Heat Transfer Questions & Answers

This is a very good question, and I am glad you are asking it. When using the relationship for radiation heat transfer (i.e. $q^{\prime\prime}=\sigma T^4$), you must use units of absolute temperature like Kelvin or Rankine. But, when using the relationship for convective heat transfer (i.e. $q^{\prime\prime}=h(T_{\rm s}-T_\infty)$) or conduction heat transfer (i.e. $q^{\prime\prime}_x=-k \partial T/\partial x$), you can use either units of Kelvin or of Celcius. This is because convective and conduction heat transfer depend on a simple difference of temperatures, and the difference in Kelvin is exactly equal to the difference in Celcius. Thus, you can change the unit Kelvin to Celcius as you wish. But, you can't do this for radiation thus because the radiative heat transfer can not be expressed as a simple difference of temperatures: it corresponds to a difference of $T^4$, and a difference of K$^4$ is not equal to a difference of $^\circ$C$^4$. ​ I'll give you 1.5 points bonus boost for your question. I would have given more if your LATEX typesetting would have been better. You should write “W/m$^2$$^\circ$C” not “W/m $\wedge$(2)$^\circ$C”. To learn LATEX quickly, just right click on any mathematical expression on my website. Then select “Show Math As” and then “TEX command”. ​ Also, don't attach your solutions to your post (I deleted them). It's too time consuming for me to go through your solutions. Your question by itself was fine.

03.19.16

Question by Student 201027112 Hello professeur, I have a question about derivation of integral form of heat equation. Following the steps, it starts from differential form of heat equation. $$ \frac{\partial}{\partial t}(\rho cT) =-\frac{\partial}{\partial x}q_x^{"} -\frac{\partial}{\partial y}q_y^{"} -\frac{\partial}{\partial z}q_z^{"}+S $$ And multiply by $dV=dxdydz$ and integrating both sides. I can understand up to this step. $$ \int_V\frac{\partial}{\partial t}(\rho cT)dV =-\int_z\int_y\partial q_x^{"}dydz -\int_z\int_x\partial q_y^{"}dxdz -\int_y\int_x\partial q_z^{"}dxdy+\int_V SdV $$ But I've little confused from next step. $$ \frac{\partial}{\partial t}\int_V(\rho cT)dV =-\int_S \overrightarrow{q^{"}}\cdot\widehat{n} dS+\int_V SdV $$ I understand that this step is expressing each directions of heat fluxes (x, y, z direction) to single vecter form. But suddenly partial operator in front of each heat flux is gone. I've wondered this point. $$ \frac{\partial}{\partial t}\int_V(\rho cT)dV =q_{in}-q_{out}+\int_V SdV $$ Equation above is conclusion of integral form of heat equation. But for derivation of $q_{in}$ and $q_{out}$, maybe partial operator should remain. Then partial operator and integral operator can be elimiated. I think the step that I wondered should be below. $$ \frac{\partial}{\partial t}\int_V(\rho cT)dV =-\int_S \partial\overrightarrow{q^{"}}\cdot\widehat{n} dS+\int_V SdV $$ Am I wrong or I miswrote in your lecture? If I'm wrong, where am I missed something in process of derivation? I'll wait your answer. Thank you. :)

03.20.16

This is a good point. The problem is in the second equation (maybe you made a mistake when writing down what I wrote, or maybe I made a mistake on the blackboard — this happens sometimes): $$ \int_V\frac{\partial}{\partial t}(\rho cT)dV =-\int_z\int_y\partial q_x^{"}dydz -\int_z\int_x\partial q_y^{"}dxdz -\int_y\int_x\partial q_z^{"}dxdy+\int_V SdV $$ There is an integral missing. This equation should be: $$ \int_V\frac{\partial}{\partial t}(\rho cT)dV = - \int_z\int_y\int_{x=0}^{x=L}\partial q_x^{"}dydz - \int_z\int_x\int_{y=0}^{y=H} \partial q_y^{"}dxdz - \int_y\int_x\int_{z=0}^{z=D} \partial q_z^{"}dxdy + \int_V SdV $$ Notice the missing integrals in front of the heat fluxes, and note that $L$, $H$, $D$ are the length, height, and depth of the domain we are integrating. The integrals of the heat fluxes can be easily done and this will yield: $$ \int_{x=0}^{x=L}\partial q_x^{"} = q_{x=L}^{"}-q_{x=0}^{"} $$ and similarly for the other dimensions. I'll give you a 2 point bonus boost for your question.

Question by Student 201427130 When I solve assignment #1, question #3. I find something strange. when I solve that, i define $q_{\rm rad.in}$ and $q_{\rm rad.out}$. I only know energy moving occur that high temperture move to low temperture. But $q_{\rm rad.out}$. it is moved to high temperture to low temperture. $q_{\rm rad.out}$ is energy that $T_{\rm 1}$ to $T_{\rm 2}$ . But $T_{\rm 1}$ is colder than $T_{\rm 2}$. why we define that energy moving?

