or Vy, ; Gngmilcal ae ) Chemie [Gs ea THE re. READAMAGAZINE IN THE CHEMICAL INDUSTRY NS) D9 yy Fluted plow ftps) ine Fluid Flow Air & Steam Equipment Design —_ Fluid InstrumentsDiiVtniT a i a aaa Control valve sizing subjects relevant to sizing and selecting the right control valve for a particular appication, followed by simpli- fied step-by-step procedures for perform- ing sizing calculations by hand along with worked examples. Te guide briefly discusses several Selection of control valve style The choice of control valve style (globe, bal, butter, etc) is often based on tra: dition or plant preference. For example, a majority of the control valves in a pulp ‘and paper mils are usually ball or seg- ‘mented ball valves. Refineries tradtional- ly use a high percentage of globe valves for control Globe valves offer the widest range of options for flow characteristics, pressure, temperature and noise and cavitation reduction. Globe valves also tend to be the most expensive. ‘Segment ball valves tend to have a higher rangeability, and size for size, nearly twice the flow capacity of globe valves and in adkition are less expensive than globe valves. On the other hand, segment ball valves are limited in avail abiity for extremes of temperature and Pressure and are more prone to noise and cavitation than are globe vaives. High-performance buttery valves are even less expensive than ball valves, especially in larger sizes (say 8-in and larger). They also have less rangeabilty than the ball valves and are more prone to cavitation. The eccentric rotary plug valves combine features of rotary valves, such as high cycle life stem seals and compact construction with the rugged construction of glove valves. Unlike the other rotary velves whose flow capacity is approximately double that of lobe valves, the flow capacity of eccen- tric rotary plug valves is on a par with globe valves. Table 1 is generalized comparison of the various styles of con- trol vaives. Although an extensive discussion of the selection of the proper valve flow characteristics for a particular applcation is beyond the scope of this article, as a general rule, systems with a significant amount of pipe and fitings, andlor with EEA 005 riio row annua centrifugal pumps (the most common case) are best suited to equal percent. age valves. Systems with very little pipe tend to be better suited to linear charac- teristics valves. For the final design of more rtical systems the valve manufac- turer should calculate the installed char- acteristic of the selected valve in a partic- ular system to ensure a good match between system and valve. Process data ‘Avvalve sizing calculation wil only be reli- able if the process data used in the cal= culation accurately represents the true process. Perhaps the most misunder- stood area of control valve sizing is the ‘selection of the pressure drop, Ap, to use in the sizing calculation. The Ap cannot be arbitrarily specified without regard for the actual system into which the valve will be installed. The correct procedure for determining the pressure drop across a control valve is to start upstream of the valve at a point where the pressure is. known (for example, a pump where the pressure can be determined from the head curve) and subtract the pressure losses due to the upstream pipe and fit tings. When the inlet to the valve p, is NOMENCLATURE ©, Valve flow coefcent Valve inlet diameter, in; ‘Ap Pressure differential across a control valve, py =P psi, bar; Pressure differential beyond which cavitation damage Is likely, psi, bar; Po ‘pr Pressure differential at which liquid flow becomes choked, psi, ba; Fe Liquid erica! pressure rato factor, dimensionles Fx Ratio of specific heats factor, dimensionless; FL guid pressure recovery factor, dimensionless; 74 Specific weight of gas o vapor at upstream conditions, Ib/ouf,kg/cu m; & k Ratio of spectic heats, dimensionless; M Molecular weight, atomic mass units; P; Upstream absolute static pressure, psa, bara; Pe Downstream absolute static pressure, p, ~ Ap, pla, baa; Pe Absolute thermodynamic critical