Exercise set

Industrial Emissions Capture and Control Device Exercises

Worked air-emissions control exercises for hood capture, duct velocity, fan power, baghouse, scrubber, carbon bed, oxidizer and bypass gates.

Branch: Environmental Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.0 · reviewed

These exercises practise the physical side of industrial air-emissions control: capturing a pollutant at the source, moving the gas through ducts, keeping the fan and pressure boundary credible, and proving that a baghouse, scrubber, carbon bed or oxidizer still operates inside its release envelope.

The calculations are simplified, but the evidence logic is practical. A control device is not released because it exists in a drawing. It is released when source capture, duct transport, pressure drop, treatment mechanism, bypass state, maintenance condition and monitoring evidence all support the same operating decision.

How to Use These Exercises

For each exercise, define:

the source and operating mode being captured;
whether the boundary is hood, duct, fan inlet, control-device inlet, control-device outlet or stack;
whether flow is actual, standard, dry, wet or corrected;
whether the check proves capture, transport, removal, destruction, adsorption, energy use, interlock readiness or maintenance action;
the evidence needed before a process can continue operating.

The common mistake is treating outlet concentration as proof of control. A low stack number does not prove that fugitives are captured, that bypass dampers are closed, that pressure drop is healthy or that the same condition will persist through the next production campaign.

Release Evidence Notes

Capture evidence should be recorded separately from treatment evidence. A hood face velocity, smoke visualization result, enclosure pressure, duct traverse or make-up air balance proves whether emissions enter the collection system. A baghouse outlet concentration or scrubber pH does not prove capture if the source plume escapes before the duct.

Duct and fan evidence should preserve the operating point. Flow, pressure drop, fan speed, damper position, filter loading, gas temperature, gas density and system curve determine whether the equipment can keep the required capture flow. A running fan is not a release criterion unless the measured operating point still matches the required boundary.

Control-device evidence should match the physical mechanism. Fabric filters need cloth area, differential pressure, cleaning health and broken-bag detection. Scrubbers need gas-liquid contact, liquid flow, chemistry and demister condition. Carbon beds need loading history, humidity, breakthrough criteria and changeout evidence. Oxidizers need temperature, residence time, mixing, oxygen and bypass integrity.

Bypass and interlock evidence should be explicit. Maintenance doors, relief dampers, temporary ducts and startup bypasses can defeat otherwise sound controls. A release package should state which bypass paths exist, how they are detected, and what operating state is allowed when they are open.

Scenario Map

Scenario	Exercises	Primary calculation	Engineering decision
Capture and transport	1-4	Hood flow, make-up air, duct velocity, fan power	Decide whether emissions are collected and conveyed without losing capture.
Particulate collection	5-9	Pressure drop, air-to-cloth ratio, loading trend, cyclone residence, removal efficiency	Decide whether the particulate-control train is still inside its physical envelope.
Gas and vapor treatment	10-16	Residual emissions, carbon capacity, scrubber liquid rate, reagent demand, oxidizer residence, bypass leakage, heat recovery	Decide whether treatment mechanism and energy boundary are credible.
Release governance	17-18	Interlock proof, evidence closure and guarded residual risk	Decide whether operation can continue or must be held.

Exercise 1: Hood Capture Flow

A rectangular local exhaust hood has opening dimensions:

1.2\ \text{m}\times0.8\ \text{m}

The required face velocity is:

v_f=0.65\ \text{m/s}

Calculate the required hood flow.

Solution

Hood area:

A=(1.2)(0.8)=0.96\ \text{m}^2

Required flow:

Q=A v_f

Q=(0.96)(0.65)=0.624\ \text{m}^3/\text{s}

Convert to hourly flow:

Q_h=0.624(3600)=2246\ \text{m}^3/\text{h}

Engineering Comment

This is a capture requirement, not a treatment requirement. If the downstream baghouse performs well but the hood flow falls below this value, fugitive emissions can still escape into the workplace or building.

Plausibility Check

An opening close to $1\ \text{m}^2$ at a face velocity below $1\ \text{m/s}$ should require slightly less than $1\ \text{m}^3/\text{s}$ . The result is credible.

Exercise 2: Capture Flow Deficit

The same hood requires:

Q_{req}=0.624\ \text{m}^3/\text{s}

A traverse during production measures:

Q_{meas}=0.54\ \text{m}^3/\text{s}

Calculate the deficit and decide whether the capture basis is met.

