Air Quality and Control practice questions

Answer: B) At a lower temperature, because the catalyst reduces the temperature needed for oxidation

A catalytic oxidizer uses a solid-phase catalyst to lower the temperature needed to oxidize VOCs. It runs cooler than a thermal oxidizer and uses less supplemental fuel.

• AIncorrect. The catalyst lowers the temperature needed for oxidation. Catalytic units run well below thermal oxidizer temperatures, which is why they use less supplemental fuel.
• BCorrect. The solid-phase catalyst lowers the activation energy, so VOCs are oxidized at much lower temperatures than in a thermal oxidizer. This saves fuel.
• CIncorrect. A catalyst changes reaction kinetics on its surface. It doesn't work by adding residence time, and its main advantage is a much lower operating temperature.
• DIncorrect. Mixing (turbulence) is a design factor in thermal oxidizers. A catalyst speeds up the reaction itself, which lets the unit run at a lower temperature.

Answer: B) Reaction with ammonia on a catalyst

• AIncorrect. That increases NOx.
• BCorrect. It's common on boilers and turbines.
• CIncorrect. That's for SO₂.
• DIncorrect. NOx is a gas.

Answer: A) Higher stack exit velocity and higher stack gas temperature

Plume rise comes from momentum (exit velocity) and buoyancy (stack gas hotter than ambient air). Higher wind speed bends the plume over and reduces rise.

• ACorrect. Higher exit velocity increases momentum flux, and hotter gas increases buoyancy flux. Both make the plume rise higher.
• BIncorrect. Stronger wind bends the plume over and dilutes it faster, and a lower exit velocity reduces momentum. Both reduce plume rise.
• CIncorrect. Both shrink the temperature difference between stack gas and ambient air, which reduces buoyancy and therefore plume rise.
• DIncorrect. Plume rise falls as wind speed rises, and a smaller gas flow carries less momentum and buoyancy flux. Both reduce rise.

Answer: C) 86%

Deutsch–Anderson: η = 1 − exp(−wA/Q). With wA/Q = 2, η = 1 − e^−2 ≈ 86%.

• AIncorrect. This is 1 − e^−1, which means wA/Q was taken as 1. Here wA/Q = (0.1 × 1,000)/50 = 2.
• BIncorrect. e^−2 ≈ 0.135 is the fraction of particles that gets through the ESP uncollected. The efficiency is 1 minus that value.
• CCorrect. η = 1 − exp(−wA/Q) = 1 − exp(−0.1 × 1,000/50) = 1 − e^−2 ≈ 0.865, or about 86%.
• DIncorrect. This is 1 − e^−4, which doubles the exponent. The correct exponent is wA/Q = 2.

Answer: C) Six

• AIncorrect. There are six.
• BIncorrect. There are six.
• CCorrect. CO, lead, NO₂, ozone, PM and SO₂.
• DIncorrect. That's roughly the hazardous air pollutant list.

Answer: D) 5,000 kg

• AIncorrect. That's off by a factor of 10.
• BIncorrect. Multiply.
• CIncorrect. That's off by a factor of 10.
• DCorrect. 0.5 × 10,000.

Answer: A) Lowering peak flame temperature and staging combustion

• ACorrect. Thermal NOx rises sharply with temperature.
• BIncorrect. That can raise NOx.
• CIncorrect. That reduces SO₂.
• DIncorrect. They reduce NOx formation.

Answer: B) How much light the plume blocks

• AIncorrect. That isn't opacity.
• BCorrect. It's a visual indicator of particulate emissions.
• CIncorrect. That isn't opacity.
• DIncorrect. That isn't opacity.

Answer: A) Thermal or catalytic oxidation

• ACorrect. VOCs are converted to CO₂ and water.
• BIncorrect. That removes large particles.
• CIncorrect. They remove particles.
• DIncorrect. That removes particles.

Answer: B) 2 m/min

The air-to-cloth ratio (filtering velocity) is gas flow divided by cloth area: 600 m³/min ÷ 300 m² = 2 m/min.

• AIncorrect. This is cloth area divided by gas flow (300/600). The ratio is defined as flow divided by area.
• BCorrect. Air-to-cloth ratio = volumetric gas flow ÷ cloth area = 600 m³/min ÷ 300 m² = 2 m/min, which is the filtering velocity.
• CIncorrect. 120 is the correct ratio of 2 m/min converted to m/h (2 × 60) but still labeled m/min. The units were mishandled.
• DIncorrect. This multiplies flow by area. The air-to-cloth ratio divides gas flow by cloth area.