Resolução do White cap 06, Provas de Engenharia Sanitária

Baixe Resolução do White cap 06 e outras Provas em PDF para Engenharia Sanitária, somente na Docsity! Chapter 6 • Viscous Flow in Ducts 6.1 In flow past a sphere, the boundary layer becomes turbulent at about ReD ≈ 2.5E5. To what air speed in mi/h does this correspond to a golf ball whose diameter is 1.6 in? Do the pressure, temperature, and humidity of the air make any difference in your calculation? Solution: For air at 20°C, take ρ = 1.2 kg/m3 and µ = 1.8E−5 kg/m⋅s. Convert D = 1.6 inches to 0.0406 m. The critical Reynolds number is D VD 1.2V(0.0406) m Re 2.5E5 , or V 92 1.8E 5 s Ans. ρ µ = = = = ≈ − mi 206 h Since air density and viscosity change with pressure, temperature, and humidity, the calculation does indeed depend upon the thermodynamic state of the air. 6.2 Air at approximately 1 atm flows through a horizontal 4-cm-diameter pipe. (a) Find a formula for Qmax, the maximum volume flow for which the flow remains laminar, and plot Qmax versus temperature in the range 0°C ≤ T ≤ 500°C. (b) Is your plot linear? If not, explain. Solution: (a) First convert the Reynolds number from a velocity form to a volume flow form: 2 4 , therefore Re 2300 for laminar flow ( /4) d Q Vd Q V dd ρ ρ µ πµπ = = = ≤ Maximum laminar volume flow is given by . (a)Ansmax 2300 Q 4 = π µ ρ d With d = 0.04 m = constant, get µ and ρ for air from Table A-2 and plot Qmax versus T °C: Fig. P6.2 364 Solutions Manual • Fluid Mechanics, Fifth Edition The curve is not quite linear because ν = µ/ρ is not quite linear with T for air in this range. Ans. (b) 6.3 For a thin wing moving parallel to its chord line, transition to a turbulent boundary layer occurs at a “local” Reynolds number Rex, where x is the distance from the leading edge of the wing. The critical Reynolds number depends upon the intensity of turbulent fluctuations in the stream and equals 2.8E6 if the stream is very quiet. A semiempirical correlation for this case [Ref. 3 of Ch. 6] is crit 2 1/2 1/2 2 1 (1 13.25 ) Re 0.00392x ζ ζ − + +≈ where ζ is the tunnel-turbulence intensity in percent. If V = 20 m/s in air at 20°C, use this formula to plot the transition position on the wing versus stream turbulence for ζ between 0 and 2 percent. At what value of ζ is xcrit decreased 50 percent from its value at ζ = 0? Solution: This problem is merely to illustrate the strong effect of stream turbulence on the transition point. For air at 20°C, take ρ = 1.2 kg/m3 and µ = 1.8E−5 kg/m⋅s. Compute Rex,crit from the correlation and plot xtr = µRex/[ρ(20 m/s)] versus percent turbulence: Fig. P6.3 The value of xcrit decreases by half (to 1.07 meters) at ζ ≈ 0.42%. Ans. Chapter 6 • Viscous Flow in Ducts 367 Solution: Equation (6.9b) applies in both cases, noting that τw is negative: 2 2( 72 ) (a) : (a) 0.04 wdp PaHorizontal Ans. dx R m τ −= = = − Pa3600 m 2 (b) : 3600 998(9.81) (b)w dp dz Vertical, up g Ans. dx R dx τ ρ= − = − − = Pa13, 400 m − 6.9 A light liquid (ρ = 950 kg/m3) flows at an average velocity of 10 m/s through a horizontal smooth tube of diameter 5 cm. The fluid pressure is measured at 1-m intervals along the pipe, as follows: x, m: 0 1 2 3 4 5 6 p, kPa: 304 273 255 240 226 213 200 Estimate (a) the total head loss, in meters; (b) the wall shear stress in the fully developed section of the pipe; and (c) the overall friction factor. Solution: As sketched in Fig. 6.6 of the text, the pressure drops fast in the entrance region (31 kPa in the first meter) and levels off to a linear decrease in the “fully developed” region (13 kPa/m for this data). (a) The overall head loss, for ∆z = 0, is defined by Eq. (6.8) of the text: ρ ∆ −= = = 3 2 304,000 200,000 (a) (950 / )(9.81 / ) f p Pa h Ans. g kg m m s 11.2 m (b) The wall shear stress in the fully-developed region is defined by Eq. (6.9b): 4 413000 , solve for (b) 1 0.05 w w fully developed w p Pa Ans. L m d m τ τ τ∆ = = = = ∆ | 163 Pa (c) The overall friction factor is defined by Eq. (6.10) of the text: 2 , 2 2 2 0.05 2(9.81 / ) (11.2 ) (c) 6 (10 / ) overall f overall d g m m s f h m Ans. L mV m s
= = =    0.0182 NOTE: The fully-developed friction factor is only 0.0137. 1 368 Solutions Manual • Fluid Mechanics, Fifth Edition 6.10 Water at 20°C (ρ = 998 kg/m3) flows through an inclined 8-cm-diameter pipe. At sections A and B, pA = 186 kPa, VA = 3.2 m/s, zA = 24.5 m, while pB = 260 kPa, VB = 3.2 m/s, and zB = 9.1 m. Which way is the flow going? What is the head loss? Solution: Guess that the flow is from A to B and write the steady flow energy equation: 2 2 186000 260000 , or: 24.5 9.1 , 2 2 9790 9790 : 43.50 35.66 , solve: , . . (a, b) A A B B A B f f f p V p V z z h h g g g g or h Yes flow is from A to B Ans ρ ρ + + = + + + + = + + = + fh 7.84 m= + 6.11 Water at 20°C flows upward at 4 m/s in a 6-cm-diameter pipe. The pipe length between points 1 and 2 is 5 m, and point 2 is 3 m higher. A mercury manometer, connected between 1 and 2, has a reading h = 135 mm, with p1 higher. (a) What is the pressure change (p1 − p2)? (b) What is the head loss, in meters? (c) Is the manometer reading proportional to head loss? Explain. (d) What is the friction factor of the flow? Solution: A sketch of this situation is shown at right. By moving through the manometer, we obtain the pressure change between points 1 and 2, which we compare with Eq. (6.9b): 1 2,w m wp h h z pγ γ γ+ − − ∆ = 1 2 3 3 133100 9790 (0.135 ) 9790 (3 ) 16650 29370 (a) N N p p m m m m Ans.

