Chapter 6. Fabric Filters
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Transcript of Chapter 6. Fabric Filters
![Page 1: Chapter 6. Fabric Filters](https://reader034.fdocuments.net/reader034/viewer/2022042514/55cf98e0550346d0339a3264/html5/thumbnails/1.jpg)
Chapter 6. Fabric filters
薛人瑋 Ph.D
2nd October 2012
An Introduction to Air
Pollution
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Fabric Filters
• Fabric filtration
– Air or combustion gases pass through a fabric
– Dust is trapped on the fabric
– Cleaned air exits the system
• Baghouses
– Rows of bags
– Inlet
– Exit
– Cleaning mechanism
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Baghouse application and operation
• Baghouses-
– Compartmented:
• Shaker
• Reverse flow
– Noncompartmented: Pulse Jet
• Operation:
– Dirty air enters at low velocity
– Multiple filters (bags)collect PM
– PM falls to bottom
– Cleaned air exits
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Advantages and disadvantages
• Advantages
– High efficiency for small particles
– Modular design
– Low velocity
– Low pressure drop
• Expense
– Large area
– Frequent cleaning/maintenance
– Operating temp limitations
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Types of Baghouses
• Compartmented:
– Shaker baghouse
• Bags cleaned by oscillating framework
– Reverse air
• Clean air blown through bag is opposite direction
• Non-compartmented:
– Pulse jet
• Compressed air blown down bags for cleaning
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Dust loading
• Fabric: filter material
– Woven fibers
• 100-150 micron diameter
– Interstitial holes
• 50-75 microns
– PM layer forms between fibrils
• Increased filtration efficiency
• Increased pressure drop
S= filter drag
V= filtering velocity
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Theory
Δ P Total pressure drop
Δ Pf Pressure drop due to the fabric
Δ Pp Pressure drop due to the particulate layer
Δ Ps Pressure drop due to the bag house structure
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Darcy’s equation
ΔPf Pressure drop N/m2
ΔPp Pressure drop N/m2
Df Depth of filter in the direction of flow (m)
Dp Depth of particulate layer in the direction of flow (m)
μ Gas viscosity kg/m-s
V superficial filtering velocity m/min
Kf, Kp Permeability (filter & particulate layer m2)
60 Conversion factor δ/min
V = Q/A
Q volumetric gas flow rate m3/min
A cloth area m2
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Dust Layer
L Dust loading kg/m3
t time of operation min
ρL Bulk density of the particulate layer kg/m3
ΔP = ΔPf + ΔPp
Filter Drag S = ΔP/V
Areal dust density W = LVt
S= k1+k2W
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Permeability, K
• Permeability of filter material, (K1)( Ke )
– Extrapolated from test data
• Permeability of particulate layer, (K2)( Ks )
– Slope of test plot
• Determined from test data
– Fabric, dust
• Contributes to filter drag (S) as a function of
areal dust density (W)
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Filter drag model
• Filter drag: dependent on areal dust density (W)
and fabric and dust layer permeability (K)
S= filter drag, Pa-min/m or inches of water- min/ft
W= areal dust density, kg/m2 of fabric or lb/ft2 of fabric
L= dust loading, kg/m3 or lb/ft3
T= time of operation, minutes
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Filter drag model
• Using the filter drag model to predict pressure
drop (ΔP) after 60 min of operation
– Dust loading (L) = 15 g/m3, V= 0.8 m/min
– Ke = 500 Pa-min/m, Ks= 3 Pa-min-m/g
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Filter drag model
• Using the filter drag model to predict pressure
drop (ΔP) after 60 min of operation
– Dust loading (L) = 15 g/m3, V= 0.8 m/min
– Ke = 500 Pa-min/m, Ks= 3 Pa-min-m/g
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Filter drag model
• Using the filter drag model to predict pressure
drop (ΔP) after 60 min of operation
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Dust Layer
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DESIGN OF FABRIC FILTERS
The equation for fabric filters is based on Darcy’s law for flow
through porous media.
Fabric filtration can be represented by the following equation:
S = Ke + Ksw
Where,
S = filter drag, N-min/m3 S = ∆P/V
Ke = extrapolated clean filter drag, N-min/m3
Ks = slope constant. Varies with the dust, gas and fabric, N-min/kg-m
W= Areal dust density = L V t
L = dust loading (g/m3), V = velocity (m/s)
Both Ke and Ks are determined empirically from pilot tests.