pressure, psia, bara; POC Fipe diameter correction for aerodynamic noise calculation, dBA; PSC Pipe schedule correction for aerodynamic noise calcuiaton, dBA; Py Absolute vapor pressure of liquid atinlet temperature, psa, bara; Volumetric flowrate of gas, scfh,N cu m/r; or volumetric fowrateof liquid, gpm, cu mihr SPL Sound pressure level, dBA; T, Absolute upstream temperature, °R, °X; SC Valve style correction for aerodynamic noise calculation, dBA; W Weight or mass flowrate, Ir, kg/hr, x Ratio of pressure drop to absolut inet pressure, dimensionless; %_ Terminal or limiting pressure drop rato for gas, dimensionless: Y Expansion factor, dimensionless; 2 —_Compressblty factor, dimensionless Liquid-specfic gravity at upstream conditions (rato of density of liquid at flowing ‘temperature to density of water at 60°F), dimensionless; Gas-specitic gravity (ratio of density of flowing gas to density of ir with both at standard conditions, which is equal tothe ratio ofthe molecular weight of gas to the molecular weight of air), dimensionless;known, the next step is to go to a point downstream of the control valve where the pressure is known (for example, a tank where the head is known) and then work upstream toward the control valve, adding the pressure losses of the pipe and fittings. (The pressure lossos are added by working in the direction ‘opposite to the flow.) At the valve outlet 2 is known, The actual pressure drop across the control valve Is the different between the upstream and downstream pressure, ie., Ap = py ~ Pz. To perform sizing calculations at more than one flowrate (@.g., at both maximum and ‘minimum design flows), the calculation of py and pp at each flowrate must be repeated, as the system pressure loss- es (and pump head) are dependent on the flow. Liquid choked flow, cavitation, flashing When liquid flow in a control valve pass- es through the vena contracta (the point at which the cross-sectional area of the flow stream is at a minimum), the flow velocity reaches a maximum and the pressure decreases to a minimum. The static pressure at the vena con- tracta is a function of three things—the pressure immediately upstream of the valve (p,); the pressure drop (ap) across the valve; and the valve geometry ‘expressed in the manufacturers literature as the liquid pressure recovery factor, F, (see Fig. 1 for typical values of F,) For a fixed value of p;, as the pres- sure drop across a control valve increas- es, the pressure at the vena contracta decreases. If the pressure drop across the control valve increases to a point where the vena contracta pressure decreases to siight- ly below the vapor && . Recovery Factor, F. pressuro, py, of tho |= 9.7 liquid, vapor bub- bles form in the | % 06 vena contracta Once this happens, |% 0-5 additional increas- | 9,4 es in pressure drop across the valve do rot result in addi- tional flow, and 0 10 20 30 40 50 60 70 80 90 100 Percent Open flow is choked. This limiting or choking pressure drop is called the terminal pressure drop, apy. (The same thing is also occasionally referred to as APMax OF APatonabie) The calculation of pr (see step 2 of the sizing method for liquids) is important, because when the actual pressure drop, 4p, is greater than 4pr, then Apr and not Ap must be used in the sizing equations to prevent under- sizing the valve. CChoked flow produces either flashing cr cavitation. Flashing results ifthe pres- sure downstream of the valve, Po, is less than the vapor pressure of the liquid. In this case, the vapor bubbles that formed at the vena contracta continue down- stream. Flashing conditions have the potential for erosive damage to the valve by drops of liquid entrained in high veloci- ty vapor and selection of erosion-resis- tant materials (such as stainless-steel valves. Fig. 1. Liquid pressure recovery factor (FL) of control bodies and hardened trim) is advisable. Cavitation results from choked flow ‘when pp is greater than p,. In this case, ‘as the Vapor bubbles travel downstream from the vena contracta they collapse vio- lently, resulting in vibration, noise and