In closed pipes configured such that they are parallel in fashion, we know that the total volumetric flow rate $Q$ is equal to the sum of all $Q$s in the parallel pipes.

But I have one doubt (as I'm a newbie in this thematic) - if the two parallel pipes are equal in diameter, hence area, thus we know that they manifest the same magnitude of fluid velocity. So does this fact apply too, for similar-sized pipes in parallel? Is the velocity really the same for both pipes? Please correct me if I'm wrong.

So let's say two parallel pipes of similar diameter, connecting a water tank, Pipe 1 and Pipe 2, converge and subsequently join with a single Pipe 3 of different diameter, and water discharges out of its end. So from the continuity equation, it is $Q_1 + Q_2 = Q_3$. Therefore $A_1V_1 + A_2V_2 = A_3V_3$. But since $A_1 = A_2$, I learned that $V_1 = V_2$. So in that sense, does that make $2(A_1V_1) = A_3V_3$?

Please correct me. Thanks!

  • $\begingroup$ Welcome to Physics SE. Here is a tutorial in MathJax which allows us to typeset equations and formulae. $\endgroup$
    – Sam
    Feb 17, 2020 at 10:31
  • 1
    $\begingroup$ Yes. That is a correct mathematical deduction. So what is the difficulty? What is the cause of your doubt? $\endgroup$ Feb 17, 2020 at 12:52
  • $\begingroup$ The thing that might change in going from two pipes to one is the rate of pressure loss which results from friction. $\endgroup$
    – R.W. Bird
    Feb 17, 2020 at 16:05

1 Answer 1


I believe that the equation you would get would be $A_{1}\left(V_{1}+V_{2}\right)=A_{3}V_{3}$

or $$V_{1}+V_{2}=\frac{A_{3}V_{3}}{A_{1}}$$

This equation does not require $V_{1}=V_{2}$

Things like the length of the pipe, and the roughness can affect the relative velocities. From the pressure drop equation for parallel pipes



$$f_{1}=function\left(\frac{ρV_{1}D}{μ}\right)$$ $$f_{2}=function\left(\frac{ρV_{2}D}{μ}\right)$$

as you can see the velocities need not be the same.


This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .