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Author: Subject: Pressure cooker regulator weight help...
Fulmen
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Not sure, it didn't factor in to my answer. I calculated how to get the unit it's design pressure based on a given altitude, that's all.

We're not banging rocks together here. We know how to put a man back together.
moray james
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I have access to one of those laser thermometers Can I simply shoot my pot when it is cooking and take measurements with each of the two regulators I have (2.7 ounces and 3 ounces available)? While I am asking do I shoot the top of the pot or would shooting the water vapor escaping the regulator be the best option? Thanks for the help.

[Edited on 27-9-2016 by moray james]
BillA
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Weigths

Just for kicks and giggles I did a quick calc. My vent pipe top diameter is also 3/16, or .19". However that is not the effective area that the pressure acts upon. If it were, you would need 6.7 oz to resist 15 psig. The .19" is only a lip that the weight rests on. The vent diameter under the lip is only .12" diameter. With such a diameter and 15 psig pressure underneath (@ SL) you would need (.12/2)^2*3.14159*15 *16oz/lb=2.71 oz. Exactly as supplied.
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 Sciencemadness Discussion Board » Fundamentals » Miscellaneous » Pressure cooker regulator weight help... Select A Forum Fundamentals   » Chemistry in General   » Organic Chemistry   » Reagents and Apparatus Acquisition   » Beginnings   » Miscellaneous   » The Wiki Special topics   » Technochemistry   » Energetic Materials   » Biochemistry   » Radiochemistry   » Computational Models and Techniques   » Prepublication Non-chemistry   » Forum Matters   » Legal and Societal Issues   » Detritus   » Test Forum