(Solution Manual) Fluid Mechanics

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e ⎝ ⎠⎝ ⎠ Chapter 01 • 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 Molecular weight 28.97 mol−1 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,  = ⎛1012 molecules ⎞⎛ 4.81E −23 g ⎞ | mm3 || molecule | = 4.81E −11 g mm3 = 4.81E −5 kg m3 Finally, from the perfect gas law, Eq. (1.13), at 20C = 293 K, we obtain the pressure: ⎛ kg ⎞⎛ m2 ⎞ p =  RT = | 4.81E −5 || 287 m3 s2K |(293 K) = 4.0 Pa ns. ⎝ ⎠⎝ ⎠ 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 mt = ∫  dVol = avg (Air Vol)  avg 4 R2 (Air thickness) = (0.6 kg/m3 )4 (6.377E6 m)2 (20E3 m)  6.1E18 kg Ans. 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: N = m(atmosphere) = 6.1E21 grams  1.3E44 molecules Ans. molecules m(one molecule) 4.8E−23 gm/molecule 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 p a, 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 show n, 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 , velocity V, and surface tension Y may be combined into a dimensionless group. Find the combination which is proportional to . This group has a customary name, which begins with C. Can you guess its name? Solution: The dimensions of these variables are {} = {M/LT}, {V} = {L/T}, and {Y} = {M/T2}. We must divide  by Y to cancel mass {M}, then work the velocity into the group: {  F = { M /LT F = {T F, hence multiply by {V} = { L F; { Y } { M /T 2 } { L } {T } % J % J % J % J finally obtain V = dimensionless. Y Ans. This dimensionless parameter is commonly called the Capillary Number . 1.5 A formula for estimating the mean free path of a perfect gas is: l = 1.26  = 1.26   (RT) (1)   (RT) p 3 Chapter 1 • Introduction ⎝ ⎠⎝ ⎠ where the latter form follows from the ideal -gas law,  = pRT. What are the dimensions of the constant “1.26”? Estimate the mean free path of air at 20 C and 7 kPa. Is air rarefied at this condition? Solution: We know the dimensions of every term except “1.26”: { M F { M F { L2 F {l} = {L} {} = { } LT {} = { L3 } {R} = { T2 } {T} = {} % J % J % J Therefore the above formula (first form) may be written dimensionally as {L} = {1.26?} {M/L T} = {1.26?}{L} {M/L3}[{L2 /T2  }{}] Since we have {L} on both sides, {1.26} = {unity}, that is, the constant is dimensionless. The formula is therefore dimensionally homogeneous and should hold for any unit system. For air at 20 C = 293 K and 7000 Pa, the density is  = pRT = (7000)/[(287)(293)] = 0.0832 kg m3. From Table A-2, its viscosity is 1.80E −5 N s/m2. Then the formula predict a mean free path of l = 1.26 1.80E −5  9.4E−7 m (0.0832)[(287)(293)]1/2 Ans. This is quite small. We would judge this gas to approximate a continuum if the physical scales in the flow are greater than about 100 l, that is, greater than about 94 m. 1.6 If p is pressure and y is a coordinate, state, in the {MLT} system, the dimensions of the quantities (a) py; (b) ∫ p dy; (c) 2py2; (d) p. Solution: (a) {ML−2T−2}; (b) {MT−2}; (c) {ML−3T−2}; (d) {ML−2T−2} 1.7 A small village draws 1.5 acre-foot of water per day from its reservoir. Convert this water usage into (a) gallons per minute; and (b) liters per second. Solution: One acre = (1 mi2640) = (5280 ft)2640 = 43560 ft2. Therefore 1.5 acre -ft = 65340 ft3 = 1850 m3. Meanwhile, 1 gallon = 231 in3 = 231/1728 ft3. Then 1.5 acre -ft of water per day is equivalent to ft3 ⎛ 1728 gal ⎞⎛ 1 day ⎞ gal Q = 65340 day | 231 ft3 || 1440 min |  340 min Ans. (a) 4 Solutions Manual • Fluid Mechanics, Fifth Edition % J % J Similarly, 1850 m3 = 1.85E6 liters. Then a metric unit for this water usage is: ⎛ Q = 1.85E6 L ⎞⎛ 1 day ⎞  21 L Ans. (b) | day || 86400 sec | s ⎝ ⎠⎝ ⎠ 1.8 Suppose that bending stress  in a beam depends upon bending moment M and beam area moment of inertia I and is proportional to the beam half -thickness y. Suppose also that, for the particular case M = 2900 in lbf, y = 1.5 in, and I = 0.4 in4, the predicted stress is 75 MPa. Find the only possible dimensionally homogeneous