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Physics question for all you engineers, etc.

dexterq20

dexterq20

Turbo Monkey
Mar 6, 2003
3,442
1
NorCal
Here's the problem:

"A hot dog on a grill can be thought of as a cylindrical pressure vessel. Explain in which orientation (direction) the hot dog will fail when the pressure is increased through cooking and why this is so."

My roommate and I can't figure it out. Any good explanations?

EDIT: Serious answers would be appreciated.
 
B

blue

boob hater
Jan 24, 2004
10,160
2
california
I'm confused.

I'd guess outward due to the moisture content of the meat itself and the non-moisture content of the "skin", aka sausage tubing.
 
Toshi

Toshi

butthole powerwashing evangelist
Oct 23, 2001
39,388
8,479
the pressure will be equal everywhere due to our friend pascal's law, so it must be some special property of pressurized cylinders. it must have been mentioned in class, check the notes...
 
Westy

Westy

the teste
Nov 22, 2002
55,728
21,750
Sleazattle
I can speak from years of experience in blowing things up, the hotdog will fail with a lengthwise crack.

But just think of it this way, a spherical vessel has equal forces in all directions but a cylinder has more stress radially. Imagine a very long cylinder with a very small diameter, it will have much more stress radially than along its axis.

Now excuse me while I polish my pocket protector and dust my graphing calculator.
 
Toshi

Toshi

butthole powerwashing evangelist
Oct 23, 2001
39,388
8,479
Westy said:
But just think of it this way, a spherical vessel has equal forces in all directions but a cylinder has more stress radially. Imagine a very long cylinder with a very small diameter, it will have much more stress radially than along its axis.
Click to expand...
why is this so? it seems (somewhat) intuitively obvious, but that's not the path to problem set success...
 
firetoole

firetoole

duch bag
Nov 19, 2004
1,910
0
Wooo Tulips!!!!
Ok here’s the deal, if they are talking about cylindrical pressured objects in general. Most are designed to fail at the ends an example of this is large propane tanks will almost never fail on their sides. But the ends will blow off quite easily.
 
Denny

Denny

Chimp
Aug 27, 2002
96
0
Seattle, WA
In cylindrical pressure vessels, the two stresses are "Hoop Stress" (splitting along the length), and "Axial Stress" (splitting around the circumference).

The equations are:
Hoop stress = P*D/(2*t)
Axial Stress = P*A/(pi*D*t)

P = internal pressure
D = diameter
t = thickness of wall / skin
A = cross sectional area of circular section

Since A = pi * D^2 /4
The axial stress equation simplifies to:

Axial Stress = P * D / 4*t

So, by definition, Hoop Stress = 2 * Axial Stress.

This concludes your weekly pressure vessel safety announcement.
 
N8 v2.0

N8 v2.0

Not the sharpest tool in the shed
Oct 18, 2002
11,003
149
The Cleft of Venus
Denny said:
In cylindrical pressure vessels, the two stresses are "Hoop Stress" (splitting along the length), and "Axial Stress" (splitting around the circumference).

The equations are:
Hoop stress = P*D/(2*t)
Axial Stress = P*A/(pi*D*t)

P = internal pressure
D = diameter
t = thickness of wall / skin
A = cross sectional area of circular section

Since A = pi * D^2 /4
The axial stress equation simplifies to:

Axial Stress = P * D / 4*t

So, by definition, Hoop Stress = 2 * Axial Stress.

This concludes your weekly pressure vessel safety announcement.
Click to expand...
Interesting.... I did not know this...
 
Pau11y

Pau11y

Turbo Monkey
Oct 20, 2004
1,968
0
In a web in the corner of your room
Something entirely off topic:
I bit into a hotdog last year to find a green vein. I'm sure you can picture me dry heaving (I was hungry and on an empty stomach) in the corner over a waste basket. I will NEVER eat another hotdog! :dead: Brats, however, is another story :D
 
Psychic_Pimp

Psychic_Pimp

Chimp
Sep 2, 2003
95
0
On the corner pimpin' psychics
Cool, I have seen a bunch of CO2 cylinders that blew up in ships' fire extinguisher systems. Those that didn't blow off the valves split down the side. Now I know why. They also blew open hatches and bulged steel walls, and I'm sure scared the crap out of several unsuspecting crewmen. Exploding high-pressure cylinders are scary mo-fos. Haven't seen propane cylinders go, they are lower pressure and a bit different construction than what I dealt with tho.
 
allsk8sno

allsk8sno

Turbo Monkey
Jun 6, 2002
1,153
33
Bellingham, WA
hotdogs are in a sausage like casing...the weakness/closure point is the ends... pressure vessel analysis is kinda useless in this case
though your analysis comes up with the correct answer
 
Smelly

Smelly

Turbo Monkey
Jun 17, 2004
1,254
1
out yonder, round bout a hootinany
Pau11y said:
Something entirely off topic:
I bit into a hotdog last year to find a green vein. I'm sure you can picture me dry heaving (I was hungry and on an empty stomach) in the corner over a waste basket. I will NEVER eat another hotdog! :dead: Brats, however, is another story :D
Click to expand...
grossest post of the day...
bleck!
 
SkaredShtles

SkaredShtles

Michael Bolton
Sep 21, 2003
67,305
13,869
In a van.... down by the river
Pau11y said:
Something entirely off topic:
I bit into a hotdog last year to find a green vein. I'm sure you can picture me dry heaving (I was hungry and on an empty stomach) in the corner over a waste basket. I will NEVER eat another hotdog! :dead: Brats, however, is another story :D
Click to expand...
This calls for a photo! Mmmm....... black sausage........ :drool:



Or as the German's call it: blutwurst:



-S.S.-
 
Westy

Westy

the teste
Nov 22, 2002
55,728
21,750
Sleazattle
Toshi said:
why is this so? it seems (somewhat) intuitively obvious, but that's not the path to problem set success...
Click to expand...

Denny has already filled us in with the actual formulae but the way you get them is to look at cross sections of a cylinder in both directions.

To look at the axial stress figure out the force pressed against the ends = pressure(pi)R^2. Now divide that by the cross sectional area of a cylinder. For simplicity sake we will examine a cyclinder with 0 thickness so we come up with force/distance instead of force/distance^2. We will assume the pressure to be 1force unit/distance unit^2. The resulting stress would be (1force/distance^2 unit)(pi)R^2/2(pi)R which eqyals 1R/2 (force/distance units) where R= the radius of the vessel.

Now for the axial cross section the resulting stress in similar fake units under the same fake 1force/distance^2 pressure. The stress ends up being (1 force unit/distance^2)(2RL)/(2L) which equals R(force/distance unitw) where L is the length of the pressure vessel.

From this, as Denny previously stated proves that the axial stress will be one half that of the radial or hoop stress.

The sad thing is that while sitting in traffic today I actually sat there and though about this. I was giddy with excitement when I found Denny had posted the actual formulas that agreed with my assumptions.
 
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