# Math Help - fluids and pumps definitions

1. ## fluids and pumps definitions

Hello,

I have a homework due tomorrow and I don't understand some things, or even what some of the things mean.

1) this problem is about a pump, and I have to convert a pressure in psi to ft. The hint given to us shows multiplying by 2.31 but I don't know what 2.31 stands for.
it shows: ___ (psi) x 2.31 = ___ f(t)

I have some other problems but I want to try and solve them on my own, and if I have some problems then I can post here (hopefully)

2. Originally Posted by jon_dknight
Hello,

I have a homework due tomorrow and I don't understand some things, or even what some of the things mean.

1) this problem is about a pump, and I have to convert a pressure in psi to ft. The hint given to us shows multiplying by 2.31 but I don't know what 2.31 stands for.
it shows: ___ (psi) x 2.31 = ___ f(t)

I have some other problems but I want to try and solve them on my own, and if I have some problems then I can post here (hopefully)
I don't where the 2.31 came from but I can tell you what units it must have

psi= $\frac{lbs}{(in)^2}=\frac{144lbs}{(ft)^2}$

so if we want to multiply the above by something and get ft we can solve

$\frac{lbs}{(ft)^2}\cdot x =ft \iff x=\frac{(ft)^2}{lbs}\cdot ft =\frac{(ft)^3}{lbs}$

pounds are a force not a mass but the converstion factor has similar units to density

$D=\frac{m}{V}$

or in your case

$\frac{(ft)^3}{lbs}=\frac{V}{F}=\frac{V}{m \cdot g}=\frac{1}{g}\cdot \frac{V}{m}=\frac{1}{g}\cdot \frac{1}{D}$

I hope this helps.

Good luck.

3. Thanks for the help, with what you posted I figure out that if you divide the pressure in psi by the water specific weight then you get a head in ft, so based on that, I figured that the 2.31 comes from having:
(12)^2/62.4
that gives you 2.31, so that way you don't convert the psi to psf and then divide by the specific weight, you just multiply by the conversion factor.