On Wed, 23 Jun 2010 09:48:02 -0700, Winston
wrote:

15 PSIG distributed under a 22" diameter drum lid is
~5700 lbs of force.

http://www.use-enco.com/CGI/INPDFF?P...PARTPG=INLMK32

Any of these is likely to be far less dramatic
than pressurizing the drum.

--Winston

But you don't need nearly that much pressure. A 55 gal drum is about
35" high. 35" of water is a head of 1.26 PSIG. Oils have lower
specific gravity than water so will have less head for given height.

I plan to use a scheme like this for fuel transfer. No motors, no
sparks. The air reservoir will be an old fire extinguisher
pressurized to 120 PSIG, with a propane regulator set to about 1.5
PSIG.

What happens with a 5 gallon jerrycan with 1.5 PSIG inside? Well, the
vapor pressure of "summer" gasoline varies about 3.5 PSI from 50F to
90F, so a jerrycan certainly can handle 3.5 PSIG without problem.

Don Foreman wrote:

On Wed, 23 Jun 2010 09:48:02 -0700, Winston
wrote:

15 PSIG distributed under a 22" diameter drum lid is
~5700 lbs of force.

http://www.use-enco.com/CGI/INPDFF?P...PARTPG=INLMK32

Any of these is likely to be far less dramatic
than pressurizing the drum.

--Winston

But you don't need nearly that much pressure. A 55 gal drum is about
35" high. 35" of water is a head of 1.26 PSIG. Oils have lower
specific gravity than water so will have less head for given height.

I plan to use a scheme like this for fuel transfer. No motors, no
sparks. The air reservoir will be an old fire extinguisher
pressurized to 120 PSIG, with a propane regulator set to about 1.5
PSIG.

What happens with a 5 gallon jerrycan with 1.5 PSIG inside? Well, the
vapor pressure of "summer" gasoline varies about 3.5 PSI from 50F to
90F, so a jerrycan certainly can handle 3.5 PSIG without problem.

http://www.harborfreight.com/12-gpm-...ump-93755.html

Has 10psi relief valve, so presumably a standard 55 gal drum can handle
10psi.

On 6/23/2010 12:21 PM, Pete C. wrote:

Don Foreman wrote:

(...)

What happens with a 5 gallon jerrycan with 1.5 PSIG inside? Well, the
vapor pressure of "summer" gasoline varies about 3.5 PSI from 50F to
90F, so a jerrycan certainly can handle 3.5 PSIG without problem.

A 20 l jerrycan has about 10% the internal surface area of a 55 ggl drum.

http://www.harborfreight.com/12-gpm-...ump-93755.html

Has 10psi relief valve, so presumably a standard 55 gal drum can handle
10psi.

In cursory Googling, I've seen specifications from 5 PSIG to 43.5 PSIG.

As someone mentioned, there are economical pumps available that would
place insignificant pressure on the contents, though.
I would choose that safer alternative.

--Winston -- Needs to buy a new compressor to dispense coolant to
machine a part to repair the old compressor.
No. Wait....

On Wed, 23 Jun 2010 13:50:07 -0700, Winston
wrote:

On 6/23/2010 12:21 PM, Pete C. wrote:

Don Foreman wrote:

(...)

What happens with a 5 gallon jerrycan with 1.5 PSIG inside? Well, the
vapor pressure of "summer" gasoline varies about 3.5 PSI from 50F to
90F, so a jerrycan certainly can handle 3.5 PSIG without problem.

A 20 l jerrycan has about 10% the internal surface area of a 55 ggl drum.

The flat top & bottom of a 55 gal drum have areas of about 380 in^2.
The 13" x 18.5" flat sides of a jerrycan are 240 in^2. The
cylindrical part of a 55 gal drum could probably stand upwards of 100
PSIG, it's the flat surfaces and seams that dictate.

On 6/23/2010 3:57 PM, Don Foreman wrote:
On Wed, 23 Jun 2010 13:50:07 -0700,
wrote:

On 6/23/2010 12:21 PM, Pete C. wrote:

Don Foreman wrote:

(...)

What happens with a 5 gallon jerrycan with 1.5 PSIG inside? Well, the
vapor pressure of "summer" gasoline varies about 3.5 PSI from 50F to
90F, so a jerrycan certainly can handle 3.5 PSIG without problem.

