test 1 book notes examples-1.pdf
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book notesTRANSCRIPT
L
I II
PCO2 =
76 mmHg
PCO2 = 456 mmHg
d
Molecular Diffusion in Gases
Molecular Diffusion of Helium in Nitrogen Example 6.1-1, page 413
A mixture of He and N2 gas is contained in a pipe at 298K and 1.0atm total pressure which is
constant throughout. At one end of the pipe at point 1, the partial pressure PA1 of He is 0.60atm
and at the other end 0.2m, PA2 is 0.20 atm. Calculate the flux of He at steady state if DAB is
0.687 x 10-4
m2/s.
PA1 = 0.6 atm
PA2 = 0.2 atm
P = 1 atm
R = 82.057 m3 atm/kgmol K
T = 298K
z2-z1 = 0.2m
DAB = 0.687 x 10-4
m2/s
DAB for a gas is constant; P is constant meaning C is also constant; flux is constant at steady
state.
Flux,
12
21
zz
CCDJ AAAB
AZ
if C = P/RT
12
21
zzRT
PPDJ AAAB
AZ
Substitute in values:
64
1063.52.0298057.82
2.06.010687.0
AXJ kgmol/m2s
Equimolar Counterdiffusion Example 6.2-1, page 415
Ammonia gas (A) is diffusing through a uniform tube 0.10m
long containing N2 gas (B) at 1.0132x105 Pa pressure and
298K. At point 1, PA1 = 1.013x104 Pa and at point 2,
PA2 = 0.507x104 Pa. DAB = 0.230x10
-4 m
2/s.
Calculate the flux J*
A and J*
B at steady state.
PA1 = 1.013x104 Pa
PA2 = 0.507x104 Pa
DAB = 0.230x10-4
m2/s
T = 298K
P = 1.0132x105 Pa
z2 – z1 = 0.10 m
R = 8314.3 m3 Pa/kgmol K
7444
12
21 1070.41.02983.8314
10507.010013.110230.0
zzRT
PPDJ AAAB
AZ kgmol A/m2 s
H2O
Diffusion of Water Through Stagnant, Nondiffusing Air Example 6.2-2 page 419
Water in the bottom of a narrow metal tube is held at a constant
temperature of 293K. The total pressure of air (assumed dry) is
1.01325x105 Pa (1.0atm) and the temperature is 293K.
Water evaporates and diffuses through the air in the tube,
and the diffusion path z2 – z1 is 0.1524 m long. Calculate the
rate of evaporation at steady state. The diffusivity of water vapor at 293K and 1 atm is
0.250x10-4
m2/s. Assume that the system is isothermal.
P = 1.01325x105 Pa = 1.0 atm
T = 293K
z2 – z1 = 0.1524m
DAB = 0.250x10-4
m2/s
Vapor pressure of water, PA1 = 0.0231 atm
Water pressure in dry air, PA2 = 0 atm
R = 82.057 m3 atm/kgmol K
988.0
0231.01
01ln
0231.0101
ln1
2
12
B
B
BBBM
P
P
PPP atm
21
12
AA
BM
ABA PP
PzzRT
PDN
74
10595.100231.0988.01524.0293057.82
0.110250.0
AN kgmol/m2s
Diffusion in a Tube with Change in Path Length Example 6.2-3, page 419
Water in the bottom of a narrow metal tube is held at a constant temperature of 293K. The total pressure of air
(assumed dry) is 1.01325x105 Pa (1.0atm) and the temperature is 293K. At a given time, t, the level is z meters from
the top. As water vapor diffuses through the air, the level drops slowly. Derive the equation for the time tF for the
level to drop from a starting point of zo m at t = 0 and zF at t = tF seconds.
We can assume a pseudo-steady state condition because the level drops slowly. Now, both NA
and z are variables.
21
12
AA
BM
ABA PP
PzzRT
PDN
Assuming a cross-sectional area of 1 m2, the level drops dt in dz seconds, and the leftover kgmol
of A is PAdz/MA:
dtM
dzN
A
AA
11
Rearranging and integrating:
FF t
AA
BM
ABz
zA
A dtPPRTP
PDzdz
M 021
0
21
2
0
2
2 AAABA
BMFA
FPPPDM
RTPzzt
Stagnant B
r2
Diffusing
A
r1
Diffusion Through a Varying Cross-Sectional Area – Evaporation Derivation
HW1.1 In the lecture we showed that the molar rate of material A evaporating from a spherical
drop immersed in material B could be written as
2,
1,
21
ln11
4 A
AABA
pP
pP
RT
PD
rr
N
where
AN is the molar rate of material A leaving
the drop
r1 and r2 are two radial points away from
the sphere center
DAB is the diffusion coefficient P is the total
system pressure
R is the ideal gas constant
T is the system temperature
pA,1 is the partial pressure of A at point 1
pA,2 is the partial pressure of A at point 2
Starting with this equation, derive the following approximate equation for the molar flux at the particle
surface, NA,1
2,1,
1
1,
2AA
ABA cc
D
DN
where D1 is the diameter of the spherical drop
cA,1 is the molar concentration of material A at the surface of the drop
cA,2 is the molar concentration of material A far from the drop
Assume point 1 is at the drop’s surface
Assume point 2 is very far away from the drop, so r2 >> r1
1
2
21
ln11
4 A
AABA
PP
PP
RT
PD
rr
N
1
2
1
ln4 A
AABA
PP
PP
RT
PD
r
N
Now multiply the equation by 1/r1, rewrite the radius as 2/D on the right
side, substitute for surface flux, NAs, and PBM:
1
2
21
1
2
12
lnlnA
A
AA
B
B
BBBM
PPPP
PP
PP
PPP and
2
14 r
NN A
AS
So we rewrite the equation as: BM
AAABAS
P
PP
DRT
DN 212
Assume low vapor pressure, so PA1, PA2 << P and use a Taylor Series approximation ln(x) = x-1:
1
2
A
A
PP
PPis approaching 1, so 1ln
1
2
1
2
A
A
A
A
PP
PP
PP
PP
Now,
P
PP
PP
PP
PP
PPPP AA
A
AA
A
AA 21
1
21
1
12
so PBM ≈ P.
Lastly, PA1 = CA1RT and PA2 = CA2RT:
212121
222AA
ABAA
ABAA
BM
ABAS CC
D
DRTCRTC
RTD
DPP
P
P
RTD
DN
where CA1 = concentration at surface of the drop and CA2 = concentration far from the drop ≈ 0
Evaporation of a Napthalene Sphere Example 6.2-4, page 421
A sphere of naphthalene having a radius of 2.0mm is suspended in a large volume of still air at 318K
and 1.01325x105 Pa (1.0atm). The surface temperature of the naphthalene can be assumed to be at
318K and its vapor pressure at 318K is 0.555 mmHg. The DAB of naphthalene in air at 318K is
6.92x10-6
m2/s. Calculate the rate of evaporation of naphthalene from the surface.
r = 2.0mm
T = 318K
P = 1.01325x105 Pa
DAB = 6.92x10-6
m2/s
Vapor pressure, PA1 = 0.555 mmHg = 74 Pa
PA2 = 0 Pa
R = 8314.3 m3 Pa/kgmol K
5
5
5
55
1
2
12 100129.1
74101.01325
0101.01325ln
74101.013250101.01325
ln
B
B
BBBM
P
P
PPP Pa
8
5
56
21
01
1068.9100129.1002.03183.8314
1001325.11092.6
AA
BM
ABA PP
PrrRT
PDN kgmol A/m
2s
6.2-5.) Mass Transfer from a Napthalene Sphere to Air
Mass transfer is occurring from a sphere of naphthalene having a radius of 10mm. The sphere is in a
large volume of still air at 52.6ºC and 1atm absolute pressure. The vapor pressure of naphthalene at
52.6ºC is 1.0 mmHg. The diffusivity of naphthalene in air at 0ºC is 5.16 x 10-6
m2/s. Calculate the rate
of evaporation of naphthalene from the surface in kg mol/s m2. [Note: the diffusivity can be corrected
for temperature by using the temperature correction factor from the Fuller et al. equation (6.2-45)]
Assumptions:
the system is at steady-state, so the radius of the sphere is not changing
point 1 is at the surface of the sphere and point 2 is very far away , so r2 >> r1
R = 8314 m3 Pa/kgmol K
T = 325.75 K
r = 0.01 m
PA1=(1.0mmHg)(1.01325 x 105
Pa/760 mmHg) =
133.322 Pa
PA2 = 0 Pa because the air is still
First calculate DAB at the new temperature:
DAB(52.6ºC) = DAB(0ºC)
1
1
2
75.1
1
2
P
P
T
T= (5.16 x 10
-6 m
2/s)(325.75/273.15)
1.75 = 7.023 x 10
-6
m2/s
Now solve for AN :
1
2
21
ln11
4 A
AABA
PP
PP
RT
PD
rr
N
1
2
1
ln4 A
AABA
PP
PP
RT
PD
r
N
01001325.1
322.1331001325.1ln
75.3258314
1001325.110023.7
1.04 5
556
AN
= 4.34736 x 10-11
kgmol/s
Now solve for NA = AN /A = (4.34736 x 10-11
kgmol/s)/(4πr12) = 3.46 x 10
-8 kgmol/m
2s
6.2-9 Time to Completely Evaporate a Sphere A drop of liquid toluene is kept at a uniform temperature of 25.9ºC and is suspended in air by a
fine wire. The initial radius r1 = 2.00mm. The vapor pressure of toluene at 25.9ºC is PA1 = 3.84
kPa and the density of liquid toluene is 866 kg/m3.
(a) Derive Eq. (6.2-34) to predict the time tF for the drop to evaporate completely in a large
volume of still air. Show all steps.
(b) Calculate the time in seconds for complete evaporation.
(a) Equations for flux:
BM
AAABA
P
PP
RTr
PD
r
N 21
24
and
dtM
dVN
A
AA
Volume of the sphere, 3
34 rV so the derivative of volume is: 24 r
dr
dV
Plug this volume derivative into molar flux equation for dV:
dtM
drrN
A
AA
24
Now set AN equal in each equation:
BM
AAAB
A
A
P
PP
RTr
PD
rdtM
drr 21
2
2
4
14
Separate variables and integrate both sides:
r
AAABA
BMAt
rdrPPPDM
PRTdt
021
0
21
2
2 AAABA
BMA
PPPDM
PRTrt
(b) Now use the equation to find the time in seconds:
PA1 = 3840 Pa
PA2 = 0 Pa
P = 1.01325 x 105 Pa
MA = 92.14 kg/kgmol
DAB = 0.86 x 10-4
m2/s
R = 8314 m3 Pa/kgmol K
T = 299.05 K
ρ = 866 kg/m3
r1 = 0.002 m
Solve for PBM:
1
2
21
lnA
A
AABM
PPPP
PPP = 99392.6 Pa
tevap = 21
2
2 AAABA
BMA
PPPDM
PRTr
=
0384010086.014.922
6.9939205.2998314002.08664
2
1388.23 seconds
6.2-10 Diffusion in a Nonuniform Cross-Sectional Area – Changing Pipe Diameter
The gas ammonia (A) is diffusing at steady state through N2 (B) by equimolar counterdiffusion in a conduit
1.22m long at 25ºC and a total pressure of 101.32 kPa abs. The partial pressure of ammonia at the left end
is 25.33 kPa and at the other end is 5.066 kPa. The cross section of the conduit is in the shape of an
equilateral triangle, the length of each side of the triangle being 0.0610m at the left end and tapering
1.22m
.0305m .061m
uniformly to 0.0305m at the right end. Calculate the molar flux of ammonia. The diffusivity is DAB = 0.230 x
10-4
m2/s.
