Assignment 2 (2006) Solutions

CHEE 2940 Particle Processing 2006 – Assignment #2 Due 5 PM Tuesday 28 March - Chemical Engineering Office 1. A size reduction operation involves crushing and grinding. The feed ore has 90% passing a 36 mm screen and is crushed to 90% passing 2.5 mm screen in a cone crusher. The crushed material is further ground in a ball mil to 90% passing a 160 micron screen. Use the Bond method to calculate the power required for each stage of the size reduction operation if 100 t/hr of a) iron ore and b) clay is processed. The Bond work index is given in the table. (20 marks)

Assignment #2-1

2. A granulation operation runs at 90% relative humidity. The particle radius is 10 microns. The Hamaker constant is 1 × 10-19 J. The closest separation distance between the particles is 2 Angstrom. The surface tension is 72 mN/m. Calculate the van der Waals force and the force of the liquid bridge between the particles. Describe your calculation steps. Please use the updated lecture note in Blackboard. (20 marks) 3. The results of shear cell tests on a powder for the unconfined yield stress, σy, and compacting stress, σc, are shown in the following table. σc (kN/m2)

1

1.4

1.8

2.2

2.6

σy(kN/m2)

0.7

0.93

1.05

1.12

1.2

The effective angle of internal frictions is δ = 30o. The kinematic angle of wall friction is φw = 20o. The flow factor chart is shown below. (40 marks)

δ = 30ο

ϕw a) Determine the powder flow factor, ff, and the semi-included angle of the conical hopper, ϕw, which will confidently give mass flow if the safety margin of 3o is required. The allowed error of the determination of the angle is 1o. Show your solution on the flow factor chart.

Assignment #2-2

b) Determine the stress developed, σD, as a function of the compacting stress, σc, and fill in the following table. σc(kN/m2)

σD (kN/m2)

0 0.2 … 3.0

c) Plot the stress developed, σD, and the unconfined yield stress, σy, versus the compacting stress, σc. d) Use the “Add trendline” function in Microsoft Excel to obtain the fitting function for σD and σy versus σc. The linear function with 0 intercept should be used for σD. The second order polynomial should be used for σy. Show the functions in your diagram in Question c. e) The critical stress, σcrit, developed by the powder to cause the powder flow is determined by the intercept of the curves obtained and shown in the diagram in Question c. At the intercept, the two functions obtained in Question d should give the same stress value. Use the functions and solve the equation to obtain the compacting stress of the intercept and σcrit. You can use the Solver function in Microsoft Excel. f) Determine the critical outlet dimension of the conical hopper if the solid density is 2500 kg/m3. 4. A vertical standpipe of inside radius of 1 cm is used to transport solid particles at flux 100 kg/m2 s from a lower vessel to an upper vessel. The constant voidage in the pipe is 0.4. The solid density is 2500 kg/m3. The gas density is 1.2 kg/m3. The relative velocity between the gas and solid phases is 0.1 m/s. Calculate: (20 marks) a) The actual velocities of the solid and gas phases in the pipe. b) The superficial velocity, volumetric flow rate, and mass flow rate of the gas phase.

Assignment #2-3

Solutions 1.

The power required for crushing the iron ore (5 marks):

 1   1  1 1 − − × (100t/h ) = 250.11kW P = 10Wi   × ( mass rate ) = 10 × 16.98kWh/t   3 3 D D × × 2.5 10 36 10   0    The power required for grinding the iron ore (5 marks):  1  1  1  1 − − × (100t/h ) = 1002.79kW P = 10Wi   × ( mass rate ) = 10 × 16.98kWh/t   D0  2.5 × 103   D  160 The power required for crushing the clay sample (5 marks):  1   1  1 1 − − × (100t/h ) = 115.04kW P = 10Wi   × ( mass rate ) = 10 × 7.81kWh/t   3 D0  36 ×103   2.5 ×10  D The power required for grinding the clay sample (5 marks):  1  1  1  1 − − × (100t/h ) = 461.23kW P = 10Wi   × ( mass rate ) = 10 × 7.81kWh/t   D0  2.5 ×103   160  D

Assignment #2-4

2.

