Mass Transfer Part (3)

3. MASS TRANSFER COEFFICIENT AND INTERPHASE MASS TRANSFER 3.1.

Introduction In the previous chapter, we have emphasized molecular diffusion in stagnant fluids or fluids at laminar flow. It is known that the rate of diffusion under molecular diffusion is very slow. In order to increase the fluid velocity for introducing turbulence, the fluid has to flow past a solid surface. When a fluid flows past a solid surface, three regions for mass transfer can be visualized. There is a region of laminar or thin viscous sub layer very adjacent to the surface where most of the mass transfer occurs by molecular diffusion due to which a sudden concentration drop is seen. Next a gradual change in concentration of the diffusing substance is obtained in transition region. In the third region called turbulent region, a very small variation in the concentration is observed since the eddies present which tend to make the fluid as one of more uniform concentration. The above trend of concentration is distribution with distance from the solid surface is shown in Figure.3.1

3.2.

Mass Transfer Coefficient Since turbulent flow mechanism is yet to be understood, it is better to express the turbulent diffusion in a similar manner as that of molecular diffusion. Molecular diffusion is characterized by the term DAB C/Z as in Eq. (2.14) which is modified by ‘F’, a mass transfer coefficient for binary system. Here the flux depends upon the cross sectional surface area which may vary, the diffusional path which is not specifically known and the bulk average concentration difference. Hence, flux can be written using a convective mass transfer coefficient. Flux = (coefficient) (concentration difference) Since concentration can be expressed in many ways, there are different types of equations followed: 1. For Transfer of A through stagnant B [NB = 0] For Gases: NA = kG (pA1 – pA2) = ky (yA1 – yA2) = kC (CA1 – CA2) (3.1) For Liquids: NA = kx (xA1 - xA2) = kL (CA1 – CA2)

(3.2)

where kG, ky, kC, kx and kL are individual mass transfer coefficients. 38

2. For Equimolar counter transfer [NA = − NB] For Gases: NA= kG’ (pA1 – pA2) = ky’ (yA1 – yA2) = kC’ (CA1 – CA2)

(3.3)

For Liquids: NA= kx’ (xA1 – xA2) = kL’ (CA1 – CA2)

(3.4)

Thus kC is a replacement of DAB C/Z of eq. (2.14) used for low mass transfer rates. ‘F’ can be used for high mass transfer rates and it can be related to k’s as

F=

kGpB, m. The other relations of mass transfer coefficients and flux equations are presented in Table 3.1. The mass transfer coefficient can also be correlated as a dimensionless JD factor by JD = (kC’/V) (NSC)2/3

(3.5)

where V is the mass average velocity of the fluid, and NSC is Schmidt number i.e. (µ/ρDAB) Concentration units

where ρ and µ are the density and viscosity of the mixture respectively. Laminar

CA0

Transition Turbulent CA1 CA2

0

Distance from the surface, mm

Fig. 3.1 concentration distribution flow past a solid surface 3.3.

Mass Transfer Coefficients in Laminar flow When mass transfer occurs in a fluid flowing in laminar flow, it follows the same phenomena of heat transfer by conduction in laminar flow. However, both heat and mass transfer are not always analogous since mass transfer involves multicomponent transport. Thatswhy it needs some simplications to manipulate the mathematical equations for conditions of laminar flow in many complex situations. One simplified situation is illustrated in this section.

39

3.3.1. Mass transfer from a Gas into a falling liquid film Consider the absorption of solute from a gas into a falling liquid film as in wetted wall column. This is shown in Fig 3.2. Here the laminar flow of liquid and diffusion occur in such conditions that the velocity field can be virtually unaffected by the diffusion.

Fig. 3.2 Falling liquid film The component, A from gas is slightly soluble in liquid B, and hence the viscosity of liquid is not changing appreciably. In addition to that diffusion takes place very slowly in the liquid film and A will not penetrate more into ‘B’. The penetration distance is so small when compared with film thickness. Let the concentration of A in the inlet liquid be CA0, the concentration of A at the surface of the liquid is in equilibrium with the concentration of A in the gas phase which is constant throughout at CAi and the film thickness is δ. First by solving the momentum transfer equation we obtain the velocity profile Vz(x) for the film as Vz(x) = Vmax [1 – (x/ δ)2]

(3.6)

Having seen the momentum transfer let us analyze the mass transfer. Let us make a mass balance for component ‘A’ in the elemental volume (W) (Δx ) (Δz). For steady state Rate of input = Rate of output. Hence, NAZ

WΔx ─ NAZ Z

WΔx + NAx Z + Δz

x

WΔz − NAx

WΔz = 0 (3.7) x+ Δx

40

where W is width of the film Let us divide by W. Δx Δz and let Δx and Δz → 0, we get (∂/∂x) NAx + (∂/∂z) NAz = 0

(3.8)

Now let us substitute in Eq. (3.8) the flux components in terms of diffusional and convectional fluxes of A . The one-directional molar fluxes are defined by, NAx = − DAB (∂CA/ ∂x) + xA (NA,x +NB,x)

(3.9)

NAz = − DAB (∂CA/ ∂z) + xA (NA,z +NB,z)

(3.10)

