SPI Sizing Equations

March 27, 2018 | Author: nwabukingz | Category: Statistical Mechanics, Quantity, Materials Science, Physical Sciences, Science


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SmartPlant InstrumentationSizing Equations Version 2013 March 2013 DSPI2-PE-200008B Copyright Copyright © 1995-2013 Intergraph Corporation. All Rights Reserved. Including software, file formats, and audiovisual displays; may be used pursuant to applicable software license agreement; contains confidential and proprietary information of Intergraph and/or third parties which is protected by copyright law, trade secret law, and international treaty, and may not be provided or otherwise made available without proper authorization. Restricted Rights Legend Use, duplication, or disclosure by the government is subject to restrictions as set forth below. 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Contents Preface .................................................................................................................................3 1 2 Nomenclature .....................................................................................................4 Control Valve Sizing..........................................................................................7 2.1 2.2 Liquid, Water............................................................................................................ 7 Gas, Steam ................................................................................................................ 9 3 Noise Calculation....................................................................................................11 3.1 3.2 3.3 3.4 Hydrodynamic Noise: Masoneilan Standard .......................................................... 11 Hydrodynamic Noise: IEC Standard ...................................................................... 11 Aerodynamic Noise: ISA Standard ........................................................................ 14 Aerodynamic Noise: IEC Standard ........................................................................ 17 4 Flow Meter Sizing ..................................................................................................20 4.1 4.2 4.3 Calculation of Beta Ratio ....................................................................................... 20 Calculation of Flow rate ......................................................................................... 22 Calculation of Differential range ............................................................................ 23 5 Restriction Device Sizing .......................................................................................26 5.1 5.2 5.3 Calculation of Beta Ratio ....................................................................................... 26 Calculation of Flow Rate ........................................................................................ 30 Calculation of Pressure Loss .................................................................................. 33 6 Relief Valve Sizing .................................................................................................35 6.1 6.2 6.3 6.4 6.5 Blocked Flow: Liquid Relief .................................................................................. 35 Blocked Flow: Gas Relief ...................................................................................... 36 Blocked Flow: Steam Relief ................................................................................... 38 Fire Case: Gas Expansion....................................................................................... 39 Fire Case: Liquid Filled Vessel .............................................................................. 40 7 Thermowell Calculation .........................................................................................42 Kf 7.1.1 7.1.2 7.1.3 Define Initial Value of Constant:................................................................... 42 Calculate Ratio of Frequency at Fluid Temperature to Frequency at 70 °F: ....... 42 Calculate Maximum Thermowell Length Based on Vibration Limitation: ......... 42 7.1.4 Define 7.1.5 Paragraphs 7.1.3 - 7.1.4 are repeated until current 7.1.6 7.1.7 7.1.8 43 Calculate Natural Frequency for L=L1 ................................................................ 43 Calculate Wake Frequency .................................................................................. 43 Calculate Magnification Factor for L1(R  0.8)................................................... 44 Kf Constant: ............................................................................................ 42 Kf value equals previous K f value K 7.1.9 Define K1 , K 2 , 3 constants .............................................................................. 44 7.1.10 Calculate Maximum Allowable Static Gage Pressure....................................... 44 7.1.11 Calculate Maximum thermowell length based on Steady State Stress Consideration 44 7.2.1 Calculate Mounting factor ................................................................................... 45 7.2.2 Calculate Reynolds number ................................................................................. 45 7.2.3 Calculate Strouhal number ................................................................................... 45 7.2.4 Calculate Natural frequency ................................................................................ 45 7.2.5 Calculate Vortex shedding frequency .................................................................. 46 Sizing Equations 1 ......... 58 Segmental Orifice ............................2...........................................................................7 9............................................................ 54 Flow Meter Sizing ...........2............................................4 9.....................................3 8.........................................................1 9............................................................... 53 8 Standards....10 7..............................................................6 7.......................................................... 54 Relief Valve Sizing.....................3 9..........................................7........................................2............ 57 Lo-Loss Tube .....................................................................................................Contents 7.......... 55 Thermowell Calculation ................9 7................................................................................................. 55 9 Appendix................... 52 Compliance with the Hydrostatic Pressure Limit requirement ........................................................................ 57 Venturi Tube.............5 9.............................. 47 Calculate miscellaneous parameters .........12 Calculate Scruton number ................... 57 Nozzle ...................54 8...................1 8....................... 7.................................2....................................... 54 Noise Calculation ..........5 Control Valve Sizing ..................................2..... 58 2 Sizing Equations .2................................. 57 Conical Entrance Orifice ..................................................................................9 Square Edge Orifice .................... 58 Eccentric Orifice........................ 58 Restriction Device ..........8 9..2 8.............................. 56 Quarter of Circle (Quadrant Edge) Orifice ..6 9.... 47 Calculate Frequency Limit ...................2 9.............................. 48 Compliance with the Static Stress Limit requirement ..........................56 9..................................................................................................................................................................................11 7................................................................................................................ 50 Compliance with the Dynamic Stress Limit requirement ............2............... 48 Compliance with the Frequency Limit requirement ............................4 8.........................................................8 7........................................................................................................ Preface This document is a reference for sizing equations used by the SmartPlant Instrumentation Calculation module. Send documentation comments or suggestions to PPMdoc@intergraph. Sizing Equations 3 .com. m Diameter at tip. dimensionless Piping geometry factor. dimensionless Liquid critical pressure ratio factor. bar Differential pressure at normal flow rate. dimensionless Valve style modifier. dimensionless Product of the liquid pressure recovery factor of a valve with attached fittings and the piping geometry factor. Hz Ring frequency. mm Nominal size of sensing element. Hz Wake frequency. psi-g Acoustical efficiency factor. US gallon per minute of water (60 ºF) at 1 psi pressure drop Coefficient for critical flow equation.1 1 A A1 Ad B b bs C CV Cr Cs Cp D D0 Dt D1 D2 Dd2 DP DPA DPloss DPn Dbh Dj d db d0 E Ef Eref Eps F F1 F2 Fd Fe FF Fk FL FLP Fp FR Fs fp fo fr fw Gf Nomenclature Nomenclature Required effective discharge area. dimensionless Heat Absorption Factor. or differential pressure transmitter range. 