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<p>Paul,</p>
<p>How do you intend to use this? Will this make a difference for a
cook who gathers and burns sticks to cook food, and really doesn't
care about mathematics? Would not your efforts be better used
working on a level closer to the cook? <br>
</p>
<p>This reminds me of the years of conflict on how to calculate how
much heat gets into a pot of water (thermal efficiency?). None of
those calculations told us about a wok, plancha, or a tandoori, or
about putting vegetables or meat into the water to make soup or
stew (change in heat absorption and circulation). Also, those
calculations told us more about the <b><u>interface</u></b>
between the stove and cooking vessel (pot stand, skirt, super pot,
different size or shape pot, and I will include lid for the pot
here) and its effect on getting the heat produced by the stove
into an open pot of water, then it told us about the burning
efficiencies of the stove. Is not the direction of your inquiry
of limited value for cooks in the field, but rather getting
caught-up in the fine mathematical details and minimizing value in
the overall picture?<br>
</p>
<p>I do like that you are discussing food rather than a pot of
water. Does a stew have the same heat absorbing qualities as a
stir-fry or a tortilla? Currently thermal efficiency is
determined using an open pot of water, which is different from
these.<br>
</p>
<p>With respect for your work,<br>
</p>
<p>Kirk H.</p>
<p><br>
</p>
<div class="moz-cite-prefix">On 10/15/2019 3:03 PM, Paul Arveson
wrote:<br>
</div>
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<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Thanks for those who responded to my
question, how much energy is required to cook a typical meal?</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">I need more information. The way we would
calculate the basic energy requirement is:<br>
<br>
</p>
<p class="MsoNormal">E = q m dT </p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Where q = heat capacity of food</p>
<p class="MsoNormal">m = mass of food</p>
<p class="MsoNormal">dT = temperature change </p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Example (in metric units):</p>
<p class="MsoNormal">Let's assume a meal is 1 kg of water. = m</p>
<p class="MsoNormal">Water has a higher heat capacity than solid
foods, so water is the worst case.
</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">It takes 4186 J to heat 1 kg of water 1
deg. C. = q</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Suppose we want to heat the water from 27
to 77 deg. C, or 50 deg. increase. = dT</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">So the energy required is 4186 x 50 deg. =
209,300 Joules.</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">This is only the energy required to heat
the food. But cookers are not 100% efficient, since some
energy goes to heat the pot, the stove parts, etc. Also there
is some heat loss during cooking due to conduction, convection
and radiation.
</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Apparently efficiency is low for most
stoves, so most of the energy doesn’t go into cooking the
food. So the “steam cooker” was quoted by Frans as 1.2 MJ,
which is 17% efficient (assuming he was heating 1 kg of water,
which was not stated). This in turn drives the fuel
requirement. </p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">What then are some typical numbers for
cookstove thermal efficiency?
</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Thanks,</p>
<p class="MsoNormal">Paul Arveson</p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
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<br>
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