In the above Eq. (3.9), A is transported in X- direction by diffusion and not by convective transport because of very slight solubility of A in B. Similarly in Eq. (3.10), A moves in the Z – direction because of the flow of the film, and thereby diffusive contribution is negligible. Hence, Eqs. (3.9) and (3.10) become NAx = − DAB (∂CA/ ∂x)

(3.11)

NAz = xA (NA,z + NB,z ) = CA VZ (x)

(3.12)

Substituting Eqs. (3.11) and (3.12) into (3.8) we get, VZ (∂CA/ ∂z) = ∂/∂z [VZ. CA] = DAB (∂2CA/ ∂x2)

(3.13)

In this case, solute has penetrated only a very short distance and it gives the impression that it is carried along with the film with a velocity equal to V max. Hence, at x =0, VZ is replaced by Vmax in Eq.(3.13). Vmax (∂CA/ ∂z) = DAB (∂2CA/ ∂x2)

(3.14)

The Eq. (3.14) has been solved by Laplace transform using the boundary conditions, B.C.1;

At Z = 0,

CA = 0

B.C.2;

At x = 0,

CA = CA0

B.C.3;

At x = ∞,

CA = 0

Since A does not penetrate very far, the distance, δ becomes infinite in view of A. The solution of Eq. (3.14) is given as (CA/ CA0) = erfc[x/ √ (4 DAB Z/Vmax)]=1-erf[x/√(4DAB Z/Vmax)] (3.15) where erf is the error function . The flux at the surface, x = 0 as a function of position Z is given by (N A,x)

x=0

= - DAB (∂CA/∂x) x=0 = CA0√(DAB Vmax)/ πZ

(3.16)

41

The rate of transfer of A to the fluid over the length Z = L is given by L

NA [ L.W] = W ∫ (NA,x )

x=0

dZ

(3.17)

0 L

= W ∫ CA0[√(DAB Vmax)/ πZ ]dZ

(3.18)

0

NA [ L.W] = L.W CA0 √ 4(DAB Vmax)/ πL NA = CA0 √ 4(DAB Vmax)/ πL

(3.19) (3.20)

This shows that the liquid mass transfer coefficient is proportional to D AB0.5 for short contact times. 3.4.

Mass Transfer Theories. Various theories have been used as models for explaining the turbulent mass transfer. These models can be used for predicting the mass transfer coefficients and they can be correlated with experimental data to obtain the design parameters of process equipments.

3.4.1. Film theory Consider turbulent flow of liquid over a solid surface and a simultaneous mass transfer-taking place. The film theory postulates that there is a stagnant film of thickness, Zf adjacent to the interface, where the concentration difference is attributed to molecular diffusion and is shown in Fig.3.3. As the molecular diffusion is occurring only in Zf, the flux equation can be written as NA = kC (CA1 – CA2) = (DAB/Zf) (CA1 – CA2)

(3.21)

where (CA1 – CA2) is the concentration difference Hence, kC = (DAB/Zf), the mass transfer coefficient is proportional to DAB1.0. However, the JD factor is given by, JD = (kC/V) (NSC)2/3 = (kC/V) (µ/ρDAB)2/3

(3.22)

Hence the film theory deviates from the actual turbulent mass transfer.

42

Fig. 3.3 Concentration distribution in Film theory 3.4.2.

Penetration theory This theory explains the mass transfer at fluid surface and was proposed by Higbie. In many situations, the time of exposure for mass transfer is too short and hence, there may not be sufficient time for the steady state concentration gradient of film theory to develop. This theory has been described by the following Fig. 3.4. An eddy ‘b’ rising from the turbulent liquid is exposed for a short time, θ at the interface for absorption. In this situation the exposure time is assumed to be constant for all the eddies or particles of liquid. Initially the eddy concentration is CA0 and when it comes to the surface, the interfacial concentration is CAi. Since the exposure time is less, molecules of solute from gas never reach the depth Zb, which is nothing but the thickness of eddy. The liquid particle is subjected to unsteady state diffusion and hence Fick’s II law is applicable.(i.e.) ∂CA/∂θ = DAB[∂2CA/∂Z2 ]

(3.23)

From the solute point of view, the depth Zb is considered to be infinite. The boundary conditions applied are as follows: CA =CA0

at θ = 0

CA = CAi at Z = 0 CA = CA0

for all Z

θ>0

at Z = ∞

for all θ

43

Fig. 3.4. Higbie’s theory: By solving the above Eq. (3.23), the average flux can be obtained as described in falling film Hence, NA, av = 2(CAi – CA0) kL, av =

DAB / πθ

DAB / πθ

(3.24) (3.25)

Thus in penetration theory kL is proportional to DAB0.5. However, the exponent on DAB varies from zero to 0.8 or 0.9. 3.4.3

Surface - Renewal theory: In reality the time of exposure of all eddies as proposed in Penetration theory is not constant. Hence, Danckwerts modified the penetration theory to account for varying lengths of time of exposure. If ‘S’ is the fractional rate of replacement of elements, Then, NA, av = (CAi – CA0) DAB S

(3.26)

Hence, kL, av is proportional to DAB0.5 in this theory. 3.4.4

Combination Film – surface renewal theory: Film theory is meant for steady state diffusion where kL α DAB and in surface renewal theory, kL α DAB0.5. So kL is proportional to DABn with ‘n’ dependent upon circumstances. In this theory Dobbins replaced the third boundary condition of Eq. (3.23) by CA = CA0 at Z = Zb where Zb is of finite depth. Finally he obtained kL, av =

