Constant pressure drop filtration If ðpÞ is constant then,
6.4 WORKED EXAMPLES WORKED EXAMPLE 6.1
Water flows through 3.6 kg of glass particles of density 2590 kg/m3forming a packed bed of depth 0.475 m and diameter 0.0757 m. The variation in frictional pressure drop across the bed with water flow rate in the range 200 –1200 cm3/min is shown in columns one and two in Table 6.W1.1.
(a) Demonstrate that the flow is laminar.
(b) Estimate the mean surface-volume diameter of the particles.
(c) Calculate the relevant Reynolds number.
Figure 6.3 Analysis of the pressure drop – flow relationship for a compressible cake
Solution
(a) First, convert the volumetric water flow rate values into superficial velocities and the pressure drop in millimetres of mercury into pascal. These values are shown in columns 3 and 4 of Table 6W1.1.
If the flow is laminar then the pressure gradient across the packed bed should increase linearly with superficial fluid velocity, assuming constant bed voidage and fluid viscosity. Under laminar conditions, the Ergun equation [Equation (6.15)] reduces to:
ðpÞ
H ¼ 150mU x2sv
ð1 eÞ2 e3
Hence, since the bed depthH, the water viscosity m and the packed bed voidage e may be assumed constant, then ðpÞ plotted against U should give a straight line of gradient
150mH x2sv
ð1 eÞ2 e3
This plot is shown in Figure 6W1.1. The data points fall reasonably on a straight line confirming laminar flow. The gradient of the straight line is 1:12 106Pa s/m and so:
150mH x2sv
ð1 eÞ2
e3 ¼ 1:12 106Pa s=m
(b) Knowing the mass of particles in the bed, the density of the particles and the volume of the bed, the voidage may be calculated:
mass of bed¼ AHð1 eÞrp
giving e ¼ 0:3497
Substituting e ¼ 0:3497, H ¼ 0:475 m and m ¼ 0:001 Pa s in the expression for the gradient of the straight line, we have
xsv¼ 792 mm Table 6W1.1
Water flow rate Pressure drop U Pressure
(cm3/min) (mmHg) (m/s 104) drop (Pa)
162 FLUID FLOW THROUGH A PACKED BED OF PARTICLES
(c) The relevant Reynolds number isRe¼ xUrf
mð1 eÞ[Equation (6.12)] giving Re¼ 5:4 (for the maximum velocity used). This is less than the limiting value for laminar flow (10); a further confirmation of laminar flow.
WORKED EXAMPLE 6.2
A leaf filter has an area of 0.5 m2and operates at a constant pressure drop of 500 kPa.
The following test results were obtained for a slurry in water which gave rise to a filter cake regarded as incompressible:
Volume of filtrate collected (m3) 0.1 0.2 0.3 0.4 0.5
Time (s) 140 360 660 1040 1500
Calculate:
(a) the time need to collect 0.8 m3of filtrate at a constant pressure drop of 700 kPa;
(b) the time required to wash the resulting cake with 0.3 m3of water at a pressure drop of 400 kPa.
Solution
For filtration at constant pressure drop we use Equation (6.27), which indicates that if we plott=V versus V a straight line will have a gradient
rcfm 2A2ðpÞ and an intercept rcfm
A2ðpÞVeqon thet=V axis.
0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000
5
Pressure drop (Pa)
10 15
Superficial fluid velocity (m/s x 10000)
20 25 30 35 40 45
Figure 6W1.1 Plot of packed bed pressure drop versus superficial fluid velocity
Using the data given in the question:
Vðm3Þ 0.1 0.2 0.3 0.4 0.5
t=V ðs=m3Þ 1400 1800 2200 2600 3000
This is plotted in Figure 6W2.1.
From the plot: gradient¼ 4000 s=m6 intercept¼ 1000 s=m3 hence rcfm
2A2ðpÞ¼ 4000 and rcfm
A2ðpÞVeq¼ 1000
which, withA ¼ 0:5 m2andðpÞ ¼ 500 103Pa, gives rcfm ¼ 1 109Pa s=m2 andVeq¼ 0:125 m3
Substituting in Equation (6.27):
t
V¼0:5 109
ðpÞ ð4V þ 1Þ
which applies to the filtration of the same slurry in the same filter at any pressure drop.
