Lecture No 8
Solving Example no. 2 lecture 7
Example 2
Evaporation from sphere droplet of A vaporizing on to an infinite stagnant
atmosphere, using (a) shell balance method, and (b) the general equation. Determine
the mole fraction of A in B at distance R2 from the centre.
r
Δr
Liquid A
Assumptions
1. Steady state process: Rate of accumulation is zero.
2. Constant T, P: DAB is constant.
3. No chemical reaction: Rate of generation is zero.
4. Gas B is not soluble in liquid A: NB = 0.
5. Diffusion is in r-direction : NA = f(r)
Solution using shell balance method
{𝑅𝑎𝑡𝑒 𝑜𝑓 𝐴}𝑖𝑛 − {𝑅𝑎𝑡𝑒 𝑜𝑓 𝐴}𝑜𝑢𝑡 + 𝑔𝑒𝑛𝑒 = 𝐴𝑐𝑐𝑢𝑚
4𝜋𝑟 2 𝑁𝐴 |𝑟 − 4𝜋(𝑟 + ∆𝑟)2 𝑁𝐴 |𝑟+∆𝑟 = 0 conservation of mass law
Dividing by 4𝜋∆𝑟
Take the limit as ∆𝑟 → 0
𝑟 2 𝑁𝐴 |𝑟 − (𝑟 + ∆𝑟)2 𝑁𝐴 |𝑟+∆𝑟
lim
∆𝑟→0 ∆𝑟
𝑑(𝑟 2 𝑁𝐴 )
=0
𝑑𝑟
Using the general equations for spherical coordinates
𝜕𝐶𝐴 1 𝜕(𝑟 2 𝑁𝐴𝑟 ) 1 𝜕𝑁𝐴𝜃 𝑠𝑖𝑛𝜃 1 𝜕𝑁𝐴𝜙
+[ 2 + + ] − 𝑅̇𝐴𝑣 = 0
𝜕𝑡 𝑟 𝜕𝑟 𝑟𝑠𝑖𝑛𝜃 𝜕𝜃 𝑟𝑠𝑖𝑛𝜃 𝜕𝜙
1 𝜕(𝑟 2 𝑁𝐴𝑟 )
=0
𝑟2 𝜕𝑟
𝑑(𝑟 2 𝑁𝐴𝑟 )
𝑑𝑟
=0
𝑟 2 𝑁𝐴 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ∴ 𝑅12 𝑁𝐴 |1 = 𝑅22 𝑁𝐴 |2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑑𝑦𝐴
𝑁𝐴 = −𝐶𝐷𝐴𝐵 + 𝑦𝐴 (𝑁𝐴 + 𝑁𝐵 )
𝑑𝑟
−𝐶𝐷𝐴𝐵 𝑑𝑦𝐴
𝑁𝐴 =
(1 − 𝑦𝐴 ) 𝑑𝑟
Substituting into the diff. equation
𝑑 −𝐶𝐷 𝑑𝑦
𝑑𝑟
[𝑟 2 (1−𝑦𝐴𝐵) 𝑑𝑟𝐴 ] =0
𝐴
For ideal gas mixture at constant temperature and pressure, C is a constant, and DAB is
very nearly independent of concentration. Hence C DAB can be taken outside the
derivative to get.
