Transport Phenomena II CME 3703
Nehal I Abu-Lail
04/26/2023
Recitation 4 (Chapters 21)
Problem 21.1
Problem 21.1-2
Problem 21.3-1: 21.3-1. Mass Transfer from a Flat Plate to a Liquid. Using the data and
physical properties of Example 21.3-2, 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.
Given: L=0.137 m, υ= 0.152m/s, ρ= 996.8 kg/m3, = 8.7 1×10-4 Pa.s, MWA= 122.1 g/mol
Nsc = 702, solubility = CA1= 0.02948 kg mol/m2, CA2= 0
NRe, L =
=
.
.
.
2.383
.
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Basis: 1 m3 solution
Mol has xA1 =
.
.
.
0.00053
xB1 = 1.0-0.00053 = 0.99947 xB2 = 1.0000
.
xBM=
.
0.9997
JD = 0.99 NRe, L-0.5 = 0.99(2.883 104)-0.5= 6.416 10-3
kc’= JD*v*Nsc-(2/3) = 6.416 10-3(0.152)(702)-2/3
kc’= 1.235 10-5 m/s
NA=
’
𝐶𝐴1
𝐶𝐴2
.
NA= 3.642 10-7 kgmol/s.m2
Problem 21.3-2
.
0.02948
0
Problem 21.3-9
𝑘 𝐶
1.) 𝑁 𝑑𝐴
𝑑𝐴
2.)
𝐶 𝑑𝐴
𝑉𝑑𝐶
𝐴
𝑙𝑛
3.) Use the mean driving force and material balances:
NAA= kc (𝐶
𝐶 )meanA = V(CA2- CA1)
𝐶 )mean=
4.) A(𝐶
(CA2-CA1)
5.) Solve for kc in (2) and substitute to (4)
𝐶 )mean = A
A(𝐶
6.) 𝐶
𝐶 mean
NAA=
Problem 21.5-1
Given: V= 0.11 m/s, T= 26.1 oC, L=0.40m, DAB=1.245×10-9 m/s, ρ= 996 kg/m3, = 8.71×10-4 Pa.s,
Nsc=702
(a) 𝑁
.
,
.
5.031
.
JD= 0.99NRe, L-0.5 = 4.414 10-3
4.414 10-3 =
𝑁𝑠𝑐
JD =
.
702
Kc(dilute) =kc’= 6.148 10-6 m/s
(b) Film model
kc’= DAB/f ,
6.148 10-6 = 1.245 10-9 /f
f = 2.025 10-4 m= 0.2025 mm
6.148 10-6=
C.) kc’=
tl = 41.94 seconds
d.)𝑘
𝐷 𝑠
6.148 10-6 = (1.245 10-9 s)1/2
s= 3.036 10-2 s-1
.
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