Is there a way to convert cubic feet per second to feet per second. Or in general volumetric flow to velocity?

I want to know the time taken by water to travel from point A to point B. I have the distance between them and volumetric flow.

Any suggestions?

  • 2
    $\begingroup$ divide by the cross-sectional area of the path. $\endgroup$
    – Jim
    Mar 2 '15 at 21:33
  • 1
    $\begingroup$ You can convert that if you know the surface area it's flowing through. $\endgroup$
    – ahemmetter
    Mar 2 '15 at 21:33

This is a more complicated question than you think.

Superficially, if you have a volumetric flow rate of $V$ through a pipe with constant area $A$, then the average velocity of the fluid is given by

$$v = \frac{V}{A}$$

If you use cubic feet per second and feet squared, the result will be in feet per second.


The actual velocity of a fluid in a pipe is a function of the position in the pipe. For laminar flow, the velocity follows a parabolic profile, with the points near the wall being almost stationary, and the points in the middle going fastest. In that case, there is no direct answer "how quickly will the water come out the other end" - because some water will come out much more quickly than the rest. Note - if you have a river, the flow is unlikely to be laminar (and the cross section is neither circular nor constant) so this analysis does not apply. But when I initially wrote this answer, we had no inkling that you were asking about flow in a river...

How much faster is the fastest liquid? Well, if you assume a circular cross section and a parabolic profile, then we can write the velocity as a function of radius r (for a pipe with radius R):

$$v(r) = v_0\left(1 - \left(\frac{r}{R}\right)^2\right)$$

where $v_0$ is the maximum flow velocity (at the center of the pipe).

Integrating the flow rate over the entire cross section, we find

$$V = \int_0^R 2\pi r v(r) dr\\ =2\pi v_0\int_0^R r\ \left(1 - \left(\frac{r}{R}\right)^2\right) dr\\ = 2\pi v_0 (\frac12 R^2 - \frac14 R^2)\\ = \frac12 \pi R^2 v_0$$

If we put $$A=\pi R^2$$

Then we find

$$v_{av} = \frac12 v_0$$

In other words, the mean flow velocity is half the peak velocity. Worth thinking about when you try to figure out what you are really trying to ask.

  • $\begingroup$ My friend is doing his internship. Although he has several other tasks, he wanted to figure out a way to forecast next day's flow from the one of the dams on the McClellan–Kerr Arkansas River Navigation System. This could help the company save money on the hydro-projects. I have put a link of the locations of the different hydroprojects on the map link $\endgroup$
    – Naddy bill
    Mar 3 '15 at 15:13
  • $\begingroup$ Flow in rivers is more complex than flow in pipes - if you change the flow rate (for example by opening a dam) you create a wave and the speed at which this wave travels depends on the depth of the rive (not simply the area). This is not a simple "pressure across a tube" problem because the cross sectional area is not constant if the flow rate changes. I guess you are asking "when does the wave arrive?" right? $\endgroup$
    – Floris
    Mar 3 '15 at 15:20
  • $\begingroup$ Right. And its effect on the flow rate at the turbine. Cause more the turbine spins more it generates electricity. $\endgroup$
    – Naddy bill
    Mar 3 '15 at 15:40
  • $\begingroup$ @Floris also a flow of water at any meaningful speed is likely to be turbulent and hence does not follow the analysis you put forward above. $\endgroup$
    – Dai
    Mar 4 '15 at 5:47
  • $\begingroup$ @dai - you are right, I should make that point more explicitly. When I wrote the answer, we did not know this was related to a river; so I just wrote "for laminar flow". $\endgroup$
    – Floris
    Mar 4 '15 at 5:50

Any suggestions?

Summarizing the answers in the comments:

Divide the volumetric flow rate (in cm^3/s) by the cross-sectional area of the pipe (in cm^2) to get the velocity in cm/s.


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