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ch08


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Mass flow rate is independent of x.
PROPERTIES: Table A-4, Air (300K, 1 atm): \u3c1 = 1.16 kg/m3, \u3bd = 15.9 × 10-6 m2/s. Prescribed,
Vapor (300K): pA,sat = 15 mm Hg, AM = 70 kg/kmol, DAB = 10
-5
 m
2/s.
ANALYSIS: (a) With the vapor assumed to behave as an ideal gas, pA = CA\u211c T = \u3c1A( )/ A T,\u211c M
and isothermal conditions, the vapor pressure at the outlet may be obtained from the expression
A,sat A,o A,s A,m,o m
A,sat A,i A,s A,m,i
p p D L h
exp
p p m
\u3c1 \u3c1 \u3c1\u3c0
\u3c1 \u3c1
\u2212 \u2212 \uf8eb \uf8f6
= = \u2212\uf8ec \uf8f7
\u2212 \u2212 \uf8ed \uf8f8\ufffd
where ( )3 32m cm u A 1.16 kg / m 0.5 m / s 0.05m / 4 1.14 10 kg / s.\u3c1 \u3c0 \u2212= = × × = ×\ufffd With ReD = um D/\u3bd =
0.5 m/s × 0.05m/15.9 × 10-6 m2/s = 1570, the flow is laminar and mh may be determined from the
mass transfer analog to Eq. 8.57. With Sc = \u3bd/DA = 1.59 and [ReD Sc/(L/D)]1/3 = 2.92 > 2
D AB
5 21/ 3D 3AB
m
Sh ReSc D 10 m / s
h 1.86 1.86 2.92 1.09 10 m / s
D L / D D 0.05m
\u2212
\u2212
= = = × × = ×
\uf8eb \uf8f6\uf8ec \uf8f7\uf8ed \uf8f8
Hence, with pA,i = 0
[mA,o A,sat DL hp p l exp 15mm Hg 1
m
\u3c1\u3c0\uf8ee \uf8f9\uf8eb \uf8f6
= \u2212 \u2212 = \u2212\uf8ef \uf8fa\uf8ec \uf8f7\uf8ed \uf8f8\uf8f0 \uf8fb\ufffd
3 3
3
1.16kg / m 0.05m 5m 1.09 10 m / s
exp 8.7 mm Hg
1.14 10 kg / s
\u3c0 \u2212
\u2212
\uf8f9\uf8eb \uf8f6× × × × × \uf8fa\uf8ec \uf8f7\u2212 =\uf8ec \uf8f7\uf8fa×\uf8ed \uf8f8\uf8fb
<
The corresponding mass density of the vapor is
( )( )A
A,o 3
A,m,o 3
p 8.7 mm Hg 70 kg / kmol 0.0326 kg / m
T 760 mm Hg / atm 0.082 m atm / kmol K 300K
\u3c1 ×= = =
\u211c
\u22c5 \u22c5
M <
(b) The evaporation rate is
( ) 3 2 3 5A m c A,m,o A,m,in u A 0.5m / s 1.96 10 m 0.0326 kg / m 3.20 10 kg / s\u3c1 \u3c1 \u2212 \u2212= \u2212 = × × × = × <
COMMENTS: (1) Since the evaporation rate (nA = 3.2 × 10-5 kg/s) is much less than the air flow
rate ( m\ufffd = 1.14 × 10-3 kg/s), the assumption of a fixed flow rate is reasonable. (2) The evaporation
rate is also given by nA = mh \u3c0 D L \u2206 \u3c1A,lm = - mh \u3c0 D L \u3c1A,m,o/ln [(pA,sat \u2013 pA,o)/pA,sat] = 3.22 ×
10-5 kg/s, which agrees with the calculation of part (b).
PROBLEM 8.112
KNOWN: Air flow rate through trachea of diameter D and length L.
FIND: (a) Average mass transfer convection coefficient, mh , and (b) Rate of water loss per day
(liter/day).
SCHEMATIC:
ASSUMPTIONS: (1) Trachea can be approximated as a smooth tube with uniform surface
temperature, (2) Laminar, fully developed flow, (3) Trachea inner surface is saturated with water at
body temperature, Ts = 37°C, (4) Negligible water vapor in air at 310 K during inhalation, and (5)
Heat-mass analogy is applicable.
PROPERTIES: Table A-4, Air (310 K, 1 atm): \u3c1B = 1.128 kg/m3, µ = 1.893 × 10-5 N\u22c5s/m2; Table
A-6, Water (Ts = 37°C = 310 K): \u3c1A,f = 993 kg/m3, \u3c1A,g = 0.04361 kg/m3; Table A-8, Water-vapor
air (310 K, 1 atm): DAB = 0.26 × 10-4 (310/298)3/2 = 2.76 × 10-5 m2/s.
