Tutorial Solution Chapter 8 Week 7 (Q3 -4) PDF

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Tutorial 8 solution for Groundwater Contamination Tutorial 8.3 Continuous source in one dimension. A saline solution with a concentration of 1823 mg/L is introduced into a 2-m-long sand column in which the pores are initially filled with distilled water. If the solution drains through the column at an average linear velocity of 1.43 m/day and the dynamic dispersivity of the sand column is 15 cm, what would the concentration of the effluent be 0.7 day after flow begins? Assume the effective molecular diffusion coefficient is negligible. Solution:

Step 1: Calculate Dx from Eq. (assume

D  0)

 DL  aLV  D

D L  0.15 m  1.43 m d  0.22 m2 d V x  1.43 m d L 2m Co  1823 mg L t  0.7 d Step 2: Using Eq. to calculate C: C

 2 m  (1.43 m d  0.7 d)    1823 mg L    exp 1.43 m d  2m   erfc 2 m  (1.43 m d  0.7 d)   erfc 2  0.22 m d   2  0.22 m 2 d  0.7 d   2  0.22 m 2 d  0.7 d   2        

C  911.5 [erfc(1.29)  exp(13.33) erfc(3.87)] since erfc(3.87)  0, ignore second term.

C  911.5 mg L erfc(1.29) From Appendix Table or EXCEL,

erfc(1.29)  0.0683 C  911.5 mg L  0.0683 C  62.3 mg L.

Tutorial 8 solution for Groundwater Contamination Tutorial 8.4 The one-dimensional application of contaminant transport equation An underground tank leaches an organic (benzene) continuously into a onedimensional aquifer having a hydraulic conductivity of 2.15 m/day, an effective porosity of 0.1, and a hydraulic gradient of 0.04m/m. Assuming an initial concentration of 1000mg/L and longitudinal dispersity of 6.45m, find the time taken for the contaminant concentration to reach 100 mg/L at L = 750 m. Neglect any other degradation processes. Solution: (a) Seepage velocity is calculated from Darcy's law:

V 

2.15 m / day  0.04 m / m  0.86 m / day q K dh  ( )( )  n n dl 0 .1

 Calculate Dx from Eq. (assume D  0 )

D x  a xVx  D *  6.45 m  0.86m / d  5.55m 2 / d Using Eq. and ignoring the second term,

C ( x, t ) 

 L  vx t   C 0  erfc   2  D t   x   2 

 750  0.86t   100  500erfc   ,  2 5.55t      750  0.86t   100  0.2   erfc   500  2 5.55t    750  0.86t 0.9  2 5.55t By trial and error, t=739days...