the development of poiseuille flow of a pseudoplastic fluid

THE DEVELOPMENT OF POISEUILLE FLOW OF A
PSEUDOPLASTIC FLUID
M.A.M. Al Khatib
HBCC, King Fahd University of Petroleum & Minerals
KFUPM Box 5087, Dhahran, Saudi Arabia
e-mail: alkhatibm@yahoo.com
M.A.M. Al Khatib
1. INTRODUCTION
We consider here the two-dimensional steady flow of a liquid down a semi-infinite straight
channel, a much-studied problem of obvious practical and theoretical interest. Far from the
entrance, the flow is parallel to the walls and the determination of the velocity distribution is
a simple matter, whatever the constitutive model. We consider here the region sufficiently far
downstream that the flow is almost parallel, so that the equations can be linearized about the
parallel-flow solution. We then look for the approach to fully-developed flow by way of decaying
eigenmodes. This was done for the Newtonian liquid by Wilson [1]. In that paper both large
and small Reynolds numbers were considered; here we consider only low Reynolds number. This
is reasonable in view of the generally large viscosity of the liquids.
Al Khatib and Wilson [2] extended that work to the case of viscoplastic liquids. Viscoplastic
liquids have a yield stress; little or no motion of the fluid will occur until a critical yield stress
is exceeded. Al Khatib and Wilson expected that there should be two types of regions of a
viscoplastic liquid in the channel; an unyielded region which is at the center of the channel, and
yielded regions which are close to its walls. These regions are separated by the yield surface.
For a Bingham fluid this natural assumption turns out to be kinematically impossible because
of the absolute rigidity of the unyielded region assumed by the Bingham model. To overcome
this difficulty, they used the smoothed-out form of the Bingham model; biviscosity model. They
showed that as the biviscosity model approaches the Bingham model, the approach of the entry
flow to the fully-developed parallel flow becomes infinitely delayed.
This paper extends the above work to the case of Pseudoplastic (shear thinning) liquids.
The viscosity of pseudoplastic liquids decreases as the shear rate increases. Greases, mayonnaise,
many polymers, polymer solutions, and suspensions exhibit this type of behavior (see Tanner [3]).
The next section gives more details about the power law and Carreau models which are used
to describe pseudoplastic liquids.
Key words: power law model, Pseudoplastic fluids, Carreau model, two-point boundary-value problem.
Paper Received 17 January 2004; Revised 8 October 2004; Accepted 26 March 2005.
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M.A.M. Al Khatib
2. CONSTITUTIVE MODEL
The power law model, which is simple and easy to handle, is the most well-known and widely used empiricism
in engineering work; many specific flow problems and heat transfer problems have been solved using it and the
results have proven to be useful. The constitutive equation of the power law model in simple shear flow can be
written as
τ = m .
γ
.
.
n−1 . γ (2.1)
where γ is the shear rate, n is the index number (0

3. EQUATIONS OF MOTION
M.A.M. Al Khatib
We consider a region far downstream from the entrance of a straight channel of width 2h, with the origin
in the centre of the channel. The velocity components of the flow in the downstream, x, and transverse, y,
directions are u and v respectively.
The smoothed-out power law model which is going to be used here, the Carreau Model, is given in
3-dimensional form by Equation (2.2) with · γ= 2eijeij (see Bird [4]). With η∞ = 0 for simplicity, we get
where e ∗ ij
τ ∗ ij =2η◦
is the strain-rate tensor.
1+(λ · γ) 2(n−1) /2
e ∗ ij
By neglecting the inertial effects, the equations of motion in the steady state in the x and y directions are
given respectively by
0=− ∂P∗
∂x + ∂τ∗ xx
∂x + ∂τ∗ xy
∂y ,
0=− ∂P∗
∂y + ∂τ∗ xy
∂x + ∂τ∗ yy
∂y .
First we determine the basic parallel flow which is approached far downstream. Here the pressure gradient is
uniform and we put − ∂P
∂x = Gr > 0 for convenience. Then from (3.2) we get
0=Gr + ∂τ∗ xy
∂y
January 2006 The Arabian Journal for Science and Engineering, Volume 31, Number 1A. 103
(3.1)
(3.2)
. (3.3)
Integrating (3.3) with respect to y using the symmetry condition τ ∗ xy = 0 when y =0weget
τ ∗ xy = −Gr y . (3.4)
Using Equations (3.1) and (3.4) we get
−Gr y = η◦
1+(λ (u ∗ ◦) ′ ) 2 (n−1) /2
where (u ∗ ◦) ′ = ∂u ∗ ◦/∂y and the subscript ◦ refers to the parallel flow solution.
(u ∗ ◦) ′ . (3.5)
Now we choose dimensionless variables; we put Y = y/h, X = x/h, P = P ⋆ /Gr h, τij = τ ∗ ij /Grh, and
u◦ = η◦u∗ ◦/Grh2 . Then the dimensionless form of the solution is given by the following equations
t◦ = −Y, (3.6)
−Y =
1+k 2 ( u ′
◦) 2 (n−1)/2
u ′
◦ , (3.7)
where t◦ and u ′
◦ are the stress and velocity derivative in the case of parallel flow and k =(λGrh)/η◦. Note
that when k is large, (3.7) represents a shear thinning fluid and when k is small, (3.7) represents a Newtonian
fluid. Equation (3.7) is solved numerically with the boundary condition u◦ = 0 when Y = −1, giving u ′
◦ and u◦.
Examples of velocity profiles are given in Figures 2, 3, and 4.