The van der Waals force (4 marks):

FvdW = −

A R1 R2 1× 10−19 J 10 × 10−6 m × 10 ×10−6 m = − × = −2.083 × 10−6 N 2 − − 2 6 6 6 D R1 + R2 6 × ( 2 × 10−10 m ) 10 × 10 m + 10 ×10 m

The radius of the liquid meniscus (4 marks) r =

1.34 = −12.72 nm ln ( 90 /100 )

The semi-included angle of the liquid bridge (4 marks)  R 10 ×10−6 m    = acos   = 0.050 −6 −9  R + abs ( r )  10 × 10 m + abs ( −12.72 × 10 m )  

ψ = acos 

The capillary pressure (4 marks) ∆p =

2γ 2 × 72 ×10−3 N/m = = −11322324.07N/m 2 −9 r −12.72 × 10 m

The force of the liquid bridge (4 marks)

FLBridge = ∆p π ( R sinψ ) + γ 2π R sin 2 ψ = −9.019 ×10−6 N 2

Assignment #2-5

Using the flow factor chart provided for δ = 30o with φw = 20o and a 3o safety margin we obtain a hopper flow factor, ff = 1.8, and the semi-included angle of the conical hopper case, ϕw = 26o. The solution is shown by the red lines in the diagram. (5 marks) 3a.

3o margin

δ = 30ο

ϕw The stress developed, σD, is determined using the equation σ D = σ C / ff . The results are shown below. (5 marks) 3b.

σc(kN/m2)

σD (kN/m2)

0.0

0.000

0.2

0.111

0.4

0.222

0.6

0.333

0.8

0.444

1.0

0.556

1.2

0.667

1.4

0.778

1.6

0.889

1.8

1.000

2.0

1.111

2.2

1.222

2.4

1.333

2.6

1.444

2.8

1.556

3.0

1.667

Assignment #2-6

3c. The plots for the stress developed, σD, and the unconfined yield stress, σy, versus the

compacting stress, σc, are shown below. (10 marks)

σcrit 1 Unconfined yield stress

σ d (kN/m )

2

2

Unconfined yield stress, σ y, or stress developed,

1.5

y = -0.1562x + 0.86x + 0.0083

0.5

Compacting stress y = 0.5556x

0 0

1

2

3

Compacting stress (kN/m2)

3d. The fitting functions for σD and σy versus σc were obtained using the “Add trend” function

in Microsoft Excel and are shown in the diagram in Question c. (5 marks) 2

3e. The equation to be solved is -0.1562x + 0.86x + 0.0083 = 05556 x, where x describes the

compacting stress at the intercept. Solving the equation gives σc = 1.976 kN/m2 at the intercept. The corresponding critical stress is σcrit = 1.098 kN/m2. (7 marks) 3f. Determine the critical outlet dimension of the conical hopper. (8 marks)

H (ϕ w ) = 2.0 +

ϕw 60

= 2.0 +

26 = 2.433 60

The outlet dimension B=

H (ϕ w ) σ crit 2.433 × 1.217 × 103 N/m 2 = = 0.121 m ρB g 2500kg/m3 × 9.81m/s 2

Assignment #2-7

4a. The actual velocity of the solid phase (4 marks)

Up =

solid flux 100 kg/m 2 = = 0.067 m/s (1-ε ) ρ s (1- 0.4 ) × 2500 kg/m3

The actual velocity of the gas phase (4 marks)

U f = U rel + U p = 0.167 m/s 4b. The gas superficial velocity (4 marks)

J f = U f ε = 0.167 m/s × 0.4 = 0.067 m/s The gas volumetric flow rate (4 marks) Q f = J f A = 0.067 m/s × 3.1415 × ( 0.01 m ) = 0.0000209 m3 /s 2

The gas mass flow rate (4 marks): M f = Q f ρ f = 0.0000209 m3 /s × 1.2 kg/m3 = 0.0000251 kg/s

Assignment #2-8