A thin, flat pate that is 0.2m by 0.2m on a side is oriented parallel to an atmospheric airstream having a velocity of 40m/s. The air is at a temperature of T∞ = 20 °C, while the plate is maintained at Ts = 1200 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?

This question is incomplete, the complete question is;

A thin, flat pate that is 0.2m by 0.2m on a side is oriented parallel to an atmospheric airstream having a velocity of 40m/s. The air is at a temperature of T∞ = 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?

the rate of heat transfer from both sides of the plate to the air is 236.54 W

Explanation:

Given the data in the question,

first we calculate the  Reynold's number for the flow

Re = pu∞d / Ц

Re = (1.12 × 40 × 0.2) / 1.983 × 10⁻⁵

Re = 451840

Now the Local skin friction coefficient is given as;

Cfx =  T / ( 1/2pu∞²)

Cfx = (Fd/A) / ( 1/2pu∞²)

Cfx = (0.075/(2×0.2×0.2)) / ( 1/2 × 1.12 × 40²)

= 0.9375 / 896

= 0.0010463

Cfx = 1.0463 × 10⁻³

Apply Reynold's- cOLBURN analogy

Cfx/2 = StₓPr^2/3

so

1.0463 × 10⁻³ / 2 = (h/pu∞Cp) × ( 0.711)^2/3

5.2315 × 10⁻⁴ × 1.12 × 40 × 1.005 × 1000 = h(0.711)^2/3

h = 23.554 / 0.7966

h = 29.56 W/m².K

so

The heat transfer rate from both the sides of the plate will be;

Q = 2 × 29.56 × 0.2 × 0.2 × ( 120 - 20 )

Q = 236.54 W

Therefore the rate of heat transfer from both sides of the plate to the air is 236.54 W

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explanation: