Assignment 3 (2006) Solutions

CHEE 2940 Particle Processing 2006 – Assignment #3 Due 5 PM Tuesday 11 April - Chemical Engineering Office 1. A suspension of solid particles with the average diameter of D = 100 µm and the density of ρs = 2500 kg/m3 is dewatered in a thickener with the diameter of 2m. The water viscosity and density are µ = 0.001 Pa s and ρf = 1000 kg/m3, respectively. The acceleration due to gravity is equal to g = 9.81 m/s. The suspension with the solid volume fraction of εF = 0.25 is fed into the thickener at the volumetric flow rate of Q = 5.35 L/s. The underflow product with the solid volume fraction of εu = 0.5 is withdrawn from the thickener at the volumetric flow rate of U = 2.35 L/s. The operation of the thickener is schematically shown below.

a) Calculate the terminal settling velocity of the particles using the Stokes law and the average diameter. (10 marks) b) Knowing the particle settling velocity from Question a), the particle Reynolds number, Re = ρ f DVT / µ , can be calculated. Inserting the Reynolds number into the following

general equation for the settling velocity, we see that the particle terminal velocity, VT, is smaller than the terminal velocity calculated in Question a). To obtain the correct value for the terminal velocity, the non-linear equation given below must be solved for VT. Use the Excel Solver function and solve the following equation for the actual terminal settling velocity of the particles. Describe your procedure with the Solver function or the VBA macro. (10 marks) VT =

D2 g ( ρs − ρ f

)

18µ (1 + 0.15 Re0.687 )

c) Calculate and plot the particle flux due to the particle settling versus ε. (10 marks) d) Calculate and plot the particle flux due to the underflow and the net solid flux in the thickener zone below the feed inlet. Show these two curves and the flux curve in Question c) in one diagram. (10 marks) e) Use the diagram in Question d) and estimate the solid volume fraction and the downward solid flux under the critical condition of the thickener operation. (10 marks)

Assignment #3-1

f) Establish the balance equations for the volumetric flow rates and the mass of solid particles. Solve the balance equations for the volumetric flow rate, O, and the solid volume fraction, εo, of the overflow product. (10 marks) g) Calculate and plot the particle flux due to the overflow and the net solid flux in the thickener zone above the feed inlet. Show these two curves and the flux curve in Question c) in one diagram. (10 marks) h) Determine the net solid flux in the thickener zone above the feed inlet. You can use the diagram obtained in Question g) and the solid volume fraction, εo, of the overflow product in Question f). (10 marks) i) Calculate the suspension density of the feed, underflow product and overflow product. (10 marks) 2. The particle suspension described in Question 1 can be fluidised in a vertical pipe using the liquid upward flow (against gravity). Use the Ergun equation for the pressure gradient over the particle bed (Lecture 9) and establish the force balance on the particle bed for determining the minimum (superficial) liquid velocity required to fluidise the particle bed. Solve the force balance equation for the minimum liquid velocity. The particle volume fraction at the fluidisation onset is ε = 0.4. (10 marks)

Assignment #3-2

Solutions 1a. The terminal settling velocity predicted by the Stokes law is described as VStokes =

D2 g ( ρs − ρ f ) 18µ

Inserting the numerical values gives

VStokes

(100 ×10 =

−6

m ) × 9.81m/s × ( 2500kg/m3 − 1000kg/m3 ) 2

18 × 0.001Pa s

= 0.008175m/s

1b. The non-linear equation to be solved is described as

VT

(100 ×10 =

−6

m ) × 9.81m/s × ( 2500kg/m3 − 1000kg/m3 ) 2

 1000kg/m3 × 100 ×10−6 m × VT  18 × 0.001Pa s  1+  0.001Pa s  

The initial setting in Excel is shown below. VT (Guessed) Re

0.008175 0.818

LHS

0.008175

RHS

0.00736

LHS-RHS

0.00082

The Excel Solver is used to adjust the “LHS-RHS” to zero by changing the initial guess for VT, giving the following results: VT (Solution) Re

0.00736 0.736

LHS

0.00736

RHS

0.00736

LHS-RHS

0.00000

The actual terminal settling velocity of the particles is VT = 0.00736 m/s.

