# 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

### Astronuc

Staff Emeritus
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