Consider atmospheric air at 25oC in parallel flow at 4 m/s over both surfaces of a 1-m long flatplate maintained at 75oC. (a) Determine the velocity boundary layer thickness, the surface shear stress, and the heat flux at the trailing edge. (b) Determine the drag force on the plate and the total heat transfer from the plate, each per unit width of the plate. (c)Plot each parameter of part (a) as a function of distance from the leading edge of the plate.

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:

- 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