Heat Transfer Questions & Answers  
Hm, I think you're on the right track, but some logical steps may be missing. In the question, it is stated that the steel can not be cooled faster than 22 degrees per hour. This means that the steel anywhere within the block (at the center, on the edges, and on the corners) can not be cooled faster than 22 degrees per hour. As we learned in class, you can use either the LCA (or a similar approach based on a similar derivation with the same assumptions) or the Heisler charts to solve such a problem. First, you need to determine which approach is appropriate and why. Then determine which part of the steel will be cooling fastest and focus on that from then on.. I'll give just a 0.5 point boost here because the question is not really a question but more a verification of your solution.
04.24.16
Question by Student 201327557
Sir, I have question about design problem #2. Do I have to neglect interactions such as heat convection between each fins?? If my assumption is right, do I just multiply 10 for 10 aluminum fins??
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.
Question by Student 201327557
Sir, I got question about Design problem #3 just for curious. In class you mentioned APU that located on tail wing. I searched internet but I couldn't find anything about APU. What is APU and Why does airplane need APU??
04.26.16
You can read about it here:
https://en.wikipedia.org/wiki/Auxiliary_power_unit
Question by Student 201327557
Professor I have another question about fin efficiency. In my lecture note that derived fins assumptions are steady state, k constant, h constant, 1-dimensional heat transfer in x axis and t$\gg$w. I do not know why t$\gg$w is needed.
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.
Question by Student 201227135
Hello, I have a question about Heat generation in cylinder that learned few weeks ago. While solving the Heat equation start with $$\frac{\partial}{\partial t}(\rho c T)=\frac{1}{r}\frac{\partial}{\partial r}(kr\frac{\partial T}{\partial r})+S$$ I found the note that $$(q_{in})_{rad}= \sigma\epsilon A_{r=r_o}({T_w}^4-{T_s}^4)$$ from radiation H-T in large room. As I know that radiation H-T is heat flux going out of body, but in my note $(q_{in})_{rad}$. Why is the q going in? Thank you.
04.27.16
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.
Question by Student 201027112
Professeur, I have a question about derivation of power in couette flow. For derivation of power, consider fluid element which is located on top surface. Force will be needed, and that will be the friction force at y=H. Multiply the area to the shear stress ($\tau_{yx}$), friction force can be determined. You taught that area is in xz plane because we think about the shear strees ($\tau_{yx}$) in upper surface of fluid element. I wonder about this point. Why $\tau_{yz}$ is neglected? Because multiplied area is in xz plane, I think it means that this problem regarded as the 3-Dimensional problem. Thus $\tau_{yz}$ should be also considered. Am I wrong or misunderstand something? I'll wait your answer. Thank 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.
Question by Student 201227127
I have a qustion. In cylinder, we use boundary condition $\frac{dT}{dr}=0$. But when we find shape factor of sphere, we didn't consider boundary condition $\frac{dT}{dr}=0$. Why we didn't consider this boundary condition in this problem??
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.
Question by Student 201427130
When i study radiation H-T occurs without any gas, fluid. But it doesn't mean that radiation doesn't occur between solid and gas. However when i solve assignment #1 q#2, i neglect rad H-T with air. Otherwise i used rad H-T with wall. Can i do that at another problem? I confused.
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.
Question by Student 201327557
Professor, I have a question about transient conduction. You had posed example of multidimensional transient heat trasfer by cube. But you told cube is hard to express in blackboard, so you made the cylinder example instead of the cube example. So I think about transient conduction of cube problem. Cube is in 3-dimension, put the origin of coordinate on the geometrical center. And each direction has each wall, so 3 walls interact (intersect) each other. So finally I think that if I should to use Heisler chart to solve the cube example, it needs to use the formula below in the table. $$ (\frac{T-T_{\infty}}{T_{i}-T_{\infty}})_{int} =(\frac{T-T_{\infty}}{T_{i}-T_{\infty}})_{wall_x}\times (\frac{T-T_{\infty}}{T_{i}-T_{\infty}})_{wall_y}\times (\frac{T-T_{\infty}}{T_{i}-T_{\infty}})_{wall_z} $$ My idea is right? If it is not working, which part do I miss something? Thank you.
04.28.16
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.
Question by Student 201600012
Professor, I have a question about Question 3 of Design Problem Set 1. In order to compute the heat flux generated by the copper cable, we multiply the expression $RI^2$ by the length of the cable L and we divide it by the area of the cross section A. Although I do understand that we need to divide by a unit length in order for the expression to be correct from the dimensional point of view, I do not understand why we use this combination of the length and the cross section to have our expression in Watts. Thank you in advance.
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.
Question by Student 201027112
Professeur, I have a question in your lecture about boundary layer. In momentum equation, it needs to compare convection term and viscous term when friction force is significant. $$ \mu\vert\frac{\partial^2 u}{\partial x^2}\vert\gtrsim \rho u\vert \frac{\partial u}{\partial x} \vert $$ From next part, I've little bit confused. It is the process to express characteristic length. $$ \begin{equation} \mu\frac{\Delta u}{L_c^2}\gtrsim \rho u\frac{\Delta u}{L_c} \end{equation} $$ And the result, $$ L_c \lesssim \frac{\mu}{\rho u} $$ In equation (1), why the numerator of left hand side of equation above is $\Delta u$, not $(\Delta u)^2$? I think that left hand side is second derivative of x, $\Delta u$ should be $(\Delta u)^2$. But consider the result, dimension seems to be right. What do I missing in this progress? I'll wait your answer. Thank you.
05.02.16
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|”.
05.03.16
Question by Student 201027112
Professeur, I have a question about derivation of the thermal layer. In process to derive thermal layer, valid region of Prandtl number should be exist due to the assumption. I have some problems to understand about the inequality. \begin{equation} \frac{3}{280}\left( \frac{\delta_t}{\delta} \right)^4 \ll \frac{3}{20}\left(\frac{\delta_t}{\delta} \right)^2 \end{equation} In 10% error, \begin{equation} \label{good1} \frac{3}{280}\left( \frac{\delta_t}{\delta} \right)^4 < \frac{3}{20}\left( \frac{\delta_t}{\delta} \right)^2 \end{equation} But next step below, \begin{equation} \label{good2} \left( \frac{\delta_t}{\delta} \right)^2 < \frac{7}{5} \end{equation} It is clear that if arrange the coefficients of both sides of $\ref{good1}$, right hand side of $\ref{good2}$ should be 14. Why is the right hand side of $\ref{good2}$ is $\frac{7}{5}$? I'll wait your answer. Thank you. :)
05.07.16
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.
Question by Student 201427130
Sir. I have a question for class. When we solve example at end of class, we use ideal gas equation. But turbulent 'flow' can't achieve ideal gas state. Because ideal gas don't have flow direction. Why we can use it?
05.08.16
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.
05.09.16
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