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ENGINEERING INFORMATION 12 flow data flow factor and orifice size Importance of properly sizing valves Conditions to be known The formulas necessary to determine the Kv are quite complicated and for that It is important to properly size a valve. In general, we must obtain as many of the reason ASCO/JOUCOMATIC has devel- There are undesirable affects in both conditions surrounding the application as oped a series of flow graphs which reduce undersizing and oversizing. possible. that problem to one of a simple multiplica- tion or division. Undersizing may result in: Flow required in cubic metres per hour All flow calculation work of a medium has (m3/h) as used for liquids, Normal Cubic been simplified to a basic formula: 1) inability to pass desired flow requirements Metre per hour (nm3/h) as used for gases, 2) flashing of liquids to vapours on the or Kilogram per hour (kg/h) as used for Flow required: Q outlet size of the valve steam. This can be obtained by merely Kv = --------------------- 3) lowering the outlet pressure asking the customer's requirements or Graph factors: Fgm, Fsg, Fgl 4) creating a substantial pressure loss in from nameplates on pumping equipment, a piping system boiler room charts or calculations. The graph factors Fgm, Fsg, Fgl can be easily picked out by aligning known pres- Oversizing may result in: Inlet Pressure (p1) - This is usually known sure conditions on the graphs I to X on the by knowing the source of the supply or following pages (for calculations see next 1) unnecessary cost in oversized equipment readily obtained by placing a gauge near page). 2) variable flow through the valve or er- the valve inlet. ratic control of the flow The tables below can be used to estimate 3) shorter life of some valve designs Outlet Pressure (p2) - This can be ob- a Kv if the orifice size is known or relate through oscillating of internal parts tained by gauge observations but usually the approximate orifice size if the Kv is caused by lack of flow to maintain re- is tied in with specifications regarding known. The chart is based on the ASCO/ quired internal pressure differentials allowable system pressure drop. If we JOUCOMATIC design of in-line globe type 4) erratic operation of some designs such know the inlet pressure and the pressure valves. The flow charts must be used for as failure to shift position due to lack of drop, we, of course, know the outlet pres- precise sizing and converting Kv factors required flow in 3- and 4-way valves sure. to actual flow terms and the catalogue 5) erosion or wire drawing of seats in sheet must be consulted for the actual Kv some designs because they operate at Pressure Drop (∆p) - In large or compli- of a particular valve. nearly closed position cated systems, it is desirable to keep the pressure drop across a valve to a mini- Definition of Kv mum and often the customer will have definite specifications concerning the fac- The flow coefficient Kv in cubic metres per tor. Of course, if the valve is discharging Approx. Approx. orifice approx. Kv orifice approx. Kv hour or litres per minute is a special volu- to atmosphere, the pressure drop is equal size size metric flow rate (capacity) through a valve to the inlet pressure when dealing with (mm) (m3/h) (l/min) (mm) (m3/h)(l/min) at a specified travel and at the following liquids. With gases and steam, although conditions: the valve may be discharging to atmos- 0.8 0,02 0,33 13 3 50,0 phere, when sizing a valve, only 50 per- - the static pressure loss (∆pKv) across cent of the inlet pressure can be used for 16 4 66,7 the valve is 105 Pa (1 bar) the pressure drop used in the formulas 1.2 0,05 0,83 - the fluid is water within a temperature (commonly called critical pressure drop). 