Quick math question(s?) on the 1-D Heat Equation

[Darien]

I need to make sure I have these two things down...so, I turn to ATOT for a math problem for the first time in my life to check my understanding:



Let T (temperature) be a function of x (position) and t (time) => T(x,t)

If a problem is described as adiabatic, does this mean

dT/dx of (x,t) = 0?, for all x and t?

and if a problem describes a bar insulated at x = 0 and x = L, does this mean

dT/dx of (0, t) and (L,t) = 0?

Thanks

 

[Darien]

bump.

 

[dullard]

Wow that was one of the fastest bumps I've ever seen.

An adiabatic process is one in which there is no external heat added or removed. You can have a slope in tempertature in an adiabatic process. That just means that there are internal exchanges of heat but not external exchanges. Thus dT/dx is not zero for all x and t.

However you are correct on the second half. If there is no heat exchanged with the outside world, then the outside world is the same temperature as your bar. Therefore, by definition, dT/dx=0 at the border.

 

[duke]

nvm

 

[Darien]

thanks guys

 

[duke]

Forget my answer. Dullard is correct.

 

[Darien]

Does the problem state an adiabatic surface on one end?

I have an exam soon (~2.5 hours), and I just realized this is one way they can give initial/boundary conditions -- by just describing something in a few words and not explicitely giving the initial/boundary conditions.