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高等輸送二 — 熱傳
Lecture 11Simultaneous Heat and Mass Transfer
郭修伯 助理教授
Mathematical analogies
• The diffusion of mass and the conduction of heat obey very similar equations– diffusion in one dimension - Fick‘s law
– heat conduction - Fourier‘s law
dz
dcDj 1
1
dz
dTkq
semiinfinite slab
Dt
z
cc
cc
4erf
101
101
semiinfinite slab
t
z
TT
TT
4erf
0
01
pC
kˆ
Thermal diffusivityThermal conductivity
Flat plate moved into an initially stagnant fluid
Mathematical analogies
– Momentum transfer in one dimension - Newton‘s law
dz
du t
z
V
Vu
4erf
0
viscosity plate velocity kinetic viscosity
Confusing?~~~ kD
~~~ D
dz
dcDj 1
1
Mass flux
dz
dTkq
Mass per volume
Energy flux
Not energy per volume?
dz
TCd
C
kq p
p
)ˆ(ˆ
dz
du
Momentum flux
Not momentum per volume?
dz
ud )(
Interfacial mass flux: 11 ckN
Interfacial energy flux: TCC
hThq p
p
zˆ
ˆ| 0
Interfacial momentum flux: 02
| 0 ufu
z
Table 20.1-1
Cooling metal spheresWe want to quickly quench a liquid metal to make fine powder. We plane to do this by spraying drops into an oil bath. How can we estimate the cooling speed of the drops?
No suitable heat transfer correlations! However, several mass transfer correlations for drops are given:For large drops without stirring:
213
1
2
3
42.0
D
v
v
gd
D
kd
Sherwood number, kd/D ~ Nusselt number, hd/k
Schmidt number, v/D ~ Prandtl number, v/α
213
1
2
3
42.0
v
v
gd
k
hd
This correlation will be reliable only if the Grashöf number for the cooling falls in the same range as that used to develop the mass transfer correlation.
Heat transfer from a spinning discImagine that a spinning metal disc electrically heated to 30C is immersed in 1000 cm3 of an emulsion at 18C. The disc is 3 cm in diameter and is turning at 10 rpm. The emulsion’s kinetic viscosity is 0.082 cm2/sec. After an hour, the emulsion is at 21C. What is its thermal diffusivity?
Energy balance:
qRdt
dTVC p
20
ˆ
)(| 0 TThq discz
I.C., t = 0, T = T0
tV
R
C
h
TT
TT
pdisc
disc20
0ˆ
exp
36001000
)5.1(ˆ
exp1830
21303
2
cm
cm
C
h
p
sec011.0ˆ
cmC
h
p
312
12
62.0
D
v
v
d
D
kd
α = ?Mass transfer away from a spinning disc:
312
12
62.0ˆ
v
v
dd
C
h
p
sec/102.1
ˆ62.0
1
23
23
21
61
cm
vC
h
p
111 cbuackN
TCubaTCC
hq pp
p
ˆ''ˆˆ
0''''02
uubaufu
• The situations in which mass transfer, heat transfer, and fluid flow occur at the same rate:– the rates of mass, heat, and momentum transfer can be
essentially the same for fluids in turbulent flow.– Reynolds(1874): mass or heat transfer in a flowing
fluid must involve two simultaneous processes:• Natural diffusion of the fluid at rest
• the eddies caused by visible motionAll caused by flow?
''' bbb a << bua’ << b’ua’’ << b’’u
2ˆf
uC
h
u
k
p
Reynolds analogy
Reynolds analogy
• It suggests a simple relation between different transport phenomena.
• This relation should be accurate when transport occurs by means of turbulent eddies.
• We can estimate mass transfer coefficients from heat transfer coefficients or from friction factors!
• However, experimental results show that the Reynolds analogy is accurate for gases, but not for liquids.
2ˆf
uC
h
u
k
p
The Chilton-Colburn analogy
• How to extend to liquid?
• By an analysis of experimental data:
– reduce to the Reynolds analogy for gases whose Schmidt and Prandtl numbers equal unity
– later apply theories, especially boundary layer theory, to rationalize the exponent of 2/3.
