The heat flow into the volume is due to the temperature gradient (dTdr) according to Fourier’s law
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Consider a thin spherical shell between radii r and r + dr. Its surface area at radius r is A(r) = 4π r^2 and its volume is dV = 4π r^2 dr. The material has density rho and specific heat c. Show more…
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Five 17.8 Divergence Theorem Fourier's Law of heat transfer states that the heat flow vector F of at a point is proportional to the negative gradient of the temperature; that is F = -k∇T, which means that heat energy flows from hot regions to cold regions. The constant k is called the conductivity, which has metric units of J/m-s-K. A temperature function T for a region D is given below. T(x, y, z) = 100 + x" + y" + z" D = {(x, y, z): 0 ≤ x ≤ 2, 0 ≤ y ≤ 1, 0 ≤ z ≤ 3} Find the net outward heat flux ∬ F · n dS = -k ∬ ∇T · n dS across the boundary s of D. Assume k = 1.
Sri K.
Obtain the heat flow equation (1.3) as follows: The quantity of heat $Q$ flowing across a surface is proportional to the normal component of the (negative) temperature gradient, $(-\nabla T) \cdot n$. Compare Chapter 6 , equation (10.4), and apply the discussion of flow of water given there to the flow of heat. Thus show that the rate of gain of heat per unit volume per unit time is proportional to $\nabla \cdot \nabla T$. But $\partial T / C t$ is proportional to this gain in heat; thus show that $T$ satisfies $(1.3)$.
PARTIAL DIFFERENTIAL EQUATIONS
Introduction
Obtain the heat flow equation (1.3) as follows: The quantity of heat $Q$ flowing across a surface is proportional to the normal component of the (negative) temperature gradient, $(-\nabla T) \cdot \mathbf{n} .$ Compare Chapter $6,$ equation $(10.4),$ and apply the discussion of flow of water given there to the flow of heat. Thus show that the rate of gain of heat per unit volume per unit time is proportional to $\nabla \cdot \nabla T .$ But $\partial T / \partial t$ is proportional to this gain in heat; thus show that $T$ satisfies (1.3)
Parial Differential Equations
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