For the first of the plate(characteristic length would be 400mm)
Re=$\frac{VL}{v}$=$\frac{(2×.4)}{(18.97×10^-6) }$=42.171×$10^3$
which is less than critical reynolds number
Therefore
Nu=0.322×$Re^.5$ $Pr^.333$
Nu=$0.322×(42.171×10^3)^.5×0.696^.333$
Nu=58.60= $\frac{h_x x}{k}$
k=53.8 from heat transfer table for 600℃ for air
58.60= $\frac{h×.4}{53.8}$
h=7881.7w/$m^2$ k
Q=$hA(T_s-T_a )$
Q=7881.7×.4×.4(1000-200)
Q=1.00885×1$0^6$ Watt
For entire plate we would consider …
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