Process Control and Optimization, VOLUME II - Freetwanclik.free.fr/electricity/IEPOPDF/1081ch6_15.pdf · The control valve is an important component of the total process control loop,

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6.15 Sizing

B. G. LIPTÁK (1970) C. G. LANGFORD (1985) F. M. CAIN (1995) H. D. BAUMANN (2005)

Reviewed by F. M. Cain and B. G. Lipták

INTRODUCTION

The control valve is an important component of the total processcontrol loop, and proper valve sizing is essential for this finalcontrol element to fulfill its role. There are two important pre-requisites for the control valve to be able to fulfill its role: Firstis to have the correct process data. This includes the accurateknowledge of the maximum and the minimum flow conditions,the available pressure drops across the valve at the various flowconditions, the maximum inlet pressure and inlet temperature,the viscosity of the fluid, and whether or not the fluid is two-phased. Correct selection of the valve’s pressure drop, for exam-ple, depends on the head loss in the piping system or thecharacteristic of a pump, to name just two system variables(Figure 6.1b). As the reader will notice, just the selection of theprocess conditions involves considerable uncertainties.

The second criterion is the selection of the correct valvesizing equation for the prevailing operating conditions. There-fore, before starting the process of sizing, the process controlengineer has to determine which of the following sizing basisare applicable. For liquid flow sizing, they include 1) turbulentnon-choked, 2) turbulent, choked, 3) saturated, 4) laminar(viscous), 5) non-Newtonian, and 6) two-phase. For gas flows,the sizing conditions can be: 1) turbulent non-choked, 2)choked, and 3) laminar (small flow valves).

The Standard IEC 60534-2-1 and the Instrumentation, Sys-tems, and Automation Society equivalent Standard S 75.01have improved the accuracy of valve sizing over the years,lowering the sizing errors to less than 5% except for two-phaseflow and for non-Newtonian flow conditions.

ABOUT THIS SECTION

For the definition of all the terms used in this section, refer tothe “Nomenclature” listing provided at the end of this section.For capacity testing procedures see Section 6.6. The subjectsof control valve characteristics and rangeability are discussedin Section 6.7, and the subject of control valve noise and itscalculation is covered in Section 6.14. The engineering unitsused in this section are the U.S. Customary units. Use thenumerical constants listed in Tables 6.15l and 6.15p for con-version to metric units.

This section begins with a definition of the valve coeffi-cients (Cv and Cd) and with a brief description of fluid behav-

ior during flashing, cavitation, and under turbulent or lam-inar flow conditions. After this general discussion thevarious liquid sizing factors (FL, FF , FP , FLP , and FR) aredefined, and then the liquid sizing equations are provided,as well as some examples. Some of the examples deal withviscous (laminar flow), sizing applications and pipe expanderlosses, and generalized approaches to liquid sizing. This isfollowed with a more detailed discussion of cavitation andflashing.

The second half of this section begins with a discussionof valve sizing for gas and vapor services. After the sizingequations are described, an example outlines a generalizedapproach to gas/vapor sizing. The section is concluded witha description of sizing for two-phase flow applications.

STANDARDS

The study of control valve sizing is based on the science offluid dynamics. The information in this chapter is an overviewof current knowledge, which is continually growing. Thisknow-how has been developed, in large measure, from thefollowing two standards of the Instrumentation, Systems, andAutomation Society (ISA): S75.01-2000, “Control Valve SizingEquations,” and S75.02-1996, “Control Valve Capacity TestProcedure.” The author is grateful to ISA for its permissionto copy and adopt portions of these standards for this section.

The international equivalent of these U.S. standards areIEC 60534-2-1, “Industrial Process Control Valves, Part 2-1:Flow Capacity Sizing Equations for Fluid Flow under InstalledConditions,” and IEC 60534-2-3, “Industrial Process ControlValves, Part 2-3, Flow Capacity Test Procedure.” The equationsin the international editions are essentially equivalent to thosein the U.S. standards with the exception of using metric units.

The user is cautioned that the calculations are not per-fectly accurate for all conditions. Accuracy is best for water,air, or steam applications using conventional valve designs,which are installed in straight piping. For non-Newtonianfluids or high-viscosity or low Reynolds number flows, thesecalculations are less accurate. In all cases, reliable calcula-tions require reliable information about the characteristics ofthe valve and the process fluid, including the fluid’s flow ratesand pressures at normal and unusual operating conditions.

The English units are used in the various sample calcu-lations and also in the “Nomenclature” listing at the end of

© 2006 by Béla Lipták

6.15 Sizing 1235

the section. While this was necessary because conversionconstants are embedded in the working equations, other unitscan be substituted with appropriate conversions found inTables 6.15i and 6.15n. The notations used in this section arethose used in the ISA standards.

General Principles

The earliest control valves were manual globe valves. Sizingwas easy: a 4 in. (100 mm) valve “belonged” in a 4 in. (100mm) line. Later it became obvious that at higher pressuredrops such a valve had too much capacity for good control.Thus evolved the next rule-of-thumb practice, which was tomake the control valve one size smaller than line size.

The limitations of such rule-of-thumb methods of valvesizing led to the development of a sizing coefficient basedon Bernoulli’s theorem for the conservation of energy, andthe continuity equation for the conservation of mass. Theseequations can be combined to express the ideal flow ratethrough a pipe restriction such as a control valve :

6.15(1)

Calculating the actual flow rate requires the introductionof a discharge coefficient, C, which accounts for the reallosses in the specific valve. The head loss (∆H) may also beexpressed as pressure drop divided by weight density (γ).Thus,

6.15(2)

By combining the area terms and discharge coefficient(C) with the other constants and expressing density in termsof specific gravity, Gf , the flow equation can be written as afunction of a capacity parameter, Cv , which is a generallyaccepted valve capacity parameter called the valve coefficient.

6.15(3)

It should be noted that Equation 6.15(3) is only valid forturbulent and nonchoked (low pressure drop) liquid flows insizing control valves.

It is important to note that Cv is not a dimensionlesscoefficient, and in the SI system Kv is used. The conversionbetween engineering units is covered in later paragraphs ofthis section. Table 6.15i provides the conversion constantsfor liquid and Table 6.15n for gas and vapor sizing equations.The Cv valve coefficient, which is based on English units, iswidely used in the United States, while Kv , based on the SIsystem, is the preferred unit in Europe.

The Flow Coefficient

Prior to about 1946 control valves were sized using the flowarea of the valve’s orifice in square inches together with some,usually proprietary equations in order to estimate the amountof liquid or gas passing through the valve. Thereafter, a newcoefficient was introduced called Cv, which started standard-ization in this field. This coefficient denotes the number of(U.S.) gallons of water that a valve would pass when thepressure drop across it is 1 lb/in.2

Under turbulent flow conditions the water velocity isrelated to the square root of the pressure drop (inlet pressureminus outlet pressure). Therefore, one can multiply this quan-tity with the Cv number of the valve to obtain the expectedflow of cold water in gallons per minute. Later in this sectionit will be shown how this simple definition of the Cv can beexpanded to liquids having specific gravity other than coldwater and with the use of conversion factors, for gases aswell.

Correction Factors Yet, the turbulent Cv alone turned outto be inaccurate. Using it, occasionally there appeared to besevere sizing errors (up to 50%) that nobody could explain.This problem was resolved in 1963 by the introduction of aCritical Flow Factor.1 This factor was later renamed the Liq-uid Pressure Recovery Factor and designated as FL.

What this factor describes is the amount of pressurerecovery, which is occurring downstream of the throttlingorifice. A low FL factor, which is usually the case with stream-lined valves, will promote vaporization in liquid streams orcause the development of sonic velocity in gas streams. Theseeffects tend to “choke’ the flow and consequently, the valvewill pass less than would be predicted if the FL factor wasnot considered.

In 1983 two additional factors were introduced. One iscalled Xt, the Pressure Differential Factor of a Control Valveat Choked Flow.2 The other is called the Expansion Factor Y,which enables the calculation of the correct gas density atthe valve’s orifice.

Yet another, albeit less serious, sizing error source hasbeen noticed in applications where a valve, smaller than linesize, was installed between reducers. The error was causedby the pressure drops across the reducers, which lowered thepressure drop available for the valve. In 1968, this problemwas overcome by the introduction of a Piping GeometryFactor,3 Fp.

One more factor was introduced for viscous fluid services,which is called the Reynolds Number Factor,4 FR. In 1993,this factor was redefined and its accuracy was improved.5

Finally, the Liquid Pressure Ratio Factor, FF (based onthe original work of Allen6 and others), was introduced forboiling or flashing liquid applications.

Table 6.15a tabulates the various sizing equations for theoverall orientation of the reader. The equations listed in thistable will be derived in the following paragraphs. In thissection, the equations are numbered consecutively, so the

Q Ag H

A A= ∆

−22 1

2

2

1

( )

( / )

Q CAg p

A A= ∆

−22 1

2

2

1

( / )

( / )

γ

Q Cp

Gvf

= ∆


1236 Control Valve Selection and Sizing

reader can easily find the derivation of any of the equationslisted in the table.

Units In that part of the world where the SI system is used,the valve coefficient is called Kv and it is defined as the flowof cold water in cubic meter per hour units, when the a

pressure drop across the valve is one bar. The conversionbetween Cv and Kv is

Kv = 1.17Cv 6.15(4)

The flow coefficient can be converted into an area that isthe area of the vena contracta, or the net flow area inside a

TABLE 6.15a Orientation Table: Summary of Valve Sizing Equations (for U.S. Customary Units: gpm, SCFH, psi, °F, lbm/hr, lb/ft3, etc.)

Selection Basis

Fluid State

Liquid Gas or Vapor

Nonchoked, turbulent flow in liquids:

Volumetric flow in gpmor SCFH

Eq. 6.15(35) Eq. 6.15(78)

Nonchoked vapors and gases: ∆p/P1 < xT

Mass flow in lbm/h Eq. 6.15(36) Eq. 6.15(77)

Choked flow due to cavitation or flashing liquids in liquids, or choked flow in gases

Volumetric flow in gpmor SCFM

Eq 6.15(31) Eq. 6.15(80)

Choked flow due to sonic velocity in gases or vapors, or choked flow in liquids

Mass flow in lbm/h Eq. 6.15(32) Eq. 6.15(79)

Piping effect for above equations

Not choked Eq. 6.15(22)

See Eq. 6.15(21)

Choked Eq. 6.15(23)

Ki See Eq. 6.15(20)

Nonturbulent (viscous) flow Volumetric flow in gpm Eq. 6.15(63)

where FR is from Figure 6.15m and is a function of Rev from Eq. 6.15(64 or 69)

Laminar, i.e., nonturbulent, conditions generally do not occur in gases or vapors except for small flow valves.

Mass flow in lbm/h Eq. 6.15(39)

∆ < −p F P F PL F v2

1( ) Cq

F

G

pvP

f=∆

CQ

F PY

G T Z

xvP l

g l=1360

Yx

F xx

p

PF

k

k T lk= − = ∆ =1

3 1 4; ;

.

Cw

F pv

P

=∆63 3 1. ( )γ

Cw

F Y xPv

P l l

=63 3. γ

Cq

F

G

P F PvLP

f

F v

=−

max

1

F P PF v c= −0 96 0 28 1 2. . ( / ) /

CQ

F PY

MT Z

F xvP l K T

= max

73201

Cw

F P F Pv

LP l F v l

=−

max

. ( )63 3 γC

w

F PY

T Z

F x MvP l K T

= max

.19 31

FK C

Pd= +

∑

−

1890

21 2

( )/

ΣK

FF

K CLP

L

i d= +

−1

8902

21 2/

Cq

F

G

pvR

f=∆

Cw

F pv

R l

=∆63 3. ( )γ


6.15 Sizing 1237

single stage valve orifice, in accordance with Equation 6.15(5):

Av = Cv × FL × 0.026 in square inches 6.15(5)

More refined equations are applicable for multistagevalves but are not yet standardized.7 Here the reader mayrefer to their respective vendors for assistance.

There has been substantial progress in increasing theaccuracy of valve sizing, yet additional research is needed inthe areas of mixed phase and non-Newtonian flows.

LIQUID SIZING

Fluid flowing through a control valve obeys the basic laws ofthe conservation of mass and energy. As the fluid streamapproaches the valve restriction, fluid velocity increases inorder to pass the same amount of flow through the restrictedarea. The restriction inside the valve is the result of moving aclosure member (e.g., plug, disk, ball) closer to the valve seat.

Energy to accelerate the fluid comes from a correspond-ing decrease in static pressure, as illustrated in Figure 6.15b.Maximum velocity and minimum static pressure occur imme-diately downstream from the minimum area of valve restric-tion (the throttling point), the most constricted area of theflow stream. This minimum pressure point downstream fromthe throttling point is known as the vena contracta.

Figure 6.15b shows that as the velocity decreases afterthe vena contracta, some of the kinetic energy is convertedback into static pressure. This conversion is called the pres-sure recovery of the valve, and its amount is equal to P2 –Pvc. Valves with large pressure recovery relative to P1 – P2

are called high recovery valves. Such valves include mostrotary and gate valves. The net pressure loss (P1 – P2)between the valve inlet and outlet is due to frictional effects(turbulence) and represents the permanent pressure loss.

Relative Valve Capacity Coefficient (Cd)

The valve capacity coefficient, Cv , increases as valve sizeincreases, but valve geometry is also a major factor in the

magnitude of the pressure loss for a given flow rate. It istherefore useful to define the relative valve capacity (Cd) inorder to compare the effects of geometry of different valvedesigns. Cd is defined as

6.15(6)

Representative values of Cd for various valve styles (atfull capacity) may be found in Table 6.15c.

Cd can also be useful for evaluating the relative magni-tude of pressure recovery. Generally, the valve types withlarger Cd values are also more likely to have higher pressurerecovery (P2 – Pvc); that is, they have a lower FL factor andtherefore tend to “choke” or have liquids evaporate at lowerpressure drops (P1 – P2). The significance of this will becomeclearer in the discussions of sizing factors for choked flow.

