6 134 and 6 135 n rev v v and n scv v av v dv 1 cv

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2) 1 1/ 2 4(0.930) HG = ( 0.947 - 0.0241) 0.422 3004 1/ 2 ( 850 ) -3/ 4 ( 0.36 ) -1/ 3 (2.56)(300) = 0.16 m (0.28 10-4 )(78) At the top of the column, vapor rate = V =1,495 lbmol/h and liquid rate = L = 727 lbmol/h. Therefore, mV/L = (0.42)(1,495)/(727) = 0.864 mV From Eq. (7.51), for NOR PAC, H OG = H G + H L = 0.34 + (0.864)(0.117) = 0.44 m = 1.5 ft L Exercise 7.52 (continued) Analysis: (c) at the top of the column (continued) mV For Montz, H OG = H G + H L = 0.16 + (0.864)(0.054) = 0.21 m = 0.68 ft L At the bottom of the column: Near a mole fraction, xB , of 0.005, m = 2.5 Estimate HL from Eq. (6-123), using the following properties and parameters: CL hL , m3/m3 , m3/m3 uL, ft/s a, m2/m3 ah, m2/m3 NOR PAC 1.080 0.0269 0.947 0.0228 86.8 91.7 Montz 1.165 0.0341 0.930 0.0150 300 162 Need an estimate of the diffusivity of methanol in water at low concentrations of methanol. From Example 3.7, the diffusivity of methanol in water at infinite dilution at 25oC is 1.5 x 10-5 cm2/s. Use Eq. (3-39) to correct this for...
View Full Document

This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .

Ask a homework question - tutors are online