UNIT II

2.1 EDDY DIFFUSION.

Turbulent motion of a fluid is characterized by rapid and highly irregular fluctuation of the

velocity at any point in the fluid. Experimental evidences indicate that in a turbulent medium, tiny fluid

elements move about randomly and are responsible for a high rate of transport of momentum, heat or

mass. These fluid elements, not necessarily of the same size, are called eddies. It is visualized that eddies

are continually formed and they break up by interaction among themselves or even may disappear in the

process. Eddies are, therefore, short-lived.

Mass or heat transfer in a turbulent medium is a result of the mixing process caused by the

movement of the eddies. In Figure for transport of a soluble substance from a pipe wall to a flowing fluid.

For the sake of simplicity it may be visualized that the effect of turbulence does not reach the wall.

Transport of momentum, mass or heat occurs by diffusion through a thin layer, called the laminar

sublayer, at the wall. Beyond this layer, transport occurs predominantly by eddies. However, the motion

of eddies and the phenomenon of eddy transport are not well-understood despite the extensive theoretical

and experimental research done on these phenomena. To simplify the analysis of the phenomenon, the

physical laws of transport (of heat, mass and momentum) are very often extended to the case of eddy

transport by replacing the molecular transport coefficient by an eddy transport coefficient. For example,

in the case of mass transfer from the tube wall we can write

where ED is the eddy diffusivity of mass. DAB is replaced by ED. The total flux is, therefore, obtained by

adding the fluxes due to molecular diffusion and eddy diffusion

The eddy diffusivity ED is not as simple a quantity as the molecular diffusivity DAB. It depends upon the

intensity of the local turbulence. The intensity of turbulence decreases as one approaches the wall.

Therefore, the contribution of eddy diffusion towards the rate of mass transfer is substantially less than

that by molecular diffusion very near the wall. The situation is just the opposite in the bulk of the

turbulent medium.

2.2 CONCEPT OF MASS TRANSFER COEFFICIENTS

The mass transfer coefficient is defined on the following _phenomenological basis .

Rate of mass transfer ∞ Concentration driving force (i.e. the difference in concentration)

Rate of mass transfer ∞ Area of contact between the phases

If WA is the rate of mass transfer (kmol/s) of the solute A, ΔCA is the concentration driving force

between two points, and a is the area of mass transfer,

WA ∞ a ΔCA => WA ∞ kc a ΔCA

where kc, the proportionality constant, is a phenomenological coefficientt called the `mass transfer

coefficient . If NA is the molar flux (expressed as kmol/m2 ' s, say), we may write

WA = a NA = kCa ΔCA => NA = kCΔCA

WA =NA

ΔCA=

molar flux

concentration driving force

For the purpose of comparison, we may recall teat transfer coefficient

h =heat flux

temperature driving force, Δ T

The inverse of mass transfer coefficient is a measure of the mass transfer resistance . If the driving force

is expressed as the difference in concentration (kmol/m3), the unit of mass transfer coefficient is m/s (or

cm/s, ft/s, etc. which is the same as the unit of velocity).

If the mass transfer coefficient is expressed as the ratio of the local flux and the local driving force, it is

called the local mass transfer coefficient . When it is expressed as the ratio of the average flux (over a

surface) and the average driving force, it is known as the average mass transfer coefficient .

2.2.1 TYPES OF MASS TRANSFER COEFFICIENTS

Different types of mass transfer coefficients have been defined depending upon: (i) whether mass transfer

occurs in the gas phase or in the liquid phase, (ii) the choice of the driving force, and (iii) whether it is a

case of diffusion of A through non-diffusing B or a case of counter diffusion.

Convective heat transfer is often visualized to occur through a "stagnant film" adhering to the surface.

The transport of heat through the film is assumed to occur purely by conduction. In the study of mass

transfer too, this concept is frequently used. If the transport of mass occurs through a stagnant film of

thickness d, we may write down the following expressions for the mass transfer flux as the product of a

mass transfer coefficient and the appropriate driving force.

(A) Diffusion of A Through Non-diffusing B

Mass transfer in the gas phase: NA = kG(pAl - pA2) = ky(yAl - yA2) = kc(CA1 - CA2)

Mass transfer in the liquid phase: NA = kx(x411 - xA2) - kL(CA1 - CA2)

Here kG, ky, and kc are the gas-phase mass transfer coefficients, and kx and kL are the liquid-phase mass

transfer coefficients; the subscripts 1 and 2 refer to two positions in a medium. The units of the mass

transfer coefficients can be obtained from the definitions above. For example, a unit of ky is

(kmol)/(m2)(s)(Δy), where Δy stands for me driving force in mole fraction unit. If the gas phase is ideal,

the concentration term in Eq. (3.3) is given by CA = pA/RT, where pA is the partial pressure of A.

Supposing that the distance between the two locations 1 and 2 is d (the film thickness), the expressions

for the mass transfer coefficients can be obtained by comparing above equation with NA(Gas) &

NA(Liquid).

Gas phase : 𝑘𝐺 =𝐷𝐴𝐵𝑃

𝑅𝑇𝛿𝑝𝐵𝑀 𝑘𝑦 =

𝐷𝐴𝐵𝑃2

𝑅𝑇𝛿𝑝𝐵𝑀 𝑘𝑐 =

𝐷𝐴𝐵𝑃

𝛿𝑝𝐵𝑀

Liquid phase : 𝑘𝑥 =𝐷𝐴𝐵(𝜌 𝑀)𝑎𝑣⁄

𝛿𝑥𝐵𝑀 𝑘𝐿 =

𝐷𝐴𝐵

𝛿𝑥𝐵𝑀

The relations among the three types of gas-phase mass transfer coefficients (i.e. kG, ky, and kc) can be easily

obtained. Similarly, the relation between the two types of liquid phase mass transfer coefficients, kx and kL, can be

obtained

kc = RTkG; ky = PkG; kx= (𝜌/M)av kL

(B) Equimolar Counterdiffusion of A and B

The set of notations for mass transfer coefficients are used here with a prime (') to differentiate them from

the case of diffusion of A through non-diffusing B.

Gas phase: NA = k'G (pA1 - pA2) = k'y (yA1 - yA2) = k'c(CA1 - CA2)

Liquid phase: NA = (xA1 - xA2) = (CA1 - CA2)

Comparing above gas Equations for gas-phase transport and liquid-phase transport, we can have the

following expressions for the mass transfer

coefficients in this case:

Gas phase:

Liquid phase:

If the concentration of A is expressed in the mole ratio unit (moles A per mole of A-free medium),

Conversion: NA = kY(YA1 - YA2), for the gas phase

and NA = kX(XA1 - XA2), for the liquid phase

Here YA and XA are the concentrations of A in the gas or in the liquid phase in mole ratio unit [note that similar

expressions can be written using the mass ratio (mass A/mass B) unit as well]. Note that YA = yA/(1 - yA) and

XA=xA/(I - xA). The types of mass transfer coefficients defined above and their interrelations are given in Table..

The fundamental difference between the types of mass transfer coefficients defined in Eqs. and the types

of coefficients defined has to be carefully noted. The former class of coefficients (kG, ky, kc, kx, and kL) are

inherently associated with the log mean concentration of the other species(B) which is non-diffusing

Accordingly, this type of mass transfer coefficient has a dependence on concentration because of the term

pBM or xBM (this dependence can however be ignored at low concentrations of A). On the contrary, the

coefficients k’G, k’y, k’c, k’x, k’L do not have dependence on concentration. The second type of

coefficient, k’c, is called `Colburn-Drew mass transfer coefficient.

2.3 THEORIES OF MASS TRANSFER

(A) The Film Theory {also called 'Film Model')

Let us describe this theory (Whitman, 1923) through an illustration. We consider mass transfer from a

solid surface to a flowing liquid. Even though the bulk liquid is in turbulent motion, the flow near the wall

may be considered to be laminar. The concentration of the dissolved solid (A) will decrease from CAi at

the solid-liquid interface to CAb at the bulk of the liquid. In reality the concentration profile will be very

steep near the solid surface where the effect of turbulence is practically absent. Molecular diffusion is

responsible for mass transfer near the wall while convection dominates a little away from it. The film

theory, however, visualizes a simpler picture. It is based on the following assumptions.

(a) Mass transfer occurs by purely molecular diffusion through a stagnant fluid layer at the phase

boundary (here the wall is the phase boundary). Beyond this film, the fluid is well- mixed having

a concentration which is the same as that of the bulk fluid (i.e. CAb).

(b) Mass transfer through the film occurs at steady state.

(c) The bulk flow term [i.e. (NA + NB)CA/C] in the expression for the Fick’s law is small. So the

flux can be written as NA = -DAB(dCAldz). This is valid when: (i) the flux is low and (ii) the mass

transfer occurs at low concentrations [or the mass transfer occurs by equimolar counter diffusion].

For many practical situations this assumption is satisfactory.

Wall or phase boundary of interface

CAi = cone, at the interface CAh = cone,

at the bulk

The underlying concept of the film theory of mass transfer is analogous to that of heat transfer (the heat

transfer coefficient is frequently called the ‘film coefficient’). Figure shows the stagnant film. We

consider an elementary volume of thickness Δz and of unit area normal to the z-direction (i.e. the

direction of mass transfer). We make a steady state mass balance over this element located at position z.

Rate of input+ of the solute at z = NA|z

Rate of output of the solute at z + Δz = NA|z+Az

Rate of accumulation = 0 (at steady state)

NA|z - NA|z+Az = 0

Dividing by Az throughout and taking the limit Δz —> 0, we get

-dNA/dz = 0

Putting NA= - DAB dCA/dz in the preceding equation, we have

Integrating Eq and using the following boundary conditions (i) and

(ii),

(i) z = 0 (i.e. the wall or the phase boundary or the interface), CA = CAi

(ii) Z = 𝛿 (i.e. the other end of the film of thickness, 𝛿). CA = CAb,

CA = CAi – (CAi – CAb) 𝑧

𝛿

The above equation indicates that the theoretical concentration profile, according to the film theory, is

linear as shown in Figure2 (where the ‘true’ concentration profile is also shown). The mass transfer flux

through the film is constant at steady state and is given as

(b) Concentration profiles of the solute near the phase

boundary (5 indicates the edge of the film)

Thus the mass transfer coefficient can be calculated from the film theory if the diffusivity and the film

thickness are known. Whereas the former can possibly be obtained from the literature or may be

estimated by using a suitable correlation, the latter (i.e. the film thickness 𝛿) is unknown. So this theory

does not help us in reality to predict the mass transfer coefficient. However, the film theory, like two

other theories described below, has been extremely useful in the analysis of mass transfer accompanied by

a chemical reaction.

The film thickness, however, can be attributed a physical significance. It is the thickness of the stagnant

layer of fluid that offers a mass transfer resistance equal to the actual resistance to mass transfer offered

by the fluid in motion. The film theory predicts linear dependence of kL upon the diffusivity DAB.

Unfortunately, experimental data for diverse systems show that the coefficient of mass transfer to a

turbulent fluid varies as (DAB)n where n may have any value from zero to about 0.8.

The film theory is simple indeed and that is why it has got wide acceptance. Nevertheless, it does not

visualize a realistic physical picture regarding the mechanism of mass transfer at the phase boundary

(B) The Penetration Theory

Most of the industrial processes of mass transfer is unsteady state process. In such cases, the contact time

between phases is too short to achieve a stationary state. This non stationary phenomenon is not generally

taken into account by the film model. In the absorption of gases from bubbles or absorption by wetted-

wall columns, the mass transfer surface is formed instantaneously and transient diffusion of the material

takes place. Figure 3.4 demonstrates the schematic of penetration model.

Basic assumptions of the penetration theory are as follows:

1) Unsteady state mass transfer occurs to a liquid element so long it is in contact with the bubbles or other

phase

2) Equilibrium exists at gas-liquid interface

3) Each of liquid elements stays in contact with the gas for same period of time

Under these circumstances, the convective terms in the diffusion can be neglected and the unsteady state

mass transfer of gas (penetration) to the liquid element can be written as:

The appropriate initial and boundary conditions are:

Initial condition : t = 0, z≥0 CA = CAb

Boundary condition 1 : t>0 z =0 CA = CAi

Boundary condition 2 : t>0 z= ∞ CA = CAb

The initial condition implies that in a fresh liquid element coming from the bulk, the concen-tration is

uniform and is equal to the bulk concentration. The boundary condition 1 assumes that interfacial

equilibrium exists at all time. The last condition means that if the contact time of an element with the gas

is small, the depth of penetration of the solute in the element should also be small and effectively the

element can be considered to be of infinite thickness in a relative sense. Equation subject to the initial and

boundary conditions, can be solved for the transient concentration distribution of the solute in the element

by introducing a similarity variable η, as in the case of unsteady state heat conduction (see Dutta, 2001).

The solution is

The mass flux to the element at any time t can be derived from the equation

The flux decreases with time because of a gradual build-up of the solute concentration within the element

and the resulting decrease in the driving force. At a large time, the element becomes nearly saturated and

the flux becomes vanishingly small.

The average mass flux over the contact time tc is given by

Instantaneous mass transfer coefficient, 𝑘𝐿 = √𝐷𝐴𝐵

𝜋𝑡

Average mass transfer coefficient, 𝑘𝐿,𝑎𝑣 = 2 √𝐷𝐴𝐵

𝜋𝑡𝑐

The above equations show that the mass transfer coefficient is proportional to the square root of

diffusivity. Although this is gain not in conformity with experimental observations in general, this is

definitely an improvement over the film theory for a more realistic visualization. Here the contact time tc

is the model parameter like the film thickness d in the film theory.

(C) The Surface Renewal Theory

For the mass transfer in liquid phase, Danckwert (1951) modified the Higbie’s penetration theory. He

stated that a portion of the mass transfer surface is replaced with a new surface by the motion of eddies

near the surface and proposed the following assumptions:

1) The liquid elements at the interface are being randomly swapped by fresh elements from bulk

2) At any moment, each of the liquid elements at the surface has the same probability of being substituted

by fresh element

3) Unsteady state mass transfer takes place to an element during its stay at the interface.

𝑘𝐿 = √𝐷𝐴𝐵𝑆

where s is fraction of the surface renewed in unit time, i.e., the rate of surface renewal [s-1].

(D) The Boundary Layer Theory

Boundary layer theory takes into account the hydrodynamics/flow field that characterizes a system and

gives a realistic picture of the way mass transfer at a phase boundary. A schematic of concentration

boundary layer is shown in Figure

Theoretical analysis of mass transfer in a laminar boundary can be done following the same approach as

adopted in the case of heat transfer, and the following equation for the local Sherwood number, Shx, can

be developed.

Here x is the distance of a point from the leading edge of the plate, kL,x is the local mass transfer coefficient, and Rex is the local Reynolds number. If l is the length of the plate, the average Sherwood number, Shav, can be obtained from the above equation as follows:

2.4 TWO PHASE MASS TRANSFER

Consider two phases are immiscible with each other than an inter phase is seen between the phases.

Consider a solute A which is in bulk gas phase G and diffusing into the liquid phase L. there should be a

concentration gradient within each phase to cause direction through the resistance.

YAG – concentration of A in bulk gas phase

YAi – concentration of inter phase

XAL – concentration of A in bulk liquid phase

XAi – concentration of A in inter phase

YAG, XAL are not in equilibrium

This enables diffusion to occur

At inter phase there is no resistance to transfer of solute and concentration YAi & XAi are in equilibrium

and they are related by equilibrium relation.

YAi = f(XAi)

We consider state mass transfer the rate at which molecules reach the inter phase will be the same rate at

which the molecules are transferred to the liquid. Since the inert phase has no resistaance the flux for each

phase can be expressed in terms of mass transfer coefficient.

NA= ky(yAG-yAi) = (kX(xAi-xAL)

kX, kY are local liquid and gas mass transfer coefficient.

Therefore 𝑦𝐴𝐺−𝑦𝐴𝑖

𝑥𝐴𝐿−𝑥𝐴𝑖 = - (

𝑘𝑥

𝑘𝑦)

2.5 RELATION FOR LOCAL AND OVERALL MTC

Determine the rate of mass transfer is very difficult since it is

not possible to evaluate the inter phase composition. The bulk concentration are easily measured and

measuring XAL is good as measuring YA* because both have same chemical potential. Similarly YAG is as

good as measuring XA*.

