ADVANCED TRANSPORT PHENOMENA
Assignment 1
Shinta Leonita (0906635772)
Problem 1
A viscous fluid is in laminar flow in a slit formed by two parallel walls a distance 2B apart.
Make a differential momentum balance and obtain the expression for the distributions of
momentum flux and velocity. What is the ratio of average to maximum velocity in the slit?
Obtain the analog of the Hagen-Poiseuille law of the slit.
Answer:
Laminar flow in a narrow slit:
Figure 1. Flow Through a Slit, with B << W << L
From figure above, thickness of a system is Δx, length L, and listing the various distributions
to the momentum balance in the z-direction:

Rate of momentum in across slit surface at x
(𝑊𝐿𝜏𝑥𝑧 )|𝑥

Rate of momentum out across slit surface at x + Δx
(𝑊𝐿𝜏𝑥𝑧 )|𝑥+Δx

(1)
(2)
Rate of momentum in across beam surface at z = 0
(Δx𝑊𝑣𝑧 )(𝜌𝑣𝑧 )|𝑧=0
(3)



Rate of momentum out across beam surface at z = L
(Δx𝑊𝑣𝑧 )(𝜌𝑣𝑧 )|𝑧=𝐿
(4)
(Δx𝑊𝐿)𝜌𝑔
(5)
Gravity force acting on slit shell
Pressure force acting on beam surface at z = 0
(Δx𝑊)𝑝0

