Pitot Tube experiment

  • Thread starter knight92
  • Start date
  • #1
101
0
Hi there I am new to this forum and after reading like 100s of forums I cant find a solution so here I am

So recently I conducted an experiment to measure the pressure using pitot tube and use that to find the velocity profile which would be used to find mass flow rate. The whole experiment is rather long so I will only consider the first pressure. The pressure measured in pitot is in mmH2O and atmospheric pressure is 763.25 mmHg.

Experiment Setup:
Air pump blowing air at an unknown constant velocity into a horizontal large tube which narrows down to a tube with a smaller diameter(venturi). Pitot is attached to the small tube in the middle about 15cm away from where the small tube starts. A digital display is used to display the pressure from pitot. All of the apparatus is at the same level(horizontal), no elevation. the other end of the small tube is open.

Now stagnation pressure = dynamic + static pressure. So stagnation pressure must be the pressure measured by the pitot which appears on a digital display and the static pressure must be the atmospheric as it is 90 degrees to the flow. Am I right or wrong ?

So what I am doing is taking the pitot pressure 80 mmH2O and converting it to Pascals which gives me 784.5 Pa. Now when I convert the atmospheric pressure 763.25 mmHg to Pascals which gives me 101767 Pa which I sub back into my equation to find dynamic pressure gives me 784.5 - 101767 = Dynamic Pressure. To my understanding I should now be able to find the velocity using 0.5*desnity of air*v^2 = 784.5 - 101767 but as you can see it gives me a negative number which cant be square rooted. So I am thinking I am wrong can anyone explain to me how I might go about finding the velocity using the data from the pitot tube because really all I have is the pitot pressure, temperature of the room (18 Degrees Celcius) and the atmospheric pressure.

Thank you very much for any replies.
 

Answers and Replies

  • #2
2,685
22
Now stagnation pressure = dynamic + static pressure. So stagnation pressure must be the pressure measured by the pitot which appears on a digital display and the static pressure must be the atmospheric as it is 90 degrees to the flow. Am I right or wrong ?

Stagnation pressure is static pressure (measured at a point where flow velocity is zero).

Total Pressure = Static Pressure + Dynamic Pressure
So what I am doing is taking the pitot pressure 80 mmH2O and converting it to Pascals which gives me 784.5 Pa. Now when I convert the atmospheric pressure 763.25 mmHg to Pascals which gives me 101767 Pa which I sub back into my equation to find dynamic pressure gives me 784.5 - 101767 = Dynamic Pressure.

Well, for a start the fact your total pressure is lower than atmospheric should have flagged an error to you (see above equation).

I'm also curious why your pitot reading would be in mmH20? Did you build it yourself?
 
  • #3
2,685
22
Also, doesn't the pitot tube do both readings? So the figure you receive on the digital reading would be the dynamic pressure to convert to flow velocity, not the total pressure (this would explain the excessively low reading).
 
  • #4
101
0
Also, doesn't the pitot tube do both readings? So the figure you receive on the digital reading would be the dynamic pressure to convert to flow velocity, not the total pressure (this would explain the excessively low reading).

Hey jarednjames thank you for your reply, thats what I was saying i am getting a negative number which cant be square rooted so I was wrong but like you said the reading would be the dynamic pressure so I tried doing the whole thing using just the dynamic pressure equation which is 0.5*density*v^2. Now the results look better the mass flow rate I found using the venturi is very similar to the mass flow rate I found using the pitot velocity traverse. I am not sure why its in mmH2O its just the way the equipment was set by the technician. Anyways thanks alot mate :)
 
  • #5
2,685
22
Your welcome, glad to help.
 
  • #6
1
0
Hey , can someone please explain me this experiment , exactly how to find and what do we have to calculate in it ............................ ???
 
  • #7
boneh3ad
Science Advisor
Insights Author
Gold Member
3,262
926
Don't listen to JaredJames, his answer is incorrect.

Now stagnation pressure = dynamic + static pressure. So stagnation pressure must be the pressure measured by the pitot which appears on a digital display and the static pressure must be the atmospheric as it is 90 degrees to the flow. Am I right or wrong ?

There is a difference between a Pitot and a Pitot tube and not a Pitot-static tube. You can tell the difference because if it is a Pitot-static tube, there will be a second pressure port somewhere on the tube that opens normal to the flow direction so as to measure static pressure as well.

If it is, in fact, a Pitot tube as you describe, then yes, the measurement you get with it would be the stagnation pressure (or total pressure, they are the same thing). However, since it is measuring such a low value, I would be more inclined to believe you have a Pitot-static tube and the readout is actually the differential pressure between your total pressure and your static pressure. In other words, it is reading out dynamic pressure directly.

Under that assumption, you are measuring a flow velocity of
[tex]v = \sqrt{\frac{2 q}{\rho}} = 36.1 \text{ m/s}[/tex]

I don't know much about your pump, but that is a reasonable value.

Now when I convert the atmospheric pressure 763.25 mmHg to Pascals which gives me 101767 Pa which I sub back into my equation to find dynamic pressure gives me 784.5 - 101767 = Dynamic Pressure.

Atmospheric pressure is not your static pressure in this case. I am sure you could find a pump setting that would result in your atmospheric pressure and your static pressure being equal, but this is not, in general, the case.

So I am thinking I am wrong can anyone explain to me how I might go about finding the velocity using the data from the pitot tube because really all I have is the pitot pressure, temperature of the room (18 Degrees Celcius) and the atmospheric pressure.

Like I said, you have two potential problems. One is that your may be using a Pitot-static tube, which would be reading out dynamic pressure directly. The second possibility is that you are using the wrong static pressure. In general, if you have a simple Pitot tube, you need to have a separate static pressure port somewhere in the flow to get that value. Atmospheric doesn't help you.
 

Related Threads on Pitot Tube experiment

  • Last Post
Replies
8
Views
13K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
22
Views
5K
Replies
3
Views
1K
Replies
2
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
2K
Top