104
M.A.M. Al Khatib
Figure 2. The velocity profiles of the Carreau model with k =2and for different values of n.
velocity
10
9
8
7
6
5
4
3
2
1
0
0 –0.8 –0.6 –0.4 –0.2 0 0.2 0.4 0.6 0.8 1
V
Figure 3. The velocity profiles of the Carreau model with k =10and for different values of n.
The Arabian Journal for Science and Engineering, Volume 31, Number 1A. January 2006
n=0.3
n=0.5
n=0.8
n=1

velocity
0.6
0.5
0.4
0.3
0.2
0.1
0
-1 -0.8 -0.6 -0.4 -0.2 0
y
0.2 0.4 0.6 0.8 1
M.A.M. Al Khatib
Figure 4. The velocity profiles of the Carreau model with k =0.5 and for different values of n (curves from top to bottom
are, respectively, n =0.1, 0.2, 0.3, 0.5, 0.8, 1).
The velocity profiles in Figures 2 and 3 correspond to the case k>1. They indicate that the velocity gradients
near the boundaries diverge as the index n tends to 0. In fact, it is easy to prove from Equation (3.7) for k>1
and small n that
|u ′ ◦(±1)| = k 1/n−1 .
On the other hand, this divergence does not exist if k

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M.A.M. Al Khatib
The constitutive Equation (3.1) becomes, in dimensionless form,
After linearization we find
τij =2 1+k 2 · γ 2 (n−1) /2
eij. (3.10)
∂u1
τ1 XX =2θ◦
∂X
∂v1
τ1YY =2θ◦
∂Y
τ1 XY = µ◦
∂u1
∂Y
∂v1
+
∂X
where θ◦ =(1+k 2 u ′ 2
◦ ) (n−1)/2 and µ◦ = θ◦
1+nk 2 u ′ 2
◦
1+k 2 u ′ 2
◦
.
(3.11)
From (3.9) and (3.11) with u1 =ΨY and v1 = −ΨX, where Ψ is the stream function, we have
∂2 ∂X∂Y (4θ◦ΨXY
2 ∂ ∂2
)+ −
∂Y 2 ∂X2
[µ◦(ΨYY − ΨXX)] = 0 (3.12)
and the boundary conditions on the walls Y = ±1 are
Now let
Ψ(X, ±1) = 0, ΨY (X, ±1) = 0. (3.13)
Ψ=Φ(Y ) exp(−αX) (3.14)
as in Wilson [1]. Substituting (3.14) in (3.12) we get
2 d
4α
dY
with boundary conditions
θ◦
2
2 dΦ d d Φ
+ − α2 µ◦
dY dY 2 dY 2 − α2
Φ = 0 (3.15)
Φ(±1) = Φ ′ (±1) = 0 (3.16)
and this determines the eigenvalues α. These are complex in general and we are interested in the decaying modes
for which Re(α) > 0. Note that when k =0,orn =1weget
θ◦ = µ◦ =1,
and we recover the Newtonian case
Φ iv +2α 2 Φ ′′
+ α 4 Φ = 0 (3.17)
studied by Wilson [1]. The eigenvalues satisfy sin 2α = ±2α, with four families of roots, one in each quadrant
of the complex plane. The two signs in sin 2α = ±2α correspond to odd and even solutions to Equation (3.15),
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M.A.M. Al Khatib
respectively. Examples of the eigenvalues for the Newtonian case are given in Table 1. These eigenvalues are the
same obtained by Wilson [1].
Note that if we work out the same theory for the power law model, then θ◦ and µ◦ are given by
θ◦ = k 1−1/n |y| 1−1/n and µ◦ = nθ◦,
and the equation corresponding to (3.15) is given by
Φ iv +(4r − 2)α 2 Φ ′′
+ α 4 Φ+ 2
(1 − r) Φ
Y ′′′
+(2r− 1)α 2 Φ ′
+ r
(r − 1)(Φ′′ − α
Y 2 2 Φ) = 0 (3.18)
where r =1/n. The scaling used here is Y = y/h, X = x/h, andu◦ = uQ/h where Q is the flow rate. By using
the power series method with Φ = Y σ amY m the solution of the above equation will be given by
Φ ≈ A + BY + CY r+1 + DY r+2
where A, B, C, and D are constant, and σ =0, 1, r+1, and r + 2 are the four roots of the indicial equation
associated with Equation (3.18). Since r is not an integer, in general, the third and fourth terms are weakly
singular, as noted. And in order to satisfy the boundary conditions at the center line of the channel, that the
solution is odd or even, requires that either A and C or B and D are zero respectively. But it would be very
difficult to distinguish numerically between Y r+1 and Y r+2 because they are generally high powers.
4. RESULTS AND DISCUSSION
Table 1. The Values of α for the Newtonian Case.