(6)
Pressure force acting on beam surface at z = L
−(Δx𝑊)𝑝𝐿
(7)
Momentum balance (assuming fluid is incompressible, vz is the same at z = 0 and z = L):
(𝑊𝐿𝜏𝑥𝑧 )|𝑥 − (𝑊𝐿𝜏𝑥𝑧 )|𝑥+Δx + (Δx𝑊𝑣𝑧 )(𝜌𝑣𝑧 )|𝑧=0 − (Δx𝑊𝑣𝑧 )(𝜌𝑣𝑧 )|𝑧=𝐿 + (Δx𝑊𝐿)𝜌𝑔 +
Δx𝑊(𝑝0 − 𝑝𝐿 )
(8)
Dividing eq. (8) by WL Δx and take the limit as Δx goes to zero; this gives:
𝜏𝑥𝑧 |𝑥+Δx −𝜏𝑥𝑧 |𝑥
lim (
Δx
Δx→0
)=
(𝑝0 −𝑝𝐿 )
𝐿
+ 𝜌𝑔
(9)
The expression on the left side is the definition of the first derivative, and 𝑃 = 𝑝 + 𝜌𝑔ℎ =
𝑝 − 𝜌𝑔𝑧. Hence eq. (9) may be written as:
𝑑
𝑑𝑥
𝜏𝑥𝑧 =
(𝑃0 −𝑃𝐿 )
(10)
𝐿
Integrating eq. (10) gives:
𝜏𝑥𝑧 =
(𝑃0 −𝑃𝐿 )
𝐿
𝑥 + 𝐶1
(11)
The constant C1 must be zero if the momentum flux is not to be infinite at x = 0. Hence the
momentum flux distribution is:
𝝉𝒙𝒛 =
(𝑷𝟎 −𝑷𝑳 )
𝑳
Substituting Newton’s law of viscosity 𝜏𝑥𝑧 = −𝜇
𝑑𝑣𝑧
𝑑𝑥
=−
𝑑𝑣𝑧
𝑑𝑥
𝒙
(12)
to eq. (12) gives:
(𝑃0 −𝑃𝐿 )
𝜇𝐿
𝑥
(13)
Integrating eq. (13) gives:
(𝑃0 −𝑃𝐿 )
𝑣𝑧 = −
2𝜇𝐿
𝑥 2 + 𝐶2
(14)
Because of the boundary condition that vz be zero at x = ±B, the constant C2 has the value
(𝑃0 −𝑃𝐿 )
2𝜇𝐿
𝐵 2. Hence the velocity distribution is:
𝒗𝒛 =
(𝑷𝟎 −𝑷𝑳 )
𝟐𝝁𝑳
𝒙 𝟐
𝑩𝟐 [𝟏 − ( ) ]
𝑩
(15)
The maximum velocity vz,max at the middle of the slit occurs at x = 0 and has the value:
(𝑃0 −𝑃𝐿 )
𝑣𝑧,𝑚𝑎𝑥 =
2𝜇𝐿
𝐵2
(16)
Hence the ratio of the average to the maximum velocity is then:
𝑣𝑧
𝑣𝑧,𝑚𝑎𝑥
𝑥 2
=1−( )
𝐵
𝑾 𝑩
⟨𝒗𝒛 ⟩
𝒗𝒛,𝒎𝒂𝒙
=
∫𝟎 ∫−𝑩[𝟏−(𝒙/𝑩)𝟐 ]𝒅𝒙𝒅𝒚
𝑾 𝑩
∫𝟎 ∫−𝑩 𝒅𝒙𝒅𝒚
𝟏
=
∫𝟎 (𝟏−𝝃𝟐 )𝒅𝝃
𝟏
∫𝟎 𝒅𝝃
(17)
𝟏
𝟐
𝟑
𝟑
= (𝟏 − ) =
(18)
The analog of the Hagen-Poiseuille law:
2
𝑄 = (2𝐵𝑊)⟨𝑣𝑧 ⟩ = (2𝐵𝑊) (3) 𝑣𝑧,𝑚𝑎𝑥
𝑸=
𝟐 (𝑷𝟎 −𝑷𝑳 )𝑩𝟑 𝑾
𝟑
𝝁𝑳
(19)
(20)
Problem 2
A droplet of substance A is suspended in a stream of gas B. The droplet radius is r1. We
postulate that there is a spherical stagnant gas film of radius r2. the concentration of A in the
gas phase is xA1 at r = r1 and xA2 at r = r2.
a) By a shell balance, show that for steady state diffusion r2NAr is a constant and set the
constant equal to r12NAr1 at the droplet surface.
b) Show that:
