0
$\begingroup$

Problem statement

I am solving a simple looking, but actually difficult fluid dynamics problem. As shown, a hinged door is placed at the bottom of the box-shaped duct. Here, it is possible to assume any friction at hinge. The inlet of the duct is just atmosphere, the outlet is connected to the chamber with a strong fan.

When the system is at rest, door is at completely closed ($\theta = 0$) Once the system activated, air is blown out at the outlet with controlled flow rate $Q'$. Then, air is blowing in to the duct at inlet, opening(rotating) the door with some degree theta. My goal here is to model relation between $Q'$ and $\theta = 0$. Reynolds number is big enough, so the flow is turbulent. However, the system is in the steady state for given $Q'$.

First, I listed forces (moments)that push door upside and downside.

1) Pulling up the door: - Air that hits the door; momentum exchange. - Viscous drag at the tip of the door - maybe negligible since door is very thin - Pressure difference between two side of the door (by experiment, it is found that there's pressure difference between inlet and outlet, which is not negligible. However, I am not sure the same argument can hold between two sides of the door).

2) Pushing down the door: - Weight of the door - This is only one thing that I can think of

Then, I tried to balance those forces so that moment at the hinge becomes 0. Roughly speaking upward forces have a $\cos(\theta)$ term while moment induced by door weight has a $\sin(\theta)$ term, thereby balancing at some angle theta. Get the relationship something like $\tan(\theta) \sim (V_{\mathrm{air}}^2)$. However, this relation gives much stiffer slope compared to the one from the experiment. Experimental result showed there there's a linear relationship ($R^2>0.99$) between $Q'(=V_{\mathrm{air}})$ and $\theta$, for $\theta$ between 15-30 degree.

I didn't expect I can have a very accurate analytical solution, but wonder is there any method or model that I can use for this kind of problem.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.