13.8. DERIVATION OF FINITE-DIFFERENCE NUMERICAL METHOD FOR ASYMMETRIC MEMBRANES

13.8A. Countercurrent Flow

The flow diagram for the numerical method is shown in Fig. 13.8-1. Using the method derived by McCabe et al. (M5) and taking an area ΔAm, the mass balances on both streams can be written as

Equation 13.8-1


Figure 13.8-1. Flow diagram for countercurrent flow with asymmetric membrane using finite difference method.


The value of can be calculated from Eq. (13.7-27) using xin for x and for y'. Also, can be calculated similarly from xout. Then, writing a balance on A,

Equation 13.8-2


where . Substituting Lout from Eq. (13.8-1) into (13.8-2),

Equation 13.8-3


Equations (13.7-27), (13.8-2), and (13.8-3) can be solved numerically starting at the feed xf = xin to determine the permeate ΔV and for each ΔAm and increments of x to xo at the outlet reject. Usually 10 or so increments of x from xf to xo are sufficient.

Then, to obtain the values of V and the bulk composition y as a function of x, the calculation is started at xo. Using Equations (13.8-4) and (13.8-5), the increments of ΔV are added to get V, and y is calculated for each ΔAm increment up to xf at the feed inlet:

Equation 13.8-4


Equation 13.8-5


For cocurrent or parallel flow, the only difference is in calculating V. Starting at xf inlet and using Eqs. (13.8-4) and (13.8-5), the increments of ΔV are added to get V up to xo.

To calculate the area for countercurrent or cocurrent flow, rewriting Eq. (13.7-4),

Equation 13.8-6


where r = pt/ph. The average driving force is

Equation 13.8-7


Solving for ΔAm,

Equation 13.8-8


Starting at xf, ΔAm is calculated for each increment to obtain Σ ΔλAm versus x.

13.8B. Short-Cut Numerical Method

Making an approximate material balance to obtain a or yp for the total area from to (M5),

Equation 13.8-9


An overall and component material balance gives

Equation 13.8-10


Equation 13.8-11


Substituting Eq. (13.8-10) into (13.8-11),

Equation 13.8-12


Equation (13.8-12) can be solved for Vp and (13.8-10) solved for Lo.

To calculate the approximate area for countercurrent or cocurrent flow, Eq. (13.8-8) is rewritten for the total area Am:

Equation 13.8-13


where (x - ry')lm is the ln mean of (xf - ) and (xo - ). Since plots of x and y' versus Am are approximately straight lines, this approximate area is within about 15% of the Am calculated by the finite-difference method.

EXAMPLE 13.8-1. Air Separation Using an Asymmetric Membrane for Countercurrent Flow

It is desired to design a hollow-fiber asymmetric membrane for air separation to produce a residue which contains 97.0% N2 using countercurrent flow. The dense polymer layer is in the inside of the tubes and the feed is in the tubes. Experimental values for commercial membranes give separation factors α* of O2/N2 between 3 and 7 (H2) with a typical value of 5.0 (G1). Permeance values of /t are given (H2) of 5 × 10-6 to 250 × 10-6 cm3 (STP)/(s · cm2 · cm Hg) with a typical value of 20 × 10-6 (G1). The typical values will be used. The feed or tube side pressure is 1034 kPa abs and the shell side or permeate is 103.4 kPa. The feed rate of air is 10.0 m3 (STP)/h.

  1. Using the finite-difference method, calculate the permeate composition, the fraction θ of feed permeated, and the residue and permeate flows. Assume negligible pressure drop in the tubes. Calculate the composition y of the permeate as a function of the residue composition x.

  2. Calculate the membrane area needed.

  3. Using the short-cut procedure, calculate the permeate composition, θ, flows, and membrane area.

  4. Repeat (a) for cocurrent flow. Plot x, y, and y' for countercurrent and cocurrent flow versus area.

Solution: For part (a), Lf = 10.0 m3 (STP)/h, xf = 0.209, xo = 0.03, ph = 1034 kPa, pl = 103.4 kPa, /t = 20 × 10-6 cm3 (STP)/(s · cm2 · cm Hg), α* = 5.0. Using Table 13.3-1,


Starting at the feed inlet (Fig. 13.8-2), where xf = 0.209, using an area ΔAm, going from xf = x1 = 0.209 to x2 = 0.190, and using Eq. (13.7-27) to calculate y',

Equation 13.7-27


Figure 13.8-2. Finite-difference elements for countercurrent flow in Example 13.8-1.


