12. Mass Transter - Heat and Ma= ss Transfer, 2nd Edition

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11. Experimental Hea= t Transfer

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Appendix A

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12 Mass Transfer

= 12.1 INTRODUCTION

The transfer of mass from a h= igher concentration region to a lower concentration region in a mixture is = known as mass transfer. The mass transfer is similar to heat transfer in ma= ny aspects. In heat transfer, heat is transferred toward the lower temperat= ure decreasing the temperature of the source and, in mass transfer, the mas= s is transferred toward the lower concentration matter decreasing the conce= ntration of the source. Both depend on the driving potential and the resist= ance that exist between the two sources. Mass transfer takes place from liq= uid to gas and solid, from solid to gas, and also from gas to gas. It depen= ds on the molecular spacing of the receiving mass and larger spacing gives = a higher diffusion rate. The mass transfer rate of gases is usually higher = than that of liquids and solids.

= 12.2 MODES OF MASS TRANSFER

Mass transfer occurs by conve= ction or diffusion. Mass transfer by convection is the transfer of mass bet= ween a moving fluid and a surface. It depends on the properties and charact= eristics of the fluid. If the mass transfer by diffusion (microscopic level= ) takes place due to temperature gradient in a system, then it is called as= thermal diffusion. Mass diffusion can also occur due to temperature gradie= nt within a system and that is known as thermal diffusion.

= 12.3 FICK'S LAWS OF DIFFUSION

Fick's law of diffusion descr= ibes diffusion and are used to solve for the coefficient of diffusion. The = amount of species per unit volume is known as concentration gradient of tha= t species. It states that the flow rate of the species is proportional to t= he concentration gradient and the area normal to the flow direction.

Mass transfer by diffusion,<= /p>

where m_b is the mass flux of species =E2=80=98b=E2=80=99 pe= r unit time in kg/s, D is the proportionality constant or diffusio= n coefficient in m²/s, A is the area th= rough which the mass is flowing in m², and C= _b is the mass concentration of a component = =E2=80=98b=E2=80=99 per unit volume in kg/m^{3<=
/sup>. The negative sign implies concentration of the mass decreases in the=
flow direction.}

Fouri= er law of heat conduction and Newton's law of viscosity are also in the for= m of Equation (12.2). Fourier equa= tion for conduction is

Newton's equation for shear = is

The perfect gas equation for= partial pressure is given as,

P =3D =CF=81R= T

where R =3D R_u / M in which R_u= is the universal gas constant (8314 J/kg mol.K) and M is the mole= cular weight of any gas

Mass flux is defined as the = rate of mass flow across a unit area. It can also be stated as the mass flu= x across a surface A is the total mass of particles which move acr= oss A per unit time:

We can also derive the follo= wing expression for the mass flux by integrating Equation (12.5):

= 12.4 EQUIMOLAR DIFFUSION

When a mole of one species is= replaced by another mole of another species and vice versa, it is known as= equimolar diffusion. Let B and C are two species, then t= he mole diffusion rates of =E2=80=98B=E2=80=99 and =E2=80=98C<= /em>=E2=80=99 are

Under= steady-state condition, the total pressure is

P =3D P_c+P_b

In case of equimolar steady-= state diffusion,

=E2=88=92N_=
b =3D N_c

(=E2=88=92ve sign shows the = flow of species in the opposite direction)

The diffusion coefficient ca= n be determined using the semi-empirical equation,

where D is in cm2/s, T is the absolute temperature in K, P is the total pressure in Pascal, V_c= and V_b are the molecular volume s= pecies C and B calculated from the atomic volumes, and M_c and M_b<= /em> are the molecular weights of C and B.

= 12.5 ISOTHERMAL EVAPORATION

Consider the evaporation of w= ater from a source maintained at a constant temperature and its diffusion t= hrough the stagnant air layer over it when the free surface of the water so= urce is exposed to air. The assumptions made are the following: (1) the wat= er source temperature and the total pressure remain constant; (2) the syste= m is at steady-state; (3) there is a continuous movement of air over the su= rface of water source to remove the diffused water vapor; (4) air movement = does not affect the concentration profile of air; (5) air and water vapor b= ehave as ideal gas.

