Heat Transfer Questions & Answers | |
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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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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$^4$. |
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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.
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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).
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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.
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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.
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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.
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$\pi$ |