A HEAT TRANSFER TEXTBOOK - THIRD EDITION Episode 3 Part 8 pps

Hence, nH 2O,s h fgT wet-bulb = hTe − Twet-bulb For low-rate mass transfer, nH2O,s jH 2O,s, and this equation can be written in terms of the mass transfer coefficient flow rates of the psyc

664 An introduction to mass transfer §11.9

Figure 11.20 The wet bulb of a sling psychrometer.

perature, is directly related to the amount of water in the surrounding

air.12The highest ambient air temperatures we normally encounter are fairlylow, so the rate of mass transfer should be small We can test this sug-

gestion by computing an upper bound on B m,H2O, under conditions thatshould maximize the evaporation rate: using the highest likely air tem-perature and the lowest humidity Let us set those values, say, at 120◦F(49◦ C) and zero humidity (mH2O,e = 0).

We know that the vapor pressure on the wet bulb will be less than thesaturation pressure at 120◦F, since evaporation will keep the bulb at alower temperature:

xH2O,s  psat(120 ◦ F)/patm= (11, 671 Pa)/(101, 325 Pa) = 0.115

12The wet-bulb temperature for air–water systems is very nearly the adiabatic

satu-ration temperature of the air–water mixture — the temperature reached by a mixture

if it is brought to saturation with water by adding water vapor without adding heat It

is a thermodynamic property of an air–water combination.

§11.9 Simultaneous heat and mass transfer 665

so, with eqn (11.67),

This means that under the worst normal circumstances, the low-rate

the-ory should deviate by only 4 percent from the actual rate of evaporation

We may form an energy balance on the wick by considering the u, s,

and e surfaces shown in Fig.11.20 At the steady temperature, no heat is

conducted past the u-surface (into the wet bulb), but liquid water flows

through it to the surface of the wick where it evaporates An energy

balance on the region between the u and s surfaces gives

Since mass is conserved, nH 2O,s = nH 2O,u, and because the enthalpy

change results from vaporization, ˆhH2O,s − ˆ hH2O,u = hfg Hence,

nH 2O,s h fgT

wet-bulb = h(Te − Twet-bulb)

For low-rate mass transfer, nH2O,s jH 2O,s, and this equation can be

written in terms of the mass transfer coefficient

flow rates of the psychrometer, so it would appear that Twet-bulb should

depend on the device used to measure it The two coefficients are not

in-dependent, however, owing to the analogy between heat and mass

trans-fer For forced convection in cross flow, we saw in Chapter 7 that the

heat transfer coefficient had the general form

hD

k = C Re aPrb

666 An introduction to mass transfer §11.9

where C is a constant, and typical values of a and b are a 1/2 and

b 1/3 From the analogy,

gmD

ρ D12 = C Re aScbDividing the second expression into the first, we find

h gmcp

D12

α =

PrSc

b

Both α/ D12and Sc/Pr are equal to the Lewis number, Le Hence,

h

The Lewis number for air–water systems is about 0.847 Eqn (11.108)

shows that the ratio of h to g mdepends primarily on the physical erties of the mixture, rather than the geometry or flow rate

prop-This type of relationship between h and g m was first developed by

W K Lewis in 1922 for the case in which Le = 1 [11.27] (Lewis’s mary interest was in air–water systems, so the approximation was nottoo bad.) The more general form, eqn (11.108), is another Reynolds-Colburn type of analogy, similar to eqn (6.76) It was given by Chiltonand Colburn [11.28] in 1934

pri-Equation (11.107) may now be written as

wet-bulb temperature as a function of the dry-bulb temperature, T e, and

the humidity of the ambient air, mH2O,e The psychrometric charts found

in engineering handbooks and thermodynamics texts can be generated inthis way We ask the reader to make such calculations in Problem11.49.The wet-bulb temperature is a helpful concept in many phase-changeprocesses When a small body without internal heat sources evaporates

or sublimes, it cools to a steady “wet-bulb” temperature at which vective heating is balanced by latent heat removal The body will stay atthat temperature until the phase-change process is complete Thus, thewet-bulb temperature appears in the evaporation of water droplets, thesublimation of dry ice, the combustion of fuel sprays, and so on If thebody is massive, however, steady state may not be reached very quickly

