Syvum Info Page :
http://www.syvum.com/cgi/online/serve.cgi/eng/fluid/fluid204.html
Transport Phenomena - Fluid mechanics Problem : Newtonian fluid flow in a falling film
Problem. Consider a Newtonian liquid (of viscosity μ and density ρ) in laminar flow down an inclined flat plate of length L and width W . The liquid flows as a falling film with negligible rippling under the influence of gravity. End effects may be neglected because L and W are large compared to the film thickness δ.
Figure.
Newtonian liquid flow in a falling film.
a) Determine the steady-state velocity distribution. b) Obtain the mass rate of flow and average velocity in the falling film. c) What is the force exerted by the liquid on the plate in the flow direction? d) Derive the velocity distribution and average velocity for the case where x is replaced by a coordinate x' measured away from the plate (i.e., x' = 0 at the plate and x' = δ at the liquid-gas interface).
Solution. Click here for stepwise solution a) Step. Shear stress distribution
For axial flow in rectangular Cartesian coordinates, the differential equation for the momentum flux is (click here for derivation) dτ xz dx
=
Δ P L
(1)
where P is a modified pressure, which is the sum of both the pressure and gravity terms, that is, Δ P = Δ p + ρ g L cos β . Here, β is the angle of inclination of the z -axis with the vertical. Since the flow is solely under the influence of gravity, Δ p = 0 and therefore Δ P / L = ρ g cos β . On integration, τ xz
= ρ g x cos β + C 1
(2)
On using the boundary condition at the liquid-gas interface (τ xz = 0 at x = 0), the constant of integration C 1 is found to be zero. This gives the expression for the shear stress τ xz for laminar flow in a falling film as τ xz = ρ g x cos β (3)
It must be noted that the above momentum flux expression holds for both Newtonian and non-Newtonian fluids (and does not depend on the type of fluid). The shear stress for a Newtonian fluid (as per Newton's law of viscosity) is given by τ xz
= − μ
dv z dx
(4)
Step. Velocity distribution To obtain the velocity distribution, there are two possible approaches. In the first approach, equations (3) and (4) are combined to eliminate τ xz and get a firstorder differential equation for the velocity as given below. dv z
= − ρ g cos β x
(5)
dx
μ
On integrating, v z = − ρ g x2 cos β /(2 μ) + C 2. On using the no-slip boundary condition at the solid surface (v z = 0 at x = δ ), C 2 = ρ g δ2 cos β /(2 μ). Thus, the final expression for the velocity distribution is parabolic as given below. 2 v z
2
ρ g δ
=
cos β
2 μ
1−
x δ
(6)
In the second approach, equations (1) and (4) are directly combined to eliminate τ xz and get a second-order differential equation for the velocity as given below. d 2v z − μ 2 dx
= ρ g cos β
(7)
Integration gives dv z dx
= −
ρ g cos β x + C 1 μ
(8)
On integrating again, v z =
ρ g x2 cos β − 2 μ
+ C 1 x + C 2
(9)
On using the boundary condition at the liquid-gas interface (at x = 0, τ xz = 0 dv z /dx = 0), the constant of integration C 1 is found to be zero from equation (8). On using the other boundary condition at the solid surface (at x = δ, v z = 0), equation (9) gives C 2 = ρ g δ2 cos β /(2 μ). On substituting C 1 and C 2 in equation (9), the parabolic velocity profile in equation (6) is again obtained. ⇒
b) Step. Mass flow rate The mass flow rate may be obtained by integrating the velocity profile given by equation (6) over the film thickness as shown below.
1
δ w
=
∫ ρ v z W dx 0
=
2
3
ρ g W δ 2 μ
cos β ∫
2
1−
x x d δ δ
(10)
0
w
=
ρ2 g W δ3 cos β 3 μ
(11)
When the mass flow rate w in equation (11) is divided by the density ρ and the film crosssectional area (W δ), an expression for the average velocity is obtained. Also, the maximum velocity v z, max occurs at the interface x = 0 and is readily obtained from equation (6). Thus, equations (6) and (11) give =
ρ g δ2 cos β 3 μ
=
2 v 3 z, max
(12)
c) Step. Force on the plate The z -component of the force of the fluid on the solid surface is given by the shear stress integrated over the wetted surface area. Using equation (3) gives F z
= L W τ xz | x = δ = ρ g δ L W cos β
(13)
As expected, the force exerted by the fluid on the plate is simply the z -component of the weight of the liquid film. d) Step. Choice of coordinates Consider the coordinate x' starting from the plate (i.e., x' = 0 is the plate surface and x' = δ is the liquid-gas interface). The form of the differential equation for the momentum flux
given by equation (1) will remain unchanged. Thus, equation (2) for the new choice of coordinates will be τ x'z
= ρ g x' cos β + C 1
(14)
On using the boundary condition at the liquid-gas interface (τ x'z = 0 at x' = δ), the constant of integration is now found to be given by C 1 = − ρ g δ cos β . This gives the shear stress as τ x'z = ρ g ( x' − δ) cos β (15)
Note that the momentum flux is in the negative x'-direction. On substituting Newton's law of viscosity, dv z ρ g cos β = (δ − x') (16) dx' μ
On integrating, v z = ( ρg / μ) (δ x' − ½ x'2) cos β + C 2. On using the no-slip boundary condition at the solid surface (v z = 0 at x' = 0), it is found that C 2 = 0. Thus, the final expression for the velocity distribution is v z
=
ρ g δ cos β x' μ δ 2
2
−
1 x' 2 δ
The average velocity may be obtained by integrating the velocity profile given in equation (17) over the film thickness as shown below. 1 2 δ 2 1 1 x' x ρ g δ cos β ∫ x' = − w = d δ ∫ v z dx' μ δ 2 δ δ 0 0
Integration of equation (18) yields the same expression for obtained earlier in equation (12).
(17)
(18)
It must be noted that the two coordinates x and x' are related as follows: x + x' = δ. By substituting ( x/δ)2 = [1 − ( x'/δ)]2 = 1 − 2( x'/δ) + ( x'/δ)2, equation (17) is directly obtained from equation (6). Related Problems in Transport Phenomena - Fluid Mechanics : Transport Phenomena - Fluid Mechanics Problem : Power law fluid flow in a falling film
- Determination of shear stress, velocity profile, mass flow rate, film thickness and force for non Newtonian fluid described by power law model
© 1999-2012 Syvum Technologies Inc. www.syvum.com
