Answer:
q = 4870.0524 W
Explanation:
Given that:
[tex]T_{ \infty} = 27^0C = (27+273)K = 300 K[/tex]
[tex]T_s = 523K[/tex]
[tex]T_f = \frac{T _\infty- T_s }{2}[/tex]
[tex]T_f = \frac{300-523}{2}[/tex]
[tex]T_f =412 K[/tex]
From Table A4; Thermophysical Properties of Gases at Atmospheric Temperature; Properties of air at film Temperature are as follows:
density [tex]\rho = 0.8478 \ kg/m^3[/tex]
kinematic viscosity (v) = [tex]27.845*10^{-6} \ m^2/s[/tex]
thermal conductivity K = [tex]34.64 *10^{-3 } \ W/m.K[/tex]
Prandtl number (Pr) = 0.689
Given that :
the length of the plate is = 150 cm = 1.5 m
the width of the plate = 10 cm = 0.1 m
Then the Area = W×L = 1.5 × 0.10 = 0.15 m²
Air flow velocity [tex](u_{\infty } )= 50 \ mph = 22.35 m/s[/tex]
The Reynolds number [tex]R_{eL}[/tex] is calculated by using the formula :
[tex]R_{eL} = \frac{\mu_{\infty}L}{v}[/tex]
[tex]R_{eL} = \frac{22.35*1.5}{27.845*10^{-6}}[/tex]
[tex]R_{eL} = 1.2*10^6[/tex] turbulent flow.
Thus, we can say that the turbulent flow is throughout the entire plate. Therefore, the appropriate correlation is addressed via the Nusselt number;
[tex]N \bar {u} _L = (0.037R_{eL} -A ) Pr ^{1/3}[/tex]
where;
[tex]A = 0.037 \ Re ^{4/5}_{x,e} - 0.664 \ Re ^{1/2}_{x,e}[/tex]
and [tex]Re_{x.e} = 5*10^5[/tex]
Then;
[tex]A = 0.037 \ *5*10^5 ^{(4/5)}- 0.664 \* 5*10^5 ^{(1/2)}[/tex]
A = 871
Now;
[tex]N \bar {u} _L = (0.037*1.2*10^6-871 ) *0.689 ^{1/3}[/tex]
[tex]N \bar {u} _L = 3152.24[/tex]
We can now determine the convention coefficient since the [tex]N \bar {u} _L[/tex] is known. By using the equation;
[tex]\bar {h} = \frac{N \bar {u} _L *k}{L}[/tex]
[tex]\bar {h} = \frac{3152.24*34.64*10^{-3}}{1.5}[/tex]
[tex]\bar {h} =72.796 \ W/m^2 .K[/tex]
Finally the rate of heat removal q from both cooling surface of the plate (fin) is calculated as:
[tex]q= 2*A * \bar {h} (T_s - T _ {\infty})[/tex]
[tex]q= 2*0.15 * 72.796 (523 - 300)[/tex]
q = 4870.0524 W