formula for . Solution: We are given that  = y fcn(M,I) and we are not to study up on strength of materials but only to use dimensional reasoning. For homogeneity, the right hand side must have dimensions of stress, that is, {} = {y}{fcn(M,I)}, or: { M F = {L}{fcn(M,I)} { LT2 } or: the function must have dimensions {fcn(M,I)} = { M F { L2T2 } Therefore, to achieve dimensional homogeneity, we somehow must combine bending moment, whose dimensions are {ML2T–2}, with area moment of inertia, {I} = {L4}, and end up with {ML–2T–2}. Well, it is clear that {I} contains neither mass {M} nor time {T} dimensions, but the bending moment contains both mass and time and in exactly the com- bination we need, {MT–2}. Thus it must be that  is proportional to M also. Now we have reduced the problem to:  = { M F = { ML2 F = −4 yM fcn(I), or { LT2 } {L}{ T2 }{fcn(I)}, or: {fcn(I)} {L } % J % J We need just enough I’s to give dimensions of {L–4}: we need the formula to be exactly inverse in I. The correct dimensionally homogeneous beam bending formula is thus:  = C My , I where {C} = {unity} Ans. The formula admits to an arbitrary dimensionless constant C whose value can only be obtained from known data. Convert stress into English units:  = (75 MPa) (6894.8) = 10880 lbfin2. Substitute the given data into the proposed formula:  = 10880 lbf = C My = C (2900 lbfin)(1.5 in) , or: C  1.00 Ans. in2 I 0.4 in4 The data show that C = 1, or  = My/I , our old friend from strength of materials. 5 Chapter 1 • Introduction = =   3  gL { } 1.9 The dimensionless Galileo number, Ga, expresses the ratio of gravitational effect to viscous effects in a flow. It combines the quantities density , acceleration of gravity g, length scale L, and viscosity . 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 {} = {ML3}, {g} = {LT2}, {L} = {L}, and { } = {MLT}. Divide  by  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: {  F { M /L3 F { T F {  } { M / LT } { L2 } % J % J % J while only {g} contains {T}. To keep {g} to the 1st power, we need to multiply it by {}2. Thus {}2{g} = {T2L4}{LT2} = {L−3}. We then make the combination dimensionless by multiplying the group by L3. Thus we obtain: ⎛ ⎞2 Galileo number = Ga = | | ⎝ ⎠ 2 3 (g)(L) =  2 = gL3  2 Ans. 1.10 The Stokes -Oseen formula [10] for drag on a sphere at low velocity V is: F = 3DV + 9 16 V2D2 where D = sphere diameter,  = viscosity, and  = density. Is the formula homogeneous? Solution: Write this formula in dimensional form, using Table 1-2: {F} = {3}{}{D}{V} + { 9 F{}{V}2{D}2 ? % 16 J { ML F { M F { L F { M F{ L2 F 2 or: { T2 } = {1}{ LT }{L}{ T } +{1}{ L3 }{ T2 }{L } ? % J % J % J % J% J where, hoping for homogeneity, we have assumed that all constants (3, ,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 (gh) 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: Q = 0.68D2 where g is the acceleration of gravity. What are the dimensions of the constant 0.68? Solution: Write the equation in dimensional form: { L3 F ? { L F1/ 2 { L3 F {Q} = { } = {0.68?}{L2}{ } {L}1/ 2 = {0.68} { } % T J % T2 J % T J Thus, since D2 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 Q = Cd Ao (p/ ) 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 u = B p (r2 − r2 )  o where  is viscosity and p the pressure drop. What are the dimensions of B? Solution: Using Table 1-2, write this equation in dimensional form: {p} 2 { L F {M/LT2} 2 { L2 F {u} = {B} {r }, or: { } = {B?} {L } = {B?}{ }, {} % T J {M/LT} % T J 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 u = C p (r2 − r2 ) L o 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. gh 7 Chapter 1 • Introduction 1.13 The efficiency  of a pump is defined as  = Qp Input Power where Q is volume flow and p the pressure rise produced by the pump. What is  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  is a dimensionless ratio. Then convert everything to consistent units, for example, BG: L ft2 lbf lbf ftlbf Q = 40 = 1.41 ; s s p = 35 in2 = 5040 ft2 ; Power = 16(550) = 8800 s  = (1.41 ft3s)(5040 lbfft2 )  8800 ftlbfs 0.81 or 81% Ans. 