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3 years ago

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A thin, flat plate that is 0.2 m × 0.2 m on a side is oriented parallel to an atmospheric airstream having a velocity of 40 m/s.

The air is at a temperature of T[infinity] = 20°C, while the plate is maintained at Ts = 120°C. The air flows over the top and bottom surfaces of the plate, and measurement of the drag force reveals a value of 0.075 N. What is the rate of heat transfer from both sides of the plate to the air?

2 answers:

Tamiku [17]3 years ago

7 0

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

Leto [7]3 years ago

4 0

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

$h' = \frac{C_f}{2} \rho u C_p P_r^{-2/3}\\\\\frac{C_f}{2} = \frac{\tau'_s}{2*\rho u^2/2}$

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

$\frac{C_f}{2} = \frac{(0.075/2)/(0.2)^2}{2*(1.018)(40)^2/2} = 5.756*10^{-4}\\\\Thus,\\h' = 5.756*10^{-4} (1.018*40*1009)*(0.7)^{-2/3}\\\\h' = 30 \ W/m^2.K$

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

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