A 20 l jerrycan has about 10% the internal surface area of a 55 ggl drum.

The flat top& bottom of a 55 gal drum have areas of about 380 in^2.
The 13" x 18.5" flat sides of a jerrycan are 240 in^2. The
cylindrical part of a 55 gal drum could probably stand upwards of 100
PSIG, it's the flat surfaces and seams that dictate.

Hokay but the drum still has 2.8 x the seam length of the jerrycan.

I stumbled across a claim that the flat top and bottom
of the drum go noticeably convex at only 5 PSIG.

Gives me the willies, it does.

I'd stay with a lever pump and keep the
'air over oil' pressure down around 0 PSIG.

The pump would be a great metalworking project.





--Winston

On Wed, 23 Jun 2010 22:07:50 -0700, Winston
wrote:



Hokay but the drum still has 2.8 x the seam length of the jerrycan.

I stumbled across a claim that the flat top and bottom
of the drum go noticeably convex at only 5 PSIG.

5 PSIG is about 11.5 feet of water head. Head necessary to get the
stuff out of the drum is only 35" or 1.26 PSIG. "Noticibly convex" is
markedly different from "BOOM" so about 0.252 NC (Noticably Convex)
seems safe enough to me.

Gives me the willies, it does.

Nothing wrong with being safe. Belt, suspenders and staples.

I'd stay with a lever pump and keep the
'air over oil' pressure down around 0 PSIG.

The pump would be a great metalworking project.

There is that.

I did get a bit of machining in with my fuel-transfer project, though.
Hadda make a plug to fit the defunct fire extinguisher, cut some
threads on the lathe. More later on that if interested. I have
photos back in wideband land.

On 6/23/2010 10:29 PM, Don Foreman wrote:
On Wed, 23 Jun 2010 22:07:50 -0700,
wrote:



Hokay but the drum still has 2.8 x the seam length of the jerrycan.

I stumbled across a claim that the flat top and bottom
of the drum go noticeably convex at only 5 PSIG.

5 PSIG is about 11.5 feet of water head. Head necessary to get the
stuff out of the drum is only 35" or 1.26 PSIG. "Noticibly convex" is
markedly different from "BOOM" so about 0.252 NC (Noticably Convex)
seems safe enough to me.

Gives me the willies, it does.

Nothing wrong with being safe. Belt, suspenders and staples.

I'd stay with a lever pump and keep the
'air over oil' pressure down around 0 PSIG.

The pump would be a great metalworking project.

There is that.

I did get a bit of machining in with my fuel-transfer project, though.
Hadda make a plug to fit the defunct fire extinguisher, cut some
threads on the lathe. More later on that if interested.

Yes please.
Especially your 120 PSIG pressurization scheme.

I have photos back in wideband land.

http://www.metalworking.com/dropbox/

^ Hint Hint

--Winston

On Thu, 24 Jun 2010 00:29:27 -0500, Don Foreman
wrote:

On Wed, 23 Jun 2010 22:07:50 -0700, Winston
wrote:



Hokay but the drum still has 2.8 x the seam length of the jerrycan.

That's a good thing. The total force is divided by the length of the
seam.


I stumbled across a claim that the flat top and bottom
of the drum go noticeably convex at only 5 PSIG.

5 PSIG is about 11.5 feet of water head. Head necessary to get the
stuff out of the drum is only 35" or 1.26 PSIG. "Noticibly convex" is
markedly different from "BOOM" so about 0.252 NC (Noticably Convex)
seems safe enough to me.

And the 35" of head represents the hydrostatic pressure that the
bottom head sees anyway when the drum is full. Add a few more inches
to get over the rim of the drum, and a bit more to get a reasonable
flow thru the spout, and I don't see a problem, conceptually anyway.
The real challenge is in making damn sure the drum is never
over-presssurized.

I could imagine using a tall standpipe of generous diameter as a
pressure relief. Not that I'd recommend anything as crazy as
pressurizing a 55 gallon drum. g

--
Ned Simmons

"Ned Simmons" wrote in message
news
On Thu, 24 Jun 2010 00:29:27 -0500, Don Foreman
wrote:

On Wed, 23 Jun 2010 22:07:50 -0700, Winston
wrote:



Hokay but the drum still has 2.8 x the seam length of the jerrycan.

That's a good thing. The total force is divided by the length of the
seam.


