Heat Transfer Questions & Answers  


You have to find the most accurate answer possible given the tools outlined in class. If we have seen a way to take into consideration the interaction of one fin's heat transfer on the other, then use such a way. I'll give you 0.5 points bonus boost here.




You can read about it here: https://en.wikipedia.org/wiki/Auxiliary_power_unit 



I do not understand your question.. I don't think I mentioned that $t\gg w$.. Your questions don't sound genuine or are badly formulated. To give you more time to ask a proper question, I set the minimum interval between questions per student to 2 days.


The bonus boost has been added to the scoresheet for the midterm bonus. Check if it has been added correctly.




The equation that you outline is correct. Here $(q_{\rm in})_{\rm rad}$ refers to the heat transfer by radiation going into the surface of the body from the room walls. Thus, it will be positive (going in the body) if $T_w>T_s$ and negative (coming out of the body) if $T_w<T_s$. I'll give you 1 point bonus boost for the question. Not a bad question but it's not clear what confused you.




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.




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.




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.




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.




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.




I skipped a step in class hence why it may not have been so clear. Here, we replace $\partial_x(\cdot)$ by $\Delta(\cdot)/\Delta x$ $$ \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x} \right) \approx \frac{\partial}{\partial x}\left(\frac{\Delta u}{\Delta x} \right) \approx \frac{\Delta\left({\Delta u}/{\Delta x} \right)}{\Delta x} \approx \frac{\Delta\left({\Delta u} \right)}{\Delta x^2} $$ But $\Delta(\Delta u)$ has the same units as $u$ and is typically of the same order of magnitude as $\Delta u$ except near discontinuities. Thus, we can say: $$ \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x} \right) \sim \frac{\Delta u}{\Delta x^2} \sim \frac{\Delta u}{L_{\rm c}^2} $$ I liked your question. I'll give you 1.5 point bonus boost for it. I would have given 2 points if you would have typeset the absolute operators “” properly with the command “\left” and “\right”.




For 10% error, you should divide the RHS of Eq. (3) by 10.. This will then yield a RHS of (4) of 7/5. This was mentioned in class. I'll give you 0.5 point bonus boost.




A gas can be treated as ideal when the collisions between molecules are elastic and when there are no intermolecular forces.. This has nothing to do with whether the flow is turbulent or not. I'll give you 0.5 point bonus boost.



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