Area = tan4
1 2b (60º)
To find an equation for how b changes with the length, create a linear fit based on two points:
At the first end: (0, 0.061) and at the other end: (1.22, 0.0305)
Slope = -0.0305/1.22 = -0.025
b = -0.025L + 0.061
So now Area = tan061.0025.04
1 2 L (60º) = 0.000271 L
2 – 0.001321 L + 0.001611
dL
dP
RT
D
A
N AABA
dL
dP
RT
DN AABA
0.001611 L 0.001321 - L 0.000271 2
Now separate and integrate both sides:
2
1
22.1
0 2 0.001611 L 0.001321 - L 0.000271
PA
PAA
ABA dPRT
DdL
N
PA1 = 25.33 kPa
PA2 = 5.066 kPa
DAB = 0.230 x 10-4
m2/s
R = 8314 m3 Pa/kgmol K
T = 298.15 K
Now plug into integrate equation:
15.2988314
25330506610230.08.1514
4
12
RT
PPDN AAAB
A AN = 1.24 x 10-10
kgmol/s
Now solve for NA:
NA = AN /A = 60tan0610.0025.0
41
1024.12
10
L at any point along the length, L
Molecular Diffusion in Liquids
Diffusion of Ethanol (A) Through Water (B) Example 6.3-1, page 429
An ethanol(A)-water(B) solution in the form of a stagnant film 2.0mm thick at 293K is in contact at one
surface with an organic solvent in which ethanol is soluble and water is insoluble. Hence, NB = 0. At
point 1, the concentration of ethanol is 16.8 wt % and the solution density is ρ1 = 972.8 kg/m3. At
point 2, the concentration of ethanol is 6.8 wt % and ρ2 = 988.1 kg/m3. The diffusivity of ethanol is
0.740x10-9
m2/s. Calculate the steady-state flux NA.
DAB = 0.740x10-9
m2/s
T = 293K
CA1 = 16.8
CA2 = 6.8
ρ1 = 972.8 kg/m3
ρ2 = 988.1 kg/m3
Meth =46.05
Mwater = 18.02
Calculate the mole fractions, taking a basis of 100 kg:
0732.0
02.182.83
05.468.16
05.468.16
1
AX 0277.0
02.182.93
05.468.6
05.468.6
2
AX
9268.00732.011 11 AB XX 9723.00277.011 22 AB XX
Now calculate the molecular weight:
07.20
02.182.83
05.468.16
1001
kgM kg/kgmol 75.18
02.182.93
05.468.6
1001
kgM kg/kgmol
Now calculate CAvg:
6.502
75.18/1.98807.20/8.972
2
// 2211
MM
Cavg
kgmol/m
3
949.0
9723.0
9268.0ln
9268.09723.0
ln1
2
12
B
B
BBBM
X
X
XXX
79
21
12
1099.8949.0002.0
0277.00732.06.5010740.0
AA
BM
avgAB
A xxxzz
CDN kgmol/m
2s
6.3-2 Diffusion of Ammonia in an Aqueous Solution An ammonia (A) – water (B) solution at 278K and 4.0mm thick is in contact at one surface with an organic
liquid at this interface. The concentration of ammonia in the organic phase is held constant and is such that
the equilibrium concentration of ammonia in the water at this surface is 2.0 wt % ammonia (density of
aqueous solution 991.7 kg/m3) and the concentration of ammonia in water at the other end of the film 4.0
mm away is 10 wt% (density = 961.7 kg/m3). Water and the organic are insoluble in each other. The
diffusion coefficient of ammonia in water is 1.24 x 10-9
m2/s.
(a) At steady state, calculate the flux NA in kg mol/s m2
(b) Calculate the flux NB. Explain.
(a)
ρ1 = 991.7 kg/m3
ρ2 = 961.7 kg/m3
M1 = 17.9806 kg/kgmol
M2 = 17.9031 kg/kgmol
T = 278 K
CA1 = 0.02
CA2 = 0.1
z1 = 0.004 m
z2 = 0 m
Calculate Cavg:
9031.17
7.961
9806.17
7.991
2
1
2
1
2
2
1
1
MMCavg
54.435 kgmol/m
3
Calculate XBM:
Xw1 = 1-0.02 = 0.98
Xw2 = 1-0.1 = 0.9
98.0/9.0ln
98.09.0
ln 12
12
BB
BBBM
XX
XXX 0.939432
Now calculate flux:
BM
AAavgAB
AXzz
XXCDN
12
21
=
939432.0004.00
1.002.0435.541024.1 9
= 1.437 x 10-6
kgmol/m2 s
Prediction of Diffusivities in Liquids Prediction of Liquid Diffusivity Example 6.3-2, page 432
Predict the diffusion coefficient of acetone in water at 25ºC and 50ºC using the Wilke-Chang equation. The
experimental value 1.28x10-9
m2/s at 298K.
From Appendix A.2, the viscosity of water at 25ºC is μB = 0.8937 x 10-3
Pa s and at 50ºC is 0.5494 x 10-3
Pa
s. From Table 6.3-2 for CH3COCH3 with 3 carbons + 6 hydrogens + 1 oxygen.
VA = 3(0.0148) + 6(0.0037) + 1(0.0074) = 0.0740 m3 kgmol
For the water association parameter φ = 2.6 and MB = 18.02 kg mass/kgmol. For 25ºC:
9
6.0
2/116
6.0
2/116 10277.10740.08937.0
29802.186.210173.110173.1
AB
BABV
TMD
m
2/s
For 50ºC:
9
6.0
2/116
6.0
2/116 10251.20740.05494.0
32302.186.210173.110173.1
AB
BABV
TMD
m
2/s
Prediction of Diffusivities of Electrolytes in Liquids Diffusivities of Electrolytes Example 6.3-3, page 434
Predict the diffusion coefficients of dilute electrolytes for the following cases:
(a) For KCl at 25ºC, predict DºAB and compare with the value in Table 6.3-1
(b) Predict the value of KCl at 18.5ºC. The experimental value is 1.7x10-5
cm2/s
(c) For CaCl2 predict DAB at 25ºC. Compare with the experimental value of 1.32x10-5
cm2/s; also,
predict Di of the ion Ca2+
and of Cl- and use Eq. 6.3-12
(a) From Table 6.3-3, λ+ (K+) = 73.5 and λ- (Cl
-) = 76.3:
51010 10993.1
3.76/15.73/1
1/11/129810928.8
/1/1
/1/110928.8
nnTDAB
cm2/s
(b) To change the temperature, we use a simple correction factor:
T = 18.5ºC = 291.7K
From Table A.2-4, μw = 1.042 cP
55 10671.1042.1
7.29110993.1
334255.18
w
ABAB
TDD
cm
2/s
(c) From Table 6.3-3, λ+ (Ca2+
/2) = 59.5 and λ- (Cl-) = 76.3, n+ = 2 and n- = 1:
577 10792.02
5.5910662.210662.22
nD
Ca
cm
2/s
577 10031.21
3.7610662.210662.2
nD
Cl
cm
2/s
5
5510335.1
10031.2/110792.0/2
12
//
DnDn
nnDAB
cm2/s
Molecular Diffusion in Biological Solutions and Gels
Prediction of Diffusivity of Albumin Example 6.4-1, page 438
Predict the diffusivity of bovine serum albumin at 298K in water as a dilute solution using the modified
Polson equation (6.4-1).
T = 298 K
MA = 67500 kg/kgmol from Table 6.4-1 (pg 437)
μwater = 0.8937x10-3
Pa s
We can use the equation for the prediction of diffusivities for biological solutes:
11
3/13
15
3/1
15
1070.767500108937.0
2981040.91040.9
A
ABM
TD
m
2/s
This value is different from the experimental value because the shape of the molecule differs greatly from a
sphere.
Diffusion of Urea in Agar Example 6.4-2, page 439
A tube or bridge of a gel solution of 1.05 wt% agar in water at 278K is 0.04 m long and connects two
agitated solutions of urea in water. The urea concentration in the first solution is 0.2 gmol urea per liter
solution and is 0 in the other. Calculate the flux of urea in kgmol/m2s at steady-state.
T = 278 K
DAB = 0.727x10-9
m2/s from Table 6.4-2 (page 440)
CA1 = 0.2 kgmol/m3
CA2 = 0 kgmol/m3
Because XA1 is less than 0.01, the solute is very dilute and XBM ≈1.0.
99
12
21 1063.304.0
02.010727.0
zz
CCDN AAAB
A kgmol/m2s
Molecular Diffusion in Solids
Diffusion of H2 Through Neoprene Membrane Example 6.5-1, page 442
The gas hydrogen at 17ºC and 0.010 atm partial pressure is diffusing through a membrane of vulcanized
neoprene rubber 0.5 mm thick. The pressure of H2 on the other side of the neoprene is zero. Calculate the
steady-state flux, assuming that the only resistance to diffusion is in the membrane. The solubility S of H2
gas in neoprene at 17ºC is 0.051 m3 (STP of 0ºC and 1atm)/m
3 solid atm and the diffusivity DAB is 1.03x10
-10
m2/s at 17ºC.
Solubility = 0.051 m3/m
3 solid atm
DAB = 1.03x10-10
m2/s
PA1 = 0.10 atm
PA2 = 0
z2 – z1 = 0.5 mm
The equilibrium concentration at the inside surface of the rubber is:
5
11 1028.201.0414.22
051.0
414.22
AA PS
C kgmol H2/m3 solid
Because PA2 on the other side of the rubber is zero, CA2 = 0:
12510
12
21 1069.40005.0
01028.21003.1
zz
CCDN AAAB
A kgmol H2/m2 s
Diffusion Through a Packaging Film Using Permeability Example 6.5-2, page 443
A polyethylene film 0.00015m thick is being considered for use in packaging a pharmaceutical product at
30ºC. If the partial pressure of O2 outside the package is 0.21 atm and inside it is 0.01atm, calculate the
diffusion flux of O2 at steady-state. Use permeability data from Table 6.5-1. Assume that the resistances to
diffusion outside the film and inside the film are negligible compared to the resistance of the film.
From Table 6.5-1, PM = 4.17x10-12
m3 solute/(s m
2 atm/m)
1012
12
21 10480.200015.0414.22
01.021.01017.4
414.22
zz
PPPN AAM
A kgmol/m2s
2.0 atm
N2
0 atm
6.5-5 Diffusion Through a Membrane in Series Nitrogen gas at 2.0 atm and 30ºC is diffusing through a membrane of nylon 1.0 mm thick and polyethylene
8.0 mm thick in series. The partial pressure at the other side of the two films is 0 atm. Assuming no other
resistances, calculate the flux, NA at steady state.
P1 = 2.0 atm
P2 = 0 atm
PN2/Ny = 0.152 x 10-12
m2/s atm
PN2/Poly = 1.52 x 10-12
m2/s atm
Diffusion of gas through a solid:
CA can be related to the permeability: 414.22
AA
PSC
where S is the solubility
The permeability of a gas in a solid, PM = DABS
When there are several solids with thicknesses L1, L2,…, we can write the flux as:
12
1212
2
2
1
1
21 10256.1
1052.1
008.0
100152.0
001.0
1
414.22
00.21
414.22
MM
AAA
P
L
P
L
PPN kmole/m
2s
12
21
zz
CCDN AAAB
A
Diffusion in Porous Solids That Depends on Structure
Diffusion of KCl in Porous Silica Example 6.5-3, page 445
A sintered solid of silica 2.0mm thick is porous, with a void fraction ε of 0.30 and tortuosity τ of 4.0. The
pores are filled with water at 298K. At one face the concentration of KCl is held at 0.1 gmol/liter, and fresh
water flows rapidly past the other face. Neglecting any other resistance but that in the porous solid,
calculate the diffusion of KCl at steady-state.
DAB = 1.87x10-9
m2/s from Table 6.3-1 (page 431)
CA1 = 0.1 gmol/liter = 0.1 kgmol/m3
CA2 = 0
τ = 4.0
ε = 0.30
z2 – z1 = 0.002 m
99
12
21 1001.7002.0
01.01087.1
0.4
30.0
zz
CCDN AAAB
A
kgmol/m
2s
P1 = 2.026 x 105 Pa
P2 = 0 Pa
1.25m
6.5-6 Diffusion of CO2 in a Packed Bed of Sand It is desired to calculate the rate of diffusion of CO2 gas in air at steady state through a loosely packed bed
of sand at 276K and a total pressure of 1.013 x 105 Pa. The bed depth is 1.25m and the void fraction ε is
0.30. The partial pressure of CO2 is 2.026 x 103 Pa at the top of the bed and 0 Pa at the bottom. Use a τ of
1.87. τ = 1.87 PA1 = 2.026 x 10
3 Pa
ε = 0.30 PA2 = 0 Pa
T = 276K
025.187.1276314.8
010026.230.010142. 34
12
21
zzRT
PPDN AAAB
A
610609.1 AN mol/m
2s
Unsteady-State Diffusion in Various Geometries
Unsteady-State Diffusion in a Slab or Agar Gel Example 7.1-1, page 463
A solid slab of 5.15 wt% agar gel at 278K is 10.16 mm thick and contains a uniform concentration of urea of
0.1 kgmol/m3. Diffusion is only in the x-direction through two parallel flat surfaces 10.16 mm apart. The
slab is suddenly immersed in pure turbulent water, so the surface resistance can be assumed to be
negligible; that is, the convective coefficient kc is very large. The diffusivity of urea in agar from Table 6.4-2
is 4.72x10-10
m2/s.
