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Solution: Heat transfer Fourier’s Law of heat transfer (or
Chapter 14, Problem 44E(choose chapter or problem)
Heal transfer Fourier's Law of heat transfer (or heat conduction) states that the heat flow vector F al a point is proportional To the negative gradient of the temperature: that is, \(\mathbf{F}=-k \nabla T\). which means that heal energy flows fmnt hot regions to cold regions. The constant k is called the conductivity which has metric units of J/m-s-K or W/m-K. A temperature function for a region D is given. Find the net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=-k \iint_{S} \nabla T \cdot \mathbf{n} d S\) across the boundary S of D.
In some cases it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume that k = 1.
\(T(x, y, z)=100+x^{2}+y^{2}+z^{2}\); D is the unit sphere centered at the origin.
Text Transcription:
F = -k nabla T
iint_S F cdot n dS = -k iint_S nablaT cdot n dS
T(x, y, z) = 100 + x^2 +y^2 + z^2
Questions & Answers
QUESTION:
Heal transfer Fourier's Law of heat transfer (or heat conduction) states that the heat flow vector F al a point is proportional To the negative gradient of the temperature: that is, \(\mathbf{F}=-k \nabla T\). which means that heal energy flows fmnt hot regions to cold regions. The constant k is called the conductivity which has metric units of J/m-s-K or W/m-K. A temperature function for a region D is given. Find the net outward heat flux \(\iint_{S} \mathbf{F} \cdot \mathbf{n} d S=-k \iint_{S} \nabla T \cdot \mathbf{n} d S\) across the boundary S of D.
In some cases it may be easier to use the Divergence Theorem and evaluate a triple integral. Assume that k = 1.
\(T(x, y, z)=100+x^{2}+y^{2}+z^{2}\); D is the unit sphere centered at the origin.
Text Transcription:
F = -k nabla T
iint_S F cdot n dS = -k iint_S nablaT cdot n dS
T(x, y, z) = 100 + x^2 +y^2 + z^2
ANSWER:Solution 44E