Heat Transfer Questions & Answers  
You have to use the “high-speed flow” correction when the Eckert number is higher than the inverse of the Prandtl number. Sometimes, it's not possible to know this before solving the problem. If you think the high-speed flow correction is not necessary, then solve the problem ignoring the high-speed flow correction, then calculate the Eckert number, and make sure it's low enough. If the Eckert number is too high, recalculate the problem using the high-speed flow correction. I'll give you 2 point bonus boost for this question.
06.09.14
Question by Student 201027128
Hello, Dr. parent I want to know about assignment 8 question number 4 condition of problem is 1m square vertical plate Can I use the following ???
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and then plate has two surface so I have to find 2*(convection heat transfer)???
06.13.14
Yes you can use that correlation as long as the Rayleigh number falls in the appropriate range restriction. Also, you should compute the heat loss on both sides of the plate by multiplying by 2. I'll give you 0.5 point bonus boost for this question.
Question by Student 201027128
Professor one more question about free convection first, It is about irregular solids free convection table box is written "The charaterisitic length L corresponds to the distance a fluid particle travels in a boundary layer":
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what does mean?? you solve about the irregular solids problem in class the solid has each length 0.02m, 0.05m and 0.04m and you find L =(0.4+0.25)*0.25+(0.4+0.1)*0.75=0.54 How can I do that????
06.14.14
Hm, well, the characteristic length $L_c$ is the average distance a fluid particule would travel while touching the surface of the body.. How to determine $L_c$ depends on the situation.. I explained this in class.
Question by Student 201027128
professor, I want to know about the final exam 2013 question number 6. condition is Friction Force 0.144N and you give a hint friction factor f is equal to (-dP/dx)D/[$\rho$u$_b$^2*0.5] D(Diameter) is given and also u$_b$ is found by mass flow rate. I think (-dP/dx) is getted by Friction force -dP/dx have to positive and dimenssional of -dP/dx is N/m^3 so I use Friction Force/ Volume of pipe -dP/dx$= $61N/m^3 , u$_b$$=$0.2m/s friction factor is 0.0305 almost same 0.03 this pipe is smooth e/D$=$0 We find Reynolds Number if we use moody chart but laminar flow and turblent flow each different value when laminar flow viscous is 0.001kg/ms(answer of question) but turblent flow viscous is 0.0002kg/ms But we don't know this flow is laminar flow or turblent flow but Nusselt Number is different for each flow We want also Prandtl Number so we find C$_p$ and k many value is unknowm so I think it needs to iteration but when I play iteration, i have to iterration each flow laminar and turblent ??? How can I deside sort of flow??
Please typeset the mathematics correctly using , and put proper punctuation.. Then I will answer your question..
Question by Student 201427150
Hello Professor :) I have 2 questions about your previous lecture contents. First, at first time of Heat Transfer lecture, you said there are 2 types of Heat Transfer. These are Conduction and Radiation. However, we studied about 'Convection' today. Does 'Conduction' includes 'Convection' on broad sense?? Also, I have one more question about our handwriting. This is also at first lecture, 'mean free path'. You said 'Z' is the number of collisions by one particle during time 'delta t', and 'N' is the number of particles per unit volume. What I wonder is that why Z equals to N? Actually I don't understand it well... I want to know that what relationship exists between Z and N. Thank you very much!!
03.14.16
This is a good question. “Convective” heat transfer is just an handy way of dealing with the complex conduction heat transfer that occurs within matter in movement. So in this course, we say that convective heat transfer is the heat transfer that occurs at the interface between a solid and a fluid and we use the parameter $h$ to deal with this situation: this is an engineer “trick” that makes it easier to solve engineering problems. But fundamentally in physics, there are only two types of heat transfer: conduction and radiation.
For the second question, please ask it below and delete it from your 1st post. Only 1 question per post is allowed.
I'll give you 2 points bonus boost for your question.
Question by Student 201427564
Hello. I have a question about ur 2nd lecture. Actually its the range of thermo. Anyway, i learned that work 'by the volume' on environment is positive in thermo class. But u wrote negative. Am i wrong? Please answer and have a good day. Thx.
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03.16.16
Hi, I'll answer your question after you type it better. Please use standard English spelling: “u” should be “you”, “i” should be “I”, etc. Also, Place your figure inline so that it appears within your question instead of being an attachment. This will make it easier for everyone to read your question.
Question by Student 201427564
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I'm sorry. Hello again. I have a question about your 2nd lecture. Actually it is the range of Thermodynamics. Anyway, I learned that work 'by the volume' on environment is positive in Thermodynamics class. But you wrote negative. Am I wrong? Please answer and have a good day. Thank you.
03.17.16
Your question is incomplete: you have to outline in your post the first law that the professor used in his course. Make a correction by clicking on EDIT just right of your question and add the outline of the first law using (write the precise mathematical terms he used). There are many ways the first law can be written.. I will answer your question once you do this.
Question by Student 201427115
hello professor, I think there is something wrong at our assignment. At problem 2, you wrote the answer is 9.6 $^{\circ} C$. But there is some problem to 9.6 $^{\circ} C$ become answer. Two coefficients of temperature of the plate have different units. $$\sigma = 5.67\times 10\wedge (-8)[ W/m\wedge (2)K\wedge (4)]$$ Its temperature unit is K. $$h=12[W/m\wedge (2)^{\circ} C]$$ its temperature unit is $^{\circ} C$. so it can't be calculated directly. I attached my solution. So please answer me. Thank you for reading my question.
03.19.16
This is a very good question, and I am glad you are asking it. When using the relationship for radiation heat transfer (i.e. $q^{\prime\prime}=\sigma T^4$), you must use units of absolute temperature like Kelvin or Rankine. But, when using the relationship for convective heat transfer (i.e. $q^{\prime\prime}=h(T_{\rm s}-T_\infty)$) or conduction heat transfer (i.e. $q^{\prime\prime}_x=-k \partial T/\partial x$), you can use either units of Kelvin or of Celcius. This is because convective and conduction heat transfer depend on a simple difference of temperatures, and the difference in Kelvin is exactly equal to the difference in Celcius. Thus, you can change the unit Kelvin to Celcius as you wish. But, you can't do this for radiation thus because the radiative heat transfer can not be expressed as a simple difference of temperatures: it corresponds to a difference of $T^4$, and a difference of K$^4$ is not equal to a difference of $^\circ$C$^4$.
I'll give you 1.5 points bonus boost for your question. I would have given more if your typesetting would have been better. You should write “W/m$^2$$^\circ$C” not “W/m $\wedge$(2)$^\circ$C”. To learn quickly, just right click on any mathematical expression on my website. Then select “Show Math As” and then “ 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.
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 , 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, skills can be an asset on a resume ( 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/:)
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