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'J Sizing & Selection Control Valve Sizing CONTENTS Introduction : 3-1 Nomenclature 3-1 Calculating Cvfor Liquids 3-3 Liquid Sizing Examples .,"" 3-7 CalculatingCv for Gases. 3-10 Gas SizingExamples , , 3-13 Calculating Cvfor Two Phase Flbw 3-15 Vapor Pressure Charts 3A-1 Steam Tables 3A-7 Fluid Property Correlation Constants ...3A-21 Pipa Data 3A-29 INTRODUCTION '--" Valtek uses a systematic method for selecting body types, sizes, materials, pressure ratings and trim sizes basad on flow characteristics. Valtek control valva flow capacity (C) is basad upon the industry standard, ANSI/ISA S75.01. This standard and the corresponding measuring standards contain Equations usad to predict the flow of compressible and incompressible fluids in control valves. Slightly differ- ent forms of the basic Equation are usad for liquids and gases. Basic steps for sizing and selecting the correGí valva include calculating the required Cv' Equations for calcu- lating Cv for both gases and liquids are found in ibis section. Valtek has programmed the ANSI/ISA sizing Equations and procedures, making computer-aided sizing avail- able on 18M-PC or compatible computers. These programs permit rapid control valva flow capacity calcu- lations and valva selection with minimal effort. The programs also include exit velocity, noise prediction and actuator sizing calculations. Sea Section 22 for more details on computer-aided valva selection. These instructions are designad to expose the usar to the different aspects of valva sizing. The step-by-step method outlined in ibis section is the most common method of sizing. , ". NOMENCLATURE Flow Capacity The valva sizing coefficient most commonly usad as a measure of the capacity of the body and trim of a control valva is Cv' One C)s defined as one U.S. gallon par minuta of 60 degree Fahrenheit waterthat flows through Rev. 6/94 a valva with a one psi pressuredrop.' Thegeneral Equationfor Cvis as follows: . specific gravity at flowing temperatura pressure drop When selecting a control valva for an application, the calculated C is usad to determine the valva size and the v trim size that will allow the valva to pass the desired flow rata and provide stable control of the process fluido Pressure Profile C = flow v Fluid flowing through a control valva obeys the basic laws of conservation of mass and energy, anct ;i/:1é continuity Equation. The control valva acts as a resure- tion in the flow stream. As the fluid stream approaches ibis restriction, its velocity increases in arder for the full fJow to pass through the restriction. Energy for ibis increase in velocity comes from a corresponding decrease in pressure. Maximum velocity and mínimum pressure occur imme- diately downstream from the throttling point at the narrowest constriction of the fluid stream, known as the vena contracta. Downstream from the vena contracta, the fluid slows and part of the energy (in the form of velocity) is converted back to pressure. A simplified prefije of the fluid pressure is shown in Figure 3-1. The slight pressure losses in the inlet and outlet passages are due to frictional effects. The majar excursions of pressure are due to the velocity changas in the region of the vena contracta. P1 (Valve Pressure Drop) --- Pv (VaporPressure) PVC (Pressureat VenaContracta) Figure 3-1: Pressure Profile of Fluid Passing Through a Valve 3-1

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calculo de válvulas de control Valtek

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' JSizing & Selection

Control Valve SizingCONTENTS

Introduction : 3-1Nomenclature 3-1

Calculating Cvfor Liquids 3-3Liquid Sizing Examples .,"" 3-7CalculatingCv for Gases. 3-10GasSizingExamples , , 3-13Calculating Cvfor Two Phase Flbw 3-15Vapor Pressure Charts 3A-1Steam Tables 3A-7Fluid Property Correlation Constants ...3A-21Pipa Data 3A-29

INTRODUCTION

'--"

Valtek uses a systematic method for selecting bodytypes, sizes, materials, pressure ratings and trim sizesbasad on flow characteristics.

Valtek control valva flow capacity (C) is basad upon theindustry standard, ANSI/ISA S75.01. This standardand the corresponding measuring standards containEquations usad to predict the flow of compressible andincompressible fluids in control valves. Slightly differ-ent forms of the basic Equation are usad for liquids andgases.

Basic steps for sizing and selecting the correGí valvainclude calculating the required Cv' Equations for calcu-lating Cvfor both gases and liquids are found in ibissection.

Valtek has programmed the ANSI/ISA sizing Equationsand procedures, making computer-aided sizing avail-able on 18M-PC or compatible computers. Theseprograms permit rapidcontrol valva flow capacity calcu-lations and valva selection with minimal effort. Theprograms also include exit velocity, noise predictionand actuator sizing calculations. Sea Section 22 formore details on computer-aided valva selection.

These instructions are designad to expose the usar tothe different aspects of valva sizing. The step-by-stepmethod outlined in ibis section is the most commonmethod of sizing.

, ".

NOMENCLATURE

Flow Capacity

The valva sizing coefficient most commonly usad as ameasure of the capacity of the body and trim of a controlvalva is Cv' One C)s defined as one U.S. gallon parminuta of60 degree Fahrenheit waterthat flows through

Rev. 6/94

a valva with a one psi pressuredrop.' ThegeneralEquationfor Cvis as follows: .

specific gravity at flowing temperatura

pressure drop

When selecting a control valva for an application, thecalculated C is usad to determine the valva size and thev

trim size that will allow the valva to pass the desired flowrata and provide stable control of the process fluido

Pressure Profile

C = flowv

Fluid flowing through a control valva obeys the basiclaws of conservation of mass and energy, anct ;i/:1écontinuity Equation. The control valva acts as a resure-tion in the flow stream. As the fluid stream approachesibis restriction, its velocity increases in arder for thefull fJow to pass through the restriction. Energy foribis increase in velocity comes from a correspondingdecrease in pressure.

Maximum velocity and mínimum pressure occur imme-diately downstream from the throttling point at thenarrowest constriction of the fluid stream, known as thevena contracta. Downstream from the vena contracta,the fluid slows and part of the energy (in the form ofvelocity) is converted back to pressure. A simplifiedprefije of the fluid pressure is shown in Figure 3-1. Theslight pressure losses in the inlet and outlet passagesare due to frictional effects. The majar excursions ofpressure are due to the velocity changas in the regionof the vena contracta.

P1 (ValvePressureDrop)

- - - Pv (VaporPressure)

PVC (Pressureat VenaContracta)

Figure 3-1: Pressure Profile of FluidPassing Through a Valve

3-1

,,',

F " C- ~L

~""

".

'

.

' v-q I'>P/G,- -' f,

-- '--" - --

)1q max

I'>PCh~j I'>P/Gf

Figure 3-2: Choked Pressure Drop

Allowable Pressure DropThe capacity curva shown in Figure 3-2 shows that, withconstant upstream pressure, flow rata, q, is relatad tothe square root of pressure drop through the proportion-ality constant Cv' The curva departs from a linearrelationship at the onset of "choking" described usingthe F¡factor. Th~flow rata reaches a maximum, qmax'atthe fully choked condition due to effects of cavitation forliquids or sonic velocity for compressible fluids. Thetransition to choked flow may be gradual or abrupt,depending on valva designo ANSI/ISA liquid sizingEquations use a pressure recovery factor, FL'to calcu-late the L1PChat which choked flow is assumed for sizingpurposes. For compressible fluids, a terminal pressuredrop ralla, xT' similarly describes the choked pressuredrop for a specific valva.When sizing a control valva, the smaller of the actualpressure drop or the choked pressure drop is alwaysusad to determine the correGíCv. This pressure drop isknown as the allowable pressure drop, L1Pa'

Cavitation

In liquids, when the pressure anywhere in the liquiddrops below the vapor pressure of the fluid, vaporbubblesbégin tóform in the fluid stream. As the fluid

,decelerafestpere is a resultant increase in pressure. Ifthis pressÜreis higher than the vapor pressure, thebubblescollapse (or implode) as the vapor returns to the

, liqUid phase. This two-stepmechanism - callad:cavifation - procjuces noise, vibration, and causes

"'erosion damage to the valva and downstream piping.

