Southwest Engine Failure

BobDDuck

Island Bus Driver
I would avoid reading that author again about any aviation topics...
The thing that really concerns me is that this is attributed to Allen Fuhs (http://www.nmspacemuseum.org/halloffame/detail.php?id=114), somebody who really should understand the physics involved. Not sure what exactly he was thinking, unless he was forgetting that a rocket gets a large part of it's thrust from the venturi created by the exit nozzle, and not just from the differential in pressure created by the fuel being burned off. Also, he's 99 years old, so that could play in to it as well.
 

Stone Cold

Well-Known Member
The thing that really concerns me is that this is attributed to Allen Fuhs (http://www.nmspacemuseum.org/halloffame/detail.php?id=114), somebody who really should understand the physics involved. Not sure what exactly he was thinking, unless he was forgetting that a rocket gets a large part of it's thrust from the venturi created by the exit nozzle, and not just from the differential in pressure created by the fuel being burned off. Also, he's 99 years old, so that could play in to it as well.
Yeah, I didn't see a link on the original post and didn't look up where it came from. I took a cursory glance at the article and I was laughing. Then I saw calculations happening, and I wanted to stop the shenanigans before it became ridiculous.
 

Stone Cold

Well-Known Member
I completely agree, but it's not like using equations properly does any harm...
Oh, I agree, and go for it if you want to. I just wanted to point out the physics of the leverages you're talking about.

Also, according to some Gulfstream instructors (I still don't buy it, but it's possible), the outlet valve on the G V aircraft produces 8 pounds of thrust while at cruise. (~9.6 psi is normal cruise differential, and the opening is approximately 16-25 square inches located at the copilot's side of the cockpit. A window is much bigger, especially since the valve isn't fully open normally, so the thrust would be less at a window.
 

Markf64

Well-Known Member
You don't even have to do any calculations. Where's the thrust vector from the window? I believe it was an over wing seat, so very near the vertical axis. It would take a huge amount of force to counteract the rudder which has a huge arm from that axis. Leverage is a real thing...

Debunked. The author should have paid more attention in high school physics. @Markf64, I would avoid reading that author again about any aviation topics...
@Stone Cold
Red flags went up in the back of my mind when I read that, even if the blowout occurred in the most aft/forward windows. I chose to post for the sake of conversation here on JC, and to make sure I wasn't missing something. I could have also done the google search but again I wanted to hear from smarter people than me. I should've been more clear in my original post. Disclaimer: The 'heaviest metal' I fly is a 172 at max gross.
 

Stone Cold

Well-Known Member
@Stone Cold
Red flags went up in the back of my mind when I read that, even if the blowout occurred in the most aft/forward windows. I chose to post for the sake of conversation here on JC, and to make sure I wasn't missing something. I could have also done the google search but again I wanted to hear from smarter people than me. I should've been more clear in my original post. Disclaimer: The 'heaviest metal' I fly is a 172 at max gross.
It definitely wasn't meant to be a knock on you. I was merely pointing out what I saw wrong with the theory. It was great for discussions.

That's the beauty of aviation...if you think you know it all, it's time to get out because you've become dangerous. Keep posting, as we all learn from each other.
 

Markf64

Well-Known Member
It definitely wasn't meant to be a knock on you. I was merely pointing out what I saw wrong with the theory. It was great for discussions.

That's the beauty of aviation...if you think you know it all, it's time to get out because you've become dangerous. Keep posting, as we all learn from each other.
Thanks
From your post history, I knew you you weren't knocking me.
 

Roger Roger

Paid to sleep, fly for fun
Wow.

I don't know any spherical horses, but here:

~10 PSI x ~80sq.in. (non scientific WAG) = ~~800lbs of thrust

Moment of 800lbs of thrust at unknown station? Probably not greater than rudder authority in cruise...even if we get past that wild assumption, I'm pretty sure a rolling moment would be created by the yaw starting a spiral which would have been easy to recover from in short order. Negative flat spin in cruise, Sir Newton!

