# Volume of water flowing out a cylindrical pipe

1. Sep 9, 2006

### Jacobpm64

Water is flowing down a cylindrical pipe of radius r.

(a) Write a formula for the volume, V, of water that emerges from the end of the pipe in one second if the water is flowing at a rate of

(i) 3 cm/sec (ii) k cm/sec

(i) r, assuming k is constant
(ii) k, assuming r is constant

All right, I just don't know where to start at all... I'm thinking that I'll probably need the formula for the volume of a cylinder. Besides that, I don't know what I have to do with the formula. I'm completely clueless on how to implement the rates of water flow into the volume of a cylinder formula.. if that's even what I have to do... I'm sure that if I can figure out part (a), part (b) won't be much of a problem. (just solving for different variables and graphing)

2. Sep 9, 2006

The cross area of a cylinder equals A = r^2*PI, so the volume that flows through the cylinder in one second equals V = v*A [m^3/sec], where v is the speed of water.

3. Sep 9, 2006

### Jacobpm64

you said the volume that flows through is in m^3/sec... does that mean I have to convert to m/sec, or will the rates work if i keep them as cm/sec?

Should work as cm/sec, right?

After you respond to this.. I'm gonna give it a shot, and you can let me know how i'm doing.

Thanks a lot.

4. Sep 9, 2006

Nope, you don't have to convert anything, of course it works perfectly well with cm^3/sec, too. I just forgot that the speed was given in cm/sec.

5. Sep 9, 2006

### Jacobpm64

ok, let's see if i did it correctly

(a) (i) V = 3pi*r2 (where V is in cm^3/sec)
(ii) V = kpi*r2 (where V is in cm^3/sec)

(b) (i) V on the vertical axis, r on the horizontal axis. It's just a parabola .. it'll look like x2.
(ii) V is on the vertical axis, k is on the horizontal axis. It's just a linear function with a positive slope.

How's this? (and is there any way to be more specific with the graphs, like to plot out any points, even if the points are just defined as variables?)

6. Sep 9, 2006