as ± 25%.

Also, the use of equivalent lengths or resistance coefficients is, as published, essentially an approximate correlation of acomplex problem. If pressure drop is a critical factor for safety, economic, or other considerations, consult with the Reactor andFluid Dynamics Section.

When piping details are not available, the following guidelines may be used for estimating equivalent length:

Onsite Lines - Actual pipe length can be estimated from the plot plan, tower heights, etc. For a rough estimate the equivalentlength of fittings in onsite piping adds between 200% and 500% to the actual length. Accordingly, a multiplier of 3.0 to 6.0 maybe applied to the estimated length of straight pipe. A better estimate can be obtained from Table 2 as long as the componentsin the circuit are known. If the designer wants even better accuracy, he must know all the components in the circuit, make apreliminary pipe routing on the plot plan and finally add an appropriate allowance.

Offsite Lines - For offsite lines, the approximate length of straight pipe can be estimated from the plot plan. Since fittings inoffsite lines usually have an equivalent length of 20% to 80% of the actual length, a multiplier of 1.2 to 1.8 can be applied to theestimated length of straight pipe.

EXPANSIONS AND CONTRACTIONS

The pressure drop in cross-section changes, such as exits and entrances of process vessels, reducers and diffusers, consistsof two components: one for friction and one for change in kinetic energy. Calculation of the friction loss is based on thediameter of the smaller of the two pipes with no obstructions.For pipes ending in an area of very large cross-section, such as a process vessel, the frictional pressure drop is equal to thegain in pressure caused by the change in kinetic energy. As a result, the net pressure change over the cross-section change iszero.

For a very gradual contraction, friction pressure drop is calculated based on a straight piece of pipe with inside diameter equalto the narrowest cross-section of the contraction.

In pressure drop calculations for lines containing fittings and cross-section changes, the line is first broken into sections ofconstant nominal diameter. The friction pressure drop of each change in cross-section is accounted for in the equivalent lengthof the smaller diameter pipe attached to it as defined by Eq. (16). The pressure drop due to the various changes in kineticenergy in the line is determined by calculating the overall change in kinetic energy between the inlet and outlet of the line.

NON-HORIZONTAL PIPES

In case of non-horizontal pipes, an elevation term must be added to the pressure change calculated for friction loss and kineticenergy [Eq. ( 6)].

COMBINING AND DIVIDING OF STREAMS

When a stream is split in two or more substreams there is both a friction loss and a pressure change due to the change inkinetic energy. The same applies to the combining of streams. For tees the total pressure change is given by Eq. (8). For Y’ssee Reference 8, and for manifolds see Reference 9. Further information may be obtained by consulting with the Reactor andFluid Dynamics Section.



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ORIFICES, FLOW NOZZLES AND VENTURISFor orifices, flow nozzles, and venturis, two pressure drops can be distinguished:

Flow Measurement Pressure Drop is a pressure drop used in connection with flow measurements, which does not includepressure recovery downstream of the constriction. For orifices and flow nozzles, this pressure drop is measured across flangetaps; for venturis, between an upstream tap and a tap at the narrowest cross-section. The flow coefficients presented inFigures 7A, 7B, and 8 relate to this pressure drop.

Total Pressure Drop is the pressure drop between a point upstream of the restriction and a point several pipe diametersdownstream. This pressure drop is smaller than the flange tap pressure drop because of pressure recovery (i.e., conversion ofmomentum into pressure) downstream of the restriction. The total pressure drop can be obtained by multiplying the flange tappressure drop by a pressure recovery factor (Figure 10). For determining the pressure drop along a line containing an orifice,flow nozzle, or venturi, one must use the total pressure drop. If the orifice or flow restriction is at the end of a line discharginginto a large vessel or the ambient, there is no pressure recovery and the flow measurement pressure drop should be used.For more detailed information on the flow measurement aspects of orifices, flow nozzles, and venturis see Section XII.

PERFORATED PIPE DISTRIBUTORS

In most cases, perforated pipe distributors can be designed by the short procedure given in Section III-H. These designsmainly apply to liquid-liquid settlers and fractionation towers where a relatively low velocity is more important than liquiddistribution. In these cases, some non-uniformity in liquid distribution may occur, depending on the ratio of the pressure dropacross the distributor holes to the pressure drop (or gain) along the pipe. If inertial forces predominate over friction losses inthe pipe, flow through the holes will increase in the direction of the closed end. If friction loss along the pipe is more importantthan inertial forces, the opposite will be the case. When an upstream disturbance, such as that produced by a bend, issuperimposed upon a case where inertial forces predominate, flow through the holes near the distributor inlet and near theclosed end can be greater than in the middle.The degree of maldistribution in a liquid distributor can be predicted from Eq. (15). Where less than 5% maldistribution isrequired, the design procedure given in this section must be used. In this procedure, pressure drop across the holes, (∆P)o, isset at ten times the greater of either inlet kinetic energy per unit volume of flowing fluid, Ek, or pressure drop over the length ofthe distributor pipe, (∆P)p.

The following guidelines should be followed for choosing hole diameter and number of holes:

• Minimum hole diameter ≈ 1/2-in. (13 mm) to avoid plugging and to limit the number of holes to a reasonable value. In veryclean service, smaller holes may be considered, but in severely fouling service, 1/2-in. (13 mm) holes may be too small.

• Maximum hole diameter = 0.2 times inside diameter of distributor.

• The ratio of hole diameter, do, to inside pipe diameter, d1, should be between 0.15 and 0.20 when the criterion(∆P)o = 10 Ek is used. If it is necessary to use do /d1 < 0.10, then make (∆P)o = 100 Ek.

• To provide sufficient pipe strength, the minimum distance (edge-to-edge) between adjacent holes should approximatelyequal the hole diameter.

• Within the limitations imposed by the above requirements, a larger number of small holes is preferred over a smallernumber of large holes.

• If slots are used instead of holes, the slot width should be at least 1/2-in. (13 mm).To assure optimum distribution, flow conditions upstream and downstream of the distributor should be considered. Conditionsupstream of the distributor are controlled by the piping outside of the unit. In general, this means minimizing the number andseverity of sharp turns or sudden contractions or enlargements just ahead of the distributor. Conditions downstream of thedistributor depend on the geometry of the downstream internals, which are usually designed to maintain uniform distribution forgood contacting.


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DESIGN PROCEDURES

The following design methods, equations, and guidelines must be used together with the material given above under BASICDESIGN CONSIDERATIONS. The first section below presents procedures for calculating pressure drop in single pipingcomponents. The second section should be used for calculating pressure drop in flow systems containing more than onepiping component.

PRESSURE DROP ACROSS SINGLE PIPING COMPONENTS

Use the procedures below for calculating frictional pressure drop across single piping components, such as runs of straightpipe, bends, valves, orifices, etc.

Straight Pipe

For commercial steel pipe, find the pressure drop in psi/100 ft (kPa/m) from the charts in Figures 3A and 3B or 4A - 4D. Forconditions not covered by these charts or more precise answers, use the procedure given below. [For non-circular conduits,calculate the equivalent hydraulic diameter from Eq. (2). Note: This is valid only for turbulent flow, and deq should be usedonly to calculate the Reynolds number, Eq. (3), and the frictional pressure drop, Eq. (5). It should not be used to calculatevelocity, V, from the flow rate, Q.]

dcross area

wetted perimeter in consistent unitseq =

4

- sectional Eq. (2)

Step 1 - For given diameter and flow rate, calculate the Reynolds number, Re, from the following equation:

µ

ρ=

µρ

=DV

K DV

Re3a

Eq. (3a)

µρ

d Q

K =3b

Eq. (3b)

µ d

W K =

3cEq. (3c)

where: D = Inside diameter of pipe or equivalent hydraulic diameter, ft (m)d = Inside diameter of pipe or equivalent hydraulic diameter, in. (mm)Q = Volumetric flow rate, gpm (dm3/s)Re = Reynolds number, dimensionlessV = Velocity, ft/s (m/s)W = Mass flow rate, thousands lbm/h (kg/s)ρ = Density, lbm /ft3 (kg/m3)µ = Viscosity, cP (Pa⋅s)

Customary Metric

K3a = 123.9 10-3

K3b = 50.6 1.27

K3c = 6.31 x 103 1.27 x 103



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DESIGN PROCEDURES (Cont)

Step 2 - Look up the friction factor, f, in Figures 1A, 1B or 2A, 2B. For values of Re lower than those covered bythese figures, with Re < 2100 (Laminar Flow), calculate f from the Eq. (4):

f = 16

ReEq. (4)

where: f = Friction factor, dimensionless

Step 3 - Calculate the frictional pressure drop from one of the following equations:

( )

ρ

=∆c

2

5af g 2

V

DL f 4

KP Eq. (5a)

ρd

VL f K=

2

5bEq. (5b)

ρ5

2

5c d

Q L f K= Eq. (5c)

ρ 5

2

5d d

WL f K= Eq. (5d)

where: (∆P)f= Frictional pressure drop, psi (kPa)L = Pipe length, ft (m)

Customary Metric

K5a = 1/144 10-3

K5b = 5.18 x 10-3 2

K5c = 8.63 x 10-4 3.24 x 106

K5d = 13.4 3.24 x 1012

Step 4 - In case the pipe is not horizontal, calculate the pressure drop due to the change in elevation from thefollowing equations:

( ) ( )1 2c

6ae z z g

g K=P −ρ

∆ Eq. (6a)

( )1 26b z z K −ρ= Eq. (6b)

where: (∆P)e = Pressure drop due to change in elevation, psi (kPa)z1, z2 = Elevation of beginning and end of pipe, ft (m)

Customary Metric

K6a = 1/144 10-3

K6b = 6.94 x 10-3 9.81 x 10-3

Step 5 - Find the total pressure drop by adding the frictional pressure drop (∆P)f and the pressure drop due to changein elevation (∆P)e.


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Bends - Use the following procedure:

Step 1 - Find the resistance coefficient, K, in Figure 5C or 5D.

For pipes larger than 10 in. (250 mm) I.D., use the resistance coefficient for 10 in. (250 mm) I.D. pipe. If theReynolds number is such that the flow is not in the region of complete turbulence (f is constant), the value ofK should be multiplied by the ratio:

f

f(at calculated Reynolds number)

(in range of complete turbulence)

Step 2 - Calculate the frictional pressure drop from the following equations:

( )

ρ∆

c

2

7af g 2 VK

K =P Eq. (7a)

( )2

7b VK K ρ= Eq. (7b)

ρ=4

2

7c d

Q K K Eq. (7c)

ρ=

4

2

7dd

K W K Eq. (7d)

where:

Customary Metric

K7a = 1/144 10-3

K7b = 1.08 x 10-4 5.0 x10-4

K7c = 1.8 x 10-5 810

K7d = 0.28 8.10 x 108

Step 3 - For long non-horizontal bends, add the pressure drop due to the change in elevation calculated from Eq. (6).