damage. The use of hard or erosion- resistant materials is not very effective in preventing cavitation damage and, as a rule, cavitation should be avoided. Gavitation can usually be eliminated by selecting a valve style with a higher value of F. In general, as the F, increas: es, so does the price of the valve. In addition to the standard valve styles shown in Fig. 1, there are special cavita- tion-resistant valves available. In practice, at pressure drops approaching, but below the calculated value of Apr, there is usually some for- TABLE 1. COMPARISON OF CONTROL VALVE TYPES. Top-guided Cage-guided ‘Segment Eccentric High-performance globe globe ball rotary plug buttertly Cost High High Medium Medium Low Weight High High Medium Medium Low Flow capacity (compared to globe) 1K 1K x 1K x Cavitation potential Low Low Medium Medium High Inline repairable Yes Yes No No Inherent flow characteristic linear, linear, Modified linear Modified = % Quick opening quick opening Cavitation/noise reduction options No Yes Some Some No Suitable for high pressure differential Limited Yes Limited Yes Limited Suitable for dirty service Yos No Yes Yes Yes Suitable for slurries Limited No Yes Yes Limited Suitable for pulp stock No No Yes No LimitedTABLE 2, NUMERICAL CON- STANTS FOR LIQUID SIZING EQUATIONS. pop 100 psi 69 bar Ke 1.0 pmsl 0.865 cummr bar mation of vapor bubbles and some degree of cavitation. A simplified method for approximating the pressure drop above which cavitation damage and noise are likely to be a problem is given in Equation 3, Sizing method for liquids The following seven steps are used for control valve sizing for liquids (see nomen- lature sidebar). Constants K; and Kp (Table 2) are included to make the formu- las readily adaptable to customary U.S. or SS. units, The values of p;, p2 and py must always be expressed in absolute unis. 1. Calculate the critical pressure ratio factor, Fr: .96 0.28 [Be w) 2, Calculate the terminal pressure drop, Apr: 491 =FE(01-Fep,) O Fis a function of valve opening that is yet to be determined, so it will be nec- essary to make an initial estimate of valve opening. A good initial estimate is the value of Fat 80% open. For manu- facturers using factors other than F., the following is used to convert from one fac- torto another: C; = Ky'!2 =F. 3.1f Ap > Apr, flow willbe choked. Use 4p in place of 4p in Step 7. Go to Step 4 If.4p < Apr, skip Step 4 and go to Step 5. 4. If Bp < By flow will be flashing, ero- sion-resistant valve materials are requited; go to Step 7. Otherwise contin- ve with Step 5. 5. Calculate the pressure drop at which cavitation damage is likely to begin (4p): The values of Rand S for the vari- us valve styles are found in Table 3 EE 1055 rivio riow annuat TABLE 3, VALVE STYLE COR- RECTION COEFFICIENTS FOR AB) (CAVITATION DAMAGE) CALCULATION. Valve style R Ss Globe 1.0 05 Eccentric rotary plug 19 035 ‘Segment ball o7 02 High-performance butterfly. 06 0.16 Ky, @ APD -rei(S) (P:-P,) 6. If Ap > App, there is the potential for cavitation damage. A valve with a higher Fy should be used. If Ap < App, there is limited danger of cavitation and exces- sive noise. 7. Calculate the required C,; Kat « Using the caleulated C,, an appropriate valve size is chosen from the manufactur- ers’ tables of C, vs. valve opening as rep- resented in Tables 4 and 5. The goal is to. select a valve that will be as fer open as possible without exceeding 80% open at the maximum design flow. For valves installed with reducers, selecting a valve with an opening not exceeding 75% according o the tables will ensure that, with the reducer effect, it will ‘ot be in excess of 80% open in actual operation Size 20 40 1 1 4 15 3 iT 2 5 7 3 2 39 4 19 63 6 34 115 8 55 187 10 a7 208 2 123 a8 ‘TABLE 4. TYPICAL VALVE FLOW COEFFICIENTS, Cy, FOR SEGMENT BALL CONTROL VALVES, (%) 7 100 14 45 40 110 38 82 180 88 143 420 141 208 620 258 418 4,260 418 67 2030 658 1088 3210 937 1516 4490 TABLE 5. TYPICAL VALVE FLOW COEFFICIENTS, Gy, FOR EQUAL PERCENTAGE GLOBE CONTROL VALVES. Relative opening (%) Size 20 40 60 80 100 1 oA 08 2 6 13 15 1. 