I stumbled across a claim that the flat top and bottom
of the drum go noticeably convex at only 5 PSIG.

5 PSIG is about 11.5 feet of water head. Head necessary to get the
stuff out of the drum is only 35" or 1.26 PSIG. "Noticibly convex" is
markedly different from "BOOM" so about 0.252 NC (Noticably Convex)
seems safe enough to me.

And the 35" of head represents the hydrostatic pressure that the
bottom head sees anyway when the drum is full. Add a few more inches
to get over the rim of the drum, and a bit more to get a reasonable
flow thru the spout, and I don't see a problem, conceptually anyway.
The real challenge is in making damn sure the drum is never
over-presssurized.

I could imagine using a tall standpipe of generous diameter as a
pressure relief. Not that I'd recommend anything as crazy as
pressurizing a 55 gallon drum. g

If I were to do this, I would make it so I would only hand-apply a rubber
tipped air nozzle, and eventually put a relief valve on, as well. And just
crack the ball valve on the compressor, to limit air flow.

I do something similar with a grease gun and motors.
I actually remove the grease fitting, and use a rubber air-hose-type
tip ( from a grease gun "kit" from HD, iirc), to grease the bearings.
Baldor told me to be careful about grease fittings/guns on motors, cuz they
siad it's relatively easy to rupture seals. So I use this method, where it
becomes very evident when things are packed.

Automotive joints are different, I would imagine.

So applying the air "by hand" will hopefully afford a similar safety hedge.
Unless I fall asleep.....
--
EA



--
Ned Simmons

On 6/24/2010 7:01 AM, Ned Simmons wrote:
On Thu, 24 Jun 2010 00:29:27 -0500, Don Foreman
wrote:

On Wed, 23 Jun 2010 22:07:50 -0700,
wrote:



Hokay but the drum still has 2.8 x the seam length of the jerrycan.

That's a good thing. The total force is divided by the length of the
seam.

More is really better?

I thought the seams represented the weakest link.

I stumbled across a claim that the flat top and bottom
of the drum go noticeably convex at only 5 PSIG.

5 PSIG is about 11.5 feet of water head. Head necessary to get the
stuff out of the drum is only 35" or 1.26 PSIG. "Noticibly convex" is
markedly different from "BOOM" so about 0.252 NC (Noticably Convex)
seems safe enough to me.

And the 35" of head represents the hydrostatic pressure that the
bottom head sees anyway when the drum is full. Add a few more inches
to get over the rim of the drum, and a bit more to get a reasonable
flow thru the spout, and I don't see a problem, conceptually anyway.
The real challenge is in making damn sure the drum is never
over-presssurized.

A failed gas regulator shouldn't cause anything worse than
minor disappointment. I am lazy by nature and would be
likely to leave the air connected. This approach would
make me nervous.


I could imagine using a tall standpipe of generous diameter as a
pressure relief. Not that I'd recommend anything as crazy as
pressurizing a 55 gallon drum.g

When a drum pump fails, you get no product.
When a regulator fails, you get product *everywhere*.

--Winston

On Thu, 24 Jun 2010 11:01:48 -0700, Winston
wrote:

On 6/24/2010 7:01 AM, Ned Simmons wrote:
On Thu, 24 Jun 2010 00:29:27 -0500, Don Foreman
wrote:

On Wed, 23 Jun 2010 22:07:50 -0700,
wrote:



Hokay but the drum still has 2.8 x the seam length of the jerrycan.

That's a good thing. The total force is divided by the length of the
seam.

More is really better?

In this case, yes.


I thought the seams represented the weakest link.

I don't know that for a fact, but let's assume they are. Two inches of
weak seam will resist more force than one inch.


I stumbled across a claim that the flat top and bottom
of the drum go noticeably convex at only 5 PSIG.

5 PSIG is about 11.5 feet of water head. Head necessary to get the
stuff out of the drum is only 35" or 1.26 PSIG. "Noticibly convex" is
markedly different from "BOOM" so about 0.252 NC (Noticably Convex)
seems safe enough to me.

And the 35" of head represents the hydrostatic pressure that the
bottom head sees anyway when the drum is full. Add a few more inches
to get over the rim of the drum, and a bit more to get a reasonable
flow thru the spout, and I don't see a problem, conceptually anyway.
The real challenge is in making damn sure the drum is never
over-presssurized.

