J-F APPENDIX 3 EXAMPLE CEI CALCULATIONS
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1 ~ Dow's Chemical Exposure Index Guide, First Edition by AIChE technical manual Copyright 1994 American Institute of Chemical Engineers APPENDIX 3 EXAMPLE CEI CALCULATIONS CHLORINE VAPOR RELEASE The 3/4 inch vapor connection on a 1 ton chlorine cylinder stored at ambient temperature (30 "C or 86 "F) has broken. Needed information: Pressure inside the cylinder, Pg Absolute Pressure, Pa Molecular Weight, MW Storage Temperature, T Diameter of Hole, D SI US/Brit Pa gauge psig kpa psia "C 86 OF 19 mm 0.75 in Determine Airborne Quantity. SI Units (Equation IA) MW AQ=4.751~10 D Pa -6 d(t+273) i- AQ=4.751 x 104(19)2(889.5) AQ = 0.74 kg / see USIBrit Units (Equation I B) ge AQ=3.751D2Pa / AQ = (.75)2(129.0) :9) (8:!: AQ = 98.2 lb I min Calculate the CEI. SI Units (Equation I OA) whereerpg-2 = 9mg/m3 CEI = &z J-F CEI= CEI = 188 USIBrit Units (Equation I OB) where ERPG-2 = 3 PPM /= CEI = /- CEI = CEI = 191 Differences in CEI values result from rounding the EFWG values when converting between PPM and mghn3. 27
2 Calculate Hazard Distances. SI Units (Equation I IA) USIBrit Units (Equation I I B) For ERPG-2 = 9 mdm3 JZ' HD=6551 -,/? HD = HD = 1,878 meters For ERPG-1 = 3 mg/m3 HD = G HD = 3,254 meters For ERPG-3 = 58 mdm3 HD = 655 1,/; 0.74 ForERPG-2 = 3PPM HD = 9243 /2zi HD = HD = 6,280 feet ForERPG-1 = 1 PPM /- HD = 9243 HD = 10,878 feet ForERPG-3 = 20PPM HD = 740 meters HD = 2,432 feet Differences in HD values result from rounding the ERPG values when converting between PPM and mghn3. 28
3 AMMONIA LIQUID RELEASE Ammonia is stored in a 12 ft diameter by 72 ft long horizontal vessel under its own vapor pressure at ambient temperature (30 "C or 86 OF). The largest liquid line out of the vessel is 2 inch diameter (50.8 mm). Needed infomation: Pressure inside vessel, P, Temperature inside vessel, T Normal boiling point Liquid density in vessel, p1 Ratio Cp/Hv Height of liquid in tank, Ah Diameter of hole, D Molecular weight, MW SI 1G64 kpa gauge 30 OC "C kdm E m 50.8 mm US/Brit psig 86 OF -28 O F 37.1 wft E ft 2.0 in Estimate liquid released. SI Units (Equation 2A) USIBnt Units (Equation 2B) lo0op L = 9.44 x 10-7D2 p l, + / 9.8 Ah F L= 9.44 x 10-7(50.8)2(594.5) J+9.8(3.66) L= 61.9 kg /se~ L = 2.234(2.0)2 (37.1) L = 8,200 Ib /min Estimate flash fraction. SI Units (Equation 4A) C F~ =-E(T,-T~) HV Fv = 0. oo40 1 (30 - (-3 3.4)) Fv = USIBnt Units (Equation 4B) r- Fv =-(T,-Tb) LP HV Fv = (86-(-28)) Fv = Sine F,, > 0.2 AQ = L Since Fv > 0.2 AQ = L AQ = 61.9 kg/sec AQ = 8,2001b/min 29
4 Calculate CEI. SI Units (Equation I OA) where ERPG-2 = 139 mg/m3 USIBnt Units (Equation I OB) whereerpg-2 = 200PPM CEI = i E 2 2 CEI = E CEI = 437 /z& CEI = CEI = 437 Calculate the Hazard Distances. SI Units (Equation IIA) For ERPG-2 = 139 mg/m3 USlBrit Units (Equation IIB) For ERPG-2 = 200 PPM HD = 655 1, / g HD = 6551E HD = 4,372 meters For ERPG-I = 17 mg/m3 HD = 6551E HD = 12,500 meters ForERPG-3 = 696mg/m3 d6qg69 HDz HD = 1,953 meters / HD = 9243 :?03) 2: HD = 14,342 feet For ERPG-I = 25 PPM Jzz HD = 9243 HD = 40,564 feet For ERPG-3 = 1000 PPM HD = 9243 Jzz3) HD = 6,414 feet 30
