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Heat transfer Fourier’s Law of heat transfer
Chapter 14, Problem 45E(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.
\(T(x, y, z)=100 e^{-x^{2}-y^{2}-z^{2}}\); D is the sphere of radius a 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) = 100e^-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.
\(T(x, y, z)=100 e^{-x^{2}-y^{2}-z^{2}}\); D is the sphere of radius a 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) = 100e^-x^2 -y^2 -z^2
ANSWER:Solutio