A failed gas regulator shouldn't cause anything worse than
minor disappointment. I am lazy by nature and would be
likely to leave the air connected. This approach would
make me nervous.

I think the proposal is to meter the air in, not the liquid out. The
spout is unrestricted and the liquid runs out as long as there's
enough pressure in the drum to overcome the head that results from the
highest point in the plumbing.


--
Ned Simmons

Ned Simmons wrote:

On Thu, 24 Jun 2010 11:01:48 -0700, Winston
wrote:

On 6/24/2010 7:01 AM, Ned Simmons wrote:
On Thu, 24 Jun 2010 00:29:27 -0500, Don Foreman
wrote:

On Wed, 23 Jun 2010 22:07:50 -0700,
wrote:



Hokay but the drum still has 2.8 x the seam length of the jerrycan.

That's a good thing. The total force is divided by the length of the
seam.

More is really better?

In this case, yes.


I thought the seams represented the weakest link.

I don't know that for a fact, but let's assume they are. Two inches of
weak seam will resist more force than one inch.


I stumbled across a claim that the flat top and bottom
of the drum go noticeably convex at only 5 PSIG.

5 PSIG is about 11.5 feet of water head. Head necessary to get the
stuff out of the drum is only 35" or 1.26 PSIG. "Noticibly convex" is
markedly different from "BOOM" so about 0.252 NC (Noticably Convex)
seems safe enough to me.

And the 35" of head represents the hydrostatic pressure that the
bottom head sees anyway when the drum is full. Add a few more inches
to get over the rim of the drum, and a bit more to get a reasonable
flow thru the spout, and I don't see a problem, conceptually anyway.
The real challenge is in making damn sure the drum is never
over-presssurized.

A failed gas regulator shouldn't cause anything worse than
minor disappointment. I am lazy by nature and would be
likely to leave the air connected. This approach would
make me nervous.

I think the proposal is to meter the air in, not the liquid out. The
spout is unrestricted and the liquid runs out as long as there's
enough pressure in the drum to overcome the head that results from the
highest point in the plumbing.

Just get the Harbor Freight air operated piston pump, they are designed
to be output throttled i.e. oil dispenser guns. Close off the outlet and
the air piston just stalls, the fluid line is pressurized ready to
dispense again. The drum is never pressurized.

On Wed, 23 Jun 2010 22:41:22 -0700, Winston
wrote:

On 6/23/2010 10:29 PM, Don Foreman wrote:
On Wed, 23 Jun 2010 22:07:50 -0700,
wrote:



Hokay but the drum still has 2.8 x the seam length of the jerrycan.

I stumbled across a claim that the flat top and bottom
of the drum go noticeably convex at only 5 PSIG.

5 PSIG is about 11.5 feet of water head. Head necessary to get the
stuff out of the drum is only 35" or 1.26 PSIG. "Noticibly convex" is
markedly different from "BOOM" so about 0.252 NC (Noticably Convex)
seems safe enough to me.

Gives me the willies, it does.

Nothing wrong with being safe. Belt, suspenders and staples.

I'd stay with a lever pump and keep the
'air over oil' pressure down around 0 PSIG.

The pump would be a great metalworking project.

There is that.

I did get a bit of machining in with my fuel-transfer project, though.
Hadda make a plug to fit the defunct fire extinguisher, cut some
threads on the lathe. More later on that if interested.

Yes please.
Especially your 120 PSIG pressurization scheme.

I have photos back in wideband land.

http://www.metalworking.com/dropbox/

^ Hint Hint

--Winston

Here's what I have so far:

http://members.goldengate.net/dforeman/ext_plug/

On 6/24/2010 9:07 PM, Don Foreman wrote:

Here's what I have so far:

http://members.goldengate.net/dforeman/ext_plug/

Very pretty, Don.

It looks unnatural without the obligatory
burrs and strings hanging off at odd angles though.

--Winston -- 'Way less brave than Don.

On 6/24/2010 5:39 PM, Pete C. wrote:

(...)

Just get the Harbor Freight air operated piston pump, they are designed
to be output throttled i.e. oil dispenser guns. Close off the outlet and
the air piston just stalls, the fluid line is pressurized ready to
dispense again. The drum is never pressurized.

http://www.harborfreight.com/air-ope...ipe-43575.html

That'll work!