r12 N Ar1  
cD AB 2 dx A
r
1  x A  dr
c) Integrate that equation between the limits r1 and r2 to get:
N Ar1  
cDAB  r2  xB 2
  ln
r2  r1   r1  xB1
d) If a mass transfer coefficient kv is defined by NAr1 = kv (pA1 – pA2), show that:
kv 
2cD AB / D
 p B ln
Answer:
Figure 2. Diffusion through a Spherical Film
(a) Showing that for steady state diffusion r2NAr is a constant and set the constant equal to
r12NAr1 at the droplet surface:
A mass balance on A over a spherical shell of thickness Δr is (in molar units) at a steady
state:
4𝜋𝑟 2 . 𝑁𝐴𝑟 |𝑟=𝑟 − 4𝜋(𝑟 + ∆𝑟)2 . 𝑁𝐴𝑟 |𝑟=𝑟+∆𝑟 = 0
(1)
(4𝜋𝑟 2 . 𝑁𝐴𝑟 )|𝑟=𝑟 − (4𝜋𝑟 2 . 𝑁𝐴𝑟 )|𝑟=𝑟+∆𝑟 = 0
(2)
Or, equivalently:
Dividing eq. (2) by 4πΔr and take the limit as Δr goes to zero gives:
𝒅
𝒅𝒓
(𝒓𝟐 . 𝑵𝑨𝒓 ) = 𝟎
(3)
Eq. (3) may be integrated to give 𝑟 2 . 𝑁𝐴𝑟 = 𝐶1 and we may use the boundary condition
that 𝑁𝐴𝑟 = 𝑁𝐴𝑟1 at r = r1 (the gas liquid interface) to evaluate the constant and obtain that
r2NAr = r12NAr1
(b) Equation for the radial component in spherical coordinates is:
𝑁𝐴𝑟 = −𝑐𝐷𝐴𝐵
𝑑𝑥𝐴
𝑑𝑟
+ 𝑥𝐴 (𝑁𝐴𝑟 + 𝑁𝐵𝑟 )
(4)
If gas B is not moving, then 𝑁𝐵𝑟 may be set equal to zero and the equation may be solved
for the molar flux of A:
𝑐𝐷
𝑁𝐴𝑟 = − 1−𝑥𝐴𝐵
𝐴
𝑑𝑥𝐴
(5)
𝑑𝑟
Multiplying by r2 and using the result obtained in (a), then we get:
𝒄𝑫
𝒓𝟐𝟏 𝑵𝑨𝒓𝟏 = − 𝟏−𝒙𝑨𝑩 𝒓𝟐
𝑨
𝒅𝒙𝑨
𝒅𝒓
(6)
𝑟2
𝑥
𝑟12 𝑁𝐴𝑟1 ∫𝑟1 1/𝑟 2 𝑑𝑟 = −𝑐𝐷𝐴𝐵 ∫𝑥 𝐴2
(c)
𝑑𝑥𝐴
𝐴1 (1−𝑥𝐴 )
(7)
Integrated that equation gives:
1
1
(1−𝑥
1
2
)
𝐴2
𝑟12 𝑁𝐴𝑟1 ( − ) = 𝑐𝐷𝐴𝐵 𝑙𝑛
𝑟
𝑟
(1−𝑥 )
𝑟2 −𝑟1
𝑟12 𝑁𝐴𝑟1 (
𝑵𝑨𝒓𝟏 =
1
1
1
2
𝑟2 𝑟1
𝒄𝑫𝑨𝑩
(𝑥
)
) = 𝑐𝐷𝐴𝐵 𝑙𝑛 (𝑥𝐵2)
(9)
𝐵1
𝒓
𝒓𝟐 −𝒓𝟏
(8)
𝐴1
(𝒙
)
(𝒓𝟐) 𝒍𝒏 (𝒙𝑩𝟐 )
𝟏
(𝑥
(10)
𝑩𝟏
)
(d) At 𝑟 → ∞ 𝑟12 𝑁𝐴𝑟1 (𝑟 − 𝑟 ) = 𝑐𝐷𝐴𝐵 𝑙𝑛 (𝑥𝐵2 )
𝐵1
𝑁𝐴𝑟1 =
(𝑥
𝑐𝐷𝐴𝐵
𝑟1
𝑙𝑛
(𝑥𝐵2 )
(𝑥𝐵1 )
(11)
−𝑥𝐵1 )
𝐵2 /𝑥𝐵1 )
But, (𝑥𝐵 )𝑙𝑛 = 𝑙𝑛(𝑥𝐵2
Therefore,
𝑁𝐴𝑟1 =
𝑁𝐴𝑟1 =
𝑁𝐴𝑟1 =
𝑐𝐷𝐴𝐵
1
𝑟1
(𝑥𝐵 )𝑙𝑛
𝑐𝐷𝐴𝐵
1
𝑟1
(𝑥𝐵 )𝑙𝑛
𝑐𝐷𝐴𝐵
1
𝑟1
(𝑝𝐵 )𝑙𝑛
(𝑥𝐵2 − 𝑥𝐵1 )
(12)
(𝑥𝐴1 − 𝑥𝐴2 )
(13)
(𝑝𝐴1 − 𝑝𝐴2 )
(14)
But, 𝑁𝐴𝑟1 = 𝑘𝑣 (𝑝𝐴1 − 𝑝𝐴2 )
So,
𝒌𝒗 =
𝒄𝑫𝑨𝑩
𝒓𝟏(𝒑 )
𝑩 𝒍𝒏
=
𝟐𝒄𝑫𝑨𝑩
𝑫(𝒑𝑩 )𝒍𝒏
(15)