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What is the pressure in Kilopascals on a vertical plate

  1. May 19, 2015 #1
    The plate (glass) is 1400 x 700 MM at right angles to the wind. The wind velocity is 42 Metres per second.

    The height above the ground is 6 metres.

    Your help to an 85yo would be enormously appreciated.

    Ian Richards.
  2. jcsd
  3. May 21, 2015 #2
    Hi Ian...
    I was thinking that :
    Pressure=Force /area
    If we assume that the air particles collide elastically with the plate then for a small air particle of mass dm the momentum change will be:
    $$2 (dm)v $$
    Where ##v## is the velocity of air.
    We know that force =change in momentum /change in time.
    Then we get $$F=2 v (dm/dt) $$
    And mass =volume×density=##V×\rho##
    And volume here=area of plate × small length travelled by the particle=A×dl
    So $$F=2v\rho A (dl/dt) $$
    And we can take ##dl/dt=velocity=v##
    So $$F=2\rho Av^2 $$
    And thus $$P=2\rho v^2$$
    [Thats what I think, an assumption can also be taken that the air particle stops after collision ]
  4. May 21, 2015 #3


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    Interesting derivation, but because of your elastic collision assumption, you ended up at 4x the real world worst case. If the fluid flow stops entirely, all of its kinetic energy will be used to increase the pressure, which results in a pressure increase (compared to ambient) of ##1/2\rho v^2##. This is called the stagnation pressure, and is the worst case scenario.

    Out of curiosity, where on earth are you that has a 42 m/s wind speed?
  5. May 21, 2015 #4


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    I realize what you meant here but I figured I would clarify for everyone else: the stagnation pressure is the pressure that occurs if that entire ##\rho v^2/2## is converted into pressure energy. The ##\rho v^2/2## term itself is commonly referred to as dynamic pressure.

    I suppose he never explicitly said Earth... hmm.
  6. May 21, 2015 #5


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    True enough boneh3ad, and thanks for the clarification. I was ignoring the ambient pressure, but that is a good point to make.
  7. May 21, 2015 #6

    jim mcnamara

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    Staff: Mentor

    42m/s ~= 151km/hr or about 93 mi/hr - hurricane force winds. I know there are geographically based building codes for areas subject to hurricane force winds. Don't know if they are applicable to large glass sheets. I would guess shuttering would be the only solution, if there is one.
  8. May 21, 2015 #7
    Sorry... but I didn't understand how will I get the ##1/2## part. If I assume that the particle will collide and stop then it will cause stagnation pressure. Therefore the change in momentum will be ##(dm)v## for the small particle. Which will give a ##1##. How will I get that ##1/2##?
  9. May 21, 2015 #8

    Thank you for your observations.

    To CJL - Where am I????? I am 6 metres above the level ground at Sydney Australia's Mascot Airport on two separate days with the strongest winds recorded there since 1929 - 152 Kilometres per hr

    As I understand it the Pressure in Kilopascals = p (which is ??? - the figure and the terminology?) ) x v (velocity in m/s = 42)squared all divided by 2 == ???? kps

    Can you pse fill in the figures to give a reading for the Pressure in Kilopascals.

    Is this figure a maximum? -an average ? - a???

    I cannot begin to express my gratitude for your interest and help. Ian Richards.
  10. May 21, 2015 #9
    It's ##\rho## the density of air.
  11. May 21, 2015 #10
    So pressure = 1. 274×(42)2/2=1123. 668 Pascal =0. 112 kpa.....
    that's the value I am getting..... And I ain't sure about it.
  12. May 22, 2015 #11
    For identical newtonian fluid the force to the plate must be 0. I think it is wrong to use newtonian lows for this calculations because air viscosity. For example, there is not only the front superpressure but and the back subpressure. Without calculations, I think that is safe to take as energy density the double amount $$ \Delta{p} = \frac{F}{A} = \rho v_0^2 \Leftrightarrow F=A\rho v_0^2 $$
  13. May 22, 2015 #12

    jim mcnamara

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    How about this approximation --

    Air density ρ = 1.225kg/m^3 at ground level on average
    Area of window A = 1.4m x .7m = .98m^2
    Wind velocity v = 42m/s
    We need something like a Betz limit factor,
    ... but I do not know enough to guess, so how about 1?
    This number is usually less than one AFAIK.

    So Force in Newtons = (1/2) * 1 * 1 (Betz #) * 1.225 * .98 * (42)^2

    F ~= (1.225 * .98 * 1764) / 2 == 1058.8 rounded > 1059 Newtons ( 1 Newton - 1 Pa )
    This is not too far from mooncrater's estimate. .11 Kilopascals is my sort-of answer.

    Once somebody can provide a limit factor you can get a really decent answer.
    Then multiply .11 times the Betz limit (or whatever it is called in Engineering) to get the true value note:
    42m/s wind velocity is the upper limit for a Saffir-Simpson 1 - Hurricane (typhoon)

    See: http://en.wikipedia.org/wiki/Saffir–Simpson_hurricane_wind_scale
    This is the scale for estimating destruction of a storm in the North Atlantic. Scale that applies to Australia is RMSC Tokyo - and would be "Typhoon". Correct?
    Last edited: May 22, 2015
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