Can anyone explain to me what the Velocity Area method for measuring river or water flow is?

My guess is that the product of the cross sectional area and the velocity of water flowing in a pipe is always constant. If the Cross sectional area of the pipe increases at a particular point, then the velocity decreases so that the product $AV$ is a constant. Am I right?

If so, how can we extend this to pipes where the water is accelerating & does not have a constant velocity? For example, the system may be under the action of gravity & hence the acceleration of the water is $g$, the acceleration due to gravity?


3 Answers 3


What you refer to is conservation of mass under some assumptions:

  • Constant density
  • A steady state flow

I'll bring us back to your equation by starting with the very fundamental mass accounting for a given fluid flow. To be comprehensive, we need to recognize that velocity isn't constant over the entire area, but we will assume that it is. Take the flow rate to be $\dot{m}$.

$$\dot{m} = \rho V A$$

Now, if we have a steady state flow along a single flow path, then this quantity will be constant over the entire path, $\dot{m}=const$. Water in the cases you are concerned about is sufficiently incompressible so $\rho = const$. This results in your conclusion that $VA$ is constant.

Gravity may or may not shift the balance from $V$ to $A$ or vice versa. It depends on if there are rigid boundaries to the flow. If you have a flow fall freely in air or flow downward in a trench (like a river) then the boundary of the fluid may change freely. If you have a pipe with a given flow area, then the velocity is fully determined from that. Anyway, there are laws that conserve other things - like energy. So in a rigid pipe flowing downward (absent friction) the pressure will increase as you go down in elevation, which results directly from gravity.


You are right, if you assume the velocity of the fluid is more or less constant across the pipe, then conservation of mass dictates that $AV$ is constant.

Now, if you have a pipe, with no constraints at the bottom to let the water accelerate, this will change. The issue to consider here, is that when the water is accelerating, it will not longer occupy the full cross-sectional area of the pipe. So in your expression you have to change the meaning of $A$ to the effectively occupied area.

You can see this happening when you open the tap in your kitchen. The water accelerates due to gravity, and therefor the radius of the jet gets smaller. If it gets to narrow, it will eventually break up into droplets due to surface tension.


I am assuming that you want the relation for a noncompressible fluid ($\rho$ is constant). Well, what we are doing is conserving the volume(i.e. mass) of fluid entering and leaving the pipe in unit time. Thus, $\frac{dV}{dt}=const$ (more correctly $\frac{\partial V}{\partial t}$). As $V=Ax$, $\frac{dV}{dt}=A\frac{dx}{dt}=Av$. So, The area at a point(in the pipe) times the velocity at that point is constant.

Actually there's a small flaw in this argument. This will only work if the water is not in a pipe but is instead in freefall (or something similar). Also, this will only work at steady state, when the "shape" of the flow is constant. Otherwise, there will be loss of volume of water in changing the shape too.

If the water was in a pipe with varying area, the principle would be the same (conservation of mass), but you would have to take change of density into account (i.e. an incompressible fluid is no longer possible).

  • $\begingroup$ Your last sentence is incorrect. If the area of the pipe varies and if we are still assuming an incompressible fluid (ie water to a good approximation), then velocity changes with area, not density. $\endgroup$ Commented Feb 4, 2012 at 17:47
  • $\begingroup$ Uhh, no. Take a pipe which decreases in area, while the water decelerates. We have to take compression into account. If the pipe was increasing in area, then we would have to forget about the area of the pipe, and instead worry about the area of the fluid itself. $\endgroup$ Commented Feb 5, 2012 at 2:40
  • $\begingroup$ This makes little sense. Why is the water decelerating if the pipe area decreases? We are still assuming a steady state regime right? I don't know if you are confusing compressiblity from an equation of state point of view or compressibility from a fluid mechanics point of view, but for what we were talking about, neither of them applies. $\endgroup$ Commented Feb 5, 2012 at 14:19
  • $\begingroup$ I meant if the water was in a decelerating field (like in the OPs question). For example, take a right conical pipe, with apex upwards. The water will have to compress if we assume that the acceleration of the water is independant of the pipe (which is what the OP seems to be asking for). $\endgroup$ Commented Feb 5, 2012 at 14:49
  • $\begingroup$ Your last edit about the incompressible fluid makes sense and I guess I misunderstood that you were still under that constant acceleration assumption. Sorry ! $\endgroup$ Commented Feb 6, 2012 at 6:30

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