CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
CHEE3731 Filtration Dr Simon Iveson School of Engineering & Built Environment University of Newcastle Callaghan NSW 2308, Australia • Many thanks to Dr Paul Stevenson and Professor Kevin Galvin for the use of their lecture notes in preparing this lecture material.
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Filtration • Filtration Filtration:: Cake Cake filtrati filtration on theory, theory, determ determinati ination on of the the specific cake and medium resistance, constant pressure and constant volume operations, continuous continuous filtration. • Resources: – 1 hour lectures, 1 hour tutorials. – Lecture notes will be posted on blackboard. – Tutorial solutions will will be made available. – McCabe, Smith & Harriott, Harriott, Chapter 29 (6th Ed.) • Asses sessment: – Assignment B3 worth 7 % due Monday 04/11/13. – Topic is an assessable item item in the final exam. exam. 10:15
4
Lecture Outline • Filtra Filtratio tion n defin definiti ition on and and reas reasons. ons. • Flow Flow thro through ugh pack packed ed bed beds. s. • Filt Filtrat ratio ion n equa equati tion ons: s: – Constant pressure batch filtration. – Incompressible & compressible cakes. – Constant rate batch filtration. – Continuous filtration. • Filt Filtrat ratio ion n Equi Equipm pmen ent. t.
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
What is Filtration? • Removal Removal of solid particles particles from a fluid by passing passing the fluid through a filter medium. • Drivin Driving g force force is an applied applied press pressure ure,, either either above above atmospheric to push the fluid and/or below atmospheric to suck the fluid through.
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Why Filtration? • Aim is to to concentra concentrate te a solids solids product product or or to remove remove solids from a fluid. • Usuall Usually y a pre-cur pre-curser ser to drying. drying. • Dry solids solids cheaper to transport transport and have have longer longer shelf shelf life. • Removi Removing ng water water by by drying drying is an an energy energy inten intensiv sivee process. So it is much better better to first remove as much as possible of the free moisture by mechanical means.
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Schematic of a Simple Filtration Unit
Slurry
Filter Cake
Filter Medium
Filtrate
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• How can can we incre increase ase the the rate rate of filtr filtrate ate flow flow?? 8
Nomenclature • L = length of pipe or bed (m) •
P = pressure change across pipe or bed (Pa)
• d SV ≡ 6V p/S p = Sauter-mean average particle size (m) where V p and S p are the total volume and surface area of all the particles in the bed. • D = pipe diameter (m) • e = average height of pipe surface roughness (m). • = bed porosity (volume fraction pores) (-). • f ≡ (P/L) D D/( U U2 /2) = Darcy friction factor (-) • = fluid dynamic viscosity (Pa·s). • L = liquid (fluid) density (kg/m3). 10:15
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Pressure Gradient in a Pipe • Fric Fricti tion onal al pre pressu ssure re los loss: s:
P L
f
1 2
LU 2 D
where U is the average fluid velocity over the crosssection of the pipe i.e. U = Q (m3/s) / Ax (m2) • For For lam lamin inar ar flow flow ( Re Re < 2300) f = 64/ Re Re where Re = UD UD/ . Hence: P 32U L
D
2
• For turbu turbulen lentt flow flow in smoot smooth h pipes, pipes, Blasi Blasius us correlation: f = 0.316/ Re Re1/4 (valid Re < 105). • For turbul turbulent ent flow flow in in rough rough pipes pipes f = f (e/ D D). Use Colebrook equation or Moody diagram.
