Problem 6-6
For each of the solute breakthrough curves given below, describe whatprocesses are likely causing the specific shape and location of thecurves. Assume results are for laboratory columns and give brief, concise
descriptions. The calculation were performed using STANMOD (Simuneket al., 1999), this code may be downloaded free of charge at
http://www.pc-progress.com/en/Default.aspx?stanmod

AB

CDFig. 6-3: Solute breakthrough curves for various conditions

Answer:a. The breakthrough curves represented in the top left graph arecharacteristic of non-sorbing and non-degrading solutes, when the soluteExercises in Soil Physics: Chemical fate and transport

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is continuously applied to the surface of a soil column. The observed
concentrations in these breakthrough curves would come from theeffluent exiting the bottom of the soil columns. The variation in the curves
is due to dispersive processes. When there is no dispersion (D = 0), allsolutes move at the same velocity, and the front arrives as a squarewave (see arrow), termed “piston flow”. The dispersion of a solute in aflow path can result from velocity variations within a pore, pore size differences, or path length differences. Breakthrough curves characterized byincreased dispersion will be spread out along the x-axis, resulting fromincreased variation in solute arrival times (or pore volumes) around theidealized piston flow arrival value. Nevertheless, the inflection point of all
curves, representing 50% displacement of the non-sorbing and nondegrading solute, should occur at about one pore volume.
b. The breakthrough curves represented in the top right graph arecharacteristic of sorbing solutes, when the solute is continuously applied
to the surface of a soil column. The observed concentrations in thesecolumn breakthrough curves would come from the effluent exiting thebottom of the soil columns. Retardation factors (R) in the soil may resultin breakthrough curves similar to those displayed in the top right graph.
R = 1+

b K d, where Kd is the distribution coefficient, b is bulk density,and  is volumetric water content. If there is no adsorption, then Kd = 0
and R = 1, which would represent the curve furthest to the left. As
adsorption increases (i.e. Kd > 0) the breakthrough time increases(moves right) and the curve spreads more across the x-axis.
c. The breakthrough curves represented in the bottom left graph arecharacteristic of degradable or volatile solutes such as ammonia (NH3).
The solute for these breakthrough curves was applied as a pulse ofsolution where water was applied after the pulse application wasdiscontinued. The observed concentrations come from the effluent exiting
the bottom of the soil columns. The greater the rate of degradation, thesmaller the peak of the curves.
d. The solute for these breakthrough curves was applied as a pulse ofsolution where water was applied after the pulse application was
discontinued. The observed concentrations come from the effluent exiting
the bottom of the soil columns. The more symmetrical shape of thebreakthrough curve results from a more uniform pore size distribution
and the concomitant removal of large pores. In the second case, early
breakthrough of the solute is made possible by solute moving at higher
convective velocity through more mobile areas (e.g. root holes, cracks).

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