--Winston

On 6/24/2010 5:21 PM, Ned Simmons wrote:
On Thu, 24 Jun 2010 11:01:48 -0700,
wrote:

On 6/24/2010 7:01 AM, Ned Simmons wrote:
On Thu, 24 Jun 2010 00:29:27 -0500, Don Foreman
wrote:

On Wed, 23 Jun 2010 22:07:50 -0700,
wrote:

(...)

Hokay but the drum still has 2.8 x the seam length of the jerrycan.

That's a good thing. The total force is divided by the length of the
seam.

More is really better?

In this case, yes.


I thought the seams represented the weakest link.

I don't know that for a fact, but let's assume they are. Two inches of
weak seam will resist more force than one inch.

The force per inch is the same in both cases, yes?

I don't grok.

--Winston

On Thu, 24 Jun 2010 10:01:43 -0400, Ned Simmons
wrote:

On Thu, 24 Jun 2010 00:29:27 -0500, Don Foreman
wrote:

On Wed, 23 Jun 2010 22:07:50 -0700, Winston
wrote:



Hokay but the drum still has 2.8 x the seam length of the jerrycan.

That's a good thing. The total force is divided by the length of the
seam.


I stumbled across a claim that the flat top and bottom
of the drum go noticeably convex at only 5 PSIG.

5 PSIG is about 11.5 feet of water head. Head necessary to get the
stuff out of the drum is only 35" or 1.26 PSIG. "Noticibly convex" is
markedly different from "BOOM" so about 0.252 NC (Noticably Convex)
seems safe enough to me.

And the 35" of head represents the hydrostatic pressure that the
bottom head sees anyway when the drum is full. Add a few more inches
to get over the rim of the drum, and a bit more to get a reasonable
flow thru the spout, and I don't see a problem, conceptually anyway.
The real challenge is in making damn sure the drum is never
over-presssurized.

I could imagine using a tall standpipe of generous diameter as a
pressure relief. Not that I'd recommend anything as crazy as
pressurizing a 55 gallon drum. g


I think that the most logical answer is that pressurizing the drum is
not a common solution. Which seems to say that either the rest of the
world is not bright enough to think of the idea; or perhaps the one
lonely guy that did is in error :-)

Cheers,

John D. Slocomb
(jdslocombatgmail)

On 6/25/2010 4:17 AM, J. D. Slocomb wrote:

(...)

I think that the most logical answer is that pressurizing the drum is
not a common solution. Which seems to say that either the rest of the
world is not bright enough to think of the idea; or perhaps the one
lonely guy that did is in error :-)

If we pressurize a nearly empty fuel tank with 20%
oxygen, aren't we in danger of doing an impromptu
stoichiometry experiment?:
http://en.wikipedia.org/wiki/Inerting_system

That would be an error, yes?

--Winston

On Fri, 25 Jun 2010 06:38:23 -0700, Winston
wrote:

On 6/25/2010 4:17 AM, J. D. Slocomb wrote:

(...)

I think that the most logical answer is that pressurizing the drum is
not a common solution. Which seems to say that either the rest of the
world is not bright enough to think of the idea; or perhaps the one
lonely guy that did is in error :-)

If we pressurize a nearly empty fuel tank with 20%
oxygen, aren't we in danger of doing an impromptu
stoichiometry experiment?:
http://en.wikipedia.org/wiki/Inerting_system

That would be an error, yes?

--Winston

A spark or heat is still needed to start a reaction, and if a spark
occurs then even the unpressurized tank will blow.

On 6/25/2010 2:50 PM, Don Foreman wrote:

(...)

A spark or heat is still needed to start a reaction, and if a spark
occurs then even the unpressurized tank will blow.

Not if there is a layer of CO2 floating
on the fuel, yes?

Fuel, Oxygen, Flame

I'm for limiting #2 and #3 when working with #1.

--Winston

On Thu, 24 Jun 2010 21:54:18 -0700, Winston
wrote:

On 6/24/2010 5:21 PM, Ned Simmons wrote:
On Thu, 24 Jun 2010 11:01:48 -0700,
wrote:

On 6/24/2010 7:01 AM, Ned Simmons wrote:
On Thu, 24 Jun 2010 00:29:27 -0500, Don Foreman
wrote:

On Wed, 23 Jun 2010 22:07:50 -0700,
wrote:

(...)

