[Stoves] Energy requirement

[Harris, Kirk]

Paul,

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?

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 *_interface_* 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?

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.

With respect for your work,

Kirk H.


On 10/15/2019 3:03 PM, Paul Arveson wrote:
>
> Thanks for those who responded to my question, how much energy is 
> required to cook a typical meal?
>
> I need more information.  The way we would calculate the basic energy 
> requirement is:
>
> E = q m dT
>
> Where q = heat capacity of food
>
> m = mass of food
>
> dT = temperature change
>
> Example (in metric units):
>
> Let's assume a meal is 1 kg of water.  = m
>
> Water has a higher heat capacity than solid foods, so water is the 
> worst case.
>
> It takes 4186 J to heat 1 kg of water 1 deg. C. = q
>
> Suppose we want to heat the water from 27 to 77 deg. C, or 50 deg. 
> increase.  = dT
>
> So the energy required is 4186 x 50 deg. = 209,300 Joules.
>
> 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.
>
> 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.
>
> What then are some typical numbers for cookstove thermal efficiency?
>
> Thanks,
>
> Paul Arveson
>
>
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