(a) Calculate the concentration at the midpoint of the slab (5.08mm from the surface) and 2.54mm from the
surface after 10 hours.
(b) If the thickness of the slab is halved, what would the midpoint concentration be in 10 hours?
C0 = 0.10 kgmol/m3
C1 = 0 for pure water
C = concentration at distance x from the center
line and at time t
DAB = 4.72x10-10
m2/s
t = 10 hours * 3600 seconds = 36000 seconds
1.01.00.1/0
0.1/0
/
/
01
1 cc
CKC
CKCY
(a)
X1 = 10.16mm/2 = 5.08mm = 0.00508 m
X = 0 (center)
658.000508.0
360001072.42
10
2
1
X
tDX AB
Relative position, n = X/X1 = 0
Relative Resistance, m = 0
On Figure 5.3-5 “Unsteady State Conduction in a Large Flat Plate” when X = 0.685, m = 0, n = 0
1.0275.0
cY c = 0.0275 kgmol/m
3
On Figure 5.3-5, when X = 0.658, m = 0, n = 0.00254/0.00508 = 0.5: (n is not 0 now because not at center)
1.0172.0
cY c = 0.0172 kgmol/m
3
(b)
If the thickness is halved, X becomes:
632.200254.0
360001072.42
10
2
1
X
tDX AB
Relative position, n = 0, m = 0
1.00020.0
cY c = 0.0002 kgmol/m
3
7.1-5 Unsteady State Diffusion in a Cylinder of Agar Gel – Radially and Axially A wet cylinder of agar gel at 278K containing a uniform concentration of urea of 0.1 kgmol/m
3 has a
diameter of 30.48mm and is 38.1mm long with flat parallel ends. The diffusivity is 4.72 x 10-10
m2/s.
Calculate the concentration at the midpoint of the cylinder after 100h for the following cases if the cylinder
is suddenly immersed in turbulent pure water.
(a) For radial diffusion only
(b) Diffusion occurs radially and axially
Assume surface resistance is negligible; assume k = 1.0 because properties are similar
T = 278 K
C0 = 0.1 kgmol/m3
C1 = 0 for pure water
D = 4.72 x 10-10
m/s
Time, t = 100h = 360000 seconds
1.01.00.10
0.10
01
1 cc
ckc
ckcY
(a)
r1 = 0.01524 m
r0 = 0 m (center line)
731601.0
01524.0
3600001072.4 10
2
1
r
tDX AB
Relative position at midpoint of the cylinder, n = 0/0.1524 = 0
Relative resistance, m ≈ 0 because kc is assumed to be very large
From Figure 5.3-7, “Unsteady-State Heat Conduction in a Long Cylinder”:
Using X = 0.731601, m = 0, n = 0 Y = 0.024 = c/0.1
Concentration = 0.0024 kgmol/m3
(b)
X1 = L/2 = 38.1 mm / 2 = 19.05 mm = .01905 m
468.0
01905.0
3600001072.4 10
2
1
r
tDX AB
From Figure 5.3-7, Unsteady-State Heat Conduction in a Long Cylinder:
Using X = 0.468, m = 0, n = 0 Y = 0.375
For both axial and radial diffusion, Y = YxYy = (0.375)(0.24) = 0.009
Concentration = 0.0009 kgmol/m3
Unsteady-State Diffusion in a Semi-Infinite Slab Example 7.1-2, page 465
A very thick slab has a uniform concentration of solute A of C0 = 1.0x10-2
kgmol A/m3. Suddenly, the front
face of the slab is exposed to a flowing fluid having a concentration C1 = 0.10 kgmol A/m3 and a convective
coefficient kc = 2x10-7
m/s. The equilibrium distribution coefficient K = cLi/ci = 2.0. Assuming that the slab
is a semi-infinite solid, calculate the concentration in the solid at the surface (x = 0) and x = 0.01m from the
surface after t = 30000 seconds. The diffusivity in the solid is DAB = 4x10-9
m2/s.
t = 3000 seconds
DAB = 4x10-9
m2/s
C0 = 0.01 kgmol.m3
C1 = 0.10 kgmol/m3
kc = 2x10-7
m/s
K = 2.0
095.13000104104
1020.2 9
9
7
tDD
KkAB
AB
c
For x = 0.01:
457.0
30001042
01.0
2 9-
tD
x
AB
From the chart 5.3-3 “Unsteady State Heat Conducted in a Semi-Infinite Solid with Surface Convection”:
26.001.02/1.0
01.0
/1
01
0
C
CKC
CCY C = 2.04x10
-2 kgmol/m
3
For x = 0
0
30001042
0
2 9-
tD
x
AB
From the chart 5.3-3 “Unsteady State Heat Conducted in a Semi-Infinite Solid with Surface Convection”:
62.001.02/1.0
01.0
/1
01
0
C
CKC
CCY C = 3.48x10
-2 kgmol/m
3
This is the same value as Ci. To calculate CLi:
22 1096.61048.30.2 iLi KCC kgmol/m3
7.1-6 Drying of Wood – Unsteady State Diffusion in a Flat Plate A flat slab of Douglas fir wood 50.8mm thick containing 30 wt% moisture is being dried from both sides
(neglecting ends and edges). The equilibrium moisture content at the surface of the wood due to the drying
air blown over it is held at 5 wt% moisture. The drying can be assumed to be represented by a diffusivity of
3.72 x 10-6
m2/h. Calculate the time for the center to reach 10% moisture.
Assume there is no surface resistance and kc = ∞
At t = 0, c0 = 0.3, c1 = 0.05
At t = ?, c = 0.1
DAB = 3.72 x 10-6
m2/h
Solve for Y: 2.03.00.15.0
1.00.15.0
01
1
ckc
ckcY
Solve for X for the graph:
X1 = 25.4 mm = 0.254 m
X0 (center) = 0 m
tt
x
tDX AB 005766.0
254.0
1072.32
6
2
1
Relative position, n = 0 at the center
Relative resistance, m = 0 because kC is large
From Chart 5.3-5, “Unsteady State Heat Conduction in a Flat Plate”: X = 0.75 = 0.005766t
Time, t = 130.073 hours
Convective Mass Transfer Coefficients Vaporizing A and Convective Mass Transfer Example 7.2-1, page 469
A large volume of pure gas B at 2 atm pressure is flowing over a surface from which pure A is vaporizing.
The liquid A completely wets the surface, which is a blotting paper. Hence, the partial pressure of A at the
surface is the vapor pressure of A at 298K, which is 0.2 atm. The k’y has been estimated to be 6.78x10
-5
kgmol/s m2 mol frac. Calculate NA, the vaporization rate, and also the value of ky and kG.
P = 2.0 atm
PA1 = 0.2 atm
PA2 = 0 atm
k'y = 6.78x10
-5 kgmol/s m
2 mol frac
First calculate the mole fraction of A:
1.00.2/2.0/1 PPY AA
02 AY
To calculate ky, we must relate it to k'y:
'
yBMy kyk
95.0
9.0/0.1ln
9.01
ln 12
12
BB
BBBM
yy
yyy
55'
10138.795.0
1078.6
BM
y
yy
kk kgmol/m
2 s mole fraction
Now we can calculate kG:
BMyBMG ykPyk 55
10569.30.2
10138.7
P
kk
y
G kgmol/m2 s atm
Now we can calculate the vaporization rate, NA:
65
21 10138.7010.010138.7 AAyA yykN kgmol/m2s
or
65
21 10138.702.010569.3 AAGA PPkN kgmol/m2s
7.2-1 Flux and Conversion of Mass-Transfer Coefficient A value of kG
was experimentally determined to be 1.08 lbmol/h ft
2 atm for A diffusing through stagnant B.
For the same flow and concentrations it is desired to predict kG’ and the flux of A for equimolar
counterdiffusion. The partial pressures are PA1 = 0.20 atm, PA2 = 0.05 atm, P = 1.0 atm abs total. Use
English and SI Units.
BMGG PkPk '
Solve for PBM:
872853.0
2.01
05.01ln
05.02.0
ln1
2
21
A
A
AABM
PP
PP
PPP atm
Plug into equation:
k’G (1.0 atm) = (1.08 lbmol/h ft
2 atm)(0.872853 atm)
k’G(1.01x10
5Pa) = (1.08lbmol/hft
2atm)(.872853atm)(1hr/3600sec)(1ft
2/.0929 m
2)(.453kgmol/lbmol)
k’G = 0.9427 lbmol/h ft
2 atm
k’G = 1.262 x 10
-8 kgmol/s m
2 Pa
Now solve for flux:
PA1 = 0.2 atm = 20265 Pa
PA2 = 0.05 atm = 5066.25 Pa
1414.05.02.09247.021
' AAGA PPkN
48
21
' 1092.125.50662026510262.1 AAGA PPkN
NA = 0.1414 lbmol/ft2h
NA = 1.92 x 10-4
kgmol/m2s
Mass Transfer Under High Flux Conditions
High Flux Correction Factors Example 7.2-2, page 472
Toluene A is evaporating from a wetted porous slab by having inert pure air at 1 atm flowing parallel to the
flat surface. At a certain point, the mass transfer coefficient, k’x for very low fluxes has been estimated as
0.20 lbmol/hr ft2. The gas composition at the interface at this point is XA1 = 0.65. Calculate the flux NA and
the ratios kc/ k’c or kx/ k
’x and k
0c/ k
’c or k
0x/ k
’x to correct for high flux.
First find XBM:
619.0
65.01
01ln
65.0101
ln1
2
12
B
B
BBBM
X
X
XXX
To find the flux, NA, use:
210.0065.0619.0
20.021
'
AA
BM
x
A XXX
kN lbmol/ft
2 hr
To find the ratios, set up the following:
616.1619.0
11''
BMc
c
x
x
Xk
k
k
k
So 323.02.0616.1616.1 ' xx kk lbmol/ft2 hr; and
565.0619.0
65.011 1
'
0
'
0
BM
A
c
c
x
x
X
X
k
k
k
k
So 113.02.0565.0565.0 '0 xx kk lbmol/ft2 hr
1 2
7.2-3 Absorption of H2S by Water In a wetted-wall tower an air H2S mixture is flowing by a film of water that is flowing as a thin film down a
vertical plate. The H2S is being absorbed from the air to the water at a total pressure of 1.50 atm abs and
30ºC. A value for kc’ of 9.567 x 10-4
m/s has been predicted for the gas-phase mass transfer coefficient. At a
given point, the mole fraction of H2S in the liquid at the liquid-gas interface is 2.0(10-5
) and PA of H2S in the
gas is 0.05 atm. The Henry’s law equilibrium relation is PA (atm) = 609xA (mole fraction in the liquid).
Calculate the rate of absorption of H2S. Hint: Call point 1 the interface and point 2 the gas phase. Then
calculate PA1 from Henry’s Law and the given xA. The value of PA2 is 0.05 atm.
P = 1.50 atm
PA1 = 609 (CA1) = 609(2 x 10-5
) = 0.01218 atm
PA2 = 0.05 atm
T = 30 ºC = 303.15 K
k’c = 9.567 x 10
-4 m/s
XA1 = 2 x 10-5
Henry’s Law: PA = 609XA
R = 82.057 x 10-3
m3 atm/kgmol K
PkRT
PkG
c ''
5
3
4' 10846.3
15.30310057.82
10567.9
Gk kgmol/m
2s atm
65
21 10455.105.001218.010846.3 AAGA PPkN
NA = 1.455 x 10-6
kgmol/m2s
Mass Transfer for Flow Inside Pipes
Mass Transfer Inside a Tube Example 7.3-1, page 479
A tube is coated on the inside with naphthalene and has an inside diameter of 20 mm and a length of 1.10m.
Air at 318K and an average pressure of 101.3 kPa flows through this pipe at a velocity of 0.80 m/s.
