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Old 06-01-2015, 09:30 PM   #41
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Ooops. I forgot to include the link. Here it is. https://diracseashore.wordpress.com/...ssure-cookers/

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Old 06-01-2015, 09:40 PM   #42
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Quote:
Originally Posted by Selkie View Post
Gravity has nothing to do with it. Atmospheric pressure altitude is the determining factor and is more significant than most people realize.
When compared with sea level, a person at 5,000 ft. altitude experiences a pressure loss of 15%. At 10,000 ft. you've lost 30%, so instead of your 11.5 pounds of internal pressure of your pressure cooker, you're only cooking with 8.05 pounds of pressure before the relief valve begins to open.
That makes no physical sense. In the Chief's example, it's weight pure and simple, and that weight is unaffected by changes in atmospheric pressure. The vent hole is too small and the pressure applied to that point by the controlling weight too great to be affected as you say.

Your theory would only apply to a valve controlled by a flexible diaphragm where pressure differential could be the determining factor. This would make using the cooker more difficult because of the differing way that it would cook at different elevations.
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Old 06-01-2015, 09:50 PM   #43
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This guy doesn't take into account the change in external atmospheric pressure pressing the escape value weight against the valve hole. His calculations don't compensate for change (sea level vs 10,000 ft.). They only speak to a static setup.
As both a pilot and a submariner, I have just a little experience with changing pressure differentials, not that I claim to be without fault. It has been a while since I had work a problem like this, and my aging brain is admittedly rusty.
Now it's time for me to go soak my head! Where's my scotch?
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Old 06-01-2015, 10:51 PM   #44
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Originally Posted by roadfix View Post
RPC.......But if the ambient pressure is lower then there is less pressure "weighing" that plug down against the relief hole, I would think....
At 12,000 feet it would be 5.35 psi or 36 % less.
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Old 06-01-2015, 11:09 PM   #45
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Quote:
Originally Posted by Selkie View Post
This guy doesn't take into account the change in external atmospheric pressure pressing the escape value weight against the valve hole. His calculations don't compensate for change (sea level vs 10,000 ft.). They only speak to a static setup.
As both a pilot and a submariner, I have just a little experience with changing pressure differentials, not that I claim to be without fault. It has been a while since I had work a problem like this, and my aging brain is admittedly rusty.
Now it's time for me to go soak my head! Where's my scotch?
I too worked on submersibles, the U.S. Navy DSRV. Yes the pressure exerted on the hull of a submersible does affect the inside pressure of the vessel, just as elevation requires aircraft to compensate by pressurizing the cabin. However, in both cases, the inner pressure is compensated to a degree by the rigidity of the vessel walls. The inside environment of the Steel spheres of the DSRV resist the intense pressure exerted by the outside water at great depths. This allows the occupants to stay in an atmospheric environment that will not crush them, or even cause excess gasses to dissolve in their bodily fluids. This allows them to safely travel to the surface with no ill effects.

This pressure mechanism is independent of the outside pressure forces, within allowable limits, which is determined by the strength of the materials used to make the spheres.

A scuba diver, on the other had, is subject to the full force of the water pressure. His tank remains at an internal pressure, when full, of 3000 lbs. until he starts breathing. The airflow to him through the regulator is dependant on the water pressure, which when the ambient pressure increases, it presses on a diaphragm, or valve to open the orifice more, and allow a greater pressure of air to flow to the diver. This provides sufficient pressure to allow his/her lungs to inflate properly, even though there is more pressure exerted by the body.

Like the submarine, the PC is a vessel designed to capture and maintain a pressure load, regulated by a weight that requires a certain amount of pressure to lift it and allow steam to escape. Notice that pressure cookers are not designed to work horizontally, but the valve points upward. The weight of the regulator presses down over a tiny orifice, trapping the air inside the PC. When sufficient pressure is achieved to lift the regulator from that orifice, pressure is released. The regulator weight is literally dancing above the orifice on a column of upward moving steam. If more weight is applied to the regulator cap, it takes more pressure to lift it. That's why I said that the main forces that create pressure in the pot are the excitation of liquid, and heating vapor working against the weight, which is a function of gravity and mass. I thus propose that the temperature inside the PC is independent of atmospheric pressure, as that pressure is exerted from every angle, and does not press the weight against the release orifice.

