4.10 Problem Set
- A soil profile is in hydraulic equilibrium with a water table located at a depth of 275 cm below the surface. Draw a sketch representing that soil profile, label the soil surface as point “A” and the water table as point “B”. Create a table showing the gravitational head, pressure head, and total head at points A and B.
- An initially saturated soil sample is brought into hydraulic contact with a thin porous plate connected to a hanging water column (see Fig. 3‑12a) with a vertical length of 55 cm.
- When the soil reaches hydraulic equilibrium, what is the pressure head at the base of the soil sample?
- In one complete sentence explain why this pressure head occurs.
- A cylindrical soil column of 100. cm2 cross-sectional area and 50.-cm height is filled with homogeneous soil and saturated, and 10. cm of water is kept ponded on the surface. The soil column is open to the atmosphere and freely draining at the bottom. The steady-state volumetric flow rate through the soil column is 1000. cm3 h-1.
- Draw a sketch of this soil column.
- Create a table to determine the difference in hydraulic head across the column.
- Convert the volumetric flow rate to water flux (cm h-1).
- Calculate the saturated hydraulic conductivity of the soil.
- A 1.0-mm diameter tube is pushed through the column described in problem 3 and hollowed out. Steady state flow is established, with water flowing through the soil in accordance with Darcy’s Law and through the tube in accordance with Poiseuille’s Law. Assume the viscosity of water is 1.0 x 10-3 kg m-1 s-1.
- Calculate the volumetric flow rate through the tube (cm3 h-1).
- Calculate the flux for the combined column-tube system in cm h-1. You can assume that the tube takes up a negligible portion of the cross-sectional area of the column, so the flow through the soil matrix itself is unchanged.
- Calculate the effective saturated hydraulic conductivity for the combined column/tube system.
- Write one sentence explaining the practical significance of this example.
Hint for #4: Convert your water potential difference from head units (e.g. cm) to pressure units (e.g. Pascals [Pa] which is the same as kg m-1 s-2) for use in Poiseuille’s Law. To do that, first express the water potential difference in meters and use the fact that P = ρw g H where ρw is the density of water (1000 kg m-3) and g is acceleration due to gravity (9.81 m s-2).
- A saturated soil column contains two soil layers, each 10. cm thick, with sand having Ks = 10. cm h-1 underneath loam having Ks = 5.0 cm h-1, and 10. cm of water is kept ponded on the surface. The bottom of the soil column is open to the atmosphere and drains freely.
- Draw a sketch of this soil column.
- Create a table to determine the difference in hydraulic head across the column.
- Calculate the hydraulic resistance for each layer.
- Calculate the flux of water through the column using Darcy’s Law for layered soil.
- Calculate the pressure head at the sand-loam interface.