, .,Theonset of cavitation - known as rncipient cavitation<"~ is the point when the bubblesfirst begin to form and

,;. ,coJlapse. Advanced cavitation can affect capacity and, ,°

.' "'valve performance, which begins at a L1Pdeterminadfrom the factor, F¡. The point at which full or chokedcavitation occurs (severe damage, vibration, and noise)can be determinad from Equation 3.3. Under chokedconditions, "allowable pressure ,prop," is the chokedpressure drop. .

3-2.' °,..

"; '(".

Liquid Pressure Recovery Factor, FLThe liquid pressure recovery factor, FL' predictstheamount of pressurerecoverythat will occur betweenthe vena contracta and the valva outlel. FL is anexperimentally determinad coefficient that accounts forthe influence of the valve's infernal geometry on themaximum capacity of the valva. It is determinad fromcapacity test data like that shown in Figure 3-2.

FLalso varias according to the valva type. High recov-ery valves - such as butterfly and ball valves - havesignificantly lower pressures atthe vena contracta andhence recover much farther forthe sama pressure dropthan a gleba valva. Thus theytend to choke (orcavitate)at smaller pressure drops than gleba valves.

~

Liquid Critical Pressure Ratio Factor, FF

The liquid critical pressure rallo factor, FF'multiplied bythe vapor pressure, predicts the theoretical vena con-tracta pressure at the maximum effective (choked)pressure drop across the valva.

Flashing

If the downstream pressure is equal to or less than thevapor pressure, the vapor bubbles created at the venacontracta do not collapse, resulting in a liquid-gasmixtura downstream of the valva. This is commonlycallad flashjng. When flashing of a liquid occurs, theinlet fluidis 100 percent liquid which experiences pres-sures in and downstream of the control valve which areat or below vapor pressure. The result is a two phasemixtura (vaporand liquid) at the valva outlet and in thedownstream piping. Velocity of this two phase flow isusually very high and results in the possibility for erosionof the valva and piping components.

....

Choked Flow

Choked flow occurs in gases and vapors when the fluidvelocity reaches sonic values at any point in the valvabody, trim, or pipa. As the pressure in the valva or pipais lowered, the specific volume increases to the pointwhere sonic velocity is reached. In liquids, vaporformad as the result of cavitation or flashing increasesthe specific volume of the fluid at a tastar rata than theincrease in flow due to pressure differential. Loweringthe downstream pressure beyond this point in eithercase will not increase the flow rata for a constantupstream pressure. The velocity at any point in thevalva or downstream piping is limitad to sonic(Mach = 1). As a result, the flow rata will be limitad toan amount which yields a sonic velocity in the valva trimor the pipa under the specified pressure conditions.

'--

11~

Reynolds Number Factor, FR

The Reynolds Number Factor, FR'is usad to correGíthecalculated C for non-turbulent flow conditions due tov

high viscosity fluids, very low velocities, or very smallvalva C .v

~.~

I

Piping Geometry Factor, Fp

Valva sizing coefficients are determinad from tests runwith the valva mounted in a straight run of pipa whiéh isthe sama diameter as the valva body. If the processpiping configurations are different from the standard testmanifold, the apparent valva capacity is changed. Theeffect of reducers and increasers can be approximatedbythe use of the piping geometry factor, FP'

Velocity

As a general ruja, valva outlet velocities should belimitad to the following maximum values:

.-."The above figures are guidelines for typical applica-tions. In general,smaller sized valves handle slightlyhigher velocities and larga valves handle lower veloci-tieso Special applications have particular velocity re-quirements; a few of which are provided below.

Liquid applications - where the fluid temperatura isclase to the saturation point - should be limitad to 30feet par second to avoid reducing the fluid pressurebelow the vapor pressure. This is also an appropriatelimit for applications designad to pass the full flow ratawith a mínimum pressure drop across the valva.

Valves in cavitating service should also be limitad to 30feet par second to minimiza damage to the downstreampiping. This will also localiza the pressure recoverywhich causes cavitation immediately downstream fromthe vena contracta.

In flashing services, velocities become much higherdueto the increase in volume resulting from vapor forma-tion. For most applications, it is important to keepvelocities below 500 feet par secando Expanded outletstyle valves - such as the Mark One-X - help to controloutlet velocities on such applications. Erosion damagecan be limitad by using chrome-moly body material andhardened trim. On smaller valva applications whichremain closed for mostofthe time - such as heater drainvalves - higher velocities of 800 to 1500 feet par secondmay be acceptable with appropriate mataríais.

'-'

Gas applications where special noise attenuation trimare usad should be limitad to approximately 0.33.Mach.In adclition, pipa velocities downstream fromthe valvaare critical to the overall noise level. Experimentationhas shown that velocities around 0.5 Mach can createsubstantial noise even in a straight pipa. The additionof a control valva to the line will increase the turbulencedownstream, resulting in even higher noise levels.

Expansion Factor, Y

The expansion factor, Y, accounts for the variation ofspecific weight as the gas passes from the valva inlet tothe vena contracta. It also accounts for the c~ange incross-sectional area of the vena contracta as the pres-sure drop is variad.

Ratio of Specific Heats Factor, Fk

The ratio of specific heats factor, Fk,adjusts the Equa-tion to account for different behavior of gasesother thanair.

Terminal Pressure Drop Rallo, XT

The terminal pressure drop ratio for gases, xT' isused topredict the choking point where additional pressuredrop (by lowering the downstream pressure) will notproduce additional flow due to the sonic velocity limita-tion across the vena contracta. This factor is a function

of the valva geometry and varias similarly to FL' depend-ing on the valva type.

Compressibility Factor, Z

The compressibility factor, Z, is a function of the tem-perature and the pressure of a gas. It is usad todetermine the density of a gas in relationship to itsactual temperatureand pressure conditions.

CALCULA TING Cv FO.RLlQUIDSIntroduction

The Equation for theflow coefficient(C)in non"laminar .

liquid flow is:

q~Cv= F;J~ (3.1)

Where: Cv = Valv~ sizing coefficient

Fp = Piping geometry factor

q = Flow rata, gpm

LiPa = Allowable pressure drop acrossthe valva for sizing, psi

Gj = Sp~cific gravity at flowingtemperatura

>

3-3

Liquids 50 feet par second

Gases Approaching Mach 1.0

Mixed Gases 500 feet persecondand Liquids

Table 3-1:Typical Valve Recovery Coefficient and Incipient Cavitation FactorsNOTE: Va/ues are given for full-open va/ves: See charts be/ow for part-stroke va/ues

* Typica/ ** ns = number of stages

1.00

0.90

0.80

FL 0.70

0.60

0.50

0.40O 20 40 60 80

Percent of Rated Cv

Globe Valve 'FLValues

1.00

0.90

0.80

Fi 0.70

0.60

0.50

1000.40

O 20

Percent of Rated Cv

6040 80

Globe Valve F.ValuesI

~

1.00

~

0.90

0.80

FL 0.70

0.60

0.50

1000.40

O 20 80 10040 60

Percent Open

Rotary Disc Valve FL Values

The following steps shoul~ be used to compute thecorrect Cv' body size andlrim number:

Step1: Calculate Actual Pressure Drop

The allowable pressure drop, ~p , acrossthe valve fora

calculating Cv' is the smaller of the actual ~p from

Equation 3,2and choked~P chfrom Equation 3.3.

