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