3.4.5

DAB S coth √SZb2/DAB

(3.27)

Surface – Stretch theory: Lightfoot and his coworkers explained this theory. They found that the mass transfer at the interface varies with time periodically. When mass transfer is proceeding for a particular system the central portion of the drop is thoroughly

44

turbulent and resistance to mass transfer resides in a surface layer with varying thickness and the drop is elongated as shown in Fig. 3.5. According to this theory, kL, av = [(A/Ar) DAB / ∏ θ r ] θ / θr

[√

∫

(A/Ar)2dθ]

(3.28)

0

where A is time dependent interface surface, Ar is reference value of A, defined for every situation and θr is constant with dimensions of time or drop formation time.

Surface Layer

Fig. 3.5: Surface – stretch theory 3.5.

Analogies As flow past solid surface occurs, at a uniform velocity U0, the curve ABCD separates the region of velocity U0 from a region of lower velocity. The curve ABCD which separates these two regions is called boundary layer. In the same way as mass transfer takes place, a similar concentration boundary layer also occurs. In understanding the analogies between momentum and mass transfer it is worth having a review of universal velocity distribution which is shown in Fig. 3.6

45

25 Viscose sublayer

20

Buffer layer

Turbulent core

u+=2.5lny++ 5.5

15 u+=5lny+- 3.05

u

+

10 5

0

u+=y+

1

3 5

10

30 50 100 300 500 + y Fig. 3.6 (a) : Universal velocity profile

1000

46

BOUNDARY - LAYER THICKNESS Laminar flow in boundary layer Turbulent flow in boundary layer

Buffer layer Viscous sub layer

DISTANCE FROM LEADING EDGE, x Fig. 3.6 (b) : Turbulent boundary layer The similarity between the transfer processes of momentum, heat and mass lead to the possibility of determining mass transfer characteristics for different situations from the knowledge of other two processes. Let us now deal with the various analogies and the assumptions involved in each. 3.5.1.

Reynolds Analogy In this analogy, the assumptions considered are:

(i)

Only turbulent core is present.

(ii)

Velocity, temperature and concentration profiles are perfectly matching

(iii)

All diffusivities are same. Hence, (α) = (DAB) = (μ/ρ)

(3.29)

47

When all the three diffusivities are equal, then Prandtl Number (NPr) = Schmidt number (NSc) = 1. The basic flux equations of heat, mass and momentum can be written as follows: q = h (ti – t0) = − α ∂/∂Z (ρCPt)

(3.30)

NA = kC (CAi – CA0) = - DAB ∂CA/∂Z

(3.31)

τi =( μ/gc )∂U/∂Z

(3.32)

Let us consider heat and momentum transfer and from Eq. (3.30) h (ti – t0) = −α ∂/∂Z (ρCPt) = − (K/ρCP) ( ρCP) dt/dz since α = (K/ρCP) (3.33) h (ti – t0) = - K(d/dZ) (t – ti)

(3.34)

∴ h/K = (d/dZ) [(t – ti)/ (t0 – ti)]

(3.35)

As per assumption (ii), velocity and temperature profiles match and hence, (d/dZ) (Ux/U0) = (d/dZ) [(t – ti)/ (t0 – ti)]

(3.36)

Multiplying by ‘CP μ’ on both sides of Eq. (3.36), we get [(CP μ/U0) (dUx/dZ)] = K (d/dZ) [(t – ti)/ (t0 – ti)]

(3.37)

Since K = CP μ by assumption (iii), and rearranging gives [(CP μ/KU0) (dUx/dZ)] = (d/dZ) [(t – ti)/ (t0 – ti)]

(3.38)

Combining Eqs. (3.35) and (3.38), we get (CP μ/KU0) (dUx/dZ)] = h/K

(3.39)

Therefore, μ (dUx/dZ) = h U0/ CP = (f/2) ρU02

(3.40)

⇒ (f/2) = h/ (ρ CP U0)

(3.41)

Similarly, by considering mass and momentum transfer we get (f/2) = kc/U0

(3.42)

Hence, the Reynolds Analogy equation is (f/2) = h/ (ρ CP U0) = kc/U0

(3.43)

3.5.2. Chilton – Colburn Analogy Assumptions involved are: (i) Only turbulent core is present (ii)

Velocity, temperature and concentration profiles are same.

(iii)

NPr and NSC are not equal to unity. In this Analogy, the equation obtained is (f/2) = (kc/U0) (Sc) 2/3 = [h/ (ρ CP U0)] (Pr) 2/3

(3.44)

48

3.5.3.

Taylor – Prandtl Analogy Assumptions involved are: i) Assumes the presence of turbulent core and laminar sublayer ii) NPr and NSc are not equal to unity. In this Analogy, the equation obtained is kc/U0 = h/(ρCPU0)={(f/2)/[1 +5 f/2 (SC – 1)]}={(f/2)/[1 +5

f / 2 (Pr – 1)]}(3.45)

3.5.4. Von – Karman Analogy Assumptions involved are: i)Assumes the presence of turbulent core, laminar sublayer and buffer layers ii)Universal velocity profile equations are applicable. iii) NPr and NSc are not equal to unity. In this Analogy, the equation obtained is kc/U0 = {(f/2)/ [1 +5 f/2 [(SC – 1) + ln (5SC + 1)/6)]} = {(f/2)/ [1 +5 f/2 [(Pr – 1) + ln (5Pr + 1)/6)]}

(3.46)

Eqs. (3.44), (3.45) and (3.46) lead to Reynolds Analogy when SC = Pr = 1. 3.6.

