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<p>Chapter 07.00C Physical Problem for Integration Civil Engineering</p><p>Problem Statement In a class X528 well in New York State, you are asked to find if the concentration of benzene is below the toxicity level of 0.5mg/l at a distance of 36m from the point of contamination. The contamination at the source is constant at3.5mg/l for a whole year. Solution The equations governing transport of groundwater contaminant are quite complex. They depend on several parameters such as “the molecular diffusion of the contaminant, physical and chemical isotropy of the medium, and actual direction of the groundwater flow” [1]. Here we consider that the groundwater is flowing in one-direction, x and the aquifer has isotropic and homogeneous physical and chemical characteristics. In that case, the governing equation [2] for the concentration cx,t of the contaminant is given by c c 2c u D . (1) t x x2 where u = velocity of ground water flow in the x -direction (m/s) t = time (s) x = distance from source (m) D = dispersion coefficient m 2 . To solve this partial differential equation (1) [3], we use Laplace transforms. Applying Laplace transform to both sides on the variable t , c c 2c L u L D 2 t x x dC d 2C sC cx,0 u D . (2) dx dx 2 where L[c(x,t)] Cx, s</p><p>07.00C.1 07.00C.2 Chapter 07.00C</p><p>Since the initial concentration is zero, cx,0 0 , Equation (2) becomes dc d 2C sC u D dx dx 2 d 2C dC D u sC 0 (3) dx 2 dx This is a homogenous ordinary differential equation. It is linear with fixed coefficients. The characteristics equation of the above ordinary differential equation (3) is Dm 2 um s 0 u u 2 4Ds m (4) 2D The homogeneous as well as the complete solution of Equation (3) is u u 2 4Ds u u 2 4Ds x x Cx, s Ase 2D Bse 2D (5) At x , that is far away from the source, the concentration of the pollutant is zero, that is c, s 0 , that is, C, s 0 . This forces As 0 as the exponent of the exponential term is always positive as u, D, s 0 . The exponent of the exponential term on the second term will be negative as u 2 uDs u for all values of u, D, s as u, D, s 0 . Equation (5) hence reduces to u u 2 4Ds x Cx, s Bse 2D (6) c At x 0 , c c (initial concentration, c ), then C0, s 0 . Substituting this in equation 0 0 s (6) gives c o Bs s then equation (6) is u u 2 4Ds c x Cx, s 0 e 2D s u 2 4Ds x ux e 2D c e 2D 0 s </p><p> u 2 4Ds x ux e 2D 1 1 2D L [C(x, s)] L c0e s Physical Problem for Integration: Civil Engineering 07.00C.3</p><p> u2 4Ds x ux e 2D cx,t c e 2D L1 0 s x u 2 s ux D 4D 2D 1 e C0e L (7) S Let x a D u b 2 D Then Equation (7) becomes 2 ea b s cx,t c eab L1 (8) 0 s Using the formulas a2 a L1 (e a s ) e 4t ,a 0 , and 3 2 t L1 Fs a eat f t then 2 a 2 2 a b t L1 e a b s e 4t (9) 2 t 3 Then from t Fs f d 0 s we get 2 a b2 s t a e a 2 L1 e 4 e b d s 3 0 2 t a2b 2 a e ab e 4 d 3 0 2 t a2b 2 a 2b a 2b e ab e 4 d 3 3 0 4 4 t a2b 2 t a2b 2 a 2b a 2b e ab e 4 d e ab e 4 3 3 0 4 0 4 07.00C.4 Chapter 07.00C</p><p> t a2b 2 t a2b 2 8ab 8ab a 2b a 2b e ab e 4 d e ab e 4 4 4 3 3 0 4 0 4 t a2b 2 t a2b 2 a 2b a 2b e ab e 4 d e ab e 4 d 3 3 0 4 0 4 Let a 2b p , 4 a 2b q 4 then a2bt a2bt a b2 s ab 4t ab 4t e e 2 e 2 L1 e p dp eq dq s e ab a 2bt e ab a 2bt erfc erfc (10) 2 2 t 2 2 t Using Equation (10), Equation (8) becomes ab ab ab e a 2bt e a 2bt cx,t c0e erfc erfc 2 2 t 2 2 t 2ab c0 a 2bt e a 2bt erfc erfc (11) 2 2 t 2 2 t Substituting back x a , D u b 2 D ux c0 x ut D x ut cx,t erfc e erfc (12) 2 2 Dt 2 Dt The velocity of the ground flow, u is given by k dh u (13) n ed dℓ where</p><p> ned = effective Darcian porosity, K = permeability, dh ground water gradient. dl Assuming</p><p> ned = 22% = 0.22 Physical Problem for Integration: Civil Engineering 07.00C.5</p><p> cm K = 0.002 s dh cm 0.01 . dl cm Then from Equation (13), 0.002 u (0.01) 0.22 cm 9.091105 . s So in the formula for concentration given by Equation (12), substituting cm 2 D 0.01 s mg c 3.5 0 L x 36m t 1 year 3.15107 s the concentration of benzene after 1 year at a distance of 36m is</p><p>5 5 7 9.091x10 36 5 7 3.5 3.6 9.09110 3.1510 36 9.09110 3.1510 c erfc e 0.01 erfc 2 7 7 2 0.013.1510 2 0.013.1510 1.75[erfc0.6560 e32.73erfc5.758] (14) As you can see to calculate the concentration of benzene at x 36m , we need to calculate erfc0.6560 and erfc5.758. How is erfcx defined? x 2 erfcx e z dz So 0.6560 2 erfc0.6560 ez dz 2 Since e z decays rapidly as z , we will approximate 0.6560 2 erfc0.6560 ez dz 5 So to find whether the concentration of benzene is below toxicity level, we need to find the integral numerically.</p><p>QUESTIONS 1. What is the concentration of benzene after 2 years? 2. After 1 year, at what distance does the concentration go below the toxicity level of 0.5 mg/L? 07.00C.6 Chapter 07.00C</p><p>REFERENCES 1. EPA Region 5 Water Draft UIC Class IV and V Site Assessment Guidelines, http://www.epa.gov/R5water/uic/r5guid/siteasst.htm, last accessed July 2004. 2. Fetter, C.W., Jr., Applied Hydrology, Merill Publishing Co, Columbus, 1980. 3. Li, W., Differential Equations of Hydraulic Transients, Dispersion, and Groundwater Flow, Prentice Hall, Englewood Cliffs, NJ, 1972</p><p>Topic INTEGRATION Sub Topic Physical problem Summary Find if the concentration of benzene is above or below the toxicity limit at a critical distance from its source. Authors Autar Kaw Last Revised 2018 年 4 月 9 日 Web Site http://numericalmethods.eng.usf.edu</p>
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