Baixe Soluções White Mecânica de Fluídos cap 1 e outras Exercícios em PDF para Administração Empresarial, somente na Docsity! Chapter 1 • Introduction 1.1 A gas at 20°C may be rarefied if it contains less than 1012 molecules per mm3. If Avogadro’s number is 6.023E23 molecules per mole, what air pressure does this represent? Solution: The mass of one molecule of air may be computed as 1Molecular weight 28.97 mol m 4.81E 23 g Avogadro’s number 6.023E23 molecules/g mol − = = = − ⋅ Then the density of air containing 1012 molecules per mm3 is, in SI units, ρ = − = − = − 12 3 3 3 molecules g 10 4.81E 23 moleculemm g kg 4.81E 11 4.81E 5 mm m Finally, from the perfect gas law, Eq. (1.13), at 20°C = 293 K, we obtain the pressure: ρ Α = = − = ⋅ 2 3 2 kg m p RT 4.81E 5 287 (293 K) . m s K ns4.0 Pa 1.2 The earth’s atmosphere can be modeled as a uniform layer of air of thickness 20 km and average density 0.6 kg/m3 (see Table A-6). Use these values to estimate the total mass and total number of molecules of air in the entire atmosphere of the earth. Solution: Let Re be the earth’s radius ≈ 6377 km. Then the total mass of air in the atmosphere is 2 t avg avg e 3 2 m dVol (Air Vol) 4 R (Air thickness) (0.6 kg/m )4 (6.377E6 m) (20E3 m) . Ans ρ ρ ρ π π = = ≈ = ≈ 6.1E18 kg Dividing by the mass of one molecule ≈ 4.8E−23 g (see Prob. 1.1 above), we obtain the total number of molecules in the earth’s atmosphere: molecules m(atmosphere) 6.1E21 grams N m(one molecule) 4.8E 23 gm/molecule Ans.= = ≈ − 1.3E44 molecules 2 Solutions Manual • Fluid Mechanics, Fifth Edition 1.3 For the triangular element in Fig. P1.3, show that a tilted free liquid surface, in contact with an atmosphere at pressure pa, must undergo shear stress and hence begin to flow. Solution: Assume zero shear. Due to element weight, the pressure along the lower and right sides must vary linearly as shown, to a higher value at point C. Vertical forces are presumably in balance with ele- ment weight included. But horizontal forces are out of balance, with the unbalanced force being to the left, due to the shaded excess-pressure triangle on the right side BC. Thus hydrostatic pressures cannot keep the element in balance, and shear and flow result. Fig. P1.3 1.4 The quantities viscosity µ, velocity V, and surface tension Y may be combined into a dimensionless group. Find the combination which is proportional to µ. This group has a customary name, which begins with C. Can you guess its name? Solution: The dimensions of these variables are {µ} = {M/LT}, {V} = {L/T}, and {Y} = {M/T2}. We must divide µ by Y to cancel mass {M}, then work the velocity into the group: 2 / , { } ; Y / . M LT T L hence multiply by V L TM T finally obtain Ans µ = = = µV dimensionless. Y = This dimensionless parameter is commonly called the Capillary Number. 1.5 A formula for estimating the mean free path of a perfect gas is: 1.26 1.26 (RT) p(RT) µ µ ρ = = √ √ (1) Chapter 1 • Introduction 5 1.9 The dimensionless Galileo number, Ga, expresses the ratio of gravitational effect to viscous effects in a flow. It combines the quantities density ρ, acceleration of gravity g, length scale L, and viscosity µ. Without peeking into another textbook, find the form of the Galileo number if it contains g in the numerator. Solution: The dimensions of these variables are {ρ} = {M/L3}, {g} = {L/T2}, {L} = {L}, and {µ} = {M/LT}. Divide ρ by µ to eliminate mass {M} and then combine with g and L to eliminate length {L} and time {T}, making sure that g appears only to the first power: 3 2 / / M L T M LT L ρ µ = = while only {g} contains {T}. To keep {g} to the 1st power, we need to multiply it by {ρ/µ}2. Thus {ρ/µ}2{g} = {T2/L4}{L/T2} = {L−3}. We then make the combination dimensionless by multiplying the group by L3. Thus we obtain: ρ ρ µ µ = = = = 2 2 3 3 2 ( )( ) . gL Galileo number Ga g L Ans 3 2 gL ν 1.10 The Stokes-Oseen formula [10] for drag on a sphere at low velocity V is: 2 29F 3 DV V D 16 ππµ ρ= + where D = sphere diameter, µ = viscosity, and ρ = density. Is the formula homogeneous? Solution: Write this formula in dimensional form, using Table 1-2: 2 29{F} {3 }{ }{D}{V} { }{V} {D} ? 16 ππ µ ρ = + 2 2 2 3 2 ML M L M L or: {1} {L} {1} {L } ? LT TT L T = + where, hoping for homogeneity, we have assumed that all constants (3,π,9,16) are pure, i.e., {unity}. Well, yes indeed, all terms have dimensions {ML/T2}! Therefore the Stokes- Oseen formula (derived in fact from a theory) is dimensionally homogeneous. 6 Solutions Manual • Fluid Mechanics, Fifth Edition 1.11 Test, for dimensional homogeneity, the following formula for volume flow Q through a hole of diameter D in the side of a tank whose liquid surface is a distance h above the hole position: 2Q 0.68D gh= where g is the acceleration of gravity. What are the dimensions of the constant 0.68? Solution: Write the equation in dimensional form: 1/ 2 33 ? 2 1/ 2 2 L LL {0.68?}{L } {L} {0.68}{Q} TTT = = = Thus, since 2D gh( ) has provided the correct volume-flow dimensions, {L3/T}, it follows that the constant “0.68” is indeed dimensionless Ans. The formula is dimensionally homogeneous and can be used with any system of units. [The formula is very similar to the valve-flow formula d oQ C A ( p/ )ρ= ∆ discussed at the end of Sect. 1.4, and the number “0.68” is proportional to the “discharge coefficient” Cd for the hole.] 1.12 For low-speed (laminar) flow in a tube of radius ro, the velocity u takes the form ( )2 2opu B r rµ ∆= − where µ is viscosity and ∆p the pressure drop. What are the dimensions of B? Solution: Using Table 1-2, write this equation in dimensional form: 2 2 2 2{ p} L {M/LT } L{u} {B} {r }, or: {B?} {L } {B?} , { } T {M/LT} Tµ ∆ = = = or: {B} = {L–1} Ans. The parameter B must have dimensions of inverse length. In fact, B is not a constant, it hides one of the variables in pipe flow. The proper form of the pipe flow relation is ( )2 2opu C r rLµ ∆= − where L is the length of the pipe and C is a dimensionless constant which has the theoretical laminar-flow value of (1/4)—see Sect. 6.4. Chapter 1 • Introduction 7 1.13 The efficiency η of a pump is defined as Q p Input Power η ∆= where Q is volume flow and ∆p the pressure rise produced by the pump. What is η if ∆p = 35 psi, Q = 40 L/s, and the input power is 16 horsepower? Solution: The student should perhaps verify that Q∆p has units of power, so that η is a dimensionless ratio. Then convert everything to consistent units, for example, BG: 2 2 2 L ft lbf lbf ft lbf Q 40 1.41 ; p 35 5040 ; Power 16(550) 8800 s s sin ft ⋅= = ∆ = = = = 3 2(1.41 ft s)(5040 lbf ft ) 0.81 or 8800 ft lbf s Ans.