Solution to Risk Assessments

Solution to Risk Assessments - Given: Data in Problem 5-1...

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Solution to Chapter 5/6 Homeworks (numbering based on old Ed) 5-2 CDI for sulfur dioxide Given: NAAQS = 80 μg/m 3 , lifetime (24 h/d, 365 d/y), average adult. Solution: a. Using the assumptions in Table 5-8 (80 μg/m 3 )(20 m 3 /d)(365 d/y)(70 y) CDI = ------------------------------------------------ (70 kg)(365 d/y)(70 y) CDI = 22.86 μg/kg · d or 2.29 x 10 -2 mg/kg · d 5-3 Comparison of adult and child CDI for nitrate Given: Drinking water at 10 mg/L, one year averaging time, 1 year old child. Solution: a. Using Table 5-9 values for 1 year old child (10 mg/L)(1 L/d)(365 d/y)(1 y) CDI = ------------------------------------------------ (10 kg)(365 d/y)(1 y) CDI = 1.0 mg/kg · d b. Using Table 5-9 for an adult (10 mg/L)(2 L/d)(365 d/y)(1 y) CDI = ------------------------------------------------ (70 kg)(365 d/y)(1 y) CDI = 0.29 mg/kg · d
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5-7 Hexavalent chromium risk
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Unformatted text preview: Given: Data in Problem 5-1 Solution: a. The slope factor for Cr(VI) from Table 5-5 is 42 kg · d/mg. From Problem 5-1 CDI = 2.2 x 10-3 mg/ kg · d. The risk is then: Risk = (2.2 x 10-3 mg/ kg · d)(42 kg · d/mg) = 9.66 x 10-2 5-10 Characterize risk Given: tetrachloroethylene, arsenic, dichloromethane Solution: a. These are all carcinogens so calculate risk using Eqn 5-17 and 5-20 and slope factors from Table 5-5 Risk = (1.43 x 10-4 )(0.052) + (1.43 x 10-3 )(1.5) + (1.43 x 10-4 )(0.0075) Risk = 2.15 X 10-3 5-11 Cumulative probability Given: Table 5-2. Solution: a. Lifetime risk from driving motor vehicle (2.2 x 10-4 /y)(70 y) = 1.54 x 10-2 b. Lifetime risk from falling in home (7.7 x 10-5 /y)(70 y) = 5.39 x 10-3 c. Total Risk 1.54 x 10-2 + 5.39 x 10-3 = 2.08 x 10-2...
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This note was uploaded on 08/27/2009 for the course ENVE 331 taught by Professor Vigil during the Winter '08 term at Cal Poly.

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Solution to Risk Assessments - Given: Data in Problem 5-1...

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