Solution of Fluid Mechanics 7th.pdf -

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www.mechassis.com 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 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,  molecules  g     1012 3  4.81E23   mm  molecule  g kg  4.81E11 3  4.81E5 3 mm m 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 3   287 2  (293 K)  4.0Pa ns.  m  s K  P1.2 Table A.6 lists the density of the standard atmosphere as a function of altitude. Use these values to estimate, crudely, say, within a factor of 2, the number of molecules of air in the entire atmosphere of the earth. Solution: Make a plot of density  versus altitude z in the atmosphere, from Table A.6: 1.2255 kg/m3 Density in the Atmosphere  0 z 30,000 m This writer’s approximation: The curve is approximately an exponential,    o exp(-b z), with b approximately equal to 0.00011 per meter. Integrate this over the entire atmosphere, with the radius of the earth equal to 6377 km:  b z   d (vol )  0 [ o e 2 matmosphere  ](4 Rearth dz )  2 o 4 Rearth (1.2255 kg / m3 )4 (6.377 E 6 m) 2    5.7 E18 kg b 0.00011 / m 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: m(atmosphere) 5.7E21 grams N molecules   1.2Ε44 molecules Ans. m(one molecule) 4.8E  23 gm/molecule This estimate, though crude, is within 10 per cent of the exact mass of the atmosphere. Chapter 1  Introduction 3 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 Fig. P1.3 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. www.mechassis.com P1.4 Sand, and other granular materials, definitely flow, that is, you can pour them from a container or a hopper. There are whole textbooks on the “transport” of granular materials [54]. Therefore, is sand a fluid? Explain. Solution: Granular materials do indeed flow, at a rate that can be measured by “flowmeters”. But they are not true fluids, because they can support a small shear stress without flowing. They may rest at a finite angle without flowing, which is not possible for liquids (see Prob. P1.3). The maximum such angle, above which sand begins to flow, is called the angle of repose. A familiar example is sugar, which pours easily but forms a significant angle of repose on a heaping spoonful. The physics of granular materials are complicated by effects such as particle cohesion, clumping, vibration, and size segregation. See Ref. 54 to learn more. ________________________________________________________________________ 1.5 A formula for estimating the mean free path of a perfect gas is:     1.26  1.26 RT (1)  RT p 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 M  L2  {}  {L} {}    {}   3  {R}   2  {T}  {}  LT  L  T  Therefore the above formula (first form) may be written dimensionally as {M/LT} {L}  {1.26?}  {1.26?}{L} {M/L }  [{L2 /T 2  }{}] 3 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 predicts a mean free path of 1.80E5   1.26  9.4E7 m Ans. (0.0832)[(287)(293)]1/2 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 , that is, greater than about 94 m. P1.6 The Saybolt Universal Viscometer, now obsolete but still sold in scientific catalogs, measures the kinematic viscosity of lubricants [Ref. 49, p. 40]. A container, held at constant temperature, is filled with 60 ml of fluid. Measure the time t for the fluid to drain from a small hole or short tube in the bottom. This time unit, called Saybolt universal seconds, or SUS, is correlated with kinematic viscosity , in centistokes (1 stoke = 1 cm2/s), by the following curve- fit formula: 145   0.215 t  for 40  t 100 SUS t (a) Comment on the dimensionality of this equation. (b) Is the formula physically correct? (c) Since  varies strongly with temperature, how does temperature enter into the formula? (d) Can we easily convert  from centistokes to mm2/s? Solution: (a) The formula is dimensionally inconsistent. The right-hand side does not have obvious kinematic viscosity units. The constants 0.215 and 145 must conceal (dimensional) information on temperature, gravity, fluid density, and container shape. (b) The formula correctly predicts that the time to drain increases with fluid viscosity. (c) The time t will reflect changes in , and the constants 0.215 and 145 vary slightly (1%) with temperature [Ref. 49, p. 43]. (d) Yes, no conversion necessary; the units of centistoke and mm2/s are exactly the same.. P1.7 Convert the following inappropriate quantities into SI units: (a) a velocity of 3,937 yards per hour; (b) a volume flow rate of 4,903 acre-feet of water per week; and (c) a mass flow rate of 25,616 gallons per day of SAE 30W oil at 20ºC. Solution: Most of what we need is on the inside front cover of the text. (a) One yard equals 3 ft. Thus 3937 yards = 11811 ft x (0.3048 m/ft) = 3600 m. One hour = 3600 s. Thus, finally, the velocity in SI units is (3600 m)/(3600 s) = 1.00 m/s Ans.(a) (b) One acre = 4046.9 m2, and 1 ft = 0.3048 m, thus 4903 acre-ft = (4903)(4046.9)(0.3048) = 6,048,000 m3. One week = (7 days)(24 h/day)(3600 s/h) = 604,800 s. Finally, 4903 acre-ft per week = (6,048,000 m3)/(604,800 s) = 10.0 m3/s Ans.(b) (c) From Table A.3, the density of SAE 30W oil at 20ºC is 891 kg/m3. Meanwhile, 25616 gallons x (0.0037854 m3/gal) = 96.97 m3. One day = (24 h/day) x (3600 s/h) = 86400 s. Finally, 25616 gal/day = (891 kg/m3)(96.97 m3)/(86400 s) = 1.00 kg/s Ans.