|
First, let's re-state the puzzler:
Polluted air is flowing into the filter at a rate of 20 cubic meters per minute. There is half a gram of uranium dust per cubic meter of air. 30% of the dust escapes the stack, since the filter has an efficiency rating of 70%. You need to calculate the daily uranium emissions from the stack in kilograms per day and determine if the plant is in compliance. Here's how to do it (though this is not the only way): 1. (20m3 of air) x (0.5g uranium) = 10 grams of uranium per minute (or 10gU/min) 2. (10gU/min) x (60 mins) = 600gU/hour 3. (600gU/hr.) x (24 hours) = 14,400gU/day 4. 30% escapes from stack, so 0.3 x 14,400 = 4,320 gU/day 5. Regulations for emissions are given in kilograms (kg) per day, so the 4,320 grams per day must be converted to kilograms per day. One kg = 1,000g, so by converting you get: 6. 4,320 / 1,000 = 4.32kgU/day 7. Regulations say emissions should be below 1.5kgU/day, so the plant is out of compliance, as Gamma suspected! Advanced Question: This question required that you calculate the radioactivity in the uranium that has escaped the stack to see if it violates the standard for air quality at the plant boundary. The standard is expressed in picocuries per liter, and your information is expressed in grams per cubic meter of air. Here's how Gamma figured it: 1. We know that 30% of the 0.5gU per cubic meter is escaping the stack, so:
0.15gU / m3 (0.15 grams of uranium per cubic meter of air is escaping the stack.)
0.000015gU / m3 (or: 1.5 x 10-5 grams of uranium per m3)
1.5 x 10-5gU / 103 liters (or: 1.5x10-8gU / liters)
(NOTE: For more information about working
1.005 x 10-8µCi / liter (For more information on specific activity, see Dr. Egghead in SDA Vol.4 No.1.)
1.005x10-2pCi / liter or: 0.01pCi / liter (rounded)
So Gamma was right again, the plant does violate the air quality standard!
|
Institute for Energy and Environmental Research
Comments to:
Outreach Coordinator: ieer@ieer.org
Takoma Park, Maryland, USA
March, 1997