Caution: The published Cv data from manufacturers isbased on standard testing methods that include the frictionlosses associated with the test manifold pipe between thepressure taps (Section 6.6). When the Cd is under 20, thefriction loss from eight diameters of this pipe adjacent tothe valve is relatively insignificant. However, as the Cd

increases above 20, the manifold pressure drop rises expo-nentially relative to the losses from the valve alone. There-fore, for valves with Cd > 20, one must also consider thepressure drop in the valve test manifold piping to avoidsignificant errors.

In the majority of valve applications the flow is turbulent,and the velocity profile across the restrictive (throttling) areaof the valve is relatively uniform. Turbulent flow occurs whenthe Reynolds number is high (typically above 10,000). Thisoccurs when the velocity is high and the viscosity is low [Re= (γUd/µ)]. In the turbulent range, the incompressible fluidflow is proportional to the square root of the pressure dropacross the valve, as was shown in Equation 6.15(3). Thisproportionality is shown by the straight line on the left sideof Figure 6.15d.

Factors FL, FF, FP , and FLP

Pressure Recovery Factor (FL) FL is the pressure recoveryfactor, which indicates the size of the pressure recovery rela-tive to the valve pressure drop (Figure 6.15b). FL is defined as

6.15(7)

Note: FL can also be used to calculate the pressure dropthat causes sonic velocity in the valve orifice. As it has beendiscussed in Section 6.14, this is important for noise calcu-lations, where

∆psonc = 0.5 × FL2 × P1 6.15(8)

The values of FL for various valve designs are listed inTable 6.15c.

FIG. 6.15b Pressure profile through a valve.

Pressurerecovery(P2 − Pvc)

Pvc

P2

P1

P1 − Pvc

P1 − P2

Increasereffect

Inletreducereffect

C C dd v= / 2

F P P P PL vc= − −[( )/( )] /1 2 1

1 2



Liquid Critical Pressure Ratio Factor (FF) The complexgeometry of most valves makes experimental measurementof the pressure at the vena contracta (Pvc) impossible. TheISA sizing equations9 use the liquid critical pressure ratiofactor, FF, for approximating Pvc used for saturated liquidflow conditions.

6.15(9)

where FF can be approximated by

6.15(10)

A graph of this relationship is plotted in Figure 6.15e.Equation 6.15(10) is based on the assumption that the fluid

is always in thermodynamic equilibrium. However, a liq-uid does not remain in thermodynamic equilibrium as itflashes across the valve restriction; therefore, the actualflow rate can be greater than that predicted by Equation6.15(10). Estimating Pvc by using FF is not fully agreedupon by experts but it is accepted by the valve sizingstandard (ISA S75.01) and is in common use in the controlvalve industry.

The pressure drop required to cause choked flow in liq-uids is given by the following equation:

∆Pchoked = FL2 (P1 − FFPv) 6.15(11)

Piping Geometry Factor (FP and FLP) By convention, valvetests and calculations have included a portion of piping adja-cent to the valve. P1 is measured upstream of the pipe reducer,

TABLE 6.15c Representative Values of Relative Valve Capacity Coefficients (Cd) and of Other Sizing Factors for a Variety of Valve Designs.The Cd Values Listed are for Valves with Full Area Trims, When the Valve is Fully Open

Valve Type Trim TypeFlow

Direction* XT FL Fd Fs Cd = Cv/d2** Kc

GLOBESingle-port Ported plug, 4 port Either 0.70 0.90 0.48 1.0 9.5 0.65

Contoured plug Open 0.72 0.90 0.46 1.1 11 0.65

Close 0.55 0.80 1.0 1.1 11 0.58

Characterized cage, 4 port Open 0.75 0.90 0.45 1.1 14 0.65

Close 0.75 0.85 0.41 1.1 16 0.60

Wing-guided, 3 wings Either 0.75 0.90 0.58 1.1 11 0.60

Double-port Ported plug Either 0.75 0.90 0.28 0.84 12.5 0.80

Contoured plug Either 0.70 0.85 0.32 0.85 13 0.70

Wing-guided Either 0.75 0.90 0.41 0.84 14 0.80

Rotary Eccentric spherical plug Open 0.60 0.85 0.42 1.1 12 0.60

Close 0.40 0.68 0.42 1.2 13.5 0.35

ANGLE Contoured plug Open 0.72 0.90 0.46 1.1 17 0.65

Close 0.65 0.80 1.0 1.1 20 0.55

Characterized cage, 4 port Open 0.65 0.85 0.45 1.1 12 0.60

Close 0.60 0.80 1.0 1.1 12 0.55

Venturi Close 0.20 0.50 1.0 1.3 22 0.21

BALL Segmented (throttling) Open 0.30 0.80 0.98 1.2 25 0.25

Standard port(diameter ≈ 0.8d)

Either 0.42 0.74 0.99 1.3 30 0.20

BUTTERFLY 60°, no offset seat Either 0.42 0.70 0.5 0.95 17.5 0.39

90°, offset seat Either 0.35 0.60 0.45 0.98 29 0.32

90°, no offset seat Either 0.08 0.53 0.45 1.2 40 0.12

*Flow direction tends to open or close the valve, i.e., push the closure member away from or toward the seat.**In this table, d may be taken as the nominal valve size, in inches.

P F PF vvc =

F P PF v c≅ −0 96 0 28 1 2. . ( / ) /


6.15 Sizing 1239

if used, and P2 is downstream of the increaser. Some manu-facturers supply tables for Cv for valves with standard pipereducers. Where this data is based on actual test data, it shouldbe used in preference to the approximations shown below.Elbows and block valves installed near the valve will upset

the flow profile and cause some reduction in capacity. Noprocedure is available to predict these losses.

The standard reducer and increaser fittings result in anabrupt change in size. As a result of the change in cross-section, first there is an irreversible energy loss due to turbu-lence (friction loss), and second there is the conversionbetween pressure energy and velocity energy. If the controlvalve is smaller than the line size, the pressure at the valveinlet is reduced by the friction loss and is also reducedbecause the smaller flow area requires acceleration to a highervelocity.

If the downstream piping is the same size as the inletpiping, then the velocity-energy/pressure-energy exchange atthe outlet is reversed and it cancels the effect at the inlet.Friction losses are always additives. For a reducer at the valveinlet, the ISA standard determines the inlet fitting frictionloss coefficient:

6.15(12)

For an increaser at the valve outlet the ISA standarddefines the outlet fitting friction loss coefficient:

6.15(13)

Although not explicitly stated by ISA S75.01, an analysisof the references shows that for the unusual case where anincrease might exist at the valve inlet, K1 is calculated as

6.15(14)

and for a reducer at the outlet, K2 is found as

6.15(15)

For the velocity-pressure exchange (Bernoulli effect) forthe usual case:

6.15(16)

6.15(17)

It can also be shown from the same basic fluid mechanicsfor the unusual case:

6.15(18)

6.15(19)

Certain combinations of these coefficients are used incalculations:

6.15(20)

6.15(21)

FIG. 6.15d Typical flow rate vs. curve for liquid at constant upstreampressure and vapor pressure. The straight line segment on the leftshows the behavior of turbulent liquid flow. As the pressure droprises the pressure at the vena contracta drops and when it reachedthe vapor pressure of the liquid, vaporization starts. From this pointon (right side of figure) the relationship is no longer linear, until atqmax the choked flow condition is reached.

FIG. 6.15e Liquid critical pressure ratio factor FF, plotted as a function of thePv /Pc ratio.

Flow

rate

(q)

∆P ∆Pch

qmax

∆p

1.0

00 0.2 0.4 0.6 0.8 1.0

0.6

0.7

0.8

0.9

Pv/Pc

FF

FF = 0.96 − 0.28 (Pv/Pc)1/2 Equation 6.15 (10)

K d D1 12 20 5 1= −. ( ( / ) )

K d D2 22 21= −( ( / ) )

K D d1 12 21= −( ( / ) )

K D d2 22 20 5 1= −. ( ( / ) )

reducer at inlet K d DB1 141= − ( / )

increaser at outlet K d DB2 24 1= −( / )

increaser at inlet K D dB1 14 1= −( / )

reducer at outlet K D dB2 241= − ( / )

K K Ki B= +1 1 ( )inlet combination

K K K Ki B= + +2 2



A piping geometry factor is defined as:

6.15(22)

Valve flow capacity is directly proportional to this FP

factor (see Figure 6.15f ). Note that capacity is reduced for high values of

combined with high Cd. Because the inlet fittings also affectthe valve inlet pressure, a combination of FL and FP is usedto predict choking or cavitation (see Figure 6.15g):

6.15(23)

Again note the large loss in capacity caused by reducerson valves with high Cd. This equation provides only an esti-mate. The actual point where choking starts is preferred tobe determined by test.

6.15(24)

For all practical purposes, the flow at rated valve capacityis always turbulent (otherwise the outlet velocity would betoo high). This makes correcting for Fp much simpler andavoids most iterations. Figure 6.15h provides the piping con-figuration factor (Fp ) for two pipe-to-valve diameter ratios.The curve on the left is for a D/d = 2, and on the right for aD/d = 1.5.

Installing a valve between reducers and expanders canlead to the distortion of the valve’s inherent flow characteris-tic. This in turn can alter the gain of the valve (the percentagechange in flow resulting from 1% change in control signal).

Fp Correction Shortcut The correction to the required Cv

for reducer and expander losses can also be obtained by thefollowing steps: 1) Calculate the Cv for nonchoked liquid orgas flows. 2) Estimate the valve size d (inches) by determin-ing the required Cd (dividing the calculated Cv by d2). 3)Make sure that this required Cd is at least 40% smaller thanthe typical values listed in Table 6.15c. 4) Divide the pipediameter by the chosen valve diameter to obtain the D/d ratio.5) Select the applicable curve on Figure 6.15h. 6) Read Fp

from the applicable graph using the calculated Cd (requiredCv /valve size squared). 7) Divide the required system Cv byFp. This then gives the required and corrected valve Cv . 8)

FIG. 6.15f Piping geometry factor as a function of Cd and K.

1.0

0.9

0.8

0.7

0.6

0.5

0.40 10 20 30 40

FP

FP

1.61.41.21.00.80.6

0.4

0.2ΣK

ΣK

Cd

CdFP = [1 + (ΣK) (Cd

2)/890]−1/2 Ref. 6.15 (22)

F K CP d= + ∑ −[ ( ) ( ) / ] /1 8902 1 2

ΣK

F F K CLP L i d= + −[ / ( ) ( )/ ] /1 8902 2 1 2

( ) ( / )P P F F pvc LP P12− = ∆ critical

FIG. 6.15g Combined factor (FLP) for liquids as a function of Cd, Ki, and FL.

FIG. 6.15h Correction factor for pipe reducers and expanders. The curve onthe left is for a pipe to valve diameter ratio (D/d) of 2 and on theright, the curve is for D/d = 1.5.

Cd

FLP

0.3 0.4 0.5 0.6 0.7 0.8 1.00.9

0 10 20 30 40

0.5 0.6 0.7FL −

0.8 0.9 1.0

FLP = [1/FL2 + (Ki) (Cd )2/890]−1/2 Ref. 6.15 (23)

Ki

Ki

Cd

FL

FLP

1.61.41.21.00.80.60.4

0.20.1

0.05

25

1.00

0.82

0.84

0.86

0.88

0.90

0.92

0.94

0.96

0.98

0 5 10 15 20

Relationship Fp v.s. required CdRelationship Fp v.s. required Cd

Required Cd

F p fo

r D/d

= 1

.5

Required Cd

15 20 25 30

1.0

0.9

0.8

0.7

0.60 5 10

F p fo

r D/d

= 2


6.15 Sizing 1241

Check the manufacturer’s literature to make sure that the Cv

of the chosen valve size is sufficient.

EXAMPLE

Assume that the required valve Cv was calculated from theflow data as 70 and the pipe size is 4 in. If a line size valveis assumed, the corresponding Cd = 70/42 = 4.4. This thencould result in the selection of a 3− or 4-in. globe valve.

However, one might consider a less costly 2-in. ballvalve. In that case the required Cd = 70/22 = 17.5. ConsultingFigure 6.15h for a D/d = 2 (curve on the left), the factor Fp

for a Cd of 17.5 is 0.84. Therefore, the required operatingCv = 70/0.84 = 83.3. For such a requirement, one needs toselect a valve with a catalog Cv of 100 or more. Based onTable 6.15c, a 2-in. segmented ball valve can provide that,so the choice is acceptable.

Units Used in Valve Sizing

USA Units The valve sizing equations described here arebased on Instrumentation, Systems, and Automation SocietyStandard S75.01, “Flow Equations for Sizing ControlValves.”9 For turbulent, nonvaporizing flow conditions, theflow rate in gallons per minute is calculated in accordancewith Equation 6.15(25):

6.15(25)

The valve coefficient may be calculated as

6.15(26)

where

P1 and P2 are in units of lb/in.2 absolute.

If flow rate, w, is given in lbm/hr and specific weight, γ1,is in lbm/ft3, then

6.15(27)

or if solved for Cv:

6.15(28)

Under choked flow conditions, the maximum flow rateis no longer dependent on the pressure drop but rather on thepressure drop at the onset of choking (∆pch in Figure 6.15d).See Equation 6.15(9). Therefore, the maximum volumetricflow rate, q, under choked conditions, in gallons/minute is

6.15(29)

The maximum mass flow rate, w, in lbm/hr is

6.15(30)

When reducers, increasers, or other fittings are installedbetween the pipe and the valve, a combined liquid pressurerecovery factor, FLP, is used. See the discussion on liquidsizing factors in connection with Equation 6.15(23).The vol-umetric flow rate becomes

or if solved for Cv: 6.15(31)

The mass flow rate is calculated by

or if solved for Cv:

6.15(32)

The pressure drop at the onset of choking (∆pch) mustalways be calculated to check if choking will occur. It isconvenient to write the above Equations 6.15(25–32) usingthe allowable sizing pressure drop, ∆pa , which is defined asthe smaller value of the actual pressure drop, P1 – P2, andthe pressure drop at the onset choking, ∆pch.

