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?
Answer:
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