Flux for gases

NA = KY(YAG – YA*)

NA = kY(yAG – yAi)

KY is overall MTC

YAG – YA* = (YAG – YAi) + (YAi – YA*)

= (YAG – YAi) +m’ (XAi – XAL)

m’ – slope of chord SN

𝑁𝐴

𝐾𝑌=

𝑁𝐴

𝑘𝑦+

𝑚′𝑁𝐴

𝑘𝑥

1

𝐾𝑌=

1

𝑘𝑦+

𝑚′

𝑘𝑥

Flux for liquids

NA = KX(XA*– XAL)

NA = kx(xAi – xAL)

KX is overall MTC

XA* - XAL = (XA* - XAi) + (XAi - XAL)

= (YAG – YAi)

𝑚" + (XAi – XAL)

m” – slope of chord NT

𝑁𝐴

𝐾𝑋=

𝑁𝐴

𝑚"𝑘𝑦+

𝑁𝐴

𝑘𝑥

1

𝐾𝑋=

1

𝑚"𝑘𝑦+

1

𝑘𝑥

Since KY is overall MTC its inverse 1

𝐾𝑌 can be considered as overall MT resistance in gas phase. It

is the sum of individual MT resistance of two phases. 1

𝑘𝑦 - individual gas phase MT resistance

𝑚′

𝑘𝑥 - individual liquid phase MT resistance

Since KX is overall MTC its inverse 1

𝐾𝑋 can be considered as overall MT resistance in liquid phase.

It is the sum of individual MT resistance of two phases. 1

𝑚"𝑘𝑦 - individual gas phase MT resistance

1

𝑘𝑥 - individual liquid phase MT resistance

The fractional resistance offered by gas phase

= resistance offered by the gas−phase

total resistance of two phases=

1/ky

1/KY

The fractional resistance offered by liquid phase

= resistance offered by the liquid−phase

total resistance of two phases=

m′/kx

1/KY

UNIT III PART A

1. What is humid heat?

Humid heat is the heat required to raise the temperature of unit mass of gas and its accompanying vapor one degree at constant pressure.

2. What is natural draft? It ensures more positive air movement even in calm weather by depending upon the displacement of warm air inside the tower by the cooler outside the air.

3. Define Humidity.(Dec 2009)

It is the mass of vapor carried by a unit mass of vapor free gas. It is denoted by H or y

4. What is Saturated gas?

It is a gas in which vapor is equilibrium with the liquid at gas temperature.

5. Define Relative humidity. (MAY 2015)

It is defined as the ratio of the partial pressure of vapor to the vapor pressure of the liquid at gas temperature.

6. What is Percentage humidity?

It is the ratio of actual humidity to the saturation humidity.

7. What is Humidification?

In the processing of materials the operation of increasing the amount of vapor present in a gas stream is known as humidification.

8. What is dry bulb temperature?

This is the temperature of a vapor free gas mixture as ordinarily determined by immersion of thermometer in a mixture.

9. What is wet bulb temperature?(Dec 2012) (May 2016)

It is the steady state temperature reached by a small amount of liquid evaporating in to a large amount of unsaturated vapor gas mixture.

10. What is bound water?(MAY14) (MAY 2015) (May 2016)

If the equilibrium moisture content of a given material is continued to its intersection with 100% humidity line the moisture is called bound water

11. What is Unbound water?(MAY14) (MAY 2015) (May 2016)

If a material contains more water than indicated by intersection with 100%humidity time. It exerts only a vapor pressure as high as ordinary water at same temperature.

12. What are the types of cooling towers?

Natural draft towers , Induced draft towers and forced draft towers

13. What are Induced draft towers?

In this case the fan is placed at the top avoids the dis advantages of previous and permits more uniform internal distribution of air.

14. What are Tray towers?

These are very effective Not commonly used in humidific , dehumidific or gas cooling operation for reasons of cost relatively high pressure drop.

15. What are Spray chambers?

These are horizontal spray towers.T hese are used for adiabatic humidification cooling operations with recirculating liquid.

16. What are the methods of determination of humidity?(MAY14)

Chemical methods, Dew point, WBT, Piezo electric hygrometer.

17. What are the methods of increasing humidity?

Live steam addition

Water spray

The gas mixed with a gas of higher humidity.

18. What is humid volume? (MAY 2015)

Humid volume is defined as volume occupied by unit mass of dry gas and its associated vapor .

19. What are the characteristics of good refrigerants?

It should be non toxic and non flammable,Cost must be less,Must have low viscosities.

20. What is Air conditioning?

It is the simultaneous control of temperature humidity motion and purity of the atmosphere in confined space.

UNIT III PART B: 1. (a) Describe the methods available for estimating humidity of a sample of air. (b) What are the different types of cooling towers used in industries? Briefly explain them. 2. The temperature of air in a room is 40.2OC/50 OC and the total pressure is 101.3 kPa abs. The air contains water ( NOV14) vapor with a partial pressure PA = 3.74 kPa. Calculate: (i) the humidity

(ii) the saturation humidity and percentage humidity (iii) the percentage relative humidity. 3. Explain the theory of adiabatic saturation curves and wet bulb temperature theory (Dec12) (May 2016) 4. Water is to be cooled in a cooling tower counter currently from 55OC to 30OC, using air counter- currently. The entering air is at 20OC and an absolute humidity of 0.007 kg water / kg dry air. The value of Kya is 0.3 kg/m3.sec and the liquid flow rate is 1500 kg/hr.m2. Estimate the minimum air requirement and the height of the cooling tower, if 1.3 times the minimum air flow rate is used

Temperature, to OC 20 30 40 50 55 Enthalpy, H' kJ/kg 60 102 166 279 355 5. a) Explain the principle and theory of a cooling tower. b) With a neat sketch, explain the operation of a counter-current induced draft tower. 6. Explain how the height of a cooling tower can be determined. 7. A mixture of oxygen-acetone vapor at a total pressure of 1050 mm. Hg. at 25OC has a percentage humidity of 70%. The saturation vapor pressure of acetone at 25OC is 290 mm. Hg. The specific heats of oxygen and acetone vapor are 0.25 and 0.352 Kcal/Kg OC respectively. Calculate its molal humidity,

absolute humidity, relative humidity, molal humid volume and molal humid heat.

8. Fresh air at 21.2OC, in which the partial pressure of water vapor is 0.0118 atm. is blown at the rate of 214 m3/hr., first through a preheater and then adiabatically saturated in a spray chamber to 100% saturation and then again reheated. The reheated air has a humidity of 0.024 kg water vapor/kg. dryair. It is assumed that the fresh air and air leaving the preheater have the same % humidity. Determine:

a) The temp. of the preheater, spray chamber and reheater. b) Heat requirement for preheating and reheating.

9.a) Write short notes on absolute humidity, relative humidity and dew point. b) Discuss Interphase mass transfer(MAY14) (MAY 2015)

C) Write a note on induced and forced draft cooling tower (NOV 14)

10. a) Discuss briefly about spray chamber with a neat sketch.(MAY14)(Dec 12)

b) Write a note on two industrial cooling towers with a neat sketch (MAY14) (MAY 2015)

C) write briefly about adiabatic saturator (MAY14)

D) Discuss Air conditioning process (NOV14)

Humidification

Humidification is the generic name of operations where an insoluble gas is

brought into contact with a liquid. This gas-liquid contact results in heat and mass

transport of vaporized liquid between the two phases.

L���� G, it involves

vaporization of liquid. The operation is called Humidification (as it increases humidity

of the gas).

If the transport of vapor(A) is from gas (B) to liquid, ie., G���� L, it involves

condensation of vapor. The operation is called Dehumidification (as it decreases

humidity of the gas).

:

: It is the vapor content of one mole of dry gas.

Y = At

A

B

A

B

A

pp

p

p

p

y

y

−== = moles of A / moles of B =

gasmolesofdry

ormolesofvap …..(1)

A – Vapor- Water vapor

B – dry gas- Air

It is the vapor content

of one unit mass of dry gas. It is generally called absolute humidity.

Y’ = B

A

M

MY = -

B

A

At

A

M

M

pp

p

− = Mass A / Mass B ……(2)

At saturation, partial pressure of the G-V mixture, AA pp = , vapor pressure

of the the liquid,A at the prevailing temperature.

Y Y’

----- x 100 or ----- x 100 …(3) YS & & Y‘S are

YS Y’s molal & mass abs. humidities

at saturation computed at the

dry bulb temperature of the mixture.

3.1 Introduction:

If the transport of vapor(A) is from liquid to gas (B), ie

Absolute Humidity

(A)Molal absolute humidity

(B)Mass absolute humidity (Grosvenor humidity) :

(C)Saturated vapor$Gas mixtures:

(D)Percentage Saturation or Percentage Absolute Humidity:

---

pA ----

----- x 100 …(4) pA = partial pressure of A

pA pA = vapor pressure of a at the

given temperature

pt = total pressure

constant total pressure out of contact with a liquid.

H : vH of a vapor-gas mixture is the volume of unit mass of dry gas and its accompanying

vapor at the prevailing temperature and pressure.

t

G

ABt

G

AB

Hp

t

M

Y

Mp

xt

M

Y

Mv

273'18315

10013.1

273

27341.22

'1 5 +

+=

+

+= ….. (5)

where

MA = Molecular weight of A MB= Molecular weight of B

C

It is the heat required to raise the temperature of unit mass of dry gas and its

For a given mass of V-G mixture of abs. humidity,Y’,

CS = CB + Y’CA …(6) CA & CB = specific heat of A & B

Q = WB CS ∆t = the heat in Btu …. (7) WB = Mass of dry gas

Sum of the enthalpies of the dry gas and the vapor content.

H’ = CB(tG - t0 ) + Y’ [ CA ( tG - tDP) + λDP + CA,L (tDP- t0) ] …(8)

H’ = CB(tG - t0 ) + Y’ [ CA ( tG - t 0) + λ0 ] …(9)

=CS(t-t0 ) + Y’ λ0 …(10)

tG = dry bulb temperature

tDP = dew point temperature

tw = wet bulb temperature

t0 = reference temperature

Hint: Eqn (9) is obtained by replacing tDP by t0 in eqn.(8).

S

(E)Relative Saturation or Percentage Relative humidity:

(F)Dew Point :This is the temperature at which a vaporgas mixture becomes saturated when cooled at

(G)Humid volume , v

(H)Humid heat, C

accompanying vapor one degree at constant pressure.

(I)Enthalpy,H’:

When a vapor-gas mixture has to be humidified, it can be done by bringing it in contact

with the liquid. The liquid is sprayed into the entering vapor-gas mixture at humidity,

Y’1 and dry bulb temperature, (DBT ) tG. The gas mixture gets humidified as a result of

the vaporization of the liquid sprayed. The latent heat for this humidification is supplied

by the sensible heat of gas mixture and hence its temperature (originally at tG ) drops.

The mass balance for the substance,A:

L’ = G’S (Y’2 – Y’1) ….(1)

The enthalpy balance:

G’S H’1 + L’HL = G’S H’2 ….(2)

Where L’= mass liquid/(time)(area)

G’S = mass dry gas /(time)(area)

H’ = Enthalpy of gas at the given condition

H’L = Enthalpy of liquid at the given condition

1, 2 refer to the inlet and outlet conditions respectively.

Eqn. (2) can be rewritten as

H’1 + (Y’2 –Y’1)HL =H’2 …(3)

If the liquid enters the process at tas and the leaving gas mixture is also at tas with

humidity, Y’as ( saturated condition), eqn. (3) becomes

H’1 + (Y’as –Y’1)HL =H’as (4)

H’1 = sensible heat of (air + vapor) at point 1+ latent heat of vaporization of Y’1

= ( ) ( ) 0

'

10

'

10 λYttCYttC GAGB +−+− ….(5)

(Y’as –Y’1)HL = Sensible heat of liquid

= (Y’as –Y’1) CA,L (tas – t 0)

H’2 = sensible heat of )air + vapor) at point 2 + latent heat of vaporization of Y’as

= ( ) ( ) 000 '' λasasAasasB YttCYttC +−+− …(6)

Substituting eqns (4),(5) &(6) in eqn.(3), and simplifying,

CS1 (tG - tas) = (Y’as –Y’1) λas

i.e., Sensible heat drop in Gas-vapor mixture = latent heat of vaporization of (Y’as –Y’1)

amount of moisture at tas

(tG –tas) = [ (Y’as –Y’1) λas] / CS1 …(7)

Adiabatic saturation curve:

CS1 ( tG1 – tas ) = (Y’as – Y’1)λas

Or

(Y’as – Y’1)λas

( tG1 – tas ) = ------------------

CS1

Detailed derivation:

Enthalpy in = G’S H’1 + L’HL

Enthalpy in per Kg dry air = H’1 + (Y’2 –Y’1)HL = H’1 + (Y’as –Y’1)HL

= ( ) ( ) 0

'

10

'

10 λYttCYttC GAGB +−+− + (Y’as –Y’1) CA,L (tas – t 0)

3.2 Adiabatic Saturation :

H’1 = ( ) ( ) 0

'

10

'

10 λYttCYttC GAGB +−+−

In the above Picture, t1 = tG

t2 = tas

t4 = t0

Line A’E represents CA, specific heat of vapor A.

CA( tG- t0) = Enthalpy at A’ – Enthalpy at E = H1 - HE

λ 0 = λ at t0 = Enthalpy at E – Enthalpy at D = HE - H4

CA( tG- t0) + λ 0 = H1 - HE + HE - H4 = H1 - H4

But (H1 - H4 ) is also equal to (H1 – H2) + (H2 -H3) + (H3 – H4)

= (H1 – HB) + (HB -HC) + (HC – H4)

=[CA( tG – t as)]+ λas +[CA,L(tas – t0)]

Hence

CA( tG- t0) + λ 0 = [CA( tG – t as)]+ λas +[CA,L(tas – t0)]

Now

H’1 = CB( tG- t0) + Y’1 [CA( tG – t as)+ λas + CA,L(tas – t0)] …(5’)

Enthalpy in per Kg dry air = H’1 + (Y’as –Y’1)HL

= CB( tG- t0) + Y’1 [CA( tG – t as) + λas + CA,L(tas – t0)] +(Y’as –Y’1) CA,L (tas – t 0)

= CB( tG- t0) + Y’1 [CA( tG – t as) + λas] + Y’as CA,L(tas – t 0) …. (5.1)

Enthalpy out per Kg dry air = H’2

= ( ) ( ) 000 '' λasasAasasB YttCYttC +−+− …(6)

But CA(tas – t0 ) + λ0 = CA,L (tas – t0 ) + λas ( See the picture)

(H2 - HE ) + (HE – HD) = (HC – HD) + (HB – HC)

H2 – H4 = ( H3 –H4) + (H2 – H3)

Hence

Enthalpy out per Kg dry air = H’2

= CB (tas –t0) + Y’as [CA,L (tas – t0 ) + λas ] …….(6.1)

Enthalpy in = Enthalpy out

(5.1) = (6.1)

CB( tG- t0) + Y’1 [CA( tG – t as) + λas] + Y’as CA,L(tas – t 0)

= CB (tas –t0) + Y’as [CA,L (tas – t0 ) + λas ] ….(7.1)

Cancelling common terms and bringing CB term to LHS,

CB( tG- t0) - CB (tas –t0) + Y’1 CA( tG – t as) + Y’1 λas = Y’as λas

CB( tG- tas) + Y’1 CA( tG – t as) = (Y’as - Y’1) λas

CS ( tG- tas) = (Y’as - Y’1) λas

( tG- tas) = (Y’as - Y’1) λas / CS

thus by replacing CA and λ 0 terms with CA,L and λ as terms, we get

(Y’as – Y’1)λas

( tG1 – tas ) = ------------------

CS1

It is the steady state temperature reached by a small amount of liquid evaporating into a

large amount of unsaturated V-G mixture.

This is measured by a thermometer covered with a wick that is wetted by the liquid A.

When this thermometer is immersed in a rapidly moving stream of vapor-gas mixture

with a humidity Y’ and dry bulb temperature, t G, the liquid vaporizes if its vapor

pressure (pA at t G ) is greater than the partial pressure of vapor A in the gas mixture.The

liquid temperature fluctuates as there is a loss in sensible heat on account of vaporization.

Finally it reaches a steady state temperature, tW which is greater than the dew point of the

vapor-gas mixture. This (tW) is the wet bulb temperature of the vapor-gas mixture.

Wet-Bulb Thermometry: Consider a drop of liquid immersed in a rapidly moving stream of vapor-gas mixture

with a humidity Y’ and dry bulb temperature, t G. If the liquid temperature,t L higher than

the dew point of vapor, t D , the vapor pressure of the liquid at the drop surface will be

higher than the partial pressure of vapor in the gas and hence the liquid vaporizes and

3.3 Wet$Bulb Temperature:

diffuses into the gas. Thus mass transfer occurs due to the difference in pressures of

liquid & gas phases (pA- Ap ). But the humidity of the gas phase is not significantly

changed as the flow rate of gas is high. Hence, Y’ remains constant.

The latent heat of vaporization required for this mass transferred is supplied by the

sensible heat of the liquid drop due to which the temperature of the liquid drop reduces.