The Reynolds number factor, FR, may also be includedin the equation. Refer to the discussion on liquid sizingfactors for nonturbulent in the later paragraphs discussinglaminar and viscous flow. Thus, the following set of simpli-fied equations can be written for volumetric flow rate (gpm):

6.15(33)

and if solved for Cv:

6.15(34)

Other Engineering Units The equations presented previ-ously were derived using U.S. Customary Units. In theseequations Cv is not a dimensionless coefficient. Therefore, itis helpful to develop units conversion constants that willpermit the use of these equations with other systems of units.ISA has chosen to write the equations using numerical con-stants (N1, N2, etc.) that take on unique values for specific

q F CP P

GP vf

=−1 2

Cq

F

G

P PvP

f=−1 2

w F C P PP v= −63 3 1 2 1. ( )γ

Cw

F P Pv

P

=−63 3 1 2 1. ( )γ

q F CP F P

GL vF v

fmax =

−1

w F C P F PL v F vmax . ( )= −63 3 1 1γ

q F CP F P

G

Cq

F

G

P F P

LP vF v

f

vLP

f

F v

max

max

=−

=−

1

1

w F C P F P

Cw

F P F

L v F v

v

LP

max

max

. ( )

. (

= −

=−

63 3

63 3

1 1

1

γ

FF vP )γ 1

q F F Cp

GP R va

f

=∆

Cq

F F

G

pvR P

f

a

=∆



combinations of units. Equations 6.15(35–41) and Table 6.15iare from ISA S75.01.*

or 6.15(35)

or 6.15(36)

where:

6.15(37)

The corresponding nonturbulent equations* are

or 6.15(38)

or 6.15(39)

The valve Reynolds number is defined* as

6.15(40)

The equations* for determining maximum flow rate of aliquid under choked conditions for valves in straight pipes ofthe same size as the valve are as follows:

or 6.15(41)

Metric Capacity Coefficients, Kv and Av Valve capacity co-efficients, which have been derived in a similar manner aswas the Cv but which have their basis in the metric or SIsystem of units, can be easily converted to their equivalentCv values. If units for volumetric flow rate (q) are in cubicmeters per hour (m3/h) and the pressure drop (∆p), is givenin bars (1 bar = 100 kPa), then the capacity coefficient Kv is

6.15(42)

TABLE 6.15i Numerical Constants for the Conversion of Liquid Flow Equations*

Constant Units Used in Equations

N w Q P, ∆P d, D γ1 ν (nu)

N1 0.08650.8651.00

———

m3/hm3/hgpm

KPaBarpsia

———

———

———

N2 0.00214890

——

——

——

mmin.

——

——

N4 76,00017,300

——

m3/hgpm

——

mmin.

——

Centistokes**Centistokes**

N6 2.7327.363.3

kg/hkg/hlb/h

———

KPaBarpsia

———

kg/m3

kg/m3

lb/ft3

———

*Reprinted by permission. Copyright © 1985, Instrumentation, Systems, and AutomationSociety. From ANSI/ISA-S75.01, “Flow Equations for Sizing Control Valves.”**To convert m2/s to centistokes, multiply m2/s by 106. To convert centipoise to centistokes,divide centipoise by Gf.

* Reprinted by permission. Copyright © 1985, Instrumentation, Systems,and Automation Society. From ANSI/ISA-S75.01, “Flow Equations for Siz-ing Control Valves.”

q N F CP P

G

Cq

N F

G

P P

P vf

vp

f

=−

=−

11 2

1 1 2

w N F C P P

Cw

N F P P

P v

v

P

= −

=−

6 1 2 1

6 1 2 1

( )

( )

γ

γ

FK C

N dPv=

∑+

−( )

/2

24

1 2

1

q N F CP P

G

Cq

N F

G

P P

R vf

vR

f

=−

=−

11 2

1 1 2

w N F C P P

Cw

N F P P

R v

v

R

= −

=−

6 1 2 1

6 1 2 1

( )

( )

γ

γ

Re/ /

/

vd

L v

L vN F q

F C

F C

N d= +

4

1 2 1 2

2 2

24

1 4

1ν

q N F CP F P

G

Cq

N F

G

P F P

L vF v

f

vL

f

F v

max

max

=−

=−

11

1 1

K q G pv f= ∆( / ) /1 2


6.15 Sizing 1243

Simple units conversions lead to the following relation-ship between Kv and Cv:

Kv = 0.865Cv 6.15(43)

The derivation of the capacity coefficient, Av, yields anequivalent area in square meters (m2) that is related to Cv bythe conversion

6.15(44)

It should be noted that valve capacity or sizing coeffi-cients are defined in terms of a flow test with pressure tapslocated in straight pipe of the same nominal size as the valveand at specific distances upstream and downstream of thevalve. The test configuration for determining Cv is welldefined in ISA Standard S75.0210 and is discussed in Section6.6. The above conversions are valid only if Kv or Av aredetermined using the same test manifold configuration andmethod as those specified in ISA-S75.02.

Sizing Example for Liquids

The above working equations can be used in a logicalsequence to determine the required valve capacity for a givenapplication. The following example illustrates the basic steps.However, there are other considerations when selecting avalve, such as noise, cavitation, corrosion, and actuator siz-ing, that are discussed in other sections of this chapter.

Example 1 Determine the required valve capacity coeffi-cient (Cv) and valve size for the following application:

Liquid: WaterCritical pressure (Pc): 3206.2 psiaTemperature: 250°FUpstream pressure (P1): 314.7 psiaDownstream pressure (P2): 204.7 psiaFlow direction (relative to valve plug): Flow-under-to-

openLine size/class: 4 in., ANSI Class 600Flow rate (q): 500 gpmLiquid vapor pressure (Pv): 30 psiaKinematic viscosity (v): 0.014 centistokesValve characteristics: Equal percentage

Step 1: Calculate actual ∆p.

Step 2: Calculate choked pressure drop ∆pch and deter-mine allowable sizing pressure drop ∆pa. From Equation6.15(10):

In order to determine the ∆pch , a preliminary valve-typeselection must be made in order to establish the liquid pres-sure recovery coefficient FL to be used in the calculation.Assume the use of a single-ported globe valve, which has anFL equal to 0.9 (Table 6.15c). Solving for ∆pch from Equation6.15(11):

The allowable sizing drop ∆pa is the smaller of ∆p and∆pch. Therefore, we have determined that ∆pa = 110 psi, andthe flow is not choked.

Step 3: Determine the specific gravity, Gf. The Instru-mentation, Systems, and Automation Society sizing equationsare based on water at 60°F with a density of 62.37 lb/ft3.From water properties data we find the density of water at250°F is 58.8 lb/ft3.

Step 4: Calculate the approximate required Cv assumingFP and FR equal 1.0. From Equation 6.15(34):

Step 5: Select the approximate body size based on theapproximate Cv from Step 4. From manufacturers’ catalogsit can be determined that the smallest globe valve body sizethat will accommodate Cv 46.2 is a 2 in. size with a maximumvalve capacity coefficient of Cv51.

Step 6: Determine the piping geometry factor FP. If man-ufacturer’s test data are not available for the selected valvetype, Fp may be estimated using Equations 6.15(12–22) orFigure 6.15f.

Therefore, in Figure 6.15h, the left side (D/d = 2) curveshould be used. For a reducer on the inlet and an increaseron the valve outlet the following values are calculated:

A Cv v= × −24 10 6

∆ = − = − =p P P1 2 314 7 204 7 110. . psi

F P PF v c= −= −

0 96 0 28

0 96 0 28 30 3206 2

1 2. . ( / )

. . ( / .

/

)) ./1 2

0 93=

∆ = − = − =p F P F PL F vch2

120 9 314 7 0 93 30 23[ ] . [ . ( . )( )] 22 3. psi

Gf = =58 80 62 37 0 94. / . .

Cq

F F

G

pvR P

f

a

=∆

= =500 0 94 110 46 21 2( . / ) ./

d D D d D= = = =2 0 4 0 0 501 2 1. , . , / .and

K

K

12 2

22 2

0 5 1 0 5 0 2813

1 0 5 0 56

= − =

= − =

. [ ( . ) ] .

[ ( . ) ] . 225

1 0 5 0 9375

0 5 1 0 9375

14

24

K

K

B

B

= − =

= − = −

∑

( . ) .

( . ) .

KK

Cd

= + + − =

=

0 2813 0 5625 0 9375 0 9375 0 8438

46

. . . . .

.22 2 11 55

1 890

1 0

2

2 1 2

/( ) .

[ ( )( ) / ]

[ (

/

=

= + ∑

= +

−F K CP d

.. )( . ) / ] ./8438 11 55 890 0 942 1 2− =



Step 7: Calculate the Reynolds number factor, FR. Formost process applications, this step is not necessary, becausein most cases turbulent flow (i.e., Rev > 10,000) conditionsexist. The calculation is shown here for illustration purposes.From Equation 6.15(64):

Referring to Figure 6.15m, FR = 1.Step 8: Calculate the final required Cv using Equation

6.15(34) and the values for Fp and FR that were calculatedabove:

Note: It was noted in Step 5 that the maximum Cv fora 2-in. globe valve was found to be Cv(max) = 51. Therequired Cv(req) = 49.1 is too close to the maximum, becauseit exceeds 85% of it. Therefore, a 3-in. globe valve shouldbe selected.

Step 9: Calculate the fluid velocity at the 3-in. valve outletusing Equation 6.15(45) with appropriate units conversions.

6.15(45)

whereU = velocityq = volumetric flow rateAo = area of the body outlet port

If q is given in gallons per minute (gpm) and Ao is insquare inches, then U can be calculated in feet per second:

6.15(46)

For this example, the outlet velocity is

U = 0.321(500 gpm/7.07 sq. in.) = 22.7 fps

Evaluation: The valve sizing and outlet velocity of 22.7fps is of consequence only under flashing (liquid/vapor) con-ditions, or where the fluid contains erosive particles. In suchcases, the valve manufacturer should also be asked to double-check the sizing for the particular service.

The Cavitation Phenomenon

Cavitation occurs when static pressure anywhere in the valvedrops to or below the vapor pressure of the process liquid.

Vaporization begins around microscopic gaseous nuclei. Thecavitation process includes the vapor cavity formation andsudden condensation (collapse) of the vapor bubble drivenby pressure changes. The basic process of cavitation is relatedto the conservation of energy and Bernoulli’s theorem, whichdescribes the pressure profile of a liquid flowing through arestriction or orifice (Figure 6.15b).

In order to accelerate the fluid through the restriction,some of the pressure head is converted into velocity head.This transfer of static energy is needed to maintain the samemass flow through the reduced passage. The fluid acceleratesto its maximum velocity, which corresponds to the point ofminimum pressure (vena contracta). The fluid velocity grad-ually slows down as it expands back to the full pipe area.The static pressure also recovers somewhat, but part of it islost due to turbulence and friction.

If the static pressure at any point drops below the liquidvapor pressure (Pv) corresponding to the process temperature,then vapor bubbles will form. If enough energy is impartedto the growing vapor bubble to overcome surface tensioneffects, the bubble will reach a critical diameter and expandrapidly. As the static pressure recovers to a point greater thanthe vapor pressure, the vapor will condense, causing thebubbles to collapse violently back into their liquid phase. Thegrowth and collapse of the bubbles produce high-energyshock waves in the fluid.

The collapse stage of the process (the bubble implosion)produces the more severe shock waves. Shock waves andliquid microjets radiate for short distances from implodingcavities and erode nearby surfaces. Cavitation can cause ero-sion, noise, and vibration in piping systems and thereforemust be avoided. Extensive cavitation also causes chokedflow conditions in the valve.

Predicting and Mitigating Cavitation Sizing a valve in liquidservice for choked flow allows one to determine its maximumflow capacity, but this is of limited value, because mostliquid-service valves should not be operated under chokedconditions. Special trim designs with multiple stages or mul-tiple flow paths like those in Figure 6.1y are typically usedto prevent severe cavitation and are better able to operate ator near choked conditions without damage.

Metal erosion from cavitation damage has a very distinc-tive appearance, like that of cinder block or sandblasting. Noknown material will withstand continuous, severe cavitationwithout damage and eventual failure. The length of time itwill take is a function of the fluid, metal type, and severityof the cavitation. Without special trim geometry, mild orintermittent cavitation can be mitigated in standard valves bythe use of extremely hard trim materials or overlays.

Sometimes it is feasible to increase the downstream back-pressure, or limit the pressure drop by installing control valvesin series to reduce the pressure drop in each valve. Anothermitigating effect in some processes is a result of the fluidthermodynamic properties and operating conditions. As liquidoperating temperature approaches its critical temperature, the

Re( )

R

/ /

/

vd

L v

L dN F q

F C

F C

N= +

41 2 1 2

2

2

1 4

1ν

ee( )( )( )

( . ) ( . )( . )

[( . )v = 17300 1 500

0 014 0 9 46 2

0 9 (( . )]

.

/

Re

11 55890

1

98 6 10

2

6

1 4+

= ×v

Cq

F F

G

pvR P

f

a

=∆

= ( )/[( . )( )]( . / ) /500 0 94 1 0 94 110 1 22 49 1= .

U q Ao= /

U q Ao( ) ( ) ( .). /fps gpm sq.in= 0 321


6.15 Sizing 1245

growth and collapse rates of cavities slow down, as heat trans-fer effects become increasingly significant relative to the dom-inant inertial effects. This can greatly reduce the energy ofcavity collapse and surface damage to the valve parts.

Some cryogenic and hydrocarbon applications are thoughtto behave this way, which may partly explain why cavitationdamage in these cases is minimal or absent even when cavi-tation is present in the valve. Further information about cavi-tation and predicting its effects on valve performance can befound in the ISA Recommended Practice RP75.23.01, “Con-siderations for Evaluating Control Valve Cavitation.”

Various methods are in use to establish the pressure dropat which cavitation starts. Two of the common techniques aredescribed here. The first is based on a cavitation index, Kc,determined from flow capacity test data similar to that shownin Figure 6.15d. The second is based on a cavitation index,σ, and vibration test data.

