Answer:
The rate of heat transfer from both sides of the plate to the air is 240 W
Explanation:
Given;
area of the flat plate = 0.2 m × 0.2 m = 0.04 m²
velocity of atmospheric air stream, v = 40 m/s
drag force, F = 0.075 N
The rate of heat transfer from both sides of the plate to the air:
q = 2 [h'(A)(Ts -T∞)]
where;
h' is heat transfer coefficient obtained from Chilton-Colburn analogy
[tex]h' = \frac{C_f}{2} \rho u C_p P_r^{-2/3}\\\\\frac{C_f}{2} = \frac{\tau'_s}{2*\rho u^2/2}[/tex]
Properties of air at 70°C and 1 atm:
ρ = 1.018 kg/m³, cp = 1009 J/kg.K, Pr = 0.7, v = 20.22 x 10⁻⁶ m²/s
[tex]\frac{C_f}{2} = \frac{(0.075/2)/(0.2)^2}{2*(1.018)(40)^2/2} = 5.756*10^{-4}\\\\Thus,\\h' = 5.756*10^{-4} (1.018*40*1009)*(0.7)^{-2/3}\\\\h' = 30 \ W/m^2.K[/tex]
Finally;
q = 2 [ 30(0.04)(120 - 20) ]
q = 240 W
Therefore, the rate of heat transfer from both sides of the plate to the air is 240 W
Answer:
1. q = 240 W
Explanation:
Let's begin by listing out the parameters given unto us:
Length = 0.2 m, Width = 0.2 m, T∞ = 20 °C, Ts = 120 °C, velocity = 40 m/s, τ = 0.075 N
At 70 °C and 1 atm, air has these thermodynamic properties (we can get these from the thermodynamic tables)
ρ = 1.018 kg/m³, Cp = 1009 J/kgK, v = 20.22 * 10^(-6) m²/s, Pr = 0.70
The formula for heat transfer rate is given as:
q = 2hL² ΔT
f ÷ 2 = 0.5τ ÷ ρ u² = (0.5 * 0.075/0.2²) ÷ (1.018 * 40²)
f = 5.76 * 10 ^(-4)
h = f ÷ 2 * [ρ u CP * Pr ^(-⅔)]
h = 5.76 * 10^(-4) * 1.018 * 40 * 1009 * (0.7^(-⅔))
h = 30 W/m²K
q = 2 * 30 * 0.2 * (120 - 20)
q = 240 W