Solution

Flow deficit:

\Delta Q=Q_{req}-Q_{meas}

\Delta Q=0.624-0.54=0.084\ \text{m}^3/\text{s}

Percentage deficit:

\displaystyle \text{deficit}=\frac{0.084}{0.624}(100)=13.5\%

The measured flow is:

\displaystyle \frac{0.54}{0.624}=0.865

or $86.5\%$ of the requirement. The capture basis is not met.

Engineering Comment

A $13.5\%$ shortfall is large enough to justify a hold, derate or retest depending on the source hazard and permit basis. The next diagnostic step is not only replacing filters; it is checking damper position, fan speed, duct deposits, make-up air, cross-drafts and hood obstruction.

Plausibility Check

Because measured flow is visibly below the required value, the deficit should be positive and the pass fraction should be below $100\%$ . Both are true.

Exercise 3: Duct Velocity and Transport Screen

A round duct has diameter:

D=0.45\ \text{m}

The operating flow is:

Q=0.82\ \text{m}^3/\text{s}

Calculate duct velocity and compare it with a minimum transport velocity of:

v_{min}=12\ \text{m/s}

Solution

Duct area:

\displaystyle A=\frac{\pi D^2}{4}

\displaystyle A=\frac{\pi(0.45)^2}{4}=0.159\ \text{m}^2

Velocity:

\displaystyle v=\frac{Q}{A}

\displaystyle v=\frac{0.82}{0.159}=5.16\ \text{m/s}

Transport margin:

M=v-v_{min}=5.16-12=-6.84\ \text{m/s}

The duct velocity fails the transport screen.

Engineering Comment

For particulate streams, low velocity can allow settling, duct plugging and later re-entrainment. A low velocity also invalidates any downstream assumption that the control device inlet represents the source mass flow.

Plausibility Check

A $0.45\ \text{m}$ duct has a relatively large area. Less than $1\ \text{m}^3/\text{s}$ through that area should produce only a few metres per second, so the failure is plausible.

Exercise 4: Fan Power from Pressure Drop

An exhaust system moves:

Q=1.6\ \text{m}^3/\text{s}

The total pressure drop is:

\Delta p=1850\ \text{Pa}

The fan plus motor efficiency is:

\eta=0.58

Calculate required electrical input power.

Solution

Air power:

P_{air}=Q\Delta p

P_{air}=(1.6)(1850)=2960\ \text{W}

Input power:

\displaystyle P_{in}=\frac{P_{air}}{\eta}

\displaystyle P_{in}=\frac{2960}{0.58}=5103\ \text{W}

Therefore:

P_{in}=5.10\ \text{kW}

Engineering Comment

The power estimate is a useful operating check. If measured power is much lower than expected, the system may not be moving the assumed flow. If it is much higher, filters, dampers or ducts may be restricted.

Plausibility Check

A few kilopascals at a few cubic metres per second normally gives air power of a few kilowatts. Dividing by efficiency raises the electrical value, so $5.1\ \text{kW}$ is reasonable.

Exercise 5: Filter Loading Pressure-Drop Gate

A baghouse has a clean differential pressure:

\Delta p_{clean}=850\ \text{Pa}

The maintenance trigger is:

\Delta p_{trigger}=1800\ \text{Pa}

The current differential pressure is:

\Delta p_{now}=2050\ \text{Pa}

Calculate the exceedance above the trigger.

Solution

Exceedance:

\Delta p_{excess}=2050-1800=250\ \text{Pa}

Percentage above trigger:

\displaystyle \frac{250}{1800}(100)=13.9\%

Increase relative to clean condition:

\displaystyle \frac{2050-850}{850}(100)=141\%

The baghouse exceeds the maintenance trigger.

Engineering Comment

High pressure drop can reduce capture flow even when outlet particulate appears acceptable. A release decision should pair this result with measured hood or duct flow, not just with stack concentration.

Plausibility Check

The pressure drop is slightly above the trigger but much higher than clean operation. That combination is typical of filter loading or cleaning failure.

Exercise 6: Baghouse Air-to-Cloth Ratio

A pulse-jet baghouse receives:

Q=3.4\ \text{m}^3/\text{s}

The installed cloth area is:

A_c=245\ \text{m}^2

Calculate air-to-cloth ratio in $\text{m/min}$ and compare with a limit of:

0.90\ \text{m/min}

Solution

Convert flow to cubic metres per minute:

Q=3.4(60)=204\ \text{m}^3/\text{min}

Air-to-cloth ratio:

\displaystyle G=\frac{Q}{A_c}

\displaystyle G=\frac{204}{245}=0.833\ \text{m/min}

Margin:

M=0.90-0.833=0.067\ \text{m/min}

The baghouse passes the simplified air-to-cloth screen.