η / /= ≈ ⋅ / 81% Similarly, one could convert to SI units: Q = 0.04 m3/s, ∆p = 241300 Pa, and input power = 16(745.7) = 11930 W, thus h = (0.04)(241300)/(11930) = 0.81. Ans. 1.14 The volume flow Q over a dam is proportional to dam width B and also varies with gravity g and excess water height H upstream, as shown in Fig. P1.14. What is the only possible dimensionally homo- geneous relation for this flow rate? Solution: So far we know that Q = B fcn(H,g). Write this in dimensional form: 3 2 L {Q} {B}{f(H,g)} {L}{f(H,g)}, T L or: {f(H,g)} T = = = = Fig. P1.14 So the function fcn(H,g) must provide dimensions of {L2/T}, but only g contains time. Therefore g must enter in the form g1/2 to accomplish this. The relation is now Q = Bg1/2fcn(H), or: {L3/T} = {L}{L1/2/T}{fcn(H)}, or: {fcn(H)} = {L3/2} 10 Solutions Manual • Fluid Mechanics, Fifth Edition Solution: List the dimensions: {α} = {L2/T}, {L} = {L}, {µ} = {M/LT}, {δY} = {M/T2}. We divide δ Y by µ to get rid of mass dimensions, then divide by α to eliminate time: { }2 2Y Y 1 1, thenM LT L L TM T T LT Lδ δµ µ α = = = = Multiply by L and we obtain the Marangoni number: .Ans L M = δ µα Y 1.20C (“C” means computer-oriented, although this one can be done analytically.) A baseball, with m = 145 g, is thrown directly upward from the initial position z = 0 and Vo = 45 m/s. The air drag on the ball is CV 2, where C ≈ 0.0010 N ⋅ s2/m2. Set up a differential equation for the ball motion and solve for the instantaneous velocity V(t) and position z(t). Find the maximum height zmax reached by the ball and compare your results with the elementary-physics case of zero air drag. Solution: For this problem, we include the weight of the ball, for upward motion z: = − − − = = − = − + o V t 2 z z 2 V 0 dV dV F ma , or: CV mg m , solve dt t dt g CV /m φφ φ − √= − = mg Cg m cos( t (gC/m) Thus V tan t and z ln C m C cos where –1 otan [V (C/mg)]φ = √ . This is cumbersome, so one might also expect some students simply to program the differential equation, m(dV/dt) + CV2 = −mg, with a numerical method such as Runge-Kutta. For the given data m = 0.145 kg, Vo = 45 m/s, and C = 0.0010 N⋅s 2/m2, we compute 1mg m Cg m0.8732 radians, 37.72 , 0.2601 s , 145 m C s m C φ −= = = = Hence the final analytical formulas are: = − − = m V in 37.72 tan(0.8732 .2601t) s cos(0.8732 0.2601t) and z(in meters) 145 ln cos(0.8732) The velocity equals zero when t = 0.8732/0.2601 ≈ 3.36 s, whence we evaluate the maximum height of the baseball as zmax = 145 ln[sec(0.8734)] ≈ 64.2 meters. Ans. Chapter 1 • Introduction 11 For zero drag, from elementary physics formulas, V = Vo − gt and z = Vot − gt 2/2, we calculate that 2 2 o o max height max V V45 (45) t and z g 9.81 2g 2(9.81) = = ≈ = = ≈4.59 s 103.2 m Thus drag on the baseball reduces the maximum height by 38%. [For this problem I assumed a baseball of diameter 7.62 cm, with a drag coefficient CD ≈ 0.36.] 1.21 The dimensionless Grashof number, Gr, is a combination of density ρ, viscosity µ, temperature difference ∆T, length scale L, the acceleration of gravity g, and the coefficient of volume expansion β, defined as β = (−1/ρ)(∂ρ/∂T)p. If Gr contains both g and β in the numerator, what is its proper form? Solution: Recall that {µ/ρ} = {L2/T} and eliminates mass dimensions. To eliminate tem- perature, we need the product {β∆Τ} = {1}. Then {g} eliminates {T}, and L3 cleans it all up: 2 3 2Thus the dimensionless g / .Gr TL Ansρ β µ= ∆ 1.22* According to the theory of Chap. 8, as a uniform stream approaches a cylinder of radius R along the line AB shown in Fig. P1.22, –∞ < x < –R, the velocities are 2 2u U (1 R /x ); v w 0∞= − = = Fig. P1.22 Using the concepts from Ex. 1.5, find (a) the maximum flow deceleration along AB; and (b) its location. Solution: We see that u slows down monotonically from U∞ at A to zero at point B, x = −R, which is a flow “stagnation point.” From Example 1.5, the acceleration (du/dt) is 2 2 2 2 3 3 5 du u u R 2R U 2 2 x u 0 U 1 U , dt t x R Rx x ∂ ∂ − ζ ∂ ∂ ζ ζ ∞ ∞ ∞ = + = + − + = = This acceleration is negative, as expected, and reaches a minimum near point B, which is found by differentiating the acceleration with respect to x: 2 max decel. min d du 5 x 0 if , or . (b) dx dt 3 R du Substituting 1.291 into (du/dt) gives . (a) dt Ans Ans ζ ζ = = ≈ = − = | | ∞ − − 2 1.291 U 0.372 R 12 Solutions Manual • Fluid Mechanics, Fifth Edition A plot of the flow deceleration along line AB is shown as follows. 1.23E This is an experimental home project, finding the flow rate from a faucet. 1.24 Consider carbon dioxide at 10 atm and 400°C. Calculate ρ and cp at this state and then estimate the new pressure when the gas is cooled isentropically to 100°C. Use two methods: (a) an ideal gas; and (b) the Gas Tables or EES. Solution: From Table A.4, for CO2, k ≈ 1.30, and R ≈ 189 m 2/(s2⋅K). Convert pressure from p1 = 10 atm = 1,013,250 Pa, and T1 = 400°C = 673 K. (a) Then use the ideal gas laws: 1 1 2 2 3 1 1,013,250 ; (189 / )(673 ) 1.3(189) . (a) 1 1.3 1p p Pa kg RT m s K K m kR J c Ans k kg K ρ = = = = = = − − ⋅ 7.97 819 For an ideal gas cooled isentropically to T2 = 100°C = 373 K, the formula is /( 1) 1.3 /(1.3 1) 2 2 2 2 1 1 373 0.0775, . (a) 1013 673 k k p T p K p Ans p T kPa K − − = = = = = or: 79 kPa For EES or the Gas Tables, just program the properties for carbon dioxide or look them up: 3 1 2 kg/m ; J/(kg K); kPa . (b)pc p Ansρ = = ⋅ =7.98 1119 43 (NOTE: The large errors in “ideal” cp and “ideal” final pressure are due to the sharp drop- off in k of CO2 with temperature, as seen in Fig. 1.3 of the text.) Chapter 1 • Introduction 15 1.28 Wet air, at 100% relative humidity, is at 40°C and 1 atm. Using Dalton’s law of partial pressures, compute the density of this wet air and compare with dry air. Solution: Change T from 40°C to 313 K. Dalton’s law of partial pressures is υ υ = = + = +a wtot air water a w m m p 1 atm p p R T R T υ υ= + = +a wtot a w a w p p or: m m m for an ideal gas R T R T where, from Table A-4, Rair = 287 and Rwater = 461 m 2/(s2⋅K). Meanwhile, from Table A-5, at 40°C, the vapor pressure of saturated (100% humid) water is 7375 Pa, whence the partial pressure of the air is pa = 1 atm − pw = 101350 − 7375 = 93975 Pa. Solving for the mixture density, we obtain a w a w a w m m p p 93975 7375 1.046 0.051 R T R T 287(313) 461(313) Ans.ρ υ += = + = + = + ≈ 3 kg 1.10 m By comparison, the density of dry air for the same conditions is dry air 3 p 101350 kg 1.13 RT 287(313) m ρ = = = Thus, at 40°C, wet, 100% humidity, air is lighter than dry air, by about 2.7%. 1.29 A tank holds 5 ft3 of air at 20°C and 120 psi (gage). Estimate the energy in ft-lbf required to compress this air isothermally from one atmosphere (14.7 psia = 2116 psfa). Solution: Integrate the work of compression, assuming an ideal gas: 2 2 2 2 1-2 2 2 1 11 1 mRT p W p d d mRT ln p ln p υυ υ υ υ υ = − = − = − = where the latter form follows from the ideal gas law for isothermal changes. For the given numerical data, we obtain the quantitative work done: 32 1-2 2 2 2 1 p lbf 134.7 W p ln 134.7 144 (5 ft ) ln . p 14.7ft Ansυ = = × ≈ 215,000 ft lbf⋅ 16 Solutions Manual • Fluid Mechanics, Fifth Edition 1.30 Repeat Prob. 1.29 if the tank is filled with compressed water rather than air. Why is the result thousands of times less than the result of 215,000 ft⋅lbf in Prob. 1.29? Solution: First evaluate the density change of water. At 1 atm, ρ o ≈ 1.94 slug/ft3. At 120 psi(gage) = 134.7 psia, the density would rise slightly according to Eq. (1.22): 7 3 o p 134.7 3001 3000, solve 1.940753 slug/ft , p 14.7 1.94 ρ ρ = ≈ − ≈ 3 waterHence m (1.940753)(5 ft ) 9.704 slugρυ= = ≈ The density change is extremely small. Now the work done, as in Prob. 1.29 above, is 2 2 2 1-2 avg2 2 avg1 1 1 m m d W pd pd p p m ρ ρυ ρ ρ ρ ∆= − = = ≈ for a linear pressure rise 3 1-2 2 2 14.7 134.7 lbf 0.000753 ft Hence W 144 (9.704 slug) 2 slugft 1.9404 Ans. + ≈ × ≈ 21 ft lbf⋅ [Exact integration of Eq. (1.22) would give the same numerical result.] Compressing water (extremely small ∆ρ) takes ten thousand times less energy than compressing air, which is why it is safe to test high-pressure systems with water but