# Basic Dimensional Analysis Question

• armolinasf
In summary, the SI units of K, the hydraulic conductivity of an aquifer, are [L/T], indicating a speed. The volume of water that moves through the aquifer is given by V/t=KA(H/L), where H is the vertical drop and L is the horizontal distance. The formula can be simplified to L^3t^-1=K*L^2, with H/L being dimensionless.

## 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

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.

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?

armolinasf said:
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.