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8/3/2019 FM 301 B6(b)
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FM 301
Flow of Solids through Bins
andPneumatic Conveying
Date of Experiment : 28th
September
Date of Presentation : 3rd
October
Lab Group: B-6 (b)
Rupak Kumar 09D02033 (Report)
Akshansh Gupta 09002039 (Presentation)
Sourabh Biswas 09002044 (PPT)
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PART A: Flow of solids through Bins
OBJECTIVE:
The objective of our experiment is to study the effect of the powder level above the exit and
the effect of particle diameter and orifice diameter on the mass flow rate.
MOTIVATION:
Several chemical processes involve solid raw materials or products in the form of powders
(e.g. cement manufacture, pharmaceuticals, etc). The powders need to be transported during
their processing, hence their flow properties are of considerable importance. An application,
which is very common, is the flow of powders from storage bins & hoppers, under the
influence of gravity.
THEORY:
Particulate flow consists of both liquid-like and solid like characteristics. They occupy the
volume of the container, and exert pressure on the side walls of the containers, which is liquid
like behaviour. Unlike liquids, the shearing stress is proportional to the normal load ratherthan to the rate of deformation. Like solids, they can sustain a shear stress, though the
magnitude of the shearing stress at a point is generally indeterminate.
The discharge flow rate depends mainly on the orifice geometry, the nature of the powder and
is nearly independent of the height of the powder above the exit, and the vessel diameter. If
the effect of particle diameter is ignored, then based on the dimensional analysis, the mass
flow rate is given by
Where Do is the orifice diameter, g is the acceleration due to gravity, is the bulk density of
the solids and C is constant.
Beverloo et al. (1981) have shown that the mass flow rate is given by
where d is particle diameter, k and C are constants.
By plotting W2/5
versus Do, we can check whether straight lines are obtained for the two
different particle sizes. The constants C and k can be obtained from the slopes and intercept
of the lines respectively.
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EXPERIMENTAL PROCEDURE:
Figure 1 : Flowchart showing the procedure followed in the experiment
Fix the plate havingorifice diameter as 7.9mm at the bottom of
the bin
Fill the bin with thematerials of particle size
1.1 mm size upto 300mm height in the bin
Release the cork toallow the sand to flow
out and record the timetaken till the sand stops
coming out
Weigh the sandcollected
Repeat the above 4steps for different orifice
plate
Follow the same aboveprocedure for particle
size 0.7 mm
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SCHEMATIC DIAGRAM:
Figure 2 : Schematic diagram of the apparatus
CALCULATION PROCEDURE:
Figure 3 : Calculation procedure followed
Calculate the mass flow rate
Draw the graph between W2/5 and orificediameter Do and obtain equation of lineobtained
From the slope of graph calculate the constant Cby comparing it with Beverloos equation
From the intercept and value of C calculate valueof constant k
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OBSERVATIONS:
Weight of the tub = 189 gm
Average particle diameter = 1.1 mm
Orifice
Diameter
(mm)
Height
(mm)
Mass
collected
(kg)
Actual Mass
collected (kg)
Time Taken (sec) Mean Mass
flow rate W
(gm/s)
7.9 300 2.302 2.113 241.4 8.753
7.9 200 1.608 1.419 162.77 8.717
10.8 300 2.288 2.099 94.5 22.211
10.8 200 1.598 1.409 68.16 20.671
15.8 300 2.403 2.214 36.9 60
15.8 200 1.634 1.445 35.5 40.704
20.5 300 2.332 2.143 16.21 132.202
20.5 200 1.666 1.477 11.4 129.561
Average particle diameter = 0.7 mm
Orifice
Diameter
(mm)
Height
(mm)
Mass
collected
(kg)
Actual Mass
collected (kg)
Time taken (sec) Mean Mass
flow rate W
(gm/s)
7.9 300 2.186 1.997 210.5 9.486
7.9 200 1.545 1.356 142.56 9.511
10.8 300 2.273 2.084 84.58 24.639
10.8 200 1.564 1.375 59.01 23.301
15.8 300 2.283 2.094 32.69 64.056
15.8 200 1.592 1.403 22.07 63.570
20.5 300 2.273 2.084 14.5 143.724
20.5 200 1.646 1.457 10.8 134.907
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GRAPHS:
From the above graph, slope = 0.367
Intercept = -0.54
From the above graph, slope = 0.350
Intercept = -0.529
y = 0.3672x - 0.54
R = 0.9984
0
1
2
3
4
5
6
7
8
0 5 10 15 20 25
Mass
flowrate^(2/5)(gm/sec)
Orifice Diameter D (mm)
W^(2/5) Vs D for grains of 1.1 mm
with Height 300 mm
Series1
Linear (Series1)
y = 0.3501x - 0.529
R = 0.9569
0
1
2
3
4
5
6
7
8
0 5 10 15 20 25
Massflowrate^(2/5)(gm/sec)
Orifice Diameter D (mm)
W^(2/5) Vs D for grains of 1.1 mmwith Height 200 mm
Series1
Linear (Series1)
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From the above graph, slope = 0.378
Intercept = -0.539
From the above graph, slope = 0.366
Intercept = -0.453
y = 0.3781x - 0.5397
R = 0.9974
0
1
2
3
4
5
6
7
8
0 5 10 15 20 25
Massflowrate^(2/5)(gm/sec)
Orifice Diameter D (mm)
W^(2/5) Vs D for grains of 0.7 mmwith Height 300 mm
Series1
Linear (Series1)
y = 0.3668x - 0.4536
R = 0.9993
0
1
2
3
4
5
6
7
8
0 5 10 15 20 25
Mass
flowrate^(2/5)(gm/sec)
Orifice Diameter D (mm)
