# Flow velocity in parallel pipes confusion

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 $$2(A_1V_1) = A_3V_3$$?

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

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

$$V_{1}^{2}f_{1}L_{1}=V_{2}^{2}f_{2}L_{2}$$

and

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

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