CEE 350 - Korshin - Winter 2012 - Homework 6(1)

CEE 350 - Korshin - Winter 2012 - Homework 6(1) - 1 CEE 350...

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Unformatted text preview: 1 CEE 350 Week 6 Homework Solutions Problem 1 (textbook 5.35) An often used chemical representation of algae is C 106 H 263 O 110 N 16 P. (a) Determine the mass (mg) of each element in 1 g of algae. First calculate the molar weight of algae molar wt = (106 × 12)+(263 × 1)+(110 × 16)+(16 × 14)+31 = 3550 g/mol Mass (mg) of each element: C = (106 × 12)/3550 = 358 mg H = 263/3550 = 74 mg O = (16 × 110)/3550 = 496 mg N = (16 × 14)/3550 = 63 mg P = 31/3550 = 8.7 mg (b) Suppose there are 0.10 mg of N and 0.04 mg of P available for algal production per liter of water. Assuming adequate amount of other nutrients, which is the limiting nutrient? N allows: 0.1 mg N/L × (1000 mg algae/ 63 mg N) = 1.6 mg/L algae P allows: 0.04 mg P/L × (1000 mg algae/8.7 mg P) = 4.6 mg/L algae Therefore N is the limiting nutrient (c) What mass of algae could be produced (milligram algae per liter of water)? The limiting nutrient, N, provides for 1.6 mg/L of algal growth (d) If the nitrogen source could be cut by 50%, how much algae could be produced? Cutting N by 50% would half the amount of algal production to 0.8 mg/L (e) If the phosphorus sources could be cut by 50%, how much algae could be produced? Cutting P sources by 50% would allow for 4.6/2 = 2.3 mg/L, however N would still be the limiting nutrient in this case so only 1.6 mg algae/L can be produced under this scenario Problem 2 (textbook 5.36) Suppose the N and P content of some algae is shown in the table below, The third column shows milligrams of nutrient available per liter of water Nutrient Milligram per gram of algae Milligram available per liter Nitrogen 60 0.12 2 Phosphorus 10 0.03 (a) What percent reduction in nitrogen is needed to control algal production to 1.0 mg/L? (0.12 mg N/L) × (1000 mg algae / 60 mg N) = 2.0 mg algae/L (0.03 mg P/L) × (1000 mg algae / 10 mg P) = 3.0 mg algae/L Therefore, N is the limiting nutrient To control algal production down to 1.0 mg/L, available N must be cut by 50% down to 0.06 mg N/L (b) What percent reduction in phosphorus is needed to control algal production to 1.0 mg/?...
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This note was uploaded on 03/31/2012 for the course CEE 350 taught by Professor Korshin during the Spring '10 term at University of Washington.

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CEE 350 - Korshin - Winter 2012 - Homework 6(1) - 1 CEE 350...

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