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{SECT 0 {PARA 260 "" 0 "" {TEXT -1 25 "Passage Across Membranes:" }}
{PARA 261 "" 0 "" {TEXT -1 34 "Concurrent or Counter-Current Flow" }}
{PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT 256 9 "Jim H
erod" }}{PARA 257 "" 0 "" {TEXT 257 21 "School of Mathematics" }}
{PARA 258 "" 0 "" {TEXT 258 12 "Georgia Tech" }}{PARA 259 "" 0 ""
{TEXT 259 21 "herod@math.gatech.edu" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 532 " We model a procedure for cleaning \+
a solute from a fluid. Imagine two tubes; fluid flows through both of \+
them. Imagine now that the inner tube contains an expensive fluid whic
h contains the solute we wish to remove. Think of the outer fluid as a
cleaning agent in which the solute is soluble and from which the solu
te can be removed inexpensively, without damage to the fluid in the in
ner tube. The wall for the inner tube is a membrane through which the \+
solute will pass, but neither of the fluids will pass. We draw a pictu
re." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 192 "J:=tubeplot([t,t+1,0],t=0..1,radius=1,orientation=[75,80],\n \+
color=BLUE):\nK:=tubeplot([t,t+1,0],t=0..1,radius=0.5,orientatio
n=[75,80],\n tubepoints=20,color=RED):\ndisplay(\{J,K\});" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 509 " We suppose that the fluid
is flowing in both tubes and that the solute passes through the membr
ane by osmosis. We ask: is it more efficient for the two fluids to flo
w in the same direction, or to flow in opposite directions? In the fir
st case, the cleaning fluid gets progressively dirtier as it moves in \+
the same direction as the inner fluid. In the latter case, the cleanin
g fluid is cleanest at the exit for the inner fluid and is dirtiest at
the end where the inner fluid begins the cleaning process." }}{PARA
0 "" 0 "" {TEXT -1 103 " A first understanding of the mathematics \+
of such a problem is contained in the previous worksheet " }{TEXT 268
21 "Flow Across Membranes" }{TEXT -1 111 ". This worksheet can be unde
rstood independent of that one, it does give intuition for the situati
on, however. " }}{PARA 0 "" 0 "" {TEXT -1 127 " We derive equation
s for this model. The equations are only a little more complicated tha
n those of the previous worksheet." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 53 "Derivation of the equations for a
same direction flow" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 123 "Suppose the two tubes have length L. Suppose also that
the rate of diffusion across the walls of the inner tube depends on \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 15 "" 0 "" {TEXT -1 79 "the \+
area of the wall of the vessel across which the diffusion takes place,
and " }}{PARA 15 "" 0 "" {TEXT -1 125 "the difference between the con
centration of the solute inside the inner cylinder and the concentrati
on outside this cylinder." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 58 " Thus, take a small portion of the cylinder of
length " }{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT -1 65 "x. We appr
oximate the amount A of diffusion out of this portion:" }}{PARA 0 ""
0 "" {TEXT -1 22 " A(x + " }{XPPEDIT 18 0 "Delta" "6#%&
DeltaG" }{TEXT -1 21 "x) - A(x) = k 2 " }{TEXT 261 1 "a" }{TEXT
-1 1 " " }{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT -1 18 "x ( S(x) - \+
C(x) )," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9
"where 2 " }{TEXT 262 1 "a" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Delta" "6
#%&DeltaG" }{TEXT -1 594 "x is the surface area of the small portion o
f the cylinder under consideration, and k is a constant characteristic
of the properties of the wall of the vessel and the particular diffus
ing substance that is being considered. The difference S(x) - C(x) is
the difference in the concentration of the solute across the membrane
at distance x from the end where the inner fluid enters. If S(x) < C(
x), the fluid in this section of the inner tube will decrease. At the \+
same time a volume V of the solution passes through the small section \+
and as it does, the concentration will change by the amount " }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 " \+
C(x + " }{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT -1 14 "x) - C(x) =
