The last topic in hydraulics and hydrologic systems concerns groundwater flows. And here, first of all, we look at some definitions and basic concepts, and then look at Darcy's law. There are only a few topics referred to groundwater in the reference handbook. And that is Darcy's law. And drawdown in aquifers, which may be either unconfined or confined. An aquifer, of course, contains groundwater or subsurface water. And it can be unconfined or free, in which case the water surface is at atmospheric pressure. And the water surface can rise or fall. As contrasted to a confined aquifer, which is bounded on all sides and may be under pressure. So, the first major topic is Darcy's law, and here is the section from from the reference handbook. Darcy's law relates flow through a porous medium to the head and the property of the medium. And it can be most easily understood by this simple construction here, where I suppose that we have a flow of water through a pipeline here. And the pipeline has cross-sectional area A and length L. And I'll suppose that the hydraulic head, or the hydrostatic head, at the left-hand side is h and drops by an amount delta h in flowing through this pipe. So, Darcy's law for this particular sho, flow, is very simple. It says that Q is equal to KA, delta h, over L, where A is the area and K is the hydraulic conductivity of the porous medium. We can generalize this expression because delta h over L is the hydraulic gradient. So generalizing, we have Darcy's law, which is Q, is minus KA dh by dx. Where dh by dx is the slope of the hydrostatic head and the negative sign comes in because the flow is down the gradient from high head to low head. In this equation, K, which they denote by upper case K, is the hydraulic conductivity. But also, later on it's also the permeability or the coefficient of permeability, which is usually given the symbol lowercase k. So, also from this, we can define a variable q, lowercase q, which is the volume flow rate divided by the area, which is therefore equal to minus k, dh by dx. And it would seem like this is a velocity, because it's a volume flow rate divided by a cross-sectional area. However, the actual area is less than that. And the actual velocity is given by q over n, where n is the porosity or the effective porosity of the media. And we note that this equation only applies for very low velocities, in other words very laminar flow, where the Reynolds number is less than about 1. To give an example, we have a soil sample in horizontal tube 1 foot, and it has a cross-sectional area of 0.08 feet squared. The upstream head is 3 feet, 1 feet, and 1 foot downstream. The flow rate is 1 cubic foot per day. The permeability is most nearly which of these? So, we start from that basic Darcy's law again. But in this case, we want to calculate the permeabil, ability, or the conductivity, so I recast it in terms of K, as given here. So, that is equal to minus Q of A, dh by dx. The flow rate is 1 cubic foot per second. Divide by 86400, which is the number of seconds in a day to change to cubic feet per second. Area is 0.08, hydraulic gradient is 1 minus 3 over 1. So that gives me 7.23 times 10 to the minus 5 feet per second, so the answer is c. Another example, a saturated soil column to measure permeability is 2 feet in diameter and 2 feet tall. The head is 5 feet at the top and 1 foot at the bottom. After 24 hours, 5 gallons have drained through the column. The hydraulic conductivity of the soil is most nearly which of these. So, again, we'll have the same equations. But we have to be a little careful with the, the units, here. We're given five gallons in 24 hours. So, to convert that, I show the conversion here. It's 5 gallons times 1 cubic foot, divided by 7.48 gallons, divided by 24 hours. Multiplied by 36 seconds over 1 hour. And when converting units, I recommend going through it in in detail like this because you're less likely to make a mistake, from which you see that gallons and gallons cancels hours and hours cancels. Leaves me with 7.74 times 10 to the minus 6 units of cubic feet per second. Now, I'm ready to apply Darcy's law. And again recalculating or recasting that in terms of conductivity, we have this equation. And the cross-sectional area is pi d squared over 4. So, substituting in the numbers, we have 1.23 times 10 to the minus 6 feet per second. So, the closest answer is B, and this concludes our discussion of Darcy's law.