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A hole of diameter D = 0.25 m is drilled through the center of a solid block of square cross section with w = 1 m on a side. The hole is drilled along the length, l = 2 m, of the block, which has a thermal conductivity of k = 150 W/m·K. The four outer surfaces are exposed to ambient air, with T[infinity],2 = 25°C and h2 = 4 W/m2·K, while hot oil flowing through the hole is characterized by T[infinity],1 = 330°C and h1 = 50 W/m2·K. Determine the corresponding heat rate and surface temperatures.

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Answer:

[tex]q=4.013\:\:kW\\\\T_1=278.91\:\:^{\circ}C\\\\T_2=275.82\:\:^{\circ}C[/tex]

Explanation:

[tex]R_{conv,1}=(h_1\pi D_1L)^{-1}=(50*0.25*2)^{-1}=0.01273\:\:K/W\\\\R_{conv,2}=(h^2*4wL)^{-1}=(4*4*1)^{-1}=0.0625\:\:K/W\\\\R_{cond(2D)}=(Sk)^{-1}=(8.59*150)^{-1}=0.00078\:\:K/W[/tex]

So, heat rate can be calculated as follows:

[tex]q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{conv,1}+R_{conv,2}+R_{cond(2D)}} =\frac{330-25}{0.076} =4.013\:\:kW[/tex]

Surface temperatures can be calculated as follows:

[tex]T_1=T_{\infty,1}-qR_{conv,1}=330-51.09=278.91\:\:^{\circ}C\\\\T_2=T_{\infty,2}+qR_{conv,2}=25+250.82=275.82\:\:^{\circ}C[/tex]

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