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Pressure Drop through Packed Bed • Ergu Ergun n (1952 (1952)) equa equati tion on::
P L
150
1 2 U s 2 3 d SV
1.75
1 LU s2 3d SV
where U s = volumetric flowrate per bed crosssectional area (superficial velocity). • First term is is viscous viscous term; term; second term is inertia inertial. l. • In most most filtratio filtration n situation situations, s, inertia inertiall term term negligibl negligible. e. 2 Hence: 1 U s P 150 2 L 3d SV • This This is same same as Darcy’ Darcy’ss law law (1856 (1856): ): where is the permeability (m 2). 10:15
P L
U s 11
10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Plot of Ergun Equation
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Source: Levenspiel (1984)
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Packed Bed Flow Regimes • Fully Fully lamina laminarr condi conditio tions ns exist exist for for Re Re* < 10, and fully turbulent/inertial flow occurs for Re Re* > 2000, where: Re
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*
d LU fs 1
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Example: Packed Bed find
P
• What What press pressure ure drop drop is is needed needed to to produc producee a flowr flowrate ate of 100 kg/(m2·s) of water at 20 ºC ( = 10-3 Pa·s; L = 1000 kg/m3) through a filter cake of cross-sectional area 1 m2 that is 10 cm thick, has a porosity of 0.4 and consists of particles with a Sauter-mean diameter diameter 100 m. Solution:
• Strate Strategy: gy: Use the Ergun Ergun equat equation ion::
P H
150
1 2 U 3
2
d SV
1.75
1 LU 2 3
d SV
• First First deter determin minee superfi superficia ciall veloci velocity: ty: U fs = G/ f = (100 kg/(s·m2))/(1000 kg/m3) = 0.10 m/s 10:15
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Solution (cont). P H
150
1 2 U 3
2 SV
d
1.75
1 LU 2 3
d SV
1 0.42 0.001 kg/(m s)0.10 m/s 150 2 0.43 0.0001 m 1 0.41000 kg/m 3 0.10 m/s2 1.75 0.43 0.0001 m 8.44 106 kg/(m 2 s 2 ) 1.64 106 kg/(m 2 s 2 ) 10.08 106 Pa/m • Hence -P = (0.1 m)(10.08×106 Pa/m) = 1008 kPa 10:15
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Check Solution CHECK: • The lamin laminar ar term is bigger bigger than the turbulent turbulent term, although in this case the inertial term (~ 15 %) is not negligible negligible.. Hence it is expected expected that Re* will be between 10 and 2000, and closer to 10 than 2000. Re
*
d LU 1
(10 4 m)(103 kg/m 3 )(0.1 m/s) (10
3
kg/(m s))(1 0.4)
25
• NOTE: 100 m is fairly large and even for these particles the laminar term is the major one. In most practical cases the laminar laminar term is dominate and the the inertial term can often be neglected. 10:15
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Pressure Profile in Filtration Feed L
P3 = P0
LS
LC L M
Slurry
P2
Filter Cake
dL P1 P0
Filter Medium
P0
dP
P
PC = P2 – P1 PM = P1 – P0 Filtrate Q (m3/s) 10:15
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Filtration Theory • Total Total applie applied d pressure pressure drop drop due due to hydrost hydrostati aticc pressure:
Ph = P3 – P0 = g L = g( L LS + LC + LM) • At steady steady state the hydrostatic hydrostatic pressure pressure must must match match the frictional pressure drop:
P f = PS + PC + P M • The fric frictio tional nal pressure pressure loss loss in the the slurry slurry PS is usually negligible, so:
PC + P M = g( L LS + LC + LM)
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NOTE: If using a pump, then it it will supply a pressure drop that is usually much greater than hydrostatic. A vacuum pump will suck out at less than atmospheric.
