Question

ff" +2f" = 0 with the boundary conditions n = 0 f = f' = 0 n + f' = 1

convert boundary condition problem to initial condition problem and solve with using classical R-K 4.

Answer 1

) ff"+2+"=0 boundary layer appront This was done by transforming the two partial differential eans into single ordinary differential ean new independent variable called by introducing the similarity Variable. av du u do Jx tudu and? ean Momentum ean- g (1) - La 1 8xgy Pastral Ty by 2 differentice continuity cans n similarity variable = y 97€ to get rid of continuity ean. Blasius golved the above by differentiating and obtaining single ordinary differential equation, la, f d²R dne dn3 nag run a + where f(n) = f g (n) dr e df 2 ! ff"+202" ian= dy go to Physical coordmates Uzo at y=0 similarity Coodmaites df dn f n=0 at R=0 O OU at y=0 at y -1.0 at no dn The value of n corresponding to /uo = 0.992 is 50. (taken from table below) n=y > 5=8 Vx so 5 5x va Je Shen (28x U 5.* (-6 ) dy (-elon * 0.5*(1-ö to dy Sci-f'}{"} 2) St 1.7208 ✓ Rex 2ux ū dn 0.664 o 2 Rex Similarity function f and its derivatives for laminar boundary layer along a flat plate. df d2f ຕ dn Uz U dna 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 0 0.042 0.166 0.370 0.650 0.996 1.397 1.838 2.306 2.790 3.283 3.781 4.280 0 0.166 0.330 0.487 0.630 0.751 0.846 0.913 0.956 0.980 0.992 0.997 0.999 0.332 0.331 0.323 0.303 0.267 0.217 0.161 0.108 0.064 0.034 0.016 0.007 0.002 0 8 1 Scanned with CamScanner

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