air speed above airplane wings physics question?

An airplane flies on a level path. There is a pressure difference of 525 Pa between the lower and upper surfaces of the wings. The area of each wing surface is about 112 m^2. The air moves below the wings at a speed of 80 m/s. Find the air speed above the wings if the weight of the plane is 117600 N.

3 Answers

  • 1 decade ago
    Favorite Answer

    The A/S above the wing is faster than the A/S below the wing. This comes from the cross sectional shape of the wing, which is called camber. Camber is that shape where the top of the wing sort-of bulges up near the leading edge and then trails away toward the trailing edge. Thus, the air above the wing has to go around that bulge and travel farther than air below the wing where there is no bulge.

    We all know that S = VT; where S is distance traveled above the wing, V is velocity, and T is time traveled. Assume S > s; where S is the distance traveled around the camber bulge and s = vt is the distance traveled under the wing at velocity v in time t. Then we have S/s = VT/vt. So when T = t, meaning the air above and below the wing must reach the trailing edge at the same time, we have S/s = V/v; so that V = (S/s) v, which clearly shows the air above a wing travels faster than air below.

    Now, here's the Bernoulli equation Bekki was talking about, when air travels along a surface, the static pressure on that surface reduces proportionately to the square of the velocity of that air (e.g., V and v) along the surface. This results because total pressure TP = SP + DP + PP is constant around the wing; where SP is static pressure, PP is potential pressure (based on height), and DP is dynamic pressure based on velocity. As the DP increases because of increasing velocity, SP has to come down for TP to remain fixed.

    Thus TP(A) = TP(B) = constant; where A and B mean above and below the wing. So invoking Bernoulli's equation for total pressure, we have SP(A) + DP(A) = SP(A) + 1/2 rho V^2 = SP(B) + 1/2 rho v^2 = SP(B) + DP(B); where S(B) - SP(A) = 525 Newton/m^2, which was given. The potential pressures above and below the wing are essentially the same value PP(A) = PP(B); so they cancel out.

    If you have rho, the air mass density, you can solve for V from V^2 = 2*525/rho + v^2; where v = 80 mps.

    Source(s): Physics and engineering degrees.
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  • 4 years ago

    As you assert, the decrease floor is shorter front to lower back than the better floor, yet you acquire the speeds the incorrect way around. If there are 2 neighbouring molecules on the front, they are going to be reunited on the lower back. the only decrease than the wing won't would desire to run so speedy, whilst the only above the wing will would desire to seize up after working an prolonged distance, so will would desire to run swifter.

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  • Anonymous
    1 decade ago

    Is this a trick question. They ask for air speed and they give you air speed.

    And here I was about to pull out Bernoulli's equation.

    EDIT: Oh, okay, I should read better. I guess if it looks enough like a Bernoulli problem, it must be a Bernoulli problem.

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