Heat Transfer Questions & Answers


B. Parent • AE23815 Heat Transfer

Heat Transfer Questions & Answers

Question by Student 201527110

Professor, I wonder the difference of conductivties between that of gas and solid. The definiton of conductivity is $K\equiv\frac{m5R\sqrt{3RT}}{\sigma4\sqrt{2}}$ and is it valid only for gases? If that definition is also valid for both gas and solid, why solid doesn't exactly proporsional to $\sqrt{T}$?

03.12.18

This is valid for gases assuming negligible intermolecular collisions and assuming that all the particules going towards the interface have the same average energy independently of speed. For liquids, the expression will be different because of intermolecular attraction forces. For solids, the expression for $k$ will be totally different and depend on how the phonons propagate (different from solid to solid). 1 point bonus boost.

Question by Student 201427104

Professor. I tried to sort out the problem that you asked us to explain yesterday. The solution of the given problem is the same as that of the $T_1>T_2$ graph. Assume $q_x^"=const$ and $\beta>0$. $q_x^"=const=-k\frac{\partial T}{\partial x}$ and $k=k_0(1+\beta T)\frac{\partial T}{\partial x}$. Therefore $q_x^"=-k_0(1+\beta T)\frac{\partial T}{\partial x}$. In other words $q_x^"=k_0(1+\beta T)(-\frac{\partial T}{\partial x})$. That is, since $q_x^"=const$, $(-\frac{\partial T}{\partial x})$ must be small if $k_0(1 +\beta T)$ is large. The reverse is also true. So $(-\frac{\partial T}{\partial x})$ is low for $T_2$, and $(-\frac{\partial T}{\partial x})$ is high for $T_1$. Therefore, the shape of the graph is a symmetrical shape of the graph when $T_1>T_2$. It is convex shape.

03.13.18

This is a better explanation but builds on the one of the previous student: 0.5 point bonus boost. I deleted the second question from your post — please ask only 1 question per post.

Question by Student 201427111

In the last lesson, I thought about the problem that professor aksed me to explain why. As the professor explained about T profile in wall that $q_x^"=constant=-k\frac{\partial T}{\partial x}$, $k=(k_0+k_0 \beta T)$. So $q_x^"=constant=(k_0+k_0 \beta T)(-\frac{\partial T}{\partial x})$. And if $\beta>0$ because of $T_1>T_2$ when $T_1$, $(k_0+k_0 \beta T)$ is high and $(-\frac{\partial T}{\partial x})$ is low. similarly when $T_2$ , $(k_0+k_0 \beta T)$ is low and $(-\frac{\partial T}{\partial x})$ is high. As a result slope is decrease in positive direction along x. that is convex up shape In the same case if $T_2>T_1$ when $T_2$, $(k_0+k_0 \beta T)$ is high and $(-\frac{\partial T}{\partial x})$ is low when $T_1$ , $(k_0+k_0 \beta T)$ is low and $(-\frac{\partial T}{\partial x})$ is high. As a result slope is increase in negative direction along x. that is convex up shape.

This is a repetition of the explanations given by the previous students. No bonus.

Question by Student 201327128

Dear professor, I would like to answer about your question(this monday).

I'm sorry, I'm not good at writing with a formula on the computer. So I attached a image.

You need to typeset your post using LATEX. Only attach images for figures/schematics. Also, other students have provided good explanations already — we need to move on now.

Question by Student 201527110

Professor, I have question about the differences between heat flux and energy density. At one glance, they have exactly same unit ($W/m^2$) and similar form ($\sigma T^2$). So is there any differences between those or assumptions (conditons) of those?

03.15.18

Here you mean that the energy density in a room is the same as the heat flux due to radiation coming out of a black body. They may have a similar form but this doesn't mean they are subject to the same assumptions.. You can determine the assumptions from my explanations in class.

Question by Student 201327139

Professor, I have a question about Assignment #1, Problem 3. I used heat eqs,
$ \frac{\partial}{\partial t} \left( \int_{V}^{} \rho cT dV \right) =-\int_{S}^{} q^{\prime\prime} \centerdot \vec{n}dS +\int_{V}^{}SdV$, $ q^{\prime\prime}_{conv.1} = -q^{\prime\prime}_{conv.2} $, and $ h_1(T_{P_1}-T_{\infty_1}) = h_2(T_{\infty_2}-T_{P_2}) $, therefore I found a expression that $ T_{P_2}=\frac {20}{3} \left( 80'C - T_{P_1} \right) +20'C $. I want to know another eqs for solving $ T_{P_1}, T_{P_2} $. But I can't find it. Where can I get it? Thank you.

03.21.18

Good question. You can get a second equation by applying the heat equation in integral form to one of the plates. Then, you'll have 2 equations for 2 unknowns. 2 points bonus.

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