A HEAT TRANSFER TEXTBOOK - THIRD EDITION Episode 3 Part 7 doc

Figure 11.14 Concentration boundary layer on a flat plate.Convective mass transfer at low rates Convective mass transfer is analogous to convective heat transfer whentwo conditions apply:

§11.6 Mass transfer at low rates 639

Figure 11.13 Mass diffusion into a

semi-infinite stationary medium

with

m i = mi,0 for t = 0 (all x)

m i = mi,u for x = 0 (t > 0)

mi → mi,0 for x → ∞ (t > 0)

This math problem is identical to that for transient heat conduction

into a semi-infinite region (Section5.6), and its solution is completely

The reader can solve all sorts of unsteady mass diffusion problems

by direct analogy to the methods of Chapters4and5when the

concen-tration of the diffusing species is low At higher concenconcen-trations of the

diffusing species, however, counterdiffusion velocities can be induced,

as in Example11.8 Counterdiffusion may be significant in concentrated

metallic alloys, as, for example, during annealing of a butt-welded

junc-tion between two dissimilar metals In those situajunc-tions, eqn (11.72) is

sometimes modified to use a concentration-dependent, spatially varying

interdiffusion coefficient (see [11.6])

Figure 11.14 Concentration boundary layer on a flat plate.

Convective mass transfer at low rates

Convective mass transfer is analogous to convective heat transfer whentwo conditions apply:

1 The mass flux normal to the surface, n i,s, must be essentially equal

to the diffusional mass flux, j i,s from the surface In general, this

requires that the concentration of the diffusing species, m i, be low.9

2 The diffusional mass flux must be low enough that it does not affectthe imposed velocity field

The first condition ensures that mass flow from the wall is diffusional,

as is the heat flow in a convective heat transfer problem The secondcondition ensures that the flow field will be the same as for the heattransfer problem

As a concrete example, consider a laminar flat-plate boundary layer in

which species i is transferred from the wall to the free stream, as shown

in Fig.11.14 Free stream values, at the edge of the b.l., are labeled with

the subscript e, and values at the wall (the s-surface) are labeled with the subscript s The mass fraction of species i varies from m i,s to m i,e

across a concentration boundary layer on the wall If the mass fraction

of species i at the wall, m i,s , is small, then n i,s ≈ ji,s, as we saw earlier inthis section Mass transfer from the wall will be essentially diffusional.This is the first condition

In regard to the second condition, when the concentration difference,

mi,s − mi,e , is small, then the diffusional mass flux of species i through the wall, j i,s, will be small compared to the bulk mass flow in the stream-

9 In a few situations, such as catalysis, there is no net mass flow through the wall, and convective transport will be identically zero irrespective of the concentration (see Problems 11.9 and 11.44).

§11.6 Mass transfer at low rates 641

wise direction, and it will have little influence on the velocity field Hence,

we would expect that
v is essentially that for the Blasius boundary layer.

These two conditions can be combined into a single requirement for

low-rate mass transfer, as will be described in Section11.8 Specifically,

low-rate mass transfer can be assumed if

 0.2 condition for low-ratemass transfer (11.74)

The quantity B m,i is called the mass transfer driving force It is

writ-ten here in the form that applies when only one species is transferred

through the s-surface The evaporation of water into air is typical

exam-ple of single-species transfer: only water vapor crosses the s-surface.

The mass transfer coefficient In convective heat transfer problems,

we have found it useful to express the heat flux from a surface, q, as

the product of a heat transfer coefficient, h, and a driving force for heat

transfer,∆T Thus, in the notation of Fig.11.14,

qs = h (Ts − Te) (1.17)

In convective mass transfer problems, we would therefore like to

ex-press the diffusional mass flux from a surface, j i,s, as the product of a

mass transfer coefficient and the concentration difference between the

s-surface and the free stream Hence, we define the mass transfer

coeffi-cient for species i, g m,i (kg/m2·s), as follows:

j i,s ≡ gm,im i,s − mi,e (11.75)