Interphase mass transfer:

3.6.1. Equilibrium

y

x To generalize the equilibrium Fig.: 3.7 Equilibrium curve characteristics, consider that an amount of solute from a gaseous mixture is dissolved in solvent. After sufficient time, the system will attain equilibrium with respect to a particular temperature and pressure. The concentration of solute in both gas and liquid phase may not be equal but the chemical potential of solute will be equal at equilibrium. At the

49

same temperature and pressure, if some more amount of solute is introduced, then once again a new equilibrium will be attained in the same system. The equilibrium curve can be represented for any system as shown in Fig. 3.7. The net rate of diffusion is zero at equilibrium. Conventionally, the concentration of solute in liquid phase is expressed by mole fraction, x and the concentration of solute in gas phase is by mole fraction y. 3.6.2. Two phase mass transfer Generally the two phase systems occur in most

of

the

transfer

mass

operations.

Suppose the two phases are each

immiscible other,

interface Fig.3.8 Two phase mass transfer

then is

with an seen

between the two 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 diffusion through the resistances and is shown in Fig. 3.8. In the Fig.:3.8 yAG is the concentration of A in bulk gas phase, y Ai is the concentration at interface, xAL is concentration of A in bulk liquid phase, xAi is concentration at interface. The bulk phase concentrations, yAG and xAL are certainly not at equilibrium, which enables diffusion to occur. At the interface, there is no resistance to transfer of solute and the concentrations yAi and xAi are in equilibrium and they are related by the equilibrium distribution relation as yAi = f (xAi) (3.47) The concentration driving forces can be shown graphically as in 50 x Fig.: 3.9 concentration driving force

Fig.3.9. If we consider a steady state mass transfer, the rate at which molecules reach the interface will be the same rate at which the molecules are transferred to the liquid phase. Since interface has no resistance, the flux for each phase can be expressed in terms of mass transfer coefficient. NA = ky (yAG – yAi) = kx (xAi – xAL)

(3.48)

Where ky and kx are local gas and liquid mass transfer coefficients. (i.e.) (yAG – yAi)/ (xAL – xAi) = − (kx/ ky)

(3.49)

Hence, the interface compositions can be determined if kx, ky, yAG and xAL values are known. 3.6.3. Overall Mass Transfer coefficient Experimentally determining the rate of mass transfer is very difficult since it is not possible to evaluate the interface compositions. However, bulk concentrations are easily measured and measuring xAL is as good as measuring yA* because both have the same chemical potential. Similarly yAG is as good as measuring xA*.The concentration driving forces can be shown as in Fig. 3.10. P

yAG

D Slope m''

(-kx/ky) yAi

M

y Slope m’ yA*

C xAL

xAi

xA*

x Fig.3.10. Concentration Driving Force The flux can be written in terms of overall mass transfer coefficient for each phase. NA = Ky (yAG – yA*)

(3.50)

where Ky is overall mass transfer coefficient 51

From the geometry of the Fig. 3.10 (yAG – yA*) = (yAG – yAi) + (yAi – yA*) = (yAG – yAi) + m’ (xAi – xAL)

(3.51)

where m’ is the slope of the chord CM in Fig. 3.10. Substituting for the concentration differences as given by Eqs. (3.48) and (3.50) (NA/Ky) = [(NA/ky) + (m’NA/kx)]

(3.52)

1 1 m’ (i.e.) --- = ---- + ---Ky ky kx Similarly, for the liquid side

(3.53)

NA = Kx (xA* – xAL)

(3.54)

On simplification, we get 1 1 1 --- = ------- + ---(3.55) Kx m”ky kx Where m” is the slope of the chord MD in Fig. 3.10. The above two Eqs. (3.53) and (3.55) show the relationship between the individual and overall mass transfer coefficients. These equations also lead to the following relationships between the mass transfer resistances. Resistance in gas phase/Total resistance = (1/ky)/ (1/Ky)

(3.58)

Resistance in liquid phase/Total resistance = (1/kx)/ (1/Kx)

(3.59)

Assuming that kx and ky are same and if m’ is small so that solute A is highly soluble in liquid (i.e. equilibrium curve will be flat), the term m’/kx will be negligible when compares to 1/ky. Hence, 1/Ky ≈ 1/ky

(3.60)

This condition says that overall resistance lies only in the gas phase, conversely when m” is very large, then solute A is relatively insoluble in liquid. Under this condition, the term (1/m”ky) will be negligible compared to 1/kx. Then 1/Kx ≈ 1/kx

(3.61)

In this case the entire rate of mass transfer is controlled by liquid phase. For cases where kx and ky are not nearly equal, then it will be the relative size of the ratio (kx/ky) and of m’ or m” which will determine the location of the controlling mass transfer resistance. 3.7

Types of operations

52

The mass transfer operations take place both on batch and continuous basis. However, due to various demands in an industry most of the operations are carried out on continuous basis. These continuous operations are carried on cocurrent and countercurrent basis. In these operations the concentration of each phase changes with position whereas in batch process the concentration changes with time. 3.7.1

Co-current process Schematic diagram for a co-current process is shown in Fig. 3.11.