As tL is reduced below tG , the dry bulb temperature of the gas, heat will flow from gas to

liquid at an increasing rate due to the large temperature difference. Ultimately, the rate of

heat transfer from the gas will be equal to the rate of heat required for the evaporation

and tL reaches a steady value, tW which is the wet bulb temperature of the vapor-gas

mixture.

Flux of mass transfer = kY (Y’W –Y’)

Heat flux for mass transfer = flux required for vaporization of (Y’W –Y’)

=kY(Y’W –Y’)λW

Heat flux for heat transfer (from gas side) = hG ( tG – tW )

At equilibrium, both heat flxes are equal.

Hence, kY(Y’W –Y’)λW = hG ( tG – tW )

i.e.,( )

YG

WWWG

kh

YYtt

/

'' λ−=−

( tG – tW )is called the wet bulb depression.

tG - tw = wet-bulb depression

hG / k y = psychrometric ratio

Detailed derivation:

Flux of liquid evaporated = NA = ( )AGAwG ppk − moles/(time)(area)

Mass flux of liquid evaporated =NAMA = MA ( )AGAwG ppk −

( )B

A

Bm

AGAw

M

M

p

pp − =(Y’W – Y’)

MA ( )AGAwG ppk − = MAkG Bmp (MB /MA) (Y’W – Y’)

= kG Bmp MB(Y’W – Y’)

But kG Bmp MB = kY

Hence

MA ( )AGAwG ppk − = kY (Y’W – Y’) = mass flux evaporated

Heat flux required for latent heat of vaporization = mass flux (latent heat of vapn)

= λW kY (Y’W – Y’)

At equilibrium,

Flux of sensible heat supplied by the gas = Flux of latent heat required by the liquid

hG (tG – tW) = λW kY (Y’W – Y’)

(tG – tW) = λW (Y’W – Y’) / (hG /kY)

(tG – tW) is called the wet bulb depression

( )

Y

G

WW

WG

kh

YYtt

'' −=−λ

hG/kY is called the psychrometric ratio.

It is the ratio of heat transfer coefficient to mass transfer coefficient.

Estimation of hG / k y :

Henry-Epstein equation:

For flow of gases past cylinders and single spheres,

hG /(kyCS) = (Sc /Pr)0.567

= Le0.567

For dilute mixtures, where CB = CS, and with Pr for air taken as 0.707,

hG /ky = 1223 Sc0.567

For Air-Water vapor systems,

hG /ky = 950 N.m/Kg.K

Lewis relation:

For air-water vapor system, hG /ky = CS

Or apprxly. hG /(kyCS) = 1 = (Sc /Pr)0.567

= Le0.567

i.e., Le=1

or Sc/Pr = 1 which means ν /DAB = ν /α = 1

i.e., ν = α = DAB ; The molecular diffusivities of momentum, heat & mass transfers are

equal. This results in

tw = tas

Le is Lewis No. which is equal to 1 for air-water vapor system.

As wet bulb temperature = adiabatic saturation temperature (apprxly) for air-water vapor

system, the adiabatic saturation curves of psychrometric chart can be used to find wet

bulb temperature of the given mixture.

3.4 PSYCHROMETRIC CHART

Plotting the given air-water vapor system: Here the data are given in FPS system. Given:

Air-Water Vapor system with tG = 800 F and Y’ = 0.004 lb water vapor/ lb dry air Plot this at point A on

the chart.

1. Finding Dew Point:

Draw a horizontal line from A towards left to intersect the 100 % humidity line at B. The temperature corresponding to B is the dew point of the given mixture.

Dew point, t DP : 330 F.

2. Finding Saturation Humidity,Y’S:

Draw a vertical line from point A upwards to intersect the 100 % humidity line at C. The humidity

corresponding to C is the Saturation humidity of the given mixture.

Saturation Humidity,Y’S : 0.0218 lb water vapor /lb dry air.

3. Finding Wet Bulb Temperature, tW:

From A draw a line parallel to the adjacent constant wet bulb temperature line (adiabatic saturation

line) to intersect the 100 % humidity line at D.

The temperature corresponding to D is the wet bulb temperature of the given mixture.

4. Wet bulb temperature, tW =56 0 F.

Similarly, for a mixture at 90 0 F and Y’ = 0.004, tW = 60 0 F.

5. Percentage saturation:

The line representing 20% saturation is passing adjacent to point A. The Percentage saturation of the given mixture is approximately 20% or say, 19.7%.

3.5 COOLING TOWERS

Cooling towers are a very important part of many chemical plants. The primary task of a cooling

tower is to reject heat into the atmosphere. They represent a relatively inexpensive and

dependable means of removing low-grade heat from cooling water. The make-up water source is

used to replenish water lost to evaporation. Hot water from heat exchangers is sent to the cooling

tower. The water exits the cooling tower and is sent back to the exchangers or to other units for

further cooling.

3.5.1 Components of Cooling Tower

The basic components of an evaporative tower are: Frame and casing, fill, cold water basin, drift

eliminators, air inlet, louvers, nozzles and fans.

Frame and casing: Most towers have structural frames that support the exterior enclosures

(casings), motors, fans, and other components. With some smaller designs, such as some glass

fiber units, the casing may essentially be the frame.

Fill: Most towers employ fills (made of plastic or wood) to facilitate heat transfer by maximizing

water and air contact. Fill can either be splash or film type.

With splash fill, water falls over successive layers of horizontal splash bars, continuously

breaking into smaller droplets, while also wetting the fill surface. Plastic splash fill promotes

better heat transfer than the wood splash fill.

Film fill consists of thin, closely spaced plastic surfaces over which the water spreads, forming a

thin film in contact with the air. These surfaces may be flat, corrugated, honeycombed, or other

patterns. The film type of fill is the more efficient and provides same heat transfer in a smaller

volume than the splash fill.

Cold water basin: The cold water basin, located at or near the bottom of the tower, receives the

cooled water that flows down through the tower and fill. The basin usually has a sump or low

point for the cold water discharge connection. In many tower designs, the cold water basin is

beneath the entire fill.

Drift eliminators: These capture water droplets entrapped in the air stream that otherwise would

be lost to the atmosphere.

Air inlet: This is the point of entry for the air entering a tower. The inlet may take up an entire

side of a tower–cross flow design– or be located low on the side or the bottom of counter flow

designs.

Louvers: Generally, cross-flow towers have inlet louvers. The purpose of louvers is to equalize

air flow into the fill and retain the water within the tower. Many counter flow tower designs do

not require louvers.

Nozzles: These provide the water sprays to wet the fill. Uniform water distribution at the top of

the fill is essential to achieve proper wetting of the entire fill surface. Nozzles can either be fixed

in place and have either round or square spray patterns or can be part of a rotating assembly as

found in some circular cross-section towers.

Fans: Both axial (propeller type) and centrifugal fans are used in towers. Generally, propeller

fans are used in induced draft towers and both propeller and centrifugal fans are found in forced

draft towers. Depending upon their size, propeller fans can either be fixed or variable pitch. A

fan having non-automatic adjustable pitch blades permits the same fan to be used over a wide

range of kW with the fan adjusted to deliver the desired air flow at the lowest power

consumption.

3.5.2 Classification of cooling towers

Natural draft towers

Natural draft towers use very large concrete chimneys to introduce air through the media. Due to

the large size of these towers, they are generally used for water flow rates above 45,000 m3/hr.

These types of towers are used only by utility power stations.

Mechanical draft towers

Mechanical draft towers utilize large fans to force or suck air through circulated water. The water

falls downward over fill surfaces, which help increase the contact time between the water and the

air - this helps maximize heat transfer between the two. Cooling rates of Mechanical draft towers

depend upon their fan diameter and speed of operation.

Mechanical draft towers are available in the following airflow arrangements:

1. Counter flows induced draft.

2. Counter flow forced draft.

3. Cross flow induced draft.

In the counter flow induced draft design, hot water enters at the top, while the air is introduced at

the bottom and exits at the top. Both forced and induced draft fans are used. In cross flow

induced draft towers, the water enters at the top and passes over the fill. The air, however, is

introduced at the side either on one side (single-flow tower) or opposite sides (double-flow

tower). An induced draft fan draws the air across the wetted fill and expels it through the top of

the structure.

Mechanical draft towers are available in a large range of capacities. Normal capacities range

from approximately 10 tons, 2.5 m3/hr flow to several thousand tons and m3/hr. Towers can be

either factory built or field erected - for example concrete towers are only field erected.

Counter flow Natural draft tower

Counter flow induced draft tower

Cross flow induced draft tower

Forced draft cooling tower

UNIT IV PART A

1.What is meant by equilibrium moisture? (Dec 2012)

This is the moisture content of a substance when at equilibrium with a given partial pressure of the

vapor.

2.What is bound water?

This refers to the moisture contained by a substance which exerts an equilibrium vapor pressure less than

that of the pure liquid at the same temperature.

3.What is drying?

Drying refers to the removal of moisture from a substance.

4.What are the types of moisture? Equilibrium moisture , Bound moisture , Unbound moisture and free moisture.

5.What is meant by moisture content?

The moisture content of a solid or solution is described in terms of weight % moisture and unless

otherwise qualified this is ordinarily understood to be expressed on wet basis

6.What are hygroscopic substance?

Substance containing bound water are called hygroscopic substance

7.What are fiber saturation point? The condition at which wood textiles and other cellular materials is in equilibrium with saturated air is

called fiber saturation point.

8.What are the types of drying?

Batch drying ,Continuous drying ,freeze drying ,Cross circulation drying , Through circulation drying.

9.What is free moisture? (Dec 2012) (Dec 2010)

It is the moisture contained by a substance in excess of the equilibrium moisture only free moisture can be evaporated. free moisture content of the solid depends upon the vapor concentration in the gas.

10.What are special drying methods?

IR radiation ,Dielectric heating and vaporization from ice.

11.What are the disadvantages of batch drying?

High labour cost

Expensive operation and temperature of the interior falls at the time of loading and unloading.

12.What are the advantages of batch drying? Low maintenance cost

13.What are the advantages of tunnel driers?

Convenience of continuous operation It is used on brick , ceramic , and other materials must be dried .

14.What are the advantages of Rotoloure?

It does not lift the material and drop it down through the shell but merely allows it to roll along the bottom.There is less tendency for size degradation and fragile materials.

Hot air passes through the material the air comes more nearly in equilibrium with the material .

15.what are funicular state?

It is that condition in drying a porous body when capillary suction results in air being sucked in to the pores.

16.What is pendular state?

It is that state of a liquid in a porous solid when a continuous film of liquid no longer exists around and between discrete particles so that flow by capillary cannot occurs

This state succeeds the funicular state.

17.What is shrinkage? An important factor in controlling the drying rate is the shrinkage of the solid as the moisture content is

lowered, rigid, porous or non porous solids do not shrink appreciably during drying.But colloidal and

fibrous materials undergo severe shrinkage as the moisture is removed from them.

18.What are the effects of shrinkage? (May 2016) It alters the surface of the material per unit weight.

This is true for vegetables and food stuffs.

19. What is meant by case hardening?(Nov 2012) (May 2016)

There may be developed hardened layer on the surface. It occurs in clay and soap. Change in gross structure. It occurs in clay and soap. This slows down the drying.

20.What are freeze drying ?

Substance which cannot be heated even to moderate temp such as food stuffs .Moderate temperatures such as food stuffs and pharmaceuticals can be dried by this method.

PART B

UNIT IV: 1. a) Explain the mechanism of dying during constant rate and falling rate period. (MAY14)

b) A porous solid is to be batch dried under constant drying conditions. A trial shows that it requires 6

hours to reduce the moisture content from 30 to 10%. The critical moisture content is 16% and the equilibrium moisture content is 2.5%. Assuming that the rate of drying during the falling rate period is

proportional to the free moisture content. How long will it take to dry a sample of the same solid from

35% to 6% under the same drying conditions? All moisture contents are on a wet basis.

2. With neat diagrams, explain the working of:

a) Spray dryer(MAY14) b) Rotary dryer( Dec 12) (May 2016)

c) Mechanically agitated dryers

3. (i) Write short notes on:

a) Critical Moisture content. Bound and unbound moisture. (MAY 2015)

(ii) Discuss any one industrial dryer for continuous operation. Give a neat sketch.

4. Explain how time of drying can be calculated. 5. 1400 kg. of bone dry granular solid is to be dried under constant drying conditions. From a moisture

content of 0.2 kg water/kg. dry solid to a final value of 0.02 kg. water/kg. dry solid. The material has an

effective area of 0.0616 ,2/kg. dry solid. Calculate the time of drying

Flux (kg/hr.m2) 1.71 1.71 1.71 1.46 1.29 0.88 0.54 0.376 Moisture content 0.3 0.2 0.14 0.096 0.056 0.046 0.026 0.016

(dry basis)

6. Find an expression for the determination of total time of drying of a wet solid material under constant drying conditions to a final moisture content well below the critical moisture content.

7. In a textile mill, wet cloth passes through a hot air drier. The cloth enters with 90% moisture regain and

leaves at 6% moisture regain at a speed of 1.15 m/sec. The width of the cloth is 120 cm and its density on bone-dry basis is 0.095 kg/m2. The temperature of the cloth leaving the drier is 368 K. The ambient

air enters the dryer at 303 K DB and 298 K WB while the hot air leaves the dryer at 393 K DB and 328 K

WB. Calculate:

(i) the bone-dry production of the dryer (ii) the evaporation taking place in the dryer and

(iii) the air circulation rate.

8. A time of 5 hrs. was taken to dry a material from an initial moisture of 30% to a final moisture of 7%. Critical and equilibrium moisture are found to be 15% and 2% respectively. How much further time

would be required to dry the material to a final moisture of 4%. All moisture contents are on wet basis.

9. Explain about tray dryer and fluidized bed dryer with neat sketches. 10. Discuss the temp. patterns in batch and continuous countercurrent adiabatic dryers.

Drying

Drying refers to the removal of relatively small amounts of moisture from a substance

which is generally a solid. Eg. Wet cloth dried by evaporation of its moisture into a

stream of air. In some specific cases, it includes the removal of moisture from liquids

and gases as well. Eg. Benzene liquid dried of its moisture content that was present in

traces.

Difference between drying and other similar operations:

Evaporation : Here, moisture is boiled off without the need of a gas stream carrying

it off. In drying, moisture of a wet substance is converted into vapor and taken off

by a gas stream.

Humidification: Here, moisture is transported from a liquid to gas stream in the form

of vapor. In drying, moisture is removed from a wet solid into a gas stream.

Adsorption : Moisture from a fluid is adsorbed on a solid (fluid � solid). Drying is a

desorption operation where moisture is leaving a solid and enters a gas phase

(Solid� Gas).

Squeezing: Here moisture is removed in large quantity from a wet solid by mechanical

operation. Drying removes only traces of moisture by diffusion.

In drying, as moisture from the wet material is converted into vapor, it involves latent

heat of vaporization. Hence, drying is a simultaneous heat and mass transfer operation.

Equilibrium:

The moisture contained in a wet solid or liquid solutions exerts a vapor pressure, pA to an

extent depending upon the nature of the moisture, the nature of the solid, and the

temperature. If the wet solid is exposed to a continuous supply of fresh air, containing a

partial pressure of the vapor p ,

the solid will lose moisture by evaporation (if pA > p ) …. drying

or gain moisture from the gas (if pA < p ) ….. adsorption

until the vapor pressure of the moisture of the solid equals partial pressure of vapor in air,

(pA = p ).

When pA = p , the solid and the gas are in equilibrium, and the moisture content of the

solid is termed its equilibrium- moisture content at the prevailing conditions.

Definitions of Different Types of Moisture Contents:

(i) Moisture Content- Wet Basis: x:

This is defined as the weight of moisture per unit weight of wet substance.

(ii) Moisture Content- Dry Basis: X: This is defined as the weight of moisture per unit weight of bone dry substance.

(iii) Equilibrium- moisture, X*:

This is the moisture content of a substance when at equilibrium with a given partial

pressure of the vapor. It is the limiting moisture content to which a material can be dried

under specific conditions of gas temperature and humidity.

(iv) Free Moisture: X – X*:

This is the moisture contained by a substance in excess of equilibrium moisture. Only

free moisture can be removed with air of given temperature and humidity. It may include

both bound and unbound moisture.

(v) Bound Moisture:

This refers to the moisture contained by a substance which exerts an equilibrium

vapor pressure less than that of the pure liquid (p0) at the same temperature. Liquid may

become bound by retention in small capillaries, by solution in cell or fiber walls or by

adsorption on solid surface.

(vi) Unbound Moisture:

This refers to the moisture contained by a substance which exerts an equilibrium vapor

pressure equal to that of the pure liquid (p0) at the same temperature.

Relationship between wet and dry basis: x and X:

Formula: x = Kg moisture / (Kg moisture + Kg dry solid)

=X / (1+X)

Or

X = x / ( 1-x)

Problem to understand the moisture contents on wet & dry basis:

A wet solid is to be dried from 80 to 5% moisture, wet basis. Compute the moisture to be

evaporated per 1000 kg of dried product.