1/K. dimensionless Peak frequency. Hz Coincidence or Natural frequency. inch2 Root diameter of thermowell. m Valve inlet diameter or orifice (throat) diameter at flow temperature. bar Bleed/Vent hole diameter. in Fillet radius at root. inch2 Exposed surface area. m/s Internal pipe diameter (upstream) at flow temperature. dimensionless Environmental factor for thermowell calculation. mm Internal pipe diameter (upstream) at ambient temperature. m Pressure differential. mm Bore diameter of thermowell. bar Maximum allowable pressure drop. m Fillet radius at base for stepped thermowell. P1-P2 . dimensionless Coefficient for subcritical flow equation. mm Outlet internal diameter of the pipe. dimensionless Correction factor for steam quality. ratio of flow coefficient for a gas to that for a liquid at the same Reynolds number. m Orifice (throat) diameter at ambient temperature. m/s Speed of sound in pipe wall. dimensionless Speed of sound. m Discharge coefficient Valve flow coefficient. dimensionless Liquid pressure recovery factor. Pa-g Gas expansion factor.6 C). Hz Liquid specific gravity (ratio of density of liquid at flowing temperature to density of water at 15. mm Diameter of the jet. in Inlet internal diameter of the pipe. dimensionless 4 Sizing Equations . mm Outlet outside diameter of the pipe. mm Module of Elasticity at 70 °F. dimensionless Ratio of specific heats factor. dimensionless Reynolds number factor. dimensionless Environment Factor. dimensionless Reference value for module of elasticity. bar Pressure loss. dimensionless Number of flow passages . Cp/Cv dimensionless Effective velocity head coefficients. PVCC P2c. dimensionless Superheat steam correction factor. m Molecular weight. dimensionless Sizing Equations 5 . psi-g Upstream relieving pressure. Pa Total back pressure. in “A”-weighted sound level. dimensionless Combination capacity factor for rupture disk at the relief valve inlet. which is equal to the ratio of molecular weight of gas to the molecular weight of air). kg/m3 Density of the thermowell sensor at ambient temperature. dimensionless Density at flow conditions (upstream). dimensionless Correction factor for Napier equation. psi-g Critical flow throat pressure. dimensionless Correction factor for viscosity. J/kg Specific Heat Ratio. dimensionless Correction factor for overpressure. mm Heat of vaporization. dB Internal sound pressure level. dimensionless Segmental height at flow temperature.Kb1. % Set pressure. dimensionless Constant for thermowell calculation. bar Absolute pressure in vena contracts at subsonic and critical flow conditions. Pa Border pressures for different noise regimes.Kb2 KC KCD KCV Kb Kcc Kd Kf Kn Kp Ksh Kv Kw L La Lg Lpi Ls M MN MN0 Mj No Ns Nsc P PC PV PVC. kg/m3 Ratio of frequency at fluid temperature to frequency at 70 °F. psi-g Variable back pressure. bar Absolute vapor pressure of liquid at inlet temperature. dimensionless Effective coefficient of discharge. mm Segmental height at ambient temperature. dimensionless Maximum thermowell length or unsupported length of thermowell. kg/m3 Density of well material at ambient temperature. psi-a Limitation on Wake to Natural frequency ratio.K2. CV/d2 Valve cavitation index. dB(A) Correction for pipe Mach number. dimensionless Segmental radius at ambient temperature. dimensionless Mach number in valve outlet. dimensionless Mounting factor. mm Valve (pipe) Reynolds number. dimensionless Pipe Reynolds number at maximum and normal flow rate. dimensionless Strouhal number Scruton number Absolute static pressure. dimensionless Mach number in the jet. dimensionless Capacity correction factor due to back pressure. Pa Absolute outside pipe pressure. kg/m3 Density of pipe material. P2ce Pa Pb Pb1 Pb2 Pb3 Pcf Pover Ps Pup R RO ROm ROp ROs Rf RS0 Re Remax. P2b. dimensionless Correction factor for back pressure. dimensionless Cavitation index. dB Length of reduced-diameter shank for stepped thermowell. dimensionless Relative capacity. psi-a Overpressure. psi-g Constant back pressure. atomic mass units Mach number in pipe. bar Absolute thermodynamic critical pressure. psi-g Built-up back pressure. Ren Nomenclature Gas specific gravity (ratio of density of flowing gas to density of air with both at standard conditions.Preface1 Gg Hc HS HS0 HV K K1. m Ratio of pressure drop to absolute inlet pressure (DP/P 1). 10Wn/Wmax Water in steam. dimensionless Absolute (dynamic) viscosity. m Velocity. dB Vessel Wall Temperature. dimensionless First estimate Upstream conditions Downstream conditions Vena Contracta 6 Sizing Equations . Pa×s Kinematic viscosity. ft/s Flow rate. Wn Wa Wm. dimensionless Value of XT for valve-fitting assembly. centistokes Damping factor. K Thermodynamic critical temperature. dB Transmission loss across the pipe wall at the ring frequency.  Transmission loss. °R Ambient temperature.1 Nomenclature Rq S Sf SRF St T T1 T0 Tc TL TLfo TLfr Tw t U V W Wmax. Pa-g Fatigue endurance limit. dimensionless Compressibility at operating conditions. pe  Lf TLfp    ξ    Subscripts 0 1 2 VC Radius of upstream profile of Quarter of Circle (Quadrant Edge) orifice. Wms wall X XT XTP Z Zb p. kg/h Maximum and normal flow rate. dimensionless Pressure drop ratio factor. dimensionless Linear expansion coefficient of pipe and primary element material. 1/ºC Diameter ratio of orifice or throat and inside diameter of line Correction value for cavitating flow. kg/s Sound power. m/s Velocity. dB Acoustic efficiency factor. mm Maximum allowable working stress. K Gas temperature at relieving conditions. W Pipe wall thickness. W Stream powers. dB Transmission loss at coincidence frequency. dimensionless Compressibility at base conditions. °R Minimum tip thickness of thermowell. Pa-g Scale reading factor. dB Correction for ratio of peak frequency and coincidence frequency. %wt Flow temperature. 6 2.25 Calculate DPA : F  DPA   LP   FP  2  P1  FF PV  Sizing Equations 7 .1.  0.5  1  d 2 / D12 K b1  1  d / D1   2. FL  2.96  0.3 Calculate K C : P1  P2 P1  PV KC  2.5  4 Ki  K1  Kb1 W 27.28 2.2 Calculate :   10 6   / RO 2.5   FL  CV0  2     1  0.1.8 K 2   1  d 2 / D2 2  2 76000  FD  W   RO   FL  CV0  0.1.1.428571  1869048 .Preface2 2 Control Valve Sizing Control Valve Sizing 2.4 Calculate effective velocity head coefficients:  K1  0.7   K  CV0 2   FP   1  0.10 2 Calculate first value of CV: 2.1.1.1.1 Liquid.00214  d  1/ 2 Re : Re  2.00214 d 4    Calculate KCV : 1/ 2   KCV  FL FL 166667 .3   P1  P2   RO FP : Calculate 2.1 Calculate FF : PV PC FF  0.9 Kb 2  1   d / D2  4 K  K1  K 2  K b1  K b 2 CV0  2. Water 2.00214  d 4    0.3077143 Calculate FLP : FLP Calculate  Ki  FL 2  CV0 2   FL    1 4  0.1.1.1. 121  10  Az  6.  Az  0.2.11.67 Case 56  Re  40000 .Turbulent flow FR  1 2.1 FR  1 W CV  B1.11.1.99 2.2 2.11.Cavitation and Flashing A1 Cavitation case takes place if flashing case takes place.11.1.3 2470  Re  10200 3 2 FR  9184 .2. the following cases for different Re are possible 2.844539  10 2  Az 2  01708764 .4 10200  Re  20000 FR  0.1.1.3  FR  FP   P1  P2   RO .212891  10 3  Az 3  2.11.019  Re 0.2 Case A .1.43  10 2  Az  088 .97 20000  Re  30000 2.11 Control Valve Sizing Calculate FR .2. A2 Calculate new CV .3  FP   P1  P2   RO IEC standard: CV  8 Sizing Equations  P1  P2   RO FR  1 CV  B2 27.1.3  FR  FLP  ( P1  FF  PV )  RO Case B ( P1  P2  DPA ) .Transitional flow 56  Re  620 Az  Re/ 56  1 5 4 FR  6.2925969 Az  Re/ 620  1 620  Re  2470 3 2 FR  9.5 FR  0.1. A2.2.1 FR  0.2 W 27.1. At this stage.12 The condition P1  P2  DPA determines: 2.11.1.11.3  FLP  ( P1  FF  PV )  RO IEC standard: CV  W 27.2 P2  PV .684  10 2  Az  0.3  FR  W 27.1.7614 Az  Re/ 2470  1 2.3 Case Re  40000 . otherwise 27.2.11.1 ISA standard: W CV  A2.Laminar flow 2.082774  10  Az  2.98 30000  Re  40000 2.6 FR  0.1.1 Case Re  56 .2.Usual B1 ISA standard: B1.2 2.11. 2.  10  Az  5. 13 Control Valve Sizing Calculate the relative change in the CV: CV  2.2. CV0  CV Paragraphs 2.001.1 Calculate XT : X T  0.16 CV0  CV CV0 .1.1.3 2.5  K 2   1  d 2 / D2 2  2 Kb 2  1   d / D2  4 2 4 Ki  K1  Kb1 K  K1  K 2  K b1  K b 2 Calculate FK : FK  K / 14 . Define “Incipient” case.5  1  d 2 / D12 Kb1  1  d / D1   2. Calculate X : X   P1  P2  / P1 Calculate first value of CV: *d [in] CV0  KCD  d 2 2.13 are repeated until the relative change in the CV is less than 0.  (D2 / 2 ) 2  RO 2.6 Calculate FP:   K  CV0 2   FP    1  0. D2 [m] U2  W 31416 .3  FP  Eps  Fk  X TP  P1  RO Sizing Equations 9 .2. The “Incipient” case takes place if P1  P2  DPA and KC  KCV Calculate outlet pipe velocity: *W [kg/s].1.2.6 .Critical flow A1 Eps  0.2.667 .1.00241  d 4  The condition X  FK  X TP determines: 1 Case A .2 Gas.1.4 2.8 2  X T  X T  Ki  CV0  X TP  2   1 FP  0.7 Calculate XTP: 2.2. Steam 2.1.15 2.2.2 Calculate effective velocity head coefficients:  K1  0.Preface2 2.2.00214 d 4    1/ 2 2.84  FL2 2.14 2.2.2. A2 Calculate new CV  CV : W 27. 2.  ( D2 / 2) 2  RO2 Calculate outlet Max number: Cs2  8314  K  T / M MN  U 2 / Cs2 10 Sizing Equations .001.10 Paragraphs 2. 2.2.9 W 27. 3  FK  X TP Calculate new CV : Eps  1  B1 B2 CV  2.2. CV0  CV CV0 2.2.11 Calculate outlet pipe velocity: *W [kg/s].12 P2 P1 W 31416 .3  FP  Eps  X  P1  RO Calculate the relative change in CV: CV  CV0  CV . D2 [m] RO2  RO  U2  2.Usual X .2.2.2.2 Control Valve Sizing Case B ( X  FK  X TP ) .6 .9 are repeated until the relative change in CV is less than 0. 07 . Pv [Pa-a].1 Hydrodynamic Noise: Masoneilan Standard 3. W[kg/s] Calculate the downstream speed of sound: Cs2  3. P1 [Pa-a].2.1 Case DP  dPi .1 *D2 [m].1.2 Hydrodynamic Noise: IEC Standard 3.3 Case DP  dPC and P2  PV .07  PV   70.Cavitation Noise: La  10  log CV  20  log DP  30  log( wall )   DP   P  P  K CV    log 14.Preface3 Noise Calculation 3 Noise Calculation 3.dPC ) + 6] 3.3 2 Calculate Differential pressure ratio Xf  P1  P2 P1  Pv Sizing Equations 11 .2. Pc [Pa-a]. 3.[5  log(DP + 0.2.5   P2  0.2.2 Case dPi  DP  dPC .7 .2. 3.84  FL 3.1 Calculate dPi .1.4 Case P2  PV .07  PV  + 5  1 2 V  FL  K CV      70. P2 [Pa-a].1. dPC : dPi  KCV  ( P1  PV ) dPC  FL  ( P1  PV ) 2 3.5   P2  0. 3.2 8314  K  T1 M Calculate Characteristic pressure ratio X fz  0.Flow Noise: La  10  log CV  20  log DP  30  log( wall )  705 .2.Flashing Noise: There is no method to predict Noise for flashing case.1.1.1.2 Calculate hydrodynamic noise.5 5  1 2 V  FL  KCV      3.2.Incipient Cavitation Noise: La  10  log CV  20  log DP  30  log(wall )   DP   P  P  KCV    log 14. 5.5 Cp 31416 .2.1 Calculate Liquid critical pressure ratio factor Ff  0.2.8  log 1  X fz 1 X f .2 Calculate Differential pressure 3.2.2.0625 Xf X fz 12 Sizing Equations   1 X f  0.5.  Dd 2 Calculate Internal sound power level 3.2.1 Case Xf < Xfz .Cavitating flow 3.3 Calculate Lwi.2.2.96  0.5.5.2 Case P1  P2  FL   P1  FF  Pv  DP  P1  P2 2 P1  P2  FL   P1  FF  Pv  2 DP  FL   P1  FF  Pv  2 3.2.2.5.4 Calculate Ring frequency F fz  3.28  ( Pv / Pc ) 3.2.2.1 Case 3.2 Case Xf => Xfz .2. Lwi  120  10  log( E f )  10  log(W )  10  log( P1  P2 )  10  log( RO)  L f  180  X fz 0.2.5.2.5.2.Non-cavitating flow Lwi  120  10  log( E f )  10  log(W )  10  log( P1  P2)  10  log( RO) 3.2.3 Noise Calculation 3. 1TL 3  TL 3  10  log Dd 2 Dd 2 C p  RO p  wall 1. 12 Lwe5  Lwi 5  17. 12 Lwe 4  Lwi 4  17.7 C p  RO p  wall Cs 2  RO  Dd 2 1.0)  10 0.Preface3 Noise Calculation 3.37   10 0.37   10 0.8  Calculate External A-weighted sound pressure level   D  La  Lwae  10  log  31416 .1( Lwe 4 1. 12 Lwe 2  Lwi 2  17.1( Lwe 3 1.2.5  F  8000   fr    10  10  log  20  log   Cs 2  RO  Dd 2  8000  F fr     Lwi 5  Lwi  10  log16  2.37   10 0.1) 3.9 15 .6 Calculate Un-weighted external sound power levels TL1  10  10  log Lwe1 TL 2 TL 3 TL 4 TL5 3.1TL 4  TL 4  10  log Dd 2 Dd 2 C p  RO p  wall 1.9 15 .1TL 2  TL 2  10  log Dd 2 Dd 2 C p  RO p  wall 1.9 15 .1( Lwe1  3.5  F  2000   fr    10  10  log  20  log   Cs 2  RO  Dd 2  2000  F fr     Lwi 3  Lwi  10  log4  2.2.9 15 .5  F  4000   fr    10  10  log  20  log   Cs 2  RO  Dd 2  4000  F fr     Lwi 4  Lwi  10  log8  2.1TL1  TL1  10  log Dd 2 Dd 2 1.37   10 0.9 15 .  3   2  1   Dd 2   Sizing Equations 13 .2 )  10 0.2 )  10 0.1TL 5  TL5  10  log Dd 2 Dd 2 C p  RO p  wall Calculate External A-weighted sound power level  Lwae  10  log 10 0.1( Lwe 5 1.5 F  500   fr    20  log    500  F fr     Lwi1  Lwi  2.0)  10 0.2.5  F  1000   fr    10  10  log  20  log   Cs 2  RO  Dd 2 1000  F fr     Lwi 2  Lwi  10  log2  2. 12 Lwe 3  Lwi 3  17.1( Lwe 2  0.37   10 0. 12  Lwi1  17. 1416  RO2  D2 / 2  Cs 2  MN 2  U 2 / C s 2 3. P2 [Pa-a].3 Calculate Fd.3.3 Noise Calculation 3. Dj.3.3 Aerodynamic Noise: ISA Standard *D2 [m]. Mj: Fd  N o1/2 D j  0.FL2(P1-Pvcc)    = Pvcc/P2c  K    P  1  K 1  P2b  1     K P2ce = P1/(22) 3.(P1-P2)/FL2  K    PVCC  2  K 1    P1    K  1 P2c = P1 . P1 [Pa-a]. W[kg/s] *FL=FLP/Fp 3.3.0046  Fd CV  FL K 1    K   P 2    1 M j    1         K  1    P2      14 Sizing Equations 1/2 .2 Calculate the following pressures: Pvc = P1 .1 Calculate the downstream parameters: RO2  RO  P2 / P1 8314  K  T1 M W U2  2 3. 4.5  K 1         2 K   PVC   K   P1   1     RO   K  1   P1       W  U VC Wm  2 U MN  VC C sVC 1/ 2 2   0.P1  P2  P2 c P  TVC  T1  VC   P1  C sVC U VC  K 1     K   8314  K  TVC      M 0.0001  M 6j .3.2 M j C sVCC Dj Sizing Equations 15 .Preface3 Noise Calculation 3.3.6 Wa    Wm  FL f P  0.1 Regime I .4.  P1  PVCC  fP  0.2 Common calculations for II-V Regimes: TVCC  2T1 K 1  8314 KTVCC  C sVCC      M 0.4.0001 MN 3.3 Regime II .3.4 Now 5 regimes are possible: 3.P2 c  P2  Pvcc :   0.5  K 1         2 K    PVCC   K   P1  U VCC     1    RO   K  1   P1       2 W  U VCC Wms  2 1/ 2 3.3.6 F 2 L  P  P2  Wa    Wms  1 .2  U VC / D j 2 3. 3.35C sVCC fP  1/ 2 2 125 .3 fp>4fo   TLfp = 20log(fp/4fo) + 7.Pvcc  P2  P2 b :   0.0001  Mj 2 2 1/ 2 6.3.3.8    3.6 FL 2 2 Wa    Wms 0.TLfp 16 Sizing Equations .P2b  P2  P2 ce :   0.5664D2) Three cases are possible when calculating TLfp: 3. Dj M j  1  3.1 fpfo   TLfp = 20log(fo/fp) 3.3.3.3 Noise Calculation 3.6 FL 2 Wa    Wms 0.4.2 fo<fp4fo   TLfp = 13log(fp/fo) 3.6 F Wa    Wms 0.4 Regime III .3.0001  Mj 2 2 2 6.8 Calculate transmission loss: TL = TLfo .9 Calculate fo : fo = 5000/(12. Dj M j  1   3.4.8.8.3.3.6 Regime V .3.5 Calculate Lpi: :   L pi  10  log 8  10 8  Wa  RO2  Cs2 / D2 3.6 2  Calculate TLfo :  11  .7 3.P2 ce  P2  0 :  2    K 1    K  M j   22  1      K  1    0.0001  M 6j .4.8.3.3.35C sVCC fP  1/ 2 2 125 .2 M j C sVCC fP  Dj 2 L 3.5 Regime IV .  10 7  D23 Pa TLfo  10 log   2  1  D2 / 2   wall  ( P2  101325)  3. P1 [Pa-a]. d [m].4.0046  Fd CV  FL K 1     2   P1  K  M j    1       K  1    P2      1/2 Sizing Equations 17 .4 Aerodynamic Noise: IEC Standard 3. P2 [Pa-a].4.1416  C s 2  RO2  d 2 Calculate the following pressures: Pvc = P1 .1416  C s 2  RO2  D2 Cs 2  MN 0  3. W[kg/s] *FL=FLP/Fp Calculate the downstream parameters: RO2  RO  P2 / P1 8314  K  T1 M 4 W MN 2  2 3. Mj: Fd  N o1/2 D j  0.1 *D2 [m].2 4 W 3.Preface3 Noise Calculation 3.3 Calculate Fd.10 Calculate Lg :   1  L g  16 log 5 1.3.(P1-P2)/FL2  K    PVCC  2  K 1    P1    K  1 P2c = P1 .4. Dj.FL2(P1-Pvcc)    = Pvcc/P2c  K    P  1  K 1  P2b  1     K P2ce = P1/(22) 3.11       Calculate sound level La = 5 + Lpi +TL +Lg 3.3.3  10  P1  CV  FL 1    D2  2  P2 3. 4.4.4.1 Regime I .4 Regime III .5  K 1         2 K   PVC   K   P1   1     RO   K  1   P1       W  U VC Wm  2 U MN  VC C sVC 2   0.5 1/ 2 .2 M j C sVCC Dj 3.3 Noise Calculation 3.2 M j C sVCC fP  Dj L 18 Sizing Equations 2 0.4.6F 2 L  P  P2  Wa    Wms  1 .0001  M j6.4.2  U VC / D j 2 3.4 Now 5 regimes are possible: 3.3 Regime II .2 Common calculations for II-V Regimes: TVCC  2  T1 K 1  8314  K  TVCC  C sVCC      M 2 W  C sVCC Wms  2 3.P2 c  P2  Pvcc :   0.Pvcc  P2  P2 b :   0.6 F Wa    Wms 0.0001  M 6j .4.4.  P1  PVCC  fP  0.6 Wa    Wm  FL f P  0.0001 MN 3.P1  P2  P2 c P  TVC  T1  VC   P1   8314  K  TVC      M C sVC U VC  K 1     K  0.4.4. 4.6 Regime V .8.4.4.4.11 3.Preface3 Noise Calculation 3.4.4.35  C sVCC fP  1/ 2 2 125 .TLfp Calculate Lg :  1  Lg  16  log   1  M 2  3.10 Calculate fr and fo : fr = 5000/(3.3 fp>fr   TLfp = 20log(fp/fr) Calculate transmission loss: TL = TLfr .0001  Mj 2 2 6.4.12 Calculate sound level at the outside diameter of the pipe La0 = 5 + Lpi +TL +Lg Calculate sound level at a distance of 1 m from the pipe wall La  La 0  10  log D2  2 D2 Sizing Equations 19 .4. 6 FL 2 2 Wa    Wms 0.8.5 Calculate Lpi: :   L pi  10  log 8  10 8  Wa  RO2  Cs2 / D2 3.7 3.6 FL 2 Wa    Wms 0.4.P2 ce  P2  0 :  2    K 1     M j    22 K  1     K  1    0.  Dj  M j  1  3.9 3.  Dj  M j  1   3.5 Regime IV .4.8    3.1 fp<fo   TLfp = 20log(fo/fp) + 13log(fo/fr) 3.4.6 2  Calculate TLfr: 2 2  3  10 13  Cs 2  D2 Pa  TLfr  10  log   2  1  Cs 2  RO2 / 415   wall  101325  3.P2b  P2  P2 ce :   0.8.4.1416D2) fo = fr Cs2/1372 Three cases are possible when calculating TLfp: 3.4.35  C sVCC fP  1/ 2 2 125 .4.2 fofpfr   TLfp = 13log(fp/fr) 3.4.0001  Mj 2 2 2 1/ 2 6. 4 Calculate Differential pressure at normal flow rate: SRF  10  Wn / Wmax DPn  DP   SRF / 10 2 4.6.2 For Lo-Loss tube.1.   0  0. 4.0074  S t 4.256   0  0.1.4 For square-edge orifices specified in ISO 5167 – 2003 (flange tappings.25 4. Water.1. Eps  1 ). Venturi tubes and Nozzles: 2   K   Q K  Eps    K  1  where Q  4 1 0  2   4   1   0   Q K K 1     1   Q K     1  Q      0.5 For other Flow meters:  Eps  1  0.7.333  1145 . Fs=1): Fs  1  0.6.1.7.1 4.9675   0  4.35   0 20 Sizing Equations 4 8   ( DP n / P1 )  K 1 .3 Calculate Scale reading factor: 4. corner tappings.3 For Eccentric orifices:  Eps  1  01926 .  