Simple Heat Transfer Question

Hmmm doesn't the table you are referring to give a dimension of the sample?

It seems odd it gives a value of 205 Watts per Kelvin without stating the volume or shape of the sample...

usually it will say for a 1" cube or 1 cm cube and assumes both sides of the cube are making good thermal connection - although if the temperature measurements are made on the cube surface it doesn't matter about any temperature gradients at the interfaces.

As you say if the table is for a 1 cm cube then if the cube dimensions are changed the thermal characteristics are changed.

John.

The conduction coefficient is a material property, and is given in Watts per meter per degree Kelvin.

The formula is Q = k*A*T/L

where

Q is in watts (a watt is one Joule per second)

k is in watt/metre/Kelvin, and is a property of the material

T is temperature difference between ends in Kelvins

A is cross sectional area of cylinder in metre squared

L is the length in metres

Greg

Yes, if the impediment to heat travel is 50% less, then the heat transfer rate will double.

http://en.wikipedia.org/wiki/Thermal_conductivity

You are correct with the exception of the fact that the heat transfer rate is measured in Watts = Joules per second, not Watts per second.

Hi Scott! Thermal conductivity, k, is mathematically expressed thus:

k = Q/A gradΘ, where Q is in Watts (Joules/second)

A, in m² is heat transfer area perpendicular to the heat flux

gradΘ in K (kelvin) is the temperature gradient or differential between the hot and cold surfaces.

From thence, it implies that thermal conductivity is a thermodynamic property of a material and it is defined as the quantity of heat in Watts (Joules/second) passing through a unit isothermal surface (in m²) when the temperature gradient is 1 degree (in this case 1 K) along every metre in the direction of normal to the surface.

k = ƒ( t °C, structure, density of the material, Pressure - for gases and liquids, humidity etc)

Therefore, the driving force in heat transfer process is temperature gradient. Using Aluminium as the heat conductor, in 1 second and at a temperature gradient/differential of 1 degree, 205 Joules of energy will be transferred through 1 metre along the normal.

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