5 STYRENE UQUID RELEASE Styrene is stored in a 40 ft x 40 ft API tank at ambient temperature (25 "C or 77 OF). The tank has a closed vent system but is essentially at ambient pressure. The outlet is a 6-inch Schedule 40 nozzle. Needed information: Pressure inside the tank, P, Temperature inside the tank, T Normal boiling point Vapor pressure, ambient temperature Liquid density, pi Height of liquid, Ah Molecular weight, MW SI 0.0 Wa 25 OC "C kpa kg/m m USBrit 0.0 psig 77 O F OF psi 56.3 lb/ft ft Scenario selection - For greater than 4-inch diameter, use 20% of the cross sectional area (CSA). For 6-inch Schedule 40, CSA = in2 0.20(28.89)=5.78 in2 =fi D = = 2.71 in or 68.9 mm Estimate liquid released. SI Units (Equation 2A) USIBrit Units (Equation 2B) L = 9.44 x 10-7D2 p lo0op 1 i F - 4 L=9.44 x 10-7(68.9)2(901.6) L = 2.234D2 + Ah L= 2.234(2.71)2(56.3) JF L = 44.2 kg /sec L = 5,842 lb / min Compare operating temperature to normal boiling point: 25 OC < 145 OC 77 OF < OF Therefore, Flash Fraction = 0 Estimate pool size. SI Units (Equation 3A) w' = m(l) WT = 39,800 kg = wp USIBrit Units (Equation 3B) WT = 15(L) WT =87,6001b=Wp 31
6 SI Units (Equation 7A) A, = 100- WP PI A, = A, = 4,410 m2 USlBnt Units (Equation 7B) WP A, = PI A, = A, = 47,460 ft2 Assume no dike. SI Units (Equation 8A) AQ, =9.OxlO4(A, 0.95) (Mw)pv T+273 USIBnt Units (Equation 8B) AQ, = 0.154( A, 0.95) (Mw) pv T+459 Characteristic pool temperature equals ambient (0.841) AQ, =9.0x1O4(4410 ) AQ, =0.767kg/se~ SI Units (Equation 9A) AQ= AQf + AQ, AQ = AQ = kg 1 sec (0.122) AQ, = (47460 ) AQ, =lollb/min USIBrit Units (Equation 9B) AQ= AQf + AQ, AQ = AQ = 101 lb / min Calculate CEP. SI Units (Equation I OA) where ERE-2 = 1065 mg/m3 USIBrit Units (Equation 1 OB) whereerpg-2 = 250PPM CEI = 655.1,/",/% EFWG-2 CEI = CEI=18 CEI = 281.8,/z CEI=18 32
7 Calculate the Hazard Distances. SI Units (Equation I IA) For ERPG-2 = 1065 mg/m3 HD = J"" EFWG E HD= HD = 176 meters ForERPG-1 = 213mg/m3 HD = 393 meters ForERPG-3 = 4259mg/m3 JZ HD= HD = 87.9 meters USIBrit Units (Equation 1 I B) ForERPG-2 = 250PPM HD = 9243 / = ' HD = 9243 J25o:pd1.15) HD = 576 feet ForERPG-I = SOPPM HD = 9243 /zz HD = 1,287 feet ForERPG-3 = 1OOOPPM JizL HD = 9243 HD = 288 feet 33
8 CHLORINE LIQUID RELEASE Chlorine is stored in a sphere at 5 OC (41 OF). liquid to escape. A 2-inch nozzle fails on the bottom of the vessel allowing Needed information: Pressure inside the cylinder, P, Molecular weight, MW Storage temperature, T Liquid density, p1 Height of liquid in the sphere, Ah Diameter of hole, D Capacity of sphere Estimate liquid released. SI 332 kpa gauge "C 1458 kg/m3 6m 50.8 mm x 106 kg UsBrits 48.2 psig O F lb/ft ft 2 in 2.5 x 106 lb SI Units (Equation 2A) USfBrit Units (Equation 2B) Ah L = 2.234(2) JF L = 60.1 kg / sec L = 7,967 lb/min Determine the total liquid released. For 