The document provides solutions to several exercises related to slurry transport. For Exercise 4.1, the solution analyzes shear stress and shear rate data for a phosphate slurry and determines it follows a power-law relationship with a flow index of 0.15 and consistency index of 23.4 Ns0.15/m2. Exercise 4.2 verifies an equation for pressure drop in pipe flow of a power-law fluid. Exercise 4.3 similarly verifies an equation incorporating a yield stress. Subsequent exercises provide solutions for pressure drop, slurry concentration, and rheological properties calculations using data given.

1. 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，γ& ( Shear Stress,
)
1
sec−
τ ( )
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).
n
k
τ γ
= &
log log log
k n
τ γ
= + &
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

2. Slope = 0.15
Intercept = 1.37
Hence,
Slope 0.15
n
= =
Intercept log 1.37
k
= =
k = 23.4 Ns0.15
/m2
0.15
23.4
τ γ
∴ = & with γ& in 1
sec−
and τ in Pa
and
0.85
app 23.4
τ
μ γ
γ
−
= = &
&
Problem 4.1
0
10
20
30
40
50
60
70
0 200 400 600 800 1000
Shear Rate
Shear
Stress
Problem 4.1
1.55
1.6
1.65
1.7
1.75
1.8
1.85
0 1 2 3 4
Log (Shear Rate)
Log
(Shear
Stress)
2

3. 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
rz
p
O r
z r r
τ
∂ ∂
= − +
∂ ∂
For a power-law fluid
d
d
n
z
rz
v
k
r
τ
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
Hence,
d
d
d d
n
z
v
P k
O r
L r r r
⎡ ⎤
Δ ⎛ ⎞
= + + ⎢ ⎥
⎜ ⎟
⎝ ⎠
⎢ ⎥
⎣ ⎦
Here
p
z
∂
−
∂
has been replaced by
P
L
Δ
+ since the pressure gradient is constant.
Integrating once,
1 d
2 d
n
z
C v
r P
k L r r
Δ ⎛ ⎞
− + = ⎜ ⎟
⎝ ⎠
must equal to zero since the velocity gradient is zero at
1
C 0
r = .
Integrating again,
1
1 1
2
1
2
1
n
n
z
P r
C v
Lk
n
+
Δ
⎛ ⎞
− +
⎜ ⎟
⎛ ⎞
⎝ ⎠ +
⎜ ⎟
⎝ ⎠
=
Applying the boundary condition that 0
z
v = at r R
=
1 1
1 1
1
1
2
n n
n
z
P R r
v
n
kL
n
+ +
⎡ ⎤
⎢ ⎥
Δ −
⎛ ⎞
⎢ ⎥
= +⎜ ⎟ +
⎛ ⎞
⎝ ⎠ ⎢ ⎥
⎜ ⎟
⎢ ⎥
⎝ ⎠
⎣ ⎦
3

4. The average velocity is equal to
AV
v
AV 2
0
2
R
z
v rv
R
= ∫ dr
Hence
1 1
1 3
1
2
AV 2
0
2
n+1 3 1 1
2
2
n
R
n n
n
P R r r
v
n n
R kL
n n
+ +
⎡ ⎤
⎢ ⎥
Δ
⎛ ⎞
⎢ ⎥
= −
⎜ ⎟ + +
⎛ ⎞ ⎛ ⎞⎛ ⎞
⎝ ⎠ ⎢ ⎥
⎜ ⎟ ⎜ ⎟⎜ ⎟
⎢ ⎥
⎝ ⎠ ⎝ ⎠⎝ ⎠
⎣ ⎦
and
( )
1 1
3
AV 2
2
2 2 3
n
n
P n
v R
R kL n
+
⎡ ⎤
⎛ ⎞
Δ
⎛ ⎞
= ⎢ ⎥
⎜ ⎟
⎜ ⎟ ⎜ ⎟
+
⎝ ⎠ ⎢ ⎥
⎝ ⎠
⎣ ⎦
1
Simplifying with
2
D
R
=
( )
1
AV
4 2 3
n
PD Dn
v
kL n
Δ
⎛ ⎞
= ⎜ ⎟
+
⎝ ⎠ 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
rz
P
O r
z r r
τ
∂ ∂
= − +
∂ ∂
For any fluid
4

