Consider airflow over a plate surface maintained at a temperature of 220°C. The temperature profile of the airflow is given as fluid ()()expsV TyTTTy     The airflow at 1 atm has a free stream velocity and temperature of 0.08 m/s and 20°C, respectively. Determine the heat flux on the plate surface and the convection heat transfer coefficient of the airflow.

Consider airflow over a plate surface maintained at a temperature of 220°C. The temperature profile of the airflow is given as T(y) = T∞ − (T∞−Ts)exp((−V/α_fluid)y).

The airflow at 1 atm has a free stream velocity and temperature of 0.08 m/s and 20°C, respectively. Determine the heat flux on the plate surface and the convection heat transfer coefficient of the airflow

A) heat flux on the plate is;q_o = 11737.34 W/m²

B) convection heat transfer coefficient of the airflow is;h = 58.67 W/m².k

Explanation:

The temperature profile of the airflow is given as;

T(y) = T∞ − (T∞−Ts)exp((−V/α_fluid)y)

Let's differentiate with respect to y;

dT/dy = [[(T∞−Ts)V]/α](e^(-vy/α)

Where;

T∞ = 20°C

Ts = 220°C

V = 0.08 m/s

α is thermal diffusivity of air and from the table i attached at a temperature of 220°C, by interpolation it has a value of;

α = 5.33 x 10^(-5) m²/s

Thus, at y =0;

dT/dy = [[(20 − 220)0.08]/(5.33 x 10^(-5))](e^(0))

dT/dy = -300187.62 °C/m

A) Now, heat flux at y = 0 would be given by;

q_o = -k(dT/dy)

Where k is thermal conductivity

from the table attached at 220°C and by interpolation, the thermal conductivity k = 0.0391 W/m.k

Thus,

q_o = -0.0391(-300187.62)

q_o = 11737.34 W/m²

B) the convection heat transfer coefficient of the airflow is gotten from;

q_o = h(Ts - T∞).

Where h is the convection heat transfer coefficient of the airflow

Thus making h the formula, we have;

h = q_o/(Ts - T∞)

h = 11737.34/(220 - 20)

h = 58.67 W/m².k

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explanation:

answer:

yes that is right have a good day