E1, ES, Y1, y1

E, ES, Y, y ② II ① R, RS, X, x Fig. 3.11 Co-current process

R1, RS, X1, x1

E2, ES, Y2, y2 R2, RS, X2, x2

Where E1, E2 are mass or molar flow rates of ‘E’ stream at ① and ② position respectively, R1, R2 are mass or molar flow rates of ‘R’ stream at ① and ② position respectively, ES, RS are solute free flow rates of streams, x1, x2,y1,y2 are concentration of solute in mass or mole fraction of streams at ① and ② position respectively and X1,X2,Y1,Y2 are mass or mole ratio of solute in streams at ①and ② position respectively Making a component balance for solute, we get R1x1 + E1y1 = R2x2 + E2y2

(3.62)

RSX1 + ESY1 = RSX2 + ESY2

(3.63)

(i.e) RS (X1 – X2) = ES (Y2 – Y1)

(3.64)

(- RS/ES) = (Y2 – Y1)/(X2 – X1)

(3.65)

This indicates a line passing through the points (X 1,Y1) and (X2,Y2) which is called as operating line in the X vs. Y plot. The operating line also indicates the material balance in the operation. Also,

RSX1 + ESY1 = RSX + ESY

(3.66)

RS (X1 – X) = ES (Y – Y1)

(3.67)

53

Eq.Curve

Y2*

Ye Y2

(Transfer from R to E)

Y1

X2* X2 Xe

X1

Fig. : 3.12 Equilibrium curve and operating line for a co-current process. This represents the general equation of operating line in a co-current process. Graphically the operation can be represented for transfer from R to E as shown in Fig.3.12. 3.7.2

Counter-current process Schematic diagram for a counter-current process is shown in Fig.3.13. E1, ES, Y1, y1

E, ES, Y, y

R1, RS, X1, x1

①

② II R, RS,X,x

E2, ES, Y2, y2 R2, RS, X2, x2

Fig. :3.13 Counter current process Making a component balance for solute yields E2y2 + R1x1 = E1y1 + R2x2

(3.68)

ESY2 + RS X1 = ESY1 + RS X2

(3.69)

RS ( X1 – X2) = ES (Y1 – Y2)

(3.70)

(i.e) RS/ES = (Y1 – Y2)/ (X1 – X2)

(3.71)

This represents the equation of a line passing through the coordinates (X 1,Y1) and (X2, Y2) with a slope of RS/ES in a plot of X vs. Y. Similarly another balance gives, ESY + RSX1 = ESY1 + RS X

(3.72)

54

(i.e) RS/ES = (Y1 – Y)/(X1 – X)

(3.73)

This equation is a generalized equation representing the operating line in a countercurrent process. The following Fig.3.14 shows the operating line and equilibrium in a counter current process. (Operating lineTransfer fromE to R) Equilibrium Curve Y (Operating line - Transfer from R to E)

X Fig : 3.14 Equilibrium curve and operating line for a counter current process The advantage of the counter current over the co-current is the higher driving force, which results in a smaller equipment for a specified transfer condition or lesser flow rates for a given equipment. 3.7.3 Stages A stage is defined as any device or combination of devices in which two insoluble phases are brought into intimate contact, where mass transfer occurs between phases leading them to equilibrium and subsequently the phases are separated. A process carried out in this manner is a single stage process. An ideal theoretical or equilibrium stage is one in which the leaving streams are in equilibrium. However, in reality there is a shortfall in reaching the equilibrium and the actual stages needed to effect a specified separation is more . 3.7.4

Stage efficiency It is defined as the fractional approach to equilibrium, which a real stage produces . Multiphase stage efficiency: It is defined as the fractional approach of one leaving stream to equilibrium with the actual concentration in the other leaving stream

(i.e) E ME = (Y2 - Y 1) / (Y2 – Y1)

and

E MR = (X1 - X2) / (X1 – X2)

(3.74)

55

3.7.5. Cascade It is one which has a group of interconnected stages, in which the streams from one stage flows to the other .We have both cross flow and counter flow cascades. A typical cross current cascade is shown in Fig. 3.15. E1Y1

R0, X0

1

R1, X1

ES1 Y0

E2Y2

E3Y3

2

3 R2, X2

ES2 Y0

E4Y4

R3, X3

ES3 Y0

4

R4, X4

ES4 Y0

Fig. 3.15 :Cross current cascade A typical countercurrent cascade is shown in Fig. 3.16. E2,Y2 E1Y1 R0, X0

R1, X1

E3,Y3 R2, X2

E4,Y4 R3, X3

E5,Y5 R4, X4

Fig. 3.16 :Counter current cascade The number of stages ‘N’, required for a countercurrent cascade can be estimated analytically for cases where both the equilibrium relationship and operating line are linear. If ‘m’ is the slope of the equilibrium curve and ‘A’ (RS/mES) the absorption factor For absorption, (transfer from E to R) For A ≠ 1 (YN+1 – Y1)/ (YN+1 – mX0) = (AN+1 – A)/ (AN+1 – 1)