Flow Curve Cavitation Index (Kc) Figure 6.15j illustratesthe effect that cavitation has on the linear relationshipbetween flow rate and the square root of pressure drop. Theinflection point, the point at which the linear slope of the Cv

curve marks the ∆p, at which measurable amounts of vapor-ization exist is measured by a flow test.10

The calculation of the index Kc is based on the assumptionthat a valve may operate cavitation-free at any pressure dropthat is less than that associated with Kc, which is defined as15

6.15(47)

where ∆pi is the pressure drop at the inflection in the tested Cv

slope. Kc can then be compared to the index describing the

actual service conditions of an application in question, Ksc:

6.15(48)

If Ksc for the service is less than the Kc for the valve, thevalve is assumed to be adequate for the service. EstimatedKc values for various valve types are shown in Table 6.15c.These values of Kc are only representative and should alwaysbe verified with the valve manufacturer. However, care mustbe exercised with this method, because it has not consistentlyindicated damaging cavitation for some valve designs.2,15,18

Vibration Curve Cavitation Index (�) Noise and vibrationmeasurements have identified four recurring regimes of cav-itation in many valves.19–22 Figure 6.15k illustrates the rela-tionship between acceleration measurements taken on thedownstream pipe and a cavitation index called sigma (σ).

There are several forms of sigma, all of which representa ratio between the forces resisting cavitation and the forcespromoting cavitation. The following forms of sigma havebeen used for valves.

6.15(49)

6.15(50)

Upstream and downstream pressures should be correctedfor piping and fitting losses to obtain net values of P1 and P2

at the valve for calculating σ when Cv /d2 > 20. The σ indexes

FIG. 6.15jThe flow rate curve showing the effect of choking and the methodshow FL and Kc are determined. (Copyright © 1991, Instrumentation,Systems, and Automation Society. From ISA Paper No. 91-0462,“Solving the Problem of Cavitation in Control Valves.”)

KP

P Pci

v

=−

∆

1

Kc

∆Pch

∆Pi

FL2

Cv = Q/ ∆P/Gf

∆P/Gf ( psi )

Q (g

pm)

FIG. 6.15kClassical cavitation level plot (log-log). (Copyright © 1991, Instru-mentation, Systems, and Automation Society. From ISA Paper No.91-0462, “Solving the Problem of Cavitation in Control Valves.”)

KP

P Pscv

=−

∆

1

σ11

1 2

=−−

P P

P Pv

σ 22

1 2

=−−

P P

P Pv

Acce

lera

tion

s

si

sc

smv

Super-cavitation

Fullcavitation

Incipientcavitation

Nocavitation



can be related to the K-type indexes by using the relationship:

6.15(51)

Although Kc can be converted to a σ value for subjectivecomparison, the meaning of the value does not change. Thereis no proven correlation between specific points on the flowcurve Figure 6.15j and the inflection points on the vibrationcurve in Figure 6.15k. Various cavitation regimes can be iden-tified for many (but not all) valves with the type of plot shownin Figure 6.15k.22 The ISA Recommended Practice RP75.23.01uses σ defined by σ1 in equation 6.15(49), so where σ is usedin this section without a subscript, σ1 is implied.

The various regimes of cavitation identified in Figure6.15k are determined in a test defined in ISA RP75.23.01,which begins at a maximum attainable pressure drop in thetest valve. The pressure drop is then reduced in small stepswhile maintaining a constant upstream pressure and temper-ature; vibration, pressures, and flow rates are measured ateach step until the noise, vibration, and ∆P indicate that anoncavitating flow regime has been attained. The results areplotted showing vibration as a function of σ.

The noncavitating regime in Figure 6.15k is distinguishedby turbulent flow noise and mild vibration. The incipientcavitation regime begins at a level called incipient cavitationand progresses to another level called constant22 (or critical18)cavitation. Incipient cavitation begins with intermittent cavityformation and collapse. The resulting noise may be scarcelyaudible above the background noise. Constant cavitation pro-duces a steady process of cavity growth and collapse, so thatthe sound is no longer intermittent but is constant.

Cavitation in the incipient regime is not generally harm-ful to valve hardware,18 and limiting valve operation belowa level of constant cavitation is a conservative applicationapproach—although sometimes overly conservative. In thefull cavitation regime, vibration or sound intensity increasesto the maximum vibration level. When damage testing canbe done, the onset of cavitation damage, called incipientdamage, is often found somewhere in this regime.22

In many cases, the value of 1/Kc plotted on this graph willalso appear in this regime. Thus, Kc is sometimes referred to as“incipient choking.” Noise and vibration in the super-cavitationregime18 are less than at maximum vibration as a result of theway that large volumes of vapor in the flow affect the fluiddensity, compressibility, and vibration transmission properties.Super-cavitation may exhibit a vapor pocket extending fromthe valve downstream into the pipe, where it will collapse backinto liquid as the static pressure recovers. This differs from trueflashing, in which the downstream pressure remains less thanthe vapor pressure of the flowing fluid. Damage to downstreampiping and fittings is common in this regime.

The cavitation index σ describing the service conditionsis useful if it can be compared with a valve cavitation indexthat represents the resistance of the particular valve to poten-tial cavitation. The coefficient σmr (“manufacturer’s recom-

mended σ ”) is defined in ISA RP75.23.01 to designate thelimit for σ above which the valve may be operated.

If the operating pressures for the valve are very close tothe pressures used to test the valve, a direct comparison couldbe made, such that σ > σmr indicates safe operating condi-tions. However, research has shown that valve σ coefficientsfor incipient, constant, and incipient-damage cavitation donot remain constant for all pressures or in different sizes ofgeometrically similar valves. Scaling equations have beendeveloped that take these effects into consideration, and theyhave been adopted in the ISA recommended practice.

Pressure Scale Effect (PSE) The scaling factors PSE andsize scale effect (SSE) are used to adjust the recommendedvalve coefficient, σmr, when the reference size and referencepressure are known for which σmr was determined. Equation6.15(52) defines a scaled coefficient for the valve, σv .

6.15(52)

σv is then compared to the service σ calculated by Equa-tion 6.15(49). The service σ ≥ σv means that the valve willoperate at a level of cavitation less severe than the levelcorresponding to the manufacturer’s σmr.

The pressure scale effect, PSE, is

6.15(53)

The subscript R refers to reference pressures used intesting the valve. Typically, valves are tested at about (P1 –Pv)R = 100 psi; this value may be used as an estimate unlessotherwise stated by the valve manufacturer. The exponent, a,is a constant determined for a specific valve. Table 6.15lshows representative values of a for different valve types anddifferent levels of cavitation.18,22 When the exponent a is 0,PSE reduces to a value of 1, indicating that there is nopressure scale effect for those conditions.

Size Scale Effect The size scale effect, SSE, can be esti-mated from Equation 6.15(54). The exponent b was derivedfrom limited testing.18,20 It should be noted that testing underchoked conditions showed no size scale effect; i.e., SSE forchoked flow has a value of 1.

6.15(54)

6.15(55)

where d is the inside diameter of the valve port in inchesdR refers to the reference (tested) valve port diameter in inches

σ σ1 2 11= + =K

σ σv = − +[ ( ) ]( )mr SSE PSE1 1

PSE =−

−

P P

P Pv

v R

a

1

1( )

SSE =

ddR

b

bC

dv=

0 0682

1 4

.

/


6.15 Sizing 1247

Size scale equations are valid only when the valves arethe same style and have approximately the same relativecapacity; i.e., Cv /d

2 must be about the same for both valves.There are additional considerations if the valve is larger orsmaller than the pipe size with attached fittings (reducers,increasing). RP75.23.01 recommends additional calculationswhen Cv /d

2 > 20. The following example demonstrates howthis method can be applied.

Example 2. Rotary Valve Service data:

Fluid: WaterT = 74°FLine size = 10 in. Sch. 40P1 = 82 psiaQ = 3500 gpmGf = 0.998Pv = 0.41 psiaP2 = 70 psia

Results of valve sizing: Cv = 1010; 8 in., ANSI Class 150throttling rotary disk valve; body outlet velocity = 22 fps;approximately 75% open with Cv/d

2 = 1010/(8)2 = 15.8, soeffects of fittings may be ignored.

Calculate σ using Equation 6.15(49).

Compare with manufacturer’s cavitation data for the but-terfly valve, which are given as:

σmr = 5.1 (P1 − Pv)R = 100 psi, a = 0.12dR = 6 in.

Calculate PSE, b, SSE from Equations 6.15(53–55).

PSE = [(82 − 0.41)/100] 0.12 = 0.976b = 0.068(15.8)1/4 = 0.136SSE = (8/6)0.136 = 1.040

Calculate σv from Equation 6.15(52).

Evaluation: Since σ is 6.8 and is greater than σv of 5.2,the valve can operate without significant cavitation.

Rule of Thumb for Incipient Cavitation One definition ofthe state of incipient cavitation is when the pressure dropincreases to the point where it is first audible that somecavitation bubbles burst. This condition is usually not a causefor concern, although the pressure ratio factor XFzi is impor-tant in noise calculations.16

XFzi = 0.9(P1 ref /P1actual)0.125/[1 + 3Fd(Cv/1.17FL)0.5]0.5

6.15(56)

In Equation 6.15(38), if P1actual is in psia, the value ofP1 ref is 87 psia. If P1actual is in bars, P1actual is 6 bars absolute.Using this pressure ration factor XFzi, the pressure drop caus-ing incipient cavitation can be calculated using Equation6.15(39):

∆Pinc = XFzi (P1 − Pv) 6.15(57)

As a rule of thumb that can be used as a guide only inan attempt to avoid damage or noise, one should make surethat the valve’s pressure drop does not exceed the followinglimit:

∆Plimit = 0.5(XFzi + FL0.5)(P1 − Pv) 6.15(58)

For more accurate cavitation calculations one should useEquations 6.15(47–55), while for accurate noise calculations,should consult Section 6.14.

Example 3 Assume that a 2 in. globe valve was selectedfor a particular flow condition where the inlet pressure was195 psia and the outlet pressure 87 psia. Let us also assumethat using Equation 6.15(25), the required Cv was calculated

TABLE 6.15l Exponent “a” for Calculating Pressure Scale Effect (PSE)*

Valve Type Cavitation Level Exponent a

Quarter-turn valves (i.e., ball, butterfly)

IncipientConstantIncipient damageChoking

0.22–0.300.22–0.300.10–0.180

Segmented ball and eccentric plug

IncipientConstantIncipient damageChoking

0.30–0.400.30–0.40N/A0

Single-stage globe IncipientConstantIncipient damageChoking

0.12–0.160.12–0.160.10–0.140

Multistage globe IncipientConstantIncipient damageChoking

0.00–0.100.00–0.10N/A0

Orifice IncipientConstantIncipient damageChoking

000.200

*Reprinted by permission. Copyright © 1991, Instrumentation,Systems, and Automation Society. From ISA Paper No. 91-0462,“Solving the Problem of Cavitation in Control Valves.” Single-stage globe data modified per Tullis, Reference 18.

σ = −−

=82 0 4182 70

6 8.

.

σv = − + =[ . ( . ) ]( . ) .5 1 1 040 1 0 976 1 5 2



to be 38, and that the supplier’s catalog for the particularvalve showed an FL of 0.87 and an Fd of 0.4 at the givenvalve travel. Finally, assume that the vapor pressure of theflowing fluid is 8.5 psia.

Using Equation 6.15(56), we can calculate the pressuredifferential ratio for incipient cavitation as

XFzi = 0.9 (87/195)0.125/[1 + 3 × 0.4(38/1.17 × 0.87)0.5]0.5 = 0.282

Using this XFzi value, the limiting pressure drop to avoidcavitation damage or noise, using Equation 6.15(58), is

∆Plimit = 0.5(0.282 + 0.872)(195 − 8.5) = 96.8 psi.

Because the actual pressure drop is only 80 psi, we shouldnot be concerned.

Flashing

Flashing begins in the same manner as cavitation. However,in case of flashing, the pressure downstream of the cavitygrowth region remains at or below vapor pressure of theprocess fluid. This causes the vapor to stay in the vapor stateas it leaves the valve and enters the downstream piping. Thespecific volume increases as liquid changes to vapor, whichin turn causes an increase in the fluid velocity. If enoughvapor is formed, the resulting high valve outlet, or pipe, theresulting velocities can erode metals.

In many cases, flashing is a normal part of the process;it cannot be avoided, and special system and valve designsare required to accommodate it. The heat of vaporizationcomes from the process liquid, causing its temperature todecrease. The relative masses of liquid and vapor will therebyapproach thermodynamic equilibrium. The amount of flash-ing can be calculated from an energy balance. Even smallamounts of flashing (e.g., where the vapor equals 1–3% byweight) can significantly affect a valve’s capacity, sizing, andselection; therefore, flashing should be stated in the valvespecification data sheets.

Large amounts of flashing (e.g., 10–15% by weight)require special valve designs, such as oversized outlets, anda larger downstream pipe. Note, choking can occur in thedownstream pipe due to the large volume. This in turn willincrease the back-pressure, P2, which in turn then causes thevalve to undergo cavitation again.

When the valve outlet pressure, P2, is less than or equalto the vapor pressure of the process liquid, some of the liquid“flashes” into vapor and stays in the vapor phase as it entersthe downstream piping. The phase change from liquid tovapor may cause high velocities and erosion of metals at theoutlet. Even small amounts of flashing can significantly affectvalve sizing and selection. Large amounts of flashing (e.g.,where the vapor equals 10–15% by weight) require specialvalve designs and materials.

In order to select the right valve, it is necessary to knowthe fraction of the liquid that will flash to vapor and the

flowing velocity of the resulting vaporized mixture. If thevapor content and therefore the velocity is high, the resultingback-pressure at the valve outlet can cause “choked flow” inthe downstream pipe. This can cause cavitation in the valveand can cause severe erosion of the valve trim.

Calculating the Flash Fraction (X) The method for calcu-lating the vapor fraction is essentially the same for all liquids.Water is the most common liquid that is likely to flash, anddata for water are readily available in the steam tables. Datafor other fluids may be found in other references.11

The vapor fraction by weight can be calculated fromEquation 6.15(59).

6.15(59)

where hf1, hf2, and hfg2 are the enthalpy of saturated liquidupstream, the enthalpy of saturated liquid downstream, and theenthalpy of evaporation at downstream pressure, respectively.