10 3  Wn Re n  D Calculate first estimate for  .   0  0.1.1.1.1.1.24   0  3.7.1. 4.1 For square edge orifice with 2½D & 8D pipe taps:   Eps  1  0.7   Wn2 0   8 4 2   4.25 4.  0574 .1 Calculation of Beta Ratio 4.  10  D  DPn  RO  Wn   123 0.351  0.1.2 For other Flow meters: 0.6 12732 .1 For Venturi tubes:   Wn2 0   7 4 2  .4 Flow Meter Sizing 4 Flow Meter Sizing 4.7.1.7 0  12 0 2 5 13    DP n P1   K 1 4.62   0 2 3 4   ( DP n / P1 )  K 1 4.41  0.6  10  D  DPn  RO  Wn  Calculate Gas expansion factor (for Liquid and Water.5 P1  DPn P1 4.5 Calculate Pipe Reynolds number at normal flow rate: 4.93   0  (1  ( P2 / P1 )1 / K ) 4 4.1.2 Calculate Internal pipe diameter at flow temperature:   D  D0  1   p T  T0  Calculate Correction factor for Steam quality (For Liquid.7. and radius tappings):   Eps  1  0.ratio.1. and Gas. 1.12 0   0 0. 5  .12.0001.8 Calculate Discharge coefficient.1. See 9 Appendix.13.1 4.11 4.1.13 Calculate Orifice diameter at ambient temperature.0   Paragraphs 4.ratio is less than 0. 4.1.2   1    Segmental orifices.2 1   pe For other orifices: d0  d  4.1. 4.55   Dbh d  2 1   pe  T  T0  For Quarter of Circle (Quadrant Edge) orifice. Nozzles.07571    0.05  W  1   4  n 0   2  D  Eps  Fs  C  DPn  RO    4.2.1.ratio: .12.68286   Sizing Equations 21 .1.1 For all Flow Meters except Segmental orifices: d  D 4.1.1.49  D0 Calculate Segmental height at flow temperature using the equation: 2     arccos 1  2  HS   2   1  2  HS    HS   HS       D  D   D  D           0. calculate Radius of upstream profile:  0.10 Calculate the relative change in   4.5 For further calculation of Segmental orifices.14 d  T  T0  1  0.06253 Rq  d      0. HS and HS0 are analogous to d and d0 respectively.Preface4 Flow Meter Sizing 4.1.1.2 4.12.12.1.13.ratio:  2847. Calculate Segmental radius: RS 0  0.10 are repeated until the relative change in the  .1 For Lo-Loss tube.4. and Venturi tubes: d0  4.7 .9 Calculate new  .2. Calculate Orifice diameter at flow temperature.1.1. 4.1. 4. 24   3  3.2.ratio.4 Flow Meter Sizing 4.1 For square edge orifice with 2½D & 8D pipe taps:      Eps  1  0.2.5.4 For square-edge orifices specified in ISO 5167 – 2003 (flange tappings.256   4  0. and radius tappings):   Eps  1  0.574    0. Nozzles.4.2.2. Eps  1 ).  0. corner tappings.5 Note: HS is calculated with the help of HS0 in accordance with paragraph 4. Venturi tubes and Nozzles: 2  K K  Q    Eps    K  1  where Q   1 4  2   4   1     Q K K 1   K 1  Q       1  Q      0.2.2.1 For all Flow Meters except Segmental orifices: 4.1 Calculate Segmental radius: RS 0  0.2. Fs=1): Fs  1  0.3.49  D0 4.3.4  d D 4. 4.2. Water.3.2.3 Calculate Orifice diameter at flow temperature.2.2 Calculate Internal pipe diameter at flow temperature:   D  D0  1   p T  T0  Calculate Correction factor for Steam quality (For Liquid.2 Calculate .62   4  ( DP / P1 )  K 1 4. Calculate Gas expansion factor (for Liquid and Water.2.2.ratio:   1    4.4.   2  0.5 P1  DP P1 4.2.333  1145 .3 For Eccentric orifices:   Eps  1  01926 .4.2 Calculation of Flow rate 4.351  0.2.2 For Lo-Loss tube. and Gas.2.1 4.93   8  (1  ( P2 / P1 )1 / K ) 22 Sizing Equations .5. and Venturi tubes:   d  d 0  1   pe  T  T0  4. 4.5 2     arccos 1  2  HS   2   1  2  HS    HS   HS       D  D   D  D           0.2 Segmental orifices.2 For other orifices:    d  d 0  1  0.5.1 For Lo-Loss tube.9675   2  4.2.5.2.4.2.0074  S t 4.2.7 5  12 13  DP P1  K 1 4. 4. 4.55   Dbh d 0   1   pe  T  T0  2 Calculate . 3.3 Calculate Orifice diameter at flow temperature.13. Nozzles.4. Water. Lo-loss tube.7 Calculate Pipe Reynolds number: *W0 [kg/s] Re  4.3.2.7 . 4.W 0  W Paragraphs 4.2.2.5 For other Flow meters:   Eps  1  0.3. Calculate new Flow rate: *W [kg/s] W  35124 .06253 Rq  d      0. For Quarter of Circle (Quadrant Edge) orifice. calculate Radius of upstream profile:  0.11 Calculate the relative change in Flow rate: W  4.9 12732 . Fs=1): Fs  1  0.68286   4.2.2.35   4  ( DP / P1 )  K 1 4.3. and Segmental orifices.2 For other orifices:    d  d 0  1  0.12 4.3.1 For Lo-Loss tube.2.8 4.2.1 4.5. and Gas.55   Dbh d 0   1   pe  T  T0  2 Sizing Equations 23 .3. Venturi tubes.2.11 are repeated until the relative change in the flow rate is less than 0.0074  S t 4. go to the paragraph 4.10 For Quarter of Circle (Quadrant Edge) orifices. and Venturi tubes:   d  d 0  1   pe  T  T0  4.13 W 0 W W0 .0001.2.107  10 4  Eps  d 2  4.2.3 Calculation of Differential range 4. 4. See 9 Appendix.6 Calculate first estimate for Flow rate: *W0 [kg/s] DP  RO 1 4 W0  2.  10 4  C  Fs  Eps  d 2  DP  RO 1 4 4.3.2.  10 3  W0 D Calculate Discharge coefficient.41  0.2 Calculate Internal pipe diameter at flow temperature:   D  D0  1   p T  T0  Calculate Correction factor for Steam quality (For Liquid.2.07571    0.Preface4 Flow Meter Sizing 4. 2 Segmental orifices.574    0.3.3.4 Calculate Pipe Reynolds number: *W [kg/s] Re  4.3.9675   2  4.5 For other Flow meters:   Eps  1  0.ratio:   1     arccos 1  2  HS   2   1  2  HS     D  D      HS  HS  2           D  D        0. and radius tappings):   Eps  1  0.8. 4.   2  0.  0.2 For Lo-Loss tube.3.7 First estimate for Differential range: DP0  P1 4.1 For all Flow Meters except Segmental orifices:  d D 4.4 For square-edge orifices specified in ISO 5167 – 2003 (flange tappings.5 P1  DP P1 4. 4.3 For Eccentric orifices:   Eps  1  01926 .3.1 For square edge orifice with 2½D & 8D pipe taps:      Eps  1  0.62   4  ( DP / P1 )  K 1 4. 4.333  1145 .93   8  (1  ( P2 / P1 )1 / K ) 4.3. See 9 Appendix.3.8.8 Calculate Gas expansion factor (for Liquid and Water.8.3.351  0. corner tappings.35   4  ( DP / P1 )  K 1 4.4 Flow Meter Sizing 4.9 Calculate new Differential range: *W [kg/s] 2   W  1 4   RO 1 DP   2 4  3.2.2 Calculate .256   4  0.5.3.5.5 Note: HS is calculated with the help of HS0 in accordance with paragraph 4.3.6 Calculate Discharge coefficient.24   3  3.3. Eps  1 ).3.49  D0 4.8.3.7 5  12 13  DP P1  K 1 4.3.3.  10 3  W D Calculate .5. 4.5 12732 .2.5124  10  C  Eps     D  Fs  24 Sizing Equations .3.5.1 Calculate Segmental radius: RS 0  0.8. 4.ratio.2.41  0.3. Venturi tubes and Nozzles: 2   K   Q K  Eps    K  1  where Q   1 4  2   4   1     Q K K 1     1   Q K     1  Q      0. 3.8 .12 4.3.11 For Liquid and Water.3.Preface4 Flow Meter Sizing 4.10 4.13. calculate Radius of upstream profile:  0.3.3.3.68286   Sizing Equations 25 .0001.3. go to paragraph 4. DP 0  DP Paragraphs 4.13 DP 0  DP DP 0 .4.07571    0. For Quarter of Circle (Quadrant Edge) orifice.06253 Rq  d      0.11 are repeated until the relative change in the differential range is less than 0. Calculate the relative change in Differential range: DP  4. 6.25 . Water.2 Calculate Correction factor for Steam quality (For Liquid.25 For other Restriction Devices:   Wn2 0   8 4 2   4.  10  D  DPn  RO  Wn   123 0.2.2732 103 Wn D Checking for Critical flow. Fs=1): 5.   Wn2 0   7 4 2  .2 Gas/Steam case 5.1 Calculate first estimate for For Venturi tubes:  .1.1.1.1.6 1.1.1.1 Calculation of Beta Ratio 5.1. 5.5 Calculate Pipe Reynolds number at normal flow rate: Re n  5.0074  S t SRF  10  Wn / Wmax DPn  DPloss  SRF / 10 2 5.3 Calculate Scale reading factor: 5.1 Calculate Internal pipe diameter at flow temperature: 5.1 Liquid / Water case P1  DPloss If  Pv then the flow is critical.ratio. 5.4 Calculate Differential pressure at normal flow rate:   D  D0  1   p T  T0  Fs  1  0.1.6  10  D  DPn  RO  Wn  26 Sizing Equations 0.5 Restriction Device Sizing 5 Restriction Device Sizing 5. and Steam.6.6.1. If not. 0   0.ratio:  2847.2.Preface5 Restriction Device Sizing 5. Case B.6  10  D  DPn  RO  Wn A3 A4 A5 A6    0.ratio: 0     . Calculate  . the flow is Sub-Critical.1. See 9 Appendix.  10  D  DPn  RO  Wn   123 0.05  W  1   4  n 0   2  D  Eps  Fs  C  DPn  RO    Calculate the relative change in  . Gas / Steam – Critical flow B1 Calculate Gas expansion factor  k Eps     Z  B2   2      k 1  k 1 /  k 1 1/ 2     Calculate Discharge coefficient. Sizing Equations 27 .2 Calculate Critical Pressure Ratio and define whether the flow is critical.7 then the flow is critical. See 9 Appendix.ratio is less than 0.ratio   Wn2 0   7 4 2  .25 Eps = 1. 