03.22.16

Please typeset your question correctly using LATEX, and I will answer it afterwards: T1 should be $T_1$, q[radiation in] should be $q_{\rm rad,in}$, etc. This will make your question easier to read for me and for the class, and I can then answer it using the same notation as you used. Besides, LATEX skills can be an asset on a resume (LATEX is not only used on my website: it's commonly used to typeset mathematics within google docs, wikipedia, scholarpedia, scholarly publications and books, etc).

Question by Student 201427130 I'm sorry for my mistake for not use latex. So I ask it again sir. When I solve assignment #1, question #3. I find something strange. when I solve that, i define $q_{\rm rad.in}$ and $q_{\rm rad.out}$. I only know energy moving occur that high temperture move to low temperture. But $q_{\rm rad.out}$. it is moved to high temperture to low temperture. $q_{\rm rad.out}$ is energy that $T_{\rm 1}$ to $T_{\rm 2}$ . But $T_{\rm 1}$ is colder than $T_{\rm 2}$. why we define that energy moving?

The net energy transfer must be from a hot body to a cold body. But this doesn't mean a cold body can not give energy to a hot body. If a cold body gives some energy to a hotter body, then the hotter body must give in return more energy to the cold body so that the net energy transfer is from hot to cold. I'll give you 1 point bonus boost for your question. Your question was good and I would have given more if you had typeset your mathematics correctly and if the spelling, grammar, and formulation of the sentences would be better.

Question by Student 201227123 Hello professor. In heat transfer tables, There is chart of efficiencies of retangular fins. I think this chart is relatvie with fins that isn't insulated at tip because $L_c$ = L + $\frac{t}{2}$. If tip of fin is insulated $L_c$ = L I wonder if the tip of fin is insulated and $L_c$ = L. Can I use chart of efficiencies of retangular fins? If I can't, how can I get a fin efficieny of retangular fins with respect to insulated tip.

04.05.16

Yes, you are right, the chart is not for a fin with an insulated tip (because such fins are not commonly used). Thus, if the tip is insulated, you need to remember that $L_{\rm c}=L$ instead of $L_{\rm c}=L+\frac{1}{2}t$. I think I mentioned something about this in class... But I feel generous so I'll give you 1.5 point bonus boost for this question.	04.06.16

Question by Student 201327557 Hello sir. During doing assignment 3. #1. I just wonder about efficiencies of circumferential fins. What is the geometrical meaning of characteristic length $L_c$ and $r_{2c}$ and why is $L_c$ = L + $\frac{t}{2}$ and $r_{2c}$ = $r_1$ + $L_c$. I heard in class that $L_c$ = L + $\frac{t}{2}$ is because $L_c$ = $\frac{volume}{area}$ but it didn't working.

Hmm, there is no “geometrical meaning” I can think of.. As for the normal fins, the circumferential fins were derived from the heat equation. Thus, the equations listed in the tables were simply derived making the standard fin assumptions. The question is not so clear, so I'll give you just 1 point bonus boost for it.

Question by Student 201027112 Hello professeur, I have a question about problem #2 of assignment #3. To solve this problem, I used resistance analogy between surface of pipe and ground surface. Therefore, I had to find shape factor when considering conductive resistance over ground. But I found the table, there is three shape factors about problem's condition. $$ \frac{2\pi L}{\cosh^{-1}(D/r)} $$ $$ \frac{2\pi L}{\ln(2D/r)} $$ $$ \frac{2\pi L}{\ln\frac{L}{r}[1-\frac{\ln(L/2D)}{\ln(L/r)}]} $$ Of course, each shape factor has each restrictions. In these restrictions, I'm not clear that how much bigger or lower value makes the sign " $\ll$ " or " $\gg$ ". For example, A$\gg$B means that A is 10 times bigger than B ? or 100 times? If it has some typical references or criteria? I'll wait your answer/:)

This is a good question. When they derived these shape factors, they had to make some assumptions and those are indicated in the rightmost column. When they indicate $L \gg D$ it means that $L$ should be infinitely greater than $D$ for the given exact solution to be valid. I'll give you 2 points bonus boost for your question.

Question by Student 201427130 Sir, I want to ask a question. In today class, we didn't use assumption in water jet problem. We just solve it without any assumption. In that problem '$\frac{\partial pct}{\partial x}=\frac{\partial }{\partial x}(\frac{K\partial T}{\partial x})$'. I think it is width and we must think about radiation. But Professer didn't use assumption. Why? I want to know.