Assuming that the absolute pressure remains essentially constant, calculate the concentration of naphthalene
in the exit air.
v = 0.80 m/s
D = 0.02 m
T = 318 K
z2 – z1 = 1.10 m
DAB = 6.92x10-6
m2/s
PAi = 74.0 Pa
μair = 1.932x10-5
Pa s from Appendix A
ρ = 1.114 kg/m3
R = 8314 m3 Pa/kgmol K
Calculate the concentration, CAi:
510799.23188314
74 RT
PC Ai
Ai kgmol/m3
Calculate the Schmidt number:
506.2
1092.6114.1
10932.16
5
AB
ScD
N
Calculate the Reynolds number:
6.922
10932.1
114.180.002.05Re
DN
Hence, the flow is laminar. We will use Figure 7.3-2 for laminar flow streamlines:
02.33410.1
02.0506.26.922
4Re
L
DNN Sc
Using Figure 7.3-2 with rodlike flow:
55.00
0
AAi
AA
CC
CC
If CA0 = 0,
010799.2
055.0
5
0
0
A
AAi
AA C
CC
CC CA = 1.539 x 10
-5 kgmol/m
3
Pure Water
T = 26.1ºC
v = 3.05m/s
0.00635 m
1.22 m
1.829 m
7.3-7 Mass Transfer from a Pipe and Log Mean Driving Force Use the same physical conditions as in problem 7.3-2, but the velocity in the pipe is now 3.05
m/s. Do as follows:
(a) Predict the mass transfer coefficient kc (Is this turbulent flow?).
(b) Calculate the average benzoic acid concentration at the outlet. [ Note: in this case,
Eqs. (7.3-42) and (7.3-43) must be used with the log mean driving force, where A is
the surface area of the pipe.]
(c) Calculate the total kgmol of benzoic acid dissolved per second.
ρ = 996 kg/m3
μ = 8.71 x 10-4
Pa s
T = 26.1ºC
DAB = 1.245x10-9
m2/s
Solubility = 0.2948 kgmol/m3
D = 0.00635 m
As = 2rπL
V = Aυ
Calculate the Schmidt number:
408.702
10245.1996
1071.89
4
AB
ScD
N
Calculate the Reynolds number:
22147
1071.8
99605.300635.04Re
DN
Calculate the Sherwood number:
000049487.023.033.083.0
Re ScSh NNN
Now calculate k’c:
AB
c
ShD
DkN
'
00635.0
10245.1000049487.0
9' ck
4' 1059.1 ck m/s
(b)
CA1 = 0
CAi = 0.02948 (solubility)
CA2 = ?
So now we plug this into the log mean driving force equation:
2
1
1212
lnAAi
AAi
AACAA
CC
CC
CCKACCv
CA1 = 0, so it cancels out of the equation:
2
22
lnAAi
Ai
ACA
CC
C
CKACv
Now plug in:
2
242
2
02948.0
02948.0ln
1059.1003175.005.3
A
AA
C
CC
CA2 = 0.001151 kgmol/m3
(c)
Rate of benzoic acid dissolved = NA:
7
2
12 1011.1003175.0
0001151.005.3
A
CCvN AA
A
Mass Transfer for Flow Outside Solid Surfaces Mass Transfer From a Flat Plate
Example 7.3-2, page 481
A large volume of pure water at 26.1ºC is flowing parallel to a flat plate of solid benzoic acid, where L =
0.24 m in the direction of flow. The water velocity is 0.061 m/s. The solubility of benzoic acid in water is
0.02948 kgmol/m3. The diffusivity of benzoic acid is 1.245x10
-9 m
2/s. Calculate the mass-transfer coefficient
kL and the flux NA.
Solubility = 0.02948 kgmol/m3
T = 26.1ºC
L = 0.24 m
DAB = 1.245x10-9
m2/s
v = 0.061 m/s
Because the solution is dilute, we can use the properties of water for the solution:
ρ = 996 kg/m3
μ = 8.71 x 10-4
Pa s
Calculate the Schmidt Number:
702
10245.1996
1071.89
4
AB
ScD
N
Calculate the Reynolds number:
17000
1071.8
996061.024.04Re
DN
Now calculate JD:
00758.01700099.099.05.05.0
Re,
LD NJ
Now solve for k’c:
3/2'
Sc
c
D Nk
J
6' 1085.5 ck m/s
In this case, A is diffusing through stagnant B. We use the solubility for CA1 and CA2 = 0. Also, since the
solution is dilute, xBM ≈ 1:
76
21
'
10726.1002948.00.1
1085.5
AA
BM
c
A CCX
kN kgmol/m
2s
Mass Transfer From a Sphere Example 7.3-3, page 482
Calculate the value of the mass-transfer coefficient and the flux for mass transfer from a sphere of
naphthalene to air at 45ºC and 1 atm abs flowing at a velocity of 0.305 m/s. The diameter of the sphere is
0.0254m. The diffusivity of naphthalene in air at 45ºC is 6.92x10-6
m2/s and the vapor pressure of solid
naphthalene is 0.555 mmHg.
D = 0.0254m
PA1 = 0.555 mmHg = 74 Pa
T = 45ºC
P = 1 atm
v = 0.305 m/s
DAB = 6.92x10-6
m/s
ρ = 1.113 kg/m3
μ = 1.93 x 10-5
Pa s
Calculate the Schmidt number:
505.2
1092.6113.1
1093.16
5
AB
ScD
N
Calculate the Reynolds number:
446
1093.1
113.1305.00254.05Re
DN
Now find the Sherwood number:
0.21552.023/153.0
Re ScSh NNN
Now solve for k’c:
AB
cshD
DkN ' 3
6' 1072.5
0254.0
1092.60.21
D
DNk AB
shc m/s
Now solve for k’G:
9
3'
' 10163.23188314
1072.5
RT
kk c
G kgmol/s m2 Pa
Since the gas is dilute, k’G ≈ kG. Using that we can solve for the flux:
79
21 10599.107310163.2 AAGA PPkN kgmol/m2 s
7.3-1 Mass Transfer from a Flat Plate to a Liquid Using the data and physical properties from Example 7.3-2, calculate the flux for a water velocity of 0.152
m/s and a plate length of L = 0.137 m. Do not assume that xBM = 1.0 but actually calculate its value.
DAB = 1.245 x 10
-9 m
2/s
Solubility, S = 0.02948 kgmol/m3
ρ = 996 kg/m3
Mw = 18 kg/kgmol
CA1 = 0.02948 kgmol/m3
μ = 8.71 x 10-9
Pa s
Solve for XBM:
XB1 = 11
1
AB
B
CC
C
= [(996 kg/m
3)/(18kg/kgmol)]/[(996 kg/m
3)/(18kg/kgmol) + 0.02948 kgmol/m
3]
XB1 = 0.999468
XB2 = 1.0
999734.0
999468.0
0.1ln
999468.00.1
ln1
1ln
1
2
12
1
2
21
B
B
BB
A
A
AABM
X
X
XX
X
X
XXX
Calculate Schmidt Number:
70210245.1996
1071.89
4
AB
SCD
N
Calculate Reynolds Number:
5.23812
1071.8
996152.0137.09Re
LN
Calculate JD:
06416.05.2381299.099.05.05.0
Re
NJ D
Calculate K’c:
53/23/2' 102346.1702152.006416.0 SCDC NJK m/s
Calculate Flux:
75
21
'
10641.3999734.0
02948.0102346.1
BM
AAC
AX
CCKN kgmol/m
2s
7.3-4 Mass Transfer to Definite Shapes – Flat Plate and a Sphere Estimate the value of the mass-transfer coefficient in a stream of air at 325.6 K flowing in a duct past the
following shapes made of solid naphthalene. The velocity of the air is 1.524 m/s at 325.6K and 202.6 kPa.
The DAB of naphthalene in air is 5.16 x 10-6
m2/s at 273K and 101.3 kPa.
(a) For air flowing parallel to a flat plate 0.152 m in length
(b) For air flowing past a single sphere 12.7 mm in diameter
CA1
0.137 m
Benzoic Acid
Water at 26.1ºC, velocity, v = 0.152 m/s
P = 202600 Pa
R = 8314.34 m3 Pa/kgmol K
T = 325.6 K
Mw = 28.8 kg/kgmol
Correct DAB:
6
175.1175.1
1051182.33.101
6.202
273
6.3256.202,6.325
ooABP
P
T
TkPaKD m
2/s
Calculate ρ and μ:
1958.0273
6.3250171.0
768.0
cP
T
Tn
o
o cP = 1.958 x 10-5
kg/m s
155.26.32534.8314
2026008.28
RT
PM w kg/m3
(a)
Calculate NRe:
5.25495
10958.1
155.2524.1152.05Re
LN
Because NRe > 15000:
004732.05.25495036.0036.02.02.0
Re NJ D
58722.2
1051182.3155.2
10958.16
5
AB
SCD
N
3/2'
SC
C
D NK
J
3/2'
58722.2524.1
004732.0 CK K
’C = 0.003827 m/s
9
'
1041366.16.32534.8314
003827.0 RT
KK C
G kgmol/m2 Pa s
(b) Dp = 0.0127 m
Calculate Reynolds and Schmidt Numbers:
21.2130
10958.1
155.2524.10127.05Re
pDN
58722.2
1051182.3155.2
10958.16
5
AB
SCD
N
Because 0.6 < NSC < 2.7:
0164.4658722.221.2130552.02552.023/153.03/153.0
Re SCSH NNN
Air at 325.6K, 202.6 kPa, v = 1.524 m/s
0.152 m
AB
pC
SHD
DKN
'
6
'
1051182.3
0127.00164.46
CK
K’C = 0.012725 m/s
9
'
1070.46.32534.8314
012735.0 RT
KK C
G kgmol/m2 Pa s
Mass Transfer of a Liquid in a Packed Bed Example 7.3-4, page 485
Pure water at 26.1ºC flows at the rate of 5.514x10-7
m3/s through a packed bed of benzoic acid spheres
having diameters of 6.375mm. The total surface area of the spheres in the bed is 0.01198 m2 and the void
fraction is 0.436. The tower diameter is 0.0667m. The solubility of benzoic acid in water is 2.948x10-2
kgmol/m3.
(a) Predict the mass transfer coefficient kc.
(b) Using the experimental value of kc, predict the outlet concentration of benzoic acid in the water.
Because the solution is dilute, we can use the properties of water for the solution:
ρ = 996 kg/m3
μ26.1ºC = 8.71 x 10-4
Pa s
μ25ºC = 8.94 x 10-4
Pa s
V = 5.514x10-7
m3/s
Dp = 6.375mm
AS = 0.01198 m2
ε = 0.436
D = 0.0667m
S = 2.948x10-2
kgmol/m3
DAB(25ºC) = 1.21x10-9
m2/s from Table 6.3-1
Correct DAB:
9
3
39 10254.1
108718.0
108940.0
298
1.2991021.1251.26
new
old
old
new
ABABT
TDD
m
2/s
Calculate the area of the column:
32 10494.30667.044
DA m2
Now use the area and volumetric flow to find the velocity:
437 10578.110494.3/10514.5/ AVv m/s
Calculate the Schmidt Number:
6.702
10245.17.996
1071.89
4
AB
ScD
N
Calculate Reynolds number:
150.1
1071.8
7.99610578.1006375.04
4
Re
pDN
Calculate JD:
277.2150.1436.
09.109.1 3/23/2
Re
NJ D
Then, assuming k’c = kc for dilute solutions,
3/2'
Sc
c
D Nv
kJ 3/2
4
'
6.70210578.1
277.2
ck 6' 10447.4 ck m/s
Now set the log mean driving force equation can be set equal to the material balance equation on
the bulk stream:
12
2
1
21
ln
AA
AAi
AAi
AAiAAi
c CCV
CC
CC
CCCCAk
where CAi = 0.02948 (the solubility)
CA1 = 0
A = 0.01198 m2 (the surface area of the bed)
V = 5.514x10-7
m3/s
CA2 = 2.842 x 10-3
kgmol/m3
7.3-5 Mass Transfer to Packed Bed and Driving Force Pure water at 26.1ºC is flowing at a rate of 0.0701ft
3/h through a packed bed of 0.251-in.
benzoic acid spheres having a total surface area of 0.129 ft2. The solubility of benzoic acid in
water is 0.00184 lbmol benzoic acid/ft3 solution. The outlet concentration is cA2 is 1.80 x 10
-4
lbmol/ft3. Calculate the mass transfer coefficient kc. Assume dilute solution.