Hokay but the drum still has 2.8 x the seam length of the jerrycan.

That's a good thing. The total force is divided by the length of the
seam.

More is really better?

In this case, yes.


I thought the seams represented the weakest link.

I don't know that for a fact, but let's assume they are. Two inches of
weak seam will resist more force than one inch.

The force per inch is the same in both cases, yes?

I don't grok.


Perhaps we have different cases in mind. My point is only that the if
the pressure and area remain constant while the seam length increases,
the force per unit length of seam will decrease. Clearly this is
meaningless if we're limited to cylindrical drums, because there's a
fixed relationship between head diameter and circumference.

force/length of seam = pressure x area / seam length
or
lb/in = lb/in^2 x in^2 / in


--
Ned Simmons

Here's what I was on about:

Envision a barrel 22" diameter by 35" high. Let's agree I pressurized
that barrel to 10 PSIG.

Let's arbitrarily say that all seams measure 0.05" in width.
We have a total seam length of ~173" (Two circumferences plus the side
seam). So 173" of seam length x 0.05" of seam width is 8.66 square inches
of total seam area. If we multiply that by our 10 PSIG we get
a force of 86.6 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?

Envision that same barrel only this time, we now have two side
seams and four circumferential seams (instead of just two).
Our new total seam length is now 346.5" for a total seam area
of ~17.3 sq. inches. If we multiply that by our 10 PSIG we get
a force of ~173.2 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?



Where did I go wrong?


--Winston -- I mean, concerning 'seam pressure'.

On Fri, 25 Jun 2010 22:48:29 -0700, Winston
wrote:

Here's what I was on about:

Envision a barrel 22" diameter by 35" high. Let's agree I pressurized
that barrel to 10 PSIG.

Let's arbitrarily say that all seams measure 0.05" in width.
We have a total seam length of ~173" (Two circumferences plus the side
seam). So 173" of seam length x 0.05" of seam width is 8.66 square inches
of total seam area. If we multiply that by our 10 PSIG we get
a force of 86.6 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?

Envision that same barrel only this time, we now have two side
seams and four circumferential seams (instead of just two).
Our new total seam length is now 346.5" for a total seam area
of ~17.3 sq. inches. If we multiply that by our 10 PSIG we get
a force of ~173.2 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?



Where did I go wrong?


--Winston -- I mean, concerning 'seam pressure'.

In a closed cylinder pressure is applied to all surfaces equally. so
you have an area of 2,419 sq. inches with a pressure of 10 PSIG =
24,190 pounds of force applied to the inside the vessel.

How much of this force is trying to rip the seams apart?

Cheers,

John D. Slocomb
(jdslocombatgmail)

On 6/26/2010 5:00 AM, J. D. Slocomb wrote:
On Fri, 25 Jun 2010 22:48:29 -0700,
wrote:

Here's what I was on about:

Envision a barrel 22" diameter by 35" high. Let's agree I pressurized
that barrel to 10 PSIG.

Let's arbitrarily say that all seams measure 0.05" in width.
We have a total seam length of ~173" (Two circumferences plus the side
seam). So 173" of seam length x 0.05" of seam width is 8.66 square inches
of total seam area. If we multiply that by our 10 PSIG we get
a force of 86.6 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?

Envision that same barrel only this time, we now have two side
seams and four circumferential seams (instead of just two).
Our new total seam length is now 346.5" for a total seam area
of ~17.3 sq. inches. If we multiply that by our 10 PSIG we get
a force of ~173.2 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?



Where did I go wrong?


--Winston-- I mean, concerning 'seam pressure'.

In a closed cylinder pressure is applied to all surfaces equally. so
you have an area of 2,419 sq. inches with a pressure of 10 PSIG =
24,190 pounds of force applied to the inside the vessel.

How much of this force is trying to rip the seams apart?

I agree the problem is more complicated than my arithmetic
suggests because I did not consider the most important aspect
and that is the leverage against the top and bottom seams
by the bowed flat surfaces.

The point of my arithmetic was to discover how much normal
force was placed on a seam of a given length and how that
force changed as seam length increased. Apparently it's
linear (for the side seam anyway!)

--Winston

On Fri, 25 Jun 2010 22:48:29 -0700, Winston
wrote:

Here's what I was on about:

Envision a barrel 22" diameter by 35" high. Let's agree I pressurized
that barrel to 10 PSIG.

