If you any good at Math, could you help me please?

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.

You seem to have a typo in this sentence. Could you fix it to make the question more clear?

"The water in your neighbor's well rises to 10 meters below the top of cross-sectional area 0.12 cubic meters of water goes by per second."

You seem to have a typo in this sentence. Could you fix it to make the question more clear?

 

"The water in your neighbor's well rises to 10 meters below the top of
cross-sectional area 0.12 cubic meters of water goes by per second."

DiaperMan2

 There you go homie. Yeah, I didnt realize I missed an entire sentence. Sorry. :D

So you want to solve for k.

k = QL/A(delta)h

delta = triangle

Q = 0.12 cubic metres/s, we're given that

L = 15 m I believe. That's where the water can seep into the well.

A = pi*r^2 assuming the well is circular, which is a total guess. What the radius is, there's no clue given. Also I'm not sure if you need to multiply this by 2 since there are 2 wells.

(delta)h = k this is a toughy since your explanation was vague. It's the change in height of what? Is it the difference in height between your neighbor's well and yours? Or is it the height between the aquifer and the surface, or what?

hope this kinda helps.

On another note, this isn't exactly Darcy's law. You should be dealing with pressure gradients, not just heights, but whatever.

Blah, well that is tough, but here I'll try to help as much as I can and maybe guess.

So you want to solve for k.

k = QL/A(delta)h

delta = triangle

Q = 0.12 cubic metres/s, we're given that

L = 15 m I believe. That's where the water can seep into the well.

A = pi*r^2 assuming the well is circular, which is a total guess. What the radius is, there's no clue given. Also I'm not sure if you need to multiply this by 2 since there are 2 wells.

(delta)h = k this is a toughy since your explanation was vague. It's the change in height of what? Is it the difference in height between your neighbor's well and yours? Or is it the height between the aquifer and the surface, or what?

 

hope this kinda helps.

On another note, this isn't exactly Darcy's law. You should be dealing with pressure gradients, not just heights, but whatever.

DiaperMan2

I really have no idea holmes. I copied that word for word from the EC paper the teacher gave me so I dunno. 

Okay, I understand the question a bit better now. Q is 0.12 for every 1 m^2 cross-sectional area. Therefore Q = 0.12 A.

 The equation therefore becomes:

0.12*A = kA(delta)h/L

The A's cancel out, so the radius is irrelevant. That's why we weren't given it.

0.12 = k(delta)h/L

k = 0.12*L/(delta)h 

L I still think is 15m...just that pesky delta h. I'm not sure what it's referring to. I also can't figure out the significance of knowing your neighbor's well is 100m away. 

Also this pic may help you understand

Also this pic may help you understand

DiaperMan2

Okay, I understand the question a bit better now. Q is 0.12 for every 1 m^2 cross-sectional area. Therefore Q = 0.12 A.

The equation therefore becomes:

0.12*A = kA(delta)h/L

The A's cancel out, so the radius is irrelevant. That's why we weren't given it.

0.12 = k(delta)h/L

k = 0.12*L/(delta)h

L I still think is 15m...just that pesky delta h. I'm not sure what it's referring to. I also can't figure out the significance of knowing your neighbor's well is 100m away.

DiaperMan2

Dang homie, I thank you sooooooooooooo much for takin time to help me with that. It makes a hella lot more sense than it did Friday when I first saw it. Haha, thanks alot cuz. 

Ya know, although I don't know what delta h is, I'd probably guess it was 5 m if I was forced to choose. Reason I chose 5 is because that's the difference in water heights between you and your neighbor, and it is also the difference between the aquifer water height and the height of the water in your well. If you can ask your teacher for clarification, then that'd probably be a lot better, but if you can't figure it out, use 5m. for delta h. Then you can solve the question.

As for the second part, I can't really figure it out cause the hint doesn't make proper sense. Could there be another typo?

 "(Hint: it is A that changes--now we have a cross sectional area 1 meter wide and the height of the aquifer layer)"

What about the height of the aquifer layer? 

Ya know, although I don't know what delta h is, I'd probably guess it was 5 m if I was forced to choose. Reason I chose 5 is because that's the difference in water heights between you and your neighbor, and it is also the difference between the aquifer water height and the height of the water in your well. If you can ask your teacher for clarification, then that'd probably be a lot better, but if you can't figure it out, use 5m. for delta h. Then you can solve the question.

As for the second part, I can't really figure it out cause the hint doesn't make proper sense. Could there be another typo?

"(Hint: it is A that changes--now we have a cross sectional area 1 meter wide and the height of the aquifer layer)"

What about the height of the aquifer layer?

DiaperMan2

Nah homles, it matches word for word for what's on the E.C. paper.