Step 4 - For 90° mitered bends, the curves and table in Figure 5E may be used. If minimizing pressure drop is criticaland the design is based on the use of smooth bends or mitered bends with many segments, care must betaken during detailed design and construction to make sure that miters with few segments are not installed.

Tees and Y’s - For blanked-off tees and Y’s, use Eq. (7) and the resistance coefficients for tees in Figure 5C or 5D. For teesin which streams are split or joined, the pressure drop should be calculated from the following equations (Reference 8):

1. Split Flow

1 2

3

( ) ( . ) ( . . . )∆P x V V V V1 24

22

12

1 2108 10 136 0 64 0 72−−= − −ρ Eq. (8a)



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1 2

3

( ) ( . ) ( . . )∆P x V V V1 34

32

1 3108 10 18 0 368−−= −ρ Eq. (8b)

1 2

3

( ) ( . ) ( . . )∆P x V V V3 14

12

1 3108 10 18 0 368−−= −ρ Eq. (8c)

2. Join Flow

1 2

3

( ) ( . ) . .∆ P x V V V VQQ

VQQ1 2

422

12

2 33

21

1

2108 10 2 0 05 2 0 205−

−= − − +

ρ Eq. (8d)

1 2

3

( ) ( . ) . .∆ P x V V V VQ

QV

Q

Q1 34

32

12

3 11

32

2

3108 10 2 0 4 0 41−

−= − − +

ρ Eq. (8e)

1 2

3

( ) ( . ) . .∆ P x V V V VQ

QV

Q

Q3 14

12

32

1 33

12

2

1108 10 2 0 4 2 0 205−

−= − − +

ρ Eq. (8f)

Equations (8a-f) account for both frictional pressure drop and pressure drop due to change in kinetic energy. A multiplyingfactor of 1.25 has been used in these equations to allow for entrance and exit effects when the length of the inlet leading line isshort (L/D < 10). For Y’s, equations similar to Equations (8a-f) can be derived with a method presented in Reference 8.Reference 9 contains procedures for the pressure drop in manifolds. When more accurate calculation of pressure drop for Y’sor in manifolds is required, consult the Reactor and Fluid Dynamic Section of Exxon Engineering.

Valves - Find the resistance coefficient, K, by using L/D values in Table 3 and Figure 5A or 5B. Use the same procedure asused for bends. Figure 9 can be used to determine flow coefficient Cv from K.

Orifices - For calculating the “measurement” pressure drop (as measured across flange taps), use the following equations:


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ρ∆

4o

2

2

9a d C

Q K =P Eq. (9a)

ρ=

4o

2

2

9b d C

W K Eq. (9b)

where: C = Flow coefficient, dimensionless (Figure 7A or 7B)do = Orifice diameter, in. (mm)

Customary Metric

K9a = 1.8 x 10-5 810

K9b = 0.28 8.10 x 108

To obtain the total pressure drop (including pressure recovery downstream of the orifice), multiply ∆P from Eq. (9) by thepressure recovery factor, r, of Figure 10.

Flow Nozzles - Use same procedures as for orifices, except with a flow coefficient from Figure 8.

Venturis - For calculating pressure drop as measured across venturi taps (one upstream and one at the narrowest crosssection - diameter do), use Eq. (9) with the following flow coefficient:

( )41o d/d1

0.98 = C

−Eq. (10)

where: d1 = Inside diameter of upstream pipe, in. (mm)

To obtain the total pressure drop, multiply ∆P from Eq. (9) by the pressure recovery factor of Figure 10.

Contractions and Expansions - Use the following procedure:

Step 1 - Look up the appropriate resistance coefficient, K, in Figure 6.Step 2 - Calculate the frictional pressure drop from the following equations:

( ) g 2 VK

K Pc

2

7af

ρ=∆ from Eq. (7a)

( )27b V K K ρ= from Eq. (7b)

ρ=

4

2

7cd

Q K K from Eq. (7c)

ρ=

4

2

7d d

K W K from Eq. (7d)

where: d = Inside diameter or equivalent hydraulic diameter of the smaller diameter pipe, in. (mm)V = Velocity in smaller-diameter pipe, ft/s (m/s), and K7a-d given on Page 10

Calculate the frictional pressure drop in a gradual contraction as if it were a pipe with diameter equal tothe smallest diameter in the contraction.



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Step 3 - Calculate the pressure drop due to change in kinetic energy of the flow from the following equation:

( ) ( )21

22

c11aVV

g 2 K kP −

ρ=∆ Eq. (11a)

( )21

22

11bVVK −ρ= Eq. (11b)

−ρ=

41

42

211c d

1

d

1 Q K Eq. (11c)

−

ρ=

41

42

2

11d d

1

d

1W K Eq. (11d)

where: d1, d2 = Upstream and downstream inside diameters or equivalent hydraulic diameters, in.(mm)

V1, V2 = Upstream and downstream velocities, ft/s (m/s)

Customary Metric

K11a = 1/144 10-3

K11b = 1.08 x 10-4 5.0 x 10-4

K11c = 1.8 x 10-5 810

K11d = 0.28 8.10 x 108

Step 4 - For non-horizontal gradual contractions and expansions, calculate the pressure drop due to change inelevation from Eq. (6).

Step 5 - Calculate the total pressure drop by adding the pressure drops obtained from Equations (6), (7) and (11).

Perforated Pipe Distributors

Use the following procedure for designing perforated pipe distributors with less than 5% maldistribution. (Note: In liquid-liquidsettlers and fractionation towers low velocity is more important than liquid distribution and the procedures in Section III-Hshould be followed.)

Step 1 - For the first trial, set the distributor pipe diameter, d, equal to that of the inlet line.

Step 2 - Calculate the Reynolds number, Rei of the inlet stream from Eq. (3).Step 3 - Find the friction factor, f, from Figures 1A, 1B or 2A, 2B.

Step 4 - Calculate the kinetic energy per unit volume of the inlet stream, Ek, in psi (kPa ) from the following equations:

ρα

c

2i

12ak g 2 V

K =E Eq. (12a)

2ib12

V K ρα= Eq. (12b)

ρα=4

2

12c d

Q K Eq. (12c)

ρ

α=

4

2

12d d

W K Eq. (12d)


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where: α = Velocity correction factor, dimensionless (Use α = 1.1 for turbulent flow, and α = 2.0 forlaminar flow)

Customary Metric

K12a = 1/144 10-3

K12b = 1.08 x 10-4 5.0 x 10-4

K12c = 1.8 x 10-5 810

K12d = 0.28 8.10 x 108

Step 5 - Calculate the pressure change (∆P)p along the pipe due to friction and momentum recovery from thefollowing equations:

( )

ρ

α−=∆c

2i

13a21a13

g 2 V

Kd

J Lf KpP Eq. (13a)

k 13b E 1d

JLfK

−

α= Eq. (13b)

where: J = Dimensionless factor from Figure 11 (Use J = 0.35 for first trial)

Customary Metric

K13a1 = 48.0 4 x 103

K13a2 = 1/144 10-3

K13b = 48.0 4 x 103

Step 6 - Find the required pressure drop, (∆P)o, across the outlet holes by multiplying the greater of Ek or (∆P)p by 10.If the calculated value of (∆P)o is less than 0.25 psi (1.75 kPa), make (∆P)o equal to 0.25 psi (1.75 kPa).

Step 7 - Calculate the required total area of the outlet holes from the following equations:

( )o14ao P

CQ

K A∆

ρ= Eq. (14a)

( )

∆ρ=

o14b P

1CW

K Eq. (14b)

where: Ao = Total required hole area, in.2 (mm2)

Customary Metric

K14a = 3.32 x 10-3 22.3

K14b = 0.415 22.3 x 103

For the first trial, take the flow coefficient C equal to 0.60.



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Step 8 - Choose a hole diameter and number of holes to obtain the desired Ao based on the guidelines presentedunder “Basic Design Considerations.”

Step 9 - Calculate Rei / n, where n is the number of holes of the distributor. If Rei / n < 4,000, look up a new flowcoefficient in Figure 7, taking Re in this figure equal to Rei / n.

Step 10 - Using the calculated number of holes, find the factor J from Figure 11 and compare this with the assumedvalue of 0.35. If this revised value of J affects the value of (∆P)o by more than 10%, substitute the revisedvalue of J in Eq. (13) and repeat Steps 5 through 10.The maldistribution in a pipe distributor can be calculated from the following equation:

% Maldistribution = 100 ( ) ( )

( )∆ ∆

∆

P - P

Po p

o

−

1 Eq. (15)

INTEGRATED PRESSURE DROP CALCULATION FOR PIPING SYSTEMS

Use the procedure below for calculating pressure drop in any flow system containing more than one piping component.

Step 1 - Constant Flow and Nominal-Diameter Sections - Break the system in question into sections of constantflow rate and constant nominal diameter. Apply Steps 2 through 6 to each of the sections.

Step 2 - Equivalent Hydraulic Diameter - For any section having a non-circular cross-section, calculate theequivalent hydraulic diameter, deq, from Eq. (2).

Step 3 - Reynolds Number (not needed for rough estimates) - Find the Reynolds number, Re, for each section fromEq. (3).

Step 4 - Friction Factor (not needed for rough estimates) - Find the friction factor, f, for commercial steel pipe fromFigure 2. For other materials use Figure 1 or the correction factors in Figure 2.

For Reynolds numbers smaller than 2100, find the friction factor from Eq. (4).Step 5 - Equivalent Length of Fittings - If piping details are not available, assume for offsite lines that the equivalent

length of fittings lies between 20 and 80% of the actual pipe length and for onsite lines 200 to 500%.Estimate pipe length from the plot plan, tower heights, etc.When the fittings are known or can be estimated, find their total equivalent length, Leq, from the followingequation:

L = K4 f

Keq16d

∑ Eq. (16)

where: Leq = Equivalent length of all fittings, ft (m)∑K = Sum of resistance coefficients of all fittings, dimensionlessK16 = 1/12 (Customary)

K16 = 10-3 (Metric)

The resistance coefficient, K, of bends, blanked-off tees, and valves is found in Figures 5A to 5E as functionof nominal pipe diameter. For fittings larger than 10 in. I.D., use the resistance coefficient for 10 in. (250 mm)I.D. fittings in Figures 5C or 5D.

