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Determine the thickness of the viscous sublayer, 6', at a distance I m downstream from the leading edge. Would a roughness element 100 pm high afTect the local skin friction coefficient? Wlry? and assumptions , tr- t _ H u _ 1 0 _ 6 m 2 l s problem statement 1 m downstream from leading edge. : cf pU2 12 r0lp : Re, : : Ru|'' : r0l p ,ol p u* 6' :5v/u* (0.058/ xe]'21ufr 1z Usrf u (5)(1) i10-6 :5 x 106 2L.87 (0.05s/21 .87)(2512) 0.0332 m2ls2 (ro ldo ' ' :0 .1822 mls 6' :5vlu. - (b)(10-6)/(o.rB22) 6t :27 .4 x 10-o m microns is about 4 times greater than the thickness of the skin friction coefficient. viscous sublayer; 9.73 Consider the boundary layer next to the smooth hull of a ship. The ship is cruising at a speed of 30 ft,/s in 60'F fresh water. Assuming that the boundary layer on the ship hull develops the same as on a flat plate, deter- mine a. The thickness of the boundary layer at a distance of 100 ft downstream from the bow. b. The velocity of the water at a point in the boundary layer at y/ 6 -- 0.50. c. The shear stress, rp, adjacent to the hull at this posi- t ion. 9.73 Information and assumptions B,J Flom Table H u : I.22 x 10-5 ft,1" and p - I.g4 slug/ft3. provided in problem statement Find a) thickness of boundary layer at l00faownstream, b) velocity of water at yl6 - 0.5 and c) shear stress on hull. Solution Re* : Ur lu : (30) (100) / (1 .22 x t0 -5) :2 .46x 10s 0.455 ' In2(0.06 Re*; : 0.00167 r0 : cyprJfr lz: (0.00167)(1.94)(30')12 - 1.4b6_IbVft1.) ' t .L* : (roldo'' : (I.456fl.94)0'5 : 0.866 ft/s 6l* - 0.L6 Re, 1/7 : 0.010 6 -- (0.010)(100) : l0-ft (") 612 : 0.5 ft F}om Fig. 9-12 at y 16: 0.50, (Uo - ulu* x 3.0. Then ( 3 0 - r ) / 0 . 8 6 6 : 3 . 0 uo/z-27.4ft/s (b) g,Se A llat plate is oriented parallel to a 15 rn,/s air flow at 20"C and atmospheric pressure. The plate is I nt long in the flow direction and 0.5 rn wide. On one side of the plate. the boundary layer is tripped at the leading edge and on the other side there is no tripping device. Find the total drag force on the plate. PROBLEM 9.66 9.66 Information and assumptions provided in problem statement Find total drag force on plate. Solution The force due to shea.r' stress is The density N-s/m2, respectively. and kinematic viscosity The Relrrolds number 1 l% : C1 ,pffint 4 of air at 20"C and atmospheric pressure based on the plate length is is 1.2 kg/*S and 1.5x10-5 Rer : 15j!1 l . b x l o - 5 - 1 o o The average shear stress coefficient on the "bripped." side of the plate is cr : *Ftr :0.0047 The average sheal stress on the "untripped" side is C f : 0.523 $24 tn2(0.06 x 106) 106 :0 .0028 " ' -1 , s - 2 Trip str ip The total force is x l .2 x 152 x 1x 0 .5 x (0 .0047+0.0028) 9.76 A supertanker has length, breadth, and draught (ful ly loaded) dimensions of 325 m, 48 m, and 19 m, respectively. In open seas the tanker normally operates at a speed of 15 k t ( l k t : 0 .515 m,zs) . For these con- dit ions, and assuming that f lat-plate boundary-layer condit ions are approximated, est irnate the skin-f i ict ion drag of such a ship steaming in l0oC water. What power is required to overcome the skin-friction drag? What is the boundary-layer thickness at 300 m from the bow? 9.76 Information and assumptions Fbom Table'& , : i.4x 10-o provided in problem statement Find skin friction drag, power required and boundary layer thickness 300 m from bow. Solution Re.r . - UsLlu: (15 x 0.515) x3251$.a x tO-6) :L.79 x 10s Fhom rig. 9-|$ Cf :0.00153. Then F" : CyApufr lz 0.00153 x 325(48 + 33) x 1,026 x (rr x 0.51s)2/2 : 1.309 MN 1.309 x 106 x 15 x 0.515 - 10.1 MW T o f i n d 6 a t r - 3 0 0 m Fr s P ^2 ls and p = L026 kg/*3 Ugrf u: 15 x 0.515 x 0.515 x 300/(1.a x 10-6) 1.66 x 10e 0.16lRe]/? : 0.0077 300 m x .0077 2.31 m Resoo : : 6l* 6 6
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