CHE425: Problem set #2
1. A 3-ft depth of stagnant water at 25oC lies on top of a 0.10-in. thickness of NaCl. At time t
< 0, the water is pure. At time t = 0, the salt begins to dissolve into the water. If the
concentration of salt in the water at the solid-liquid interface is maintained at saturation
(0.00544 mol NaCl/cm3) and the diffusivity of NaCl is 1.2×10-5 cm2/s, independent of
concentration, estimate, by assuming the water to act as a semi-infinite medium, the time and
of salt in the water when: (a) 10% of the salt has dissolved; (b) 50% of the salt has dissolved;
and (c) 90% of the salt has dissolved. Density of NaCl = 2.165 g/cm3. Molecular weight of
NaCl = 58.44 g/mol.
Part
(a) 10% transferred
(b) 50% transferred
(c) 90% transferred
Moles transferred
0.000941
0.004705
0.008469
Time, s
1,960
48,900
159,000
Time, h
0.544
13.6
44.0
2. A slab of dry wood of 4-inch thickness and sealed edges is exposed to air of 40% relative
humidity. Assuming that the two unsealed surfaces of the wood immediately jump to an
equilibrium moisture content of 10 lb H2O per 100 lb of dry wood, determine the time for the
moisture content at the center of the slab to reach 1% of the equilibrium value. Assume a
diffusivity of water of 8.3×10-6 cm2/s.

0.01
x
1.82
t, s
234,000
Time, h
65
3) Dry air at 1 atm flows at 2 m/s across the surface of a 2-inch-long surface that is covered
with a thin film of water. If the air and water are at 25oC and the diffusivity of water in air is
0.25 cm2/s, estimate the water flux in kmol/sm2 for the evaporation of water at the middle of
the surface, assuming laminar boundary-layer flow. Air at 25oC,  = 0.018 cP = 1.8 x 10-5
kg/m-s.
nH2O
A
 (0.016)[0.00128  0]  2.05 105 kmol/s-m2
4. Air at 1 atm and 100oC flows across a thin, flat plate of subliming naphthalene that is 1 m
long. The Reynolds number at the trailing edge of the plate is at the upper limit for a laminar
boundary layer. Estimate: (a) the average rate of sublimation in kmol/sm2; and (b) the local
rate of sublimation 0.5 m from the leading edge. Vapor pressure of naphthalene = 10 torr;
viscosity of air = 0.0215 cP; molar density of air = 0.0327 kmol/m3; and diffusivity of
naphthalene in air = 0.94×10-5 m2/s.
Solution
c
h
nA,B / A  0.0059 4.3  102  0  2.54  104 kmol / s - m2
nA,B / A  kcx  cA i   cA o   0.00417  4.3 102  0   1.79 104 kmol/s-m 2
5. Water at 25oC flows turbulently at 5 ft/s through a straight, cylindrical tube cast from
benzoic acid, of 2-in inside diameter. If the tube is 10 ft long, and fully developed, turbulent
is assumed, estimate the average concentration of acid in the water leaving the tube. The
properties are: solubility of benzoic acid in water = 0.0034 g/cm3; viscosity of water = 0.89
cP = 0.0089 g/cms; and diffusivity of benzoic acid in water at infinite dilution = 9.18×10-6
cm2/s.

 (0.00367)(4,870)  
3
cAout =0.0034 1  exp  
  = 0.000020 g/cm
(152)(20.2
5)



Therefore, the concentration of benzoic acid in the exiting water is way below the solubility
value.
6.2 Air at 1 atm flows at a Reynolds number of 50,000 normal to a long, circular, 1-in
diameter made of naphthalene. Calculate the average sublimation flux in kmol/sm2. Vapor
pressure of naphthalene = 10 torr; viscosity of air = 0.0215 cP; molar density of air = 0.0327
kmol/m3; and diffusivity of naphthalene in air = 0.94×10-5 m2/s.


N A  kc cAs  cA  (0.080)(4.3 104  0)  3.44 105 kmol/s-m2
7. Carbon dioxide is stripped from water by air in a wetted-wall tube. At a location where
pressure is 10 atm and temperature 25oC, the flux of CO2 is 1.62 lbmol/hft2. The partial
pressure of CO2 is 8.2 atm at the interface and 0.1 atm in the bulk gas. The diffusivity of CO2
in air at these conditions is 1.6×10-2 cm2/s. Assuming turbulent flow, calculate by film theory
the mass-transfer coefficient kc for the gas phase and the film thickness.
kc 
(0.00022)(0.475)
 0.315 cm/s
(0.000409)  0.82  0.01
 = DAB / kc = 1.6 x 10-2/0.315 = 0.051 cm
8. Water is used to remove CO2 from air by absorption in a column packed with Pall rings.
At a region of the column where the partial pressure of CO2 at the interface is 150 psia and
the concentration in the bulk liquid is negligible, the absorption rate is 0.017 lbmol/hft2. The
CO2 diffusivity in water is 2.0×10-5 cm2/s. Henry’s law for CO2 is p = Hx, where H = 9,000
psia. Calculate by film theory the mass-transfer coefficient kc in cm/s for the gas phase and
the film thickness in cm.
kc 
2.31 106
 0.0025 cm/s
(0.0556)  0.0167  0 
 = DAB /kc = 2 x 10-5/0.0025 = 0.0080 cm
Ref: J. D. Seader and E. J. Henley, Separation Process Principles , Wiley, 2011