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Author: Subject: Pressure cooker regulator weight help...
Fulmen
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[*] posted on 26-9-2016 at 15:31


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.



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moray james
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[*] posted on 26-9-2016 at 19:08


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]
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BillA
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[*] posted on 5-12-2017 at 14:51
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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