[Stoves] stove test

[Crispin Pemberton-Pigott]

Dear Dieter

>From a thermodynamic viewpoint the efficiency in the simmering phase is always zero, if vaporizing of water is not targeted. 

Well that is the main point: what was the task assigned? To evaporate water (which is what they measure) or to keep a hot pot hot, while allowing it to cool a little.

If you use any energy at all and the pot cools overall, what is the performance?

The problem with ‘specific’ performance numbers is you have to be very careful what you chose as the denominator. If you use as a denominator a number that has no bearing on the thing you are trying to report, then the result is meaningless.

Take the hot pot off the stove and put it into a haybox cooker at the beginning of the low power test and store it in another room.  Keep the fire going, it doesn’t matter what the firepower because it isn’t doing anything anyway. 

After 45 minutes, check to see if the temperature in the pot is still high enough to qualify as a ‘successful simmer’.

Having weighed the mass of fuel consumed in the fire during that portion of the test, divide that mass by the number of litres of water remaining in the pot in the other room. Of what value is the resulting number?

Repeat the test but leave the pot on the stove this time. Perform the same calculation at the end of the test. What will the “difference in performance” tell you about the stove that is somehow important for rating, regulation and trade?

Regards

Crispin

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