GENERALIZED MASS BALANCES

GENERALIZED MASS BALANCES 1. THEORY: Equations that describe both steady and non-steady state Mass balance for species 1 in volume ∆x ∆y ∆z  mass of ...
Author: Frank Welch
1 downloads 0 Views 835KB Size
GENERALIZED MASS BALANCES 1. THEORY: Equations that describe both steady and non-steady state Mass balance for species 1 in volume ∆x ∆y ∆z  mass of species 1  mass flux of   mass of species 1         accumulati ng in    species 1 in    produced or destroyed     min us that out   by chemical reaction  x y z      

 n1x yz   n1x yz   x x  x    c1 xyz    n1y xz y  n1y xz yy   r xyz t        n1z xy z  n1z xy z z 

Divide by the volume x y z:

 c1 t



 n1x x



 n1y y



 n1z z

 r1 1

Mass balance for species “1” for different coordinate systems from Cussler “Diffusion”:

2 5‐2

Mass balance for species “1” in vector notation:  c1 t

   n1  r1

Express contributions of diffusion and convection to the species mass flux:

n1  j1  c1 v 0  D  c1  c1 v 0

(1)

where v0 is the volume average velocity.

 c1 t

 D  2 c1   (c1 v 0 )  r1 3 5‐3

Overall (total) mass balance, considering all species

  v x y z    v x y z   x x  x    x y z     v y x z y   v y x z yy  t     v z x y z   v z x y z z  where vx, vy, vz are the components of the mass average velocity. Note: The overall mass balance has no reaction term since no total mass is generated or destroyed. Dividing by x y z yields the continuity equation:

     vx  vy  vz t x y z

or

    v  t

(2) 4 5‐4

Overall (total) mass balances for different coordinate systems from Cussler “Diffusion”:

5 5‐5

Simplify the species mass balance (1) with the help of the continuity equation (2)! Problem: Different reference velocities used. For constant density, the mass and volume average velocities are the same and 

t

 0   ( v )   ( v 0 )   v 0

Divide by ρ, multiply by c1 and note that





→ 0  c 1  v 0   c 1 v 0  v 0 c 1 so eqn. (1) becomes

 c1 t

so

 (c1 v 0 )  v 0c1  c1v 0





 c 1 v 0  v 0 c 1

 v 0  c1  D  2 c1  r1 6 5‐6

Mass balance for species “1” with diffusive and convective terms for constant density systems (Cussler “Diffusion”):

0 0 v v c 1   c 1 c c   v r0 1   1   t r r  r  sin  

 1   2 c 1     2c 1  c 1  1 1  D   2 r  r1  2  2  sin   2  2            r r r r sin r sin       7 5‐7

2 Examples Example : Fast diffusion through a stagnant film

Remember few weeks back….

1. Step: Select appropriate mass balance  c1 t



n n 1 r  n1r   1 1  1z  r1 r  r r z

or for const. density:  c1 t

v

0 r

 c1

 c1 v 0  c 1   v 0z  r r  z

 1    c 1  1  2c 1  2c 1   r   2 D  r1  2 2  r r r r z          8 5‐8

2. Step: Simplify mass balance

 c1

=0

=0

=0

no flow in r-direction

symmetry

1 1  n1  n1z   r n1r     r1 t rr z r 

steady state

gives:

 n1z z

0

or:

v 0z

=0

no reaction

c1  2 c1 D z z 2

Then solve as before: 9 5‐9

Example 5.2.2: Fast diffusion into a semi-infinite slab

Remember again: 1. Step: Select appropriate mass balance For rectangular geometry:

 c1 t



 n1,x x



 n1,y y



 n1,z z

 r1

Here: capillary  c1

1  n1  n1z 1 r n1r   r     z  r1  r r t 10 5‐10

2. Step: Simplify mass balance • No flow in x-direction • No flow in y-direction • No chemical reaction Thus: Or:

 c1 t

 c1 t



 n1,z z

 v 0z

0

 c1 z

D

 2 c1  z2

Then solve as before (Ex. 3.2.4).

11 5‐11

Example 5.2.3: The flux near a spinning disk

A solvent flow approaches a spinning disk made out of a sparingly soluble solute, as shown below. Calculate the diffusion-controlled rate at which the disk slowly dissolves at steady state. 1. Step: Select appropriate mass balance  c1 t

v

0 r

 c1

 c1 v 0  c1   v 0z  r r  z

 1    c 1  1  2c 1  2c 1   2  r  D 2   z 2 r r r r     

   r1 

12 5‐12

2. Step: Simplify mass balance

=0  c1 t

v

steady state

see flow pattern =0

see flow pattern =0 =0 0 r

=0

=0

1    c 1  1  2c 1  2c 1  c1 v 0  c 1 0   2  r   D   vz 2 r  z 2 r z  r  r   r  r 

 c1

angular symmetry

angular symmetry

   r1 

no reaction

c1  2c 1 v D z z 2 0 z

13 5‐13

Velocity over a spinning disk From: Levich, “Physiochemical Hydrodynamics”, Prentice-Hall, 1962.

The velocity profile suggests that there is no gradient in radial direction! This is true for sufficiently large disks where edge effects are negligible.

