cee291_hw2

# cee291_hw2 - January 25, 2010 CEE 291 Problem Solving Using...

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January 25, 2010 CEE 291 – Problem Solving Using Computer Tools Homework 2 Solve these problems using Excel. Copy and Paste solutions and work into a Word document with problems and solutions clearly identified. Assignment is due on Monday (2/1) in class. 1. Environmental Engineering (DO – BOD in a stream)- The consumption of oxygen in the degradation of anthropogenic materials (e.g., treated municipal and industrial waste water) in rivers and streams in its simplest form is represented by the mass balance, differential equations below. These equations assume that river flow velocity and cross sectional shape are constant, concentrations of dissolved oxygen (expressed as O ) and of waste (expressed as L = BOD = biological oxygen demand) are uniformly distributed over the river cross-section, dissolved oxygen consumption by bottom deposits is negligible, and longitudinal transport by diffusion and dispersion is negligible. L k x L U t L dt dL r - = + = for BOD or L ( 29 L k O O k x O U t O dt dO d s a + - - = + = for O or L k D k x D U t D dt dD d a - = + = for O deficit with L,O = concentrations in mg/l; D = O s – O = dissolved oxygen deficit; O s = saturated O concentration; U = river velocity; x, t = distance, time; k r = k d + k s = BOD removal rate in time -1 , k d = BOD degradation rate in time -1 , k s = BOD settling rate in time -1 , k a = re-aeration rate in time -1 . For steady conditions ( 29 = 0 t with L 0 ,D 0 = conditions at x = 0 , and U, k r , k d , k s , k a = constant along the river, the solution to these equations for L, D are: U x k r e L L - = 0 and - - + = - - - U x k U x k r a d U x k a r a e e k k L k e D D 0 0 . The figure below shows a sketch of the dissolved oxygen deficit with distance along the river. Of concern for the health and maintenance of aquatic life is the magnitude and location of the minimum O or maximum D. These quantities can be obtained by setting the derivative of D (or O ) with x = 0. Thus, ( 29 - - - = d r a r a r a c k L k k D k k k k U x 0 0 1 ln = location of minimum allowable O or maximum allowable D ( 29 U x k a d k k k d r a r a a d c s c c r r a r e k k L k L k k D k k k k L O O D - - - = - - = - = 0 0 0 0 1 = critical

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## This note was uploaded on 05/04/2010 for the course CEE 291 taught by Professor Hoopes during the Spring '10 term at Wisconsin.

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cee291_hw2 - January 25, 2010 CEE 291 Problem Solving Using...

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