~P:;:: P1 -1?2 . (3.2)

Step2:Check for Choked Flow, Cavitationand Flashing

Use Equation 3.3 to check for choked flow:

áPCh :;:: FL2 (P1 - FFPV) ,

3A

(3.3)

Where: FL :;:: LiquidpressurerecoveryfactorFF :;:: Liquid critical pressure ratio factorPv :;:: Vaporpressureof the liquidat inlet

temperature, psiaP1 :;:: Upstreampressure,psia

See Table 3-1for FLfactors for both full-open and part-stroke values.

FF can be estimated by the following relationship:

~vFF:;:: 0.96 - 0.28 -

Pe

(3.4 )

Where: FF :;:: Liquid critical pressure ratioPv :;:: Vapor pressure of the liquid, psiaPe :;:: Critical pressure of the liquid, psia

(see Table 3-11)

<-

~

ValveType Flow Direction Trim Size FL F. XT FdI

Globe'.'.

Over Seat Full Area 0.85 0.75 .70 1.0Over Seat Reduced Area 0.80 0.72 .70 1.0Under Seat Full Area 0.90 0.81 .75 1.0Under Seat Reduced Area 0.90 0.81 .75 1.0

Yaldisk 60° Open Full 0.76 0.65 .36 .71

Rotary Disc 90° Open Full 0.56 0.49 .26 .71

ShearStream 60° Open Full 0.78 0.65 .51 1.0

Rotary Ball 90° Open.' Full 0.66 0.44 .30 1.0

CavControl Over Seat AII 0.92 0.90 N/A - .2/Jd

MegaStream Under Seat AII -1.0 N/A -1.0 (ni25df/3**ChannelStream Over Seat AII -1.0 0.87 to 0.999 N/A .040*

Tiger-Tooth Under Seat AII -1.0 0.84 to 0.999 -1.0 .035*

..- F. Flowto Open

J1

I

I F Flowto CloseI

i-'"

I1

... Fi Flowto Ope.!!.-

J11"

J F. Flow to Close-IJ

""

Valdisk*...-

I '" ,1" ...\

J ,ShearStream*

\."i

If ~Pch (Equation 3.3) is less than the actual ~P

(Equation 3.2) , use ~P ch for ~P a in Equation3.1..

~

O. 10 .20.30.40.50.60.70.80.901.00

P v =Vapor PressureP = Critical Pressuree

Figure 3-3: Liquid Critical Pressure RatioFactor Curve

Table 3-11:Critical Pressures

(.,

e

It may also be useful to determine the point at whichsubstantial cavitation begins. The following Equationdefines the pressure drop at which substantial cavita-tion begins:

~P (cavitation) = F¡2 (P1 - Py) (3.5)

In high pressure applications, alternate analysis may berequired; verify analysis with factory if~P ~ ~P (cavita-tion) ~ 300 psi (globe valves) or 100 psi (rotary valves).

Where:F¡ =Liquidcavitation factor

(Typical values for F¡are given in Table 3-1)

P1 = Upstream pressure, psia

Py = Vaporpressureof the liquid,psia

The required Cvfor flashing applications is determinadby using the appropriate ~P allowable [~PChcalculatedfrom Equation 3.3].

Step 3: Determine Specific Gravity

Specific gravity is generally available for the flowingfluid at the operating temperatura. The appendix pro-vides fluid property data for 268 chemical com"pounds,from which the specific;gravity, Gf can be calculated.

Step 4: Calculate Approximate Cy

Generally the effects of nonturbulent flow can be ig-nored, provided the valva is not operatjng in a laminarortransitional flowregion due to high viscosity, very lowvelocity, orsmall Cv.lnthe eventthere is some question,calculate the Cy,from Equation 3.1, a9suming Fp=1,andthen proceed to steps 5-7. If the ReYl3olds numbercalculatedin Equation3.6a is greaterthan 40,000, FR

can be ignorad (proceed to step 8 after step 5.)

Step 5: Select Approximate Body SizeBásed on C

y

From the Cv tablas in section 4, select the smallest bodysize that will handle the calculated C .v

Step 6: Calculate Valve Reynolds NumberReyand Reynolds Number Factor FR

Use Equation 3.6a to calculátevalveReynolds NumberFactor:

Re =v

N4F

.

dq ..

(0 FL2C

o

v'

o

0)

1/1

. 0-+1v J FLC v N2 d4

(3.6a)

Use Equation 3.6b to calculate valva Reynolds NumberFactorFR if Rev< 40,000,otherwiseFR =1.0:

(

C

j

O0655

FR=1.044-.358 cvs o," . vIo

(3.6b)

Where: C = Laminar flow C .w v

C = ~(~

)

2/3

vs F N ~Ps s

(3.6c)

Cvt = Turbulent flow Cv (Equation 3.1)

Fs = streamline flow factor

3-5

"-LL 1.0o

0.9

0.8::JenenQ) 0.7C.C5

0.6<.J:;:::"5

0.5""O"5o-

::::¡

\."""""

""""'....... -

Critical CriticalLiquid Press. Liquid Press.

(psia) (psia)

Ammonia 1636.1 HydrogenChloride 1205.4

Argon 707.0 Isobutane 529.2Benzene 710.0 Isobutylene 529.2Butane 551.2 Kerosene 350.0Carbon Dioxide 1070.2 Methane 667.3Carbon Nitrogen 492.4Monoxide 507.1 Nitrous Oxide 1051.1Chlorine 1117.2 Oxygen 732.0Dowtherm A 547.0 Phosgene 823.2Ethane 708.5 Propane 615.9Ethylene 730.5 Propylene 670.3FuelOil 330.0 Refrigerant 11 639.4Fluorine 757.0 Refrigerant 12 598.2Gasoline 410.0 Refrigerant 22 749.7Helium 32.9 Sea Water 3200.0Hydrogen 188.1 Water 3208.2

Table 3-111:Piping Geometry Factors torValves with Reducer and Increaser,

F versus C Id2P v

NOTE: The maximum effective pressure drop (!1Pchoked) mar be affected by the use of reducers andincreasers. This is especiallytrue of Va/disk va/ves.Contact factory for critica/ applications.

Table 3-IV: Piping Geometry Factors torIncreaser Only on Valve Outlet,

F versus C I d2P v ~

Where:d =Valve port inside diameter in inchesD =Infernal diameter of the piping in inches(See Tables 3-VII and 3-VIII)

F 2/3

~

F2C2

)1/6

d L v

Fs= fm N d4 +1L 2(3.6d)

Where: d = Valve inJet díameter, ihc~es

Fd =Valv8:stylem()difier(Tablé 3,-1)'F\$::::Lamiriár, or streamline, flow factor"q =Flow rate; gpm

N2 = 890 when d is in,lnchesN4 =17,300, when q is in gpm and d ,In

inches .

Ns = 47 when q is in gpm and !1Pin psi¡.t= absoluteViscosity,centipoise' .

v ==kinematicviscosity,centistokes=¡l/Gf

Step 7: RecalculateCv,Using ReynoldsNUmber Factor

Ifthe calculated,vaIÜ~úfERisless than 0.48, the flowisconsideredJaminar;'andtl1eCis equál to C calculated.' ,v. '" ". vs

fror:nEquation 3.6c. IfFR ís greátE;j(thaq0,98, turbulentflp\IYcan be assumed (F~'= 1.0);'iandCv is calculatedfrbm Equation 3.1. Do not use the'pipiríg geometryfactor Fp if F R is less than 0.98. For values of FR betWeen0.48 and 0.98, the flow is considered trahsitional; andthe Cvis calculatedfromEquation3.6e:

3-6

q¡;:Cv =F;J ~2

~(3.6e)

For laminar and transitional flow,note the!1P is alwaystaken as P1- P2.

Step 8: Calculate Piping Geometry Factor

If the pipe size is not given, use the approximate bodysize (from step 5) to choose the corresponding pipesize. The pipe diameter is used to calculate the pipinggeometry factor, Fp' which can be determined byTables 3-111and 3-IV. If the pipe diameter is the sameas the valvesize, Fp is 1 and does notaffect Cv'

Step 9: Calculate the Final Cv

Using the value of Fp' calculate the required Cv fromEquation 3.1 .