And yes, I dove as well as working on the DSRV. Neither activity qualifies me as a pressure expert. Selkie, if you can produce recognized documents to support your theory, I will step back and state that I was wrong and learned something new.

As to the original question posted by the OP, Check your pressure cooker to verify that all seals, and valves are working correctly. You may have to turn in your new pressure cooker as defective, and obtain a replacement. Again, if you know anyone with a pressure cooker in your area, that would be where the best advice would come from.

One more bit of factual info to support my argument. The weight of air is greater the closer you get to the center of the Earth. The heavier air has greater density that the air above it. When a ballon is filled with hot air, the density of the air is less than that of the air around it. If air pressure pushed downward, the balloon would stay down, as the air around it is heavier. Instead, the more dense air at the bottom pushes the balloon upward. This phenomenon is called bouancy. The air at the bottom of the weight is actually pushing the regulator upward, but with negligible force. The lesser pressure exerted at higher altitudes still works the same way as the air at the bottom of the weight is more dense than the air above the weight.

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Old 06-01-2015, 11:54 PM   #46
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Originally Posted by powerplantop View Post
At 12,000 feet it would be 5.35 psi or 36 % less.
After rereading this I need to clarify The pressure goes to 9.35 PSI or 5.35 PSI less.

Where this pressure pushing down on the relief will be lower, I think that gravity will be the main thing causing the pressure.
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Old 06-02-2015, 01:16 AM   #47
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Originally Posted by powerplantop View Post
After rereading this I need to clarify The pressure goes to 9.35 PSI or 5.35 PSI less.

Where this pressure pushing down on the relief will be lower, I think that gravity will be the main thing causing the pressure.
While the pressure is in pounds per square inch in the US (atmospheric pressure is 14.7 psi at sea level to 10.5 psi at 10000 feet), the actual surface area that is affected is the area of the relief hole in the lid. That is a tiny fraction of the one square inch that atmospheric pressure is figured in. That tiny area on the top of the pressure controller which is affected by any pressure change is insignificant when compared to the gravitational effect pulling down on the heavy controller. It would result in an effective change of no more than a gram or two, maybe a lot less than that - I don't know how to do the math involved. Certainly not enough to have any effect on the food being cooked.
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Old 06-02-2015, 01:33 AM   #48
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I think the pressure inside the pressure cooker has to be the differential between it and the outside pressure. At sea level air pressure ≈14.7 pounds-force per square inch (psi). Since most pressure cookers can be set to 5 pressure, this would make no sense if it were not the differential. It's has to mean 5 lbs more than the external pressure.
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Old 06-02-2015, 01:45 AM   #49
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Quote:
Originally Posted by RPCookin View Post
While the pressure is in pounds per square inch in the US (atmospheric pressure is 14.7 psi at sea level to 10.5 psi at 10000 feet), the actual surface area that is affected is the area of the relief hole in the lid. That is a tiny fraction of the one square inch that atmospheric pressure is figured in. That tiny area on the top of the pressure controller which is affected by any pressure change is insignificant when compared to the gravitational effect pulling down on the heavy controller. It would result in an effective change of no more than a gram or two, maybe a lot less than that - I don't know how to do the math involved. Certainly not enough to have any effect on the food being cooked.
The surface area affected by atmospheric pressure would be on the surface of the weight not the hole. But that's still only about a 1 inch area.

The size of the hole only affects how much internal pressure is applied to the weight to overcome the total forces holding it down.

All that said when you look at the temp of the water on a steam table 1 or 2 psi change inside the vessel will only give a small change in temp.

But the vessel will start at a lower pressure and the water will boil at a lower temp. But once the weight starts to lift it should be very close to the same pressure / temp as sea level.
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Old 06-02-2015, 01:58 AM   #50
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Originally Posted by taxlady View Post
I think the pressure inside the pressure cooker has to be the differential between it and the outside pressure. At sea level air pressure ≈14.7 pounds-force per square inch (psi). Since most pressure cookers can be set to 5 pressure, this would make no sense if it were not the differential. It's has to mean 5 lbs more than the external pressure.
When discussing PSI we normally use PSIG or PSIA. PSI is taken as PSIG.

PSIG means pounds per square inch gage. These gages show 0 with only atmospheric pressure.

PSIA gages at sea level will show +14.7 with with only atmospheric pressure.
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