Calculations for raising temperature in specific amount of time

When talking differentials, it is better / customary to swing the degree symbol across the units... 20 C=B0, means a "difference of 20 centigrade degrees"... whether than be between 0=B0C and 20=B0C and 235=B0C and 255=B0C...

q =3D "heat energy", assumed joules p =3D power =3D "heat energy per second", assumed watts /\t =3D period of time, seconds q =3D p * /\t

May require some units conversions on your part

David A. Smith

Never heard of this convention.

Also incorporate formula for radiant heat transfer to find probable heat flux at the surface of the sphere.

I wouldn't expect the temperature to be uniform with depth in the sphere. If the radiant heater has high output, the outer surface might melt before the core reaches the desired temperature.

An equation for a time-varying temperature profile in the sphere due to conduction should be doable at a given surface heat flux. Start by assuming the heat flux is constant for all sphere temperatures. How much heat flux can the rubber sphere stand before melting the surface?

... Exercise 16.1. Look at how the "o" jumps across the temperautre unit, when forming differential temperature.

... which involves the "emissivity / absorptivity" of that rubber surface.

Good point.

Not much. Starts altering its cross-linking... even if it does not boil off the additives used to keep it pliable.

David A. Smith

I guess everyone is assuming infra-red radiation. I don't think he would be able to heat rubber to the center in 10 seconds. If he is talking about doing it in a microwave then maybe it could be done in 10 secs.

-jim

Jim,

Based on my research it does seem the infra-red heating/radiation is the most effective way of heating rubber.

You are indeed correct that I have tried to heat the rubber in the microwave, and it did succeed but only after 30 seconds with a 1000W microwave.

--pokkie

Microwaves in a microwave oven are tuned to be resonant with water molecules. So it'll get the job done, the energy does all end up as heat, but more of it ends up heating the oven itself, than if you'd immersed the ball in water (which limits the temp to 212=B0F, and affects the surface of the rubber to some extent).

David A. Smith

Hi,

Am Thu, 10 Feb 2011 08:15:26 -0800 schrieb pokkie:

well, the question is not how much energy you need, the question is what

*power* you need to achieve the requested temperature/energy rise in the given amount of time.

Unfortunately, there is no easy answer to this. The point is that while there actually is an analytical solution to the problem, it is awkward to apply. Temperature within the ball is a function of time, distance from the centre and surface temperature. The governing differntial equation is easy to formulate

dT/dt = a * d/dr (r^2 * dT/dr)

where a denotes conductivity/(Density * heat capacity). It is *not* easy to solve, but does nevertheless have an analytical solution involving, for the given type of boundary conditions where temperature on the surface is prescribed (or where ambient temperature and a heat transfer coefficient is described - your boundary conditions are a special case for that), a Fourier Series. This gives you the temperature at any time in the sphere, at any radius. Unfortunately working out the series is rather cumbersome, since it involves solving some non-linear equations. Approximations for solutions "after a lot time has passed" and "after very little time has passed" are available - in your case the latter helps: you can get an approximation of the temperature within the sphere by

(T(r)-T,wall)/(T,initial-T,wall) = erf(x/(2*sqrt(a*t))

where erf denotes the error function and x denotes the distance from the surface. This is called the "semi infinite body approximation" and will do nicely for your 20 s heating period. BTW, "a" for rubber is about 0.1 *

10-6 m^2/s. Off you go.

By the way, if you consider microwaving the ball it would be either a) much easier or b) much more difficult: a microwave would act as an infinite number of internal heat sources in the sphere. If you assume that the microwaves penetrate equally in the centre and the outer region, heat pickup is as easy as Q = T * cp * t * m. Piece of cake. If you however raise your ambition level and assume that the intensity of the microwave is a function of penetration depth, it becomes much *more* complicated, since then the differential equation becomes

dT/dt = a * d/dr (r^2 * dT/dr) + dq(r)/dt / (Density * Conductivity)

where I doubt an analytical solution is available. You could solve this numerically of course, but if you haven't done this before it may hold a few unwelcome surprises. Consult a book on heat transfer if you want to know the gory details.

Hope this made some sense to you, regards Andreas