The solution to this is given by the quadratic equation (13.4-7):

Equation 13.4-7


where a = 1 - α*, b = -1 + α* + 1/r + x/r(α* - 1), and c = -α*x/r. Substituting and solving for where x1 = 0.209,


Again, for x2 = 0.190, = 0.4830 is obtained from Eq. (13.7-27).

Then, = (0.5148 + 0.4830)/2 = 0.4989. Substituting into Eq. (13.8-3),


Then by Eq. (13.8-1),


For the second increment for ΔAm2 in Fig. 13.8-2, solving for with x3 = 0.170,


Using Eq. (13.4-7), = 0.4471. Then = (0.4830 + 0.4471)/2 = 0.4651


Also,


This is continued, and for the eighth increment, x8 = 0.07, x9 = 0.050, = 0.2198, = 0.1629, = 0.1914, ΔV8 = 0.7757, and L9 = 4.7088. For the final or ninth increment, x10 = xo = 0.030, x9 = 0.050, = 0.1629, = 0.1014, = 0.1322, ΔV9 = 0.9215, and L10 = Lo = 3.7873 m3/h. The permeate flow rate Vp = Lf - Lo = 10.00 - 3.7873 = 6.2127 m3/h. The stage cut θ = 6.2127/10.00 = 0.6213.

To calculate the bulk composition y as a function of x for countercurrent flow, it is necessary to start at xo. Using Eq. (13.8-4),


For calculation of y9 using Eq. (13.8-5),


For the eighth plus ninth increments,


This calculation is continued up to the feed entrance. Values of y, x, and y' are given in Table 13.8-1.

Table 13.8-1. Compositions and Areas for Numerical Method in Example 13.8-1
XCountercurrent yCocurrent yCounter and Cocurrent y'ΔA, Area (m2)A, Area (m2)
0.2090.31810.51480.51480.0000.000
0.1900.29830.49890.48303.6743.674
0.1700.27690.48170.44713.9687.642
0.1500.25490.46370.40844.11711.759
0.1300.23230.44460.36654.32316.082
0.1100.20900.42420.32144.60720.689
0.0900.18470.40210.27265.00625.695
0.0700.15930.37790.21985.59231.287
0.0500.13220.35050.16296.50837.795
0.0300.10140.31810.10148.14445.939

To calculate y as a function of x for cocurrent flow, it is necessary to start at xf. For the first increment, from Eq. (13.8-4),


For the second increment,


This calculation is continued up to the outlet at xo.

For part (b), Eq. (13.8-8) is used starting at xf for both countercurrent and cocurrent flows. For the first increment, the flow rates in m3/h are converted to m3/s by multiplying by (1/3600). Also, xf = 0.209, x2 = 0.190, = 0.5148, = 0.4830, = 0.4989, ΔV1 = 0.6151 m3/h, and ph = 1034 kPa. Calculating x - ry' at points 1 and 2 and the average (xry') from Eq. (13.8-7),


Using Eq. (13.8-8),


For the second increment, x3 = 0.170, = 0.4471, = 0.4651, and ΔV2 = 0.6360. Then,


This is continued, and the results are given in Table 13.8-1. The final total area is 45.94 m2.

For part (c), using the shortcut procedure, Eq. (13.8-9) gives the value of yp:


This value of 0.3081 compares to 0.3181 for the numerical method. Making a component balance in Eq. (13.8-12),


Solving, Vp = 6.437 and Lo = (10.00 - 6.437) = 3.563. This gives a stage cut θ = 6.437/10.00 = 0.6437, as compared with 0.6213 for the numerical method. To calculate the area, the driving forces are


The area from Eq. (13.8-13) is


This value of 53.43 is about 16% greater than the numerical method. A plot of these data for the numerical method is given in Fig. 13.8-3.

Figure 13.8-3. Plot of composition versus area for numerical method in Example 13.8-1.



13.8C. Use of Spreadsheet for Finite-Difference Numerical Method

The numerical method can easily be adapted so that a spreadsheet can be used to solve the problem for asymmetric membranes. Using Excel® or other programs, basic formulas can be entered into cells.

