assignment 2 fluid particles and processes

CURTIN UNIVERSITY OF TECHNOLOGY 2015 0

Table of Contents Question 1 ..................................................................................................................................................... 1

Question 2 ..................................................................................................................................................... 7

Question 3 ................................................................................................................................................... 14

Question 4 ................................................................................................................................................... 16

Question 5 ................................................................................................................................................... 21

References .................................................................................................................................................. 24

Appendix. .................................................................................................................................................... 25

Appendix question 4 ............................................................................................................................... 25

Table of figures Table 2 1 Filtrate volume over time .............................................................................................................. 7

Table 2 2 Basic data for calculation used in question 2 ................................................................................ 7

Table 2 3 Calculated values for the plotting of B vs t/B ................................................................................ 9

Table 4 1 Time and interface height for tank thickener 16

Table 4 2 Tabulated and calculated values for height vs time .................................................................... 18

Table 4 3 Tangent lines generated with Matlab 2nd m-file (Martina and Delli 2009) ............................... 20

Graph 2 1 t/b vs b to find alpha and betta for resistance calculation ........................................................ 13

Graph 4 1 H vs T graph ................................................................................................................................ 19

ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015


Question 1

Oxygen is supplied to an astronaut through an umbilical hose that is 7 m long. The pressure in

the oxygen tank is 200 kPa at a temperature of 10 ºC, and the pressure in the space suit is 20 kPa. If

the umbilical hose has an equivalent roughness of 0.01 mm, what should the hose diameter be to supply

oxygen at a rate of 0.05 kg/s?

The general assumptions:

o The flow is taken to be under isothermal condition

o Behaviour of oxygen as an ideal gas.

o Pipe length is assumed long, hence average density is taken for calculation

First, calculate the upstream (𝜌 1) and downstream (𝜌2) density:

Upstream pressure: 200 kPa≈ 200 × 103𝑃𝑎

Molecular weight of oxygen, Mw: 32 g/mol

The ideal gas law is used to calculate the density:

𝜌 =𝑃𝑀

𝑅𝑇

1.1

𝜌1 =𝑃𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚𝑀𝑤

𝑅𝑇=

200 × 103(32

1000)

8.314(10 + 273.15)= 𝟐. 𝟕𝟐 𝒌𝒈/𝒎𝟑

𝜌2 =𝑃𝑑𝑜𝑤𝑛𝑠𝑡𝑟𝑒𝑎𝑚𝑀𝑤

𝑅𝑇=

20 × 103(32

1000)

8.314(10 + 273.15)= 𝟎. 𝟐𝟕𝟐 𝒌𝒈/𝒎𝟑

�̅� =𝜌1 + 𝜌2

2=

2.7187 + 0.2719

2= 𝟏. 𝟒𝟗𝟔 𝒌𝒈/𝒎𝟑



As velocity is not given, it needs to be substituted.

�̇� = 𝜌𝑣𝐴

1.2

𝑣 =�̇�

�̅�𝐴

𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒 𝑖𝑛 ∶ 𝑣 = 0.0234

𝐷2

The isothermal equation is:

(𝑃12 − 𝑃2

2) = 𝑓𝜌1𝑣2𝑋𝑃1

𝐷

1.3

Substituting value in, we will obtain:

(2000002 − 200002) = 𝑓 × 2.72 × (0.0234

𝐷2)2 × 7 × 200000 ×

1

𝐷

𝐷 = 0.035(𝑓)1/5

Where D = diameter of pipe, X = length of pipe and f = friction factor.

Reynolds number equation is:

𝑅𝑒 =𝜌𝑢𝑝𝑠𝑡𝑟𝑒𝑎𝑚𝑣𝐷

𝜇=

3230.9

𝐷

The viscosity of oxygen , 𝜇 = 1.995 × 10−5 𝑃𝑎. 𝑠(The Engineering Toolbox n.d).