μ = 0.8718 x 10-3
Pa s
ρ = 996.7 kg/m3
υ = 0.0701 ft3/hr
A = 0.129
CAi = 0.00184 lbmol/ft3
CA1 ≈ 0
CA2 = 1.80 x 10-4
lbmol/ft3
Log mean driving force equation:
2
1
12
lnAAi
AAi
AACA
CC
CC
CCKAAN
where CA1 is the inlet bulk flow concentration, CA2 is the outlet bulk flow concentration, and CAi is the
concentration at the surface of the solid (which in this case equals the solubility)
Material Balance on the bulk stream: 12 AAA CCvAN
So now we plug this into the log mean driving force equation:
2
1
1212
lnAAi
AAi
AACAA
CC
CC
CCKACCv
CA1 = 0, so it cancels out of the equation:
2
22
lnAAi
Ai
ACA
CC
C
CKACv
We can now plug in given numbers:
4
44
1080.100184.0
00184.0ln
1080.1129.01080.10701.0 CK KC = 0.0559 ft/hr
Mass Transfer to Suspensions of Small Particles Mass Transfer from Air Bubbles in Fermentation Example 7.4-1, page 488
Calculate the maximum rate of absorption of O2 in a fermenter from air bubbles at 1 atm abs pressure
having diameters of 100 μm at 37ºC into water having a zero concentration of dissolved oxygen. The
solubility of O2 from air in water at 37ºC is 2.26x10-7
kgmol O2/m3 liquid. The diffusivity of O2 in water at
37ºC is 3.25x10-9
m2/s. Agitation is used to produce the air bubbles.
Dp = 1 x 10-4
m
DAB = 3.25x10-9
m2/s
Solubility = 2.26x10-7
kgmol O2/m3 liquid
μc,water = 6.947 x 10-4
Pa s = 6.947 x 10-4
kg/ m s
ρc,water = 994 kg/m3
ρp,air = 1.13 kg/m3
Calculate NSc:
215
1025.3994
10947.69
4
ABc
c
ScD
N
Now calculate k’L:
3/1
2
3/2' 31.02
c
c
Sc
p
ABL
gN
D
Dk
4
3/1
2
43/2
4
9' 1029.2
994
806.910947.613.199421531.0
101
1025.32
Lk m/s
Assuming k’L = kL for dilute solutions,
874
21 1018.501026.21029.2 AALA CCkN kgmol O2/m2 s
Molecular Diffusion Plus Convection and Chemical Reaction Proof of Mass Flux Equation Example 7.5-1, page 491
Table 7.5-1 gives the following relation: 0 BA jj
Prove this relationship using the definition of the fluxes in terms of velocities.
From Table 7.5-1 (page 490), substituting AA for jA and BB for jB, and rearranging:
0 BBBAAA 0 BABBAA
BB
AA
and BA
0
B
BA
ABBAA 0 BBAABBAA
Thus the identity is proved.
Diffusion and Chemical Reaction at a Boundary Example 7.5-2, page 495
Pure gas A diffuses from point 1 at a partial pressure 101.32 kPa to point a distance 2.00mm away. At point
2, it undergoes a chemical reaction at the catalyst surface and A 2B. Component B diffuses back at
steady state. The total pressure is P = 101.32 kPa. The temperature is 300K and DAB = 0.15 x 10-4
m2/s.
(a) For instantaneous rate of reaction, calculate xA2 and NA.
(b) For a slow reaction where k’1 = 5.63x10
-3 m/s, calculate xA2 and NA.
PA2 = XA2 = 0 because no A can exist next to the
catalyst surface
XA1 = PA1/P = 1.0
δ = 0.002 m
T = 300 K
C = P/RT = 101320/(8314*300) = 4.062 x 10-2
kgmol/m3
NB = -2NA
DAB = 0.15 x 10-4
m2/s
k’1 = 5.63x10
-3 m/s
(a)
442
2
1 10112.201
0.11ln
002.0
1015.010062.4
1
1ln
A
AABA
X
XCDN
kgmol A/m
2s
(b)
SubstituteCk
NX A
A '
1
2 in for XA2:
Ck
N
XCDN
A
AABA
'
1
1
1
1ln
4
23
42
10004.1
10062.41063.51
0.11ln
002.0
1015.010062.4
A
A NN kgmol A/m
2s
7.5-7 Unsteady State Diffusion and Reaction Solute A is diffusing at unsteady state into a semi-infinite medium of pure B and undergoes a first-order
reaction with B. Solute A is dilute. Calculate the concentration CA at points z = 0, 4, and 10 mm from the
surface for t = 1 x 105 seconds. Physical property data are DAB = 1 x 10
-9 m
2/s, k’ = 1 x 10
-4 s
-1, CA0 = 1.0 kg
mol/m3. Also calculate the kg mol absorbed/m
2. Find CA and Q.
For un-steady state diffusion and homogenous reaction in a semi-infinite medium,, the general solution is:
tk
tD
zerf
D
kztk
tD
zerf
D
kz
C
Ci
ABAB
i
ABAB
A
2exp
2exp
'
12
1
'
12
1
For z1=0m:
542
1542
1 1011010exp1011010exp0.1
erferfCA
0.100exp20exp 21
21 AC kgmol/m
3
For z2=0.004m
54
959
4
21
54
959
4
21
1011011011012
004.0
101
101004.0exp
1011011011012
004.0
101
101004.0exp
0.1
erf
erfCA
28226.02282264.00101
101004.0exp2
101
101004.0exp 2
19
4
21
9
4
21
AC kgmol/m3
For z2=0.01m
54
959
4
21
54
959
4
21
1011011011012
01.0
101
10101.0exp
1011011011012
01.0
101
10101.0exp
0.1
erf
erfCA
9998.1042329.00101
10101.0exp9998.1
101
10101.0exp 2
19
4
21
9
4
21
AC
0423.0.0AC kgmol/m3
(b)
The total amount, Q, of A absorbed up to time t is represented by:
tkAB
A etk
tkerftkk
DcQ
''
'
21'
'0
Plugging in and solving for Q, we get:
54 101101
5454
2154
'4
9 101101101101101101
101
1011 eerfQ
0332.0Q kgmol/m2
7.5-11 Effect of Slow Reaction Rate on Diffusion Gas A diffuses from point 1 to a catalyst surface at point 2, where it reacts as follows: 2A B. Gas B
diffuses back a distance δ to point 1.
(c) Derive an equation for NA for a slow first-order reaction where k1’ is the reaction velocity constant.
(d) Calculate NA and xA2 for part (c) where k1’ – 0.53 x 10-2
m/s.
(c)
Begin with the convective diffusion equation:
BAAA
ABA NNC
C
dz
dCDN where: AB
AA NN
C
Cx
21,
Plugging in:
21
21ln
2
21
1
2
0
21
2
1
A
AABA
x
x A
ABAAAA
ABA
x
xCDN
x
dxCDdzNNx
dz
dxCDN
A
A
For a slow reaction, Ck
Nx A
A '
1
2 . Plugging into the flux equation and replacing C with P/RT:
21
21ln
2
1
'
1
A
AABA
x
PkRTN
RT
PDN
(d)
Plugging in the values, we get:
297.01
1001325.11053.02298314.81ln
0013.0298314.8
102.01001325.125245
AA
NN
41076.1 AN kgmol/m2 s
Solving for the mole fraction:
81479.0
298314.8
1001325.10053.0
1076.15
4
'
1
2
Ck
Nx A
A
7.5-10 Diffusion and Chemical Reaction of Molten Iron in Process Metallurgy In a steel-making process using molten pig iron containing carbon, a spray of molten iron particles
containing 4.0 wt % carbon falls through a pure oxygen atmosphere. The carbon diffuses through the
molten iron to the surface of the drop, where it is assume that it reacts instantly at the surface because of the
high temperature, as follows, according to a first order reaction:
Calculate the maximum drop size allowable so that the final drop after a 2.0 second fall contains an average
of 0.1 wt % carbon. Assume that the mass transfer rate of gases at the surface is very great, so there is no
outside resistance. Assume no internal circulation of the liquid. Hence, the decarburization rate is
controlled by the rate of diffusion of carbon to the surface of the droplet. The diffusivity of the carbon in
iron is 7.5 x 10-9
m2/s (S7). Hint: use figure 5.3-13
For unsteady-state diffusion through a spherical geometry where we are looking at average temperature and
time, Figure 5.3-13 (page 377) can be used. So, to find the radius, r, we need to find Y, and X, where:
025.00040
001.00
01
1
CC
CCY ave
Because A reacts instantly, we know that C1=0. From the graph for a sphere at Y, X=0.32
Solving for r:
2
1
9
2
1
2105.732.0
rr
tDX AB
000217.0r m
Unsteady State Diffusion and Reaction in a Semi-Infinite Medium Reaction and Unsteady State Diffusion Example 7.5-3, page 498
Pure CO2 gas at 101.32 kPa pressure is absorbed into a dilute alkaline buffer solution containing a catalyst.
The dilute, absorbed solute CO2 undergoes a first-order reaction, with k’= 35 s
-1 and DAB = 1.5x10
-9 m
2/s.
The solubility of CO2 is 2.961x10-7
kgmol/m3 Pa. The surface is exposed to the gas for 0.010 seconds.
Calculate the kgmol CO2 absorbed/m2 surface.
P = 101.32 kPa
k’= 35 s
-1
DAB = 1.5x10-9
m2/s
Solubility of CO2 is 2.961x10-7
kgmol/m3 Pa
Time, t = 0.010 seconds
The total amount, Q, of A absorbed in time, t is:
tk
ABA etktkerftkkDCQ'
// ''
21''
0
k’t = (35s
-1)(0.010s) = 0.350
CA0 = Solubility*Pressure = (2.961x10-7
kgmol/m3 Pa)( 101.32 kPa) = 3.00x10
-2 kgmol CO2/m
3
7350.0
2192 10458.1/350.0350.0350.35/105.11000.3 eerfQ kgmol CO2/m
2
Multicomponent Diffusion of Gases
Diffusion of A Through Nondiffusing B and C Example 7.5-4, page 498
At 298K and 1 atm total pressure, methane (A) is diffusing at steady state through nondiffusing argon (B)
and helium (C). At z1 = 0, the partial pressures in atm are PA1 = 0.4, PB1 = 0.4, PC1 = 0.2, and at z2 =
0.005m, PA2 = 0.1, PB2 = 0.6, PC2 = 0.3. The binary diffusivities from Table 6.2-1 are DAB = 2.02x10-5
m2/s,
DAC = 6.75x10-5
m2/s, and DBC = 7.29x10
-5 m
2/s. Calculate NA.
PA1 = 0.4,
PB1 = 0.4,
PC1 = 0.2
PA2 = 0.1
PB2 = 0.6
PC2 = 0.3
z2 = 0.005m
z1 = 0
T = 298K
P = 1 atm
DAB = 2.02x10-5
m2/s
DAC = 6.75x10-5
m2/s
DBC = 7.29x10-5
m2/s
R = 82.057 m3 atm/kgmol K
Find DAM:
At point 1:
667.04.01
4.0
1
'
A
BB
X
XX
At point 2:
667.01.01
6.0
1
'
A
BB
X
XX
333.0
4.01
2.0
1
'
A
C
CX
XX
5
55
''10635.2
1075.6333.0
1002.2667.0
11
AC
C
AB
B
AM
DX
DX
D m2/s
Calculate PiM:
6.04.00.111 Ai PPP atm
9.01.00.122 Ai PPP atm
740.0
4.09.0ln
4.09.0
ln1
2
12
i
i
ii
iM
PP
PPP atm
Now solve for NA:
55
21
12
1074.86.09.0740.0005.0298057.82
0.110635.2
AA
iM
AMA PP
PzzRT
PDN kgmol A/m
2s
7.5-8 Multicomponent Diffusion At a total pressure of 202.6 kPa and 358 K, ammonia gas (A) is diffusing at steady state through an inert,
nondiffusing mixture of nitrogen (B) and hydrogen (C). The mole fractions at z1 = 0 are xA1 = 0.8, xB1 =
0.15, and xC1 = 0.05; and at z2 = 4.0 mm, xA1 = 0.2, xB1 = 0.6, and xC1 = 0.2. The diffusivities at 358 K and
101.3 kPa are DAB = 3.28 x 10-5
m2/s and DAC = 1.093 x 10
-4 m
2/s. Calculate the flux of ammonia.
