Fig. P1.51P1.21 In 1908, Prandtl’s student Heinrich Blasius proposed the following formula for the wall shear stress w at a position x in viscous flow at velocity V past a flat surface:Determine the dimensions of the constant 0.332.Solution: From Table 1.2 we find the dimensions of each term in the equation:Use these dimensions in the equation to determine {0.332}:The constant 0.332 is dimensionless. Blasius was one of the first workers to deduce dimensionally consistent viscous-flow formulas without empirical constants.1.32 A blimp is approximated by a prolate spheroid 90 m long and 30 m in diameter. Estimatethe weight of 20°C gas within the blimp for (a) helium at 1.1 atm; and (b) air at 1.0 atm. Whatmight the difference between these two values represent (Chap. 2)?Solution: Find a handbook. The volume of a prolate spheroid is, for our data, 2 2 32 2LR (90 m)(15 m) 42412 m3 3Estimate, from the ideal-gas law, the respective densities of helium and air: Hehelium3Hep 1.1(101350) kg(a) 0.1832 ;R T 2077(293)m airair3airp101350 kg(b) 1.205 .R T 287(293)mThen the respective gas weights are 3He He3 2kg mW g 0.1832 9.81 (42412 m ) (a)m sAns.76000 N air airW g (1.205)(9.81)(42412) (b)Ans.501000 N2/12/32/12/1332.0 xVw}{}{;}{}{;}{}{;}{}{;}{}{111321LxLTVTMLMLTMLw.:or,}{}332.0{}{:upClean}{}{}{}{}332.0{}{222/12/32/12/132AnsLTMLTMLTLLTMLMLTM{1}{0.332} The difference between these two, 425000 N, is the buoyancy, or lifting ability, of the blimp. [SeeSection 2.8 for the principles of buoyancy.]P1.51 An approximation for the boundary-layer shape in Figs. 1.6b and P1.51 is the formulawhere U is the stream velocity far from the walland is the boundary layer thickness, as in Fig. P.151.If the fluid is helium at 20°C and 1 atm, and if U = 10.8 m/s and = 3 mm, use the formula to (a) estimatethe wall shear stress w in Pa; and (b) find the position in the boundary layer where is one-half of w.Solution: From Table A.4, for helium, take R = 2077 m2/(s2-K) and = 1.97E-5 kg/m-s. (a) Then the wall shear stress is calculated asA very small shear stress, but it has a profound effect on the flow pattern.(b) The variation of shear stress across the boundary layer is a cosine wave, = (du/dy):_______________________________________________________________________1.67 A vertical concentric annulus, with outer radius ro and inner radius ri, is lowered into fluidof surface tension Y and contact angle 90. Derive an expression for the capillary rise h inthe annular gap, if the gap is very narrow. yyUyu 0,)2sin()(u(y)Uy0y = Fig. P1.510 0| ( cos )2 2 2(1.97 5 / )(10.8 / )Numerical values: 0 .( )2(0.03 )w y ywu y UUyE kg m s m sAns amp p p mt m md d dpt= =�= = =�- -= = 0. 11 Pa).(:or,32when2)2cos()2cos(2)( bAnsyyyUyww32δy Solution: For the figure above, the force balance on the annular fluid is 2 2o i o icos (2 2 ) r rY r r g hCancel where possible and the result is .Anso i2 cos /{ (r r )} h Y g1.73 A small submersible moves at velocity V in 20C water at 2-m depth, where ambientpressure is 131 kPa. Its critical cavitation number is Ca 0.25. At what velocity will cavitationbubbles form? Will the body cavitate if V 30 m/s and the water is cold (5C)?Solution: From Table A-5 at 20C read pv 2.337 kPa. By definition, a vcrit crit2 3 22(p p ) 2(131000 2337)Ca 0.25 , solve V aV (998 kg/m )VAns32.1 m/sIf we decrease water temperature to 5C, the vapor pressure reduces to 863 Pa, and the densitychanges slightly, to 1000 kg/m3. For this condition, if V 30 m/s, we compute: 22(131000 863)Ca 0.289(1000)(30)This is greater than 0.25, therefore the body will not cavitate for these conditions. Ans. (b)P1.79 From Table A.3, the density of glycerin at standard conditions is about 1260 kg/m3. At avery high pressure of 8000 lb/in2, its density increases to approximately 1275 kg/m3. Use this data to estimate the speed of sound of glycerin, in ft/s.Solution: For a liquid, we simplify Eq. (1.38) to a pressure-density ratio, without knowing if the process is isentropic or not. This should give satisfactory accuracy:2223 2(8000 15 / )(6895 / )| 3.67 6(1275 1260) /3.67 6 1920 / / .glycerinlb in Pa psip ma Ekg m sHence a E m s ft s Ansr-D� = =D-� � � 6300The accepted value, in Table 9.1, is 6100 ft/s. This accuracy (3%) is very good, considering the small change in density (1.2%).C1.3 Two thin flat plates are tilted at an angle and placed in a tank of known surface tensionY and contact angle , as shown. At the free surface of the liquid in the tank, the two plates are adistance L apart, and of width b into the paper. (a) What is the totalz-directed force, due to surface tension, acting on the liquid column between plates? (b) If theliquid density is , find an expression for Y in terms of the other variables.Solution: (a) Considering the right side of the liquid column, the surface tension acts tangent tothe local surface, that is, along the dashed line at right. This force has magnitude F Yb, as shown.Its vertical component is F cos(), as shown. There are two plates. Therefore, the total z-directed force on the liquid column isFvertical 2Yb cos(– ) Ans. (a)(b) The vertical force in (a) above holds up the entire weight of the liquid column between plates,which is W g{bh(L h tan)}. Set W equal to F and solve for [gh(L h tan)]/[2 cos( )]
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