The water evaporates diffuse= upward from the surface and this upward water vapor will be balanced by th= e downward diffusion of air. So that concentration at any =E2=80=98x=E2=80=99 will remain the same. At the surface of water, there is no mass= movement of air downward. Therefore, there must be a bulk mass movement up= ward with a velocity just enough to balance the diffusion of air downward.<= /p>
Diffusion of air downward is=

where = =E2=80=98A=E2=80=99 is the cross-sectional area of the tank.

This is balanced by the bulk= mass transfer of air upward:

where =E2=80=98V=E2=80=99= is the upward bulk mass velocity.

Using the above equations,

The upward mass diffusion of= water vapor is

Bulk water vapor transport i= s given by

The total transport of the m= ass of water vapor is

Substituting the value of = =E2=80=98V=E2=80=99 in Equat= ion (12.18), we get

Using the Dalton's law of pa= rtial pressure,

P_t =3D P_a +P_v=

This = relation is known as Stefan's law.

Integrating Equation (12.21), we get

= 12.6 MASS TRANSFER COEFFICIENT IN CONVECTION

Convective mass transfer is t= he transfer of mass between a surface and a moving fluid. It involves both = mass diffusion and bulk fluid motion.

The rate of mass transfer by= convection is

m_a =3D =C4=A7_DaA(C_{a1 - C_a2}) (= 12.24)

where m_a is the diffusive mass flux of species =E2=80=98A=E2=80=99, = =C4=A7_Da is the mass transfer coefficie= nt of =E2=80=98A=E2=80=99, and (C_a₁ and C_a₂) are the concentration through which diffusion occurs.<= /p>
If a steady-state diffusion = across a layer of thickness =E2=80=98=E2=88=86x=E2=80=99 is considered, the= n

For t= he process of water vaporization,

From Equation (12.22),

= 12.7 MASS TRANSFER THROUGH BOUNDARY LAYER

If the concentration profile = exists on the surface as shown in Fig= . 12.1, C_a > C_=E2=88=9E, then the component =E2=80=98A=E2=80=99 diffuse= s in to the fluid through concentration boundary layer.

F= igure 12.1 Concentration Boundary Layer

For a laminar boundary layer= :

Energy equation is given as<= /p>

Momentum transfer equation i= s given as

Diffusion on a flat plate in= a laminar boundary layer gives the mass equation as

where = C_a is the concentration of the species,= which diffuses through the boundary layer.

From Equations (12.30) and (12.31)

and the Schmit number plays a= n important role where both convection and mass transfer are important.

Combing Equations (12.31) and (12.32) we get,

The temperature and concentr= ation profiles will be similar when =CE=B1/ D=3D1.

The concentration and veloci= ty profiles will be similar when =CE=B3/D=3D1.

The governing equations for = heat, mass, and momentum transfer give an empirical conclusion for mass tra= nsfer coefficient which is similar to heat transfer coefficient. For liquid= s diffuse into air inside the columns in which the liquid wets the surface = and air is forced, we get

The ratio is called the Sherwood number.

The mass transfer coefficien= t in terms of a friction factor is

Compared with the heat trans= fer analogy, we get

For laminar flow over a smoo= th flat surface, the Reynolds analogy of mass transfer gives

For turbulent flow over a fl= at surface, the Reynolds analogy of mass transfer gives

Mass = and heat transfer coefficient may be expressed as

= 12.8 EVAPORATION PROCESSES IN THE ATMOSPHERE

The atmospheric air receives = water from water bodies of the earth by evaporation and by condensation it = gives back some water. These processes are complicated and in practice gove= rned by convection currents. The diffusion of water vapor from the surface = into atmospheric air can be estimated. With the partial pressure of vapor (= p_s), the vapor pressure drops with the = increase in elevation =E2=80=98z =E2=80=99to the free atmospheric pressure = of =E2=80=98p_=E2=88=9E, =E2=80=99the di= ffusion of water vapor is given by

The total pressure is

where =E2=80=98=CF=81=E2=80= =99 is the density and =E2=80=98M=E2=80=99is the molecular weight = of the moisture air.