We expect g m,i , like h, to be determined mainly by the flow field, fluid,

and geometry of the problem

The analogy to convective heat transfer We saw in Sect. 11.5 that

the equation of species conservation and the energy equation were quite

similar in an incompressible flow If there are no reactions and no heat

generation, then eqns (11.61) and (6.37) can be written as

These conservation equations describe changes in, respectively, the amount

of mass or energy per unit volume that results from convection by a givenvelocity field and from diffusion under either Fick’s or Fourier’s law

We may identify the analogous quantities in these equations For thecapacity of mass or energy per unit volume, we see that

or, in terms of the mass fraction,

ρ dmi is analogous to ρcp dT (11.76b)The flux laws may be rewritten to show the capacities explicitly

ji,s = gm,imi,s − mi,e =



g m,i ρ

§11.6 Mass transfer at low rates 643

From these comparisons, we conclude that the solution of a heat

convec-tion problem becomes the soluconvec-tion of a low-rate mass convecconvec-tion

prob-lem upon replacing the variables in the heat transfer probprob-lem with the

analogous mass transfer variables given by eqns (11.76)

Convective heat transfer coefficients are usually expressed in terms

of the Nusselt number as a function of Reynolds and Prandtl number

Nux = hx

k = (h/c p )x ρ(k/ρcp) = fn (Rex , Pr) (11.77)For convective mass transfer problems, we expect the same functional de-

pendence after we make the substitutions indicated above Specifically,

if we replace h/c p by g m,i , k/ρc p byDi,m, and Pr by Sc, we obtain

Num,x ≡ g m,i x

ρ Dim = fn (Rex , Sc) (11.78)

where Num,x, the Nusselt number for mass transfer, is defined as

indi-cated Num is sometimes called the Sherwood number10, Sh

Example 11.10

A napthalene model of a printed circuit board (PCB) is placed in a

wind tunnel The napthalene sublimates slowly as a result of forced

convective mass transfer If the first 5 cm of the napthalene model

is a flat plate, calculate the average rate of loss of napthalene from

that part of the model Assume that conditions are isothermal at

303 K and that the air speed is 5 m/s Also, explain how napthalene

sublimation might be used to determine heat transfer coefficients

Solution. Let us first find the mass fraction of napthalene just

above the model surface A relationship for the vapor pressure of

napthalene (in mmHg) is log10pv = 11.450−3729.3 (T K) At 303 K,

this gives p v = 0.1387 mmHg = 18.49 Pa The mole fraction of

napthalene is thus x nap,s = 18.49/101325 = 1.825 × 10 −4, and with

10 Thomas K Sherwood (1903–1976) obtained his doctoral degree at M.I.T under

War-ren K Lewis in 1929 and was a professor of Chemical Engineering there from 1930 to

1969 He served as Dean of Engineering from 1946 to 1952 His research dealt with

mass transfer and related industrial processes Sherwood was also the author of very

influential textbooks on mass transfer.

eqn (11.9), the mass fraction is, with Mnap= 128.2 kg/kmol,

The analogy therefore applies

The convective heat transfer coefficient for this situation is thatfor a flat plate boundary layer The Reynolds number is

ReL = u ∞ L

ν = (5)(0.05)

1.867 × 10 −5 = 1.339 × 104

where we have used the viscosity of pure air, since the concentration

of napthalene is very low The flow is laminar, so the applicable heattransfer relationship is eqn (6.68)

NuL = hL

k = 0.664 Re 1/2

L Pr1/3 (6.68)Under the analogy, the Nusselt number for mass transfer is

Num,L = g m,i L

ρ Dim = 0.664 Re

1/2

L Sc1/3

The diffusion coefficient for napthalene in air, from Table 11.1, is

Dnap,air = 0.86 × 10 −5m/s, and thus Sc= 1.867 × 10 −5 /0.86 × 10 −5 =

§11.6 Mass transfer at low rates 645

The average mass flux from this part of the model is then

nnap,s = gm,napmnap,s − mnap,e

= (0.0200)(8.074 × 10 −4 − 0)

