PROBLEM 6.2 KNOWN: Form of the velocity and temperature profiles for flow over a surface. FIND: Expressions for the friction and convection coefficients. SCHEMATIC: ANALYSIS: The shear stress at the wall is 2sy=0y=0 u A2By3Cy A. y∂τμμ μ∂⎤⎡⎤==+−=⎥⎢⎥⎣⎦⎦ Hence, the friction coefficient has the form, sf222AC u/2 uτμρρ∞∞== f22AC.uν∞= < The convection coefficient is ()2ffy=0 y=0skE2Fy3Gyk T/ yhTT DT∂∂∞∞⎡⎤−+−−⎢⎥⎣⎦==−− fkEh.DT∞−=− < COMMENTS: It is a simple matter to obtain the important surface parameters from knowledge of the corresponding boundary layer profiles. However, it is rarely a simple matter to determine the form of the profile.PROBLEM 6.5 KNOWN: Variation of hx with x for laminar flow over a flat plate. FIND: Ratio of average coefficient, xh , to local coefficient, hx, at x. SCHEMATIC: ANALYSIS: The average value of hx between 0 and x is xx00-1/2xx1/2 -1/2xxx1ChhdxxdxxxCh2x2Cxxh2h.=∫ =∫=== Hence, xxh2.h= < COMMENTS: Both the local and average coefficients decrease with increasing distance x from the leading edge, as shown in the sketch below.PROBLEM 6.19 KNOWN: Air speed and temperature in a wind tunnel. FIND: (a) Minimum plate length to achieve a Reynolds number of 108, (b) Distance from leading edge at which transition would occur. SCHEMATIC: ASSUMPTIONS: (1) Isothermal conditions, Ts = T∞. PROPERTIES: Table A-4, Air (25°C = 298K): ν = 15.71 × 10-6m2/s. ANALYSIS: (a) The Reynolds number is x ux uxRe .ρμν∞∞== To achieve a Reynolds number of 1 × 108, the minimum plate length is then ()862xmin1 10 15.71 10 m /sReLu 50 m/sν−∞××== minL 31.4 m.= < (b) For a transition Reynolds number of 5 × 105 ()5-62x,cc5 10 15.71 10 m /sRexu 50 m/sν∞××== cx 0.157 m.= < COMMENTS: Note that x,ccLRexLRe= This expression may be used to quickly establish the location of transition from knowledge of Re .x,c L and