For multi-component diffusion, the flux can be calculated as:
21
12
AA
AM
AMA CC
Czz
CDN
Calculate DAM in m2/s:
5
55
1
1
1
1
109878.1
8.0110456.5
05.0
0811064.1
15.0
1
11
1
AAC
C
AAB
B
AM
xD
x
xD
xD
Calculate CAM:
0647.68
3588314
106.202 3
RT
PC mol/m
3
45.548.00647.6811 AA CxC mol/m3
61.132.00647.6822 AA CxC mol/m3
2
1
21
lnA
A
AAAM
CC
CC
CCCCC
4605.29
61.130647.68
45.540647.68ln
61.130647.6845.540647.68
mol/m
3
Substituting back in and solving for flux:
4335
21
12
1068.461.1345.544605.290004.0
0647.68109878.1
molmmolmCC
Czz
CDN AA
AM
AMA
41068.4 AN kgmole/m2s
Knudsen Diffusion of Gases
Knudsen Diffusion of Hydrogen Example 7.6-1, page 500
A H2(A) – C2H6(B) gas mixture is diffusing in a pore of a nickel catalyst used for hydrogenation at
1.01325x105 Pa and 373K. The pore radius is 60
A (angstrom). Calculate the Knudsen diffusivity, DKA, of
H2.
P = 1.01325x105 Pa
T = 373K
r = 60 angstroms = 6.0x10-9
MA(H2) = 2.016
62
1
92
1
1092.7016.2
37310600.970.97
A
KAM
TrD m
2/s
Flux Ratios for Diffusion of Gases In Capillaries
Transition-Region Diffusion of He and N2 A gas mixture at a total pressure of 0.10 atm abs and 298K is composed of N2 (A) and He (B). The mixture
is diffusing through an open capillary 0.010m long having a diameter of 5x10-6
m. The mole fraction of N2
at one end is XA1 = 0.8 and at the other is XA2 = 0.2. The molecular diffusivity DAB is 6.98x10-5
m2/s at 1 atm,
which is an average value based on several investigations.
(a) Calculate the flux NA at steady state.
(b) Use the approximate equations for this case.
L = 0.010m
r = 2.5x10-6
m
P = 0.10 atm
T = 298 K
XA1 = 0.8
XA2 = 0.2
DAB = 6.98x10-5
m2/s
MA = 28.02 kg/kgmol
MB = 4.003 kg/kgmol
R = 8314.3 m3 Pa/kgmol K
(a)
42
1
62
1
1091.702.28
273105.20.970.97
A
KAM
TrD m
2/s
In an open system with no chemical reaction, the ratio of NA/NB is constant:
645.2003.4
02.28
B
A
A
B
M
M
N
N
Now we can solve for the flux factor, α:
645.1645.211 A
B
N
N
Now we can solve for the transition region NA:
KAABA
KAABAABA
DDx
DDx
RTL
PDN
/1
/1ln
1
2
5
45
4545
1040.61091.7/1098.68.0645.11
1091.7/1098.62.0645.11ln
010.02983.8314645.1
10013.11098.6
AN kgmol/m
2s
(b) If we estimate that equimolar counterdiffusion is taking place at steady state, α = 1-1 = 0. We can
estimate the diffusivity, D’NA:
4
45
' 10708.3
1091.71
1098.61
1
11
1
KAAB
NA
DD
D m2/s
Now we can solve for the approximate flux:
5
44
21
'
1010.92.08.0010.02988314
10013.110708.3
AA
NA
A XXRTL
PDN kgmol/m
2s
7.6-4 Transition-Region Diffusion in Capillary A mixture of nitrogen gas (A) and helium (B) at 298 K is diffusing through a capillary 0.10 m long in an
open system with a diameter of 10 μm. The mole fractions are constant at XA1 = 1.0 and XA2 = 0.0. DAB =
6.98 x 10-5
m2/s at 1 atm. MA = 28.02 kg/kg mol, MB = 4.003.
(a) Calculate the Knudsen diffusivity DKA and DKB at the total pressures of 0.001, 0.1 and 10 atm.
(b) Calculate the flux NA at steady state at the pressures.
(c) Plot NA versus P on log-log paper. What are the limiting lines at lower pressures and very high
pressures? Calculate and plot these lines.
(a)
The Knudsen diffusivity for A and B will be the same at all pressures (it is independent of pressure): 2
1
0.97
A
KM
TrD
001582.002.28
2981050.97
21
6
KAD m2/s
004185.0003.4
2981050.97
21
6
KBD m2/s
(b)
For the transition region, flux is calculated using:
KAABA
KAABAABA
DDx
DDx
RTL
PDN
1
2
1
1ln
For P = 0.001:
7
2
22
1021.6001582.01098.616457.11
001582.01098.606457.11ln
1.0298082157.06457.1
001.01098.6
AN kgmol/m
2s
For P = 0.1:
5
2
22
10321.1001582.01098.616457.11
001582.01098.606457.11ln
1.0298082157.06457.1
1.01098.6
AN kgmol/m
2s
For P = 10:
5
2
22
1068.1001582.01098.616457.11
001582.01098.606457.11ln
1.0298082157.06457.1
101098.6
AN kgmol/m
2s
(c)
Lower limit is representative of Knudsen diffusion and the upper limit is representative of molecular
diffusion.
9.3 Vapor Pressure of Water and Humidity
9.3-1 Humidity from Vapor-Pressure Data The air in a room is at 26.7ºC and a pressure of 101.325 kPa and contains water vapor with a partial
pressure pA = 2.76 kPa. Calculate the following:
(a) Humidity, H
(b) Saturation humidity, HS, and percentage humidity, HP.
(c) Percentage relative humidity, HR.
From the steam tables at 26.7ºC, the vapor pressure of water is pAs = 3.50 kPa. Also pA = 2.76 kPa and P =
101.3 kPa. For part (a):
01742.0
76.23.10197.28
76.202.18
97.28
02.18
A
A
pP
pH kg H2O/ky dry air
For part (b)
02226.0
76.23.10197.28
50.302.18
97.28
02.18
As
As
pP
pH
%3.7802226.0
01742.0100100
s
pH
HH
For part (c)
%9.7850.3
76.2100100
As
AR
p
pH
9.3-2 Use of Humidity Chart Air entering a dryer has a temperature (dry bulb) of 60ºC and a dew point of 26.7ºC. Using the humidity
chart, determine the actual humidity H, percentage humidity HP, humid heat cs, and humid volume vH.
The dew point of 26.7ºC is the temperature when the given mixture is at 100% saturation. Starting at 26.7ºC,
and drawing a vertical line until it intersects the line for 100% humidity, a humidity of H = 0.0225 kg
H2O/kg dry air is read off the plot. This is the actual humidity of the air at 60ºC. Stated in another way, if
air at 60ºC and having a humidity of H = 0.0225 kg H2O/kg dry air is cooled, its dew point will be 26.7ºC.
Locating this point where H = 0.0225 and T = 60ºC on the chart, the percentage humidity HP is found to be
14% by linear interpolation vertically between 10 and 20% lines. The humid heat for H = 0.0225 is from Eq.
(9.3-6):
047.10225.088.1005.1 sc kJ/kg dry air K
The humid volume at 60ºC from Eq. (9.3-7) is
977.0273600225.01056.41083.2 33
Hv m3/kg dry air
9.3-3 Adiabatic Saturation of Air An air stream at 87.8ºC having a humidity H = 0.030 kg H2O/kg dry air is contacted in an adiabatic
saturator with water. It is cooled and humidified to 90% saturation.
(a) What are the final values of H and T?
(b) For 100% saturation, what would be the values of H and T?
For part (a), the point H = 0.030 and T = 87.8ºC is located on the humidity chart. The adiabatic saturation
curve through this point is followed upward to the left until it intersects with the 90% line at 42.5ºC and H =
0.0500 kg H2O/kg dry air.
For part (b), the same line is followed to 100% saturation where T = 40.5ºC and H = 0.0505 kg H2O/kg dry
air.
9.3-4 Wet Bulb Temperature and Humidity A water vapo-air mixture having a dry bulb temperature of T = 60ºC is passed over a wet bulb, as shown in
Figure 9.3-4, and the wet bulb temperature obtained is TW = 29.5ºC. What is the humidity of the mixture?
The wet bulb temperature of 29.5ºC can be assumed to be the same as the adiabatic saturation temperature
TS, as discussed. Following the adiabatic saturation curve of 29.5ºC until it reaches the dry bulb temperature
of 60ºC, the humidity is H = 0.0135 kg H2O/kg dry air.
9.6 Calculation Methods for Constant-Rate Drying Period
9.6-1 Time of Drying from Drying Curve A solid whose drying curve is represented by Figure 9.5-1a is to be dried from a free moisture content X1 =
0.38 kg H2O/kg dry solid to X2 = 0.25. Estimate the time required.
From Figure 9.5-1a for X1 = 0.38, t1 is read off as 1.28 hours. For X2 = 0.25, t2 = 3.08 hours. Hence the time
required is t = t2 – t1 = 3.08 – 1.28 = 1.80 hours.
9.6-2 Drying Time from Rate-of-Drying Curve A solid whose drying curve is represented by Figure 9.5-1b is to be dried from a free moisture content X1 =
0.38 kg H2O/kg dry solid to X2 = 0.25 in the constant rate drying period. Estimate the time required.
For Figure 9.5-1b, a value of 21.5 for LS/A was used to prepare the graph. From the figure, RC = 1.51 kg
H2O/m2hours. Substituting into Eq. 9.6-2:
85.125.038.051.1
5.2121 XX
AR
Lt
C
S hours
9.6-3 Prediction of Constant Rate Drying An insoluble wet granular material is to be dried in a pan 0.457 by 0.457 m and 25.4 mm deep. The material
is 25.4 mm deep in the pan, and the sides and bottom can be considered insulated. Heat transfer is by
convection from an air stream flowing parallel to the surface at a velocity of 6.1 m/s. The air is at 65.6ºC
and has a humidity of 0.010 kg H2O/kg dry air. Estimate the rate of drying for the constant rate period.
For a humidity of H = 0.010 and a dry bulb temperature of 65.6ºC, using the humidity chart, the wet bulb
temperature TW = 28.9ºC and HW = 0.026 by following the adiabatic saturation line to the saturated humidity.
Using Eq 9.3-7 to calculate the humid volume,
974.02736.6501.1056.41083.21056.41083.2 3333 THvH m3/kg dry air
The density for 1.0 kg dry air + 0.010 kg H2O is
037.1974.0
010.00.1
kg/m
3
The mass velocity G is
22770037.136001.6 vG kg/h
Using Eq. 9.6-9
45.62227700204.00204.08.08.0 Gh W/m
2K
At TW = 28.9ºC, λW = 2433 kJ/kg from the steam tables. Substituting into Eq. 9.6-8:
39.336009.286.6510002433
45.623600
W
W
C TTh
R
kg/m2h
The total evaporation rate for a surface area of 0.457 by 0.457 m2 is
708.0457.0457.039.3 ARC kg H2O / h
9.7 Calculation Methods for Falling-Rate Drying Period
Example 9.7-1 Numerical Integration in Falling-Rate Period A batch of wet solid whose drying rate is represented by Figure 9.5-1b is to be dried from a free moisture
content of X1 = 0.38 kg H2O/kg dry solid to X2 = 0.04. The weight of the dry solid is LS = 399 kg solid and A
= 18.58 m2 of top drying surface. Calculate the time for drying. Note that the LS/A = 399/18.58 = 21.5
kg/m2.
From Figure 9.5-1b, the critical free moisture content is XC = 0.195 kg H2O/kg dry solid. Hence the drying
occurs in the constant-rate and falling-rate periods.
For the constant-rate period, X1 = 0.38 and X2 = XC = 0.195. From Figure 9.5-1b, RC = 1.51 kg H2O/m2h.
Substituting Eq. 9.6-2:
63.2195.038.051.158.18
39921 XX
AR
Lt
C
S hours
For the falling-rate period, reading values of R for
various values of X from Figure 9.5-1b, the
following table was prepared. To determine this
area by numerical integration using a spreadsheet,
the calculations below are provided. The area of the
first rectangle is the average height (0.663+0.826)/2
= 0.745, times the width ΔX = 0.045, giving 0.0335.
Other values are similarly calculated and all the
values are summed to give a total of 0.1889.
Substituting into Eq. 9.6-1
06.41889.058.18
3991
2
X
X
S
R
dX
A
Lt hours
The total time is 2.63 + 4.06 = 6.69 hours.
Approximation of Straight Line for Falling-Rate Period A batch of wet solid whose drying rate is represented by Figure 9.5-1b is to be dried from a free moisture
content of X1 = 0.38 kg H2O/kg dry solid to X2 = 0.04. The weight of the dry solid is LS = 399 kg solid and A
= 18.58 m2 of top drying surface. Calculate the time for drying. Note that the LS/A = 399/18.58 = 21.5
kg/m2. As an approximation, assume a straight line for the rate R versus X through the origin from point XC
to X = 0 for the falling-rate period.
RC = 1.51 kg H2O/m2h and XC = 0.195. Drying in the falling rate region is from XC to X2 = 0.040.