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Pseudo Steady State Assumption • A batch batch filter filter is NOT NOT a steady steady state system since the thickness of the cake is increasing with time. • Howeve However, r, the rate rate at at which which pressu pressure re change changess propagate through the system (speed of sound in in the liquid) is much faster than the rate of change of cake thickness. • Hence, Hence, at any any particular particular point in time, time, it it is reasona reasonable ble to assume that the rate of filtrate flow can be calculated based on the steady state fluid flow equations for the thickness of filter cake c ake at that point in time. 10:15
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Incremental Analysis • Assum Assuming ing lamin laminar ar flow, flow, then then across across a thickne thickness ss dL of the filter cake:
dP dL
150
1 2 U s 2 3 d SV
where d SV ≡ 6V p/S p. Hence:
dP 150
1 2 U s 2 3d SV
dL
• The mass dm of solids in the layer of thickness dL is: dm = p AdL(1 – )
• Hence:
dP 150
1
U s
2 SV
3
d p A
dm
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Incompressible Filter Cakes • If the the cake cake is is approxim approximately ately incompressi incompressible ble (i.e. is constant), then: P1
dP 150 P2
P2
P1 150
1 U s m
c
1
0
U s mc
2 p A 3 d SV
• This This is often often writ writte ten: n: Pc where is the specific cake resistance (m/kg). 10:15
dm
2 p A 3d SV
Pc
U s mc A
Pc A U s mc
1501 2
3
d SV p 21
10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Filter Medium Resistance • As well well as as the specif specific ic cake cake resis resistan tance ce , the filter medium also has a resistance Rm (m-1): Rm
• Hence: P Pc
P1 P0 U s
Pm U s
m Pm U s c Rm A
• Rm may slowly increase due to fine particle inclusion. • Rm is significant during the early stages but may become negligible once the the cake becomes thick. Hence = f (PC) becomes f (P). ≈
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Filtrate Velocity & Cake Mass • If V is the volume of filtrate collected from the start of the filtration time t , then: dV dt U s A • This value is constant constant at all all points points through through the cake. • Let C be be the mass of solids deposited mc per unit volume of filtrate i.e. C = mc/V so mc = CV : dV / dt CV • Subs Substi titu tuti ting ng:: P
Rm A • Re-ar e-arra rang ngiing: ng: dt 1 CV Rm dV Q PA A A
where Q is the volumetric volumetric flowrate flowrate of filtrate filtrate (m3/s). 10:15
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Constant Pressure Filtration • If P is constant and the cake is incompressible, then only V varies with time. When t = 0, V = 0 and P = Pm. Hence: dt 1 PA 0 Rm Rm dV t 0 PA Q0 Q0 dt 1 1 • So: wh e r e C dV
• Integrate:
t
0
Q
K cV
K c
Q0
V
dt K cVdV 0
V
1
Q 0
PA2
dV
0
t K cV 2 / 2 V / Q0
t V K c 2 V 1 Q0
• Hence t /V vs V should be straight line of slope K c/2. 10:15
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Example • A slu slurry rry cont contai aini ning ng 23.5 23.5 g CaCO3 per L filtrate (H2O) was filtered at constant pressure of 6.7 psia (46.2 kPa) across a filter of area 440 cm2 at a temperature of 25°C ( water = 0.886 cP). The volume of filtrate as a function of time is tabulated below. Evaluate and Rm (McCSH, Example 29.1a). 10:15
Filtrate Volume, V Time, t
t /V
(L)
(s)
(s/L)
0.5
17.3
34.6
1
41.3
41.3
1.5
72.0
48.0
2
108.3
54.2
2.5
152.1
60.8
3
201.7
67.2
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Solution: Crude Method (1 of 2) 3.5
• Plot V vs. t .
Slope at t = 200 s is: (3 L - 1 L)/(200 s) = 0.01 L/s
3
• Assu Assume me that that at end end of of experiment Rm is negligible. At this point Q = 0.01 L/s.
2.5
) L ( 2 e m u 1.5 l o V 1
dt dV
1 Q
CV R m PA A
Slope at t = 41.3 s is: (2.5 - 0.25 L)/(125 - 0 s) = 0.018 L/s
0.5 0 0
50
100
150
200
250
Time (s)
• Hence:
PA
2
CV Q
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Solution: Crude Method (2 of 2) 46171 Pa0.044 m2 23.5 kg/m 3 0.003 m3 8.86 104 Pa s105 m3 /s 1.43 1011 m/kg 2
• Substi Substitut tutee this this into into earlie earlierr data data at t = 41.3 s, V = 1 L and Q = 0.018 L/s: Rm
PA Q
46171 Pa0.044 m2 8.86 104 Pa s1.8 105 m3 /s
CV A
1.43 10
11
m/kg 23.5 kg/m 3 0.001 m3 0.044 m2
5.10 1010 1/m 10:15
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Solution: Better Method (1 of 2) t V K c 2V 1 Q0 80.0
• Plot t /V versus V and fit straight line.
