Unformatted text preview: cumference = 35.3/0.487 = 72.5
Call it 72 turns Radial distance of the turns = (0.3  0.01)/2 = 0.145 m
Center to center distance between turns = 0.145/72 = 0.002 m or 2 mm. Exercise 14.6
Subject: Module volume and packing density of a monolithic membrane element.
Given: Element of 0.85 m length, with 19 flow channels (see Fig. 14.4d) of 0.5 cm inside
diameter each. 9 elements per module.
Assumptions: Center to center triangular spacing of holes = l = 0.5 cm.
Find: (a) Module volume in m3.
(b) Packing density in m2/m3, with comparison to Table 14.4.
Analysis: Each equilateral triangle of 0.5 cm includes onesixth of a hole at each apex or 50%
of the area of one hole. The area of each triangle = 0.433 l2 = 0.433(0.5)2 = 0.1083 cm2. The
number of triangles within the area covered by the holes (that is, excluding the borders) = 24.
For the borders we might assume that the triangular spacing extends to the edges of the monolith.
If so, we add 30 more triangles, giving a total of 54. Then, the total crosssectional area of one
element = 54(0.1083) = 5.85 cm2.
(a) Module volume for 9 elements = 9(5.85)(0.85)/(104) = 0.00448 m2
(b) Membrane area of one element = 3.14(0.5)(19)(0.85)/(102) = 0.253 m2
Module membrane area for 9 elements = 9(0.253) = 2.28 m2
Packing density = 2.28/0.00448 = 508 m2/m3. This is higher than given in Table 14.4. Exercise 14.7
Subject: Passage of water at 70oC through a porous polyethylene membrane by a pressure
differential
Given: Membrane of 25% porosity, with 0.3 µm diameter pores and a tortuosity of 1.3.
Pressures on either side of the membrane are 500 kPa and 125 kPa.
Assumptions: Assume membrane is 1 micron thick and flow is fully developed laminar flow.
Find: Flow rate of water in m3/m2day = Q/AM
Analysis: Provided that the flow is laminar and fully developed, Eq. (14) applies, with a
correction for the tortuosity:
2
Q
N εD P0 − PL
=
=
AM
ρ
32µl M τ (1) At 70oC, µ of water = 0.42 cP or (0.42)(0.001) = 0.00042 Pas and ρ = 1000 kg/m3
P0 = 500 kPa = 500,000 Pa and PL = 125 kPa = 125,000 Pa
ε = 0.25, D = 0.3 µm = 3 x 107 m, lM = 1 x 106 m, and τ = 1.3.
Substitution into Eq. (1) gives:
−7 2
Q
N (0.25)(3 × 10 ) ( 500, 000 − 125, 000 )
==
(3600)(24) = 41, 700 m3 /m 2 day
−3
−6
AM ρ
32(0.42 × 10 )(1× 10 )(1.3) Now check the Reynolds number.
Velocity in the pore from Eq. (14.2) =
2 3 × 10 −7 500,000 − 125,000
D2
υ=
P0 − PL =
= 2.51 m / s
32µl M
32 0.42 × 10 −3 1 × 10−6
N Re −7
Dυρ 3 × 10 2.51 1000
=
=
= 1.78
µ
0.42 × 10 −3 Since the Reynolds number is < 2100, the flow is laminar, but it probably is not fully developed. Exercise 14.8
Subject: Knudsen flow through a porous glass membrane
Given: Porous glass membrane with an average pore diameter of 40 Angstroms. Pressures are
not more than 120 psia upstream and 15 psia downstream. At 25oC and this pressure range,
permeability of helium = 117,000 barrer. Experimental permeability of CO2 = 68,000 barrer.
Assumptions: Knudsen flow is dominant.
Find: The...
View
Full Document
 Spring '11
 Levicky
 The Land

Click to edit the document details