(c) ________________________________________________________________________ 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,  M  { }  {y}{fcn(M,I)}, or:  2   {L}{fcn(M,I)}  LT   M  or: the function must have dimensions {fcn(M,I)}   2 2  L T  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   ML2  4   yM fcn(I), or  2  {L}  2 {fcn(I)}, or: {fcn(I)}  {L }  LT   T  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: My  C , where {C}  {unity} Ans. I 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: lbf My (2900 lbf in)(1.5 in)   10880 2 C C , or: C  1.00 Ans. in I 0.4 in 4 The data show that C  1, or   My/I, our old friend from strength of materials. P1.9 An inverted conical container, 26 inches in diameter and 44 inches high, is filled with a liquid at 20C and weighed. The liquid weight is found to be 5030 ounces. (a) What is the density of the fluid, in kg/m3? (b) What fluid might this be? Assume standard gravity, g = 9.807 m/s2. Solution: First find the volume of the liquid in m3, from our high school cone-volume formula:   Liquid Vol.  R2 h  (13 in) 2 (44 in)  7787 in3  4.506 ft 3  0.1276 m3 3 3 Then find the mass of liquid in kilograms: Liquid mass  5030 oz  16  314.4 lbm  0.45359  142.6 kg mass 142.6 kg kg Then liquid density   3  1117 3 Ans.(a ) volume 0.1276 m m (b) From Appendix Table A.3, this could very well be ethylene glycol. Ans.(b) ________________________________________________________________________ 1.10 The Stokes-Oseen formula [10] for drag on a sphere at low velocity V is: 9 F  3 DV   V 2 D2 16 where D  sphere diameter,   viscosity, and   density. Is the formula homogeneous? Solution: Write this formula in dimensional form, using Table 1-2:  9  {F}  {3 }{}{D}{V}   {}{V}2 {D}2 ?  16  M  L  2  ML  M L  or:  2   {1}  {L}    {1}  3   2 {L2} ? T   LT  T  L T  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. P1.11 In English Engineering units, the specific heat c p of air at room temperature is approximately 0.24 Btu/(lbm-F). When working with kinetic energy relations, it is more appropriate to express c p as a velocity-squared per absolute degree. Give the numerical value, in this form, of c p for air in (a) SI units, and (b) BG units. Solution: From Appendix C, Conversion Factors, 1 Btu = 1055.056 J (or N-m) = 778.17 ft-lbf, and 1 lbm = 0.4536 kg = (1/32.174) slug. Thus the conversions are: Btu 1055.056 N  m N m m2 SI units : 0.24  0.24  1005  1005 2 Ans.(a ) lbm  F (0.4536 kg )(1K /1.8) kg  K s K Btu 778.17 ft  lbf ft  lbf ft 2 BG units : 0.24  0.24  6009 Ans.(b)  6009 lbm  F [(1/ 32.174) slug ](1  R) slug  R s2  R _______________________________________________________________________ 1.12 For low-speed (laminar) flow in a tube of radius r o , the velocity u takes the form p uB  r 2 o  r2  where  is viscosity and p the pressure drop. What are the dimensions of B? 8 Solutions Manual  Fluid Mechanics, Sixth Edition Solution: Using Table 1-2, write this equation in dimensional form: {p} 2 L  {M/LT 2} 2  L2  {u}  {B} {r }, or:    {B?} {L }  {B?}   , {} T  {M/LT} T 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 p 2 2 uC L  ro  r  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. 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 ft 2 lbf lbf ftlbf Q  40  1.41 ; p  35 2  5040 2 ; Power  16(550)  8800 s s in ft s (1.41 ft 3 s)(5040 lbf ft 2 )   0.81 or 81% Ans. 8800 ftlbf s 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: Fig. P1.14  L3  {Q}     {B}{f(H,g)}  {L}{f(H,g)}, T  L2  or: {f(H,g)}    T 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}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  C B g1/2 H3/2, where C is a dimensionless constant Ans. P1.15 Mott [49] recommends the following formula for the friction head loss h f , in ft, for flow through a pipe of length L o and diameter D (both in ft): h f  Lo ( )1.852 Q 0.63 0.551AC h D where Q is the volume flow rate in ft3/s, A is the pipe cross-section area in ft2, and C h is a dimensionless coefficient whose value is approximately 100. Determine the dimensions of the constant 0.551. Solution: Write out the dimensions of each of the terms in the formula: {h f }  {L} ; {Lo }  {L} ; {Q}  {L3 / T } ; { A}  {L2 } ; {C h }  {1} ; {D}  {L} Use these dimensions in the equation to determine {0.551}. Since h f and L o have the same dimensions {L}, it follows that the quantity in parentheses must be dimensionless: L3 / T L0.37 {( Q )}  {1}  { }  { }  {1} 0.551AC h D 0.63 {0.551}{L2 }{1}{L}0.63 {0.551) T It follows that {0.551}  { L0.37 /T} Ans. The constant has dimensions; therefore beware. The formula is valid only for water flow at high (turbulent) velocities. The density and viscosity of water are hidden in the constant 0.551, and the wall roughness is hidden (approximately) in the numerical value of C h . 1.16 Test the dimensional homogeneity of the boundary-layer x-momentum equation: u u p  u  v    gx  x y x y Solution: This equation, like all theoretical partial differential equations in mechanics, is dimensionally homogeneous. Test each term in sequence:   u    u  M L L/T  M    p  M/LT 2  M   u    v  3   2 2 ;     2 2   x   y L T L  L T   x  L L T  M L  M     M/LT 2  M  { g x }   ;     2 2 L3 T 2  L2 T 2    x  L L T  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  p  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? Chapter 1  Introduction 11 Solution: Write out the dimensions of each side of the equation: 0.54 0.54  L3  ?  p   M /LT 2  {Q}     {61.9}{D2.63}    {61.9}{L2.63}   T L  L  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 than 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  V o . Show (a) that its velocity 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: V t dV dV m  Fx  Drag  ma x , or: KV  m , integrate    dt dt V V 0 K o t Solve V  Vo e  mtK and x   V dt  mVo K  1  e  mtK  Ans. (a,b) 0 Vo Thus, as asked, V drops off exponentially with time, and, as t  , x  K m P1.19 In his study of the circular hydraulic jump formed by a faucet flowing into a sink, Watson [53] proposes a parameter combining volume flow rate Q, density  and viscosity  of the fluid, and depth h of the water in the sink. He claims that the grouping is dimensionless, with Q in the numerator. Can you verify this? Solution: Check the dimensions of these four variables, from Table 1.2: {Q}  {L3 / T } ; { }  {M / L3 } ; {}  {M / LT } ; {h}  {L} Can we make this dimensionless? First eliminate mass {M} by dividing density by viscosity, that is, / has units {T/L2}. (I am pretending that kinematic viscosity is unfamiliar to the students in this introductory chapter.) Then combine and Q to eliminate time: (Q has units {L}. Finally, divide that by a single depth h to form a dimensionless group: Q {M / L3 }{L3 / T } { }   {1}  dimensionless Ans. Watson is correct. h {M / LT }{L} P1.20 Books on porous media and atomization claim that the viscosity  and surface tension  of a fluid can be combined with a characteristic velocity U to form an important dimensionless parameter. (a) Verify that this is so. (b) Evaluate this parameter for water at 20C and a velocity of 3.5 cm/s. NOTE: Extra credit if you know the name of this parameter. Solution: We know from Table 1.2 that {}= {ML-1T-1}, {U} = {LT-1}, and {  }= {FL-1} = {MT-2}. To eliminate mass {M}, we must divide  by  , giving {/  } = {TL-1}. Multiplying by the velocity will thus cancel all dimensions: U  is dimensionless, as is its inverse, Ans.(a)  U The grouping is called the Capillary Number. (b) For water at 20C and a velocity of 3.5 cm/s, use Table A.3 to find  = 0.001 kg/m-s and  = 0.0728 N/m. Evaluate U (0.001 kg / m  s )(0.035m / s )    0.00048 ,  2080 Ans.(b)  2 (0.0728 kg / s ) U _______________________________________________________________________ Chapter 1  Introduction 13 P1.21 In 1908, Prandtl’s student Heinrich Blasius proposed the following formula for the wall shear stress  w at a position x in viscous flow at velocity V past a flat surface:  w  0.332  1/ 2  1/ 2 V 3 / 2 x 1/ 2 Determine the dimensions of the constant 0.332. Solution: From Table 1.2 we find the dimensions of each term in the equation: { w }  {ML1T 2 } ; {}  {ML3 } ; {}  {ML1T 1} ; {V }  {LT 1} ; {x}  {L} Use these dimensions in the equation to determine {0.332}: M M M 1/ 2 L 3 / 2 { 2 }  {0.332} { 3 } { } {L}1 / 2 }1 / 2 { LT L LT T M M Clean up : { 2 }  {0.332} { 2 } , or : {0.332}  {1} Ans. LT LT The constant 0.332 is dimensionless. Blasius was one of the first workers to deduce dimensionally consistent viscous-flow formulas without empirical constants. P1.22 The Ekman number, Ek, arises in geophysical fluid dynamics. It is a dimensionless parameter combining seawater density , a characteristic length L, seawater viscosity , and the Coriolis frequency sin , where  is the rotation rate of the earth and  is the latitude angle. Determine the correct form of Ek if the viscosity is in the numerator. Solution : First list the dimensions of the various quantities: { }  {ML-3 } ; {L}  {L} ; {}  {ML-1T -1} ; { sin }  {T -1} Note that sinis itself dimensionless, so the Coriolis frequency has the dimensions of . Only and  contain mass {M}, so if  is in the numerator,  must be in the denominator. That combination / we know to be the kinematic viscosity, with units {L2T-1}. Of the two remaining variables, only sin contains time {T-1}, so it must be in the denominator. So far, we have the grouping /(sin, which has the dimensions {L2}. So we put the length-squared into the denominator and we are finished:  Dimensionless Ekman number: Ek  Ans.  L2  sin  ________________________________________________________________________ P1.23 During World War II, Sir Geoffrey Taylor, a British fluid dynamicist, used dimensional analysis to estimate the energy released by an atomic bomb explosion. He assumed that the energy released, E, was a function of blast wave radius R, air density , and time t. Arrange these variables into a single dimensionless group, which we may term the blast wave number. Solution: These variables have the dimensions {E} = {ML2/T2}, {R} = {L}, {} = {M/L3}, and {t} = {T}. Multiplying E by t2 eliminates time, then dividing by  eliminates mass, leaving {L5} in the numerator. It becomes dimensionless when we divide by R5. Thus Et2 Blast wave number   R5 ____________________________________________________________________________ P1.24 Air, assumed to be an ideal gas with k = 1.40, flows isentropically through a nozzle. At section 1, conditions are sea level standard (see Table A.6). At section 2, the temperature is –50C. Estimate (a) the pressure, and (b) the density of the air at section 2. Solution: From Table A.6, p 1 = 101350 Pa, T 1 = 288.16 K, and  1 = 1.2255 kg/m3. Convert to absolute temperature, T 2 = -50°C = 223.26 K. Then, for a perfect gas with constant k, p2 T 223.16 1.4 /(1.41)  ( 2 )k /( k 1)  ( )  (0.7744)3.5  0.4087 p1 T1 288.16 Thus p2  (0.4087)(101350 Pa)  41, 400 Pa Ans.