Pat Naughtin wrote:
>
> Dear All,
>
> My wife, some years ago, was spoiled for modern cookery when we lived in a
> farm cottage that had only a wood fired stove. This stove was of the
> 'slow-combustion' variety that was well insulated with excellent control of
> air inflow and outflow � in short it burned slowly.
>
> One feature of this stove was its ability to hold a largish pot at a
> 'simmer', which as I understand it is a temperature at, or just below
> boiling temperature where the odd bubble breaks the surface while the soup
> or stew does not 'catch' on the bottom.
>
> The main pot we use for soups and stew is stainless steel with a laminated
> base of aluminium and copper; it is about 240 mm in diameter and we fill it
> to about 110 mm so its capacity is about 5 litres. Has anyone on the list
> ever considered energy flow into and out of a cooking pot as it simmers? My
> question is how much energy do I need to flow from an electric hotplate to
> maintain a slow simmer?
>
> Thanks.
>
> Pat Naughtin CAMS
> Geelong, Australia
One can fairly simply estimate this, Pat. If the pot is at a constant
temperature, the heat flow in equals the heat flow out. So the real
question is, at the desired temperature, "What is the rate of heat flow
to the environment from the hot pot?"
That breaks down into two parts: the losses through the sides of the
pan and the losses through the top.
To get the losses through the pan's sides, use the thermal conductivity
for the material, its thickness, along with the difference in
temperature between the contents and the atmosphere to calculate the
rate of heat loss.
For the top, consider whether the pan is lidded or not. If lidded, it
is a similar calculation except that one now has an added layer of air.
Most tables will provide information for non-convective circumstances,
so the actual thermal conductivity of this layer will be greater than
the tables' values. The true value will be somewhere between that of a
non-convective layer of air and that of a situation with no air layer at
all. If the pan is not lidded, then one must estimate the rate of
evaporation and include the losses by that process in the form of latent
heat borne away in the water vapor. You probably have a feel for how
quickly the level fell in the pot if left unlidded.
The above assumes that all the heat input goes to the contents and
disregards any conduction of heat directly from the source to the
atmosphere by conduction along the pan's sides. But, then, we're not
putting this into a tight near-Earth orbit so we don't need to use
rocket science for this problem.
Jim
--
Metric Methods(SM) "Don't be late to metricate!"
James R. Frysinger, LCAMS http://www.metricmethods.com/
10 Captiva Row e-mail: [EMAIL PROTECTED]
Charleston, SC 29407 phone: 843.225.6789