(3.75)

For A = 1 N = (YN+1 – Y1)/(Y1 – mX0) = (Yin –Yout)/ (Yout – mXin)

(3.76)

For desorption, (transfer from R to E) For A ≠ 1 (X0 – XN)/ [X0 – (YN+1/m)] = [(1/A)N+1 – (1/A)]/[(1/A)N+1 – 1]

(3.77) 56

For A = 1 N = (X0 – XN)/[XN – (YN+1/m) = (Xin – Xout)/[Xout – (Yin/m)]

(3.78)

The above four equations are called Kremser –Brown- Souders equation. Worked Examples:

1) Calculate the rate of sublimation from a cylinder of naphthalene 0.075 m i.d. by 0.6 m long into a stream of pure air flowing at a velocity of 6 m/s at 1 atm. and 100ºC. The vapour pressure of naphthalene at 100ºC and 1 atm. may be taken as 10 mm Hg and the diffusivity of naphthalene in CO2 as 5.15 x 10- 6 m2/s. Density and viscosity of air are: 0.946 kg/m3and 0.021 cp respectively at operating condition. Cf = 0.023 (Re)- 0.2. Use any two analogies. V = 6 m/s, P = 1 atm, T = 373 K, pA = 10 mm Hg, Dnapth,CO2 = 5.15 × 10 –6 m2/s, ρair = 0.946 kg/m3, μair = 0.021 Cp, Cf/2 = 0.023(Re) –0.2 NRe = DVρ = 0.075 ×6 × 0.946 = 20271.43 0.021 × 10 –3

μ

Cf/2 = 3.1648 × 10 – 3 NRe =

μ

= 0.012 × 10 –3_______ = 2.463

ρDAB

0.946 × 5.15 × 10 – 6

kc = f × u0 2(NSc) 2/3 kc = 3.1648 × 10 – 3 × 6 = 0.0104 m/s 1.8238 NA = kc (CA1 – CA2) = kc (pA1 – pA2) RT = 0.0104 × 10 5 [(10/760) – 0] 8314 × 373 NA = 4.417 × 10 – 6 kgmole/m2s Rate of sublimation = NA × 2∏rl = 4.417 × 10 –6 × 2 ×∏ × (0.075/2) × 6 57

= 6.24 × 10 –7 k mole/s.

2)

A one-meter square thin plate of solid naphthalene, is oriented parallel to a

stream of air flowing at 30 m/s. The air is at 300°K and 1 atm pressure. The plate is also at 300°K. Determine the rate of sublimation from the plate. The diffusivity of naphthalene in air at 300°K and 1 atm is 5.9 x 10- 4 m2/s. Vapour pressure of naphthalene at 300°K is 0.2 mm Hg. Ut = 30 m/s, T = 300 K, Pt = 1 atm, DAB = 5.9 × 10 – 4 m2/s, pA = 0.2 mm Hg ρair = 1.15 × 10 – 3 g/cc, μair = 0.0185 Cp, D = 1 m NSc =

μ

=

ρDAB

0.0185 × 10 –3

= 0.0273 ≠ 1

1.15 ×10 – 3 ×103 × 5.9 × 10 – 4

So we will use Chilton analogy NRe = DVρ = 1.15 ×10 - 3 × 10 3 × 1 × 30 = 1864864.865 μ

0.0185 × 10 –3

So flow is turbulent, so Chilton colburn analogy can be used. f/2 = Kc/U0 (NSC) – 0.25 f = 0.072 × (NRe) – 0.25 f = 1.9484 × 10 – 3 kc =

f × u0 2(NSc) 2/3

kc = 1.9484 × 10 – 3 ×30 = 0.3226 m/s 2 (0.0273) 2/3 NA = kc (CA1 – CA2) = kc (pA1 – pA2)/RT = 0.3226 × 10 5 [(0.2/760) – 0] 8314 × 300 NA = 3.4037 × 10 – 6 k mole/m2s

58

3)

In a wetted wall column Carbon dioxide is being absorbed from air by water flowing at 2 atmospheres pressure and 25 C. The mass transfer coefficient k’y has been estimated to be 6.78 × 10-5 k mol/m2 s mole fraction. Calculate the rate of absorption if the partial pressure of carbon dioxide at the interface is 0.2 atmospheres and the air is pure. Also determine ky and kg. pA1 = 0.2 atm, yA1 = 0.1, yB1 =0.9 pA2 = 0, yA2 =0.0 and yB2 =0.0 ( yB)ln = (yB2 - yB1) / ln (yB2/ yB1) = 0.95 Also, ky (yB)lm = ky’ = kG(yB) lm P Hence, ky = ky’/(yB)lm = 6.78 × 10-5 /0.95 = 7.138 × 10-5 k mol/m2 s mole fraction kG = ky/P = 3.569 × 10-5 k mol/m2 s atm NA = ky (yA1-yA2) = 7.138 × 10-5 (0.1-0.0) = 7.138 × 10-6 k mol/m2 s NA = kG (pA1-pA2) = 3.569 × 10-5 × (0.2-0.0) = 7.138 × 10-6 k mol/m2 s