ANALYSIS: (a) Begin by characterizing the air (B) flow in the trachea modeled as a smooth tube,
B
D
4 m 4
Re
D D
\u3c1
\u3c0 µ \u3c0 µ
\u2200
= =
\ufffd
\ufffd
3 3 3
D 5 2
4 10 liter / min 10 m / liter 1min/ 60s 1.128 kg / m
Re 632
0.020 m 1.893 10 N s / m\u3c0
\u2212
\u2212
× × × ×
= =
× × × \u22c5
Hence, the flow is laminar, and for fully developed conditions and invoking the heat-mass analogy
D D m ABNu Sh 3.66 Sh h D / D= = =
5 2
m ABh 3.66 D / D 3.66 2.76 10 m / s / 0.020 m 0.0050 m / s
\u2212
= = × × = <
(b) The species (A) transfer rate equation, Eq. 8.75, has the form
A m s A, mn h A \u3c1= \u2206 &quot;
( ) ( )
( ) ( )
A,s A,m,o A,s A,m,i
A, m
A,s A,m,o A,s A,m,im /
\u3c1 \u3c1 \u3c1 \u3c1
\u3c1
\u3c1 \u3c1 \u3c1 \u3c1
\u2212 \u2212 \u2212
\u2206 =
\u2212 \u2212\uf8ee\uf8f0&quot; &quot;
where the mean outlet species density, \u3c1A,m,o, can be determined from Eq. 8.78
mA,s A,m,o
A,s A,m,i
h P
exp
m
\u3c1\u3c1 \u3c1
\u3c1 \u3c1
\u2212
= \u2212
\u2212
\uf8eb \uf8f6\uf8ec \uf8f7\uf8ed \uf8f8\ufffd
where m c Bm / u A .\u3c1 = = \u2200\ufffd Substituting numerical values with P = \u3c0D, find
6
A,m,o A0.009233 n 1.54 10 kg / s\u3c1
\u2212
= = ×
The volumetric rate of water loss on a daily basis, assuming a 12 hour inhalation period, is
( ) ( )6 3 3 3A 1.54 10 kg / s / 993 kg / m 10 liter / m 3600s / h 12 h / day\u2212\u2200 = × × × ×\ufffd
AV 0.067 liter / day=\ufffd
PROBLEM 8.113
KNOWN: Air (species B) is in fully developed, laminar flow as it enters a circular tube wetted with
liquid A (water). Tube length and diameter. Flow rate of air and system temperature.
FIND: (a) Governing differential equation for species transfer, (b) Heat transfer analog and an
expression for ShD, (c) General expression for \u3c1A,m,o, (d) Value of \u3c1A,m,o for prescribed conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Steady, incompressible flow, (2) Flow rate is independent of x, (3) Laminar, fully
developed flow (hydrodynamically), (4) Isothermal conditions, (5) Dry air at inlet.
PROPERTIES: Table A.4, Air (298 K, 1 atm): \u3c1 = 1.1707 kg/m3, µ = 183.6 × 10-7 N\u22c5s/m2, \u3bd = 15.71 ×
10-6 m2/s; Table A.6, Water vapor (298 K): \u3c1A,sat = 1/vg = 0.0266 kg/m3; Table A.8, Air-vapor (298 K):
DAB = 26 × 10-6 m2/s, Sc = \u3bd/DAB = 0.60.
ANALYSIS: (a) The governing differential equation may be inferred by analogy to Eq. 8.48. In this
case, the dependent variable is the vapor mass density, \u3c1A (x,r), and the diffusivity is DAB. With v = 0 for
fully-developed flow, it follows that
A AB ADu r
x r r r
\u2202\u3c1 \u2202 \u2202\u3c1
\u2202 \u2202 \u2202
\uf8eb \uf8f6
= \uf8ec \uf8f7\uf8ed \uf8f8 <
The entrance condition is
( )A 0, r 0\u3c1 = <
and the boundary conditions are
( )A o A,sr , x\u3c1 \u3c1= A r 0x 0\u2202\u3c1 \u2202 = = <
(b) The foregoing conditions are analogous to those of the thermal entry length condition associated with
Eq. 8.56. Invoking this analogy the average Sherwood number for laminar, fully developed flow is
( )
( )
D
D 2/3
D
0.0668 D L Re Sc
Sh 3.66
1 0.04 D L Re Sc
=
\uf8ee \uf8f9+ \uf8f0 \uf8fb
<
(c) Applying conservation of species to the differential control volume,
A,m
A,m m c A A,m m c
d
u A dn dx u A
dx
\u3c1
\u3c1 \u3c1\uf8eb \uf8f6+ = +\uf8ec \uf8f7\uf8ed \uf8f8
or, with umAc = m \u3c1\ufffd and ( )A m A,s A,mdn h Ddx\u3c0 \u3c1 \u3c1= \u2212
( )A,m m A,s A,mdm dx h Ddxdx
\u3c1
\u3c0 \u3c1 \u3c1
\u3c1
= \u2212
\ufffd
Continued...