2ˆ
32
32
fv
uC
h
D
v
u
k
p
The wet-bulb thermometer
• measures the colder temperature caused by evaporation of the water
• applied to calculate the relative humidity in air:– mass flux:
– energy flux:
11111 yykccckN ii
TThq i
Coupling: qHN vap ~1
TThyycHk iivap 11
~
TThyycHk iivap 11
~
the Chilton-Colburn analogy
2ˆ
32
32
fv
uC
h
D
v
u
k
p
32
ˆ
D
C
hk
p
TThyycHC
hiivap
p
11
~ˆ
=1 for gases
pp CcC~ˆ
i
vap
pi TT
H
Cyy
~
~
11
Relative humidity =
i
vap
pii TT
H
CTatsaty
TatsatyTatsaty
y~
~)(
)(
1
)( 111
1
Design of cooling towersCalculate the size of a tower required to cool a given amount of water:
Fig. 20.3.2
Fig. 20.3.1
Hot water in
Cold water out
zCold air in
Hot air out
The mass balance on the water vapor in the control volume
Water accumulation = water convection in minus that out + water added by evaporation
OHiOHzzOHzOH cckzaGyGy2222 ,0
OHiOHOH cckaGydz
d222 ,0
airnG ~
OHOH cyc22
OHiOHOHair yykacydz
dn
222 ,0
The energy balance on the wet air in the control volume
airiairairpair TThaTdz
dCn ,
~0
The energy balance on both liquid water and wet air in the control volume:
Hdz
dnT
dz
dCn airOHOHpOH
~~0
222 , air
OHpOH
OH n
Cn
dT
Hd22
2
,
~~
OHiOHOHair yykacydz
dn
222 ,0
airiairairpair TThaTdz
dCn ,
~0
Coupling:
X vapH~
OHiOHvapairOHOHvapairairpair yyHkacTThayHTCdz
dn
2222 ,,
~~~
Assuming, OHi TT2
the Chilton-Colburn analogy
2ˆ
32
32
fv
uC
h
D
v
u
k
p
32
ˆ
D
C
hk
p
=1 for gasespp CcC~ˆ
HHkacHdz
dn iair
~~~ OHvapairairp yHTCH2
~~~,
iOHvapOHairpi yHTCH ,, 22
~~~
HHkacHdz
dn iair
~~~
integration
out
in
H
Hi
airl
HH
Hd
kac
ndzl
,
,
~
~0
~~
~ orair
OHpOH
OH n
Cn
dT
Hd22
2
,
~~
...
Fig 20.3-3
Fig 20.3-5
Fig 20.3-4
For kc values
Design a countercurrent cooling tower to cool water at 2150 kg/min. The water enters at 60C and is to be cooled to 25C. The air is fed at 60 g-mol/m2.sec with a dry-bulb temperature of 30C and a dew point temperature of 10C. The water flux should be 40% lower than the maximum allowed thermodynamically. Find (1) the flow rate of the water per tower cross section, (2) the tower cross section, and (3) the height of tower required.
Refer to Fig 20.3-5, the maximum water flow : AB’
(1) Slope of AB’ = Cmolg
J
n
Cn
dT
Hd
air
OHpOH
OH
230
~~22
2
,
sec180
22
2
m
OHmolgn OH
40%
Slop of actual operating line, AB = 110 x 75 / 60 = 137.5
(2) the tower cross section: Am
OHmolgOHkg
sec110
min2150
222 218 mA
sec110
22
2
m
OHmolgn OH
(3) the tower height: out
in
H
Hi
airl
HH
Hd
kac
ndzl
,
,
~
~0
~~
~ml 1.8)7.2(
3
9
Thermal diffusion and effusion
• Temperature gradient effects a solute flux
Uniform
salt solution
Heated
Cooled
Soret, 1879
Dilute salt solution
Concentrated salt solution
TxxxDcj 2111
For liquid,
Soret coefficient
T
TxxxDcj 2111
For gas,
Thermal diffusivityHeavier molecules usually will concentrate in the cooler region.
Experimental values:
Table 20.5-1
T
TxxxDcj 2111
The temperature gradient effect disappears rapidly for dilute solution and is largest when solute and solvent concentrations are similar.
Thermal diffusion is studied in a two-bulb apparatus. Each bulb is 3 cm3 in volume; the capillary is 1 cm long and has an area of 0.01 cm2. The left-hand bulb is heated to 50C, and the right-hand bulb is kept at 0C. The entire apparatus is initially filled with an equilmolar mixture, either of hydrogen-methane or of ethanol-water. How much separation is achieved? About how long does this separation take?
T
TxxxDcj 2111
Thermal diffusion:
average
coldhotcoldhot T
TTxxxx 2111
gas mixture
012.0298
50)5.0)(5.0)(29.0(11
K
Kyy coldhot 05.0
298
50)5.0)(5.0)(3.1(11
K
Kxx coldhot
ethanol-water
The separations are small; that with liquids is slightly larger but in the opposite direction.
Mass balance on the left-hand bulb:
ave
ABABi
BB T
TTxxxx
l
ADj
c
A
dt
dxV 11
21111 )(
Mass balance on the left-hand bulb:
ave
ABAB
AA T
TTxxxx
l
AD
dt
dxV 11
21111 )(
ave
ABAB
ABAB T
TTxxxx
VVl
ADxx
dt
d 11211111 )(
11 Integration…gas ~ 500secliquid ~ 180 days
Conclusions
• For gases, D and α are nearly equal, and k and are very similar.
• For liquids and solids, D is much less than α, and k is much less than
• For liquids and solids, the heat transfer is much more rapid than the mass transfer, and so proceeds as if the mass transfer did not exist. The two processes are essential uncoupled.
pC
hˆ
pC
hˆ