(a) To calculate the time required to pass 0:8 m3of filtrate at a pressure drop of 700 kPa, we substituteV ¼ 0:8 m3andðpÞ ¼ 700 103Pa in the above equation, giving
t ¼ 2400 s ðor 40 minÞ
0
t/v
0 500 1000 1500 2000 2500 3000
0.05 0.01 0.15 0.15 0.25 0.3 0.35 0.4 0.45 0.5
Volume of filtrate passed, V Figure 6W2.1 Plot oft=V versus V
164 FLUID FLOW THROUGH A PACKED BED OF PARTICLES
(b) During the filtration the cake thickness is continuously increasing and, since the pressure drop is constant, the volume flow rate of filtrate will continuously decrease.
The filtration rate is given by Equation (6.26). Substituting the volume of filtrate passed at the end of the filtration periodðV ¼ 0:8 m3Þ, rcfm ¼ 1 109Pa s=m2,Veq¼ 0:125 m3 andðpÞ ¼ 700 103Pa, we find the filtration rate at the end of the filtration period is dV=dt ¼ 1:89 104m3=s.
If we assume that the wash water has the same physical properties as the filtrate, then during a wash period at a pressure drop of 700 kPa the wash rate is also 1:89 104m3=s. However, the applied pressure drop during the wash cycle is 400 kPa.
According to Equation (6.26) the liquid flow rate is directly proportional to the applied pressure drop, and so
flow rate of wash waterðat 400 kPaÞ ¼ 1:89 104 400 103 700 103
¼ 1:08 104m3=s
Hence, the time needed to pass 0:3 m3of wash water at this rate is 2778 s (or 46.3 min).
TEST YOURSELF
6.1 For low Reynolds number (<10) flow of a fluid through a packed bed of particles how does the frictional pressure drop across the bed depend on (a) superficial fluid velocity, (b) particle size, (c) fluid density, (d) fluid viscosity and (e) voidage?
6.2 For high Reynolds number (>500) flow of a fluid through a packed bed of particles how does the frictional pressure drop across the bed depend on (a) superficial fluid velocity, (b) particle size, (c) fluid density, (d) fluid viscosity and (e) voidage?
6.3 What is the correct mean particle diameter to be used in the Ergun equation? How can this diameter be derived from a volume distribution?
6.4 During constant pressure drop filtration of an incompressible cake, how does filtrate flow rate vary with time?
EXERCISES
6.1 A packed bed of solid particles of density 2500 kg/m3occupies a depth of 1 m in a vessel of cross-sectional area 0.04 m2. The mass of solids in the bed is 50 kg and the surface-volume mean diameter of the particles is 1 mm. A liquid of density 800 kg=m3 and viscosity 0:002 Pa s flows upwards through the bed, which is restrained at its upper surface.
(a) Calculate the voidage (volume fraction occupied by voids) of the bed.
(b) Calculate the frictional pressure drop across the bed when the volume flow rate of liquid is 1:44 m3=h.
[Answer: (a) 0.50; (b) 6560 Pa (Ergun).]
6.2 A packed bed of solids of density 2000 kg/m3occupies a depth of 0:6 m in a cylindrical vessel of inside diameter 0:1 m. The mass of solids in the bed is 5 kg and the surface-volume mean diameter of the particles is 300mm. Water (density 1000 kg=m3and viscosity 0:001 Pa s) flows upwards through the bed.
(a) What is the voidage of the packed bed?
(b) Calculate the frictional superficial liquid velocity at which the pressure drop across the bed is 4130 Pa.
[Answer: (a) 0.4692; (b) 1:5 mm=s (Ergun).]
6.3 A gas absorption tower of diameter 2 m contains ceramic Raschig rings randomly packed to a height of 5 m. Air containing a small proportion of sulfur dioxide passes upwards through the absorption tower at a flow rate of 6 m3=s. The viscosity and density of the gas may be taken as 1:80 105Pa s and 1:2 kg=m3, respectively. Details of the packing are given below:
Ceramic Raschig rings
surface area per unit volume of packed bed,SB¼ 190 m2=m3 voidage of randomly packed bed¼ 0:71
(a) Calculate the diameter,dsv, of a sphere with the same surface-volume ratio as the Raschig rings.
(b) Calculate the frictional pressure drop across the packing in the tower.
(c) Discuss how this pressure drop will vary with flow rate of the gas within 10% of the quoted flow rate.
