Solution Exercise 24
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]D_{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]
Concentration is given by Equation (98):
[latex]C = \frac{C_{o}}{2}\left\lbrack erfc\left( \frac{x - \overline{v_{x}}t}{2\sqrt{D_{x}t}} \right) + \exp\left( \frac{x\overline{v_{x}}}{D_{x}} \right)erfc\left( \frac{x + \overline{v_{x}}t}{2\sqrt{D_{x}t}} \right) \right\rbrack[/latex]
a) For 25 cm the concentration is calculated as follows:
[latex]C = \frac{1000\frac{mg}{L}}{2}\left\lbrack erfc\left( \frac{25\ cm - 0.05\ \frac{cm}{s}1000\ s}{2\sqrt{0.25\frac{{cm}^{2}}{s}1000\ s}} \right) + exp\left( \frac{25\ cm\ 0.05\ \frac{cm}{s}}{0.25\frac{{cm}^{2}}{s}} \right)erfc\left( \frac{25\ cm + 0.05\ \frac{cm}{s}\ 1000\ s}{2\sqrt{0.25\frac{{cm}^{2}}{s}1000\ s}} \right) \right\rbrack[/latex]
[latex]C = 500\frac{mg}{L}\left\lbrack erfc( - 0.79057) + \exp(5)erfc(2.3717) \right\rbrack[/latex]
[latex]erfc( - 0.79057) = 1 - \left( - {erf}(0.79057) \right) = 1 + 0.7364[/latex] = 1.7364
[latex]\exp(5) = 148.4132[/latex]
[latex]erfc(2.3717) = 1 - \left( {erf}(2.3717) \right) = 1 - 0.999204[/latex] = 0.000796
[latex]C = 500\ \frac{mg}{L}\left\lbrack (1.7364) + 148.4132(0.000796) \right\rbrack[/latex]
[latex]C = 500\ \frac{mg}{L}\lbrack 1.854537\rbrack = \mathbf{927\ }\frac{\mathbf{mg}}{\mathbf{L}}[/latex]
Following similar steps:
b) at 50 cm, C=500 mg/L
c) at 75 cm, C=132 mg/L
- These values make sense. We could have deduced the concentration at 50 cm because it takes 1000 seconds to reach 50 cm so the concentration would be half of the source concentration. The concentration at 25 cm should be higher than half the source concentration because it is closer than the location the contaminant would travel in 1000 seconds, while the concentration at 75 cm should be less because it is further than the average travel distance at 10000 seconds.
- 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.