Of course, you need to list your assumptions (as for any other problem), but I went quickly to give you a hint on how to solve the problem. I'll give you 0.5 point bonus boost for this question.

Question by Student 201427130 Sir, I have a question for class. When we solve the Design problem which is find $\varepsilon$, we use $[kr^{2}\frac{\partial T}{\partial r}=-Sr^{3}/3]$. In this equation, S is term of volume. But heat generation by electric is term of surface . I confused. Because s's term is not equal. I want to ask what is my error and how we can decided S. If S's term is not equal to function term, how can we do?

04.11.16

What is given is not the heat generation per surface area but the power in W. Thus, the heat generation per unit volume $S$ is the power divided by the volume. I'll give you 0.5 point bonus boost for your question because it's not really a question but a misunderstanding of the problem statement.	04.12.16

Question by Student 201327557 Hello professor, I have a simple question about Heisler chart example for cylinder. You told that using the Heisler chart with interaction between infinite cylinder and the "plane wall". Thus Equation below, $$ (\frac{T-T_{\infty}}{T_{i}-T_{\infty}}) =(\frac{T-T_{\infty}}{T_{i}-T_{\infty}})_{Vol1} (\frac{T-T_{\infty}}{T_{i}-T_{\infty}})_{Vol2} $$ In this example, I can't understand how 2 volumes are considered. Clearly, cylinder is just "one" volume. And What the "plane wall" is designated?

04.15.16

You need to fix your question before I can answer it. Please typeset your mathematics better by making the parentheses as large as the terms within them are. Also, phrase your question better. I can not understand what you don't understand. Explain in more detail what you don't understand with additional information and example(s) if necessary.

Question by Student 201600011 Professor, I have a question about Heisler charts use. In order to compute the time needed to reach a temperature at the center of the body ( for instance a cylinder) Do we need to use the chart "temperature distribution in a cylinder" because of at centerline x=r? In the example that we have made in class, we have computed the temperature on the surface of the cylinder using "temperature distribution in a cylinder" and "mid plane temperature as a fonction of time", what if we need to compute the centerline temperature of the cylinder? Thank you,

04.22.16

If you're seeking the center temperature in a cylinder, sphere or plane wall, then there's no need to use the “temperature distributions” charts. Just use the chart “centerline temperature” or “midplane temperature”. I think I mentioned this in class, so I'll give only 1 point bonus boost for this question.

Question by Student 201227124 Professor, I have a porblem about finding $\alpha$ of concrete. To solve Assignment 4 q#1, I have to know $\alpha$ of concrete but in table there are two concrete property(concrete cinder and stone 1-2-4 mix). In two case, $\alpha$ of concrete was not certain value or not written. how can I solve this problem?	04.23.16

You should always compute $\alpha$ using more fundamental properties $k$, $c$, etc, and never use $\alpha$ directly from the tables. This is because in the tables some solids have a range of $c$, $k$, etc, and not specific values. The value you choose for $k$ and $c$ at one point in your solution must be the same as those that are used to calculate $\alpha$, or your solution will be inconsistent. I think I mentioned this in class already, so I'll give you just 0.5 point bonus boost.

Question by Student 201227127 Hi professor, I have a question about not infinite cylinder. in this case, use heisler chart with intersection between infinte cyl. and plane wall. but I think we must find biot and fourier number about cylinder. but in my lecture note, we find fourier number about infinite cyl. and plane wall, saparetely. When I calculate fourier number about total cylinder, $\mathit{Fo}$<0.2. but $\mathit{Fo}$ in infinite wall and infinite cylinder are satisfied $\mathit{Fo}$>0.2 saparately. then Can't we use heisler chart about total cylinder?

You can still use the Heisler charts if $\textrm{Fo}<0.2$. But keep in mind there will be a more substantial error on the term determined. In this case, the error is on the parameter $(T_0-T_\infty)/(T_i-T_\infty)$. Thus, if your Fourier number is much less than 0.2, and you have no choice but to use the Heisler chart to solve the problem, you need to make a statement in your solution that there may be significant error on a certain term (don't just say there is error, specify on which term the error is high). I liked your question but it was sloppily typed with no uppercases at the beginning of sentences and some wrong uppercases in the middle of sentences. I'll give you 1 point bonus boost for it.

Question by Student 201600011 I have a question about the exercise 1 in design project. The first idea that came to me to solve this problem, is to write the temperature equation of the steal as a function of time : $T=780-[\frac{22t}{3600}]$ and then substitute this equation into heat equation, substituting T in $[\frac{dρcT}{dt}]$ and with $Q_{in}=0$ ,$Q_{out}=h(T-T_∞)$ and isolate $T_∞$ which is the temperature in the hooven that we are looking for. Is this method correct to solve this kind of problem? Thank you,	04.24.16

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