4) Sulfurdioxide is absorbed from air into water in a packed absorption tower. At a certain location in the tower, the mass transfer flux is 0.027 k mol SO2/m2.h. and the liquid phase mass transfer coefficient are 0.0025 and 0.0003 respectively at the two phase interface and in the bulk liquid. If the diffusivity of SO 2 in water is 1.7 × 10-9 m2/s, determine the mass transfer coefficient, kc and film thickness. Mass flux of SO2 = 0.027 × 1000/(3600 × 100 ×100) = 7.5 × 10-7 g moles/ cm2 s C = Density/ Molecular weight = 1/18.02 = 5.55 g moles / cm3 NA = kc (CA1- CA2) = DAB × (CA1- CA2) / δ Therefore, kc = DAB / δ = NA/ C (xA1-xA2) = 7.5 × 10-7/ [(5.55 × 10-2) × (0.0025 - 0.0003) ] = 0.00614 cm/s δ = DAB / kc = 1.7 ×10-5/0.00614 = 0.0028 cm

59

5) In an experimental study of absorption of Ammonia by water in a wetted wall column the overall gas phase mass transfer coefficient , Kg was estimated as 2.72 × 10-4 k moles /m2 s atm. At one point in the column the gas contained 10 mole % ammonia and the liquid phase concentration was 6.42 × 10-2 k moles NH3/m3 of solution. Temperature is 293 K and the total pressure is 1 atm. 85% of the resistance to mass transfer lies in gas phase. If Henry’s law constant is 9.35 × 10-3 atm. m3 / k mole, calculate the individual film coefficient and the interfacial composition. Kg = 2.72 × 10-4 k moles /m2 s atm. Resistance 1/ Kg = 1/ 2.72 × 10-4 = 3676.47 m2 s atm./ k moles 1/kg = 0.85 × 1/ Kg = 3125 m2 s atm./ k moles We know that, 1/ Kg = 1/ kg + m / kl m = 9.35 × 10-3 atm . m3 / k mole Hence, kl = 1.695 × 10-5 m/s We shall now calculate the interfacial concentration from the following relation: NA = Kg(pAg – pA*) = kg (pAg – pAi) = kl (CAi- CAL) y AG = 0.1 CAL

=

6.42 × 10-2 k moles NH3 /m3 of solution.

pAg = y Ag × Pt = 0.1 × 1.0 = 0.1 atm. CAL = 6.42 × 10-2 k moles NH3/m3 of solution. pAi = m CAi NA = kg (pAg – pAi)

= kl (CAi- CAL)

kg / kl = (CAi- CAL ) / (pAg – pAi)

60

kg / kl = (CAi- CAL ) / (pAg – m CAi) 18.88 = (CAi – 6.42 × 10-2) / (0.1 - 9.35 × 10-3 CAi ) On solving, CAi = 1.6593 k mole/m3 pAi = 0.0155 atm. 6)

At 293 K the solubility of ammonia in water is given by p = 0.00127 C, where, p is in atmosphere and C is in k mole/m3. A mixture of 15% Ammonia and 85% air by volume at 1 atm is in contact with an aqueous solution containing 0.147 gmole /lit. The air velocity is such that kg/kl = 0.9. Find the concentration of ammonia and partial pressure at interface. We have, p = 0.3672 C where p is in atmosphere and C is in k mole/m3. NA = kg (pAg – pAi) = kl (CAi- CAL) where pAi and CAi indicate the interfacial pressure and composition and pAg and CAL indicate the bulk phase compositions. pAg = 1 × 0.15 = 0.15 atm. CAL = 0.147 g mole/lit = 0.147 k mole/m3 kg / kl = (CAi- CAL ) / (pAg – pAi) 0.9 = (CAi- 0.147 ) / (0.15 – 0.3672 CAi) Solving for CAi we get, CAi = 0.212 k mole/m3 pAi = 0.078 atm. 7) Pure gas is absorbed in a laminar liquid jet. The volumetric flow rate of the liquid was 4 cc/s and the diameter and length of the jet were 1 mm and 3mm respectively. The rate of absorption of A at atmospheric pressure was 0.12 cc/s at 303 K. The solubility of gas at 303 K is 0.0001 g mole/cc. atm. Estimate the diffusivity of gas. If the diameter of the jet is reduced to 0.9 mm, under otherwise the same conditions how would it affect the rate of evaporation. Assume the validity of Higbie’s penetration theory. We know, NA = kL A (CA*- CA) 61

(kL) Av = 2 (DAB/πt) 0.5 = 0.12 × 1 × 273/(22414 × 303) = 0.482 × 10-5 A = πDL = π (0.1) (3) = 0.942 cm2 CA* = 0.0001 g mol;e/cc.atm CA = 0 kL = NA/ [A (CA*- CA)] = 0.482 × 10-5 /0.942 × 10-4 = 0.051 cm/s Time of contact ‘t’ = Bubble length/linear velocity Linear velocity = Q/A = 4 / [π × (0.1) × (0.1) / 4] =509 cm/s t= 3/509 = 0.006 s (kL) Av = 2 (DAB/ πt)0.5 0.051 = 2 (DAB/ π×0.006)0.5 DAB

= 1.23 × 10-5 cm2/s

Revised diameter = 0.09 cm Area = = πDL = π (0.09) (3) = 0.848 cm2 Velocity = Q/A = 4 / [π × (0.09) × (0.09) / 4] = 628.8 cm/s Time of contact ‘t’ = Bubble length/linear velocity = 3/628.8 = 0.00477 cm/s (kL) Av = 2 (DAB/ πt)0.5 = 2 (1.23 × 10-5 / π × 0.00477)0.5 = 0.0573 cm/s. NA = kL A (CA*- CA) = 0.0573 × 0. 848 × (0.0001-0) = 4.86 × 10-6 g mole/s atm.

