Rhodes Solutions Ch4

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& ( sec −1 )

Shear Stress, τ

25 75 125 175 225 325 425 525 625 725 825

38 45 48 51 53 55.5 58 60 62 63.2 64.3

( Pa )

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).

τ = kγ& n log τ = log k + n log γ& Shear Rate 25 75 125 175 225 325 425 525 625 725 825

Shear Stress Log (Shear Stress) 38 45 48 51 53 55.5 58 60 62 63.2 64.3

1.58 1.65 1.68 1.71 1.72 1.74 1.76 1.78 1.79 1.80 1.81

Log (Shear Rate) 1.40 1.88 2.10 2.24 2.35 2.51 2.63 2.72 2.80 2.86 2.92

From this plot

1

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

∴τ = 23.4γ& 0.15 with γ& in sec −1 and τ in Pa and

μapp =

τ = 23.4γ& −0.85 γ&

Problem 4.1 70 60 50 40 30 20 10 0 0

200

400

600

800

1000

Shear Rate

Problem 4.1 Log (Shear Stress)

Shear Stress

Hence,

1.85 1.8 1.75 1.7 1.65 1.6 1.55 0

1

2

3

4

Log (Shear Rate)

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 ∂p 1 ∂ O=− + ( rτ rz ) ∂z r ∂r For a power-law fluid ⎛ dv ⎞

n

τ rz = k ⎜ z ⎟ ⎝ dr ⎠ Hence, ΔP k d ⎡ ⎛ dvz ⎞ ⎤ + ⎢r ⎜ ⎟ ⎥ L r dr ⎢⎣ ⎝ dr ⎠ ⎥⎦ ΔP ∂p has been replaced by + since the pressure gradient is constant. Here − L ∂z Integrating once, n

O=+

r ΔP C1 ⎛ dvz ⎞ − + =⎜ ⎟ 2k L r ⎝ dr ⎠

n

C1 must equal to zero since the velocity gradient is zero at r = 0 . Integrating again, 1

1 +1

⎛ ΔP ⎞ n r n −⎜ + C2 = v z ⎟ ⎝ 2 Lk ⎠ ⎛ 1 + 1⎞ ⎜ ⎟ ⎝n ⎠ Applying the boundary condition that vz = 0 at r = R ⎡ 1 1 ⎤ 1+ 1+ ⎢ n R −r n ⎥ ⎛ ΔP ⎞ ⎢ ⎥ vz = + ⎜ ⎟ ⎝ 2kL ⎠ ⎢ ⎛ n + 1 ⎞ ⎥ ⎢⎣ ⎜⎝ n ⎟⎠ ⎥⎦ 1

n

3

The average velocity vAV is equal to R

vAV =

2 rvz dr R 2 ∫0

Hence R

vAV

⎡ ⎤ 1 1 1+ 3+ 1 2 ⎥ n n n ⎢ 2 ⎛ ΔP ⎞ R r r ⎢ ⎥ = 2⎜ − ⎟ R ⎝ 2kL ⎠ ⎢ ⎛ n+1 ⎞ ⎛ 3n + 1 ⎞ ⎛ n + 1 ⎞ ⎥ 2 ⎢⎣ ⎜⎝ n ⎟⎠ ⎜⎝ n ⎟⎠ ⎜⎝ n ⎟⎠ ⎥⎦ 0

vAV

⎞⎤ 2 ⎛ ΔP ⎞ n ⎡ 3+ 1n ⎛ n = 2⎜ ⎟⎟ ⎥ ⎟ ⎢ R ⎜⎜ R ⎝ 2kL ⎠ ⎣⎢ ⎝ 2 ( 3n + 1) ⎠ ⎦⎥

and 1

Simplifying with

vAV

D =R 2

⎛ ΔPD ⎞ =⎜ ⎟ ⎝ 4kL ⎠

1

n

Dn 2 ( 3n + 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 O=−

∂P 1 ∂ + ( rτ rz ) ∂z r ∂r

For any fluid

4

−

Δ Pr d = ( rτ rz ) dr L

Integrating, Δ Pr = τ rz 2L For pipe flow −

since τ rz must be finite at r = 0

τ rz = −τ y + μ p

dvz dr

since

dvz is negative for pipe flow dr

Hence, −

dv Δ Pr = −τ y + μ p z 2L dr

Rearranging, −

Δ Pr τ y dvz + = 2 Lμ p μ p dr

Integrating, −

Δ Pr 2 τ y r + + C2 = v z 4 Lμ p μ p

Apply the boundary condition vz = 0 −

r=R

Δ P R2 τ y R + = −C2 4 Lμ p μp

Δ P ( R2 − r2 ) τ y ( r − R ) ∴ vz = + 4 Lμ p μp

This velocity profile is valid for R* ≤ r ≤ R . For the plug flow region r ≤ R* ,

dvz = 0. dr

In the plug flow region, the velocity is constant and −τ rz = τ y =

ΔPR* 2L 5

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

vz =

4 Lμ p

∴ vz =

+

Δ P ( R − R* ) 4μ p L

Δ P R* ( R* − R )