18 4,5 75,0 range of 278 K to 313 K (5°C to 40°C) In all other cases the pressure drop is, the 1.6 0,08 1,33 19 6,5 108 - the unit of the volumetric flow rate is the difference between inlet and outlet pres- cubic metre per hour or liters per minute sures. 25 11 183 2.4 0,17 2,83 The value of Kv can be obtained from test Note: It often is difficult to understand the 32 15 250 results with the help of the following equa- meaning of the term "minimum operating tion: pressure differential" (see page V1210). 3.2 0,26 4,33 38 22 366 Certain pilot operated valves operate by ∆pKv . ρ Kv = Q differential pressures created internally 51 41 683 ∆p . ρw by "pilot" and "bleed" arrangements. This 3.6 0,31 5,17 where: differential is measured as the difference 64 51 850 Q is the measured volumetric flow rate in between inlet and outlet conditions on all 4.8 0,45 7,50 cubic metres per hour or liters per minute valve construction. If pressure conditions 76 86 1433 ∆pKv is the static pressure loss of 105 Pa (see are not known, but only flow information, we can use the graphs or formulas to 6.4 0,60 10,0 80 99 1650 above) ∆p is the measured static pressure loss solve the resulting pressure drop. If the drop is less than assigned minimum 100 150 2500 across the valve in pascals 1,5 25,0 differential, the valve is oversized. In these 8 ρ is the density of the fluid in kilograms per 125 264 4400 cubic metre situations, offer a valve with a lower min. ρw is the density of water (see above) in operating pressure differential, or go to a 9 1,7 28,3 150 383 6375 kilograms per cubic metre (according to smaller sized valve with a closer Kv. IEC 534) All leaflets are available on: www.ascojoucomatic.com V1215-GB-R2 12 ENGINEERING INFORMATION SECTION 12 SAMPLE PROBLEMS LIQUIDS (tables I and III) AIR AND GASES (tables I and IV - VII) STEAM (tables VIII - X) To find Kv: What Kv is required to pass To find Kv: A valve is required to pass 14 To find Kv: A valve is required to pass 25 22 litres of oil per minute with a specific nm3/h at an inlet pressure of 4 bar and a kg/h of saturated steam at an inlet pres- gravity of 0,9 and a pressure drop of 1,5 pressure drop (∆p) of 0,5 bar. sure of 1 bar and a ∆p of 0,2 bar. bar? Find the Kv if the medium is carbon diox- What is the Kv? ide. The viscosity is less than 9° Engler. Solution: The formula is: Solution: Refer to the 1-10 bar graph. Solution: Refer to the appropriate Steam The formula used is: Graph. Use the formula: Q (m3 /h) Q (Nm3 /h) Q (kg/h) Kv (m3 /h) = Kv (Nm3 /h) = Kv (m3 /h) = Fgm . Fsg Fgm . Fsg Fgm Q (m3 /h) Q (Nm3 /h) Q (kg/h) Kv (l/min) = Kv (Nl/min) = Kv (l/min) = Fgl . Fsg Fgl . Fsg Fgl Locate Fgm at the intersection of 4 bar Locate Fg on graph corresponding to 1 To find Fg, use the Liquid Flow Graph. inlet pressure and 0,5 bar pressure drop bar inlet pressure and a ∆p of 0,2 bar ∆p (along curve). Read down to Fgm = (along curve). The Fgm factor is that which corresponds 43,5. Fgm = 13,8 and the Fgl = 0,83 to a pressure drop of 1,5 bar and equals Fgl factor is 2,61 1,25. The Fgl factor is 0,075. Locate Fsg corresponding to specific grav- ity of carbon dioxide (= 1,5) on Fsg Chart. The Fsg factor can be obtained from the Fsg = 0,81. Fsg chart and is that which corresponds to 0,09 specific gravity and equals 1,05. Insert values into formula : Insert values into formula : Insert values into formula : −3 Q (Nm3 /h) 14 Q (kg/h) Kv = 60.22.10 = 1 m3 /h Kv = = = 0, 4 Nm3 /h Kv = = 25 = 1, 8 m3 /h 1, 25.1, 05 Fgm . Fsg 43, 5.0, 81 Fgm 13, 8 −3 Q (Nm3 /h) 14 Q (kg/h) Kv = 60.22.10 = 16, 7 l/min Kv = = = 6, 62 Nl/min Kv = = 25 = 30 l/min 0, 075.1, 05 Fgl . Fsg 2, 61.0, 81 Fgl 0, 83 Formula for