5 0 A7 Paragraphs A4 – A6 are repeated until the relative change in the  . Fttp k 1 / k 1   k   2  4   1    0      2   k  1   DPrat  1 P1  DPloss Fttp  P1 k /  k 1  2  A   k 1 A  DPrat If 5.6.25 For other Restriction Devices:  Wn2  0   8 4 2  4.ratio. DP  P1  Pv A2 Calculate first estimate for For Venturi tubes:  . Calculate new  . Case A.1.0001. Calculate Discharge coefficient. Liquid / Water – Critical flow A1 Calculate DP. 35   0 C4    / P1 )  K 1 . See 9 Appendix.ratio in accordance with equations in paragraph B3. Eps = 1) For Venturi tubes and Nozzles: 2  K   Q K  Eps    K  1  where Q For Orifice Plates:  4 1 0  2   4   1   0   Q K K 1     1   Q K     1  Q       4   ( DP n Calculate Discharge coefficient.41  0.25 For other Restriction Devices:  Wn2  0   8 4 2  4.25 Calculate Gas expansion factor (for Liquids/Water. Sub-Critical flow C1 First estimate for DPn: DPn0  DPloss C2 Calculate first estimate for For Venturi tubes:  . See 9 Appendix. Case C.9625)  P1  D 4  Eps Calculate first estimate for k  2  am  aa        2   k  1  k 1 /  k 1 cm  aa  D 4 d2     1  1  4  am  cm 2  am d2 D B4 Calculate Orifice diameter at flow temperature: B5 Calculate Bore Reynolds number at normal flow rate: d  D Re n  1.ratio  Wn2  0   7 4 2  1. P1[psi-a].2732  10 3  Wn  d B6 Calculate Discharge coefficient.6  10  D  DPn0  RO  Wn C3 0.5 P1  DPn P1 Eps  1  0. D[in]): Wn  T aa  2195.5 Restriction Device Sizing B3  . 28 Sizing Equations 0. T[R].ratio (Wn[lb/h]. B7 Calculate new  .591  C  ( M / 28.23  10  D  DPn0  RO  Wn    0. ratio:  2847.06   2  1.18   3 For Nozzles: For Orifice Plates: DP  C7  DPloss  1   4   C   2 1    4 0.0001 Calculate Orifice diameter at flow temperature: C9 5.1. DPn0  DPn .1. 5 Calculate new DP For Classical Venturi Tube: DP  DPloss 0.5 C 2   Calculate new DPn:  W DPn  DP   n  Wmax C8 0.  0   DPn0 Paragraphs C3 – C8 are repeated until the relative change in the DPn is less than 0.9 d  D Calculate Orifice diameter at ambient temperature.42    0.38   2 DP  DPloss 1  0.5    2 Calculate the relative change in DPn: DPn  DPn0  DPn .8 5.Preface5 Restriction Device Sizing C5 Calculate new  .05  W  1   4  n 0   2  D  Eps  Fs  C  DPn  RO    C6 0.014    2. For Venturi tubes and Nozzles: d0  1   pe For Orifice Plates: d0  d  d  T  T0  1  0.218  0.55   Dbh d  2 1   pe  T  T0  Sizing Equations 29 . 2.38   2 DPloss DP  1  0. 5.2.1.3 Calculate Orifice diameter at flow temperature.0074  S t   d  d 0  1   pe  T  T0  For Orifice Plates:    d  d 0  1  0.42    0.2.2.5 C 2 5.2.1.2 Calculate Correction factor for Steam quality (For Liquid.4 Calculate first estimate for Flow Rate (W0 [kg/s]): W0  2.1.5.06   2  1. and Steam.5.ratio:  5.5 2 d D Checking for Critical flow and some initial calculations.2.5  4 0.18   3 DP  Nozzles: Orifice Plates: DP   DPloss  1   4   C   2 1   0.107  10 4  d 2  30 Sizing Equations DP  RO 1  4   .2. If not.1 Calculate downstream pressure: P2  P1  DPloss 5.3 Calculate Differential Pressure to be used in Flow Rate calculation.218  0.2. Water. For Critical flow: DP  P1  Pv For Sub-Critical flow: Classical Venturi Tube: DPloss 0.5.4 Calculate .014    2. the flow is Sub-Critical 5.55   Dbh d 0   1   pe  T  T0  5.2 Identify Critical flow: P2  Pv If then the flow is critical.2.5.1 Liquid / Water case 5.1 Calculate Internal pipe diameter at flow temperature: 5. For Venturi tubes and Nozzles:   D  D0  1   p T  T0  Fs  1  0.2 Calculation of Flow Rate 5. Fs=1): 5.2.5 Restriction Device Sizing 5.5.1. 5 C 2 C  2   Gas expansion factor: For Venturi tubes and Nozzles: 2  K  Q  K   Eps   K  1  where Q   1  4  2  4  1    Q  K P1  DP P1  1  Q  KK1   1  Q      0.5 Sizing Equations 31 . P1[psi-a]): M   W0  2195.3 Initial calculations for Sub-Critical flow. T[R].2.2 Gas/Steam case 5.38   2 For Nozzles: DP  DPloss 1  0.2. Fttp DPrat  k 1 /  k 1   k   2    1   4     2   k  1   1 P1  DPloss Fttp  P1 k /  k 1  2  A   k 1 A  DPrat If then the flow is critical.9625  T  0. Differential Pressure to be used in Flow Rate calculation: For Classical Venturi Tube: DP  DPloss 0.014    2.2.5.42    0.06   2  1. the flow is Sub-Critical 5.2.5.591  d     28.2.1 Calculate Critical Pressure Ratio and define whether the flow is critical. D[in]. Gas expansion factor:   k   2  k 1 / k 1   Eps         Z   k  1     1/ 2 First estimate for Flow Rate (W0 [lb/h].5.5.5  P1  Eps  Fttp 2 5.Preface5 Restriction Device Sizing 5.2 Initial calculations for Critical flow.5  0.18   3 For Orifice Plates: DP   DPloss  1   4 1    4 0. If not.2.2.218  0. 10 are repeated until the relative change in the Flow rate is less than 0.W 0  W 5. calculate DP and Gas Expansion factor: DP   DPloss  1   4   C   2 1   0.2.0001.7 5.2. For Gas / Steam Critical flow (Bore Reynolds number): Re  1.2.2732  10 3  W0 Re  D 5.591  C  d 2     28.9625  T  0.5 Restriction Device Sizing For Orifice Plates: Eps  1  0.35    ( DP / P1 )  K 1 4 5.2732 103 W0  d For Liquid / Water and Gas / Steam Sub-Critical flow (Pipe Reynolds number): 1.11 Paragraphs 5.2.5  4 0.  10 4  C  Fs  Eps  d 2  DP  RO 1 4 5.107  10 4  d 2  5.2.9 Calculate new Flow rate: For Gas / Steam at Critical flow: M   W  2195.10 Calculate the relative change in Flow rate: W  W 0 W W0 .6 – 5.2.35   4  ( DP / P1 )  K 1 First estimate for Flow Rate (W0 [kg/s]): W0  2. See 9 Appendix.6 DP  RO 1  4 Calculate Reynolds number.2.41  0.5  P1  Eps  Fttp For other cases (W [kg/s]): W  35124 .2.8 Calculate Discharge coefficient. For Orifices in the case of Gas / Steam Sub-Critical flow. 32 Sizing Equations .41  0.5 C 2   Eps  1  0. 8 0. Sizing Equations 33 .2 Calculate Correction factor for Steam quality (For Liquid.3.5124  10 4  C  Fs  Eps  d 2  DP  RO 1  4 For Gas / Steam   k   2  k 1 / k 1   Eps         Z   k  1     1/ 2 M   Wcr  2195.5  P1  Eps  Fttp Checking for Critical flow and Pressure Loss calculation.001*Wcr. For Venturi tubes and Nozzles:   D  D0  1   p T  T0  Fs  1  0.1 Calculate Internal pipe diameter at flow temperature: 5.  5.5 12732 . For Venturi tubes and Nozzles: Eps  1 ). For this case. and Steam.999*Wcr C1 First estimate for Differential range: DP0  P1  Pv C2 C3 Calculate Discharge coefficient.Preface5 Restriction Device Sizing 5.3. Sub-Critical flow: W<=0.ratio.0074  S t   d  d 0  1   pe  T  T0  For Orifice plates:    d  d 0  1  0. Case A. Flow rate cannot be achieved.001*Wcr.4 Calculate Pipe Reynolds number (W [kg/s]): Re  5. See 9 Appendix.7 2 d D Calculate Discharge coefficient.3.3. Fs=1): 5.3 Calculate Orifice diameter at flow temperature.3. Case C. Calculate Gas expansion factor (for Liquid and Water. Calculate Critical flow rate.9625  T  2 5. Critical flow: W>0.55   Dbh d 0   1   pe  T  T0  5.3. See 9 Appendix.3. For Liquid / Water: DP  P1  Pv Wcr  3. Water. Wrong data: W=>1.591  C  d     28. User has to check his data.  10 3  W D Calculate .6 5.3.999*Wcr and W<1. Pressure Loss: DPloss  P1  Pv Case B.3 Calculation of Pressure Loss 5. 5124  10  C  Eps     D  Fs  C5 Calculate the relative change in Differential range: DP  C6 C7 DP 0  DP DP 0 .0001.5 C2   .41  0. DP 0  DP Paragraphs C3 – C5 are repeated until the relative change in the DP is less than 0.5  4 0. Calculate Pressure Loss DPloss  34 Sizing Equations  DP  1   4   C   2 1   0.5 Restriction Device Sizing 2  K   Q K  Eps    K  1  where Q   1 4  2   4   1     Q K P1  DP0 P1 K 1     1   Q K     1  Q      0.35   4  ( DP0 / P1 )  K 1 C4 Calculate new Differential range (W [kg/s]): 2   W  1 4   RO 1 DP   2 4  3.5 For Orifice Plates: Eps  1  0. 1 Calculate Total back pressure: Pb  Pb1  Pb 2  Pb3 Sizing Equations 35 . 6.1.1.1.6611 Case: Case Re  50000 : 6.3.1.2 Not requiring ASME capacity certification 6. 6.4 Calculate Upstream relieving pressure: *Pup [psi-g] Kw  1152 .1.1.62838  Y  0.1.1.8.Preface6 Relief Valve Sizing 6 Relief Valve Sizing 6.1.1.1.1.1.1.1.7 Calculate Correction factor for viscosity: Y  ln Re / ln10 Case Re  35 : Case 35  Re  100 : Kv  0.2 Effective coefficient of discharge: Kd  0.5 : Case Pb / Ps  0.2 For Bellows valve: Case Pb / Ps  016 .  Pb / Ps  0.1 For Conventional and Pilot operated valves.1.1. : Kw  1 Case 016 .1 With Rupture disk: A  A0 /  Kv  K cc  6.1.1 Calculate Total back pressure: Pb  Pb1  Pb 2  Pb3 6.1 Blocked Flow: Liquid Relief 6.3 Determine Correction factor for back pressure.1.65 (if K d is not entered by user).1. Pup [psi-g] A0  W  38  K d  KW Gf pup  pb 6.  0.1.01  Pover   Ps 6.6 Calculate Reynolds number: *W [US gal/min].1.677 Pup  1  0.1 Requiring ASME capacity certification 6. 6.1. 6.1.8 Calculate Required effective discharge area. [cP] Re  2800  W  G f   A0 6.2.5 Estimate Required effective discharge area: *W [US gal/min].2 Without Rupture disk: A  A0 / Kv 6.95  Pb / Ps Kw  0.1.3. Kw  1 .1.1.5 : 6.3 Kv  0.8.  0.1.2  Y 2  0.3 Case 35  Re  100 : Kv  0.2.2 For Bellows valve: Case Pb / Ps  016 . 6.2228 Case Re  50000 : Kv  1.  Pb / Ps  0.6611 Case 100  Re  50000 : Kv  0.91334 K p  1086 .023362  Y 4  0.1 With Rupture disk: A  A0 /  Kv  Kcc  6.1.2.5 : 6.5  Y  5 : Case Y  5 : Case K p  0.4 Calculate correction factor for overpressure: Kw  1152 . 6.2.034664  Y  0.01  Pover   Ps  Pa 36 Sizing Equations .2.62838  Y  0. Kw  1 .2. 6.1.2 Effective coefficient of discharge: Kd  0.95464  Y  0.1 For Conventional and Pilot operated valves.30196  Y 2  0.5 : Case 2.35578  Y 3  2. : Kw  1 Case 016 .62 (if K d is not entered by user).037641  Y 3  0.5 : Case Pb / Ps  0.1.  ps  pb 6.1. [cP] Re  2800  W  G f   A0 6.7 Calculate Correction factor for viscosity: Y  ln Re / ln10 Case Re  35 : Kv  0.1.2.1.95  Pb / Ps Kw  0.2.2.037  Y 2  5.677 Y  Pover / 10 Case K p  0.1.2507  Y  4. 6.1.2.1 Calculate Upstream relieving pressure: *Pa [psi-a] Pup  1  0.8 Calculate Required effective discharge area.08544 K p  0.8.2. 