Solution:

Initial moisture content, wet basis= x = 80 % = 80 /100 =0.8

Initial moisture content, dry basis = X = x /(1-x)

= 0.8 / (1-0.8) = 4.0 kg water / kg dry solid

Final moisture content, wet basis = 5% = 5/100 = 0.05

Final moisture content, dry basis = 0.05 /( 1- 0.05 ) = 0.0527 kg water / kg dry solid

Dry solid in product = 1000 ( 1- 0.05) = 1000 x 0.95 = 950 Kg

Moisture to be removed= 950 ( 4- 0.0527) = 3750 kg.

Other Definitions:

Funicular State:

It is a condition that occurs in drying a porous body when capillary suction results in air

being sucked into pores. It generally indicates first falling rate period.

Pendular State:

As drying proceeds beyond funicular state, water is being progressively removed from

the solid, the fraction of pore volume that is occupied by air increases. When the fraction

reaches a certain limit, there is insufficient water left to maintain a continuous film across

the pores. The interfacial tension in the capillaries breaks and air becomes the continuous

phase filling the pores. The residual water is relegated to small isolated pores and

interstices of the pores. This state is called the pendular state and it generally refers to the

second falling rate drying period.

Drying Operations:

Classification:

(i) Batch drying (ii) Continuous drying.

Batch drying is actually a semi batch operation as the solid to be dried remains

stationary and the gas phase is in continuous movement. It is an unsteady state operation.

Continuous drying is a steady state operation where both the solid to be dried and the

drying gas are both in continuous motion. It can either be a cocurrent or countercurrent

operation.

Rate of Batch Drying:

The time required to dry a substance from one moisture content to another under

specified conditions should be known to (i) set up drying schedule (ii)determine size of

the equipment. There is also need to estimate the influence of drying conditions on drying

time. To accomplish this, experimental measurements are required.

Measurements of the rate of batch drying are relatively simple ; these data can be used

not only for batch but also for continuous drying operations.

Drying Tests:

The rate of drying can be determined for a sample of substance by suspending it in a duct

or cabinet in a stream of air, from a balance. The weight of the drying sample can be

measured as a function of time.

Precautions to be followed:

Sample should not be too small.

The following conditions should resemble as closely as possible those which prevail in

the contemplated large scale operation:

(i) The sample should be similarly supported on a tray or frame.

(ii) It should have the same ratio of drying to non drying surface.

(iii) It should be subjected to similar conditions of radiant heat transfer.

(iv) The air should have the same temperature, humidity and velocity (both speed and

direction w.r.t. the sample).

If possible, several tests can be carried for different thicknesses of the sample.

The dry weight of the sample should also be obtained.

Constant drying conditions:

The exposure of the sample to air of constant temperature, humidity and velocity

constitutes drying under constant drying conditions.

Rate of Drying Curve:

(i) Weight vs Time. i.e., W vs t (Time: x axis;Weight : y axis).

This can be converted to X vs Time.

X = (W –w ) / w = weight of moisture/weight of dry solid

where W = Weight of wet substance,

w = Weight of bone dry solid

From this, another plot of

(ii) Drying flux vs Moisture content ( dry basis) is made.

i.e., N in mass evaporated/( area) (time) vs X in Mass moisture / Mass dry solid

(N : y axis; X: x axis)

Both the plots are shown below:

The first plot is X vs Time. This shows the moisture content decreasing as drying

proceeds and finally reaching a level asymptotically. X corresponding to this level is X* ,

equilibrium moisture content. The drying time for large batches can be directly found

from this plot.

In the same plot, drying rate curve is also shown, which is obtained by plotting the

slopes, dX/dt obtained from the first curve at various instants during drying.

The second plot is N vs X obtained from drying rate curve.

The above plot consists of 4 stages:

Stage AB or A’B: Initial disturbances prevail in this stage.

Initially the solid and liquid surfaces are usually colder than the (ultimate)equilibrium

temperature attained by the surface, tS. In this case, evaporation rate increases as drying

proceeds, which is shown by curve AB.

Alternatively, some times, the surface temperature may be greater than the ultimate

temperature,tS. This gives rise to the curve,A’B. As this initial adjustment periods are

short, they can be ignored.

Stage BC: Constant rate period.

If the solid is initially very wet, its surface is covered with a film of liquid. As the liquid

film is exposed to relatively dry air, the interface is liquid-gas interface and evaporation

takes place. The vapor pressure exerted by the wet solid = vapor pressure of pure liquid at

the prevailing temperature. Hence the surface moisture is entirely unbound moisture.

Drying during this stage is called saturated surface drying.

Since evaporation of moisture absorbs latent heat , the drying surface will ultimately

come to and remain at an equilibrium temperature,tS, such that the rate of heat flow from

the surroundings to the surface = rate of absorption of heat by the surface for evaporation.

The humidity at the surface is equal to YS ,the saturation humidity at tS.

When the constant rate period ends, the solid reaches an average moisture content, XC,

called the critical moisture content. The surface film has been so reduced that further

drying caused dry spots on the surface. Saturated surface drying ends here.XC is also

called Leidenfrost point.

Stage CD: First falling rate period.

Dry spots occupy increasingly larger proportions of the exposed surface. Rate of moisture

evaporated from wet surface remains constant. However, N is calculated on the basis of

gross (wet + dry) drying surface. Hence, N decreases during this period. In most cases,

this unsaturated surface drying gives a linear relationship between N and X. Ultimately,

at point D, the wet film will have completely evaporated leaving the surface totally dry.

Stage DE: Second falling rate period.

As drying proceeds, the rate of drying becomes diffusion controlled. As the deeper parts

(farther away from the drying surface) contain more moisture than the drying surface,

concentration gradients drive the moisture towards the surface. The rate at which

moisture is evaporated depends upon the rate at which moisture moves towards the

surface, which in turn depends on the concentration gradient. Further, as the average

moisture content is lowered, internal moisture movement within the solid decreases, the

only direction of moisture movement being from the interior to the drying surface.

The drying rate plot is curvilinear, typically that of a diffusion controlled operation.

E refers to the equilibrium moisture content for the prevailing air humidity, and then

drying stops.

Batch Drying: at Constant Drying Conditions

To find time of drying:

--SS dX

The rate of drying, N = ------------ where Ss = mass of dry solid in a batch

A dθ A = drying surface

Rearranging,

SS dX

θ = d θ = ------ ----------

A N

I. Constant rate period:

X1 and X2 both > XC and N = NC

SS (X1 – X2 )

Θ = --------------------

ANC

II. Falling rate period:

(i) If X1 and X2 both < XC , and N varying,

draw a curve of 1/N vs X, and integrate graphically the area under this

curve between X1 and X2 to get the value of dX / N.

As a special case, if N is linear in X,

ie., N = mX + b,

SS dX

θ = d θ = ------ ----------

A mX + b

SS mX1 + b SS N1

= ---------ln ------------- = -------ln ---------

m A mX2 + b mA N2

SS (X1 – X2 ) N1 – N2

Θ = -------------------- where Nm = -------------

ANm ln (N1 / N2)

N1 - N2

& m = ------------------

X1 – X2

(ii) If X1 and X2 fall on either side of XC, split the time into 2 parts:

θ = θC + θf

where θC = duration for constant rate period from X1 to XC

SS (X1 – XC )

= --------------------

ANC

θf =duration for falling rate period from XC to X2

SS dX

= ---- -------

A N

(iii) If the entire falling rate curve can be taken as a st.line between

critical point and equilibrium moisture ,

NC ( X – X*)

N = m (X – X*) = -----------------

XC – X*

SS (XC – X*) ( X1 – X

* )

θ = ----------------- ln ----------------

NCA ( X2 – X* )

The Mechanisms of Batch Drying

Cross-Circulation Drying:

I. Constant rate period:

Flux at constant rate drying, Nc = kY(Ys –Y) for cross-circulation drying.

kY, the gas mass transfer coefficient remains constant as long as the speed and direction

of gas flow past the surface remains the same.

YS = Saturated humidity at the liquid surface temperature, tS.

Y = humidity at the main gas stream.

As tS is constant ( being the equilibrium surface temperature), YS , which is saturation

humidity at saturation temperature, tS is also constant.

Y is constant as drying is done at constant drying conditions.

Fir the given system of wet solid and gas, kY is constant.

Hence, NC is constant, and results in constant rate period.

Consider the section of a material drying in a stream of gas. The solid has a thickness zS

and placed on atray of thickness zM. The flux of total heat arriving at the surface = q

where

q = qc + qR + qK,

qC = convectional flux

qR = radiational flux

qK = conductional flux

The flux of evaporation = The flux of total heat arriving at the surface

NCλS = q

But NC = kY (YS-Y)

qC = h C(TG-TS)

qR= hR (TR-TS)

qK= UK (TG-TS)

NC = q / λS = [(hC+ UK) (TG-TS) + hR (TR-TS)] / λS = kY (YS-Y)

With conduction, convection & Radiation:

(YS – Y) λS Uk (TG – TS) hC (TG – TS) hR (TR – TS)

-------------- = ---------------- + ------------------+ ---------------------- = (TG – TS)

hC / kY hC hC hC

Neglecting conduction & radiation effects,

(YS – Y) λS hC (TG – TS)

-------------- = ---------------- = (TG – TS)

hC / kY hC

(a)For flow of gas parallel to a surface & confined between parallel plates:

hC kY

jH = ------- Pr2 / 3

= jD = ----------Sc2 / 3

= 0.11 Ree – 0.29

CP G G S

Where Ree = de G / µ & de = equivalent diameter of the airflow space.

This is for Ree = 2600 to 22000.

With the properties of air at 950 C, this becomes in SI units,

G0.71

hC = 5.9 -------------

de 0.29

(b) Airflow perpendicular to the surface:

For G = 1 to 5 kg/ m2 s , ie, 1 to 4.5 m/s, in the SI unit,

hC = 24.2 G 0.37

Effect of various parameters on NC:

a. Gas velocity, G:

In the absence of radiation & conduction effects,

NC G0.71

for parallel flow of gas

G0.37

for perpendicular flow

b. Gas temperature, TG :

Increase in TG increases TG – TS & hence increases NC.

In the absence of radiation effects, if λ variation over moderate temp. ranges

can be neglected,

NC ∝ ( TG – TS )

c. Gas humidity, Y:

NC ∝ YS – Y

Hence increase in gas humidity decreases NC.

In the absence of radiation & conduction effects,

(YS – Y) λS

-------------- = (TG – TS)

hC / kY

d. Thickness of drying solid, z:

If heat conduction through the solid occurs, increase in z decreases NC.

II.Falling Rate Period: Unsaturated surface drying:

N varies linearly with moisture content.

If this period constitute the whole of falling-rate drying,

_ dX kY A (X – X*) (YS-Y)

-------- = ----------------------------------

dθ SS (XC – X* )

As SS = zS A ρs & letting kY = f(G),

_ dX f(G) (X – X*) (YS-Y)

-------- = ----------------------------------

dθ zS ρs (XC – X* )

α f(G) (X – X*) (YS-Y)

= ----------------------------------

zS

where α is a constant.

Alternatively, if at time θ , the moisture content is X,

X - X* _ NC θ

ln ---------- = -------------------------

X1 – X zS ρs ( XC – X*)

Through-Circulation Drying:

The rate of drying of unbound moisture:

Nmax = GS ( Yas – Y1)

where GS = mass dry gas / (area bed cross section)(time)

N = mass moisture evaporated / (area bed cross section)(time)

= Nmax if the gas leaving the bed is saturated at tas with

humidity YS

Instantaneous rate of drying:

N = GS ( Y2 - Y1)

For a differential section of the bed, the rate of drying is

dN = GS dY = kYdS (Yas – Y)

where S = interfacial surface / unit are of bed cross section

If a is interfacial surface per unit volume of bed whose thickness is zS ,

dS = adzS

∫ −

2

1

Y

YYYas

dY = ∫

zs

Gs

kadz

0

lnYYas

YYas

−−

= NtG = Gs

kaz

(Yas - Y1 ) - ( Yas – Y 2 )

Mean driving force = ------------------------------------

( Yas – Y1 )

ln -----------------

( Yas – Y2 )

N Y2 – Y1 Yas – Y2

----- = ---------- = 1 - ------------- = SSYtG GZkNee

/11

α−−=− ………(1)

Nmax Yas – Y1 Yas – Y1

(a) For small particles, (zs / d ) > 5 , drying of unbound moisture from

Nonporous particles:

Equation can be used for both constant & falling rates.

The interfacial area varies with moisture content.

0.273 dpG 0.215

NtG = ---------- ----------- ( X ρs zs ) 0.64

dp

0.35 µ

where dp is the particle diameter & ρs is the apparent density of the bed

in mass dry solid / volume.

Through- drying of such beds involve a very high pressure drop for gas flow.

Hence they can be dried only on continuous rotary filters ( crystal- filter driers).

(b) Larger particles in shallow beds, (zs / d) < 4 , drying of unbound moisture from

porous & nonporous particles:

During the constant- rate period, the gas leaves the bed unsaturated, and the

Constant rate of drying is given by equation (1) where

j D GS

kY = ------------

Sc 2 / 3

For air drying of water from solids, Sc = 0.6.

Rotary Driers:

Holdup in Rotary driers:

It is defined as the fraction of the drier volume occupied by the solid at any instant.

The average time of retention θ is given by dividing the holdup by the volumetric

feed rate.

ФD z π TD 2 / 4 z ФD ρs

θ = -------------------------- = ---------------

(SS / ρs) (π TD 2 / 4 ) SS

where SS / ρs = volumetric feed rate / cross-sectional area of the drier

SS = mass velocity of dry solids, mass / ( area) ( time)

ρs = apparent solid density, mass dry solid / volume

Z = drier length

TD = drier diameter.

Continuous Direct-Heat Driers:

Three separate zones are distinguished in such driers,owing to the variation in

temperatures of the gas and solid in the various parts of the drier. Refer figure.

Drying at high temperatures:

Zone I: Preheat zone: the solid is heated by the gas until the rate of heat transfer

to the solid is balanced by the heat requirements for evaporation of moisture.

Little actual drying usually occurs here.

Zone II: The equilibrium temperature of the solid, tS remains constant; surface and

unbound moisture are evaporated. Point B: XC, critical moisture content is reached.

Zone III:Unsaturated surface drying and evaporation of bound moisture occur.

The decreased rate of evaporation results in increased solid temperature;

the discharge temperature of the solid � inlet temperature of the gas.

Design of the drier:

Considering only heat transfer from the gas, and neglecting other heat transfers,

for a differential length of the drier,

dq = GSCSdt’G =Ua (tG-tS)dZ

where q = heat flux transferred from the gas to the solid

U = overall heat transfer coefficient between gas and solid

tG-tS = temperature difference for heat transfer

S = interfacial surface / drier cross section

a = interfacial surface / drier volume

dt’G = temperaturedrop experienced by the gas as a result of heat transfer to solid only

CS = humid heat

dt’G Ua dZ

dNt OG = ----------- = -----------------

tG – tS GSCS

If the heat transfer coefficient is constant,

∆ t’G Z GS CS

Nt OG = ----------- = -------------- where Ht OG = -----------

∆ tm Ht OG Ua

where Nt OG = No. of heat-transfer units

Ht OG = Length of heat-transfer unit

∆ t’G = Change in gas temperature owing to heat transfer to solid only

∆ tm = average temperature difference between gas and solid

= log average of terminal temperature differences for each zone

taken separately if the temperature profile is idealized as straight line.

Rotary Driers: 237G 0.67

In commercial driers, the value of Ua in SI units is Ua = -------------

TD

In zone II, ∆ tm = average wet bulb depression of the gas.

Drying at low temperatures:

Zone I- Preheat zone can be ignored.

Zone II-unbound and surface moisture is evaporated ; X1 - X C

Zone III – unsaturated surface drying and evaporation of bound moisture; XC – X2

Gas humidity rises from Y2 - YC

SS (XC – X2 ) = GS (YC –Y2 ) From this equation, YC can be calculated.

Retention time = θ = θ II + θ III = ∫1X

XcN

dX

A

Ss + ∫

Xc

XN

dX

2

Where A / SS is the specific exposed drying surface, area / mass dry solid.

Zone II , X > XC:

N = NC = kY (YS – Y)

θ II = ∫ −

11X

XcYYs

dX

kA

Ss

As GSdY = SS dX,

θ II = ∫ −

11Y

YcYYs

dY

kA

Ss

Ss

Gs

Ys varies with Y.

If the gas temperature is held constant, and if the effects of conduction and radiation

Are negligible,

Ys is the saturation humidity corresponding to the wet bulb temperature of Y.

Integrate the above equation graphically to get θ II.