Calculating Velocity Velocity calculations of the liquid-vapor mixture downstream of the valve assume that the mix-ture is in thermodynamic equilibrium. In most cases, thisassumption is a good enough approximation when comparedto the accuracy of other valve sizing factors. The velocity, U,for the mixture is

6.15(60)

where all units are consistent and vf2 and vg2 are specificvolumes of the saturated liquid and saturated vapor, respec-tively, at downstream conditions. If mass flow rate, w, is givenin lb/hr, area A in in.2, and specific volumes in ft3/lb, Equation6.15(60) for the calculation of velocity, U, in fps becomes

6.15(61)

Similarly, if flow rate is given as q in gpm, then U in fpscan be calculated by

6.15(62)

It has been the experience that velocities exceedingapproximately 300–500 ft/s have caused erosion in down-stream pipe fittings in flashing water applications. Therefore,flashing water applications often require angle body valvesand piping made out of chromium-molybdenum-steel alloysto resist erosion from the flashing water that is moving athigh velocities.

Sizing Example 4 A flashing application must always besized on the basis of choked flow conditions. Piping down-stream of a flashing valve should generally be larger thanthe valve size because it must accommodate the largeincrease in volume. Therefore, the corresponding combined

X h h hf f fg= −( )/1 2 2

U w A X v Xvf g= − +( / )[( ) ]1 2 2

U w A X v Xvf g= − +( . / )[( ) ]0 040 1 2 2

U qG A X v Xvf f g= − +( . / )[( ) ]20 0 1 2 2


6.15 Sizing 1249

pressure recovery and piping geometry factor, FLP , must alsobe applied to the sizing equations. See Equation 6.15(23)and Figure 6.15g.

Using the same configuration and conditions as in Exam-ple 1, except that P2 in this case is 20 psia, determine thevalve size and required Cv. (The other conditions in Example1 were: process fluid was water at 250°F, Pv = 30 psia, Gf =0.94, P1 = 314.7 psia, and q = 500 gpm.)

Step 1: Calculate actual ∆p.

Step 2: Calculate ∆Pch and determine ∆Pa. From Example1, FF = 0.93. Assuming that an angle valve with a characterizedcage and flow over the seat (flow to close) is the preliminaryselection for this flashing application, from Table 6.15c onecan obtain the values FL = 0.80, Fd = 1.0, and Cv/d2 = 12.From Equation 6.15(11)

Step 3: Determine specific gravity of liquid at upstreamconditions. This was determined in Example 1; Gf = 0.94.

Step 4: Calculate the approximate required Cv assumingFP and FR = 1. From Equation 6.15(34)

Step 5: Select the approximate body size. Refer to man-ufacturers’ catalogs or, by using estimates from Table 6.15c,Cv /d2 = 12, in which case the minimum body size is

Thus, a 2-in. body size is selected.Step 6: Determine FLP; d = 2.0 in.; D1 = D2 = 4.0 in.;

d/D1 = 0.5.From Equations 6.15(12–22) and Example 1: K1 =

0.2813; KB1 = 0.9375; Ki = 0.2813 + 0.9375 = 1.2188; Cd =35.8/(2)2 = 8.95.

Using Equation 6.15(23):

Step 7. Calculate the final required Cv from Equation6.15(30).

Therefore, the 2-in. valve size is adequate and the capac-ity required for the application is within the estimated limit.

Example 5 Given the flashing conditions described inExample 4, determine the mass fraction of vapor at the valveoutlet.

Step 1: Determine the enthalpies of saturated liquid atupstream conditions and of saturated vapor and liquid atdownstream conditions using the steam tables. From satu-rated steam temperature tables at 250°F for upstream condi-tions: hf1 = 218.59 BTU/lbm.

From saturated steam pressure tables at 20 psia for down-stream conditions: hf2 = 196.27 BTU/lbm; hg2 = 1156.3BTU/lbm; hhfg2 = 960.1 BTU/lbm.

Step 2: Using the assumption that the process is adiabaticand therefore the enthalpies immediately upstream and down-stream of the valve are equal, Equation 6.15(59) can be usedto calculate the vapor fraction, X, or steam “quality” down-stream:

Therefore, the amount of fluid mass flashing to steam is2.3%.

Example 6 Given the flashing case in Examples 4 and 5,estimate the velocity of the two-phase mixture at the valveoutlet.

Step1: From the steam tables, find the specific volumesof saturated liquid and vapor at downstream conditions.

Step 2: Determine vapor fraction or steam quality atdownstream conditions. For this example, the vapor fraction,X, was found in Example 5. X = 0.023.

Step 3: Find specific gravity (or density) of the mixtureand the cross-sectional area at the valve outlet. From Example4, Gf = 0.94. Flow area is the outlet area of a 2-in. ANSIClass 300 valve:

Step 4: Calculate velocity using one of Equations6.15(60–62). Selecting Equation 6.15(62), because the flowrate is in units of gpm and area in square inches,

Evaluation: This is not a realistic velocity since the actualvelocity cannot exceed the speed of sound (about 1300 ft/s).As a result of such high velocity, the pressure at the valveoutlet will rise, which in turn will reduce the vapor content

∆p = − =314 7 20 294 7. . psi

∆ p F P F PL F vch = − = − =21

20 8 314 7 0 93 30 18( ) . [ . ( . )( )] 33 6.

,

psi

therefore =ch c∆ ∆ ∆ ∆p p p pa< hh = 183.6 psi

Cq

F F

G

pvR P

f

a

= = =∆

500 0 94 183 6 35 81 2( . / . ) ./

d ≈ ≈( . / ) ./35 8 12 1 731 2

F F K CdLP L i= +

= +

−

−

1 890

0 8 1 21

2 2 1 2

2

/ / )

[ . ( .

/

888 89 5 890 0 772 1 2)( . ) / ] ./− =

C q F G P F Pv LP f F v= −

=

( / ) /( )

/ . . /

max 1

500 0 77 0 94 (( . ( . )( )) .314 7 0 93 30 37 2− =

X h h hf f fg= − = − =( )/ ( . . )/ . .1 2 2 218 59 196 27 960 1 0 0233

v vf g12 20 016834 20 087= =. ; .ft /lbm ft /lbm3 3

A = =π ( . ) / . .2 00 4 3 142 2in

U qG A X v Xvf f g= − +

=

[ / ]( ) ]

[( )( )( .

20 1

20 500 0 9

2 2

44 3 14 1 0 023 0 016834

0 023

)/ . ][( . )( . )

( . )

−

+ (( . )]20 087 1432= ft/s



until an equilibrium is established at sonic velocity. The resultwill be choked flow in the pipe.

At that point, the back-pressure in the pipe can causecavitation in the valve and in the pipe. Wet steam at 1300 ft/scan be erosive. An angle valve with hardened trim exhaustingit into either a long, straight run of pipe or directly into areceiver vessel should be used in order to avoid erosion ofdownstream pipe and fittings.

Another alternative would be to use an expanded outletwith a venturi outlet on the valve to reduce the velocity beforeit exits into the pipe. A venturi outlet with the same exit areaas the 4-in. downstream pipe would reduce velocity to approx-imately 360 ft/s. These velocities are estimates only becausethe assumption of thermodynamic equilibrium was used todetermine fluid properties. Note: The longer the downstreampipe and the more elbows there are in your piping system, themore the valve’s back-pressure will increase, and with it cav-itation will also rise.

Laminar or Viscous Flow

The proportionality between flow rate and the square root ofpressure drop holds true only for fully turbulent Newtonianfluid behavior. Non-Newtonian fluids include most polymersand many other fluids. If no experimental sizing data exist,one might specify a line size valve and purchase a numberof reduced Cv trims for it, so that the final choice of trimwould be determined by trial and error.

Laminar or transitional flow regimes may result fromvery low flows, small valves or passages, low pressure dif-ferentials, and high viscosity. The valve Reynolds number,Rev, is calculated to determine the effect of laminar or tran-sitional flow behavior on the required valve Cv.

Reynolds Number Factor (FR ) In the majority of controlvalve applications the flow velocities inside the valve arehigh, which in turn causes the Reynolds number to be high,and therefore turbulent conditions exist. Under turbulent con-ditions pressure drop relates to the square of flow.

On the other hand, when the viscosity of the process fluidis high or when the size of the valve and the flow velocitythrough it are low, the Reynolds number will also be low andlaminar flow conditions will exist. Under laminar conditionspressure drop linearly relates to flow. Therefore, the sameincrease in pressure drop results in a larger increase in flowthan it would under turbulent conditions.

Here the flow coefficient under laminar or transitionalflow becomes

Cv = q/FR (∆P/Gf)0.5 6.15(63)

The valve equivalent Reynolds number, Rev , is defined as

6.15(64)

(See Table 6.15i for the values of constants N2 and N4

for various units.)

FR = 0.7 × 77 (Rev)0.5/FL (Cv /d2) 6.15(65)

When the valve operates in the laminar regime and thediameter (d) is in inches,

FR = 1 + (0.06 (Cv /d2)0.5FL0.5) log (Rev /10,000)

6.15(66)

Note: When operating in the transitional regime, use thesmaller of the two FR values calculated by using Equations6.15(65) and 6.15(66).

If the diameter (d) is given in millimeters, the FR numbersare determined by using Equations 6.15(67) and 6.15(68).For the laminar regime use Equation 6.15(67):

FR = 0.0076(Rev)0.5/FL (Cv/d

2) 6.15(67)

For the transitional regime use Equation 6.15(68):

FR = 1+ (1.53(Cv /d2)0.5 FL0.5) log (Rev/10,000) 6.15(68)

The curves provided in Figure 6.15m relate the valveReynolds number (Rev ) to the Reynolds number factor (FR ).

Example 7 Assuming that in a particular application Cv =10, d = 25 mm, FL = 0.9, Rev = 500, the FR in the transitional

Re( )

/ /

/

vd

L v

L dN F q

F C

F C

N= +

4

1 2 1 2

2

2

1 4

1ν

FIG. 6.15m The valve Reynolds number (Rev ) can be corrected as a function ofthe Reynolds number factor (FR ) and of the type of valve used (ifthe diameter d is in inches).8 Curve No. 1 applicable for globe style valves Cd ≤ 10. Curve No. 2 applicable for globe valves and eccentric rotary plug

valves, Cd = 10–15. Curve No. 3 applicable for butterfly valves, Cd = 15–25. Curve No. 4, applicable for ball valves Cd ≥ 25.

1

2

3

4

1.00

0.01

0.10

Valv

e Rey

nold

s num

ber f

acto

r-F R

Valve Reynolds number-Rev

1 10 100 1,000 10,000


6.15 Sizing 1251

regime is calculated as follows:

1 + (1.53(10/625)0.5 0.90.5) log (500/10,000)= 1 + 0.193 × −1.3 = 0.748

In Figure 6.15m the straight diagonal lines extending down-ward from right to left, starting at an FR value of approximately0.5, are in the laminar region and operate under the conditionswhere laminar flow exists. At a valve Reynolds number above10,000, all three curves in Figure 6.15m reach an FR value of1.0. At this number and at all higher values of Rev, fullyturbulent flow conditions exist. Between the laminar region(straight diagonal lines in Figure 6.15m), and the turbulentregion (over Rev = 10,000, where FR = 1.0), the flow regime istransitional (i.e., neither laminar nor turbulent).

Equation 6.15(69) can be used for determining the valveReynolds number Rev for globe valves and small flow valves.In these cases the Bernoulli correction for reducers (thebracketed term in Equation 6.15[64]) is not required.

Rev = N4Fd q/FL0.5Cv

0.5 6.15(69)

The valve-style modifier in Equation 6.15(69) is desig-nated Fd. Baumann17 defined Fd as the ratio of the equivalenthydraulic diameter(s) of a valve flow orifice to that of a circularone. For representative values please consult Table 6.15c.

Valve Sizing for Viscous (Laminar) Services The follow-ing treatment is applicable to valves with or without attachedfittings. The Fp factor (Table 6.15c) is assumed to be 1.0,because of the absence of turbulence at the valve outlet. Thefirst step in the valve sizing procedure is to calculate a pseudovalve flow coefficient Cvt, assuming turbulent flow and usingEquation 6.15(70):

6.15(70)

One may calculate the valve’s Reynolds number by usingeither Equation 6.15(64) or 6.15(69) and then determine FR

from Figure 6.15m. The required Cv for the laminar or tran-sitional flow conditions is then found by CvL = Cvt /FR.

The problem with this method is that it requires iteration,because one first has to guess the laminar valve Cv in orderto calculate the Reynolds number, and later one has to repeatthis procedure if the actual CvL is too far off from the estimate.

In the sizing approach below a simplified method is pre-sented for two typical valve classes. This approach, whileignoring transitional flow (a minor error), saves the timeassociated with the above-mentioned iteration.

Globe Valve (Example 8) For the calculations below, thevalues of FL = 0.9 and Fd = 0.46 are assumed. In this simpli-fied approach one would first calculate a pseudo Revi bymodifying Equation 6.15(69) by also using Equation6.15(70) for Cvt, in order to arrive at Equation 6.15(71) below:

Revi = N4q/0.95νCvt0.5 6.15(71)

Next, the required Cv is calculated under laminar flowconditions:

CvL = Cvt /0.022 (Revi)0.666 6.15(72)

Assuming that q = 4.13 gpm, ν = 31,729 cSt, ∆P = 8.41 psi,and Gf = 0.977, one can, using Equation 6.15(70), calculate theCvt of 1.407. Using that value in Equation 6.15(71), the Reynoldsnumber comes out as 2.

Based on Equation 6.15(72), the Cv under laminar flowconditions is calculated as:

CvL = 1.407/0.022 (2)0.666 = 40.3

This Cv requirement can be met with a 2-in. globe-stylecontrol valve.

Ball or Butterfly Valves (Example 9) For the calculationsbelow, the values of FL = 0.6 and Fd = 0.5 are assumed. Herethe required Cv under laminar flow conditions is

CvL = Cvt /0.04 (Revi)0.666 6.15(73)

Using the same data as in Example 8 above, but usingEquation 6.15(73), we get

CvL = 1.407/0.04(2) 0.666 = 22.

