Heat Transfer Questions & Answers | |
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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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$\pi$ |