SOLUTION TO CHAPTER 4 EXERCISES: SLURRY TRANSPORT EXERCISE 4.1 Samples of a phosphate slurry mixture are analyzed in a lab. The following data describe the relationship between the shear stress and the shear rate: Shear Rate& ( Shear Stress, )1sec ( )Pa 25 38 75 45 125 48 175 51 225 53 325 55.5 425 58 525 60 625 62 725 63.2 825 64.3 The slurry mixture is non-Newtonian. If it is considered a power-law slurry, what is the relationship of the viscosity to the shear rate? SOLUTION TO EXERCISE 4.1: First prepare a plot of log(shear stress) versus log (shear rate).

nk = & log log logk n = + &

Shear Rate Shear Stress Log (Shear Stress) Log (Shear Rate)

25 38 1.58 1.40 75 45 1.65 1.88 125 48 1.68 2.10 175 51 1.71 2.24 225 53 1.72 2.35 325 55.5 1.74 2.51 425 58 1.76 2.63 525 60 1.78 2.72 625 62 1.79 2.80 725 63.2 1.80 2.86 825 64.3 1.81 2.92

From this plot

1

Slope = 0.15 Intercept = 1.37

Hence,

Slope 0.15n= =Intercept log 1.37k= = k = 23.4 Ns0.15/m2

0.1523.4 = & with & in 1sec and in Pa and

0.85app 23.4

= = &&

Problem 4.1

010203040506070

0 200 400 600 800 1000

Shear Rate

Shea

r Str

ess

Problem 4.1

1.551.6

1.651.7

1.751.8

1.85

0 1 2 3 4

Log (Shear Rate)

Log

(She

ar S

tres

s)

2

EXERCISE 4.2: Verify equation 4.7 SOLUTION TO EXERCISE 4.2:

r

z

For one-dimensional, fully-developed, laminar flow in a pipe, the z component of the momentum balance simplifies to the differential equation

( )1 rzpO rz r r = +

For a power-law fluid

dd

nz

rzvkr

= Hence,

ddd d

nzvP kO r

L r r r = + +

Here pz

has been replaced by PL+ since the pressure gradient is constant.

Integrating once,

1 d2 d

nzC vr P

k L r r + =

must equal to zero since the velocity gradient is zero at 1C 0r = . Integrating again,

11 1

212 1

nn

zP r C vLk

n

+ + + =

Applying the boundary condition that 0zv = at r R=

1 11 11

12

n nn

zP R rv

nkLn

+ + = + +

3

The average velocity is equal to AVv

AV 20

2 R

zv rvR= dr

Hence

1 11 31 2

AV 2

0

2 n+1 3 1 12 2n

R

n nnP R r rvn nR kLn n

+ + = + +

and

( )1 13

AV 2

22 2 3

nnP nv R

R kL n+ = + 1

Simplifying with 2D R=

( )1

AV 4 2 3

nPD DnvkL n

= + 1 EXERCISE 4.3: Verify equation 4.28 SOLUTION TO EXERCISE 4.3:

r

z

For one-dimensional, fully-developed laminar flow in a pipe, the z component of the momentum balance simplifies to the differential equation

( )1 rzPO rz r r = +

For any fluid

4

( )Pr ddr rzr

L =

Integrating,

Pr2 rzL

= since rz must be finite at 0r = For pipe flow

ddz

rz y pvr

= + since ddzvr

is negative for pipe flow

Hence,

dPr2 d

zy p

vL r

= + Rearranging,

dPr2 d

y z

p p

vL r

+ = Integrating,

2

2Pr

4y

zp p

rC v

L

+ + =

Apply the boundary condition 0zv = r R=

2

2P

4y

p p

RR CL

+ = ( ) ( )2 2P r

4

yz

p p

R r Rv

L

= +

This velocity profile is valid for *R r R . For the plug flow region , *r R d 0dzvr= .

