sup6 014

Prévia do material em texto

PROBLEM 6S.14 
KNOWN: Thickness and inclination of a liquid film. Mass density of gas in solution at free surface 
of liquid. 
FIND: (a) Liquid momentum equation and velocity distribution for the x-direction. Maximum 
velocity, (b) Continuity equation and density distribution of the gas in the liquid, (c) Expression for 
the local Sherwood number, (d) Total gas absorption rate for the film, (e) Mass rate of NH3 removal 
by a water film for prescribed conditions. 
SCHEMATIC: 
 NH3 (A) – Water (B) 
 L = 2m 
 δ = 1 mm 
 D = 0.05m 
 W = πD = 0.157m 
 ρA,o = 25 kg/m3 
 DAB = 2 × 10-9 m2/s 
 φ = 0° 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) The film is in fully developed, laminar flow, (3) 
Negligible shear stress at the liquid-gas interface, (4) Constant properties, (5) Negligible gas 
concentration at x = 0 and y = δ, (6) No chemical reactions in the liquid, (7) Total mass density is 
constant, (8) Liquid may be approximated as semi-infinite to gas transport. 
PROPERTIES: Table A-6, Water, liquid (300K): ρf = 1/vf = 997 kg/m3, μ = 855 × 10-6 N⋅s/m2, ν 
= μ/ρf = 0.855 × 10-6m2/s. 
ANALYSIS: (a) For fully developed flow (v = w = 0, ∂u/∂x = 0), the x-momentum equation is 
 ( ) ( )yx yx0 / y X where u/ y and X g cos .∂τ ∂ τ μ ∂ ∂ ρ φ= + = = 
That is, the momentum equation reduces to a balance between gravitational and shear forces. Hence, 
 ( ) ( )2 2u/ y g cos .μ ∂ ∂ ρ φ= − 
Integrating, ( ) ( ) 21 1 2 u/ y g cos / y C u g cos /2 y C y C .∂ ∂ φ ν φ ν= − + = − + + 
Applying the boundary conditions, 
 
)
( )
1y=0
2
2
 u/ y 0 C 0
u 0 C g cos .
2
∂ ∂
δδ φ ν
= → =
= → =
 
Hence, ( ) ( )2 22 2g cos g cos u y 1 y/2 2φ φ δδ δν ν= − = −⎡ ⎤⎢ ⎥⎣ ⎦ < 
and the maximum velocity exists at y = 0, 
 ( ) ( )2maxu u 0 g cos / 2 .φ δ ν= = < 
(b) Species transport within the liquid is influenced by diffusion in the y-direction and advection in the 
x-direction. Hence, the species continuity equation with u assumed equal to umax throughout the 
region of gas penetration is 
 Continued ….. 
PROBLEM 6S.14 (Cont.) 
 
2 2
maxA A A A
AB 2 2 AB
u 
u D .
 x D x y y
∂ρ ∂ ρ ∂ ρ ∂ρ
∂ ∂∂ ∂
= = 
Appropriate boundary conditions are: ρA(x,0) = ρA,o and ρA(x,∞) = 0 and the entrance condition is: 
ρA(0,y) = 0. The problem is therefore analogous to transient conduction in a semi-infinite medium 
due to a sudden change in surface temperature. From Section 5.7, the solution is then 
 ( ) ( )
A A,o
A A,o1/ 2 1/ 2A,o AB max AB max
y y
erf erfc
0 2 D x/u 2 D x/u
ρ ρ ρ ρρ
− = =− < 
(c) The Sherwood number is defined as 
 y=0m,x A,xx m,x
AB A,o A,o
h x n AB A
Sh where h
D
D / y′′= ≡ =
−
ρ ρ
∂ρ ∂
 
( ) ( )
1/ 22
max maxA
A,o A,o1/ 2 1/ 2AB ABy=0 AB max y=0
y u u2 1
exp .
 y 4 D x D x2 D x/u
∂ρ ρ ρ∂ ππ
= − − = −
⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎣ ⎦
 
Hence, 
 ( ) ( )
1/ 2 1/ 21/ 2 1/ 2
max AB max max
m,x x 1/ 2 1/ 2AB AB
u D u x u x1 1
h Sh
 x D D
ν
π νπ π
= = =⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦ 
and with Rex ≡ umax x/ν, 
 ( )1/ 2 1/2 1/2 1/2 1/2x x xSh 1/ Re Sc 0.564 Re Sc .π= =⎡ ⎤⎢ ⎥⎣ ⎦ < 
(d) The total gas absorption rate may be expressed as 
 ( )A m,x A,on h W L ρ= ⋅ 
where the average mass transfer convection coefficient is 
 
L
0 0
1/ 2 1/ 2
Lmax AB max AB
m,x m,x 1/2
u D 4u D1 1 dx
h h dx .
L L Lxπ π
= ∫ = ∫ =⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ 
Hence, the absorption rate per unit width is 
 ( )1/ 2A max AB A,on / W 4u D L / .π ρ= < 
(e) From the foregoing results, it follows that the ammonia absorption rate is 
 
1/ 21/ 2 2
max AB AB
A A,o A,o
4u D L 4 g cos D L
n W W .
2
φδρ ρπ πν= =
⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥⎣ ⎦
 
Substituting numerical values, 
( ) ( ) ( )
1/ 222 3 -9 2
3 4
A -6 2
4 9.8 m/s 1 10 m 2 10 m /s 2m
n 0.157m 25 kg/m 6.71 10 kg/s.
2 0.855 10 m / sπ
−
−× × × ×= = ×
× ×
⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
 < 
COMMENTS: Note that ρA,o ≠ ρA,∞, where ρA,∞ is the mass density of the gas phase. The value 
of ρA,o depends upon the pressure of the gas and the solubility of the gas in the liquid.