thanks for the video, its a good refresher. the velocity equation can also be found on page 188 of the new handbook under Jet propulsion, if anyone forgets it
Great Video. One question. Why is the pressure at the outlet equal to zero? I can see that the outlet is flowing under some pressure from the ponded water. Thanks
The outlet is a free jet that is open to the atmosphere. The pressure term is for atmospheric pressure, and both points are under the same atmospheric pressure.
Thanks for the video! For the first problem, it seems like an orifice discharging freely into the atmosphere with the orifice coefficient C = 1. In a problem like this, if they don't say anything about the orifice (i.e sharp-edged, rounded), should I just solve it like you did? Also the FE handbook 10.1 (updated June 2021) provides H = v^2/2g for the pitot tube, so we don't have to derive it :)
Hi Ryan, yeah if doesn't specify anything then that approach would be appropriate. But! Make sure to stick to the FE Handbook equations. Namely, "Submerged Orifice Operating under Steady-Flow Conditions" and "Orifice Discharging Freely into Atmosphere" as they will likely use those on the FE exam. If not, you can always use the approach discussed in the video. Thank you for telling me about that pitot tube equation! I just noticed it and realized it's a little tricky because they don't' say H = v^2/2g, they just give v^2/2g lol. Still very helpful! Thanks Ryan
Hi! Please check out the image below: snipboard.io/Qc9G4u.jpg Remember, pressure is always the unit weight times the depth to the point of interest. For point 2 (stagnation pressure point), the water streamlines inside the tube is ultimately causes the pressure, therefore, the depth will be H + h.
@@directhubfeexam currently is like this -->"J" --> change to like this --> "L" -->, how to calculate the pressure in Pitot tube, as water velocity is still 0, but..
@@directhubfeexam I mean equation fundamental engineering fe mechanical engineering questions like design mechanical, thermodynamic fluid mechanics, etc
@@محمدالمعاني-ظ5ش You can download 2020 NCEES reference handbook following this link. You just have to register an account. account.ncees.org/reference-handbooks/
thanks for the video, its a good refresher. the velocity equation can also be found on page 188 of the new handbook under Jet propulsion, if anyone forgets it
ruclips.net/video/KZNMYV23sj4/видео.html
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Great video and explanation. Couldn’t have done it any better! Thank you!
Thank you!! I'm happy to know you find this helpful! Keep up the good work!
You can one-step plug n chug the pitot question using the pitot tube section of the fluids section of the handbook.
Yes! Just showing a few concepts in this one.
The equation is given on page 195. On the Pitot Tube
ruclips.net/video/KZNMYV23sj4/видео.html
...
Great Video.
One question. Why is the pressure at the outlet equal to zero? I can see that the outlet is flowing under some pressure from the ponded water.
Thanks
The outlet is a free jet that is open to the atmosphere. The pressure term is for atmospheric pressure, and both points are under the same atmospheric pressure.
Can you please make a video on orifices..
ruclips.net/video/KZNMYV23sj4/видео.html
Thanks for the video! For the first problem, it seems like an orifice discharging freely into the atmosphere with the orifice coefficient C = 1. In a problem like this, if they don't say anything about the orifice (i.e sharp-edged, rounded), should I just solve it like you did? Also the FE handbook 10.1 (updated June 2021) provides H = v^2/2g for the pitot tube, so we don't have to derive it :)
Hi Ryan, yeah if doesn't specify anything then that approach would be appropriate. But! Make sure to stick to the FE Handbook equations. Namely, "Submerged Orifice Operating under Steady-Flow Conditions" and "Orifice Discharging Freely into Atmosphere" as they will likely use those on the FE exam. If not, you can always use the approach discussed in the video. Thank you for telling me about that pitot tube equation! I just noticed it and realized it's a little tricky because they don't' say H = v^2/2g, they just give v^2/2g lol. Still very helpful! Thanks Ryan
@@directhubfeexam Thanks!
@@ryanlakner8950 ruclips.net/video/KZNMYV23sj4/видео.html
can you explain how the pressure at point 2 is equal to the specific weight * (h+H)? still a bit confused about it
Hi! Please check out the image below:
snipboard.io/Qc9G4u.jpg
Remember, pressure is always the unit weight times the depth to the point of interest. For point 2 (stagnation pressure point), the water streamlines inside the tube is ultimately causes the pressure, therefore, the depth will be H + h.
@@directhubfeexam thank you so much!
How about turn the Pitot tube 180degree, then point 2's velocity is still 0, but pressure definitely is lower, how to calculate that pressure?
Would need to see a picture, turn it 180 degrees Clockwise or Counterclockwise?
@@directhubfeexam currently the Pitot tube is facing the flow, by saying "turn 180 degree", I mean, make the Pitot tube point to the flow direction.
@@directhubfeexam currently is like this -->"J" --> change to like this --> "L" -->, how to calculate the pressure in Pitot tube, as water velocity is still 0, but..
Can you send me all the mechanical engineering questions
I'm sorry I don't know what you are referring to. I have none of those, I just create and cover FE type questions on here to help students.
@@directhubfeexam
I mean equation fundamental engineering fe mechanical engineering questions like design mechanical, thermodynamic fluid mechanics, etc
@@محمدالمعاني-ظ5ش You can download 2020 NCEES reference handbook following this link. You just have to register an account. account.ncees.org/reference-handbooks/
@@directhubfeexam
Please can I communicate with
you
Facebook or what's up
DirectHUBfh@gmail.com