6.8.5 Estimate Required effective discharge area: *W [US gal/min] A0  Gf W  38  Kd  KW  K p 125 .2.8  Y Y  1: 1  Y  2.3.2.1.1. 6.3 Determine Correction factor for back pressure.2 Without Rupture disk: A  A0 / Kv 6.3.6 Calculate Reynolds number: *W [US gal/min].6 Relief Valve Sizing 6.2 Blocked Flow: Gas Relief 6. Preface6 Relief Valve Sizing 6.2.2 Calculate Total back pressure: Pb  Pb1  Pb 2  Pb3 6.2.3 Calculate Critical flow throat pressure:  2  Pcf  Pup    K  1 K /  K 1 6.2.4 Effective coefficient of discharge: Kd  0.975 . 6.2.5 The condition Pb  Pcf (Pb [psi-a] ) determines: Case A - Critical flow. A1 Calculate Coefficient for critical flow equation: Case K  101 . : Cr  317 K 1 Case 101 .  K  2:  2  K 1 Cr  520  K     K  1 Cr  400 Case K  2 : A2 Calculate Capacity correction factor due to back pressure (if Kb is not entered by user). A2.1 For Conventional and Pilot operated valves, Kb  1 . A2.2 For Bellows valve: A2.2.1 Y  Pb / Ps Y  0.315 : Case Y  0.315 : A2.2.3 Case Y  0.325 : Case Y  0.325 : A2.2.4 Case Pover  20 : Case 10  Pover  20 : Case Pover  10 : A2.2.2 Case A3 Estimate Required effective discharge area: *W [lb/h], T [°R] A0  Case B ( Pb B1 Kb10  153 .  168 . Y Kb10  1. Kb20  114 .  0.43  Y Kb20  1. Kb  Kb 20 Kb  Kb10  01 .   Pover  10   Kb20  Kb10  Kb  Kb10 W TZ  Cr  K d  Pup  Kb M  Pcf ) - Subcritical flow: Bellows valve. Calculate Cr: . : Case K  101 Cr  317 K 1 Case 101 .  K  2: B2  2  K 1 Cr  520  K     K  1 Cr  400 Case K  2 : If Capacity correction factor due to back pressure is not entered by user, Kb  1 . Sizing Equations 37 6 Relief Valve Sizing B3 Estimate Required effective discharge area: *W [lb/h], T [°R] A0  Case C ( Pb  Pcf ) - Subcritical flow: Conventional and Pilot operated valves. C1 Calculate Coefficient for subcritical flow equation: *Pb [psi-a] C1.1 R  Pb / Pup C1.2  K  2/ K F2    R  K  1 C2 1  R  K 1 / K     1 R  Estimate Required effective discharge area: *W [lb/h], Pb [psi-a], T [°R] A0  6.2.6 W TZ  Cr  K d  Pup  Kb M W  735  F2  K d TZ  M  Pup  Pup  Pb  For Rupture disk application, calculate Required effective discharge area: A  A0 / K cc 6.3 Blocked Flow: Steam Relief 6.3.1 Calculate Upstream relieving pressure: *Pa [psi-a] Pup  1  0.01  Pover   Ps  Pa 6.3.2 6.3.3 Calculate Correction factor for Napier equation: Case Pup  1515 : Kn  1     Case 1515  Pup  3215 : K n  01916 .  Pup  1000 / 0.2292  Pup  1061 Case Pup  3215 : Kn  1 Calculate Superheat steam correction factor: Y1   T  27316 .   18 .  32 / 1000 6.3.3.2 Y 2  Pup / 1000 6.3.3.1 6.3.3.3 K sh  0.201  Y12  Y 2 2  0168 .  Y12  Y 2  0.291  Y12  0.389  Y1  Y 2 2  0.256  Y1  Y 2  6.3.4 6.3.5 0.838  Y1  0164 .  Y 2 2  0.025Y 2  1276 . Effective coefficient of discharge: Kd  0.975 . Estimate Required effective discharge area: *W [lb/h] A0  6.3.6 W 515 .  Pup  Kd  K n  K sh For Rupture disk application, calculate Required effective discharge area: A  A0 / K cc 38 Sizing Equations Preface6 Relief Valve Sizing 6.4 Fire Case: Gas Expansion 6.4.1 Calculate Upstream relieving pressure: *Pa [psi-a] Pup  1  0.01  Pover   Ps  Pa 6.4.2 Calculate Coefficient for critical flow equation: Case K  101 . : Cr  317 K 1  2  K 1 Cr  520  K     K  1 Cr  400 Case 101 .  K  2: 6.4.3 6.4.4 6.4.5 Case K  2 : If Heat absorption factor is entered, go to paragraph 6.4.8. If Normal pressure is not entered or Normal temperature is not entered, a message appears: “Please enter Heat absorption factor or Normal pressure and Normal temperature”. Calculate Gas temperature at relieving conditions: *Pa [psi-a], T [°R] T1  T  Pup P 6.4.6 If Vessel Wall temperature is not entered, the following message appears: “Please enter Heat absorption factor or Vessel wall temperature”. 6.4.7 Calculate Heat absorption factor: T [°R] F1  01406 .  6.4.8  Tw  T11.25 Cr  K d  T10.6506 Estimate Required effective discharge area: *W [lb/h], T [°R] A0  F1  A1 Pup 6.4.9 If Normal pressure is not entered or Normal temperature is not entered or Molecular Mass is not entered or Vessel wall temperature is not entered, a message appears: “Maximum discharge cannot be calculated because Normal pressure or Normal temperature or Molecular Mass or Vessel wall temperature is not entered”. Then go to paragraph 6.4.11. 6.4.10 Calculate Maximum discharge: *Pa [psi-a], W[lb/h] W  01406 .  A1  6.4.11 M  P   up  Tw  T11.25 T11.1506 For Rupture disk application, calculate Required effective discharge area: A  A0 / K cc Sizing Equations 39 6 Relief Valve Sizing 6.5 Effective coefficient of discharge: Kd  0.2 F   A1 0.1 Y  Pb / Ps A2 Y  0.  168 . A1 Calculate Coefficient for critical flow equation: Case K  101 .5 Fire Case: Liquid Filled Vessel 6.Critical flow. A2.1 For Conventional and Pilot operated valves.6 The condition Pb  Pcf (Pb [psi-a] ) determines: Case A .82 HV Calculate Upstream relieving pressure: *Pa [psi-a] Pup  1  0. Kb  Kb 20 Kb  Kb10  01 . T [°R] A0  Case B ( Pb Kb10  153 .315 : A2. : Cr  317 .2.  K  2:  2  K 1 Cr  520  K     K  1 Cr  400 Case K  2 : Calculate Capacity correction factor due to back pressure (if Kb is not entered by user).01  Pover   Ps  Pa 6.325 : A2.   Pover  10   Kb20  Kb10  Kb  Kb10 W TZ  Cr  K d  Pup  Kb M  Pcf ) .2 For Bellows valve: A2.5.2.5.43  Y Kb20  1. : Cr  317 K 1 Case 101 .975 .5.2.5. A2.1 Calculate Maximum discharge: *HV [Btu/lb].4 Calculate Critical flow throat pressure:  2  Pcf  Pup    K  1 K /  K 1 6. 6. B1 40 Sizing Equations Calculate Cr: Case K  101 .3 Case Y  0.315 : Case Y  0. Kb  1 .5.  0.2 Case A3 Estimate Required effective discharge area: *W [lb/h].Subcritical flow: Bellows valve. Kb20  114 .4 Case Pover  20 : Case 10  Pover  20 : Case Pover  10 : A2.3 Calculate Total back pressure: Pb  Pb1  Pb 2  Pb3 6. Y Kb10  1. W[lb/h] W  21000  6.2.325 : Case Y  0.5. T [°R] A0  6.5. Kb  1 . T [°R] A0  Case C ( Pb Calculate Coefficient for subcritical flow equation: *Pb [psi-a] C1. C1 C2  2  K 1 Cr  520  K     K  1 Cr  400 W  735  F2  K d TZ  M  Pup  Pup  Pb  For Rupture disk application.Preface6 Relief Valve Sizing K 1 Case 101 .2  K  2/ K F2    R  K  1 1  R  K 1 / K     1 R  Estimate Required effective discharge area: *W [lb/h].Subcritical flow: Conventional and Pilot operated valves.7 W TZ  Cr  K d  Pup  Kb M  Pcf ) .  K  2: Case K  2 : If Capacity correction factor due to back pressure is not entered by user. Pb [psi-a].1 R  Pb / Pup C1. calculate Required effective discharge area: A  A0 / K cc Sizing Equations 41 . B2 B3 Estimate Required effective discharge area: *W [lb/h]. 000000000336  T 3 Calculate Maximum Thermowell Length Based on Vibration Limitation: * ROmaterial[lb/in3] L1  ( R  K f  E  B  R f ) /( 2.1 7.32 =3.1.000138  T + 0.96 K f =2.97 =3.1 ASME PTC 1974 7.0101 .0.4 Define Kf Constant: Case L  3.0000000742  T 2 For Ferrite steel.000000267  T 2 .5<L  6 If Dt =1/4 then If Dt =3/8 then If Dt =9/16 then If Dt =11/16 then If Dt =7/8 then Case 6<L  9 If Dt =1/4 then If Dt =3/8 then 42 Sizing Equations Kf Kf Kf Kf Kf =2.39 =3.5 If Dt =1/4 then If Dt =3/8 then If Dt =9/16 then If Dt =11/16 then If Dt =7/8 then Case 3.7 Thermowell Calculation 7 Thermowell Calculation 7.64  ROmaterial V ) 7. R f  0.075 =2.01 =3.02 =3.99996 .06 Kf Kf Kf Kf Kf =2.0.46 .1.965 Calculate Ratio of Frequency at Fluid Temperature to Frequency at 70 °F: * T[°F] For Austenitic steel.0.07 =2. 7.08 K f =2.1.0.3 R f  1.40 =3.1.445 =3.42 =2.2 Define Initial Value of Kf If Dt =1/4 then If Dt =3/8 then If Dt =9/16 then If Dt =11/16 then If Dt =7/8 then Constant: Kf Kf Kf Kf Kf =2.84 =2.0000036111  T .45 =3. 03 Kf Kf Kf Kf Kf =2.3 .46 =4.44 K f =4.06 =2.47 =3.09 =2.07 =3.1.48 =4.6 Calculate Natural Frequency for L=L1 Kf value equals previous K f value * ROmaterial[lb/in3] fn  7.1.7.47 =4.09 Kf Kf Kf Kf Kf =2.25 If Dt =1/4 then If Dt =3/8 then If Dt =9/16 then If Dt =11/16 then If Dt =7/8 then Case 13.47 =3.47 =3.25<L  20 If Dt =1/4 then If Dt =3/8 then If Dt =9/16 then If Dt =11/16 then If Dt =7/8 Case 20<L If Dt =1/4 then If Dt =3/8 then If Dt =9/16 then If Dt =11/16 then If Dt =7/8 then then K f =3.06 =3.09 Kf Kf Kf Kf Kf =2.4 are repeated until current 7.08 =2.1.Preface7 Thermowell Calculation If Dt =9/16 then If Dt =11/16 then If Dt =7/8 then Case 9<L  13.07 =3.05 K f =3.5 Paragraphs 7.1.64  V B Sizing Equations 43 .7 L 2 1 K f  E  Rf  ROmaterial  Calculate Wake Frequency fW  2.09 7.1. { K1 =0.334.1. K 2 . K 3 =0.5.223.1. *S [psi-g] P max  K1  S 7.3. Convert from UOM used for pressure. K 2 =42.116} K 3 =0.9 7.548} K 3 =0.8. { K1 =0.864} Calculate Maximum Allowable Static Gage Pressure *Pmax [psi-g].412.1. { K1 =0. K 3 constants If Dt =1/4 then If Dt =3/8 then If Dt =9/16 then If Dt =11/16 then If Dt =7/8 then { K1 =0.8) FM 7.8 Calculate Magnification Factor for L1(R  0. K 2 =46. K 2 =50. K 2 =48.7.11 Calculate Maximum thermowell length based on Steady State Stress Consideration *L2 [inch] *ROfluid [lb/ft3] *P [psi-g] – operating pressure L2  K2  V 44 Sizing Equations S  K 3 P  ROfluid  1  FM  .10  fw     fn  2 R2   2 1 R 2  fw   1   f n   Define K1 .1.389} K 3 =0.7 Thermowell Calculation 7.202. K 2 =37.205} K 3 =0.1. { K1 =0.155. 