Question

# Darcy's law gives us the volume flow rate-volume per time (for example, cubic meters per day,...

Darcy's law gives us the volume flow rate-volume per time (for example, cubic meters per day, m^3/day). Another way to express volume is on a per-unit-area basis. For example, if we have a 1-m^3 cube, we know its base is 1 m^2 the cross-sectional area of the cube-and its height is 1m. If we fill the cube with water, we can say we have 1m of water per unit area. Darcy's law can be rearranged to calculate the volume flow rate per unit area. This is called the specific discharge, which has units of length per time (for example, meters per day, m/day). Completing parts A through C will illustrate how specific discharge relates to the volume flow rate. (Hint: Assume that the units for hydraulic conductivity are m/day and the units for cross-sectional area are m^3.)

Part A:  Suppose 1 m^3 of water is pumped from a well into an empty cylindrical tank. If the water level is 3.0 m above the base, what is the cross-sectional area of the tank? Hint: Volume of a cylinder = πr^2h = area of base × height.

A=

Part B: If it takes half a day to pump the 1.0 m^3 of water into the tank, what is the flow rate in term of volume per time?

Part C: What is the flow rate in term of specific discharge?

Part D: Write Darcy's law so that it calculates the specific discharge.

A:

The volume puped out = volume occupied by water in the tank.

i.e. 1m3 = volume of tank = Area of base x height

1 m3 = (cross section area) x (3m)

Thus, cross-sectional area of the cylindrical tank is (1/3) m2 = 0.33 m2.

B:

Volumetric flow rate = 1m3 / 0.5 day = 2 m3 / day.

C:

Specific discharge = flow rate per unit area of the tank

i.e. Specific discharge = (volumetric flow rate)/(cross-sectional area of the tank)

= (2 m3 / day) / (1/3 m2)

Therefore, Specific discharge = 6 m/day.

D:

Darcy's Law is given by (in terms of volumetric flow):

where;

Q = rate of water flow, K = hydraulic conductivity, A = tank cross section area, dh/dl = hydraulic gradient.

Converting it in terms of specific discharge (Q/A):

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