Pressure Cooker PSI/Cooking Time Conversion Tables
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I received as an Xmas gift a Nesco Model PC625P electric pressure cooker (PC) that I originally thought produced 15 pounds per square inch (PSI) of pressure in the “High” pressure mode. I later discovered that it was actually 10 PSI on the “High” setting.
I subsequently purchased the Great Food Fast book by Bob Warden. Since it was written specifically for electric PC’s, I thought it would reference the lower PSI as most electric PC’s are between 9 and 11 PSI. I then noted the paragraph that stated, “…Most of the recipes in this book are written to be cooked under the highest pressure—HIGH on some models of pressure cookers and 15psi on others.”
I attempted to find a conversion table or other reference that described how to convert the cooking time for recipes written for a 15 PSI PC to a PC using 10 PSI. The only source I found was in a Miss Vicky’s cookbook and she evidently assumed a linear relationship between PSI/Temp and cooking time and indicated that the cooking time should be increased by 1/3 since the PSI is decreased by 1/3. I’m not sure that this is a correct assumption since the temperature difference is only 10°, i.e. 240° for 10 PSI versus 250° for 15 PSI.
A couple of people that I contacted said that they followed the 15 PSI recipes with their 10 PSI PC’s, didn’t alter the cooking time at all and had successful results.
Any help in clarifying this issue would be greatly appreciated.

<The only source I found was in a Miss Vicky’s cookbook and she evidently assumed a linear relationship between PSI/Temp and cooking time and indicated that the cooking time should be increased by 1/3 since the PSI is decreased by 1/3. I’m not sure that this is a correct assumption since the temperature difference is only 10°, i.e. 240° for 10 PSI versus 250° for 15 PSI.>
I agree with your general thinking. At one atmosphere, the boiling point for water is 212°F or 100°C. A 10 psi pressure cooker will bring the boiling point to 240°F, but a 15 psi pressure cooker will top at 250°F. In other words, the boiling point increases at a decreasing rate against increaing pressure.
http://science.jrank.org/article_imag...
Keep in mind that our atmosphere (1 atmosphere) is already 15 psi (14.6 psi to be accurate). Therefore, a 10 psi pressure cooker is providing a 10 psi on top of 15 psi, which is 25 psi total, and a 15 psi pressure cooker is really cooking at 30 psi, so the actual difference in pressure is really 25 psi vs 30 psi. I briefly read MissVickie's website, it seems that she suggests the "total cooking time" to be increased by 33%, but that won't even be linear against the pressure difference, let's alone the temperature difference.
On top of this, the food cooking is much more complex than merely temperature, or should I say the relationship is not linear. We cannot say I cook this lamb stew at 80°C for 1 hour, and therefore I can make the lamb stew at 40°C by doubling the time to 2 hours  because 80 X 1 = 40 X 2. That is purely wrong. Surely you will also notice that the temperature ratio will be different if you use °C vs °F vs K.
I have a pressure cooker which offers two modes at 10 and 15 psi. In my expereince, the 15 psi does feel to work a bit faster than 10 psi, but certainly not 1/3. I would say about a 10% increase in time. In other words, you may not notice a whole lot of difference even if you use the same time. At the end, you will have to try it to optimize your cooking time, but I would start by increasing the time by ~10% give or take.
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re: Chemicalkinetics
I think you’re correct about the smaller increase in time. If the cooking time is dependent solely on temperature and not pressure then the delta of 10° between 240° and 250° is an increase of a little over 4%. Based on that assumption and rounding off the time to the nearest minute I would increase the cooking time as follows:
1 to 12 mins. = 0 increase
13 to 37 mins. = 1 min. increase
38 to 62 mins. = 2 min. increase
63 to 87 mins. = 3 min. increaseSince most PC recipes seem to call for a cooking time of 15 minutes or less, increasing the time is probably not significant or terribly important.
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re: TomDel
<the delta of 10° between 240° and 250° is an increase of a little over 4%>
It get complicate because that "percentage" difference depends on your temperature units. You were using Fahrenheit. Using Kelvin, the temperatures are 389K (240°F) and 394K (250°F), which then the percentage difference becomes 1%. As you can see, the so called percentage in temperature change can be problematic depending the "zero" point of the temperature. Regardless, I think MissVicky's 33% increase in time is probably a bit high. Good luck.
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re: Chemicalkinetics
With a name like Chemicalkinetics, can't you figure out a way for us to use the Arrhenius equation with a pseudo "activation energy" (which would presumably differ for meat vs. potato, etc) ??????????????????
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re: drongo
:) Sometime empirical determination is easier. :P
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