Iteration 1: By assuming f= 0.01:

Substitute the variables inside to find diameter first:

𝐷 = 0.035(0.01)1/5 = 0.0139𝑚

Then, use the calculated D to find the Reynolds number:

𝑅𝑒 =3230.9

0.0139= 232438.8

𝜀

𝐷= 0.000719

𝑓 = 0.0198

Iteration 2: f=0.0198:

𝐷 = 0.035(0.0198)1/5 = 0.01597𝑚

𝑅𝑒 =3230.9

0.01597= 202265.8

𝜀

𝐷= 0.0005

𝑓 = 0.0189

Iteration 3: f=0.0189:

𝐷 = 0.035(0.0189)1/5 = 0.0158𝑚



𝑅𝑒 =3230.9

0.0158= 204156.5

𝜀

𝐷= 0.00053

𝑓 = 0.0189

Since the difference between the friction factor is insignificant, f = 0.0189 is acceptable.

Thus, the diameter of the umbilical hose = 0.0158m.

Checking the magnitudes of the terms:

𝐺 =𝑚

𝐴= 255 𝑘𝑔/𝑚2𝑠

Term 1:

𝐺2𝑙𝑛 (𝜌1

𝜌2) = 2552 × 𝑙𝑛 (

2.72

0.272) = 199725

Term 2:

�̅�(𝑃2 − 𝑃1) = 1.496 × (20,000 − 200,000) = −269280

Term 3:

𝑓𝐺2𝑋

2𝐷=

0.0189 × 2552 × 7

2 × 0.158= 272240.7

Term 1 (kinetic energy term) is not negligible; therefore it needed to be considered in calculating the

diameter of the umbilical hose.



The general equation for isothermal flow is shown as:

𝐺2𝑙𝑛 (𝜌1

𝜌2) + �̅�(𝑃2 − 𝑃1) +

𝑓𝐺2𝑋

2𝐷= 0

It is known that 𝜌1

𝜌2=

𝑃1

𝑃2:

𝐺2𝑙𝑛 (𝑃1

𝑃2) +

𝑀𝑤

2𝑅𝑇(𝑃2

2 − 𝑃12) +

𝑓𝐺2𝑋

2𝐷= 0

(𝜌𝑣)2𝑙𝑛 (𝑃1

𝑃2) +

𝑀𝑤

2𝑅𝑇(𝑃2

2 − 𝑃12) +

𝑓𝐺2𝑋

2𝐷= 0

After substituting value:

(9.33 × 10−3 × 𝐷) − (269289 × 𝐷5) + 0.014175 × 𝑓 = 0

1st iteration, f = 0.0189

(9.33 × 10−3 × 𝐷) − (269289 × 𝐷5) + 0.000268 = 0

After using try and error method, we obtain:

𝐷 = 0.0174 𝑚

𝑅𝑒 =3230.9

0.0174= 185683.9

𝜀

𝐷= 0.000575

𝑓 = 0.01932

2nd iteration, f = 0.01932

(9.33 × 10−3 × 𝐷) − (269289 × 𝐷5) + 0.000274 = 0

𝐷 = 0.0174 𝑚

𝑅𝑒 =3230.9

0.0174= 185684



𝜀

𝐷= 0.00055

𝑓 = 0.0193

Since the difference between f is insignificant, D = 0.0174 m.



Question 2

A filter press with a plate area of 1.5 m2 operates at a constant differential pressure of 330 kPa is

fed with a slurry of 16.5 mass percent CaCO3 in water.

Filtrate volume x103 4 6.1 7.7 9.65 11.47 13.6 15.5

Time(min) 10 20 30 45 60 80 100 Table 2 1 Filtrate volume over time

i) Estimate the individual resistance of the cake and filter medium and also the total combined specific resistance of the filtration equipment.

