This document contains the work of Steven Brandon in answering 6 calculation questions regarding the flow properties of a non-Newtonian fluid flowing through a pipe. Brandon calculates the velocity profile, volumetric flow rate, average velocity, Reynolds number, friction factor, and viscosity at 5°C. He determines that the flow is laminar based on the Reynolds number. The velocity profile is flattened compared to Newtonian flow. The volumetric flow rate is 1.71 L/s. The average velocity is 1.78 m/s. The Reynolds number is 288.6. The friction factor is 0.222. The viscosity at 5°C is calculated to be 1100 mPa·s.

1. Preliminary Exam
Dr. Terry Walker
Steven Brandon May 4, 2004
1
A non-Newtonian fluid is flowing in a 10-m long pipe. The inside diameter is
3.5 cm. The pressure drop is measured at 100 kPa. The consistency
coefficient is 5.2 Pa•sn
and flow behavior index is 0.45.
1. Calculate and plot the velocity profile (velocity vs. radius)
2. Determine volumetric flow rate
3. Determine average velocity
4. Determine generalized Reynolds number
5. Determine the friction factor
6. Determine the viscosity at 5°C, if the Arrhenius coefficient, A = 4.3E-9
Pa•s and the activation energy, Ea = 10.7 Kcal/g-mol.
All the formulas used in these calculations are from Rheological Methods in
Food Process Engineering, 2nd
Ed. by James F. Steffe.1
Numbers given in
brackets beside each formula correspond to those given in this text. If no
number is given, I have indicated the page on which the formula may be
found.
For your convenience, I have shown my calculations here. I performed all of
the calculations using Excel.
1. Calculate and plot the velocity profile (velocity vs. radius).
The formula for the velocity profile of Power-Law fluids, such as the fluid
described above, flowing in laminar flow inside a circular conduit (pipe) is:
   
 nnnn
n
rR
n
n
LK
P
rfu /1/1
1
12
)( 












 
 [2.40]
where K is the consistency index, n is the flow behavior index, R is the
inside radius of the pipe and r is a variable radius measured from the
centerline of the pipe out to R = D/2. Since this formula assumes laminar
flow, I must first determine the Reynolds Number to verify that the flow is
laminar.
Steffe gives the following formula for power law fluids:
  n
n
nn
PL
n
n
K
uD
N 













 

13
4
8 1
2
Re,
 [2.52]
L = 10 m
D = 3.5 cm
K = 5.2 Pa•sn
n = 0.45ΔP = 100 kPa
m

2. Preliminary Exam
Dr. Terry Walker
Steven Brandon May 4, 2004
2
where D is the inside diameter of the pipe, ρ is the density of the fluid and
u is the average velocity in the pipe.
The critical Reynolds Number for Power Law fluids, below which the flow
may be assumed to be laminar, is determined from:
2
)1/()2(
Re,
)31(
)2(6464
)(
n
nn
N
nn
CRITICALPL




[2.51]
Before proceeding, the average velocity must be found using the following
equation:
 2
/ RQu 
p. 104
Since the average velocity is calculated from the volumetric flow rate, Q,
the flow rate must be found first.
 
  nn
n
R
n
n
LK
P
Q /13
1
132













 
 
[2.31]
Q =  [(100,000 Pa)/(2(10 m)(5.2 Pa•s0.45
))](1/0.45)
x
[0.45/(1.35+1)](0.0175 m)(1.35+1)/0.45
Q = 0.00171 m³/s = 1.71 L/s = 27.1 gpm.
The average velocity may now be calculated.
u = Q/ R2
= (0.00171 m³/s)/ (0.0175 m)2
= 1.78 m/s.
Now, the Reynolds Number can be found using Equation 2.52, above.
Since no density was given for this fluid, I will assume that it has the
density of water at room temperature, 1000 kg/m3
.
NRe,PL = [(0.035 m)0.45
(1.78 m/s)(2-0.45)
(1000 kg/m3
) ÷
(8-0.55
(5.2 Pa•s0.45
))[1.8/(1.35+1)]0.45
NRe,PL = 288.6.
The critical Reynolds Number is determined from Equation 2.51, above.
(NRe,PL)CRITICAL = 6464(0.45)(2.45)(2.45/1.45)
/(1+1.35)2
(NRe,PL)CRITICAL = 2,394.

3. Preliminary Exam
Dr. Terry Walker
Steven Brandon May 4, 2004
3
Since NRe,PL < (NRe,PL)CRITICAL, the flow can be considered to be laminar.
Therefore, the velocity profile may be determined using Equation 2.40,
above. To determine the entire profile, the velocity must be calculated
over a range of radii from the centerline (r = 0 m) to r = R. By way of
example, I present here the determination of the centerline (maximum)
velocity.
u(r = 0 m) = [(100,000 Pa)/(2(10 m)(5.2 Pa•sn
)]1/0.45
x [0.45/(1.35)] x
[(0.0175 m)(1.35/0.45)
– 0]
u(r = 0 m) = 2.88 m/s
The no-slip boundary condition is accurately predicted by this formula,
since the velocity goes to zero when r = R.
I used Excel to calculate velocities at r = 0 m and at radii in Δr = 0.05 cm
increments up to r = R (1.75 cm). Figure 1, below, is a plot of the
velocity profile for this case. Note that this profile is flattened in contrast
to the parabolic velocity profile of a Newtonian fluid in laminar flow in a
pipe, where the velocity at the center line of the pipe, which is also the
maximum velocity, uMAX, is twice the average velocity, u . In this profile,
the ratio
u
uMAX
is only 1.62.
Velocity Profile
0
1
2
3
4
0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75
Radius, cm
Velocity,m/s
Velocity Profile
Figure 1. Velocity Profile

4. Preliminary Exam
Dr. Terry Walker
Steven Brandon May 4, 2004
4
2. Determine volumetric flow rate, Q.
The volumetric flow rate was determined in the calculations for Question
1, above. Q = 0.00171 m³/s = 1.71 L/s = 27.1 gpm.
3. Determine average velocity, u .
The average velocity in the pipe was determined in the calculations for
Question 1, above. u = 1.78 m/s.
4. Determine generalized Reynolds number.
The Reynolds number (generalized to apply to non-Newtonian fluids, as
well as Newtonian fluids) was determined in the calculations for Question
1, above. NRe,PL = 288.6.
5. Determine the friction factor, f.
The Fanning Friction Factor is determined using the following formula:
PL
FANNING
N
f
Re,
16
 [2.113]
fFANNING = 16/288.6 = 0.0554.
The Darcy Friction factor (which I favor) is simply four times the Fanning
Friction Factor.
FANNINGDARCY ff 4
fDARCY = 4(0.0554) = 0.222.
6. Determine the viscosity at 5°C, if the Arrhenius coefficient, A = 4.3E-9
Pa•s and the activation energy, Ea = 10.7 Kcal/gmol.
  






RT
E
ATf a
exp p. 85
where T is the absolute temperature in kelvins; R, the Universal Gas
Constant, = 1.987 calories/(g-mole•K); Ea is the activation energy and A
is the Arrhenius Coefficient.
μ = (4.3E-9 Pa•s) x
exp[(10,700 cal./g-mole)/(1.987 cal./(g-mole•K))((273.15 +5) K)]
μ = 1.10 Pa•s = 1100 mPa•s = 1100 cP.

5. Preliminary Exam
Dr. Terry Walker
Steven Brandon May 4, 2004
5
1
Steffe, James F., Rheological Methods in Food Process Engineering, 2nd
ed.,
Freeman Press, 2807 Still Valley Dr., East Lansing, MI 48823 USA, 1996