con-§11.9 Simultaneous heat and mass transfer 667

Stagnant film model of heat transfer at high mass transfer rates

The multicomponent energy equation Each species in a mixture

car-ries its own enthalpy, ˆhi In a flow with mass transfer, different species

move with different velocities, so that enthalpy transport by

individ-ual species must enter the energy equation along with heat conduction

through the fluid mixture For steady, low-speed flow without internal

heat generation or chemical reactions, we may rewrite the energy balance,

eqn (6.36), as

−



S ( −k∇T ) · d
S −

where the second term accounts for the enthalpy transport by each species

in the mixture The usual procedure of applying Gauss’s theorem and

re-quiring the integrand to vanish identically gives

This equation shows that the total energy flux—the sum of heat

conduc-tion and enthalpy transport—is conserved in steady flow.13

The stagnant film model Let us restrict attention to the transport of a

single species, i, across a boundary layer We again use the stagnant film

model for the thermal boundary layer and consider the one-dimensional

flow of energy through it (see Fig.11.21) Equation (11.110) simplifies to

d dy

13 The multicomponent energy equation becomes substantially more complex when

kinetic energy, body forces, and thermal or pressure diffusion are taken into account.

The complexities are such that most published derivations of the multicomponent

energy equation are incorrect, as shown by Mills in 1998 [ 11.29 ] The main source

of error has been the assignment of an independent kinetic energy to the ordinary

diffusion velocity This leads to such inconsistencies as a mechanical work term in the

thermal energy equation.

668 An introduction to mass transfer §11.9

Figure 11.21 Energy transport in a stagnant film.

If we neglect pressure variations and assume a constant specific heatcapacity (as in Sect.6.3), the enthalpy may be written as ˆhi = cp,i(T −Tref),

and eqn (11.111) becomes

d dy

§11.9 Simultaneous heat and mass transfer 669

Substitution of eqn (11.114) into eqn (11.113) yields

h = n i,s c p,i

exp(n i,s c p,i /h ∗ ) − 1 (11.115)

To use this result, one first calculates the heat transfer coefficient as if

there were no mass transfer, using the methods of Chapters6through8

The value obtained is h ∗, which is then placed in eqn (11.115) to

de-termine h in the presence of mass transfer Note that h ∗ defines the

effective film thickness δ t through eqn (11.114)

Equation (11.115) shows the primary effects of mass transfer on h.

When n i,s is large and positive—the blowing case—h becomes smaller

than h ∗ Thus, blowing decreases the heat transfer coefficient, just as it

decreases the mass transfer coefficient Likewise, when n i,s is large and

negative—the suction case—h becomes very large relative to h ∗:

suc-tion increases the heat transfer coefficient just as it increases the mass

transfer coefficient

Condition for the low-rate approximation When the rate of mass

trans-fer is small, we may approximate h by h ∗ , just as we approximated g m

by g ∗ m at low mass transfer rates The approximation h = h ∗ may be

tested by considering the ratio n i,s cp,i/h ∗in eqn (11.115) For example,

if n i,scp,i/h ∗ = 0.2, then h/h ∗ = 0.90, and h = h ∗ within an error of

only 10 percent This is within the uncertainty to which h ∗ can be

pre-dicted in most flows In gases, if B m,i is small, n i,scp,i/h ∗will usually be

small as well

Property reference state In Section11.8, we calculated g ∗ m,i(and thus

gm,i ) at the film temperature and film composition, as though mass

transfer were occurring at the mean mixture composition and

tempera-ture We may evaluate h ∗ and g ∗ m,iin the same way when heat and mass

transfer occur simultaneously If composition variations are not large,

as in many low-rate problems, it may be adequate to use the freestream

composition and film temperature When large properties variations are

present, other schemes may be required [11.30]

670 An introduction to mass transfer §11.9

Figure 11.22 Transpiration cooling.

Energy balances in simultaneous heat and mass transfer

Transpiration cooling To calculate simultaneous heat and mass

trans-fer rates, one must generally look at the energy balance below the wall aswell as those at the surface and across the boundary layer Consider, for

example, the process of transpiration cooling, shown in Fig.11.22 Here awall exposed to high temperature gases is protected by injecting a coolergas into the flow through a porous section of the surface A portion ofthe heat transfer to the wall is taken up in raising the temperature of thetranspired gas Blowing serves to thicken the boundary layer and reduce

h, as well This process is frequently used to cool turbine blades and

combustion chamber walls

Let us construct an energy balance for a steady state in which the wall

has reached a temperature T s The enthalpy and heat fluxes are as shown

in Fig.11.22 We take the coolant reservoir to be far enough back from

the surface that temperature gradients at the r -surface are negligible and the conductive heat flux, q r , is zero An energy balance between the r - and u-surfaces gives

n i,r hˆi.r = ni,u hˆi,u − qu (11.116)

and between the u- and s-surfaces,

n i,u hˆi,u − qu = ni,s hˆi,s − qs (11.117)

