My pressure cooker takes too long to cook the grained rice

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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.
 
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. :wacko:
Now it's time for me to go soak my head! Where's my scotch?
 
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. :wacko:
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.

Seeeeeeeya; Chief Longwind of the North
 
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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.
 
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.
 
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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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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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.
 
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.
So what will the PSIG gauge show at 12,000 ft.? -5.35?

Most pressure cookers use weights, not a gauge.
 
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.Seeeeeeeya; Chief Longwind of the North

A scuba regulator first stage is adjusted to deliver 120 psi to 140 psi to the second stage. Whether the first stage is a diaphragm or piston design, it depends upon a spring to control the pressure (intermediate) to the second stage. Ports in the first stage, open to the ambient environment, allow the first stage to deliver intermediate pressure to the second stage above the ambient pressure. The second stage has a diaphragm that depresses a lever when the diver draws breath. It is also exposed to the ambient pressure. The second stage is "tuned" to the intermediate pressure via a spring and seal which will only allow flow when the lever is depressed either by the diver drawing breath or manually depressing the purge button. Whether the diver is drawing breath at the surface or at depth, the regulator performance will remain the same.

Sorry Chief, your description of operation was not quite accurate.
 
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The only pressure cooker I've ever seen in action was my mother's. It was an old Presto (had to be pre 1950), and the relief valve was nothing more than a hole in a stem in the center of the lid, and you used a weighted plug sort of thing that just rested on top of the stem. It was the weight of that plug which regulated the pressure, and when the cooking was over, you just took that off and it depressurized through the relief hole in a few seconds.

That one would have cooked to the same internal pressure no matter the ambient atmospheric pressure because it was controlled by gravity.

Your mother and I had the same PC. It made the greatest Yankee Pot Roast for Sunday dinner. You never had to worry about getting the meal on the table on time. You knew you could turn to your trusty Presto PC.

The regulator was removed when the food was cooked. It got washed under running hot water, dried and placed right back in the top drawer where it belonged. Nobody dared to remove it. Along with the scissors in that drawer. Small items had a specified place and it never changed. Including my Presto regulator. :angel:
 
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.

Think about what you are saying. Sure there is 14.7 psi on the top (at sea level), but there is also 14.7 psi pushing on the bottom and sides, everywhere except for the tiny spot where the weight rests in the hole.

Before you apply the heat, there is equal pressure on every surface of the weight including the point on the hole. Pressure is balanced over the entire surface, no matter what the elevation. As heat builds inside the pot, so does the pressure on the point of the weight that rests in the hole, pushing back against gravity and atmosphere. The ambient pressure doesn't change anywhere else, so the only point where an imbalance in pressure can have an effect is the tiny surface area that is directly opposite the point affected by the internal pressure. Since that infinitesimal difference is mostly negated by the weight, it becomes irrelevant in any practical sense.

Water starts to boil at a lower temperature, but as the pressure builds that boiling point has to rise too, mostly negating the higher elevation. I may be all messed up, but this is how my logic sees it. Since I don't own a PC, it is really just a discussion for me.
 
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So what will the PSIG gauge show at 12,000 ft.? -5.35?

Most pressure cookers use weights, not a gauge.

Most gages showing PSIG are vented to atmospheric presure. For those at 12,000 ft alttitude they will still show 0.

The one in the OP loks to be a weight and not a valve.
 
Are you following the instruction manual that came with the pressure cooker?


I've never heard of taking it off the heat completely when it comes to temp.
 
I am unclear of the OP's cooking method. The correct method for using a pressure cooker is to place the food into the cooker, add the appropriate amount of liquid, secure the lid and regulator and turn on the heat source. When the regulator starts lifting, as is shown by the escape of steam, and movement of the regulator, you set the timer for the recommended period of time. When that time has elapsed, the quickest way to reduce the internal pressure is to place the PC into the sink and run cold water over it. There is usually an indicator valve that will open when the PC is safe to open. If not, simply remove the regulator while running water over the PC. When steam no longer emerges from the valve, pressure is equalized inside and out, and the PC lid can be safely removed.

The key to PC timing is to start timing when the pot comes up to the proper internal pressure, then reduce the heat until the regulator just barely moves. Cook for the recommended time.

Seeeeeeeya; Chief Longwind of the North
 
Selkie, this discussion has me taking in one more bit of info than I had initially considered. I still believe that the weight of the regulator is the primary force that controls the internal pressure. However, Before the cover is placed on the pot, the air at the higher elevation is thinner, with less particulate mass per square inch. It exerts less pressure at any given temperature than air at lower elevations as there are not as many molecules striking the inner surfaces of the pot. When the lid is first secured to the pot, the liquid would indeed boil at a lower temperature, thus the ineternal temperature of the pot would be lower. Because of the more rarefied atmosphere, it would take longer for the water vapor to create sufficient pressure to raise the regulator weight. Once reached however, the internal temperature would be the same as at lower elevations. It would also require the application of more eternal energy to maintain that internal pressure for the same reason.

I suspect that at any elevation, over time, as vapor escapes through the steam valve, the temperature would cool if a constant external energy was applied to the pot because again, less molecules means less internal pressure, to a point. So yes, your were correct, but not because any pressure was exerted on the regulator by ambient atmospheric pressure, but because the internal pressure was lower to start with.

Still, without experimentation, I couldn't say how much longer the pot would take to come to temperature at varying elevations.

Seeeeeeeya; Chief Longwind of the North
 

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