Example 4.1 Air at 20C and 0.5 105 N/m (0.5 bar) flows over a flat plate with a velocity of 60 m/s. The length of the plate is 25 cm. Calculate the boundary-layer thickness at various loca- tions along the plate. Solution From Appendix B, the kinematic viscosity of air at 20 C and atmospheric pressure is Vo=15.1110 m/s. Since the variation of the dynamic viscosity with pressure is negligible, the kinematic viscosity of air at 0.5 105 N/m and 20 C can be found from the equation of state, that is, v=Vo Po P -=2v = 215.1110 m/s=30.2210 m/s. The thickness of the boundary layer can now be calculated from Equation 4.34a: 8=5.0 (30.2210)x 60 = 3.55 103xm. The following table gives the boundary-layer thickness at various points along the plate: X (cm) 0 24 8 (mm) 0 0.502 0.710 10 25 G5 1.122 15 1.374 1.775 The local shear stress at the surface of the plate is given by Tw(x)=| = uu... u df(0) vx dn y=0 (4.36a) From Table 4.1, it is seen that f"(0) = 0.332. Hence, Equation 4.36a can be rewritten as u Tw(x)=0.332U VX (4.36b) A dimensionless wall shear stress, which is usually referred to as the drag coefficient or, since drag is due only to wall shear, the friction coefficient, may be defined as 0.664 () (1/2)puRex' where Re, is the local Reynolds number defined by Equation 4.35. (4.37) This solution results in an infinite shear stress at the leading edge of the plate. This is due to the fact that the boundary-layer assumptions are not valid in the vicinity of the leading edge where the derivative ou/x is not small compared to du/dx. Because the region in which this is true is usually very small compared to the size of the entire plate, 4.2 Plot the distribution of the velocity u in the boundary layer of Example 4.1 at a dis- tance 15 cm from the leading edge of the plate.