𝑑 𝑟2 𝑑𝑦𝐴
[ ]=0
𝑑𝑟 (1 − 𝑦𝐴 ) 𝑑𝑟
Integrate:
𝑟2 𝑑𝑦𝐴
= 𝐾1
(1 − 𝑦𝐴 ) 𝑑𝑟
𝑑𝑦𝐴 𝑑𝑟
= 𝐾1 2
(1 − 𝑦𝐴 ) 𝑟
−𝑲1
−𝒍𝒏(1 − 𝒚𝑨 ) = + 𝑲2 General Solution
𝒓
Boundary Condition
𝑦𝐴 = 𝑦𝐴1 𝑎𝑡 𝑟 = 𝑅1
𝑦𝐴 = 𝑦𝐴2 𝑎𝑡 𝑟 = 𝑅2
−𝐾1
−𝑙𝑛(1 − 𝑦𝐴1 ) = 𝑅1
+ 𝐾2
−𝐾1
−𝑙𝑛(1 − 𝑦𝐴2 ) = + 𝐾2
𝑅2
1 − 𝑦𝐴1 1 1
𝐾1 = 𝑙𝑛 [ ]⁄[ − ]
1 − 𝑦𝐴2 𝑅2 𝑅1
1 1 − 𝑦𝐴1 1 1
𝐾2 = 𝑙𝑛[1 − 𝑦𝐴1 ] + 𝑙𝑛 [ ] ⁄[ − ]
𝑅1 1 − 𝑦𝐴2 𝑅2 𝑅1
1 1
−
𝑅 𝑟
[ 11 1 ]
1 − 𝑦𝐴 1− 𝑦𝐴1 𝑅 −𝑅
=[ ] 2 1
(1 − 𝑦𝐴1 ) 1 − 𝑦𝐴2
1 1
( − )
𝑹 𝒓
[ 11 1 ]
𝒚𝑩 𝒚𝑩1 (𝑹 −𝑹 )
𝒚𝑩1
= [𝒚 ] 2 1 Particular Solution
𝑩2
Example (3)
Gas A diffuses through a stagnant gas film to the surface of a non-porous cylindrical
𝑘1
catalyst. Gas A then undergoes the following chemical reaction 2𝐴 → 𝐵.
Gas B diffuses from the catalyst surface and swept away. Neglecting diffusion and
reaction on the ends of the particle. Derive an equation for the molar flux of A, using
(a) shell balance method, and (b) the N-stocks equation, derive an expression for the
concentration profile of A for (a) a slow reaction (b) fast reaction.
Assumptions
1. Steady state process➔ Rate of accumulation is
A
zero.
B
2. Constant T, P, and DAB.
3. Chemical reaction takes places on the surface.
catalyst
4. 1st order chemical reaction 𝑁𝐴 = 𝑘 𝑠 𝐶𝐴 ,
5. Diffusion is only in the r-direction : NA = f(r)
6. Chemical traction at the boundary.
7. 2 moles of A diffuses to the catalyst for every 1 mole of B diffuses away
𝑁𝐴 = −2𝑁𝐵
Using the general equations for Cylindrical coordinate
Δr
𝜕𝐶𝐴 1 𝜕(𝑟𝑁𝐴𝑟 ) 1 𝜕𝑁𝐴𝜃 𝜕𝑁𝐴𝑧
+[ + + ] − 𝑅̇𝐴𝑣 = 0
𝜕𝑡 𝑟 𝜕𝑟 𝑟 𝜕𝜃 𝜕𝑧
1 𝜕(𝑟𝑁𝐴𝑟 ) 𝑑(𝑟𝑁𝐴 )
=0 ⇒ =0
𝑟 𝜕𝑟 ⬚ 𝑑𝑟
Solution using Shell Balance Method
Take the element of thickness ∆𝑟 and length L.
2𝜋𝑟𝐿𝑁𝐴 |𝑟 − 2𝜋𝐿(𝑟 + ∆𝑟)𝑁𝐴 |𝑟+∆𝑟 = 0
Divide by 2𝜋𝐿∆𝑟 and take the limit as ∆𝑟 → 0
𝑑(𝑟𝑁𝐴 )
𝑑𝑟
=0 (1)
𝑑𝑦𝐴
𝑁𝐴 = −𝐶𝐷𝐴𝐵 + 𝑦𝐴 (𝑁𝐴 + 𝑁𝐵 )
𝑑𝑟
1
𝑁𝐵 = − 𝑁𝐴
2
kinetics of the problem
𝑑𝑦𝐴 1
𝑁𝐴 = −𝐶𝐷𝐴𝐵 + 𝑦𝐴 (𝑁𝐴 − 𝑁𝐴 )
𝑑𝑟 2
𝑑𝑦𝐴 1
𝑁𝐴 = −𝐶𝐷𝐴𝐵 + 𝑦 𝑁
𝑑𝑟 2 𝐴 𝐴
1 𝑑𝑦𝐴
𝑁𝐴 (1 − 𝑦𝐴 ) = −𝐶𝐷𝐴𝐵
2 𝑑𝑟
−𝐶𝐷𝐴𝐵 𝑑𝑦𝐴
𝑁𝐴 =
1
(1 − 2 𝑦𝐴 ) 𝑑𝑟
)1(Substituting into equation
𝑑 −𝐶𝐷𝐴𝐵 𝑟 𝑑𝑦𝐴
[ ]=0
𝑑𝑟 (1 − 1 𝑦 ) 𝑑𝑟
2 𝐴
For ideal gas mixture at constant temperature and pressure, C is a constant, and DAB is
very nearly independent of concentration. Hence C DAB can be taken outside the
derivative to get
Integrating the previous equation:
𝑑 𝑑𝑦𝐴 𝑟
[ ]=0
𝑑𝑟 (1 − 1 𝑦 ) 𝑑𝑟
2 𝐴
𝑑𝑦𝐴 𝑟
= 𝐾1
1
(1 − 2 𝑦𝐴 ) 𝑑𝑟
1
−2𝒍𝒏 (1 − 𝒚𝑨 ) = 𝑲1 𝒍𝒏𝒓 + 𝑲2
2
General solution
Case I
The reaction is very fast, therefore the concentration of gas A on the surface
of the catalyst is zero.