Step 10: Calculate Valve Exit VelocityThe following Equation is used to calculate entrance orexit velocities for liquids:

'JJJv = 0.321 q

Av

(3.7)

e Id2 d/Dv

0.50 0.60 0.70 0.80 0.904 0.99 0.99 1.00 1.00 1.006 '0.98 0.99 0.99 1.00 1.008 0.97 0.98 0.99 0.99 1.0010 0.96 0.97 0.98 0.99 1.0012 0.94 0.95 0.97 0.98 1.0014

.,

0.92 0.94 0.96 0.98 0.9916 0.90 0.92 0.95 0.97 0.99

18 0.87 0.90 0.94 0.97 0.9920 0.85 0.89 0.92 0.96 0.99

25 0.79 0.84 0.89 0.94 0.9830 0.73 0.79 0.85 0.91 0.9735 0.68 0.74 0.81 0.89 0.9640 0.63 0.69 0.77 0.86 0.95

e I d2 d/Dv

0.50 0.60 0.70 0.80 0.904 1.00 1.00 1.00 1.00 1.006 1.01 1.01 1.01 1.01 1.018 1.01 1.02 1.02 1.02 1.0110 1.02 1.03 1.03 1.03 1.02

12 1.03 1.04 1.04 1.04 1.0314 1.04 1.05 1.06 1.05 1.04

16 1.06 1.07 1.08 1.07 1.05

18 1.08 1.10 1.11 1.10 1.06

20 1.10 1.12 1.12 1.12 1.08

25 1.17 1.22 1.24 1.22 1.13

30 1.27 1.37 1.42 1.37 1.2035 1.44 1.65 1.79 1.65 1.3240 1.75 2.41 3.14 2.41 1.50

Where:

~v = Velocity, ft/sec

q = Liquid flow rata, gpm

Av = Applicable flow area, in2of body port (Table 3-VIII)

After calculating the exit velocity, compare that numberto the acceptable velocity for that application. It may benecessary to go to a largar valva size.

Step 11: Recalculate Cy 11Body SizeChanged

RecalculateCvif the Fp has been changed due toselection of a largerbody size.

Step 12: Select Trim NumberFirst identify if the valva will be usad for on/off orthrottling service. Using the Cv tablas in Section 4,select the appropriate trim number for the calculated Cvand body size selected. The trim number and flowcharacteristic (Section 9) may be affected by how thevalva will be throttled. When cavitaiton is indicated,retar to Section 14 to evaluate special trims for cavita-tion protection.

LlQUID SIZING EXAMPLES

~Example OneGiven:

Liquid Water'

Critical Pressure (Pe) 3206.2 psiaTemperature... 2500 F

Upstream Pressure (P1) 314.7 psia

Oownstream Pressure (P2) 104.7 psiaSpecific Gravity 0.94

Valva Action Flow-to-open

Line Size 4-inch (Class600)

Flow Rate 500 gpm

Vapor Pressure (Py) 30 psiaKinematic Viscosity (v) 0.014 centistokes

Flow Characteristic Equal Percentage.

Step 1: Calculate actual pressure drop using Equation(3.2).

~

AP = 314.7 psia - 104.7 psia = 210 psi

Step 2: Check for choked flow. Find FLusing Table 3-1. Looking under "globe, flow-under," find FLas 0.90.Next,estímateFF usingEquation3.4:

~ oFF = 0.96 - 0.28 = 0.93

3206.2

Insert FL and FF into Equation 3.3:

APch = (0.90)2 [314.7 - (0.93)(30)] = 232.3 psi

SinGathe actual AP is less than APch' the flow is notchoked; therefore, use the smaller (or actual AP) to sizethe valva.

At this point, also check for incipient cavitation usingEquation 3.5 and Table 3-1: "

AP (cavitation) = (0.81)2(314.7-30)= 187 psi

SinGaAP (actual) exceeds AP (cavitation), substantialcavitation is occurring, but flow is not choked. Specialattention should be paid to material and trini sel~ction.

Step 3: The specific gravity for water is given as 0.94

Step 4: CalculatetheapproximateCvFp usingEquation3.1 and assumingFp is 1.0:

e =soi Ó.94 ,

= 33.4v , 210

Step 5: From the Gvtablas (Mark One, flow-under,equal percentage, Class 600) select the smallest bodysize for a Cvof 33.4, which is a 2-inch body.

\$tep 6: Calcul¡:¡.tethe Reynolds Number Factor, FR'using Equations 3.6a and 3.6e as required.

,

r ~

1/4

(1

",

7

,

'

"

.

,

,3(),

O,

) (1)

,

:(5"

'

,.

0,,, ,

Q

,

'

,

')"

(0'

,',

9

,,

O

,

),,

2

,,

(33

,

.4

,,

)2

,

Re= .'.,." +1=114x106

. " v:'(0.014Ú(0.90)(33A) ,', (890)(2)4

Step 7: SinGa Rev > 40,000, FR= 1.0andthe recalcu-lated CvFp remains as 33.4. '

Step 8:Using the 2-inch body from step 5, determinetheFpusing Table 3-111,where:

diO = 2/4 = 0.5 aMC Id2 = 33.4/22 = 8.35, v

Therefore according to Table 3-111,theF pis 0.97.

Step 9: Recalculate the'fin'al(::,,:

500 ~0.94",

:;,;C = - - = 34.5

v 0.97 210

Step 10: Using Equation 3.7, the velocity for a 2-inchbody is found to be nearly 51 ft(sec.Since this applica-tion is cavitating, damage may result in a 2-inch valva.A 3-inch body reduces velocity to about 23 ft/sec which

3-7

is a more aqceptable value. How8ver, special trim maybe tequired to eliminate cavitation damage.

NOTE: In this example, a 2 x 4-inch Mark One -X mightalso be chÓsen. It is less costly than a 3-inch valve andtl)e larger outlets willlower the velocities. It will also beless costly fo install in a 4-inch line.

Step 11: SinGathe body size ha,schanged, recalculatethe Cyby following steps8and 9. The Fp for a 3-inchbody is nearly 1.0, and the final C"is 3304.

i

Step 12: Referringtothe C~tablas, a Cy33,3-inch valvawould requireat least a Him number of L2q. Trimnumber 2,Omayalso suffice andhave no reduced trimprice adder. Refer to Section 14 on special trims forcavitation protection. '

Example TwoGiven:

Liquid Ammonia

Critical Pressure (Pc) 1638.2 psiaTemperature , 20° F

Upstream Pressure (P1) 149.7 psia

Downstream Pressure (P2) 64.7 psia

Specific Gravity 0.65Valva Action Flow-to-close

Line Size 3-inch (Class 600)

Flow Rate 850 gpm

Vapor Pressure (Pv) 45.6 psiaKinematic Viscosity (v) 0.02 centistokes

FlowC,haractetistic , :' : Linear

Step 1: Calculate áctual pressuredropusing Equation3.2.

~p = 149.7 psia - 64.7 psia ==85 psid

Step 2: Check fer choked flow. Find FL using Table3-1. Lookirig under "globe,flow-over," find FLas 0.85.Next, estimate F¡::usingEquatioIl3.4:

F =0.96 - 0.28 j 4,

.

,

5,6"

..

",

.' ,,'.

,

=0.91

~ " 1638.2 "

Insert FL and F~>into Equation 3.3:'

~P9h (GhOké'd{=(O~8~Jd'[149.{'- (0.91 )(15.6)] =78.2 ps;, .-'; ,.. "

;sfnce'the "aduaf~P,)s morethan ~p Ch" the flow is

chPke,dand cavitating; therefore, use the ~Pch'for ~Pa, to sLzethe valva. SinGa the service is cavitating, special .'attentipn should be made to material and trim selection.GavC.oritrolor ChannelStream\$hould be considerad.

