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If carbon monoxide is released at the rate of 0.03m^3/min from a gasoline engine and 50 ppm is the threshold limit for an 8 hr exposure the quantity of air which dilutes the contaminant to a safe level will be A) 60m^3/min B) 600m^3/min C)60m^3/s D)600m^3/s?
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If carbon monoxide is released at the rate of 0.03m^3/min from a gasol...
Given:
Rate of carbon monoxide release from a gasoline engine = 0.03 m^3/min
Threshold limit for an 8-hour exposure = 50 ppm
To find:
The quantity of air that dilutes the contaminant to a safe level
Solution:
To find the quantity of air required to dilute the contaminant, we need to determine the concentration of carbon monoxide in the air.
Step 1: Convert m^3/min to m^3/s
Given rate of carbon monoxide release = 0.03 m^3/min
To convert minutes to seconds, we multiply by 60 (1 minute = 60 seconds).
0.03 m^3/min × (1 min/60 s) = 0.0005 m^3/s
Therefore, the rate of carbon monoxide release from the gasoline engine is 0.0005 m^3/s.
Step 2: Determine the concentration of carbon monoxide in the air
The threshold limit for an 8-hour exposure is 50 ppm.
To convert ppm to a ratio, we divide by 1,000,000 (1 ppm = 1/1,000,000).
Threshold limit in ratio = 50 ppm ÷ 1,000,000 = 0.00005
Therefore, the concentration of carbon monoxide in the air should be 0.00005.
Step 3: Calculate the quantity of air
To find the quantity of air required to dilute the contaminant, we can use the formula:
Quantity of air = Quantity of contaminant / Concentration of contaminant in air
Quantity of contaminant = 0.0005 m^3/s (given)
Concentration of contaminant in air = 0.00005 (calculated)
Quantity of air = 0.0005 m^3/s ÷ 0.00005 = 10 m^3/s
Therefore, the quantity of air required to dilute the carbon monoxide contaminant to a safe level is 10 m^3/s.
Conclusion:
The quantity of air required to dilute the carbon monoxide contaminant to a safe level is 10 m^3/s.
Rate of carbon monoxide release from a gasoline engine = 0.03 m^3/min
Threshold limit for an 8-hour exposure = 50 ppm
To find:
The quantity of air that dilutes the contaminant to a safe level
Solution:
To find the quantity of air required to dilute the contaminant, we need to determine the concentration of carbon monoxide in the air.
Step 1: Convert m^3/min to m^3/s
Given rate of carbon monoxide release = 0.03 m^3/min
To convert minutes to seconds, we multiply by 60 (1 minute = 60 seconds).
0.03 m^3/min × (1 min/60 s) = 0.0005 m^3/s
Therefore, the rate of carbon monoxide release from the gasoline engine is 0.0005 m^3/s.
Step 2: Determine the concentration of carbon monoxide in the air
The threshold limit for an 8-hour exposure is 50 ppm.
To convert ppm to a ratio, we divide by 1,000,000 (1 ppm = 1/1,000,000).
Threshold limit in ratio = 50 ppm ÷ 1,000,000 = 0.00005
Therefore, the concentration of carbon monoxide in the air should be 0.00005.
Step 3: Calculate the quantity of air
To find the quantity of air required to dilute the contaminant, we can use the formula:
Quantity of air = Quantity of contaminant / Concentration of contaminant in air
Quantity of contaminant = 0.0005 m^3/s (given)
Concentration of contaminant in air = 0.00005 (calculated)
Quantity of air = 0.0005 m^3/s ÷ 0.00005 = 10 m^3/s
Therefore, the quantity of air required to dilute the carbon monoxide contaminant to a safe level is 10 m^3/s.
Conclusion:
The quantity of air required to dilute the carbon monoxide contaminant to a safe level is 10 m^3/s.
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If carbon monoxide is released at the rate of 0.03m^3/min from a gasoline engine and 50 ppm is the threshold limit for an 8 hr exposure the quantity of air which dilutes the contaminant to a safe level will be A) 60m^3/min B) 600m^3/min C)60m^3/s D)600m^3/s?
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If carbon monoxide is released at the rate of 0.03m^3/min from a gasoline engine and 50 ppm is the threshold limit for an 8 hr exposure the quantity of air which dilutes the contaminant to a safe level will be A) 60m^3/min B) 600m^3/min C)60m^3/s D)600m^3/s? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about If carbon monoxide is released at the rate of 0.03m^3/min from a gasoline engine and 50 ppm is the threshold limit for an 8 hr exposure the quantity of air which dilutes the contaminant to a safe level will be A) 60m^3/min B) 600m^3/min C)60m^3/s D)600m^3/s? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If carbon monoxide is released at the rate of 0.03m^3/min from a gasoline engine and 50 ppm is the threshold limit for an 8 hr exposure the quantity of air which dilutes the contaminant to a safe level will be A) 60m^3/min B) 600m^3/min C)60m^3/s D)600m^3/s?.
If carbon monoxide is released at the rate of 0.03m^3/min from a gasoline engine and 50 ppm is the threshold limit for an 8 hr exposure the quantity of air which dilutes the contaminant to a safe level will be A) 60m^3/min B) 600m^3/min C)60m^3/s D)600m^3/s? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about If carbon monoxide is released at the rate of 0.03m^3/min from a gasoline engine and 50 ppm is the threshold limit for an 8 hr exposure the quantity of air which dilutes the contaminant to a safe level will be A) 60m^3/min B) 600m^3/min C)60m^3/s D)600m^3/s? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If carbon monoxide is released at the rate of 0.03m^3/min from a gasoline engine and 50 ppm is the threshold limit for an 8 hr exposure the quantity of air which dilutes the contaminant to a safe level will be A) 60m^3/min B) 600m^3/min C)60m^3/s D)600m^3/s?.
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