What is the net pressure force on a pressurized jumbo jet flying at 10 km?

[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?

 

[SteamKing]

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.

 

[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?

 

[rude man]

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?

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.

 

[haruspex]

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.

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.

 

[rude man]

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.

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.

 

[haruspex]

The problem asks for the net force. Force = pressure times area.

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.

 

[rude man]

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.

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.

 

[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.

 

[Von Neumann]

Which is to say, 2x10^7N

 

[haruspex]

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

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.

 

[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.

 

[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.

 

[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?

 

[haruspex]

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.

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.

 

[Von Neumann]

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.

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

 

[haruspex]

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

The half cylinder, length*πr²/2.

 

[SteamKing]

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.

 

[tms]

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.

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

 

[haruspex]

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

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.)

 

[Von Neumann]

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

That is precisely what I found most confusing: the ambiguity in the problem.

 

[rude man]

That is precisely what I found most confusing: the ambiguity in the problem.

I agree. A goofy problem.

 

[haruspex]

That is precisely what I found most confusing: the ambiguity in the problem.

I think we can safely discount oblique planes. Most likely it means a plane containing the axis of the cylinder (because that's the interesting case), but it won't hurt to solve both that and the only other reasonable guess, a plane perpendicular to the axis.