One Dimensional Slab Heat Transfer Taylor Expansion in Glasstone

1. Mar 23, 2014

terryphi

Hi There,

I came across the following passage in Sam Glasstone's 'Nuclear Reactor Engineering'

See where I underlined in red that taylor series expansion? I don't understand how (dt/dx)_(x+dx) is equal to that.

I know it's a Taylor series expansion, but where did the x+dx go?

2. May 4, 2014

Greg Bernhardt

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

3. May 5, 2014

terryphi

Well,

I know it's the incremental form of the Taylor series:

But I have no clue how to derive this incremental form.

Last edited by a moderator: May 6, 2017
4. May 5, 2014

Staff: Mentor

5. May 12, 2014

caldweab

The differential is supposed to be small, so we can neglect higher order terms in the Taylor series expansion since squaring a small number will give you an even smaller number. So you have heat entering and exiting your differential control volume,

That is Net Heat Flow:

qx- qx+dx.

See the heat enters at position x and leaves at a position x + a very small amount (dx). Now if we do a Taylor series expansion

qx+dx= qx +(dqx/dx)*dx

Take this expression and substitute into the above expression for net heat flow.

qx-qx+dx=qx-qx+(dqx/dx)*dx

Then:
qx-qx+dx=(-dqx/dx)*dx

No you can't cancel those two dx terms because one is a partial differential and the other is total. The one in parentheses is the partial differential.

Now we can describe qx using Fourier's law

qx=-kA(dT/dx).

Substitute this into our above expression

-d/dx*(-k *dydz*dT/dx)*dx

Net heat flow in x direction is now

d/dx*(kdT/dx)*dxdydz

The area in the x direction is given by y times z, similarly in the y direction area will be given by x and z

If you do those equations for the y and z directions and then divide the whole equation by dxdydz you will obtain the heat equation.

I assume you were ok with the other terms in the heat equation. If not I'm happy to explain the derivation of those as well

6. May 12, 2014

Morbius

terryphi,

I looks like a Taylor series expansion for the first derivative instead of the function itself.

The Taylor series expansion for function "f" about x = 0 is:

f(x) = f(0) + df/dx dx + 1/2 d2f/dx2 (dx)^2....

In this case, the function f is dT/dx; so all the derivative are one degree higher.

Greg