In the plug flow region, the velocity is constant and

*

2rz yPRL

= =

5

The plug flow velocity can be found by substituting this expression for y into the velocity profile. ( ) ( )2 *2 * *P P

4 2z p p

R R R Rv

L L = + R

( )2*P

4z p

R Rv

L = for *0 r R

Both velocity regions must be integrated to determine the average velocity . AVv

( ) ( ) ( )**

2* 2 2AV 2 2

0

2 2 d 4 4

R Ry

p pR

P Pv R R r r R r rR L R L

= + + dp R r r Integrating,

42 * *

AV4 11

8 3 3p

PR R RvL R R = +

or

4

0AV

0 0

4 114 3 3

y y

p

Rv

= +

where 0 is the wall shear stress

0 2R PL

= EXERCISE 4.4 A slurry behaving as a pseudoplastic fluid is flowing through a smooth round tube having an inside diameter of 5 cm at an average velocity of 8.5 m/s. The density of the slurry is 900 kg/m3 and its flow index and consistency index are n = 0.3 and k = 3.0 Ns0.3/m2. Calculate the pressure drop for a) 50 m length of horizontal pipe and b) 50 m length of vertical pipe with the flow moving against gravity.

6

SOLUTION TO EXERCISE 4.4: First check if the flow is laminar or turbulent

( )( )

2.31.71.3

*0.3

*

6464 0.3 2.3 1Re1.91.9

Re 2345

transition

transition

=

=

*Re at flow conditions,

( ) 1.70.3 0.33

*0.3

2

kg m8 900 0.05 m 8.5 0.3m sRe

Ns 3.83.0 m

=

* *Re 17340 Re turbulenttransition= > The friction factor ff is given by

( )* 20.751 4 0log Re nff

fn

.4f n

= Solving for ff with and

*Re 17340= 0.3n = , yields 0.0026ff = a) the pressure drop for horizontal flow is then

22 f m AL

Vp f vD =

( ) ( )23kg 50m m2 0.0026 900 8.5 sm 0.05mp =

7

2kN338m

p = b) the pressure drop for vertical flow is given by

fm

ph zg

= + or

22m f m f m AV mLp gh g z f v g zD

= + = +

( ) ( )23 3kg 50m kg m2 0.0026 900 8.5 900 9.8 50mm 0.05m m smps

= + 2

2

kN779m

p = EXERCISE 4.5 The concentration of a water-based slurry sample is to be found by drying the slurry in an oven. Determine the slurry weight concentration given the following data:

Weight of container plus dry solids 0.31kg Weight of container plus slurry 0.48kg Weight of container 0.12kg

Determine the density of the slurry if the solid specific gravity is 3.0. SOLUTION TO EXERCISE 4.5: Weight of dry solids = 0.31-0.12 = 0.19kg Weight of slurry = 0.48-0.12 = 0.36kg

Concentration of solids by weight = 0.19 0.530.36

= The density of the slurry is found by

8

3 3

3

1 0.53 0.47kg kg3000 1000m m

kg1546m

m

m

= +

=

EXERCISE 4.6 A coal-water slurry has a specific gravity of 1.3. If the specific gravity of coal is 1.65, what is the weight percent of coal in the slurry? What is the volume percent coal? SOLUTION TO EXERCISE 4.6:

( )11 ww

m s f

CC

= +

( )11

1.3 1.65 1.0ww CC = +

Solving for , wC

0.586wC = mass coal

mass slurrywC =

Volume fraction coal = w ms

C

Volume fraction coal = ( )( )0.586 1.3 0.46

1.65=

EXERCISE 4.7 The following rheology test results were obtained for a mineral slurry containing 60 percent solids by weight. Which rheological model describes the slurry and what are the appropriate rheological properties for this slurry?

9

Rate of Shear (1/s) Shear Stress (Pa) 0 4.0

0.1 4.03 1 4.2 10 5.3 15 5.8 25 6.7 40 7.8 45 8.2

SOLUTION TO EXERCISE 4.7: The shear stress at zero shear rate is 4.0 Pa. Hence, this slurry exhibits yield stress equal to 4.0 Pa. In order to determine whether the slurry behaves as a Bingham fluid or if it follows the Herschel-Bulkley model, we need to plot y versus shear rate.

