Pressure forces on a jet

1. Jan 19, 2013

Von Neumann

Problem:

The fuselage (body) of a 747 jumbo jet is roughly a cylinder 60m long and 6m in diameter. if the interior of the plane is pressurized to .75 atm, what is the net pressure force tending to separate half the cylinder from the other half when the plane is flying at 10 km, where air pressure is about 0.25 atm?

Possible solution:

I understand that forces exist here because a difference in pressure is present. Namely, since the interior of the plane is at 3 times the pressure of that outside the plane, the forces will be directed outward. Any advice on setting this up?

2. Jan 19, 2013

SteamKing

Staff Emeritus
Draw a picture of the pressurized fuselage.

use the net pressure to calculate the force acting on one half of the fuselage.

Use the pressure to calculate the separation force.

3. Jan 19, 2013

haruspex

The easiest way to approach this is to think instead of a half cylinder, the other half being replaced by a flat rectangular plate. What is the force on the plate? What does that tell you about the force on the half cylinder?

4. Jan 19, 2013

rude man

Pretend the outside pressure is zero and the inside is .75 - .25 = 0.50 at. And realize the pressure on the airplane's inside surface is equal everywhere and normal to the surface at every point on the surface. And of course force = pressure x area.

5. Jan 19, 2013

haruspex

Normal force isn't what's wanted here, ultimately. (SteamKing seems to have headed down the same path.) Von N, please try my earlier hint.

6. Jan 19, 2013

rude man

But it is.

The problem asks for the net force. Force = pressure times area. Area = πR2h.
Pressure = 0.50 at. directed normally everywhere on the cylindrical surface. The answer is
πR2h*p.
h = 60m
R = 3m
p = 0.5e5 Pa.

This does neglect the force on the plane's two end-points, to be sure. These surfaces would add 2πR2p to the total force on the plane's inside.

7. Jan 19, 2013

haruspex

Force is a vector. You cannot integrate it over an area and get a meaningful answer if the force is acting in different directions at different points.

8. Jan 19, 2013

rude man

Oh really? What about Gauss' theorm in electrostatics? ∫D*da = q? It's exactly the same idea.

Actually, I acknowledge there is a semantic issue with this problem. The force tending to "pull the two sides apart" can be interpreted to be the projection of the pressure on the normal plane you suggest.

9. Jan 19, 2013

Von Neumann

Hey guys, thanks for all help. I'm going to see if I can figure it out when I get home from work. I recall the answer being 2000 tons of force if I recall correctly.

10. Jan 19, 2013

Von Neumann

Which is to say, 2x10^7N

11. Jan 19, 2013

haruspex

Let me word my statement more accurately: you cannot sum the forces acting in different directions as though they were scalars. I was wrong to say 'integrate' because what's actually being integrated is pressure, a scalar: dF = P.dA. The integral is therefore a vector and the summation may involve some cancellation.
This is quite different from the integration of flux: $d\Phi = \textbf{D.dA}$, which sums scalars.

12. Jan 20, 2013

rude man

haruspex then seems to have taken the correct tack. The "horizontal" component of the force would be the area 2*3m*60m = 360 m^2 and times 5e4 Pa would give 1.8e7 N, pretty close to 2e7 N. Again, ignoring the end surfaces.

13. Jan 20, 2013

Von Neumann

I think I may have gone somewhere. I calculated my answer to be 1.43e6N. First I found the net pressure by simply taking the difference between the two acting pressures. Then, I used P = F/A, and rearranged it to F=P_net*A. A=pi*d^2/4 being the area on which the force is acting.

14. Jan 20, 2013

Von Neumann

Or rather, should I assume the cylindrical surface to not be composed of two half-cylinders, but instead two half-rectangular plates? Thus the area A through which the force is acting is the following,

A = 2*(Area of one plate)
= 2*(60*3)
= 360 m^2

So, plugging in much the same way rude man has, I get approximately 2e7 N. I think my difficulty has come from the way I visualized the situation. I understand that forces are present due to the pressure difference, which are directed away from the body. However, I was originally imaging the fuselage being stretched along its horizontal axis, therefore the forces acting through its circular ends, hence making the effective area A = 2*(area of one circular end) = 2*(pi*d^2/4) (or 2*(pi*r^2)). Is there a way to solve this problem that way, or is it stupid?

15. Jan 20, 2013

haruspex

As I explained, that's wrong. You cannot simply multiply the pressure by the half-cylinder area. If you think about the ways the forces act on different parts of that area, you see that they're not all acting in the same direction. This means they'll partly cancel each other. You want the net force tending to separate the two halves.
Use the method I described in the third post.

16. Jan 20, 2013

Von Neumann

Specifically, what do you mean by "that" area?

17. Jan 20, 2013

haruspex

The half cylinder, length*πr2/2.

18. Jan 21, 2013

SteamKing

Staff Emeritus
Re-reading the problem statement in the OP, it is not clear which force is being sought. When the fuselage is pressurized internally, the forward (circular) end wants to separate from the after (circular) end, while at the same time, the left half of the fuselage wants to separate from the right half, where the line of separation runs longitudinally between the forward and after ends.

19. Jan 21, 2013

tms

There are also infinitely many other planes of separation possible. The problem is not well-defined.

20. Jan 21, 2013

haruspex

True, but two semicircular sections defined by a plane along the axis of the cylinder is surely the most likely. (It's a common enough question, in various guises.)