(a ) 2 T 223.16 1/(1.41)  ( 2 )1/( k 1)  ( )  (0.7744) 2.5  0.5278 1 T1 288.16 Thus  2  (0.5278)(1.2255kg / m3 )  0.647 kg / m3 Ans.(b) Alternately, once p 2 was known, we could have simply computed  2 from the ideal-gas law.  2 = p 2 /RT 2 = (41400)/[287(223.16)] = 0.647 kg/m3 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/m 2 He    0.3286 kg/m 3 R He T (2077 J/kgK)(293 K) Now mass is mass, no matter where you are. Therefore, on the moon or wherever, m He  He  (0.3286 kg/m 3)(0.9 m 3)  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  mc v T  0 since T2  T1 (isothermal) Then the heat added equals the work of expansion. Estimate the work done: 2 2 2 m 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 (above atmospheric 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/in 2 )(144 in 2 /ft 2 )  2116 lbf/ft 2  4608  2116  6724 lbf/ft 2 From this compute the density of the air in the tire: p 6724 lbf/ft 2 air    0.00732 slug/ft 3 RT (1717 ftlbf/slug R)(535R) Then the total weight of air in the tire is Wair   g  (0.00732 slug/ft 3 )(32.2 ft/s2 )(3.0 ft 3 )  0.707 lbf Ans. 1.27 Given temperature and specific volume data for steam at 40 psia [Ref. 23]: 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/in 2 )(144 in 2 /ft 2 )(12.624 ft 3 /lbm)(32.2 lbm/slug) ftlbf “R”400 F    2721 T (400  459.6)R 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, 455 457 459 460 460 m2/(s2K): 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: R steam   R i  458  0.6% Ans. 5 i=1 kg  K 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: 2  pV  5 E 5  pV  Minimize E    R  i  by differentiating  0  2R  i  i 1  Ti  R i 1  Ti  p 5 Vi 40(144) 12.624 18.699  Thus R least-squares       (32.2) 5 i 1 Ti 5  860R 1260R  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: R steam  2739/5.9798  458  0.6% J/kgK Ans. 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 ma mw p tot  1 atm  pair  p water  Ra T  RwT   pa p w or: m tot  m a  m w   for an ideal gas Ra T R w T where, from Table A-4, Rair 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  p w  101350  7375  93975 Pa. Solving for the mixture density, we obtain ma  m w pa p 93975 7375 kg    w    1.046  0.051  1.10 3 Ans.  R a T R w T 287(313) 461(313) m By comparison, the density of dry air for the same conditions is p 101350 kg dry air    1.13 3 RT 287(313) m 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: p   lbf   134.7  W1-2  p22 ln  2    134.7  144 2  (5 ft 3 ) ln    215,000 ftlbf Ans.  p1   ft   14.7  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): 7 p 134.7      3001   3000, solve   1.940753 slug/ft 3 , po 14.7  1.94  Hence m water    (1.940753)(5 ft 3 )  9.704 slug The density change is extremely small. Now the work done, as in Prob. 1.29 above, is 2 2 2  m m d  W1-2   p d   pd    p  pavg m 2 for a linear pressure rise 1 1   1  2 avg  14.7  134.7 lbf   0.000753 ft 3  Hence W1-2    144 2  (9.704 slug)   21 ftlbf Ans.  2 ft   1.94042 slug  [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. P1.31 One cubic foot of argon gas at 10C and 1 atm is compressed isentropically to a new pressure of 600 kPa. (a) What will be its new density and temperature? (b) If allowed to cool, at this new volume, back to 10C, what will be the final pressure? Assume constant specific heats. Solution: This is an exercise in having students recall their thermodynamics. From Table A.4, for argon gas, R = 208 m2/(s2-K) and k = 1.67. Note T 1 = 283K. First compute the initial density: p1 101350 N / m 2 1   2 2  1.72 kg / m3 RT1 (208 m / s  K )(283K ) For an isentropic process at constant k, p2 600, 000 Pa     5.92  ( 2 ) k  ( 2 )1.67 , Solve  2  4.99 kg/m 3 Ans.(a) p1 101,350 Pa 1 1.72 p2 T2 k /( k 1) T  5.92  ( )  ( 2 )1.67 / 0.67 , Solve T2  578 K  305 C Ans.(a) p1 T1 283K (b) Cooling at constant volume means  stays the same and the new temperature is 283K. Thus kg m2 p3   3 R T3  (5.00 3 )(208 2 )(283K )  294,000 Pa  294 kPa Ans.(b) m s K 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    LR 2   (90 m)(15 m)2  42412 m 3 3 3 Estimate, from the ideal-gas law, the respective densities of helium and air: pHe 1.1(101350) kg (a)  helium    0.1832 3 ; R He T 2077(293) m pair 101350 kg (b)  air    1.205 3 . Rair T 287(293) m Then the respective gas weights are  kg   m WHe   He g   0.1832 3   9.81 2  (42412 m3 )  76000 N Ans. (a)  m  s  Wair   air g  (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.] P1.33 Experimental data [55] for the density of n-pentane liquid for high pressures, at 50C, are listed as follows: Pressure, MPa 0.01 10.23 20.70 34.31 Density, kg/m3 586.3 604.1 617.8 632.8 Interestingly, this data does not fit the author’s suggested liquid state relation, Eq. (1.19), very well. Therefore (a) fit the data, as best you can, to a second order polynomial. Use your curve-fit to estimate (b) the isentropic bulk modulus of n-pentane at 1 atm, and (c) the speed of sound of n-pentane at a pressure of 25 MPa. Solution: The writer used his Excel spreadsheet, which has a least-squares polynomial feature for tabulated data. The writer hit “Add Polynomial Trend Line” and received this 2nd-order curve-fit. p  1.702 E 9  6.278 E 6   5755  2 Ans.