− = − +       = + = or: 46,000 Pa 3 46000 . , 3 4.7 3.0 (b) 9790 / f w p Pa From Eq (6.9b) h z m Ans. N mγ ∆= − ∆ = − = − = 1.7 m 2 2 2 2 0.06 2(9.81 / ) (1.7 ) (d) 5 (4 / ) f d g m m s The friction factor is f h m Ans. L mV m s
= = =   0.025 By comparing the manometer relation to the head-loss relation above, we find that: ( ) (c)m wf w h h Ans. γ γ γ − = isand thus head loss proportional to manometer reading. Chapter 6 • Viscous Flow in Ducts 369 NOTE: IN PROBLEMS 6.12 TO 6.99, MINOR LOSSES ARE NEGLECTED. 6.12 A 5-mm-diameter capillary tube is used as a viscometer for oils. When the flow rate is 0.071 m3/h, the measured pressure drop per unit length is 375 kPa/m. Estimate the viscosity of the fluid. Is the flow laminar? Can you also estimate the density of the fluid? Solution: Assume laminar flow and use the pressure drop formula (6.12): 4 4 ? ?p 8Q Pa 8(0.071/3600) , or: 375000 , solve . L mR (0.0025) Ans µ µ µ π π ∆ = = ≈ kg0.292 m s⋅ oil 3 kg Guessing 900 , m 4 Q 4(900)(0.071/3600) check Re . d (0.292)(0.005) Ans ρ ρ πµ π ≈ = = ≈ 16 OK, laminar It is not possible to find density from this data, laminar pipe flow is independent of density. 6.13 A soda straw is 20 cm long and 2 mm in diameter. It delivers cold cola, approximated as water at 10°C, at a rate of 3 cm3/s. (a) What is the head loss through the straw? What is the axial pressure gradient ∂p/∂x if the flow is (b) vertically up or (c) horizontal? Can the human lung deliver this much flow? Solution: For water at 10°C, take ρ = 1000 kg/m3 and µ = 1.307E−3 kg/m⋅s. Check Re: 34 Q 4(1000)(3E 6 m /s) Re 1460 (OK, laminar flow) d (1.307E 3)(0.002) ρ πµ π −= = = − f 4 4 128 LQ 128(1.307E 3)(0.2)(3E 6) Then, from Eq. (6.12), h (a) gd (1000)(9.81)(0.002) Ans. µ πρ π − −= = ≈ 0.204 m If the straw is horizontal, then the pressure gradient is simply due to the head loss: f horiz p gh 1000(9.81)(0.204 m) (c) L L 0.2 m Ans. ρ∆ = = ≈| Pa9980 m If the straw is vertical, with flow up, the head loss and elevation change add together: f vertical p g(h z) 1000(9.81)(0.204 0.2) (b) L L 0.2 Ans. ρ∆ + ∆ += = ≈| Pa19800 m The human lung can certainly deliver case (c) and strong lungs can develop case (b) also. 372 Solutions Manual • Fluid Mechanics, Fifth Edition Solution: (a) Assume no pressure drop and neglect velocity heads. The energy equation reduces to: 2 2 1 1 2 2 1 20 0 ( ) 0 0 0 , :2 2 f f f p V p V z L l z h h h L l g g g gρ ρ + + = + + + = + + + = + + + ≈ +or 4 128 , ,f LQ For laminar flow h and, for uniform draining Q tgd µ υ πρ = = ∆ (a)Solve for Ans. L t gd L l ∆ = +4 128 ( ) µ υ πρ (b) Apply to ∆t = 6 s. For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Formula (a) predicts: 3 3 2 4 128(0.001 / )(0.12 )(8 6 ) 6 , (998 / )(9.81 / ) (0.12 0.02 ) Solve for (b) kg m s m E m t s kg m m s d m Ans. π ⋅ −∆ = = + d 0.0015 m≈ 6.18 To determine the viscosity of a liquid of specific gravity 0.95, you fill, to a depth of 12 cm, a large container which drains through a 30-cm-long vertical tube attached to the bottom. The tube diameter is 2 mm, and the rate of draining is found to be 1.9 cm3/s. What is your estimate of the fluid viscosity? Is the tube flow laminar? Fig. P6.18 Solution: The known flow rate and diameter enable us to find the velocity in the tube: 3 2 1.9 6 / 0.605 ( /4)(0.002 ) Q E m s m V A smπ −= = = Evaluate ρ liquid = 0.95(998) = 948 kg/m3. Write the energy equation between the top surface and the tube exit: 2 2 2 2 2 2 0 , 2 2 32 (0.605) 32 (0.3)(0.605) : 0.42 2 2(9.81) 948(9.81)(0.002) topa a top f Vp p V z h g g g g V LV or g gd ρ ρ µ µ ρ = + = + + + = + = + Chapter 6 • Viscous Flow in Ducts 373 Note that “L” in this expression is the tube length only (L = 30 cm). Solve for ( ) 948(0.605)(0.002) 446 ( ) 0.00257d laminar flow Ans. Vd Re laminar µ ρ µ = = = = kg 0.00257 m s⋅ 6.19 An oil (SG = 0.9) issues from the pipe in Fig. P6.19 at Q = 35 ft3/h. What is the kinematic viscosity of the oil in ft3/s? Is the flow laminar? Solution: Apply steady-flow energy: 2 2 atm atm 2 1 2 f p p0 V z z h , g 2g g 2gρ ρ + + = + + + Fig. P6.19 2 2 Q 35/3600 ft where V 7.13 A s(0.25 /12)π = = ≈ 2 2 2 f 1 2 V (7.13) Solve h z z 10 9.21 ft 2g 2(32.2) = − − = − = Assuming laminar pipe flow, use Eq. (6.12) to relate head loss to viscosity: f 4 4 128 LQ 128(6)(35/3600) h 9.21 ft , solve gd (32.2)(0.5/12) Ans. ν ν µν ρπ π = = = = ≈ 2ft 3.76E 4 s − Check Re 4Q/( d) 4(35/3600)/[ (3.76E 4)(0.5/12)] 790 (OK, laminar)πν π= = − ≈ 6.20 In Prob. 6.19 what will the flow rate be, in m3/h, if the fluid is SAE 10 oil at 20°C? Solution: For SAE 10 oil at 20°C, take ρ = 1.69 slug/ft3 and µ =2.17E−3 slug/ft⋅s. The steady flow energy analysis above gives, for laminar flow, 2 f 2 2 V 32 LV 32(2.17E 3)(6.0)V h 10 4.41V (quadratic equation) 2(32.2) gd (1.69)(32.2)(0.5/12) µ ρ −= − = = = 2 3ft 0.5 ft Solve for V 2.25 , Q (2.25) 0.00307 s 4 12 s Ans. π
≈ = =   3m 0.31 h ≈ 374 Solutions Manual • Fluid Mechanics, Fifth Edition 6.21 In Tinyland, houses are less than a foot high! The rainfall is laminar! The drainpipe in Fig. P6.21 is only 2 mm in diameter. (a) When the gutter is full, what is the rate of draining? (b) The gutter is designed for a sudden rainstorm of up to 5 mm per hour. For this condition, what is the maximum roof area that can be drained successfully? (c) What is Red? Solution: If the velocity at the gutter surface is neglected, the energy equation reduces to 2 ,laminar 2 32 , where 2 f f V LV z h h g gd µ ρ ∆ = + = Fig. P6.21 For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. (a) With ∆z known, this is a quadratic equation for the pipe velocity V: 2 2 3 2 2 32(0.001 / )(0.2 ) 0.2 , 2(9.81 / ) (998 / )(9.81 / )(0.002 ) V kg m s m V m m s kg m m s m ⋅= + or 2: 0.051 0.1634 0.2 0, 0.945 ,mV V Solve for V s + − = = 3 2(0.002 ) 0.945 2.97 6 (a) 4 m m Q m E Ans. s s π
= = − =   3 0.0107 m h (b) The roof area needed for maximum rainfall is 0.0107 m3/h ÷ 0.005 m/h = 2.14 m2. Ans. (b) (c) The Reynolds number of the gutter is Red = (998)(0.945)(0.002)/(0.001) = 1890 laminar. Ans. (c) 6.22 A steady push on the piston in Fig. P6.22 causes a flow rate Q = 0.15 cm3/s through the needle. The fluid has ρ = 900 kg/m3 and µ = 0.002 kg/(m⋅s). What force F is required to maintain the flow? Fig. P6.22 Chapter 6 • Viscous Flow in Ducts 377 6.26 For the system in Fig. P6.25, if the fluid has density of 920 kg/m3 and the flow rate is unknown, for what value of viscosity will the capillary Reynolds number exactly equal the critical value of 2300? Solution: Add to the energy analysis of Prob. 6.25 above that the Reynolds number must equal 2300: f 4 4 128 L 128 (1.2) 2300 (0.002) h 0.9 m Q 4(920)gd (920)(9.81)(0.002) µ µ πµ πρ π
= = =     Solve for Ans.µ = 0.000823 kg/m s⋅ 6.27 Let us attack Prob. 6.25 in symbolic fashion, using Fig. P6.27. All parameters are constant except the upper tank depth Z(t). Find an expression for the flow rate Q(t) as a function of Z(t). Set up a differential equation, and solve for the time t0 to drain the upper tank completely. Assume quasi- steady laminar flow. Solution: The energy equation of Prob. 6.25, using symbols only, is combined with a control-volume mass balance for the tank to give the basic differential equation for Z(t): Fig. P6.27 2 2 2 f 2 32 LV d energy: h h Z; mass balance: D Z d L Q d V, dt 4 4 4gd µ π π π ρ
= = + + = − = −   2 2 2dZ gdor: D d V, where V (h Z) 4 dt 4 32 L π π ρ µ = − = + Separate the variables and integrate, combining all the constants into a single “C”: o Z t Z 0 dZ C dt, or: , where h Z Ans.= − +

4 Ct o 2 gd Z (h Z )e h C 32 LD −= + − = ρ µ Tank drains completely when Z 0, at Ans.
=    o 0 Z1 t ln 1 C h = + 378 Solutions Manual • Fluid Mechanics, Fifth Edition 6.28 For straightening and smoothing an airflow in a 50-cm-diameter duct, the duct is packed with a “honeycomb” of thin straws of length 30 cm and diameter 4 mm, as in Fig. P6.28. The inlet flow is air at 110 kPa and 20°C, moving at an average velocity of 6 m/s. Estimate the pressure drop across the honeycomb. Solution: For air at 20°C, take µ ≈ 1.8E−5 kg/m⋅s and ρ = 1.31 kg/m3. There would be approximately 12000 straws, but each one would see the average velocity of 6 m/s. Thus Fig. P6.28 laminar 2 2 32 LV 32(1.8E 5)(0.3)(6.0) p d (0.004) Ans. µ −∆ = = ≈ 65 Pa Check Re = ρVd/µ = (1.31)(6.0)(0.004)/(1.8E−5) ≈ 1750 OK, laminar flow. 6.29 Oil, with ρ = 890 kg/m3 and µ = 0.07 kg/m⋅s, flows through a horizontal pipe 15 m long. The power delivered to the flow is 1 hp. (a) What is the appropriate pipe diameter if the flow is at the laminar transition point? For this condition, what are (b) Q in m3/h; and (c) τw in kPa? Solution: (a, b) Set the Reynolds number equal to 2300 and the (laminar) power equal to 1 hp: 3 2(890 / )Re 2300 or 0.181 / 0.07 /d kg m Vd Vd m s kg m s = = = ⋅ 2 2 laminar 2 32 1 745.7 32(0.07)(15) 4 4 LV Power hp W Q p d V V d π µ π


= = = ∆ = =           Solve for 5.32 and (a) m V s = d 0.034 m Ans.= It follows that Q = (π/4)d2V = (π/4)(0.034 m)2(5.32 m/s) = 0.00484 m3/s = 17.4 m3/h Ans. (b) (c) From Eq. (6.12), the wall shear stress is 8 8(0.07 / )(5.32 / ) 88 (c) (0.034 )w V kg m s m s Pa Ans. d m µτ ⋅= = = = 0.088 kPa Chapter 6 • Viscous Flow in Ducts 379 6.30 SAE 10 oil at 20°C flows through the 4-cm-diameter vertical pipe of Fig. P6.30. For the mercury manometer reading h = 42 cm shown, (a) calculate the volume flow rate in m3/h, and (b) state the direction of flow. Solution: For SAE 10 oil, take ρ = 870 kg/m3 and µ = 0.104 kg/m⋅s. The pressure at the lower point (1) is considerably higher than p2 according to the manometer reading: Fig. P6.30 1 2 Hg oilp p ( )g h (13550 870)(9.81)(0.42) 52200 Paρ ρ− = − ∆ = − ≈ oilp/( g) 52200/[870(9.81)] 6.12 mρ∆ = ≈ This is more than 3 m of oil, therefore it must include a friction loss: flow is up. Ans. (b) The energy equation between (1) and (2), with V1 = V2, gives 1 2 2 1 f f f 4 p p 128 LQ z z h , or 6.12 m 3 m h , or: h 3.12 m g gd µ ρ πρ − = − + = + ≈ = 4 3(6.12 3) (870)(9.81)(0.04) m Compute Q 0.00536 (a) 128(0.104)(3.0) s Ans. π−= = ≈ 3m 19.3 h Check Re 4 Q/( d) 4(870)(0.00536)/[ (0.104)(0.04)] 1430 (OK, laminar flow).ρ πµ π= = ≈ 6.31 Light oil, ρ = 880 kg/m3 and µ = 0.015 kg/(m⋅s), flows down a vertical 6-mm- diameter tube due to gravity only. Estimate the volume flow rate in m3/h if (a) L = 1 m and (b) L = 2 m. (c) Verify that the flow is laminar. Solution: If the flow is due to gravity only, the head loss matches the elevation change: 4 f 4 128 LQ gd h z L , or Q 128gd independent of pipe length µ πρ µπρ = ∆ = = = For this case, 3 4 mQ (880)(9.81)(0.006) /[128(0.015)] 1.83E 5 (a, b) s Ans. π= ≈ − = 3m 0.066 h Check Re 4 Q/( d) 4(880)(1.83E 5)/[ (0.015)(0.006)] . (c)Ans. ρ πµ π= = − ≈ 228 (laminar) 382 Solutions Manual • Fluid Mechanics, Fifth Edition 6.35 By analogy with Eq. (6.21) write the turbulent mean-momentum differential equation for (a) the y direction and (b) the z direction. How many turbulent stress terms appear in each equation? How many unique turbulent stresses are there for the total of three directions? Solution: You can re-derive, as in Prob. 6.34, or just permute the axes: y dv p v v (a) : g u’v’ v’v’ dt y x x y y v v’w’ z z ∂ ∂ ∂ ∂ ∂ρ ρ µ ρ µ ρ ∂ ∂ ∂ ∂ ∂ ∂ ∂µ ρ ∂ ∂

= − + + − + −       
+ −    y z dw p w w (b) : g u’w’ v’w’ dt z x x y y w w’w’ z z ∂ ∂ ∂ ∂ ∂ρ ρ µ ρ µ ρ ∂ ∂ ∂ ∂ ∂ ∂ ∂µ ρ ∂ ∂

= − + + − + −       
+ −    z 6.36 The following turbulent-flow velocity data u(y), for air at 75°F and 1 atm near a smooth flat wall, were taken in the University of Rhode Island wind tunnel: y, in: 0.025 0.035 0.047 0.055 0.065 u, ft/s: 51.2 54.2 56.8 57.6 59.1 Estimate (a) the wall shear stress and (b) the velocity u at y = 0.22 in. Solution: For air at 75°F and 1 atm, take ρ = 0.00230 slug/ft3 and µ = 3.80E−7 slug/ft⋅s. We fit each data point to the logarithmic-overlap law, Eq. (6.28): w u 1 u*y 1 0.0023u*y ln B ln 5.0, u* / u* 0.41 3.80E 7 ρ τ ρ κ µ   ≈ + ≈ + = − Enter each value of u and y from the data and estimate the friction velocity u*: y, in: 0.025 0.035 0.047 0.055 0.065 u*, ft/s: 3.58 3.58 3.59 3.56 3.56 yu*/ν (approx): 45 63 85 99 117 Each point gives a good estimate of u*, because each point is within the logarithmic layer in Fig. 6.10 of the text. The overall average friction velocity is avg 2 2 w,avg ft*u 3.57 1%, u* (0.0023)(3.57) (a) s Ans.τ ρ≈ ± = = ≈ 2 lbf 0.0293 ft Chapter 6 • Viscous Flow in Ducts 383 Out at y = 0.22 inches, we may estimate that the log-law still holds: u*y 0.0023(3.57)(0.22/12) 1 396, u u* ln(396) 5.0 3.80E 7 0.41 ρ µ
= ≈ ≈ + −   or: u (3.57)(19.59) (b)Ans.≈ ≈ ft70 s Figure 6.10 shows that this point (y+ ≈ 396) seems also to be within the logarithmic layer. 6.37 Two infinite plates a distance h apart are parallel to the xz plane with the upper plate moving at speed V, as in Fig. P6.37. There is a fluid of viscosity µ and constant pressure between the plates. Neglecting gravity and assuming incompres- sible turbulent flow u(y) between the plates, use the logarithmic law and appropriate Fig. P6.37 boundary conditions to derive a formula for dimensionless wall shear stress versus dimensionless plate velocity. Sketch a typical shape of the profile u(y). Solution: The shear stress between parallel plates is constant, so the centerline velocity must be exactly u = V/2 at y = h/2. Anti-symmetric log-laws form, one with increasing velocity for 0 < y < h/2, and one with decreasing velocity for h/2 < y < h, as shown below: 384 Solutions Manual • Fluid Mechanics, Fifth Edition The match-point at the center gives us a log-law estimate of the shear stress: 1 * ln B, 0.41, B 5.0, 2 * 2 V hu Ans. u κ κ ν
 ≈ + ≈ ≈   u 1 2w* ( ) /= /τ ρ This is one form of “dimensionless shear stress.” The more normal form is friction coefficient versus Reynolds number. Calculations from the log-law fit a Power-law curve-fit expression in the range 2000 < Reh < 1E5: w 2 1/4 0.018 (1/2) ( / ) Ans. V Vh τ ρ ρ ν = ≈ =f 1 4 h 0.018 C Re / 6.38 Suppose in Fig. P6.37 that h = 3 cm, the fluid is water at 20°C (ρ = 998 kg/m3, µ = 0.001 kg/m⋅s), and the flow is turbulent, so that the logarithmic law is valid. If the shear stress in the fluid is 15 Pa, estimate V in m/s. Solution: Just as in Prob. 6.37, apply the log-law at the center between the wall, that is, y = h/2, u = V/2. With τw known, we can evaluate u* immediately: 15 /2 1 * /2 * 0.123 , ln , 998 * w m V u hu B s u τ ρ κ ν
= = = ≈ +   /2 1 0.123(0.03/2) or: ln 5.0 23.3, . 0.123 0.41 0.001/998 V Solve for Ans
 = + =    m V 5.72 s ≈ 6.39 By analogy with laminar shear, τ = µ du/dy. T. V. Boussinesq in 1877 postulated that turbulent shear could also be related to the mean-velocity gradient τturb = ε du/dy, where ε is called the eddy viscosity and is much larger than µ. If the logarithmic-overlap law, Eq. (6.28), is valid with τ ≈ τw, show that ε ≈ κρu*y. Solution: Differentiate the log-law, Eq. (6.28), to find du/dy, then introduce the eddy viscosity into the turbulent stress relation: 1 * ln , * u yu du u If B then u dy yκ ν κ ∗
= + =   2 ** ,w du u Then, if u solve for Ans. dy y τ τ ρ ε ε κ ≈ ≡ = = ε κρ= u * y Chapter 6 • Viscous Flow in Ducts 387 With f known, we compute the head loss and (horizontal) pressure drop as 2 2 f L V 5280 1(11.35) h f (0.0189) (a) d 2g 3/12 2(32.2) Ans.
= = ≈    800 ft fand p gh (1.94)(32.2)(800) (b)Ans.ρ∆ = = ≈ 2 lbf 49900 ft 6.44 Mercury at 20°C flows through 4 meters of 7-mm-diameter glass tubing at an average velocity of 5 m/s. Estimate the head loss in meters and the pressure drop in kPa. Solution: For mercury at 20°C, take ρ = 13550 kg/m3 and µ = 0.00156 kg/m⋅s. Glass tubing is considered hydraulically “smooth,” ε/d = 0. Compute the Reynolds number: .13550(5)(0 007) 304,000; Moody chart smooth: 0.0143 0.00156d Vd f ρ µ = = = ≈Re
= = =    10.4 m 2 24.0 5 0.0143 (a) 2 0.007 2(9.81)f L V h f Ans. d g (13550)(9.81)(10.4) 1,380,000 (b)fp gh Pa Ans. ρ∆ = = = = 1380 kPa 6.45 Oil, SG = 0.88 and ν = 4E−5 m2/s, flows at 400 gal/min through a 6-inch asphalted cast-iron pipe. The pipe is 0.5 miles long (2640 ft) and slopes upward at 8° in the flow direction. Compute the head loss in feet and the pressure change. Solution: First convert 400 gal/min = 0.891 ft3/s and ν = 0.000431 ft2/s. For asphalted cast-iron, ε = 0.0004 ft, hence ε/d = 0.0004/0.5 = 0.0008. Compute V, Red, and f: 2 0.891 4.54(0.5) 4.54 ; 5271; calculate 0.0377 0.000431(0.25) d Moody ft V f sπ = = = = =Re 2 22640 (4.54) then 0.0377 (a) 2 0.5 2(32.2)f L V h f Ans. d g
= = =   63.8 ft If the pipe slopes upward at 8°, the pressure drop must balance both friction and gravity: ( ) 0.88(62.4)[63.8 2640sin8 ] (b)fp g h z Ans.ρ∆ = + ∆ = + ° = 2 lbf 23700 ft 388 Solutions Manual • Fluid Mechanics, Fifth Edition 6.46 Kerosene at 20°C is pumped at 0.15 m3/s through 20 km of 16-cm-diameter cast- iron horizontal pipe. Compute the input power in kW required if the pumps are 85 percent efficient. Solution: For kerosene at 20°C, take ρ = 804 kg/m3 and µ = 1.92E−3 kg/m⋅s. For cast iron take ε ≈ 0.26 mm, hence ε/d = 0.26/160 ≈ 0.001625. Compute V, Re, and f: 2 0.15 m 4 Q 4(804)(0.15) V 7.46 ; Re 500,000 s d (0.00192)(0.16)( /4)(0.16) ρ πµ ππ = = = = ≈ / 0.001625: Moody chart: f 0.0226dε ≈ ≈ 2 2 f L V 20000 (7.46) Then h f (0.0226) 8020 m d 2g 0.16 2(9.81)
= = ≈   At 85% efficiency, the pumping power required is: fgQh 804(9.81)(0.15)(8020)P 11.2E 6 W 0.85 Ans. ρ η = = ≈ + = 11.2 MW 6.47 The gutter and smooth drainpipe in Fig. P6.47 remove rainwater from the roof of a building. The smooth drainpipe is 7 cm in diameter. (a) When the gutter is full, estimate the rate of draining. (b) The gutter is designed for a sudden rainstorm of up to 5 inches per hour. For this condition, what is the maximum roof area that can be drained successfully? Solution: If the velocity at the gutter surface is neglected, the energy equation reduces to Fig. P6.47 2 2 2 2 2(9.81)(4.2), , solve 2 2 1 / 1 (4.2/0.07)f f V L V g z z h h f V g d g fL d f ∆∆ = + = = = + + For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Guess f ≈ 0.02 to obtain the velocity estimate V ≈ 6 m/s above. Then Red ≈ ρVd/µ ≈ (998)(6)(0.07)/(0.001) ≈ 428,000 (turbulent). Then, for a smooth pipe, f ≈ 0.0135, and V is changed slightly to 6.74 m/s. After convergence, we obtain 26.77 m/s, ( /4)(0.07) . (a)V Q V Ansπ= = = 30.026 m /s Chapter 6 • Viscous Flow in Ducts 389 A rainfall of 5 in/h = (5/12 ft/h)(0.3048 m/ft)/(3600 s/h) = 0.0000353 m/s. The required roof area is 3 roof drain rain/ (0.026 m /s)/0.0000353 m/s (b)A Q V Ans.= = ≈ 2740 m 6.48 Show that if Eq. (6.33) is accurate, the position in a turbulent pipe flow where local velocity u equals average velocity V occurs exactly at r = 0.777R, independent of the Reynolds number. Solution: Simply find the log-law position y+ where u+ exactly equals V/u*: ?1 Ru* 3 1 yu* 1 y 3 V u* ln B – u* ln B if ln 2 R 2κ ν κ κ ν κ κ

= + = + = −       3/2rSince y R – r, this is equivalent to 1– e 1– 0.223 R Ans.−= = = ≈ 0.777 6.49 The tank-pipe system of Fig. P6.49 is to deliver at least 11 m3/h of water at 20°C to the reservoir. What is the maximum roughness height ε allowable for the pipe? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Evaluate V and Re for the expected flow rate: Fig. P6.49 2 Q 11/3600 m Vd 998(4.32)(0.03) V 4.32 ; Re 129000 A s 0.001( /4)(0.03) ρ µπ = = = = = = The energy equation yields the value of the head loss: 2 2 2 atm atm1 2 1 2 f f p pV V (4.32) z z h or h 4 3.05 m g 2g g 2g 2(9.81)ρ ρ + + = + + + = − = 2 2 f L V 5.0 (4.32) But also h f , or: 3.05 f , solve for f 0.0192 d 2g 0.03 2(9.81)
= = ≈   With f and Re known, we can find ε/d from the Moody chart or from Eq. (6.48): 101/2 1/2 1 / 2.51 2.0 log , solve for 0.000394 3.7(0.0192) 129000(0.0192) d d ε ε
= − + ≈    Then 0.000394(0.03) 1.2E 5 m (very smooth) Ans.ε = ≈ − ≈ 0.012 mm 392 Solutions Manual • Fluid Mechanics, Fifth Edition Solution: Evaluate ρ = 0.68(998) = 679 kg/m3. Evaluate V = Q/A = (27/3600)/[π (0.025)2] = 3.82 m/s. The energy analysis of the previous problem now has f as the unknown: 2 2 2 1 700000 (3.82) 17070 1 , solve 0.0136 679(9.81) 2 2 2(9.81) 0.05 p V L V z f f f g g d gρ
= = ∆ + + = + + =   679(3.82)(0.05) Smooth pipe: 0.0136, 416000 , Solve df Ans. µ µ = = = = Re kg 0.00031 m s⋅ The density and viscosity are close to the likely suspect, gasoline. Ans. 6.54* A swimming pool W by Y by h deep is to be emptied by gravity through the long pipe shown in Fig. P6.54. Assuming an average pipe friction factor fav and neglecting minor losses, derive a formula for the time to empty the tank from an initial level ho. Fig. P6.54 Solution: With no driving pressure and negligible tank surface velocity, the energy equation can be combined with a control-volume mass conservation: 2 2 2 2( ) , : 2 2 4 1 /av out pipe av V L V gh dh h t f or Q A V D WY g D g f L D dt π= + = = = − + We can separate the variables and integrate for time to drain: ( ) 0 2 0 2 0 2 4 1 / o t o av h g dh D dt WY WY h f L D h π = − = − − +

:Clean this up to obtain Ans.o avdrain h f L DWY t gD + /≈ 1/2 2 2 (1 )4 π Chapter 6 • Viscous Flow in Ducts 393 6.55 The reservoirs in Fig. P6.55 contain water at 20°C. If the pipe is smooth with L = 4500 m and d = 4 cm, what will the flow rate in m3/h be for ∆z = 100 m? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The energy equation from surface 1 to surface 2 gives 1 2 1 2p p and V V ,= = f 1 2thus h z z 100 m= − = Fig. P6.55 2 24500 VThen 100 m f , or fV 0.01744 0.04 2(9.81)
= ≈   Iterate with an initial guess of f ≈ 0.02, calculating V and Re and improving the guess: 1/2 smooth 0.01744 m 998(0.934)(0.04) V 0.934 , Re 37300, f 0.0224 0.02 s 0.001
≈ ≈ ≈ ≈ ≈   1/2 better better better 0.01744 m V 0.883 , Re 35300, f 0.0226, etc...... 0.0224 s
≈ ≈ ≈ ≈   This process converges to 3f 0.0227, Re 35000, V 0.877 m/s, Q 0.0011 m /s / . Ans.= = = ≈ ≈ 34.0 m h 6.56 Consider a horizontal 4-ft-diameter galvanized-iron pipe simulating the Alaska Pipeline. The oil flow is 70 million U.S. gallons per day, at a density of 910 kg/m3 and viscosity of 0.01 kg/m⋅s (see Fig. A.1 for SAE 30 oil at 100°C). Each pump along the line raises the oil pressure to 8 MPa, which then drops, due to head loss, to 400 kPa at the entrance to the next pump. Estimate (a) the appropriate distance between pumping stations; and (b) the power required if the pumps are 88% efficient. Solution: For galvanized iron take ε = 0.15 mm. Convert d = 4 ft = 1.22 m. Convert Q = 7E7 gal/day = 3.07 m3/s. The flow rate gives the velocity and Reynolds number: 2 3.07 910(2.63)(1.22) 2.63 ; 292,500 0.01(1.22) /4 d Q m Vd V A s ρ µπ = = = = = =Re 0.15 0.000123, 0.0157 1220 Moody mm f d mm ε = = ≈ 394 Solutions Manual • Fluid Mechanics, Fifth Edition Relating the known pressure drop to friction factor yields the unknown pipe length: 910 2 2 28,000,00 400,000 0.0157 (2.63) , 2 1.22 L L p Pa f V d ρ
∆ = − = =     Solve 117 miles (a)L Ans.= =188, 000 m The pumping power required follows from the pressure drop and flow rate: 3.07(8 6 4 5) 2.65 7 watts 0.88 (35,500 hp) (b) Q p E E Power E Efficiency Ans. ∆ −= = = = 26.5 MW 6.57 Apply the analysis of Prob. 6.54 to the following data. Let W = 5 m, Y = 8 m, ho = 2 m, L = 15 m, D = 5 cm, and ε = 0. (a) By letting h = 1.5 m and 0.5 m as representative depths, estimate the average friction factor. Then (b) estimate the time to drain the pool. Solution: For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The velocity in Prob. 6.54 is calculated from the energy equation: 2 with (Re ) and Re , / 300 1 / D smooth pipe D gh VD V f fcn L D fL D ρ µ = = = = + (a) With a bit of iteration for the Moody chart, we obtain ReD = 108,000 and f ≈ 0.0177 at h = 1.5 m, and ReD = 59,000 and f ≈ .0202 at h = 0.5 m; thus the average value fav ≈ 0.019. Ans. (a) The drain formula from Prob. 6.54 then predicts: 2 2 2 (1 / )4 4(5)(8) 2(2)[1 0.019(300)] 9.81(0.05) 33700 (b) o av drain h f L DWY t gD s Ans. π π + +≈ ≈ = = 9.4 h 6.58 In Fig. P6.55 assume that the pipe is cast iron with L = 550 m, d = 7 cm, and ∆z = 100 m. If an 80 percent efficient pump is placed at point B, what input power is required to deliver 160 m3/h of water upward from reservoir 2 to 1? Fig. P6.55 Chapter 6 • Viscous Flow in Ducts 397 Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. For commercial steel, take ε ≈ 0.00015 ft, or ε/d = 0.00015/(0.5/12) ≈ 0.0036. Compute 2 Q 0.015 ft V 11.0 ; A s( /4)(0.5/12)π = = = Moody Vd 1.94(11.0)(0.5/12) Re 42500 / 0.0036, f 0.0301 2.09E 5 d ρ ε µ = = ≈ = ≈ − The energy equation, with p1 = p2 and V1 ≈ 0, yields an expression for surface elevation: 2 2 2 f V V L (11.0) 80 h h 1 f 1 0.0301 2g 2g d 2(32.2) 0.5/12 Ans.
  = + = + = + ≈     111 ft 6.62 Water at 20°C is to be pumped through 2000 ft of pipe from reservoir 1 to 2 at a rate of 3 ft3/s, as shown in Fig. P6.62. If the pipe is cast iron of diameter 6 in and the pump is 75 percent efficient, what horsepower pump is needed? Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. For cast iron, take ε ≈ 0.00085 ft, or ε/d = 0.00085/(6/12) ≈ 0.0017. Compute V, Re, and f: Fig. P6.62 2 Q 3 ft V 15.3 ; A s( /4)(6/12)π = = = Moody Vd 1.94(15.3)(6/12) Re 709000 / 0.0017, f 0.0227 2.09E 5 d ρ ε µ = = ≈ = ≈ − The energy equation, with p1 = p2 and V1 ≈ V2 ≈ 0, yields an expression for pump head: 2 2 pump L V 2000 (15.3) h z f 120 ft 0.0227 120 330 450 ft d 2g 6/12 2(32.2)
= ∆ + = + = + ≈    pgQh 1.94(32.2)(3.0)(450)Power: P 112200 550 0.75 Ans. ρ η = = = ÷ ≈ 204 hp 398 Solutions Manual • Fluid Mechanics, Fifth Edition 6.63 A tank contains 1 m3 of water at 20°C and has a drawn-capillary outlet tube at the bottom, as in Fig. P6.63. Find the outlet volume flux Q in m3/h at this instant. Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For drawn tubing, take ε ≈ 0.0015 mm, or ε/d = 0.0015/40 ≈ 0.0000375. The steady-flow energy equation, with p1 = p2 and V1 ≈ 0, gives Fig. P6.63 2 2 2 2 f L V V V 0.8 35.32 h f z , or: 1 f 1.8 m, V d 2g 2g 2g 0.04 1 20f
= = ∆ − + ≈ ≈   + 1/2 better better 35.32 m 998(5.21)(0.04) Guess f 0.015, V 5.21 , Re 208000 1 20(0.015) s 0.001 f 0.0158, V 5.18 m/s, Re 207000 (converged)
≈ = ≈ = ≈ +  ≈ ≈ ≈ 2 3Thus V 5.18 m/s, Q ( /4)(0.04) (5.18) 0.00651 m /s / . Ans.π≈ = = ≈ 323.4 m h 6.64 Repeat Prob. 6.63 to find the flow rate if the fluid is SAE 10 oil. Is the flow laminar or turbulent? Solution: For SAE 10 oil at 20°C, take ρ = 870 kg/m3 and µ = 0.104 kg/m⋅s. For drawn tubing, take ε ≈ 0.0015 mm, or ε/d = 0.0015/40 ≈ 0.0000375. Guess laminar flow: 2 2 f 2 2 ?V 32 LV V 32(0.104)(0.8)V h 1.8 m , or: 1.8 0.195V 2g 2(9.81)gd 870(9.81)(0.04) µ ρ = − = − = = 2Quadratic equation: V 3.83V 35.32 0, solve V 4.33 m/s Check Re (870)(4.33)(0.04)/(0.104) + − = = = ≈ 1450 (OK, laminar) So it is laminar flow, and Q = (π/4)(0.04)2(4.33) = 0.00544 m3/s = 19.6 m3/h. Ans. 6.65 In Prob. 6.63 the initial flow is turbulent. As the water drains out of the tank, will the flow revert to laminar motion as the tank becomes nearly empty? If so, at what tank depth? Estimate the time, in h, to drain the tank completely. Chapter 6 • Viscous Flow in Ducts 399 Solution: Recall that ρ = 998 kg/m3, µ = 0.001 kg/m⋅s, and ε/d ≈ 0.0000375. Let Z be the depth of water in the tank (Z = 1 m in Fig. P6.63). When Z = 0, find the flow rate: 2 f 2(9.81)(0.8) Z 0, h 0.8 m, V converges to f 0.0171, Re 136000 1 20f = = ≈ = = + 3V 3.42 m/s, Q 12.2 m /h (Z 0)≈ ≈ = So even when the tank is empty, the flow is still turbulent. Ans. The time to drain the tank is 2tank tank d d dZ ( ) Q (A Z) (1 m ) Q, dt dt dt υ = − = = = − 0m drain avg1m dZ 1 or t (1 m) Q Q
= − =   
So all we need is the average value of (1/Q) during the draining period. We know Q at Z = 0 and Z = 1 m, let’s check it also at Z = 0.5 m: Calculate Qmidway ≈ 19.8 m 3/h. Then
≈ + + ≈ = ≈   |avg drain3 1 1 1 4 1 h 0.0544 , t Q 6 23.4 19.8 12.2 m Ans.0.0544 h 3.3 min 6.66 Ethyl alcohol at 20°C flows through a 10-cm horizontal drawn tube 100 m long. The fully developed wall shear stress is 14 Pa. Estimate (a) the pressure drop, (b) the volume flow rate, and (c) the velocity u at r = 1 cm. Solution: For ethyl alcohol at 20°C, ρ = 789 kg/m3, µ = 0.0012 kg/m⋅s. For drawn tubing, take ε ≈ 0.0015 mm, or ε/d = 0.0015/100 ≈ 0.000015. From Eq. (6.12), w L 100 p 4 4(14) (a) d 0.1 Ans.τ  ∆ = = ≈ 56000 Pa The wall shear is directly related to f, and we may iterate to find V and Q: 2 2 w f 8(14) V , or: fV 0.142 with 0.000015 8 789 d ετ ρ= = = = 1/2 better better better 0.142 m 789(3.08)(0.1) Guess f 0.015, V 3.08 , Re 202000 0.015 s 0.0012 f 0.0158, V 3.00 m/s, Re 197000 (converged)
≈ = ≈ = ≈   ≈ ≈ ≈ 402 Solutions Manual • Fluid Mechanics, Fifth Edition Guess f ≈ 0.02, calculate d, ε/d and Re and get a better f and iterate: 1/5 4 Q 4(1.94)(3.0)f 0.020, d [3.94(0.02)] 0.602 ft, Re , d (2.09E 5)(0.602) ρ πµ π ≈ ≈ ≈ = = − better 0.00085 or Re 589000, 0.00141, Moody chart: f 0.0218 (repeat) 0.602d ε≈ = ≈ ≈ We are nearly converged. The final solution is f ≈ 0.0217, d ≈ 0.612 ft ≈ 7.3 in Ans. 6.70 In Prob. 6.62 suppose the pipe is 6-inch-diameter cast iron and the pump delivers 75 hp to the flow. What flow rate Q in ft3/s results? Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. For cast iron, take ε ≈ 0.00085 ft, or ε/d = 0.00085/(6/12) ≈ 0.0017. The energy analysis holds: 2 2 p 2 3 better 2000 [4Q/ (6/12) ] ft lbf Power gQh 62.4Q 120 f 75(550) 6/12 2(32.2) s Clean up: 661 Q(120 1611Q f), solve by iteration using / and Re Guess f 0.02, Q 2.29 ft /s, Re 4 Q/ d 541000, then f d πρ ε ρ πµ
  ⋅= = + =   ≈ + ≈ ≈ = ≈ ≈ 0.0228 Convergence is near: f 0.02284, Re 522000, / . Ans.≈ ≈ 3Q 2.21 ft s≈ 6.71 It is desired to solve Prob. 6.62 for the most economical pump and cast-iron pipe system. If the pump costs $125 per horsepower delivered to the fluid and the pipe costs $7000 per inch of diameter, what are the minimum cost and the pipe and pump size to maintain the 3 ft3/s flow rate? Make some simplifying assumptions. Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. For cast iron, take ε ≈ 0.00085 ft. Write the energy equation (from Prob. 6.62) in terms of Q and d: 2 2 in hp f 5 gQ 62.4(3.0) 2000 [4(3.0)/ d ] 154.2f P ( z h ) 120 f 40.84 550 550 d 2(32.2) d ρ π
 = ∆ + = + = +   5 hp inches 5 Cost $125P $7000d 125(40.84 154.2f/d ) 7000(12d), with d in ft. Clean up: Cost $5105 19278f/d 84000d = + = + + ≈ + + Chapter 6 • Viscous Flow in Ducts 403 Regardless of the (unknown) value of f, this Cost relation does show a minimum. If we assume for simplicity that f is constant, we may use the differential calculus: 1/6 f const best6 d(Cost) 5(19278)f 84000, or d (1.148 f) d( ) dd ≈ −= + ≈| 1/6 better better 4 Q Guess f 0.02, d [1.148(0.02)] 0.533 ft, Re 665000, 0.00159 d d Then f 0.0224, d 0.543 ft (converged) ρ ε πµ ≈ ≈ ≈ = ≈ ≈ ≈ ≈ Result: dbest ≈ 0.543 ft ≈ 6.5 in, Costmin ≈ $14300pump + $45600pipe ≈ $60000. Ans. 6.72 Modify Prob. P6.57 by letting the diameter be unknown. Find the proper pipe diameter for which the pool will drain in about 2 hours flat. Solution: Recall the data: Let W = 5 m, Y = 8 m, ho = 2 m, L = 15 m, and ε = 0, with water, ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. We apply the same theory as Prob. 6.57: 2 2 (1 / )2 4 , , (Re ) for a smooth pipe. 1 / o av drain av D h f L Dgh WY V t f fcn fL D gDπ += ≈ = + For the present problem, tdrain = 2 hours and D is the unknown. Use an average value h = 1 m to find fav. Enter these equations on EES (or you can iterate by hand) and the final results are av2.36 m/s; Re 217,000; 0.0154; 0.092 mDV f D Ans.= = ≈ = ≈ 9.2 cm 6.73 The Moody chart, Fig. 6.13, is best for finding head loss (or ∆p) when Q, V, d, and L are known. It is awkward for the “3rd” type of problem, finding d when hf (or ∆p) and Q are known (see Ex. 6.11). Prepare a modified Moody chart whose abscissa is independent of d, using as a parameter ε non-dimensionalized without d, from which one can immediately read the (dimensionless) ordinate to find d. Use your chart to solve Ex. 6.11. Solution: An appropriate Pi group which does not contain d is β = (ghfQ3)/(Lν5). Similarly, an appropriate roughness parameter without d is σ = (εν/Q). After a lot of algebra, the Colebrook friction factor formula (6.48) becomes 1/2 3/2 5/2 103 3 1/2 2.51128 2.0 log 14.8 (128 / ) d d d Re Re Re πσβ π β π
 = − +   404 Solutions Manual • Fluid Mechanics, Fifth Edition A plot of this messy relation is given below. To solve Example 6.11, a 100-m-long, unknown-diameter pipe with a head loss of 8 m, flow rate of 0.342 m3/s, and ε = 0.06 mm, we use that data to compute β = 9.8E21 and σ = 3.5E−6. The oil properties are ρ = 950 kg/m3 and ν = 2E−5 m2/s. Enter the chart above: let’s face it, the scale is very hard to read, but we estimate, at β = 9.8E21 and σ = 3.5E−6, that 6E4 < Red < 8E4, which translates to a diameter of 0.27 < d < 0.36 m. Ans. (Example 6.11 gave d = 0.3 m.) 6.74 In Fig. P6.67 suppose the fluid is gasoline at 20°C and h = 90 ft. What commercial-steel pipe diameter is required for the flow rate to be 0.015 ft3/s? Solution: For commercial steel, take ε ≈ 0.00015 ft. For gasoline, take ρ = 1.32 slug/ft3 and µ = 6.1E−6 slug/ft⋅s. From Prob. 6.61 the energy relation gives 2 2 2 2 2 f V (4Q/ d ) L [4(0.015)/ d ] 80 h h 1 f , or: 90 ft 1 f 2g 2g d 2(32.2) d π π
  = + = + = +   1/4 0.00015Clean up: d 0.0158(1 80f/d) with f determined from Re and dd ε≈ + = Because d appears implicitly in two places, considerable iteration is required: 1/4Guess f 0.02, d 0.0158[1 80(0.02)/d] 0.0401 ft,≈ ≈ + ≈ better 4 Q 4(1.32)(0.015) Re 103000, 0.00374, f 0.0290 d (6.1E 6)(0.0401) d ρ ε πµ π = = ≈ ≈ ≈ − Converges to f 0.0290, Re 96000, d 0.0431 ft Ans.≈ ≈ ≈ ≈ 0.52 inches Chapter 6 • Viscous Flow in Ducts 407 6.78 In Fig. P6.78 the connecting pipe is commercial steel 6 cm in diameter. Estimate the flow rate, in m3/h, if the fluid is water at 20°C. Which way is the flow? Solution: For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For commercial steel, take ε ≈ 0.046 mm, hence ε/d = 0.046/60 ≈ 0.000767. With p1, V1, and V2 all ≈ 0, the energy equation between surfaces (1) and (2) yields Fig. P6.78 2 1 2 f f p 200000 0 0 z 0 z h , or h 15 5.43 m g 998(9.81)ρ + + ≈ + + + = − ≈ − ←(flow to left) 2 2 2 f L V 50 V Guess turbulent flow: h f f 5.43, or: fV 0.1278 d 2g 0.06 2(9.81) = = = ≈ 1/2 fully rough 0.1278 m 0.00767, guess f 0.0184, V 2.64 , Re 158000 0.0184 sd ε
= ≈ ≈ ≈ =   better better better 3rd iteration m f 0.0204, V 2.50 , Re 149700, f 0.0205 (converged) s ≈ = ≈ ≈ The iteration converges to f ≈ 0.0205, V ≈ 2.49 m/s, Q = (π/4)(0.06)2(2.49) = 0.00705 m3/s = 25 m3/h ← Ans. 6.79 A garden hose is used as the return line in a waterfall display at the mall. In order to select the proper pump, you need to know the hose wall roughness, which is not supplied by the manufacturer. You devise a simple experiment: attach the hose to the drain of an above-ground pool whose surface is 3 m above the hose outlet. You estimate the minor loss coefficient in the entrance region as 0.5, and the drain valve has a minor- loss equivalent length of 200 diameters when fully open. Using a bucket and stopwatch, you open the valve and measure a flow rate of 2.0E−4 m3/s for a hose of inside diameter 1.5 cm and length 10 m. Estimate the roughness height of the hose inside surface. Solution: First evaluate the average velocity in the hose and its Reynolds number: 2 2.0 4 998(1.13)(0.015) 1.13 , 16940 ( ) 0.001( /4)(0.015) Q E m Vd V turbulent A s ρ µπ −= = = = = =dRe 408 Solutions Manual • Fluid Mechanics, Fifth Edition Write the energy equation from surface (point 1) to outlet (point 2), assuming an energy correction factor α = 1.05: 2 2 2 1 1 1 2 2 2 2 2 1 2 ,2 2 2 eq f loss loss e Lp V p V V z z h h where h K f g g g g d g α α α ρ ρ
+ + = + + + +   = +   The unknown is the friction factor: 21 22 2 3 1.05 0.5 /2 (1.13) /2(9.81) 0.0514 ( )/ (10/0.015 200) e eq z z m K V g f L L d α − − − − − = = = + + For f = 0.0514 and Red = 16940, the Moody chart (Eq. 6.48) predicts ε/d ≈ 0.0206. Therefore the estimated hose-wall roughness is ε = 0.0206(1.5 cm) = 0.031 cm Ans. 6.80 The head-versus-flow-rate characteri- stics of a centrifugal pump are shown in Fig. P6.80. If this pump drives water at 20°C through 120 m of 30-cm-diameter cast-iron pipe, what will be the resulting flow rate, in m3/s? Solution: For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For cast iron, take ε ≈ 0.26 mm, hence ε/d = 0.26/300 ≈ 0.000867. The head loss must match the pump head: Fig. P6.80 2 2 2 3 pumpf 2 5 L V 8fLQ h f h 80 20Q , with Q in m /s d 2g gdπ = = = ≈ − 2 2 2 f 2 5 8f(120)Q 80 Evaluate h 80 20Q , or: Q 20 4080f(9.81)(0.3)π = = − ≈ + 1/2 380 m 4 Q Guess f 0.02, Q 0.887 , Re 3.76E6 20 4080(0.02) s d ρ πµ
≈ = ≈ = ≈ +  better better 0.000867, f 0.0191, Re 3.83E6, converges to Ans. d ε = ≈ ≈ 3m Q 0.905 s ≈ Chapter 6 • Viscous Flow in Ducts 409 6.81 The pump in Fig. P6.80 is used to deliver gasoline at 20°C through 350 m of 30-cm-diameter galvanized iron pipe. Estimate the resulting flow rate, in m3/s. (Note that the pump head is now in meters of gasoline.) Solution: For gasoline, take ρ = 680 kg/m3 and µ = 2.92E−4 kg/m⋅s. For galvanized iron, take ε ≈ 0.15 mm, hence ε/d = 0.15/300 ≈ 0.0005. Head loss matches pump head: 2 2 2 2 2 f pump2 5 2 5 8fLQ 8f(350)Q 80 h 11901fQ h 80 20Q , Q 20 11901fgd (9.81)(0.3)π π = = = = ≈ − = + 3 rough better better m Guess f 0.017, Q 0.600 , s Re 5.93E6, 0.0005, f 0.0168 d ε ≈ ≈ ≈ = ≈ This converges to f ≈ 0.0168, Re ≈ 5.96E6, Q ≈ 0.603 m3/s. Ans. 6.82 The pump in Fig. P6.80 has its maximum efficiency at a head of 45 m. If it is used to pump ethanol at 20°C through 200 m of commercial-steel pipe, what is the proper pipe diameter for maximum pump efficiency? Solution: For ethanol, take ρ = 789 kg/m3 and µ = 1.2E−3 kg/m⋅s. For commercial steel, take ε ≈ 0.046 mm, hence ε/d = 0.046/(1000d). We know the head and flow rate: 2 3 pumph 45 m 80 20Q , solve for Q 1.323 m /s.= ≈ − ≈ 2 2 1/5 p f 2 5 2 5 5 8fLQ 8f(200)(1.323) 28.92f Then h h 45 m, or: d 0.915f gd (9.81)d dπ π = = = = = ≈ 1/5Guess f 0.02, d 0.915(0.02) 0.419 m, 4 Q Re 2.6E6, 0.000110 d d ρ ε πµ ≈ ≈ ≈ = ≈ ≈ better better better betterThen f 0.0130, d 0.384 m, Re 2.89E6, 0.000120d ε≈ ≈ ≈ ≈| This converges to f 0.0129, Re 2.89E6, . Ans.≈ ≈ d 0.384 m≈ 412 Solutions Manual • Fluid Mechanics, Fifth Edition Solution: For water, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. For commercial steel, take ε ≈ 0.00015 ft. Compute the hydraulic diameter of the annulus: h 2 2 2 f h 4A D 2(a b) 1 inch; P L V 40 V h 20 ft f f , or: fV 2.683 D 2g 1/12 2(32.2) = = − =
= = = ≈   We can make a reasonable estimate by simply relating the Moody chart to Dh, rather than the more complicated “effective diameter” method of Eq. (6.77). Thus 1/2 rough h 0.00015 ft 0.0018, Guess f 0.023, V (2.683/0.023) 10.8 D 1/12 s ε = ≈ ≈ = ≈ h better better VD 1.94(10.8)(1/12) ft Re 83550, f 0.0249, V 10.4 2.09E 5 s ρ µ = = ≈ ≈ ≈ − This converges to f ≈ 0.0250, V ≈ 10.37 ft/s, Q = π(a2 − b2)V = 0.17 ft3/s. Ans. 6.88 An oil cooler consists of multiple parallel-plate passages, as shown in Fig. P6.88. The available pressure drop is 6 kPa, and the fluid is SAE 10W oil at 20°C. If the desired total flow rate is 900 m3/h, estimate the appropriate number of passages. The plate walls are hydraulically smooth. Fig. P6.88 Solution: For SAE 10W oil, ρ = 870 kg/m3 and µ = 0.104 kg/m⋅s. The pressure drop remains 6 kPa no matter how many passages there are (ducts in parallel). Guess laminar flow, Eq. (6.63), 3 3one passage bh p Q Lµ ∆= Chapter 6 • Viscous Flow in Ducts 413 where h is the half-thickness between plates. If there are N passages, then b = 50 cm for all and h = 0.5 m/(2N). We find h and N such that NQ = 900 m3/h for the full set of passages. The problem is ideal for EES, but one can iterate with a calculator also. We find that 18 passages are one too many—Q only equals 835 m3/h. The better solution is: h 3 D, 935 m /h, 1.47 cm, Re 512 (laminar flow)NQ h= = =N = 17 passages 6.89 An annulus of narrow clearance causes a very large pressure drop and is useful as an accurate measurement of viscosity. If a smooth annulus 1 m long with a = 50 mm and b = 49 mm carries an oil flow at 0.001 m3/s, what is the oil viscosity if the pressure drop is 250 kPa? Solution: Assuming laminar flow, use Eq. (6.73) for the pressure drop and flow rate: π µ
∆ −= − −    2 2 2 4 4p (a b )Q a b , or, for the given data: 8 L ln(a/b) 2 2 2 3 4 4250000 {(0.05) (0.049) }0.001 m /s (0.05) (0.049) 8 1 m ln(0.05/0.049) π µ
− = − −   Solve for / Ans.µ ≈ 0.0065 kg m s⋅ 6.90 A 90-ft-long sheet-steel duct carries air at approximately 20°C and 1 atm. The duct cross section is an equilateral triangle whose side measures 9 in. If a blower can supply 1 hp to the flow, what flow rate, in ft3/s, will result? Fig. P6.90 Solution: For air at 20°C and 1 atm, take ρ ≈ 0.00234 slug/ft3 and µ = 3.76E−7 slug/ft⋅s. Compute the hydraulic diameter, and express the head loss in terms of Q: h 4A 4(1/2)(9)(9sin60 ) D 5.2 0.433 ft P 3(9) °= = = =′′ 2 2 2 2 f h L (Q/A) 90 {Q/[0.5(9/12) sin60 ]} h f f 54.4fQ D 2g 0.433 2(32.2) °
= = ≈   414 Solutions Manual • Fluid Mechanics, Fifth Edition For sheet steel, take ε ≈ 0.00015 ft, hence ε/Dh ≈ 0.000346. Now relate everything to the input power: 2 f 3 3 3 1/3 h ft lbf Power 1 hp 550 gQh (0.00234)(32.2)Q[54.4fQ ], s or: fQ 134 with Q in ft /s ft (Q/A)D Guess f 0.02, Q (134/0.02) 18.9 , Re 209000 s ρ ρ µ ⋅= = = = ≈ ≈ = ≈ = ≈ Iterate: fbetter ≈ 0.0179, Qbetter ≈ 19.6 ft 3/s, Rebetter ≈ 216500. The process converges to f ≈ 0.01784, V ≈ 80.4 ft/s, Q ≈ 19.6 ft3/s. Ans. 6.91 Heat exchangers often consist of many triangular passages. Typical is Fig. P6.91, with L = 60 cm and an isosceles- triangle cross section of side length a = 2 cm and included angle β = 80°. If the average velocity is V = 2 m/s and the fluid is SAE 10 oil at 20°C, estimate the pressure drop. Fig. P6.91 Solution: For SAE 10 oil, take ρ = 870 kg/m3 and µ = 0.104 kg/m⋅s. The Reynolds number based on side length a is Re = ρVa/µ ≈ 335, so the flow is laminar. The bottom side of the triangle is 2(2 cm)sin40° ≈ 2.57 cm. Calculate hydraulic diameter: 2 h 1 4A (2.57)(2 cos40 ) 1.97 cm ; P 6.57 cm; D 1.20 cm 2 P A = ° ≈ = = ≈ h h D VD 870(2.0)(0.0120) Re 201; from Table 6.4, 40 , fRe 52.9 0.104 ρ θ µ = = ≈ = ° ≈ 2 2 h 52.9 L 0.6 870 Then f 0.263, p f V (0.263) (2) 201 D 2 0.012 2 Ans. ρ

= ≈ ∆ = =        ≈ 23000 Pa Chapter 6 • Viscous Flow in Ducts 417 6.95 A wind tunnel is made of wood and is 28 m long, with a rectangular section 50 cm by 80 cm. It draws in sea-level standard air with a fan. If the fan delivers 7 kW of power to the air, estimate (a) the average velocity; and (b) the pressure drop in the wind tunnel. Solution: For sea-level air, ρ = 1.22 kg/m3 and µ = 1.81E−5 kg/m⋅s. The hydraulic diameter is: 4 4(50 )(80 ) 61.54 0.6154 2(50 80 )h A cm cm D cm m P cm = = = = + (a, b) The known power is related to both the flow rate and the pressure drop: 2[ ] 2h L Power Q p HWV f V D ρ
= ∆ =     3 2 328 1.22 /[(0.5 )(0.8 ) ] 11.1 7000 0.6154 2 m kg m m m V f V fV W m
 = = =    Thus we need to find V such that fV3 = 631 m3/s3. For wood, take roughness ε = 0.5 mm. Then ε/Dh = 0.0005 m/0.6154 m = 0.000813. Use the Moody chart to find V and the Reynolds number. Guess f ≈ 0.02 to start, or use EES. The iteration converges to: 0.0189, Re 1.33E6, / (a, b) hD f Ans.= = ∆V p= =32 m s, 540 Pa 6.96 Water at 20°C is flowing through a 20-cm-square smooth duct at a (turbulent) Reynolds number of 100,000. For a “laminar flow element” measurement, it is desired to pack the pipe with a honeycomb array of small square passages (see Fig. P6.28 for an example). What passage width h will ensure that the flow in each tube will be laminar (Reynolds number less than 2000)? Fig. P6.96 Solution: The hydraulic diameter of a square is the side length h (or a). For water, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The Reynolds number establishes flow velocity: 998 (0.2) 100,000 , Solve for 0.501 0.001hD Vd V m Re V s ρ µ = = = = 418 Solutions Manual • Fluid Mechanics, Fifth Edition This velocity is the same when we introduce small passages, if we neglect the blockage of the thin passage walls. Thus we merely set the passage Reynolds number = 2000: 998(0.501) 2000 0.001h Vh h Re if h Ans. ρ µ = ≤ ≤ 0.004 m 4 mm= 6.97 A heat exchanger consists of multiple parallel-plate passages, as shown in Fig. P6.97. The available pressure drop is 2 kPa, and the fluid is water at 20°C. If the desired total flow rate is 900 m3/h, estimate the appropriate number of passages. The plate walls are hydraulically smooth. Fig. P6.97 Solution: For water, ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. Unlike Prob. 6.88, here we expect turbulent flow. If there are N passages, then b = 50 cm for all N and the passage thickness is H = 0.5 m/N. The hydraulic diameter is Dh = 2H. The velocity in each passage is related to the pressure drop by Eq. (6.58): 2 where 2 h smooth h VDL p f V f f fcn D ρρ µ
 ∆ = = =    3 22.0 998 /For the given data, 2000 2(0.5 / ) 2 m kg m Pa f V m N = Select N, find H and V and Qtotal = AV = b 2V and compare to the desired flow of 900 m3/h. For example, guess N = 20, calculate f = 0.0173 and Qtotal = 2165 m 3/h. The converged result is 3 total 908 m /h, 0.028, Re 14400, 7.14 mm, hD Q f H Ans. = = = = N = 70 passages Chapter 6 • Viscous Flow in Ducts 419 6.98 A rectangular heat exchanger is to be divided into smaller sections using sheets of commercial steel 0.4 mm thick, as sketched in Fig. P6.98. The flow rate is 20 kg/s of water at 20°C. Basic dimensions are L = 1 m, W = 20 cm, and H = 10 cm. What is the proper number of square sections if the overall pressure drop is to be no more than 1600 Pa? Fig. P6.98 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For commercial steel, ε ≈ 0.046 mm. Let the short side (10 cm) be divided into “J” squares. Then the long (20 cm) side divides into “2J” squares and altogether there are N = 2J2 squares. Denote the side length of the square as “a,” which equals (10 cm)/J minus the wall thickness. The hydraulic diameter of a square exactly equals its side length, Dh = a. Then the pressure drop relation becomes 21600 Pa, where N 2J 2 2 2 h L 1.0 998 Q 0.1 p f V f and a 0.0004 D 2 a 2 JNa ρ

∆ = = ≤ = = −       As a first estimate, neglect the 0.4-mm wall thickness, so a ≈ 0.1/J. Then the relation for ∆p above reduces to fJ ≈ 0.32. Since f ≈ 0.036 for this turbulent Reynolds number (Re ≈ 1E4) we estimate that J ≈ 9 and in fact this is not bad even including wall thickness: 2 2 0.1 20/998 m J 9, N 2(9) 162, a 0.0004 0.0107 m, V 1.078 9 s162(0.0107) = = = = − = = ≈ Moody Va 998(1.078)(0.0107) 0.046 Re 11526, 0.00429, f 0.0360 0.001 a 10.7 ρ ε µ = = ≈ = ≈ ≈ 21.0 998Then p (0.036) (1.078) 1950 Pa 0.0107 2

∆ = ≈       So the wall thickness increases V and decreases a so ∆p is too large. Try J = 8: m J 8, N 128, a 0.0121 m, V 1.069 , s Re 12913, 0.0038, f 0.0347 a ε = = = = = = ≈ 2Then p f(L/a)( /2)V . Close enough, J 8, Ans.ρ∆ = ≈ =1636 Pa N 128= [I suppose a practical person would specify J = 7, N = 98, to keep ∆p < 1600 Pa.] 422 Solutions Manual • Fluid Mechanics, Fifth Edition Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. For galvanized iron, ε ≈ 0.0005 ft, whence ε/d = 0.0005/(2/12 ft) ≈ 0.003. Without the 6° cone, the minor losses are: reentrant elbows gate valve sharp exitK 1.0; K 2(0.41); K 0.16; K 1.0≈ ≈ ≈ ≈ 2 Q 0.4 ft Vd 1.94(18.3)(2/12) Evaluate V 18.3 ; Re 284000 A s 2.09E 5(2/12) /4 ρ µπ = = = = = ≈ − At this Re and roughness ratio, we find from the Moody chart that f ≈ 0.0266. Then 2 2 pump V L (18.3) 60 (a) h z f K 20 0.0266 1.0 0.82 0.16 1.0 2g d 2(32.2) 2/12
  = ∆ + +  = + + + + +   p pump gQh (62.4)(0.4)(85.6) or h 85.6 ft, Power 0.70 3052 550 (a)Ans. ρ η ≈ = = = ÷ ≈ 5.55 hp (b) If we replace the sharp exit by a 6° conical diffuser, from Fig. 6.23, Kexit ≈ 0.3. Then 2 p (18.3) 60 h 20 0.0266 1.0 .82 .16 0.3 81.95 ft 2(32.2) 2/12
  = + + + + + =   then Power (62.4)(0.4)(81.95)/0.7 550 (4% less) (b)Ans.= ÷ ≈ 5.31 hp 6.103 The reservoirs in Fig. P6.103 are connected by cast-iron pipes joined abruptly, with sharp-edged entrance and exit. Including minor losses, estimate the flow of water at 20°C if the surface of reservoir 1 is 45 ft higher than that of reservoir 2. Fig. P6.103 Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. Let “a” be the small pipe and “b” the larger. For wrought iron, ε ≈ 0.00015 ft, whence ε/da = 0.0018 and ε/db = 0.0009. From the continuity relation, 2 2 a a b b b a b a 1 Q V d V d or, since d 2d , we obtain V V 4 4 4 π π= = = = Chapter 6 • Viscous Flow in Ducts 423 For pipe “a” there are two minor losses: a sharp entrance, K1 = 0.5, and a sudden expansion, Fig. 6.22, Eq. (6.101), K2 = [1 − (1/2) 2]2 ≈ 0.56. For pipe “b” there is one minor loss, the submerged exit, K3 ≈ 1.0. The energy equation, with equal pressures at (1) and (2) and near zero velocities at (1) and (2), yields 2 2 a a b b f-a m-a f-b m-b a b a b V L V L z h h h h f 0.5 0.56 f 1.0 , 2g d 2g d