2 Lμ p

2

for 0 ≤ r ≤ R*

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

2 = 2 R

R*

∫ 0

2 2 ΔP R − R* ) rdr + 2 ( 4μ p L R

R

∫

R*

⎡ ΔP ⎤ τ R 2 − r 2 ) + y ( r − R ) ⎥ rdr ( ⎢ μp ⎣⎢ 4 μ p L ⎦⎥

Integrating, vAV

4 ΔPR 2 ⎡ 4 ⎛ R* ⎞ 1 ⎛ R* ⎞ ⎤ = ⎢1 − ⎜ ⎟ + ⎜ ⎟ ⎥ 8μ p L ⎢ 3 ⎝ R ⎠ 3 ⎝ R ⎠ ⎥ ⎣ ⎦

vAV

4 Rτ 0 ⎡ 4 τ y 1 ⎛ τ y ⎞ ⎤ ⎢1 − = + ⎜ ⎟ ⎥ 4μ p ⎢ 3 τ 0 3 ⎝ τ 0 ⎠ ⎥ ⎣ ⎦

or

where τ 0 is the wall shear stress

τ0 =

RΔP 2L

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

* transition

Re

=

6464 × 0.3 ( 2.3)

(1.9 )

2.3 1.3

0.3

1.7

⎛ 1 ⎞ ⎜ ⎟ ⎝ 1.9 ⎠

Re*transition = 2345 Re* at flow conditions, 1.7

kg ⎞ m⎞ 0.3 ⎛ ⎛ 8 × ⎜ 900 3 ⎟ × ( 0.05 m ) × ⎜ 8.5 ⎟ 0.3 m ⎠ s ⎠ ⎡ 0.3 ⎤ * ⎝ ⎝ Re = ⎢⎣ 3.8 ⎥⎦ Ns 0.3 3.0 m2 Re* = 17340 > Re*transition ⇒ turbulent The friction factor f f is given by

(

1 4 = 0.75 log Re* ff n

)

f f 2−n −

0.4 n

Solving for f f with Re = 17340 and n = 0.3 , yields *

f f = 0.0026 a) the pressure drop for horizontal flow is then

Δp = 2 f f ρ m

L 2 v AV D

(

kg ⎞⎛ 50m ⎞ ⎛ ⎛ m Δp = 2 ( 0.0026 ) ⎜ 900 3 ⎟⎜ ⎟ ⎜ 8.5 s m 0.05m ⎝ ⎝ ⎠⎝ ⎠

) ⎞⎟⎠ 2

7

Δp = 338

kN m2

b) the pressure drop for vertical flow is given by

−h f = −

Δp + Δz ρm g

or

⎛L⎞ 2 Δp = ρ m gh f + ρ m g Δz = 2 f f ρ m ⎜ ⎟ v AV + ρ m g Δz ⎝D⎠

2

kg ⎞⎛ 50m ⎞⎛ m ⎞ ⎛ kg ⎞⎛ m⎞ ⎛ Δp = 2 ( 0.0026 ) ⎜ 900 3 ⎟⎜ ⎟⎜ 8.5 ⎟ + ⎜ 900 3 ⎟⎜ 9.8 2 ⎟ ( 50m ) m ⎠⎝ 0.05m ⎠⎝ s⎠ ⎝ m ⎠⎝ s ⎠ ⎝ Δp = 779

kN m2

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

1

ρm

=

0.53 0.47 + kg kg 3000 3 1000 3 m m

ρ m = 1546

kg m3

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

ρm

=

Cw

ρs

+

(1 − Cw ) ρf

(1 − Cw ) 1 C = w + 1.3 1.65 1.0 Solving for Cw ,

Cw = 0.586 mass coal Cw = mass slurry Cw ρ m

Volume fraction coal =

Volume fraction coal =

ρs

( 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 0.1 1 10 15 25 40 45

4.0 4.03 4.2 5.3 5.8 6.7 7.8 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 τ − τ y versus shear rate.

