# Thermal conductivity calculation.xls

### Description

KNOWN: Heat flux and surface temperatures associated with a slab of prescribed thickness.

FIND: Thermal conductivity, k, of the wood.

ASSUMPTIONS: 1. One-dimensional conduction in the x direction.

2. Steady-state conditions.

3. Constant properties.

**Calculation Reference**

Fundamentals of Heat and Mass Transfer - Frank P. Incropera

To calculate the thermal conductivity (k) of a wood slab, given the known heat flux and surface temperatures associated with the slab and the given assumptions of one-dimensional conduction in the x direction, steady-state conditions, and constant properties, you can use Fourier's Law of heat conduction.

Fourier's Law states that the heat flux (q) is proportional to the temperature gradient (ΔT/Δx) in the direction of heat flow and is given by:

q = -k * (ΔT/Δx)

Where q is the heat flux, k is the thermal conductivity, and (ΔT/Δx) is the temperature gradient.

For a slab with one-dimensional conduction in the x direction, the temperature gradient can be approximated as:

(ΔT/Δx) = (T2 - T1) / L

Where T1 and T2 are the surface temperatures of the slab and L is the thickness of the slab.

Rearranging the equation, we have:

k = -q * L / (T2 - T1)

Using this equation, you can calculate the thermal conductivity (k) of the wood slab by substituting the known values of the heat flux (q), slab thickness (L), and surface temperatures (T1 and T2). Make sure to use consistent units for the heat flux, thickness, and temperature difference to obtain the correct units for the thermal conductivity.

By following these steps and applying Fourier's Law, you can calculate the thermal conductivity (k) of the wood slab based on the given heat flux and surface temperatures, while considering the assumptions of one-dimensional conduction, steady-state conditions, and constant properties.

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