4.3.0095  [ Log10 (Re/ 1300]3 7.61 7.2.2.22  (1  22 / Re) 7.3 1.000 N s  0.1.2. B[m].2.1.1416 m L 7.1416  ROm  ( Da  d b ) / 4 2 2 1.3 Calculate Strouhal number 7.2 Other connection types: H c  1  0.2.0248  [ Log10 (Re/ 1300]2  0.213  0.1.1 Calculate Mounting factor *L [m] 7.3.4.1 Tapered thermowell: Da  ( Ad  B) / 2 7.5  (b / Ad )]2 Calculate Reynolds number *B [m] Re  U  B  RO /  7.2. E[Pag] 7.4.3 Calculate correction factor Hf 7.2.2.4.4.867 2 E  I 0.1 Tapered and straight thermowell: Sizing Equations 45 .2.9  Ad L 7.1 Threaded connection: H c  1  0.3.2 Ad / L [1  1.2.1 Re < 1.3.1 Calculate average outside diameter 7.2.1416  ( Da  d b ) / 64 4 4 m  3.22 7.000 N s  0.300 N s  0.2 Calculate approximate natural frequency I  3.Preface7 Thermowell Calculation 7.1.5 1 fa  ( )  2 2  3.2.2.2.2 ASME PTC 2010 7.2 Other thermowells: Da  Ad 7.2.300 ≤ Re ≤ 500.2 Re > 500.4 Calculate Natural frequency *L [m].4. 407. c7=1.99  [1  (1  B / Ad )  (1  B / Ad ) 2 ] 1  1.949.714.228.66 cm Case A1: B = 2. c5=-2. c9=9.1 ( Da / L) 3[10.4.2.3. c2=-0. c3=-0. c8=0.27cm: c1=1.132.466 Case A2: B = 1. c2=-0.594. c9=8. c10=-7.22cm: c1=1. c10=-5.4.4.2 Stepped thermowell: Case A: db = 0.4.2.2.6 Calculate natural frequency with ideal support f n 0  f a  H f  H af  H as 7.5 Calculate correction factor Has H as  1  ROs 1 [ ] 2  ROm ( Da / d b ) 2  1 7.7 Calculate natural frequency of the mounted thermowell f n  f n0  H c 7.8( db / Da )] 7.022.865.362.839.7 Thermowell Calculation Hf  0. c7=0.5 Calculate Vortex shedding frequency *B[m] f s  N s U / B 46 Sizing Equations .00. c8=1.022. c3=-0.091. c5=-1.2.313. c4=1.4.861.275.299.4 Calculate correction factor Haf H af  1  RO 2  ROm 7.2. c6=1. c6=0.2. c4=1.376 For both A1 and A2: y1  [c1  ( Ad / B)  c2 ]  ( Ls / L)  [c3  ( Ad / B)  c4 ] y2  [c5  ( Ad / B)  c6 ]  ( Ls / L)  [c7  ( Ad / B)  c8 ] Bet  [c9  ( Ad / B)  c10 ] H f  ( y1  Bet  y2  Bet ) 1/ Bet Case A3: Other values of db: H f 1 Case B: Other values of B: H f 1 7.41. 1416 2    ( ROm / RO)  [1  (d b / B) 2 ] 7.7.Preface7 Thermowell Calculation 7.7.1  0.7.2.6 Calculate Scruton number *B [m] N sc  3.3.1 Parameter Gsp 7.1416  B 2  [1  (d b / B) 4 ] 2 Grd 7.2.1.2.7.7.7.2.033  Ad / b 7.2 Case B: b is known and Ad/b ≤ 33: K t  1.1416  Ad  [1  (d b / Ad ) 4 ] 7.7.2.2.1.2 Other connections Case A: b is unknown or Ad/b > 33: K t  2.3 7.2 Parameter Grd (for stepped thermowell only): 16  Ls  3.7. E[Pag] 7.2.3.2 Sizing Equations 47 .1 Tapered and straight thermowell: Gsp  16  L2  (1  2  B / Ad ) 2 3  3. L[m].2. Calculate miscellaneous parameters *B [m].3 Stress concentration factor 7.2.2.7.1 Threaded connections: K t  2.4 Stress concentration factor at support plane (for stepped thermowell only) Case A: bs is unknown or B/bs > 33: K t1  2.1416  Ad  [1  (d b / Ad ) 4 ] 7.2 Stepped thermowell: Gsp  16  L2  [ B / Ad  (1  B / Ad )  (1  Ls / L) 2 ] 2 3. 2.8.2.2 Re < 1.1.8.5 Temperature de-rating factor: Ft  E / Eref 7.2.2.1.2.0496  R VR  48 Sizing Equations B  fn B  fn aR  [1   Log10 ( )] 2  Ns Ns 2  N s U .8  f n 7.5)and (Re  100000) Go to 7.9.1 Calculate velocity for in-line resonance 7.0285  R 2  0.2.9.2.2.9.1 Low density gases: Nsc > 2.2.1  0.5 and Re < 100000 FrL  0.2.1.1 VR  7.8 Calculate Frequency Limit 7.2 General case: other values of Nsc and Re FrL  0.9.2.2.300 ≤ Re < 500.2 The thermowell is in non-compliance with the Frequency Limit requirement if f s  FrL and ( N sc  2.9.2.2.1.1 The thermowell is in compliance with the Frequency Limit requirement.2 Evaluate compliance with Cyclic Stress Limit for General case 7.2.2.7 Thermowell Calculation Case B: bs is known and B/bs ≤ 33: K t1  1.300 B  fn 22   22    (1  ) 2  Ns B  RO U B  RO 1.9.9.10 7.1 Initial evaluation of compliance 7.10 7. if f s  FrL Go to 7.000 R  Log10 (Re/ 1300) aR  0.2.9 Compliance with the Frequency Limit requirement 7.033  B / bs 7.4  f n 7.7. 9.2 Stepped thermowell 7.5.2.2.2.2.2.2.4 Calculate fatigue stress limit Limit  Ft  Fe  S f 7.2.2.2 Complied with the cyclic stress limit if Set new frequency limit: FrL1 S max  Limit  0.9.Preface7 Thermowell Calculation 7.2.9.05  RO VR 2 Sd  Gsp  Fm  Pd 7.1 Not complied with the cyclic stress limit and frequency limit if S max  Limit Go to 7.2.2.8  f n Go to 7.2.9.1.2.9.5.9.3 7.10 7.5 Compliance with cyclic stress limit 7.2.5.9.2.2.2 Case 7.3 VR  Re ≥ 500.2.9.2.2.5.2.9.1 Not complied with the cyclic stress limit and frequency limit if S max  Limit Go to 7.2.2 Calculate cyclic drag stress at the root Fm  1/( 2   ) Pd  0.1.10 S max  Limit 7.2.1 Tapered and straight thermowell 7.2.5.3 Calculate combined drag and lift stresses S max  K t  Sd 7.1 Calculate: Sd1  Grd  Fm  Pd S max 1  K t1  Sd1 Sizing Equations 49 .2.2.5.2.2.5.9.1.9.2.000 B  fn 2 Ns 7.9.2.2.2.9. 9.2.8  f n Go to 7.5  [(S max  Sr ) 2  ( S max  St ) 2  ( St  Sr ) 2 7.9.2.2.3 Complied with the cyclic stress limit if Set new frequency limit: FrL1 S max 1  Limit  0.2.2.2.10. and axial stresses due to the external pressure: Sr  P St  P  Sa  1  (d b / Ad ) 2 1  (d b / Ad ) 2 P 1  (d b / Ad ) 2 7.2.5 Calculate left-hand side of the Von Mises criteria: LHS  0.3 Re-evaluate compliance with the frequency limit requirement if cyclic stress is acceptable 7.10.10.3. tangential. S[Pag] 7.2.9.2.2.2 The thermowell is in non-compliance with the Frequency Limit requirement if f s  FrL1 7.6 Calculate right-hand side of the Von Mises criteria: RHS  1. P[Pag].5  S 50 Sizing Equations .3.2.2.1 Calculate radial.3 7.2.2.10.2.2 Not complied with the cyclic stress limit and frequency limit if S max 1  Limit Go to 7.10 7.2.9.10 Compliance with the Static Stress Limit requirement *B [m].9.2.2. if f s  FrL1 7.10.2 Calculate steady-state drag stress at the root: PD  0.2.4 Calculate maximum stress S max  Sd  Sa 7.5.2.1 The thermowell is in compliance with the Frequency Limit requirement.10.7 Thermowell Calculation 7.7  RO U 2 7.9.5.3 Calculate steady-state stress due to the drag force: Sd  Gsp  PD 7. 10.2.7.10.7.2.2.10.7.7.2 Case 7.2.2.2 Stepped thermowell 7.1.1 Not complied with the static stress limit if LHS  RHS Go to 7.1 Tapered and straight thermowell 7.2 0.7.1 LHS  RHS Calculate: St1  P  Sa1  1  (d b / B) 2 1  (d b / B) 2 P 1  (d b / B) 2 Sd1  Grd  PD S max 1  Sd1  Sa1 LHS1  7.2.5  [(S max 1  Sr ) 2  ( S max 1  St1) 2  ( St1  Sr ) 2 Not complied with the static stress limit if LHS1  RHS 7.10.7.2.10.2.7 Compliance 7.10.10.2.2.Preface7 Thermowell Calculation 7.2.2.7.7.2.7.3 Complied with the static stress limit if LHS1  RHS Sizing Equations 51 .2.1 Not complied with the static stress limit if LHS  RHS 7.1.10.2.11 7.10.2.2 Complied with the static stress limit if LHS  RHS 7.2.10.2.2. 2.1 Not complied with the dynamic stress limit if S max  Limit Go to 7.1 Tapered and straight thermowell 7.11.11.5.11 Compliance with the Dynamic Stress Limit requirement 7.5 Compliance 7.5.11.2.11.5.3 Calculate combined drag and lift stresses: S max  K t  Sd 2  S12 7.1 Calculate parameters: r  fs / fn Fm1  1 1 r 2 r1  2  f s / f n Fm 2  1 1  r12 7.2.7 Thermowell Calculation 7.2 Complied with the dynamic stress limit if S max  Limit 7.11.2.2.2.2.4 Calculate fatigue stress limit: Limit  Ft  Fe  S f 7.2 Calculate drag and lift stresses at the root: PD1  0.1 Not complied with the dynamic stress limit if S max  Limit 7.2 Stepped thermowell 7.11.5  RO U 2 Sd  Gsp  Fm2  PD1 S1  Gsp  Fm1 P1 7.12 52 Sizing Equations .11.2.2.11.5.2.2.05  RO U 2 P1  0.11.11.5.2. Pt ) 7. P[Pag].5.2.0833] 2  B /( B  d b ) 7.2.1 Not complied with the hydrostatic pressure limit if Pr ating  P 7.2.11.2.12.2 Calculate external pressure rating for the tip: Pt  S t  ( )2 0.2.2 Case 7.5.5.11.12.11.2.2.2 Not complied with the dynamic stress limit if S max 1  Limit 7.66  S  [ 2.1 Calculate pressure rating for the shank: Pc  0.4.2.2 Complied with the hydrostatic pressure limit if Pr ating  P Sizing Equations 53 .2.5.3 Calculate pressure rating of the thermowell: Pr ating  MIN ( Pc.12.12.4.13 d b 7.2.11.2.Preface7 Thermowell Calculation 7.12.1 S max  Limit Calculate: Sd1  Grd  Fm2  PD1 S11  Grd  Fm1 P1 S max 1  K t1  Sd12  S112 7.12 Compliance with the Hydrostatic Pressure Limit requirement *B [m].12.167  0.2.2. S[Pag] 7.2.3 Complied with the dynamic stress limit if S max 1  Limit 7.4 Compliance 7. 