The K of contractions and expansions is found in Figure 6, based on the smaller diameter pipe that isattached to them.For orifices, flow nozzles and venturis, K should be calculated from the following equation:

K = r

C

dd2

1

o

4

Eq. (17)

where: r = Pressure recovery factor (Figure 10), dimensionless


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For orifices and flow nozzles obtain C from Figure 7 or 8.

For venturis:

( )C =

0.98

1 d / do 14−

from Eq. (10)

Step 6 - Frictional Pressure Drop, (∆∆P)f - Find this for commercial steel pipe from Figure 3A or 3B or Figure 4 bymultiplying the pressure drop in psi per 100 ft (kPa per m) by the sum of the actual pipe lengths and theequivalent lengths of all fittings divided by 100 (1 for metric). For more precise answers, or conditions notcovered by the charts, calculate the frictional pressure drop in each section of the system from Eq. (5), butuse for L the sum of the actual pipe lengths and the equivalent lengths of all fittings.

Step 7 - Overall Kinetic Energy Change, (∆∆P)k - For each constant-flow section, check the flow cross-sections atbeginning and end. If they are not equal, calculate the pressure change (∆P)k due to the change in kineticenergy from Eq. (11). Note that (∆P)k can be either positive or negative.

Step 8 - Overall Elevation Change (∆∆P)e - For each constant-flow section, check the elevation at beginning and end.If it is not equal, calculate the resulting pressure change (∆P)e from Eq. (6). Note that (∆P)e can be eitherpositive or negative.

Step 9 - Total Pressure Drop per Constant-Flow Section - Find the total pressure drop in each constant-flowsection from the following equation:

( P) = ( P) + ( P) + ( P)t f k e∆ ∆ ∆ ∆∑ Eq. (18)

where: (∆P)t = Total pressure drop, psi (kPa)∑ (∆P)f = Sum of frictional pressure drops in all constant nominal diameter sections, psi (kPa)

Step 10 - Stream Junctions - For tees, calculate the pressure drop from Eq. (8). For Y’s or manifolds see Reference8 or Reference 9, respectively, or contact the Reactor & Fluid Dynamics Section of Exxon Engineering.

The pressure drop over the entire system is obtained by combining the pressure drops in the various streamjunctions with the pressure drops across the various constant-flow sections calculated in Step 9.



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SAMPLE PROBLEMS

PROBLEM 1 - INTEGRATED PRESSURE DROP CALCULATION

Given: The following flow system, with a constant flow rate through the piping system, from the process vessel to the pump.

DP14B

Gate Valve

50 psig(345kPa)

20' (

6.1m

)

15' (

4.6m

)

90o Bend

NPS 3in. (75mm)Orifice, do = 2" (50mm)

25' (7.6m) 10' (3m)

NPS 4 in. (100mm)

Diffuser, θ = 40o

Customary Metric

Liquid flow rate = 200 gpm 12.6 dm3/s

Liquid density = 50 lbm/ft3 800 kg/m3

Liquid viscosity = 0.30 cP 0.3 x 10-3 Pa•s

Find: The suction pressure of the pump.

Solution:Step 1 - Divide the system into three sections of constant nominal diameter:

a. Section I includes the vessel.

b. Section II includes the contraction at the bottom of the vessel, the NPS 3 (75 mm) line and the diffuser.

c. Section III includes the NPS 4 (100 mm) line, from the diffuser to the pump.

Steps 2 through 6, where applicable, will now be carried out for each of these three sections to find the friction pressure drops.


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SAMPLE PROBLEMS (Cont)

Section I - Because of low velocity in the vessel, friction pressure drop can be neglected.

Section IIStep 2 - Pipe is round.

Step 3 - From Section XIV-A, Table 1: Inside pipe diameter d = 3.068 in. (77.9 mm). Reynolds number (Eq. 3b):

Re = (0.30)(3.068)

(50)(200)(50.6) =

dQ50.6µ

ρ = 550,000 (rounded) (Customary)

= )10x(0.3(77.9)

(800)(12.6)(1.27) =

dQ1.27

3-µρ

= 550,000 (rounded) (Metric)

Step 4 - From Figure 2A or 2B: Friction factor, f = 0.0046, essentially fully turbulent

Step 5 - First determine the various resistance coefficients:

a. Resistance coefficient of the contraction at the vessel bottom: K = 0.5 (Figure 6).

b. Resistance coefficient for the gate valve: L/D = 13 (Table 3), thus K = 0.23 approximately (Figure 5A or5B).

c. Resistance coefficient for the 90o long radius ell: K = 0.25 (Figure 5C or 5D).

d. The orifice has an orifice diameter do of 2 in. (50 mm); therefore:

1

o

dd

= 0.65 = 3.068

2.0(Customary)

= 0.65 = 9.77

50(Metric)

Flow coefficient of orifice: C = 0.67 (Figure 7B).

Pressure recovery factor: r = 0.58 (Figure 10).

Use Eq. (17) to calculate the resistance coefficient of the orifice:

( )( )

7.15 =2

3.068

0.67

0.58 =

dd

C

r =K

4

2

4

o

12

(Customary)

( )( )

7.15 = 50

77.9

0.67

0.58 =

4

2

(Metric)

e. Resistance coefficient of the diffuser: β = 3.026/4.068 = 0.76; K = 0.2 approximately (Figure 6).

Sum of resistance coefficients: ∑ K = 0.5 + 0.23 + 0.25 + 7.15 + 0.2 = 8.33

Obtain the equivalent length of valve and other fittings from Eq. (16):

( ) ( ) ( )

ft116= 8.33 0.00464

(3.068)12

1

=K f48

d =Leq ∑

(Customary)

( ) ( ) ( )∑

m35.3 = 8.33

0.00464(77.9)

10=K f4d10

= 3--3

(Metric)



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Step 6 - Sum of actual pipe length and equivalent lengths of all fittings:

L = 40 + 116 = 156 ft (Customary)

L = 12.2 + 35.3 = 47.5 m (Metric)

Specific gravity, S, of the liquid relative to water at 60°F:

S = 50

62.37 = 0.80

Kinematic viscosity νµ

= = =S

cSt0 30

0 800 37

.

.. (Customary)

Kinematic viscosity ν = =−

100 30 10

8000 376

32.

. /x

mm s (Metric)

Use Eq. (5c) to calculate the frictional pressure drop:

( ) psi 4.56(3.068)

(50)(200)(156)(0.0046))10x (8.63P

5

24

f ==∆−

(Customary)

kPa31.4(77.9)

(800)(12.6)(47.5)(0.0046))10x (3.24P)(

5

26

f ==∆ (Metric)

Alternatively, from Figure 3A or 3B, at a flow rate of 200 gpm (12.6 dm3/s) and a kinematic viscosity of0.37 cSt (0.37 mm2/s), ∆P/L is 3.7 psi/100 ft (0.84 kPa/m).

The friction pressure drop in Section II is therefore:

( ) ( )

psi 4.63 = 0.8023.7 100156

= P

∆(Customary)

( ) ( ) ( ) kPa 32.0 = 0.8020.84 417.6= P∆ (Metric)

Generally, Eq. (5) will give more precise answers than Figure 3A or 3B and is the recommended method.

Section IIIStep 2 - Pipe is round. From Section XIV-A, Table 1: Inside diameter of 4-in. (100 mm) Pipe = 4.026 in.

(102.3 mm).

Since Section III does not contain any fittings, Steps 3 through 5 may be omitted.

Step 6 - Length of Section III: L = 10 ft (3.05 m)

Use Eq. (3b) to calculate the Reynolds number in Section III:

(rounded)420,000(0.30)(4.026)

(50)(200)(50.6)d

Q 50.6 = Re ==

µρ

(Customary)

(rounded)420,000)10x(0.3(102.3)

(800)(12.6)(1.27)d

Q 1.27 = Re

3-==

µρ

(Metric)

From Figure 2: Friction factor, f = 0.0044

Use Eq. (5c) to calculate the frictional pressure drop:

psi 0.072(4.026)

(50)(200)(10)(0.0044))10x (8.63P)(

5

24

f ==∆−

(Customary)

kPa0.49(102.3)

(800)(12.6)(3.05)(0.0044))10 x (3.24P)(

5

24

f ==∆−

(Metric)


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Alternatively, from Figure 3A or 3B, at a flow rate of 200 gpm (12.6 dm3/s) and a kinematic viscosity of0.37 cSt (0.37 mm2/s), ∆P/L is 0.9 psi/100 ft (0.20 kPa/m).

The friction pressure drop in Section III is therefore:

psi 0.07 = (0.802) (0.9) 10010

=P

∆ (Customary)

( ) kPa 0.49 = (0.802) (0.20) 3.05 = (Metric)

Step 7 - Assume that the diameter of the vessel is very large compared to the pipe diameter.

Eq. (11c) now gives the overall pressure change due to change in kinetic energy from the liquid surface inthe vessel to the outlet of the 4-in. (100 mm) pipe:

( ) ( )

−ρ=∆

41

42

25k

d

1

d

1 Q -10 x 1.8 P (Customary)

= ( )( ) ( )( )

psi 0.14 04.026

1 200 50 -10 x 1.8

4

25 =

− (Customary)

( )

−ρ=∆

41

42

2k

d

1

d

1 Q 810 P (Metric)

= ( ) ( )( )

kPa 0.94 0102.3

1 12.6 800 810

4

2 =

− (Metric)

Step 8 - The pressure change due to change in elevation is given by Eq. (6b) taken from the liquid surface in thevessel to the outlet of the 4-in. (100 mm) pipe:

( ) ( )123

e zz 10x94.6P −ρ=∆ − (Customary)

( )( ) ( ) psi 6.9- 20050 10 x 6.94 -3 =−= (Customary)

( )1 2-3 zz 10x 9.81 −ρ= (Metric)

( ) ( ) kPa 47.9- 6.10800 10 x 9.81 -3 =−= (Metric)

Step 9 - The total pressure drop can be calculated with Eq. (18):

∑ ∆∆∆∆ ekft P)( + P)( + P)( = P)(

psi2.136.90.140.072)4.56(0P)( t −=−+++=∆ (Customary)

kPa 15.1=47.90.94+0.49)+31.4+(0 = P)( t −−∆ (Metric)

Therefore, the suction pressure of the pump is:

tP)(50 ∆− = 50 − (− 2.13) = 52.1 psig (Customary)

tP)(345 ∆− = 345 − (− 15.1) = 360 kPa gage (Metric)

Answer: P = 52.1 psig (360 kPa gage)



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PROBLEM 2 - ORIFICE PRESSURE DROP CALCULATION

Find: Determine the measurement pressure drop across the flange taps of the orifice in Problem 1 and compare this to thecontribution of the orifice plate to the pressure drop of the NPS 3 (75 mm) pipe section.

