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Stagnation Pressure

  1. Oct 17, 2012 #1
    First, let me say I am not a engineering student but just a average person trying to get his pilots license and enjoy working with aerospace problems.

    My problem is:

    “An aircraft is flying at 150 mph. The atomspheric pressure is 14.1 lbs./sq in and the temperature is 50 degrees F. Find the total or stagnation pressure at the point on the airfoil. If the pressure at a point on the upper surface of the airfoil is measured and found to be 13.9 lbs/sq in., what is the local air speed at this point?”

    I am trying to use the Bernoulli principle to solve this problem

    P1 + ½ρV^2 = P2 + ½ρV^2

    I have converted a few things to help work with the equation:

    150 mph x 1.4667 = 220 ft/sec
    50 F + 459.6 = 509.6 degrees Rankine.
    14.1 x 144 = 2030 lbs/sq. ft.

    I understand that as the velocity increases the pressure decreases in the venturi tube and that as velocity decreases the pressure will increase but I am still having trouble solving the entire equation and I guess understanding how to calculate everything in the problem.

    Here is what I have though and please correct me if I am wrong or have mathematical errors or failed reasoning 101.

    Solving for P1 (dynamic pressure)

    2030 + ½ (.0023769 slugs)(220)^2
    =2030 + (.00118845)(48400)
    =2030 +57.52
    =2087 lbs/sq ft.

    How do I go about solving the the second part of the word problem if the pressure is measured at 13.9 lbs/sq in., what is the local air speed?

    Also, let me apologize for the ole school units. My book came from a old bookstore and it by a professor at University of Kansas (Vincent Muirhead). I would like to have the Intro to Flight by Anderson but I refuse to give 180.00 for a book.
     
  2. jcsd
  3. Oct 17, 2012 #2

    rcgldr

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    You can probably assume the 13.9 lbs / in^2 pressure is sensed using a static port, so the 13.9 lbs is the static pressure of the moving air.
     
  4. Oct 24, 2012 #3
    I think I might have it figured out now.

    150 mph x 1.4667 = 220 ft/sec
    50 F + 459.6 = 509.6 degrees Rankine.
    14.1 x 144 = 2030 lbs/sq. ft.
    13.9 X 144 = 2001.6
    R = 1716

    ρ = P/RT

    2030 / (1716 * 509.6) = .002321
    2001.6 / (1716 * 509.6) = .002289

    P1 + ½ρV^2 = P2 + ½ρV^2


    2030 + 1/2 * .002321 * (220)^2 = 2001.6 +1/2 * .002289 * V^2

    2086 -2001.6 / (.001145) = V^2
    sqrt 73,743.99 = V
    271.55 ft./sec = V
    or 185.15 mph.
     
  5. Oct 24, 2012 #4

    rcgldr

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    One issue not clear to me. It seems that static pressure of the air should be independent of the frame of reference. So I tried to calculate the speed relative to the air, with the ambient air at 14.1 psi and not moving (0 mph), then tried to calculate the speed corresponding to 13.9 psi, then adding the 150 mph back to get the local speed, but I'm not getting the correct answer. Perhaps my assumption about frame independence is wrong because of the v^2 factors in Bernoulli's equation.
     
  6. Oct 25, 2012 #5

    boneh3ad

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    You are right, static pressure is frame independent. However, total pressure is not, and the total pressure is not conserved in the frame of reference you are trying to work in. Unfortunately, when you look at it in the frame of a stagnant fluid with an object moving through it, Bernoulli's equation no longer applies because you can no longer assume a steady flow. If you place a probe somewhere in that flow, Imagine the reading you would get on something, say, velocity. It would change with time, so you can't use inherently steady concepts like Bernoulli.
     
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