[Stoves] Energy requirement

Crispin Pemberton-Pigott crispinpigott at outlook.com
Wed Oct 16 09:10:36 CDT 2019


Dear Paul

There was a lot of work done on this matter at Eindhoven University in the early 80's. They published a list of foods and the energy requirement to cook them.  It is not obvious but as chemical transformations take place in the food, the temperature drops as the reactions are endothermic. So even then energy is going in, they do not rise in temperature some of the time.

You can take foods to be wet, so using water mass is not far wrong. Cabbage requires one of the greatest inputs of energy to cook.

Regards
Crispin


From: Stoves <stoves-bounces at lists.bioenergylists.org> On Behalf Of Paul Arveson
Sent: Wednesday, October 16, 2019 6:04
To: stoves at lists.bioenergylists.org
Subject: Re: [Stoves] Energy requirement


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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