Fluid particle process

8/10/2019 Fluid particle process

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FLUID & PARTICLE PROCESSASSIGNMENT 2

Mahsa Lotfi 14863624

Danise

Mohammed


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Figure 1.1

From the gradient it becomes possible to calculate and from the y-intercept it is possible tocalculate .

y = 6.6518x + 18106R = 0.9976

0100002000030000400005000060000700008000090000

100000

0 2000 4000 6000 8000 10000 12000 14000

D P

B-Bo

Series1

Linear (Series1)


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Question 2.

DataA=

Volume of filtrate collected (litres) 280 430 540 680 800Time (minutes) 10 20 30 45 60Table 2.1a. Volume of filtrate collected (litres) at time t.

a- 1200 L of filtrateConstant pressure drop of 400 kPa

b- Washing with 500L of waterConstant pressure drop of 200 kPa

Calculations

A)In this case the basic filtration equation can be written as the equation represented below.

If is plotted against , it would result in a linear graph with a gradient of

and a y-intercept of .Time (s)

600 280000 535.71431200 430000 697.6744

1800 540000 833.33332700 680000 992.64713600 800000 1125

Table2.2a. Graph values


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From the graph we

obtain the linear relationship:

y = 0.0011x + 213.54

Hence;

Time taken to collect 1200 liters of filtrate is seconds.

Figure 2.1a.


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B)We can again use the same equation used in part A in order to find a linear relationshipbetween against

Time (min)

600 280000 857.14285711200 430000 1116.279071800 540000 1333.3333332700 680000 1588.2352943600 800000 1800

Figure 2

Figure 3

Using the same method as part A, we can calculate the time taken to wash the resulted cakewith 500 liters of water. The equation for the resulted linear relationship is,

y = 0.0018x + 341.66

The time taken to wash the resulted cake with 500 liters of water is 3104 seconds.

y = 0.0018x + 341.66R = 0.9997

0200400600

800100012001400160018002000

0 200000 400000 600000 800000 1000000

Linear ()


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Question 3.

Data

Time (mins) Interface height (cm) Time (mins) Interface height (cm)0 48.3 21 19.32 46.2 24 16.44 13.2 30 13.66 40.3 39 10.98 37.5 59 7.5110 34.6 77 6.2112 31.7 102 5.2314 28.8 4.416 26.2

CalculationsFrom this we are able to calculate the volume fractions.At feed:

At under flow:

Therefore the volume fraction of limestone at feed and at underflow is:


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Using the values calculated above and table 3.1, a curve of interface height against time isconstructed. Using the tangential lines gradient could be obtained. This gradient is equal tothe velocity of the flow (cm/min).

.Figure 3.1 Graph of Interface Height Vs Time

Using the equation below, we can construct tabulated results for volume fraction, solid fluxand batch flux.

(cm) Gradient, v(cm/min)

Volumefraction, c

cm/min (cm/min)

48.3 1.45 0.00937 0.01345 1.16649 0.0156945 1.216 0.01006 0.01223 1.18091 0.0144530 0.988 0.0132 0.01118 1.20823 0.0135135 0.777 0.01293 0.01005 1.24527 0.0125230 0.0577 0.01509 0.00871 1.29835 0.0113025 0.385 0.01811 0.00697 1.38073 0.00963

20 0.232 0.02264 0.00525 1.52598 0.008015 0.125 0.0302 0.00377 1.85041 0.0069812 0.0789 0.0377 0.00298 2.35004 0.00710 0.0526 0.04527 0.00238 3.21926 0.007678 0.034 0.0566 0.00192 7.23127 0.01391

Table 3.2. Gradient, Volume fraction, , and etc.


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Using and the volume fraction we can construct the following graph:

Figure 3.2. Flux rate Vs Volume Fraction.

Therefore flux rate of the solid phase at minimum is Also the feed rate is,

Since,

Therefore,


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Question4.

Data

Calculations

A)

Since the particles are uniformly spherical,

Assume slow flow; thus:

( )

This assumption can be validated by checking the Reynolds number:

Using Richarsdson-Zaki correlation and Reynold number relationship, we can iterate to getthe correct velocity.

And,

Re nReReRe500 2.39Table 4.1. Range of n values over for different Re values.


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Iteration Re N 1 4.65 2 4.65 3 4.65

Table 4.2. Iterations

The minimum fluidization velocity is m/s. To calculate bed height:


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B)

To calculate superficial velocity;

( )

( ) =1.06Using Khan and Richardson correlation:

N=4.72

To calculate porosity:

Therefore mean bed height is:


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Question 5.

Data

Size Range ( 6.6-

9.49.4-13.3

13.3-18.7

18.7-27

27.0-37.0

37.0-53.0

Feed Size Distribution 0.05 0.2 0.35 0.25 0.1 0.05Coarse Production Size Distribution 0.016 0.139 0.366 0.3 0.12 0.06Table 5.1. Given Data

Calculations

A)

B)The size distribution of the fine product can be calculated using the equation below.

( ) Tabulated values for size distributions for fine products are displayed below.

Size Range ( Inlet Mass CoarseMass

6.6-9.4 0.05 0.016 10 2.66 0.2199.4-13.3 0.2 0.139 40 23.14 0.50313.3-18.7 0.35 0.366 70 60.94 0.27118.7-27 0.25 0.3 50 49.95 0.001527-37 0.1 0.12 20 19.98 0.000637-53 0.05 0.06 10 9.99 0.0003

C)

Grade efficiency is can be calculated by the equation shown below.

Tabulated values for grade efficiency are displayed below.Size Range ( Average Size ( G(x)6.6-9.4 8 26.649.4-13.3 11.35 57.8613.3-18.7 16 87.0618.7-27 22.85 99.927-37 32 99.937-53 45 99.9Table 3


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Plotting the results,

From the graph we get 10.5 is the equiprobable size where g(x) = 50%.

D)

Size Range ( 6.6-9.4

9.4-13.3

13.3-18.7

18.7-27

27.0-37.0

37.0-53.0

Feed Size Distribution 0.08 0.13 0.27 0.36 0.14 0.02

The coarse product size distribution can be determined from the equation below.

Tabulated values are shown below.Size Range ( Feed Distribution Size G(x) Coarse Distribution6.6-9.4 0.08 0.266 0.02569.4-13.3 0.13 0.579 0.090413.3-18.7 0.27 0.871 0.282318.7-27 0.36 0.999 0.43227-37 0.14 0.999 0.16837-53 0.02 0.999 0.024

Fluid particle process

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