Output data from a spreadsheet are given in Table 13.8-2 for Example 13.8-1. Input data for the calculations are briefly as follows: In cells D3-D10 known values are entered. In cell D11 the equation for r = pl/ph is given by the formula $D$8/$D$7.

Table 13.8-2. Output from Spreadsheet for Finite-Difference Method for Example 13.8-1
Col/ RowABCDEFGHIJK
1Example 13.8-1          
2           
3Input Variables:          
4Lf (feed rate)  10m**3(STP)/hr      
5xf (O2 mole fraction— feed)  0.209       
6xo (O2 mole fraction— outlet)  0.03       
7Pressure (feed), ph  1034kPa      
8Pressure (permeate), pl  103.4kPa      
9/t  1.50E-10m**3(STP)/      
10α*(O2/N2 separation factor)  5s · m**2 · Pa      
11r = pl/ph  0.1       
12           
13I12345678910
14x(I)0.20900.19000.170.150.130.110.090.070.050.03
15A-4.0000-4.0000-4.0000-4.0000-4.0000-4.0000-4.0000-4.0000-4.0000-4.0000
16B22.360021.600020.800020.000019.200018.400017.600016.800016.000015.2000
17c-10.4500-9.5000-8.5000-7.5000-6.5000-5.5000-4.5000-3.5000-2.5000-1.5000
18y'(I)0.51480.48300.44710.40830.36650.32140.27260.21980.16290.1014
19y(avg) 0.49890.46510.42770.38740.34390.29700.24620.19140.1321
20ΔV 0.61510.63610.63000.63070.64010.66170.70220.77590.9219
21L(I)10.00009.38498.74878.11877.48806.84786.18615.48394.70813.7862
22           
23I 123456789
24V(counter)6.21386.21385.59874.96264.33263.70183.06172.40001.69780.9219
25yavgV1.97641.97641.66951.37371.10420.85990.63970.44320.27030.1218
26y(I)0.31810.31810.29820.27680.25490.23230.20890.18470.15920.1321
27           
28Checks:          
29Mass(in)10.0000         
30Mass(out)10.0000         
31[O2]in2.0900         
32[O2]out2.0900         
33           
34I 123456789
35V(cocurr) 0.61511.25131.88132.51203.15223.81394.51615.29196.2138
36yavg*ΔV 0.30690.60270.87221.11661.33671.53331.70611.85461.9764
37y(I) 0.49890.48170.46360.44450.42410.40200.37780.35050.3181
38           
39Calculate the Areas for each increment:          
40           
41(x - ry')avg0.14960.13350.11720.10130.08560.07030.05540.04090.02680.0099
42(/t)*ph0.5584         
43ΔA3.67353.96904.11704.32234.60635.00615.59076.50728.1449 
44Total Area45.9370         

Row 13 is used to keep track of the indexes used in this part of the calculation. Row 14 gives the input values of x1, x2, and so on. In rows 15–18 the values of a, b, c, and y' are calculated using Eqs. (13.4-7) and (13.7-27). For example, in cell B18 the formula entered is =(-B16+SQRT(B16*B16-4*B15*B17))/2*B15). In row 19, . In row 20, ΔV is calculated from Eq. (13.8-3). The formula in cell C20 is =B21*(B14-C14)/ (C19-C14) and in D20 is =C21*(C14-D14)/(D19-D14). In row 21, L is calculated from Eq. (13.8-1), where L2 = L1 - ΔV1 by using the formula in cell C21, which is = -(C20-B21), for C22, =-(D20-B21), and so on.

Row 23 is a new set of indexes for countercurrent flow. To calculate V in row 24 using Eq. (13.8-4), where V = ∑ ΔV, and starting at the outlet xo, the formula in cell K24 is =K20, in J24, = J20+K24, and so on. For row 25, to calculate ∑ ΔV, the formula in cell K25 is =K19*K20, in J25, J19*J20+K25, and so on. For the bulk permeate composition, y is calculated in row 26 from Eq. (13.8-5), where y= ΔV/V. In cell B26, the formula is =B25/B24.