Neglecting the molecular wei= ghts of water vapor, we get

By using boundary conditions= , p_w =3D p_s,= at z =3D 0 and p_w =3D p_=E2=88=9E, at z =3D z₁:

= 12.9 CONCLUSIONS

The amount of commodity per u= nit volume is known as concentration. It can be represented based on mass b= asis as kg/m³ and based on mole basis as Kmol/m<= sup class=3D"EBsup">3. Mass transfer is the movement of a chemical sp= ecies from a higher concentration region toward a lower concentration one (= due to concentration difference). Bulk fluid motion occurs on a macroscopic= , as a fluid is transported from one location to another due to temperature= difference.

Fick's law of diffusion stat= es that the rate of mass flow is directly proportional to the concentration= gradient and area normal to the flow direction. Mass transfer involves the= diffusion of a species through a plane =E2=80=93 pa= rallel medium that does not involve any chemical reaction. Similarly, heat = is transferred by conduction through a plane wall without heat generation.<= /p>
The growth of velocity and c= oncentration boundary layer is governed by Schmidt number. It states the ma= gnitude of thermal and concentration boundary layer. Mass transfer takes pl= ace by diffusion, convection, and phase change.

Mass transfer by convection = involves the transfer of mass between a moving fluid and a surface. It depe= nds on the transport properties and dynamic characteristics of the flowing = fluid. Mass transfer by phase change occurs whenever there is a change from= one phase to another. Velocity of a component relative to the mass average= velocity of the mixture is known as mass diffusion velocity. Velocity of a= component relative to the molar average velocity of the mixture is known a= s molar diffusion velocity.

Evaporation of water, dissol= ution of sugar in water, diffusion of smoke into atmosphere, and neutron di= ffusion within nuclear reactor are examples for mass transfer. The ratio of= mass of a component to the total mass of the mixture is known as mass frac= tion. The ratio of number of moles of a component to the total number of mo= les of the mixture is known as mole fraction. Equimolar diffusion is define= d as the isothermal diffusion process in which one mole of a component is r= eplaced by one mole of another component in a binary gas mixture and vice v= ersa.

Fick's diffusion coefficient= is a function of pressure, temperature, and nature of the system component= . But it is assumed as constant for ideal gases and dilute liquids. The rat= e of mass transfer can be represented similar to heat transfer.

Stefan's law of diffusion gi= ves the diffusion of an ideal gaseous component through a practically stagn= ant and ideal constituent of the binary system.

Molar concentration is defin= ed as the ratio between the number of moles species in a mixture to the vol= ume of the mixture. Mass fraction is defined as the ratio between the mass = of species to the total mass of the mixture.

Mass average velocity flow i= s the sum of products of specific gravity and velocity of species in a mixt= ure. It can be measured by pitot tube and hot wire anemometer.

EXAMPLE PROBLEMS

12.1 Calculate the= diffusion coefficient for CO₂ in air at a atmos= pheric pressure of 298.5 K and compare with table value.

V_COi= =3D34.0 M_COi=3D44

V_air=3D2= 9.9 M_air =3D 28.9

From the table, D=3D0.16= 4 cm²/s.

12.2 Calculate the= diffusion rate of water vapor from the bottom of a test tube 5 mm in radiu= s and 20 cm long into dry atmospheric air at 298 K.

The partial pressure at = the bottom of a tube surface (P_w1)=3Dsaturated p= ressure at 298 K

=3D 0.03169 bar

Water vapour pressure at= the top of the tube (P_W2) is zero beca= use it is dry air.

P_A₁=3D P_t =E2=80=93P_w₁=3D 9.815x10⁴ Pa

P_A2<= /em>=3D pt -P_W2=3D 1.0132x10⁵ Pa

From the table, D=3D= 0.256 cm²/s=3D0.256 x 10⁴ m²/s

12.3 Dry air at 30= =C2=B0C and one atmospheric pressure flows over a wet flat plate of 0.5 m l= ong and at a velocity of 60 m/s. Determine the mass transfer coefficient of= water vapor in air at the end of the plate.

