F-L-I-P's forum posts

Avatar image for F-L-I-P
F-L-I-P

161

Forum Posts

0

Wiki Points

0

Followers

Reviews: 0

User Lists: 0

#1 F-L-I-P
Member since 2006 • 161 Posts

Washappnin?


Alright, Im jes barely pullin a C- off in Math and the teacher gave use an extra credit oppurtunity, but I don't understand wtf to do. I really need the E.C. 'cuz I dun wanna risk droppin to a D-F. Any help would be sooooo greatly appreciated. Here goes:

 

DARCY'S LAW:

Darcy's Law deals with flow rates of water in aquifers and other porous materials The equation is Q=KA(_)h/L where Q is the flow rate (volume over time), K is a term known as hyrdaulic conductivity (units of lenght over time such as mm/h. a product of the aquifer material) A is the cross of sectional area being considered (units of lenght squared), (_)h is the change in hydrologic head (for our purposes, this is the change in height) and L is the distance in length.

(I dont have no triangle button on my computer so (_) is a triangle.)

And the question is: Let's say you and your neighbor live above a confined aquifer with really good water in it. At the bottom of your aquifer is impermeable bedrock. Then there's a layer of sand and gravel that is 15 meters thick. On top of that is a clay layer 10 meters thick that keeps surface water from draining in or water from the aquifer from escaping. And finally, there is another 10 meters of silt on top. You and your neighbor both have wells that go down to the surface of the bedrock. Both of your wells are screened through the entire 15 meters of good aquifer. Your wells are 100 meters apart and the tops of your wells are at the same elevation. The water in your neighbor's well rises to 10 meters below the top of the well. In your well, the water surface is 15 meters down. For every square meter of cross-sectional area 0.12 cubic meters of water goes by per second.

- Identify the parameters and write them out with units

- Rearrange Darcy's Law to solve for K.

- Insert the known parameters and solve for K.

 

Then, using the same situation as before and the K you calculated, imagine you laid a meter stick on the ground perpendicular to the direction of the flow. How much water flows under the meter stick per second? (Hint: it is A that changes--now we have a cross sectional area 1 meter wide and the height of the aquifer layer)

 

Yeah, I dont understand wtf to do? Any help would be soo gladly appreciated.

 

Then it says to draw it out and label.