Step 1 - Calculate the pressure drop as measured across the flange taps using Eq. (9).

∆P = K9a (2) (0.67)

(200) (50) )-10 x (1.8 =

d C

Q 42

25

4o

2

2ρ= 5.01 psi (Customary)

= 42

2

(50.8)(0.67)

(12.6)(800) (810) = 34.4 kPa (Metric)

Step 2 - Calculate the contribution of the orifice plate to the total pressure drop. Recovery factor r = 0.58.

∆P = (5.01) (0.58) = 2.91 psi (Customary)

= (34.4) (0.58) = 20 kPa (Metric)

The total pressure drop across the NPS 3 pipe section was 4.56 psi (31.4 kPa).

PROBLEM 3 - PERFORATED - PIPE DISTRIBUTOR

Given: A 12.00 in. (305 mm) ID pumparound return line carries 2,000 gpm (126 dm3/s) of a hydrocarbon stream with a densityof 40 lbm/ ft3 (640 kg/m3) and a viscosity of 0.8 cP (0.8 x 10-3 Pa•s).

Find: For an 8-ft (2.43 m) perforated pipe distributor, determine the pipe diameter, and the number and size of holes that willprovide good distribution.

Solution:Step 1 - For the first trial, set the distributor pipe diameter equal to the line size:

d = 12.00 in. ID (305 mm)

Step 2 - Obtain the Reynolds number from Eq. (3).

µρ

= d

Q 50.6 Rei (Customary)

( ) ( ) ( )( ) ( )0.8 12.00

40200050.6 = (Customary)

(rounded) 420,000 = (Customary)

µρ

= d

Q 1.27 Rei (Metric)

( ) ( ) ( )( ) ( )310x 0.8 305

640 126 1.27 = − (Metric)

(rounded)420,000 = (Metric)

Step 3 - The friction factor in a 12.00 in. (305 mm) ID steel pipe at Rei = 420,000:

f = 0.0033 (Figure 2A or 2B)


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Step 4 - Kinetic energy per unit volume of inlet stream [Eq. (12c)]:

Ek = 4

25-

d

Q 10x1.8 ρα (Customary)

= ( ) ( ) ( )( )

psi0.153 = 12.00

2,000401.1 )10x(1.8

4

25-

(Customary)

Ek = 4

2

d

Q 810 ρα (Metric)

= ( ) ( ) ( )( )

kPa 1.05 = 305

1266401.1 810

4

2

(Metric)

Step 5 - For first trial, use J = 0.35 in Eq. (13b):

( ) k13b

p E 1d

J L f K P

−α

=∆ (Customary)

( ) ( ) ( ) ( ) ( ) (0.153) 1 12.001.1

0.35 8 0.0033 48

−

= (Customary)

= -0.148 psi (Customary)

( ) k13b

p E 1d

J L f K P

−α

=∆ (Metric)

( )( ) ( ) ( ) ( ) (1.05) 1 3051.1

0.35 2.43 0.0033 10x4 3

−

= (Metric)

= -1.01 kPa (Metric)

Step 6 - Required pressure drop across the holes: (∆P)o. Here EK >(∆P)p , so (∆P)o = 10 Ek

(∆P)o = (10) (0.153) = 1.53 psi (Customary)

(∆P)o = 10 (1.05) = 10.5 kPa (Metric)

This is larger than 0.25 psi (1.75 kPa). Therefore take (∆P)o = 1.53 psi (10.5 kPa).

Step 7 - Obtain the required total hole area from Eq. (14a):

( ) ( ) o

3-o P

CQ

10 x 3.32 A∆ρ

= (Customary)

( ) 23- in. 56.6 = 1.5340

0.602000

10 x 3.32

= (Customary)

oo )P(C

Q3.22A

∆ρ

= (Metric)

( ) 2mm 36,600 = 10.5640

0.60126

22.3

= (Metric)



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Step 8 - Choose a hole diameter by following steps on page 7 under Basic Design Considerations for Perforated-PipeDistributors:

a) Minimum hole diameter

do = 0.5 in. (13 mm)

b) Maximum hole diameter

do = (0.2) (di) = (0.2) (12) = 2.40 in. (Customary)

= 0.2 (di) = 0.2 (305 mm) = 61.0 mm (Metric)

c) Prefer do between (0.15) (di) = (0.15) (12) = 1.8 in. and maximum do = 2.4 in.

Therefore, 1.8 in. = do = 2.40 in. (Customary)

Prefer do between 0.15(di) = (0.15) (305) = 46 mm and maximum do = 61 mm.

Therefore, 46 mm = do = 61 mm (Metric)

d) Find the maximum number of 1.8 in. (46 mm) holes - hole spacing = do (1.8 in./46 mm):

( ) ( )( ) ( ) holes 27 26.7 =

1.8 212 8

= n ≈ (Customary)

( ) ( )( ) ( ) holes 27

46 2305 8

= n ≈ (Metric)

Similarly, find the maximum number of 2.4 in. (61 mm) holes:

( ) ( )( ) ( ) 20 =

2.4 212 8

= n (Customary)

( ) ( )( ) ( ) 20 =

61 2305 8

= n (Metric)

e. Try 1-7/8 in. holes: (Customary) - to bracket maximum and minimum hole size

( ) 22 in. 2.76 = 1.875 4

= hole per Areaπ

(Customary)

holes 21 20.5 = 2.7656.6

= holes of Number ≈

Try 48 mm holes: (Metric)

( ) 22 mm 1810 = 48 4

= hole per Areaπ

(Metric)

holes 20 = 1810

36,600 = holes of Number

Step 9 - Check the Reynolds number criterion:

4,000 > 20,000 = 21

420,000 =

n

Rei (Customary)

4,000 > 21,000 = 20

420,000 =

n

Rei (Metric)

Since Re

n > 4,000 this solution is acceptable.i


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Step 10 - Check the value of J used in Step 1: From Figure 11, for 20-21 holes, J = 0.357. Substituting 0.357 insteadof 0.35 in Eq. (13) would have little effect on (∆P)p. Ek is still greater than (∆P)p and still controls. Thecalculations do not have to be repeated.

Answer: Use 12 in. pipe with 21 holes at 1-7/8 in. diameter. (Customary)

Use 305 mm pipe with 20 holes at 48 mm diameter. (Metric)

NOMENCLATURE

Ao = Total required hole area in perforated pipe distributor, in.2 (mm2)C = Flow coefficient for orifices, flow nozzles and venturis, dimensionlessCv = Flow coefficient for valves (Figure 9), dimensionlessD = Inside diameter of pipe or equivalent hydraulic diameter, ft (m)d = Inside diameter of pipe or equivalent hydraulic diameter, in. (mm)Ek = Kinetic energy per unit volume, psi (kPa)F = Friction or head loss, ft lbf /lbm (kPa m3/kg)f = Fanning fraction factor, dimensionless (Figures 1A, 1B and 2A, 2B)fD = Darcy friction factor = 4f, dimensionlessg = Acceleration of gravity, ft/s2 (m/s2)gc = Dimensional constant, 32.174 ft lbm /lbf s2 (1.0 kg m/N s2)J = Factor for calculating head loss along a perforated pipe distributor, dimensionlessK = Resistance coefficient of valves, fittings, and cross-section changes, dimensionlessL = Length of pipe; actual length plus equivalent length of fittings, ft (m)Leq = Equivalent length of pipe or fitting, ft (m)n = Number of holes in perforated pipe distributorP = Pressure, lbf/in.2 or psi, (kPa) absolute∆P = Pressure drop, psi (kPa)Q = Quantity (volume) flow rate at conditions, gpm (dm3/s)r = Pressure recovery factor of orifices, flow nozzles and venturis, dimensionless (Figure 10)Re = Reynolds number, dimensionlessS = Specific gravity relative to water at 60°F, dimensionlessV = Linear fluid velocity, averaged over pipe cross-section, ft/s (m/s)Vi = Average inlet velocity in a perforated pipe distributor, ft/s (m/s)W = Mass flow rate, thousands lbm/h (kg/s)z = Elevation, ft (m)α = Velocity correction factor for calculating the kinetic energy of a stream, dimensionless

(α = 1.1 for turbulent flow; α = 2.0 for laminar flow)

ε = Pipe roughness, in. (mm)θ = Angle of divergence of diffuser, degreesµ = Viscosity, cP (Pa•s)

ρ = Density, lbm/ft3 (kg/m3)



XIV-B

Page

25 of 61


December, 1998


EXXONENGINEERING

NOMENCLATURE (Cont)

Subscripts (Unless indicated otherwise)e = By elevation changeeq = Equivalent (for equivalent hydraulic diameter)f = Frictional, forcei = Inletk = Kineticl = Linem = Masso = Hole, orificep = Distributor pipet = Total1 = Upstream location or conditions2 = Downstream location or conditions

COMPUTER PROGRAMS

GUIDANCE AND CONSULTING

For up-to-date information on available programs and how to use them, affiliate personnel should get in touch with their AffiliateLibrary Contact. Florham Park personnel should consult either the Exxon Engineering Section responsible for the technologyinvolved and/or the Computer Utilization Group of Exxon Engineering Technology Department (EETD).

LITERATURE

The following references are available:

Exxon Engineering Computer Users’ Manual, published and maintained by the Computer Utilization Group of EETD.

AVAILABLE PROGRAMS

The calculations methods in this section are available under the PEGASYS PC programs. (See EE.108E.96, September 1996,or contact the Plant Computing Division of Exxon Engineering Technology Department.)


Section

XIV-B

Page

26 of 61




EXXONENGINEERING

TABLE 1ADESIGN BASIS FOR “AVERAGE” CARBON STEEL LINES (CUSTOMARY)

LINEAVERAGE ∆∆P

psi/100 ftMAXIMUM ∆∆P

psi 100 ftMAXIMUM TOTAL ∆∆P

psi (APPROX.)

