Question

The velocity profile for a turbulent boundary layer over a flat plate is to be approximated by the expression и an"* +b7072 where n=y/8 U a) (10P) Evaluate the coefficients a and b b) (20P) Obtain an expression for 8/x c) (5P) Obtain an expression for shear stress coefficient Cf. d) (5P) Draw velocity profile precisely.

Answer 1

Given for turbulent B.L :- Velocity profile 0.8 an 0.72 + bn where n U 0.8 0.72 u (3** + using boundary conditions. at y=S a+b = ci) - At y=8 du dy 0.8 0.72 ua (5 bu ) + 0.8-1 du dy ua. 0.8 Y 0.72-1 bU, 0.72 y + (6) 0.8 80.72 -0.28 bu X 0.72 X S -0.2 U a x 0.8 X SI + O 0.72 8º.8 - 0.8 a XS + 0.72 6 5 cii) 0.89 + 0.72 6 = 0 solving equation ci) and (ii) we get a = 9 b = 10

and b a] coefficients b=10 a = -9 e of zlos b] Expression To Sw? өe 2 ē inst To I ve [1 (-4) dy 1-(35*+- (** ol u = -9mm * + 10 20:22 S = (1-4) dy y = 8 y=0 y S dy=s.dn y = 0 n = 0 y=1 2 = 1

s (10 monz 994-8)(1+ gnu - 10 Muz?). sdn 0 = S Como z goto) (1+90 - 10 mºnte) do 0.8 0.72 2:52)dn] 1.52 1-44 1.6 1.52 on +90 n - 100 m - 9 m - 81 + gon 0 = [ścion o 2.52 1.8 1.72 2.52 L 10 m gon gr 2.6 8) n + 2.6 2.44 100 m2 2.44 gon 8 + 1.72 2.52 1.8 2.52 O 0.105072 ] 0.1050728 To = SU? 9 C0.1050728) 22 su ²x To 0.105042 x as - To = M. du deye yao dy

0.72-1 0.8-1 10 0.72 y 9 x 0.8 . du dy 89.72 89.8 the dy dy lyzo This dibbiculty is overcome by considering velocity in the viscous laminar sublayer to be linear and tangential to the velocity profile at the point where the laminar sublayer merges with the turbulent part of the boundary layer. Blasius suggested the following relation for shear stoess. Sub viscous 1/4 To = 0.0226 SU? • (505) M SUS we b equating get and egna egn. 14 0.0226304 시 SUS 0,050 72 304. as 14 1/4 0.2151 = S '.28 SU on integrating we get 4 0.215 + c 5/4 8 5 4 ( colonit SU

5/4 4 s 1/4 el 0.2151 ( x + c SU condition for turbulent How using boundary at S=0 . C=O 5/4 1/4 0.215) el 4 8 5 re 514 114 = 0.26887 M Su Y5 415 8 0-3497 SU 415 15 a 1/5 a S- 0.3497 ll sua 15 8 0.3497 ( ㅗ Rea 0.3497 (Rex)'s 14 2 (To ) 0.0225 SU 4 SUS Shear S Hess "4 u : To x 0.0225 SU? su X 0.3497 (Rex) 15

To = 0.0225 (Rex) (0.3497) 14 Vs (Rex) To = 0.02926 SU? Ren 0.0 5852 To = X 2 (Rex)'5 * С. of drag Local coefficient suz х 0.05852 X To 2 2 175 (Real Ju LSU? 0.05852 15 (Rex)" Drag force (fo) Fo= Što To x B x da 0.05852 SU 2 x x B x da S $? 2 115 (Rex)' L -115 0.05852 SU В de X 2 15 4

SU? 0.05852 Fo Х B 5 2 ***] ( sus sul Fo= su² 4/5 0.058 52 X B x 5 XL 2 leurs 2 0.07315 Fo = su² B XL x Х iis 2 (Rey) coepticient of drag (CD) : For co xh sau 2 suz 0-07315 x B x L = Cox Į SAU² (Re)'s 0.07315 x B xx = Cox Cox Xxxxx xu- 2 115 (Re) 0.07315 CD = (Re)'s