Data Given:

CaCO3 slurry feed Filter aid

Solid density 2700 kg/m3 -

Feed solid loading 50 kg/m3 -

Filtrate viscosity 10 cP -

Sphericity 0.9 0.7

Average particle size 3.33mm 1mm

Porosity 0.25 0.75 Table 2 2 Basic data for calculation used in question 2

Question:

- Resistance of the cake and the combine resistance of the filter medium and equipment

Assumptions:

Initial filtrate volume, 𝐵𝑜 = 0

Initial time, 𝑡𝑜 = 0

Answer:

Area=1.5m2

Constant pressure: 330𝑘𝑃𝑎 × 103 = 330000𝑃𝑎

Filter aid=medium

Slurry feed=cake



Slurry of 16.5% kg CaCO3 in water.

𝑋 = (30|1𝑚

100𝑐𝑚 × 20) = 6𝑚

Step 1: Volume fraction of solids

𝒄 =𝒎

𝝆𝒔

4.1

50

2700= 0.0185 ≪ 1

For filter aid, specific surface 𝑺𝒇

𝑆𝑓 =6

∅𝐷

4.2

6

0.75 × 1 × 10−3= 8000𝑚2𝑚−3

Since constant ∆𝑃; the values of the resistance 𝛼 𝑎𝑛𝑑 𝛽 can be found from a t/B vs. B graph. The

graph is plotted from the values in table 2.3. Graph 2.2 shows the plotted values. A line if best fit

is drawn. The equation for the linear line was determined to be:

𝑦 = 22.8𝑥 + 57.9 4.3

The following equation is to be used to solve this problem:

∆𝑃(𝑡 − 𝑡0) =𝜇𝑚𝑠𝛼

2𝐴2(𝐵2 − 𝐵0

2) +𝜇

𝐴𝛽(𝐵 − 𝐵0)

4.4

Since 𝐵0 and 𝑡0 are both equal to zero, the equation can be simplified as follow:

∆𝑃𝑡 =𝜇𝑚𝑠𝛼

2𝐴2𝐵2 +

𝜇

𝐴𝛽𝐵



𝑡

𝐵=

𝜇𝑚𝑠𝛼

2𝐴2∆𝑃𝐵 +

𝜇

𝐴∆𝑃𝛽

Filtrate volume x103 dm3 4 6.1 7.7 9.65 11.47 13.6 15.5

Filtrate volume m3 (B) 4 6.1 7.7 9.65 11.47 13.6 15.5

Time(min) 10 20 30 45 60 80 100

Time(s) 600 1200 1800 2700 3600 4800 6000

t/B (s/m3) 150 196.7213 233.7662 279.7927 313.8622 352.9412 387.0968

Table 2 3 Calculated values for the plotting of B vs t/B

From the graph, it is obtained that the gradient of the line, m = 22.8 and the intercept = 57.9.

(i) Resistance of cake

𝜇𝑚𝑠𝛼

2𝐴2∆𝑃= 22.8

∴ 𝛼 =(22.8)(2𝐴2∆𝑃)

𝜇𝑚𝑠=

22.8 × 2 × 1.52 × 330 × 103

0.001 × 50

𝜶 = 𝟔. 𝟕𝟕 × 𝟏𝟎𝟖𝒎

𝒌𝒈

(ii) Resistance of medium

𝜇

𝐴∆𝑃𝛽 = 57.9

∴ 𝛽 =(57.9)(𝐴∆𝑃)

𝜇=

57.9 × 330 × 103 × 1.5

0.001

𝜷 = 𝟐. 𝟖𝟔 × 𝟏𝟎𝟏𝟎𝒎−𝟏

iii)Specific resistance of cake 𝒓𝒎:

𝒓𝒎 =𝛽

𝑋

4.5

𝒓𝒎 =423.18 × 106

6= 𝟕𝟎. 𝟓𝟑 × 𝟏𝟎𝟔𝒎−𝟐



iv)Specific resistance of cake, 𝒓𝒄

𝒓𝒄 = 𝜌𝑠(1 − 𝜀)𝛼 4.6

𝒓𝒄 = 2700(1 − 0.25)(299.26 × 103) = 𝟔𝟎𝟔. 𝟎 × 𝟏𝟎𝟔 𝒎−𝟐

Total specific resistance:

𝒓𝒎 + 𝒓𝒄 = 70.53 × 106𝑚−2 + 606.0 × 106 𝑚−2 = 𝟒. 𝟐𝟕𝟒 × 𝟏𝟎𝟏𝟔𝒎−𝟐

2) For 75% slurry

Filtrate volume 𝑖𝑠 11.47 × 103 𝑑𝑚3 at 60 minutes.