Rows 28-32 are optional and are used to check the mass balances for the inlet and outlet streams. In cell B29, the mass in is =$D$4; the mass out in B30 is =B24+K21. Similarly, cell B31 is =$D$4*$D$5 and B32 is =B24*B26+K21*K14.

For cocurrent flow, starting at xf, Eq. (13.8-4), where V = ∑ ΔV, is calculated in row 35. In row 36, ∑ ΔV is calculated using the formula in cell C36 of =19*C20, in D36, =D19*D20+C36, and so on. Then Eq. (13.8-5) is used to calculate the bulk composition y from ∑ ΔV/V in row 37.

To calculate the areas, in row 41 (x - ry')av is calculated for each increment, using Eq. (13.8-7). The formula in cell B41 is =((B14-$D$11*B18) + (C14-$D$11*C18))/2. In cell B42 the conversion of V m3/h to m3/s and kPa to Pa is performed using = ($D$9*$D$7*1000*3600). In row 43, using Eq. (13.8-8) to calculate ΔAm, the formula in cell B43 is C19*C20/($B$42*B41). In row 44 all of the ΔAm values in row 43 are added to get the total Am using =SUM(B43:J43).

13.8D. Calculation of Pressure-Drop Effects on Permeation

1. Estimation of number of tubes and length

In order to calculate the pressure drop inside the hollow-fiber tubes, the dimensions, number of tubes, and area Am of the membrane must be known. The following procedure can be employed. The shortcut procedure is first used with Eqs. (13.8-9)-(13.8-13). This gives Lf, Lo, Vp, , and Am. Then the average flow rate in the tubes is calculated from the equation Lav = (Lf + Lo)/2.

The number of tubes in parallel NT must be estimated from typical laboratory or commercial data. For example, for air with a flow rate of 3 m3/h, 3.8 × 104 fibers (NT) are used with an ID of 95 μm, and a length (z2 - z1) of 48.26 cm. For a flow rate Lf of 10.0 m3/hr in Example 13.8-1 and to keep a similar and reasonable velocity in the tubes, the NT needed will be directly proportional to the inlet Lf. Then, knowing the Am, an estimated total length (z2 - z1) of the tubes can be calculated from Eq. (13.8-14):

Equation 13.8-14


2. Finite-difference method to include effect of pressure drop

For the finite-difference numerical method, Eq. (13.8-14) can be rewritten as

Equation 13.8-15


Also, Eq. (13.7-38) can be rewritten for a finite increment of Δz length:

Equation 13.8-16


The following steps, which may require several iterations, can be used in the finite-difference numerical method to include the effect of pressure drop in the tubes:

1.
For the first increment shown in Fig. 13.8-2, a constant inlet pressure ph1 (ph) is used for phav to calculate and , ΔV1, L2, and ΔAm1 from Eqs. (13.7-27), (13.8-1), (13.8-3), (13.8-7), and (13.8-8).

2.
The value of NT is obtained from typical data as discussed above in Subsection 1. Using Eq. (13.8-15), the incremental length Δz is obtained.

3.
Using Lav = (L1 + L2)/2 and the inlet ph1 for ph av, the Δph for the first increment is calculated from Eq. (13.8-16).

4.
For the second iteration, the new ph av to use is calculated from ph av = ph1 - Δph/2. A new value of the outlet ph2 = ph1 - Δph is used to recalculate a new from Eq. (13.7-27). The value of remains the same. Then by repeating steps 1 through 4, a new value of Δph is obtained. Usually only three or so iterations are needed to obtain a constant value of Δph, phav, and ΔAm1. Using a spreadsheet as in Ex. 13.8-1, this procedure can be repeated for each finite section.

3. Determination of effect of pressure drop using shortcut method

The values of Lav, NT, Am, and overall length (z2 - z1) are obtained from the shortcut calculation in Subsection 1 above. The shortcut value of the overall pressure drop (ph - pho) can be obtained from Eq. (13.8-16) using ph at the inlet as ph av and the overall length (z2 - z1). Then for a second and final iteration, the new and corrected pressures ph av = ph - Δph/2 and pho = ph - Δph are used in Eqs. (13.7-27) and (13.8-9)-(13.8-13) to recalculate , , Vp, and Am again. This recalculated Am will include an approximate effect of pressure drop in the tubes.

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