3-8

Step 3: The specific gravity for ammonia is given as0.65

Step 4: Calculate the approximate Cy using Equation3.1 :

~

Cy = 850 J 0.6578.2= 77.5

Step 5: From the Cytablas (Mark One, flow-over, linear,Class 600) select the smallest body Bize for a Cyof 77 .5,which is a 3-inch body.

Steps 6 and 7: Turbulent flow is assumed, so ReynoldsNumberFactoris ignorad,FR = 1.0.

Step 8: With the 3-inch body and 3-inch line, Fp = 1.

Step 9: SinGaFp = 1, the final Cyremains as 77.5.

Step 10: Using Equation 3.7, the velocity for a 3-inchbody is found to be ovar 38 ft/sec. SinGathis applicationis cavitating, this velocity may damage a 3-inch valva.However, sinGathe Bize is restricted to a 3-inch line, alargar valva size cannot be chosen to lower the velocity.Damage problems may result from such a system. Acavitation control style trim should be suggested; seaSection 14.

~

Step 11: If cavitation control trim is not selected, Cyrecalculation is not necessary sinGathe body size ortrill1 style did not changa.

Step 12: Heferring to the Cytablas, a Cy77.5, 3-inchvalva would use atrim numberof2.000rthefull sizetrimn!1mber 2.62. Use of this trim, however, could result incavitationdamage to body and trim; sea Section 14.

Flashing Liquids Velocity CalculationsWhen the valva outlet pressure is lower than or aquel toth,~saturation pressure for the fluid temperatura, part ofthe fluid flashes into vapor. When flashing exists, thefollowing calculations must be usad to determine veloc-ity. Flashing requires special trim designs and/orhardened materials. Flashingvelocity greater than 500ftlsec requires special body designs. If flow rafe is in lb/hr:

. V = 0,040.,,

'W

[(1- ~ ) V¡2+ ~ Vg

j.. A, 100% 100%y ,

...-

'-'(3.8)

"-'

'"

J.,

if the flow rata is given in gpm, the following Equationcan be usad:

v= ~ q~ 1-1~O%)Where:

V = Velocity, ft/sec

w = Liquid flow rata, Ib/hr

q = Liquid flow rata, gpm

Av = Valva outlet flow area, in2,sea Table 3-VIII.

V'2 = Saturated liquid specific volume (fP/lb atoutlet pressure)

Vg2= Saturated vapor specific volume (fP/lb atoutlet pressure)

x = % of liquid mass flashed to vapor

v. +10;% vj(3.9)

Calculating Percentage Flash

The % flash (x) can be calculated as follows:

(h -h

)x = f1 f2 X 100%

h'92

(3.10)

Where:

x = % of liquid mass flashed to vapor

h'1= Enthalpy of saturated liquid at inlettemperatura

h'2= Enthalpy of saturated liquid at outletpressure

hfg2= Enthalpy of evaporation at outlet:pressure

For water, the enthalpies (hf1' hf2and hfg2)and specificvolumes (Vf2and Vg2)can be found in the saturationtemperatura and pressure tablas of any set of steamtablas.

Flashing Liquid Example

Assume the sama conditions exigí as in Example OIl.É3,

except that the temperatura is 3500 F rather than 2500F. By referring to the saturated steam temperaturatablas, you find that the saturation pressure of wateí tlt3500 F is 134.5 psia, which is greater than the outletpressure of 105 psia (90 psia). Therefore, the fluid isflashing. SinGa a portion- of the liquid is flashing,Equations 3.9 and 3.1Omustbe usad. x (%flashed) canbe determinad by using Equation 3.10:

h'1 = 321.8 Btu/lb at 3500 F(from saturation temperatura tablas)

h'2 = 302.3 Btu/lb at 105 psia(from saturatioh pressure tablas)

hfg2= 886.4 Btu/lb ~t 105 psia(from saturatlon pressure tablas)

(321.8 -302;3

) x 1909/0= 2.2%x = 886.4

Therefore, the velocity in a 3-inch valva can be deter-minad by using Equation 3.9:

Vf2 = 0.0178 fP/lb at 105 psia(from saturation pressure tablas)

Vg2 = 4.234 fP/lb at 105 psia(from saturation pressure tablas)

(20)(5.

00)

.[.(.

6._02%

).

.(2.2

.

0

.

Yc

..

o

. ..J .. ~V = - 1 -~ 0.0178 + - - - 4.234

7,07 100% - -:'10P%"

V = 156 ft/sec

Flashing velocity is less than 500 ft/sec, which is accept-able for Mark One bodies. Hardened trim andCavControl should algo be considerad.

3-9

CALCULAiJNGC FOA GASES,'.'" v,

", ij1trQduction, ,

aecél:Useofcompressibility, gases and vapors expandas thepressuredrops at the venacontracta, decreasingtheir specificweight. To account for the change inspecific weight, anexpansion factor, V, is introducedinto the valve sizing fOrmula. The forro of the Equationused is one of thefollowing, depending en the processvariables available:

w =63.3 FpCvV JxPl 'Y1 (3.11)

Q = 1360 FpC P1VJ,

xv G T Z

9 1

w ==19".3.FCP V¡ xMw',p v 1 T Z, 1

Q = 7320 FpC Plyj xv M T Zw 1

(3.12)

(3.13)

(3.14 )

Where:

w = Gas flow rate,lb/hr "

Fp = Piping geometry factorCv = Valve sizing coefficientV = Expansion factorx = Pressure droo ratio

'Y1= Specific welght at inlet conditions, Ib/fP ,

Q = Gas flow in standard fP/hr (SCFH)G = Specific gravllY ur gas relative to air at9

standard conditions

T1' :::;Absolute upstream temperature," °A = (OE+ 460°)

Z = Compressibility fa<;:tor

Mw = Molecular INeight

P1 = Upstream absOlute pressure, psia,

'NOTE:The numerícal constants in Equations 3.11-3.14 aré unit conversion factors.

The following steps should be used to compute thecorrectCv' body size and trim number:

Step 1: Select the Appropriate EquationBased on the information available, select one of thetour Equations: 3.11, 3.12, 3.13 or 3.14.

~

Step 2: Check tor Choked Flow

Determine the terminal pressure drop ratio, xT' for thatparticular valve by referring to Table 3-V.

Next, determine the ratio of specific heats factor, Fk' byusing the Equation below:

k

Fk=1.40 (3.15)

Where:

Fk = Ratio ofspecific heats factork = Ratio of specific heats (taken from

Table 3-VI).

Calculate the ratio of actual pressure drop to absoluteinlet pressure, x, by using Equation 3.16:

x = APP1

(3.16)

Where: ~x = RatiOof pressure drop to absolute inlet

pressure

AP = Pressure drop (P1 - P2)

P1 = Inletpressure, psiaP::, = Outlet pressure, psia

Choked flow occurs when x reaches the value of Fkxr

Therefore, if x is lessthan FkxT'the flow is not choked.If x is greater, the flowis choked. If flow is choked, then

FkxTshould be used in place of x (whenever it applies)in the gas sizing Equations.

Table 3-V: Pressure Drop Ratios, XT

3-,10"...'"

~

Valve Type Flow,Direction Trim Size XT

Globe Flow-to-close Full Area 0.70Flow-to-close Reduced Area 0.70

Flow-to-open Full Area 0.75

Flow-to-open Reduced Area 0.75

High Performance 60° apeA Full 0.36

Butterfly , 90° Open .Full 0.26

Multi-stage Under Seat AII -1.00"

Ball 90° Open Full 0.30,

. .

Figure 3-4: Compressibility Factors tor Gases with Reduced Pressures tr()m O to 4()~..