(a ) with p in Pa and  in kg/m3. The error, for the four given data points, is less than 0.5%. (b) The isentropic bulk modulus can be estimated by differentiating the p- relation at the atmospheric density of 586.3 kg/m3. p B  ( ) s   (11510   6.278E6) | 586.3  2.76E8 Pa Ans.(b)  (c) First examine Ans.(a) to see what density corresponds to p = 25 MPa. We find, approximately, by iteration or EES, that   624 kg/m3. Then the speed of sound at this pressure is p a |25MPa  ( ) s  11510(624)  6.278E 6  904240  950 m / s Ans.(c)  We are assuming, reasonably, that the given p- relation is approximately isentropic. ______________________________________________________________________ 1.34 Consider steam at the following state near the saturation line: (p 1 , T 1 )  (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 T 1  563 K. Then, p1 1,310,000 Pa kg 1  2 2  5.05 3 Ans. (a) RT1 (461 m s K )(563 K ) m 1/k 1/1.33 2  p   414 kPa  kg  2  2   0.421, or: 2  2.12 Ans. (b) 1 5.05  p1   1310 kPa  m3 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 [21], 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. Most tables list k = 1.32 for NH 3 at about 20C, implying N  6. 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. 22 Solutions Manual  Fluid Mechanics, Sixth Edition P1.36 The isentropic bulk modulus B of a fluid is defined in Eq. (1.39). (a) What are its dimensions? Using theoretical p-relations for a gas or liquid, estimate the bulk modulus, in Pa, of (b) chlorine at 100C and 10 atm; and (c) water, at 20C and 1000 atm. Solution: (a) Given B = (p/) s , note that the density dimensions cancel. Therefore the dimensions of bulk modulus B are the same as pressure: {ML-1T-2}. Ans.(a) (b) For an ideal gas, we may differentiate to find a formula for bulk modulus: p p Bgas   ( ) s   (kRT )   (k )  k p   Chlorine : k  1.34 , p 10 atm , B  (1.34)(10)  13.4 atm  1.36E6 Pa Ans.(b) (c) For water, differentiate the suggested curve-fit, Eq. (1.19): p      ( B  1)( ) n  B ; Water : B  3000 , n  7 po o  dp n( B  1) po  n Bulk modulus    [ ]( ) ; Now find  at 1000 atm : d o o  7  1000  (3001)( )  3000 , solve ( )  1.0419 ; then, finally, o o Bwater  (1.0419)(7)(3001)(1 atm)(1.0419)7  29200 atm  2.96E9 Pa Ans.(c) This is 35% more than the low-pressure bulk modulus listed for water in Table A.3. We apologize for using the same symbol B for bulk modulus and also for the curve-fit constant in Eq. (1.19). It won’t happen again. ________________________________________________________________________ 1.37 A near-ideal gas has M  44 and c v  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 N2 O] With k and R known, the speed of sound at 100ºC  373 K is estimated by a  kRT  1.31[189 m2 /(s2  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,   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 Ans. (a) h (0.006 m) 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) Re L    28 Ans. (b)  1.5 kg/m  s 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), 0.7  773 kg   o (T/To )  (1.80E 5)  n  3.55E 5 Ans. (a)  293  ms 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    (1.80E 5)    (T  S)   (773  110)  kg  3.52E 5 Ans. (b) ms 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: kg   1.8c  (1.8)(1.93E5)  3.5E5 Ans. (c) 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   where T is in °K and A and B are curve-fit constants.  T 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 E E Minimize E   [i  A exp(B/Ti )]2 , then set  0 and 0 i 1 A B The result of this minimization is: A  0.0016 kg/ms, B  1903K. Ans. 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). P1.41 An aluminum cylinder weighing 30 N, 6 cm in diameter and 40 cm long, is falling concentrically through a long vertical sleeve of diameter 6.04 cm. The clearance is filled with SAE 50 oil at 20C. Estimate the terminal (zero acceleration) fall velocity. Neglect air drag and assume a linear velocity distribution in the oil. [HINT: You are given diameters, not radii.] Solution: From Table A.3 for SAE 50 oil,  = 0.86 kg/m-s. The clearance is the difference in radii: 3.02 – 3.0 cm = 0.02 cm = 0.0002 m. At terminal velocity, the cylinder weight must balance the viscous drag on the cylinder surface: V W   wall Awall  ( )( DL) , where C  clearance  rsleeve  rcylinder C kg V or : 30 N  [0.86 )( )]  (0.06 m)(0.40 m) m  s 0.0002 m Solve for V  0.0925 m / s Ans. ________________________________________________________________________ 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, an Excel graph, take logarithms, plot them, and make a linear curve-fit. The result is: 0.68  T K  Power-law curve-fit: He  1.505E 5  Ans. (a)  200 K  