τ − τ y (Pa) 0 0.03 0.2 1.3 1.8 2.7 3.8 4.2

Shear Rate (1/s) 0 0.1 1 10 15 25 40 45

A plot of τ − τ y versus shear rate on an arithmetric scale is not linear. However, a plot of τ − τ y versus shear rate on a log-log scale is linear (the data for zero shear rate is excluded)

τ − τ y = kγ& n ln (τ − τ y ) = ln k + n ln γ&

10

ln (τ − τ y )

ln γ&

-3.51 -1.61 +0.26 +0.59 +0.99 +1.34 +1.44

-2.30 0 +2.30 +2.71 +3.22 +3.69 +3.81

Shear Rate 0.1 1 10 15 25 40 45

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

Ns 0.81 ∴ k = 0.20 m2 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

2 ⎧ vAV ⎫ 0 = + h f + Δz + Δ ⎨ ⎬ ⎩ 2g ⎭

or, 2 vAV 0 = + h f + z2 − z1 + 2g

where h f =

2 ff ⎛ L ⎞ 2 ⎜ ⎟ vAV and z2 − z1 = −3.048 m g ⎝ Dh ⎠ 2

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

Dh =

4 cross-sectional area wetted perimeter

2b

2a

( 4 )( πab )

Dh =

2π

Dh =

(a

2

+ b2 ) 2

( 4 )( 2 in )(1 in ) = 2.53 in = 0.0643 m 2

4 in 2 + 1 in 2 2

Plugging in numbers (SI units), 2 2 f f ⎛ 15.24 m ⎞ 2 vAV 0= v − 3.048 m + m ⎜⎝ 0.0643 m ⎟⎠ AV m⎞ ⎛ 9.8 2 2 ) ⎜ 9.8 2 ⎟ ( s s ⎠ ⎝ 2 2 0 = 0.051vAV − 3.048 + 48.3 f f vAV

(*)

12

Solution Procedure:

1) Calculate He 2) Guess velocity v AV 3) Calculate Re 4) Find f f from Figure 6. 5) Check governing equation (*) 6) If governing equation is not satisfied, guess a new velocity vAV kg ⎞ kg ⎞ 2⎛ ⎛ 1400 3 ⎟ ( 0.0643 m ) ⎜10 ⎜ ⎟ ρm D τ m ⎠ m ⋅ s2 ⎠ ⎝ ⎝ He = = 2 μ kg ⎞ ⎛ ⎜ 0.050 ⎟ m⋅s ⎠ ⎝ 2 h y 2 p

He = 23,153 Solution via iterative procedure for water:

kg ⎞ m⎞ ⎛ ⎛ 1000 3 ⎟ ( 0.0643 m ) ⎜ 3.65 ⎟ ⎜ ρ Dv ⎝ m ⎠ s ⎠ ⎝ = = 2.3 × 105 Re = kg ⎞ ⎛ μ ⎜ 0.001 ⎟ m⋅s ⎠ ⎝

vAV = 3.65 m s

f f (Re = 2.3 × 105 ;smooth tube) = 0.0037 Solution via iterative procedure for mud:

vAV = 3.2

m s

kg ⎞⎛ m⎞ ⎛ ⎜1400 3 ⎟⎜ 3.2 ⎟ ( 0.0643 m ) m ⎠⎝ s ⎠ = 5.7 × 103 Re = ⎝ kg 0.05 m ⋅s

f f (Re = 5.7 × 103 ; He = 2.3 × 104 ) ≅ 0.005

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 Re transition for n = 0.3

* transition

Re

=

6464 ( 0.3)

(1 + 3( 0.3) )

0.3

( 2 + 0.3)

⎛ 2+ 0.3 ⎞ ⎜ ⎟ ⎝ 1+ 0.3 ⎠

⎛ ⎞ 1 ⎜ ⎟ + 1 3 0.3 ( ) ⎝ ⎠

2−0.3

Re*transition = 2340 Also, need to calculate k , the consistency index

μapp = kγ& n−1 kg ⎛ 100 ⎞ = k⎜ 0.07 ⎟ m ⋅s ⎝ s ⎠ ∴ k = 1.76

−0.7

kg ms1.7

Applying equation 4.19, the average velocity in the smooth pipe can be found. 2− n ⎡ n ⎤ 8 ρ m D nvAV Re = ⎢ ⎥ k ⎣ ( 6n + 2 ) ⎦

