I Equalizing airflow through multi-hole orifices

  • I
  • Thread starter Thread starter Anachronist
  • Start date Start date
  • Tags Tags
    Airflow
AI Thread Summary
The discussion centers on the airflow dynamics through multi-hole orifices compared to a single hole of equivalent area, particularly in the context of a 3D-printed siren whistle inspired by a Francis turbine. While the Bernoulli equation suggests no distinction between the two configurations, the presence of friction and the perimeter-to-area ratio indicates that multiple smaller holes may behave differently due to increased friction. Observations show that the side with more holes (8) yields higher rotation speeds than the side with fewer holes (6), prompting speculation that the blade area may significantly influence airflow dynamics. The difference in performance could also be attributed to varying pressure drops when driving the turbine with constant flow. Ultimately, the interaction between hole configuration and blade design plays a critical role in the device's efficiency.
Anachronist
Gold Member
Messages
123
Reaction score
59
TL;DR Summary
One pressure vessel has 6 holes from which air escapes, another has 8 holes, total open area of the holes is same on each vessel. Should the volume flow be the same?
I know the Bernoulli equation for calculating the air velocity through a pressurized hole, but I am wondering if one hole of a given area behaves the same as multiple holes with the same total area. The Bernoulli equation doesn't make a distinction.

Background: As a personal 3D design and printing project I designed a siren whistle patterned after a Francis turbine, in which a dual runner (two runners back to back) chops the air coming out of exit holes on each side. The number of blades on each side of the runner is different (6 and 8) with a corresponding number of exit holes. As the runner rotates, the turbine produces two tones, with the airflow interrupted 6 times per rotation on one side, and 8 times per rotation on the other. The holes on each side have the same total open area.

The device works pretty well, but if I cover the holes on one side or the other, to allow only one side of the runner to drive its rotation, the rotation speed is clearly different for the same input pressure (by my best guess by breathing into it). It's slower when air exits only the 6-hole side.

That made me wonder if several small holes behave differently than fewer holes with the same open area.

Hmm, as I write this, I realize the difference could also be due to the different number of blades. The side with the higher number has a higher surface area on which the air can impart force.
 
Physics news on Phys.org
To your summary question:

No - the flow shouldn't be the same. One of the factors controlling the flow is friction. The implementation with more holes has a higher 'perimeter/area' ratio. More 'perimeter' results in more friction. How much actual difference that makes depends on how large a factor friction is in your specific case - higher velocities result in more significant friction terms.
 
That's what I thought. However, the effect I observe is the opposite of what would be expected (higher rotation when exhausting through the 8 hole side instead of the 6 hole side) so maybe the runner blade area is having a greater effect, which may be mitigated only slightly by the greater surface area of the holes.
 
If you drive the turbine with a constant flow, then it may have greater pressure drop with the 8 hole side, which may explain the higher velocity.
 
Thread ''splain this hydrostatic paradox in tiny words'
This is (ostensibly) not a trick shot or video*. The scale was balanced before any blue water was added. 550mL of blue water was added to the left side. only 60mL of water needed to be added to the right side to re-balance the scale. Apparently, the scale will balance when the height of the two columns is equal. The left side of the scale only feels the weight of the column above the lower "tail" of the funnel (i.e. 60mL). So where does the weight of the remaining (550-60=) 490mL go...
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Back
Top