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Boundary Layer Theory and Turbulent Flow in Pipes.

a) Calculate the power required to overcome the skin friction drag on the keel of a sailing dingy moving through water at 2m/s. The keel projects 0.8m into the water and it is 0.4m wide. Assume that the keel is a smooth flat rectangular plate and that transition from laminar to turbulent flow occurs instantaneously at Re = 4*10^5.

For water: v = 10^-6 m^2 / s p = 1000kg/m^3

b) In reality, the keel in (a) is not a thin flat plate but a slim ellipse with C_D = 0.1 for turbulent flow and a frontal area of 0.04m^2. Assuming that the entire boundary layer is turbulent, determine the contributions of each component of drag (skin friction drag and pressure drag) to the total drag force and explain your result. You may take the skin friction area to be the same as in (a).

The velocity profile for turbulent flow in a pipe is well described by the relationship:

u/u_max = (y/R)^(1/n)

n varies according to the relationship:

Re n
4*10^3 6
2.3*10^4 6.6
1.1*10^5 7
1.1*10^6 8.8
2*10^6 10
3.2*10^6 10

Demonstrate that this profile cannot apply all the way to the wall. Do not describe the actual flow in the near-wall region; merely consider the mathematics of the above equation.

Solution Preview

1) First, we need to know what kind of flow we are dealing with.
In order to do that, we need to compute the Re number:
Re = (V*L)/nu = (2*0.4)/(10^-6)=0.8*10^6=8*10^5 > 4*10^5 --> turbulent flow
For the turbulent flow, the formula of drag coefficient is
Cf = 0.074/Re^0.2
By replacing with the value of Re obtained before, we will have
Cf = 0.074/(8*10^5)^0.2 = 0.0049
This coefficient applies to one face of the keel. The total drag force is given by skin friction on both sides (faces), that is
F1 = (ro/2)*V^2*(2A)*Cf
A = 0.8*0.4 = 0.32 m2
By replacing with numerical values, we get

Solution Summary

Solution includes calculations and explanation with 500 words.