# T: Transient Conduction - A "Lumped" Approach

 Consider a "hot" piece of iron that we throw into a "cold" oil bath. Our problem is greatly simplified if we assume that the internal resistance is low (relative to what?!).

What will be true in this case?

• the temperature is constant (in time) at the boundaries
• the temperature within the solid is spatially uniform
• the flux out of the solid is constant (in time)
• there is no flux out of the solid

So, what we are doing here is making an assumption that makes the problem easier, because the full problem (where the temperature varies in both space and time) is too hard (don't worry, we will do the hard version too!).

First, we need to figure out what our governing equation is...

 rate of change of heat in the material = net rate of heat flow into the material

or in equation form

$\displaystyle{\frac{d(\rho cVT)}{dt} = \rho cV\frac{dT}{dt} = -hA(T-T_{\infty})}$

We started looking at this problem by making the assumption that internal resistance is negligible. We will now look at the question of: how do we validate this assumption? and: what are the other cases that may present themselves if this isn't true?

Early in the course, we spent a lot of time considering thermal resistances. What do we mean by internal resistance in this case?!

Conduction, of course! And what is the resistance due to conduction?

$\displaystyle{R_{cond} = R_{internal} = \frac{L}{kA}}$

##### NOTE:

It is important to note, that we need to be careful hear in our choice of L. A good rule of thumb is that L should be the ratio of V/A.

So,

$\displaystyle{R_{internal} = \frac{V/A}{kA} = \frac{L}{kA}}$

In contrast to internal resistance, what other resistance is important?

External. And in this case, what is the external resistance?

Convection, of course! And what is the resistance due to convection?

$\displaystyle{R_{conv} = R_{external} = \frac{1}{hA}}$

Finally, how do we determine the importance of the internal resistance? By comparing it to the external resistance! This yields a new dimensionless quantity, the Biot (Bi) number.

$\displaystyle{Bi = \frac{R_{internal}}{R_{external}} = \frac{\frac{V/A}{kA}}{\frac{1}{hA}} = \frac{Vh/A}{k} = \frac{hL}{k}}$

For internal resistance to be negligible what does the Bi have to be like? How small?

A good rule of thumb is that for Bi<0.1, most geometries yield a full solution where the center-line temperature is less than 5% different from the surface temperature (in other words a "lumped" solution would be pretty good). Before we move on to looking at $\displaystyle{Bi /ge 0.1}$, how would this problem change is we had radiation occurring?

If we had radiation, we would first have to recalculate our Bi. Since radiation is also an external resistance, and it is in parallel with the convection, we would modify our external resistance to be

$\displaystyle{\frac{1}{R_{external}} = \frac{1}{\frac{1}{ha}}+\frac{1}{\frac{1}{h_rA}} = (h+h_r)A}$

or

$\displaystyle{R_{external} = \frac{1}{(h+h_r)A}}$

so our Bi is

$\displaystyle{Bi = \frac{(h+h_r)L}{k}}$

If we find that this Bi is sufficiently small, we then need to modify our governing equation. If the temperature difference is sufficiently small (so that our hr approximation works), we get

$\displaystyle{\rho cV\frac{dT}{dt} = -(h+h_r)A(T-T_\infty)}$

If, however, the temperature difference is large (so hr is not valid), we get something similar to

$\displaystyle{\rho cV\frac{dT}{dt} = -hA(T-T_\infty)-F\epsilon \sigma A(T^4-T_\infty^4)}$

which we cannot solve analytically (we need to use a numerical solution). (Note: it is also somewhat questionable to use hr in our Bi in this case).

##### OUTCOME:

Explain the utility of the Biot number