§11.9 Simultaneous heat and mass transfer 671

Since there is no change in the enthalpy of the transpired species when

it passes out of the wall,

ˆ

and, because the process is steady, conservation of mass gives

n i,r = ni,u = ni,s (11.119)Thus, eqn (11.117) reduces to

The flux q u is the conductive heat flux into the wall, while q s is the

con-vective heat transfer from the gas stream,

Combining eqns (11.116) through (11.121), we find

ni,sˆ

hi,s − ˆ hi,r

= h(Te − Ts ) (11.122)This equation shows that, at steady state, the heat convection to the

wall is absorbed by the enthalpy rise of the transpired gas Writing the

enthalpy as ˆh i = cp,i (Ts − Tref), we obtain

ni,scp,i(Ts − Tr ) = h(Te − Ts ) (11.123)or

Ts = hTe + ni,scp,iTr

It is left as an exercise (Problem11.47) to show that

Ts = Tr + (Te − Tr ) exp( −ni,scp,i/h ∗ ) (11.125)

The wall temperature decreases exponentially to T r as the mass flux of

the transpired gas increases Transpiration cooling may be enhanced by

injecting a gas with a high specific heat

672 An introduction to mass transfer §11.9

Sweat Cooling A common variation on transpiration cooling is sweat

cooling, in which a liquid is bled through a porous wall The liquid is

vaporized by convective heat flow to the wall, and the latent heat ofvaporization acts as a sink Figure11.22 also represents this process.The balances, eqns (11.116) and (11.117), as well as mass conservation,eqn (11.119), still apply, but the enthalpies at the interface now differ bythe latent heat of vaporization:

ˆ

h i,u + hfg = ˆ h i,s (11.126)Thus, eqn (11.120) becomes

where c p,i f is the specific heat of liquid i Since the latent heat is generally

much larger than the sensible heat, a comparison of eqn (11.127) toeqn (11.123) exposes the greater efficiency per unit mass flow of sweatcooling relative to transpiration cooling

Thermal radiation When thermal radiation falls on the surface through

which mass is transferred, the additional heat flux must enter the energybalances For example, suppose that thermal radiation were present dur-

ing transpiration cooling Radiant heat flux, q rad,e, originating above the

e-surface would be absorbed below the u-surface.14 Thus, eqn (11.116)becomes

ni,r hi,rˆ = ni,u hi,uˆ − qu − αq rad,e (11.128)

where α is the radiation absorptance Equation (11.117) is unchanged.Similarly, thermal radiation emitted by the wall is taken to originate be-

low the u-surface, so eqn (11.128) is now

n i,r hˆi,r = ni,u hˆi,u − qu − αq rad,e + q rad,u (11.129)

or, in terms of radiosity and irradiation (see Section10.4)

n i,r hˆi,r = ni,u hˆi,u − qu − (H − B) (11.130)for an opaque surface

14Remember that the s- and u-surfaces are fictitious elements of the enthalpy

bal-ances at the phase interface The apparent space between them need be only a few

molecules thick Thermal radiation therefore passes through the u-surface and is

ab-sorbed below it.

Problems 673

Chemical Reactions The heat and mass transfer analyses in this

sec-tion and Secsec-tion 11.8 assume that the transferred species undergo no

homogeneous reactions If reactions do occur, the mass balances of

Sec-tion11.8are invalid, because the mass flux of a reacting species will vary

across the region of reaction Likewise, the energy balance of this section

will fail because it does not include the heat of reaction

For heterogeneous reactions, the complications are not so severe

Re-actions at the boundaries release the heat of reaction released between

the s- and u-surfaces, altering the boundary conditions The proper

sto-ichiometry of the mole fluxes to and from the surface must be taken into

account, and the heat transfer coefficient [eqn (11.115)] must be

modi-fied to account for the transfer of more than one species [11.30]

Problems

11.1 Derive: (a) eqns (11.8); (b) eqns (11.9)

11.2 A 1000 liter cylinder at 300 K contains a gaseous mixture

com-posed of 0.10 kmol of NH3, 0.04 kmol of CO2, and 0.06 kmol of

He (a) Find the mass fraction for each species and the pressure

in the cylinder (b) After the cylinder is heated to 600 K, what

are the new mole fractions, mass fractions, and molar

concen-trations? (c) The cylinder is now compressed isothermally to a

volume of 600 liters What are the molar concentrations, mass

fractions, and partial densities? (d) If 0.40 kg of gaseous N2

is injected into the cylinder while the temperature remains at

600 K, find the mole fractions, mass fractions, and molar

con-centrations [(a) mCO2 = 0.475; (c) cCO2 = 0.0667 kmol/m3;

(d) xCO2= 0.187.]

