Basic Dimensional Analysis Question

[armolinasf]

Homework Statement

Porous rock through which groundwater can move is called an aquifer. The volume V of water that, in time t, moves through a cross section of area A of the aquifer is given by:

V/t=KA(H/L)

where H is the vertical drop of the aquifer over the horizontal distance L. The quantity K is the hydraulic conductivity of the aquifer. What are the SI units of K

The Attempt at a Solution

So since this is supposed to be a dimensional analysis question, I figured that I'd start by figuring out what the dimensions of the problem are and I came up with:

[V/t]=[K^a][A^b][(H/L)^c] ===> L^3*t^-1=[K^a][L^2b][L^c-c]

Since there is a time dimension on the left side of the equation I'm guessing that K has a time component but I'm not really too sure how to approach the problem...Thanks for the help

 

[Dick]

Why are you writing [A]=[L^2b]? Area is just plain length squared, right? [A]=L^2. I don't know why you are writing [K^a] either. There's just a K in the formula. And H/L is dimensionless.

 

[armolinasf]

Alright that clarifies some things for me...so then it would be L^3t^-1=K*L^2 (H/L is dimensionless because L/L is one right?) So for the dimensions to balance out K must be [L/T]? and that would be a speed right?

 

[Dick]

Alright that clarifies some things for me...so then it would be L^3t^-1=K*L^2 (H/L is dimensionless because L/L is one right?) So for the dimensions to balance out K must be [L/T]? and that would be a speed right?

Yes.

 

[rwisz]

The SI units of K would be m/s. This can be determined by setting up the equation in terms of its base units:

[V/t] = [K] * [A] * [H/L]

L^3 * T^-1 = L^2 * T^-1 * [K]

L^3 * T^-1 / L^2 * T^-1 = [K]

K = L/T

Therefore, K has units of length per time, which in SI units is m/s. This makes sense because hydraulic conductivity is a measure of how easily water can flow through the aquifer, so it would have units of distance per time.