Pump suction and gravity rundown lines 0.25 0.4 -

Pump discharge lines (except high-pressure) 1.5 2.0 -

High-pressure pump discharge lines (700 psig and up) 3.0 4.0 -

Vapor lines (atm. & press., tower overhead lines) 0.2 0.5 0.5-1.0

Gas lines (inside battery limits) 0.2 0.5 4-5

Gas lines (tie-in lines) - - 5-10% of available pressure

Compressor suction lines 0.1 0.3 0.5-1.0

Compressor discharge lines 0.2 0.5 4-5

High-pressure steam lines (short) 0.5 1.0 2

High pressure steam lines (long) 0.1 0.4 5

Exhaust steam lines (short) 0.2 0.4 1

Exhaust steam lines (long) 0.05 0.1 1.5-2.0

Water lines (short) 1.0 1.5 2

Water lines (long) 0.25 0.5 5

Liquid transfer and tie-in lines - - 25

TABLE 1BDESIGN BASIS FOR “AVERAGE” CARBON STEEL LINES (METRIC)

LINEAVERAGE ∆∆P

kPa/mMAXIMUM ∆∆P

kPa/mMAXIMUM TOTAL ∆∆P

kPa (APPROX.)

Pump suction and gravity rundown lines 0.06 0.09 -

Pump discharge lines (except high-pressure) 0.34 0.45 -

High-pressure pump discharge lines (700 psig and up) 0.68 0.90 -

Vapor lines (atm. & press., tower overhead lines) 0.05 0.11 3.5-7

Gas lines (inside battery limits) 0.05 0.11 28-35

Gas lines (tie-in lines) - - 5-10% of available pressure

Compressor suction lines 0.02 0.07 0.4-7

Compressor discharge lines 0.04 0.11 28-35

High-pressure steam lines (short) 0.11 0.23 14

High pressure steam lines (long) 0.02 0.09 35

Exhaust steam lines (short) 0.04 0.09 7

Exhaust steam lines (long) 0.01 0.02 10.5-14

Water lines (short) 0.23 0.34 14

Water lines (long) 0.06 0.11 35

Liquid transfer and tie-in lines - - 175



XIV-B

Page

27 of 61


December, 1998


EXXONENGINEERING

TABLE 2TYPICAL PROCESS LINE EQUIVALENT LENGTHS

LINE RUN EQUIVALENT LENGTH(1)

Pump Suction Vessel/Rack/Pump 32 D + 200(2)

PA Pump Suction Vessel/Rack/Pump 42 D + 250(2)

Pump Discharge Pump/Rack - Line Size 30 D + 50

- Reduced Size 45 D + 50(3)

Compressor Suction Vess/Rack/Compressor 25 D + 150

Compressor Discharge Compressor/Rack 12 D + 150

Exchanger Rack/Exch./Rack 20 D + 150

Add for each bundle 10 D + 50

Control Valve Station Rack/CV/Rack - Line Size 20 D + 100

- Reduced Size 30 D + 100(3)

Furnace Rack/Furnace 10 D + 100

Furnace/Rack 15 D + 50

Vessel Rack/Vessel 10 D + 100

Vessel/Rack 10 D + 100

Expansion Loop In Rack(4) 10 D + 100

Transfer line Unit/Unit 10 D + 1000(5)

Notes:

(1) D is nominal pipe size in inches.

(2) Add 1 psi (7 kPa) for permanent strainer.

(3) Normally used for estimates.

(4) Use when process temperature > 400°F (200°C).

(5) Check plot plan to firm up length since they can readily exceed 1000 ft (300 m).


Section

XIV-B

Page

28 of 61




EXXONENGINEERING

TABLE 3REPRESENTATIVE EQUIVALENT LENGTH OF VARIOUS VALVES, IN PIPE DIAMETERS (L/D)

DESCRIPTION OF PRODUCT

EQUIVALENTLENGTH IN PIPE

DIAMETERS(L/D)

GlobeStem Perpendicular to Run With no obstruction in flat, bevel, or plug type seat Fully open

With wing or pin guided disc Fully open

340

450

Valves Y-Pattern (No obstruction in flat, bevel, or plug type seat)

- With stem 60 degrees from run of pipe line Fully open

- With stem 45 degrees from run of pipe line Fully open

175

145

Angle ValvesWith no obstruction in flat, bevel, or plug type seat Fully open

With wing or pin guided discFully open

145

200

Gate

Wedge, Disc,

Double Disc,

or Plug Disc

Fully open

Three-quarters open

One-half open

One-quarter open

13

35

160

900

Valves

Pulp Stock

Fully open

Three-quarters open

One-half open

One-quarter open

17

50

260

1200

Conduit Pipe Line Gate, Ball, and Plug Valves Fully open 3**

CheckValves

Conventional Swing

Clearway Swing

Globe Lift or Stop; SternPerpendicular to Run or Y-Pattern

Angle Lift or Stop

In-Line Ball

0.5 (3.5)† Fully open




2.5 (17.5) vertical and 0.25 (1.75) horizontal† Fully open

135

50

Same as Globe

Same as Angle

150

Foot Valves with Strainer With poppet lift-type disc 0.3 (2.1)† Fully open

With leather disc 0.4 (2.8)† Fully open

420

75

Butterfly Valves [8-in. (200 mm) and larger] Fully open 40

Straight-Through Rectangular plug port area equal to 100% of pipe area Fully open 18

Cocks Three Way Rectangular plug port area equal to Flow straight through

80% of pipe area (fully open) Flow through branch

44

140

Notes:

** Exact equivalent length is equal to the length † Minimum calculated pressure drop psi (kPa)between flange faces or welding ends. across valve to provide sufficient flow to lift disc fully.

From Crane Co. Technical Paper No. 410. Reprinted by permission.



XIV-B

Page

29 of 61


December, 1998


EXXONENGINEERING

FIGURE 1FRICTION FACTORS FOR ANY TYPE OF COMMERCIAL PIPE (CUSTOMARY) (METRIC)

0.025

0.020

0.015

0.010

0.009

0.008

0.007

0.006

0.005

0.004

0.003

Turbulent Zone

104 105 106 107 108103

Complete Turbulence

Trans-itionZone

LaminarZone

f = 16/Re

Smooth Pipes

Drawn TubingCommercial Steel or Wrought IronAsphalted Cast IronGalvanized IronCast IronConcrete

0.000 060.002 (2)

0.0050.0060.010.01-0.1

MaterialRoughness

ε, Inches

0.002

Fric

tion

Fac

tor,

f

Rel

ativ

e R

ough

ness

,ε d

0.00001

0.00005

0.0001

0.0002

0.0004

0.00060.00080.001

0.002

0.004

0.006

0.008

0.01

0.015

0.02

0.03

0.04

0.05

= 0.000001

εd

εd= 0.000005

DP14BF1

0.00150.051 (2)

0.1270.1520.250.25 - 2.5

Roughnessε, mm

5 5 5 5 5

Reynolds No., Re


Section

XIV-B

Page

30 of 61




EXXONENGINEERING

ç FIGURE 2AFRICTION FACTORS FOR CLEAN COMMERCIAL STEEL AND WROUGHT IRON PIPE (CUSTOMARY)

For

pip

e ot

her

than

com

mer

cial

ste

el, a

djus

t for

roug

hnes

s by

mul

tiply

ing

the

actu

al d

iam

eter

by

the

follo

win

g fa

ctor

to fi

nd th

e pr

oper

cha

rt p

aram

eter

.*A

spha

lted

Cas

t Iro

nG

alva

nize

d Ir

onC

ast I

ron

Con

cret

e

0.37

0.30

0.18

0.01

5 -

0.15

* U

se a

ctua

l I.D

. in

all o

ther

cal

cs.

Insi

deD

iam

eter

,In

ches

Nom

inal

Pip

eS

ize,

Inch

es1/

8

1/4

3/8

1/2

3/4

1 1 1/

41

1/2

2 2 1/

23 3

1/2

4 5 6 8 10 12 142

46

81

11

10

00

00

24

60

00

0S

ched

ule

Num

ber

Tra

ns-

ition

Zon

e

Com

plet

eT

urbu

lenc

e

Tur

bule

ntZ

one

Lam

inar

Zon

e f = 16/Re

0.01

5

0.01

0

0.00

9

0.00

8

0.00

7

0.00

6

0.00

5

0.00

4

0.00

3

0.00

2

Friction Factor, f

Rey

nold

s N

o. R

e [E

q. (

3)]

No

tes:

(1)

Dat

a ex

trac

ted

from

Cra

ne C

o. T

echn

ical

Pap

er N

o. 4

10, b

y pe

rmis

sion

(2)

See

als

o F

igur

e 1.

103

104

105

106

107

108

23

45

68

12

34

56

81

23

45

68

12

34

56

81

23

45

68

16

81

0.20

0.25

0.30

0.40

0.50

0.75 1.0

1.5 2 3 4 5 6 8 10 12 16 20 24 36 48

DP

14B

F2A



XIV-B

Page

31 of 61


December, 1998


EXXONENGINEERING

ç FIGURE 2BFRICTION FACTORS FOR CLEAN COMMERCIAL STEEL AND WROUGHT IRON PIPE (METRIC)

For

pip

e ot

her

than

com

mer

cial

ste

el, a

djus

t for

roug

hnes

s by

mul

tiply

ing

the

actu

al d

iam

eter

by

the

follo

win

g fa

ctor

to fi

nd th

e pr

oper

cha

rt p

aram

eter

.*A

spha

lted

Cas

t Iro

nG

alva

nize

d Ir

onC

ast I

ron

Con

cret

e

0.37

0.30

0.18

0.01

5 -

0.15

* U

se a

ctua

l I.D

. in

all o

ther

cal

cs.

Insi

deD

iam

eter

,m

m

Nom

inal

Pip

eS

ize,

mm

6 8 10 15 20 25 32 40 50 65 80 90 100

125

150

200

250

300

350

24

68

11

11

00

00

02

46

00

00

Sch

edul

e N

umbe

r

Tra

ns-

ition

Zon

e

Com

plet

eT

urbu

lenc

e

Tur

bule

ntZ

one

Lam

inar

Zon

e f = 16/Re

0.01

5

0.01

0

0.00

9

0.00

8

0.00

7

0.00

6

0.00

5

0.00

4

0.00

3

0.00

2

Friction Factor, f

Rey

nold

s N

o. R

e [E

q. (

3)]

No

tes:

(1)

Dat

a ex

trac

ted

from

Cra

ne C

o. T

echn

ical

Pap

er N

o. 4

10, b

y pe

rmis

sion

(2)

See

als

o F

igur

e 1.

103

104

105

106

107

108

23

45

68

12

34

56

81

23

45

68

12

34

56

81

23

45

68

16

81

DP

14B

F2B

5.08

6.35

7.62

10.1

6

12.7

19.0

5

25.4

38.1

50.8

76.2

101.

612

715

2.4

203.

2

304.

8

508.

060

9.6

1219

.2

254

914.

