Unformatted text preview: drop at the end of stage 1 = 25 psia = 172,400 Pa
Rearrange (14103) to compute the time, t, in seconds to reach the pressure drop of 25 psia, using
SI units.
t= ∆P
Rm
172, 400
1.43 × 1010
−
=
−
J 2µK 2 cF K 2 JcF ( 9.63 × 10−5 )2 ( 0.001) ( 3.78 × 1012 ) ( 4.3) ( 3.78 × 1012 )( 9.63 ×10 −5 ) ( 4.3) = 1,138 – 9 = 1,129 s = 18.82 min
During stage 1, the cumulative volume, mL, as a function of time in minutes is given by,
V (in mL) = 10 t (in minutes) up to 18.82 minutes Exercise 14.27 (continued) Cumulative Permeate Volume, mL Analysis: (continued)
For the stage 2 with operation at constant pressure drop of 25 psia, (14110) is used to
compute the cumulative permeate volume V, starting from VCF = volume at the end of stage 1 =
10(18.82) = 188.2 mL and tCF = 18.82 minutes = 1,129 s. The decrease in permeate flux, J, in
stage 2 is obtained from (14111). The calculations are best carried out with a spreadsheet.
The results shown in the following two plots show that stage 2 ends at 2,850 seconds or
47.5 minutes.
400
350
300
250
200
150
100
50
0
0 10 20 30 40 50 Time, minutes Permeate Flux, mL/minsq cm 0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
0 5 10 15 20 25 30 Time, minutes 35 40 45 50 Exercise 15.1
Subject: Estimation of adsorbent characteristics. Given: Porous particles of activated alumina. BET area = 310 m2/g = Sg. Particle porosity =
0.48 = εp. Particle density = 1.30 g/cm3 = ρp.
Assumptions: Straight pores of circular crosssection with uniform diameter.
Find: (a) Vp = specific pore volume in cm3/g.
(b) ρs = true solid density, g/cm3.
(c) dp = average pore diameter, angstoms.
Analysis:
(a) From Eq. (153), Vp = εp/ρp = 0.48/1.30 = 0.369 cm3/g
(b) From a rearrangement of Eq. (155), ρs = ρp/(1  εp) = 1.30/(1  0.48) = 2.50 g/cm3
(c) From a rearrangement of Eq. (152), dp = 4 εp/Sgρg = 4(0.48)/[310 x 104(1.30)]
= 4.76 x 107 cm or 47.6 angstroms. Subject: Exercise 15.2
Estimation of surface area of two forms of molecular sieves. Given: Form A with Vp = 0.18 cm3/g and average dp = 5 angstroms.
Form B with Vp = 0.38 cm3/g and average dp = 2.0 microns.
Assumptions: Straight pores of circular crosssection with uniform diameter.
Find: Surface area, Sg, of each form.
Analysis: By substitution of Eq. (153) into (152), Sg = 4 Vg/dp (1) Form A: From Eq. (1), Sg = 4(0.18)/(5 x 108) = 14.4 x 106 cm2/g or 1,440 m2/g Form B: From Eq. (1), Sg = 4(0.38)/(2 x 104) = 7,600 cm2/g or 0.760 m2/g This is a very large difference in surface area. Exercise 15.3
Subject: Consistency of the characteristics of a smallpore silica gel. Given: Smallpore silica gel. Pore diameter = 24 angstroms = dp. Particle porosity = 0.47 = εp.
Particle density = 1.09 g/cm3 = ρp. Specific surface area = 800 m2/g = Sg.
Assumption: Straight pores of circular crosssection with uniform diameter.
Find: (a) Reasonableness of the above values.
(b) Fraction of a monolayer adsorbed if the adsorption capacity for water vapor at 25oC
and 6 mmHg partial pressure is 18% by weight.
Anal...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details