Answer:
a) 15.37 mm
b)
c) 5.7186 W/m². K
d) 0.60 m
Explanation:
Given that :
The surface temperature = 130°C = ( 130+ 273 ) K = 473 K
suspended in quiescent air at 25°C = ( 25 + 273 ) K = 298 K
[tex]T_f = \frac{T_s + T \infty}{2 } \\ \\ T_f = {403+298}{2} \\ \\ T_f = \frac{701}{2} \\ \\ T_f = 350 \ K[/tex]
Atmospheric Pressure = 1 atm
The properties obtained from Table A - 4 include :
v = [tex]20.92 *10^{-6} \ \ m^2 /s[/tex]
k = 0.03 W/m K
Pr = 0.700
η = 5
[tex]Gr_x = 9.8[T_s - T \infty] \frac{x^3}{v^2}[/tex]
[tex]Gr_x = 9.8*\frac{1}{350}[130-25] \frac{x^3}{20.92*10^{-6}^2}[/tex]
[tex]Gr_x =6.718*10^9 \ x^3[/tex]
[tex]\gamma = 5(0.15)((6.718*10^9)(0.15)^{3}/4)^{-1/4}[/tex]
[tex]\gamma = 0.01537 \ m[/tex]
[tex]\gamma = 15.37 \ mm[/tex]
Hence, the boundary layer thickness at a location 0.15 m measured from the lower edge is 15.37 mm
b) The maximum velocity in the boundary layer [tex]\eta x_1[/tex] with f'(n) = 0.275
[tex]u = \frac{2v}{x} Gr_x^{1/2} f'(n)[/tex]
[tex]u= \frac{2*20.92*10^{-6}}{0.75} [6.718*10^9(0.15)^3]^{1/2} * 0.275[/tex]
u = 0.3659 m/s
the maximum velocity in the boundary layer at this location is 0.3659 m/s
the position in the boundary layer where the maximum occur is calculated as:
[tex]y_{max} = \frac{1}{5}(15.37 \ mm)[/tex]
[tex]y_{max} =[/tex] 3.074 mm
c) Using the similarity solution result, , determine the heat transfer coefficient 0.15 m from the lower edge.
we know that:
[tex]Nu_x = \frac{h_x *x }{K} = (Gr_x/4)^{1/4} \ g (pr)[/tex]
[tex](Gr_x/4)^{1/4} \ g (pr)[/tex] = [tex](6.718*10^9(0.15)^3/4)^{1/4} *0.586[/tex]
[tex](Gr_x/4)^{1/4} \ g (pr)[/tex] = 28.593
Making [tex]h_x[/tex] the subject from the above formula:
[tex]h_x = \frac{Nu_xK}{x}[/tex]
[tex]h_x= \frac{28.593*0.03}{0.15}[/tex]
[tex]h_x[/tex] = 5.7186 W/m². K
d) to determine the location on the plate that the boundary layer we become turbulent ; we have the following:
[tex]Ra _ {x,c} = Gr_x,c^{pr} \approx 10^9[/tex]
[tex]x_c = [10^9/6.718*10^9(0.7)]^{1/3}[/tex]
[tex]x_c =[/tex] 0.60 m
