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
Answer:
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
Step-by-step 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