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 f ar we know that Q = B fcn(H,g). Write this in dimensional form: { L3 F {Q} = { } = {B}{f(H,g)} = {L}{f(H,g)}, % T J Fig. P1.14 { L2 F or: {f(H,g)} = { } % T J 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} 8 Solutions Manual • Fluid Mechanics, Fifth Edition p⎛ ⎞ % J % J % J % J % J In order for fcn(H) to provide dimensions of {L3/2}, the function must be a 3/2 power. Thus the final desired homogeneous relation for dam flow is: Q = CBg1/2H3/2, where C is a dimensionless constant Ans. 1.15 As a practical application of Fig. P1.14, often termed a sharp -crested weir, civil engineers use the following formula for flow rate: Q  3.3 BH3/2, with Q in ft3/s and B and H in feet. Is this formula dimensionally homogeneous? If not, try to explain the difficulty and how it might be converted to a more homogeneous form. Solution: Clearly the formula cannot be dimensionally homogeneous, because B and H do not contain the dimension time. The formula would be invalid for anything except English units (ft, sec). By comparing with the answer to Prob. 1.14 just above, we see that the constant “3.3” hides the square root of the acceleration of gravity. 1.16 Test the dimensional homogeneity of the boundary -layer x-momentum equation:  u  u + v  u = −  p + g +   x  y  x x  y Solution: This equation, like all theoretical partial differential equations in mechanics, is dimensionally homogeneous. Test each term in sequence: {  u F {  u F M L L/T { M F { p F M/LT2 { M F { u  x } = { v  y } = L3 T L = { L2T2 }; { x } = L = { L2T2 } M L { M F {  F M/LT2 { M F { gx} = L3 T2 = { L2T2 }; { x } = L = { L2T2 } % J % J % J All terms have dimension {ML–2T–2}. This equation may use any consistent units. 1.17 Investigate the consistency of the Hazen -Williams formula from hydraulics: 0.54 Q = 61.9D2.63 | | ⎝ L ⎠ What are the dimensions of the constant “61.9”? Can this equation be used with confidence for a variety of liquids and gases? 9 Chapter 1 • Introduction ( t Solution: Write out the dimensions of each side of the equation: { L3 F ? { p F0.54 { M/LT2 F0.54 {Q} = { } = {61.9}{D2.63}{ } = {61.9}{L2.63}{ } % T J % L J % L J The constant 61.9 has fractional dimensions: {61.9} = {L1.45T0.08M–0.54} Ans. Clearly, the formula is extremely inconsistent and cannot be used with confidence for any given fluid or condition or units. Actually, the Hazen -Williams formula, still in common use in the watersupply industry, is valid only for water flow in smooth pipes larger than 2 -in. diameter and turbulent velocities less tha n 10 ft/s and (certain) English units. This formula should be held at arm’s length and given a vote of “No Confidence.” 1.18 * (“*” means “difficult” —not just a plug-and-chug, that is) For small particles at low velocities, the first (linear) term in Stokes’ drag law, Prob. 1.10, is dominant, hence F = KV, where K is a constant. Suppose a particle of mass m is constrained to move horizontally from the initial position x = 0 with initial velocity V = Vo. Show (a) that its velo city will decrease exponentially with time; and (b) that it will stop after travelling a distance x = mV o/K. Solution: Set up and solve the differential equation for forces in the x-direction: Σ Fx = −Drag = max , or: −KV = m dV dt V , integrate ∫ Vo dV = −∫ V 0 m dt K Solve V = Voe −mtK and t x = ∫ 0 V dt = mVo 1 − e K −mtK ) Ans. (a,b) Thus, as asked, V drops off exponentially with time, and, as t → , x = mVo K. 1.19 Marangoni convection arises when a surface has a difference in surface tension along its length. The dimensionless Marangoni number M is a combination of thermal diffusivity  = k/(cp) (where k is the thermal conductivity), length scale L, viscosity , and surface tension difference Y. If M is proportional to L, find its form. 10 Solutions Manual • Fluid Mechanics, Fifth Edition Cg 0 | {  } T 2 M { } { T   } {T L2 } { L } % J % J % J % J % J ∫ Solution: List the dimensions: {} = {L2/T}, {L} = {L}, {} = {M/LT}, {Y} = {M/T2}. We divide  Y by  to get rid of mass dimensions, then divide by  to eliminate time: { YF = M LT = { L F, then { Y 1 F = { L T F = { 1 F Multiply by L and we obtain the Marangoni number: M =  YL  Ans. 1.20 C (“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 CV2, 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 z max reached by the ball and compare your resu lts with the elementary -physics case of zero air drag. Solution: For this problem, we include the weight of the ball, for upward motion z: 2 dV V dV t ΣFz = −maz , or: −CV − mg = m , solve dt Vo g + CV2 /m = −∫ dt = −t Thus V = ⎛ tan  − t ⎞ and z = m ln 「cos( − t  (gC/m) $ | | C | cos | ⎝ ⎠ ) ] where  = tan–1[Vo  (C/mg)] . 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s2/m2, we compute  = 0.8732 radians, = 37.72 m , s = 0.2601 s−1, m = 145 m C Hence the final analytical formulas are: ⎛ V| in m ⎞ = 37.72 tan(0.8732 −.2601t) s ⎝ ⎠ and z(in meters) = 145 ln 「cos(0.8732 − 0.2601t) ( | 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. mg mg C Cg m 11 Chapter 1 • Introduction | |  U= −  For zero drag, from elementary physics formulas, V = Vo − gt and z = Vot − gt2/2, we calculate that Vo 45 V2 (45)2 tmax height = =  4.59 s and zmax = o =  103.2 m g 9.81 2g 2(9.81) 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 , viscosity , temperature difference T, length scale L, the acceleration of gravity g, and the coefficient of volume expansion , defined as  = (−1/)(/T)p. If Gr contains both g and  in the numerator, what is its proper form? Solution: Recall that {/} = {L2/T} and eliminates mass dimensions. To eliminate tem- perature, we need the product {} = {1}. Then {g} eliminates {T}, and L3 cleans it all up: Thus the dimensionless Gr =  2gTL3/ 2 Ans. 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 u = U (1 − R2/x2 ); 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 du  u  u ⎛ R2 ⎞ 「 ⎛ 2R2 ⎞" U2 ⎛ 2 2 ⎞ x = + u = 0 + U |1 − 2 | |U | + || =  | −   5 |,  =  dt  t  x ⎝ x ⎠ |+ ⎝ x ⎠|] R ⎝   ⎠ R This acceleration is negative, as expected, and reaches a minimum near point B, which is found by differentiating the acceleration with respect to x: d ⎛ du ⎞ = 0 if  2 = 5 , or x max decel.  −1.291 Ans. (b) dx ⎝ dt ⎠ Substituting  3 R = −1.291 into (du/dt) gives du 2 min 0.372 dt R Ans. (a) 3 3  12 Solutions Manual • Fluid Mechanics, Fifth Edition ⎝ A plot of the flow deceleration along line AB is shown as follows. 1.23 E 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  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 CO 2, k  1.30, and R  189 m2/(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 = p1 RT1 = 1,013,250 Pa (189 m2/s2 K )(673 K ) = 7.97 kg ; m3 c = kR = 1.3(189) = 819 J Ans. (a) p k − 1 1.3 − 1 kgK For an ideal gas cooled isentropically to T2 = 100C = 373 K, the formula is p ⎛ T ⎞ k/(k −1) p ⎛ 373K ⎞1.3 /(1.3−1) 2 = | 2 | p1 ⎝ T1 ⎠ = 2 = | 1013 kPa 673K |⎠ = 0.0775, or: p2 = 79 kPa Ans. (a) For EES or the Gas Tables, just program the properties for carbon dioxide or look them up: 1 = 7.98 kg/m3; cp = 1119 J/(kg K); p2 = 43 kPa Ans. (b) (NOTE: The large errors in “ideal” cp and “ideal” final pressure are due to the sharp drop - off in k of CO 2 with temperature, as seen in Fig. 1.3 of the text.) 13 Chapter 1 • Introduction 1.25 A tank contains 0.9 m3 of helium at 200 kPa and 20 C. Estimate the total mass of this gas, in kg, (a) on earth; and (b) on the moon. Also, (c) how much heat transfer, in MJ, is required to expand this gas at constant temperature to a new volume of 1.5 m3? Solution: First find the density of helium for this condition, given R = 2077 m2/(s2K) from Table A-4. Change 20C to 293 K:  = p = 200000 N/m2  3 RHe T (2077 J/kgK)(293 K) 0.3286 kg/m Now mass is mass , no matter where you are. Therefore, on the moon or wherever, mHe = He = (0.3286 kg/m3)(0.9 m3)  0.296 kg Ans. (a,b) For part (c), we expand a constant mass