dangerous with air. 1.31 The density of water for 0°C < T < 100°C is given in Table A-1. Fit this data to a least-squares parabola, ρ = a + bT + cT2, and test its accuracy vis-a-vis Table A-1. Finally, compute ρ at T = 45°C and compare your result with the accepted value of ρ ≈ 990.1 kg/m3. Solution: The least-squares parabola which fits the data of Table A-1 is: ρ (kg/m3) ≈ 1000.6 – 0.06986T – 0.0036014T2, T in °C Ans. When compared with the data, the accuracy is less than ±1%. When evaluated at the particular temperature of 45°C, we obtain ρ45°C ≈ 1000.6 – 0.06986(45) – 0.003601(45)2 ≈ 990.2 kg/m3 Ans. This is excellent accuracya good fit to good smooth data. The data and the parabolic curve-fit are shown plotted on the next page. The curve-fit does not display the known fact that ρ for fresh water is a maximum at T = +4°C. Chapter 1 • Introduction 17 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 32 2LR (90 m)(15 m) 42412 m 3 3 υ π π= = ≈ Estimate, from the ideal-gas law, the respective densities of helium and air: He helium 3 He p 1.1(101350) kg (a) 0.1832 ; R T 2077(293) m ρ = = ≈ air air 3 air p 101350 kg (b) 1.205 . R T 287(293) m ρ = = ≈ Then the respective gas weights are 3 He He 3 2 kg m W g 0.1832 9.81 (42412 m ) (a) m s Ans.ρ υ = = ≈ 76000 N air airW g (1.205)(9.81)(42412) (b)Ans.ρ υ= = ≈ 501000 N 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.] 20 Solutions Manual • Fluid Mechanics, Fifth Edition For water at 20°C, we could just look it up in Table A-3, but we more usefully try to estimate B from the state relation (1-22). Thus, for a liquid, approximately, n n o o o o o d B [p {(B 1)( / ) B}] n(B 1)p ( / ) n(B 1)p at 1 atm d ρ ρ ρ ρ ρ ρ ≈ + − = + = + For water, B ≈ 3000 and n ≈ 7, so our estimate is water oB 7(3001)p 21007 atm≈ = ≈ 2.13E9 Pa Ans. (b) This is 2.7% less than the value B = 2.19E9 Pa listed in Table A-3. 1.37 A near-ideal gas has M = 44 and cv = 610 J/(kg⋅K). At 100°C, what are (a) its specific heat ratio, and (b) its speed of sound? Solution: The gas constant is R = Λ/Μ = 8314/44 ≈ 189 J/(kg⋅K). Then v v 2c R/(k 1), or: k 1 R/c 1 189/610 (a) [It is probably N O]Ans.= − = + = + ≈ 1.31 With k and R known, the speed of sound at 100ºC = 373 K is estimated by 2 2a kRT 1.31[189 m /(s K)](373 K)= = ⋅ ≈ 304 m/s Ans. (b) 1.38 In Fig. P1.38, if the fluid is glycerin at 20°C and the width between plates is 6 mm, what shear stress (in Pa) is required to move the upper plate at V = 5.5 m/s? What is the flow Reynolds number if “L” is taken to be the distance between plates? Fig. P1.38 Solution: (a) For glycerin at 20°C, from Table 1.4, µ ≈ 1.5 N · s/m2. The shear stress is found from Eq. (1) of Ex. 1.8: V (1.5 Pa s)(5.5 m/s) . (a) h (0.006 m) Ans µτ ⋅= = ≈ 1380 Pa The density of glycerin at 20°C is 1264 kg/m3. Then the Reynolds number is defined by Eq. (1.24), with L = h, and is found to be decidedly laminar, Re < 1500: 3 L VL (1264 kg/m )(5.5 m/s)(0.006 m) Re . (b) 1.5 kg/m s Ans ρ µ = = ≈ ⋅ 28 Chapter 1 • Introduction 21 1.39 Knowing µ ≈ 1.80E−5 Pa · s for air at 20°C from Table 1-4, estimate its viscosity at 500°C by (a) the Power-law, (b) the Sutherland law, and (c) the Law of Corresponding States, Fig. 1.5. Compare with the accepted value µ(500°C) ≈ 3.58E−5 Pa · s. Solution: First change T from 500°C to 773 K. (a) For the Power-law for air, n ≈ 0.7, and from Eq. (1.30a), 0.7 n o o 773 (T/T ) (1.80E 5) . (a) 293 Ansµ µ = ≈ − ≈ kg 3.55E 5 m s − ⋅ This is less than 1% low. (b) For the Sutherland law, for air, S ≈ 110 K, and from Eq. (1.30b), 1.5 1.5 o o o (T/T ) (T S) (773/293) (293 110) (1.80E 5) (T S) (773 110) . (b)Ans µ µ + += ≈ − + + = kg3.52E 5 m s − ⋅ This is only 1.7% low. (c) Finally use Fig. 1.5. Critical values for air from Ref. 3 are: c cAir: 1.93E 5 Pa s T 132 Kµ ≈ − ⋅ ≈ (“mixture” estimates) At 773 K, the temperature ratio is T/Tc = 773/132 ≈ 5.9. From Fig. 1.5, read µ/µc ≈ 1.8. Then our critical-point-correlation estimate of air viscosity is only 3% low: c1.8 (1.8)(1.93E 5) . (c)Ansµ µ≈ = − ≈ kg 3.5E 5 m s − ⋅ 1.40 Curve-fit the viscosity data for water in Table A-1 in the form of Andrade’s equation, B A exp T µ ≈ where T is in °K and A and B are curve-fit constants. Solution: This is an alternative formula to the log-quadratic law of Eq. (1.31). We have eleven data points for water from Table A-1 and can perform a least-squares fit to Andrade’s equation: 11 2 i i i 1 E E Minimize E [ A exp(B/T )] , then set 0 and 0 A B ∂ ∂µ ∂ ∂= = − = = The result of this minimization is: A ≈ 0.0016 kg/m⋅s, B ≈ 1903°K. Ans. 22 Solutions Manual • Fluid Mechanics, Fifth Edition The data and the Andrade’s curve-fit are plotted. The error is ±7%, so Andrade’s equation is not as accurate as the log-quadratic correlation of Eq. (1.31). 1.41 Some experimental values of µ for argon gas at 1 atm are as follows: T, °K: 300 400 500 600 700 800 µ, kg/m · s: 2.27E–5 2.85E–5 3.37E–5 3.83E–5 4.25E–5 4.64E–5 Fit these values to either (a) a Power-law, or (b) a Sutherland law, Eq. (1.30a,b). Solution: (a) The Power-law is straightforward: put the values of µ and T into, say, “Cricket Graph”, take logarithms, plot them, and make a linear curve-fit. The result is: Power-law fit: . (a)Ansµ ≈ 0.73 T K 2.29E 5 300 K °− Note that the constant “2.29E–5” is slightly higher than the actual viscosity “2.27E–5” at T = 300 K. The accuracy is ±1% and would be poorer if we replaced 2.29E–5 by 2.27E–5. (b) For the Sutherland law, unless we rewrite the law (1.30b) drastically, we don’t have a simple way to perform a linear least-squares correlation. However, it is no trouble to perform the least-squares summation, E = Σ[µi – µo(Ti/300)1.5(300 + S)/(Ti + S)]2 and minimize by setting ∂ E/∂ S = 0. We can try µo = 2.27E–5 kg/m⋅s for starters, and it works fine. The best-fit value of S ≈ 143°K with negligible error. Thus the result is: Sutherland law: . (b) / Ans − µ 1.5(T/300) (300 143 K) 2.27E 5 kg m s (T 143 K) +≈ ⋅ + Chapter 1 • Introduction 25 These results (#1) are pretty terrible, errors of ±50%, even though they are “least- squares.” The reason is that µ varies over three orders of magnitude, so the fit is biased to higher µ. An alternate fit to Andrade’s equation would be to plot ln(µ) versus 1/T (°K) on, say, “Cricket Graph,” and then fit the resulting near straight line by least squares. The result is: 1 Least-squares of ln( ) versus : . (#2) T Ansµ µ ≈ − ⋅ ° kg 5476 K 3.31E 9 exp m s T K The accuracy is somewhat better, but not great, as follows: T, °C: 0 20 40 60 80 100 µSAE30, kg/m ⋅ s: 2.00 0.40 0.11 0.042 0.017 0.0095 Curve-fit #1: 2.00 0.42 0.108 0.033 0.011 0.0044 Curve-fit #2: 1.68 0.43 0.13 0.046 0.018 0.0078 Neither fit is worth writing home about. Andrade’s equation is not accurate for SAE 30 oil. 1.45 A block of weight W slides down an inclined plane on a thin film of oil, as in Fig. P1.45 at right. The film contact area is A and its thickness h. Assuming a linear velocity distribution in the film, derive an analytic expression for the terminal velocity V of the block. Fig. P1.45 Solution: Let “x” be down the incline, in the direction of V. By “terminal” velocity we mean that there is no acceleration. Assume a linear viscous velocity distribution in the film below the block. Then a force balance in the x direction gives: x x terminal V F W sin A W sin A ma 0, h or: V .Ans θ τ θ µ = − = − = = = hW sin A θ µ 1.46 Find the terminal velocity