n

*

kg ⎞ 0.3 ⎛ 0.3 8 ⎜1500 3 ⎟ ( 0.0127 m ) v1.7 AV ⎛ ⎞ 0.3 m ⎠ 2340= ⎝ ⎜ ⎟ kg 6 0.3 2 + ( ) ⎝ ⎠ 1.76 ms1.7 Solving for vAV 14

vAV = 1.61

m s

The volumetric flow rate Q is then

2

m⎞ 2 ⎛ 0.0254 m ⎞ ⎛ ⎜1.61 ⎟ ( 0.5 in ) ⎜ ⎟ π s ⎠ 1 in ⎠ ⎝ ⎝ Q = vAV ⋅ (Cross-Sectional Area, A C ) = 4 = 2.0 × 10

−4

m3 s

b)

⎛L⎞ 2 Δp = 2 f f ρ m ⎜ ⎟ vAV ⎝D⎠ ff =

16 = 0.007 Re*

kg ⎞⎛ 4.572 m ⎞⎛ m⎞ ⎛ Δp = 2 ( 0.007 ) ⎜1500 3 ⎟⎜ ⎟⎜1.61 ⎟ m ⎠⎝ 0.0127 m ⎠⎝ s ⎠ ⎝

2

Δp = 19600 Pa

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

SOLUTION TO EXERCISE 4.10:

Applying the modified Bernoulli equation to the system above

−h f = z2 − z1 +

2 vAV 2

2g

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

hf =

p3 − p2 ρm g

Applying equation 4.30,

32 μ p vAV2 L hf =

D

2

+

ρm g

16τ y L 3D

kg ⎞ N ⎞ ⎛ ⎛ 32 ⎜ 0.04 ⎟ vAV2 (1 m ) 16 ⎜10 2 ⎟ (1 m ) m ⋅s ⎠ m ⎠ ⎝ + ⎝ 2 3 ( 0.01 m ) ( 0.01 m ) hf = kg ⎞⎛ m⎞ ⎛ ⎜1500 3 ⎟⎜ 9.8 2 ⎟ m ⎠⎝ s ⎠ ⎝

(

)

h f = 0.87vAV + 0.36

16

Plugging this head loss back into the modified Bernoulli equation, 2 vAV −0.87vAV − 0.36 = −4 + 2 ( 9.8 )

Rearranging, 2 0 = −71.3 + 17.1 vAV + vAV

vAV =

−17.1 ±

vAV = 3.5 m

(17.1)

2

− 4 ( −71.3)

2 s

(other solution yields a negative velocity)

Check original assumption to see if flow is laminar

kg ⎞⎛ m⎞ ⎛ 1500 3 ⎟⎜ 3.5 ⎟ ( 0.01 m ) ⎜ ρ v D m ⎠⎝ s ⎠ Re = m AV = ⎝ = 1300 kg ⎞ μp ⎛ ⎜ 0.04 ⎟ m ⋅s ⎠ ⎝ ∴ 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 2 vAV −h f = z2 − z1 + 2g

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

hf =

p3 − p2 = ρm g

32 μ p vAV2 L D2

+

ρm g

16τ y L 3D

kg ⎞ N ⎞ ⎛ ⎛ 32 ⎜ 0.06 ⎟ vAV2 ( 5 m ) 16 ⎜15 2 ⎟ ( 5 m ) m ⋅s ⎠ m ⎠ ⎝ + ⎝ 2 3 ( 0.01 m ) ( 0.01 m ) hf = kg ⎞⎛ m⎞ ⎛ ⎜ 2000 3 ⎟⎜ 9.8 2 ⎟ m ⎠⎝ s ⎠ ⎝

(

)

h f = 4.90 vAV2 + 2.04

18

Plugging this head loss back into the modified Bernoulli equation,

−4.9vAV2 − 2.04 = −5 +

2 vAV

m⎞ ⎛ 2 ⎜ 9.8 2 ⎟ s ⎠ ⎝

Rearranging, 2 vAV + 96.0 vAV − 58 = 0

vAV =

−96 ±

vAV = 0.6

( 96 )

2

− 4 ( −58 )

2 m s

(other solution yields a negative velocity)

Check original assumption to see if flow is laminar

kg ⎞⎛ m⎞ ⎛ 2000 3 ⎟⎜ 0.6 ⎟ ( 0.01 m ) ⎜ ρ v D m ⎠⎝ s ⎠ Re = m AV = ⎝ kg ⎞ ⎛ μp ⎜ 0.06 ⎟ m ⋅s ⎠ ⎝ Re = 200 ⇒ flow is laminar

19