Let's arbitrarily say that all seams measure 0.05" in width.
We have a total seam length of ~173" (Two circumferences plus the side
seam). So 173" of seam length x 0.05" of seam width is 8.66 square inches
of total seam area. If we multiply that by our 10 PSIG we get
a force of 86.6 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?

Envision that same barrel only this time, we now have two side
seams and four circumferential seams (instead of just two).
Our new total seam length is now 346.5" for a total seam area
of ~17.3 sq. inches. If we multiply that by our 10 PSIG we get
a force of ~173.2 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?



Where did I go wrong?

The force per unit length resisted by a head seam is pressure times
the area of the *head* divided by the length of the seam. In the
example above:
10 lb/in^2 x 380 in^2 / 69 in = 55 lb/in

The stress in the cylindrical part of the barrel is equal to pressure
x radius / wall thickness -- the classical formula for stress in a
cylindrical shell. Remove the wall thickness and you get the force per
unit length in that seam:
10 lb/in^2 x 11 in = 110 lb/in


--
Ned Simmons

On Sat, 26 Jun 2010 07:47:41 -0700, Winston
wrote:

On 6/26/2010 5:00 AM, J. D. Slocomb wrote:
On Fri, 25 Jun 2010 22:48:29 -0700,
wrote:

Here's what I was on about:

Envision a barrel 22" diameter by 35" high. Let's agree I pressurized
that barrel to 10 PSIG.

Let's arbitrarily say that all seams measure 0.05" in width.
We have a total seam length of ~173" (Two circumferences plus the side
seam). So 173" of seam length x 0.05" of seam width is 8.66 square inches
of total seam area. If we multiply that by our 10 PSIG we get
a force of 86.6 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?

Envision that same barrel only this time, we now have two side
seams and four circumferential seams (instead of just two).
Our new total seam length is now 346.5" for a total seam area
of ~17.3 sq. inches. If we multiply that by our 10 PSIG we get
a force of ~173.2 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?



Where did I go wrong?


--Winston-- I mean, concerning 'seam pressure'.

In a closed cylinder pressure is applied to all surfaces equally. so
you have an area of 2,419 sq. inches with a pressure of 10 PSIG =
24,190 pounds of force applied to the inside the vessel.

How much of this force is trying to rip the seams apart?

I agree the problem is more complicated than my arithmetic
suggests because I did not consider the most important aspect
and that is the leverage against the top and bottom seams
by the bowed flat surfaces.

The point of my arithmetic was to discover how much normal
force was placed on a seam of a given length and how that
force changed as seam length increased. Apparently it's
linear (for the side seam anyway!)

--Winston


I was only trying to illustrate that the force trying to separate the
seam is not simply the force applied to the seam itself by the 10 PSI.

Years ago, in the USAF I watched some fuel cell guys build a box of
several cubic feet with a circular cut-out in one side. They mounted a
pressure relief valve in one cut-out side of the box and screwed an
air fitting in the other. I asked them what they were doing and they
said that they needed to actually see the valve operate when pressure
was only a few inches of water pressure high. I wouldn't let them use
my air system to do it so they went over to an unused hanger and set
up there.

A bit later the Sergeant in charge of the test came back and said,
"well I didn't believe you but you made so much noise about it that we
put the box out in the middle of the hanger and all went outside.
Reached around the door and started to open the air valve, just a tiny
bit. the box blew up before the air gage even started to wiggle.

Low PSI in big boxes (or cylinders) is sometimes more potent then it
appears simply from looking at the gage.

Cheers,

John D. Slocomb
(jdslocombatgmail)

On 6/26/2010 8:57 PM, Ned Simmons wrote:
On Fri, 25 Jun 2010 22:48:29 -0700,
wrote:

Here's what I was on about:

Envision a barrel 22" diameter by 35" high. Let's agree I pressurized
that barrel to 10 PSIG.

Let's arbitrarily say that all seams measure 0.05" in width.
We have a total seam length of ~173" (Two circumferences plus the side
seam). So 173" of seam length x 0.05" of seam width is 8.66 square inches
of total seam area. If we multiply that by our 10 PSIG we get
a force of 86.6 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?

