Answer:
[tex]At 10^5 = 17270W/m[/tex]
[tex] At 5*10^5 = 13127.84 W/m[/tex]
[tex] At 10^6 = 8472.2 W/m[/tex]
Explanation:
We are given:
T_o = 20°C
Flat plate temp, T_s = 230°C
v = 30m/s
Let's find the mean:
[tex] T_f = \frac{20+130}{2} [/tex]
= 75°C
Using air table at 75°C at 1 atm pressure, we have:
[tex]Thermal conductivity, k = 29.66*10^-^3 W/m•K[/tex]
Prandtl number, Pr = 0.7087
[tex]Viscosity, v= 20.82*10^-^6 m^2/s [/tex]
Calculating Reynolds number, we have:
[tex]Re_L = \frac{VL}{v}[/tex]
[tex] Re_L = \frac{30*1}{20.82*10^-^6}[/tex]
[tex] = 1.44*10^6[/tex]
Since the Reynolds number is higher than the critical Reynolds number, we can say there is a turbulent flow.
Therefore we'll now use the formula:
[tex] Nu_L = (0.037Re_L^0^.^8 - A)Pr^0^.^3^3[/tex]
[tex]For Re_x = 10^5[/tex]
To get A using the formula, we have:
[tex]A = 0.037(10^5) - 0.664(10^5)^0^.^5[/tex]
= 160.024
[tex]To find Nu_L[/tex]
[tex]Nu_L =(0.037(1.44*10^6)^0^.^8-160.024)(0.7087)^0^.^3^3[/tex]
= 2646.78
Connective heat transfer:
[tex]h_L = \frac{Nu_L K}{L}[/tex]
[tex]=\frac{(2646.78)(2966*10^-^3)}{1}[/tex]
[tex]=78.50W/m^2•K[/tex]
Lets calc for heat transfer:
[tex] q_1_0_^_5 = 2h_L L(T_s-T_o)[/tex]
= 2(78.50)(1)(130-20)
= 1720W/m
[tex]For Re_x = 5*10^5[/tex]
[tex]A = 0.037Re_x^0^.^8-0.664Re_x^0^.^5[/tex]
[tex] = 0.037(5*10^5)^0^.^8-0.664(5*10^5)^0^.^5[/tex]
= 871.323
Calculating Nu_L we have:
[tex] h_L =\frac{Nu_L K}{L}[/tex]
[tex] = \frac{2011.88*29.66*10^-^3}{1}[/tex]
= 59.672W/m²•K
For the heat transfer:
[tex] q_5_*_1_0_^5 = 2(59.672)(1)(130-20)[/tex]
= 13127.84W/m
[tex]For Re_x = 10^6[/tex]
[tex]A= 0.037(10^6)^0^.^8-0.664(10^6)^0^.^5[/tex]
= 1670.54
Calculating Nu_L:
[tex] Nu_L = (0.037(1.44*10^6)^0^.^8-1670.54)(0.7087)^0^.^3^3[/tex]
= 1298.51
Calculating the convective heat transfer coefficient:
[tex] h_L = \frac{Nu_L*k}{L}[/tex]
[tex]\frac{(1298.51)(29.66*10^-^3)}{1}[/tex]
=38.51 W/m²•K
To calculte for rate of heat transfer, we have:
[tex] q_1_0_^6=2h_LL(T_s-T_o)[/tex]
=2(38.51)(1)(130-20)
= 8472.2W/m
