The EPA requires hazardous waste incinerators to meet a standard of 99.99% destruction and removal of hazardous constituents injected into the incinerator. Determine the allowable quantity of contaminant in the exit stream if the incinerator is burning 1.0000 g/s of hazardous constituent.
If the incinerator is 90% efficient in destroying the hazardous constituent, what scrubber (which injects water into the process) efficiency is required to meet the standard?
Any help is appreciated, thank you!
To determine the allowable quantity of contaminant in the exit stream of the incinerator, we first need to calculate the amount of hazardous constituent that is destroyed during the incineration process.
Given that the incinerator is burning 1.0000 g/s of hazardous constituent and is 90% efficient in destroying it, we can calculate the amount of hazardous constituent that is destroyed per second:
Amount destroyed per second = 1.0000 g/s * 0.9 = 0.9000 g/s
Now, we need to determine the maximum allowable quantity of contaminant in the exit stream to meet the EPA standard of 99.99% destruction and removal.
To achieve 99.99% destruction and removal, the remaining quantity of contaminant in the exit stream should be:
0.0001% of the hazardous constituent burnt
Therefore, the maximum allowable quantity of contaminant in the exit stream is:
Allowable quantity of contaminant = 0.0001/100 * 0.9000 g/s = 0.00000009 g/s
Now, let's move on to determining the scrubber efficiency required to meet this standard. The scrubber injects water into the process to remove contaminants.
To calculate the required scrubber efficiency, we need to consider the mass balance equation:
Mass of contaminant in the exit stream = Mass of contaminant in the input stream - Mass of contaminant removed by the scrubber
We know that the mass of contaminant in the exit stream should be 0.00000009 g/s. Assuming that the scrubber removes a fraction 'x' of the contaminant, the mass balance equation becomes:
0.00000009 g/s = 1.0000 g/s - x * 1.0000 g/s
Simplifying the equation:
x * 1.0000 g/s = 1.0000 g/s - 0.00000009 g/s
x = (1.0000 g/s - 0.00000009 g/s) / 1.0000 g/s
Calculating x:
x ≈ 0.99999991
Therefore, the scrubber efficiency required to meet the standard is approximately 99.999991%.
Please note that these calculations are based on the given information and assumptions. Different conditions or requirements may lead to different results.