(Reprodueed from eharts of L.C. Nelson and E.F. Obert, Northwestern Teehnologieallnstitute)

~ 4.0

3.0N

15"tílOLI.

~:s 2.0.¡¡;U>!!?

Q.Eoo

1.0

oo 10 15

Reduced Pressure, Pr

5

Step 3: Calculate the Expansion FactorThe expansion factor, Y, may be expressed as:

xY = 1--

3FkxT(3.17)

~ NOTE: If the fIow is choked, use FkxT for x.

Step 4: Determine the CompressibilityFactor

To obtain the compressibility factor, Z, first calculate thereduced pressure, P , and the reduced temperature, T :r r

P,

Pr= pe(3.18)

Where:

P = Reduced pressurer

P1 = Upstream pressure, psia

Pe = Critical Pressure, psia (fromTable 3-VI)

T1T =-

r Te(3.19)

¡,

Where:

T = Reduced temperaturer

T, = Absolute upstream temperature

Te = Critical absolute temperature(from Table VI)

Using the factors P and T , find Z in Figures 3-4 orr r

3-5.

20 25

1.02

1.00

0.98

0.96

0.94

0.92N.: 0.90o-~ 0.88LL>- 0.86-~ 0.84

.~ 0.82Q)

Q. O 80Eo 0.78U

0.76

0.74

0.72

0.70

0.68

0.66

0.64O

Tr-1.001.051.101.151.201.301.401:501.60

1.80

2.00

2.50

3.003.504.005.006.008.0010.0015.00

30 35 40

1 2 63 54

Reduced Pressure, Pr

Figure 3-5: Compressibility Factors for Gase\$

with Reduced Pressures from O to 6.. (Reprodueed .fromeharts of LC. Nelson and E.F. Obert,

Northwestern Teehnologieallnstitute)

3-11

,/ ;...-

vV.J""

V

v/. f....-

v.

¡...... ¡......

L--"..... 1-" -

8U¡.- ¡.-

¡.--

'

r

V\ "f.&

T= .00

A"I

1 1 I

l."'.'0'

,/

/'

...... , .....

\.' ... .......... ,- -""" ./

\\\\1 \

...... .....

'T. I

\ \ """¡....¡U'

1,.....-

\ I\. /./

\ \1\ ./\ ./

\ \ V I\ , 11=,1.5(\1\ ...... ... I\ 1\ :.'/

I\ \ "'" I 1

1\ \ ; ..... -\ 1\ I\ \ /

\ 1 II

", .," In\ \ " I /

",1/ /

\\ J\ 1\ .'

T, - 1 j /'\ ;; 11

L\ '"I = I ¿:'¡J\ ........ I

/

Table 3-VI: Gas Physical Data

~

Step 5: Calculat.e Cv

Using the above calculations, use one of the tour gassizing Equations to determine Cv (assuming Fp is 1).

Step 6: Select Approximate Body SizeBased on Cv

From the Cvtablas in the appendix, select the smallestbodysizethatwill handlethe calculatedCv'

Step 7:Calculate Piping Geometry FactorIf the pipa size is not given, use the approximate bodysize (fromstep 6) to choose the corresponding pipasize. The pipa size is usad to calculate the pipinggeometry factor, Fp' which can be determinad byTables 3-111or 3-IV. If the pipa diameter is the sama asthevalvasize, Fp ís 1 and is nota factor.

Step 8: Calculate the Final Cv

With the calculation of the Fp' figure the final Cv' ,

Step 9: Calculate Valve Exit Mach NumberEquations 3.20, 3.21,3.22 or 3.23 are usad to calculateentrance or exit velocities (in terms of the approximateMach number). Use Equations 3.20 or 3.21 for gases,Equation 3.22 for air and Equation 3.23 for steam. Usedownstream temperature if it is known, otherwise use

, upstream temperatura as an approximation.

3-12

°aM (air) = (3.22)

1225 Av jTwv

M (steam) = (3.23)1514 AvfT

Where:M = Mach number

°a ,=FActual flow rafe, fP/hr. (CFH, not SCFH; sea paga 3-13)

A = Applicable flow area, in2,ofv body port (Table 3-VIII)

TJp,= Absolute temperatureo R, (OF + 460°)w = Massflow rafe, Ib/hr

v = Specific volume at flow conditions, fP/lb

G = Specific gravity at standard conditions9 relativato air

Mw = Molecular weight

k = Ratio of specific heats

'*

..Critical Critical Ratio of

Pressure Temperature Molecular SpecificGas (psia) (OR) Weight (Mw) Heats (k)

Air 492.4 227.1 28.97 1.40Ammonia 1636.1 729.8 17.0 1.31

Argon 707.0 271.1 39.9 1.67Carbon Dioxide 1070.2 547.2 44.0 1.29Carbon Monoxide 507.1 238.9 28.0 1.40Ethane 708.5 549.4 30.1 1.19

Ethylene 730.6 508.0 28.1 1.24Helium 32.9 9.01 4.00 1.66

Hydrogen 188.2 59.4 2.02 1.40Methane 667.4 342.8 16.04 1.31Natural Gas 667.4 342.8 16.04 1.31

Nitrogen 492.4. 226.8 28.0 1.40

Oxygen 732.0 278.0 32.0 1.40

Propane..

665.3 44.1 1.13-615.9Steam . 3208.2 1165.1 18.02 1.33

°a (3.20)M (gas)=

5574A:) kT ..Mw

°a(3.21)

M (gas) = 1036 Av jY;Gg

NOTE: To convert SCFH to CFH use the Equation.'

..~(Pa)(Oa)

Ta

(Ps)(O)(3.24)=

Ts

Where:

Pa = Actual operating pressure°a = Actual volume flow rata, CFHTa = Actualtemperatura,°R (OF + 460°)Ps = Standard pressure (14.7 psi)O = Standard volume flow rata, SCFH

Ts = Standard temperatura (520° Rankine)

After calculating the exit velocity, compare that numberto the acceptable velocity for that application. Select alargar size valva if necessary. Refer to section 13 topredict noise level.

Caution: Noise levels in excess of 110 dBA may causevibration invalves/pipingresulting inequipmentdamage.

Step 10: Recalculate Cyif Body SizeChanged

Recalculate Cv if Fp has changed due to the selectionof a largar body size.

Step 11: Select Trim Number

'-., Identify if the valva is for en/off or throttling service.Using the Cvtablas in Section 4, select the appropriatetrim number for the calculated Cv and body size se-lected. The trim number and flow characteristic (Sec-tion 9) may be affected by how the valva is throttled.

GAS SIZING EXAMPLES

Example One

..~

Given:Gas Steam

Temperature 450° F

Upstream Pressure (P1) 140 psiaDownstream Pressure (P2) 50 psiaFlow Rate 10,000 Ib/hr

Valva Action ,... Flow-to-open

Critical Pressure (Pe) 3206.2 psiaCritical Temperature (Te) 705.5° FMolecular Weight (Mw) 18.026Ratio of Specific Heats (k) 1.33

Flow Characteristic Equal percentageUne Size 2-inch (Class 600)Specific Volume 10.41

Step 1: Given the above information, Equation 3.13 canbe usad to salve for C .

v

,. .. 'u' .,".

. ." .. '.' ~

Step 2: Referring to Table 3-V, the pressure drop'ratlO,

XI' is 0.75. Calculate Fkusing Equation 3.15 and x using...Equation 3.16:

Fk= 1.33 = 0.951.40

140 - 50 = 0.64x = 140

Therefore, FkxTis(0.95)(O.75) or 0.71. Since x is lessthan FkxT,flow isnot choked. Use x in all Equations.

Step 3: Determine Y using Equation 3.17:

y = 1 - 0.643 (0.71)

~ 0.70

"....

