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# ch06

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```5 2
evapm 2.76 10 kg/s m .
\u2212
\u2032\u2032 = × \u22c5\ufffd <
Because evaporation is a convection mass transfer process, it also follows that
evap Am n\u2032\u2032 \u2032\u2032=\ufffd
or in terms of the rate equation,
( ) ( ) ( )
( ) ( )
evap m A,s A, m A,sat s A,sat
evap m A,sat
m h h T T
m h 305K 1 ,
\u3c1 \u3c1 \u3c1 \u3c6 \u3c1
\u3c1 \u3c6
\u221e \u221e \u221e
\u221e
\uf8ee \uf8f9\u2032\u2032 = \u2212 = \u2212\uf8f0 \uf8fb
\u2032\u2032 = \u2212
\ufffd
\ufffd
and solving for the convection mass transfer coefficient,
( ) ( ) ( )
5 2
evap
m 3A,sat
m 2.76 10 kg/s mh
305K 1 0.0336 kg/m 1 0.4\u3c1 \u3c6
\u2212
\u221e
\u2032\u2032 × \u22c5
= =
\u2212
\u2212
\ufffd
3
mh 1.37 10 m/s.
\u2212
= × <
COMMENTS: Conservation of species has been applied in exactly the same way as a
conservation of energy. Note the sign convention.
PROBLEM 6.48
KNOWN: CO2 concentration in air and at the surface of a green leaf. Convection mass
transfer coefficient.
FIND: Rate of photosynthesis per unit area of leaf.
SCHEMATIC:
ANALYSIS: Assuming that the CO2 (species A) is consumed as a reactant in photosynthesis
at the same rate that it is transferred across the atmospheric boundary layer, the rate of
photosynthesis per unit leaf surface area is given by the rate equation,
( )A m A, A,sn h .\u3c1 \u3c1\u221e\u2032\u2032 = \u2212
Substituting numerical values, find
( )2 -4 4 3An 10 m/s 6 10 5 10 kg/m\u2212 \u2212\u2032\u2032 = × \u2212 ×
6 2
An 10 kg/s m .
\u2212
\u2032\u2032 = \u22c5 <
COMMENTS: (1) It is recognized that CO2 transport is from the air to the leaf, and (\u3c1A,s -
\u3c1A,\u221e) in the rate equation has been replaced by (\u3c1A,\u221e - \u3c1A,s).
(2) The atmospheric concentration of CO2 is known to be increasing by approximately 0.3%
per year. This increase in \u3c1A,\u221e will have the effect of increasing the photosynthesis rate and
hence plant biomass production.
PROBLEM 6.49
KNOWN: Species concentration profile, CA(y), in a boundary layer at a particular location
for flow over a surface.
FIND: Expression for the mass transfer coefficient, hm, in terms of the profile constants,
CA,\u221e and DAB. Expression for the molar convection flux, \u2032\u2032NA.
SCHEMATIC:
ASSUMPTIONS: (1) Parameters D, E, and F are constants at any location x, (2) DAB, the
mass diffusion coefficient of A through B, is known.
ANALYSIS: The convection mass transfer coefficient is defined in terms of the
( ) )( )
A y=0
m AB
A,s A,
C / y
h x D .
C C
\u2202 \u2202
\u221e
= \u2212
\u2212
The gradient at the surface follows from the profile, CA(y),
( )2A y=0y=0 C Dy Ey F E. y y\u2202 \u2202\u2202 \u2202
\uf8f9
= + + = +\uf8fa\uf8fb
Hence,
( ) ( ) ( )AB ABm A,s A, A,
D E D Eh x .
C C F C
\u221e \u221e
\u2212
= \u2212 =
\u2212 \u2212
<
The molar flux follows from the rate equation,
( ) ( ) ( )ABA m A,s A, A,s A,A,s A,
D EN h C C C C .
C C\u221e \u221e
\u221e
\u2212
\u2032\u2032 = \u2212 = \u22c5 \u2212
\u2212
A ABN D E.\u2032\u2032 = \u2212 <
COMMENTS: It is important to recognize that the influence of species B is present in the
property DAB. Otherwise, all the parameters relate to species A.
PROBLEM 6.50
KNOWN: Steady, incompressible flow of binary mixture between infinite parallel plates with
different species concentrations.
FIND: Form of species continuity equation and concentration distribution. Species flux at upper
surface.
SCHEMATIC:
ASSUMPTIONS: (1) Two-dimensional flow, (2) No chemical reactions, (3) Constant properties.
