Rewrite the DE for the concentration of pollutant as dc t dt f t V c t f t p t

Rewrite the de for the concentration of pollutant as

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Rewrite the DE for the concentration of pollutant as dc ( t ) dt + f ( t ) V c ( t ) = f ( t ) p ( t ) V with c (0) = 0 This DE has the integrating factor μ ( t ) = e R ( f ( t ) /V ) dt With the integrating factor, the DE becomes d dt ( μ ( t ) c ( t )) = μ ( t ) f ( t ) p ( t ) V Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (27/64)
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Introduction Falling Cat 1 st Order Linear DEs Examples Pollution in a Lake Example 2 Mercury in Fish Modeling Mercury in Fish Pollution in a Lake 7 Solution of the DE (cont): The DE is integrated to produce μ ( t ) c ( t ) = Z ( μ ( t ) f ( t ) p ( t ) /V ) dt + C With the initial condition, c (0) = 0, we have c ( t ) = μ - 1 ( t ) Z t 0 ( μ ( s ) f ( s ) p ( s ) /V ) ds or c ( t ) = e - R ( f ( t ) /V ) dt Z t 0 e R ( f ( s ) /V ) ds f ( s ) p ( s ) /V ds Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (28/64)
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Introduction Falling Cat 1 st Order Linear DEs Examples Pollution in a Lake Example 2 Mercury in Fish Modeling Mercury in Fish Basic Example: Pollution in a Lake 1 Basic Example: Pollution in a Lake Part 1 Suppose that you begin with a 100,000 m 3 clean lake Assume the river entering (and flowing out) has a constant flow, f = 100 m 3 /day Assume the concentration of some pesticide in the river is constant at p = 5 ppm (parts per million) Form the differential equation describing the concentration of pollutant in the lake at any time t and solve it Find out how long it takes for this lake to have a concentration of 2 ppm Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (29/64)
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Introduction Falling Cat 1 st Order Linear DEs Examples Pollution in a Lake Example 2 Mercury in Fish Modeling Mercury in Fish Basic Example: Pollution in a Lake 2 Solution: This example follows the model derived above with V = 10 5 , f = 100, and p = 5, so the differential equation for the concentration of pollutant is dc ( t ) dt = - f V ( c ( t ) - p ) = - 0 . 001( c ( t ) - 5) with c (0) = 0 This can be solved like we did by substitution for Newton’s Law of Cooling. Alternately, we use an integrating factor dc ( t ) dt + 0 . 001 c = 0 . 005 with μ ( t ) = e R 0 . 001 dt = e 0 . 001 t so d dt ( e 0 . 001 t c ( t ) ) = 0 . 005 e 0 . 001 t or e 0 . 001 t c ( t ) = 5 e 0 . 001 t + C Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (30/64)
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Introduction Falling Cat 1 st Order Linear DEs Examples Pollution in a Lake Example 2 Mercury in Fish Modeling Mercury in Fish Basic Example: Pollution in a Lake 3 Solution: From the integration before and multiplying by μ - 1 ( t ), c ( t ) = 5 + Ce - 0 . 001 t with c (0) = 0 Thus, the solution is c ( t ) = 5 - 5 e - 0 . 001 t . Solving c ( t ) = 2 = 5 - 5 e - 0 . 001 t gives e 0 . 001 t = 5 3 It follows that the concentration reaches 2 ppm when t = 1000 ln ( 5 3 ) 510 . 8 days. Joseph M. Mahaffy, h [email protected] i Lecture Notes – Linear Differential Equations — (31/64)
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Introduction Falling Cat 1 st Order Linear DEs Examples Pollution in a Lake Example 2 Mercury in Fish Modeling Mercury in Fish Example 2: Pollution in a Lake 1 Example 2: Pollution in a Lake Varying flow, f ( t ), and pollutant entering p ( t ) Again start with a constant volume, V = 100 , 000 m 3 clean lake Assume the river entering (and flowing out) has a seasonal flow, f ( t ) = 100 + 60 sin(0 . 0172 t ) m 3 /day If there is a point source pollutant dumped at t
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