" }{XPPEDIT 18 0 "(A(x+Delta*x)-A(x))/V" "6#*&,&-%\"AG6#,&%\"xG\"\"
\"*&%&DeltaGF*F)F*F*F*-F&6#F)!\"\"F*%\"VGF/" }{TEXT -1 12 " = k 2 \+
" }{TEXT 260 1 "a" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Delta" "6#%&DeltaG
" }{TEXT -1 20 "x ( S(x) - C(x) )/V." }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 26 "In the limit, we may write" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 " \+
" }{XPPEDIT 18 0 "diff(C(x),x) = (2*Pi*a*k/V*(S(x)-C(x))" "6#
/-%%diffG6$-%\"CG6#%\"xGF**.\"\"#\"\"\"%#PiGF-%\"aGF-%\"kGF-%\"VG!\"\"
,&-%\"SG6#F*F--F(6#F*F2F-" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 147 "The derivation of the equation fo
r the concentration in the outer tube is about the same. The change in
the amount of diffusion into the portion is" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 " A(x + " }
{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT -1 21 "x) - A(x) = k 2 \+
" }{TEXT 263 1 "a" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Delta" "6#%&DeltaG
" }{TEXT -1 18 "x ( C(x) - S(x) )," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 173 "The equation reflects that if the concen
tration in the inner tube is larger than in the outer tube, then the a
mount of diffusion increases. This change leads to the equation" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 " \+
" }{XPPEDIT 18 0 "diff(S(x),x) = (2*Pi*a*k/V*(C(x)-S(
x))" "6#/-%%diffG6$-%\"SG6#%\"xGF**.\"\"#\"\"\"%#PiGF-%\"aGF-%\"kGF-%
\"VG!\"\",&-%\"CG6#F*F--F(6#F*F2F-" }{TEXT -1 1 "." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "We solve these two equat
ions with C(0) and S(0) specified. We expect that, C(L) < C(0) and S(L
) > S(0)." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 ""
{TEXT -1 47 "Solution of the equations for con-current flow." }}{PARA
0 "" 0 "" {TEXT -1 222 " We take the flow in both tubes to be the \+
same: V = 3, take the diffusivity constant to be 1/5, and the radius o
f the inner tube to be 1/2. Assume the initial value of C to be 5 and \+
of S to be 0: C(0) = 5 and S(0) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 21 "a:=1/2: V:=3: k:=1/5:" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 126 "dsolve(\{diff(C(x),x)=2*Pi*a*k*(S(x)-C(x))
/V,\n diff(S(x),x)=2*Pi*a*k*(C(x)-S(x))/V,C(0)=5,S(0)=0\},\n \+
\{C(x),S(x)\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "C:=
x->5/2*exp(-2/15*Pi*x)+5/2;\nS:=x->5/2-5/2*exp(-2/15*Pi*x);" }}}{PARA
0 "" 0 "" {TEXT -1 27 "We graph the two solutions." }}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 26 "plot([C(x),S(x)],x=0..10);" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 52 "Derivation of the equations for c
ounter-current flow" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 133 "We continue to suppose that the two tubes have length \+
L and that the rate of diffusion across the walls of the inner tube de
pends on " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 15 "" 0 "" {TEXT
-1 79 "the area of the wall of the vessel across which the diffusion t
akes place, and " }}{PARA 15 "" 0 "" {TEXT -1 124 "the difference betw
een the concentration of the solute inside the inner cylinder and the \+
concentration outside the cylinder." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 69 " As before, take a small portion of \+
the inner cylinder of length " }{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }
{TEXT -1 61 "x. The amount A diffused out of this portion is approxima
tely" }}{PARA 0 "" 0 "" {TEXT -1 22 " A(x + " }
{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT -1 21 "x) - A(x) = k 2 \+
" }{TEXT 265 1 "a" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Delta" "6#%&DeltaG
" }{TEXT -1 18 "x ( S(x) - C(x) )," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 9 "where 2 " }{TEXT 266 1 "a" }{TEXT -1 1 " \+
" }{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT -1 496 "x is the surface \+
area of the small portion of the cylinder under consideration, and k i
s the diffusivity constant. As before, the difference S(x) - C(x) is \+
the difference in the concentration of the solute across the membrane \+
at distance x from the end where the inner fluid enters. If S(x) < C(x
), the fluid in this section of the inner tube will decrease. At the s