70.0 60.0
• NOTE NOTE:: Dat Dataa is is a good good 50.0 ) L / match to theory theory – cake s 40.0 ( V t really was incompressible. / 30.0 • Slope K c/2 = 13.03 s/L2.
20.0
• Int Interce rcept 1/Q0 = 28.2 s/L.
10.0
y = 13.025x + 28.227 2
R = 0.9999
0.0
• Hence:
0
K c = 26.1 s/L2
0. 5
1
1. 5
2
2. 5
3
3. 5
V (L)
Q0 = 0.0354 L/s. 10:15
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Solution: Better Method (2 of 2) • Hence:
Rm
PA Q0
46171 Pa 0.044 m 2
8.86 10
4
Pa s 3.54 10 5 m 3 /s
6.48 1010 1/m
Crude method: 5.10×10 10 m-1
K c PA2 C
26.110 s/m 46171Pa0.044 m 8.86 10 Pa s23.5 kg/m 6
4
1.12 1011 m/kg 10:15
2 2
6
3
Crude method: 1.43×10 11 m/kg
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
CF ≠ C • The conce concentr ntrati ation on of solid solidss in the the feed feed slurry slurry C F (kg solid/m3 H20 in feed) is different to C (kg solid/m3 H20 in filtrate) since some of the feed liquid remains in the cake at the end of the filtration process.
Feed Slurry
Water Vapouor
Wet Cake, mwc
Dry Cake, mdc Filtrate, V
C
V
C F
mdc V
C F
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CF
versus C
• If the the mas masss rati ratio o of wet wet cake cake mwc to dry cake mdc is known, then the mass of water in the cake per unit mass of solids is: (mwc/mdc – 1) kg solid water kg solid
1
kg solid water kg solid kg solid
kg solid
kg water kg solid
• Hence the volume volume of water in the the cake cake per volume volume of water in the feed is: C F(mwc/mdc – 1)/ L kg solid kg water in cake m 3 water in cake m 3 water in feed kg solids kg water in cake 10:15
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
• Hence:
CF
versus C (cont.) C F
C
1 kg solid m 3 water in feed m 3 water in cake
1
Dr Simon Iveson University of Newcastle
C F mwc L
1 mdc
kg solid
m 3 water in feed
m 3 water in feed m 3 water in filtrate
kg solid m 3 water in filtrate
m 3 water in feed
• Unde Underr what what cond condit itio ions ns is C ≈ C F ? – Dilute slurries with thin cake so little water retained. • When is C >> C F ? 10:15
– Concentrated slurries, or ones that form porous cakes.
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Example: Particle Size • Examining Examining the cake cake afterwar afterwards, ds, it it was was found found that that its its dry density was 1700 kg/m3, whereas the density of pure CaCO3 is 2700 kg/m3. Estimate the size of the particles in the cake.
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Solution Calculate cake porosity assuming a basis 1 m 3 cake: • (1 m3 cake)×(1700 kg solid/m3 cake) /(2700 kg solid/m3 solid) = 0.63 m3 solid • Hence ence poro porosi sitty = (1 m3 – 0.63 m3)/(1 m3) = 0.37 • From From the the defi defini niti tion on of : 1501 d SV
3 p
1501 0.37 3 11 3 1.12 10 m/kg 0.37 2700 kg/m 2.48 10 6 m 10:15
0.5
• Very Very fine fine par parti ticl cles es – only only 2.5 2.5 micr micron ons! s!
