Mass Transfer 1phenomenon of one-dimensional conduction.pptx
1. TRANSPORT PHENOMENA: 3RD SECTION
MASS TRANSFER
DIAH SUSANTI, PH.D
UNDERGRADUATE PROGRAM
MATERIALS AND METALLURGICAL ENGINEERING DEPARTMENT
INSTITUT TEKNOLOGI SEPULUH NOPEMBER (ITS)
APRIL 19, 2020
4. MASS TRANSFER (MASS DIFFUSION): FICK'S LAW
𝐽𝐴
𝑨 𝑥
= −𝐷
𝜕𝐶𝐴
𝜕𝑥
(1)
at T, P is constant
The minus sign indicates that the direction of the flux is from the high to low concentration side, while the
direction of the concentration increase is from the low to high side
(JA/A) is the molar flux, which is the number of mole A transferred per unit of time per area (kmol m-2s-1)
(JA) is the molar transfer rate that is the number of mole A transferred per unit time (kmols-1)
A is the area where there is no volume flow (stationary) (m2)
Subscript x shows that the direction of the transfer is only in the x direction
D is proportionality constant = diffusivity coefficient (m2s -1) specific for each material
CA is concentration (kmol m-3)
x is the distance (m)
So that:
Rate = JA
Driving force = CA
Resistance = x/(DA)
N2 CO2
x = 0
CA = 0
x = L
CA
L
Direction of flux
Direction of concentration increase
5. FICK’S LAW:
𝐽𝐴
𝑨
= −𝐷𝜵𝐶𝐴 at T, P is constant (2)
in which 𝛁CA = 𝐢
𝜕C𝐴
𝜕𝑥
+ 𝐣
𝜕C𝐴
𝜕𝑦
+ k
𝜕C𝐴
𝜕𝑧
(3)
i, j, k are the unit vectors in the x, y, z directions, respectively, and (del) is an operator which may operate on any
scalar.
Mass transfer is more complex than heat or momentum transfers because at least two species are needed. If there
is one particular molecule in a mixture, the molecule must diffuse through other molecules.
For example, the second component is called B. There are two possibilities for B; diffuse or not. If it does not
participate to diffuse, then it is called A diffuses on a stationary film (diffusion through stagnant film).
The mass transfer equation for B in the x direction is:
𝐽𝐵
𝑨 𝑥
= −𝐷
𝜕𝐶𝐵
𝜕𝑥
at T, P is constant (4)
For 3-dimensional directions:
𝐽𝐵
𝑨
= −𝐷𝜵𝐶𝐵 at T, P is constant (5)
The diffusion coefficients for A and B will be the same if they are ideal gases. In reality, the two are not the same.
Therefore it can be written with DAB and DBA.
DAB is the diffusion coefficient of component A which diffuses through A + B
DBA is the diffusion coefficient of component B which diffuses through A + B
6. NOTATION
The picture on the left shows gas fluxes A
and B through the room between two
stationary walls.
Therefore, the calculation of flux A or B is
relative to the stationary coordinates, not
the coordinates that moves. Thus we can
use the NA/A and NB/A notation to replace
the JA/A and JB/A notations used to express
molar flux relative to volumetric velocity.
wall
wall
1 Fluxes 2
(JA/A)x
(JB/A)x
1 x 2
Concentration Profile
CA,1
CB,1
CB,2
CA,2
7. In the picture above it can be seen that gas A diffuses from left to right, and therefore the molecules
must be replaced by gas molecules B so that the pressure is constant.
Because at each position x the number of molecules is constant, there is no volume flow. The diffusion is
called ‘equimolar counter diffusion’.
For ‘equimolar counter diffusion’ NA = -NB (minus sign shows the opposite direction)
(NA/A)x = -(NB/A) x (6)
Because there is no volumetric flow, so:
(NA/A)x =(JA/A)x=-DAB(CA/x) (7)
(NB/A)x =(JB/A)x=-DBA(CB/x) (8)
The total concentration is constant because pressure is kept constant:
CT = CA + CB = constant (9)
(CT/x) = (CA/x) + (CB/x) = 0 (10)
Hence, (CA/x) = - (CB/x) (11)
Therefore, DAB = DBA (12)
This equation (11) is restricted to the binary diffusion of ideal gas at constant T and P.
EQUIMOLAR COUNTER DIFFUSION
8. MASS FLUX
Mass flux is defined as mass per area per time. It is written mathematically as:
(jA/A)x = [(JA/A)x]MA (13)
(nA/A)x = [(NA/A)x]MA (14)
Mass flux unit is kgm-2s-1,while the unit of mass flowrate is kgs-1.
The driving force is the mass concentration =
(CA)MA = A (15)
where MA is molecular weight of A (kg kmol-1) and A is the density of A (kg m-3)
So that the Fick’s law could be written as:
(jA/A)x=-D (A/x) (16)
9. PARTIAL PRESSURE
Ideal gas equation:
PV = nRT (17)
So that CA =
𝑛
𝑉
=
𝑃𝐴
𝑅𝑇
(18)
where 𝑃𝐴 is the partial pressure of A, yA is the mole fraction of A.