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 (T i /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% Ans. (b) 1.50E5 kg/m s (T  95.1K) 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. P1.43 For the flow between two parallel plates of Fig. 1.8, reanalyze for the case of slip flow at both walls. Use the simple slip condition, u wall = l (du/dy) wall , where l is the mean free path of the fluid. (a) Sketch the expected velocity profile and (b) find an expression for the shear stress at each wall. Solution: As in Fig. 1.8, the shear stress remains constant between the two plates. The analysis is correct up to the relation u = a + b y . There would be equal slip velocities, u, at both walls, as shown in the following sketch U y u Ans. (a) h u(y) u x Fixed plate Because of slip at the walls, the boundary conditions are different for u = a + b y : du At y  0 : u   u  a  ( ) y 0   b dy U At y  h : u  U   u  a  b h  U   b   b  b h , or : b  h  2 Chapter 1  Introduction 27 One way to write the final solution is U  u  u  ( )y , where  u  ( )U , h  2 h  2 du U and  wall   ( ) wall  Ans.(b) dy h  2 ______________________________________________________________________ P1.44 SAE 50 oil at 20C fills the concentric annular space between an inner cylinder, r i = 5 cm, and an outer cylinder, r o = 6 cm. The length of the cylinders is 120 cm. If the outer cylinder is fixed and the inner cylinder rotates at 900 rev/min, use the linear profile approximation to estimate the power, in watts, required to maintain the rotation. Neglect any temperature change of the oil. Solution: Convert  i = 900 rpm x (2/60) = 94.25 rad/s. For SAE 50 oil, from Table A.3,  = 0.86 kg/m-s. Then the rotational velocity of the inner cylinder, and its related shear stress, are: Vi  i ri  (94.25 rad / s )(0.05 m)  4.71 m / s V kg 4.71  0 m / s kg i    (0.86 )( )  405  405 Pa r m  s 0.06  0.05 m m  s2 The total moment of this stress about the centerline is 2 M   ri dF   ri ( i ri d L)   i 2 ri 2 L  (405)(2 )(0.05) 2 (1.2)  7.64 N  m 0 Then the power required is P =  i M = (94.25 rad/s)(7.64 N-m) = 720 watts Ans. 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:  V  Fx  W sin   A  W sin     A  ma x  0,  h hW sin  or: Vterminal  Ans. A P1.46 A simple and popular model for two non-newtonian fluids in Fig. 1.9a is the power-law: du n   C( ) dy where C and n are constants fit to the fluid [15]. From Fig. 1.9a, deduce the values of the exponent n for which the fluid is (a) newtonian; (b) dilatant; and (c) pseudoplastic. (d) Consider the specific model constant C = 0.4 N-sn/m2, with the fluid being sheared between two parallel plates as in Fig. 1.8. If the shear stress in the fluid is 1200 Pa, find the velocity V of the upper plate for the cases (d) n = 1.0; (e) n = 1.2; and (f) n = 0.8. Solution: By comparing the behavior of the model law with Fig. 1.9a, we see that (a) Newtonian: n = 1 ; (b) Dilatant: n > 1 ; (c) Pseudoplastic: n < 1 Ans.(a,b,c) From the discussion of Fig. 1.8, it is clear that the shear stress is constant in a fluid sheared between two plates. The velocity profile remains a straight line (if the flow is laminar), and the strain rate duIdy = V/h . Thus, for this flow, the model becomes  = C(V/h)n. For the three given numerical cases, we calculate: N N  s1 V m (d ) n  1 :   1200 2  C (V / h)  (0.4 n 2 )( )1 , solve V  3.0 Ans.(d ) m m 0.001m s N N  s1.2 V m (e) n  1.2 :   1200 2  C (V / h) n  (0.4 2 )( )1.2 , solve V  0.79 Ans.(e) m m 0.001m s N N  s 0. 8 V m ( f ) n  0.8 :   1200  C (V / h) n  (0.4 )( ) 0.8 , solve V  22 Ans.( f ) m2 m2 0.001m s A small change in the exponent n can sharply change the numerical values. P1.47 Data for the apparent viscosity of average human blood, at normal body temperature of 37C, varies with shear strain rate, as shown in the following table. Shear strain rate, s-1 1 10 100 1000 Apparent viscosity, 0.011 0.009 0.006 0.004 kg/(ms) (a) Is blood a nonnewtonian fluid? (b) If so, what type of fluid is it? (c) How do these viscosities compare with plain water at 37C? Solution: (a) By definition, since viscosity varies with strain rate, blood is a nonnewtonian fluid. (b) Since the apparent viscosity decreases with strain rate, it must be a pseudoplastic fluid, as in Fig. 1.9(a). The decrease is too slight to call this a “plastic” fluid. (c) These viscosity values are from six to fifteen times the viscosity of pure water at 37C, which is about 0.00070 kg/m-s. The viscosity of the liquid part of blood, called plasma, is about 1.8 times that of water. Then there is a sharp increase of blood viscosity due to hematocrit, which is the percentage, by volume, of red cells and platelets in the blood. For normal human beings, the hematocrit varies from 40% to 60%, which makes this blood about six times the viscosity of plasma. _______________________________________________________________________ 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   V  V F  1A   2 A   1  2  A Ans. (a )  h1 h2  The formula is of course valid only for laminar (nonturbulent) steady viscous flow. (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, fixed axially and rotated inside the sleeve. Let the inner and outer cylinders have radii r i and r o , respectively, with