17 (1991) .4 Lo-Loss Tube Miller 8.8 Standards 8 Standards 8.3.1 Control Valve Sizing ISA S75.3.3 Conical Entrance Orifice Miller 8.2 Quarter of Circle (Quadrant Edge) Orifice Flange Tappings BS 1042**** Corner Tappings BS 1042 Radius Tappings BS 1042 8.3 Flow Meter Sizing 8.01 (1995) IEC 60534-2-1 (1998) 8.3.7 Eccentric Orifice Diametrically Opposite Flange Tappings Miller Side (90) Flange Tappings Miller Diametrically Opposite Vena Contracta Tappings Miller Side (90) Vena Contracta Tappings Miller 54 Sizing Equations .5D & 8D) Miller AGA Report 3 (Flange Tappings) AGA Rep.3.Aerodynamic Noise IEC 534-8-3 (1995) IEC 534-8-4 (1994) 8.3.3.2 Noise Calculation Masoneilan OZ 3000E (1984) .1 Square Edge Orifice Standard Flange Tappings ISO*/ Miller** Corner Tappings ISO / Miller Radius Tappings ISO / Miller Small-Bore Honed-Orifice Meter Run with Corner Tappings Miller Small-Bore Honed-Orifice Meter Run with Flange Tappings Miller Pipe Tappings (2.5 Venturi Tube Classical venturi tube with an “as cast” convergent section ISO / Miller Classical venturi tube with a machined convergent section ISO / Miller Classical venturi tube with a rough-welded sheet-iron convergent section ISO / Miller Universal Venturi Tube Miller 8.3.6 Nozzle ISO Long Radius Nozzle ISO ASME Long Radius Nozzle Miller ISA 1932 Nozzle ISO / Miller Venturi Nozzle ISO / Miller 8.3*** AGA Report 3 (Pipe Tappings) AGA Rep.3 8.Hydrodynamic Noise ISA S75. 8 Segmental Orifice Flange Tappings Miller Vena Contracta Tappings Miller Note: * ISO 5167-1 (1991). ISO 5167-1 (1998). seventh edition (January 2000) 8.3.Preface8 Standards 8. or ISO 5167 (2003) ** Flow Measurement Engineering Handbook by R. 3rd edition (1996) *** ANSI/API 2530 (1991) [AGA Report 3] **** British standard 1042.3 TW (2010) Sizing Equations 55 . Section 1.2 (1989) 8.4 Relief Valve Sizing API RP 520.3 (1974) ASME PTC 19. Miller.5 Thermowell Calculation ASME PTC 19.W. 47/(1B))^1.011*(0.4/D)-0.8*(2*0/(1-B))^1.000521*(10^6*B/Re)^0.75 9.8))*(B^4/(1B^4))-0.5*(10^6/Re)^0.080*exp(-10*25.5959+0.5961+0.0312*B^2.216B^8+0.0261*B^2-0.3+0.123)*(1.4/D/(1B))^1.039*B^4/(1-B^4)0.2.75-B)*(2.0029*B^2.3+0.1-0.0188+0.2ISO 5167-1 (1998) and ISO 5167 (2003) If D<71.000521*(10^6*B/Re)^0.5*(10^6/Re)^0.216B^8+0.0261*B^2-0.5961+0.0063*(19000*B/ Re)^0.5*(10^6/Re)^0.8)*B^3.8))*(B^4/(1-B^4))-0.1 Square Edge Orifice 9.123)*(1.1-0.1.4/D))*(10.0312*B^2.123*exp(-7*1))*(10.5*(10^6/Re)^0.000521*(10^6*B/Re)^0.216B^8+0.8-D/25.0188+0.1.080-0.8*(2*0.216B^8+0.0029*B^2.1 ISO 5167-1 (1991) and Miller If D<58.0261*B^2-0.1.0188+0.031*(2*0/(1-B)-0.043+0.4/D/(1-B)-0.031*(2*25.8-D/25.123*exp(-7*25. inch Re Pipe Reynolds Number at Normal Flow Rate Rd Bore Reynolds Number at Normal Flow Rate 9.5961+0.0261*B^2-0.11*(19000*B/Re)^0.1)*B^1.043+0.0261*B^2-0.0312*B^2.3+(0.0063*(19000*B/ Re)^0.1.3 9.0029*B^2.043+0.8))*(B^4/(1-B^4))-0.12 C = 0.216B^8+0.7+(0.4) Else C = 0.5961+0.3 9.0063*(19000*B/ Re)^0.8))*(B^4/(1-B^4))-0.1-0.5*(10^6/Re)^0.0063*(19000*B/ Re)^0.031*(2*0/(1-B)-0.1)*B^1.3 Radius Tappings 9.000521*(10^6*B/Re)^0.5*(10^6/Re)^0.2 Corner Tappings 9.123*exp(-7*25.031*(2*0.62 C = 0.080-0.75-B)*(2.5*(10^6/Re)^0.1 Flange Tappings 9.4/D))*(10.0188+0.5959+0.2.ratio C Discharge Coefficient D Internal Pipe Diameter.1)*B^1.85598*B^3/D+ 2.75-0.8*(2*0/(1-B))^1.8*(2*25.1.7+(0.5*(10^6/Re)^0.85598*B^3/D+ 0.11*(19000*B/Re)^0.75-B)*(2.1.75-0.9 Appendix 9 Appendix Formulae for Discharge Coefficient Notation B .080*exp(-10*1)-0.184^B^8+0.8)*B^3.184^B^8+0.11*(19000*B/Re)^0.11*(19000*B/Re)^0.011*(0.8)*B^3.015839*B^3 9.0063*(19000*B/ Re)^0.1)*B^1.011*(0.7+(0.4) 56 Sizing Equations .184^B^8+0. mm Di Internal Pipe Diameter.1-0.3.031*(2*25.0.3+(0.5959+0.1 ISO 5167-1 (1991) and Miller C = 0.11*(19000*B/Re)^0.039*B^4/(1-B^4) Else C = 0.000521*(10^6*B/Re)^0.5*(10^6/Re)^0.7+(0.184^B^8+0.4) Else C = 0.1.4/D/(1-B)-0.8*(2*25.8)*B^3.12 C = 0.0188+0.0312*B^2.1.2 ISO 5167-1 (1998) and ISO 5167 (2003) If D<71.1.4/D/(1B))^1.3+0.0029*B^2.286*B^4/(D*(1-B^4)) 9.12 C = 0.5959+0.1.3+(0.8))*(B^4/(1B^4))-0.2 ISO 5167-1 (1998) and ISO 5167 (2003) If D<71.8-D/25.75+0.1 ISO 5167-1 (1991) and Miller C = 0.3+(0.3+(0.4/D)-0.47/(1-B)-0.5961+0.1)*B^1.8)*B^3.043+0.3.7+(0.0.080*exp(-10*25.1.043+0. 6-1/Di)^5*(0.730 9.0188+0.3 Conical Entrance Orifice If Re5000*B Else C = 0.5.25 A1 = (0.06/Di)*B^2+(0.076/Di^0.009+0.123*exp(-7*1))*(10.48-1.734 C = 0.5D & 8D) C = 0.1615^B^2+1.9797 Sizing Equations 57 .4*(1.564^B^2-0.5.5)))*((1-B^4)^0.5)))*((1-B^4)^0.1.47/(1-B)-0.364+0.1+0.5)*B^4 A2 = 0 If B<(0.5*(10^6/Re)^0.007/Di+(0.5*K0*(1+EP*B/Re) 9.5993+0.8 C = 0.1 9.75 AGA Report 3 (Flange Tappings) A0 = 830-5000*B+9000*B^2-4200*B^3+530/Di^0.468*(B^4+10^B^12)+(0.5-B)^(3/2)*(0.5 A3 = (0.0044/Di+(0.985 Universal Venturi Tube C = 0.5) K0 = 0.25-B)^(5/2)*(1.3 9.5 9.1*B^4)*(Re^(-0.8))*(B^4/(1-B^4))-0.4 9.6 9.1)*B^1.5*K0*(1+EF*B/Re) AGA Report 3 (Pipe Tappings) A0 = 905-5000*B+9000*B^2-4200*B^3+875/Di A1 = 0 If B<0.4 Classical venturi tube with an “as cast” convergent section C = 0.5.0182/Di+(0.7 A4 = (B-0.1.461*B^2.000015*A0) EF = A0*Di*B C = (1-B^4)^0.0063*(19000*B/ Re)^0.5/Di) A2 = 0.0261*B^2-0.16/Di)*(B^4+4*B^16))*(Re^(-0.5980+0.440-0.031*(2*0.984 Classical venturi tube with a machined convergent section C = 0.07+0.5*(10^6/Re)^0.5925+0.7 9.8*(2*0.5 Venturi Tube 9.080*exp(-10*1)-0.471*B+0.7+(0.11*(19000*B/Re)^0.07+0.216B^8+0.2 Quarter of Circle (Quadrant Edge) Orifice C = 0.7)^(5/2)*(65/Di^2+3) K0 = (A1+A2-A3+A4)/(1+0.5.5) Small-Bore Honed-Orifice Meter Run with Flange Tappings C = (0.5) Pipe Tappings (2.995 Classical venturi tube with a rough-welded sheet-iron convergent section C = 0.3155+0.5 A1 = 0.0175/Di)*(B^4+2^B^16)+(0.35*B^14+A1 EP = A0*Di*B C = (1-B^4)^0.52/Di-0.4 Lo-Loss Tube C = 1.192+ (16.73823+0.5/Di-B)^(5/2) A3 = 0 If B<0.0029*B^2.935+0.514*B^3 9.000521*(10^6*B/Re)^0.Preface9 Appendix Else 9.3 Small-Bore Honed-Orifice Meter Run with Corner Tappings C = (0.5959+0.043+0.47/(1-B))^1.3309*B-1.8)*B^3.1.034/Di) A4 = 0 If B>0.43/Di^0.1.5991+0.2 9.005-0.5084*B^3 9.039*B^4/(1-B^4)+0.5961+0.87+8.48^B^8+0.225/Di)*B^5+1.1.3+(0. 0934*B^4/(1-B^4)-0.00175*B^2-0.8 Segmental Orifice 9.1132*Bet0^3+( 55.4 ISO Long Radius Nozzle C = 0.1+0.6*Bet0^3+900.1 Critical Flow 58 Sizing Equations .6510*Bet0^4/(1-Bet0^4)0.3+556.0934*Bet0^4/(1-Bet0^4)0.7.15 Venturi Nozzle C = 0.2 Flange Tappings If D 100 C = 0.8*Bet0^4)/(Re1)^0.8.7*Bet0^3394.8*Bet0^2-2722.9 Restriction Device 9.0033*B^4.75 Else DC = 0.8^B^2-399.3312*Bet0^2.2739^B^8-0.2879*Bet0^8+0.4078*B^2.1 9.2^B^2+5691.7.2886*B^3+ (69.1462*B^2.3*B-2557.3813*B^2.3406*Bet0^8-0.5925+0.1+0.5917+0.3-15.9965-0.1019*Bet0^4/(1-Bet0^4)0.75 Side (90) Flange Tappings If D 100 C = 0.0.1+0.4628*B^3 9.1+0.7586^B^8-0.6016+0.0244*B^4/(1-B^4)-0.6^B^2-1287.2603*B^4/(1-B^4)-0.2715*Bet0^3+(-69.8.1+0.6276+0.75 Else C = 0.3932*B^2.196*B^4.6.2*B^4)/(Re)^0.5875+0.7+1328.3380*B^2.8-434.1170*Bet0^4/(1Bet0^4).3*Bet0^4)/(Re1)^0.4016^B^8-0.6276+0.75 Else C = 0.3^B^2+2977*B^3-1131.7-471.5866+0.3343*B^3 Else C = 0.4*B+1245.1598*B^2.5)*(10^6/Re)^0.1-469.3*B^4)/(Re)^0.0569*B^4/(1-B^4)-0.15)*(10^6/Re)^1.6.7*B^3+332.5 ISA 1932 Nozzle C = 0.75 Diametrically Opposite Vena Contracta Tappings If D 100 Dp1 100 DC = 0.1-1.1046*B^4/(1-B^4)-0.2918^B^8+0.0790*B^3 Vena Contracta Tappings If D 100 C = 0.4 Diametrically Opposite Flange Tappings If D 100 C = 0.3 .2*Bet0^2-2460.7.2262*B^4.6*Bet0^3+ 1569.7308*Bet0^3+( 52.3 9.7 Eccentric Orifice 9.53*B^0.2 9.9*B^3-2710.5 9.7.3212*B^3 Else C = 0.6.6284+0.3917*B^2.75 Else DC = 0.2*B^4)/(Re)^0.0547^B^8+0.2845*Bet0^3+(23.6037+0.6261+0.4*Bet0^4)/(Re1)^0.0828*B^2.207*Bet0+821.8*B-4228.3366*B^3+ (7.1 9.5581*Bet0^8+0.7*B+170.6.4*B^4)/(Re)^0.9558-0.75 Side (90) Vena Contracta Tappings If D 100 DC = 0.99-0.4*Bet0+1721.0955*B^4/(1-B^4)-0.2*Bet0^2+1303.5*Re^-0.6898^B^8-0.9 Appendix 9.3412^B^8-0.1-( 0.1+0.2*Bet0+1571.1-0.1132*B^3+ (-103.3 9.5 ASME Long Radius Nozzle C = 0.3061*Bet0^2.5608*B^3+ (-139.1963*B^4/(1-B^4)-0.1 9.1+0.5*B^3+486.9.5*Bet0^21388.1-0.1+0.2739*Bet0^8-0.1-0.6 Nozzle 9.75 9.5949+0.5922+0.0828*Bet0^2.2*Bet0^4)/(Re1)^0.9*Bet0-1332.00653*(B^0.9*Bet0^3+ 1420.8464^B^8+0.2 9.2273*B^4/(1-B^4)-0.1851*Bet0^2.2+898.9975-6. 080*exp(-10*L1) 0.4 9.5 If 3.9.0261*B^2 .1 9.3 9.0.5 C = 0.11*A)*(B^4/(1 .196*B^4.0063*A) * B^3.9.Di) 9.0.216 *B^8 + 0.5*10^5 C = 1 .1.21/Rd^0.4 9. Thick Plate Square Edge Orifice (Liquid/Water) .1 Square Edge Orifice (Liquid/Water).3 9.75 .2 9.5 If 3.3.5*10^5 C = 1 .2.0.5*10^6 C = 0.0.000521*(1000000*B/Re)^0.9975-6.9.5961 + 0.222/Rd^0.8*M^1.5)*(1/Re)^0.53*(B^0.985 Square Edge Orifice (Gas/Steam) C = 0.0.2.B) * (2.83932 Toroidal Throat Venturi Nozzle C = 0.2 9.2.0.043+0.5 * (1000000/Re)^0.5*10^5<=Rd<2.9.5 Sizing Equations 59 .5*10^6<Rd C = 1 .5*10^5<=Rd<2.5*10^6 C = 0.2 9.9.9.011*(0.0188+0.196*B^4.1)*B^1.2.2 9.1.9.525/Rd^0.7.6 Square Edge Orifice (Liquid/Water) C = 0.9.4/D L2 = 25.5 Cylindrical Throat Venturi Nozzle If Rd<3.2 Sub-Critical Flow 9.7.6 Thick Plate Square Edge Orifice (Liquid/Water) C = 0.7 + (0.9558-0.7 ASME Long Radius Nozzle If Rd<3.21/Rd^0.5 C = 0.1.B^4)) .8 .Preface9 Appendix 9.9887 If 2.5*10^6<Rd C = 1 .1.5 Classical Venturi Tube Toroidal Throat Venturi Nozzle Cylindrical Throat Venturi Nozzle ASME Long Radius Nozzle C = 0.99354 .899 Classical Venturi Tube C = 0.B) C = 0.222/Rd^0.8 M = 2 * L2 / (1 .4/D A = (19000 * B/Re)^0.031*(M .1.9. IF D < 71.984 C = 0.3 + (0.1.9558-0.12 THEN C = C +0.9.9.2.9887 If 2.5 9. and Square Edge Orifice (Gas/Steam) L1 = 25.9.9.1.123*exp(-7*L1)) * (1 .1.
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