Answer:
Heat flux on the plate surface is 1.45 x 10⁴ W/m², Heat transfer coefficient is 72.6 W/m². K
Explanation:
<u>The temperature profile of the airflow is given as</u>
<u>T(y) = T∞ – (T∞ - Ts)exp[(-V/</u>α<u> fluid)y]</u>
<u>Calculate the mean film temperature of air</u>
T(f) = T(s) + T∞/2, where T(s) is the surface temperature of the plate and T∞ is the temperature of free stream
Substitute Ts = 220 C and T∞ = 20 C
T(f) = 220 +20/2 = 120 C
Obtain the properties of air from Table A- 15 (Properties of air at 1 atm pressure) in the text book of from the internet at T(f) = 120 C, which would give us the quantities of thermal diffusivity (α fluid) and the thermal conductivity (k) of air
α fluid = 3.565 x 10⁻⁵ m²/s
k = 0.03235 W/m.k
<u>Calculate the temperature gradient for the given profile</u>
T(y) = T∞ – (T∞ - Ts)exp({-V/α fluid}y)
Where, T∞ is the ambient temperate, V is the free stream velocity of air and y is the temperature displacement thickness
<u>Differentiate the expression with respect to y</u>
dT/dY = - (T∞ - Ts){-V/α fluid}e[(-V/α fluid) x 0]
dT/dy = -(T∞ - Ts){-V/α fluid}
<u>Substitute T∞ = 20 C, Ts = 220 C, V = 0.08 m/s and </u>α<u> = 3.565 x 10⁻⁵ m²/s</u>
dT/dy = - (20 - 220)(-0.08/3.565 x 10⁻⁵)
dT/dy = -448807.8541
<u>Calculate the heat flux on the plate surface</u>
Q = -kdT/dy
K = 0.03235 W/m.k, dT/dy = -448807.8541
Q = -(0.03235)(-448807.8541)
Q = 1.45 x 10⁴ W/m²
Therefore, the heat flux on the plate surface is 1.45 x 10⁴ W/m²
<u>Calculate the heat transfer coefficient</u>
Q = h(Ts - T∞), where h is the transfer coefficient
1.45 x 10⁴ = h(220 - 20)
H = 72.6 W/m2.k
Therefore, the heat transfer coefficient is 72.6 W/m². K