36 92 23 7 2 2 5 7 45 60 3 5 12 38 1 128 4 6 3 40 106 170 6 9 2B 89 253 362 8 4 47 102 420 650 10 29 62 170 566 950TABLE 6. RATIO OF SPECIFIC HEATS FACTORS, Fy. Gas. Fi Air 10 ‘Ammonia og2 Butane 0.96 Carbon dioxide 093 Ethane 085 Freon 0.98 Hydrogen 10 Methane a9 Natural ges 094 Nitrogen 10 Oxygen 10 Steam 093 Aer selecting a valve size and deter- mining the percent of opening corre: sponding to the calculated C,, the Fi at that opening is determined, Ifthe actual F, is less than the value used in Steps 1 and 5, all the steps are repeated because the calculated values of 4p; and App wil be larger than ther real values and there isthe posskilly of over- looking choked flow and potential cavta- tion problems. Ifthe actual Fis greater than the value used in the calculation, the calculation is conservative To check for potential cavitation prob- lems, the Gy atthe minimum design flow may be calculated. Making the inital est: mate of valve opening for the purpose of obtaining the initial value of Fy, may require some tral and erro. Estimating an opening of 40% is a good starting point, but p; and 4p will most tkely be higher at the minimum design rate than they are at the maximum design flowrate. uid sizing example ‘The stated problem is to select a properly sized segment ball control valve for the | following process conditions: Fluid: water; Pipe size: 6 in Maximum design flow, q: 680 gom: Pressure upstream of the valve, p; 42 psig; Local atmospheric pressure: 14.7 psia; Critical pressure, pg: 3208 psia: Vapor pressure, py: 1.1 psiar Pressure drop, Ap: 20 psid Specific gravity, Gy: 1.0. 4. Caleulate the critical pressure ratio | ENVIRONMENTAL DTU Rael \a ag Excellence in design, fabrication and customer service has been our standard for over 70 years. Individualized CeCe Cec ruse ues ast of these industries has made us a world leader in heat transfer technology and custom coil design. UA Copper ter) Sec) ene ce Ys Pree etd Yo eens y CM eel ee T RADIATOR ry 104 Peavey Road » Chaska, MN, USA 55318-7324 Ce ere ee ee TSE a en Stee on eae RS 451 Southlake Boulevard + Richmond, VA, USA 23236-3091 Phone 804-794-2887 » Fax 804-379-2118 + E-Mail superrad@rich sreolls.com Parr ere) ete Mer eRe eS eter oneal cuemicat processing IM CIRCLE 214HP. Butterfly © 10 20 30 40 50 60 70 80 90 100 Percent Open Fig. 2. Terminal pressure drop ratio (X;) of control valves. factor, Fe; from Equation 1 F, =0.96-028 | Hr Fj =0.955 2, Calculate the terminal pressure drop, Apr, from Equation 2, using an inital estimate for F, of 0.72 (from Fig. 1) for a seg ‘ment ball vaive operating at 89% open: Apr =0.72"[(42+14.7)-(0.955\1.1)] ‘py =28.85psid 3, Because Ap (20 psid) is less than Apr (28.85 psid), flow Will not be choked. Proceed to Step 5. 4, Skipped per Step 3 5. Calculate the pressure drop at which cavitation damage is likely to begin (App) in Equation 3, using the values of Rand S for segment ball valves from Table 3 and the same value of F as used in Step 2, Pp ~orjor2y, i 2 eran [(42+14.7)-1.4] pp =22.6psid 6. Because Ap (20 psid) is less than App (22.6 psi), there is ‘no potential for cavitation damage. 7. Caloulate the required C, using Equation 4: CLT E ULERY LTE TIAL Le Whether you need vacuum, pressure or tubing pumps, Barnant builds chemical and corrosion-resistant models for most applications or pumping systems. Our pumps can handle virtually any medium: * Solvents + Etchants Inks * Acids + Toxins * Caustics * Corrosive Gases + Alkalines and More. » Call us today. Let us provide you with the oN S180 9001 > CUSTOM DESIGN 1.0 ©, =(1.0)630) 5 Cy =140.9 A in segment ball valve is selected from Table 4 that will be slightly less than 75% open, ignoring the effect of pipe reducers, which means that in actu: al operation with reducers attached to the valve will be less than 80% open. It can be seen in Fig. 1 that when a segment ball valve's opening decreas- es, the value of F increases, meaning that the intial estimate of F, (80% open) was on the conservative side and there is no need to repeal the Apy and App calculations, Gas flow, aerodynamic noise ‘As with liquids, when a gas flow stream in ‘a control valve reaches the vena contrac- ta, the flow velocity increases to a maxi- mum. Assuming a constant pressure | upstream of the valve, p;, increasing the pressure drop by decreasing the down- stream pressure, p2, results in incroasod flow through the valve until a point is ' ei perfect solution. PRIVATE LABEL Barnant Company 28W092 Commercial Avenue * Barrington, IL 60010 Toll Free: 800-637-3739 - 847-381-7050 Fax: 847-381-7053 + Web Sit: http://www.barnant.com cORCLE 215, reached where the velocity at the vena contracta becomes sonic. Any further increase in pressure drop has no effect on increasing flow. Hence, the flow has become choked, Unlike liquids where the formulas use pressure drop, Ap, for gases it is moreEE ee Convenient o use the pressure drop ratio, x. Pr © The pressure drop ratio at which flow becomes choked when the medium is air is defined as the terminal pressure drop ratio, x, and is dependent on valve geometry and the degree of opening. Control valve manufacturers test their valves to determine the values of xr and publish the results in their valve sizing literature. Some valve manufacturers use other factors (Cy, C,) for the same thing and are related by the following formula: 0.84? = CF /1,600 =x; Because x; is determined by tests with air, a correction called the ratio of specific heats factor, Fy, (Table 6), is added to account for gases with a sonic velocity that differs from that of air. For any gas, flow will choke when: X=FiXy, Sizing method for gases Constants Ks through Kg (Table 7) are included to make the for- mulas readily adaptable to either customary U.S. or Sil. units. “The values of p,, Pe and T; must be expressed in absolute units. 4. Caleulate the pressure drop ratio, x, using Equation 5, xe) Pt 2, Determine the choked pressure drop ratio from Equation 7. © a Fr Because xr is a function of valve opening that is yet to be determined, it will be necessary to make an inal estimate of valve opening, A good initial estimate is the value of xy at 80% open, Equation 6 is used when a manufacturer publishes @ fac- tor other than x7. 3.2 Fy Xp flow willbe choked and FX i used in place of x in the following calculations: "a. Calculate the expansion factor, : x ® Sor Under choked flow conditions when Fx is substituted for x, Equation 8 reduces to Y = 215. '5, Calculate C, from one of the following equations: For gas using volumetric flow units: Y= y= Se ® y ico For gas using mass ow units: (10) cy maw Kap Vx For vapor or steam using mass flow units: w Qe KV Derr ‘The compressiilly factor, Z, appears in the above equations where density is implied from specific gravity or molecular weight along with pressure and temperature and compensates for the degree to which a particular gas deviates from perfect gas behavior. Using the assumption that Z = 1.0 is usually satisfacto- 1 for valve sizing purposes for most industrial gases at the pres- sures and temperatures at which they are normally encountered in chemical plants, ‘An appropriate valve size is selected from valve manufactur- ers tables of C, versus opening (Tables 4 and 5 are examples of this). The goal is to select a vaivo that will be as far open as pos- sible without exceeding 80% at the maximum design flow. For valves installed with reducers, selecting a valve with an ‘opening not exceeding 75% according to the tables will ensure ay TABLE 7. NUMERICAL CONSTANTS FOR GAS SIZING EQUATIONS. ee ae ky 1,360 sch psi a 47 Nou bar *« mr Ky 193° ote psi R 48 igh bar *« Ks 633 b/hr psi Ibfoutt 273 koh bar kgloum Ke 404 psi 613 bar ‘TABLE 8. PIPE DIAMETER CORRECTION, PDC, FOR AERODYNAMIC NOISE CALCULATION. Downstream pipe PDC, dBA. diameter, in +40 ° 12 1.0 16 4.0 20 70 isa a aa TABLE 9. PIPE SCHEDULE CORRECTION, PSC, FOR AERODYNAMIC NOISE CALCULATION. Downstream pipe PSC, dBA, schedule 40 0 80 “4.0 160 10.0 cuemicat processineMii i TABLE 10. VALVE STYLE CORRECTION, VSC, FOR AERODYNAMIC NOISE CALCULATION. | Vabestyies 0 _VSGrdBA\ Globe 0 Eccentric rotary plug ea Ball ean Butterfly ace that it wil not be in