5. ( )
Pr d
dr
rz
r
L
τ
Δ
− =
Integrating,
Pr
2
rz
L
τ
Δ
− = since rz
τ must be finite at 0
r =
For pipe flow
d
d
z
rz y p
v
r
τ τ μ
= − + since
d
d
z
v
r
is negative for pipe flow
Hence,
d
Pr
2 d
z
y p
v
L r
τ μ
Δ
− = − +
Rearranging,
d
Pr
2 d
y z
p p
v
L r
τ
μ μ
Δ
− + =
Integrating,
2
2
Pr
4
y
z
p p
r
C v
L
τ
μ μ
Δ
− + + =
Apply the boundary condition 0
z
v = r R
=
2
2
P
4
y
p p
R
R
C
L
τ
μ μ
Δ
− + = −
( ) ( )
2 2
P r
4
y
z
p p
R r R
v
L
τ
μ μ
Δ − −
∴ = +
This velocity profile is valid for *
R r R
≤ ≤ . For the plug flow region ,
*
r R
≤
d
0
d
z
v
r
= .
In the plug flow region, the velocity is constant and
*
2
rz y
PR
L
τ τ
Δ
− = =
5

6. The plug flow velocity can be found by substituting this expression for y
τ into the
velocity profile.
( ) ( )
2 *2 * *
P P
4 2
z
p p
R R R R
v
L L
μ μ
Δ − Δ −
= +
R
( )
2
*
P
4
z
p
R R
v
L
μ
Δ −
∴ = for *
0 r R
≤ ≤
Both velocity regions must be integrated to determine the average velocity .
AV
v
( ) ( ) ( )
*
*
2
* 2 2
AV 2 2
0
2 2
d
4 4
R R
y
p p
R
P P
v R R r r R r r
R L R L
τ
μ μ μ
⎡ ⎤
Δ Δ
= − + − + −
⎢ ⎥
⎢ ⎥
⎣ ⎦
∫ ∫ d
p
R r r
Integrating,
4
2 * *
AV
4 1
1
8 3 3
p
PR R R
v
L R R
μ
⎡ ⎤
⎛ ⎞ ⎛ ⎞
Δ
= − +
⎢ ⎥
⎜ ⎟ ⎜ ⎟
⎢ ⎥
⎝ ⎠ ⎝ ⎠
⎣ ⎦
or
4
0
AV
0 0
4 1
1
4 3 3
y y
p
R
v
τ τ
τ
μ τ τ
⎡ ⎤
⎛ ⎞
⎢ ⎥
= − + ⎜ ⎟
⎢ ⎥
⎝ ⎠
⎣ ⎦
where 0
τ is the wall shear stress
0
2
R P
L
τ
Δ
=
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

7. SOLUTION TO EXERCISE 4.4:
First check if the flow is laminar or turbulent
( )
( )
2.3
1.7
1.3
*
0.3
*
6464 0.3 2.3 1
Re
1.9
1.9
Re 2345
transition
transition
× ⎛ ⎞
= ⎜ ⎟
⎝ ⎠
=
*
Re at flow conditions,
( )
1.7
0.3
0.3
3
*
0.3
2
kg m
8 900 0.05 m 8.5
0.3
m s
Re
Ns 3.8
3.0
m
⎛ ⎞ ⎛ ⎞
× × ×
⎜ ⎟ ⎜ ⎟
⎡ ⎤
⎝ ⎠ ⎝ ⎠
= ⎢ ⎥
⎣ ⎦
* *
Re 17340 Re turbulent
transition
= > ⇒
The friction factor f
f is given by
( )
* 2
0.75
1 4 0
log Re n
f
f
f
n
.4
f n
−
= −
Solving for f
f with and
*
Re 17340
= 0.3
n = , yields
0.0026
f
f =
a) the pressure drop for horizontal flow is then
2
2 f m A
L
V
p f v
D
ρ
Δ =
( ) ( )
2
3
kg 50m m
2 0.0026 900 8.5
s
m 0.05m
p
⎛ ⎞⎛ ⎞⎛ ⎞
Δ = ⎜ ⎟
⎜ ⎟⎜ ⎟
⎝ ⎠
⎝ ⎠⎝ ⎠
7

8. 2
kN
338
m
p
Δ =
b) the pressure drop for vertical flow is given by
f
m
p
h z
g
ρ
Δ
− = − + Δ
or
2
2
m f m f m AV m
L
p gh g z f v g z
D
ρ ρ ρ ρ
⎛ ⎞
Δ = + Δ = + Δ
⎜ ⎟
⎝ ⎠
( ) ( )
2
3 3
kg 50m kg m
2 0.0026 900 8.5 900 9.8 50m
m 0.05m m s
m
p
s
⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞
Δ = +
⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠
2
2
kN
779
m
p
Δ =
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
0.53
0.36
=
The density of the slurry is found by
8