(Pa)y Shear Rate (1/s) 0 0 0.03 0.1 0.2 1 1.3 10 1.8 15 2.7 25 3.8 40 4.2 45 A plot of y versus shear rate on an arithmetric scale is not linear. However, a plot of y versus shear rate on a log-log scale is linear (the data for zero shear rate is excluded) ny k = & ( )ln ln lny k n = + &

10

( )ln y ln& Shear Rate

-3.51 -2.30 0.1 -1.61 0 1 +0.26 +2.30 10 +0.59 +2.71 15 +0.99 +3.22 25 +1.34 +3.69 40 +1.44 +3.81 45

Slope = 0.81 = n Intercept = -1.62 = ln k

0.81

2

Ns 0.20m

k = EXERCISE 4.8 A mud slurry is drained from a tank through a 50 ft. long horizontal plastic hose. The hose has an elliptical cross-section, with a major axis of 4 inches and a minor axis of 2 inches. The open end of the hose is 10 feet below the level in the tank. The mud is a Bingham plastic with a yield stress of 100 dynes/cm2, a plastic viscosity of 50cp, and a density of 1.4 g/cm3. a) At what velocity will water drain from the hose? b) At what velocity will the mud drain from the hose? SOLUTION TO EXERCISE 4.8:

Applying the modified Bernoulli equation to the system between points 1 and 2,

11

2AV0

2fvh zg

= + + +

or,

2AV

2 10 2fvh z zg

= + + +

where 2

2AV

2 ff

h

f Lh vg D

= and 2 1 3.048 mz z =

Need to determine the hydraulic diameter of the pipe with the elliptical cross-section:

4 cross-sectional area

wetted perimeterhD =

2a2b

( )( )( )

( )( )( )

2 2

2 2

4

22

4 2 in 1 in2.53 in 0.0643 m

4 in 1 in22

h

h

abD

a b

D

=

+

= = =+

Plugging in numbers (SI units),

( )

2 2 AV

AV

2 2

2 15.24 m0 3.048 mm m0.0643 m9.8 2 9.8 s s

ff vv = +

(*) 2AV AV0 0.051 3.048 48.3 fv= + 2f v

12

Solution Procedure:

AV

1) Calculate 2) Guess velocity 3) Calculate Re4) Find from Figure 6.

5) Check governing equation (*)6) If governing equation is not satisfied, guess a new velocity

AV

f

Hev

f

v

( )22 3 222

kg kg1400 0.0643 m 10 m m

kg0.050 m s

23,153

m h y

p

DHe

He

s = =

=

Solution via iterative procedure for water:

( )35

AV

kg m1000 0.0643 m 3.65 m s3.65 m s Re 2.3 10

kg0.001 m s

Dvv = = = =

5(Re 2.3 10 ;smooth tube) 0.0037ff = = Solution via iterative procedure for mud:

( )3

3AV

kg m1400 3.2 0.0643 mm m s3.2 Re 5.7 10kgs 0.05

m s

v

= = =

3 4(Re 5.7 10 ; 2.3 10 ) 0.005ff He= =

13

EXERCISE 4.9 A coal slurry is found to behave as a power-law fluid with a flow index 0.3, a specific gravity 1.5, and an apparent viscosity of 70cp at a shear rate 100s-1. a) What volumetric flow rate of this fluid would be required to reach turbulent flow in a 1/2 in. I.D. smooth pipe which is 15 ft. long? b) What is the pressure drop (in Pa) in the pipe under these conditions? SOLUTION TO EXERCISE 4.9: a) First, calculate *transition for n = 0.3 Re

( )( )( ) ( ) ( )

2 0.32 0.3

* 1 0.3transition 0.3

*transition

6464 0.3 1Re 2 0.31 3 0.31 3 0.3

Re 2340

+ + = + ++

=

Also, need to calculate , the consistency index k 1app

nk = &

0.7kg 1000.07

m s sk

=

1.7kg 1.76

msk =

Applying equation 4.19, the average velocity in the smooth pipe can be found.

