The final topic in hydraulics and hydrologic systems, groundwater, concerns aquifer draw-, drawdown. And in this, we'll look at the formulas for computing drawdown in unconfined and confined aquifers. So firstly, an aquifer of course, contains groundwater or subsurface water and it can be unconfined or confined, so an unconfined aquifer, such as shown here, has a water surface which can be drawn down like this. It has a free surface, and the free surface will, generally speaking, be at atmospheric pressure. We, for a, a drawdown here, we have a cone of depression here where the depression here is given by D and the drawdown or depression at the well here is denoted by Dw, and that is called the drawdown. To predict the drawdown, we have Dupuit's Formula shown here in the extract from the reference handbook. And in this formula reproduced here, the variables are k is the coefficient of permeability, which is the same as the hydraulic conductivity, and unfortunately in the handbook they use the symbol lowercase k here for coefficient of permeability, although, in reality, it's exactly the same as the conductivity or the hydraulic conductivity which was previously denoted by K, uppercase K, in the previous segment. h two here is the head at some radius r two and h one is the head at the well which is of radius r one. And a couple of notes about this, that this equation, strictly speaking, only applies if the well drawn down, drawdown Dw, is much less than the undisturbed depth, capital D, and also if the well completely penetrates the aquifer, such as shown in this diagram here. And furthermore, this equation is an equilibrium equation. In other words, it only applies a very long time after pumping commences. Another variable are parameters called the specific capacity, which is defined as the volume flow rate divided by the well drawdown. And the units of that would be either, for example, meters squared per day or gallons per minute per foot would be typical units of specific capacity. So here's an example. We have an unconfined aquifer 200 feet deep. In other words, this depth here well away from the well is 200 feet. Pumping at a rate of 50 gallons per minute is begun from a well with a radius of 1.5 feet. Hydraulic conductivity is two feet per day, and after a very long time the drawdown is zero, essentially zero at a radial distance of 2,000 feet. So 2,000 feet from the well here, the free surface is at the undisturbed depth of 200 feet. The water depth of the well, which is h1 here, is then most nearly which of these alternatives? So, here again we have to be a little bit careful with the, the units. The flow rate is given in 50 gal in gallons per minute. And the conductivity is given in feet per day. So to convert the units to cubic feet per day, here is the conversion. It's 9,626 cubic feet per day. So using our degree formula, q is equal to pi k etc. But in this case we want to calculate the hydraulic conductivity so recasting, I mean, I'm sorry, we want to calculate the depth of the well. So recasting the equation in terms of h one, the depth of the well, we get this equation here. And now we're ready to substitute in the values, h two, the undisturbed depth is 200 feet. The flow rate is 9,626 cubic feet per day and the radii are as given so the answer is 170 feet. So, the closest answer is D. A confined aquifer is bounded on all sides and it may be under pressure. So, for example, the top of the aquifer here is bounded by an impermeable layer which is sometimes called the aquitard. And similarly, we have a number, another impermeable layer at the bottom. The equation which governs the drawdown in the flow rate here is the so-called Thiem equation, which is given by this expression right here. And in this expression, the quantity capital T is called the transmissivity and it's equal to k times D, the conductivity multiplied by the depth, or the height, of the confined aquifer. The transmissivity would have units of, for example, feet squared per second or meters squared per second. So, here is an example on that. We have a confined aquifer 200 feet thick, so this height here is 200 feet. In an experiment to determine its permeability, or conductivity, water is pumped from the well at a rate of 50 gallons per minute. The water table at a radius of 100 feet is observed to be at a depth of 40 feet and at a radius of 200 feet, it is 30 feet. The permeability is most nearly which of these alternatives? So in this case we are given the depths here. So the depth here below some reference point is 40 feet, but the depth out here is 30 feet at radii of 100 feet and 200 feet, respectively. So, from the previous question we know that Q is 9,626 cubic feet per day. So then we apply the theme equation, which is given here. But in this case we want to calculate the permeability, so we first need to compute t. So recasting the equation we have t and substituting in the numbers, we have this. So h two minus h one here is the difference between these two heights, but that's exactly equal to the difference in the do- water depths, which is just 40 minus 30. All we're interested here is the difference between those two heights. So, plugging that in, we get 106 feet squared per day, but we want to get the permeability or the conductivity, so we remember that T is equal to k times D, or k is equal to T, the transmissivity divided by the depth, which is therefore equal to 106 divided by 200, which is 0.53 feet per day. So the closest answer is B. And this finishes our discussion of knowledge area 11, hydraulics and hydrologic systems.
