- #1

Soumalya

- 183

- 2

*Considering one dimensional heat conduction we may write the Fourier's law of heat conduction in the x-direction as,*

##\dot{Q}_{x}=-kA\Big(\frac{∂T} {∂x}\Big)##

*where, ##\frac{∂T} {∂x}## is the temperature gradient which is basically the slope of the temperature curve on a T-x diagram.*

*Referring to the first attachment, if we consider a differential volume element at a distance 'x' from the origin of length 'dx' and cross sectional area 'A' normal to the direction of heat transfer, the rate of heat conduction through the left face of the element (the surface at 'x') is given by Fourier's law as,*

##\dot{Q}_{x}=-kA \Big(\frac{∂T} {∂x}\Big)##

*where,##\frac{∂T} {∂x}## is the temperature gradient 'ahead' of the surface at 'x' or it is the rate of change of temperature with the spatial coordinate 'x' for an infinitesimal change in the spatial coordinate as 'dx',the change being measured from the right of the surface at 'x' (the right hand derivative of 'T' with respect to 'x' at location 'x').*

*So we observe the rate of heat conduction through a surface is proportional to the temperature gradient immediately 'ahead' of the surface.*

*I have been observing some of the boundary conditions for heat transfer problems where they are applying Fourier's law to calculate the heat conduction rate (or sometimes heat flux) at the boundaries of a body (a slab of a material).*

*Referring to the second attachment, we again consider a differential volume element immediately adjacent to the boundary surface at x=l of thickness 'dx' and area 'A' normal to the direction of heat transfer.As we might observe if we wish to calculate the rate of heat conduction through the boundary surface at x=l, it might be written according to Fourier's law as,*

*##\dot{Q}_{x=l}=-kA \Big(\frac{∂T} {∂x}\Big)_{x=l}##*

*where,*

*##\Big(\frac{∂T} {∂x}\Big)_{x=l}## should be the temperature gradient 'ahead' of the surface at 'x=l'.But in this case the temperature curve ends at the boundary i.e, at x=l within the medium itself.*

*So how are we supposed to visualize the slope of the temperature curve on the T-x diagram i.e, the temperature gradient at location 'x=l' ?*

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