15 minutes (900 seconds), the total liquid leaving the tank is: W, = 900 (60.1) = 54,090 kg WT = 15 (7,967) = 1 19,505 lb The capacity of the tank when full is x 106 kg. Since LT = kg is less than the capacity of the tank. W, =54,090 kg WT = 119,505 lb Calculate the flash fraction. Needed information: Normal boiling point temperature = -34 OC Normal boiling point temperature = OF Heat of vaporization = 275,030 J/kg Heat capacity of liquid (at average temperature) = J/kgJ'C 34
9 A technically correct solution for evaluating the flash fraction quires the heat capacity (C,) to be evaluated at the average temperature (storage and boiling point) and the heat of vaporization at the boiling point. For example: and C, -15 OC or 5 OF) = J/kg/OC = BTU/lbPF H, (BP) = 285,457 J/kg = BTU/lb SI Units (Equation 4A) USlBrit Units (Equation 4B) F, = 943* (5 - (-34.0)) 285,457 F, = F, =- 0'2254 (41 - (-29.2)) Fv = Calculate vapor source strength from the flash. AQf=5(Fv)(L)=5(0.129)(6O.1)=38.8 kg /se~ AQf =5(Fv)(L)=5(0.129)(7967)=5,l39lb/min (SI) (UsBrit) Calculate the total liquid entering the pool. wp =w~(~-~f,)=54,~0(1-(5)(0.129))=19,202 kg wp = WT ( 1-5FV) = 1 19,505( 1 - (5)(0.129)) = 42,424 lb (SI) (USIBrits) Liquid density of chlorine at its boiling point = 1,562 kghn3 SI Units (Equation 7A) USIBnt Units (Equation 7B) A, = 100- wp P A, = A, =30.5- wp P A, = A, = 1,229 m2 A, =13,271ft2 35
10 Calculate the vapor flow rate from the pool. Since chlorine is boiling in the pool, P, = kpa = psi Molecular weight of chlorine = SI Units (Equation 8A) USlBrit Units (Equation 8B) AQ, = ~. O X ~ O ~ ( A 0.95), (Mw)pv T l(101.3) AQ, =9.Ox1O4(1229 ) (-34.0) AQ, = 0.154(AP 0.95) (Mw) pv T (14.70) AQ, =0.154(13271 ) (-29.2)+459 AQ, = 23.3 kg I sec AQ, = 3,083 lb 1 min Calculate source strength of release. SI Units (Equation 9A) AQ = AQ f + AQp USIBrit Units (Equation 9B) AQ = AQf + AQp AQ= AQ = AQ =62.1 kg Isec AQ=8,222lb/min Compare to the liquid release: 62.1 kdsec is greater than 60.1 kg/sec and 8,222 lblmin is greater than 7,967 lb/min: AQ = 60.1 kg 1 sec AQ = 7,967 lb 1 min Calculate the CEI. SI Units (Equation I OA) whereerfg-2 = 9mg/m3 i2z CEI = USIBrit Units (Equation I OB) where ERPG-2 = 3 PPM CEI = E CEI = 281.8JT 3(70.9 1) CEI = 1,963 CEI = This is greater than 1OOO; thus CEI = 1,ooO This is greater than 1OOO; thus CEI =
11 Calculate the Hazard Distances. SI Units (Equation IIA) ForERPG-2 = 9mg/m3 USIBnt Units (Equation I I B) ForERPG-2 = 3PPM HD = 6551/* EFWG HD = 6551E HD = 16,929 meters HD is greater than 10,ooO meters, thus HD = 10,OOO meters For ERPG-I = 3 mgim3 HD = 6551E HD = 29,321 meters HD is greater than 10,ooO meters, thus HD = 10,OOO meters ForERPG-3 = 58mg/m3 HD = 6551c HD = 6,668 meters J- HD = 9243 HD = 56,525 feet HD is greater than 32,800 feet, thus HD = 32,800 feet For ERPG-I = 1 PPM HD = 9243 J- HD = 97,973 feet HD is greater than 32,800 feet, thus HD = feet ForERPG-3 = 20PPM HD = 9243 J- HD = 21,907 feet 37
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