4

406.

4


Section

XIV-B

Page

32 of 61




EXXONENGINEERING

ç FIGURE 3ALIQUID PRESSURE DROP IN COMMERCIAL STEEL PIPE (CUSTOMARY)

1-Inch Extra Strong Pipe(I.D. = 0.957 Inches)

For Standard Pipe (I.D. = 1.049 Inches),Multiply Chart ∆P by 0.65

20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

∆P

/S, p

si/1

00ft.

µ/S

(CS)

10

52

1

.50

Flow Rate, gpm

1 2 3 4 6 8 10 20

Flow Rate, gpm

3 4 6 8 10 20 30 6040

20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

∆P

/S, p

si/1

00ft.

µ/S

(CS)

1 1/2-Inch Extra Strong Pipe(I.D. = 1.500 Inches)


10

5

2

1.5

0

20

DP14BF3A



XIV-B

Page

33 of 61


December, 1998


EXXONENGINEERING

ç FIGURE 3A (Cont)LIQUID PRESSURE DROP IN COMMERCIAL STEEL PIPE (CUSTOMARY)

2-Inch Extra Strong Pipe(I.D. = 1.939 Inches)


20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

∆P

/S, p

si/1

00ft.

µ/S

(CS)

10

52

1.5

0

Flow Rate, gpm

6 10 20 30 40 60 80 100

Flow Rate, gpm

20 30 40 60 80 100 200 400300

20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

∆P

/S, p

si/1

00ft.

µ/S

(CS)

3-Inch Standard Pipe(I.D. = 3.068 Inches)

For Extra Strong Pipe (I.D. = 2.900 Inches),Multiply Chart ∆P by 1.35

105

21

20

0

20

50

DP14BF3Aa

8


Section

XIV-B

Page

34 of 61




EXXONENGINEERING




20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

∆P

/S, p

si/1

00ft.

µ/S

(CS)

100

50

21

5

0

Flow Rate, gpm

40 80 100 200 300 400 600 800

Flow Rate, gpm

100 200 300 400 600 800 20001000

20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

∆P

/S, p

si/1

00ft.

µ/S

(CS)



105

2

1

200

0

20

60

10

20

5010

0

DP14BF3Ab



XIV-B

Page

35 of 61


December, 1998


EXXONENGINEERING



For Extra Strong Pipe (I.D. = 7.625 Inches),Multiply Chart ∆ P by 1.25

20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

∆P/S

, psi

/100

ft.

µ /S

(CS)

100

5020

15

0

Flow Rate, 1000 gpm

0.2 0.4 0.6 0.8 1 2 3 4

Flow Rate, 1000 gpm

0.4 0.6 0.8 1 2 3 4 86

20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

∆P

/S, p

si/1

00ft.

µ /S

(CS)



100

5020

1

5

0

500

0.3

10

200

200

10

DP14BF3Ac


Section

XIV-B

Page

36 of 61




EXXONENGINEERING


DP14BF3Ad



∆P

/S, p

si/1

00ft.

µ/S

(CS)

100

5020

15

0

Flow Rate, 1000 gpm Flow Rate, 1000 gpm

∆P

/S, p

si/1

00ft.

µ/S

(CS)



100

5020

1

5

0

500

10

200

200

10

1 2 3 4 6 8 10 200.6 1 2 3 4 6 8 100.8

20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

500



XIV-B

Page

37 of 61


December, 1998


EXXONENGINEERING




∆P

/S, p

si/1

00ft.

µ /S

(CS)

100

5020

15

0

Flow Rate, 1000 gpm Flow Rate, 1000 gpm

∆P/S

, psi

/100

ft.

µ /S

(CS)



100

5020

1

5

0

500

10

200

200

10

2 3 4 6 8 10 20 301 2 3 4 6 8 10 20

20

10

8.0

6.0

4.0

3.0

2.0

0.8

0.6

0.4

0.3

0.2

0.1

20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

500

DP14BF3Ae

1.0


Section

XIV-B

Page

38 of 61




EXXONENGINEERING


8



∆P/S

, psi

/100

ft.

µ /S

(CS)

100

5020

15

0

Flow Rate, 1000 gpm DP14BF3Af

∆P/S

, psi

/100

ft.

µ /S

(CS)



100

5020

15

0

500

10

200

200

10

3 4 6 8 10 20 30 602 4 6 8 10 20 30 403

20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

20

10

8.0

6.0

4.0

3.0

2.0

1.0

0.8

0.6

0.4

0.3

0.2

0.1

500

40

Flow Rate, 1000 gpm

1000

1000



XIV-B

Page

39 of 61


December, 1998


EXXONENGINEERING

ç FIGURE 3BLIQUID PRESSURE DROP IN COMMERCIAL STEEL PIPE (METRIC)

Flow Rate, dm 3/s

25-mm Extra Strong Pipe(I.D. = 24.3 mm)

For Standard Pipe (I.D. = 26.6 mm),Multiply Chart ∆P by 0.65

10

52

1

.50

∆P/s

, kP

a/m



10

5

2

1.5

0

20

∆P/s

, kP

a/m

DP14BF3B

.10 0.15 0.2 0.3 0.4 0.6 1.30.8 1.0.06 .08 0.2 0.4 0.6 0.8 1.0 1.5 2.0 4.03.00.3 0.5

4.0

3.0

2.0

1.5

1.0

0.7

0.5

0.3

0.2

0.15

0.10

0.07

0.05

0.03

0.04

0.06

0.08

0.4

0.6

0.8

5.0

3.0

2.0

1.5

1.0

0.7

0.5

0.3

0.2

0.15

0.10

0.07

0.05

0.03

0.04

0.06

0.08

0.4

0.6

0.8

4.0ρµ

106

(m

m2 /s

)

ρµ10

6 (

mm

2 /s)

Flow Rate, dm 3/s


Section

XIV-B

Page

40 of 61




EXXONENGINEERING

ç FIGURE 3B (Cont)LIQUID PRESSURE DROP IN COMMERCIAL STEEL PIPE (METRIC)

DP14BF3Ba



∆P/S

, kP

a/m

21

50

75-mm Standard Pipe(I.D. = 77.9 mm)

For Extra Strong Pipe (I.D. = 73.7 mm),Multiply Chart ∆P by 1.35

10

5

2 0

50

1

20

10

∆P

/S, k

Pa/

m

.5

20

1.3 3 4 5 6 8 2072 9 100.4 0.6 0.8 1.0 2.0 3.0 4.0 6.00.5 1.5 5.0

4.0

3.0

2.0

1.5

1.0

0.80

0.50

0.30

0.20

0.15

0.10

0.05

0.04

0.03

0.06

0.070.08

0.700.60

0.40

4.0

3.0

2.0

1.0

0.7

0.5

0.4

0.2

0.3

0.10

0.07

0.05

0.03

0.04

0.06

0.08

0.6

0.8

ρµ10

6 (

mm

2 /s)

ρµ10

6 (

mm

2 /s)

Flow Rate, dm 3/sFlow Rate, dm 3/s



XIV-B

Page

41 of 61


December, 1998


EXXONENGINEERING



For Extra Strong Pipe (I.D. = 97.2 mm),Multiply Chart ∆ P by 1.30

∆P/S

, kP

a/m

21

5

0150-mm Standard Pipe

(I.D. = 154.1 mm)For Extra Strong Pipe (I.D. = 146.3 mm),

Multiply Chart ∆ P by 1.30

105

2

0

50

1

2050

10

∆P/S

, kP

a/m

20

100

200

DP14BF3Bb

100

7 20 30 40 60 80 10010 7050983 5 6 10 20 30 40 504 7 8 9

4.0

3.0

2.0

1.0

0.80.7

0.5

0.4

0.3

0.2

0.10

0.05

0.04

0.03

0.060.070.08

0.6

4.0

3.0

2.0

1.0

0.80.7

0.5

0.4

0.3

0.2

0.10

0.05

0.04

0.03

0.060.070.08

0.6

ρµ10

6 (

mm

2 /s)

ρµ10

6 (

mm

2 /s)



Section

XIV-B

Page

42 of 61




EXXONENGINEERING




∆P/S

, kP

a/m

20

15

0



105

50

0

500

1

20

200

10

∆P/S

, kP

a/m

DP14BF3Bc

100

200

100

50

30 50 60 70 200 300 400 50040 10080

4.0

3.0

2.0

1.0

0.5

0.3

0.2

0.080.07

0.06

0.05

0.04

0.03

0.10

0.4

0.6

0.70.8

4.0

3.0

2.0

1.0

0.5

0.3

0.2

0.080.07

0.06

0.05

0.04

0.03

0.10

0.4

0.6

0.70.8

20 40 50 60 80 100 150 20030 70

ρµ10

6 (

mm

2 /s)

ρµ10

6 (

mm

2 /s)




XIV-B

Page

43 of 61


December, 1998


EXXONENGINEERING




∆P/S

, kP

a/m

500

15



Multiply Chart ∆P by 1.10

105

0

500

1

20

200

10

∆P/S

, kP

a/m10

0

2050 10

0

200

50

DP14BF3Bd

60 100 150 200 300 400500 60080705040

4.0

3.0

2.0

1.0

0.80.7

0.4

0.3

0.2

0.10

0.08

0.06

0.05

0.03

0.04

0.07

0.6

0.5

4.0

3.0

2.0

1.0

0.80.7

0.4

0.3

0.2

0.10

0.08

0.06

0.05

0.03

0.04

0.07

0.6

0.5

60 100 200 300 400 600 100050080 800

ρµ10

6 (

mm

2 /s)

ρµ10

6 (

mm

2 /s)



Section

XIV-B

Page

44 of 61




EXXONENGINEERING



For Extra Strong Pipe (I.D. = 381 mm),Multiply Chart ∆ P by 1.10

∆P

/S, k

Pa/

m

20

1

50




105

20

0

50

1

100

500

10

∆P

/S, k

Pa/

m

5

100

200

200

500

DP14BF3Be

60 200 300 400 500 600 800 100015010080

4.0

3.0

2.0

1.0

0.80.7

0.5

0.4

0.3

0.2

0.10

0.07

0.05

0.03

0.04

0.06

0.08

0.6

4.0

3.0

2.0

1.0

0.80.7

0.5

0.4

0.3

0.2

0.10

0.07

0.05

0.03

0.04

0.06

0.08

0.6

150 300 400 600 800 1000 1500200 2000

ρµ10

6 (

mm

2 /s)

ρµ10

6 (

mm

2 /s)

Flow Rate, dm 3/s Flow Rate, dm 3/s



XIV-B

Page

45 of 61


December, 1998


EXXONENGINEERING


500-mm Standard Pipe(I.D. = 489 mm)


∆P

/S, k

Pa/

m

20

15




10

5

100

0

5020

200

10

∆P/S

, kP

a/m

100

200

500

1000

50

500

1000

DP14BF3Bf

4.0

3.0

2.0

1.0

0.80.7

0.4

0.2

0.08

0.06

0.04

0.03

0.05

0.07

0.10

0.6

0.5

0.3

4.0

3.0

2.0

1.0

0.80.7

0.4

0.2

0.08

0.06

0.04

0.03

0.05

0.07

0.10

0.6

0.5

0.3

300 400 600 800 1000 2000200 1500 300 600 800 1000 1500 2000 3000400200

ρµ10

6 (

mm

2 /s)

ρµ10

6 (

mm

2 /s)

1

Flow Rate, dm 3/s Flow Rate, dm 3/s


Section

XIV-B

Page

46 of 61




EXXONENGINEERING

ç FIGURE 4AAPPROXIMATE LIQUID PRESSURE DROP IN COMMERCIAL PIPE (CUSTOMARY)

5252525252

103 104 105 106 107 108

W2

ρ

52525252520.01

0.05

0.1

0.2

0.5

1.0

2.0

5.0

10.0

Pre

ssur

e D

rop,

psi

/100

ft.