Engineering Comment

Passing this screen does not prove bag integrity. It only shows that the gross gas loading is within the selected design criterion. Dust properties, moisture, cleaning air, bag age and leak detection still matter.

Plausibility Check

The cloth area is large compared with flow, so the ratio should be below $1\ \text{m/min}$ . The result matches that expectation.

Exercise 7: Pulse-Cleaning Trend Rate

Baghouse differential pressure rose from:

1100\ \text{Pa}

to:

1700\ \text{Pa}

over:

6\ \text{h}

Calculate the average rise rate and compare it with an alarm rate of:

80\ \text{Pa/h}

Solution

Pressure rise:

\Delta p=1700-1100=600\ \text{Pa}

Rise rate:

\displaystyle r=\frac{600}{6}=100\ \text{Pa/h}

Alarm margin:

M=100-80=20\ \text{Pa/h}

The trend exceeds the alarm rate.

Engineering Comment

A fast pressure-drop rise can signal wet or sticky dust, failed pulse cleaning, high dust loading, collapsed bags or a change in process condition. Waiting for an absolute high-high alarm may allow capture flow to deteriorate first.

Plausibility Check

A $600\ \text{Pa}$ rise over a quarter day is fast for many dust collectors. The calculated alarm exceedance is therefore credible.

Exercise 8: Cyclone Residence-Time Screen

A small cyclone has effective gas volume:

V_c=0.42\ \text{m}^3

Gas flow through the cyclone is:

Q=1.8\ \text{m}^3/\text{s}

Estimate nominal residence time and compare it with a screening minimum of:

t_{min}=0.20\ \text{s}

Solution

Residence time:

\displaystyle t=\frac{V_c}{Q}

\displaystyle t=\frac{0.42}{1.8}=0.233\ \text{s}

Margin:

M=0.233-0.20=0.033\ \text{s}

The screen passes with a small margin.

Engineering Comment

This is not a detailed cyclone design equation. It is a quick check that flow has not increased so much that gas residence and separation opportunity become obviously weak.

Plausibility Check

Sub-second gas residence is normal in compact particulate devices. A result near $0.23\ \text{s}$ is plausible.

Exercise 9: Control-Device Removal Efficiency

A particulate control device has inlet concentration:

C_{in}=780\ \text{mg/m}^3

and outlet concentration:

C_{out}=24\ \text{mg/m}^3

Calculate removal efficiency.

Solution

Removal efficiency:

\displaystyle \eta=\frac{C_{in}-C_{out}}{C_{in}}

\displaystyle \eta=\frac{780-24}{780}=0.969

Therefore:

\eta=96.9\%

Engineering Comment

The device removes most of the inlet particulate, but release still depends on the outlet limit, uncertainty, flow basis and whether inlet and outlet samples represent the same operating state.

Plausibility Check

The outlet concentration is much smaller than the inlet concentration, so efficiency should be high but below $100\%$ . The result meets that scale check.

Exercise 10: Residual Emissions After Control

A source emits:

\dot M_{in}=11.5\ \text{kg/h}

A control device removes:

\eta=96.9\%

Calculate residual outlet mass rate.

Solution

Residual fraction:

1-\eta=1-0.969=0.031

Outlet mass rate:

\dot M_{out}=11.5(0.031)=0.3565\ \text{kg/h}

Therefore:

\dot M_{out}=0.357\ \text{kg/h}

Engineering Comment

Small efficiency changes can matter when inlet mass rate is high. A device that appears to remove almost everything may still exceed an hourly mass limit if process throughput increases.

Plausibility Check

Three percent of about $12\ \text{kg/h}$ should be a few tenths of a kilogram per hour. The answer is consistent.

Exercise 11: Activated Carbon Bed Service Time

A carbon bed has usable adsorption capacity:

m_{cap}=18\ \text{kg}

The inlet VOC mass rate reaching the bed is:

\dot M=0.42\ \text{kg/h}

A safety factor of $0.70$ is applied to avoid breakthrough. Estimate allowed service time.