Assignment #3-3

1c. The particle flux due to settling is described as

J set = VT ε (1 − ε )

n

The Richardson-Zaki index, n, as a function of the particle Archimedes number is described by n=

4.8 + 0.103 Ar 0.57 1 + 0.043 Ar 0.57

The Archimedes number is described as Ar =

D3 ρ f ( ρ s − ρ f ) g

µ2

= 14.715 , giving n = 4.401.

Knowing VT and n, Jset can be calculated as a function of the solid volume fraction, ε, and is plotted in the flowing diagram.

Solid flux due to settling (mm/s)

0.6

0.4

0.2

0 0

0.2

0.4

0.6

0.8

1

Solid volume fraction

Note: The unit of the particle flux on the vertical axis is given in mm/s. The volume fraction must be refined to obtain the continuous smooth curve.

Assignment #3-4

1d. The solid underflow flux is described as

J u = J set +

Uε A

where A is the cross sectional area of the thickener. The units for U and A must be converted to produce the consistent unit for the flux (in mm/s in this solution). The plot for Ju versus ε is given below.

0.8

Downward solid flux (mm/s)

Net flux below feed inlet 0.6

Solid flux due to settling

0.4

0.2 Solid flux due to underflow

ε crit

0 0

0.2

0.4

0.6

0.8

1

Solid volume fraction

1e. The critical condition of the thickener operation is taken at the minimum of the net flux below the feed inlet. Based on the diagram shown in Question 1d, we obtain εcrit = 0.6 and

Ju, crit = 0.53 mm/s. 1f. The balance equations can be described as

Q = O +U Qε F = Oε o + U ε u Knowing Q, U, εF and εu, these two equation can be solved for O and εo, giving

Assignment #3-5

O = P − U = 3.0 L/s and ε o =

Qε F − U ε u = 0.0542 . O

1g. The solid overflow flux is described as

J o = J set −

Oε A

where A is the cross sectional area of the thickener. The units for O and A must be converted to produce the consistent unit for the flux (in mm/s in this solution). The plot for Jo versus ε is given below.

0.8

Upward solid flux (mm/s)

0.6 0.4

Solid flux due to settling

0.2 0 0

0.2

0.4

0.6

0.8

1

-0.2 Solid flux due to underflow

Net flux above feed inlet

-0.4 -0.6 Solid volume fraction

1h. Using the solid volume fraction εo = 0.0542 from Question f) and the diagram from Question h) gives Jo = 0.26 mm/s for the net solid flux in the thickener zone above the feed inlet. 1i. Using the following equation for the suspension density

ρ sus = ερ s + (1 − ε ) ρ f

Assignment #3-6

the suspension density of the feed, underflow product and overflow product can be determined, giving

ρ feed = 1375 kg/m3 ρunderflow = 1750 kg/m3 ρ overflow = 1081 kg/m3 . 2. The force balance on the particle bed for determining the minimum (superficial) liquid velocity required to fluidise the particle bed is described as

µU mf (1 − ε ) ρ f U mf 2 (1 − ε ) g (1 − ε ) ( ρ s − ρ f ) = 153.344 + 1.568 ε 3D2 Dε 3 2

The quadratic equation can be solved for Umf. The Excel Solver function can be used and the procedure is shown below. - The initial setting in Excel: Umf (m/s) ε

0.01 0.4

LHS

8829

RHS

877260

LHS-RHS

-868431

- The Excel Solver is used to adjust the “LHS-RHS” to zero by changing the initial guess 0.01 m/s for Umf, giving the following results: Umf (m/s) ε

0.000102 0.4

LHS

8829

RHS

8829

LHS-RHS

8.18E-08

The actual minimum (superficial) liquid velocity required to fluidise the particle bed s is Umf = 0.000102 m/s. Assignment #3-7