Answer:
a) δ = 0.01066 m , τ = 0.01229 N/m^2 , q'' = 194.118 W/m^2
b) D = 0.04916 N / m , 776.472 W / m
Explanation:
Given:-
- Temperature of air, T∞ = 25°C
- The speed of air flow, U∞ = 4 m/s
- Surface Temperature, Ts = 75°C
- The Length of plate, L = 1m
Solution:-
- We will evaluate the following properties of air at (T∞) and 1 atm pressure.
Density: ρ = 1.085 kg/m^3
Dynamic viscosity: v = 18.2*10^-6 m^2/s
Prandlt Number: Pr = 0.707
Thermal conductivity: k = 0.028 W/mK
- Calculate the Reynold's number for the flow over the length (L) of the flat plate to deterimine the nature of flow as follows:
Re = U∞*L / v
= (4)*(1) / 18.2*10^-6
= 219780.21978 ...... ( Laminar Flow )
a) Determine the velocity boundary layer thickness ( δ ) , the surface shear stress (τ), and the heat flux at the trailing edge (q'').
- Use the appropriate velocity boundary layer thickness relations - Laminar flow:
δ = 5L / √Re
δ = 5*1 / √219780.21978
δ = 0.01066 m
- Similarly for the shear stress ( τ ):
τ = [0.5*ρ*(U∞)^2]*0.664/√Re
τ = [0.5*1.085*(4)^2]*0.664/√219780.21978
τ = 0.01229 N/m^2
- Determine the Nusselt number (Nu) and convection heat transfer coefficient (h) associated with the flow conditions given:
[tex]Nu = 0.332\sqrt{Re}*(Pr)^\frac{1}{3} = \frac{h*L}{k} \\\\Nu = 0.332\sqrt{Re}*(Pr)^\frac{1}{3}\\\\Nu = 138.65606 = \frac{h*L}{k} \\\\h = 138.65606*(0.028)\\\\h = 3.88236 W/m^2K[/tex]
- Use the calculated convection heat transfer coefficient (h) to determine the heat flux (q'') at trailing edge (L) using Newton's Law of cooling:
q'' = h*L*( Ts - T∞ )
q'' = (3.88236)*(1)*( 75 - 25 )
q'' = 194.118 W/m^2
(b) Determine the drag force on the plate and the total heat transfer from the plate, each per unit width of the plate.
- Use the appropriate relationship for the shear stress (τ) on both sides plate surface.
τ = 2*[0.5*ρ*(U∞)^2]*0.664/√Re
τ = 2*( 0.01229 )
τ = 0.02458 N/m^2
- The drag force per unit width (D) on the plate is given by the relationship:
D = 2*τ * L
D = 2*0.02458*1
D = 0.04916 N / m
- The total heat transfer from plate per unit length (q'):
h' = 2*h
h' = 2*(3.88236) = 7.76472 W/m^2K
q' = 2*h'*L*( Ts - T∞ )
q' = 2*(7.76472)*(1)*(75-25)
q' = 776.472 W / m
