Flowrate through an elliptical cross section of an oblique liquid jet

[rdemyan]

TL;DR Summary

Determine the flowrate through an elliptical cross section of an oblique liquid jet. This is different than the typical method of determining the flowrate through a circular cross section normal to the direction of flow.

I'm trying to solve a problem of a liquid jet impinging at an oblique angle, $\beta$. This post only involves a portion of that problem. I'm thinking that I should look at the flow within the jet based on an elliptical cross section of the jet that is coplaner with the x-z axis (see attached Fig. a). Fig. b is a top detailed view of the elliptical cross section through the jet. Now, this is different from the typical way of looking at the flow through a cylindrical jet which is based on the circular cross section that is normal to the direction of flow. Based on the typical method the flow rate is given by

$$Q = \pi R^2u_j$$

First, I want to confirm that if I change to the elliptical cross section, the flow rate Q remains the same. I can't see how it wouldn't, but I just want to confirm.

Then, based on a separation point, M, in the elliptical cross section (which corresponds to a stagnation point where the jet strikes a horizontal surface), the flowrate would be given by the following equation,

$$Q =\int_0^{2\pi} \frac{p^2}{2}u_j\sin\beta \, d\phi$$

Now plugging in Eq. 1 for Q yields

$$\int_0^{2\pi} \frac{p^2}{2}u_j\sin\beta \,d\phi = \pi R^2u_j$$

or finally,

$$\int_0^{2\pi} {p^2} \,d\phi = \frac{2 \pi R^2}{\sin\beta}$$

Am I looking at this correctly?

BTW: The preview button doesn't seem to render the latex. The only way I could check to see if I entered the latex correctly is to submit the post. Is there a way that I can check my latex before submitting the post?

 

[berkeman]

BTW: The preview button doesn't seem to render the latex. The only way I could check to see if I entered the latex correctly is to submit the post. Is there a way that I can check my latex before submitting the post?

We have been dealing with LaTeX issues since the software update yesterday. Check out the update thread in the Feedback forum for more information. In the mean time...

If you have trouble using the Preview feature to check your LaTeX, you can use a website such as https://mathb.in/ or https://latexeditor.lagrida.com/ to preview your LaTeX before you post it.

 

[erobz]

Apply continuity for a control volume:

$$ 0 = \frac{d}{dt} \int_{cv} \rho~ dV\llap{-} + \int_{cs} \rho ( \boldsymbol {u} \cdot d \boldsymbol {A} ) $$

So what is the velocity distribution over the elliptical area portions and what is its direction?

 

[rdemyan]

The velocity profile for the jet, with respect to the circular cross sectional area normal to flow, is flat (i.e. no variation with radius within the jet; same velocity everywhere in the direction of flow). In Figure b, the flow is going into the hatched slit so it is flowing in the y direction; hence the need for sin\beta. I guess, in that direction the velocity would also be constant everywhere within the slit even if the slit is swept out to be the entire ellipse.

 

[erobz]

The velocity profile for the jet, with respect to the circular cross sectional area normal to flow, is flat (i.e. no variation with radius within the jet; same velocity everywhere in the direction of flow). In Figure b, the flow is going into the hatched slit so it is flowing in the y direction; hence the need for sin\beta. I guess, in that direction the velocity would also be constant everywhere within the slit even if the slit is swept out to be the entire ellipse.

If you are saying it's a uniform velocity distribution like shown:

It will work out to be through that surface:

$$ \dot m_{out} = \int \rho \left( \boldsymbol {u} \cdot d \boldsymbol {A} \right) = \int \rho \left( u \sin \theta \boldsymbol {i} + u \cos \theta \boldsymbol {j} \right) \cdot \left( 0 \boldsymbol {i} + 1\boldsymbol {j} \right) dA = u \cos \theta \int \rho ~dA $$

for constant density $\rho$ it just $\dot m_{out} = \rho u \pi R^2$.

If you are splitting it up at $M$ it will get whatever mass fraction you decide based on your arbitrary(at this point) choice of $M$ and the boundary of the division.