9. 3 3
3
1 0.53 0.47
kg kg
3000 1000
m m
kg
1546
m
m
m
ρ
ρ
= +
=
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:
( )
1
1 w
w
m s f
C
C
ρ ρ ρ
−
= +
( )
1
1
1.3 1.65 1.0
w
w
C
C −
= +
Solving for ,
w
C
0.586
w
C =
mass coal
mass slurry
w
C =
Volume fraction coal = w m
s
C ρ
ρ
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

10. 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
τ τ
− versus shear rate.
(Pa)
y
τ τ
− 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
τ τ
− versus shear rate on an arithmetric scale is not linear. However, a plot
of y
τ τ
− versus shear rate on a log-log scale is linear (the data for zero shear rate is
excluded)
n
y k
τ τ γ
− = &
( )
ln ln ln
y k n
τ τ γ
− = + &
10

11. ( )
ln y
τ τ
− lnγ& 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 =lnk
0.81
2
Ns
0.20
m
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

12. 2
AV
0
2
f
v
h z
g
⎧ ⎫
= + + Δ + Δ⎨ ⎬
⎩ ⎭
or,
2
AV
2 1
0
2
f
v
h z z
g
= + + − +
where 2
2
AV
2 f
f
h
f L
h v
g D
⎛ ⎞
= ⎜ ⎟
⎝ ⎠
and 2 1 3.048 m
z z
− = −
Need to determine the hydraulic diameter of the pipe with the elliptical cross-section:
4 cross-sectional area
wetted perimeter
h
D =
2a
2b
( )( )
( )
( )( )( )
2 2
2 2
4
2
2
4 2 in 1 in
2.53 in 0.0643 m
4 in 1 in
2
2
h
h
πab
D
a b
D
π
=
+
= = =
+
Plugging in numbers (SI units),
( )
2
2 AV
AV
2 2
2 15.24 m
0 3.048 m
m m
0.0643 m
9.8 2 9.8
s s
f
f v
v
⎛ ⎞
= − +
⎜ ⎟
⎛ ⎞
⎝ ⎠
⎜ ⎟
⎝ ⎠
(*)
2
AV AV
0 0.051 3.048 48.3 f
v
= − + 2
f v
12

13. Solution Procedure:
AV
1) Calculate
2) Guess velocity
3) Calculate Re
4) Find from Figure 6.
5) Check governing equation (*)
6) If governing equation is not satisfied, guess a new velocity
AV
f
He
v
f
v
( )
2
2 3 2
2
2
kg kg
1400 0.0643 m 10
m m
kg
0.050
m s
23,153
m h y
p
D
He
He
ρ τ
μ
⎛ ⎞ ⎛
⎜ ⎟ ⎜
s
⎞
⎟
⋅
⎝ ⎠ ⎝
= =
⎛ ⎞
⎜ ⎟
⋅
⎝ ⎠
=
⎠
Solution via iterative procedure for water:
( )
3
5
AV
kg m
1000 0.0643 m 3.65
m s
3.65 m s Re 2.3 10
kg
0.001
m s
Dv
v
ρ
μ
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= = = =
⎛ ⎞
⎜ ⎟
⋅
⎝ ⎠
×
5
(Re 2.3 10 ;smooth tube) 0.0037
f
f = × =
Solution via iterative procedure for mud:
( )
3
3
AV
kg m
1400 3.2 0.0643 m
m m s
3.2 Re 5.7 10
kg
s 0.05
m s
v
⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
= = = ×
⋅
3 4
(Re 5.7 10 ; 2.3 10 ) 0.005
f
f He
= × = × ≅
13

14. 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 0.3
2 0.3
*
1 0.3
transition 0.3
*
transition
6464 0.3 1
Re 2 0.3
1 3 0.3
1 3 0.3
Re 2340
−
+
⎛ ⎞
⎜ ⎟
+
⎝ ⎠
⎛ ⎞
= + ⎜ ⎟
+
+ ⎝ ⎠
=
Also, need to calculate , the consistency index
k
1
app
n
k
μ γ −
= &
0.7
kg 100
0.07
m s s
k
−
⎛ ⎞
= ⎜ ⎟
⋅ ⎝ ⎠
1.7
kg
1.76
ms
k
∴ =
Applying equation 4.19, the average velocity in the smooth pipe can be found.
( )
2
* AV
8
Re
6 2
n
n n
mD v n
k n
ρ −
⎡ ⎤
= ⎢ ⎥
+
⎣ ⎦
( )
( )
0.3 1.7
0.3
AV
3
1.7
kg
8 1500 0.0127 m
0.3
m
2340=
kg 6 0.3 2
1.76
ms
v
⎛ ⎞
⎜ ⎟ ⎛ ⎞
⎝ ⎠
⎜ ⎟
+
⎝ ⎠
Solving for AV
v
14

15. AV
m
1.61
s
v =
The volumetric flow rate Q is then
( )
2
2
AV C
3
4
m 0.0254 m
1.61 0.5 in
s 1 in
(Cross-Sectional Area, A )
4
m
2.0 10
s
Q v
π
−
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
= ⋅ =
= ×
b)
2
AV
2 f m
L
p f v
D
ρ
⎛ ⎞
Δ = ⎜ ⎟
⎝ ⎠
*
16
0.007
Re
f
f = =
( )
2
3
kg 4.572 m m
2 0.007 1500 1.61
m 0.0127 m s
19600 Pa
p
p
⎛ ⎞⎛ ⎞⎛
Δ = ⎜ ⎟⎜ ⎟⎜
⎝ ⎠⎝ ⎠⎝
Δ =
⎞
⎟
⎠
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/m⋅s, and a density of 1500 kg/m3
. At what velocity will the mud slurry drain from the
hose?
15

16. SOLUTION TO EXERCISE 4.10:
Applying the modified Bernoulli equation to the system above
2
2
AV
2 1
2
f
v
h z z
g
− = − +
Assuming the flow is laminar, the head loss due to pipe friction is given by
3 2
f
m
p p
h
g
ρ
−
=
Applying equation 4.30,
2
AV
2
32 16
3
p y
f
m
v L L
D D
h
g
μ τ
ρ
+
=
( )( )
( )
( )
( )
2
AV 2
2
3 2
AV
kg N
32 0.04 1 m 16 10 1 m
m s m
3 0.01 m
0.01 m
kg m
1500 9.8
m s
0.87 0.36
f
f
v
h
h v
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⋅
⎝ ⎠ ⎝ ⎠
+
=
⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
= +
16

17. Plugging this head loss back into the modified Bernoulli equation,
( )
2
AV
AV
0.87 0.36 4
2 9.8
v
v
− − = − +
Rearranging,
( ) ( )
2
AV AV
2
AV
0 71.3 17.1
17.1 17.1 4 71.3
2
v v
v
= − + +
− ± − −
=
AV
m
3.5
s
v = (other solution yields a negative velocity)
Check original assumption to see if flow is laminar
( )
3
AV
kg m
1500 3.5 0.01 m
m s
Re 1300
kg
0.04
m s
m
p
v D
ρ
μ
⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
= = =
⎛ ⎞
⎜ ⎟
⋅
⎝ ⎠
∴ 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

18. SOLUTION TO EXERCISE 4.11:
Applying the modified Bernoulli equation to the system above
2
AV
2 1
2
f
v
h z z
g
− = − +
Assuming the flow is laminar, the head loss due to pipe friction is given by
( )( )
( )
( )
( )
2
2
2
AV
2
3 2
AV 2
2
3 2
AV
32 16
3
kg N
32 0.06 5 m 16 15 5 m
m s m
3 0.01 m
0.01 m
kg m
2000 9.8
m s
4.90 2.04
p y
f
m m
f
f
v L L
p p D D
h
g g
v
h
h v
μ τ
ρ ρ
+
−
= =
⎛ ⎞ ⎛ ⎞
⎜ ⎟ ⎜ ⎟
⋅
⎝ ⎠ ⎝ ⎠
+
=
⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
= +
18

19. Plugging this head loss back into the modified Bernoulli equation,
2
2
AV
AV
2
4.9 2.04 5
m
2 9.8
s
v
v
− − = − +
⎛ ⎞
⎜ ⎟
⎝ ⎠
Rearranging,
( ) ( )
2
AV AV
2
AV
96.0 58 0
96 96 4 58
2
v v
v
+ − =
− ± − −
=
AV
m
0.6
s
v = (other solution yields a negative velocity)
Check original assumption to see if flow is laminar
( )
3
AV
kg m
2000 0.6 0.01 m
m s
Re
kg
0.06
m s
m
p
v D
ρ
μ
⎛ ⎞⎛ ⎞
⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠
= =
⎛ ⎞
⎜ ⎟
⋅
⎝ ⎠
⇒ flow is laminar
Re 200
=
19