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T.P Presentation (Article 18.6)

This document discusses diffusion of a solid material A into a falling liquid film of material B. It presents equations to model the diffusion process assuming laminar flow, a parabolic velocity profile for the liquid, and that A is slightly soluble in B. The key equation derived is that the concentration of A, CA, at any point in the film depends on the vertical coordinate z, the distance from the wall y, the saturation concentration CA0, and involves the gamma function.

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Sohail Aziz Ahmad Malik
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87 views20 pages

T.P Presentation (Article 18.6)

This document discusses diffusion of a solid material A into a falling liquid film of material B. It presents equations to model the diffusion process assuming laminar flow, a parabolic velocity profile for the liquid, and that A is slightly soluble in B. The key equation derived is that the concentration of A, CA, at any point in the film depends on the vertical coordinate z, the distance from the wall y, the saturation concentration CA0, and involves the gamma function.

Uploaded by

Sohail Aziz Ahmad Malik
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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TRANSPORT PHENOMENON

[ ] Article 18.6

Presented to : Dr. Syed Hassan Javaid Naqvi

L/O/G/O
Presented By:

Mohsin Ali – 2011-MS-CH-68


Sohail Aziz – 2011-MS-CH-65
Syed Yasir Hussain - 2011-MS-CH-66
L/O/G/O
§ 18.6 Diffusion into A Falling
Liquid Film (Solid Dissolution)
δ
Parabolic velocity profile
of fluid B
Near Wall

vz = (ρgδ/µ)y

Insoluble
wall

y
z Slightly soluble wall
made of A

CAo = Saturation concentration

L
CA(y , z)
CA = 0
Assumptions
• Laminar Flow
• Vz only depends on “y” for z≥0
• “A” is slightly soluble in “B”
• In short interval of time, “A” will be present in a very thin
boundary layer
• “A” molecules experience the a velocity distribution that is the
characteristic of the falling film right next to the wall, y = 0

• Velocity Distribution is given by equation

• In the Present Situation


Cos θ =1 & x = δ - y
so,

At adjacent to the wall;


So approximately velocity is;

Equation 18.5-6 in the previous section becomes;

(A)

B.C. 1 : at z = 0 , CA = 0
B.C. 2: at y = 0 , CA = CA0
B.C. 3: at y = ∞ , CA = 0

Let,
(i)
By putting in (i) we get,

(B)

(ii)
By putting in (ii), we get

Diff it w.r.t “y”

Since,
from eq

(C)
Using “B” and “C” equation “A” becomes
Put,

𝑑𝑝
= +η2 𝑝 = 0
𝑑η
𝑑𝑝
= −3η2 𝑝
𝑑η
𝑑𝑝
= −3η. 𝑑η
𝑝
Taking exponential on both sides
𝑝 = 𝑒 (−η
3+ 𝐶 )
1

𝑝 = 𝑒 −η + 𝑒 𝐶1
3

3
𝑝 = 𝐶1 . 𝑒 (−η )

𝑑𝑓 3
= 𝐶1 . 𝑒 (−η )
𝑑η
(−η3 )
𝑑𝑓 = 𝐶1 . 𝑒 𝑑η

(−η3 )
𝑓 = 𝐶1 𝑒 𝑑η + 𝐶2
By Applying B.C (2)
Where Γ(4/3)=0.8930 is the gamma function of (4/3)

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