62

8) In an apparatus for the absorption of SO2 in water at one point in the column the concentration of SO2 in gas phase was 10% SO2 by volume and was in contact with a liquid containing 0.4% SO2 by weight . Pressure and temperature are 1 atm and 323 K respectively. The overall gas phase mass transfer coefficient is 7.36 × 10-10 k mole/m2. s. (N/m2). Of the total resistance 45% lies in gas phase and 55% in the liquid phase.. Equilibrium data: kg SO2/100 kg water Partial pressure of SO2, mm Hg. (i) (ii)

0.2 29

0.3 46

0.5 83

0.7 119

Estimate the film coefficients and overall mass transfer coefficient based on liquid phase. Estimate the molar flux based on film coefficients and overall transfer coefficients.

kg SO2/100 kg water Partial pressure of SO2, mm Hg. x(mole fraction of SO2 in liquid phase)×104 y(mole fraction of SO2 in gas phase) × 102

0.2 29 5.63 3.82

0.3 46 8.46 6.05

0.5 83 14.11 10.92

0.7 119 19.79 15.66

KG = 7.36 × 10-10 k mole/m2. s. (N/m2). Ky = KG P (Based on Eq. 3.3) =7.36 × 10-10 × 1.013×105 = 7.456 × 10-5 k mole/m2. s. (mole fraction). Resistance in gas phase (1/ky) is 0.45 of total resistance Resistance in liquid phase is 0.55 of total resistance We know that, 1/ Ky = 1/ ky + m’/ kx Resistance in gas phase = 0.45 × ( 1/7.456 × 10-5 ) Therefore, ky

= 6035.4 m2. s. (mole fraction)./ k mole = 1.657 × 10-4 k mole/m2. s. (mole fraction).

Resistance in liquid phase ( m’/ kx) = 0.55 × ( 1/7.456 × 10-5 ) =7376.6 m2. s. (mole fraction)./ k mole The equilibrium relationship indicates a linear behaviour in the range of ‘x’ from 0.0008 to 0.0012 and the value of slope of the curve line ( m’) is 86.45.

63

Therefore, kx

=

0.0117 k mole/m2. s. (mole fraction).

y A,G = 0.1 The liquid phase composition is 0.4 wt% of SO2 x A,L = ( 0.4/64) / (99.6/18 +0.4/64) = 0.001128 Slope of the line (to determine the interfacial compositions) [- kx /ky ]

= -70.61

It is also clear from the graph that the slope , m” is same as m’ in the range under consideration. Hence, m’=m” = 86.45 1/ Kx = 1/ kx + 1/m” ky = 1/0.0117 + 1/(86.45 ×1.657 × 10-4) Kx =6.44× 10-3 k mole/m2. s. (mole fraction). yA* = 0.083, xA* = 0.00132, yA,i = 0.0925 and xA,i = 0.00123 Flux based on Overall coefficient: Flux based on gas phase

= Ky(yA,G – yA*) = 7.456 × 10-5 (0.1 – 0.083) = 1.268 × 10-6

Flux based on liquid phase

k mole/m2. s.

= Kx(xA* – x A,L) = 6.44× 10-3 (0.00132-0.001128) =1.236 × 10-6 k mole/m2. s.

Flux based on film coefficient: Flux based on gas phase

= ky(yAG – yA,i) = 1.657 × 10-4 (0.1-0.0925) = 1.243 × 10-6 k mole/m2. s.

Flux based on liquid phase

= kx(xA,i – x A,L) =0.0117 (0.00123-0.001128) =1.193 × 10-6 k mole/m2. s.

64

Exercise : 1) Air at 25o C and 50 % relative humidity flows over water surface measuring 12m × 6 m at a velocity of 2m/s. Determine the water loss per day considering flow direction 65

is along the 12 m side. Diffusivity of water in air is 0.26 × 10-4 m2/s . Sc = 0.6 and Kinematic viscosity is 15.7 × 10-6 m2/s.

(Ans: 361.84 kgs/day)

2) The absorption of solute A from a mixture is done in a wetted wall column by a solvent at 1 atm and 25oC. The value of mass transfer coefficient is 9.0 × 10-4 m/s. At a point, the mole fraction of A in the liquid- gas interface is 2.0 × 10-5 in the liquid phase. Partial pressure of A in the gas phase is 0.08 atm. Henry’s law relation is pA = (600) xA in atm. Calculate the rate of absorption of A.

(Ans : - 2.5 × 10-6 k mole/m2 s)

3) Pure water at 27oC is flowing at a velocity of 3.5 cm/s through a tube coated with benzoic acid of 6 mm in diameter. The length of the tube is 1.25 m. Calculate the concentration of benzoic acid at the outlet. Take µ = 0.871 cP ; ρ = 1 g/cc DAB = 1.3 × 10-5 cm2/s

66