= 1.61 × 10 −5 kg/m2s= 58.0 g/m2h

Napthalene sublimation can be used to infer heat transfer

coeffi-cients by measuring the loss of napthalene from a model over some

length of time Experiments are run at several Reynolds numbers

The lost mass fixes the sublimation rate and the mass transfer

coeffi-cient The mass transfer coefficient is then substituted in the analogy

to heat transfer to determine a heat transfer Nusselt number at each

Reynolds number Since the Schmidt number of napthalene is not

generally equal to the Prandtl number under the conditions of

inter-est, some assumption about the dependence of the Nusselt number

on the Prandtl number must usually be introduced

Boundary conditions When we apply the analogy between heat

trans-fer and mass transtrans-fer to calculate g m,i, we must consider the boundary

condition at the wall We have dealt with two common types of wall

con-dition in the study of heat transfer: uniform temperature and uniform

heat flux The analogous mass transfer wall conditions are uniform

con-centration and uniform mass flux We used the mass transfer analog of

the uniform wall temperature solution in the preceding example, since

the mass fraction of napthalene was uniform over the entire model Had

the mass flux been uniform at the wall, we would have used the analog

of a uniform heat flux solution

Natural convection in mass transfer In Chapter 8, we saw that the

density differences produced by temperature variations can lead to flow

and convection in a fluid Variations in fluid composition can also

pro-duce density variations that result in natural convection mass transfer

This type of natural convection flow is still governed by eqn (8.3),

u ∂u

∂x + v ∂u

∂y = (1 − ρ∞ /ρ)g + ν ∂2u

but the species equation is now used in place of the energy equation in

determining the variation of density Rather than solving eqn (8.3) and

the species equation for specific mass transfer problems, we again turn

to the analogy between heat and mass transfer

In analyzing natural convection heat transfer, we eliminated ρ from

eqn (8.3) using (1 − ρ∞ /ρ) = β(T − T∞ ), and the resulting Grashof and

Rayleigh numbers came out in terms of an appropriate β ∆T instead of

∆ρ/ρ These groups could just as well have been written for the heat

although∆ρ would still have had to have been evaluated from ∆T

With Gr and Pr expressed in terms of density differences instead oftemperature differences, the analogy between heat transfer and low-ratemass transfer may be used directly to adapt natural convection heattransfer predictions to natural convection mass transfer As before, wereplace Nu by Numand Pr by Sc But this time we also write

Helium is bled through a porous vertical wall, 40 cm high, into

sur-rounding air at a rate of 87.0 mg/m2·s Both the helium and the air

are at 300 K, and the environment is at 1 atm What is the averageconcentration of helium at the wall,mHe,s?

Solution. This is a uniform flux natural convection problem Here

gm,He and∆ρ depend on m He,s, so the calculation is not as forward as it was for thermally driven natural convection

straight-To begin, let us assume that the concentration of helium at the wall

will be small enough that the mass transfer rate is low Since m He,e =

0, if m He,s  1, then mHe,s − mHe,e  1 as well Both conditions for

the analogy to heat transfer will be met

The mass flux of helium at the wall, n He,s, is known, and becauselow rates prevail,

nHe,s ≈ jHe,s = gm,HemHe,s − mHe,e

§11.6 Mass transfer at low rates 647

Hence,

Num,L = gm,He L

ρ DHe,air =

nHe,s L

ρ DHe,airmHe,s − mHe,e

The appropriate Nusselt number is obtained from the mass

trans-fer analog of eqn (8.44b):

The Rayleigh number cannot easily be evaluated without assuming a

value of the mass fraction of helium at the wall As a first guess, we

pickm He,s = 0.010 Then the film composition is

m He,f = (0.010 + 0)/2 = 0.005

From eqn (11.8) and the ideal gas law, we obtain estimates for the

film density (at the film composition) and the wall density

ρ f = 1.141 kg/m3 and ρ s = 1.107 kg/m3

From eqn (11.42) the diffusion coefficient is

DHe,air = 7.119 × 10 −5m2/s.