Solution

Protected usable capacity:

m_{use}=0.70(18)=12.6\ \text{kg}

Service time:

\displaystyle t=\frac{m_{use}}{\dot M}

\displaystyle t=\frac{12.6}{0.42}=30.0\ \text{h}

Engineering Comment

Carbon changeout should not rely only on theoretical capacity. Humidity, solvent mixture, temperature, concentration spikes and desorption during shutdown can shorten useful life.

Plausibility Check

At roughly half a kilogram per hour, a protected capacity a little above $12\ \text{kg}$ should last about one day plus part of another shift. Thirty hours is plausible.

Exercise 12: Scrubber Liquid-to-Gas Ratio

A packed scrubber treats gas flow:

Q_g=2.4\ \text{m}^3/\text{s}

The recirculating liquid flow is:

Q_l=0.018\ \text{m}^3/\text{s}

Calculate the liquid-to-gas ratio and compare with a minimum:

\displaystyle \left(\frac{L}{G}\right)_{min}=0.006

Solution

Liquid-to-gas ratio:

\displaystyle \frac{L}{G}=\frac{Q_l}{Q_g}

\displaystyle \frac{L}{G}=\frac{0.018}{2.4}=0.0075

Margin:

M=0.0075-0.006=0.0015

The liquid-flow screen passes.

Engineering Comment

The ratio is a hydraulic screen. It does not prove chemical absorption if pH, reagent strength, packing wetting, demister condition or gas distribution are poor.

Plausibility Check

Scrubber liquid flow is much smaller than gas volumetric flow, so a dimensionless ratio below $0.01$ is reasonable.

Exercise 13: Acid-Gas Reagent Feed

An acid-gas stream sends:

\dot n_{acid}=0.085\ \text{kmol/h}

to a caustic scrubber. The neutralization requirement is one mole of caustic per mole of acid gas. A $20\%$ excess is required. Calculate caustic feed in $\text{kmol/h}$ .

Solution

Stoichiometric feed:

\dot n_{stoich}=0.085\ \text{kmol/h}

With excess:

\dot n_{feed}=1.20(0.085)=0.102\ \text{kmol/h}

Engineering Comment

Reagent feed should be checked against pH control, liquid circulation, concentration, pump capacity and spent-liquid handling. A stoichiometric calculation alone cannot prove scrubber performance.

Plausibility Check

A $20\%$ excess should multiply the base need by $1.2$ . The calculated value is therefore slightly above the incoming acid rate, as expected.

Exercise 14: Thermal Oxidizer Residence Time

A thermal oxidizer has effective hot-zone volume:

V=5.8\ \text{m}^3

Hot gas flow through the zone is:

Q=9.1\ \text{m}^3/\text{s}

Calculate residence time and compare with a required value of:

t_{req}=0.60\ \text{s}

Solution

Residence time:

\displaystyle t=\frac{V}{Q}

\displaystyle t=\frac{5.8}{9.1}=0.637\ \text{s}

Margin:

M=0.637-0.60=0.037\ \text{s}

The residence-time screen passes with a narrow margin.

Engineering Comment

Residence time is only one oxidizer condition. Temperature, mixing, oxygen, catalyst health if applicable, LEL control, startup state and bypass position must also be proven.

Plausibility Check

An oxidizer with several cubic metres of volume and nearly $10\ \text{m}^3/\text{s}$ gas flow should have residence time well below one second. The value is credible.

Exercise 15: Oxidizer Destruction with Bypass Leakage

An oxidizer has ideal destruction efficiency:

\eta_d=98.5\%

A damper leakage path bypasses:

f_b=3.0\%

of the inlet mass. Calculate effective overall destruction efficiency.

Solution

The treated fraction is:

1-f_b=0.970

Destroyed fraction of total inlet mass:

\eta_{eff}=(1-f_b)\eta_d

\eta_{eff}=(0.970)(0.985)=0.955

Therefore:

\eta_{eff}=95.5\%

Engineering Comment

A small bypass can erase much of the benefit of a high-efficiency oxidizer. Bypass position and seal condition should be part of the release evidence, not an optional maintenance note.

Plausibility Check

The effective efficiency must be lower than $98.5\%$ because some gas is untreated. A value near $95.5\%$ is consistent with $3\%$ direct bypass.

Exercise 16: Heat-Recovery Fuel Saving

An oxidizer exhaust heat-recovery unit reduces burner demand by:

P_s=310\ \text{kW}

during:

10\ \text{h/day}

for:

250\ \text{days/year}

Calculate annual saved energy in $\text{MWh}$ .