Substituting into Eq. 9.7-8
39.4040.0
195.0ln
51.158.18
040.0399ln
2
X
X
AR
XLt C
C
CS hours
This is comparable to the value of 4.06 hours obtained in Example 9.7-1 by numerical integration
X R 1/R ΔX (1/R)avg (ΔX)( 1/R)avg
0.195 1.51 0.663 0.045 0.745 0.0335
0.150 1.21 0.826 0.050 0.969 0.0485
0.100 0.9 1.11 0.035 1.260 0.0441
0.065 0.71 1.41 0.015 2.055 0.0308
0.050 0.37 2.70 0.010 3.203 0.0320
0.040 0.27 3.70 -- Total = 0.1889
10.2 Equilibrium Relations between Phases
10.2-1 Dissolved Oxygen Concentration in Water What will be the concentration of oxygen dissolved in water at 298K when the solution is in equilibrium with
air at 1 atm total pressure? The Henry’s law constant is 4.38x104 atm/molfrac.
Given: atmpA 21.0
molfracxxHxp AAAA
64 1080.41038.321.0 or 0.000835 parts O2 to 100 parts water
10.3 Single and Multiple Equilibrium Contact Stages
10.3-1 Equilibrium Stage Contact for CO2-Air-Water A gas mixture at 1.0 atm pressure abs containing air and CO2 is contacted in a single-stage mixer
continuously with pure water at 293K. The two exit gas and liquid streams reach equilibrium. The inlet gas
flow rate is 100kgmol/hr, with a mole fraction of CO2 of yA2=0.20. The liquid flow rate entering is 300kgmol
water/hr. Calculate the amounts and compositions of the two outlet phases. Assume that water does not
vaporize to the gas phase.
Given: hkgmolLL /300'0
hrkgmolyVV A /80)20.01(100)1('
From Appendix A.3 at 293K -Henry’s Law Constant: molfracatmH /10142.0 4
liquidfracmolgasfracmolPHH /10142.00.1/10142.0/' 44
1
4
11 10142.0' AAA xxHy
Total Material Balance:
20.0,1041.1
10142.01
10142.080
1300
2.01
2.080
01
0300
1'
1'
1'
1'
1
4
1
1
4
1
4
1
1
2
2
1
1
1
1
2
2
AA
A
A
A
A
yx
x
x
x
x
y
yV
x
xL
y
yV
x
xL
Total Flow Rates:
hrkgmolx
LL
A
/3001041.11
300
1
'4
1
1
hrkgmoly
VV
A
/10020.01
80
1
'
1
1
10.3-2 Absorption of Acetone in a Countercurrent Stage Tower It is desired to absorb 90% of the acetone in a gas containing 1.0 mol% acetone in air in a countercurrent
stage tower. The total inlet gas flow to the tower is 30.0kgmol/h, and the total inlet pure water flow to be
used to absorb the acetone is 90kgmolH2O/h. The process is to operate isothermally at 300K and a total
pressure of 101.3kPa. The equilibrium relation for the acetone (A) in the gas-liquid is yA=2.53xA. Determine
the number of theoretical stages required for this separation.
Given: hrkgmolLhkgmolVxy NAAN /0.90/0.30001.0 0101
Acetone Material Balance:
Amount of entering acetone= hrkgmolVy NAN /30.0)0.30(01.011
Entering Air = hairkgmolVy NAN /7.29)0.30)(01.01()1( 11
Acetone Leaving in V1= hrkgmol /030.0)30(10.0
Acetone Leaving in LN= hrkgmol /27.0)30.0(90.0
V1=29.7+0.03=29.73 kg mol air + acetone/hr
00101.073.29
030.01 Ay
LN=0.90+0.27=90.27 kg mol water + acetone/h
00300.027.90
27.0ANx
*Since the liquid flow does not vary significantly, the operating line is assumed to be straight.
Plot the operating line and equilibrium line and step off stages. 5.2 theoretical stages required.
10.3-3 Number of Stages by Analytical Equations It is desired to absorb 90% of the acetone in a gas containing 1.0 mol% acetone in air in a countercurrent
stage tower. The total inlet gas flow to the tower is 30.0kgmol/h, and the total inlet pure water flow to be
used to absorb the acetone is 90kgmolH2O/h. The process is to operate isothermally at 300K and a total
pressure of 101.3kPa. The equilibrium relation for the acetone (A) in the gas-liquid is yA=2.53xA. Determine
the number of theoretical stages required for this separation. Solve using the Kremser analytical equations
for countercurrent stage process.
Given: hrkgmolLhkgmolVxy NAAN /0.90/0.30001.0 0101
Acetone Material Balance:
Amount of entering acetone= hrkgmolVy NAN /30.0)0.30(01.011
Entering Air = hairkgmolVy NAN /7.29)0.30)(01.01()1( 11
Acetone Leaving in V1= hrkgmol /030.0)30(10.0
Acetone Leaving in LN= hrkgmol /27.0)30.0(90.0
V1=29.7+0.03=29.73 kg mol air + acetone/hr
00101.073.29
030.01 Ay
LN=0.90+0.27=90.27 kg mol water + acetone/h
00300.027.90
27.0ANx
The equilibrium relation is: y=2.53xA so m=2.53
195.1)19.1(20.1
19.10.3053.2
27.90
20.173.2953.2
0.90
21
1
1
0
1
AAA
mV
LA
mV
L
mV
LA
N
N
N
stagesA
AAmxy
mxy
N
N
04.5195.1ln
195.1
1
195.1
11
)0(53.200101.0
)0(53.201.0ln
ln
111ln
01
01
10.4 Mass Transfer between Phases
10.4-1 Interface Compositions in Interphase Mass Transfer The solute A is being absorbed from a gas mixture of A and B in a wetted-wall tower with the liquid flowing
as a film downward along the wall. At a certain point in the tower the bulk gas concentration yAG=0.380 mol
fraction and the bulk liquid concentration is xAL=0.100. The tower is operating at 298K and 1.013x105Pa
and the equilibrium data are as follows (x,y data given). The solute A diffuses through stagnant B in the gas
phase and then through a nondiffusing liquid.
Using correlations for dilute solutions in wetted-wall towers, the film mass-transfer coefficients for A in the
gas phase is predicted as ky=1.465x10-3
kgmolA/sm2molfrac and for the liquid phase as kx=1.967x10
-3kgmolA
/sm2molfrac. Calculate the interface concentrations yAi and xAi and the flux NA.
Since system is dilute: 1)1(1)1( iMAiMA xy
Plot the equilibrium data and point P at (0.1,0.38)
Estimate the slope: 342.10.1/10465.1
0.1/10967.1
)1/(
)1/(3
3
'
'
iMAy
iMAx
yk
xkslope
Plotting this, we get yAi=0.183 and xAi=0.247. Use these new values to re-estimate slope:
715.0)]380.01/()183.01ln[(
)380.01()183.01(
)]1/()1ln[(
)1()1()1(
AGAi
AGAi
iMAyy
yyy
825.0)]247.01/()100.01ln[(
)247.01()100.01(
)]1/()1ln[(
)1()1()1(
AiAL
AiAL
iMAxx
xxx
163.1715.0/10465.1
825.0/10967.1
)1/(
)1/(3
3
'
'
iMAy
iMAx
yk
xkslope
Plotting this, we get yAi=0.197 and xAi=0.257. Use these new values to re-estimate slope:
709.0)]380.01/()197.01ln[(
)380.01()197.01(
)]1/()1ln[(
)1()1()1(
AGAi
AGAi
iMAyy
yyy
820.0)]257.01/()100.01ln[(
)257.01()100.01(
)]1/()1ln[(
)1()1()1(
AiAL
AiAL
iMAxx
xxx
160.1709.0/10465.1
820.0/10967.1
)1/(
)1/(3
3
'
'
iMAy
iMAx
yk
xkslope
This is essentially the same slope so use the interfacial values obtained before.
Calculate flux through each phase:
243'
/1078.3)197.0380.0(820.0
10465.1)(
)1(smkgmolyy
y
kN AiAG
iMA
y
A
243'
/1078.3)100.0257.0(820.0
10967.1)(
)1(smkgmolxx
x
kN ALAi
iMA
x
A
10.4-2 Overall Mass-Transfer Coefficients from Film Coefficients Using the same data as in Example 10.4-1, calculate the overall mass-transfer coefficient Ky’, the flux, and
the percent resistance in the gas and liquid films. Do this for the case of A diffusing through stagnant B.
From the graph: yA*=0.052, xAL=0.10, yAG=0.380, yAi=0.197, xAi=0.257
923.0100.0257.0
052.0197.0'
*
ALAi
AAi
xx
yym
From the previous example:
709.0
10465.1
1
3'
iMA
y
y
k and
820.0
10967.1
1
3'
iMA
x
x
k
733.0)]380.01/()052.01ln[(
)380.01()052.01(
)]1/()1ln[(
)1()1()1(
*
*
*
AGA
AGA
MAyy
yyy
4'
33'
''
*
'
1090.8820.0/10967.1
923.0
709.0/10465.1
1
773.0/
1
)1/(
1
)1/(
1
)1/(
1
y
y
iMAxiMAyMAy
KK
xkykyK
The percent resistance:
Gas film: (484/868.8)(100)=55.7% Liquid Film: 44.3%
Calculate the Flux: 244
*
*
'
/1078.3)052.0380.0(773.0
1090.8)(
)1(smkgmolyy
y
KN AAG
MA
y
A
10.6-1 Pressure Drop and Tower Diameter for Ammonia Absorption
Ammonia is being absorbed in a tower using pure water at 25C and 1.0 atm abs pressure. The feed rate is
1140 lbm/h (653.2 kg/h) and contains 3.0mol% ammonia in air. The process design specifies a liquid-to-gas
mass flow rate ratio GL/GG of 2/1 and the use of 1-in metal Pall rings.Calculate the pressure drop in the
packing and gas mass velocity at flooding. Using 50% of the flooding velocity, calculate the pressure drop,
gas and liquid flows, and tower diameter.
Size the tower for the largest flows (at the bottom of the tower). Assume all ammonia is absorbed.
61.28)03.0(0.17)97.0(97.28 gasMW and 01783.0)61.28/()17(03.0 ammoniax
3/07309.0359
64.28
77460
492ftlbmG
From Appendix A.2-4 for water: cp8937.0
From Appendix A.2-3 for water: 3/99708.0 cmgm
3/25.62)43.62(99708.0 ftlbmL
scentistokev 8963.099708.0/8937.0/
From Table 10.6-1 for 1-in Pall Rings: Fp=56ft-1
.
heightpackingftOHFP pflood /925.1)56(115.0115.0 2
7.07.0
For Figure 10.6-5 for Random Packing: 06853.0)25.62/07309.0)(0.2()/)(/( 5.05.0 GGGL GG
From Figure 10.6-5 for 0.06853 and 1.925, the ordinate=1.7.
Capacity parameter equation from ordinate: (To find volumetric flow)
05.05.0
5.0
05.05.05.0 )8963.0()56()07309.025.62(
07309.0)]/([7.1
GpGLGG vvFv
sftvG /663.6
)/(4870.0)07309.0(663.6 2sftlbvG mGGG
50% Flooding:
mGG lbGG 2435.05.0 and for the liquid: 2/4870.0)2435.0(0.2 sftlbG mL
New capacity parameter: 85.0)7.1(5.0
From Figure 10.6-5 for Random Packing at 0.85 and 0.06853: 0.18 inches of water/ft
Solving for Diameter: 22 6427.1)/(2435.0
1/
3600
1140ftsftlbslb
rateflowgas
feedrateA mC
ftDDftAC 446.1)4/(6427.1 22
Ammonia in outlet water: lb68.25)1440(01783.0
Total liquid flow rate: hrlbm /2880)1440(2
Pure liquid flow rate: slbm /3.285868.252880
10.6-2 Absorption of SO2 in a Tray Tower A tray tower is to be designed to absorb SO2 from an air stream by using pure water at 293K. The entering
gas contains 20 mol% SO2 and that leaving 2mol% at a total pressure of 101.3kPa. the inert air flow rate is
150kg air/hm2, and the entering water flow rate is 6000kgwater/hm
2. Assuming an overall tray efficiency of
25%, how many theoretical trays and actual trays are needed? Assume that the tower operates at 293K.