in Prob. P1.45 if m = 6 kg, A = 35 cm2, θ = 15°, and the film is 1-mm thick SAE 30 oil at 20°C. 26 Solutions Manual • Fluid Mechanics, Fifth Edition Solution: From Table A-3 for SAE 30 oil, µ ≈ 0.29 kg/m · s. We simply substitute these values into the analytical formula derived in Prob. 1.45: 2 hW sin (0.001 m)(6 9.81 N)sin(15 ) V . A (0.29 kg/m s)(0.0035 m ) Ans θ µ × °= = ≈ ⋅ m 15 s 1.47 A shaft 6.00 cm in diameter and 40 cm long is pulled steadily at V = 0.4 m/s through a sleeve 6.02 cm in diameter. The clearance is filled with oil, ν = 0.003 m2/s and SG = 0.88. Estimate the force required to pull the shaft. Solution: Assuming a linear velocity distribution in the clearance, the force is balanced by resisting shear stress in the oil: i wall i o i V V D L F A ( D L) R R R µ πτ µ π = = = ∆ − For the given oil, µ = ρν = (0.88 × 998 kg/m3)(0.003 m2/s) ≈ 2.63 N · s/m (or kg/m · s). Then we substitute the given numerical values to obtain the force: 2 i o i V D L (2.63 N s/m )(0.4 m/s) (0.06 m)(0.4 m) F . R R (0.0301 0.0300 m) Ans µ π π⋅= = ≈ − − 795 N 1.48 A thin moving plate is separated from two fixed plates by two fluids of unequal viscosity and unequal spacing, as shown below. The contact area is A. Determine (a) the force required, and (b) is there a necessary relation between the two viscosity values? Solution: (a) Assuming a linear velocity distribution on each side of the plate, we obtain 1 2F A A . aAnsτ τ = + = 1 2 1 2 V V A ( ) h h µ µ+ The formula is of course valid only for laminar (nonturbulent) steady viscous flow. Chapter 1 • Introduction 27 (b) Since the center plate separates the two fluids, they may have separate, unrelated shear stresses, and there is no necessary relation between the two viscosities. 1.49 An amazing number of commercial and laboratory devices have been developed to measure fluid viscosity, as described in Ref. 27. Consider a concentric shaft, as in Prob. 1.47, but now fixed axially and rotated inside the sleeve. Let the inner and outer cylinders have radii ri and ro, respectively, with total sleeve length L. Let the rotational rate be Ω (rad/s) and the applied torque be M. Using these parameters, derive a theoretical relation for the viscosity µ of the fluid between the cylinders. Solution: Assuming a linear velocity distribution in the annular clearance, the shear stress is i o i rV r r r τ µ µ Ω∆= ≈ ∆ − This stress causes a force dF = τ dA = τ (ri dθ)L on each element of surface area of the inner shaft. The moment of this force about the shaft axis is dM = ri dF. Put all this together: π πµµ θΩ Ω= = = − − 2 3 0 2i i i i i o i o i r r L M r dF r r L d r r r r { }Solve for the viscosity: .Ansοµ Μ π≈ − 2 Ω 3( )i ir r r L 1.50 A simple viscometer measures the time t for a solid sphere to fall a distance L through a test fluid of density ρ. The fluid viscosity µ is then given by ρµ π µ ≈ ≥ 2 3 netW t DLif t DL where D is the sphere diameter and Wnet is the sphere net weight in the fluid. (a) Show that both of these formulas are dimensionally homogeneous. (b) Suppose that a 2.5 mm diameter aluminum sphere (density 2700 kg/m3) falls in an oil of density 875 kg/m3. If the time to fall 50 cm is 32 s, estimate the oil viscosity and verify that the inequality is valid. Solution: (a) Test the dimensions of each term in the two equations: 2 net 3 W ( / )( ) { } and , dimensions OK. (3 ) (1)( )( ) 2 (1)( / )( )( ) { } { } and { } , dimensions OK. . (a) / tM ML T T M LT DL L L LT DL M L L L t T T Ans M LT µ π ρ µ = = = = = = Yes Yes 30 Solutions Manual • Fluid Mechanics, Fifth Edition πµωµω π θ θ = = or 4 3 o 0 r or: Torque M 2 r dr h sin 2hsin We may compute the cone’s slowing down from the angular momentum relation: 2 o o o d 3 M I , where I (cone) mr , m cone mass dt 10 ω= − = = Separating the variables, we may integrate: o w t4 o o 0 rd dt, or: . 2hI sin Ans ω πµω ω θ = − πµω ω θ = − 2 o o 5 r t exp 3mh sin 1.54* A disk of radius R rotates at angular velocity Ω inside an oil container of viscosity µ, as in Fig. P1.54. Assuming a linear velocity profile and neglecting shear on the outer disk edges, derive an expres- sion for the viscous torque on the disk. Fig. P1.54 Solution: At any r ≤ R, the viscous shear τ ≈ µΩr/h on both sides of the disk. Thus, w R 3 0 r d(torque) dM 2r dA 2r 2 r dr, h or: M 4 r dr h Ans µτ µπ Ω= = = π Ω= = . 4R h Ωπµ 1.55 Apply the rotating-disk viscometer of Prob. 1.54, to the particular case R = 5 cm, h = 1 mm, rotation rate 900 rev/min, measured torque M = 0.537 N·m. What is the fluid viscosity? If each parameter (M,R,h,Ω) has uncertainty of ±1%, what is the overall uncertainty of the measured viscosity? Solution: The analytical formula M = πµΩR4/h was derived in Prob. 1.54. Convert the rotation rate to rad/s: Ω = (900 rev/min)(2π rad/rev ÷ 60 s/min) = 94.25 rad/s. Then, 4 4 hM (0.001 m)(0.537 N m) kg or m sR (94.25 rad/s)(0.05 m) Ansµ π π ⋅ = = = . ⋅Ω 2 N s 0.29 m ⋅ Chapter 1 • Introduction 31 For uncertainty, looking at the formula for µ, we have first powers in h, M, and Ω and a fourth power in R. The overall uncertainty estimate [see Eq. (1.44) and Ref. 31] would be 1/ 22 2 2 2 h M R 2 2 2 2 1/ 2 S S S S (4S ) [(0.01) (0.01) (0.01) {4(0.01)} ] 0.044 or: Ans µ Ω ≈ + + + ≈ + + + ≈ .4.4%± The uncertainty is dominated by the 4% error due to radius measurement. We might report the measured viscosity as µ ≈ 0.29 ± 4.4% kg/m·s or 0.29 ± 0.013 kg/m·s. 1.56* For the cone-plate viscometer in Fig. P1.56, the angle is very small, and the gap is filled with test liquid µ. Assuming a linear velocity profile, derive a formula for the viscosity µ in terms of the torque M and cone parameters. Fig. P1.56 Solution: For any radius r ≤ R, the liquid gap is h = r tanθ. Then w r dr d(Torque) dM dA r 2 r r, or r tan cos τ µ π θ θ Ω = = = R 3 2 0 2 2 R M r dr , or: . sin 3sin Ans π µ π µ µ θ θ Ω Ω= = = 3 3M sin 2 R θ πΩ 1.57 Apply the cone-plate viscometer of Prob. 1.56 above to the special case R = 6 cm, θ = 3°, M = 0.157 N ⋅ m, and a rotation rate of 600 rev/min. What is the fluid viscosity? If each parameter (M,R,Ω,θ) has an uncertainty of ±1%, what is the uncertainty of µ? Solution: We derived a suitable linear-velocity-profile formula in Prob. 1.56. Convert the rotation rate to rad/s: Ω = (600 rev/min)(2π rad/rev ÷ 60 s/min) = 62.83 rad/s. Then, 3 3 3Msin 3(0.157 N m)sin(3 ) kg or . m s2 R 2 (62.83 rad/s)(0.06 m) Ans θµ π π ⋅ ° = = = ⋅Ω 2 N s 0.29 m ⋅ For uncertainty, looking at the formula for µ, we have first powers in θ, M, and Ω and a third power in R. The overall uncertainty estimate [see Eq. (1.44) and Ref. 31] would be 1/ 22 2 2 2 M R 2 2 2 2 1/ 2 S S S S (3S ) [(0.01) (0.01) (0.01) {3(0.01)} ] 0.035, or: Ans µ θ Ω = + + + ≈ + + + = . .3 5± % The uncertainty is dominated by the 3% error due to radius measurement. We might report the measured viscosity as µ ≈ 0.29 ± 3.5% kg/m·s or 0.29 ± 0.01 kg/m·s. 32 Solutions Manual • Fluid Mechanics, Fifth Edition 1.58 The laminar-pipe-flow example of Prob. 1.14 leads to a capillary viscometer [27], using the formula µ = πro4∆p/(8LQ). Given ro = 2 mm and L = 25 cm. The data are Q, m3/hr: 0.36 0.72 1.08 1.44 1.80 ∆p, kPa: 159 318 477 1274 1851 Estimate the fluid viscosity. What is wrong with the last two data points? Solution: Apply our formula, with consistent units, to the first data point: 4 4 2 o 3 2 r p (0.002 m) (159000 N/m ) N s p 159 kPa: 0.040 8LQ 8(0.25 m)(0.36/3600 m /s) m π πµ ∆ ⋅∆ = ≈ = ≈ Do the same thing for all five data points: ∆p, kPa: 159 318 477 1274 1851 µ, N·s/m2: 0.040 0.040 0.040 0.080(?) 