Baghouse

Environmental Engineering · FE Reference Handbook section

Environmental Engineering

0 formulas

10 exam-style examples

~45 min

All Environmental Engineering lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Baghouse within Environmental Engineering. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what baghouse describes physically and when it applies.
State every one of the 0 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: mg/L × MGD × 8.34 = lb/day is the single most used conversion.

Lecture

Why this section exists. Baghouse is the part of Environmental Engineering that lets you connect a treatment unit or receiving water body to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a mass balance across one reactor or one unit process. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. mg/L × MGD × 8.34 = lb/day is the single most used conversion. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Aeration basin at a wastewater treatment plant with churning aerated water and walkways. — Photo 1. Where this shows up in practice: baghouse.
Capstone Studio instructional photograph

Environmental Engineering — Baghouse: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a treatment unit or receiving water body. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 0 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Air-to-Cloth Rao for Baghouses
Shaker/Woven
Reverse Pulse
Air/Woven Jet/Felt
Dust [m3/(min•m2)] [m3/(min•m2 )]
alumina 0.8 2.4
asbestos 0.9 3.0
bauxite 0.8 2.4
carbon black 0.5 1.5
coal 0.8 2.4
cocoa 0.8 3.7
clay 0.8 2.7
cement 0.6 2.4
cosmetics 0.5 3.0
enamel frit 0.8 2.7
feeds, grain 1.1 4.3
feldspar 0.7 2.7
fertilizer 0.9 2.4
flour 0.9 3.7
fly ash 0.8 1.5
graphite 0.6 1.5
gypsum 0.6 3.0
iron ore 0.9 3.4
iron oxide 0.8 2.1

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Series particulate control: overall collection efficiency and emission rate — Baghouse

A stack gas stream of 23 m³/s carries 19 g/m³ of particulate. It passes a cyclone at 91% efficiency followed by a fabric filter at 94.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 23 m³/s
C_in = 19 g/m³
η₁ = 0.91
η₂ = 0.945

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 19 (1 - 0.91) = 1.710 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 1.710 (1 - 0.945) = 0.0941 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.090 \times 0.0550 = 0.99505 = 99.505$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.0941 g / m^{3} \times 23 m^{3} / s \times 3600 s / h \div 1000 = 7.787 k g / h$

Answer: C_out = 0.0941 g/m³, η = 99.51%, emission = 7.79 kg/h

Why the other options are there

η = 185.5% (efficiencies added)
1,573 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

Example 2

Series particulate control: overall collection efficiency and emission rate — Baghouse (2)

A stack gas stream of 19 m³/s carries 23 g/m³ of particulate. It passes a cyclone at 75% efficiency followed by a fabric filter at 97.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 19 m³/s
C_in = 23 g/m³
η₁ = 0.75
η₂ = 0.970

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 23 (1 - 0.75) = 5.750 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 5.750 (1 - 0.970) = 0.1725 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.250 \times 0.0300 = 0.99250 = 99.250$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.1725 g / m^{3} \times 19 m^{3} / s \times 3600 s / h \div 1000 = 11.799 k g / h$

Answer: C_out = 0.1725 g/m³, η = 99.25%, emission = 11.80 kg/h

Why the other options are there

η = 172.0% (efficiencies added)
1,573 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

Example 3

Series particulate control: overall collection efficiency and emission rate — Baghouse (3)

A stack gas stream of 48 m³/s carries 9 g/m³ of particulate. It passes a cyclone at 85% efficiency followed by a fabric filter at 97.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 48 m³/s
C_in = 9 g/m³
η₁ = 0.85
η₂ = 0.975

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 9 (1 - 0.85) = 1.350 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 1.350 (1 - 0.975) = 0.0338 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.150 \times 0.0250 = 0.99625 = 99.625$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.0338 g / m^{3} \times 48 m^{3} / s \times 3600 s / h \div 1000 = 5.832 k g / h$

Answer: C_out = 0.0338 g/m³, η = 99.63%, emission = 5.83 kg/h

Why the other options are there

η = 182.5% (efficiencies added)
1,555 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

Example 4

Series particulate control: overall collection efficiency and emission rate — Baghouse (4)

A stack gas stream of 41 m³/s carries 17 g/m³ of particulate. It passes a cyclone at 81% efficiency followed by a fabric filter at 97.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 41 m³/s
C_in = 17 g/m³
η₁ = 0.81
η₂ = 0.970

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 17 (1 - 0.81) = 3.230 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 3.230 (1 - 0.970) = 0.0969 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.190 \times 0.0300 = 0.99430 = 99.430$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.0969 g / m^{3} \times 41 m^{3} / s \times 3600 s / h \div 1000 = 14.302 k g / h$

Answer: C_out = 0.0969 g/m³, η = 99.43%, emission = 14.30 kg/h

Why the other options are there

η = 178.0% (efficiencies added)
2,509 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

Example 5

Series particulate control: overall collection efficiency and emission rate — Baghouse (5)

A stack gas stream of 10 m³/s carries 16 g/m³ of particulate. It passes a cyclone at 78% efficiency followed by a fabric filter at 99.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 10 m³/s
C_in = 16 g/m³
η₁ = 0.78
η₂ = 0.990

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 16 (1 - 0.78) = 3.520 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 3.520 (1 - 0.990) = 0.0352 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.220 \times 0.0100 = 0.99780 = 99.780$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.0352 g / m^{3} \times 10 m^{3} / s \times 3600 s / h \div 1000 = 1.267 k g / h$

Answer: C_out = 0.0352 g/m³, η = 99.78%, emission = 1.27 kg/h

Why the other options are there

η = 177.0% (efficiencies added)
576.0 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

Example 6

Series particulate control: overall collection efficiency and emission rate — Baghouse (6)

A stack gas stream of 46 m³/s carries 22 g/m³ of particulate. It passes a cyclone at 84% efficiency followed by a fabric filter at 91.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 46 m³/s
C_in = 22 g/m³
η₁ = 0.84
η₂ = 0.910

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 22 (1 - 0.84) = 3.520 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 3.520 (1 - 0.910) = 0.3168 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.160 \times 0.0900 = 0.98560 = 98.560$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.3168 g / m^{3} \times 46 m^{3} / s \times 3600 s / h \div 1000 = 52.462 k g / h$

Answer: C_out = 0.3168 g/m³, η = 98.56%, emission = 52.46 kg/h

Why the other options are there

η = 175.0% (efficiencies added)
3,643 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

Example 7

Series particulate control: overall collection efficiency and emission rate — Baghouse (7)

A stack gas stream of 40 m³/s carries 26 g/m³ of particulate. It passes a cyclone at 75% efficiency followed by a fabric filter at 98.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 40 m³/s
C_in = 26 g/m³
η₁ = 0.75
η₂ = 0.985

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 26 (1 - 0.75) = 6.500 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 6.500 (1 - 0.985) = 0.0975 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.250 \times 0.0150 = 0.99625 = 99.625$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.0975 g / m^{3} \times 40 m^{3} / s \times 3600 s / h \div 1000 = 14.040 k g / h$

Answer: C_out = 0.0975 g/m³, η = 99.63%, emission = 14.04 kg/h

Why the other options are there

η = 173.5% (efficiencies added)
3,744 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

Example 8

Series particulate control: overall collection efficiency and emission rate — Baghouse (8)

A stack gas stream of 50 m³/s carries 15 g/m³ of particulate. It passes a cyclone at 85% efficiency followed by a fabric filter at 91.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 50 m³/s
C_in = 15 g/m³
η₁ = 0.85
η₂ = 0.910

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 15 (1 - 0.85) = 2.250 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 2.250 (1 - 0.910) = 0.2025 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.150 \times 0.0900 = 0.98650 = 98.650$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.2025 g / m^{3} \times 50 m^{3} / s \times 3600 s / h \div 1000 = 36.450 k g / h$

Answer: C_out = 0.2025 g/m³, η = 98.65%, emission = 36.45 kg/h

Why the other options are there

η = 176.0% (efficiencies added)
2,700 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

Example 9

Series particulate control: overall collection efficiency and emission rate — Baghouse (9)

A stack gas stream of 46 m³/s carries 30 g/m³ of particulate. It passes a cyclone at 86% efficiency followed by a fabric filter at 93.5% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 46 m³/s
C_in = 30 g/m³
η₁ = 0.86
η₂ = 0.935

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 30 (1 - 0.86) = 4.200 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 4.200 (1 - 0.935) = 0.2730 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.140 \times 0.0650 = 0.99090 = 99.090$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.2730 g / m^{3} \times 46 m^{3} / s \times 3600 s / h \div 1000 = 45.209 k g / h$

Answer: C_out = 0.2730 g/m³, η = 99.09%, emission = 45.21 kg/h

Why the other options are there

η = 179.5% (efficiencies added)
4,968 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

Example 10

Series particulate control: overall collection efficiency and emission rate — Baghouse (10)

A stack gas stream of 49 m³/s carries 26 g/m³ of particulate. It passes a cyclone at 91% efficiency followed by a fabric filter at 90.0% efficiency. Compute the concentration after each device, the overall efficiency, and the emission rate in kg/h.

Given

Q = 49 m³/s
C_in = 26 g/m³
η₁ = 0.91
η₂ = 0.900

Find

Intermediate and final concentrations, overall η and kg/h emitted

Start with the thinking

Efficiencies in series multiply as penetrations (1 − η), they never simply add.
The overall penetration is the product of the individual penetrations.

Step-by-step solution

Formula
$C_{1} = C_{i} n (1 - η_{1})$
Substituting
$C_{1} = 26 (1 - 0.91) = 2.340 g / m^{3}$
Formula
$C_{2} = C_{1} (1 - η_{2})$
Substituting
$C_{2} = 2.340 (1 - 0.900) = 0.2340 g / m^{3}$
Formula
$η_{o} v er a l l = 1 - (1 - η_{1}) (1 - η_{2})$
Substituting
$η = 1 - 0.090 \times 0.1000 = 0.99100 = 99.100$
Formula
$e mi ss i o n = C_{2} Q$
Substituting
$0.2340 g / m^{3} \times 49 m^{3} / s \times 3600 s / h \div 1000 = 41.278 k g / h$

Answer: C_out = 0.2340 g/m³, η = 99.10%, emission = 41.28 kg/h

Why the other options are there

η = 181.0% (efficiencies added)
4,586 kg/h (uncontrolled rate)

Reference: FE Reference Handbook — Environmental Engineering → Baghouse

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a treatment unit or receiving water body, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Baghouse contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: a mass balance across one reactor or one unit process.
Unit rule: mg/L × MGD × 8.34 = lb/day is the single most used conversion.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

mg/L × MGD × 8.34 = lb/day is the single most used conversion
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.