excess of 0% open in actual operation with the reducer effect. ‘After selecting a valve size and determining the percent of, ‘opening corresponding to the calculated ,, the x; at that open ing is determined. If the actual x; is less than the value used in the preceding steps, repeat the process from Step 2, because the value used for F, xr is larger than its real value and there is the possibilty of overtooking choked flow. Even ifthe smaller and correct valuo of xy does not predict choked flow, the new value of x; should be used to repeat the calculations, because calculated Cy will increase with decreased x and the valve opening in actual operation will be greater than that just predicted. Ifthe actual x; is greater than the value used inthe calculation, you have made a conservative calculation. 6, Calculate the noise level: ert «4g, +18040) 20 Kg +PDC+PSC-+VSC (12) Tables 8 through 10 contain values for POC, PSC and VSG. ‘Aerodynamic noise levels of 110 dBA or greater when calculated for uninsulated Schedule 40 pipe will most likely result in severe vibration damage. Also, most plant standards limit noise to 80 GBA orless. C, and the noise level should also be calculated at the mint mum design flow. Making the initial estimate of valve opening for the purpose of obtaining a starting value for x may require some tial and error. Estimating an opening of 40% is @ good starting Point. Itis worth noting that p, and Ap will most likely be higher at the minimum design rate than they are at the maximum design flowrate. Gas sizing example The stated problem is to select a properly sized equal-percent- ‘age globe control valve forthe following process concitions: ay Bolt-On Heating Jackets Swe ei Jeti Aig ae For four rious years at sea the works largest ulphurhander. Freeport MeMeRan, Inc, has depended on our Boon eating ott more than S milion tons of ma oTrace, and 30 ContoHeat jackets have main ed process temperature onthe piping, vales and Tings withou een one reaintenance problem. The éhoice was easy. Our Bol-On system eliminates the chance for cross contamination witho sacrificing ‘ermal performance. andif thats not enough, Conirtteat and CntroTrace were able to save more than 25 in investment costs versus tration ack ted pipe and aicaed valve Jackel ur Bolton refered tech iat processes. Pease cll us at 03 or emalus at salesecsheat.com, Rese ios ae cone 225, EE 50 rium riow annuatFluid: nitrogen; Pipe size: 3-in Schedule 40, Maximum design flow, q: 190,000 sefh; Upstream pressure, py: 105 psig Local atmospheric pressure: 14.7 pia; Upstream temperature, T,: 100°F; Gompressibiity factor, Z: 1.0, Pressure drop, Ap: 40 psid; Specific gravity, Ga; 0.97; Ratio of specifics neats factor, Fy: 1.0. 1. Calculate the pressure drop ratio, x, from Equation 5. ae 4054147 x=033 2, Caleulate the choked pressure drop ratio, Fy, xr, using Equation 7. An intial estimate for x; of 0.65 1s used for a globe valve operating at 80% open (Fig. 2) Fry = (1.0)(0.65) Fir = 0.65, 8, Because x < F, xr (0.33 < 0.65), flow will not be choked. Therefore the actual value of x = 0.33 is used in the following calculations. 4, Calculate the expansion factor, Y, using Equation 8: ya1- 038 (3)(0.65) Y=083 5. Caleulate the required Cy using Equation 9 for volumetric flow units because it corresponds tothe given process data: 130,000 _|(0.97)(100+460)(1.0) 7, 360(105 +14.7)(0.83) \ 0.33 cy =39 From Table §, it can be seen that a 3:in globe valve will be about 60% open and a 2-in globe valve will be about 75% open. The 2-in valve is selected because itis the closest to 80% open. Fig. 2 shows that x; at the actual opening of about 75% is for all practical purposes the same as the 80% open value of 0.65 that was used in the calculations. 6. Calculate the noise level using Equation 12: SPL=14lo9(38) +18l09(105+14.7)+ 20 og SEZ 404+0.0+0.0 SB +T47 SPL=85dBA To receive a NELPROF™ computer program, a “Flow Control Manual” and a “Control Valve Sizing Coetticients” booklet— Neles Controls Inc., Shrewsbury, MA. cmcte as 1By Jon F. 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