0.093(?) Ans. The last two estimates, though measured properly, are incorrect. The Reynolds number of the capillary has risen above 2000 and the flow is turbulent, which requires a different formula. 1.59 A solid cylinder of diameter D, length L, density ρs falls due to gravity inside a tube of diameter Do. The clearance, o(D D) D,− is filled with a film of viscous fluid (ρ,µ). Derive a formula for terminal fall velocity and apply to SAE 30 oil at 20°C for a steel cylinder with D = 2 cm, Do = 2.04 cm, and L = 15 cm. Neglect the effect of any air in the tube. Solution: The geometry is similar to Prob. 1.47, only vertical instead of horizontal. At terminal velocity, the cylinder weight should equal the viscous drag: 2 z z s o V a 0: F W Drag g D L DL, 4 (D D)/2 πρ µ π = = − + = − + − Σ or: V .Ans= s o gD(D D) 8 −ρ µ For the particular numerical case given, ρsteel ≈ 7850 kg/m3. For SAE 30 oil at 20°C, µ ≈ 0.29 kg/m·s from Table 1.4. Then the formula predicts 3 2 s o terminal gD(D D) (7850 kg/m )(9.81 m/s )(0.02 m)(0.0204 0.02 m) V 8 8(0.29 kg/m s) .Ans ρ µ − −= = ⋅ ≈ 0.265 m/s Chapter 1 • Introduction 35 1.63 Derive Eq. (1.37) by making a force balance on the fluid interface in Fig. 1.9c. Solution: The surface tension forces YdL1 and YdL2 have a slight vertical component. Thus summation of forces in the vertical gives the result z 2 1 1 2 F 0 2YdL sin(d /2) 2YdL sin(d /2) pdA θ θ = = + − ∆ Fig. 1.9c But dA = dL1dL2 and sin(dθ/2) ≈ dθ/2, so we may solve for the pressure difference: 2 1 1 2 1 2 1 2 1 2 dL d dL d d d p Y Y dL dL dL dL θ θ θ θ +∆ = = + = 1 2 1 1 Y R R + Ans. 1.64 A shower head emits a cylindrical jet of clean 20°C water into air. The pressure inside the jet is approximately 200 Pa greater than the air pressure. Estimate the jet diameter, in mm. Solution: From Table A.5 the surface tension of water at 20°C is 0.0728 N/m. For a liquid cylinder, the internal excess pressure from Eq. (1.31) is ∆p = Y/R. Thus, for our data, 2/ 200 N/m (0.0728 N/m)/ , solve 0.000364 m, p Y R R R Ans. ∆ = = = = 0.00073 m=D 1.65 The system in Fig. P1.65 is used to estimate the pressure p1 in the tank by measuring the 15-cm height of liquid in the 1-mm-diameter tube. The fluid is at 60°C. Calculate the true fluid height in the tube and the percent error due to capillarity if the fluid is (a) water; and (b) mercury. Fig. P1.65 36 Solutions Manual • Fluid Mechanics, Fifth Edition Solution: This is a somewhat more realistic variation of Ex. 1.9. Use values from that example for contact angle θ: (a) Water at 60°C: γ ≈ 9640 N/m3, θ ≈ 0°: 3 4Y cos 4(0.0662 N/m)cos(0 ) h 0.0275 m, D (9640 N/m )(0.001 m) θ γ °= = = or: ∆htrue = 15.0 – 2.75 cm ≈ 12.25 cm (+22% error) Ans. (a) (b) Mercury at 60°C: γ ≈ 132200 N/m3, θ ≈ 130°: 3 4Y cos 4(0.47 N/m)cos130 h 0.0091 m, D (132200 N/m )(0.001 m) θ γ °= = = − trueor: h 15.0 0.91∆ = + ≈ 15.91cm( 6%error)− Ans. (b) 1.66 A thin wire ring, 3 cm in diameter, is lifted from a water surface at 20°C. What is the lift force required? Is this a good method? Suggest a ring material. Solution: In the literature this ring-pull device is called a DuNouy Tensiometer. The forces are very small and may be measured by a calibrated soft-spring balance. Platinum-iridium is recommended for the ring, being noncorrosive and highly wetting to most liquids. There are two surfaces, inside and outside the ring, so the total force measured is F 2(Y D) 2Y Dπ π= = This is crude—commercial devices recommend multiplying this relation by a correction factor f = O(1) which accounts for wire diameter and the distorted surface shape. For the given data, Y ≈ 0.0728 N/m (20°C water/air) and the estimated pull force is F 2 (0.0728 N/m)(0.03 m) . π= ≈ 0 0137 N Ans. For further details, see, e.g., F. Daniels et al., Experimental Physical Chemistry, 7th ed., McGraw-Hill Book Co., New York, 1970. 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 θ < 90°. Derive an expression for the capillary rise h in the annular gap, if the gap is very narrow. Chapter 1 • Introduction 37 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 hθ π π ρ π+ = − Cancel where possible and the result is .Ans= −o i2 cos /{ (r r )}θ ρh Y g 1.68* Analyze the shape η(x) of the water-air interface near a wall, as shown. Assume small slope, R−1 ≈ d2η/dx2. The pressure difference across the interface is ∆p ≈ ρgη, with a contact angle θ at x = 0 and a horizontal surface at x = ∞. Find an expression for the maximum height h. Fig. P1.68 Solution: This is a two-dimensional surface-tension problem, with single curvature. The surface tension rise is balanced by the weight of the film. Therefore the differential equation is 2 2 Y d d p g Y 1 R dxdx η ηρ η ∆ = = ≈ This is a second-order differential equation with the well-known solution, 1 2C exp[Kx] C exp[ Kx], K ( g/Y)η ρ= + − = To keep η from going infinite as x = ∞, it must be that C1 = 0. The constant C2 is found from the maximum height at the wall: x 0 2 2h C exp(0), hence C hη =| = = = Meanwhile, the contact angle shown above must be such that, x 0 d cot cot ) hK, thus h dx K η θθ= = − ( = − =| 40 Solutions Manual • Fluid Mechanics, Fifth Edition If we decrease water temperature to 5°C, the vapor pressure reduces to 863 Pa, and the density changes slightly, to 1000 kg/m3. For this condition, if V = 30 m/s, we compute: 2 2(131000 863) Ca 0.289 (1000)(30) −= ≈ This is greater than 0.25, therefore the body will not cavitate for these conditions. Ans. (b) 1.74 A propeller is tested in a water tunnel at 20°C (similar to Fig. 1.12a). The lowest pressure on the body can be estimated by a Bernoulli-type relation, pmin = po − ρV2/2, where po = 1.5 atm and V is the tunnel average velocity. If V = 18 m/s, will there be cavitation? If so, can we change the water temperature and avoid cavitation? Solution: At 20°C, from Table A-5, pv = 2.337 kPa. Compute the minimum pressure: 2 2 min o 3 1 1 kg m p p V 1.5(101350 Pa) 998 18 9650 Pa (??) 2 2 sm ρ = − = − = − The predicted pressure is less than the vapor pressure, therefore the body will cavitate. [The actual pressure would not be negative; a cavitation bubble would form.] Since the predicted pressure is negative; no amount of cooling—even to T = 0°C, where the vapor pressure is zero, will keep the body from cavitating at 18 m/s. 1.75 Oil, with a vapor pressure of 20 kPa, is delivered through a pipeline by equally- spaced pumps, each of which increases the oil pressure by 1.3 MPa. Friction losses in the pipe are 150 Pa per meter of pipe. What is the maximum possible pump spacing to avoid cavitation of the oil? Solution: The absolute maximum length L occurs when the pump inlet pressure is slightly greater than 20 kPa. The pump increases this by 1.3 MPa and friction drops the pressure over a distance L until it again reaches 20 kPa. In other words, quite simply, max1.3 MPa 1,300,000 Pa (150 Pa/m)L, or L Ans.