𝑃𝐴=𝑦𝐴𝑃𝑡𝑜𝑡𝑎𝑙 (19)
Hence
𝐽𝐴
𝐴
=
𝐷
𝑅𝑇
(𝜵𝑃𝐴) at T,P constant (20)
10. MASS TRANSFER EXAMPLE PROBLEMS
Example 2.6.
Two gas streams of CO2 and air are flowing in a channel. The channel is
divided into equal volumes by a piece of iron 4 cm thick as shown in the
figure below. At the A-A plane there is a hole 1.2 cm in diameter drilled in the
iron so that CO2 diffuses from left to right and air from right to left. At the
plane A-A, the two gases are at a pressure of 2 atm and a temperature of 20
oC. Upstream of the hole, both gases are pure. Under the conditions given,
the concentration of CO2 is 0.083 kmol m-3, i.e the concentration of CO2 on
the left at the point A. At the right end of the hole, the concentration of CO2
is considered 0 because air flows quickly. The diffusion coefficient of CO2 in
the air is 1.56x10-3 m2s-1.
Please find:
a). CO2 Molar Flux
b). How many pounds of CO2 flow through the hole for 1 hour
-A---------A-
iron
iron
x
air
CO2
4 cm
d=1,2
cm
11. ANSWER:
Known:
P = 2 atm, T = 20 C, CA0 = 0.083 kmol m-3. CAx = 0 kmol m-3. DAB = 1.56x10-3 m2s-1.
L = 4 cm = 4x10-2 m = 0.04 m, d = 1.2 cm = 0.012 m A = 1/4d2 = 1/4(0.012)2 = 1.131x10-4 m2.
Asked:
a). CO2 Molar Flux
b). How many pounds of CO2 flow through the hole for 1 hour
Answer:
a) The CO2 diffuses through the hole in the iron at steady-state if the flow rate of CO2 and air are constant.
Since both gases are at the same temperature and pressure, eq. (6)
(NA/A)x =(JA/A)x=-DAB(CA/x) (6)
At steady state, the molar flux (NA/A)x must be constant in the hole throughout its length because all the
CO2 entering from the left exits into the air stream. Hence eq. (6) rearranges to:
1
DAB
NA
A
x
0
0.04
x=- 0.083
0
CA (6)
1
1.56x10−3
NA
A
x
0.04 − 0 = −(0 − 0.083)
NA
A
x
= 3.237X10-3 kmol m-2 s-1.
So, the CO2 molar flux is 3.237X10-3 kmol m-2 s-1 the molar flux with respect to the stationary
coordinates (the apparatus)
12. b). The mass flux : (nA/A)x = [(NA/A)x]MA (14)
MA = MCO2 = 44 kg mol-1. The mass flow rate:
(nA)x = [(NA/A)x] AMA = 3.237X10-3 kmol m-2 s-1 x1.131x10-4 m2 x 44 kg mol-1=1.61x10-5
kg/s.
Converted to lbm/hour: 1.61x10-5 kg/s x 3600 s/hour x 1 lbm/0.4539 kg = 0.128 lbm/hour.
The mass flow rate is positive because the CO2 flows from left to right with a mass flow rate of 0.128
lbm/hour.
13. EXAMPLE 2.7.
Air and CO2 mix together in a simple T pipe. Air at pressures of 3 atm and 30 C enters at the T end with a
flow rate of 2 kmol min-1. CO2 at 3 atm and temperature at 30 C enters at the other end of T pipe with a
flow rate of 4 kmol min-1. The two gases flow out from the middle T, still at 3 atm and 30 oC.
a. Calculate the incoming CO2 concentration (kg m-3)
b. Calculate the concentration of N2 that enters the air stream (lbmol ft-3)
c. Calculate the concentration of CO2 out (mol cm-3)
d. Calculate the concentration of O2 out (kmol m-3)
Air
N2 = 79%
O2 = 21%
3 atm, 30 oC
2 kmol min-1
CO2
3 atm, 30 oC
4 kmol min-1
Air & CO2
3 atm, 30 oC
14. ANSWER:
Known:
Air at P = 3 atm and T = 30 C = 303 K, Nair = 2 kmol/min.
CO2 at P = 3 atm and T = 30 C = 303 K, NCO2 = 4 kmol/min.
The two gas streams exit still at 3 atm and 303 K.
