# What if the lid of a pressure cooker was suddenly released?

My dad and I have tried to calculate the strength of the explosion if the lid was suddenly freed. We took some measures:

• Lid mass: $0.7 \textrm{kg}$
• Lid surface: $0.415 \textrm{m}^2$
• Internal pressure (above external): $1\textrm{atm}\approx 100 \textrm{kPa}$

Now we are kind of struggling about what to do with these. We'd like to know how to calculate the initial speed, for instance, or any other value that could get us started.

I hope I was clear enough. :)

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You could probably make a reasonable model (at zeroeth order and at short times) by assuming that the gas inside flows out at the same 'rate' of the lid escaping. (So imagine the gas being fixed inside of a closed cylinder with the lid on top, moving upwards.) –  Vibert Mar 15 '13 at 22:45
Pressure is force per area, so you can get the force acting on the lid. To zeroth order you can assume that the largest accelleration happens right at the start with the initial force. That'll tell you how long it'll take for the lid to reach a certain velocity –  Lagerbaer Mar 16 '13 at 1:31
@Cranium BLEVE? –  drN Mar 16 '13 at 2:00
The problem will be with the food all over the kitchen Happened and just with a handle and valve failure youtube.com/watch?v=_THHP4j0oa0 –  anna v Mar 16 '13 at 6:27
I can tell you from my own experience that the lid from my grandmother's cooker blew off and hit the roof two or so metres above the stove such that a considerable chunk was ripped out of a structural wooden beam. I'm guessing from my memory of the hole in the beam, at least with the impulse of a pretty lustily wielded sledge hammer. Even in those days they had safety valves, so I don't know how it failed - I was four years at the time. If you google pressure cooker explosion (as opposed to bomb) 100-200kPa at failure seems a pretty realistic figure. –  WetSavannaAnimal aka Rod Vance Aug 2 '13 at 6:03

Take the internal volume of the cooker and multiply by $\Delta P=1 ~atm$. This is the theoretical maximum energy that can be imparted into the lid assuming adiabatic expansion of the pressurised gas. With the kinetic energy you can calculate the maximum height attainable by the lid assuming it captures all the work done by the expanding gas.

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Pressure equals force per unit area ($P = F/A$), so force equals pressure times area: $F = P \times A$. And force equals mass times acceleration, $F = m \times a$. We have $m \times a = P \times A$, so the acceleration is
$a = P \times A / m = 60\,000\, \mathrm{m}/\mathrm{s}^2$