(d) Discuss how the pressure drop across the packing would vary with gas pressure and temperature.
[Answer: (a) 9:16 mm; (b) 3460 Pa; for (c), (d) use the hint that turbulence dominates.]
6.4 A solution of density 1100 kg=m3and viscosity 2 103Pa s is flowing under gravity at a rate of 0:24 kg=s through a bed of catalyst particles. The bed diameter is 0:2 m and the depth is 0:5 m. The particles are cylindrical, with a diameter of 1 mm and length of 2 mm.
They are loosely packed to give a voidage of 0.3. Calculate the depth of liquid above the top of the bed. (Hint: apply the mechanical energy equation between the bottom of the bed and the surface of the liquid.)
[Answer: 0:716 m.]
6.5 In the regeneration of an ion exchange resin, hydrochloric acid of density 1200 kg=m3 and viscosity 2 103Pa s flows upwards through a bed of resin particles of density 2500 kg=m3 resting on a porous support in a tube 4 cm in diameter. The particles are
166 FLUID FLOW THROUGH A PACKED BED OF PARTICLES
spherical, have a diameter 0:2 mm and form a bed of void fraction 0.5. The bed is 60 cm deep and is unrestrained at its upper surface. Plot the frictional pressure drop across the bed as function of acid flow rate up to a value of 0.1 litres/min.
[Answer: Pressure drop increases linearly up to a value of 3826 Pa beyond which point the bed will fluidize and maintain this pressure drop (see Chapter 7).]
6.6 The reactor of a catalytic reformer contains spherical catalyst particles of diameter 1:46 mm. The packed volume of the reactor is to be 3:4 m3and the void fraction is 0.45. The reactor feed is a gas of density 30 kg=m3and viscosity 2 105Pa s flowing at a rate of 11 320 m3=h. The gas properties may be assumed constant. The pressure loss through the reactor is restricted to 68:95 kPa. Calculate the cross-sectional area for flow and the bed depth required.
[Answer: area¼ 4:78 m2; depth¼ 0:711 m.]
6.7 A leaf filter has an area of 2 m2and operates at a constant pressure drop of 250 kPa.
The following results were obtained during a test with an incompressible cake:
Volume of filtrate collected (litre) 280 430 540 680 800
Time (min) 10 20 30 45 60
Calculate:
(a) the time required to collect 1200 litre of filtrate at a constant pressure drop of 400 kPa with the same feed slurry;
(b) the time required to wash the resulting filter cake with 500 litre of water (same properties as the filtrate) at a pressure drop of 200 kPa.
[Answer: (a) 79:4 min; (b) 124 min.]
6.8 A laboratory leaf filter has an area of 0:1 m2, operates at a constant pressure drop of 400 kPa and produces the following results during a test on filtration of a slurry:
Volume of filtrate collected (litre) 19 31 41 49 56 63
Time (s) 300 600 900 1200 1500 1800
(a) Calculate the time required to collect 1:5 m3 of filtrate during filtration of the same slurry at a constant pressure drop of 300 kPa on a similar full-scale filter with an area of 2 m2.
(b) Calculate the rate of passage of filtrate at the end of the filtration in (a).
(c) Calculate the time required to wash the resulting filter cake with 0:5 m3of water at a constant pressure drop of 200 kPa.
(Assume the cake is incompressible and that the flow properties of the filtrate are the same as those of the wash solution.)
[Answer: (a) 37:2 min; (b) 20:4 litre=min; (c) 36:7 min.]
6.9 A leaf filter has an area of 1:73 m2, operates at a constant pressure drop of 300 kPa and produces the following results during a test on filtration of a slurry:
Volume of filtrate collectedðm3Þ 0.19 0.31 0.41 0.49 0.56 0.63
Time (s) 300 600 900 1200 1500 1800
(a) Calculate the time required to collect 1 m3of filtrate during filtration of the same slurry at a constant pressure drop of 400 kPa.
(b) Calculate the time required to wash the resulting filter cake with 0:8 m3of water at a constant pressure drop of 250 kPa.
(Assume the cake is incompressible and that the flow properties of the filtrate are the same as those of the wash solution.)
[Answer: (a) 49:5 min; (b) 110:9 min.]