Solution

Annual operating hours:

t=10(250)=2500\ \text{h/year}

Energy saved:

E=P_s t

E=310(2500)=775000\ \text{kWh}

Convert:

E=775\ \text{MWh/year}

Engineering Comment

Energy recovery is valuable, but it must not compromise destruction performance through condensation, corrosion, low chamber temperature or unstable draft. Energy and compliance boundaries need separate release checks.

Plausibility Check

Hundreds of kilowatts over thousands of hours should produce hundreds of megawatt-hours per year. The result has the correct order of magnitude.

Exercise 17: Bypass Damper Interlock Release Gate

A process may run only if three controls are proven:

Control	Required state	Verified?
capture fan above minimum speed	yes	yes
oxidizer above minimum temperature	yes	yes
bypass damper closed proof switch	yes	no

Assign one point for each verified required state and calculate the readiness fraction. The release rule requires all three states.

Solution

Verified states:

N_{verified}=2

Required states:

N_{required}=3

Readiness fraction:

\displaystyle R=\frac{2}{3}=0.667

or:

R=66.7\%

Because the rule requires all three states, release fails.

Engineering Comment

Interlock logic is not an average score when any single missing proof can create uncontrolled emissions. The missing bypass proof is a blocker even though the fan and oxidizer are both healthy.

Plausibility Check

Two of three checks produce exactly two thirds. The numerical score is less important than the failed all-required gate.

Exercise 18: Control Device Release Gate

A dust-control train has the following review evidence:

Item	Result
hood flow margin	$+6\%$
duct velocity margin	$+1.5\ \text{m/s}$
baghouse pressure drop margin to trigger	$-4\%$
outlet concentration margin to limit	$+18\%$
bypass proof	closed
maintenance record	overdue

Positive margin means the item passes. Decide release status.

Solution

The passing items are hood flow, duct velocity, outlet concentration and bypass proof. The pressure-drop margin is negative:

M_{\Delta p}=-4\%

The maintenance record is also not current. The release decision is therefore:

\text{status}=\text{hold or restricted operation}

The immediate action is to correct the pressure-drop and maintenance evidence before claiming normal operation.

Engineering Comment

Outlet concentration cannot override physical control evidence. A dust collector with acceptable stack concentration but rising pressure drop and overdue maintenance may be close to losing capture or breakthrough control.

Plausibility Check

The evidence is mixed: several checks pass, but two release-critical records fail. A hold or restricted decision is more defensible than a full release.

Engineering Boundary Notes

Capture flow, duct velocity, control-device efficiency and stack compliance answer different questions. Do not use a high removal efficiency to excuse poor capture, and do not use a good fan power reading as proof that the filter, scrubber, carbon bed or oxidizer is chemically effective.

Pressure-drop evidence is directional. High pressure drop can mean blinding or restriction; unexpectedly low pressure drop can mean bypass, broken bags, duct leakage or low process flow. The same number can be good or bad depending on the baseline and operating mode.

Energy recovery and emission control should be reviewed together but released separately. A heat exchanger, fan-speed reduction or pressure-drop optimization can save energy while weakening capture, changing condensation risk or pushing the control device outside the validated envelope.

Common Release Mistakes

treating a running fan as proof of source capture;
using outlet concentration without confirming hood flow;
ignoring make-up air, cross-drafts or enclosure leakage;
comparing wet, dry, actual and standard flows as if they were the same basis;
using clean-filter pressure drop after the system has loaded;
treating baghouse air-to-cloth ratio as proof of bag integrity;
overlooking a startup, maintenance or relief bypass path;
assuming scrubber liquid flow proves chemistry;
using carbon-bed theoretical capacity without humidity and breakthrough evidence;
releasing an oxidizer from temperature alone without residence time and bypass proof;
improving energy recovery without checking condensation or corrosion;
accepting a pass because most checklist items passed when one required interlock failed.

Validation Package Checklist

source and operating mode identified;
hood, enclosure or capture requirement stated;
measured flow and pressure basis recorded;
duct velocity or transport screen checked where particulate settling is possible;
fan operating point compared with required flow and pressure;
control-device mechanism identified;
inlet and outlet sampling basis documented;
pressure-drop trend reviewed against clean and trigger values;
bypass paths listed and proof state recorded;
maintenance and inspection records current;
monitoring evidence linked to the same operating mode;
release decision states whether operation is normal, restricted, held or retest-required.

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