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Obtain Pilot Data to Determine
ΔP versus loading
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Problem
• Estimate the values of Ke and Ks for the
filter drag model:
Limestone dust loading L = 1.00 g/m3
Fabric Area A = 1.00 m2
Air flow rate Q = 0.80 m3/min
Time (min) 5 10 15 20 25 30
Filter ∆P (Pa) 330 490 550 600 640 700
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Solution
• Step 1:
Calculate the air velocity
Air velocity = 0.80 (m3/min)/1.00 m2
= 0.80 m/min
• Step 2:
• Step 3:
Determine Ke and Ks graphically
Ke = 470 N-min/m3 Ks = 0.563 N-min/g-m
S = ∆P/V 412.5 612.5 687.5 750 800 875
W = LVt 4 8 12 16 20 24
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Reverse Air Fabric Filter
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Example Problem
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Solution
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Solution
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Pulse Jet Fabric Filter
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Pulse jet design considerations
• Different filtering velocities
• No compartments
• Compressed air for bag cleaning
• Compressor power
• Pressure drop
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Compressor power
• Major operating expense of pulse jet systems
• Compressor power (ω), kW:
η = compressor efficiency
γ = 1.4 (ratio of heat capacities Cp/Cv)
P1, P2 = initial and final pressures (abs), kPa
Q1 = volumetric flow rate at compressor inlet, m3/s
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Compressor power
• Example: Find compressor power (ω), kW
Flow rate (Q) = 20,000 cfm (9.5 m3/s)
T = 50˚C (323 K)
P1= 1 atm (101.3 kPa)
Air pulse (P2) 100 psig (790 kPa) abs.
Compressed/filtered air ratio = 0.6%
Compressor efficiency (η)= 50%
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Compressor power
• Compressor power (ω), kW:
• Flow rate (Q) = 20,000 cfm (9.5 m3/s)
• T = 50˚C (323 K)
• Compressed/filtered air ratio = 0.6%
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Compressor power
• Compressor power (ω), kW:
Compressor efficiency (η)= 50%
P1= 1 atm (101.3 kPa)
Air pulse (P2) 100 psig (790 kPa)
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Fan Horsepower
• Flow rate (Q) = 20,000 cfm
• Assume 60% efficiency (η) for motor
•For ΔP = 17 inches w.g. → BHP = 90 hp
•For ΔP = 3.4 inches w.g. → BHP = 18 hp
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Problem
• Calculate the number of bags required for an 8-compartment
pulse-jet baghouse with the following process information and
bag dimensions.
• Q, process gas exhaust rate 100,000 ft3/min
• A/C, gross air-to-cloth ratio 4 (ft3/min)/ft2
• Bag dimensions:
bag diameter 6 in.
bag height 12 ft
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Solution
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Solution
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Solution
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Solution
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Shaker Baghouse
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Hopper
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Filtration time, tf
• Shaker and reverse-air baghouses
– Several compartments
– One compartment off-line for cleaning
tf= filtration time, min
N= number of compartments
tr= run time, min
tc= cleaning time, min
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Filtration time, tf
• N=5, N-1=4
tf
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Number of bags
• Example: Net cloth area = 8,000 ft2
– Select 3 (N) compartments
• N-1 = 2 (1 off-line for cleaning)
• 2 compartments on line to meet NCA
• Each compartment = 4,000 ft2
• 4,000 ft2 x 3 compartments = 12,000 ft2
– Bag size: 6 inch diameter, 8 feet long
• Bag area: πdh= π(0.5)(8)= 12.6 ft2
– 12,000/12.6 = 952 bags
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Filtering Velocity
• All (N) compartments on-line
– for Q = 20,000 cfm
– flow rate (QN)through one compartment:
• N-1 compartments on line during cleaning
– flow rate through on-line compartments:
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Filtering Velocity• All (N) compartments on-line
– Filtering velocity (VN) in one compartment (C):
• N-1 compartments on line during cleaning
– Design Filtering velocity (VN-1) in on-line
compartments:
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tf
Pressure drop
• Max pressure drop (ΔPm) occurs
– before next compartment to be cleaned (j)
– end of cleaning time for last compartment (j-1)
– at time tj (the time compartment j is on-line)
tj
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Pressure drop
• Calculating Max pressure drop (ΔPm)
– tf= 60 min, tc = 4 min, tr = ?
tf
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Pressure drop
• Calculating Max pressure drop (ΔPm)
– During tj, the cloth in compartment j has
accumulated areal dust density (Wj)
– Given dust loading (L) of 10gr/ft3
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Pressure drop
• Calculating Max pressure drop (ΔPm)
– Given Ke= 1.00 in wg-min/ft, Ks= 0.003 in wg-min-ft/gr
– During tj, the filter drag (Sj) in compartment j is
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Pressure drop
• Calculating Max pressure drop (ΔPm)
– During tj, the actual filtering velocity (Vj) in
compartment j is calculated
– Ratio of Vj to VN-1 Total Number of Compartments, N
fN = Vj/VN-1
3 0.87
4 0.80
5 0.76
7 0.71
10 0.67
12 0.65
15 0.64
20 0.62
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Pressure drop
• Calculating Max pressure drop (ΔPm)
– Finally, the maximun pressure drop can be
calculated
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A Compartment of Bags
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Use Pilot Data to Design a Multi-
compartment Baghouse
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Use Pilot Data to Design a Multi-
compartment Baghouse