R (gas constant) = 8.2057x10-2 atm m3kmol-1K-1 = 0.082057 atm m3kmol-1K-1
Asked:
a. Calculate the incoming CO2 concentration (kg m-3)
b. Calculate the concentration of N2 that enters the air stream (lbmol ft-3)
c. Calculate the concentration of CO2 out (mol cm-3)
d. Calculate the concentration of O2 out (kmol m-3)
15. Answer:
a). CCO2 entering = CA =
𝑛
𝑉
=
𝑃𝐴
𝑅𝑇
(18)
CCO2 =
𝑛
𝑉
=
3 𝑎𝑡𝑚
0.082057 atm m3kmol−1K−1 3 0 3 𝐾
= 0.121 kmol m-3. To convert into kg m-3, we should multiply with MCO2.
So, concentration of CO2 = 0.121 kmol m-3 x 44 kg/mol = 5.31 kg m-3.
b). CN2 entering = CN2 =
𝑛
𝑉
=
𝑃𝑁2
𝑅𝑇
=
𝑦𝑁2𝑃𝑡𝑜𝑡𝑎𝑙
𝑅𝑇
=
0.79𝑥3𝑎𝑡𝑚
0.082057 atm m3kmol−1K−1 3 0 3 𝐾
= 0.0953 kmol m-3
Converted to lbmol ft-3: 0.0953
𝒌𝒎𝒐𝒍
𝒎𝟑 x
𝟏 𝒍𝒃 𝒎𝒐𝒍
𝟎.𝟒𝟓𝟑𝟗 𝒌𝒎𝒐𝒍
x
𝟏 𝒎𝟑
𝟑𝟓.𝟑𝟏 𝒇𝒕𝟑 = 5.945x10-3 lbmol/ft3.
So the concentration of N2 entering is 5.945x10-3 lbmol/ft3
(Note: Air is 79% nitrogen and 21% oxygen, so yN2 = 0.79 and yO2 = 0.21).
c). The exiting gases are comprised of CO2=4 kmol, N2=0.79x2 kmol=1.58 kmol and O2 = 0.21x 2 kmol = 0.42
kmol.
Total mole is 6 kmol. Mole fraction of gases: yCO2 = 4/6 = 0.667, yN2 = 1.58/6 = 0.263, and yO2 = 0.42/6 = 0.07.
So the concentration of CO2 in the exit stream is
0.667𝑥3𝑎𝑡𝑚
0.082057 atm m3kmol−1K−1 3 0 3 𝐾
= 0.08044
𝑘𝑚𝑜𝑙
𝑚3 =
𝟖. 𝟎𝟒𝟒𝒙𝟏𝟎−𝟓 𝒎𝒐𝒍
𝒄𝒎𝟑.
d). The concentration of O2 exiting is:
0.07𝑥3𝑎𝑡𝑚
0.082057 atm m3kmol−1K−1 3 0 3 𝐾
= 0.008446
𝑘𝑚𝑜𝑙
𝑚3 . Hence the
concentration of O2 is 0.008446
𝑘𝑚𝑜𝑙
𝑚3 .
16. EXAMPLE 2.8.
A tank containing 15% CO2 in the air is connected to a second tank that
contains only air. The connecting pipe has a diameter of 5 cm and a length of
30 cm. Both tanks are at a pressure of 1 atm and a temperature of 298.15 K. The
volume of each tank is very large compared to the volume of the pipe, so
changes in concentration in the tank can be ignored for a very long time after
the experiment begins. The diffusion coefficient of CO2 in the air at 1 atm and
25 C is 0.164 x 10-4 m2s-1. Calculate the initial mass transfer rate of CO2. Does
the air transfer?
15%
CO2
air
30 cm
5 cm
1 2
17. Known:
yCO2 in air = 0.15. Pipe d = 5 cm, L = 30 cm. The pressures and temperatures of both tanks are 1 atm and
298.15 K. DCO2 in air at T and P is 0.164 x 10-4 m2s-1
x = 30 – 0 = 30 cm = 0.3 m.
A = 1/4d2 = 1/4(0.05)2 = 1.96x10-3 m2.
R = 8.314x103 Nm kmol-1 K-1.
Asked:
Calculate the initial mass transfer rate of CO2. Does the air transfer?
Answer:
Assume that both gases are ideal. The diffusion transfer is equimolar counter diffusion. So,
(NA/A)x =(JA/A)x=-DAB(CA/x) (7)
(NA/A)x = −
𝐷
𝑅𝑇
(
∆𝑃𝐴
𝑑𝑥
)
NA = −
𝐷
𝑅𝑇
(
∆𝑦𝐶𝑂2𝑃𝑡𝑜𝑡𝑎𝑙
∆𝑥
)Ax = −
0.164 x 10−4m2s−1
8.314x103Nm kmol−1 K−1x298.15 K
(
0−0.15 𝑥1 𝑎𝑡𝑚𝑥101325 𝑁/𝑚2
0.3 𝑚
) 1.96x10-3 m2
= + 6.57x10-10 kmol/s. The plus sign indicates that the diffusion is from 1 to 2 ( to the right). In order to keep
the pressure 1 atm, the air should flow from right to left (in the opposite direction) with a flow rate is
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