Eigenvalue α
1 2.106196 + i1.125364
2 3.748838 + i1.384339
3 5.356269 + i1.551574
4 6.949978 + i1.671049
5 8.536682 + i1.775544
6 10.119259 + i1.85838
The problem of this paper has been reduced to a linear two-point boundary-value problem for a fourthorder
differential equation which is linearized on the assumption of small disturbance from the fully-developed
parallel flow, leading to the eigenvalue equation given in (3.15). Several numerical methods have previously
been used to solve eigenvalue equations. For example, Bramley and Dennis [5] calculated the eigenvalues for
the stationary perturbation of Poiseuille flow (Newtonian case) by expanding Φ and its derivatives in terms of
Chebyshev polynomials. In fact, the same method is used by Orszag [6] to get an accurate solution of the Orr–
Sommerfeld stability equation. Stocker and Duck [7] have also used the method to investigate the eigenvalues for
the stationary perturbation of Couette–Poiseuille flow. While this method has proven to be accurate we adopt
a more straightforward and suitable one for our equations which are more complex than those investigated by
the previously mentioned authors.
Equation (3.15) is solved numerically using the NAG routines D02CAF (Adams method) and C02NBF (Powell
hybrid method) with the boundary conditions in (3.16), giving the eigenvalues α for different values of k.
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M.A.M. Al Khatib
Examples of the eigenvalues are given in Tables 2, 3, and 4 for the same values of k used to draw the velocity
profiles in Figures 2, 3, and 4.
Table 2. The Leading Eigenvalue α for the Case k =2and n → 0.
n α
1.000 2.106196 + i1.125364
0.900 2.063433 + i1.116431
0.800 2.014212 + i1.106579
0.700 1.956469 + i1.095871
0.600 1.886985 + i1.084694
0.500 1.800307 + i1.074300
0.400 1.685923 + i1.068200
0.300 1.519094 + i1.074583
0.200 1.225893 + i1.094100
0.150 0.975289 + i1.068689
0.100 0.626237 + i 0.902539
0.050 0.218521 + i 0.380135
0.040 0.132731 + i 0.230975
0.030 0.059276 + i 0.109136
0.025 0.022121 + i 0.048524
0.024 0.014812 + i 0.036458
0.023 0.007509 + i 0.025245
0.022 0.000353 + i 0.014025
Table 3. The Leading Eigenvalue α for the Case k =10and n → 0.
n α
1.00 2.106196 + i1.125364
0.90 2.035593 + i1.128023
0.80 1.955879 + i1.135971
0.70 1.863159 + i1.152236
0.60 1.748825 + i1.180960
0.50 1.592574 + i1.223862
0.40 1.353831 + i1.260822
0.30 0.992884 + i1.205290
0.20 0.544458 + i 0.881442
0.15 0.309815 + i 0.565653
0.12 0.155244 + i 0.392127
0.11 0.105398 + i 0.332444
0.10 0.054936 + i 0.272450
0.09 0.004784 + i 0.218245
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Table 4. The Leading Eigenvalue α for the Case k =0.5 and n → 0.
n α
1.00 2.106196 + i1.125364
0.90 2.098909 + i1.122195
0.80 2.091326 + i1.118841
0.70 2.083419 + i1.115280
0.60 2.075158 + i1.111489
0.50 2.066506 + i1.107441
0.40 2.057417 + i1.103102
0.30 2.048824 + i1.098916
0.20 2.037720 + i1.093385
0.10 2.026973 + i1.087899
0.00 2.015508 + i1.081899
M.A.M. Al Khatib
From Tables 2, 3, and 4 and the corresponding graph given in Figure 5, it appears that when k = 2 and
k = 10, Re(α) → 0 as the power law index, n → 0.0220 and 0.0892 respectively. This implies that, when k =2
and k = 10, the parallel flow is unstable for 0

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M.A.M. Al Khatib
From the above results and the velocity profiles in Figures 2, 3, and 4, we conclude that when k>1, Re(α)
goes to a non-zero value as n → 0. On the other hand, when k>1, Re(α) → 0asn → nmin > 0, which implies
that the parallel flow is unstable for 0