From: neufeld@helios.physics.utoronto.ca (Christopher Neufeld)
Subject: Re: Microwave Death
Organization: University of Toronto Physics/Astronomy/CITA
Date: Thu, 12 Nov 1992 23:50:05 GMT

In article <1992Nov12.182911.861@hubcap.clemson.edu> michaet@hubcap.clemson.edu (michael dale townsend) writes:
>>Microwaves do NOT cook things "from the inside out". [ ... ]
>>Microwaves cook from the outside in. 
>>-Cindy Kandolf
>> cindy@solan.unit.no
>> Trondheim, Norway
>
>Moreover, since the skin depth at microwave frequencies for most foods is
>the same order of magnitude of their dimensions, the entire volume of the
>food is heated directly and uniformly by the microwave radiation.  . . . "
>
>Rizzi, Peter A.
>Microwave Engineering
>1988 Prentice-Hall
>TK7876.R59
>1988
>621.381'3
>87-25493
>ISBN  0-13-586702-9
>(page 10)
>
>I would speculate that ice doesn't melt easily since the water molecules
>(which are polar) are confined in a crystal lattice so they can't
>oscillate.  If the rapidly reversing electric field of a microwave can't
>physically oscillate the water molecules because they are confined, then
>no friction will be generated and thermal heating is a slower process.
>
   Actually, Cindy is right for a homogeneous food. There is more energy
per unit volume dropped into the surface than into the deeper parts of
the food. The skin depth, as you quote, is comparable to the dimensions
of the food (at 2.45 GHz the skin depth in pure water is of the order of
several centimetres). This means that there isn't a huge difference
between the energy dropped per unit volume at the surface and that
dropped deep in the bulk. The difference, in fact, is of the order of a
few tens of percent. It remains true, however, that the surface is more
strongly heated than the bulk.
   To put a bit of spin on the problem, though, and one which isn't
relevant to the issue of heating up a person, the thermal conductivity of
the food may be low. If that's the case the inside will get significantly
hotter than the outside, simply because you have to have a large
temperature gradient to support the heat flow out of the inside. Let's
create a mechanical analogy:
      Think of the food as a flat plate surrounded on the edges by a long
   drop. Now, start sprinkling sand down onto the plate in such a way that
   the sand per unit area falling at the edge is about twice the sand per
   unit area falling at the centre of the plate. The sand represents the
   heat dropped into the food, and the height of sand above the plate
   represents the local temperature at that point. Almost immediately the
   sand starts to fall off the edges of the plate and builds up a
   critical angle (the angle of repose). As the sand keeps falling the
   height profile goes from being initially higher at the edges to being
   higher in the middle. Eventually I have a cone-shaped pile of sand on
   the plate and the sand (heat) gradient is so high that no additional
   sand accumulation (heating) occurs, for each grain (joule) which falls
   on the plate (food) a grain (joule) is lost to the rim (surface).

   In this model the thermal conductivity is inversely proportional to
the angle of repose. The higher the thermal conductivity the flatter the
final temperature distribution (and the lower the average temperature at
the steady state). The model I gave doesn't quite handle the boundary
conditions properly, but hey, it's just an illustration.
   So, if the food is heated for a long time (measured on the scale of the
thermal diffusion time constant for the material) the inside will be
hotter than the outside, unlike a conventional oven where the inside will
always stay at least a bit cooler than the surface. If the food is heated
briefly then the outside will be hotter because it will have absorbed
more energy than the inside and conduction effects wouldn't have had a
chance to make themselves known.
   For the mathematicians in the audience, the microwave oven corresponds
to solving the heat equation for a cool boundary and a non-uniform source
density throughout the bulk. The conventional oven corresponds to a hot
boundary and no source density (the Laplace, rather than Poisson
equation).

   Aside: A human being has active temperature control and an effectively
high thermal conductivity because of the circulating fluids.

   Now, about this "physically oscillate the water molecules" stuff,
that's not how microwave ovens work. There is no natural resonant
frequency of water molecules anywhere near 2.45 GHz (the mandated
frequency for domestic microwave ovens). The stuff heats by "dielectric
heating" which, I imagine, will be mentioned in your Microwave
Engineering book, though probably a bit past page 10.

>My point:  this story may or may not be true, but I believe that it is
>possible.  Remember that commercial microwaves (if this store had one)
>are more powerful than the average domestic models.
>
   Commercial microwaves also work at lower frequencies (because they
expect to have to cook thicker things). This results in a more uniform
heating profile. In the sand analogy, the sprinkling is closer to uniform
over the face of the disk.

-- 
 Christopher Neufeld....Just a graduate student  | Don't let your mind
 neufeld@helios.physics.utoronto.ca    Ad astra  | wander -- it's too little
 utzoo.utoronto.ca!generic!cneufeld              | to be let out alone.
 "Don't edit reality for the sake of simplicity" |
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