Compute the drinking water equivalent level (DWEL) for methylene chloride based on a 106 risk for a lifetime consumption of 2 L of water per day for a 70-kg individual.

Bioremediation question

[Pin It]

Please complete all following questions.

Questions 1 through 6 are related to Chapter 4 on Risk Assessment from the Masters and Ela text:

Compute the drinking water equivalent level (DWEL) for methylene chloride based on a 106 risk

for a lifetime consumption of 2 L of water per day for a 70-kg individual. Use the Potency Factor

(PF) for oral route listed on Table 4.9: PF = 7.5 × 103 (mg/kg-d)1.

How would the DWEL change if we considered the more appropriate weight of 80-kg for the

individual? Use the same values for the other assumptions.

Considering methylene chloride is a fairly volatile solvent, list the properties for the compound

from your text book.

Now considering the inhalation route for exposure to methylene chloride, and considering that the

average adult is again 70-kg and breathes 20 m3/d of air, determine the corresponding methylene

chloride concentration in air that produces a 106 lifetime cancer risk.

Assuming that your calculations were for 1 atm pressure and 20°C, what would you recommend

as the corresponding air quality standard in ppm?

A man works in an aluminum smelter for 10 years. The drinking water at the smelter contains

0.0700 mg/L arsenic and 0.560 mg/L methylene chloride. His only exposure to these chemicals is at work.

(a) What is the hazard index (HI) associated with this exposure?

(b) Does the HI indicate this is a safe level of exposure?

(c) What is the incremental lifetime risk for the man due solely to the water he drinks at

work and does is seem to be an acceptable risk according to EPA?

Questions 7 through 10 are related to BOD and the DO Sag Curve:

A BOD test is run using 100 mL of treated wastewater mixed with 200 mL of dilution water

(containing no BOD). The initial DO of the mix is 9.0 mg/L. After 5 days, the DO is 4.0 mg/L. After a long period of time, the DO is 2.0 mg/L, and it no longer seems to be dropping. Assume nitrification has been inhibited so that only carbonaceous BOD is measured.

(a) What is the 5-day BOD of the wastewater?

(b) Assuming no nitrification effects, estimate the ultimate carbonaceous BOD. (c) What would be the remaining BOD after five days have elapsed?

(d) Estimate the reaction rate constant k (day1).

A standard BOD test is run using seeded dilution water. In one bottle, the water sample is mixed with seeded dilution water giving a dilution of 1:30. Another bottle, the blank, contains just seeded dilution water. Both bottles begin the test with DO at the saturation value of 9.2 mg/L. After five days, the bottle containing waste has DO equal to 2.5 mg/L while that containing just seeded dilution water has DO equal to 8 mg/L. Find the five-day BOD of the waste (BOD5).

Re-do Example 5.8 on p. 215 with a temperature of 30°C.

Consider a lake with 108 m2 of surface area for which the only source of phosphorus is the

effluent from a wastewater treatment plant. The effluent flow rate is 0.4 m3/s, and its phosphorus concentration is 10 mg/L. The lake is also fed by a stream have 20 m3/s flow with no phosphorus. If the phosphorus settling rate is estimated to be 10 m/yr, estimate the average phosphorus concentration is the lake. What level of phosphorus removal at the treatment plant would be required to keep the average lake concentration below 0.01 mg/L phosphorus?

Are you looking for a similar paper or any other quality academic essay? Then look no further. Our research paper writing service is what you require. Our team of experienced writers is on standby to deliver to you an original paper as per your specified instructions with zero plagiarism guaranteed. This is the perfect way you can prepare your own unique academic paper and score the grades you deserve.

Use the order calculator below and get started! Contact our live support team for any assistance or inquiry.

[order_calculator]