isothermally from 0.9 to 1.5 m3. The first law of thermodynamics gives dQadded − dWby gas = dE = mcvT = 0 since T2 = T1 (isothermal) Then the heat added equals the work of expansion. Estimate the work done: 2 2 m 2 d W1-2 = ∫ p d = ∫  RT d = mRT ∫  = mRT ln(2/1), 1 1 1 or: W1-2 = (0.296 kg)(2077 J/kgK)(293 K)ln(1.5/0.9) = Q1-2  92000 J Ans. (c) 1.26 A tire has a volume of 3.0 ft3 and a ‘gage’ pressure of 32 psi at 75F. If the ambient pressure is sea-level standard, what is the weight of air in the tire? Solution: Convert the temperature from 75F to 535R. Convert the pressure to psf: p = (32 lbf/in2 )(144 in2/ft2 ) + 2116 lbf/ft2 = 4608 + 2116  6724 lbf/ft2 From this compute the density of the air in the tire:  = p = 6724 lbf/ft2 = 3 air RT (1717 ftlbf/slug R)(535 R) 0.00732 slug/ft Then the total weight of air in the tire is Wair = g = (0.00732 slug/ft3 )(32.2 ft/s2 )(3.0 ft3 )  0.707 lbf Ans. He 14 Solutions Manual • Fluid Mechanics, Fifth Edition 5 1.27 Given temperature and specific volume data for steam at 40 psia [Ref. 13]: T, F: 400 500 600 700 800 v, ft3/lbm: 12.624 14.165 15.685 17.195 18.699 Is the ideal gas law reasonable for this data? If so, find a least-squares value for the gas constant R in m2/(s2K) and compare with Table A-4. Solution: The units are awkward but we can compute R from the data. At 400F, pV (40 lbf/in2 )(144 in2/ft2 )(12.624 ft3/lbm)(32.2 lbm/slug) ftlbf “R”400F = T = (400 + 459.6) R  2721 slugR The metric conversion factor, from the inside cover of the text, is “5.9798”: R metric = 2721/5.9798 = 455.1 m2/(s2K). Not bad! This is only 1.3% less than the ideal -gas approxi - mation for steam in Table A-4: 461 m2/(s2K). Let’s try all the five data points: T, F: 400 500 600 700 800 R, m2/(s2K): 455 457 459 460 460 The total variation in the data is only 0.6%. Therefore steam is nearly an ideal gas in this (high) temperature range and for this (low) pressure. We can take an average value: 1 5 J p = 40 psia, 400F  T  800F: Rsteam  ΣRi  458 i=1  0.6% kg  K Ans. With such a small uncertainty, we don’t really need to perform a least -squares analysis, but if we wanted to, it would go like this: We wish to minimize, for all data, the sum of the squares of the deviations from the perfect -gas law: 5 ⎛ pV ⎞2  E 5 ⎛ pV ⎞ Minimize E = Σ| R − i | by differentiating = 0 = Σ2 | R − i | i=1 ⎝ Ti ⎠  R i=1 ⎝ Ti ⎠ p 5 Vi 40(144) 「12.624 18.699 " Thus Rleast-squares = 5 Σ T = 5 860R +⋯+ 1260 R ]| (32.2) i=1 i For this example, then, least-squares amounts to summing the (V/T) values and converting the units. The English result shown above gives R least-squares  2739 ft lbf/slug R. Convert this to metric units for our (highly accurate) least-squares estimate: Rsteam  2739/5.9798  458  0.6% J/kgK Ans. 15 Chapter 1 • Introduction | 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 p = 1 atm = p + p = ma R T + mw R T tot air water  a  w or: m = m + m = pa + pw for an ideal gas tot a w Ra T RwT where, from Table A -4, R air = 287 and R water = 461 m2/(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 p a = 1 atm − pw = 101350 − 7375 = 93975 Pa. Solving for the mixture density, we obtain  = ma + mw = pa + pw = 93975 + 7375 = 1.046 + 0.051  1.10 kg Ans.  RaT RwT 287(313) 461(313) m3 By comparison, the density of dry air for the same conditions is  = p = 101350 = 1.13 kg dry air RT 287(313) m3 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 mRT ⎛  ⎞ ⎛ p ⎞ W1-2 = −∫ p d = −∫  d = −mRT ln| 2 | = p22 ln| 2 | 1 1 ⎝ 1 ⎠ ⎝ p1 ⎠ where the latter form follows from the ideal gas law for isothermal changes. For the given numerical data, we obtain the quantitative work done: W = p  ln ⎛ p2 ⎞ = ⎛134.7  144 lbf ⎞ (5 ft3 ) ln ⎛ 134.7⎞  215,000 ftlbf Ans. 1-2 2 2 |⎝ p1 |⎠ ⎝ ft2 |⎠ |⎝ 14.7 |⎠ 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, o  1.94 slug/ft3. At 120 psi(gage) = 134.7 psia, the density would rise slightly according to Eq. (1.22): p 134.7 ⎛  ⎞7 =  3001| | − 3000, solve   1.940753 slug/ft3, po 14.7 ⎝ 1.94 ⎠ Hence