Answer:
given:
[tex]T_{s} =130C\\T_{0} =20C\\L=1m\\u_{infinity} =30m/s\\[/tex]
to determine avg temperature:
[tex]T_{avg} =(T_{s}+ T_{0} )/2=(130+20)/2=75C=348K[/tex]
from the properties of air table A-4 corresponding to T avg=348 K approximately equal to 350 K so we can find:
[tex]k=10.10^-^3W/mK\\v=20.92.10^-^6m^2/s\\p_{r} =0.7[/tex]
A) case in which critical Reynolds number with value [tex]Re_{c} =10^5\\[/tex]
Reynolds number is equal to :
[tex]Re_{L} =(v_{infinity} *L)/v=(30*1)/20.92*10^-^6=1.4*10^6[/tex]
calculating local nusselt number :
[tex]Nu_{L} =0.037.Re_{L} ^3^/^4.Pr_{L} ^1^/^3-0.037.Re_{c} ^4^/^5.Pr_{} ^1^/^3+0.664.Re_{L} ^1^/^2.P_{r} ^1^/^3[/tex]
insert values and we get
[tex]2256[/tex]
now calculating the heat transfer coefficient:
[tex]h_{L} =(Nu_{L} .k)/L=2256.30.10^6/1=67.7W/Km^2[/tex]
calculating the rate of heat transfer per unit width of plate:
[tex]q=2.h_{L} .L(T_{s}- T_{0} )=2*67.7*1(130-20)=14.8kW/m[/tex]