Calculate the molar flow rates:
22 /3330.18
6000'/18.5
29
150' hmwaterinertkgmolLhmairinertkgmolV
From Figure 10.6-7 (drawing of tower trays): 2.01 Ny , 02.01 y , 00 x
Overall Material Balance:
00355.0
02.01
02.018.5
1333
2.01
2.018.5
01
0333
1'
1'
1'
1'
2
2
1
1
1
1
2
2
N
N
N
x
x
x
y
yV
x
xL
y
yV
x
xL
For a portion of the tower:
02.01
02.018.5
1333
118.5
01
0333
1
1
n
n
n
n
x
x
y
y
Choose several y points arbitrarily and plot the operating line.
Plot the equilibrium data from Appendix A.3.
Step off trays from bottom to top: 2.4 theoretical trays
Number of trays with 25% efficiency =0.25(2.4)= 9.6 trays
10.6-3 Minimum Liquid Flow Rate and Analytical Determination of Number of Trays A tray tower is absorbing ethyl alcohol from an inert gas stream using pure water at 303K and 101.3 kPa.
The inlet gas stream flow rate is 100.0 kgmol/h and it contains 2.2 mol% alcohol. It is desired to recover
90% of the alcohol. The equilibrium relationship is y=mx=0.68x for this dilute stream. Using 1.5 times the
minimum liquid flow rate, determine the number of trays needed. Do this graphically and also using the
analytical equations.
Given:
68.00
/00.100022.0
2
11
mx
hkgmolVy
hrinertkgmolyVV /8.97)022.01(0.100)1(' 11
Moles alcohol/h in V1 : 100-97.8=2.20.
Removing 90% moles/hr in outlet gas V2:
0.1(2.20)=0.220
02.9822.08.97220.0'2 VV
002244.002.98/22.02 y
Plot the equilibrium line and points y1, y2, and x2.
Draw the operating line from y2, and x2 to point P for Lmin. At point P on the equilibrium line, xmax is:
03235.068.0/022.0/1max1 myx
Substitute into the operating line equation to find Lmin:
hrkgmolLLL
y
yV
x
xL
y
yV
x
xL
/24.59002244.01
002244.08.97
03235.01
03235.0
22.01
22.08.97
01
0
1'
11'
1
'
min
'
min
'
min
2
2
max.1
max,1'
min
1
1
2
2
min
Solve for L and x:
86.88)24.59)(5.1(5.1' '
min LL
0218.0002244.01
002244.08.97
186.88
22.01
22.08.97
01
086.88 1
1
1
x
x
x
Plot the operating line with several points. The solution is dilute, so the line is straight.
Step off trays: 4 trays.
Flow Rates:
84.90)0218.01/(86.88/'02.98)002244.01/(8.97)1/('
8886'0.100
1122
21
mVLLyVV
LLV
Calculating A for analytical solution:
355.1
333.1)02.98)(068.0/(86.88/336.1)100)(68.0/(84.90/
21
222111
AAA
mVLAmVLA
Calculate N:
04.4335.1/1)335.1/11(0002244.0
0022.0ln
335.1ln
1/1)/11(ln
ln
1
22
21
AA
mxy
mxy
AN
10.6-4 Absorption of Acetone in a Packed Tower Acetone is being absorbed by water in a packed tower having a cross-sectional area of 0.186 m
2 at 293K
and 101.32 kPa (1 atm). The inlet air contains 2.6 mol% acetone and outlet 0.5%. The gas flow is
13.65kgmol inert air/h. The pure water inlet flow is 45.36 kgmol water/h. Film coefficients for the given
flows in the tower are: 2' 1078.3 ak y kgmol/sm3molfrac and 2' 1016.6 ak x kgmol/sm
3molfrac.
Equilibrium data are given in Appendix A.3.
a) Calculate the tower height using ky’a.
b) Repeat using kx’a
c) Calculate Ky’a and the tower height.
Given:
hkgmolVy
xhkgmolLy
/65.13'005.0
0/36.45'026.0
2
21
From Appendix A.3 for acetone-water and xA=0.0333 mol frac: 0395.00395.0760.30 yatmpA
Equilibrium line: xymmxy AA 186.1)0333.0(0395.0
Plot:
Overall Material Balance:
00648.00055.01
0055.065.13
136.45
026.01
026.065.13
01
036.45
1'
1'
1'
1'
1
1
1
2
2
1
,1
1
1
2
2
xx
x
y
yV
x
xL
y
yV
x
xL
Plot the operating line.
Approximate the slope at y1 and x1 using trial and error:
60.1)026.01/(1078.3
)00648.01/(1016.6
)1/(
)1/(
)1/(
)1/(2
2
1
'
1
'
'
'
i
y
ix
iMy
iMx
yak
xak
yak
xakslope
y1*=0.0077
Using new points yi1=0.0154 and xi1=0.0130 continue trial and error:
979.0)]026.01/()0154.01ln[(
)026.01()0154.01(
)]1/()1ln[(
)1()1()1(
AGAi
AGAi
iMAyy
yyy
993.0)]0130.01/()00648.01ln[(
)0130.01()00648.01(
)]1/()1ln[(
)1()1()1(
AiAL
AiAL
iMAxx
xxx
Recalculate the slope: 61.1929.0/1078.3
993.0/1016.6
)1/(
)1/(2
2
'
'
i
iMy
iMx
yak
xakslope
Repeat this process for y2 and x2:
62.1)005.01/(1078.3
)01/(1016.6
)1/(
)1/(
)1/(
)1/(2
2
1
'
1
'
'
'
i
y
ix
iMy
iMx
yak
xak
yak
xakslope
The slope changes very little. Plotting this line we get: y2*=0.0, yi2=0.0020, and xi2=0.0018
Using the interfacial concentrations:
00602.0)]002.0005.0/()0154.0026.0ln[(
)002.0005.0()0154.0026.0(
)]/()ln[(
)()()(
2211
221
ii
ii
Miyyyy
yyyyyy
Calculate the flow rates:
2/10811.3
005.01
3600/65.13
1
'
/10893.3026.01
3600/65.13
1
'
21
3
2
2
3
1
1VV
V
skgmoly
VV
skgmoly
VV
Av
skgmolVAv /10852.32
10811.310893.3 333
skgmolLLLL Av /10260.13600
36.45' 2
21
a)
mzzyyazkyyS
VMiy 911.1)00602.0(1078.3)005.00260.0(
186.0
10852.3)()( 2
3'
21
b) 00368.0)]00018.0/()00648.00130.0ln[(
)00018.0()00648.00130.0(
)]/()ln[(
)()()(
2211
2211
xxxx
xxxxxx
ii
ii
Mi
mzzxxazkxxS
LMix 936.1)00368.0(1016.6)000648.0(
186.0
10260.1)()( 2
2'
21
c) 983.0)]026.01/()0077.01ln[(
)026.01()0077.01(
)]1/()1ln[(
)1()1()1(
1
*
1
1
*
1
*
yy
yyy M
molfracsmkgmolaK
aKxak
m
yakyaK
y
yiMxiMyMy
32'
22'''
*
'
/10183.2
993.0/1016.1
186.1
979.0/1078.3
1
983.0/
1
)1/(
'
)1/(
1
)1/(
1
01025.0)]0005.0/()0077.0026.0ln[(
)0005.0()0077.0026.0(
)]/()ln[(
)()()(
*
22
*
11
*
22
*
11*
yyyy
yyyyyy M
mzzyyazKyyS
VMy 944.1)01025.0(10183.2)005.00260.0(
186.0
10852.3)()( 2
3*'
21
10.7 Absorbtion of Concentrated Mixtures in Packed Towers
10.7-1 Design of an Absorption Tower with a Concentrated Gas Mixture A tower packed with 25.4-mm ceramic rings is to be designed to absorb SO2 from air by using pure water at
293K and 1.013X105Pa abs pressure. The entering gas contains 20mol% SO2 and that leaving 2 mol%. The
inert air flow is 6.53x10-4
kgmol air/s and the inert water flow is 4.20x10-2
kgmol water/s. The tower cross-
sectional area is 0.0929 m2. For dilute SO2, the film mass-transfer coefficients at 293K are, for 25.44 rings
are: 25.07.0' 0594.0 xyy GGak and 82.0' 152.0 xx Gak where ky’a is kgmol/sm
3molfrac, kx
’a is kgmol/sm
3molfrac,
and Gx and Gy are kg total liquid or gas, respectively, per sec per m2 tower cross section. Calculate the
tower height.
Given:
skgmolairV /1053.6' 4 20.01 y 02 x
skgmolL /1020.4' 2 02.02 y
Overall Material Balance:
00355.002.01
02.01053.6
11020.4
2.01
2.01053.6
01
0102.4
1'
1'
1'
1'
1
4
1
1242
2
2
1
1
1
1
2
2
xx
x
y
yV
x
xL
y
yV
x
xL
Operating Line:
y
y
x
x
11053.6
00355.01
00355.01020.4
2.01
2.01053.6
1102.4 4242
Set y=0.04, x=0.000332. Choose other values of y and plug in for x.
Plot equilibrium data from Appendix A.3 and the operating data.
Repeat the following calculations for different y values. For y=0.2:
44
1
1 1016.82.01
'1053.6
1
'
y
VV
C
yA
SOairofflowmassG
'2
2
2
2
4
/3164.00929.0
/1.642.01
2.0/291053.6
smkgm
skgSOskgair
G y
Similarly for liquid flow:
04215.000355.01
'102.4
1
' 2
1
1
x
LL 241.8
0929.0
1.6400355.01
00355.0181020.4
2
2
mGx
857.0)241.8(152.0152.0 82.082.0' xx Gak , 04496.0)241.8()3164.0(0594.00594.0 0257.025.07.0' xyy GGak
Interface Compositions: Estimate the slope of PM by trial and error: Final yi=0.1685 and xi =000565
816.0)]1/()1ln[(
)1()1()1(
AGAi
AGAi
iMAyy
yyy 995.0
)]1/()1ln[(
)1()1()1(
AiAL
AiAL
iMAxx
xxx
6.15)1/(
)1/('
'
iMy
iMx
yak
xakslope
Calculate (1-y), (1-y)iM, and (y-yi)
Integrate: 33.6
)0315.0)(8.0(816.0
)0929.0(04496.0
1016.8
))(1()1(
)(4
'
i
iM
yyyy
y
aSk
Vyf
After repeating for each y value from y1 to y2, numerically integrate to get z=1.588m.
10.8 Estimation of Mass-Transfer Coefficients for Packed Towers 10.8-1 Prediction of Film Coefficients for CO2 Absorption Predict HG, HL, and HOL for absorption of CO2 from air by water in a dilute solution in a packed tower with
1 ½ in metal Pall rings at 303K (30°C) and 101.32kPa pressure. The flow rates are Gx=4.069 kg/sm2 (3000
lbm/h ft2).
From Appendix A.3-18 for CO2 at 1atm: AAAAA xpyxp 33 1086.10.1/1086.1
From Appendix A.3-3 for air at 303K: mskg /10866.1 5 and 3/166.1 mkg
From Table 6.2-1 for CO2 at 276.2K: smDAB /10142.0 24
Correct the diffusivity to 303K: smDAB /10167.02.276
30310142.0 24
75.1
4
Calculate the Schmidt No: 958.0)101670.0)(166.1(
10866.14
5
DN Sc
From Table 10.6-1 for 1 ½ in metal Pall rings to 1 ½ in. Raschig rings: 34.1pf
35.05.05.0
35.05.05.0
678.0782.6660.0
226.0
678.0782.6660.0
226.0
yxSc
p
yG
yxSc
p
yG
GGN
fHH
GGN
fHH
From Appendix A.2-4 for water at 303K: mskg /108007.0 3 and 3/68.995 mkg
At 298K: mskg /108937.0 3
From Table 6.3-1 for CO2 in water at 25C: smDAB /102 29
Adjust to 303K: smDAB /1027.22.276
303
108007.0
108937.01000.2 29
3
39
Calculate the Schmidt No: 3.354)1027.2)(68.995(
108007.04
5
DN Sc
mHH
GN
fHH
xL
xSc
p
xL
2306.0108937.0/782.6
108007.0/069.4
372
3.354
34.1
357.0
108937.0/782.6
/
372
357.0
3.0
3
35.0
3.0
3
5.0
For dilute air: 2/01872.097.28/5424.0/ smkgmolMWGV y
For water: 2/2261.00.18/069.4/ smkgmolMWGL x
mHH
HH
HmVLHHH
OxOL
OxOL
GLOxOL
2322.0001575.02306.0
2426.0))01872.0(1086.1/2261.0(2306.0
)/(
3