Chapter 6. Fabric Filters

Chapter 6. Fabric filters

薛人瑋 Ph.D

2nd October 2012

An Introduction to Air

Pollution

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

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

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

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

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

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

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

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

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)

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

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

Filter drag model



– Dust loading (L) = 15 g/m3, V= 0.8 m/min

– Ke = 500 Pa-min/m, Ks= 3 Pa-min-m/g

Filter drag model



Dust Layer

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.

Obtain Pilot Data to Determine

ΔP versus loading

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

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

Reverse Air Fabric Filter

Example Problem

Solution

Solution

Pulse Jet Fabric Filter

Pulse jet design considerations

• Different filtering velocities

• No compartments

• Compressed air for bag cleaning

• Compressor power

• Pressure drop

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

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%

Compressor power


• Flow rate (Q) = 20,000 cfm (9.5 m3/s)

• T = 50˚C (323 K)

• Compressed/filtered air ratio = 0.6%

Compressor power


Compressor efficiency (η)= 50%

P1= 1 atm (101.3 kPa)

Air pulse (P2) 100 psig (790 kPa)

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

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

Solution

Solution

Solution

Solution

Shaker Baghouse

Hopper

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

Filtration time, tf

• N=5, N-1=4

tf

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

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:

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:

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

Pressure drop

• Calculating Max pressure drop (ΔPm)

– tf= 60 min, tc = 4 min, tr = ?

tf

Pressure drop


– During tj, the cloth in compartment j has

accumulated areal dust density (Wj)

– Given dust loading (L) of 10gr/ft3

Pressure drop


– 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

Pressure drop


– 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

Pressure drop


– Finally, the maximun pressure drop can be

calculated

A Compartment of Bags

Use Pilot Data to Design a Multi-

compartment Baghouse

Use Pilot Data to Design a Multi-

compartment Baghouse

Documents

Chapter 6. Fabric Filters