Couette Flow Analytical and Numerical Solution.docx

COUETTE FLOW SOLUTION USING ANALYTICAL AND CRANKNICOLSON NUMERICAL TECHNIQUE RASIKH TARIQ

Introduction Flow of a viscous fluid between one stationary and one moving plate is a Couette Flow.

Shear stresses

∂τyy ∂y

=

∂τzy ∂z

=

∂τxy ∂x

= 0. Using this

information, the y-direction momentum equation becomes: ∂p =0 ∂y The governing x-direction momentum equation is:

FIGURE 1 FLUID MOVEMENT BETWEEN A FIXED AND A MOVING PLATE - COUETTE FLOW The report covers the following methods of finding the Couette flow solution:  

Exact Analytical Approach Implicit Numerical Approach using CrankNicolson Technique

ρ

Du ∂p ∂τxx ∂τyx ∂τzx =− + + + + ρfx Dt ∂x ∂x ∂y ∂z

Flow is flowing only in x-direction. The only shear stresses will be in x-direction on a plane perpendicular to y-axis and as a result shear stresses τxx = τzx = 0. Ignoring all the body forces, the x-direction momentum equation becomes:

Exact Analytical Approach ρ It is a viscous flow between two parallel plates separated by the vertical distance D. The upper plate is moving at a velocity ue , and the lower plate is stationary i.e. u = 0. The flow between the two plates is driven exclusively by the shear stress exerted on the fluid by the moving upper plate resulting in a velocity profile. The velocity profile is linear for steady conditions.

∂u ∂u ∂u ∂u ∂τyx + ρu + ρv + ρw = ∂t ∂x ∂y ∂z ∂y

Using the steady state assumption, the equation becomes: ρu

∂u ∂u ∂u ∂τyx + ρv + ρw = ∂x ∂y ∂z ∂y ∂u

The velocity gradient ∂x = 0 as a result ∂u

The governing y-direction momentum equation is: ρ

Dv ∂p ∂τxy ∂τyy ∂τzy =− + + + + ρfy Dt ∂y ∂x ∂y ∂z

Whereas: ρ

Dv ∂v ∂v ∂v ∂v = ρ + ρu + ρv + ρw = 0 Dt ∂t ∂x ∂y ∂z

term ρu ∂x = 0. There is no velocity in y-direction ∂u

i.e. v = 0, as a result the term ρv ∂y = 0. The equation is left with: ∂τyx =0 ∂y Using the viscous Newtonian fluid assumption, the above equation can be expressed as:

∂ ∂u (μ ) = 0 ∂y ∂y

Following is the result plotted by Matlab: -3

12

Or

x 10

Couette Flow Velocity Profile

The above equation is the exact solution for incompressible, Newtonian and steady flow. The exact solution is given by: u = c1 y + c2 Using boundary condition at y = 0 we have u = 0, we get c2 = 0. Using boundary condition at y = D we have u = ue, we get c1 = ue /D. The solution becomes: u y = ue D

10

Velocity Profile (m/s)

𝛛𝟐 𝒖 =𝟎 𝛛𝐲 𝟐

8

6

4

2

0 -0.01

%y is distance from lower plate to upper plate y = linspace(0,D,500); for i=1:1:500 U(i)=(ue*y(i))/D; end plot (y,U) legend ('Couette Flow Velocity Profile','Fontsize',12') xlabel ('Width Between Plates (m)',' Fontsize ',12') ylabel ('Velocity Profile (m/s)',' Fontsize ',12') axis ([-0.01 0.06 -0.001 0.012])

0.01

0.02

0.03

0.04

0.05

Width Between Plates (m)

Numerical Implicit Approach using Crank-Nicolson Technique A time marching procedure is chosen. For this purpose, an unsteady, incompressible, Couette flow x-direction momentum equation will be:

The equation is solved using Matlab. Following is the Matlab Code: clc clear %Workspace and Command History Cleared ue = 0.01 %Velocity of Upper Plate moving with 0.01m/s D = 0.05 %Separation Distance between plate. 5cm

0

𝛒

𝛛𝐮 𝛛𝟐 𝐮 =𝛍 𝟐 𝛛𝐭 𝛛𝐲

Eq. (1)

Nature of Partial Differential Equation The general classification of a second order partial differential equation with two variables is: A

∂2 u ∂2 u ∂2 u ∂u ∂u + B + C + D + E + Fu 2 2 ∂t ∂t ∂y ∂y ∂t ∂y = G(t, y)

By comparison, we get: A = 0,

B = 0,

C=μ

Since, B 2 − 4AC = 0 the equation is parabolic in nature.

The Numerical Formulation The numerical formation must be developed for any dimension and velocity of Couette Flow so Eq. (1) must be in a dimensionless format. Defining

0.06

u′ =

u ue

y′ = =

t

y D

t′

D⁄ ue

Eq. (1) now becomes:

un+1 − unj j Δt

1 n+1 1 1 n+1 n n+1 n n 1 2 (uj+1 + uj+1 ) + 2 (−2uj + 2uj ) + 2 (uj−1 + uj−1) = (Δ𝑦)2 ReD Or

u ∂ (u ) e

u ∂2 (u ) u ue 2 e e ρ ( )=μ 2 (D2 ) y D t ∂ (D) ∂ (D ) ⁄ue

un+1 = unj + j

Δt (un+1 + unj+1 − 2un+1 j 2(Δ𝑦)2 ReD j+1 n − 2unj + un+1 j−1 + uj−1 )

un+1 = unj + j

Or ∂u′ μ ∂2 u ′ = ∂t ′ ρue D ∂y ′ 2 Here the term

μ ρue D

is the reciprocal of ReD. We

get: ∂u′ 1 ∂2 u ′ = ∂t ′ ReD ∂y ′ 2 For simplicity and ease of writing the mathematical formulas consider: ∂u 1 ∂2 u = ∂t ReD ∂y 2 However u, t and y are the representative of dimensionless forms of u′, t ′ and y ′ . ∂u