For this application, one could select a 1.5-in. ball valveinstead of the 2-in. globe valve and thus obtain some costsaving. Note: If Rev is greater than 10,000, the flow may betaken as turbulent and it can therefore be assumed that FR =1.0. In this case CvL = Cvt.

For butterfly valves refer to Section 6.17.

Small Flow Valves (Example 10) The sizing of small flowvalves is more complicated, because their Fd factor varieswidely depending on the type of their trim. Laminar flow islikely to occur in small valves on liquid service, but at Cv

values under 0.1, it is also possible to have laminar or tran-sitional flow with gases.

The Fd values drastically change with valve type. ForV-notch-type plugs Fd = 0.7, for flat seated trim Fd = 0.3,and for tapered needle trim Fd = 0.09 × (CvL/do)

0.5, wheredo = orifice diameter in inches.

A simplified sizing equation for liquid service in smallvalves is given below:

CvL = 0.192(N11Gfνq/∆PFd)0.666 6.15(74)

Assuming the following process conditions and the useof a flat seated trim: q = 0.0035 gpm, Gf = 0.87, ∆P = 50 psi,ν = 16 cSt, Fd = 0.3 (flat seat). Having this information, onewould first calculate the turbulent Cv using CvT = q (Gf /∆P)0.5,which results in CvT = 0.00046.

Cq

Nvt P P

G f

=−

11 2



Next CvL is calculated from 6.15(74) as CvL = 0.192 (1 ×0.87 × 16 × 0.0035/50 × 0.3)0.666 = 0.004.

Note that this CvL value is substantially higher than theturbulent Cv of 0.00046, hence it indicates that laminar flowwill occur and that a valve trim with a rated Cv that is above0.004 has to be selected.

For small flow valves having tapered needle trims thesizing is even simpler. Here,

CvL = 0.973(N12 qGfνdo /∆P)0.5 6.15(75)

Using the previous example but choosing a needle trimwith an orifice diameter of do = 0.055 in., CvL is calculatedto be 0.0071 (a substantial difference from the 0.004 withthe flat seat trim).

Sizing for Laminar Gas Flow In case of gases one shoulduse absolute viscosity µ, because, in contrast with liquids,the kinematic viscosity varies with the absolute pressure ofthe gas. Sizing for the most common trim, the tapered needle,can be done as follows:

One should first calculate the turbulent Cv from theappropriate equation from the orientation table, Table 6.15a.Next, the laminar valve coefficient for gas service (CvLgas )is to be determined by using the following equation, wheredo, valve orifice diameter in inches, has to be obtained fromthe manufacturer:

CvLgas = CvT /0.0127(NgqGs /µdo)0.5 6.15(76)

whereµ = absolute viscosity, centipoisesNl1 = a constant for liquids = 1 if q = gpm, P = psi

for misc. trim 30.7 if q = m3/h, p = kPaNl2 = a constant for liquids = 1 if q = gpm, P = psi, and

for tapered trim 1204 if q = m3/h, P = kPa, and do = m, do = inches

Ng = a constant for gas = 1 if q = gpm, and do = inchesfor tapered trim 0.9 if q = m3/h, and do = m

Example 11 Assuming the following process conditions:q = 1.6 scfh and the gas has a specific gravity, Gs, of 1.34.The absolute viscosity µ is 0.0215 cP, P1 = 190 psia, P2 =170 psia. Based on these conditions, the calculated Cvt isfound to be 0.0005.

If one learns from the manufacturer that the orifice diam-eter do = 0.197 in., one can proceed to Equation 6.15(76) andcalculate the required laminar CvL as:

CvL = 0.0005/0.0127(1 × 1.6 × 1.34/0.0215 × 0.197)0.5

= 0.0017

This is 3.5 times the turbulent Cv requirement! Note: Oneshould always calculate Cvt and CvL for small valves, espe-cially when Cv is below 0.01. After having made both calcu-lations, one should select the larger of the two values.

GAS AND VAPOR SIZING*

The flow rate of a compressible fluid varies as a function ofthe ratio of the pressure differential to the absolute inletpressure (∆p/P1), designated by the symbol “x.” At values ofx near 0, the equations in this section can be traced to thebasic Bernoulli equation for Newtonian incompressible flu-ids. However, increasing values of x result in expansion andcompressibility effects that require the use of appropriatecorrection factors.

Equations for Turbulent Flow

The flow rate of a gas or vapor through a valve or the requiredCv may be calculated by using any of the following equivalentforms of the equations, which differ only in the units used.If the flow is in mass units and the density is in specificweight units,

or 6.15(77)

For volumetric flow and specific gravity units:

or 6.15(78)

For mass flow and molecular weight units:

or 6.15(79)

For volumetric flow and molecular weight units:

or 6.15(80)

Note that the numerical value of x used in these equationsmust not exceed the choking limit (FkxTP), regardless of theactual value of x.

* Reprinted by permission. Copyright © 1985, Instrumentation, Systems,and Automation Society. From ANSI/ISA-S75.01, “Flow Equations for SizingControl Valves.”

w N F C Y xP

Cw

N F Y xP

P v

v

P

=

=

6 1 1

6 1 1

γ

γ

Q N F C Yx

G T Z

CQ

N F PY

G T Z

x

P vg

vP

g

=

=

71

7 1

1

Q N F C PYxMT Z

Cw

N F PY

T Z

xM

P v

vP

=

=

8 11

8 1

1

Q N F C PYx

MT Z

Cq

N F PY

MT Z

x

P v

vP

=

=

9 11

9 1

1


6.15 Sizing 1253

Constants for Engineering Units

The numerical constants, N, are chosen to suit the engineer-ing units used in the equations. Values for N are listed inTable 6.15n.

Expansion Factor (Y )*

The expansion factor Y accounts for the change in density ofa compressible fluid as it passes from the valve inlet to thevena contracta and for the change in area of the vena contractaas the pressure drop is varied (Contraction coefficient). The-oretically, Y is affected by all of the following:

1. Ratio of port area to body inlet area2. Internal geometry of the valve3. Pressure drop ratio, x = ∆p/P1

4. Reynolds number5. Ratio of specific heats, k

The influence of items 1, 2, and 3 are defined by thefactor xT. Test data indicate that Y may be taken as a linearfunction of x, as shown in the following equation (and inFigure 6.15n for a valve without attached reducers or fittings):

6.15(81)

For a valve with attached fittings, xTP should be substi-tuted for xT. For the overwhelming majority of applications,Reynolds number effects may be disregarded when the pro-cess fluid is compressible except for small flow valves; i.e.,if Cv is less than 0.01. The effect of the ratio of specific heatsis discussed later.

Choked Flow*

If all inlet conditions are held constant and the differentialpressure ratio (x) is increased by lowering the downstreampressure (P2), the mass flow rate will increase till it reachesa maximum. Flow conditions where the value of x exceedsthis limit are described as choked flow.

Choking occurs when the jet stream across the cross-sectional area of the vena contracta attains sonic velocity andwhen the vena contracta ceases to expand. Sonic velocityalready occurs when P1/P2 exceeds about 0.5 × FL

2 × P1,while choked flow occurs at P1/P2 ratios, which are higherthan the ones causing sonic velocity. Therefore, xT is alwayshigher than the sonic pressure ratio.

The value of x at the inception of choked flow conditions(xT) varies with the type of valve (Table 6.15c). It also varieswith the piping geometry and with the thermodynamic proper-ties of the flowing fluid. The factors involved are xT, xTP, and Fk.

Choking affects the use of Equations 6.15(77–81) in thefollowing manner: The value of x used in the equations mustnot exceed FkxT or FkxTP , regardless of the actual value of x.The expansion factor Y at choked flow (x ≥ FkxTP) is at itsminimum value of 2/3.

TABLE 6.15n Numerical Constants to be Used in Gas and Vapor Flow Equations, which Reflect the Units Being Used*

Constant Units Used in Equations

N W Q** P, ∆P γ1 T1 d, D

N5 0.002411000

——

——

——

——

——

mmin.

N6 2.7327.363.3

kg/hkg/hlb/h

———

kPaBarpsia

kg/m3

kg/m3

lb/ft3

———

———

N7 4.174171360

———

m3/hm3/hscfh

kPaBarpsia

———

KK°R

———

N8 0.94894.819.3

kg/hkg/hlb/h

———

kPaBarpsia

———

KK°R

———

N9 22.522507320

———

m3/hm3/hscfh

kPaBarpsia

———

KK°R

———

*Reprinted by permission. Copyright © 1985, Instrumentation, Systems, and Automation Society. FromANSI/ISA-S75.01, “Flow Equations for Sizing Control Valves.”**Q is in cubic feet per hour measured at 14.73 psia and 60°F, or cubic meters per hour measured at 101.3kPa and 15.6°C.

* Reprinted by permission. Copyright© 1985, Instrumentation, Systems, andAutomation Society. From ANSI/ISA-S75.01, “Flow Equations for SizingControl Valves.”

Yx

F xY

k T

= − ≥ ≥13

1 0 0 67(lim : . . )its



Critical Pressure Drop Ratio Factor (xT) For maximum acc-uracy, the pressure drop ratio factor, xT, must be established byusing the test procedures specified in Reference 10. Represen-tative xT values for valves are tabulated in Table 6.15c. Theserepresentative values are not to be taken as actual. Actualvalues must be obtained from the given valve manufacturer.

When a valve is installed with reducers or other fittings,the critical pressure drop ratio factor of the assembly (xTP) isdifferent from that of the valve alone (xT). For maximum accu-racy, xTP must be determined by test.10 When estimated valuesare permissible, the following equation (or Figure 6.15p) maybe used to determine xTP:

6.15(82)

In this equation, xT is the pressure drop ratio factor for agiven valve installed without reducers or other fittings, andKi is the sum of the inlet velocity head coefficients (K1 + KB1)of the reducer or other fittings attached to the valve inlet.

Ratio of Specific Heats Factor (Fk) The ratio of specificheats of a compressible fluid affects the flow rate through avalve. The factor Fk accounts for this effect. Fk has a value of1.0 for air at moderate temperatures and pressures, where itsspecific heat ratio is about 1.40. Both theoretical and experi-mental evidence indicates that for valve sizing purposes, Fk

may be taken as having a linear relationship to k. Therefore,

6.15(83)

Compressibility Factor (Z) Equations 6.15(78–80) do notcontain a term for the actual specific weight of the fluid atupstream conditions. Instead, this term is inferred from theinlet pressure and temperature based on the laws of idealgases. Under some conditions, real gas behavior can deviatemarkedly from the ideal. In these cases, the compressibilityfactor Z shall be introduced to compensate for the discrepancy.

Z is a function of both reduced pressure and reducedtemperature. In this section, reduced pressure Pr is definedas the ratio of the actual inlet absolute pressure to the absolutethermodynamic critical pressure for the fluid in question:

6.15(84)

The reduced temperature is defined similarly. That is,

6.15(85)

Absolute thermodynamic critical pressures and temper-atures for most fluids and curves from which the compress-ibility factor, Z may be determined can be found in manyreference handbooks of physical data.11

Velocity of Compressible Fluids

The velocity of sound in a compressible fluid is an importantparameter. Pressure changes travel through an ideal gas atthe velocity of sound. Most valve and piping systems arerestricted to fluid velocities less than the speed of sound, dueto the nature of the geometric discontinuities (supersonicwind tunnels being a notable exception). In valves, the throt-tling element and seat produce the vena contracta.

As flow rate and fluid velocity increase, the effects ofcompressibility cause the relationship of mass flow (x)1/2 todepart from a straight line. Once sonic velocity is attained inthe vena contracta and the pressure ratio of xT is reached,choking occurs. Reducing the downstream pressure will notincrease the flow rate or the velocity at the vena contracta.

However, if local downstream geometry permits (e.g., sim-ilar to diverging nozzles), it is possible for velocity to increasedownstream of the vena contracta to supersonic levels. Shockwaves from supersonic velocities can produce undesirable andeven dangerous vibration in process systems; use of supersonicvelocities in system or valve applications should be avoided.The ratio of the actual fluid velocity, U, to the sonic velocity,c, is the Mach number, MN. For valves with standard trim MN

should not exceed 0.3 at the outlet of the valve body. In orderto avoid excessive noise. (See Section 6.14 for details.)

6.15(86)

The sonic velocity can be determined from the first lawof thermodynamics and the ideal gas law. Gas constants andratios of specific heat are required in the calculation.

6.15(87)

Flow rate for gases is usually given in terms of mass flowrate, w, or volumetric flow rate, Q. The velocity is

6.15(88)

From these equations and from appropriate units conver-sions, the Mach number can be derived as follows:

For gases (using molecular weight, M):

6.15(89)

For gases (using specific gravity, Gg):

6.15(90)

For air:

6.15(91)

xx

F

x K C

N dTPT

P

T i v= +

−

2

2

54

1

1

Fk

k =1 40.

P P Pr c= 1 /

T T Tr c= 1 /

M U cN = /

cg kRT

Mc= 1

U wv A Q A= =/ /

MQ

AN kT

Ma

=5574

MQ

AN kT

Ga

g

=1035

MQ

A TN

a

=1225


6.15 Sizing 1255

For steam:

6.15(92)

where

Q = actual volumetric flow, ft3/hr (not scfh)

A = flow area of the valve outlet, in.2

Ta = absolute temperature, °R (°F + 460)w = mass flow rate, lbm/hr

v = specific volume, ft3/lbmGg = specific gravity of the fluid at standard conditions

relative to air at standard conditionsM = molecular weightk = ratio of specific heats

The calculated exit velocity or Mach number should becompared to the limits set by the process or the valve man-ufacturer for the specific valve and application. Mach num-bers exceeding 0.3 can cause significant noise problems.

Sizing for Compressible Fluids (Example 12)

The above working equations can be used in a logicalsequence to determine the required valve capacity for a givenapplication. The following example illustrates the basic steps.However, there are other considerations when selecting avalve, such as noise, erosion, corrosion, and actuator sizing,that are discussed in other sections of this chapter and in thereferences.