For the case where Ys is constant, as for adiabatic drying of water into air,

θ II = 1

ln1

YYs

YcYs

kA

Ss

Ss

Gs

−−

zone III: X< XC

Unsaturated drying occurs and the drying rate depends only upon the prevailing

Conditions:

XX

XXYYk

XX

XXNcN

c

SY

c−

−−=

−

−=

))(()(*

*

*

θ III = ∫ −−

− CX

X SY

cS

XXYY

dX

k

XX

A

S

2))(( *

*

The relationship between X, X*, YS & Y should be determined and the above equation

can be evaluated graphically. For this purpose, the material balance can be written as

S

S

G

SXXYY )( 22 −+=

For a special case where bound moisture is negligible (X*=0) and YS is constant

(adiabatic drying), the above 2 equations combine with G SdY = L SdX to give

θ III = ( ) ( )[ ]∫ +−−

1

222 /

Y

Y SSSY

CS

S

S

XSGYYYY

dY

k

X

A

S

S

G

θ III = ( )

( )( )CS

SC

SSSY

CS

S

S

YYX

YYX

XSGYYk

X

A

S

S

G

−

−

+− 2

2

22

ln/

1

If drying takes place only above or only below the critical moisture content, appropriate

changes in the moisture content and humidities must be made. In parallel flow driers,

where gas enters at humidity Y1 and leaves at Y2 while solid enters at moisture content

X1 and leaves at X2 ,

θ II = A

S

S

G S

S

S ∫ −

CY

Y SY YY

dY

k1

1 =

A

S

S

G S

S

S

CS

S

Y YY

YY

k −

− 1ln1

θ III = ( )( )∫ −−

− CX

X SY

CS

XXYY

dX

k

XX

A

S

2

*

*

= A

S

S

G S

S

S

( )( )( )CS

SC

SSCSY

C

YYX

YYX

XSGYYk

X

−

−

−− 2

2

2

ln/

1

Movement of moisture within the solid:

When surface evaporation occurs, there must be a movement of moisture from the depths

of the solid to the surface. The nature of the movement influences the drying during

falling rate periods.

Three types of moisture movement can be considered:

Liquid diffusion: This occurs in cases where single phase liquid solutions are formed

with the moisture: eg.: drying of soap, glue and gelatin and also when the last portions of

water (bound moisture) from clays, paper, flour and wood is being dried.

During drying from the surface, the solid is rich in moisture content in deep portions and

lean at the surface. As a result, concentration gradients are set up, and the drying becomes

diffusion controlled. Fick’s second law for unsteady state diffusion is applicable.

During constant rate drying period, moisture moves to the surface as fast as it can be

evaporated and the rate remains constant. As dry spots appear, unsaturated surface

evaporation begins. Further drying occurs at rates which are entirely controlled by

diffusion. This results in a curvilinear falling rate plot.

Capillary movement: This occurs in granular and porous solids such as clays, sand.

paint pigments. Unbound moisture moves through the capillaries and interstices of the

solids by a mechanism involving surface tension, the way oil moves through a lamp wick.

The liquid is drawn to the surface by the tension in the capillaries which is caused by the

air entering through some openings and replacing the liquid. As drying proceeds, at first

moisture moves through the capillaries to the surface. This results in a uniformly wetted

surface and the rate of drying is constant. The surface moisture is eventually drawn to

spaces between the granules of the surface; the wetted area of the surface decreases and

the unsaturated surface drying follows. The liquid surface recedes into the capillaries.

Evaporation occurs below the surface in a zone which gradually recedes deeper into the

solid, and a second falling rate period results. During this period, diffusion of vapor will

occur within the solid.

If air can not enter the solid and replace the liquid, as in the case a wet cake adhering to

the tray or pan, liquid can not be drawn to the surface. The drying rate falls and the solid

is eventually crumpled. Now, air can enter to replace the liquid, initiating capillary

movement of liquid and drying rate again increases.(fig.)

Vapor diffusion: If heat is supplied to one surface of the solid while drying proceeds from

another, the moisture may evaporate beneath the surface and diffuse outward as a vapor.

Liquid Diffusion of moisture in drying:

When liquid diffusion controls the rate of drying in the falling rate period,Fick’s II law

can be used. Using the concentration as X,

2

2

x

XD

t

XL ∂

∂=

∂∂

where DL is the liquid diffusion coefficient in m2 /h and x is the distance

in solid in m.

During diffusion type drying, the resistance to mass transfer of water vapor from the

surface is usually very small, and the diffusion in the solid controls the rate of drying.

Then the moisture content at the surface is X*. This means the free moisture content at

the surface is essentially zero.

Assuming that the initial moisture distribution is uniform at t=0, the above equation can

be integrated as

( ) ( ) ( )

+++==−

− −−−..

25

1

9

18 21

21

21 2/252/92/

2

1

*

1

*xtDxtDxtD

t

t LLL eeeX

X

XX

XX πππ

π

Where X = average free moisture content at time t h, X1 = initial free moisture content at

time t =0, X* =equilibrium free moisture content at time, x1=1/2 thethickness of the slab

when drying occurs from the top and bottom parallel faces, and x1 =total thickness of

slab if drying occurs only from top face.

.

UNIT V PART A

1.What is Crystallisation?

It is the formation of solid particle within a homogeneous phase clarity ,good yield ,and high purity are

important objectives in crystallization.

2.Explain Delta L law of crystal growth.(MAY 2012) (Dec 10) (Dec 12) (MAY 2015) (May 2016)

If all crystals in magma grow in a uniform supersaturation field and at the same temperature and if all

crystals grow from birth at a rate governed by the super saturation

Then all the crystals are not only variant but also have same growth rate ie. independent of size.

3.Give an example of batch crystallizer and continuous crystallizer:

Agitated batch crystallizer, Swenson walker crystallizer.

4.What are the Sequence of stages

cluster ,embryo ,nucleus ,crystal.

5.What are the three techniques to produce super saturation? (Dec 2012) (Dec 2009) (NOV14)

Solutes like potassium nitrate and sodium sulphite are less soluble at low temp than high

temperature.supersaturation is produced by cooling .When solubility is independent of temperature as

with common salt supersaturation is developed by evaporation.In intermediate cases a combination of

cooling and evaporation is effective.

6. What is Magma?

In industrial crystallization from solution ,two phase mixture of mother liquor and crystals of all sizes

which occupies the crystallizer and is with drawn as product is called magma.

7.What is the purpose of Agitators? (MAY 2015)

for uniform formation of crystals.

8.What are ionic crystals?

Combination of highly electro negative and highly electro positive ions.

9.How do the mother liquid separated from crystals?

By filtration

10 What is Salting evaporator?

The crystallizer build in multiple effect is called salting

11.What is Caking? (Nov2014)

Crystalline product is having the tendency to cake or bind to gether .This is of ten trouble some in bulk

storage ,or in baralled products.

12.What is Crystal geometry?

It is charecterised by the fact that its constituents particle which may be atoms ,molecules or ions are

arranged in orderly three dimensional array called space lattices.

13.What are Invariant crystals?

Under ideal conditions a growing crystal maintains geometric similarity during growth .such a crystal is

called invariant.

14.What is Needle breeding?

Spikes are imperfect crystals which are bound to the parent crystal by weak forces and which break off to

give crystals and poor quality ,this is called needle breeding.

15.What is the rate of nucleation?

It is the number of new particles formed / unit time /unit volume of magma or solids free mother liquor.

16.What is the Critical humidity of a solid salt?

This is the humidity above which it will always become damp and below which it will stay dry.

17.What are Clusters?

Due to random motion of the particles may associate to form what is called cluster.

18.Give an example of Continuous crystallizer?

Swenson walker crystallizer and krystal crystallizer

19.What is CSD?

Crystal size distribution

20.What are the classification of nucleation?(MAY14)

Spurious nucleation , primary nucleation , secondary nucleation.

PART B:

1. (i) Classify industrial crystallizers. Discuss the working of a continuous vacuum crystallizer with the

help of a neat sketch.(Dec 12) (May 2016)

(ii) What are the parameters controlling the crystal size distribution in a crystallizer? Explain them

briefly.

2. A solution of sodium sulfate in water is saturated at a temperature of 40OC. Calculate the weight of

crystals and the percentage yield obtained by cooling this solution to a temperature of 5OC.The

solubilities are as follows:

at 40OC: 32.6% Na2SO4

at 5OC: 5.75% Na2SO4

Note: At a temperature of 5OC the decahydrate will be the stable crystalline form.

3. (i) Explain the classification of crystallizers.

(ii) Explain with neat sketch, the working of a Swenson-Walker crystallizer. (MAY14) (NOV14)

4. 5000 kg of KCl solution at a temperature of 80OC is cooled to 20OC in an open tank. The solubilities of

KCl at 80OC and 20OC are 55 parts and 35 parts per 100 parts of water. Estimate the yield of KCl

crystals by:

(i) Assuming 5% water is lost by evaporation

(ii) Assuming no loss of water by evaporation.

5. (i) Explain Mier’s supersaturation theory.(MAY 2012)

(ii) 1000 kg. of Na2CO3 solution containing 30% Na2CO3 is subjected to evaporative cooling, during

which 15% of the water present in the solution is evaporated from the concentrated solution and

Na2CO3.10H2O crystallizes out.Calculate the quantity of crystals that would be produced if the solubility

of Na2CO3 at the reduced temp. is 21.5 grams per 100 grams of water.

6. Explain with a neat sketch the working of a Draft tube Baffle crystallizer.

7. (a) Discuss the various methods of supersaturation. (MAY14)

(b) With a neat diagram, explain the agitated batch crystallizer.

8. Discuss the mechanism of nucleation and the factors influencing nucleation. (MAY14)

9. (a) Write a short note about homogeneous nucleation.

(b) Briefly discuss the following:

(i) Secondary Nucleation.

(ii) Importance of crystal size.

(iii) Kelvin equation. (MAY14)

10. A saturated solution of MgSO4 at 353K is cooled to 303K in a crystallizer. During cooling, 4% of the

solution is lost by evaporation of water. Calculate the quantity of the original saturated solution to be fed

to the crystallizer per 1000 kg. crystals of MgSO4.7H2O Solubilities of MgSO4 at 303K and 353K are

40.8 and 64.2 kg. per 100 kg. of water respectively.

CRYSTALLIZATION

Crystallization is a process where solid particles are formed from a homogeneous phase.

This process can occur in the freezing of water to form ice, in the formation of snow particles from vapor, in the formation of solid particles from a liquid melt, or in the formation of solid crystals from a liquid solution. The last process is the most important one commercially.

In crystallization the solution is concentrated and usually cooled until the solute concentration becomes greater than its solubility at that temperature. Then the solute comes out of that solution, forming crystals of approximately pure solute.

In commercial crystallization not only are the yield and purity of the crystals Important but also the sizes and the shape of the crystals. Size uniformity in crystals are desired to minimize caking in the package , for the ease of pouring, for ease In washing and filtering, and for uniform behaviour when used. Specific sizes and shapes are required according to the usage.

What is a crystal?

A crystal is an ordered, three-dimensional periodic array. This is a reversible equilibrium whose particular thermodynamic and kinetic properties are specific to the system under study.

The first step is primary nucleation, which is the first formation of the molecules in a crystalline form. This consists of molecules binding together in a very specific and repeating pattern in three dimensions.

XXXXXX

This set of Xs may be thought of as a single dimension crystal. There are two “crystal” contacts at the tops of the Xs. The symmetry that is present is a simple translation of the X.

XXXXXX

XXXXXX Here is a two dimensional crystal. Now in addition to the lateral crystal contacts there are vertical contacts between the Xs. Project this up out of the page and you have a three dimensional crystal.

Types of crystal Geometry.

In a crystal, atoms, ions or molecules are arranged in a regular, repeating pattern in space lattices. Their interatomic distance and angles between their planes are constant and characteristic of that component. On a macroscopic scale, crystals are polyhedrons having flat faces and sharp corners. The crystals of a compound may be formed in different sizes. The relative sizes of the faces and edges of these crystals of the same compound differ, but the Interfacial angles are same.

Thus the angles between corresponding faces of all crystals of the same material are equal and characteristic of the material.

5.1 Classification Of Crystals:

There are many ways to approach crystallization and the physiochemical properties of the system are different in each case. Therefore, the type of arrangement formed (space group) is likely to be different in each condition.

When crystallization of a solute occurs in a given condition, in the absence of hindrance from outside bodies, crystals of same structure but varying sizes are formed. They appear as polyhedrons having flat sides and sharp corners. Although the relative sizes of the faces and edges of various crystals of the same material may be widely different, the angles made by corresponding faces of all crystals (Interfacial angle) of the same material are equal and are characteristic of that material. Since all crystals of a definite substance have the same Interfacial angle, crystal forms are

classified on the basis of these angles. There are 7 classes of crystals:

Cubic : 3 equal axes at right angles to each other

Tetragonal : 2 axes at right angles to each other, 2 axes equal and the third longer Orthorhombic : 3 axes at right angles to each other, all of different length

Hexagonal : 3 equal axex at 600 to each other In one plane , 4

th axis at right angles to this plane

Monoclinic : 3 unequal axes , 2 at right angles In a plane , 3 rd

at some angle to this plane Triclinic : 3 unequal axes at unequal angles to each other

Trigonal : 3 equal and equally Inclined axes

Thus each crystal system consists of a set of 3 axes in a particular geometric arrangement.

Invariant crystals: A growing crystal maintains geometric similarity during its growth. Such a crystal is called invariant crystal.

These polygons are similar and the lines connecting the corners of the polygon with the centre of the crystal are straight. The rate of growth of the crystal is measured by velocity of translation of the face away from the centre of the crystal in a direction perpendicular to the face. Unless the crystal is a regular polygon, rates of growth of the various faces of an invariant crystal are not equal.

5.2 Objectives in Industrial Crystallization:

• Yield • Purity

• Crystal size and Crystal Size Distribution (CSD)

Yield: Yield of required crystals should be high with minimum loss of solute.

Purity of crystal: A well formed crystal is nearly pure. But it retains mother liquor when removed from the final magma. If the crop contains aggregates, considerable amounts of mother liquor may be occluded within the solid mass. When retained mother liquor of low purity is dried on the product, contamination results, the extent of which depends on the amount and degree of impurity of the mother liquor retained by the crystals.

Magma: In industrial crystallization from solution, the two phase mixture of mother liquor and crystals of all sizes , which occupies the crystallizer and is withdrawn as product, is called magma. Separation of retained mother liquor from crystals: This is done by filtration or centrifuging, and the balance is removed by washing with fresh

solvent.

Crystal size and Crystal Size Distribution (CSD):

Importance of crystal size: The appearance and size range of a crystalline produce are important qualities. Reasonable size and size uniformity are desirable for further processing such as filtering, washing, reacting with other chemicals, transporting and storing the crystals. If the crystals are to be marketed as the final product, the Individual crystals have to be strong, nonaggregated, uniform in size and noncaking In the package. For these reasons, crystal size distribution must be under control.

Hydrated crystals: Many important inorganic substances crystallize with water of crystallization. In some systems, several different hydrates are formed, depending on concentration and temperature. Phase equilibria in such systems can be quite complicated.

Principles of Crystallization

Equilibrium in crystallization process is reached when the solution is saturated. Solubility curve is the equilibrium relationship for bulk crystals. Phase diagrams are also known as Ostwald-Mier diagrams or sometimes just Mier diagrams. At supersaturation the solute partitions into the solid and soluble phases. Although entropy is reduced, there are a high number of specific contacts made to make the free energy favorable. For small molecules, these properties may be predicted. For macromolecular systems they cannot. Therefore, macromolecular crystallization is still empirical.

What is supersaturation? Supersaturation is the concentration difference between that of the supersaturated solution (in which the crystal is growing) and that of a solution in equilibrium with the crystal. The 2 phases are almost at the same temperature.

Units of supersaturation:

Concentrations may be defined either in mole fraction or in concentration of the solute in the

solution:

In mole fraction of the solute: moles of solute / total moles of the solution = y

In concentration of the solute: moles of solute / volume of the solution = c Supersaturation in mole fraction :

y = y - ys where y is the mole fraction of solute in solution and ys is mole fraction of solute in saturated solution

Supersaturation in concentration : Molar supersaturation

c = c – cs where c is the molar concentration of solute in solution and cs is molar concentration of solute in saturated solution

Fractional supersaturation, s : y c

s = =

y s cs Concentration ratio:

y c α = =

y s cs

Relationship between s and α : y

c

α = 1 1 = 1 + s

ys cs

Percentage of supersaturation: 100 s

Methods of supersaturation: From the graph below, it can be seen that there are three types of systems, classified on the basis of their solubility dependence on temperature:

Type I : Solubility temperature.

Here the solubility of the solute increases with increase in temperature. Most of the systems fall under this category. Eg.: KCl, KNO3 CaCl2 etc.

