Answer:
Explanation:
Given that:
V = 12.5m/s
L= 2.70m
b= 0.65m
[tex]T_{ \infty} = 27^0C= 273+27 = 300K[/tex]
[tex]T_s= 127^0C = (127+273)= 400K[/tex]
P = 1atm
Film temperature
[tex]T_f = \frac{T_s + T_{\infty}}{2} \\\\=\frac{400+300}{2} \\\\=350K[/tex]
dynamic viscosity =
[tex]\mu =20.9096\times 10^{-6} m^2/sec[/tex]
density = 0.9946kg/m³
Pr = 0.708564
K= 229.7984 * 10⁻³w/mk
Reynolds number,
[tex]Re = \frac{SUD}{\mu} =\frac{\ SUl}{\mu}[/tex]
[tex]=\frac{0.9946 \times 12.5\times 2.7}{20.9096\times 10^-^6} \\\\Re=1605375.043[/tex]
we have,
[tex]Nu=\frac{hL}{k} =0.037Re^{4/5}Pr^{1/3}\\\\\frac{h\times2.7}{29.79\times 10^-63} =0.037(1605375.043)^{4/5}(0.7085)^{1/3}\\\\h=33.53w/m^2k[/tex]
we have,
heat transfer rate from top plate
[tex]\theta _1 =hA(T_s-T_{\infty})\\\\A=Lb\\\\=2.7*0.655\\\\ \theta_1=33.53*2.7*0.65(127/27)\\\\ \theta_1=5884.51w[/tex]