Disclaimer: the type I fly only allows +6.35 PSI, so I don't know what the normal operating parameters would have been in this case, unless someone can tell me the cabin pressure altitude and pressure altitude at the time of the incident.
Yeah, but the force across the window is not the direct thrust, that will be found by the mass flow of air through the hole.
 

BobDDuck

Island Bus Driver
Yeah, but the force across the window is not the direct thrust, that will be found by the mass flow of air through the hole.
That's over time though, right? The the mass flow would account for the entirety of volume in the cabin and would be (using nibake's WAGs) 10psi x 80 sq inches x cabin in sq inches. Which is practical for thrust in zero drag or non corrective scenarios, but in this case, as long as the single moment force (the 800lbs from his wag) is less than the corrective force (the flight control input), the mass flow doesn't really matter too much.

I think... It's been forever since I cracked a fluid dynamics book.

Or did I totally misunderstand what you were saying?
 

inigo88

Composite-lover
I started to write something about Bernoulli, then remembered this is compressible flow which is hard. Much consternation over fluid mechanics can be avoided by following this guy's advice about assuming choked flow (exit velocity = Mach 1) at the window, because your P/P0 will be less than the critical value 0.528 (assuming P=3.98 psi at 32,000 ft from a standard atmosphere calculator and P0 = 8 psig in the cabin based on the differential pressure gauge from a 737 overhead panel).
https://www.physicsforums.com/threads/velocity-of-air-coming-out-of-a-nozzle.694656/

Solving for mass flow rate and then applying the General Thrust Equation for a constant mass flow rate...
https://www.grc.nasa.gov/www/k-12/airplane/thrsteq.html

...(which may or may not be a totally wrong assumption :) ) with V0=0 m/s, Ve=Mach 1 at 32,000 ft on a standard day and the intake and exit pressures above, I get a thrust force of 1885.65 lbf. To be honest I'm surprised I'm not several orders of magnitude off, especially since I converted back and forth between SI and imperial units in between a couple steps because some of the thermo constants confuse me in imperial. I also used a bigger window area of 113.9 in^2 based on some 737 window googling, which would get me to within ~700 lbs of @nibake 's pressure = force/area assumption.

That being said, you are flying in a 133,000 lb airplane. Keeping things in perspective, the point load from that "window thrust" is only 1% of the total lift force on the wings. You almost certainly get more side-load flying through turbulence. And since it's acting near the center of gravity (and thus the yaw axis) the yawing moment contribution is virtually nil.
 
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jtrain609

I'm a carnal, organic anagram.
I started to write something about Bernoulli, then remembered this is compressible flow which is hard. Much consternation over fluid mechanics can be avoided by following this guy's advice about assuming choked flow (exit velocity = Mach 1) at the window, because your P/P0 will be less than the critical value 0.528 (assuming P=3.98 psi at 32,000 ft from a standard atmosphere calculator and P0 = 8 psig in the cabin based on the differential pressure gauge from a 737 overhead panel).
https://www.physicsforums.com/threads/velocity-of-air-coming-out-of-a-nozzle.694656/

Solving for mass flow rate and then applying the General Thrust Equation for a constant mass flow rate...
https://www.grc.nasa.gov/www/k-12/airplane/thrsteq.html

...(which may or may not be a totally wrong assumption :) ) with V0=0 m/s, Ve=Mach 1 at 32,000 ft on a standard day and the intake and exit pressures above, I get a thrust force of 1885.65 lbf. To be honest I'm surprised I'm not several orders of magnitude off, especially since I converted back and forth between SI and imperial units in between a couple steps because some of the thermo constants confuse me in imperial. I also used a bigger window area of 113.9 in^2 based on some 737 window googling, which would get me to within ~700 lbs of @nibake 's pressure = force/area assumption.

That being said, you are flying in a 133,000 lb airplane. Keeping things in perspective, the point load from that "window thrust" is only 1% of the total lift force on the wings. You almost certainly get more side-load flying through turbulence. And since it's acting near the center of gravity (and thus the yaw axis) the yawing moment contribution is virtually nil.
 
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