For the example below, assume that the process condi-tions are the following:

Fluid: SteamTemperature: 450°FUpstream pressure (P1): 150 psiaDownstream pressure (P2): 65 psiaSpecific volume downstream (v): 7.98 ft3/lbFlow rate (w): 15,000 lb/hrFlow characteristic: Equal percentageLine size: 3 in., Class 600

Step 1: Select the appropriate equation based on the avail-able process information and its units. As the information isgiven in mass flow units, Equation 6.15(79) can be used andN8 can be selected from Table 6.15n as 19.3. Thus, the work-ing equation becomes

6.15(93)

In order to be able to use this equation, one must obtainother physical data for critical pressure and temperature, formolecular weight, and for the ratio of specific heats. Table 6.15qlists such data for a variety of gases and vapors.

For steam:

Critical pressure (Pc): 3208.2 psiaCritical temperature (Tc): 1165.5°RMolecular weight (M): 18.02Ratio of specific heats (k): 1.33

Step 2: Check for choked flow. Determine the criticalpressure drop ratio, xT , from manufacturer’s data or from theestimates in Table 6.16c. In order to do that, a preliminaryvalve-type selection is required. Assuming the use of a single-seated globe valve with contoured plug for this type of appli-cation and referring to Table 6.15c, xT = 0.72. The ratio ofspecific heats factor, Fk, must be calculated from Equation6.15(83):

Calculate the actual pressure drop ratio, x = ∆p/P1: x =(150 − 65)/150 = 0.57.

Choked flow occurs when x equals or exceeds the valueof FkxT or FkxTP (when the valve is installed with reducers orother fittings). The flow is not choked when x is less thanFkxT or FkxTP,

Note: If the value of x had been greater than FkxT, thenthe value of FkxT would replace the value of x in the calcu-lation of Y and Cv below.

Step 3: Calculate the expansion factor, Y, from Equation6.15(81) or obtain it from Figure 6.15o.

Step 4: Determine the reduced pressure, Pr , and thereduced temperature, Tr using Equations 6.15(84–85). Thecompressibility factor, Z, is found from generalized com-pressibility charts available in reference books such as Perryand Chilton’s Chemical Engineer’s Handbook.11

Step 5: Calculate a preliminary Cv using the equationidentified in Step 1 and assuming a line size valve with FP = 1.

Step 6: Select the approximate body size based on thecalculated preliminary Cv . Manufacturers’ tables are thebest source for Cv values, but estimates can be made basedon Cd(Cd = Cv/d

2) values from Table 6.15c. Observing that

Mwv

A TN

a

=1515

Cw

F PY

T Z

xMvP

=19 3 1

1

.

F kk = = =/ . . / . .1 40 1 331 40 0 95

F xk T = =( . ) ( . ) .0 95 0 72 0 684

Y x F xk T= − = − =1 3 1 57 3 0 684 0 72/( ) /[( )( . )] .

P P P

T T T

r c

r c

= = =

= =

1

1

150 3208 2 0 04

910 1165 5

/ / . .

/ / . ==

=

0 78

1 0

.

.Z

Cv = 15 00019 3 1 150 0 72

910 10 57 1

,. ( ) ( ) ( . )

( ) ( )( . ) ( 88 02

67 7. )

.=



a single-ported, plug-type globe valve has a Cd = 11, d =(67.7/11) = 2.48. Thus, a 3-in. valve is selected for furtherevaluation.

Step 7: Determine the effect of piping geometry. If thevalve had been smaller than line size and installed with reduc-ers, Fp and xTP would need to be known or estimated. Whereappropriate, Fp can be estimated using Equations 6.15(12–22)or Figure 6.15f. The factor, xTP, can be estimated from Equation6.15(82) or Figure 6.15p. Y and Cv values may require recal-culation using the value of FkxTP, if this product is less than x,to replace x in calculating Y and Cv. In this case, the line sizeand the valve size are 3 in. NPS, so this step is not required.

Step 8: Calculate the final required Cv using the actual orestimated FP. In this example, FP = 1, so the required Cv remains67.7. The body size must be verified if the Cv changes at this step.

Note: The Reynolds number factor, FR, is generally notconsidered when sizing for gases and vapors. However, in caseof very small valves, with Cv < 0.01, FR should be applied(when it becomes significant) in the same manner as for liquids.

Step 9: Calculate valve exit velocity to evaluate the like-lihood of velocity-related problems in the valve or in the down-stream piping. Excessive velocity can result in noise, structuralvibration, erosion, equipment damage, and personal injuries.Using Equation 6.15(92), with a flow area A = 7.07 in.2

This is a high Mach number for the application; itsacceptability should be verified with the manufacturer andwith the process system engineer. Valve noise should bepredicted and appropriate noise reduction measures taken tosafeguard against hazards of excessive noise, as discussed inSection 6.14.

TWO-PHASE FLOW

Valve sizing for two-phase flow is approximate at best,because the effects of thermal, mass, and energy transferbetween phases of the mixture are not easily quantified. Themethods commonly used by the control valve industry12,13,14

will be explained here. More elaborate experimental andnumerical methods may produce more accurate results forspecific cases.

Flow through a control valve is considered isenthalpic(adiabatic) overall and isentropic between the valve inlet and

FIG. 6.15o Graph for determining the expansion factor Y.

0.1 1.00.90.80.70.60.50.40.30.2

0.6 0.7 0.8 0.9 1.00.667

Y = 1− X/(3FK) (XTP) Ref. 6.15(81)

Fk0.6

XT1.0 0.8

0.8

0.6

0.4

0.3

0.2

0.91.0

1.2

X = ∆P/P1

XT

Fk

X

Y

FIG. 6.15p Graph for determining the critical pressure drop ratio factor for thecombined assembly of valve and fittings (xTP).

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

00 10 20 30 40

XTP

Cd

1.00.90.80.70.60.50.4

Ki

0.5 0.6 0.7 0.8 0.91.0

FP

FP

Ki

XT

XTP

Cd

XTP = (XT/FP2)/[(XT Ki Cd

2/1000) + 1] Ref. 6.15(82)

0.2

0.4

0.60.8

1.0

XT

MN = =( )( . )/[( )( ) ( . )] ./15000 7 98 1515 910 7 07 0 31 2 77


6.15 Sizing 1257

the vena contracta. It is also assumed for this analysis thatthe velocities of the liquid and gas or vapor are essentiallyequal at the throttling point or at the vena contracta. Thisdiscussion considers two cases of mixed-phase flow: (1) aliquid and a gas of different chemical species, and (2) a liquidand vapor of the same chemical species.

Liquid-Gas Mixtures

The procedure for sizing valves for liquid-gas mixturesapplies to a homogeneous mixture of a nonvaporizing liquidand a noncondensible gas. As the mixture passes through thevalve, the specific volume of the liquid remains essentiallyconstant, but the volume of the gas expands, thereby increas-ing the effective specific volume, ve, of the mixture.

6.15(94)

where fg and ff are the weight or mass fractions of gas andliquid, respectively, and Y is the expansion factor from Equa-tion 6.15(81). Equation 6.15(94) expresses ve as the specificvolume of an equivalent incompressible fluid, allowing useof ve in equations for incompressible fluids.

The required valve capacity, FPCv , assuming FR = 1, canbe obtained from rearranging Equation 6.15(36):

6.15(95)

By substituting 1/ve for γ1 and generalizing the equationfor both choked and non-choked flow using ∆pa as the smallerof P1 – P2 and ∆pch (see Equation 6.15[11]), the valve capacityis

6.15(96)

These equations have provided the required Cv valueswithin an error of about ±10% based on limited testing onair and water mixtures in small globe valves in horizontalpipe.15

Example 13 The goal in this example is to find therequired valve capacity (FpCv) for the conditions listed below:

Air flow rate: 600 lb/hrWater flow rate: 26,000 lb/hrUpstream pressure, P1: 150 psiaPressure drop, ∆p: 50 psiTemperature: 90°F (550°R)Line size: 3 in. schedule 40

Step 1: Make a preliminary selection of valve type anddetermine the critical pressure drop ratio factor (xT) for thevalve. Assume a single-seated globe valve with a contouredplug with flow under the plug (to open). Using Table 6.15b,obtain an estimate of xT = 0.72. (Manufacturers’ catalogs mayalso be used.)

TABLE 6.15q Physical Data for a Number of Common Gases and Vapors

GasCritical Pressure,

Pc (psia)Critical Temperature,

Tc (°R) Molecular Weight, MRatio of Specific

Heats, k

Air 492.4 227.1 28.97 1.40

Ammonia 1636.1 729.8 17.0 1.31

Argon 707.0 271.4 39.9 1.67

Butane 551.0 765.2 58.1 1.09

Carbon dioxide 1070.2 547.5 44.0 1.29

Carbon monoxide 507.1 239.2 28.01 1.40

Chlorine 1120.0 751.0 70.91 1.34

Ethane 708.5 549.7 30.0 1.18

Ethylene 730.5 508.3 28.05 1.21

Helium 33.2 9.45 4.00 1.67

Hydrogen 188.1 59.9 2.02 1.40

Methane 673.0 343.9 16.04 1.32

Natural gas 673.0 343.9 16.04 1.32

Nitrogen 492.4 227.1 28.02 1.40

Oxygen 736.0 278.6 32.0 1.40

Propane 615.9 665.9 44.1 1.12

Steam 3208.2 1165.5 18.02 1.33

v f v Y f ve g g f f= +/ 2

F Cw

N P PP v =

−6 1 2 1( )γ

F Cw

N

v

pp ve

a

=6 ∆



Step 2: Determine the relative mass fractions of gas andliquid, fg and ff . The total mass flow rate is w = 600 + 26,000= 26,600 lb/hr.

Step 3: Calculate the pressure drop ratio, (x), the ratio ofspecific heat factor (FK), and the expansion factor (Y) ⋅x =∆p/P1 = 50/150 = 0.333. Because x < xT, and the gas flow isnot choked, ∆pa = ∆p = 50 psi.

From Equation 6.15(83), using k = 1.40 for air, Fk =1.40/1.40 = 1.0. Using Equation 6.15(81):

Step 4: Determine the effective specific volume of themixture at upstream conditions using Equation 6.15(94). Thespecific volume of the air can be calculated from the gas lawequation of state:

6.15(97)

From saturated steam temperature tables at 90°F, thesaturated liquid specific volume is vf = 0.016099 ft3/lbm. Themixture effective specific volume from Equation 6.15(94) isve = (0.0226)(1.357)/(0.846)2 + (0.9774)(0.016099) = 0.0586ft3/lbm.

Step 5: Calculate valve capacity from Equation 6.15(96).

Liquid-Vapor Mixtures

Liquid-vapor mixtures are a more difficult, if not impossible,problem owing to the difficulty in determining the inlet vaporfraction and because of the nonequilibrium conditions pre-vailing as the fluid passes through the valve.

Tests have shown that the flow rate of low-quality (vaporfraction X < 0.03) steam-water mixtures can be as much asten times the amount predicted using calculations based onthermodynamic equilibrium, and about twice the flow pre-dicted assuming no change of state between valve inlet andvena contracta.15

Calculations produced better predictions for very high-quality (vapor fraction X > 0.9) steam-water mixtures. Forhigh vapor fraction mixtures, the same procedure is used as

in Equations 6.15(94–96). For low vapor mixtures, whilerecognizing the possibility of large errors, use the effectivespecific volume at inlet conditions, ve, and Equation 6.15(98).

6.15(98)

FLP is the combined piping and pressure recovery factorfrom Equation 6.15(23).

Example 14 Let us assume that the following conditionsare given for a liquid-vapor mixture in near-equilibrium con-ditions. Find the required valve capacity, FLPCv.

Fluid: AmmoniaVapor flow rate: 200 lb/hrLiquid flow rate: 9000 lb/hrUpstream pressure, P1: 110 psiaPressure drop, ∆p: 5 psiTemperature: 60°FLine size: 2 in. schedule 40

Step 1: Make a preliminary selection of valve type, anddetermine the valve xT. Assume a throttling (segmented) ballvalve. Using Table 6.15c for an estimate, xT = 0.25.

Step 2: Determine the relative mass fraction of gas andliquid. The total mass flow rate is w = 200 + 9000 = 9200lb/hr.

Step 3: Calculate pressure drop ratio (x), ratio of specificheat factor (FK), and expansion factor (Y)⋅x = ∆p/P1 = 5/110= 0.0455. Since x < xT , the vapor flow is not choked. FromEquation 6.15(83) and Table 4.17q, using k = 1.40 for air and1.31 for ammonia, Fk = 1.31/1.40 = 0.936. Using Equation6.15(81):

Step 4: Calculate FF from Equation 6.15(10). The criticalpressure of ammonia from Table 6.15q is Pc = 1636.1 psia.The vapor pressure from ammonia tables11 at 60°F is Pv =107.6 psia.

Step 5: Determine the effective upstream specific volumeat upstream conditions using Equation 6.15(94). The specificvolume of the ammonia vapor taken from thermodynamic

f

f

g

f

= =

= =

600 26 600 0 0226

26 000 26 600 0 9774

/ , .

, / , .

Yx

F xk T

= − = − =13

1 0 333 3 1 0 72 0 846. /[( )( )( . )] .

vRT

MPg = 1

1

vg =( ) °

(⋅

⋅ ⋅

⋅

1545 550

29

ft lbflb mol °R

lbmlb mol

R( )

)) ( ) ( ) =150 144

1 357lbfin.

in.ft

3

2

2

2

ftlbm

.

F Cw v

pP ve

a

= = =63 3

26 60063 3

0 058650

14 4.

,.

..

∆

Cw

N F

v

P FvLP

e

F

=−6 1 1( )

f

f

g

f

= =

= =

200 900 0 0217

9000 9200 0 9783

/ .

/ .

Yx

F xk T

= − = − =13

1 0 0455 3 0 936 0 25 0 93. /[( )( . )( . )] . 55

F P PF v c= −

= −

0 96 0 28

0 96 0 28 107 6 1

1 2. . ( / )

. . ( . / 6636 1 0 891 2. ) .=


6.15 Sizing 1259

tables11 is only approximate, because the vapor is not in truethermodynamic equilibrium. However, using the values forsaturated ammonia provides sufficient accuracy due to therelative uncertainty of the capacity estimate. From saturatedammonia tables,11 vg = 2.751 ft3/lb and vf = 0.02597 ft3/lb.Therefore, one can calculate ve from Equation 6.15(94).

