Mechanical Engineering

# Vacuum heating calculation

## Recommended Posts

Hello, everyone!

I am calculating the time of substrates' heating in the vacuum chamber.

It is a simple mathematical model to calculate roughly how much time does it take to heat polycor substrates in the vacuum chamber.

At least all the tube heater`s power heats the substrates (this is because the geometry of the vacuum chamber and beacuse all inner surface is polished).

So our goal is to get the time when conductive heating will be equal to the radiative heat losses.

Look at the scheme.

So the differential equation is .

And dividing the variables we get , where T_0 is the Temperature when conductive heating equals radiative losses.

If P=800 W, I have this dependence...

and as you can see, it is not physically right beacuse of two reasons. First - why there is a horisontal asymptote (it must be vertical!!!) ? Second - the heating is very-very fast...

For instance, I have calculated the heating time if there are no radiative losses.

t = c*m*(T_0-293)/W

and for W=800 W, T_0=623 K (technological process` temperature) we get t=0,205 seconds - and it is rubbish too I think...

##### Share on other sites

You're close but need to fix a couple of things. First, the radiation losses should have a minus sign (leaving the system). Second, you don't want to separate and integrate over T (temperature), you want to separate this ODE and integrate over t (time). I put it all into an Excel spreadsheet so you can see what the solution looks like. It's slow enough (or we can use time steps small enough) to easily solve this ODE with the forward Euler method (Tnew=Told+dt*dT/dt). As expected, the temperature asymptotically approaches the equilibrium value in about an hour or so.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×