Step 4: Determine Z by calculáting P.and T using.. r r

Equations 3.18 and 3.19: .

140 =0.04P = .r 3208.2

T =r

450 + 460

705.5 + 460= 0;78,

Using Figure 3-4, Z is found to be 1.0

Step 5: Determine Cvusiflg Equat'ion3.13 and assum-ing Fp is 1:

10,000C =

v (19.3) (140) (0.70)

(910) (1.0) .= 47.0

(0.64) (18.02)

Step 6: From the Cvtablas (Mark arTe, flow-under,equal percentage, Class 600), select the smallestbody size for a Cvof 47, which is a 2-inch body.

Steps 7 and 8: Since the pipa size is the sama as thebody, Fpis 1 and is not a factor. Therefore, the Cvis 47.

. Step 9: The gas is steam, calculate the Mach numberusing Equation 3.23. Assume a constant enthalpyprocess to find specific volume at downstream condi-tions; from steam tablas, v = 10.41 ft3/lbat T2= 414°F:

(10,000) (10.41)M=

1515(3.14)j 414+460

This is greater than Mach 0.5 and should be re-viewed for excessive noise and use of noise reduc-

ing trim.

= 0.74

3-13

NOTE: To convert SCFH to CFH use the Equatíon:

~ (Pa)(Oa) (P5)(0)= (3.24)

Ta T5

Where:

Pa = Actual operating pressure°a = Actual volume flow rata, CFHTa = Actual temperaturé,OR (OF+ 460°)P5 = Standardpressure(14.7psi)O = Standard volume flow rata, SCFH

T5 = Standardtemperatura(520°Rankine)

After calculating the exit velocity, comparethat numberto the acceptable velocity for that application. Select alargar size valva if necessary. Refer to section 13 topredict noise level.

Caution: Noise levels in excess of 110 dBA may causevibration invalves/pipingresultingin equipmentdamage.

Step 10: Recalculate Cv if Body SizeChanged

Recalculate Cv if Fp haschangeddueto the selectionof a largar body size.

(.Step 11: Select Trim NumberIdentify if the valva is for en/off or throttling service.Using the Cvtablas in Section 4, select the appropriatetrimnumber for the calculated Cv and body size se-lected. The trim number and flow characteristic (Sec-tion 9) may be affected by how the valva is throttled.

GAS SIZING EXAMPLES

Example One

I~

Given:

Gas Steam

Temperature 450° F

Upstream Pressure (P1) 140 psiaDownstream Pressure (P) 50 psiaFlow Rate 10,000 Ib/hr

Valva Action ,... Flow-to-open

Critical Pressure (Pe) 3206.2 psiaCritical Temperature(Tc) 705.5° FMolecular Weight (Mw) 18.026Ratio of Specific Heats (k) 1.33

Flow Characteristic Equal percentageUne Size 2-inch (Class 600)Specific Volume 10.41e

Step 1: Given the above information, Equation 3.13 canbe usadto salvefor C . -v

Step 2: Referrillg to Table 3-V, the pressure drop':¡itfo,

XI' is 0.75. CalculateFkusing Equation 3..15 and x using-_-Equation 3.16:

Fk = 1.33 = 0.951.40

140 - 50 = 0.64x = 140

Therefore, FkxTis(0.95)(0.75) or 0.71. SinGa x is lessthan FkxT, flow isnotchoked. Use x in all Equations.

Step 3: Determine Y using Equation3.17:

0.64 ~0.70y = 1- 3(0.71)

c'"

Step 4: De¡termineZ bycalculatin"Q Pründ Tr usingEquations3.18 and3.19:

P = 140 .~"0.04r3208.2

T =r

450 + 460

705.5 + 460= 0,78,

Using Figure 3-4, Z is found to be 1.0

Step 5: Determine Cvusing Equation 3.13 and assum-ing Fp is 1:

10,000 (910)(1.0) -=47.0

(0.64) (18.02)C =

v (19.3) (140) (0.70)

Step 6: From the Cvtablas (Mark One, flow-under,equal percentage, Class 600), select the smallestbody size for a Cvof 47, which is a 2-inch body.

Steps 7 and 8: SinGathe pipa size is the sama as thebody, Fpis 1and is not a factor. Therefore, the Cvis 47.

Step 9: The gas is steam, calculate the Mach numberusing Equation 3.23. Assume a constant enthalpyprocess to find specific volume at downstream condi-tions; from steam tablas, v = 10.41 fP/lb at T2= 414°F:

~.-

(10,000) (10.41)M=

1515(3.14)) 414+460

This is greater than Mach 0.5 and should be re-viewed for excessive noise and use of noise reduc-

ing trim.

= 0.74

3-13

Step 10: If bady size daes nat change, there is no.impact an Cycalculatian.

Step 11: Referring ta the Cytables, a Cy47, 2-inch MarkOne waulduse atrim number af 1.62. If naise is acansideratian, see Sectians 13 and 14.

Example Two

Given:

Gas Natural Gas

Temperature 65° F. .

U~streamPressure (P1) 1314.7 psiaDQwnstreamPressure (P) 99.7 psiaFlaw Rate """""""""',""""""'" 2,000,000 SCFH

Valve Actian """""""',"""""""""'" Flaw-ta-apen

Critical Pressure (Pe) , 672.92 psia

CriticalTemperature(Te)""""C"""""""" 342.8°R

Malecular Weight (MJ " 16.042Ratia af Specific Heats (k) 1.31Flow Characteristic ,0 Linear. ,Line Size Unknawn (Class 600)

Step 1: Given the abave informatian, Equatian 3.14 canbe used ta salve far Cv'

Step 2: Heferring ta Table 3-V, the pressure drop ratia,Xv is 0.75 byassuming a Mark One flaw-under. Calcu-late Fk using Equatian 3.15 and X using Equatian 3.16:

F = ~ = 0.936k 1.40

1314.7 - 99.7 = 0.92x=1314.7

Therefare,FkxTis (0.94)(0.75) ar 0.70. Since x is greaterthan FkxT'flaw is chaked. Use FkxTin place af x in allEquatians.

Step 3: Determine Y using Equatian 3.17:

y = 1 - 0.70

3 (0.70) '\

= 0.667

Step 4: Determine Z by calculating Pr and T: usingEquatians 3.18 and 3.19:

1314.7P =

r 667.4= 1.97

3-14

T =r

65 + 460

342.8

~= 1.53

Using Figure 3-5, Z is faund ta be abaut 0.86.

Step 5: Determine Cv using Equatian 3.14 and assum-ing Fp is 1:

(2,000,00)C =

v (7320)(1314.7)(.667)

(16.04)(525)(0.86)

0.7031.7

Step 6: Fram the Cy tables (Mark One, flaw-under,linear, Class 600), select the smallest bady size far a Cyaf 31.7, which is a 1 1/2-inch bady.

Steps 7 and 8: Since the pipe size is unknawn, use 1as the Fpfactar. Therefare, the Cv is 31.7.

Step 9: Since the gas is natural gas, calculate the Machnumber using Equatian 3.20:

M=(297,720*)

= 6.61

5574 (1.77)

~(1.31)(65 + 460)

16.04

*NOTE: TacanvertSCFHfa CFH,use Equatían3.24.

Step 10: Mach numbers in excess af sanic velacity atthe autlet af the valve are nat passible. A larger valvesize shauld be selected ta bring the velacity belaw thesanic leve!. Ta praperly size the valve, select a size tareduce the velacity ta less than 1.0 Mach.

Step 11: Using Equatian 3.20, salve far the recam-mended valve afea required far 0.5 Mach velacity:

297,720 CFH0.5 M = A = 16.3 in2

557411.31 (65 + 460) y16.04

Salve far the valve diameter fram the afea by:

#-Ay ~(16.3)

A.

= 1td2 ar d ="

'

= = 4.6 in.y . 1t 1t

Thus a 6-inch valve is required.

