Heat Transfer Questions & Answers  
Question by Student 201427103
Dear Professor.
I am writing to ask you a question while solving Question # 5 in Assignment 7. The condition given is Constant Heat Flux. So I referred to the given table.

According to this table, I see "Constant heat flux, local h " . So I did the same process as the attached picture to prove the assumption that h is also constant.

The results show satisfaction when the m is 1/4.
However, if Gr * is greater than 10^5 and less than 10^11, then the last equation attached to the picture will not be formed.
How do I interpret this?

Additionally, I am having difficulty solving this Question # 5 in Assingment 7. Do you have any suggestions for our?

Thank you.
KakaoTalk_20180609_155626471.jpg    
06.09.18
You have to use for the math. Use an attached figure only for a schematic. Also, avoid breaking lines. One question is one paragraph (one idea). Breaking lines makes it hard for me to read your question.
Question by Student 201427103
Dear Professor. I am writing to ask you a question while solving Question # 5 in Assignment 7. The condition given is Constant Heat Flux. So I referred to the given table. According to this table, I see "Constant heat flux, local h " . So I did the process as the attached picture to prove the assumption that h is also constant.(The reason for this is that if h is the constant, the average value is the same value h.) The results show satisfaction when the $$ m=\frac{1}{4} $$ However, if Gr * \begin{equation} 10^5<Gr^{*}<10^{11} \end{equation} then the last equation attached to the picture will not be adapt. because according to Table this condition, $$ m=\frac{1}{5} $$ so How do I interpret this?
KakaoTalk_20180609_155626471.jpg    
Again, all the math should be written in . Don't attach a picture with mathematics. If there's a derivation you wish to discuss, then write it all here using .
Question by Student 201427103
I'm sorry, but I've got the exact meaning now. Let's recreate the attached expression.I am writing to ask you a question while solving Question 5 in Assignment 7. The condition given is Constant Heat Flux. So I referred to the given table. According to this table, I see "Constant heat flux, local h " . So I did the process to prove the assumption that h is also constant.(The reason for this is that if h is the constant, the average value is the same value h.) first \begin{equation} Gr^* = \frac{{g}{\rho}^{2}{\beta}{q_w}{x^4}}{{k}{\mu}^{2}} \\ Nu_x = \frac{{h}{x}}{k} = C(Gr^{*}Pr)^{m}=C(\frac{{g}{\rho}^{2}{\beta}{q_w}{x^4}}{{k}{\mu}^{2}}Pr)^{m}\\ \end{equation} In reference to this equation (1) . The following is a summary of h. \begin{equation} h= \frac{Ck}{x}(Gr^{*}Pr)^{m}=Ck(\frac{{g}{\rho}^{2}{\beta}{q_w}}{{k}{\mu}^{2}}Pr)^{m}(x^{4m-1}) \\ \end{equation} Let's now measure the average value of h. \begin{equation} \bar{h} = \frac{{\int_0^Lh\,{\rm d}x}}{L} = \frac{{\int_0^LCk(\frac{{g}{\rho}^{2}{\beta}{q_w}}{{k}{\mu}^{2}}Pr)^{m}(x^{4m-1}) \,{\rm d}x}}{L} = \frac {Ck(\frac{{g}{\rho}^{2}{\beta}{q_w}}{{k}{\mu}^{2}}Pr)^{m}}{L}{\frac {1}{4m}}x^{4m}=\frac {k(C(\frac{{g}{\rho}^{2}{\beta}{q_w}{x^4}}{{k}{\mu}^{2}}Pr)^{m})}{4mL} \\ \end{equation} Therefore, the average value of the final Nu : \begin{equation} \bar {Nu_L} = \frac {{\bar h}{L}}{k} =\frac {1}{4m} C(\frac{{g}{\rho}^{2}{\beta}{q_w}{x^4}}{{k}{\mu}^{2}}Pr)^{m}\\ \end{equation} To satisfy the first assumption here , m value have to be $$m = \frac {1}{4}$$ but If you look at the table, when the Gr value are: \begin{equation} 10^5<Gr^{*}_x<10^{11} \end{equation} The m value is $$m = \frac {1}{5}$$ Therefore, under (5) conditions, we were able to confirm that we were not satisfied.Why is this so? I want to know.
Thank you for your patience Professor.
Hmm, I am not sure if I am following you correctly. Why do you say the value for $m$ is $1/4$ to satisfy assumption (1) for an average $h$? This doesn't make sense. The value for $m$ is $1/5$ if ${\rm Gr}^*_x\lessapprox 10^{12}$ and $1/4$ otherwise. If you wish to integrate $h$ over a large Grashoff number range with a lower limit less than $10^{11}$ and an upper limit greater than $10^{13}$, then you need to split the integral in 2 and use two different $m$s. 0.5 point bonus for the effort.
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