Type II : Solubility is not affected much by change In temperature. Eg.; NaCl

Type III : Solubility (1 / temperature) .

Solubility decreases with increase in temperature. This kind of inverted solubility curve is shown by salts like MgSO4 , Ce2 (SO4) 3. The solubility versus temperature curve of such systems is called inverted solubility curve.

Methods to Effect Supersaturation:

Supersaturation can be effected by the following methods: 1. Cooling

2. Evaporating a part of the solvent

3. Addition of a third substance:

(i) Drowning (ii) Reacting

Cooling: For type I systems, cooling ( reduction in temperature ) makes the unsaturated solution saturated and further cooling makes it supersaturated. This is because decrease in temperature decreases solubility of this type solute.

Evaporating a part of the solvent : For type II systems, cooling does not effect supersaturation. Hence, a portion of the solvent is evaporated; This reduces the volume of the solution thereby increasing solute concentration. This results in supersaturation ( when concentration increases beyond solubility limit of the system). Thus NaCl salt crystals are formed in sea water on solar evaporation of water.

Addition of a third substance: If neither cooling nor evaporation is desirable, as when the solubility Is very high, supersaturation may be generated by adding a third component. Drowning: The third component added combines with the solvent forming a mixed solvent in which the solubility of the solute is sharply reduced. This generates supersaturation. This is

called salting. Reacting:. A chemical reaction can be used to alter the dissolved solid to decrease its solubility in the solvent, thus working toward supersaturation. The third component added combines with the original solute forming an insoluble precipitate. This is called precipitation.

Each method of achieving supersaturation has its own benefits. For cooling and evaporative crystallization, supersaturation can be generated near a heat transfer surface and usually at moderate rates. Drowning or reactive crystallization allows for localized, rapid crystallization where the mixing mechanism can exert significant influence on the product characteristics.

5.5 Procedure for Crystal Growth: So how do you grow crystals? Let's consider an example that is fairly easy to envision. Take a pot of boiling water and add table salt while stirring to make a water-salt solution. Continue adding salt until no more salt will dissolve in the solution (this is a saturated solution). Now add one final teaspoon of salt. The salt that will not dissolve will help the first step in crystallization begin. This first step is called "nucleation" or primary nucleation. The salt resting at the bottom of the pot will provide a site for nucleation to occur.

In industrial crystallization, the formation of crystals requires 2 steps: (1) the birth of a new particle and (2) its growth to macroscopic size. The first step is called nucleation and the second is crystal growth.

Nucleation: Nucleation refers to the birth of very small bodies of a new phase within a supersaturated homogeneous existing phase. It is a consequence of rapid local fluctuations on a molecular scale in a homogeneous phase that is in a state of meta stable equilibrium. The rate of nucleation is the no. of new particles formed per unit time per unit volume of magma or solute free mother liquor.

Primary nucleation: Primary nucleation is the first step in crystallization. Simply defined, it's the growth of a new crystal. The basic phenomenon is called homogeneous nucleation which is the formation of new particles within a phase uninfluenced in any way by solids of any sort, including the walls of the container or even the most minute particles of foreign substances.

On an industrial scale, a large supersaturation driving force is necessary to initiate primary nucleation. The initiation of primary nucleation via this driving force is not fully understood. Usually, the instantaneous formation of many nuclei can be observed "crashing out" of the solution. You can think of the supersaturation driving force as being created by a combination of high solute concentration and rapid cooling In continuous crystallization, once primary nucleation has begun, the crystal size distribution begins to take shape.

Secondary nucleation:

The second chief mechanism in crystallization is called secondary nucleation. In this phase of crystallization, the formation of nuclei is attributable to the existing macroscopic crystals in the magma. Crystal growth is initiated with contact. The contact can be between the solution

and other crystals, a mixer blade, a pipe, a vessel wall, etc. This phase of crystallization occurs at lower supersaturation (than primary nucleation) where crystal growth is optimal. In the salt example, cooling will be gradual so we need to provide a "seed" for the crystals to grow on. This seeding initiates secondary nucleation.

No complete theory is available to model secondary nucleation. Correlating experimental data to model crystallization is time consuming and not feasible for batch operations, but can easily be justified for continuous processes where larger capital expenditures are necessary.

Secondary nucleation requires "seeds" or existing crystals to perpetuate crystal growth. In our salt example, we bypassed primary nucleation by "seeding" the solution with a final teaspoon of salt. Secondary nucleation can be thought of as the workhorse of crystallization.

Two kinds of secondary nucleation are : (a) Fluid shear nucleation: When supersaturated solution moves past the surface of a growing crystal at high velocity, the shear stresses in the boundary layer sweep away embryo or nuclei which later form new crystals. Thus fluid shear helps formation of new crystals and retards growth of already formed crystal. (b) Contact nucleation: Nucleation arises as a result of collisions of existing crystals with each other or with the solid surface of the equipment. Thus secondary nucleation is influenced by agitation and collisions of existing crystals.

Write short notes on Mier’s Supersaturation Theory:

Figure 2: Progression of Crystallization

According to Mier’s theory there is a definite relationship between the concentration and temperature at which crystals will spontaneously form in pure solution. This relationship is represented by the supersolubility curve which is approximately parallel to the solubility curve (refer figure 2 above: Here the solubility curve is denoted as saturation curve).

The solubility (or saturation) curve represents the maximum concentration of a solution that can be achieved by bringing a solid solute into equilibrium with pure solvent. If a solution having the composition and temperature indicated by the point 1 is cooled it first crosses the solubility curve at point 2 and we would expect here crystallization to start. Actually if we start with initially unseeded solution, crystal formation will not begin until the solution is supercooled considerably past the point 3. According to Mier’s theory, crystallization will start in the neighborhood of point 3 ( ie point 4) and the concentration of the solution then follows roughly along the curve 5-6-7. For an initially unseeded solution, the supersolubility curve (dotted curve) represents the limit at which spontaneous nuclei formation begins and consequently crystal formation starts. Mier’s theory states that under normal conditions nuclei cannot form and crystallization cannot then occur in the area between the solubility curve and the supersolubility curve, ie., at any position short of point 3 along the line 1-3 in the figure 2.

Mier’s theory is useful in the discussion of qualitative aspects of nucleation from seeded and unseeded solutions.

Homogeneous nucleation: In crystallization from solution, homogeneous nucleation almost never happens. However, its fundamentals are important in understanding the more useful types of nucleation.

Crystal nuclei may form from molecules, atoms or ions. In aqueous solutions these may be

hydrated.

Because of their random motion in restricted volume, several of them may associate to form

clusters. A cluster is a loose aggregation which usually disappears quickly. Occasionally enough particles associate to form an embryo, in which there are the beginnings of a lattice arrangement and the formation of a new and separate phase. For the most part, embryos have short lives and revert to clusters or individual particles. But if the supersaturation is large enough, an embryo may grow to such a size that it is in thermodynamic equilibrium with the solution. It is then called the nucleus, which is the smallest assemblage of particles that will not redissolve and therefore can grow to form a crystal. The number of particles needed for a stable nucleus ranges from a few to several hundreds. The evolution of a crystal can be represented in the following sequence of stages:

Cluster → embryo → nucleus → crystal

Solubility:

Ostwald ripening: Many small crystals form in a system initially but slowly disappear except for a few that grow larger, at the expense of the small crystals. The smaller crystals act as "nutrients" for the bigger crystals. As the larger crystals grow, the area around them is depleted of smaller crystals. This process is known as Ostwald ripening. This is a spontaneous process that occurs because larger crystals are more energetically favored than smaller crystals. While the formation of many small crystals is kinetically favored, (i.e. they nucleate more easily) large crystals are thermodynamically favored. Molecules on the surface of a crystal are energetically less stable than the ones already well ordered and packed in the interior, and small crystals have a larger surface area to volume ratio than large crystals. Large crystals, with their lesser surface area to volume ratio, represent a lower energy state. Thus, many small crystals will attain a lower energy state if transformed into large crystals.

In short, a small particle has more surface energy per unit mass than a large particle at the same temperature. Consequently, a small crystal less than a micron size has more solubility than that of a large crystal. Ordinary solubility data apply only to moderately large crystals. A small crystal can be in equilibrium with a supersaturated solution. If a large crystal is also present in the solution the smaller crystal will dissolve and the larger will grow until the small crystal disappears. Thus small crystals act as nutrients to large crystals which grow at the expense of small crystals. This phenomenon is called Ostwald ripening.

5.6 Effect of Particle Size on Solubility:

The effect of particle size on solubility is a key factor on nucleation. The solubility of a

particle is related to particle size by the Kelvin equation:

ln α = 4VMσvRTL

where L = crystal size α = ratio of concentrations of supersaturated and saturated solutions VM = molar volume of crystals σ = average interfacial tension between solid and liquid v = no. of ions per molecule of solute (for molecular crystals v = 1)

Since α = 1 + s , this eqn. shows that a very small crystal of size L can exist in equilibrium with a solution having a supersaturation of s, relative to a saturated solution in equilibrium with large crystals. Rate of Nucleation

Where B’ = nucleation rate, number / cm2-s

σ = solid-liquid interfacial tension, ergs / cm2 Na = Avagadro constant, 6.0222 x 1023 molecules / gmol

C = frequency factor

C is a statistical measure of the rate of formation of embryos that reach the critical size.

For B ' = 1, ie., formation of one nucleus per cm2 per second, the calculations give a very large value of s which is impossible for a material of normal solubility. This is one reason why homogeneous nucleation never occurs in ordinary crystallization.

Heterogeneous Nucleation: The catalytic effect of solid particles on nucleation rate is the reduction in energy required for nucleation. The work of nucleus formation is reduced by a factor that is a function of the angle of wetting between the nucleus and catalyst. Rate of nucleation

Different terms of nucleation:

Primary nucleation : This is equivalent to homogeneous nucleation.

Secondary nucleation: Seeding causes nucleation and contact attributes to crystal growth.

Crystallization occurs at lower supersaturation.

Homogeneous

nucleation: Nucleation occurs by the presence of submicroscopic species (ions

and molecules ) and not by macroscopic particles.

Heterogeneous

nucleation:

Nucleation rate is increased by solid particles.

Contact nucleation: Most common type of nucleation in industrial crystallizers.

Supersaturation and the impact energy of crystal-solid contact are

the

factors deciding the no. of nuclei formed.

If N = no. of nuclei formed per contact and E = contact energy,

For Inorganic crystals, N s

For some organic crystals, ln N s

For organic and hydrated inorganic crystals, N E

For anhydrous inorganic crystals, energy requires is larger than for

other crystals.

The mechanism of primary nucleation is homogeneous nucleation which happens only by virtue of supersaturation. .

The mechanism of secondary nucleation is contact nucleation.

Difference between primary and contact nucleations: Growth rate of crystals s n where s = supersaturation For primary nucleation, n ≥ 20 For secondary nucleation, n = 1

Crystal growth:

It is diffusional process consisting of 2 steps, both proceeding only when the solution is supersaturated. (1) Diffusional step: The solute molecules or ions reach the growing faces of a crystal by

diffusion through the liquid. Mass transfer coefficient, ky applies. (2) Interfacial step: The diffused molecules or ions must be accepted by the crystal and

organised into the space lattice.

Individual and Overall Growth Coefficients: For the diffusion step, driving force for mass transfer is (y - ys ) where y = mole fraction of the solute away from the crystal face and y s is the mole fraction at the interface which is the saturation value. For the interfacial step, driving force for mass transfer is (y - y' ) where y' = mole fraction of the solute at the crystal face. This is the actual driving force for crystal growth. Mass transfer flux = NA = m / s p = ky (y - y') ….. (4) where NA = molar flux, moles per time per unit area

m = rate of mass transfer, moles per time sp = surface area of a crystal ky = mass transfer coefficient, moles per time per unit area

Overall Coefficient = K

….(5) where ks is given by the equation m / sp = ks ( y' - ys ) …(6)

∆L Law of Crystal Growth:

If all crystals in a magma grow in a uniform supersaturation field and at the same

temperature and if all crystals grow from birth at a rate governed by the supersaturation, then

all crystals are not only invariant but also have the same growth rate that Is Independent of

size. This generalization Is called the ∆L Law. When applicable, G ≠ f(L) , the total growth of each crystal in the magma during the same time interval Δt is the same and L = G t. Though it is an idealized model (realistic only for particle sizes In the range 50 to 500 µm In size), it is sufficiently precise in many situations and greatly simplifies industrial process calculations.

How to predict product size distribution from seed size distribution using ∆L Law? According to ∆L Law, the product size distribution is same as the seed size distribution provided all the crystals grow in a uniform supersaturation at the same temperature and over the same time interval.

CSD: In crystallizers, CSD is determined by the interactions of the rates of nucleation and growth, and the overall process is complicated kinetically. The driving potentials for both rates is supersaturation and neither crystal growth nor formation of nuclei from the solution can occur in a saturated or unsaturated solution. The rate of nucleation is the no. of new particles formed per unit time per unit volume of magma or solute free mother liquor. This quantity (B’) is the first kinetic parameter controlling the CSD. The second one is G which can be calculated from mass transfer coefficients and concentrations. ∆L Law is used as a simplified model to calculate CSD.

Origin of Crystals:

(i) By supersaturation. (ii) By attrition: Circulating magma crystallizers have internal propeller agitator or external rotary circulating pumps. On impact with these moving parts, soft or week crystals can break into fragments, form rounded corners and edges, and so give new crystals, both large and small. Such effects degrade the quality of the product. (iii) Seeding: Seed crystals originating in previous crystallization are added to the

crystallizing systems.

Initial breeding: Seeds usually carry on their surfaces many small crystals which were

formed during the drying and storage of the seeds after their own manufacture. The small crystals soon wash off and subsequently grow in the supersaturated solution. This is called initial breeding.

Initial breeding can be prevented by curing the seed crystals before use, either by (a) contact with undersaturated solution or solvent or by (b) a period of preliminary growth in a supersaturated solution.

Growth related imperfections: (i) Needle breeding: Growth related spurious nucleation occurs at large supersaturations or accompanies poor magma circulations. It is characterized by abnormal needlelike and whiskerlike growths from the ends of the crystals, which under these conditions grow much faster than the sides. The spikes are imperfect crystals, which are bound to the parent crystal by weak forces. These break off to give crystals of poor quality. This is called needle breeding

(ii)Veiled growth: This is unrelated to nucleation and occurs at moderate supersaturations. It is the result of occlusion of mother liquor into the crystal face, giving a milky surface and impure product. Its cause is too rapid crystal growth, which traps mother liquor into crystal faces.

5.8 Principles of Design:

1. Find the theoretical yield: from mass and energy balance.

2. Estimate CSD of the product : from the kinetics of nucleation and growth. 3.Use MSMPR model to identify the kinetic parameters and utilize them in calculating

the performance of the crystallizer.

MSMPR Crystallizer:

Mixed Suspension-Mixed Product Removal Crystallizer:

Consider a continuous crystallizer that operates with the following requirements:

(i) The operation is steady state.

(ii) At all times the crystallizer contains a mixed suspension magma, with no product

classification. (iii) At all times uniform supersaturation exists throughout the magma.

(iv) The ∆L Law of crystal growth applies.

(v) No size classified withdrawal system is used.

(vi) There are no crystals in the feed. (vii) The product magma leaves the crystallizer in equilibrium, so the mother liquor in

the product magma is saturated. (viii) No crystal breakage into finite particle size occurs.

Because of the above restraints, the following are constant and independent of location.:

Nucleation rate, no.of nuclei generated in unit time in unit volume of the mother liquor.

Growth rate, length per unit time

Particles in mother liquor ranges in size from nuclei to large crystals. Particle size distribution = product size distribution.

MSMPR model can be extended to account for unsteady state operation, classified product removal, crystals in the feed, crystal fracture, variation in magma volume and time dependent growth rate. This can be done with the use of a generalized population balance (population-density function). Population Density Function The crystal number density is defined as the no. of crystals of size L and smaller in the magma per unit volume of the mother liquor. Hence if N Is the number of crystals of size L or less in the magma, V is the volume of the mother liquor In the magma and L is the size of crystals, then N/V is the crystal number density. A plot of crystal density vs crystal size is shown. At L = 0, N= 0 and at L = LT, the largest of the crystal in the magma, then N = NT. The population density n Is defined as the slope of the cumulative distribution curve at size L or it can be expressed as

At L + 0, the population density Is the maximum, I.e., n = n0 and L = LT, the population density Is zero. In MSMPR model, both n and N/V are invariant with time and location. Consider N dL crystals between sizes L and L+dL per unit volume of magma in the crystallizer. In MSMPR model, each crystal of length L has same age tm. L = G tm …….(11)

Let (Δn)ΔL be the amount of product wit5hdrawn from the above magma during a time Interval of Δt. Since the operation Is in steady state, withdrawal of product does not affect size distribution in the product. If Q is the volumetric flow rate of liquid in the product and VC is the total volume of liquid in the crystalliser, then

where the negative sign indicates the withdrawn product. The growth of each crystal Is given by ΔL = GΔt …..(13)

Combining equations (12) & (13) for

eliminating Δt,

The quantity L / (Gτ ) is the dimensionless length and Is denoted as Z.