liquids Formula for gases (with temperature correction) ∆p ∆p(2P1 − ∆p) Q (m3 /h) = Kv Q (Nm3 /h) = Kv.18, 9 ⋅ 293 S.G. (S.G.) (273 + t 2 ) ∆p ∆p(2P1 − ∆p) Q (dm3 /min) = Kv1 Q (Ndm3 /h) = Kv.18, 9 ⋅ 293 S.G. (S.G.) (273 + t 2 ) S.G. : specific gravity related to water for liquids and to air for gases t2 : fluid temperature (in °C) All leaflets are available on: www.ascojoucomatic.com V1215-2 TECHNICAL INFORMATION SECTION 12 Table I : Calculation factor Fsg Table II : Calculation factor Ft for temperature correction Factor Fsg Factor Ft FLUID TEMPERATURE t2 (°C) Specific gravity (S.G.) The correction for temperature in OTHER GRAVITIES OTHER TEMPERATURES the range of -7°C to 65°C is very small and, therefore, can be ignored specific gravity (for 1 bar absolute and 15°C) in ordinary applications Table III : Flow calculation Fgm and Fgl for liquids 0,54 Factor Fgm (m3/h) Factor Fgl (l/min) 0,48 0,42 0,36 0,30 0,24 0,18 0,12 0,06 0,03 0 Pressure drop ∆p (bar) All leaflets are available on: www.ascojoucomatic.com V1215-3 12 ENGINEERING INFORMATION SECTION 12 Table IV : Flow calculation Fgm and Fgl for air/gas Pressure drop ∆p (bar) Inlet pressure 0,01 to 0,1 bar gauge) Limiting flow curve - Do not read beyond this curve Factor Fgm (m3/h) 0,17 0,18 0,21 0,24 0,27 0,30 0,36 0,42 0,48 0,54 Factor Fgl (l/min) Table V : Flow calculation Fgm and Fgl for air/gas Pressure drop ∆p (bar) Limiting flow curve - Do not read Inlet pressure of 0,1 to 1 bar (gauge) beyond this curve Factor Fgm (m3/h) 0,24 0,30 0,36 0,42 0,48 0,6 0,72 0,84 0,96 1,08 1,2 1,32 1,44 1,56 1,68 1,8 1,92 2,04 0,54 0,66 0,78 0,9 1,02 1,14 1,26 1,38 1,5 1,62 1,74 1,86 1,98 2,1 Factor Fgl (l/min) All leaflets are available on: www.ascojoucomatic.com V1215-4 ENGINEERING INFORMATION SECTION 12 Table VI : Flow calculation Fgm and Fgl for air/gas Pressure drop ∆p (bar) Inlet pressure of 1 to 10 bar (gauge) Limiting flow curve - Dot not read beyond this curve Factor Fgm (m3/h) 0,6 1,2 1,8 2,4 3,0 3,6 4,2 4,8 5,4 6 6,6 7,2 7,8 8,4 9 9,6 1,02 1,08 Factor Fgl (l/min) Table VII : Flow calculation Fgm and Fgl for air/gas Pressure drop ∆p (bar) Inlet pressure 10 to 100 bar (gauge) Limiting flow curve - Do not read beyond this curve Factor Fgm (m3/h) 0 6 12 18 24 30 36 42 48 54 60 66 72 78 84 90 96 102 Factor Fgl (l/min) All leaflets are available on: www.ascojoucomatic.com V1215-5 12 ENGINEERING INFORMATION SECTION 12 Table VIII : Flow calculation Fgm and Fgl for steam Pressure drop ∆p (bar) Inlet pressure of 0,1 to 1 bar (gauge) Limiting flow curve - Do not read beyond this curve Factor Fgm (m3/h) 0,18 0,3 0,42 0,54 0,66 0,78 0,9 1,02 1,08 1,14 1,2 1,26 1,32 1,38 1,44 1,5 1,56 1,62 1,68 0,24 0,36 0,48 0,6 0,72 0,84 0,96 Factor Fgl (l/min) Table IX : Flow calculation Fgm and Fgl for steam Pressure drop ∆p (bar) Inlet pressure 1 to 10 bar (gauge) Limiting flow curve - Do not read beyond this curve Factor Fgm (m3/h) 0 0,6 1,2 1,8 2,4 3,0 3,6 4,2 4,8 5,4 6,0 6,6 7,2 7,8 8,4 9,6 Factor Fgl (l/min) Table X : Flow calculation Fgm and Fgl for steam Pressure drop ∆p (bar) Inlet pressure of 10 to 100 bar (gauge) Limiting flow curve - Do not read beyond this curve Factor Fgm (m3/h) 0 6 12 18 24 30 36 42 48 54 60 66 72 78 84 Factor Fgl (l/min) All leaflets are available on: www.ascojoucomatic.com V1215-6 ENGINEERING INFORMATION SECTION 12 ADDITIONAL FLOW FORMULAS AND Definition of Kv- (or Cv-) coefficient expressed in units volume "A" per time PHYSICAL DATA unit "B", that will pass through a valve Valve flow coefficient Kv (or Cv) is the with a pressure drop equal to pressure flow of water (specific gravity = 1), unit "C". (See table below) Kv- and Cv-conversion table units symbol conversion formulas volume "A" time "B" press. "C" litre min. bar