| Heat Transfer Questions & Answers | |
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Please ask your questions related to Heat Transfer in this thread, I will answer them as soon as possible. To insert mathematics use LATEX. For instance, let's say we wish to insert math within a sentence such as $q_{\rm rad}$ is equal to $\epsilon A \sigma T^4$. This can be done by typing \$q_{\rm rad}\$ is equal to \$\epsilon A \sigma T^4\$. Or, if you wish to display an equation by itself out of a sentence such as: $$ q=-\frac{\Delta T}{\sum R} $$ The latter can be accomplished by typing \$\$q=-\frac{\Delta T}{\sum R}\$\$ . You can learn more about LATEX on tug.org. If the mathematics don't show up as they should in the text above, use Chrome or Firefox or upgrade MSIE to version 9 or above. Ask your question by scrolling down and clicking on the link “Ask Question” at the bottom of the page.
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No, you don't need to register again if you already have an account. I can't give you a bonus boost for this question thus because it doesn't involve heat transfer theory ;)
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Even if $T_\infty$ is much higher, the maximum temperature in the composite wall will still be higher than $T_\infty$. This is because the left side is insulated and the heat must come out to the environment on the right side. In order for the heat to go out from the wall to the environment, the wall must have a maximum temperature higher than the environment temperature (recall $q^{\prime\prime} \propto -\partial T/\partial x$). Of course, if the environment temperature is a thousand degrees Kelvin or more and the wall is made in plastic, the wall would melt and the problem wouldn't make sense ;) But for this question, we can assume that the wall does not melt at the temperature encountered.. I'll give you 1.5 point for this question. I would have given 2 points if you wouldn't have made a mistake in typing $T_{\rm inside}$ — check carefully with the “preview” command that your post is well typeset.
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Hmm, I am not sure what could be wrong with your solution.. Make sure to impose the boundary conditions correctly (fixed heat transfer at both ends) and if the arithmetic is done without mistake, you'll get 291$^\circ$C in the center of the rod.
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Using the chart and analyzing the intervals between the curves $r_{2c}/r_1=1$, $r_{2c}/r_1=2$, $r_{2c}/r_1=5$, etc, you can take a guess of where the value $r_{2c}/r_1=11$ would be. Alternately, you could simply state that the closest answer in this case is $r_{2c}/r_1=5$ and mention how much error you would expect this to yield on the fin efficiency (give an estimate). I'll give you 1.5 point bonus for this question. I would have given more if you would have written your paragraph better with proper punctuation.
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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.
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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.
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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.
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Please typeset the mathematics correctly using LATEX, and put proper punctuation.. Then I will answer your question..
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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. |
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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.
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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 LATEX (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.
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| $\pi$ |