= = ≈ 8660 m It makes more sense to have the pump inlet at 1 atm, not 20 kPa, dropping L to about 8 km. 1.76 Estimate the speed of sound of steam at 200°C and 400 kPa, (a) by an ideal-gas approximation (Table A.4); and (b) using EES (or the Steam Tables) and making small isentropic changes in pressure and density and approximating Eq. (1.38). Chapter 1 • Introduction 41 Solution: (a) For steam, k ≈ 1.33 and R = 461 m2/s2·K. The ideal gas formula predicts: 2 2(kRT) {1.33(461 m /s K)(200 273 K)} (a)a Ans. ≈ √ = √ ⋅ + ≈ 539 m/s (b) We use the formula a = √(∂p/∂ρ)s ≈ √{∆p|s/∆ρ|s} for small isentropic changes in p and ρ. From EES, at 200°C and 400 kPa, the entropy is s = 1.872 kJ/kg·K. Raise and lower the pressure 1 kPa at the same entropy. At p = 401 kPa, ρ = 1.87565 kg/m3. At p = 399 kPa, ρ = 1.86849 kg/m3. Thus ∆ρ = 0.00716 kg/m3, and the formula for sound speed predicts: 2 3 s s{ / } {(2000 N/m )/(0.00358 kg/m )} . (b)a p Ansρ| |≈ √ ∆ ∆ = √ = 529 m/s Again, as in Prob. 1.34, the ideal gas approximation is within 2% of a Steam-Table solution. 1.77 The density of gasoline varies with pressure approximately as follows: p, atm: 1 500 1000 1500 ρ, lbm/ft3: 42.45 44.85 46.60 47.98 Estimate (a) its speed of sound, and (b) its bulk modulus at 1 atm. Solution: For a crude estimate, we could just take differences of the first two points: ρ −≈ ∆ ∆ ≈ ≈ ≈ − 2 3 (500 1)(2116) lbf/ft ft a ( p/ ) 3760 . (a) s(44.85 42.45)/32.2 slug/ft Ans m 1150 s ρ≈ = ≈ ≈2 3 2 2 lbf B a [42.45/32.2 slug/ft ](3760 ft/s) 1.87E7 . (b) ft Ans895 MPa For more accuracy, we could fit the data to the nonlinear equation of state for liquids, Eq. (1.22). The best-fit result for gasoline (data above) is n ≈ 8.0 and B ≈ 900. Equation (1.22) is too simplified to show temperature or entropy effects, so we assume that it approximates “isentropic” conditions and thus differentiate: n 2 n 1a a a a a n(B 1)pp dp (B 1)( / ) B, or: a ( / ) p d ρ ρ ρ ρ ρ ρ −+≈ + − = ≈ liquid a aor, at 1 atm, a n(B 1)p /ρ≈ + The bulk modulus of gasoline is thus approximately: 1 atm a dp “ ” n(B 1)p (8.0)(901)(101350 Pa) (b) d Ans.ρ ρ Β = = + = ≈| 731 MPa 42 Solutions Manual • Fluid Mechanics, Fifth Edition And the speed of sound in gasoline is approximately, 3 1/2 1 atma [(8.0)(901)(101350 Pa)/(680 kg/m )] . (a)Ans= ≈ m 1040 s 1.78 Sir Isaac Newton measured sound speed by timing the difference between seeing a cannon’s puff of smoke and hearing its boom. If the cannon is on a mountain 5.2 miles away, estimate the air temperature in °C if the time difference is (a) 24.2 s; (b) 25.1 s. Solution: Cannon booms are finite (shock) waves and travel slightly faster than sound waves, but what the heck, assume it’s close enough to sound speed: (a) x 5.2(5280)(0.3048) m a 345.8 1.4(287)T, T 298 K . (a) t 24.2 s Ans ∆≈ = = = ≈ ≈ ∆ 25 C° (b) x 5.2(5280)(0.3048) m a 333.4 1.4(287)T, T 277 K . (b) t 25.1 s Ans ∆≈ = = = ≈ ≈ ∆ 4 C° 1.79 Even a tiny amount of dissolved gas can drastically change the speed of sound of a gas-liquid mixture. By estimating the pressure-volume change of the mixture, Olson [40] gives the following approximate formula: [ (1 ) ][ (1 ) ] g l mixture g l l g p K a x x xK x pρ ρ ≈ + − + − where x is the volume fraction of gas, K is the bulk modulus, and subscripts and g denote the liquid and gas, respectively. (a) Show that the formula is dimensionally homogeneous. (b) For the special case of air bubbles (density 1.7 kg/m3 and pressure 150 kPa) in water (density 998 kg/m3 and bulk modulus 2.2 GPa), plot the mixture speed of sound in the range 0 ≤ x ≤ 0.002 and discuss. Solution: (a) Since x is dimensionless and K dimensions cancel between the numerator and denominator, the remaining dimensions are pressure divided by density: ρ= = = = 1/2 2 3 1/ 2 2 2 1/ 2 mixture{ } [{ }/{ }] [(M/LT )/(M/L )] [L /T ] (a) a p Yes, homogeneous Ans.L/T Chapter 1 • Introduction 45 1.84* Modify Prob. 1.83 to find the equation of the pathline which passes through the point (xo, yo) at t = 0. Sketch this pathline. Solution: The pathline is computed by integration, over time, of the velocities: 2t t o t o dx dx u x(1 2t), or: (1 2t)dt, or: x x e dt x dy dy v y, or: dt, or: y y e dt y += = + = + = = = = = We have implemented the initial conditions (x, y) = (xo, yo) at t = 0. [We were very lucky, as planned for this problem, that u did not depend upon y and v did not depend upon x.] Now eliminate t between these two to get a geometric expression for this particular pathline: = + 2o o ox x exp{ln(y/y ) ln (y/y )} This pathline is shown in the sketch below. 46 Solutions Manual • Fluid Mechanics, Fifth Edition 1.85-a Report to the class on the achievements of Evangelista Torricelli. Solution: Torricelli’s biography is taken from a goldmine of information which I did not put in the references, preferring to let the students find it themselves: C. C. Gillespie (ed.), Dictionary of Scientific Biography, 15 vols., Charles Scribner’s Sons, New York, 1976. Torricelli (1608–1647) was born in Faenza, Italy, to poor parents who recognized his genius and arranged through Jesuit priests to have him study mathematics, philosophy, and (later) hydraulic engineering under Benedetto Castelli. His work on dynamics of projectiles attracted the attention of Galileo himself, who took on Torricelli as an assistant in 1641. Galileo died one year later, and Torricelli was appointed in his place as “mathematician and philosopher” by Duke Ferdinando II of Tuscany. He then took up residence in Florence, where he spent his five happiest years, until his death in 1647. In 1644 he published his only known printed work, Opera Geometrica, which made him famous as a mathematician and geometer. In addition to many contributions to geometry and calculus, Torricelli was the first to show that a zero-drag projectile formed a parabolic trajectory. His tables of trajectories for various angles and initial velocities were used by Italian artillerymen. He was an excellent machinist and constructed—and sold—the very finest telescope lenses in Italy. Torricelli’s hydraulic studies were brief but stunning, leading Ernst Mach to proclaim him the ‘founder of hydrodynamics.’ He deduced his theorem that the velocity of efflux from a hole in a tank was equal to √(2gh), where h is the height of the free surface above the hole. He also showed that the efflux jet was parabolic and even commented on water- droplet breakup and the effect of air resistance. By experimenting with various liquids in closed tubes—including mercury (from mines in Tuscany)—he thereby invented the barometer. From barometric pressure (about 30 feet of water) he was able to explain why siphons did not work if the elevation change was too large. He also was the first to explain that winds were produced by temperature and density differences in the atmo- sphere and not by “evaporation.” 1.85-b Report to the class on the achievements of Henri de Pitot. Solution: The following notes are abstracted from the Dictionary of Scientific Biography (see Prob. 1.85-a). Pitot (1695–1771) was born in Aramon, France, to patrician parents. He hated to study and entered the military instead, but only for a short time. Chance reading of a textbook obtained in Grenoble led him back to academic studies of mathematics, astronomy, and engineering. In 1723 he became assistant to Réamur at the French Academy of Sciences and in 1740 became a civil engineer upon his appointment as a director of public works in Languedoc Province. He retired in 1756 and returned to Aramon until his death in 1771. Pitot’s research was apparently mediocre, described as “competent solutions to minor problems without lasting significance”not a good recommendation for tenure nowadays! His lasting contribution was the invention, in 1735, of the instrument which Chapter 1 • Introduction 47 bears his name: a glass tube bent at right angles and inserted into a moving stream with the opening facing upstream. The water level in the tube rises a distance h above the surface, and Pitot correctly deduced that the stream velocity ≈ √(2gh). This is still a basic instrument in fluid mechanics. 1.85-c Report to the class on the achievements of Antoine Chézy. Solution: The following notes are from Rouse and Ince [Ref. 23]. Chézy (1718–1798) was born in Châlons-sur-Marne, France, studied engineering at the Ecole des Ponts et Chaussées and then spent his entire career working for this school, finally being appointed Director one year before his death. His chief contribution was to study the flow in open channels and rivers, resulting in a famous formula, used even today, for the average velocity: V const AS/P≈ where A is the cross-section area, S the bottom slope, and P the wetted perimeter, i.e., the length of the bottom and sides of the cross-section. The “constant” depends primarily on the roughness of the channel bottom and sides. [See Chap. 10 for further details.] 1.85-d Report to the class on the achievements of Gotthilf Heinrich Ludwig Hagen. Solution: The following notes are from Rouse and Ince [Ref. 23]. Hagen (1884) was born in Königsberg, East Prussia, and studied there, having among his teachers the famous mathematician Bessel. He became an engineer, teacher, and writer and published a handbook on hydraulic engineering in 1841. He is best known for his study in 1839 of pipe-flow resistance, for water flow at heads of 0.7 to 40 cm, diameters of 2.5 to 6 mm, and lengths of 47 to 110 cm. The measurements indicated that the pressure drop was proportional to Q at low heads and proportional (approximately) to Q2 at higher heads, where “strong movements” occurred—turbulence. He also showed that ∆p was approximately proportional to D−4. Later, in an 1854 paper, Hagen noted that the difference between laminar and turbulent flow was clearly visible in the efflux jet, which was either “smooth or fluctuating,” and in glass tubes, where sawdust particles either “moved axially” or, at higher Q, “came into whirling motion.” Thus Hagen was a true pioneer in fluid mechanics experimentation. Unfortunately, his achievements were somewhat overshadowed by the more widely publicized 1840 tube-flow studies of J. L. M. Poiseuille, the French physician. 1.85-e Report to the class on the achievements of Julius Weisbach. Solution: The following notes are abstracted from the Dictionary of Scientific Biography (see Prob. 1.85-a) and also from Rouse and Ince [Ref. 23]. 50 Solutions Manual • Fluid Mechanics, Fifth Edition developed CalTech into a premier research center for aeronautics. His leadership spurred the growth of the aerospace industry in southern California. He helped found the Jet Propulsion Laboratory and the Aerojet General Corporation. After World War II, Kármán founded a research arm for NATO, the Advisory Group for Aeronautical Research and Development, whose renowned educational institute in Brussels is now called the Von Kármán Center. Kármán was uniquely skilled in integrating physics, mathematics, and fluid mechanics into a variety of phenomena. His most famous paper was written in 1912 to explain the puzzling alternating vortices shed behind cylinders in a steady-flow experiment conducted by K. Hiemenz, one of Kármán’s students—these are now called Kármán vortex streets [see Fig. 5.2a]. Shed vortices are thought to have caused the destruction by winds of the Tacoma Narrows Bridge in 1940 in Washington State. Kármán wrote 171 articles and 5 books and his methods had a profound influence on fluid mechanics education in the 20th century. 1.85-i Report to the class on the achievements of Paul Richard Heinrich Blasius. Solution: The following notes are from Rouse and Ince [Ref. 23]. Blasius (1883–1970) was Ludwig Prandtl’s first graduate student at Göttingen. His 1908 dissertation gave the analytic solution for the laminar boundary layer on a flat plate [see Sect. 7.4]. Then, in two papers in 1911 and 1913, he gave the first demonstration that pipe-flow resistance could be nondimensionalized as a plot of friction factor versus Reynolds number—the first “Moody-type” chart. His correlation, 1/4df 0.316 Re , −≈ is still is use today. He later worked on analytical solutions of boundary layers with variable pressure gradients. 1.85-j Report to the class on the achievements of Ludwig Prandtl. Solution: The following notes are from Rouse and Ince [Ref. 23]. Ludwig Prandtl (1875–1953) is described by Rouse and Ince [23] as the father of modern fluid mechanics. Born in Munich, the son of a professor, Prandtl studied engineering and received a doctorate in elasticity. But his first job as an engineer made him aware of the lack of correlation between theory and experiment in fluid mechanics. He conducted research from 1901–1904 at the Polytechnic Institute of Hanover and presented a seminal paper in 1904, outlining the new concept of “boundary layer theory.” He was promptly hired as professor and director of applied mechanics at the University of Gottingen, where he remained throughout his career. He, and his dozens of famous students, started a new “engineering science” of fluid mechanics, emphasizing (1) mathematical analysis based upon by physical reasoning; (2) new experimental techniques; and (3) new and inspired flow- visualization schemes which greatly increased our understanding of flow phenomena. Chapter 1 • Introduction 51 In addition to boundary-layer theory, Prandtl made important contributions to (1) wing theory; (2) turbulence modeling; (3) supersonic flow; (4) dimensional analysis; and (5) instability and transition of laminar flow. He was a legendary engineering professor. 1.85-k Report to the class on the achievements of Osborne Reynolds. Solution: The following notes are from Rouse and Ince [Ref. 23]. Osborne Reynolds (1842–1912) was born in Belfast, Ireland, to a clerical family and studied mathematics at Cambridge University. In 1868 he was appointed chair of engineering at a college which is now known as the University of Manchester Institute of Science and Technology (UMIST). He wrote on wide-ranging topics—mechanics, electricity, navigation—and developed a new hydraulics laboratory at UMIST. He was the first person to demonstrate cavitation, that is, formation of vapor bubbles due to high velocity and low pressure. His most famous experiment, still performed in the undergraduate laboratory at UMIST (see Fig. 6.5 in the text) demonstrated transition of laminar pipe flow into turbulence. He also showed in this experiment that the viscosity was very important and led him to the dimensionless stability parameter ρVD/µ now called the Reynolds number in his honor. Perhaps his most important paper, in 1894, extended the Navier-Stokes equations (see Eqs. 4.38 of the text) to time-averaged randomly fluctuating turbulent flow, with a result now called the Reynolds equations of turbulence. Reynolds also contributed to the concept of the control volume which forms the basis of integral analysis of flow (Chap. 3). 