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Empirical Equations for • By conduc conductin ting g constant constant pressu pressure re drop filtr filtrati ation on experiments at different pressure drops, the relationship between and P may be determined. • Usually increases with P, and this may be fitted f itted to an empirical correlation, such as one of the form: 0(P)s
where 0 and s are empirical constants. s is called the compressibility coefficient, which is zero for incompressible cakes and usually lies between 0.2 and 0.8 for compressible ones. • NOTE: NOTE: this is not not a totally totally sound analysis, analysis, since since the the derivation of assumed incompressible cakes! 10:15
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Compressible Cakes • In some some cakes, cakes, as the the pressure pressure across the cake rises the cake compresses. As porosity decreases the cake resistance increases. This is particularly an issue with biological and/or flocculated materials. • Cake Cake can exhibi exhibitt either either revers reversibl iblee (elasti (elastic) c) or irreversible (plastic) behaviour, depending on the effect of pressure on the cake structure. • Irreversible Irreversible behaviour behaviour gives gives cake with a ‘memory ‘memory’’ of the largest pressure applied: Pmaxn
where n = 0 for incompressible systems and n > 0 for compressible systems. 10:15
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Cake Variability due to Irreversible Compressibility • The The first first laye layerr of mat materi erial al deposited experiences the full pressure drop and so is compressed the most. • As the the cake cake become becomess thick thicker, er, the pressure drop is spread out more, and so the fresh material is not as tightly packed as the early material. • For irreve irreversi rsibly bly compres compressib sible le materials, this results in a cake with variable porosity. 10:15
Source: McCabe et al. (2001)
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Example: Compressibility Coefficient • The previo previous us experi experimen ments ts with with CaCO CaCO3 were repeated at a number of different pressure drops and the following values of and Rm were obtained. Determine the compressibility coefficient (McCSH Example 29.1).
P
Rm
(psi)
(m-1) ×10-10
(m/kg) ×10-11
6.7
6.5
1.11
16.2
6.7
1.50
28.2
9.1
1.63
36.3
9.3
1.77
49.1
10.7
1.88
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Solution • Compres Compressib sibili ility ty coeffi coefficie cient nt is slop slopee s of empirical correlation: 0(P)s • Plot vs P on log-log scales. Slope = 0.26 = s. 0 = 6.96×1010 m psi0.26/kg
(NOTE strange units)
10 e c n a t s 1 i 1 s ^ e 0 R 1 e x k ) g a k C / c m i f ( i c e p S
0.2587
y = 0.6956x 2
R = 0.9828
1 1
10:15
10 Pressure Drop (psig)
10 0 39
10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Constant Rate Filtration • Much less common. common. For For constant constant filtrate filtrate flowrate: flowrate: dV / dt
U
A
V At
V mc At A 2 2 • Hence: PC Vmc mc V C V 2 A t t V A t A • From 0(P)s we can substitute P/ = P1-s/ • Subs Substi titu tuti tion on::
PC
1 s C
P
0 C V t
2
A
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Constant Rate Filtration (2) 1 s C
P
0 C V t
2
A
• Hence Hence as as the the cake cake beco becomes mes thicke thickerr (V increasing), the pressure must be increased to maintain maintain constant flow. • Note Note that incr increasi easing ng pressur pressuree may also also increa increase se for compressible cakes, which makes it even harder to maintain a constant flowrate.
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Continuous Filtration • Batch operations operations are inconven inconvenient, ient, so most most plants plants use continuous filters. • A simple simple exam example ple is is the drum drum filte filterr which which was invented 1872 by the Hart brothers. • Vacu Vacuum um appl applie ied d insi inside de drum drum to to
Wash water wate r
suck material through porous wall. • Drum Drum rot rotat ates es at at 0.1 0.1 to to 3 rpm. rpm.
Vacuum
• Cake Cake remove removed d by by str strin ing g or or knife edge.
