Heat Transfer Questions & Answers | |
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You are right that $\tau_{yz}$ should be considered for the general case. But in this case, the power is equal to the dot product of the velocity vector and the force vector applied on the $xz$ top surface. Because the problem is 2D there is no contribution to the power coming from $\tau_{yz} \cdot \vec{v}_z$ because $\vec{v}_z=0$. Plus, $\tau_{yz}$ is also zero here because $\tau_{yz}=\mu \partial \vec{v}_z/\partial y$ and $\vec{v}_z$ is zero everywhere. I'll give you 1.5 points bonus boost for this question.
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You are confusing two different situations. The shape factor for the sphere was derived for the matter in between the sphere and infinity, not for the matter inside the sphere. Thus, we can not apply $dT/dr=0$ at $r=0$ because the integral goes from $r=R$ to $r=\infty$, and can not be applied at $r=0$. I'll give you 1.5 point for the question.
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Yes you are right: radiation heat transfer can occur between a gas and a solid, or between a gas region and another gas region. However, calculations involving radiation heat transfer to and from gases are difficult and beyond the scope of this course. So we neglect those. I'll give you 1.5 points bonus boost here. I would have given more if your question would be free of obvious typos and spelling mistakes.
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Hmm, this is not a genuine question again.. I'm not sure what to answer you, since you're just outlining the intersection formula from the tables. I can not give a bonus boost here.
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This is due to how the electrical resistance given in the tables (in $\rm n\Omega\cdot m$) is defined. The electrical resistance is given in the tables for copper, not for a cable in particular. But it's easy to remember: For a smaller area cross section or a longer length of cable, the current will encounter more resistance. Thus, you should divide the copper resistance by area and multiply by the length of the cable to get the cable resistance. I think I mentioned it in class. I'll give you 1 point bonus boost for the question.
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$\pi$ |