W^(2/5) Vs D for grains of 0.7 mmwith Height 200 mm
Series1
Linear (Series1)
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RESULTS:
Particle size of diameter 1.1 mm Particle size of diameter 0.7 mm
Error Analysis
Standard deviation (s) = For particle size of diameter = 1.1 mm
C = 0.002 or (10.9%)
k = 0.037 or (2.69%)
For particle size of diameter = 0.7 mm
C = 0.017 or (35.27%)
k = 0.153 or (13.25%)
For particle size of diameter = 1.1 mm
C = 0.0172 0.002
k = 1.3555 0.037
For particle size of diameter = 0.7 mm
C = 0.0411 0.017
k = 1.0775 0.153
Height of
sand level(mm)
Value of
constant C
Value of
constant k
Height of
sand level(mm)
Value of
constant C
Value of
constant k
3000.0182 1.337
3000.0499 1.001
2000.0162 1.374
2000.0323 1.154
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CONCLUSION:
The graph between W2/5
vs Do is a straight line for both the samples of the sand and the slope
and the intercept in both the graphs are nearly equal. The Particle flow depends only on the
orifice and the particle diameter while the fluid flow depends on the height of the column and
the diameter of the orifice and the vessel. The mass flow rate is independent of height
because the stress is mostly borne by the walls of the container and the stress at the orifice is
hence independent of H. Also, the mass flow rate increases with increasing orifice diameter
and decreases with increasing particle size. Unlike liquids, the shearing stress in particulate
solids is proportional to normal load rather than the rate of deformation.
The value of constant k appearing in the Beverloos equation decrease with decrease in the
particle size. The term (DOkd) accounts for the effective diameter that is available to the
particles to exit from the vessel. Further, k signifies the resistance that is offered by the walls
of the opening of exit to the particle flow. Thus, as the particle size decrease, resistance
decreases & so does the value of k. This is proved experimentally too as we obtain lesservalue of k for diameter 0.7 mm.
SOURCES OF ERROR AND PRECAUTIONS:
While filling the sand in the bin, the container should not be shaken in order to level the sand
as it will then resemble a packed bed, and not the free flowing sand that is required for the
experiment.
The height level on top should be levelled out by filling the last 10 mm of sand by hand, sothat the height level of the sand is uniform throughout the cross section of the bin. The orifice
should be placed such that it is inclining downwards like the inner side of a funnel.
The stopwatch should be started as soon as the stop cork is taken out, and this is a source of
error that will propagate into mass flow rate calculations. Also, care should be taken that the
sand does not fall out of the container.
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PART B : Pneumatic Conveying
OBJECTIVE:
The objective is to study qualitatively the different flow regimes for different gas and solids
flow rate for horizontal pneumatic conveying and also to measure pressure drops for gas-
solid flow for different gas and solids flow rates.
MOTIVATION:
Pipeline transport of particulate solids has a number of advantages over conventional
transportation systems such as conveyor belts, the former systems are compact and require
much less maintenance. They are commonly used for in-plant transports of solids (e.g. in
cement plants) as well as for loading and unloading materials from trucks and ships (e.g. food
grains).
THEORY:
Dilute Phase Transport:
When the gas volume flow rates are much higher than that of the solids, the transport of
solids is said to be in the dilute phase. In this case there is a nearly uniform distribution of
solids across the pipe cross-section of the pipe and along its axis. In this case the pressure
drop for the flow is given by
where Pfis the pressure drop due to flow of gas alone and d is the extra pressure drop
due to the pressure of the particles. fcan easily be evaluated from friction factor charts. An
expression for calculating d is
Where, = density
u = velocity
= volume fraction of gas in the pipe and the
Subscripts f and p refer to fluid and particle respectively, L is the length of the pipe, and
m = Gp /Gfis the ratio of solid to gas mass flow rates, Cds is the drag coefficient for a single
particle moving through the fluid at a velocity (uf- up).
Dense Phase Transport:
As the air flow rate is reduced below a threshold limit, the particles start to fall out of the gas
stream, and settle to the bottom of the pipe. This decreases the actual flow area, thus
increasing the pressure drop across the pipe. Further reduction in flow rate leads to more and
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more pressure drop. This phase is known as the Dense Phase, and is undesirable for solid
transport.
The velocity at which the solid particles first begin to settle is known as the Saltation
Velocity. In designing, the saltation velocity is used as a basis for choosing the design gas
velocity in a pneumatic conveying system. Usually, the saltation gas velocity is multiplied bya factor, which is dependent on the nature of the solids, to arrive at a design gas velocity.
