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

rRn

n

LK

Prfu /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

PLn

n

K

uDN

13

4

8 1

2

Re,

[2.52]

L = 10 m

D = 3.5 cm

K = 5.2 Pa•sn n = 0.45 ΔP = 100 kPa m

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

nnN

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

Rn

n

LK

PQ /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.

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

Ve

loc

ity

, m

/s

Velocity Profile

Figure 1. Velocity Profile

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

FANNINGN

fRe,

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

EATf aexp

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.

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