mwater =  = (1.940753)(5 ft3 )  9.704 slug The density change is extremely small. Now the work done, as in Prob. 1.29 above, is 2 2 ⎛ m ⎞ 2 m d  W1-2 = −∫ p d = ∫ pd |  | = ∫ p  2  pavgm  2 for a linear pressure rise 1 1 ⎝ ⎠ 1 avg ⎛ 14.7 + 134.7 lbf ⎞ ⎛ 0.000753 ft3 ⎞ Hence W1-2  |⎝ 2  144 ft2 |⎠ (9.704 slug) | 1.94042 |  21 ftlbf slug Ans. ⎝ ⎠ [Exact integration of Eq. (1.22) would give the same numerical result.] Compressing water (extremely small ) 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,  = a + bT + cT2, and test its accuracy vis-a-vis Table A-1. Finally, compute  at T = 45°C and compare your result with the accepted value of   990.1 kg/m3. Solution: The least-squares parabola which fits the data of Table A-1 is:  (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 45°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  for fresh water is a maximum at T = +4°C. 17 Chapter 1 • Introduction 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  LR2 = 2  (90 m)(15 m)2  42412 m3 3 3 Estimate, from the ideal -gas law, the respective densities of helium and air: (a)  = pHe = 1.1(101350)  0.1832 kg ; helium RHeT 2077(293) m3 (b)  = pair = 101350  1.205 kg . air RairT 287(293) m3 Then the respective gas weights are W =  g = ⎛ 0.1832 kg ⎞⎛ 9.81 m ⎞ (42412 m3 )  76000 N Ans. (a) He He | m3 || s2 | ⎝ ⎠⎝ ⎠ Wair =  airg = (1.205)(9.81)(42412)  501000 N Ans. (b) 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.] 18 Solutions Manual • Fluid Mechanics, Fifth Edition 1.33 Experimental data for density of mercury versus pressure at 20°C are as follows: p, atm: 1 500 1000 1500 2000 , kg/m3: 13545 13573 13600 13625 13653 Fit this data to the empirical state relation for liquids, Eq. (1.19), to find the best values of B and n for mercury. Then, assuming the data are nearly isentropic, use these values to estimate the speed of sound of mercury at 1 atm and compare with Table 9.1. Solution: This can be done (laboriously) by the method of least -squares, but we can also do it o n a spreadsheet by guessing, say, n  4,5,6,7,8 and finding the average B for each case. For this data, almost any value of n > 1 is reasonably accurate. We select: Mercury: n  7, B  35000  2% Ans. The speed of sound is found by differentiating Eq. (1.19) and then taking the square root: dp p ⎛  ⎞n−1 「n(B+ 1)p $1 / 2  o n(B + 1) | | , henc e a =   | o | d o ⎝ o ⎠ o + o ] it being assumed here that this equation of state is “isentropic.” Evaluating this relation for mercury’s values of B and n, we find the speed of sound at 1 atm: 「(7)(35001)(101350 N/m2 ) "1/ 2 amercury  | 13545 kg/m3 |  1355 m/s Ans. % ] This is about 7% less than the value of 1450 m/s listed in Table 9.1 for mercury. 1.34 Consider steam at the following state near the saturation line: ( p1, T1) = (1.31 MPa, 290°C). Calculate and compare, for an ideal gas (Table A.4) and the Steam Tables (or the EES software), (a) the density 1; and (b) the density 2 if the steam expands isentropically to a new pressure of 414 kPa. Discuss your results. Solution: From Table A.4, for steam, k  1.33, and R  461 m2/(s2K). Convert T1 = 563 K. Then,   1= ⎛ p p1 RT1 ⎞1/k = 1,310,000 Pa (461 m2s2K )(563 K ) ⎛ 414 kPa ⎞1/1.33 = 5.05 kg Ans. (a) m3 kg 2 = 2 = | 2 | = |⎝ ⎠| = 0.421, or: 2 = 2.12 3 Ans. (b) 1 5.05 ⎝ p1 ⎠ 1310 kPa m 19 Chapter 1 • Introduction % J For EES or the Steam Tables, just program the properties for steam or look it up: EES real steam: 1 = 5.23 kg/m3 Ans. (a), 2 = 2.16 kg/m3 Ans. (b) The ideal-gas error is only about 3%, even though the expansion approached the saturation line. 1.35 In Table A-4, most common gases (air, nitrogen, oxygen, hydrogen, CO, NO) have a specific heat ratio k = 1.40. Why do argon and helium have such high values? Why does NH 3 have such a low value? What is the lowest k for any gas that you know? Solution: In elementary kinetic theory of gases [8], k is related to the number of “degrees of freedom” of the gas: k  1 + 2/N, where N is the number of different modes