ANALYSIS: For fully developed conditions, ¶CA/¶x = 0. Hence with v = 0, the species
conservation equation reduces to
2
A
2
d C
0.
dy
= <
Integrating twice, the general form of the species concentration distribution is
( )A 1 2C y C y C .= +
Using appropriate boundary conditions and evaluating the constants,
( )
( ) ( )
A A,2 2 A,2
A A,1 1 A,1 A,2
C 0 C C =C
C L C C C C /L,
= ®
= ® = -
the concentration distribution is
( ) ( ) ( )A A,2 A,1 A,2C y C y/L C C .= + - <
From Fick\u2019s law, the species flux is
( ) AA AB
y=L
dC
N L D
dy
¢¢ = -
( ) ( )ABA A,2 A,1DN L C C .L¢¢ = - <
COMMENTS: An analogy between heat and mass transfer exists if viscous dissipation is
negligible. The energy equation is then d
2
T/dy
2
= 0. Hence, both heat and species transfer are
influenced only by diffusion. Expressions for T(y) and ( )q L¢¢ are analogous to those for CA(y) and
( )AN L .¢¢
PROBLEM 6.51
KNOWN: Flow conditions between two parallel plates, across which vapor transfer occurs.
FIND: (a) Variation of vapor molar concentration between the plates and mass rate of water
production per unit area, (b) Heat required to sustain the process.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Fully developed, incompressible flow with
constant properties, (3) Negligible body forces, (4) No chemical reactions, (5) All work
interactions, including viscous dissipation, are negligible.
ANALYSIS: (a) The flow will be fully developed in terms of the vapor concentration field,
as well as the velocity and temperature fields. Hence
( ) ( )A A A C 0 or C x,y C y .
x
\u2202
\u2202 = =
Also, with \u2202CA/\u2202t = 0, AN 0,=\ufffd v = 0 and constant DAB, the species conservation equation
reduces to
2
A
2
d C 0.
dy
=
Separating and integrating twice,
( ) ( )A 1 2C y C y C .= +
Applying the boundary conditions,
( )
( )
A A,0 2 A,0
A,0 A,L
A A,L A,L 1 2 1
C 0 C C C
C C
C L C C C L C C
L
= \u2192 =
\u2212
= \u2192 = + = \u2212
find the species concentration distribution,
( ) ( ) ( )A A,0 A,0 A,LC y C C C y/L .= \u2212 \u2212 <
From Fick\u2019s law, Eq. 6.19, the species transfer rate is
A,0 A,LA
A A,s AB AB
y=0
C C
CN N D D .
y L
\u2202
\u2202
\u2212\uf8f9
\u2032\u2032 \u2032\u2032= = \u2212 =\uf8fa\uf8fb
Continued \u2026..
PROBLEM 6.51 (Cont.)
Multiplying by the molecular weight of water vapor, \u39c A, the mass rate of water production
per unit area is
A,0 A,L
A A A A AB
C C
n N D .
L
\u2212
\u2032\u2032 \u2032\u2032= =\u39c \u39c <
(b) Heat must be supplied to the bottom surface in an amount equal to the latent and sensible
heat transfer from the surface,
lat sen
A,s fg
y=0
q q q
dTq n h k .
dy
\u2032\u2032 \u2032\u2032 \u2032\u2032= +
\uf8ee \uf8f9
\u2032\u2032 \u2032\u2032= + \u2212\uf8ef \uf8fa\uf8f0 \uf8fb
The temperature distribution may be obtained by solving the energy equation, which, for the
prescribed conditions, reduces to
2
2
d T 0.
dy
=
Separating and integrating twice,
( ) 1 2T y C y C .= +
Applying the boundary conditions,
( )
( ) ( )0 2 0L 1 1 0
T 0 T C T
T L T C T T / L
= \u2192 =
= \u2192 = \u2212
find the temperature distribution,
( ) ( )0 0 LT y T T T y/L.= \u2212 \u2212
Hence,
( )0 L
y=0
T TdTk k .
dy L
\u2212\uf8f9
\u2212 =\uf8fa\uf8fb
Accordingly,
( )A,0 A,L 0 L
A AB fg
C C T T
q D h k .
L L
\u2212 \u2212
\u2032\u2032 = +\u39c <
COMMENTS: Despite the existence of the flow, species and energy transfer across the air
are uninfluenced by advection and transfer is only by diffusion. If the flow were not fully
developed, advection would have a significant influence on the species concentration and
temperature fields and hence on the rate of species and energy transfer. The foregoing results
would, of course, apply in the case of no air flow. The physical condition is an example of
Poiseuille flow with heat and mass transfer.
PROBLEM 6.52
KNOWN: The conservation equations, Eqs. E.24 and E.31.
FIND: (a) Describe physical significance of terms in these equations, (b) Identify approximations
and special conditions used to reduce these equations to the boundary layer equations, Eqs. 6.33
and 6.34, (c) Identify the conditions under which these two boundary layer equations have the same
form and, hence, an analogy will exist.
ANALYSIS: (a) The energy conservation equation, Eq. E.24, has the form
i i T T p p
u v k k u v q.
x y x x y y x y
1a 1b 2a 2b 3 4 5
¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶r r m
¶ ¶ ¶ ¶ ¶ ¶ ¶ ¶
é ù é ùé ù+ = + + + + F +ê ú ê úê úë û ë û ë û
&
The terms, as identified, have the following phnysical significance:
1. Change```