Slurry
• Can Can wash wash the the cak cakee if desi desire red. d. 10:15
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Drum Filters
10:15
Source: Figure 18.3, 18.3, Kelly & Spottiswood (1995)
43
10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Video Clip (Drum Filtration)
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Drum Filtration Analysis • During During cake cake format formation ion phase phase,, assume assume a consta constant nt P. • Let submerge submerged d fracti fraction on of drum drum area area be f . Drum diameter D D, length L, drum speed n (revs per second). • Subm Submer erge ged d are areaa is: is: A = fAT = DLf • Subm Submer ersi sion on tim timee is: is: t sub = f/n 2 • From From prev previo ious usly ly:: t K cV / 2 V / Q0
• Quadr uadrat atiic eqn eqn:: V 2 (2 / Q0 K c )V 2t / K c 0 • ax2 + bx + c = 0 → x = [-b (b2 – 4ac)0.5]/(2a) • Solution:
10:15
2 2 2 2 t 41 V 0.5 Q0 K c Q0 K c K c 45
10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Drum Filtration Analysis (2) • Simplify:
V
Q0
• Substitute:
0.5 1 1 2 2 K ct Q0 Q0
1 K c
PA Rm
K c
C PA2
0.5 2 • Gives: V PA Rm 2 C t Rm C PA PA2 PA 0.5 2 VC 1 Rm 2 PC Rm • Simplify: tA t t t 2
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46
Drum Filtration Analysis (3) • Substitute t = f /n, A = fAT and mass rate of cake formation: V
c m
c m AT
C t
0.5 2 fn PC 2 nRm nRm
1
• Hence if R Rm is negligible:
2 fnPC AT c m
10:15
0.5
47
10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Example: Drum Filtration • A rotary rotary drum drum 30 30 % subme submerge rged d is used used to to filter filter a slurry containing 14.7 lb CaCO 3 per ft3 water. The pressure drop is 20 in.Hg. The filter cake contains 50 wt% moisture on a wet basis. What is the filter area to filter 10 gal/min of slurry with a cycle time of 5 min. 0 = 2.90×1010 (ft/lb)(ft2/lbf )0.26 and s = 0.26. Density of CaCO3 is 168.8 lb/ft3. T = 20 °C (water: = 6.72×10-4 lbm/(ft s); = 62.3 lb/ft3). Assume Rm = 0. (McCSH (McCSH Example Example 29.2) Basic unit conversions: •
P = 20 in.Hg in.Hg = 9.824 psi psi = 1414 1414 lbf /ft2
• n = 1/(300 s) 10:15•
10 gal/ gal/mi min n = 0.0 0.022 223 3 ft ft3/s
48
Solution Step 1: Calculate C , lb solids per m3 filtrate. • Feed Feed slurry slurry conta contains ins 14.7 14.7 lb solid solidss per ft ft3 water. • Take ake bas basis is:: 1 ft ft3 water = 62.4 lb water + 14.7 lb solid. • Wet filter filter cake is 50 wt% water i.e. 14.7 lb water water = 0.236 ft3 per 14.7 lb solids. • Henc Hencee filt filtra rate te vol volum umee is (1 (1 ft3 – 0.236 ft3) = 0.764 ft3 • Hence C = 14.7 lb/0.764 ft 3 = 19.24 lb/ft3 • (14. (14.7 7 lb soli solids ds)/ )/(16 (168. 8.8 8 lb/f lb/ftt3) = 0.0871 ft3 solids. • Tota Totall slur slurry ry volu volume me = 1.0 1.087 871 1 ft3 per 14.7 lb solids. • If slu slurry rry rate rate is is 0.02 0.0223 23 ft3/s, then solids rate is: dmc/dt = (14.7 lb)(0.0223 ft 3/s)/(1.0871 ft3) = 0.302 lb/s 10:15
49
10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Solution (2) c • For R m negligible: m AT
P
0 P
P
s
0 1 s
P
2 fnPC
2.90 1010
0.5
ft ft 2
c AT m 2 fnPC
0.5
0.26
lb lb f 1 0.26
1414 lbf ft 2
1.35 10
ft 3
8
lb lbf
1.35 108 ft 3 / lb lb f 6.72 10 4 lb/(ft s) AT 0.302 lb/s 20.31 / 300 s 19.2 lb/ft 3 0.5 5 ft = 463 ft2.5/s ?? 0.302 lb/s 2.36 106 lb lb f Units?