of translation, rotation, and vibration possible for the gas molecule. Example: Monotomic gas, N = 3 (translation only), thus k  5/3 This explains why helium and argon, which are monatomic gases, have k  1.67. Example: Diatomic gas, N = 5 (translation plus 2 rotations), thus k  7/5 This explains why air, nitrogen, oxygen, NO, CO and hydrogen have k  1.40. But NH 3 has four atoms and therefore more than 5 degrees of freedom, hence k will be less than 1.40 (the theory is not too clear what “N” is for such complex molecules). The lowest k known to this writer is for uranium hexafluoride , 238UF 6, which is a very complex, heavy molecule with many degrees of freedom. The estimated value of k for this heavy gas is k  1.06. 1.36 The bulk modulus of a fluid is defined as B = ( p/)S. What are the dimensions of B? Estimate B (in Pa) for (a) N2O, and (b) water, at 20°C and 1 atm. Solution: The density units cancel in the definition of B and thus its dimensions are the same as pressure or stress: {B} = {p} = {F /L2} = { M F Ans. { LT2 } (a) For an ideal gas, p = C k for an isentropic process, thus the bulk modulus is: Ideal gas: B =  d (C k ) =  kC k−1 = kC k = kp d For N2O, from Table A-4, k  1.31, so BN2O = 1.31 atm = 1.33E5 Pa Ans. (a) 20 Solutions Manual • Fluid Mechanics, Fifth Edition kRT 1.31[189 m2/(s2 K)](373 K) 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, B   d [p {(B +1)( /  )n − B}] = n(B+1)p ( /  )n = n(B +1)p at 1 atm d o o o o o For water, B  3000 and n  7, so our estimate is Bwater  7(3001)po = 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 =  =   189 J/(kg K). Then cv = R/(k −1), or: k = 1+ R/cv = 1+ 189/610  1.31 Ans. (a) [It is probably N2O] With k and R known, the speed of sound at 100ºC = 373 K is estimated by a = =  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,   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)  1380 Pa h (0.006 m) Ans. (a) 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: VL (1264 kg/m3 )(5.5 m/s)(0.006 m) ReL =  = 1.5 kg/m s  28 Ans. (b) 21 Chapter 1 • Introduction ⎝ ⎠ | T | 1.39 Knowing   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 (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), ⎛ 773⎞ 0.7  = o (T/To )n  (1.80E −5) | | 293  3.55E −5 kg ms Ans. (a) This is less than 1% low. (b) For the Sutherland law, for air, S  110 K, and from Eq. (1.30b),  =  「(T/To )1.5(To + S)"  − 「(773/293)1.5(293 + 110)" o | (T + S) | (1.80E 5) | (773 + 110) | % ] % ] = 3.52E −5 kg ms Ans. (b) This is only 1.7% low. (c) Finally use Fig. 1.5. Critical values for air from Ref. 3 are: Air: c  1.93E −5 Pas Tc  132 K (“mixture” estimates) At 773 K, the temperature ratio is T/T c = 773/132  5.9. From Fig. 1.5, read c  1.8. Then our critical -point -correlation estimate of air viscosity is only 3% low:   1.8c = (1.8)(1.93E −5)  3.5E −5 kg ms Ans. (c) 1.40 Curve -fit the viscosity data for water in Table A-1 in the form of Andrade’s equation ,   A exp⎛ B ⎞ ⎝ ⎠ 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 Minimize E = Σ[i − A exp(B/Ti )]2, then set i=1  E = 0 and  A  E = 0  B 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  for argon gas at 1 atm are as follows: T, °K: 300 400 500 600 700 800  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  and T into, say, “Cricket Graph”, take logarithms, plot them, and make a linear curve -fit. The result is: ⎛ T K ⎞0.73 Power -law fit:   2.29E −5 | 300 K | Ans. (a) 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 -square s summation, E = [i – o(Ti/300)1.5(300 + S)/(T i + S)]2 and minimize by setting  E/ S = 0. We can try o = 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:   (T/300)1.5 (300 + 143 K) Ans. (b) 2.27E −5 kg/ms (T + 143 K) 23 Chapter 1 • Introduction ⎝ ⎠ We may tabulate the data and the two curve -fits as follows: T, K: 300 400 500 600 700 800   E5, data: 2.27 2.85 3.37 3.83 4.25 4.64   E5, Power -law: 2.29 2.83 3.33 3.80 4.24 4.68   E5, Sutherland: 2.27 2.85 3.37 3.83 4.25 4.64 1.42 Some