DATOS DUKLER (1)

PROCEDIMIENTO METODO DUKLER DATOS: Vsg = 4.09 ft/s N LV = 3.02 Vs L = 2.65 ft/s. µ o = 18 cp D = 0.249 ft. µ g = 0.018 cp Y L = 56.6 lbm/ft. F TP = 0.464 Y g = 2.84 lbm/ft. H L = 0.49748 A = 0.048695 ft 2 PROCEDIMIENTO. H L = 0.49748 P K = P LλL 2 HL + P g ( 1−λ ) 2 1−H g P K = ( 56.6)( 0.3931 ) 2 0.49748 + ( 2.84 )( 1−0.3931) 2 ( 1−0.49748 ) P K = 23.4929 lbm/ft 3 V m = V SL + V SG = (2.65)+(4.09) V m = 6.74 ft/s µ n =µ L λ L + µ g λ g = [18(0.3931) + (0.018)(1-0.3931)] µ n = 7.0867 cp

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Transcript of DATOS DUKLER (1)

PROCEDIMIENTO METODO DUKLERDATOS:

Vsg = 4.09 ft/s NLV= 3.02VsL = 2.65 ft/s. o = 18 cpD = 0.249 ft. g = 0.018 cpYL = 56.6 lbm/ft. FTP = 0.464Yg = 2.84 lbm/ft. HL = 0.49748A = 0.048695 ft2

PROCEDIMIENTO.HL = 0.49748PK = + PK = PK = 23.4929 lbm/ft3Vm = VSL + VSG = (2.65)+(4.09)Vm = 6.74 ft/sn =L L + gg = [18(0.3931) + (0.018)(1-0.3931)]n = 7.0867 cpNRek = 1488NRek = 8278.5327

De la figura 4.4 Beggs and Brill

HL es aproximado a 0.49748Fn = 0.0056 +0.5NRek-0.32Fn = 0.0056 + 0.5(8278.5327)-0.32 = 0.00334

De la figura 4.3 (Beggs and Brill) tenemos que:

Cuando:y= -In(L)Entonces: