P1 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 w 0 332 1 2 1 2 V 3 2 x 1 2 Determine the dimensions of the constant 0 332 Solution From Table 1 2 we find the dimensions of each term in the equation w ML 1T 2 ML 3 ML 1T 1 V LT 1 x L Use these dimensions in the equation to determine 0 332 M LT 2 M 1 2 L 3 2 L 1 2 LT T L M M 2 0 332 2 or 0 332 1 LT LT 0 332 Clean up M 3 1 2 Ans 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 Estimate the weight of 20 C gas within the blimp for a helium at 1 1 atm and b air at 1 0 atm What might 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 LR2 90 m 15 m 2 42412 m3 3 3 Estimate from the ideal gas law the respective densities of helium and air pHe 1 1 101350 kg 0 1832 3 RHe T 2077 293 m pair 101350 kg b air 1 205 3 Rair T 287 293 m a helium Then the respective gas weights are kg m WHe He g 0 1832 3 9 81 2 42412 m 3 76000 N m s Wair air g 1 205 9 81 42412 501000 N Ans b Ans a The difference between these two 425000 N is the buoyancy or lifting ability of the blimp See Section 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 formula u y U sin y 2 y U 0 y y where U is the stream velocity far from the wall and is the boundary layer thickness as in Fig P 151 u y 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 estimate the wall shear stress w in Pa and b find the position in the boundary layer where is one half of w 0 Fig P1 51 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 as u p py p mU t w m y 0 m U cos y 0 y 2d 2d 2d p 1 97 E 5 kg m s 10 8 m s Numerical values t w 0 011 Pa Ans a 2 0 03 m A 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 U y y y 2 cos w cos w when or y Ans b 2 2 2 2 2 3 3 1 67 A vertical concentric annulus with outer radius ro and inner radius ri is lowered into fluid of surface tension Y and contact angle 90 Derive an expression for the capillary rise h in the annular gap if the gap is very narrow y Solution For the figure above the force balance on the annular fluid is h 2Y cos g ro ri Ans Y cos 2 ro 2 ri g ro2 ri2 h Cancel where possible and the result is 1 73 A small submersible moves at velocity V in 20 C water at 2 m depth where ambient pressure is 131 kPa Its critical cavitation number is Ca 0 25 At what velocity will cavitation bubbles form Will the body cavitate if V 30 m s and the water is cold 5 C Solution From Table A 5 at 20 C read pv 2 337 kPa By definition Ca crit 0 25 2 pa pv 2 131000 2337 solve Vcrit 32 1 m s V2 998 kg m 3 V2 Ans a If we decrease water temperature to 5 C the vapor pressure reduces to 863 Pa and the density 3 changes slightly to 1000 kg m For this condition if V 30 m s we compute Ca 2 131000 863 0 289 1000 30 2 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 a very 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 a 2 glycerin 8000 15 lb in 2 6895 Pa psi Dp m2 3 67 E 6 Dr 1275 1260 kg m3 s2 Hence a 3 67 E 6 1920 m s 6300 ft s Ans The 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 tension Y and contact angle as shown At the free surface of the liquid in the tank the two plates are a distance L apart and of width b into the paper a What is the total z directed force due to surface tension acting on the liquid column between plates b If the liquid 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 to the 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 zdirected force on the liquid column is Fvertical 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 Ans b
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