experimental values of  of helium at 1 atm are as follows: T, °K: 200 400 600 800 1000 1200 , kg/m  s: 1.50E –5 2.43E –5 3.20E –5 3.88E –5 4.50E –5 5.08E –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  and T into, say, “Cricket Graph,” take logarithms, plot them, and make a linear curve -fit. The result is: ⎛ T K ⎞0.68 Power -law curve -fit: He  1.505E −5 | 200 K | Ans. (a) The accuracy is less than 1%. (b) For the Sutherland fit, we can emulate Prob. 1.41 and perform the least -squares summation, E = [i – o(Ti/200)1.5(200 + S)/(T i + S)]2 and minimize by setting  E/ S = 0. We can try o = 1.50E –5 kg/m·s and T o = 200°K for starters, and it works OK. The best-fit value of S  95.1°K. Thus the result is: Helium (T/ 200)1.5 (200 + 95.1K) Sutherland law:   4% 1.50E −5 kg/m  s (T + 95.1K) Ans. (b) For the complete range 200–1200 K, the Power -law is a better fit. The Sutherland law improves to 1% if we drop the data point at 200°K. 1.43 Yaws et al. [ref. 34] suggest a 4-constant curve -fit formula for liquid viscosity: log10   A + B/T + CT + DT2, with T in absolute units. (a) Can this formula be criticized on dimensional grounds? (b) If we use the formula anyway, how do we evaluate A,B,C,D in the least-squares sense for a set of N data points? 24 Solutions Manual • Fluid Mechanics, Fifth Edition i i i N Solution: (a) Yes, if you’re a purist: A is dimensionless, but B,C,D are not. It would be more comfortable to this writer to write the formula in terms of some reference temperature To: log10   A + B(To /T) + C(T/To ) + D(T/To )2, (dimensionless A,B,C,D) (b) For least squares, express the square error as a summation of data-vs-formula differences: 「 2 "2 2 E = Σ#A + B/Ti + CTi + DTi − log10 i ] = Σfi for short. i=1 i=1 Then evaluate  E /A = 0,  E / B = 0,  E / C = 0, and  E / D = 0, to give four simultaneous linear algebraic equations for (A,B,C,D): Σfi = 0; Σfi /Ti = 0; Σfi Ti = 0; Σf T2 = 0, where fi = A + B/Ti + CTi + DT2 − log10 i Presumably this was how Yaws et al. [34] computed (A,B,C,D) for 355 organic liquids. 1.44 The viscosity of SAE 30 oil may vary considerably, according to industry -agreed specifications [ SAE Handbook , Ref. 26]. Comment on the following data and fit the data to Andrade’s equation from Prob. 1.41. T, C: 0 20 40 60 80 100 SAE30 , kg/m ·s: 2.00 0.40 0.11 0.042 0.017 0.0095 Solution: At lower temperatures, 0 C < T < 60 C, these values are up to fifty per cent higher than the curve labelled “SAE 30 Oil” in Fig. A-1 of the Appendix. However, at 100C, the value 0.0095 is within the range specified by SAE for this oil: 9.3 <  < 12.5 mm2/s, if its density lies in the range 760 <  < 1020 kg/m3, which it surely must. Therefore a surprisingly wide difference in viscosity -versus -temperature still makes an oil “ SAE 30. ” To fit Andrade’s law,   A exp(B/T), we must make a least-squares fit for the 6 data points above (just as we did in Prob. 1.41): 6 「 ⎛ B ⎞"2  E  E Andrade fit: With E = Σ|i − A exp | T || , then set  A = 0 and  B = 0 i=1 + ⎝ i ⎠] This formulation produces the following results: Least -squares of  versus T:   2.35E −10 kg exp ⎛ 6245 K ⎞ Ans. (#1) ms ⎝| TK |⎠ N 25 Chapter 1 • Introduction These results (#1) are pretty terrible , errors of 50%, even though they are “least - squares.” The reason is that  varies over three orders of magnitude, so the fit is biased to higher . An alternate fit to Andrade’s equation would be to plot ln( ) versus 1/T (°K) on, say, “Cricket Graph,” and then fit the resulting near straight line by least squares. The result is: Least -squares of ln( 1   3.31E −9 kg exp ⎛ 5476 K ⎞ Ans. (#2) ) versus : T ms ⎝| TK |⎠ The accuracy is somewhat better, but not great, as follows: T, C: 0 20 40 60 80 100 SAE30, 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: Σ F = W sin − A = W sin − ⎛  V ⎞ A = ma = 0, x | h | x ⎝ ⎠ or: Vterminal = hW sin  A Ans. 1.46 Find the terminal velocity in Prob. P1.45 if m = 6 kg, A = 35 cm2,  = 15, and the film is 1-mm thick SAE 30 oil at 20 C.