Envision that same barrel only this time, we now have two side
seams and four circumferential seams (instead of just two).
Our new total seam length is now 346.5" for a total seam area
of ~17.3 sq. inches. If we multiply that by our 10 PSIG we get
a force of ~173.2 lbs acting against all the seams collectively.
If we divide that force by the seam length, we decide that the force
per inch of seam is about 0.5 lb, yes?



Where did I go wrong?

The force per unit length resisted by a head seam is pressure times
the area of the *head* divided by the length of the seam. In the
example above:
10 lb/in^2 x 380 in^2 / 69 in = 55 lb/in

The stress in the cylindrical part of the barrel is equal to pressure
x radius / wall thickness -- the classical formula for stress in a
cylindrical shell. Remove the wall thickness and you get the force per
unit length in that seam:
10 lb/in^2 x 11 in = 110 lb/in

Sounds like the issue is much more serious than I thought.

Thanks!

--Winston

On 6/26/2010 9:41 PM, J. D. Slocomb wrote:

(...)

I was only trying to illustrate that the force trying to separate the
seam is not simply the force applied to the seam itself by the 10 PSI.

(...)

Low PSI in big boxes (or cylinders) is sometimes more potent then it
appears simply from looking at the gage.

I *knew* there was a reason for the hairs standing
on the back of my neck when we started talking about
pressurizing that drum.

Thanks!

--Winston

On Sat, 26 Jun 2010 22:06:49 -0700, Winston
wrote:

On 6/26/2010 8:57 PM, Ned Simmons wrote:


The force per unit length resisted by a head seam is pressure times
the area of the *head* divided by the length of the seam. In the
example above:
10 lb/in^2 x 380 in^2 / 69 in = 55 lb/in

The stress in the cylindrical part of the barrel is equal to pressure
x radius / wall thickness -- the classical formula for stress in a
cylindrical shell. Remove the wall thickness and you get the force per
unit length in that seam:
10 lb/in^2 x 11 in = 110 lb/in

Sounds like the issue is much more serious than I thought.


There may be more stress on the welds than you thought, but it's
unlikely EA's 5 psi came anywhere close to overstressing them. The
vapor pressure of gasoline at room temp varies from 4 to almost 10 psi
at 70F, depending on the formulation. So the pressure in a sealed drum
containing gasoline could easily be twice EA's 5 psi.

--
Ned Simmons

On 6/27/2010 8:47 PM, Ned Simmons wrote:

(...)

There may be more stress on the welds than you thought, but it's
unlikely EA's 5 psi came anywhere close to overstressing them. The
vapor pressure of gasoline at room temp varies from 4 to almost 10 psi
at 70F, depending on the formulation. So the pressure in a sealed drum
containing gasoline could easily be twice EA's 5 psi.

Doubtlessly, *some* newsgroup'ers will misinterpret this thread
to mean that it is OK to pressurize a 55 ggl drum ca. 5 PSIG
so that they dispense via a valve the output side. (Try precisely
metering a fluid using a compressible gas sometime.)

When in reality, that is exactly the corner you do not want to
be in. I know that in electronics, we look 'hard and long' at
a potential design solution that 'burns up safety margin'.

Instead, we can normally accomplish all the design goals without
decreasing safety or reliability.

In this case, a barrel pump or 'vacuuum-start siphon' are both
inexpensive, effective ways to safely handle liquids in these
large drums.

That's my story and I'm stick'n to it.

--Winston

On Mon, 28 Jun 2010 06:38:05 -0700, Winston
wrote:
In this case, a barrel pump or 'vacuuum-start siphon' are both
inexpensive, effective ways to safely handle liquids in these
large drums.

That's my story and I'm stick'n to it.

Even simpler is to put a tap in the small bung, tip the drum on its
side on some blocks to raise it above ground level so a bucket will
fit under the tap. You may have to crack the large bung to allow air
into the drum. Did it all the time at work.

Alan

On 6/29/2010 7:58 AM, wrote:
On Mon, 28 Jun 2010 06:38:05 -0700,
wrote:
In this case, a barrel pump or 'vacuuum-start siphon' are both
inexpensive, effective ways to safely handle liquids in these
large drums.

That's my story and I'm stick'n to it.

Even simpler is to put a tap in the small bung, tip the drum on its
side on some blocks to raise it above ground level so a bucket will
fit under the tap. You may have to crack the large bung to allow air
into the drum. Did it all the time at work.

Yup!

A barrel cart would make that easy.

--Winston