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 * $$10^5
 Question by Student 201427103 I'm sorry, but I've got the exact meaning now. Let's recreate the attached expression.I am writing to ask you a question while solving Question 5 in Assignment 7. The condition given is Constant Heat Flux. So I referred to the given table. According to this table, I see "Constant heat flux, local h " . So I did the process to prove the assumption that h is also constant.(The reason for this is that if h is the constant, the average value is the same value h.) first $$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}\\$$ In reference to this equation (1) . The following is a summary of h. $$h= \frac{Ck}{x}(Gr^{*}Pr)^{m}=Ck(\frac{{g}{\rho}^{2}{\beta}{q_w}}{{k}{\mu}^{2}}Pr)^{m}(x^{4m-1}) \\$$ Let's now measure the average value of h. $$\bar{h} = \frac{{\int_0^Lh\,{\rm d}x}}{L} = \frac{{\int_0^LCk(\frac{{g}{\rho}^{2}{\beta}{q_w}}{{k}{\mu}^{2}}Pr)^{m}(x^{4m-1}) \,{\rm d}x}}{L} = \frac {Ck(\frac{{g}{\rho}^{2}{\beta}{q_w}}{{k}{\mu}^{2}}Pr)^{m}}{L}{\frac {1}{4m}}x^{4m}=\frac {k(C(\frac{{g}{\rho}^{2}{\beta}{q_w}{x^4}}{{k}{\mu}^{2}}Pr)^{m})}{4mL} \\$$ Therefore, the average value of the final Nu : $$\bar {Nu_L} = \frac {{\bar h}{L}}{k} =\frac {1}{4m} C(\frac{{g}{\rho}^{2}{\beta}{q_w}{x^4}}{{k}{\mu}^{2}}Pr)^{m}\\$$ 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: $$10^5
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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