11.3 Planetary atmospheres show significant variations of

temper-ature and pressure in the vertical direction Observations

sug-gest that the atmosphere of Jupiter has the following

compo-sition at the tropopause level:

number density of H2 = 5.7 × 1021 (molecules/m3)

number density of He = 7.2 × 1020 (molecules/m3)

number density of CH4 = 6.5 × 1018 (molecules/m3)

number density of NH3 = 1.3 × 1018 (molecules/m3)

674 Chapter 11: An introduction to mass transfer

Find the mole fraction and partial density of each species at

this level if p = 0.1 atm and T = 113 K Estimate the

num-ber densities at the level where p = 10 atm and T = 400 K,

deeper within the Jovian troposphere (Deeper in the Jupiter’satmosphere, the pressure may exceed 105 atm.)

11.4 Using the definitions of the fluxes, velocities, and

concentra-tions, derive eqn (11.34) from eqn (11.27) for binary diffusion.

11.5 Show that D12= D21 in a binary mixture

11.6 Fill in the details involved in obtaining eqn (11.31) from eqn

(11.30)

11.7 Batteries commonly contain an aqueous solution of sulfuric

acid with lead plates as electrodes Current is generated bythe reaction of the electrolyte with the electrode material Atthe negative electrode, the reaction is

Pb(s) + SO2−

4  PbSO4(s) + 2e −

where the (s) denotes a solid phase component and the charge

of an electron is−1.609 × 10 −19coulombs If the current

den-sity at such an electrode is J = 5 milliamperes/cm2, what isthe mole flux of SO24− to the electrode? (1 amp =1 coulomb/s.)What is the mass flux of SO24−? At what mass rate is PbSO4produced? If the electrolyte is to remain electrically neutral,

at what rate does H+ flow toward the electrode? Hydrogendoes not react at the negative electrode [ ˙m PbSO

4 = 7.83 ×

10−5 kg/m2·s.]

11.8 The salt concentration in the ocean increases with increasing

depth, z A model for the concentration distribution in the

upper ocean is

S = 33.25 + 0.75 tanh(0.026z − 3.7)

where S is the salinity in grams of salt per kilogram of ocean water and z is the distance below the surface in meters (a) Plot the mass fraction of salt as a function of z (The region of rapid transition of msalt(z) is called the halocline.) (b) Ignoring the

effects of waves or currents, compute jsalt(z) Use a value of

Problems 675

D salt,water = 1.5 × 10 −5 cm2/s Indicate the position of

maxi-mum diffusion on your plot of the salt concentration (c) The

upper region of the ocean is well mixed by wind-driven waves

and turbulence, while the lower region and halocline tend to

be calmer Using jsalt(z) from part (b), make a simple estimate

of the amount of salt carried upward in one week in a 5 km2

horizontal area of the sea

11.9 In catalysis, one gaseous species reacts with another on a

pas-sive surface (the catalyst) to form a gaseous product For

ex-ample, butane reacts with hydrogen on the surface of a nickel

catalyst to form methane and propane This heterogeneous

reaction, referred to as hydrogenolysis, is

C4H10+ H2

Ni

→ C3H8+ CH4

The molar rate of consumption of C4H10 per unit area in the

reaction is ˙RC 4 H 10 = A(e −∆E/R ◦ T

)pC 4 H 10pH−2.42 , where A = 6.3 ×

1010 kmol/m2·s, ∆E = 1.9 × 108 J/kmol, and p is in atm.

(a) If pC4H10,s = pC3H8,s = 0.2 atm, pCH4,s = 0.17 atm, and

pH 2,s = 0.3 atm at a nickel surface with conditions of 440 ◦C

and 0.87 atm total pressure, what is the rate of consumption of

butane? (b) What are the mole fluxes of butane and hydrogen

to the surface? What are the mass fluxes of propane and ethane

away from the surface? (c) What is ˙m  ? What are v, v ∗, and

vC4H10? (d) What is the diffusional mole flux of butane? What

is the diffusional mass flux of propane? What is the flux of Ni?

[(b) nCH4,s = 0.0441 kg/m2·s; (d) jC3H8 = 0.121 kg/m2·s.]

11.10 Consider two chambers held at temperatures T1 and T2,

re-spectively, and joined by a small insulated tube The chambers

are filled with a binary gas mixture, with the tube open, and

allowed to come to steady state If the Soret effect is taken

into account, what is the concentration difference between the

two chambers? Assume that an effective mean value of the

thermal diffusion ratio is known

11.11 Compute D12 for oxygen gas diffusing through nitrogen gas

at p = 1 atm, using eqns (11.39) and (11.42), for T = 200 K,

500 K, and 1000 K Observe that eqn (11.39) shows large

de-viations from eqn (11.42), even for such simple and similar

molecules

666 An introduction to mass transfer< /i> §11.9

where C is a constant, and typical values of a and b are a. ..