1 𝑚3 = 1000 𝑑𝑚3

∴ {1 𝑚3

1000 𝑑𝑚3 |11.47 × 103𝑑𝑚3} = 11.47𝑚3

Total volume of cake is:

11.47

0.75= 15.29𝑚3

The volume of the filtrate is 11.47𝑚3 and a filter aid with the area of 3.82 𝑚2 are used.

Total surface area in cake:

(11.47 × 2002) + (3.82 × 8000) = 53522.94𝑚2

∴ 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑡𝑜𝑡𝑎𝑙 𝑠𝑜𝑙𝑖𝑑𝑠, 𝑆 =𝐴

𝑉

53522.94

15.29= 3500.52𝑚2/𝑚3



Average porosity= (0.75 × 0.25) + (0.25 × 0.7) = 0.3625

∴Total cake resistance:

(1 − 𝜀)2𝑆2

𝜀3

4.7

(1 − 0.3625)23500.522

0.36253= 104.54 × 106

∴Change of resistance:

15.29 × 104.54 × 106 = 1.69 × 109

The factor for original resistance:

Equation 4.7 is reused

=(1 − 0.25)220022

0.253

= 144.3 × 106

Relative resistance:

𝟏. 𝟔 × 𝟏𝟎𝟗

𝟏𝟒𝟒. 𝟑 × 𝟏𝟎𝟔= 𝟏𝟏. 𝟎𝟗


Graph 2 1 t/b vs b to find alpha and betta for resistance calculation



Question 3

Water percolates downwards through a san filter of thickness 15 mm, consisting of sand grains of

effective diameter 0.3 mm and void fraction 0.42. The depth of the effectively stagnant clear water above

the filter is 20 mm and the pressure at the base of the sand filter is atmospheric. Calculate the volumetric

flow rate per m2 of packed bed. Given fluid density and viscosity are 0.89x10-3 N s/m2 and 997 kg/m3

respectively.

Assumptions:

o Sand particles are taken as spherical in shape and hence the sphericity is 1.

o Kozeny constant is taken as 4.2 because the sand particles are assumed to be spherical in shape.

o Atmospheric pressure acts on the water surface.

o Flow of water through the sand filter is laminar.

Given Data:

Sand Feed (water) Diameter 0.3 mm - pressure 101325 Pa Height 15mm 20 mm Porosity 0.42 - Density - 997 kg/m3 viscosity - 0.89 x 10-3 Ns/m2 Ac

Solution:

Calculating the specific surface,

S = 6

𝜑𝐷 =

6

1 × 0.3 ×10−3 = 20000 m2/m3

P1 = Pw+ PA is taken as the pressure at the bottom of the water level. The pressure equals to the

atmospheric pressure acting on the water surface and the pressure due to the water depth (pressure

head)

P2 = 101325 Pa is the pressure at the bottom of the sand filter.



Pw = ρ𝑤.g.h = 997 x 9.81 x 20 x 10-3 = 195.6114

ΔP = P1 – P2 = (195.6114 + 101325) – 101325 = 195.6114 Pa (change in pressure is the pressure head)

Assuming the flow of water is laminar through the sand medium,

Application of Kozeny equation (K = 4.2),

−ΔP

ΔX =

𝐾𝑣𝜇(1−𝜀2)𝑆2

𝜀3 = 4.2 .𝑣.(1−0.422)200002

0.423 (where ΔX = 15 × 10−3 )

𝑣 =195.6114 × 0.423

4.2 × 0.89 × 10−3 × (1 − 0.42)2 × 200002 × 15 × 10−3

𝑣 = 0.00192 m/s (superficial fluid velocity)

Recheck the flow type using Reynolds number with the velocity value,

Calculating modified Reynolds number for porous flow through sand medium,

Re = 𝜌 𝑣𝑝𝐷𝐻

𝜇

Actual velocity of fluid through the interstices vp,

vp = v/𝜀 = 0.00192/0.42 = 0.00457 m/s

Hydraulic diameter DH = 𝜀

(1−𝜀)𝑆 =

0.42

(1−0.42) ×20000 = 3.6207 x 10-5 m

Hence, the modified Reynolds number calculation is performed,

Re = 𝜌 𝑣𝑝𝐷𝐻

𝜇 =

997 ×0.00457 × 3.6207 x 10−5

0.89 ×10−3

Re = 0.1853

This confirms the assumption of laminar flow since the modified Reynolds number is less than 2.

(Note: Turbulent flow for porous flow is for a modified Reynolds number value above 2)

Hence, the volumetric flowrate per m2 is taken from the superficial velocity v,

Q (volumetric flowrate m3/s)= vA

Q (volumetric flowrate per m2 = m/s) = v = 0.00192 m/s



Question 4 A continuous cylindrical-tank thickener is to be designed to handle 5.5 x103 pounds of solids per

hour. The feed contains 0.035 mass fraction of limestone; the required underflow concentration is 0.21

mass fraction. Solid SG: 2.71 and solid size 5 μm. Estimate the minimum area of the thickener required

(base on 10 points plot). Data: batch settling test result collect is as follow:

Time(min) interface height(cm)

0 102.8

5 91.2

10 80.9

15 71.7

20 63.6

25 56.4

30 50.1

40 39.4

50 30.9

70 19.2

90 13.88

110 10.2

130 9.6

150 9.2 Table 4 1 Time and interface height for tank thickener

Given data:

Data Values

Mass of solid =5.5 × 103 𝑝𝑜𝑢𝑛𝑑𝑠 𝑝𝑒𝑟 ℎ𝑜𝑢𝑟

Mass fraction of limestone =0.035

Underflow mass fraction =0.21

Mass fraction 𝐶𝐴𝐶𝑂3 =0.0291

Solid S.G. =2.71

Solid size(diameter) = 5 μm



Method :

1. Convert all the mass fraction to volume fraction

2. Tabulate the table in order to get the 𝜑𝑚𝑖𝑛

3. Get the area at 𝜑𝑚𝑖𝑛

Assumptions:

1. Water and 𝐶𝑎𝐶𝑂3 is assumed to be one solution.

2. The minimum area is determined using 𝜑𝑚𝑖𝑛

3. The specific gravity of the water is 1.

Convert all the mass fraction to volume fraction

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑎𝐶𝑂3 =0.0291

2.71= 0.010738 𝑙

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = (1 − 0.0291)

1= 0.9709 𝑙

Volume fraction of C0 :

0.010738

0.010738 + 0.9709= 0.0109389

𝐶𝑜𝐻𝑜 = 0.0109389 × 102.8 = 1.1245

For underflow system

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐶𝑎𝐶𝑂3 =0.21

2.71= 0.0775 𝑙

𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 = (1 − 0.21)

1= 0.79 𝑙

Volume fraction of 𝐶𝑂3 :

0.0775

0.0775 + 0.79= 0.0893



The results and the graph were obtained by using MATLAB. Two m.files were used to plot the graph. The

procedure is outlined below

1) Graph of H vs T is plotted. The first m-file (h_vs_t.m) is utilized for these. Only the curve line is obtained.

2) The tangent of the lines to the time values are determined using a separate m-file (getthetangent.m)

(Martina and Delli 2009). The slope of the lines and the gradients are recorded as shown in Table 4.3. The

gradient and the intersections are converted to y=mx+c form.12 tangent lines are taken.

3) Modifications were done to the first m-file to display the tangent lines.

4) Results after tabulation in order to get the 𝜑𝑚𝑖𝑛

H1 (cm) V (cm/min) Co Ho (cm) C=(CoHo)/H1 𝜑𝑏𝑎𝑡𝑐ℎ 𝒄𝒖 𝒄𝒖

𝒄𝒖 − 𝟏 𝜑𝑠

(kg/m2s)

102 2.19 0.010939 102.8 0.011025 0.024144 0.0893 1.140845 0.027545

100 1.95 0.010939 102.8 0.011245 0.021928 0.0893 1.144068 0.025087

97.7 1.73 0.010939 102.8 0.01151 0.019912 0.0893 1.147961 0.022858

94.2 1.53 0.010939 102.8 0.011938 0.018264 0.0893 1.154307 0.021083

90.2 1.35 0.010939 102.8 0.012467 0.01683 0.0893 1.16226 0.019561

86 1.2 0.010939 102.8 0.013076 0.015691 0.0893 1.171544 0.018383

77.8 0.96 0.010939 102.8 0.014454 0.013876 0.0893 1.193116 0.016555

69 0.762 0.010939 102.8 0.016297 0.012419 0.0893 1.223244 0.015191

49 0.426 0.010939 102.8 0.022949 0.009776 0.0893 1.34588 0.013158

34.1 0.225 0.010939 102.8 0.032977 0.00742 0.0893 1.585501 0.011764

22 0.107 0.010939 102.8 0.051114 0.005469 0.0893 2.338584 0.01279

12.9 0.025 0.010939 102.8 0.087172 0.002179 0.0893 41.96447 0.091453

Table 4 2 Tabulated and calculated values for height vs time

The formulas below were used to calculate the values above.

𝝋𝒃𝒂𝒕𝒄𝒉 = 𝑪 × 𝑽 4.1

𝝋𝒔 = 𝝋𝒃𝒂𝒕𝒄𝒉

𝒄𝒖

𝒄𝒖 − 𝟏

4.2



Graph 4 1 H vs T graph



Tangent lines from MATLAB

y1 -2.19x+102

y2 -1.95x+100

y3 -1.73x+97.7

y4 -1.53x+94.2

y5 -1.35x+90.2

y6 -1.2x+86

y7 -0.96x+77.8

y8 -0.762x+69

y9 -0.426x+49

y10 -0.225x+34.1

y11 -0.107x+22

y12 -0.0s5x+12.9 Table 4 3 Tangent lines generated with Matlab 2nd m-file (Martina and Delli 2009)

The 𝜑𝑚𝑖𝑛is calculated by first converting the mass flow rate to volumetric flow rate. The

calculation is then continued.

�̇� =5.5 × 103𝑝𝑜𝑢𝑛𝑑𝑠

1 ℎ𝑟|

1 ℎ𝑟

60 𝑚𝑖𝑛|

0.453592 𝑘𝑔

1 𝑝𝑜𝑢𝑛𝑑= 41.58 𝑘𝑔/𝑚𝑖𝑛

𝑄 =�̇�

𝜌=

41.58 𝑘𝑔/𝑚𝑖𝑛

2710𝑘𝑔𝑚3

= 0.0153𝑚3

𝑚𝑖𝑛= 1.53

𝑐𝑚3

𝑚𝑖𝑛

The minimum area is then determined:

𝑄 = 𝐴. 𝜑𝑠 4.3

𝐴 =1.53 × 104 𝑐𝑚3

𝑚𝑖𝑛𝟎. 𝟎𝟏𝟏𝟕𝟔𝟒 𝑐𝑚/𝑚𝑖𝑛

= 𝟏𝟑𝟎𝟎𝟓𝟕𝟖𝒄𝒎𝟐 = 𝟏𝟑𝟎. 𝟎𝟔𝒎𝟐



Question 5

150 kg of uniform spherical particles with a diameter of 60 μm and particle density 2000

kg/m3 are fluidized by water (density 1000 kg/m3, viscosity 0.001 Pa s) in a circular bed of cross-

sectional area 0.5 m2. The voidage at incipient fluidization is known to be 0.47. Calculate:

a. the minimum fluidized velocity (m/min) and the bed height at incipient fluidization

b. the mean fluidized bed voidage and height when the liquid flow rate is 2.25x10-5 m3/s.

Slow flow is assumed.

𝑆 =6

𝐷

5.1

𝑆 =6

60 × 10−6

𝑆 = 105

𝑉𝑚𝑓 =(𝜌𝑠 − 𝜌)𝑔𝜀3

𝐾. 𝜇. (1 − 𝜀)𝑆2

5.2

𝑉𝑚𝑓 =(2000 − 1000) × 9.81 × 0.473

4.2 × 0.001 × (1 − 0.47)(105)2

𝑉𝑚𝑓 = 4.575 × 10−5𝑚/𝑠

𝑅𝑒 =𝜌𝑉𝑚𝑓𝐷

𝜇

5.3

𝑅𝑒 =2000 × 4.575 × 10−5 × 60 × 10−6

0.001

𝑅𝑒 = 5.49 × 10−3

𝟓. 𝟒𝟗 × 𝟏𝟎−𝟑 < 𝟏.



Assumption confirmed. Flow is slow

Bed height

𝑀𝐵 = (1 − 𝜀). 𝜌𝑠. 𝐴. 𝐻

5.4

150 = (1 − 0.47). (2000). (0.5). 𝐻

𝑯 = 𝟎. 𝟐𝟖𝟑𝟎 𝒎

Voidage; 𝑉𝑓

𝑉𝑓 =𝑄

𝐴

5.5

2.25 × 10−5 𝑚3

𝑠

0.5 𝑚2= 4.5 × 10−5

𝑚

𝑠

Find Vo

𝑉𝑜 =𝐷2(𝜌𝑠 − 𝜌)𝑔

18𝜇

𝑉𝑜 =(60 × 10−6)2(2000 − 1000)(9.81)

18(0.001)

𝑉𝑜 = 1.962 × 10−3 𝑚

𝑠

Find the voidage

Since Re< 0.3, The Richardson –Zaki equation is used, n=4.65

𝜀𝑛 =𝑉𝑓

𝑉𝑜

5.6

𝜀 = √𝑉𝑓

𝑉𝑜

𝑛



𝜀 = √4.5 × 10−5

1.962 × 10−3

4.65

𝜺 = 𝟎. 𝟒𝟒𝟕𝟐

a. Find the Bed height. Reusing equation 5.4;

𝑀𝐵 = (1 − 𝜀). 𝜌𝑠. 𝐴. 𝐻

130 = (1 − 𝟎. 𝟒𝟒𝟕𝟐). (2000). (0.5). 𝐻

𝑯 = 𝟎. 𝟐𝟑𝟓𝟐𝒎



References Martina, Sandra, and Jean Luc Delli. 2009. Matlab License Universe Picarde Jules Verne. April. Accessed

May 23, 2015. https://www.u-picardie.fr/~dellis/Matlab_Licence/getthetangent.m.

Rhodes, Martin. 2008. Introduction to Particle Technology Second edition. England: John Wiley and Sons,

Ltd.

RICHARDSON, J. F., J. H. HARKER, and J. R. BACKHURST. 2002. "Chapter 16 Motions of Particle in a fluid."

In CHEMICAL ENGINEERING Volume 2 5th edition Particle technology and Separation Process,

146-187. Great Britian: Butterworth Heinemann.

Wischnewski, Berndt , and Bernhard Spang. n.d. CalcSteam.

http://www.peacesoftware.de/einigewerte/wasser_dampf_e.html.



Appendix.

Appendix question 4

assignment 2 fluid particles and processes

Documents

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tb vs b

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