CENG 411 Homework #2 Solutions
3.2. The unimolecular diffusion benzene from the tank is described by (1 − 𝑥! )𝑁! = −𝐷! 𝑐
(#$%)
"#!
"$
&!
%! &
è 𝑁! = − '() 𝑙𝑛[1 − 𝑥! (𝑧 = 0)] where 𝑥! (𝑧 = 0) = & is the mole fraction benzene at the
surface, and h is the thickness of the stagnant layer. The rate of evaporation is then 𝑚̇! =
𝑁! (𝑀𝑊)! 𝐴 = −
%! *(,-)! &
'()
𝑙𝑛 51 −
(#$%)
&!
&
6 = 987 𝑙𝑏/𝑑𝑎𝑦.
3.5. a) The equimolar counter-diffusion of Ar (in Xe) is governed by N Ar = −DArXe c
dx Ar
è
dz
DArXe P
x Ar (L) − x Ar (0) = 0.00159mol / (m2 s) è n& Ar = N Ar A = 5.00 ×10−9 mol / s . Due to
RTL
equimolar counter-diffusion, n& Xe = −n& Ar = −5.00 ×10−9 mol / s . b)
(
N Ar = −
v Ar =
)
N
N Ar N Ar RT 4.95×10−5 m / s
4.95×10−5 m / s
=
=
and v Xe = Xe = −
. c) Due to equimolar
cAr
Py Ar
y Ar
c Xe
y Xe
counter-diffusion, the average molar velocity is zero.
3.22. As derived in class, the concentration profile, assuming an infinite medium, is
⎡
⎢
2
c(z,t) = c0 ⎢1−
π
⎢
⎣
z
2 Dt
∫
0
e −ζ
2
⎤
⎥
d ζ ⎥ , where c0 is the concentration at the surface (z=0), and the flux
⎥
⎦
away from the surface is N = −D
t
∫ N dt = 2
0
∂c
∂z
=
z=0
D
c . The total moles per area leaving the surface is
πt 0
Dt
c = ρ df , where r is the salt density, d is the salt layer thickness, and f is the
π 0
π ⎛ ρ df
fraction of salt dissolved. è t( f ) = ⎜⎜
D ⎝ 2c0
1/2
⎞
⎟⎟ . The surface salt concentration is .265 wt%, and
⎠
assuming a solution density of 1.2 g/cm3, the surface concentration is 0.318 g/cm3. The time to
dissolve the salt is as follows:
fraction
time (s)
time (h)
0.1
1953
0.54
0.5
48813
13.56
0.9
158153
43.93
1
195251
54.24
3.25. The flux of water through the air is 𝑁- = 𝑘/,# 𝑐-1 =
-.
3
34.6789
;3<.4<=×<42 4;
(,-)$+, &
/01
/
3
()
37.=<?
;(687.<@ A)
5∙/01
/
-.
(4.46@? C)36 ;3<.<7= 4;
#
/
-.
<.7×<482
/∙#
=
= 1.183
2B
C4
(#$%)
2',) &*
()
. The air density is 𝜌 =
, so the Reynolds number is 𝑁(D,# =
#E7 F
G
=
= 3340 < 5 ∙ 10@ . The flow is thus laminar, and the mass transfer
9
coefficient may be expressed as 𝑘/,# =
/;
-.
3<.<7= 4 ;H4.6@×<48<
I
#
/
0.332 K
<.7×<482
𝑁- = 𝑘/,# 𝑐-1 =
-.
/∙#
(#$%)
2',) &*
()
9
:
H4.6@×<48<
L K
/;
/
I36 ;
#
#
34.4<9 ;(4.?9 &1K)L
=
9
;
=
C
L = 0.016 J . The water flux through air is then
4.46@? C
/
#
9
F% : %E ;
0.332 I G J I # 7 J
-.
/∙#;
?@A
:=><
3
37.=<?
;(687.<@ A)
5∙/01
M
CNO
= 0.0205 C; ∙J.