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Module 5: Flowcharts and Ventilation Systems - Hoods - Practice Problems
Introduction
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Instructions:
- Work these problems on a sheet of paper and check your answers against those provided below.
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#1
- Explain why maintaining a high hood capture efficiency is important.
- Answer: Small declines in hood capture efficiencies can result in significant rises in fugitive emissions. Fugitive emissions escape directly into the plant air, bypassing the air pollution control equipment, and eventually pass through roof vents and doors into the atmosphere. Even when hood capture efficiencies approach 100%, fugitive emissions can be higher than emissions leaving the stack.
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#2
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Process equipment generates 80 pounds per hour of VOCs. The hood capture efficiency is 92% and the catalytic oxidizer has a removal efficiency of 95%.
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Calculate the fugitive emissions.
- Calculate the stack emissions.
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Calculate the fugitive emissions.
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Answer: i. 6.4 lbm/hr
- Solution:
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- Answer: ii. 3.7 lbm/hr
- Solution:
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- Solution:
Practice Problems
Operating Principles
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Instructions:
- Work these problems on a sheet of paper and check your answers against those provided below.
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#1
- Why is hood placement such an important factor in the capture of fugitive emissions?
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Answer: Hood placement is so important in the capture of fugitive emissions because the volumetric gas flow rate required to pull the pollutant into the ductwork varies with the square of the distance between the hood and the pollutant source (see equation below for a freely suspended hood without a flange).
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Where:
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#2
- What factors affect the determination of the recommended hood capture velocity for a contaminant?
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- Answer: The following factors should be considered when determining the recommended hood capture velocity for a contaminant:
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The surrounding air currents
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The level of toxicity of the pollutants
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The amount of pollutant
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The area of the hood opening
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#3
- Describe three hood designs that improve capture efficiency.
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Answer: Side baffles or flanges on hoods block the movement of clean air into the hood, which allows the hood capture velocity zones to extend outward slightly. Gas flow rates for hoods with side baffles or flanges can be slightly lower than for hoods without these structures. Baffles and flanges on hoods help eliminate cross drafts, which can prevent the intended movement of pollutant-laden gas into the hood. Hood capture is further improved when the enclosure comprised of the hood and side baffles encompasses the point of pollutant generation.
- Some process equipment such as coal-fired boilers, inherently create an enclosed area for pollutant capture. In this case, the boiler walls serve as the hood.
- The push-pull hood uses a clean, high-velocity air stream that flows toward the hood and traps the pollutant-laden gas in its cross draft. This type of hood is often used on open tanks but is not appropriate in situations where the cross draft could significantly increase the quantities of vaporized pollutants.
- Some process equipment such as coal-fired boilers, inherently create an enclosed area for pollutant capture. In this case, the boiler walls serve as the hood.
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#4
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Find the furthest distance that a flanged hood, with dimensions of 6 in. by 12 in., can be placed from the contaminant source and still maintain the capture velocity of 300 fpm and a volumetric flow rate of 2000 ACFM. The equation for a flanged hood is provided below.
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Where:
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Answer: 11 in.
- Solution:
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Solve for X using the following equation.
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#5
- The recommended capture velocity for a certain pollutant is 400 fpm entering a 14-inch diameter hood (without flange). What is the required volumetric flow rate if the farthest distance from the hood face to the released contaminant is 12 inches?
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Answer: 4,428 ACFM
- Solution:
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- Solution:
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Calculate the area of the hood opening.
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Calculate the volumetric flow rate, Q, required to attain the recommended capture velocity of 400 fpm at a distance of 12 inches from the hood.
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Practice Problems
Monitoring Performance
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Instructions:
- Work these problems on a sheet of paper and check your answers against those provided below.
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#1
- Calculate the velocity pressure of a duct with an entry loss coefficient of 0.93 and a static pressure of -0.8. in. W.C.
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Answer: 1.415 in W.C.
- Solution:
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- Solution:
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#2
- Using the data provided in Problem 1, calculate the gas flow rate for a duct that has a 10-inch diameter. Assume the gas is at standard conditions.
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Answer: 1,406 ft3/min
- Solution:
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Calculate the gas velocity from the velocity pressure. At standard conditions,
Actual = 0.075 lbm/ft3.
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Calculate the cross-sectional area of the duct.
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Calculate the gas flow rate, Q.
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#3
- What would happen to the hood static pressure and fugitive emissions if the ductwork developed a hole?
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- Answer: The hole would allow air from outside the duct to enter the system, which would in effect short-circuit the hood. The hood static pressure would drop (become less negative) and the hood would draw less contaminated air into the duct. The capture efficiency of the hood would decrease causing an increase in fugitive emissions.
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