consider a thin one-dimensional rod without sources of thermal energy whose late
ID: 1889658 • Letter: C
Question
consider a thin one-dimensional rod without sources of thermal energy whose lateral surface area is not insulated.(a) Assume that the heat energy flowing out of the lateral sides per unit surface area per unit time is w(x,t). Derive the partial differential equation for the temperature u(x,t).
(b) Assume that w(x,t) is proportional to the temperature difference between the rod u(x,t) and a known outside temperature y(x,t). Derive that
cp PDE(u)/PDE(t) = DE/DE(x)(KoPDE(u)/PDE(x)-P/A{u(x,t)-y(x,t)}h(x),
where h(x) is a positive x-dependent proportionality, P is the lateral perimeter, and A is the cross-sectional area.
(c) Compare the equation above with a one-dimensional rod whose lateral surfaces are insulated, but with heat sources.
(d) Specialize the above equation to a rod of circular cross section with constant thermal properties and zero degree outside temperature.
Explanation / Answer
Thermal energy density e(x,t) = the amount of thermal energy per unit volume.
Heat ux (x,t) = the amount of thermal energy owing across boundaries per unit surface area per unit time.
Temperature u(x,t).
Specic heat c = the heat energy that must be supplied to a unit mass of a substance to raise its temperature one unit.
Mass density (x) = mass per unit volume.
Fourier’s Law: the heat ux is proportional to the temperature gradient
= K0u
Conservation of heat energy: Rate of change of heat energy in time = Heat energy owing across boundaries per unit time + Heat energy generated insider per unit time
heat energy = e(x,t)Ax.
Heat energy owing across boundaries per unit time = (x,t)A (x + x,t)A.
Heat energy owing out of the lateral sides per unit time = w(x,t)Px = [u(x,t) (x,t)]h(x)Px, w
here h(x) is a proportionality.
Then
t [e(x,t)A(x)x] = (x,t)A(x) (x + x,t)A(x + x) [u(x,t) (x,t)]h(x)Px.
Dividing it by Ax and letting x go to zero give
e t = x P/A[u(x,t) (x,t)]h(x)
Heat energy per unit mass = c(x)u(x,t)Ax.
So
e(x,t)Ax = c(x)u(x,t)Ax,
and then
e(x,t) = c(x)u(x,t). I
t then follows from Fourier’s law that
cu/ t
= /x( µK0u/x )P/A[u(x,t) (x,t)]h(x)
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