Step 12: Referring ta the Cvtables, a Cv af 31.7, 6-inchMark One wauld use a trim number af 1.62. Since theflaw is chaked, naise shauld be calculated fram Sectian13, and special trim may be selected fram Sectian 14.

~

\.CALCULATING C FOR TWO PHASE FLQ,W

Y,,'" .

tI

Introduction

The method of Cv calculation for two ph¡3.sefJow as-sumes that the gas and liquid passJhrough the valvaorífice at the sama velocity. The requrred Cv is deter-minad by using an equivaJentdénsity fer theHquid g'asmixtura. This method is intended for use with mixturesof a liquid and a non-condensable gas. To size valVeswith liquids and their own vapor at the valva ihlet willrequire good engineering judgement.Nomenclature:

Av= flow area of body port (Table 3-VIII)

~pa = allowablepressuredropqf= volumetric flow rafe of liquid, ff3/hr

q = volumetric flow rafe of gas, ff3/hr9

wf = liquid flow rafe, Ib/hr

w = gas flow rafe, Ib/hr9

Gf = liquid specific gravity at upstream conditions

G = gas specific gravity at upstream conditions9

T1 = upstream temperatura (O R)

'-'

Step1: Calculate theLimiting PressureDropFirst it must be determinad whether liquid or gas is thecontinuous phase at the vena contracta. This is done bycomparing the volumetric flow rafe of the liquid and gas.Whichever is greater will be the limiting factor:

If qf > qg'then ~Pa= ~Pafor liquidIf q > qf' then ~P = ~Pfor gasg. a a

The ~P afor liquid or gas is either P1- P2or the chokedpressure drop of the dominating phase if the valva ischoked. (See the gas and liquid choked pressureEquations.)

Step 2: Calculate the Equivalent SpecificVolume of the Liquid-gas Mixture

Where: (f v )v = ~ + fv

e Y2 f f

~i

Wf=--119 (Wg+ w¡).

f = wff(Wg + wf)

Tv = 1

9 (2.7 P1G9)

vf = 1(62.4 Gf)

Y = gas expansion factor (Equation 3.17)

~

.Sfep3: C~lculate the Required Cyof the Valve

(w+w

)F;v

C F - gf =--v p - 63 3 ~P. a

Use the smaller OfP1- P2 anq ÁPChfor Pa'

Step 4rSelect Body Size Based on CyFromthe e tablesin the appendix,selectthe smallest, v

body size that will handle the calculated Cv'

Step 5: Calculate Piping Geometry Factor. ,

If the pipasize is notgiven, usethe approximatebody size '

(from step 6) to chúosethe corresponding pipe size. Thepipa size is usad to calculate the piping geometry'factor,Fp,which can be determinad by Tables 3-111or 3-IV. Ifthe

pipa diameter is the sama as the valva size, Fp is 1.

Step 6: Caículate Final C" ,y

With the calculati?n ofthe 5p,'figure the final Cv'

, ".' "

Step 7: Calculate U"teValve Exit Velocity

Where: ,', ,(q+q)VeIocity" "= "",

,

~g, 'A v

wf

qf = 62.4 Gf .''.<'~

,,~.~.

W9T1

qg= 27G P. 9 2

Area = applicable flow area

After calculating the exit velocity, compare that numberto the acceptable velocity for that application. Select alargar valva size if necessary. '

Recommended two phase flow velocity limits are simi-lar to those for flashing when the gaseoús phase is '

dominant. If liquid is the dominant phase, velocity of th~mixtura should be less than 50 ft/sec in the body.

Step 8: Recalculate Cy if Body Size ChangedRecalc~late Cvif Fp has been changed due to the selectionof a largar body size.

Ste~9: Select Trim Number

Identify if the valva will be usad for en/off or throttlingservice. Using the Cvtablas in Section 4, select theappropriate trim number for the calculated Cvand bodysize selected. The trim number and flow characteristic(Section 9) may be affectedby how the valva is throttled.

"Special trim and mataríais may be required if high noiselevels or cavitation are indicated.

3-15

Table 3-VII: Pipe Flow Areas, A (Square Inches)p

~

Table 3-VIII: Valve Outlet Areas ~

NOTE: To find approximate fluid velocity in the pipe, use the Equation Vp= VvA/ Ap where:

Vp= Velocity in pipe Av= Valve Outlet area Irom Table 3-VIII

Vv= Velocity in valve outlet Ap= Pipe area Irom Table 3-VII

To lind equivalent diameters 01the valve or pipe inside diameter use: d =J 4AJrc, O=J 4A/rc

~

3-16 ,,' .'

Nominal Schedule

PipeDiameter 10 20 30, 40 60 80 100 120 140 160 STD XS XXS

'/2 0.30 0.23 0.17 0.30 0.23 0.05

3/4 0.53 0.43 0.30 0.53 0.43 0.15

1 0.86 0.72 0.52 0.86 0.72 0.28

1'/2 2.04 1.77 1.41 2.04 1.77 0.95

2 3.36 2.95 2.24 3.36 2.95 1.77

3 7.39 6.61 5.41 7.39 6.61 4.16

4 12.73 11.50 10.32 9.28 12.73 11.50 7.80

6 28.89 26.07 23.77 21.15 28.89 26.07 18.83

8 51.8 51.2 50.0 47.9 45.7 43.5 40.6 38,5 36.5 50.0 45.7 37.1

10 82.5 80.7 78.9 74.7 71.8 68.1 64.5 60.1 56.7 78.9 74.7

12 117.9 114.8 111.9 106.2 101.6 96.1 90.8 86.6 80.5 113.1 108.4

14 143.1 140.5 137.9 135.3 129.0 122.7 115.5 109.6 103.9 98.3 137.9 132.7

16 188.7 185.7 182.6 176.7 169.4 160.9 152.6 144.5 135.3 129.0 182.6 176.7

18 240.5 237.1 230.4 223.7 213.8 204.2 193.3 182.7 173.8 163.7 233.7 227.0

20 298.6 291.0 283.5 278.0 265.2 252.7 238.8 227.0 213.8 202.7 298.0 283.5,.

?4 434 425 411 402 382 365 344 326 310 293 425 415

30 678 661 649 663 602 574 542 513i .', .

975 956 938 914 870 830 78236

42 1328 1302 1282 1255 1187 1132 1064

Valve Valve Outlet Area, AvSize (Square Inches)

(inches) Class Class Class Class Class Class Class150 300 600 900 1500 2500 4500

'/2 0.20 0.20 0.20 0.20 0.20 0.15 0.11

3/4 0.44 0.44 0.44 0.37 0.37 0.25 0.20

1 0.79 0.79 0.79 0.61 0.61 0.44 0.37

1'/2 1.77 1.77 1.77 1.50 1.50 0.99 0.79

2 3.14 3.14 3.14 2.78 2.78 1.77 1.23

3 7.07 7.07 7.07 6.51 5.94 3.98 2.78

4 12.57 12.57 12.57 11.82 10.29 6.51 3.98

6 28.27 28.27 28.27 25.97 22.73 15.07 10.29

8 50.27 50.27 48.77 44.18 38.48 25.97 19.63

10 78.54 78.54 74.66 69.10 60.13 41.28 28.27

12 113.10 113.10 108.43 97.12 84.62 58.36 41.28

14 137.89 137.89 130.29 117.86 101.71 70.88 50.27

16 182.65 182.65 170.87 153.94 132.73 92.80 63.62

18 233.70 226.98 213.82 194.83 167.87 117.86 84.46

20 291.04 283.53 261.59 240.53 210.73 143.14 101.53

24 424.56 415.48 380.13 346.36 302.33 207.39 143.14

30 671.96 660.52 588.35 541.19 476.06 325.89

36 962.11 907.92 855.30

42 1320.25 1194.59