PROBLEM 8.113 (Cont.)
A,m
A,m,i
xA,m m
oA,s A,m
d Dh dx
m
\u3c1
\u3c1
\u3c1 \u3c1\u3c0
\u3c1 \u3c1
=
\u2212
\u222b \u222b
\ufffd
or
A,s A,m m
A,s A,m,i
(x) Dxh (x)
exp
m
\u3c1 \u3c1 \u3c1\u3c0
\u3c1 \u3c1
\u2212 \uf8eb \uf8f6
= \u2212\uf8ec \uf8f7
\u2212 \uf8ed \uf8f8\ufffd
at x = L,
A,s A,m,o m
A,s A,m,i
DLh
exp
m
\u3c1 \u3c1 \u3c1\u3c0
\u3c1 \u3c1
\u2212 \uf8eb \uf8f6
= \u2212\uf8ec \uf8f7
\u2212 \uf8ed \uf8f8\ufffd <
(d) For the prescribed conditions, ReD = 4m D\u3c0 µ\ufffd = 4 2 5 10 0 01 183 6 104 7 2. . .× × \u22c5\u2212 \u2212kg s m N s m\ufffd \ufffd \ufffd \ufffd\u3c0 =
1734 and (D/L)ReDSc = (0.01 m/1 m)1734(0.6) = 10.4. Hence,
( )
( )D 2/3
0.0668 10.4
Sh 3.66 4.24
1 0.04 10.4
= + =
+
( ) ( )6 2Dm ABh Sh D D 4.24 26 10 m s 0.01m 0.011m s\u2212= = × =
Hence,
3A,s A,m,o
4A,s A,m,i
1.1707 kg m 0.01m 1m 0.011m s
exp 0.198
2.5 10 kg s
\u3c1 \u3c1 \u3c0
\u3c1 \u3c1 \u2212
\uf8eb \uf8f6
\u2212 × × × ×\uf8ec \uf8f7= \u2212 =\uf8ec \uf8f7
\u2212 ×\uf8ed \uf8f8
( ) ( )3 3A,m,o A,s A,s A,m,i0.198 0.0226 kg m 1 0.198 0.0181kg m\u3c1 \u3c1 \u3c1 \u3c1= \u2212 \u2212 = \u2212 = <
COMMENTS: Due to evaporation, m\ufffd actually increases with increasing x. However, the increase is
small, and the assumption of fixed m\ufffd is good.
	Problems 8.1-8.25
	8.1
	8.2
	8.3
	8.4
	8.5
	8.6
	8.7
	8.8
	8.9
	8.10
	8.11
	8.12
	8.13
	8.14
	8.15
	8.16
	8.17
	8.18
	8.19
	8.20
	8.21
	8.22
	8.23
	8.24
	8.25
	Problem 8.26-8.50
	8.26
	8.27
	8.28
	8.29
	8.30
	8.31
	8.32
	8.33
	8.34
	8.35
	8.36
	8.37
	8.38
	8.39
	8.40
	8.41
	8.42
	8.43
	8.44
	8.45
	8.46
	8.47
	8.48
	8.49
	8.50
	Problems 8.51-8.75
	8.51
	8.52
	8.53
	8.54
	8.55
	8.56
	8.57
	8.58
	8.59
	8.60
	8.61
	8.62
	8.63
	8.64
	8.65
	8.66
	8.67
	8.68
	8.69
	8.70
	8.71
	8.72
	8.73
	8.74
	8.75
	Problems 8.76-8.100
	8.76
	8.77
	8.78
	8.79
	8.80
	8.81
	8.82
	8.83
	8.84
	8.85
	8.86
	8.87
	8.88
	8.89
	8.90
	8.91
	8.92
	8.93
	8.94
	8.95
	8.96
	8.97
	8.98
	8.99
	8.100
	Problems 8.101-8.113
	8.101
	8.102
	8.103
	8.104
	8.105
	8.106
	8.107
	8.108
	8.109
	8.110
	8.111
	8.112
	8.113