Boundary Conditions
𝑦𝐴 = 0 𝑎𝑡 𝑟=𝑅
𝑦𝐴 = 𝑦𝐴𝛿 𝑎𝑡 𝑟 = 𝛿𝑅
Particular Solution
Case II
The reaction is very slow, therefore the concentration of gas A on the surface
of the catalyst is expressed as:
𝑁𝐴
𝑁𝐴 = 𝑘 𝑠 𝐶𝑦𝐴 𝑎𝑡 𝑟=𝑅 → 𝑦𝐴 =
𝐶𝑘 𝑠
Boundary Conditions
𝑦𝐴 = 𝑦𝐴𝛿 𝑎𝑡 𝑟 = 𝛿𝑅
𝑁𝐴
(1 − 𝑠)
𝑙𝑛 [ 2𝐶𝑘 ]
1
(1 − 𝑦𝐴𝛿 )
𝐾1 = 2 2
𝛿𝑅
𝑙𝑛 ( 𝑅 )
is the Particular Solution
𝐶𝐷𝐴𝐵 𝑑𝑦𝐴
𝑁𝐴 |𝑟=𝑅 = |
1
(1 − 2 𝑦𝐴 ) 𝑑𝑟
𝑟=𝑅
The Molar Flux at The Surface
Example 4
A spherical catalyst particle in which species A diffuses into the catalyst and under
goes an irreversible reaction
𝑘𝑣
𝐴→ 𝐵
Assume the reaction in distributed homogeneously throughout the particle and the
reaction is first order, derive an expression for the concentration profile of A using (a)
shell balance method, and (b) the general equation.
Assumptions
Δr
1. Steady state process: Rate of accumulation is
zero.
2. Constant T, P: DAB is constant.
3. Chemical reaction takes places at the surface. R
4. 1st order chemical reaction 𝑁𝐴 = 𝑘 𝑣 𝐶𝐴 ,
5. Diffusion is only in the r-direction : NA = f(r)
6. Diffusion in solids 𝑁𝐴 = −𝐷𝐴𝑒𝑓𝑓 𝑑𝐶
𝑑𝑟
𝐴
Using General Equation
𝜕𝐶𝐴 1 𝜕(𝑟 2 𝑁𝐴𝑟 ) 1 𝜕𝑁𝐴𝜃 𝑠𝑖𝑛𝜃 1 𝜕𝑁𝐴𝜙
+[ 2 + + ] − 𝑅̇𝐴𝑣 = 0
