Answer:
Tc = 424.85 K
Explanation:
Given that,
[tex]D = 60 mm = 0.06 m[/tex]
[tex]\rho = 8000 kg/m^3[/tex]
[tex]k = 50 w/m . kc = 500 j/kg.k[/tex]
[tex]h_{\infty} = 1000 w/m^2t_{\infity} = 750 kt_w = 500 K[/tex]
[tex]surface area = As = \pi dL \\\frac{As}{L} = \pi D = \pi \timeS 0.06[/tex]
HEAT FLOW Q is
[tex]Q = h_{\infty} As (T_[\infty} - Tw) = 1000 \pi\times 0.06 (750-500)[/tex]
= 47123.88 w per unit length of rod
volumetric heat rate
[tex]q = \frac{Q}{LAs}[/tex]
[tex]= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}[/tex]
[tex]q = 1.66\times 10^{7} w/m^3[/tex]
[tex]Tc = \frac{- qR^2}{4K} + Tw[/tex]
[tex]= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 50} + 500[/tex]
= 424.85 K