At this low concentration of helium, we expect the film viscosity to

be close to that of pure air From AppendixA, for air at 300 K

We may now evaluate the mass transfer Nusselt number

= 0.01136

We have already noted that

mHe,s − mHe,e = mHe,s, so we have tained an average wall concentration 14% higher than our initial guess

ob-of 0.010

Using mHe,s = 0.01136 as our second guess, we repeat the

pre-ceding calculations with revised values of the densities to obtain

m He,s = 0.01142

Since this result is within 0.5% of our second guess, a third iteration

is not needed

In the preceding example, concentration variations alone gave rise

to buoyancy If both temperature and density vary in a natural

convec-tion problem, the appropriate Gr or Ra may be calculated using density

differences based on both the local m i and the local T , provided that

the Prandtl and Schmidt numbers are approximately equal (that is, if theLewis number 1) This is usually true in gases.

If the Lewis number is far from unity, the analogy between heat andmass transfer breaks down in those natural convection problems that in-volve both heat and mass transfer, because the concentration and ther-mal boundary layers may take on very different thicknesses, complicatingthe density distributions that drive the velocity field

In 1874, Josef Stefan presented his solution for evaporation from a liquidpool at the bottom of a vertical tube over which a gas flows (Fig.11.15)

This configuration, often called a Stefan tube, is has often been used to

§11.7 Steady mass transfer with counterdiffusion 649

Figure 11.15 The Stefan tube.

measure diffusion coefficients Vapor leaving the liquid surface diffuses

through the gas in the tube and is carried away by the gas flow across

top of the tube If the gas stream itself has a low concentration of the

vapor, then diffusion is driven by the higher concentration of vapor over

the liquid pool that arises from the vapor pressure of the liquid

A typical Stefan tube is 5 to 10 mm in diameter and 10 to 20 cm long

If the air flow at the top is not too vigorous, and if density variations

in the tube do not give rise to natural convection, then the transport of

vapor from the liquid pool to the top of the tube will be a one-dimensional

upflow

The other gas in the tube is stationary if it is not being absorbed by the

liquid (e.g., if it is insoluble in the liquid or if the liquid is saturated with

it) Yet, because there is a concentration gradient of vapor, there must

also be an opposing concentration gradient of gas and an associated

dif-fusional mass flux of gas, similar to what we found in Example11.8 For

the gas in the tube to have a net diffusion flux when it is stationary,

there must be an induced upward convective velocity — a

counterdiffu-sion velocity — against which the gas diffuses As in Example 11.8, the

counterdiffusion velocity can be found in terms of the diffusional mass

fluxes:

v = −jgas ρgas= jvapor ρgas

Figure 11.16 Mass flow across a

one-dimensional layer

In this section, we determine the mass transfer rate and tion profiles in the tube, treating it as the one-dimensional layer shown

concentra-in Fig 11.16 The s-surface lies above the liquid and the e-surface lies

at the top end of the tube We allow for the possibility that the terdiffusion velocity may not be negligible, so that both diffusion andvertical convection may occur We also allow for the possibility that the

coun-gas passes through the liquid surface (N 2,s ≠ 0) The results obtainedhere form an important prototype for our subsequent analyses of con-vective mass transfer at high rates

The solution of the mass transfer problem begins with an appropriateform of the equation of species conservation Since the mixture compo-sition varies along the length of the tube, the density may vary as well

If the temperature and pressure are constant, however, the molar centration of the mixture does not change through the tube [cf (11.14)].The system is therefore most accurately analyzed using the molar form

con-of species conservation

For one-dimensional steady mass transfer, the mole fluxes N1and N2

have only vertical components and depend only on the vertical

coordi-nate, y Using eqn (11.69), we get, with n i = MiNi,

These results are a straightforward consequence of steady-state speciesconservation

Recalling the general expression for N i, eqn (11.25), and introducing

If the Lewis number is far from unity, the analogy between heat andmass transfer breaks down in those natural convection problems that in-volve both heat and mass... saturated with

it) Yet, because there is a concentration gradient of vapor, there must

also be an opposing concentration gradient of gas and an associated

dif-fusional mass... ρgas= jvapor ρgas

Figure 11.16 Mass flow across a< /b>

one-dimensional