D for water vap= or in air =3D 0.26 x 10^-4 m^=
2/s

Properties of air at 30= =C2=B0C are =CF=81=3D1.16 kg/m³; =CE=BC=3D18.46 = x 10^-6 Ns/m²; Pr<= /em> =3D 0.71

Reynolds number

Schmit number

12.4 A well is 50 = m height and 5 m radius and the atmospheric temperature is 25=C2=B0C. The a= ir is having a relative humidity of 0.5 at the top of the well. Find the ra= te of diffusion of water vapor from the well. Assume D =3D 2.58 x 10^-5m²/s and P_t =3D 1 bar.

Partial pressure of wate= r vapor (P_w1) at 25=C2=B0C =3D 0.03169 bar

At the top of well, RH = =3D 0.5

P_w2 =3D RH =C3=97 Saturated pressure at 25=C2=B0C.

Mass of water vapor diff= used

12.5 The water in a 5 m =C3=97 20 m outdoor swimming pool is maintaine= d at a temperature of 30=C2=B0C. The wind speed is 2 m/s in the direction o= f the long side of the pool. The average ambient temperature and relative h= umidity are 40=C2=B0C and 0.5, respectively. Find the mass transfe= r coefficient for the evaporation of water from the pool surface and rate o= f evaporation of water per day.

For air at 30=C2=B0C: = =CE=B3 =3D 15.89 =C3=97 10^-6 m²/s

Density of saturated wat= er vapor at 30=C2=B0C

For water vapor-air: D =3D 0.26 =C3=97 10 ⁴ m²/s

Thus, a mixed boundary l= ayer condition exists.

Partial pressure of wate= r vapor at 25=C2=B0C at water surface =3D 3531 N/m²

Saturated pressure of wa= ter at 33=C2=B0C =3D 6221 N/m²

Therefore, partial press= ure of water vapor at 33=C2=B0C and 40% RH is

p_{=E2=88=9E =3D 0.4 =C3=97 6221 =3D 2488.4 N/m²}

Total pressure of air = =3D p_w + p_{air =3D 101.325 =C3=97 10³ N/m²}

Therefore, =CF=81AS at the water surface

R= ate of evaporation

Hence, total heat lost f= rom the swimming pool

=3D 3.081 =C3=97 5 =C3=97 15

=3D 231.1 kg/day

REVIEW QUESTIONS

What is meant by =E2=80=98concent= ration=E2=80=99?

Write the difference between mass tran= sfer and the bulk flow of fluid.

State the Fick's law of diffusion.
Write the analogous between one-dimens= ional heat and mass transfer.

Define Schmidt number and Lewis number= .

Define Sherwood number.

What are the dimensionless parameters = used in mass transfer?

What are the modes of mass transfer?
What is meant by mass transfer by conv= ection and phase change?

What are the mass diffusion velocity a= nd molar diffusion velocity?

Give examples for mass transfer.

Define mass fraction and mole fraction= .

What is meant by =E2=80=98equimolar di= ffusion=E2=80=99?

Write the significance of Fick's diffu= sion.

What is meant by mass transfer coeffic= ient?

Write the Stefan's law of diffusion.
Define molar concentration and mass fr= action.

What is meant by mass average velo= city flow?

OBJECTIVE QUESTIONS

The one-dimensionless parameter n= ot used in mass transfer is

Schmidt number

Lewis number

Sherwood number

Biot number

The modes of mass transfer are

by diffusion

by convection

by phase change

all

Which statement is not connected with = mass transfer?

Boiling of water

Dissolution of sugar in water

Diffusion of smoke into atmosphere
Neutron diffusion within nuclear r= eactor

Which are the statements correct with = respect to the significance of Fick's diffusion?

It is based on experimental princ= iple not from first principle.

It is valid for all matter irrespectiv= e of the state.

Mass transfer due to temperature gradi= ent, pressure gradient is negligibly small.