The term ∂t is written using forward difference equation, and the term

∂2 u ∂y2

is written using central

second difference. Finally the finite difference equation becomes: un+1 − unj j Δt

1 unj+1 − 2unj + unj−1 = (Δ𝑦)2 ReD

Where j represents space marching direction and n represents time marching direction. Rewriting the equation using Crank-Nicolson technique by using the averaging of terms:

Δt un+1 2(Δ𝑦)2 ReD j+1 Δt + un 2(Δ𝑦)2 ReD j+1 Δt − 2un+1 2(Δ𝑦)2 ReD j Δt − 2un 2(Δ𝑦)2 ReD j Δt + un+1 2(Δ𝑦)2 ReD j−1 Δt + un 2(Δ𝑦)2 ReD j−1

Further rearranging by taking n + 1 terms on the left side of the above equation: Further rearranging by taking n + 1 terms on the left side of the above equation: Δt Δt un+1 −[ ] un+1 ] 2un+1 j j+1 + [ j 2 2(Δ𝑦) ReD 2(Δ𝑦)2 ReD Δt −[ ] un+1 2(Δ𝑦)2 ReD j−1 Δt = unj + [ ] un 2(Δ𝑦)2 ReD j+1 Δt −[ ] 2unj 2(Δ𝑦)2 ReD Δt +[ ] un 2(Δ𝑦)2 ReD j−1

Taking certain mathematical terms common: 𝚫𝐭 𝚫𝐭 −[ ] 𝐮𝐧+𝟏 ] 𝐮𝐧+𝟏 𝐣−𝟏 + [𝟏 + 𝟐 (𝚫𝐲)𝟐 𝐑𝐞𝐃 𝐣 𝟐(𝚫𝐲) 𝐑𝐞𝐃 𝚫𝐭 −[ ] 𝐮𝐧+𝟏 𝟐(𝚫𝐲)𝟐 𝐑𝐞𝐃 𝐣+𝟏 𝚫𝐭 = [𝟏 − ] 𝐮𝐧 (𝚫𝐲)𝟐 𝐑𝐞𝐃 𝐣 𝚫𝐭 +[ ] (𝐮𝐧𝐣+𝟏 + 𝐮𝐧𝐣−𝟏 ) 𝟐(𝚫𝐲)𝟐 𝐑𝐞𝐃 The above equation becomes: 𝐧+𝟏 𝐀𝐮𝐧+𝟏 + 𝐀𝐮𝐧+𝟏 𝐣−𝟏 + 𝐁𝐮𝐣 𝐣+𝟏 = 𝐊 𝐣

The first equation for this system (j = 2) is: Au1n+1 + Bun+1 + Aun+1 = K2 2 3 Modified format is: Bun+1 + Aun+1 = K 2 − Au1n+1 2 3 The second equation for this system (j = 3) is: Aun+1 + Bun+1 + Aun+1 = K3 2 3 4 The third equation for this system (j = 4) is:

Eq. (2)

Aun+1 + Bun+1 + Aun+1 = K4 3 4 5 The fourth equation for this system (j = 5) is:

Provided: A=−

Δt 2(Δy)2 ReD

Δt B=1+ (Δy)2 ReD And

Aun+1 + Bun+1 + Aun+1 = K5 4 6 5 The fifth equation for this system (j = 6) is: Aun+1 + Bun+1 + Aun+1 = K6 6 7 5 Or

Δt K𝑗 = [1 − ] un (Δy)2 ReD j Δt +[ ] (unj+1 + unj−1 ) 2(Δy)2 ReD

Aun+1 + Bun+1 = K 6 − Aun+1 6 7 5 The results (velocity distribution) can be attained by solving the resulting matrix. 𝐁 𝐀 𝟎 𝟎 [𝟎

𝐀 𝐁 𝐀 𝟎 𝟎

𝟎 𝐀 𝐁 𝐀 𝟎

𝟎 𝟎 𝐀 𝐁 𝐀

𝐊 𝟐 − 𝐀𝐮𝐧+𝟏 𝐮𝐧+𝟏 𝟏 𝟐 𝟎 𝐊 𝐧+𝟏 𝟑 𝟎 𝐮𝟑 𝐊𝟒 = 𝟎 𝐮𝐧+𝟏 𝟒 𝐊𝟓 𝐀 𝐮𝐧+𝟏 𝟓 𝐊𝟔 𝐁] 𝐧+𝟏 [𝐮𝟔 ] [𝐊 𝟔 − 𝐀𝐮𝐧+𝟏 𝟕 ]

Solving the above set of matrices will eventually give the velocity profile.

References Figure 2 Labeling of Points for the Grid



I divided the total domain into 7 nodes. All the equations for the 7 nodes will be written. It can be seen in the boundary condition that:



u1 = 0

and

uN+1 = 0

John D. Anderson, JR (1995). Computational Fluid Dynamics: The Basics with Applications. McGraw Hill, Inc. Munson, Young, Okiishi, Huebsch (2008). Fundamentals of Fluid Mechanics. John Wiley & Sons, Inc.