Step 6: Calculate valve capacity from Equation 6.15(98).

CONCLUSIONS

Due to the uncertainties in correctly determining the processconditions, there is a tendency to be overcautious; i.e., tooverestimate pressure losses in pipes and to increase thenormal flow rates by safety factors. This tends to oversizethe control valves. This in turn affects the installed gain ofthe valve and greatly reduces the turn-down, or rangeabilityof the valve.

It is recommended that the pump heads be selected onthe basis that the control valve will add no more than 10%to the pressure drop in the piping system at maximum flow.This will not only reduce the size of the pump, but will alsosave on the required operating energy. If this approach istaken to determine the required pump head, the maximumdifferential pressure across the control valve will be the pumphead at maximum flow minus the sum of the pressure dropacross the piping system and any altitude changes, plus 10%of the pipe friction losses.

There are instances when the flow conditions are notaccurately known. This might be the case if an old valve isto be replaced in a piping system and its serial number orsource is not known. In such cases, the “best guess” approachwould be to select the same type of valve having the samepressure rating as the surrounding pipe and having a flowcoefficient (Cv) one size smaller than that listed for a full sizeorifice of a line size valve.

Another important rule is to make sure that the maximumrequired flow capacity for the valve does not exceed 85% ofthe valve’s maximum rated capacity. This is because there isa typical tolerance of ±10% between what is advertised asthe maximum Cv number and what actually is the Cv of theparticular valve.

Similar rules apply to the minimum required flow coef-ficient, which is obtained by considering a process conditionwhen the flow is the minimum and the valve pressure dropis the maximum. Once this Cv(min) is determined, one has tomake sure that the valve travel at the required minimum flow

coefficient is not less than 10% of the rated travel (or twicethe expected dead band of the valve, whatever is higher) awayfrom touching the valve seat. This safety margin is necessaryto make sure that the plug does not touch the seat during anyexcursion caused by a disturbance in the control loop.

If the minimum flow capacity of the larger valve is insuf-ficient to provide the required rangeability for the loop, usetwo valves in parallel by split-ranging the control signalamong them, as shown in Figure 6.1i.

NOMENCLATURE

ve =

+

( . )( . )/( . )

( . )( .

0 0217 2 751 0 935

0 9783 0 0259

2

77 0 0937) .= ft /lb3

F Cw v

P FLP ve

F

=−

=−63 3 1

920063 3

0 0937110 1 01. ( ) .

.( .. )

.89

12 8= A Area

A1 Cross-sectional area of pipe, in.2

A2 Area of valve opening, in.2

a Pressure scale exponent

Av Metric capacity coefficient, m2

b Size scale exponent

Cd Relative valve capacity or valve discharge coefficients,Cd = Cv/d2 (Table 6.15c)

Cv Valve capacity coefficient. Its value is 1.0 when the valve passes 1 gpm of cold water, having a specific gravity of 1.0, at a pressure drop of 1 psid.

Cvt Pseudo valve coefficient

d Valve body port diameter, inches

d/D Valve to pipe diameter ratio

D1 Upstream pipe inside diameter, inches

D2 Downstream pipe inside diameter, inches

Fd Valve-style modifier

FF Liquid critical pressure ratio factor

Ff Mass fraction of liquid in a liquid-gas mixture

Fg Mass fraction of gas in a liquid-gas mixture

Fk Ratio of specific heats factor; k/1.4

FL Liquid pressure recovery factor of a valve without attached fittings

FLP Combined liquid pressure recovery factor and piping geometry factor of a valve with attached fittings

FP Piping geometry factor of a valve with attached fittings

FR Reynolds number factor

Fs Laminar (streamline) flow factor

g Gravitational acceleration, 32.174 ft/s2

gc Units conversion constant, 32.174 (lbm-ft)(lbf-s2) or 1.0 (kg-m)/(N-s2)

Gf Specific gravity relative to water at 60°F

Gg Specific gravity relative to air 1 atm. and 60°F

H Enthalpy

K Ratio of specific heats for gases; 1.4 for air

KB1 Inlet fitting Bernoulli coefficient (static pressure to velocity exchange)

KB2 Outlet fitting Bernoulli coefficient



References

1. Baumann, H. D., “The Introduction of a Critical Flow Factor for ValveSizing,” ISA Transactions, Vol. 2, 1963.

2. Driskell, L. R., Control Valve Selection and Sizing, Research TrianglePark, NC: Instrumentation, Systems, and Automation Society, 1983.

3. Baumann, H. D., “Effect of Pipe Reducers on Control Valve Capacity,”Instruments and Control Systems, December 1968.

4. McCutcheon, E. B., “A Reynolds Number for Control Valves,” Sym-posium on Flow, Its Measurement and Control in Science and Indus-try, Vol. 1, Part 3, 1974.

5. Baumann, H. D., “A Unifying Method for Sizing Throttling Valvesunder Laminar or Transitional Flow Conditions,” Journal of FluidsEngineering, Vol. 115, No. 1., pp. 166–168, March 1993.

6. Allen, W. F., Jr., “Flow of a Flashing Mixture of Water and Steamthrough Pipes and Valves,” Journal of Basic Engineering, April 1951.

7. Singleton, E. W., “Multistage Valve Sizing Made Easy,” InTech,August 1997, pp. 47–51.

8. George, J. A., “Evolution and Status of Nonturbulent Flow Sizing forControl Valves,” Research Triangle Park, NC: Instrumentation, Sys-tems, and Automation Society, 2002.

9. ANSI/ISA S75.01, “Flow Equations for Sizing Control Valves,”Research Triangle Park, NC: Instrumentation, Systems, and Automa-tion Society, Revised 1995.

10. ANSI/ISA S75.02-1988, “Control Valve Capacity Test Procedure,”Research Triangle Park, NC: Instrumentation, Systems, and Automa-tion Society, 1996.

11. Perry, R. H. and Chilton, C. H., Chemical Engineer’s Handbook, 5thedition, New York: McGraw-Hill Book Company.

12. Masoneilan Handbook for Control Valve Sizing, Masoneilan-Dresser,2000.

13. Catalog 10, Technical Information, Fisher Controls Company.14. Monsen, J. F., “Spreadsheet Sizes Control Valves for Liquid/Gas

Mixtures,” InTech, December 1990.15. Hutchison, J. W. (ed.), ISA Handbook of Control Valves, 2nd edition,

Research Triangle Park, NC: Instrumentation, Systems, and AutomationSociety, 1976.

Kc Valve cavitation coefficient

KI Inlet fitting combined loss coefficient

Ksc Cavitation index of service conditions

Kv Metric valve sizing coefficient, (m3/h)/(bar)1/2

K1 Inlet fitting friction loss coefficient

K2 Outlet fitting friction loss coefficient

m Mass, lbm

M Molecular weight

MN Mach number

N1 to N9 Numerical constants

P Pressure, psia

P1 Valve inlet pressure, psia

P2 Valve outlet pressure, psia

Pc Critical pressure, psia

PSE Pressure scale effect for cavitation

Pvc Pressure at the vena contracta, psia

Pv or Pvp Vapor pressure at flowing conditions, psia

Pr Reduced pressure (used with compressibility curves); defined as P1/Pc

q Volumetric flow rate, gpm

Q Volumetric flow rate, ft3/hr

R Universal gas constant, 1545 (ft-lbf)/(lb-mol-°R) or 1.987 cal/(gm-mol-°K)

Re Pipeline Reynolds number

Rev Valve Reynolds number

SSE Size scale effect for cavitation

T Temperature, °F or °C

Ta Absolute temperature, °R

Tc Critical temperature, °R or °K

Tr Reduced temperature; defined as T1/Tc

T1 Absolute temperature at valve inlet, °R or °K

U Velocity, ft/s

v Specific volume, ft3/lbm

V Volume, ft3

ve Effective specific volume of a liquid-gas mixture, ft3/lbm

vf Specific volume of saturated liquid, ft3/lbm

vg Specific volume of gas or saturated vapor, ft3/lbm

W Mass flow, rate, lbm/hr

X Vapor portion of a liquid-vapor mixture, related to steam quality

XFzi Differential pressure ratio for incipient cavitation at actual inlet pressure.

XT Critical pressure drop ratio factor (inception of choked flows)

XTP Critical pressure drop ratio factor for a valve and fitting assembly

Y Expansion factor for gases

Z Compressibility factor

∆H Head loss, ft.

∆p Pressure difference (drop), psi

∆pa Allowable sizing pressure drop (the smaller of P1 − P2 and ∆pch), psi

∆pch Pressure drop at onset of choked flow defined in terms of FL, psi

∆pI Pressure drop at flow curve inflection point where Cv begins to decrease due to cavitation

γ (gamma) Weight density, lbm/ft3

γ1 Specific weight at upstream conditions, lbm/ft3

µ (mu) Absolute viscosity, centipoise = 10−3 N·sec/m2

ν (nu) Kinematic viscosity, centistokes = 10− 6 m2/sec

σ (sigma) Cavitation index at flowing conditions

σc Valve coefficient for constant cavitation

σI Valve coefficient for incipient cavitation

σid Valve coefficient for incipient damage cavitation

σmv Valve coefficient at maximum vibration cavitation

σR Recommended operating cavitation coefficient for a valve at a reference pressure

σv Valve cavitation coefficient adjusted for pressure and size scale effects


6.15 Sizing 1261

16. Kiesbauer, J. and Baumann, H. D., “Recent developments in theprediction of hydrodynamic noise of control valves,” Valve-World,February 2004.

17. Baumann, H. D., “Determination of Peak Internal Sound FrequencyGenerated by Throttling Valves for the Calculation of Pipe Transmis-sion Losses,” Noise Control Engineering Journal, Vol. 22, No. 1, pp.75–82, March–April, 1991.

18. Cain, F. M. and Barnes, R. W., “Testing for Cavitation in Low PressureRecovery Control Valves,” ISA, Volume 25, No. 2, 1986.

19. Tullis, J. P., Hydraulics of Pipelines: Pumps, Valves, Cavitation, Tran-sients, New York: John Wiley, 1989.

20. Rahmeyer, W., “Cavitation Testing of Control Valves,” Instrumenta-tion, Systems, and Automation Society, Paper No. C.I. 83-R931, pre-sented at International Conference in Houston, TX, October 1983.

21. Riveland, M. L., “The Industrial Detection and Evaluation of Con-trol Valve Cavitation,” Instrumentation, Systems, and AutomationSociety, Paper No. C.I. 82-909, presented at International Confer-ence in Philadelphia, PA, October 1982.

22. Cain, F. M., “Solving the Problem of Cavitation in Control Valves,”Advances in Instrumentation and Control—Proceedings of theISA/91 International Conference, Vol. 46, Part 2, 1991, Paper #91-0462, pp. 1133–49.

Bibliography

Baumann, H. D., Control Valve Primer, 3rd edition, Research Triangle Park,NC: Instrumentation, Systems, and Automation Society, 1998.

Cain, F. M., “Solving the Problem of Cavitation in Control Valves,”Advances in Instrumentation and Control—Proceedings of the ISA/91International Conference, Vol. 46, Part 2, 1991, Paper #91-0462, pp.1133–49.

Catalog 10, Technical Information, Fisher Controls Company.ANSI/ISA S75.02-1988, “Control Valve Capacity Test Procedure,” Research

Triangle Park, NC: Instrumentation, Systems, and Automation Society,1996.

Driskell, L., Control-Valve Selection and Sizing, Research Triangle Park,NC: Instrumentation, Systems, and Automation Society, Revised 1995.

ANSI/ISA S75.01, “Flow Equations for Sizing Control Valves,” ResearchTriangle Park, NC: Instrumentation, Systems, and Automation Society,1985.

Handbook on Control Valve Sizing, Baumann Inc., Portsmouth, NH.Hutchison, J. W. (ed.), ISA Handbook of Control Valves, 2nd edition,

Research Triangle Park, NC: Instrumentation, Systems, and AutomationSociety, 1976.

ISA S20 Specification Forms for Process Measurement and Control Instru-ment, Primary Elements and Control Valves, Research Triangle Park,NC: Instrumentation, Systems, and Automation Society, 1981.

IEC Standard 60534-2-1, “Flow Capacity: Sizing Equations for Fluid Flowunder Installed Cconditions,” Geneva, Switzerland: International Elec-tronical Commission, September 1998.

Masoneilan Handbook for Control Valve Sizing, 7th edition, Masoneilan-Dresser, 1987.

Monsen, J. F., “Spreadsheet Sizes Control Valves for Liquid/Gas Mixtures,”InTech, December 1990.

Morse, P. M. Thermal Physics, W.A. Benjamin, Inc., 1964.Perry, R. H. and Chilton, C. H., Chemical Engineer’s Handbook, 5th edition,

New York: McGraw-Hill Book Company, 1973.Rahmeyer, W., “Cavitation Testing of Control Valves,” Instrumentation, Sys-

tems, and Automation Society, Paper No. C.I. 83-R931, presented atInternational Conference in Houston, TX, October 1983.

Rahmeyer, W. and Driskell, L., “Control Valve Flow Coefficients,” Journalof Transportation Engineering, Vol. 111, No. 4, ASCE Pipeline Divi-sion Paper No. 19882, 1985.

Riveland, M. L., “The Industrial Detection and Evaluation of Control ValveCavitation,” Instrumentation, Systems, and Automation Society, PaperNo. C.I. 82-909, presented at International Conference in Philadelphia,PA, October 1982.

Shinskey, F. G., “When You Have the Wrong Valve Characteristic,” Instru-ments and Control Systems, October 1971.

Technical paper No. 410, Crane Company, 2001.Tullis, J. P., Hydraulics of Pipelines: Pumps, Valves, Cavitation, Transients,

New York: John Wiley, 1989.Valtek Control Valve Sizing and Selection, Flowserve Inc. (Valtek Inc.), 1992.


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Process Control and Optimization, VOLUME II - Freetwanclik.free.fr/electricity/IEPOPDF/1081ch6_15.pdf · The control valve is an important component of the total process control loop,