I would be very grateful if someone helps me with my issue.

I have a pump pumping water into a tube. My goal is to find velocity of the fluid flow in another tube that will be connected to the first one.

To determine the flow rate in the first tube, I used a ~$1000~ml$ container and a stopwatch. I obtained the mean volumetric flow rate, let it be about $100~ml/s$. I use the following relation for laminar flow as a fisrt approximation:

$$ V_1=\frac{Q_1}{S_1}, $$

where $V_1$ is flow velocity ($m/s$), $Q_1$ is volumetric flow rate ($\text{m}^3/s$), and $S_1$ is inner area of the first tube ($\text{m}^2$).

Let the inner radius of the first tube be $4~mm$, so $S_1=\pi \cdot 0.004^2 \approx 5 \cdot 10^{-5}~\text{m}^2$. Thus, we have

$$ V_1=\frac{100~ml/s}{5 \cdot 10^{-5}~\text{m}^2}=\frac{1 \cdot 10^{-4}~\text{m}^3/s}{5 \cdot 10^{-5}~\text{m}^2}=2~m/s. $$

Then I connect the second tube, which has $6~mm$ inner diameter, to the first one ($8~mm$ inner diameter). Outer diameter of the second tube is equal to the inner diameter of the first tube. I use a pump, so I assume flow rates $Q_1$ and $Q_2$ to be equal. From $Q_1=Q_2$ I get $V_1S_1=V_2S_2$, and

$$ V_2=\frac{V_1S_1}{S_2}=\frac{2\cdot0.004^2}{0.003^2}\approx3.6~m/s. $$

The result looks close to what I have seen, but I want to know the Reynolds number to determine if the flow is laminar or turbulent:

$$ \textrm{Re}=\frac{\rho V_2 D_H}{\eta}, $$

where $\rho$ is density of the fluid ($kg/\text{m}^3$), $D_H$ is hydraulic diameter of the second tube ($m$; we let $D_H$ be equal to geometric diameter of the tube $D_2$), and $\eta$ is dynamic viscosity of the fluid ($N\cdot s/\text{m}^2$). In our case, $\rho\approx974~kg/\text{m}^3$ and $\eta\approx 422\cdot10^{-6}~N\cdot s/\text{m}^2$, so

$$ \textrm{Re}\approx\frac{974 \cdot 3.6 \cdot 0.008}{422\cdot10^{-6}}\approx66500, $$

which is way more than $2000$, so the flow is turbulent, and the first approximation of the flow velocity should be corrected.

If I have experimental data of the volumetric flow rate in the first tube, then how can I find the flow velocity in the second tube in the case of turbulent flow? In the literature, they usually use volumetric flow rate in the case of laminar flow only. In some sources, equation $Q_1=Q_2$ ($V_1S_1=V_2S_2$) is used for pumped flows without indication of the flow type (laminar or turbulent). I am a bit confused.


1 Answer 1


Starting from the conservation of mass:

$$ \dot m_{1}=\dot m_{2} $$

This translates to $$ \rho_{1} S_{1} V_{1}=\rho_{2} S_{2} V_{2} $$

Assuming incompressible flow, thus $\rho_{1}=\rho_{2}$

gives: $$S_{1} V_{1}= S_{2} V_{2} $$ With $S_{1} V_{1} = Q_{1}$ , the formula you are using.

This formula follows directly from the mass balance, with only the assumption of incompressible flow. There is no assumption on turbulent or laminar flow, thus this equation holds for both flow types.

With the balance given above you can calculate the speed in at location 2 for the given value of $V_1$ and the ratio of Areas, just as you did. There is no need to account for the flow type.

However, it should be noted here that these are average speeds. If you want to go into further detail, you could include friction forces in the pipe. These friction forces depend on the flow type, and determine the shape in velocity profile, and the resulting velocity.

However, this is significantly more difficult to do than just solving a couple equations. I know there are some rules of thumb to estimate losses in pipes, but you have to check if their assumptions are valid for your case. Perhaps you can have a look at Pipe Flow Fluid Mechanics Course


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