( )2

* AV8Re6 2

nn nmD v nk n

= +

( )

( )0.3 1.7 0.3

AV3

1.7

kg8 1500 0.0127 m0.3m2340= kg 6 0.3 21.76

ms

v +

Solving for AVv

14

AVm1.61 s

v = The volumetric flow rate Q is then

( ) 22AV C

34

m 0.0254 m1.61 0.5 ins 1 in(Cross-Sectional Area, A )

4

m 2.0 10 s

Q v

= =

=

b)

2AV2 f mLp f vD

=

*16 0.007Ref

f = =

( ) 23kg 4.572 m m2 0.007 1500 1.61m 0.0127 m s19600 Pa

p

p

=

=

EXERCISE 4.10 A mud slurry is draining from the bottom of a large tank through a 1 m long vertical pipe that is 1 cm I.D. The open end of the pipe is 4 m below the level in the tank. The mud behaves as a Bingham plastic with a yield stress of 10 N/m2, an apparent viscosity of 0.04 kg/ms, and a density of 1500 kg/m3. At what velocity will the mud slurry drain from the hose?

15

SOLUTION TO EXERCISE 4.10:

Applying the modified Bernoulli equation to the system above

22AV

2 1 2fv

h z zg

= + Assuming the flow is laminar, the head loss due to pipe friction is given by

3 2fm

p phg

= Applying equation 4.30,

2AV2

32 163

p y

fm

v L LD Dh

g

+=

( )( )( )

( )( )

2AV 2

2

3 2

AV

kg N32 0.04 1 m 16 10 1 mm s m

3 0.01 m0.01 mkg m1500 9.8 m s

0.87 0.36

f

f

v

h

h v

+=

= +

16

Plugging this head loss back into the modified Bernoulli equation,

( )2AV

AV0.87 0.36 4 2 9.8vv = +

Rearranging,

( ) ( )

2AV AV

2

AV

0 71.3 17.1

17.1 17.1 4 71.3

2

v v

v

= + +

=

AV m3.5 sv = (other solution yields a negative velocity) Check original assumption to see if flow is laminar

( )3

AV

kg m1500 3.5 0.01 mm sRe 1300

kg0.04 m s

m

p

v D

= = =

Flow is laminar and original assumption is correct. EXERCISE 4.11 A mud slurry is draining in laminar flow from the bottom of a large tank through a 5 m long horizontal pipe that is 1 cm inside diameter. The open end of the pipe is 5 m below the level in the tank. The mud is a Bingham plastic with a yield stress of 15 N/m2, an apparent viscosity of 0.06 kg/ms, and a density of 2000 kg/m3. At what velocity will the mud slurry drain from the hose?

17

SOLUTION TO EXERCISE 4.11:

Applying the modified Bernoulli equation to the system above

2AV

2 1 2fvh z zg

= + Assuming the flow is laminar, the head loss due to pipe friction is given by

( )( )( )

( )( )

2

2

2

AV2

3 2

AV 2

2

3 2

AV

32 163

kg N32 0.06 5 m 16 15 5 mm s m

3 0.01 m0.01 mkg m2000 9.8 m s

4.90 2.04

p y

fm m

f

f

v L Lp p D Dh

g g

v

h

h v

+= =

+=

= +

18

Plugging this head loss back into the modified Bernoulli equation,

2

2AV

AV

2

4.9 2.04 5m2 9.8 s

vv = +

Rearranging,

( ) ( )

2AV AV

2

AV

96.0 58 0

96 96 4 582

v v

v

+ =

=

AVm0.6 s

v = (other solution yields a negative velocity) Check original assumption to see if flow is laminar

( )3

AV

kg m2000 0.6 0.01 mm sRe

kg0.06 m s

m

p

v D

= =

flow is laminar Re 200=

19