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I attempted to solve this problem as a tutor for a student and struggled, but want to be convince the professor didn't provide enough information.

The problem is essentially: We wish to maintain a plane in flight. The plane has a mass of 1.9E6 kg, the wings have a surface area of 1500 m^2, and the velocity of the air underneath the wing is 97 m/s.

I setup:

P1 + 1/2*density*velocity1^2 + density*gravity*y1 = P2 + 1/2*density*velocity2^2 + density*gravity*y2

where the 1 sub terms are beneath the wing, and the 2 sub terms are above

we are essentially looking for velocity2

I realized that without a thickness of the wing, the professor is probably wanting us to recognize that (y2-y1) ~ 0, thus

P1 + 1/2*density*velocity1^2 = P2 + 1/2*density*velocity2^2

Recognizing that the upward and downward forces must be equal to the pressure exerted downward by the force of gravity on the mass of the plane only, and thus

     1.9E6 kg * g
P2 = ------------
         Area

P1 is different, however the force is the same, thus

     1.96E6 kg * g
P1 = -------------  since area is given, and I can only assume is the bottom area
       1500 m^2

We now have everything we need except the top area of the wing, which given the equation

A1*v1 = A2*v2

allows us to equate A of the top to area of the bottom, this results in

     A1*v1
A2 = -----
       v2

Putting all of this together results in a quadratic equation that results in essentially the same velocity over the top as was given for the bottom (our result was 97.02 m/s).

This of course was not the answer expected which is why I am asking for help? What did I do incorrectly here, or is there truly not enough information given?

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    $\begingroup$ Note that we use MathJaX to typeset formulae here, and also note our homework policy. $\endgroup$ – ACuriousMind Apr 8 '15 at 19:53
  • $\begingroup$ This isn't my homework, I explained that very clearly. I am a tutor, and will not explicitly give him this answer. In terms of MathJaX, I have never used it, but will try to reformat :-) $\endgroup$ – trumpetlicks Apr 8 '15 at 19:56
  • $\begingroup$ Please note that the homework policy (as explicitly stated within it) is for any homework-like questions; not specifically those that the OP has for an assignment, etc. It is a broad type of question that we refer to as "homework" because it is the type that one might find in a homework assignment. Not necessarily one that is actually from someone's homework $\endgroup$ – Jim Apr 8 '15 at 20:10
  • $\begingroup$ I have to admit, that seems a bit bothersome to me. The site markets itself as being for professionals, researchers and academic students alike. An academic student is almost always going to ask questions related to their homework. How can you market this site to academics when you don't really mean that. Further I could understand (just as with stack) had I not tried anything and was asking for a blanket answer, thats NOT what Ive done here, I am clearly showing work and that I have attempted the problem. $\endgroup$ – trumpetlicks Apr 8 '15 at 20:25
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    $\begingroup$ Your use of the continuity equation ($A_1v_1 = A_2v_2$) is not valid here. This applies only for 1D incompressible flow in a conduit. $\endgroup$ – Dai Apr 8 '15 at 20:42
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If the plane is just flying at constant altitude then a vertical force balance requires that lift from the wings be equal to the plane's weight. The lift force, $L$ comes from a pressure difference above and below the wing so that $$ L = (p_1-p_2)A = mg $$

You can use the Bernoulli equation assuming a negligible difference in height to express the pressure difference as $$ p_1-p_2 = \rho/2 (v_2^2-v_1^2)$$

You should then be able to rearrange for $v_2$ and solve.

As @CarlWitthroft pointed out, this ain't how planes actually fly, but it does seem to answer your question.

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  • $\begingroup$ Thanks SOOOO much for the help. I am looking through a ton of material, and I have never seen the equation "L=(p1−p2)A=mg". Where is this derived from? How in that equation is the area of the upper portion of the wing not somehow incorporated? Thanks for you explanation :-) $\endgroup$ – trumpetlicks Apr 8 '15 at 21:09
  • $\begingroup$ Aircraft lift has almost nothing to do with stream lines and everything to do with angle of attack. The original statement of the problem leaves this out entirely. The amount of lift for a stated wing area depends on both the airspeed and the angle of attack. The airspeed above the wing can easily be the same as below. There are multiple online sources which correctly explain how an airplane flies. $\endgroup$ – Carl Witthoft Apr 8 '15 at 21:48
  • $\begingroup$ @CarlWitthoft I absolutely agree but this doesn't seem to have been a consideration when the problem was set up. $\endgroup$ – Dai Apr 9 '15 at 2:16
  • $\begingroup$ @trumpetlicks This video explains things nicely: youtube.com/watch?v=Gg0TXNXgz-w $\endgroup$ – Dai Apr 9 '15 at 3:33
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Follow two stream lines, one above the wing and one below. You can ignore the height difference and assume that under the wing you have atmospheric pressure and normal stream speed. The two stream lines follow the Bernoulli principle and thus

$$ \frac{1}{2} \rho v_{under}^2 + P_{under} = \frac{1}{2} \rho v_{top}^2 + P_{top} $$

Since you know the weight $W=M g$ and the wing area, the difference in pressures should be $$P_{under} - P_{top} = \frac{M g}{A} $$

If you know the density of air at that altitude and temperature then you can solve for $v_{top}$

$$\boxed{ \frac{M g}{\rho A} = \frac{1}{2} \left( v_{top}^2 - v_{under}^2 \right) }$$

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  • $\begingroup$ Sorry but how is this any different to my answer? $\endgroup$ – Dai Apr 8 '15 at 20:43
  • $\begingroup$ I was already typing when you submitted. What was I to do, quit? $\endgroup$ – ja72 Apr 8 '15 at 20:47

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