 

[rdemyan]

So, if I understand you correctly, what you have shown with your diagram and equations is that the mass flowrate through the elliptical cross section is the same as the mass flowrate through the circular cross section normal to flow. I "tentatively" stated that in my problem statement, but you have shown it to be true. Is there any other information regarding my question that I should have learned from your response?

 

[erobz]

So, if I understand you correctly, what you have shown with your diagram and equations is that the mass flowrate through the elliptical cross section is the same as the mass flowrate through the circular cross section normal to flow. I "tentatively" stated that in my problem statement, but you have shown it to be true. Is there any other information regarding my question that I should have learned from your response?

If $M$ were a stagnation point, there has to be a velocity gradient surrounding it (by definition). So if you are planning on being consistent, better figure that out.

 

[rdemyan]

If $M$ were a stagnation point, there has to be a velocity gradient surrounding it (by definition). So if you are planning on being consistent, better figure that out.

Well, "M" is in a cross section of the jet before it impinges so it can't be a stagnation point and is instead referred to as a separation point. However, your point is well taken because I do want to project that ellipse downwards and have that be the elliptical plane of symmetry of the impingement zone (now I am talking about two equal jets impinging upon one another as opposed to a single jet impinging upon a horizontal surface; see the attached diagram). In that case, M becomes a stagnation point and lies within the plane of symmetry of the impingement zone (i.e. the plane of symmetry bisects the impingement zone in half). The impingement of two equal jets causes a liquid sheet to form and the stagnation point is considered to be the origin of that sheet. In general, the velocity in the sheet is considered to be constant and equal to the jet velocity IF the jet velocity is uniform throughout the jet (which is the assumption I am making).

I've attached a diagram from a CFD study of two equal jets impinging to form a sheet. The diagram assumes parabolic velocity profiles in the jets prior to impingement whereas I am assuming flat profiles. However, the diagram shows that at the stagnation point (zone) the sheet velocity $u_s$ is effectively zero, but then accelerates to about the mean jet velocity ($u_j$) by the time the liquid reaches the end of the impingement zone.

 

[erobz]

I've attached a diagram from a CFD study of two equal jets impinging to form a sheet. The diagram assumes parabolic velocity profiles in the jets prior to impingement whereas I am assuming flat profiles. However, the diagram shows that at the stagnation point (zone) the sheet velocity $u_s$ is effectively zero, but then accelerates to about the mean jet velocity ($u_j$) by the time the liquid reaches the end of the impingement zone.

The impinging jets are coming down, the main outgoing jet appears to be accelerating (it appears to be converging under constant pressure), is this oriented vertically? The other outflow, (moving upward in this picture) is a sheet? Are you trying to find an analytical solution that predicts this for fun?

 

[rdemyan]

The CFD diagram is rotated 90 degrees from the diagram that I showed. What I showed is a horizontal plate and if the CFD diagram had a plate it would be vertical. The outgoing stream is actually a sheet that has approximately an elliptical shape. Here's a snapshot of an animation of two equal impinging jets producing a sheet.

 

[erobz]

The CFD diagram is rotated 90 degrees from the diagram that I showed. What I showed is a horizontal plate and if the CFD diagram had a plate it would be vertical. The outgoing stream is actually a sheet that has approximately an elliptical shape. Here's a snapshot of an animation of two equal impinging jets producing a sheet.

Ok, so the CFD model is the side view in the last image, and you hope to get the "elliptical shape" of the discharge (top view) from the model you are after?

 

[rdemyan]

At this point my only question was about converting from the circular cross section to an elliptical cross section. I'm analyzing the impingement zone of equal impinging jets. The plane of symmetry of the impingement zone, I am assuming is effectively identical to the elliptical cross section through the jet.

I think you confirmed that what I am proposing to do in my original post is correct. Please let me know if that is not the case.

BTW: modeling the shape of the sheet as shown by your red ellipse has been heavily researched; however, it is quite complex and the sheet shape changes quite drastically once the velocities are high enough where aerodynamic waves on the sheet cause rapid breakup into droplets.