Answer:
[tex]T_{surface}=343.86\°c[/tex]
Explanation:
We define the constant, so
[tex]k=0.0258W/m^2K\\\upsilon_{air}=1.608*10^5m^2/s\\Pr=0.7282[/tex]
So, begin calculating the Reynolds number, so
[tex](Re)_{x=1.5} = \frac{V_{\infty}x}{\upsilon_{air}}\\(Re)_{x=1.5} = \frac{2.5*1.5}{1.608*10^{-5}}}\\(Re)_{x=1.5} = 2.33*10^{5}[/tex]
How reynolds number is greater than critical reynolds number, the flow
of the air is near about the turbulent flow,
we now calculate the local nusselt number
[tex](Nu)_x = 0.332 (Re)^{1/2}_x(Pr)^{1/3}\\(Nu)_x= 0.322(2.33*10^5)^{1/2}*(0.782)^{1/3}\\(Nu)_x = 143.1977[/tex]
calculate the local convection heat transfer coefficient
[tex]h_X = \frac{(Nu)_xk}{x}\\h_x= \frac{143.1977*0.0258}{1.5}\\h_x=2.46300 W/m^2K[/tex]
Apply the energy balance equation
[tex]q=h_x(T_{surface}-T_{\infty})\\810=2.463*(T_{surface}-15)[/tex]
[tex]T_{surface}=343.86\°c[/tex]
