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Show AFRC-JFRC 2004 Joint International Symposium Tuesday October 12 Ftofiniery Issues Flare Liquid Seal Drum Surging: Prediction, Prevention and 4 : 0 0 Proof T. Rhode&John Zink Co., LLCand J. SeeboldChevronTexacc^Relired'i (K^AFROJFRC 2004 JOINT INTERNATIONAL COMBUSTION SYMPOSIUM _______________MAUI F la re L iq u id S e a l D ru m S u rg in g : P re d ic tio n , P re v e n tio n & P r o o f f f f TomRhodes J o h n C o rre s p o n d in g Z in k C o m p a n y , L L C A u th o r: Jo h n /in k C o m p an y . JimSeebold C h e v ro n T e b a c o L L C . 1 1920 (R e t) L ast A p ach e. J u lsa . O K 7 4 1 1 6 Flare "smoke signals" due to seal drum sloshing trailing off at intervals of about 1 second P ic tu r e from J o h n Z in k C o m b u s tio n H a n d b o o k , C R C P r e s s , 2 0 0 1 , p. 6 2 0 0 4 The F lu id M echanics o f B eer Steins Formulation $ rr + ( i / r ) $ r + ( i / r 2) $ ee + $ zz = 0 u = - $ r , v = - (i/r )$ e , w = - $ z u = - $ r = 0 at r = a w = - $ z = 0 at z = - h p /p + v2/ 2 + gs = O t st = w = -O z ® tt = gst = - g®z a t z = o The F lu id M echanics o f B eer Steins Solution $(r/0 <z,t) = ^ (r/0,z)sinwt ¥ rr + (l/r)Yr + (l/r )Tee + ¥ zz = 2 0 Yr = 0 at r = a Yz = 0 at z = - h w2Y = gYz at z = 0 Y = Jm(ar)cos(m0)cosh[a(z+h)] w2 = g(Yz/Y )|z=0 = gatanh(ah) J m'(aa) = 0; for m = 1< aa = 1.8412 w2 = g(1.8412/a)tanh(1.8412h/a) T = 2n/w = [2nV(a/g)]/V[1.8421tanh(1.8421h/a)] T h e F lu id M e c h a n ic s o f B e e r S te in s D a = d / 2 = 6 "/ 2 o e s i t w o = 3 " = 0 .2 5 ' ; h r k = ? 6 " = 0 .5 ' 2 n f (0 .2 5 /3 2 .2 ) -------- :- ;-------------------- = 0.408 sec •/ / [1 .8 4 1 2 ta n h (1 .8 4 1 2 x 0 .5 /0 .2 5 )] -------------- M odification for an A n n u lar Tank 4.0 3.8 3.6 3.4 3.2 3.0 P a 2 8 2 . 6 2.4 2 .2 2 . 0 1 .8 1 .6 1.4 1 .2 1 .0 i T = 2 n /w 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 2n V (D 0/2 g ) ] V [£ ta n h (2 £ h /D 0)] 0.8 0.9 1 1stMode Sloshing (antisymmetric) V an D y k e , M ., A n A lb u m o f F lu id M o tio n , T h e P a ra b o lic 2 nd M o d e S lo s h in g (s y m m e tric ) P r e ss, S tanford, 1 9 8 2 , P la te 191. 1 st M o d e Sloshing (antisymmetric) 2 nd M o d e Sloshing (symmetric) 1stMode Sloshing (antisymmetric) 2 nd M o d e S lo s h in g ( s y m m e tric ) Liquid PendulumMode My" + F = 0, M = (p/g)(wd02/4)h, F = 2p(ndj2/4)y, y" + [(2g/h)(di/do)2]y = y" + w2y = 0, T = 2w/w = 2w(d0/dj)//(h/2g) R e a l W o rld E x a m p le ♦ \ L_ 3 v.\ 5 A. A Jk o *\T\J V v -► 3 p 8-A t5/D0 Dj/D0 = 3/8 = 0.38; aist Mode = 1.42; T = 2nV(D0/2g)]/V[£tanh(2pH/D0)] = 2wV(8/32.2)]/V[1.42tanh(2x1.42x5/8)] = 1.91 sec; R ic h m o n d , C A c a .1 9 7 2 . f = 1/2T = 1.05/sec Calculated vs. 1.1/sec Observed S Example Sloshing Calculations H (rti- 5 , 0 0 Z Di/Do = "i a 2 . 5 3 1 .0 5 St Modfl (anlisymmtmc) T< «) = 1 .9 1 2 nDMods Symmetric) 1 . 4 2 Liquid Pendulum Tjs«)T (s«| - 3 .0 0 1 . 4 2 1 4 Dj Ht| - M 1 .& 7 T 0 . 7 0 0 . 5 4 NOTE: For the antisymmetric (1st) mode there are two surges per period so f = 2/T; for the others, f = 1/T. A n o th e r R e a l W o rld E x a m p le B a r b e r s P o i n t , HI ca.2000 D/Do = 3.5/9.42 = 0.37; ^ Mode = 1.45; T = 2nV(Do/2g)]/V[ptanh(2pH/Do)] = 2wV(8/32.2)]/V[1.42tanh(2x1.42x5/8)] = 2.58 sec; . f = 1/2T = 0.78/sec Calculated vs. "45-48/min" = 0.75-0.80/sec Observed S The problems of the vibration of air and liquid are analogous. The length independent /ibration period:s are mathematically identical when a = 1.8412C2/g. One can learn a lot from beer steins... I - 1 |