168 FLUID FLOW THROUGH A PACKED BED OF PARTICLES
7
Fluidization
7.1 FUNDAMENTALS
When a fluid is passed upwards through a bed of particles the pressure loss in the fluid due to frictional resistance increases with increasing fluid flow. A point is reached when the upward drag force exerted by the fluid on the particles is equal to the apparent weight of particles in the bed. At this point the particles are lifted by the fluid, the separation of the particles increases, and the bed becomes fluidized. The force balance across the fluidized bed dictates that the fluid pressure loss across the bed of particles is equal to the apparent weight of the particles per unit area of the bed. Thus:
pressure drop¼weight of particles upthrust on particle bed cross-sectional area
For a bed of particles of density rp, fluidized by a fluid of density rfto form a bed of depthH and voidage e in a vessel of cross-sectional area A:
p ¼HAð1 eÞðrp rfÞg
A ð7:1Þ
or
p ¼ Hð1 eÞðrp rfÞg ð7:2Þ A plot of fluid pressure loss across the bed versus superficial fluid velocity through the bed would have the appearance of Figure 7.1. Referring to Figure 7.1, the straight line region OA is the packed bed region. Here the solid particles do not move relative to one another and their separation is constant. The pressure loss
Introduction to Particle Technology, 2nd Edition Martin Rhodes
© 2008 John Wiley & Sons Ltd. ISBN: 978-0-470 -01427-1
versus fluid velocity relationship in this region is described by the Carman–
Kozeny equation [Equation (6.9)] in the laminar flow regime and the Ergun equation in general [Equation (6.11)]. (See Chapter 6 for a detailed analysis of packed bed flow.)
The region BC is the fluidized bed region where Equation (7.1) applies. At point A it will be noticed that the pressure loss rises above the value predicted by Equation (7.1). This rise is more marked in small vessels and in powders which have been compacted to some extent before the test and is associated with the extra force required to overcome wall friction and adhesive forces between bed and the distributor.
The superficial fluid velocity at which the packed bed becomes a fluidized bed is known as the minimum fluidization velocity,Umf. This is also sometimes referred to as the velocity at incipient fluidization (incipient meaning beginning). Umf
increases with particle size and particle density and is affected by fluid properties.
It is possible to derive an expression for Umf by equating the expression for pressure loss in a fluidized bed [Equation (7.2)] with the expression for pressure loss across a packed bed. Thus recalling the Ergun equation [Equation (6.11)]:
ðpÞ
substituting the expression forðpÞ from Equation (7.2):
ð1 eÞðrp rfÞg ¼ 150ð1 eÞ2
Figure 7.1 Pressure drop versus fluid velocity for packed and fluidized beds
170 FLUIDIZATION
and so
ð1 eÞðrp rfÞg rfx3sv m2
¼ 150ð1 eÞ2
e3 Remfþ 1:75ð1 eÞ
e3 Re2mf ð7:6Þ or
Ar ¼ 150ð1 eÞ
e3 Remfþ 1:751
e3Re2mf ð7:7Þ
whereAr is the dimensionless number known as the Archimedes number,
Ar ¼rfðrp rfÞgx3sv m2
andRemfis the Reynolds number at incipient fluidization, Remf¼ Umfxsvrf
m
In order to obtain a value of Umf from Equation (7.7) we need to know the voidage of the bed at incipient fluidization, e ¼ emf: Taking emf as the voidage of the packed bed, we can obtain a crudeUmf: However, in practice voidage at the onset of fluidization may be considerably greater than the packed bed voidage. A typical often used value of emfis 0.4. Using this value, Equation (7.7) becomes
Ar ¼ 1406 Remfþ 27:3 Re2mf ð7:8Þ Wen and Yu (1966) produced an empirical correlation for Umf with a form similar to Equation (7.8):
Ar ¼ 1652Remfþ 24:51Re2mf ð7:9Þ The Wen and Yu correlation is often expressed in the form:
Remf¼ 33:7½ð1 þ 3:59 105ArÞ0:5 1 ð7:10Þ and is valid for spheres in the range 0:01 < Remf< 1000.
For gas fluidization the Wen and Yu correlation is often taken as being most suitable for particles larger than 100 mm, whereas the correlation of Baeyens and Geldart (1974), shown in Equation (7.11), is best for particles less than 100mm.
Umf¼ðrp rfÞ0:934g0:934x1:8p
1110m0:87r0:066f ð7:11Þ