Number of Crystals per Unit Mass The no. of crystals nC in a unit volume of liquid in either magma or product is,

5.9 Crystallizers and their classification: Crystallizers are classified basically into batch crystallizers and continuous crystallizers.

Commercial crystallizers are generally continuous crystallizers.

Basis of classification of crystallizers: (i) Method of creating supersaturation: Crystallizers create supersaturation either by (a) cooling or by (b) evaporation or (c) both.

(ii) Method of bringing contact between crystals and supersaturated liquid.

(a) Circulating liquid method : A stream of supersaturated liquid is passed through a fluidized bed of growing crystals. Nucleation and growth bring the liquid to saturation level. This is passed through a cooling or evaporating zone to generate supersaturation. This is then recycled through the crystallization zone. Thus a supersaturated liquid releases its supersaturation to grow crystals , then pumped away from the crystals and gets supersaturated again and recycled.

(b) Circulating magma method: The entire magma is circulated through both crystallization and supersaturation steps without separating the liquid from the solid. Supersaturation and crystallization occur in the presence of crystals.

In both (a0 and (b) feed is added to the circulating stream between the crystallizing and supersaturating zones.

(iii) Method of maintaining suspension:

(a) Crystallizers with size classification: In these crystallizers, small crystals are retained in the growth zone for further growth. Only crystals of a specific minimum size can leave the unit as a product.

Role of agitation in crystallizers: To improve growth rate, to prevent segregation of supersaturated solution that causes excessive nucleation and to keep crystals in suspension throughout the crystallizing zone. Internal propeller agitators may be used, often equipped with draft tubes and baffles, and external pumps are common for circulating liquid or magma crystallizers. The latter method is called forced circulation. Advantage of forced circulation with external heaters: several identical units can be connected in multiple effect by using the vapor from one unit to heat the next in line-evaporative crystallizers.

1

MASS TRANSFER - I

UNIT I: PART A: -

1. State Fick’s first law of diffusion. (JUNE 2012) (DEC 2014) (MAY 2015)

It states that “the diffusion flux JA * (molar flux relative to molar avg velocity) is proportional

to the conc. gradient & the proportionality constant is the mass diffusivity.

JA * = - DAB dCA / dZ

JA * - molar flux (kg mol/m2 s)

dCA /dZ - conc. gradient (kg mol/m3 m)

DAB - diffusivity of A in B.

2. Define the terms: Molecular diffusion & Eddy diffusion (Dec 2013) (Dec 2011) (MAY

2015)

Molecular Diffusion: is a mechanism of mass transfer in stagnant fluids / moving in laminar

flow. Molecular diffusion is concerned with the movement of individual molecules through a

substance by virtue of its thermal energy. The phenomenon of molecular diffusion leads to a

completely uniform conc. throughout a solution.

Eddy Diffusion: is a mechanism in which the mechanical agitation produces rapid movement

of relatively large chunks /eddies, of fluid characteristics of turbulent motion, which have

carried salt with them. (method of solute transfer).

3. What is pore diffusion? Give examples (DEC 2014)

The movement of fluids (gas or liquid) into the interstices of porous solids or

membranes; occurs in membrane separation, zeolite adsorption, dialysis, and reverse

osmosis.

4. What is diffusion? (Dec 2014) (May 2016)

It is the net movement of a substance (e.g., an atom, ion or molecule) from a region of

high concentration to a region of low concentration. This is also referred to as the

movement of a substance down a concentration gradient.

5. When will the mass diffusivity and the momentum diffusivity are same?

Mass diffusivity and momentum diffusivity are same for liquids at high Concentrations

6. What is the unit for mass diffusivity?( Dec-2010)

Unit for mass diffusivity - m2/s .

7. What is the relation between the diffusivities in liquids and diffusivity in gases at

atmospheric pressure?

The diffusivities in liqs are generally less than ( four to five orders of magnitude smaller )

the diffusivity at atmospheric pressure.

8. Define Mass Flux (Dec2012)

Mass Flux of a given species is the amount of that species in either mass or moles which

crosses a given area per unit time .

9. What is the basic concept of the film theory of diffusion ?

2

The basic concept of film theory is that the resistance to diffusion can be considered

equivalent to that in a stagnant film of a certain thickness.

10.State Knudson diffusion?( NOV 2011)

Knudsen diffusion is a means of diffusion that occurs when the scale length of a

system is comparable to or smaller than the mean free path of the particles involved.

For example in a long pore with a narrow diameter (2–50 nm) because molecules

frequently collide with the pore wall

11 . What are the factors influenced eddy diffusion?

Eddy diffusion depends on fluid properties .

velocity & position in the flowing stream.

12. What is the cause of molecular diffusion?

Molecular diffusion is caused by : concentration difference

13.How does the diffusivity depends?

The diffusivity depends on the temperature , pressure and concentration of the system.

14. What is the effect of temperature on diffusion coefficient? (NOV 2011)

Diffusion increases with increase in temperature because the molecules moves rapidly

15. How do turbulent diffusion occurs?

In turbulent diffusion mass transfer is affected as a result of eddies or wakes.

16.What is Sensible heating?

This process involves heating without changing the moisture content of the air.

17.What is Sensible cooling?

This involves cooling at constant moisture content or humidity ratio of air.

18.Explain the method to Produce low temperature?

Isentropic expansion, by evaporation, and by magnetic cooling

19.What is dew point temperature?

It is the temperature of air vapor mixture at which moisture will start to condense out of

the air as the air is cooled.

20.Write about diffusion coefficient of solids.(DEC12) (May 2016)

The diffusion coefficient in solids at different temperatures is generally found to

be well predicted by the Arrhenius equation:

where

D is the diffusion coefficient (m2 / s)

D0 is the maximum diffusion coefficient (at infinite temperature; m2 / s)

EA is the activation energy for diffusion in dimensions of (J mol−1)

T is the temperature (K)

k is the Boltzmann constant

3

PART B

Unit - I

1. Derive an expression for the steady state diffusion of gas A through a stagnant film of non

diffusing gas B.(MAY 2015) (May 2016)

2. Describe a method for estimation of diffusivity of a volatile solvent into air.

3. a) Derive an expression for finding the mass flux of diffusion of A through non diffusing B,

A and B are liquids.(MAY14)

b) What is the equation of continuity and how can Fick’s second law of diffusion be derived

from it?

4. Explain a method of determination of diffusivity. Derive an expression for the same.

5. (i) Derive the equations for the steady state diffusion of a component in a liquid mixture

containing components A and B:

a) for equimolal counter diffusion.

b) Through a non diffusing B.

(ii) Write briefly about the measurement and calculation of diffusivities.

6. Differentiate between equimolal counter diffusion and diffusion through a stagnant layer.

7. Ammonia gas (A) diffuses through Nitrogen gas (B) under steady state conditions with

Nitrogen non diffusing. The total pressure is 1.013 x 105 Pa. and the temp. is 298K. The

diffusion path is 0.15m. The partial pressure of Ammonia at one end is 1.5 x 104 Pa and at the

other end is 5 x 103 Pa. The diffusivity for the mixture at 1.013 x 105 Pa. and 298K is 2.3 x

10-5 m2/sec. Calculate (i) the flux of Ammonia and (ii) the equimolal counter diffusion flux ,

assuming that Nitrogen also diffuses.

8. An open circular tank 5m. in dia. contains Benzene at 22OC exposed to the atm. in such a way

that the liquid is covered with an air film of thickness 0.9 x 10-2 m. The concentration of

Benzene beyond the film is negligible. If benzene is worth Rs. 80/litre, what is the value of

the loss of benzene from this tank in rupees per day?

Data Given:

Vapor pressure of Benzene at 22OC = 100 mm. Hg.

DAB of Benzene at 0OC = 0.77 x 10-5 m2/s.

Specific gravity of Benzene = 0.88

Viscosity of Benzene at 0OC = 0.75 x 10-3 kg / m.s

Viscosity of Benzene at 22OC = 0.55 x 10-3 kg / m.s

9. Calculate the rate of diffusion of water vapor from a thin layer of water at the bottom of a

well 6m. in height to dry air flowing over the top of the well. Assume that the entire system

is at 298K and atm. pressure. If the well dia. is 3m., find out the total weight of water diffused

per day from the surface of water in the well. The diffusion coefficient of water vapor in dry

air at 298K and atm. pressure is 0.256x 10-4 m2/s. Partial pressure of water vapor at 298K is

0.0323 x 10-4 kg./m2.

10. The diffusivity of CHCl3 in air was measured by exposing pure CHCl3 to a stream of air at

25OC and 1 atm. pressure. The liquid density of CHCl3 at 25OC is 1.485 g./cc. and its vapor

pressure at 25OC = 200 mm. Hg. At time t=0, the liquid CHCl3 surface was 7.4 cm. from the

top of the tube and after 10 hours the liquid surface had dropped 0.44 cm. If the conc. of

CHCl3 is zero at the top of the tube, calculate the diffusion coefficient of CHCl3 in air.

11. Methane diffuses at steady state through a tube containing helium. At point 1, the partial

pressure of methane is pA1= 55 kPa and at point 2, 0.03 m apart pA2 = 15 kPa. The total

pressure is 101.32 kPa and the temperature is 298 K. At this pressure and temperature, the

value of diffusivity is 6.75 x 10-5 m2/sec.

(i) Calculate the flux of CH4 at steady state for equimolal counter-diffusion.

(ii) Calculate the partial pressure at a point 0.02 m apart from point 1.

12. Through an accidental opening of a valve, water has been split on the floor of an industrial

plant in a remote, difficult to reach area. Determine the time required to evaporate the water in

4

the surrounding air. The water layer is 15 mm thick and may be assumed to remain at 25oC and 1

atm pressure with an absolute humidity of 0.002 kg water / kg dry air. Evaporation is assumed to

take place by molecular diffusion through a gas film of 10 cm thick. The mass diffusivity of water

vapor in air is 2.6 x 10-5 m2/sec. Vapor pressure of water at 25oC is 24 mm. Hg.

MASS TRANSFER OPERATIONS-I

UNIT I –Molecular Diffusion- Equations

Fick’s I Law (for diffusion in z direction):

For a constituent A in solution in B:

Valid for steady state diffusion of component A: JA = - DAB z

CA

∂

∂ ….(1.1)

Where JA= molar flux of component A in z direction in kmol/m2.s

DAB = diffusivity of A in B in 2/s

CA = concentration of A in Kmol/m3

Z = distance z2 – z1 in m

z

CA

∂

∂ = concentration gradient in kmol/m

4

The negative sign indicates that diffusion occurs in the direction of drop in concentration.

Conditions: (i) steady state, molecular diffusion.

(ii) no bulk flow.

(iii) diffusivity is independent of concentration.

JA = -cDABz

xA

∂

∂ ….(1.2)

where c = total concentration in Kmol/m3 xA = mole fraction of A

Types of Flux:

J : Flux of a constituent w.r.t. average molar velocity of all the constituents

N : Flux relative to a fixed location in space

Eg: Fish swimming upstream against the flowing current

Velocity of the fish w.r.t. a fisherman’s baited hook is analogous to N

Velocity of the fish relative to the stream is analogous to J

Relationship between N and J :

In a binary system comprising of A and B,

Total flux, N is the summation of flux of A and flux of B

NA + NB = N ….(1.3)

The movement of A is made up of two parts: (i) resulting from the bulk motion, N and

the fraction xA (ii) resulting from diffusion JA

NA = NxA + JA ……(1.4)

NA = (NA + NB )c

cA - DAB z

cA

∂

∂ …..(1.5)

NB = (NA + NB )c

cB - DBA z

cB

∂

∂ …..(1.6)

Adding these gives

- DAB z

cA

∂

∂ = DBA

z

cB

∂

∂ or JA = - JB …..(1.7)

If cA + cB = constant, it follows that DAB = DBA at the prevailing concentration and temp.

Binary System:

Steady State Molecular Diffusion In Fluids At Rest And In Laminar Flow:

In eqn (1.5), separation of variables give

∫∫ =+−

− 2

1

1

2

1

)(

z

zAB

c

c BAAA

A dzcDNNccN

dcA

A

…(1.8)

where 1 indicates the beginning of diffusion path (cA high) and 2 the end of diffusion

path (cA low). Letting z2 –z1 = z, we get

ABBAAA

BAAA

BA cD

z

NNccN

NNccN

NN=

+−

+−

+ )(

)(ln

1

1

2 ……….(1.9)

or

NA =

cc

NNN

cc

NNN

z

cD

NN

N

A

BA

A

A

BA

A

AB

BA

A

1

2

)(

)(ln

−+

−+

+ ……….(1.10)

Molecular diffusion in gases:

NA DAB pt NA/(NA+NB) -- yA2

NA = ------------ ------------- ln ----------------------------------- (1.11)

(NA+NB) RT z NA/(NA+NB) -- yA1

Molecular diffusion in liquids:

NA DAB ρ NA/(NA+NB) -- xA2

NA = ------------ ------------------ ln ----------------------------- (1.12)

(NA+NB) z x B,M M av NA/(NA+NB) -- xA1

GASES:

Steady State Molecular Diffusion In Gases In Laminar Flow:

Case 1: Diffusion of A through non diffusing B:

For gases (obeying ideal-gas law):

C = n/V = pt/RT

A

t

AA yp

p

c

c== (1.13)

Converting concentrationterms to pressure terms, equation (1.10) becomes

NA = ( )

( )

( ) t

A

BA

A

t

A

BA

A

tAB

BA

A

pp

NNN

pp

NNN

RTz

pD

NN

N

1

2

ln

−+

−+

+……………(1.14)

Case 1: Diffusion of A through non diffusing B:

NA is finite, NB = 0 Hence

NA ----------------- = 1

(NA+NB)

Equation (1.14) becomes

NA =1

2lnAt

AttAB

pp

pp

RTz

pD

−

− =

1

2lnB

BtAB

p

p

RTz

pD ………………(1.15)

22

1!

BAt

BAt

ppp

ppp

=−

=−

Subtracting one from the other,

1221 BBAA pppp −=− ……………………..(1.16)

Multiplying and dividing equation (1.15) by these equal terms,

NA =( )( ) RTz

D

p

p

pp

pp

RTz

pD AB

A

B

BB

AAtAB =−

−

1

2

12

21 ln ( )21 AA

BM

t ppp

p− -------(1.17)

where ( )

1

2

12.

lnB

B

BBMB

p

p

ppp

−= = log mean partial pressure of B

Case 2: Equimolar Counterdiffusion:

Here, NA = -NB

NA + NB = N =0

NA ----------------- = ∞

(NA+NB)

Hence use the equation NA = NyA + JA or NA = N(cA /c) + JA

But N = 0

Hence NA = JA

NA = - DAB (dcA / dz) ………………………………..(1.18)

Applying equation (1.13), dcA = A

t

pdp

c

Substituting this into equation (1.18) ,

NA =RTp

pD

t

tAB−

dz

pd A ………………………………..(1.19)

integrating,

NA = ( )21 AAAB ppRTz

D− …………………………….(1.20)

LIQUIDS:

Steady State Molecular Diffusion In Liquids At Rest And In Laminar Flow:

Case 1: Diffusion of A through non diffusing B:

Here, the total concentration varies w.r.t. position significantly. ie c ≠ constant.

As c = n / V = ρ /M,

Where ρ = density of the liquid solution and M = Molecular weight of the same,

Equation (1.10) becomes,

NA =( )

( ) 1

2

ABA

A

ABA

A

Av

AB

BA

A

xNN

N

xNN

N

Mz

D

NN

N

−+

−+

+

ρ …(1.21)

where (ρ /M)Av is average of concentrations at the start and end of diffusion path.

Case 1: Diffusion of A through non diffusing B:

NA / (NA + NB ) = 1

NA =Av

AB

BA

A

Mz

D

NN

N

+

ρ ( )MB

AA

x

xx

,

21 − …………(1.22)

Where x B,M = Log average of mole fraction of xB = ( )

−

1

2

12

lnB

B

BB

xx

xx

Case 2: Equimolar Counterdiffusion:

N= 0

NA =Av

AB

BA

A

Mz

D

NN

N

+

ρ ( )21 AA xx − …………(1.23)

Steady state molecular diffusion of fluids:

A through nondiffusing B Equimolar counterdiffusion of A & B

Gases NA =

( )MB

AAAB

p

pp

RTz

D

,

21 − NA = ( )21 AAAB ppRTz

D−

Liquids

NA =Av

AB

BA

A

Mz

D

NN

N

+

ρ ( )MB

AA

x

xx

,

21 −

NA =Av

AB

BA

A

Mz

D

NN

N

+

ρ ( )21 AA xx −

Multicomponent diffusion under steady state: For gas mixtures:

Effective diffusivity, DA,m =

( )∑

∑

=

=

−

−

n

Ai

iAAi

iA

n

Ai

iAA

NyNyD

NyN

,

1 ………(1.24)

DA,I are binary diffusivities.