0.5
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Solution (3) • Use: g c
32.174
ft lb lb f s 2
ft 5 1 lb f s 2 6 AT 0.302 lb/s 2.36 10 lb lb f 32.174 ft lb 0.5 4 2 ft s 0.302 lb/s 7.34 10 4 2 81.8 ft 2 lb
0.5
e.g. if the ratio L:D were 2:1, then AT = DL = 2 D2 = 81.8 ft2 D = 3.61 3.61 ft ft = 1.1 1.1 m 10:15
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Disc Filters
Scraper
Slurry Bath
Source: Figure 15.25 Wills (1997)
• Vacuum Vacuum insi inside de each each leaf leaf sucks sucks water water thro through ugh the the porous surface. • More compact compact than than drums, but suffer wear problems problems and more difficult to wash. 10:15
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Belt Filters
10:15
Source: Figure 18.6 18.6 Kelly & Spottiswood Spottiswood (1995)
53
10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Belt Filters
Source: Figure 15.26 Wills (1997) 10:15
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Pressure Filtration • Drum, Drum, disc disc and belt belt filt filters ers are are limite limited d to 1 atm atm vacuum, so maximum pressure drop is 1 atm. • Vacuum Vacuum filtrati filtration on also also has problems problems when when liquid liquid is volatile or close to its boiling point. Why? – Boiling of fluid breaks the cake structure and thus enables bypassing. • To achiev achievee greater greater degre degreee of de-wate de-waterin ring, g, need need to apply higher pressures.
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Belt Presses • Very Very high high press pressure uress can be applied in a belt press. • A cont contin inuou uouss fil filte terr clot cloth h travels over rollers. • Press Pressur uree is is app appli lied ed to dewater the slurry. • Ofte Often n use used d in in foo food d and and sewage treatment. Less suitable for abrasive materials.
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Batch Filters • High High pres pressu sures res can can be app appli lied. ed.
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10 September 2013
CHEE3731: Modeling of Separation Processes Part B, Topic 3: Filtration
Dr Simon Iveson University of Newcastle
Other Operational Issues Medium: • Select Select pore size and porosity porosity.. Determ Determined ined by by size size and size distribution of particles. Pressure drop: • Constrained Constrained by capital capital costs, operating operating costs, costs, safety, safety, compressible filter cakes become too compact. Filter aids: • Precoa Precoatt medium medium to stop stop foulin fouling g or partic particle le clog cloggin ging g e.g. diatomaceous earth in pool filters. • Increa Increase se partic particle le size size e.g. floc floccul culati ation. on. Removal methods: • Knife Knife edge edge,, stri string, ng, backwa backwashi shing. ng. Slurry Settling • Large particles particles will will settle settle to the the cake cake surface surface first. first. 10:15
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References • Kelly, Kelly, E.G., E.G., & Spottiswoo Spottiswood, d, D.J. D.J. (1982). (1982). Adelaide: Adelaide: Australian Mineral Foundation. • Levenspiel, Levenspiel, O. (1984). (1984). Engineering Engineering Flow and Heat Heat Exchange. New York: Plenum. • McCabe, McCabe, W.L., W.L., J.C. Smith and P. Harriott, Harriott, “Unit Operations of Chemical Engineering”, 6th. Ed. (2001), McGraw-Hill, Chapter 29. • Wills, Wills, B.A. B.A. (1997). (1997). Mineral Mineral Processing Processing Technology Technology (6th ed.). Oxford: Butterworth-Heinemann.
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