𝜕𝑡 𝑟 𝜕𝑟 𝑟𝑠𝑖𝑛𝜃 𝜕𝜃 𝑟𝑠𝑖𝑛𝜃 𝜕𝜙
1 𝜕(𝑟 2 𝑁𝐴𝑟 )
[ ] − 𝑅̇𝐴𝑣 = 0
𝑟2 𝜕𝑟
Using Shell Balance
4𝜋𝑟 2 𝑁𝐴 |𝑟 − 4𝜋(𝑟 + ∆𝑟)2 𝑁𝐴 |𝑟+∆𝑟 + 4𝜋𝑟 2 ∆𝑟 𝑅𝐴𝑣 = 0
Dividing by 4𝜋∆𝑟
Take the limit as ∆𝑟 → 0
1 𝑑(𝑟 2 𝑁𝐴 )
−[ ] + 𝑅̇𝐴𝑣 = 0
𝑟 2 𝑑𝑟
𝑅𝐴𝑣 = 𝑘 𝑣 𝐶𝐴 Kinetic of the Reaction
𝑑𝐶𝐴
𝑁𝐴 = −𝐷𝐴𝑒𝑓𝑓 𝑑𝑟
Diffusion of A through a porous solid
Equation
`
𝑑(𝑟 2 𝑁𝐴𝑟 )
+ 𝑘 𝑣 𝐶𝐴 𝑟 2 = 0
𝑑𝑟
𝐷𝐴𝑒𝑓𝑓 𝑑 𝑑𝐶𝐴
2 [𝑟 2 ] = 𝑘 𝑣 𝐶𝐴
𝑟 𝑑𝑟 𝑑𝑟
𝑑2 𝐶𝐴 2 𝑑𝐶𝐴 𝑘𝑣
+ = 𝐶
𝑑𝑟 2 𝑟 𝑑𝑟 𝐷𝐴𝑒𝑓𝑓 𝐴
Using transformation of the variables as
𝐴𝐶𝑟 = )𝑟(𝑓
𝐴𝐶𝑑
𝑟 = )𝑟( 𝑓 ′ 𝐴𝐶 +
𝑟𝑑
Then,
𝐴𝐶 𝑑𝐶𝐴 𝑓́ (𝑟) −
=
𝑟𝑑 𝑟
)𝑟( ′′
𝐴𝐶𝑑 𝐴𝐶𝑑 𝐴𝐶 𝑑2
𝑓 𝑟= + +
𝑑𝑟 2 𝑟𝑑 𝑟𝑑
𝐴𝐶 𝑑2 𝐴𝐶𝑑
𝑟 = )𝑟( 𝑓 ′′ 2
+ 2
𝑟𝑑 𝑟𝑑
بالتعويض في معادلة التفاضلية باستخدام المتغيرات الجديدة نحصل على
𝑣𝑘
𝐷 𝑓 ′′ (𝑟) − 𝑓(𝑟) = 0 A second order homogenous ODE
𝑓𝑓𝑒𝐴
𝑣𝑘
=𝛼
𝑓𝑓𝑒𝐴𝐷
𝑓 ′′ (𝑟) − 𝛼 𝑓(𝑟) = 0
Integral factor can be used to solve derivative Equation.
𝑟 𝛼√𝑓(𝑟) = 𝐾1 𝑐𝑜𝑠ℎ√𝛼 𝑟 + 𝐾2 𝑠𝑖𝑛ℎ
𝑟 𝛼√𝑟 𝐶𝐴 = 𝐾1 𝑐𝑜𝑠ℎ√𝛼 𝑟 + 𝐾2 𝑠𝑖𝑛ℎ
)𝑟(𝑓 𝐾1 𝐾2
= 𝐴𝐶 𝑟
= 𝑟
𝑐𝑜𝑠ℎ√𝛼 𝑟 + 𝑟
𝛼√𝑠𝑖𝑛ℎ 𝑟 General Solution
BC’s
𝑠𝐴𝐶 = 𝐴𝐶 𝑡𝑎 𝑅=𝑟
𝐴𝐶𝑑
=0 𝑡𝑎 𝑟=0
𝑟𝑑
باستخدام BC’sتستطيع الحصول على الثوابت & 𝐾2 𝐾1
𝐶 𝑅
𝐾1 = 0 and 𝑅 𝛼 𝑠𝐴𝐾2 = 𝑠𝑖𝑛ℎ
√
في الحل العام نحصل على بالتعويض ب & 𝐾2 𝐾1
𝑘𝑣
𝑠𝑖𝑛ℎ√ 𝑟
𝐷𝐴𝑒𝑓𝑓
𝐶𝐴 𝑅
𝐶𝐴𝑠
= 𝑟
Particular Solution
𝑘𝑣
𝑠𝑖𝑛ℎ √ 𝑅
𝐷𝐴𝑒𝑓𝑓