ϴ

This special flow behavior is taken advantage of in the application of homogeneous coatings, e.g. on wafers. 14 5‐14

3. Step: Boundary conditions B.C.:

z=0 z=

c1=c1(sat) c1=0

The solution is: z exp   0 

c1  a 

1/ D 0r v z (s) ds dr  b  

Where a and b are integration constants calculated from the boundary conditions, so r      dr  exp 1 / D v ( s ) ds  0   0 z   c1  1  c1(sat )    r v (s) ds  dr  exp 1 / D  0     0 z z

15 5‐15

If we know the velocity vz we can get c1. From the literature (Levich, 1962) we know

v z  0.51 3 2 1 2  s2

and inserting in the above eqns. gives: c1  c1(sat)

  du   du

 exp  u3 0  exp  u3 0





where

 1.82 D1/ 3 1/ 6     z   1/ 2   

1

 is the kinematic viscosity of the liquid and  is the angular velocity of the disk. The diffusion flux is: c j1 z0  D 1 z

z 0

 D 2 / 3 1/ 2   c (sat)  0.62   1/ 6  1  

16 5‐16

This equation can also be written as: 1/ 2

D  d2     j1  0.62  d  

1/ 3

   D

c 1(sat )

D Re1/ 2 Sc 1/ 3 c1(sat )  j1  0.62 d    k ( mass transfer coefficient ) More in the upco min g chapters

There is an excellent agreement when plotting j1 d / D c1(sat) vs. Re1/2 justifying the strong assumption made in the analysis. The fact that this device gives uniform mass transfer over the entire surface area makes it popular in CVD and electrochemistry. 17

5‐17

Example : Dissolving Pill

In pharmacy it is important to know HOW FAST a medicine can permeate the body to act. Estimate the time it takes to start a steadystate dissolution of a drug pill. Is dissolution diffusion controlled? This can be checked by gentle stirring. If it dissolves faster than in the absence of stirring then it is diffusion-controlled!! 1. Step: Select appropriate mass balance v 0 c 1 v 0 c 1  c1 0 c 1  vr   r  r sin   t r  1   2 c 1  1   c 1   r 2 r  r r   r 2 sin    sin        r  D   1  2c 1 1  2  2 2  r sin    18

5‐18

2. Step: Simplify mass balance “Sparingly” soluble → stagnant symmetry surroundings =0 =0 =0

0 0 v  c1   c1 c v c   v r0 1   1  t r r  r sin  

=0

=0

=0

 1   2 c1    c1  1  2c 1  1  D 2  r  r1  2  sin   2 2 2    r sin     r r  r  r sin    symmetry



 c1 t



symmetry

no reaction

D   2 c 1  r  r  r2  r  19 5‐19

3. Step: Boundary conditions t=0 t>0

r r=R0 r=

c1=0 c1=c1(sat) c1=0 (meaning that the drug is consumed at the gut wall)

This is easily solved by introducing a variable =c1r and the equation reduces to that describing unsteady diffusion through a semi-infinite slab, so

R  r  R0  c1   0  1  erf c1(sat) r  4D t 20 5‐20

Thus the flux for a sparingly soluble pill: 

n1  j1  D

c 1 r

r  R0

    R0 D c 1(sat )    1  R0 D t      unsteady state 

Now you can calculate how long you need to reach steady-state by calculating when the term R0  0.1  1 Dt Say that the pill is 3 mm and the D=10-5 cm2/s (typical for D in liquids) then t80 hours. This is clearly wrong. We know that usually a pill acts within 10-20 minutes. So where is the mistake? Revisit the approximations and think looking at slide 2-24….. 21 5‐21

Example: Diffusion through a Polymer Membrane

A diaphragm-cell separates olephins (ethylene) from aliphatic hydrocarbons. Upper compartment: vacuum, lower: ethylene. Measure the ethylene concentration as a function of time. Find the diffusivity. 1. Step: Select appropriate mass balance

 c1 t

v

0 r

 c1

 c1 v 0  c 1   v 0z  r z r 

 1    c 1  1  2c 1  2c 1  r   2  D 2 2      r r r r z   

   r1  22 5‐22

2. Step: Simplify mass balance Under which other assumptions this eqn. becomes this?

c1  2c 1 D t z 2 3. Step: Boundary conditions for the polymer film: t=0 t>0

z z=0 z=L

c1=0 c1=H p0 c1=H pL0

where L is the film (membrane) thickness and H is the partition coefficient relating ethylene concentration (partial pressure) in the gas to ethylene concentration in the membrane. 23 5‐23

Now the solution to this equation is (J. Crank “The mathematics of diffusion”, 2nd ed., 1975, p.50 eqn. 4.22):

 D  n2   2  t  c1 z 2   sinn  z / L    1     exp   2 H  p0 L  n1  n L   

(1)

This is the concentration profile c1(z,t). We need to bring this result into a convenient form for experimental measurements. Make a mass balance or better, a mole balance as we are dealing with gases. Start from the ideal gas law:

dN1 V dp   dt R T dt

moles entering upper compartment at  z=L

 A j1 z L   A D

c1 z

z L

(2) 24 5‐24

Substitute (1) into (2) and integrate for p assuming that at t=0 p=0 at upper compartment.

A  R  T  p0  2 H  L2 p H  D  t  2 V L  

 D  n2   2  t    cosn       n2  1  exp L2    n 1     

At long times the exponential term  0 A  R  T  p0  H  L2     H  D  t  p V  L 6    known

Thus the intercept of the data is related to Henry’s constant while the slope is related to DH, the permeability of ethylene through the membrane. 25 5‐25

From another point of view now: As you can obtain H from the intercept, you can estimate D from the slope! This is an elegant problem combining verification (extraction) of both equilibrium (H) and transport (D) properties! The challenge of an engineer is to QUANTITATIVELY describe the phenomenon, e.g. connect the math to the physical problem.

26 5‐26

Suggest Documents