1" to 2" - Sch. 803" to 10" - Sch. 4012" to 36" - Std. Wall

4 6 8 10

1 1/21

12

2

16 20 24 36

3

30

4

6

8

10

Turbulent FlowViscosity ≤ 1.0 cP

10310210110-110-2

W = klbm/hr

ρ = lbm/ft3

W2

ρ DP14BF4A



XIV-B

Page

47 of 61


December, 1998


EXXONENGINEERING

ç FIGURE 4BAPPROXIMATE LIQUID PRESSURE DROP IN COMMERCIAL PIPE (METRIC)

525252525210-1 100 101 102 103 104

W2

ρ

52525252520.01

0.05

0.1

0.2

0.5

1.0

2.0

5.0

10.0

25 to 50 mm Sch. 8075 to 250 mm Sch. 40300 to 900 mm Std. Wall

75 100 150

25

200 250 300

38

400

600

900

100

Turbulent FlowViscosity ≤ 10-3 Pa.s

10-110-210-310-410-510-6

W = kg/s ρ = kg/m3

W2

ρ

75075

500

50

0.02

Pre

ssur

e D

rop,

kP

a/m

DP14BF4B


Section

XIV-B

Page

48 of 61




EXXONENGINEERING

ç FIGURE 4CPRESSURE DROP IN COMMERCIAL PIPE FOR WATER AT 75°°F (CUSTOMARY)

Sch. 80

Nominal LineSize, Inches

1 101 1/2 2 3 4 6 8 12 14 16 18 20 24

Sch. 40 Std. Wall

Short Lines

Short Lines(Maximum)

(Average)

Long Lines(Maximum)

Long Lines(Average)

105104103

Flow Rate, gpm

1021011

1 2 3 4 5 6 7891 2 3 4 5 6 7891 2 3 4 5 6 7 891 2 3 4 5 6 7891 2 3 4 5 6 7 891

Pre

ssur

e D

rop,

psi

/100

ft.

DP14BF4C

0.01

2

3

456789

0.1

2

3

4567891

2

3

456789

10



XIV-B

Page

49 of 61


December, 1998


EXXONENGINEERING

ç FIGURE 4DPRESSURE DROP IN COMMERCIAL PIPE FOR WATER AT 24°°C (METRIC)

0.1 1 10 100 100010-3

10-2

10-1

1.0

Pre

ssu

re D

rop

, kP

a/m

25 38 50 75 100 200 250 300150 600500450400350

Sch. 40Sch. 80 Std. Wall

Nominal LineSize, mm

Short Lines(Maximum)

Short Lines(Average)

Long Lines(Maximum)

Long Lines(Average)

DP14BF4DFlow Rate, dm3/s


Section

XIV-B

Page

50 of 61




EXXONENGINEERING

FIGURE 5AEQUIVALENT LENGTHS L AND L/D AND RESISTANCE COEFFICIENT K FOR VALVES (CUSTOMARY)

L/D L d

16

L/D

Equ

ival

ent L

engt

h, in

Pip

e D

iam

eter

s

2000

1000900800700600

500

400

300

200

10090807060

50

40

30

20

109876

5

4

3

K = 25

K = 20

K = 18

K = 16

K = 14

K = 12

K = 10

K = 9.0

K = 8.0

K = 7.0

K = 6.0

K = 5.0

K = 4.5

K = 4.0

K = 3.5

K = 3.0

K = 2.5

K = 2.0

K = 1.5

K = 1.0

K = .9

K = .8

K = .7

K = .6

K = .5

K = .4

K = .3

K = .2

K = .15

K = .1

K = .09

K = .08

K = .07

K = .06

K = .05

K = .04

Schedule 40 Pipe, Inches

3/8

1/2

3/4 1

1 1/

41

1/2 2 3 4 5 6 8 10 12 24

Inside Diameter of Pipe, Inches

.4 .6 2.8 1.0 10 20 303 4 6 8

Reprinted from Crane Co. Technical Paper No. 410, by permission

Problem: Find the equivalent length in pipe diameters and feet ofSchedule 40 pipe, and the resistance factor K for 1, 5, and12-inch fully-opened gate valves.

Solution

Equivalent length, pipe diametersValve Size

Equivalent length, feet of Sched. 40 pipeResist, factor K, based on Sched. 40 pipe

1" 5" Refer to1313

13135.51.1

0.30 0.20 0.17 on chartDotted lines

Table 312"

L -

Equ

ival

ent L

engt

h, in

Fee

t of P

ipe

10000

8000600050004000

3000

2000

1000800

600500400

300

200

10080

605040

30

20

108

654

3

2

1.00.8

0.60.50.4

0.3

0.2

0.1

Nom

inal

Sch

edul

e 40

Pip

e S

ize,

In In

ches

24

20

18

16

14

12

10

8

6

5

4

3

2

3 1/2

2 1/2

1 1/2

1 1/4

3/4

1/2

3/8

1

d -

Insi

de D

iam

eter

of P

ipe,

In In

ches

50

40

30

20

10

9

8

7

6

5

4

3

2

1.0

0.9

0.8

0.7

0.6

0.5

DP14BF5A



XIV-B

Page

51 of 61


December, 1998


EXXONENGINEERING

FIGURE 5BEQUIVALENT LENGTHS L AND L/D AND RESISTANCE COEFFICIENT K FOR VALVES (METRIC)


Problem: Find the equivalent length in pipe diameters andmeters of Schedule 40 pipe, and the resistance factor K for25, 125, and 300-mm fully-opened gate valves.

DP14BF5B

Solution

Equivalent length, pipe diametersValve Size

Resist, factor K, based on Sched. 40 pipe

25mm 125mm Refer to131313

0.30 0.20 0.17 on chart

Table 3300mm

Equivalent length, feet of Sched. 40 pipe 4.01.80.34 Dotted lines

L/D L

d

L/D

Equ

ival

ent L

engt

h, in

Pip

e D

iam

eter

s

2000

1000900800700600500

400

300

200

10090807060

50

40

30

20

1098765

4

3

K = 25

K = 20

K = 18

K = 16

K = 14

K = 12

K = 10

K = 9.0

K = 8.0

K = 7.0

K = 6.0

K = 5.0

K = 4.5

K = 4.0

K = 3.5

K = 3.0

K = 2.5

K = 2.0

K = 1.5

K = 1.0

K = .9

K = .8

K = .7

K = .6

K = .5

K = .4

K = .3

K = .2

K = .15

K = .1

K = .09

K = .08

K = .07

K = .06

K = .05K = .04

Schedule 40 Pipe Size, MM

10 15 20 25 32 40 50 80 100

125

150

200

250

300

450

Inside Diameter of Pipe, mm

L -

Equ

ival

ent L

engt

h, in

Met

ers

of P

ipe

3000

2000

1500

1000

700600500400300

200150

100807060504030

2015

10876543

2N

omin

al S

ched

ule

40 P

ipe

Siz

e, In

mill

imet

ers

550

500450400

350300

250

200

150

125

100

80

32

90

40

25

20

8

15

d -

Insi

de D

iam

eter

of P

ipe,

In m

illim

eter

s

1

0.80.60.50.40.3

0.20.15

0.70.9

0.1

0.07

30

20

15

40

50

60

708090100

300

200

150

400

500

600

7008009001000

50

65

.4 .6 2.8 1.0 10 20 303 4 6 8

500

400


Section

XIV-B

Page

52 of 61




EXXONENGINEERING

ç FIGURE 5CRESISTANCE COEFFICIENT FOR BENDS, ELLS AND TEES (CUSTOMARY)

LongRadiusFlanged orButt Welded45° ell.

RegularScrewed45° ell.

ScrewedReturnBend

Flanged orButt WeldedReturnBend


LongRadiusScrewed90° ell.

RegularFlanged orButt Welded90° ell.


ScrewedTee

Flanged orButt Welded

Tee

DP14BF5C

0.5 1 2 4

2

1

0.6

K

0.5 1 2 4

K

0.80.60.40.30.2

1 2 4 6 10 20

K

0.60.40.30.2

0.15

dia, in.

0.1

0.20.3

1 2 4 6 10 20

K

K

0.5 1 2 4

0.60.40.30.2

K

1 2 4 6 10 200.1

0.20.3

K

0.5 1 2 4

2

10.6

K

dia, in.

1 2 4 6 10 20

0.40.30.2

0.1

KLine Flow

0.5 1 2 40.30.60.8

1

KBranch Flow

0.5 1 2 41

23

KLine Flow

1 2 4 6 10 20

0.2

0.10.06

KBranch Flow

dia, in.1 2 4 6 10 20

0.40.6

1

dia, in.

dia, in.dia, in.

dia, in. dia, in.

dia, in.

Long Radius

RegularFlanged or

Butt WeldedTee

Note: For fitting larger than 10" I.D., use the resistance coefficient for 10" I.D. fitting.

ScrewedTee

dia, in.dia, in. dia, in.



XIV-B

Page

53 of 61


December, 1998


EXXONENGINEERING

FIGURE 5DRESISTANCE COEFFICIENT FOR BENDS, ELLS AND TEES (METRIC)



ScrewedReturnBend

Flanged orButt WeldedReturnBend


LongRadiusScrewed90° ell.

RegularFlanged orButt Welded90° ell.


ScrewedTee


Tee

ScrewedTee


Tee

Note: For fittings larger than 250 mm I.D., use the resistance coefficient for 250 mm I.D. fittings.

K

K

K

K K

K

K

K

K

K

K

K

3

2

1

0.30.2

0.1

0.30.2

0.1

0.80.6

0.30.4

0.2

0.80.6

0.30.2

0.15

32

1

0.6

0.80.60.4

0.2

3.0

2.0

1.0

0.6

1.00.80.6

Line flow

Line flow

Branch flow

Branch flow

d, mm

d, mm

d, mm

d, mm

d, mm

d, mm

d, mm

d, mmd, mm

d, mm

d, mm

d, mm

25 50 100 250 500

25 50 100 250 600

25 50 100 250 60025 50 100 200 600

25 50 100 200 600400

8 10 25 50 100

8 10 25 50 100 810 25 50 100

8 10 25 50 100

8 10 25 50 100

810 25 50 100

DP14BF5D

0.4

0.6

1.0

0.05

0.1

0.2

0.30.2

0.1

0.4

25 50 100 250 500

400

Long Radius

Regular


Section

XIV-B

Page

54 of 61




EXXONENGINEERING

FIGURE 5ERESISTANCE COEFFICIENTS FOR RETURN BENDS AND MITER BENDS

SIMPLE MITER BENDRETURN BEND

104

105

106

Smooth PipeBends

Reynolds No. Multiplier for K

1.48

1.00

0.676 Length of pipe in bend is included inK as additional loss. Elsewhere, lengthcontribution is excluded from K.

Rb/D

Re=105

141210864200

0.20.4

0.60.81.0

KD

Rb

θ

θ=180°

θ=45°

θ=90°

θ

K

0

0.2

0.4

0.6

0.8

1.0

1.2

θ, Degrees

0 10 20 30 40 50 60 70 80 90

K = 1.2 (1 � cos θ )

Loss Coefficient, K m

Turbulent Flow, Re > 4000

Km*n = number of individual bends

n=2 n=3 n=4R/Dθ

(deg)

--

0.750.400.300.250.200.200.20

-

456090

2.952.950.51.01.52.03.04.05.00.5

0.110.150.700.450.350.300.350.400.454.0

--

0.750.400.350.300.200.250.25

-

* For Re < 2 x 105.

* For Re ≥ 2 x 105.

180

DP14BF5EKm Re = (Km Re = 2 x 105)(2 x 105)Re

0.2

D

All bend angles are equaln = 2 bends shown

L2

R

L1, L2 >> D

θ

L1

RESISTANCE COEFFICIENTS FORCOMPOUND MITER BENDS IN CIRCULAR PIPE

Loss coefficient K = K m + f (L1 + L2)

D



XIV-B

Page

55 of 61


December, 1998


EXXONENGINEERING

FIGURE 6RESISTANCE COEFFICIENT FOR CROSS-SECTION CHANGES

K = 0.50SharpEdged

Entrance

K = 0.23Slightly

RoundedEntrance

K = 0.04Well

RoundedEntrance

K = 1.0SharpEdged

Exit

K = 1.0Rounded

Exit

K = 1.0ProjectingPipe Exit

K = 0.78Inward

Projecting PipeEntrance

K = 0.05Bellmouth

Inlet orReducer

Note: d1 is the smaller diameter.

d2

d1

Sudden Enlargement

Sudden Contraction

K =

Res

ista

nce

Coe

ffici

ent K

, Bas

ed o

n d 1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

d1/d2

d1d2

Head Loss in Conical Diffusers

1 -d2

2

d12

2

0.8

0.6

0.4

0.20

β = d1/d2

1.2

1.0

0.8

0.6

0.4

1.2

0180160140120100806040200

θ, Degrees

K, B

ased

on

V1

V1d1d2θ

Reproduced from References 2 and 3, by Permission. DP14BF6


Section

XIV-B

Page

56 of 61




EXXONENGINEERING

FIGURE 7AFLOW COEFFICIENT FOR ORIFICES WITH FLANGE TAPS

Reprinted from Crane Co. Technical Paper No. 410, by permission. D P 1 4 B F 7 A

1.3

1.2

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

= .80

= .75= .70

= .65= .60= .50

d 0d 1

d0d1

= .40

= .30= 0 to .20

3 4 6 8 10 20 40 60 80 10 2 2 4 6 8 10 3 2 4 6 8 10 4

Reynolds Number Re, Based on d1

Flo

w C

oeffi

cien

t4

C =

Cd

1-d 1d 0

FIGURE 7BFLOW COEFFICIENT FOR ORIFICES WITH FLANGE TAPS

Reprinted from Crane Co. Technical Paper No. 410, by permission. D P 1 4 B F 7 B

��

��

0.82

0.80

0.78

0.76

0.74

0.72

0.70

0.68

0.66

0.64

0.62

0.60

4 6 8 10 4 4 6 8 10 5 4 6 8 10 62 2

Reynolds Number Re, Based on d1

Flow

d1InsideDia. ofPipe

d0

Flow

Coe

ffic

ient

C =

Cd

0-0.2.30.40.45.50.55

.60

.65

.70

.75

.80

.85

d 0 / d1

Rat

io o

f N

ozzl

e D

iam

eter

to

Pip

e D

iam

eter

1 -

d 0 d 1

4



XIV-B

Page

57 of 61


December, 1998


EXXONENGINEERING

FIGURE 8FLOW COEFFICIENT FOR FLOW NOZZLES WITH FLANGE TAPS

��

��

��

��

Flow

d1InsideDia. ofPipe d 0

Reynolds Number Re, Based on d 1


4 6 8 10 4 2 4 6 8 10 5 2 4 6 8 10 6

0-0.2.30

.40

.50

.55

.60

.65

.70

.725

.75

.775

.80

d 0/d1

Rat

io o

f N

ozzl

e D

iam

eter

to

Pip

e D

iam

eter

1.18

1.16

1.14

1.12

1.10

1.08

1.06

1.04

1.02

1.00

0.98

0.96

0.94

Flow

Coe

ffic

ient

C=

d 0

4

Cd d 1

1 -

D P 1 4 B F 8


Section

XIV-B

Page

58 of 61




EXXONENGINEERING

FIGURE 9AEQUIVALENTS OF RESISTANCE COEFFICIENT K AND FLOW COEFFICIENT CV FOR VALVES (CUSTOMARY)

15

8

9

DP14BF9A

0.1

0.15

0.2

0.3

0.4

0.5

0.6

0.70.80.91.0

1.5

2

3

4

5

6

7

8910

15

20

24

K

K -

Res

ista

nce

Coe

ffici

ent

CV -

Flo

w C

oeffi

cien

t

Nom

inal

Siz

e S

ched

ule

40 P

ipe,

in In

ches

d -

Insi

de D

iam

eter

of P

ipe,

in In

ches

24

2018

16

8

6

5

4

3 1/2

3

2 1/2

2

1 1/2

1 1/4

1

3/4

1/2

3/8

14

12

10

24

20

6

5

4

.7

3

.8

2

1.5

.6

1.0

.9

.5

.4

7

10

60,000

20,000

10,000

50,00040,000

30,000

8000

600050004000

3000

2000

1000800

600500400

300

200

10080

605040

30

20

108

654

2

1

3

CVd

Problem: Find the equivalent length in pipe diameters, the resistance coefficient K, and the flow coefficient C V for an 8-inch,125-pound Y-pattern globe valve with stern 60 degrees from run of valve.

Solution: Equivalent length in pipe diameters is 175 (taken from Table 3).

Resistance factor K based on Schedule 40 pipe is 2.5 (taken from Figure 5A).Flow coefficient C V is 1200 (see dotted line shown on chart above).

Reprint from Crane Co. Technical Paper No. 410, by permission.

891 d4K =

CV2

29.9 d2CV =

K



XIV-B

Page

59 of 61


December, 1998


EXXONENGINEERING

FIGURE 9BEQUIVALENTS OF RESISTANCE COEFFICIENT K AND FLOW COEFFICIENT CV FOR VALVES (METRIC)

400

200

0.1

0.15

0.2

0.3

0.4

0.5

0.6

0.70.80.91.0

1.5

2

3

4

5

6

7

8910

15

20

24

KK

- R

esis

tanc

e C

oeffi

cien

t

CV -

Flo

w C

oeffi

cien

t

Nom

inal

Siz

e S

ched

ule

40 P

ipe,

in m

m

d -

Insi

de D

iam

eter

of P

ipe,

in m

m

600

500

450

400

200

150

125

100

90

80

65

50

40

32

25

20

15

10

350

300

250

600

500

150

100

50

90

30

80

70

15

60

40

20

10

300

60,000

20,000

10,000

50,00040,000

30,000

8000

600050004000

3000

2000

1000800

600500400

300

200

10080

605040

30

20

108

654

2

1

3

CV d

Problem: Find the equivalent length in pipe diameters, the resistance coefficient K, and the flow coefficient CV for a 200 mm,875 kPa Y-pattern globe valve with stern 60 degrees from run of valve.

Solution: Equivalent length in pipe diameters is 175 (taken from Table 3).

Resistance factor K based on Schedule 40 pipe is 2.5 (taken from Figure 5B).Flow coefficient CV is 1200 (see dotted line shown on chart above).

Reprint from Crane Co. Technical Paper No. 410, by permission.

0.047 d2CV =

K

2.25x10-3 d4K =

CV2

DP14BF9B


Section

XIV-B

Page

60 of 61




EXXONENGINEERING

FIGURE 10PRESSURE RECOVERY FACTOR FOR ORIFICES, NOZZLES AND VENTURIS

1.00

Orifice

FlowNozzle

0.90

0.80

0.70

0.60

0.50

0.40

0.30

0.20

0.10

0

Reproduced from Reference 6, by Permission

Herschel TypeVenturi Tube

Venturi Tube with 15 o Recovery Cone

0.2 0.3 0.4 0.5 0.6 0.7 0.8

Diameter Ratio, d0/d1

DP14BF10

Pre

ssur

e R

ecov

ery

Fac

tor,

r



XIV-B

Page

61 of 61


December, 1998


EXXONENGINEERING

FIGURE 11J FACTOR FOR CALCULATING DISTRIBUTOR HEAD LOSS

0.55

0.5

0.45

0.4

0.35

0.3

Hea

d Lo

ss F

acto

r, J

3 4 5 6 7 8 9 10 15 20 30 40 50 60 80 100

Number of HolesDP14BF11

XIV-B

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Transcript of XIV-B