CL: Shape Factors

How can we use the same $\Delta T/R$ approach to multi-dimensional problems?!

Using shape factors!

DEFINITION:

A shape factor is a constant or an expression that is typically dimensionless (but sometimes has units of length) and accounts for the curvature of heat flows lines in a specific geometry such that we can use a linear relationship (i.e., $\Delta T/R$) to solve for heat flows in multi-dimensional problems.

NOTE:

In the notes below, we use a dimensionless shape factor like your text. For shape factors that have units of length, simply omit the $L$ (the "depth").

$\displaystyle{q = kSL\Delta T}$

it is useful to recast this equation into our familiar notation:

$\displaystyle{\frac{\Delta T}{R_{shape}} = kSL\Delta T}$

to yield a new resistor,

$\displaystyle{R_{shape} = \frac{1}{kSL}}$

In this way, we extend the use of shape factors to two very useful classes of problems:

OUTCOME:

Use shape factors to calculate heat flow in multi-d geometries

EXAMPLE:

For an example that uses a shape factor with units of length (so that $R_{shape} = 1/\left (kS\right)$) ... see below:

Recall in this case that:

$\displaystyle{q = kS\Delta T}$

so that

$\displaystyle{R_{shape} = \frac{1}{kS}}$

We want to figure out the heat loss through our new fridge (in order to calculate how expensive it will be to run). The fridge is a square, with side lengths of W and wall thickness of L. The inside temperature is given by Tcold and the outside (surface) temperature is given by Tsurface. We know the thermal conductivity k and it is constant with respect to T.

Since we only want the heat flow, and we have a relatively complex geometry, we want to use shape factors.

The best way to approach this problem is to break it up...

We can break our problem into it flat walls, edges, and corners (6 plane walls, 8 corners, and 12 edges). Since heat flows through each of these components in parallel we simply add each contribution to obtain the total heat flow.

Doing this, our heat flow is given by

q = 6k Swall (Tcold-Tsurface) + 8k Scorner (Tcold-Tsurface) +

12k Sedge (Tcold-Tsurface)

or

q = k (6Swall + 8Scorner + 12Sedge) (Tcold-Tsurface)

we can actually figure out Swall ourselves since we know the answer for heat flow through a plane wall...

$\displaystyle{q_{wall} = kS_{wall}\Delta T = \frac{kA}{L}\Delta T}$

so,

$\displaystyle{S_{wall} = \frac{A}{L}}$

we look up the answer for corners and edges...

Scorner = 0.15L

Sedge = 0.54W

Our final answer is given by

$\displaystyle{q = k\left (6\frac{A}{L} + 8(0.15L) + 12(0.54W)\right )(T_{cold}-T_{surface})}$

or, since A=W2

q = k (6W2/L + 8(0.15L) + 12(0.54W)) (Tcold-Tsurface)
=k (6W2/L + 1.2L + 6.48W) (Tcold-Tsurface)