When the gas velocity becomes low (below the Saltation Velocity the particles settle at the
bottom of the pipe. The settled particles are transported by partly sliding along the pipe.
The distribution of solids in such cases where mass flow rate(m) is high generally non-
uniform and this results in large pressure fluctuations. An empirical correlation for the
pressure drop in dense phase transport is
Where, D is the pipe diameter and ds is the mass per unit volume of solids in the pipe.
EXPERIMENTAL PROCEDURE:
Start compressorand fill the bin with
sand
Adjust the flow rateof air to high value
Start the flow ofsand
Measure the flowrate of sand by
collecting it for a fixtime
Also meaure thepresure dropandobserve the flow
regime
Decrease the flowand repeat the
procedure
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SCHEMATIC DIAGRAM:
OBSERVATIONS:
Weight of the mug = 110 gm
Average particle diameter = 1.1 mm
Density = 2.8 gm/cc
Air
Flow(m/hr)P(mm) Mass collected
(gm)
Time taken
(sec)
Actual mass
collected
(gm)
Mass flow rate
(W) (gm/sec)
13 13 353 20.28 243 11.982
12 11 341 20.31 231 11.373
11 9 339 20.08 229 11.404
10 8 343 20.22 233 11.523
9 15 279 18.98 169 8.904
8 19 245 20.30 135 6.650
7 26 209 20.18 99 4.905
6 27 184 19.91 74 3.716
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RESULTS AND CONCLUCIONS:
The Air Mass flow rate vs Pressure drop was plotted as shown below. Initially, as the air flow
rate is very high, dilute phase is observed. There is no settling of particles at the bottom of the
pipe, and a well mixed stream of particles and air exits from the pipe.
The above graph indicates the different phases. The first decreasing part shows the dense
phase where the sand particles are accumulating in tube. The increasing part of the graph
shows the dilute phase in which there is no accumulation.
From the above graph, the optimum flow rate to minimize the pressure drop across the tube is
10 m3/hr. And the corresponding mass flow rate is 11.523 gm/s.
y = -0.005x2 - 0.2543x + 14.404
R = 0.9304
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30
Massflowrate(gm/sec)
P(mm of CCl4)
Mass flow rate vs Pressure difference
Series1
Poly. (Series1)
0
5
10
15
20
25
30
0 2 4 6 8 10 12 14Pr
essuredrop(mmofCCl4
Air flow rate (m3/hr)
Pressure drop vs Air flow rate
Series1
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From graph we see that as the flow rate was decreased, the pressure drop reached a minimum
value and then started increasing. Now in the dense-phase pneumatic conveying more and
more solids settle down at the bottom of the pipe as the flow rate of air is decreased. This is
characterised by increase in pressure drop mainly due to large amount of frictional losses
between solid particles and pipe.
In the dilute phase transport, the physical forces contributing to the pressure drop are air-to-
pipe friction, air-to-particle friction and pipe-to-particle friction.In the absence of gravity, the
particles will never settle down at the bottom of the pipe. Thus, the particles will always be
uniformly distributed across the cross section of the pipe, whether the gas flow rates are low
or high.
SOURCES OF ERROR AND PRECAUTIONS:
The readings should only be taken once steady state is achieved and the pressure difference
becomes constant. Moreover, the collection of sand and starting of stop watch should be
simultaneous and the actual time should be considered while doing the calculations. The
stopping of stop watch and the removal of mug should also be simultaneous. Also, the hopper
should be continuously refilled with sand.
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APPENDIX:
Sample calculations for flow of solids through bins.
From graph 1, W2/5
= 0.367 Do - 0.54
Comparing with W = C x x g (DOkd)
5/2
= M/ (h x x d2
x ) = 1.425 gm/ cm3
We get,
(C x x g)0.4= 0.367
(C x x g)0.4
x K x d = 0.54
Solving we get, K = 0.54/(.367 * 1.1) = 1.337
C = 0.3672.5
/( x g) = 0.0182
From graph 2, W2/5
= 0.35 Do -0.529
Comparing with W = C x x g (DOkd)5/2 = M/ (h x x d2 x ) = 1.425 gm/ cm3
We get,
(C x x g).4
= 0.35
(C x x g).4
x K x d = 0.529
Solving we get, K = 0.529/(0.35 *1.1) = 1.374
C = 0.352.5
/( x g) = 0.0162
From graph 3, W
2/5
= 0.378 Do - 0.539Comparing with W = C x x g (DOkd)
5/2
= M/ (h x x d2 x ) = 1.365 gm/ cm3
We get,
(C x x g).4
= 0.378
(C x x g).4
x K x d = 0.539
Solving we get, K = 0.378/(.539 * 0.7) = 1.001
C = 0.5392.5
/( x g) = 0.0499
From graph 4, W2/5= 0.366 Do - 0.453
Comparing with W = C x x g (DOkd)5/2
= M/ (h x x d2 x ) = 1.365 gm/ cm3
We get,
(C x x g).4
= 0.366
(C x x g).4
x K x d = 0.453
Solving we get, K = 0.366/(.453 * 0.7) = 1.154
C = 0.4532.5
/( x g) = 0.0323
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