Kvl 16,7 Kv = 17,3 Cve = 14,4 Cv cubic metre hour bar Kv 0,06 Kvl = 1,04 Cve = 0,865 Cv imp. gallon min. psi Cve 0,058 Kvl = 0,963 Kv = 0,833 Cv U.S. gallon min. psi Cv 0,069 Kvl = 1,16 Kv = 1,2 Cve Flow calculation Air and Gases Steam and vapours (e.g. refrigerants) General: Pressure drop values for which For steam: no curves are shown, may be determined Fgm = 18, 9 ∆p (2p1 − ∆p) (m3/h) by interpolation in the graphs. However, Fgm = 15, 83 ∆p(2P1 − ∆P) (m3/h) Fgl = 1, 13 ∆p (2p1 − ∆p) (l/min) more accurate results can be obtained for Fgl = 0, 95 ∆p(2P1 − ∆P) (l/min) the calculation of the required values by using the following equations (on which Example: ∆p = 0,4 bar; p1 = 3 bar the flow graphs are based): gauge or 4,013 bar absolute. Example : ∆P = 7 bar, P1 = 40 bar or 41,013 bar abs. p1 = absolute inlet pressure (bar) = Calculation: gauge pressure plus atmospheric Calculation : Fgm = 18, 9 0, 4(8, 026 − 0, 4) = 33 m3 /h pressure of 1,013 bar Fgmgm = 15, 83 (82, 026 − 7) 7) = 363 m3 /h F = 15, 83 7 7(82, 026 − = 363 m /h 3 p2 = absolute outlet pressure (bar) = Fgl = 1, 13 0, 4(8, 026 − 0, 4) = 1, 97 l/min gauge pressure plus atmospheric pressure of 1,013 bar Note: The gas equations only apply accu- gl 0, 0, 7 7(82, 026 − = 21, 8 8 l/ min FglF= = 9595 (82, 026 − 7) 7) = 21,l/ min ∆p = p1 - p2 = pressure drop across the rately to a medium temperature of 20°C valve (bar) (for the purpose of this catalogue the t = 0°C Note 1: The steam formulas apply to standard cubic meter nm3 has been de- saturated steam. For superheated steam fined at 20°C and 1,013 bar absolute or Note: In most systems it is desirable to a correction factor is required. 760 mm mercury). keep the pressure drop to a minimum. If Consult your ASCO/JOUCOMATIC sup- At a different temperature (= t2 °C) the necessary - in case of liquids - the pres- plier. determined Kv1-value must be adjusted sure drop may equal the total inlet (gauge) by the following correction factor. pressure. This also applies to air, gases Note 2: For vapours (e.g. Freon) various and steam up to 1,013 bar inlet (gauge) other factors have to be considered. pressure but for these media never us a Ft = 293 ∆p greater than 50% of the absolute inlet 273 + t 2 pressure because excessive pressure drops will cause an irregular flow. If∆p is Specific gravity of various liquids at not specified and this information is needed 20°C (related to water at 4°C) to size the valve, a rule of thumb is to take Specific gravity of various gases (at 10% of the inlet pressure as pressure Alcohol, Ethyl 0,79 20°C and atm. pressure and related to Bezene 0,88 drop. air) Carbon-tetrachloride 1,589 Castor Oil 0,95 Liquids Acetylene 0,91 Fuel Oil no. 1 0,83 Air 1,000 Fgm = ∆p 3 (m /h) Fuel Oil no. 2 0,84 Ammonia 0,596 Fuel Oil no. 3 0,89 et Butane 2,067 Fuel Oil no. 4 0,91 Fuel Oil no. 5 0,95 Carbon-dioxide 1,53 Fgl = 0, 06 ∆p (l/min) Chloride 2,486 Fuel Oil no. 6 0,99 Ethane 1,05 Gasoline (petrol) 0,75 to 0,78 Example: for ∆p = 1,7 bar will be found Ethyl-chloride 2,26 Glycerine 1,26 Fgm = 1,3 (m3/h) and Fgl = 0,08 (l/min) Helium 0,138 Linseed Oil 0,94 Methane 0,554 Olive Oil 0,98 Methyl-chloride 1,785 Note: if the medium viscosity is higher Turpentine 0,862 Nitrogen 0,971 than 300 SSU (approx. 9°E), the deter- Water 1,000 Oxygen 1,105 mined Kv-value must be adjusted. Con- Propane 1,56 sult your ASCO/JOUCOMATIC supplier. Kv1 Sulphur-dioxide 2,264 The actual flow factor is: Kv 2 = Ft All leaflets are available on: www.ascojoucomatic.com V1215-7 12 ENGINEERING INFORMATION SECTION 12 All leaflets are available on: www.ascojoucomatic.com V1215-8

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posted: | 6/9/2010 |

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