total sleeve length L. Let the rotational rate be (rad/s) and the applied torque be M. Using these parameters, derive a theoretical relation for the viscosity  of the fluid between the cylinders. Solution: Assuming a linear velocity distribution in the annular clearance, the shear stress is V ri   r ro  ri This stress causes a force dF   dA   (r i d)L on each element of surface area of the inner shaft. The moment of this force about the shaft axis is dM  r i dF. Put all this together: 2 ri 2ri3 L M   ri dF   ri  ri L d  0 ro  ri ro  ri  Solve for the viscosity:    ( r  ri ) ri3 L  Ans. 1.50 A simple viscometer measures the time t for a solid sphere to fall a distance L through a test fluid of density . The fluid viscosity  is then given by Wnet t 2 DL  if t  3 DL  where D is the sphere diameter and W net 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: M  W t   ( ML/T 2 )(T )   M  {}    and  net        Yes, dimensions OK.  LT   (3 ) DL   (1)( L )( L )   LT   2 DL   (1)( M/L3 )( L )( L )  {t}  {T } and     {T } Yes, dimensions OK. Ans. (a)     M/LT  (b) Evaluate the two equations for the data. We need the net weight of the sphere in the fluid: Wnet  (sphere  fluid )g(Vol )fluid  (2700  875 kg/m 3)(9.81 m/s2 )( /6)(0.0025 m)3  0.000146 N Wnet t (0.000146 N )(32 s) kg Then     0.40 Ans. (b) 3 DL 3 (0.0025 m)(0.5 m) ms 2DL 2(875 kg/m3)(0.0025 m)(0.5 m) Check t  32 s compared to   0.40 kg/m  s  5.5 s OK, t is greater P1.51 An approximation for the boundary-layer shape in Figs. 1.6b and P1.51 is the formula y U y u ( y )  U sin( ), 0y  2 y= where U is the stream velocity far from the wall and  is the boundary layer thickness, as in Fig. P.151. If the fluid is helium at 20C and 1 atm, and if U = u(y) 10.8 m/s and  = 3 mm, use the formula to (a) estimate 0 the wall shear stress  w in Pa; and (b) find the position in the boundary layer where  is one-half of  w . Fig. P1.51 Solution: From Table A.4, for helium, take R = 2077 m2/(s2-K) and  = 1.97E-5 kg/m-s. (a) Then the wall shear stress is calculated as 32 Solutions Manual  Fluid Mechanics, Sixth Edition u  y  U w   | y 0   (U cos ) y 0  y 2 2 2  (1.97 E  5 kg / m  s)(10.8 m / s ) Numerical values :  w   0.11 Pa Ans.(a) 2(0.003 m) A very small shear stress, but it has a profound effect on the flow pattern. (b) The variation of shear stress across the boundary layer is a cosine wave,  = (du/dy): /dy): U y y  y  2δ  ( y)  cos( )   w cos( )  w when  , or : y  Ans.(b) 2 2 2 2 2 3 3 _______________________________________________________________________ 1.52 The belt in Fig. P1.52 moves at steady velocity V and skims the top of a tank of oil of viscosity . Assuming a linear velocity profile, develop a simple formula for the belt- drive power P required as a function of (h, L, V, B, ). Neglect air drag. What power P in watts is required if the belt moves at 2.5 m/s over SAE 30W oil at 20C, with L  2 m, b  60 cm, and h  3 cm? Fig. P1.52 Solution: The power is the viscous resisting force times the belt velocity:  V L P   oil A belt Vbelt     (bL)V   V2 b Ans.  h h (b) For SAE 30W oil,   0.29 kg/m  s. Then, for the given belt parameters, 2  kg   m 2.0 m kg  m2 P  V bL/h   0.29 2  2.5  (0.6 m)  73  73 W Ans. (b)  m  s   s 0.03 m s3 1.53* A solid cone of base r o and initial angular velocity  o is rotating inside a conical seat. Neglect air drag and derive a formula for the cone’s angular velocity (t) if there is no applied torque. Solution: At any radial position r  r o on the cone surface and instantaneous rate , Fig. P1.53  r   dr  d(Torque)  r dA w  r     2 r ,  h  sin  ro   ro4 or: Torque M   2 r 3 dr  0 h sin 2h sin We may compute the cone’s slowing down from the angular momentum relation: d 3 M   Io , where Io (cone)  mro2 , m  cone mass dt 10 Separating the variables, we may integrate: w t d ro4  5 ro2 t     2hIo sin 0 dt, or:    o exp   3mh sin   Ans. o 1.54* A disk of radius R rotates at angular velocity  inside an oil container of viscosity , 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   r/h on both sides of the disk. Thus, r d(torque)  dM  2r dA w  2r 2r dr, h R  R 4 or: M  4 h  r 3 dr  h Ans 0 P1.55 A block of weight W is being pulled over a table by another weight W o , as shown in W h Fig. P1.55. Find an algebraic formula for the steady velocity U of the block if it slides on an oil film of thickness h and viscosity . The block Wo bottom area A is in contact with the oil. Fig. P1.55 Neglect the cord weight and the pulley friction. Solution: This problem is a lot easier to solve than to set up and sketch. For steady motion, U  Fx,block  0   A  Wo  (  h ) A  Wo Wo h Solve for U  Ans. A there is no acceleration, and the falling weight balances the viscous resistance of the oil film: The block weight W has no effect on steady horizontal motion except to smush the oil film. 1.56* For the cone-plate viscometer in Fig. P1.56, the angle is very small, and the gap is filled with test liquid . Assuming a linear velocity profile, derive a formula for the viscosity  in terms of the torque M and cone parameters. Fig. P1.56 Solution: For any radius r  R, the liquid gap is h  r tan. Then  r   dr  d(Torque)  dM   dA w r      2 r  r, or  r tan   cos  R 2 2 2R 3 3M sin M  sin  0 r dr  3sin  , or:   2R 3 Ans. P1.57 Extend the steady flow between a fixed lower plate and a moving upper plate, from Fig. 1.8, to the case of two immiscible liquids between the plates, as in Fig. P1.57. V h2  2 y Fig. P1.57 h1  1 x Fixed (a) Sketch the expected no-slip velocity distribution u(y) between the plates. (b) Find an analytic expression for the velocity U at the interface between the two liquid layers. (c) What is the result if the viscosities and layer thicknesses are equal? Solution: We begin with the hint, from Fig. 1.8, that the shear stress is constant between the two plates. The velocity profile would be a straight line in each layer, with different slopes: V (2) y U Ans. (a) (1) x Fixed Here we have drawn the case where  2 >  1 , hence the upper profile slope is less. (b) Set the two shear stresses equal, assuming no-slip at each wall: (U  0) (V  U ) 1  1   2  2 h1 h2 1 Solve for U  V [ ] Ans.(b)  h 1 1 2 2 h1 (c) For equal viscosities and layer thicknesses, we get the simple result U = V/2. Ans. ______________________________________________________________________ 1.58 The laminar-pipe-flow example of Prob. 1.14 leads to a capillary viscometer [29], using the formula    r o 4p/(8LQ). Given r o 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:  ro4 p  (0.002 m)4 (159000 N/m2 ) N s p  159 kPa:    3  0.040 8LQ 8(0.25 m)(0.36/3600 m /s) m2 Do the same thing for all five data points: p, kPa: 159 318 477 1274 1851 , 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  s falls due to gravity inside a tube of diameter Do . The clearance, (Do  D)  D, is filled with a film of viscous fluid (,). Derive a formula for terminal fall velocity and apply to SAE 30 oil at 20C for a steel cylinder with D  2 cm, D o  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:   V  a z  0: Fz   W  Drag  sg D2 L      DL, 4  (Do  D)/2  sgD(Do  D) or: V  Ans. 8 For the particular numerical case given,  steel  7850 kg/m3. For SAE 30 oil at 20C,   0.29 kg/m·s from Table 1.4. Then the formula predicts sgD(Do  D) (7850 kg/m3 )(9.81 m/s2 )(0.02 m)(0.0204  0.02 m) Vterminal   8 8(0.29 kg/m  s)  0.265 m/s Ans. _________________________________________________________________________ P1.60 Pipelines are cleaned by pushing through them a close-fitting cylinder called a pig. The name comes from the squealing noise it makes sliding along. Ref. 50 describes a new non-toxic pig, driven by compressed air, for cleaning cosmetic and beverage pipes. Suppose the pig diameter is 5-15/16 in and its length 26 in. It cleans a 6-in-diameter pipe at a speed of 1.2 m/s. If the clearance is filled with glycerin at 20C, what pressure difference, in pascals, is needed to drive the pig? Assume a linear velocity profile in the oil and neglect air drag. Solution: Since the problem calls for pascals, convert everything to SI units: Find the shear stress in the oil, multiply that by the cylinder wall area to get the required force, and divide the force by the area of the cylinder face to find the required pressure difference. Comment: The Reynolds number of the clearance flow, Re = VC/, is approximately 0.8. 15 m m Dcyl  (5 in)(0.0254 )  0.1508 m ; L  (26 in)(0.0254 )  0.6604 m 16 in in 15 m Clearance  C  ( D pipe  Dcyl ) / 2  (6  5 in)(0.0254 ) / 2  0.000794 m 16 in Table A.3, glycerin :   1.49 kg / m  s ,   1260 kg / m 3 (  not needed) Shear stress    V / C  (1.49 kg / m  s )(1.2 m / s) / 0.000794 m  2252 N / m 2 Shear force F   Awall   ( Dcyl L )  (2252 N / m 2 )[ (0.1508 m)(0.6604 m)]  705 N F 705 N Finally, Δp    39500 Pa Ans. Acyl face ( / 4)(0.1508 m) 2 1.61 An air-hockey puck has m  50 g and D  9 cm. When placed on a 20C air table, the blower forms a 0.12-mm-thick air film under the puck. The puck is struck with an initial velocity of 10 m/s. How long will it take the puck to (a) slow down to 1 m/s; (b) stop completely? Also (c) how far will the puck have travelled for case (a)? Solution: For air at 20C take   1.8E5 kg/m·s. Let A be the bottom area of the puck, A  D2/4. Let x be in the direction of travel. Then the only force acting in the x direction is the air drag resisting the motion, assuming a linear velocity distribution in the air: V dV  Fx   A    Am , where h  air film thickness h dt Separate the variables and integrate to find the velocity of the decelerating puck: V t dV A  V  K  dt, or V  Vo e  Kt , where K  mh Vo 0 Integrate again to find the displacement of the puck: t Vo x   V dt  [1  e  Kt ] 0 K Apply to the particular case given: air,   1.8E5 kgm·s, m  50 g, D  9 cm, h  0.12 mm, V o  10 m/s. First evaluate the time-constant K: A (1.8E5 kg/m  s)[( /4)(0.09 m)2 ] K   0.0191 s1 mh (0.050 kg)(0.00012 m) (a) When the puck slows down to 1 m/s, we obtain the time: 1 V  1 m/s  Vo e  Kt  (10 m/s) e(0.0191 s )t , or t  121 s Ans. (a) (b) The puck will stop completely only when e–Kt  0, or: t   Ans. (b) (c) For part (a), the puck will have travelled, in 121 seconds, Vo 10 m/s x (1  e  Kt )  1 [1  e (0.0191)(121) ]  472 m Ans. (c) K 0.0191 s This may perhaps be a little unrealistic. But the air-hockey puck does decelerate slowly! 1.62 The hydrogen bubbles in Fig. 1.13 have D  0.01 mm. Assume an “air-water” interface at 30C. What is the excess pressure within the bubble?