ame time a volume V of the solution passes through the small section a
nd as it does, the concentration will change by the amount " }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 " \+
C(x + " }{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT -1 14 "x) - C(x) =
" }{XPPEDIT 18 0 "(A(x+Delta*x)-A(x))/V" "6#*&,&-%\"AG6#,&%\"xG\"\"
\"*&%&DeltaGF*F)F*F*F*-F&6#F)!\"\"F*%\"VGF/" }{TEXT -1 12 " = k 2 \+
" }{TEXT 264 1 "a" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Delta" "6#%&DeltaG
" }{TEXT -1 20 "x ( S(x) - C(x) )/V." }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 22 "In the limit, we write" }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 " \+
" }{XPPEDIT 18 0 "diff(C(x),x) = (2*Pi*a*k/V*(S(x)-C(x))" "6#/-%%
diffG6$-%\"CG6#%\"xGF**.\"\"#\"\"\"%#PiGF-%\"aGF-%\"kGF-%\"VG!\"\",&-%
\"SG6#F*F--F(6#F*F2F-" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 169 "The derivation of the equation for t
he concentration in the outer tube changes to account for the counter-
flow. The change in the amount of diffusion into the portion is" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 " \+
A(x - " }{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT -1 21 "x) - \+
A(x) = k 2 " }{TEXT 267 1 "a" }{TEXT -1 1 " " }{XPPEDIT 18 0 "Del
ta" "6#%&DeltaG" }{TEXT -1 18 "x ( C(x) - S(x) )," }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "We have used x - " }
{XPPEDIT 18 0 "Delta" "6#%&DeltaG" }{TEXT -1 88 "x to reflect the coun
ter-flow. An analysis similar to those above leads to the equation" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 " \+
" }{XPPEDIT 18 0 "diff(S(x),x) = - (2*Pi*a*k/V*(C(x
)-S(x))" "6#/-%%diffG6$-%\"SG6#%\"xGF*,$*.\"\"#\"\"\"%#PiGF.%\"aGF.%\"
kGF.%\"VG!\"\",&-%\"CG6#F*F.-F(6#F*F3F.F3" }{TEXT -1 1 "." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "We solve these tw
o equations with C(0) and S(L) specified. We expect that, C(L) < C(0) \+
and S(L) < S(0)." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "
" 0 "" {TEXT -1 51 "Solution of the equations for counter-current flow
." }}{PARA 0 "" 0 "" {TEXT -1 248 " We take the flow in both tubes
to be the same: V = 3, take the diffusivity constant to be 1/5, and t
he radius of the inner tube to be 1/2. Assume the initial value of C t
o be 5. We take S to be zero at the opposite end: C(0) = 5 and S(L) = \+
0. " }}{PARA 0 "" 0 "" {TEXT -1 134 " This is not an initial value
problem. We remedy this situation as follows. Take the initial condit
ions to be C(0) = 5 and S(0) = " }{XPPEDIT 18 0 "alpha" "6#%&alphaG" }
{TEXT -1 8 ", where " }{XPPEDIT 18 0 "alpha" "6#%&alphaG" }{TEXT -1
106 " is a generic constant. We solve the system of equations with the
se initial values, and then compute what " }{XPPEDIT 18 0 "alpha" "6#%
&alphaG" }{TEXT -1 26 " must be so that S(L) = 0." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "a:=1/2: Co:=0: V:=3: k:=1/5:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 120 "dsolve(\{diff(C(x),x)=
2*Pi*a*k*(S(x)-C(x))/V,\n diff(S(x),x)=-2*Pi*a*k*(C(x)-S(x))/V,C(0)=5
,S(0)=alpha\},\n \{C(x),S(x)\});" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 71 "C:=x->1/15*Pi*x*alpha+5-1/3*Pi*x;\nS:=x->alpha+1/15*P
i*x*alpha-1/3*Pi*x;" }}}{PARA 0 "" 0 "" {TEXT -1 14 "Now we choose " }
{XPPEDIT 18 0 "alpha" "6#%&alphaG" }{TEXT -1 18 " so that S(L) = 0." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "alpha:=solve(S(10)=0,alpha)
;" }}}{PARA 0 "" 0 "" {TEXT -1 28 "Here we draw the two graphs." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "plot([C(x),S(x)],x=0..10);"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 178 " The point to make i
s that a choice on the direction for the flow of the cleaning fluid is
important in achieving the most efficient cleaning of the contaminate
d inner fluid." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 ""
0 "" {TEXT -1 23 "Exercise for the reader" }}{PARA 0 "" 0 "" {TEXT -1
0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "1. In both situations, take the l
ength of the tubes to be 15, instead of 10. How do the graphs change?
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "2. Is
it more, or less, efficient for the fluid in the outer cylinder to fl
ow faster?" }}}}{MARK "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }