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
CURTIN UNIVERSITY OF TECHNOLOGY 2015 1
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 ΒΊC, 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 (π 1) and downstream (π2) density:
Upstream pressure: 200 kPaβ 200 Γ 103ππ
Molecular weight of oxygen, Mw: 32 g/mol
The ideal gas law is used to calculate the density:
π =ππ
π π
1.1
π1 =ππ’ππ π‘ππππππ€
π π=
200 Γ 103(32
1000)
8.314(10 + 273.15)= π. ππ ππ/ππ
π2 =ππππ€ππ π‘ππππππ€
π π=
20 Γ 103(32
1000)
8.314(10 + 273.15)= π. πππ ππ/ππ
οΏ½Μ οΏ½ =π1 + π2
2=
2.7187 + 0.2719
2= π. πππ ππ/ππ
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 2
As velocity is not given, it needs to be substituted.
οΏ½ΜοΏ½ = ππ£π΄
1.2
π£ =οΏ½ΜοΏ½
οΏ½Μ οΏ½π΄
π π’ππ π‘ππ‘π’π‘πππ π£πππ’π ππ βΆ π£ = 0.0234
π·2
The isothermal equation is:
(π12 β π2
2) = ππ1π£2ππ1
π·
1.3
Substituting value in, we will obtain:
(2000002 β 200002) = π Γ 2.72 Γ (0.0234
π·2)2 Γ 7 Γ 200000 Γ
1
π·
π· = 0.035(π)1/5
Where D = diameter of pipe, X = length of pipe and f = friction factor.
Reynolds number equation is:
π π =ππ’ππ π‘πππππ£π·
π=
3230.9
π·
The viscosity of oxygen , π = 1.995 Γ 10β5 ππ. π (The Engineering Toolbox n.d).
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 3
Iteration 1: By assuming f= 0.01:
Substitute the variables inside to find diameter first:
π· = 0.035(0.01)1/5 = 0.0139π
Then, use the calculated D to find the Reynolds number:
π π =3230.9
0.0139= 232438.8
π
π·= 0.000719
π = 0.0198
Iteration 2: f=0.0198:
π· = 0.035(0.0198)1/5 = 0.01597π
π π =3230.9
0.01597= 202265.8
π
π·= 0.0005
π = 0.0189
Iteration 3: f=0.0189:
π· = 0.035(0.0189)1/5 = 0.0158π
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 4
π π =3230.9
0.0158= 204156.5
π
π·= 0.00053
π = 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:
πΊ =π
π΄= 255 ππ/π2π
Term 1:
πΊ2ππ (π1
π2) = 2552 Γ ππ (
2.72
0.272) = 199725
Term 2:
οΏ½Μ οΏ½(π2 β π1) = 1.496 Γ (20,000 β 200,000) = β269280
Term 3:
ππΊ2π
2π·=
0.0189 Γ 2552 Γ 7
2 Γ 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.
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 5
The general equation for isothermal flow is shown as:
πΊ2ππ (π1
π2) + οΏ½Μ οΏ½(π2 β π1) +
ππΊ2π
2π·= 0
It is known that π1
π2=
π1
π2:
πΊ2ππ (π1
π2) +
ππ€
2π π(π2
2 β π12) +
ππΊ2π
2π·= 0
(ππ£)2ππ (π1
π2) +
ππ€
2π π(π2
2 β π12) +
ππΊ2π
2π·= 0
After substituting value:
(9.33 Γ 10β3 Γ π·) β (269289 Γ π·5) + 0.014175 Γ π = 0
1st iteration, f = 0.0189
(9.33 Γ 10β3 Γ π·) β (269289 Γ π·5) + 0.000268 = 0
After using try and error method, we obtain:
π· = 0.0174 π
π π =3230.9
0.0174= 185683.9
π
π·= 0.000575
π = 0.01932
2nd iteration, f = 0.01932
(9.33 Γ 10β3 Γ π·) β (269289 Γ π·5) + 0.000274 = 0
π· = 0.0174 π
π π =3230.9
0.0174= 185684
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 6
π
π·= 0.00055
π = 0.0193
Since the difference between f is insignificant, D = 0.0174 m.
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 7
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, π΅π = 0
Initial time, π‘π = 0
Answer:
Area=1.5m2
Constant pressure: 330πππ Γ 103 = 330000ππ
Filter aid=medium
Slurry feed=cake
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 8
Slurry of 16.5% kg CaCO3 in water.
π = (30|1π
100ππ Γ 20) = 6π
Step 1: Volume fraction of solids
π =π
ππ
4.1
50
2700= 0.0185 βͺ 1
For filter aid, specific surface πΊπ
ππ =6
β π·
4.2
6
0.75 Γ 1 Γ 10β3= 8000π2πβ3
Since constant βπ; the values of the resistance πΌ πππ π½ 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:
π¦ = 22.8π₯ + 57.9 4.3
The following equation is to be used to solve this problem:
βπ(π‘ β π‘0) =πππ πΌ
2π΄2(π΅2 β π΅0
2) +π
π΄π½(π΅ β π΅0)
4.4
Since π΅0 and π‘0 are both equal to zero, the equation can be simplified as follow:
βππ‘ =πππ πΌ
2π΄2π΅2 +
π
π΄π½π΅
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 9
π‘
π΅=
πππ πΌ
2π΄2βππ΅ +
π
π΄βππ½
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
πππ πΌ
2π΄2βπ= 22.8
β΄ πΌ =(22.8)(2π΄2βπ)
πππ =
22.8 Γ 2 Γ 1.52 Γ 330 Γ 103
0.001 Γ 50
πΆ = π. ππ Γ ππππ
ππ
(ii) Resistance of medium
π
π΄βππ½ = 57.9
β΄ π½ =(57.9)(π΄βπ)
π=
57.9 Γ 330 Γ 103 Γ 1.5
0.001
π· = π. ππ Γ πππππβπ
iii)Specific resistance of cake ππ:
ππ =π½
π
4.5
ππ =423.18 Γ 106
6= ππ. ππ Γ ππππβπ
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 10
iv)Specific resistance of cake, ππ
ππ = ππ (1 β π)πΌ 4.6
ππ = 2700(1 β 0.25)(299.26 Γ 103) = πππ. π Γ πππ πβπ
Total specific resistance:
ππ + ππ = 70.53 Γ 106πβ2 + 606.0 Γ 106 πβ2 = π. πππ Γ πππππβπ
2) For 75% slurry
Filtrate volume ππ 11.47 Γ 103 ππ3 at 60 minutes.
1 π3 = 1000 ππ3
β΄ {1 π3
1000 ππ3 |11.47 Γ 103ππ3} = 11.47π3
Total volume of cake is:
11.47
0.75= 15.29π3
The volume of the filtrate is 11.47π3 and a filter aid with the area of 3.82 π2 are used.
Total surface area in cake:
(11.47 Γ 2002) + (3.82 Γ 8000) = 53522.94π2
β΄ ππππππππ π π’πππππ ππ π‘ππ‘ππ π πππππ , π =π΄
π
53522.94
15.29= 3500.52π2/π3
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 11
Average porosity= (0.75 Γ 0.25) + (0.25 Γ 0.7) = 0.3625
β΄Total cake resistance:
(1 β π)2π2
π3
4.7
(1 β 0.3625)23500.522
0.36253= 104.54 Γ 106
β΄Change of resistance:
15.29 Γ 104.54 Γ 106 = 1.69 Γ 109
The factor for original resistance:
Equation 4.7 is reused
=(1 β 0.25)220022
0.253
= 144.3 Γ 106
Relative resistance:
π. π Γ πππ
πππ. π Γ πππ= ππ. ππ
CURTIN UNIVERSITY OF TECHNOLOGY 2015 0
Graph 2 1 t/b vs b to find alpha and betta for resistance calculation
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 14
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
ππ· =
6
1 Γ 0.3 Γ10β3 = 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.
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 15
Pw = Οπ€.g.h = 997 x 9.81 x 20 x 10-3 = 195.6114
ΞP = P1 β P2 = (195.6114 + 101325) β 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),
βΞP
ΞX =
πΎπ£π(1βπ2)π2
π3 = 4.2 .π£.(1β0.422)200002
0.423 (where ΞX = 15 Γ 10β3 )
π£ =195.6114 Γ 0.423
4.2 Γ 0.89 Γ 10β3 Γ (1 β 0.42)2 Γ 200002 Γ 15 Γ 10β3
π£ = 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 = π π£ππ·π»
π
Actual velocity of fluid through the interstices vp,
vp = v/π = 0.00192/0.42 = 0.00457 m/s
Hydraulic diameter DH = π
(1βπ)π =
0.42
(1β0.42) Γ20000 = 3.6207 x 10-5 m
Hence, the modified Reynolds number calculation is performed,
Re = π π£ππ·π»
π =
997 Γ0.00457 Γ 3.6207 x 10β5
0.89 Γ10β3
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
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 16
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 ΞΌm. 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 Γ 103 πππ’πππ πππ βππ’π
Mass fraction of limestone =0.035
Underflow mass fraction =0.21
Mass fraction πΆπ΄πΆπ3 =0.0291
Solid S.G. =2.71
Solid size(diameter) = 5 ΞΌm
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 17
Method :
1. Convert all the mass fraction to volume fraction
2. Tabulate the table in order to get the ππππ
3. Get the area at ππππ
Assumptions:
1. Water and πΆππΆπ3 is assumed to be one solution.
2. The minimum area is determined using ππππ
3. The specific gravity of the water is 1.
Convert all the mass fraction to volume fraction
ππππ’ππ ππ πΆππΆπ3 =0.0291
2.71= 0.010738 π
ππππ’ππ ππ π€ππ‘ππ = (1 β 0.0291)
1= 0.9709 π
Volume fraction of C0 :
0.010738
0.010738 + 0.9709= 0.0109389
πΆππ»π = 0.0109389 Γ 102.8 = 1.1245
For underflow system
ππππ’ππ ππ πΆππΆπ3 =0.21
2.71= 0.0775 π
ππππ’ππ ππ π€ππ‘ππ = (1 β 0.21)
1= 0.79 π
Volume fraction of πΆπ3 :
0.0775
0.0775 + 0.79= 0.0893
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 18
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 ππππ
H1 (cm) V (cm/min) Co Ho (cm) C=(CoHo)/H1 ππππ‘πβ ππ ππ
ππ β π ππ
(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.
ππππππ = πͺ Γ π½ 4.1
ππ = ππππππ
ππ
ππ β π
4.2
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 19
Graph 4 1 H vs T graph
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 20
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 ππππis calculated by first converting the mass flow rate to volumetric flow rate. The
calculation is then continued.
οΏ½ΜοΏ½ =5.5 Γ 103πππ’πππ
1 βπ|
1 βπ
60 πππ|
0.453592 ππ
1 πππ’ππ= 41.58 ππ/πππ
π =οΏ½ΜοΏ½
π=
41.58 ππ/πππ
2710πππ3
= 0.0153π3
πππ= 1.53
ππ3
πππ
The minimum area is then determined:
π = π΄. ππ 4.3
π΄ =1.53 Γ 104 ππ3
ππππ. ππππππ ππ/πππ
= ππππππππππ = πππ. ππππ
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 21
Question 5
150 kg of uniform spherical particles with a diameter of 60 ΞΌm 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.
π =6
π·
5.1
π =6
60 Γ 10β6
π = 105
πππ =(ππ β π)ππ3
πΎ. π. (1 β π)π2
5.2
πππ =(2000 β 1000) Γ 9.81 Γ 0.473
4.2 Γ 0.001 Γ (1 β 0.47)(105)2
πππ = 4.575 Γ 10β5π/π
π π =πππππ·
π
5.3
π π =2000 Γ 4.575 Γ 10β5 Γ 60 Γ 10β6
0.001
π π = 5.49 Γ 10β3
π. ππ Γ ππβπ < π.
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 22
Assumption confirmed. Flow is slow
Bed height
ππ΅ = (1 β π). ππ . π΄. π»
5.4
150 = (1 β 0.47). (2000). (0.5). π»
π― = π. ππππ π
Voidage; ππ
ππ =π
π΄
5.5
2.25 Γ 10β5 π3
π
0.5 π2= 4.5 Γ 10β5
π
π
Find Vo
ππ =π·2(ππ β π)π
18π
ππ =(60 Γ 10β6)2(2000 β 1000)(9.81)
18(0.001)
ππ = 1.962 Γ 10β3 π
π
Find the voidage
Since Re< 0.3, The Richardson βZaki equation is used, n=4.65
ππ =ππ
ππ
5.6
π = βππ
ππ
π
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 23
π = β4.5 Γ 10β5
1.962 Γ 10β3
4.65
πΊ = π. ππππ
a. Find the Bed height. Reusing equation 5.4;
ππ΅ = (1 β π). ππ . π΄. π»
130 = (1 β π. ππππ). (2000). (0.5). π»
π― = π. πππππ
ASSIGNMENT 2 CHEN3009 FLUID AND PARTICLE PROCESSES MIRI 05/25/2015
CURTIN UNIVERSITY OF TECHNOLOGY 2015 24
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.
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Appendix.
Appendix question 4
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