All

The mass average velocity flow is the = sum of products of specific gravity and___________of species in a mixture.

mass

velocity

mixture

density

Answer:

d

d

a

d

b

EXERCISE PROBLEMS

A mixture of O_{2<=
/sub>, N₂, and H₂ wit=
h partial pressures of 0.21 bar, 0.305 bar, and 0.5 bar, respectively, is i=
nside a container at a temperature of 300 K. Estimate the molar concentrati=
on, the mass density, the mole fraction, and mass fraction of each species =
for a total pressure of 1 bar and molecular weight of the mixture.}

(Ans: 0.00= 794, 0.0119, 0.198 k mol/m³; 0.254, 0.333, 0.039= 6, kg/m³; 0.2, 0.3, 0, 0.405, 0.531, 0.0632)
=

In a system, air and H₂ experience equ= imolar counter diffusion in a circular pipe of 59 mm diameter and 1.499 m l= ength. The ends of the tube are connected to the reservoirs in which the co= ncentrations of the species are kept at contact values. The system is maint= ained at a total pres- sure of 1 bar and a temperature of 298 K. The partia= l pressure of H₂ at one side is 0.2 bar and othe= r side is 0.1 bar. Find the mass transfer rate of air and H₂ through the pipe.

(Ans: 2.25= x 10^-5 kg/s; -32.26 x 10^-5= kg/s)

A pipe line carries water at 1 bar pressure and 299 K. A vent tub= e at the top of the pipe maintains the pressure inside the pipe. The diamet= er of the vent pipe is 49.9 mm and the height is 1 m. Calculate the rate of= water diffusion into dry air at 299 K and 1 bar.

(Ans: 1.14= x 10^-4 kg/day)

A tank of size 599 cm x 15.995 m with water is kept at a temperat= ure of 302 K in the atmos- phere. The atmospheric temperature is 302 K and = relative humidity is 49%. The wind speed in the direction of the long side = of the tank is 7.19 km/hr. Find the mass transfer coefficient of evaporatio= n from the surface.

(Ans: 9.6 = x 10^-3 m/s)

A plate of 39.9 cm long in wet condition is placed in dry air str= eam. Air at 302 K, 1 bar moves at 60 m/s over it. Find the mass transfer co= efficient of water vapor at the end of the plate in air stream.

(Ans: 0.14= 3 m/s)

A stream of air at 1 bar and 299 K with small amount of iodine is= flowing with a velocity of 5.99 m/s in a pipe. Find the mass transfer coef= fi- cient of iodine. Assume the diameter of the pipe as 4 cm.

(Ans: 0.01= 3 m/s)

A vessel of 6 m diameter has 2 mm deep layer of benzene at th= e bottom. Vapour pressure of benzene in the vessel is 0.139 bar. The film t= hickness is 3 mm and the system is under 1 bar pressure and 292 K temperatu= re. Estimate the time required to evaporate the entire benzene. Assume that= there is a stagnant air film above the benzene through which the diffusion= takes place.

(Ans: 39.9= units)

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=20 =20 =20 =20 Previous Chapter
11. Experimental Hea= t Transfer
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Appendix A
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=20

Table= of Contents for =20 Heat and Mass Transfer, 2nd Edition =20

12

Mass Transfer

= 12.1 INTRODUCTION

= 12.2 MODES OF MASS TRANSFER

= 12.3 FICK'S LAWS OF DIFFUSION

= 12.4 EQUIMOLAR DIFFUSION

= 12.5 ISOTHERMAL EVAPORATION

= 12.6 MASS TRANSFER COEFFICIENT IN CONVECTION

= 12.7 MASS TRANSFER THROUGH BOUNDARY LAYER

= 12.8 EVAPORATION PROCESSES IN THE ATMOSPHERE

= 12.9 CONCLUSIONS

EXAMPLE PROBLEMS

REVIEW QUESTIONS

OBJECTIVE QUESTIONS

EXERCISE PROBLEMS

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Table= of Contents for
=20 Heat and Mass Transfer, 2nd Edition =20