Solution Exercise 25
Average linear velocity is given by Equation (93):
[latex]\overline{v_{x}} = \frac{K\frac{\partial h}{\partial x}}{\varphi_{eff}} = \frac{0.1\ \frac{cm}{s}\frac{10\ cm}{100\ cm}}{0.2} = \mathbf{0.05\ }\frac{\mathbf{cm}}{\mathbf{s}}[/latex]
Dispersion coefficient is given by Equation (96):
[latex]\mathbf{D}_{\mathbf{x}} = \alpha_{x}\overline{v_{x}} + D_{mol} = 5\ cm\ 0.05\ \frac{cm}{s} + {1x10}^{- 9}\ \frac{m^{2}}{s}\frac{{10,000\ cm}^{2}}{m^{2}} = \mathbf{0.25}\frac{\mathbf{cm}^{\mathbf{2}}}{\mathbf{s}}[/latex]
[latex]\mathbf{D}_{\mathbf{y}} = \alpha_{y}\overline{v_{x}} + D_{mol} = \frac{1}{5}5\ cm\ 0.05\ \frac{cm}{s} + {1x10}^{- 9}\ \frac{m^{2}}{s}\frac{{10,000\ cm}^{2}}{m^{2}} = \mathbf{0.05}\frac{\mathbf{cm}^{\mathbf{2}}}{\mathbf{s}}[/latex]
[latex]\mathbf{D}_{\mathbf{z}} = \alpha_{z}\overline{v_{x}} + D_{mol} = \frac{1}{10}5\ cm\ 0.05\ \frac{cm}{s} + {1x10}^{- 9}\ \frac{m^{2}}{s}\frac{{10,000\ cm}^{2}}{m^{2}} = \mathbf{0.025}\frac{\mathbf{cm}^{\mathbf{2}}}{\mathbf{s}}[/latex]
At 1000 seconds, the center of mass of the plume is at
[latex]x = \overline{v_{x}}t = 0.05\ \frac{cm}{s}\ 1000\ s = \mathbf{50\ cm}[/latex]
Concentration relative to the center of mass is given by Equation (99):
[latex]C = \frac{M}{8{(\pi t)}^{\frac{3}{2}}\ \sqrt{D_{x}D_{y}D_{z}}}\exp\left( - \frac{X^{2}}{4D_{x}t} - \frac{Y^{2}}{4D_{y}t} - \frac{Z^{2}}{4D_{z}t} \right)[/latex]
a) For C at x=25 cm, X is -25 cm because the point of interest is 25 cm behind the center of mass, while Y=0 and Z=0 because the point is on the centerline and not to the side nor below nor above:
[latex]C = \frac{500\ \frac{mg}{L}}{8(\pi\ 1000\ s)^{\frac{3}{2}}\ \sqrt{\left( 0.25\frac{{cm}^{2}}{s} \right)\left( 0.05\frac{{cm}^{2}}{s} \right)\left( 0.02\frac{{cm}^{2}}{s} \right)}}[/latex]
[latex]\exp\left( - \frac{{( - 25)}^{2}}{4\left( 0.25\frac{{cm}^{2}}{s} \right)1000\ s} - \frac{{(0)}^{2}}{4\left( 0.05\frac{{cm}^{2}}{s} \right)1000\ s} - \frac{{(0)}^{2}}{4\left( 0.02\frac{{cm}^{2}}{s} \right)1000\ s} \right) = \mathbf{6}\frac{\mathbf{mg}}{\mathbf{L}}[/latex]
b) Concentration at the center of mass (50 cm directly down gradient of the point source with no deviation up, down, or to the sides) after 1000 sec, X is 0 cm, Y is 0 cm, and Z= 0 cm.
[latex]C = \frac{500\ \frac{mg}{L}}{8(\pi\ 1000\ s)^{\frac{3}{2}}\ \sqrt{\left( 0.25\frac{{cm}^{2}}{s} \right)\left( 0.05\frac{{cm}^{2}}{s} \right)\left( 0.02\frac{{cm}^{2}}{s} \right)}}[/latex]
[latex]\exp\left( - \frac{{( - 0)}^{2}}{4\left( 0.25\frac{{cm}^{2}}{s} \right)1000\ s} - \frac{{(0)}^{2}}{4\left( 0.05\frac{{cm}^{2}}{s} \right)1000\ s} - \frac{{(0)}^{2}}{4\left( 0.02\frac{{cm}^{2}}{s} \right)1000\ s} \right) = \mathbf{11}\frac{\mathbf{mg}}{\mathbf{L}}[/latex]
c) Concentration at 10 cm to the left and 10 below the center of mass of the plume after 1000 sec, X is 0 cm, Y is 10 cm, and Z is -5 cm.
[latex]C = \frac{500\ \frac{mg}{L}}{8(\pi\ 1000\ s)^{\frac{3}{2}}\ \sqrt{\left( 0.25\frac{{cm}^{2}}{s} \right)\left( 0.05\frac{{cm}^{2}}{s} \right)\left( 0.02\frac{{cm}^{2}}{s} \right)}}[/latex]
[latex]\exp\left( - \frac{{( - 0)}^{2}}{4\left( 0.25\frac{{cm}^{2}}{s} \right)1000\ s} - \frac{{( - 10)}^{2}}{4\left( 0.05\frac{{cm}^{2}}{s} \right)1000\ s} - \frac{{( - 5)}^{2}}{4\left( 0.02\frac{{cm}^{2}}{s} \right)1000\ s} \right) = \mathbf{5}\frac{\mathbf{mg}}{\mathbf{L}}[/latex]
- Although the aquifer properties and gradient are the same for both cases, for part (a) there is a continuous input of contaminant at a concentration of 500 mg/L while for part (b) there is only one slug of 500 mg introduced to the groundwater. To compound that significant difference in mass, the mass in the system of part (b) spread not only in the direction of flow but also to the side and down.
- Field systems will deviate from the conditions described by analytical solutions. Among other reasons, deviation from values predicted by analytical solutions will occur because:
- the geologic material is heterogenous and so does no have constant hydraulic conductivity and dispersion coefficients,
- dispersivity cannot be measured in the field so it is an estimate based on the character of the material and often obtained by fitting data describing the distribution of concentration to an analytical equation,
- the rate and direction of groundwater flow in field settings varies with variations in recharge and pumping,
- source concentration varies with time,
- water samples collected in the field include water from a zone around the point of measurement and that zone is likely to have variations in concentrations,
- the sampling process requires great care so as not to alter the concentrations of a sample, and
- laboratory measurements are not precise because various aspects of the analysis process include error.