1.85-l Report to the class on the achievements of John William Strutt, Lord Rayleigh. Solution: The following notes are from Rouse and Ince [Ref. 23]. John William Strutt (1842–1919) was born in Essex, England, and inherited the title Lord Rayleigh. He studied at Cambridge University and was a traditional hydro- dynamicist in the spirit of Euler and Stokes. He taught at Cambridge most of his life and also served as president of the Royal Society. He is most famous for his work (and his textbook) on the theory of sound. In 1904 he won the Nobel Prize for the discovery of argon gas. He made at least five important contributions to hydrodynamics: (1) the equations of bubble dynamics in liquids, now known as Rayleigh-Plesset theory; (2) the theory of nonlinear surface waves; (3) the capillary (surface tension) instability of jets; (4) the “heat-transfer analogy” to laminar flow; and (5) dimensional similarity, especially related to viscosity data for argon gas and later generalized into group theory which previewed Buckingham’s Pi Theorem. He ended his career as president, in 1909, of the first British committee on aeronautics. 52 Solutions Manual • Fluid Mechanics, Fifth Edition 1.85-m Report to the class on the achievements of Daniel Bernoulli. Solution: The following notes are from Rouse and Ince [Ref. 23]. Daniel Bernoulli (1700–1782) was born in Groningen, Holland, his father, Johann, being a Dutch professor. He studied at the University of Basel, Switzerland, and taught mathematics for a few years at St. Petersburg, Russia. There he wrote, and published in 1738, his famous treatise Hydrodynamica, for which he is best known. This text contained numerous ingenious drawings illustrating various flow phenomena. Bernoulli used energy concepts to establish proportional relations between kinetic and potential energy, with pressure work added only in the abstract. Thus he never actually derived the famous equation now bearing his name (Eq. 3.77 of the text), later derived in 1755 by his friend Leonhard Euler. Daniel Bernoulli never married and thus never contributed additional members to his famous family of mathematicians. 1.85-n Report to the class on the achievements of Leonhard Euler. Solution: The following notes are from Rouse and Ince [Ref. 23]. Leonhard Euler (1707–1783) was born in Basel, Switzerland, and studied mathematics under Johann Bernoulli, Daniel’s father. He succeeded Daniel Bernoulli as professor of mathematics at the St. Petersburg Academy, leaving there in 1741 to join the faculty of Berlin University. He lost his sight in 1766 but continued to work, aided by a prodigious memory, and produced a vast output of scientific papers, dealing with mathematics, optics, mechanics, hydrodynamics, and celestial mechanics (for which he is most famous today). His famous paper of 1755 on fluid flow derived the full inviscid equations of fluid motion (Eqs. 4.36 of the text) now called Euler’s equations. He used a fixed coordinate system, now called the Eulerian frame of reference. The paper also presented, for the first time, the correct form of Bernoulli’s equation (Eq. 3.77 of the text). Separately, in 1754 he produced a seminal paper on the theory of reaction turbines, leading to Euler’s turbine equation (Eq. 11.11 of the text). Chapter 1 • Introduction 55 C1.2 When a person ice-skates, the ice surface actually melts beneath the blades, so that he or she skates on a thin film of water between the blade and the ice. (a) Find an expression for total friction force F on the bottom of the blade as a function of skater velocity V, blade length L, water film thickness h, water viscosity µ, and blade width W. (b) Suppose a skater of mass m, moving at constant speed Vo, suddenly stands stiffly with skates pointed directly forward and allows herself to coast to a stop. Neglecting air resistance, how far will she travel (on two blades) before she stops? Give the answer X as a function of (Vo, m, L, h, µ, W). (c) Compute X for the case Vo = 4 m/s, m = 100 kg, L = 30 cm, W = 5 mm, and h = 0.1 mm. Do you think our assumption of negligible air resistance was a good one? Solution: (a) The skate bottom and the melted ice are like two parallel plates: , (a) V VLW F A Ans. h h µτ µ τ= = = (b) Use F = ma to find the stopping distance: 2 x x VLW dV F F ma m h dt µ= − = − = =Σ (the ‘2’ is for two blades) Separate and integrate once to find the velocity, once again to find the distance traveled: 2 o o 0 2 , : , (b) 2 LW t mh V mhdV LW dt or V V e X V dt Ans. V mh LW µµ µ − ∞ = − = = = (c) Apply our specific numerical values to a 100-kg (!) person: (4.0 / )(100 )(0.0001 ) 2(1.788 3 / )(0.3 )(0.005 ) m s kg m X E kg m s m m = = − ⋅ 7460 m (!) Ans. (c) We could coast to the next town on ice skates! It appears that our assumption of negligible air drag was grossly incorrect. C1.3 Two thin flat plates are tilted at an angle α and placed in a tank of known surface tension Y and contact angle θ, 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 ρ, find an expression for Y in terms of the other variables. 56 Solutions Manual • Fluid Mechanics, Fifth Edition 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(θ − α), as shown. There are two plates. Therefore, the total z-directed force on the liquid column is Fvertical = 2Yb cos(θ – α) Ans. (a) (b) The vertical force in (a) above holds up the entire weight of the liquid column between plates, which is W = ρg{bh(L − h tanα)}. Set W equal to F and solve for U = [ρgbh(L − h tanα)]/[2 cos(θ − α)] Ans. (b) C1.4 Oil of viscosity µ and density ρ drains steadily down the side of a tall, wide vertical plate, as shown. The film is fully developed, that is, its thickness δ and velocity profile w(x) are independent of distance z down the plate. Assume that the atmosphere offers no shear resistance to the film surface. (a) Sketch the approximate shape of the velocity profile w(x), keeping in mind the boundary conditions. Chapter 1 • Introduction 57 (b) Suppose film thickness d is measured, along with the slope of the velocity profile at the wall, (dw/dx)wall, with a laser-Doppler anemometer (Chap. 6). Find an expression for µ as a function of ρ, δ, (dw/dx)wall, and g. Note that both w and (dw/dx)wall will be negative as shown. Solution: (a) The velocity profile must be such that there is no slip (w = 0) at the wall and no shear (dw/dx = 0) at the film surface. This is shown at right. Ans. (a) (b) Consider a freebody of any vertical length H of film, as at right. Since there is no acceleration (fully developed film), the weight of the film must exactly balance the shear force on the wall: ( ) ( ),wall wall wall dw W g H b Hb dx ρ δ τ τ µ= = = − | Solve this equality for the fluid viscosity: ρ δµ −= wall g dw dx( / ) Ans. (b) C1.5 Viscosity can be measured by flow through a thin-bore or capillary tube if the flow rate is low. For length L, (small) diameter ,D L pressure drop ∆p, and (low) volume flow rate Q, the formula for viscosity is µ = D4∆p/(CLQ), where C is a constant. (a) Verify that C is dimensionless. The following data are for water flowing through a 2-mm-diameter tube which is 1 meter long. The pressure drop is held constant at ∆p = 5 kPa. T, °C: 10.0 40.0 70.0 Q, L/min: 0.091 0.179 0.292 (b) Using proper SI units, determine an average value of C by accounting for the variation with temperature of the viscosity of water. Solution: (a) Check the dimensions of the formula and solve for {C}: 4 4 1 2 3 ( ) { } , { }{ }( )( / ) M D p L ML T M LT CLQ LT CC L L T µ − − ∆ = = = = therefore {C} = {1} Dimensionless Ans. (a)