Laminar Flow of non-Newtonian fluids in a Horizontal Pipe ​
​Authors: Housam Binous and Ahmed Bellagi

We propose to find the velocity profiles for the laminar flow of Newtonian, power-law and Carreau fluids in a pipe, shown in below.
​
We use the following values for the pressure difference [Laplacian]P, the exponent n, the Newtonian fluid viscosity η, the consistency index k,
the infinite-shear viscosity
η
∞
, the zero-shear viscosity
η
0
, the relaxation parameter λ, the pipe length L, and radius R:
​
[Laplacian]P =100; L = 50; and R = 0.02
​
Newtonian fluid:
η=8.9×
-4
10
​
​
Dilatant fluid:
n=3.39
and
κ=
-6
10
​
​
Pseudo-plastic fluid:
n=0.4
and
κ=5×
-3
10
​
​
Carreau fluid:
n=0.5
,
λ=0.2
,
η
0
=1.72×
-3
10
, and
η
∞
=0
​
​
Reference:
​
[1] H . Binous, Introducing non-Newtonian fluid mechanics computations with Mathematica® in the undergraduate curriculum, Chemical Engineering Education, 41 (1), pp . 59 - 64, 2007.
​
[2] J. O. Wilkes, Fluid Mechanics for Chemical Engineers, Upper Saddle River, NJ: Prentice Hall, 1999.
​
Acknowledgement :
​​
Since in this laminar flow problem we have split boundary conditions. Problem’s resolution is based on the Shooting Method developed in Mathematica by Professor Brian Higgins (University of California, Davis).
​

Flow of dilatant, pseudoplastic, Carreau and Newtonian fluids in a horizontal pipe.

Analytical Solution​​
v
z
=
n
n+1
1
n
ΔP
2KL
n+1
n
R
1-
n+1
n
r
R
​​
In[]:=
v
analytical
[r_]:=n/(n+1)(ΔP/(2KL))^(1/n)R^((n+1)/n)(1-(r/R)^((n+1)/n))
In[]:=
v
analytical
[r]
Out[]=
-1/n
2
n1-
1+n
n

r
R

1+n
n
R
1
n
ΔP
KL
1+n
Dilatant Fluid
In[]:=
K=110^-6;n=3.39;ΔP=100;L=50;R=0.02;
In[]:=
system[Ω_]={D[r
τ
rx
[r],{r,1}]ΔP/Lr,D[
v
x
[r],{r,1}]If[
τ
rx
[r]≥0,-(
τ
rx
[r]/K)^(1/n),(-
τ
rx
[r]/K)^(1/n)],
τ
rx
[10^-5]0,
v
x
[10^-5]Ω};​​myODEsoln[Ω_]:=NDSolve[system[Ω],{
v
x
,
τ
rx
},{r,10^-5,R}]​​yend[Ω_?NumericQ]:=Flatten[(
v
x
[r]/.myODEsoln[Ω])/.rR]​​bc=FindRoot[yend[Ω]0,{Ω,0,0.5}][[1,2]];
In[]:=
sol1=myODEsoln[bc]
Out[]=

v
x
InterpolatingFunction
Domain: {{0.00001,0.02}}
Output: scalar
,
τ
rx
InterpolatingFunction
Domain: {{0.00001,0.02}}
Output: scalar

In[]:=
plt1=Plot[
v
x
[r]/.sol1,{r,0.00001,R},PlotStyle{Thick,Darker@Green},Frame->True,AspectRatio->1,FrameLabel->{Style["position",14],Style["velocity",14]},GridLines->Automatic,ImageSize->350{1,1}]
Out[]=
Comparing numerical result (solid line) with analytical solution (dashed curve) for dilatant fluid
In[]:=
plt2=Plot[
v
analytical
[r],{r,0.00001,R},PlotStyle{Thick,Darker@Red,Dashed},Frame->True,AspectRatio->1,FrameLabel->{Style["position",14],Style["velocity",14]},GridLines->Automatic,ImageSize->350{1,1}]
Out[]=
In[]:=
Show[plt1,plt2]
Out[]=
Pseudoplastic Fluid
In[]:=
K=510^-3;n=0.4;ΔP=100;L=50;R=0.02;
In[]:=
system[Ω_]={D[r
τ
rx
[r],{r,1}]ΔP/Lr,D[
v
x
[r],{r,1}]If[
τ
rx
[r]≥0,-(
τ
rx
[r]/K)^(1/n),(-
τ
rx
[r]/K)^(1/n)],
τ
rx
[10^-5]0,
v
x
[10^-5]Ω};​​myODEsoln[Ω_]:=NDSolve[system[Ω],{
v
x
,
τ
rx
},{r,10^-5,R}]​​yend[Ω_?NumericQ]:=Flatten[(
v
x
[r]/.myODEsoln[Ω])/.rR]​​bc=FindRoot[yend[Ω]0,{Ω,0,0.5}][[1,2]];
In[]:=
sol2=myODEsoln[bc]
Out[]=

v
x
InterpolatingFunction
Domain: {{0.00001,0.02}}
Output: scalar
,
τ
rx
InterpolatingFunction
Domain: {{0.00001,0.02}}
Output: scalar

In[]:=
plt3=Plot[
v
x
[r]/.sol2,{r,0.00001,R},PlotStyle{Thick,Darker@Blue},Frame->True,AspectRatio->1,FrameLabel->{Style["position",14],Style["velocity",14]},GridLines->Automatic,ImageSize->350{1,1}]
Out[]=
Newtonian Fluid
Comparing numerical result (solid line) with analytical solution (dashed curve) for Newtonian fluid
Comparing velocity profiles obtained for the four fluids: dilatant, pseudoplastic, Carreau and Newtonian.
​
These profiles are obtained under equal volumetric flow conditions.
​
The velocity near the wall is higher for Carreau and pseudo-plastic fluids than for Newtonian and dilatant fluids. This results in higher heat transfer rates due to a higher convection.
​