If this varies linearly with diffusion path distance,

DA,m =

∑∑==

=−

n

Bi iA

i

n

Bi iA

i

A

D

y

D

y

y

,

'

,

11 …………….(1.25)

Relationship between N & C:

Average velocity = v = c

N

c

NN

cc

vcvc BA

BA

BBAA =+

=+

+

N = vc or v = N/c

NA = JA + c A v = JA + c A N/c

But c A / c = yA for gases = xA for liquids

NA = JA + NyA for gases = JA + NxA for liquids

Estimation of Diffusivity:

Diffusivity of vapour through Air:

Pseudosteady state Diffusion:

Stefan’s Equation:

Consider a volatile liquid A(eg. Acetone or CCl4 ), kept in a tube is evaporating into air

stream (B) at constant temperature and constant total pressure. The level of liquid in the

tube in the beginning ( at t=0) measured from the open end of the tube is z 0 and this level

decreases to z t at time t due to evaporation. Here, A diffuses through non diffusing B.

Hence the flux of A is

NA = ( )21 AA

BM

tAB ppp

p

RTz

D− …………………(1.26)

As the level of liquid in the tube keeps decreasing, z, which is z 0 – z t , is varying w.r.t.

time. Thus it is unsteady state diffusion. However, dz / dt is very slow making NA almost

constant w.r.t. time. Hence, it is called pseudo-steady state diffusion.

NA = dt

dz

Mdt

dz

AM

A

A

L

A

L ρρ= …………………..(1.27)

where ρ L = density of the liquid A = kg/m3 & MA = molecular weight of A = kg / kmole

A = area of inner cross section of tube, m2

.

Combining equations (1.26) & (1.27),

( )21 AA

BM

t

L

AAB ppp

pM

RTz

D

dt

dz−=

ρ

( )∫∫ −=t

AA

BM

t

L

AABz

zdtpp

p

pM

RT

Dzdz

t

021

0 ρ…………..(1.28)

Integrating and rearranging,

DAB =( )( )21

2

0

2

2 AA

t

t

BM

A

L

ppt

zz

p

p

M

RT

−

−ρ ……………(1.29)

Point 1 is at the liquid-vapor interface.

Partial pressure of A at pt.1 = vapor pressure of A at the given temperature,T = pA

Partial pressure of A at pt.2 = 0

Prediction of DAB :

Gases: Wilke-Lee Equation:

DAB = ( ) ( )ABABt

BABA

kTfrp

MMT

MM

ε2

2/34 1111249.0084.110 +

+−−

…..(1.30)

Where DAB = diffusivity, m2/s

T = absolute temperature,K

MA, MB = molecular weight of a & B respectively, kg/kmol

Pt = absolute pressure, N/m2

rAB = molecular separation at collision, nm = ( rA + rB )/2

ε AB = energy of molecular attraction = BAεε

k = Boltzmann’s constant

f (kT/ ε AB) = collision function

r = 1.18 v 1/3

ε /k = 1.21 Tb where v is the molal volume in m3/kmol of the liquid

at normal boiling point, Tb Kelvin.

Liquids :

For dilute solutions of non-electrolytes: Wilke-Chang Equation

(1.173 x 10 -18

) (φ MB)0.5

T

DAB 0 = ---------------------------------------- ………….(1.31)

µv A 0.6

Where DAB 0 = diffusivity of a in a very dilute solution in solvent B, m

2 /s

MB = molecular weight of solvent, kg/kmol

T = temperature, K

µ = solution viscosity, kg/ m.s

vA = solute molal volume at normal boiling point, m3 / kmol

= 0.0756 for water as solute

φ = association factor for solvent

= 2.26 for water as solvent

= 1.9 for methanol as solvent

= 1.5 for ethanol as solvent

= 1.0 for unassociated solvents, e.g., benzene and ethyl ether.

For concentrated solutions:

Changes in viscosity with concentrations

DA µ = ( ) ( )

+

B

Ax

BAB

x

ABAd

dDD

BA

γγ

µµlog

log100 ………………(1.32)……

where

D0 AB is the diffusivity of A at infinite dilution in B and D

0 BA is the

diffusivity of B at infinite dilution in A.

γ is the activity coefficient (obtained from vapor-liquid equilibrium)

γA = AA

tA

AA

A

px

py

px

p= where pA is the vapor pressure of A at the given temperature

Diffusion in Solids

Steady state diffusion in solids:

Fick’s I Law (for diffusion in z direction) for solids:

Valid for steady state diffusion of component A which is a fluid: JA = - DA z

CA

∂

∂

For diffusion through a flat slab of thickness z,

JA = ( )z

ccD AAA 21 − …………(1.33)

Where JA= molar flux of component A in z direction in kmol/m2.s

DA = diffusivity of A through solid in m2/s which is constant

cA1 & cA2 = concentrations of A on the opposite sides of the slab in Kmol/m3

z = distance z2 – z1 in m

(cA1 -cA2)/z = concentration gradient in kmol/m4

For other solid shapes,

W = NASav = ( )z

ccSD AAAvA 21 − …………..(1.34)

Where Sav = average cross-section for diffusion

= 2 Π (a2 –a1) …….. for radial diffusion through a solid cylinder of inner &

ln (a2/a1) outer radii a 1 & a2 and a2 –a1 = z

= 4 Π a1a2 ……. for radial diffusion through a spherical shell of inner &

outer radii a 1 & a2 and a2 –a1 = z

Conditions: (i) steady state, molecular diffusion.

(ii) no bulk flow.

(iii) diffusivity is independent of concentration.

Unsteady state diffusion in solids:

Fick’s II Law:

2

2

2

2

2

2

z

C

y

C

x

CD

C AAAAB

A

∂

∂+

∂

∂+

∂

∂=

∂

∂

θ ……………(1.35)

where θ = time, x,y & z are directions of diffusion

Conditions: (i) Unsteady state, molecular diffusion. (ii) no bulk flow.

(iii) diffusivity independent of concentreation (iv) no chemical reaction.

Diffusion from a slab with sealed edges: E = AA

AA

cc

cc

−

−

0,

,θ =

2a

Df

θ …(1.36)

where E is the fraction of A remaining unremoved in the solid at time θ.

Diffusion through:

Polymers

Crystalline solids

Porous solids

I. Diffusion through a polymers:

Two bodies of gas, say H2 at different pressuresseparated by a polumeric membrane

(eg. Polyethylene): First the gas dissolves in the solid (dissolution obeys Henry’s law,

ie c ∝ p), then diffuses from the high- to low pressure side.

D is in the order of 10-10

m2 /s.

VA = ( )z

ppsD AAAA 21 − …………(1.37)

Where VA = diffusional flux, cm3 gas (STP)/ cm

2.s

DA = diffusivity of A, cm2 /s

Ap = partial pressure of diffusing gas, cm Hg

sA = solubility coefficient or Henry’s law constant,

cm3 gas (STP) / (cm

3 solid)cm Hg

z = thickness of polymeric membrane, cm

Permeability: P = DAsA ………..(1.38)

P is in cm3 gas (STP)/cm

2.s.(cm Hg / cm)

Commercial example: Separating hydrogen from waste refinery gases in a shell and tube

device using polymeric fibre tubes of 30µm OD in a 0.4 m dia shell.

II. Diffusion through acrystalline solid:

The mechanism of diffusion:

1. Interstitial mechanism, 2. vacancy mechanism, 3.interstitialcy mechanism, 4. crowd-

ion mechanism and 5. diffusion along grain boundaries.

III. Diffusion of a gas through a porous solid:

A porous barrier or membrane separating 2 bodies of fluids:.

(i) If the absolute pressure is constant is constant across the porous solid : diffusion

(ii) If it varies, hydrodynamic flow.

(i)Total pressure constant across the porous solid:

Knudsen’s law:

If pore dia,d and gas pressure are such that the molecular mean free path,λ is relatively

large, d / λ < about 0.2 , the rate of diffusion is governed by the collisions of gas

molecules with the pore walls and follow Knudsen’s law.

In a straight circular pore of dia d & length l,

duA

NA = ------( pA1 –pA2 )

3RTl 8gc RT

where uA = mean molecular velocity of A = ------------ ……(1.39)

π MA

DK,A (pA,1 –pA,2)

NA = ------------------------

RTl d 8gc RT

Where DK,A = Knudsen diffusion coefficient = ---- ------------ …(1.40)

3 π MA

NKn = Knudsen No. = λ / d where λ = mean free path and d is average pore dia.

d / λ

d / λ Diffusivity

Knudsen diffusion < 0.2 D K,A

21

T∝

Knudsen + Molecular diffusion 0.2 to 20 DK,A & DAB

Molecular diffusion >20 DAB 2

3

T∝

Generally the pore will not have a constant diameter along the diffusion path and the true

length,l of diffusion path is also not known. Hence, l is replaced by z, the membrane

thickness and DK,A is replaced by effective Knudsen diffusivity, DK,A,eff which is

determined experimentally.

DK,A,eff is independent of pressure and varies as (T/M)1/2

.

Molecular diffusion: d / λ > 20

NA = ( )( ) 1

2

/ln

ABAA

ABA

A

tABeff

BA

A

yNNN

yNN

N

RTz

pD

NN

N

−+

−+

+ ……….(1.41)

Transition range: 0.2 < d / λ < 20

NA =

1

,,

,

2

,,

,

1

1

ln

A

effAK

effAB

BA

A

A

effAK

effAB

BA

A

tABeff

BA

A

yD

D

NN

N

yD

D

NN

N

RTz

pD

NN

N

−

+

+

−

+

+

+ …(1.42)

Knudsen diffusion: d / λ < 0.2:

NA = ( ) ( )21

,,

21

,

AA

effAK

AA

AKpp

RTz

Dpp

RTl

D−=− ……………(1.43)

(ii) Total pressure varying across porous solid:

Hydrodynamic flow of gases:

NA = ( )21, ttavt pppRTz

k− ……(1.44)

Where pt,av =( pt1 – pt2 ) /2

K = d2 gc / 32 µ l …….(1.45)

UNIT – II 1. What is lewis number? (MAY 2015)

A dimensionless number formed by dividing thermal diffusivity by mass diffusivity.

LE = SC / PR = α / D

2. What is mass transfer coefficient? (May 2015)

Mass transfer coefficient is a diffusion rate constant that relates the mass transfer rate, mass

transfer area, and concentration change as driving force: Where: kc is the mass transfer coefficient [mol/(s. · m2)/(mol/m3)], or m/s.

3. What is wet bulb depression? (Dec 2012) (May 2016) The quantity tg – tw is called wet bulb depression. Covering the thermometer with a wick and noting

down the temperature is known as wet bulb temperature.

4. What is the dimensionless number for heat transfer? (May 2016) The dimensionless group in mass transfer identical to that of prandtl number in heat transfer is

Schmidt number

5. What is Sheerwood Number? It may be defined as the ratio of total mass transfer to molecular mass transfer.

NSH = Total Mass Transfer / Molecular Mass Transfer It is the mass transfer analogy of Nusselt’s number.

SH = F*L / C*DAB

6. What are the theories of mass Transfer?

Number of theories of mass transfer proposed in mass transfer. The most accepted theory is the

kholburn’s analogy. No other theory could explain the phenomenon of mass transfer.

7. What is stage?

Any device or a combination of devices in which intimate two insoluble phases are brought into

contact,where mass transfer occurs between the phases tending to bring them to equilibrium & where the phases are then mechanically seperable. A process carried out in this manner is a single stage

operation.

8. What is stage efficiency?

It is defined as the fractional approach to equilibrium which is real stage produce or the rate of the

actual solute transfer to that if equilibrium where attained.

9. Define Absorption factor.

This can be represented by the expression

A = RS / m * ES Where, RS is the rate of flow of non diffusing solvent in phase R

ES is the rate of flow of non diffusing solvent in phase E

m equilibrium curve slope.

10. What is the expression of Schmidt number?(may14) (May 2016)

It is the ratio of molecular momentum transfer to molecular mass transfer.

NSC = µ / ℓ * DAB

11. When do the mass transfer is gas phase controlled? When m’ is small (equilibrium distribution curve is flat).Thus the major resistance is offered by 1/ ky.

Therefore the rate of mass transfer is gas phase controlled.

1 / KY = 1 / ky + m’ / kx

1 / KY = 1 / ky

12. What is the unit of mass transfer coefficient?

pressure mass transfer coefficient KY, the unit is Kg mole / m2 sec. atm

13. When do the lewis number is unity? For air- water vapour mixture the value of Lewis number is unity

14. When do the rate of mass transfer is liquid phase controlled?

When m” is very large then the rate of mass transfer is liquid phase controlled. The major resistance to mass transfer lies within the liquid.

1 / KX = 1 / kx +1 / m’’ ky

15. What is ideal stage?

An equilibrium , ideal or theoretical stage is defined as one where the effluent phases are in equilibrium,so that a longer time of contact will bring about no additional change to composition.

16. What is the relation between stripping factor and absorption factor? stripping factor is the inverse of absorption factor. It is denoted by S .

S = m * ES / RS

17. What is Peclet number?

Peclet number

It is the product of reynold’s & prandtl number. NPE = total momentum transfer / molecular heat transfer.

18. What are film coefficients? Both heat & mass transfer coefficient are called as film co-efficients..

19. What is penetration theory? The liquid- gas interface is then a mosaic of surface elements & different exposure time& since the

rate of solute penetration depends upon exposure time ,the average rate for a unit surface and must be

determined by summing the individual values.

20. What is Surface renewal theory?

This theory assumes that the surface elements to be essentially infinitely deep , the different solute

never reaches the region of constant concentration.The observed dependence KL α DAB

n

With n dependent upon circumstances , might be explained by allowing for a finite depth of the

surface elements or eddies.

UNIT II:

1. What do you understand by the term overall mass transfer coefficient? Give an account for estimating

NTU and HTU. (May 2016)

2. Give an account of the various analogies between mass, heat and momentum transfer. Explain Reynolds anology in detail (MAY 14)

3. What do you understand by JD and JH factors?

4. Write short notes on: Higbie’s and Danckwert’s theories of mass transfer. (May 2016) 5. Derive a relation for the overall mass transfer coefficient in terms of the individual film coefficients

and discuss the limiting conditions.

6. (i) Derive an expression for N to Y for the case of a dilute solution. (ii) Explain the graphical method of determination of NTU.

7. Discuss the following theories:

(i) Penetration theory

Surface renewal theory (MAY 2015) 8. (i) Explain the method of plotting the equilibrium and operating lines for a gas liquid system. Using

this plot describe how the NTU can be calculated.

(ii) Derive the relationship between individual and overall mass transfer coefficient. 9. Explain how the height of packing for an absorber can be calculated.

10. An absorption tower operating at 20OC and 1atm. pressure was used to absorb SO2 from an air

mixture into water. At one point in the equipment, the partial pressure of SO2 was 30 mm. Hg. and concentration of the contacting liquid film was 1.2148 Kmol/m3 of solution. The individual film mass

transfer coefficients at 20OC and 1 atm. were KL = 6.347 kg. moles per (hr.m2) per (Kg. mole/m3) and

KG = 1.44 Kg. moles/hr. m2. atm. The equilibrium data at 20OC are as follows:

Partial Pressure of SO2 0.5 3.2 8.5 2.6 5.9 (mm. Hg.)

Conc. of SO2 (Kmol/m3) 0.306 1.459 2.784 6.2151 10.909

(i) Evaluate the interfacial comp. CAi and PAi (ii) Find KG and KL.

(iii) What percentage of overall resistance to mass transfer lies in the gas phase?

11. In a certain apparatus used for absorption of SO2 from air by means of water, at one point in the tower

the gas contains 10% SO2, and it was in contact with a liquid containing 0.4% SO2 (density 990 kg/m3), the temp. was 50OC and the total pressure atmospheric. Overall mass transfer coefficient

based on gas phase conc. KG = 7.36 x 10-10 kg moles / m2. sec. Of the total diffusional resistance

47% was in the gas phase and 53% in the liquid phase. The equilibrium data at 50OC is as follows: Kg. SO2 / 100Kg. water: 0.2 0.3 0.5 0.7

Partial Pressure of SO2 29 46 83 119

(mm. Hg.) Calculate:

(i) The overall mass transfer coefficient based on liquid phase conc. KL.

(ii) The interfacial comp. of both the phases.

ka, ky, kc, kx and kL.