 ∆ = +  + +  = + + + +       2 a b a a b V 120 1.0 or, since V V /4, z 45 ft 240f 1.06 f 2(32.2) 16 16
= ∆ = = + + +   where fa and fb are separately related to different values of Re and ε/d. Guess to start: a b a a a a-2f f 0.02: then V 21.85 ft/s, Re 169000, /d 0.0018, f 0.0239ε≈ ≈ = ≈ = ≈ b b b b-2V 5.46 ft/s, Re 84500, /d 0.0009, f 0.0222ε= ≈ = ≈ a b a a a Converges to: f 0.024, f 0.0224, V 20.3 ft/s, Q V A . Ans. = = ≈ = ≈ 30.111 ft /s 6.104 Reconsider the air hockey table of Problem 3.162, but with inclusion of minor losses. The table is 3 ft by 6 ft in area, with 1/16-in-diameter holes spaced every inch in a rectangular grid (2592 holes total). The required jet speed from each hole is 50 ft/s. Your job is to select an appropriate blower to meet the requirements. Fig. P3.162 Hint: Assume that the air is stagnant in the manifold under the table surface, and assume sharp-edge inlets at each hole. (a) Estimate the pressure rise (in lbf/in2) required of the blower. (b) Compare your answer to the previous calculation in Prob. 3.162, where minor losses were ignored. Are minor losses significant? Solution: Write the energy equation between manifold and atmosphere: 22 2 1 1 2 2 1 1 2 2 , where2 2 2 jet losses losses inlet Vp V p V z z h h K g g g g g α α ρ ρ + + = + + + ≈ Neglect V1 ≈ 0 and z1 ≈ z2, assume α1,2 = 1.0, and solve for 2 1 2 - -(1 ), where 0.52 jet inlet sharp edge inlet p p p V K K ρ∆ = − = + ≈ Clearly, the pressure drop is about 50% greater due to the minor loss. Ans. (b) 424 Solutions Manual • Fluid Mechanics, Fifth Edition Work out ∆p, assuming ρair ≈ 0.00234 slug/ft3: 2 2 0.00234 (50) (1 0.5) 4.39 144 (a) 2 lbf p Ans. ft ∆ = + = ÷ = 2 lbf 0.0305 in (Again, this is 50% higher than Prob. 3.162.) 6.105 The system in Fig. P6.105 consists of 1200 m of 5 cm cast-iron pipe, two 45° and four 90° flanged long-radius elbows, a fully open flanged globe valve, and a sharp exit into a reservoir. If the elevation at point 1 is 400 m, what gage pressure is required at point 1 to deliver 0.005 m3/s of water at 20°C into the reservoir? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For cast iron, take ε ≈ 0.26 mm, hence ε/d = 0.0052. With the flow rate known, we can compute V, Re: Fig. P6.105 Moody2 Q 0.005 m 998(2.55)(0.05) V 2.55 ; Re 127000, f 0.0315 A s 0.001( /4)(0.05)π = = = = ≈ ≈ The minor losses may be listed as follows: 45 long-radius elbow: K 0.2; 90 long-radius elbow: K 0.3 Open flanged globe valve: K 8.5; submerged exit: K 1.0 ° ≈ ° ≈ ≈ ≈ Then the energy equation between (1) and (2—the reservoir surface) yields 2 1 1 1 2 f m p V z 0 0 z h h , g 2gρ + + = + + + +
2 1 (2.55) 1200 or: p /( g) 500 400 0.0315 0.5 2(0.2) 4(0.3) 8.5 1 1 2(9.81) 0.05 ρ
 = − + + + + + + −   100 253 353 m, or: (998)(9.81)(353) Ans.= + = = ≈1p 3.46 MPa Chapter 6 • Viscous Flow in Ducts 427 Fig. P6.108 Solution: For water, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. The energy equation is written from point 1 to the surface of the tank: 2 2 1 1 2 2 1 2 22 2 f valve filter elbow exit p V p V z z h K K K K g g g gρ ρ + + = + + + + + + + (a) From the flow rate, V1 = Q/A = (0.4 ft 3/s)/[(π /4)(4/12 ft)2] = 4.58 ft/s. Look up minor losses and enter into the energy equation: + +
 = + + + + + + +    2 2 3 2 2 (6.5)(144) lbf/ft (4.58 ft/s) 0 62.4 lbf/ft 2(32.2 ft/s ) (4.58) 80 ft 0 0 9 ft 2.8 2(0.64) 1 2(32.2) (4/12 ft) filter f K We can solve for Kfilter if we evaluate f. Compute ReD = (1.94)(4.58)(4/12)/(2.09E−5) = 141,700. For commercial steel, ε/D = 0.00015 ft/0.333 ft = 0.00045. From the Moody chart, f ≈ 0.0193, and fL/D = 4.62. The energy equation above becomes: filter15.0 ft 0.326 ft 9 ft 0.326(4.62 2.8 1.28 1) ft, Solve (a) K Ans. + = + + + + + Kfilter 9.7≈ (b) If Kfilter = 7.0 and V is unknown, we must iterate for the velocity and flow rate. The energy equation becomes, with the disk valve wide open (KValve ≈ 0):
+ = + + + + +   2 2 80 15.0 ft 9 ft 0 7.0 1.28 1 2(32.2) 2(32.2) 1/3 V V f ≈ = =0.0189, Re 169,000, 5.49 ft/s, (b) DIterate to find f V Ans.3Q AV 0.48 ft /s= = 428 Solutions Manual • Fluid Mechanics, Fifth Edition 6.109 In Fig. P6.109 there are 125 ft of 2-in pipe, 75 ft of 6-in pipe, and 150 ft of 3-in pipe, all cast iron. There are three 90° elbows and an open globe valve, all flanged. If the exit elevation is zero, what horsepower is extracted by the turbine when the flow rate is 0.16 ft3/s of water at 20°C? Fig. P6.109 Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. For cast iron, ε ≈ 0.00085 ft. The 2″, 6″, and 3″ pipes have, respectively, (a) L/d = 750, ε/d = 0.0051; (b) L/d = 150, ε/d = 0.0017; (c) L/d = 600, ε/d = 0.0034 The flow rate is known, so each velocity, Reynolds number, and f can be calculated: a a a2 0.16 ft 1.94(7.33)(2/12) V 7.33 ; Re 113500, f 0.0314 s 2.09E 5(2/12) /4π = = = = ≈ − b b c c c cAlso, V 0.82 ft/s, Re 37800, f 0.0266; V 3.26, Re 75600, f 0.0287= = ≈ = = ≈ Finally, the minor loss coefficients may be tabulated: sharp 2″ entrance: K = 0.5; three 2″ 90° elbows: K = 3(0.95) 2″ sudden expansion: K ≈ 0.79; 3″ open globe valve: K ≈ 6.3 The turbine head equals the elevation difference minus losses and the exit velocity head: 2 t f m c 2 2 2 h z h h V /(2g) (7.33) 100 [0.0314(750) 0.5 3(0.95) 0.79] 2(32.2) (0.82) (3.26) (0.0266)(150) [0.0287(600) 6.3 1] 2(32.2) 2(32.2) = ∆ −
−
− = − + + + − − + + ≈ 72.8 ft The resulting turbine power = ρgQht = (62.4)(0.16)(72.8) ÷ 550 ≈ 1.32 hp. Ans. Chapter 6 • Viscous Flow in Ducts 429 6.110 In Fig. P6.110 the pipe entrance is sharp-edged. If the flow rate is 0.004 m3/s, what power, in W, is extracted by the turbine? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For cast Fig. P6.110 iron, ε ≈ 0.26 mm, hence ε/d = 0.26/50 ≈ 0.0052. The minor loss coefficients are Entrance: K ≈ 0.5; 5-cm(≈2″) open globe valve: K ≈ 6.9 The flow rate is known, hence we can compute V, Re, and f: 2 Q 0.004 m 998(2.04)(0.05) V 2.04 , Re 102000, f 0.0316 A s 0.001( /4)(0.05)π = = = = ≈ ≈ The turbine head equals the elevation difference minus losses and exit velocity head:
 = ∆ − − − = − + + + ≈  2 2 t f m V (2.04) 125 h z h h 40 (0.0316) 0.5 6.9 1 21.5 m 2g 2(9.81) 0.05 tPower gQh (998)(9.81)(0.004)(21.5) Ans.ρ= = ≈ 840 W 6.111 For the parallel-pipe system of Fig. P6.111, each pipe is cast iron, and the pressure drop p1 − p2 = 3 lbf/in 2. Compute the total flow rate between 1 and 2 if the fluid is SAE 10 oil at 20°C. Fig. P6.111 Solution: For SAE 10 oil at 20°C, take ρ = 1.69 slug/ft3 and µ = 0.00217 slug/ft⋅s. For cast iron, ε ≈ 0.00085 ft. Convert ∆p = 3 psi = 432 psf and guess laminar flow in each: 3ft0.0763 Check Re (OK) a a a a 4 4 a a ? 128 L Q 128(0.00217)(250)Q p 432 , d (3/12) .Q 300 s µ π π ∆ = = = ≈ ≈ 3ft0.0188 Re b b b b 4 4 b b ? 128 L Q 128(0.00217)(200)Q p 432 , d (2/12) Q . Check 112 (OK) s µ π π ∆ = = = ≈ ≈ a bThe total flow rate is Q Q 0.0763 0.0188 / . Ans.= + = + ≈ 3Q 0.095 ft s 432 Solutions Manual • Fluid Mechanics, Fifth Edition The pressure drop is the same in either leg: 2 2 1 1 2 2 1 2 1 2 2 L V L V 45000 p f f (b) d 2 d 2 Q Ans. ρ ρ∆ = = − ≈ 774, 000 Pa 6.115 In Fig. P6.115 all pipes are 8-cm-diameter cast iron. Determine the flow rate from reservoir (1) if valve C is (a) closed; and (b) open, with Kvalve = 0.5. Fig. P6.115 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For cast iron, ε ≈ 0.26 mm, hence ε/d = 0.26/80 ≈ 0.00325 for all three pipes. Note p1 = p2, V1 = V2 ≈ 0. These are long pipes, but we might wish to account for minor losses anyway: sharp entrance at A: K1 ≈ 0.5; line junction from A to B: K2 ≈ 0.9 (Table 6.5) branch junction from A to C: K3 ≈ 1.3; two submerged exits: KB = KC ≈ 1.0 If valve C is closed, we have a straight series path through A and B, with the same flow rate Q, velocity V, and friction factor f in each. The energy equation yields 1 2 fA mA fB mBz z h h h h ,− = +
+ +
2V 100 50 or: 25 m f 0.5 0.9 f 1.0 , where fcn Re, 2(9.81) 0.08 0.08 f d ε
  = + + + + =    Guess f ≈ ffully rough ≈ 0.027, then V ≈ 3.04 m/s, Re ≈ 998(3.04)(0.08)/(0.001) ≈ 243000, ε/d = 0.00325, then f ≈ 0.0273 (converged). Then the velocity through A and B is V = 3.03 m/s, and Q = (π /4)(0.08)2(3.03) ≈ 0.0152 m3/s. Ans. (a). Chapter 6 • Viscous Flow in Ducts 433 If valve C is open, we have parallel flow through B and C, with QA = QB + QC and, with d constant, VA = VB + VC. The total head loss is the same for paths A-B and A-C: 1 2 fA mA-B fB mB fA mA-C fC mCz z h h h h h h h h ,− = +
+ +
= +
+ +
2 2 A B A B 22 CA A C V 100 V 50 or: 25 f 0.5 0.9 f 1.0 2(9.81) 0.08 2(9.81) 0.08 VV 100 70 f 0.5 1.3 f 1.0 2(9.81) 0.08 2(9.81) 0.08

= + + + +      

= + + + +       plus the additional relation VA = VB + VC. Guess f ≈ ffully rough ≈ 0.027 for all three pipes and begin. The initial numbers work out to 2 2 2 2 A A B B A A C C2g(25) 490.5 V (1250f 1.4) V (625f 1) V (1250f 1.8) V (875f 1)= = + + + = + + + A B CIf f 0.027, solve (laboriously) V 3.48 m/s, V 1.91 m/s, V 1.57 m/s.≈ ≈ ≈ ≈ A A B B C C Compute Re 278000, f 0.0272, Re 153000, f 0.0276, Re 125000, f 0.0278 = ≈ = = = = Repeat once for convergence: VA ≈ 3.46 m/s, VB ≈ 1.90 m/s, VC ≈ 1.56 m/s. The flow rate from reservoir (1) is QA = (π/4)(0.08)2(3.46) ≈ 0.0174 m3/s. (14% more) Ans. (b) 6.116 For the series-parallel system of Fig. P6.116, all pipes are 8-cm-diameter asphalted cast iron. If the total pressure drop p1 − p2 = 750 kPa, find the resulting flow rate Q m3/h for water at 20°C. Neglect minor losses. Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For asphalted cast iron, ε ≈ 0.12 mm, hence ε/d = 0.12/80 ≈ 0.0015 for all three pipes. The head loss is the same through AC and BC: Fig. P6.116 2 2 2 2 fA fC fB fC A C B C p L V L V L V L V h h h h f f f f g d 2g d 2g d 2g d 2gρ



∆ = + = + = + = +                434 Solutions Manual • Fluid Mechanics, Fifth Edition Since d is the same, VA + VB = VC and fA, fB, fC are found from the Moody chart. Cancel g and introduce the given data: 2 22 2 C CA B A C B C A B C V V750000 250 V 150 100 V 150 f f f f , V V V 998 0.08 2 0.08 2 0.08 2 0.08 2 = + = + + = rough A B C m m m Guess f 0.022 and solve laboriously: V 2.09 , V 3.31 , V 5.40 s s s ≈ ≈ ≈ ≈ Now compute ReA ≈ 167000, fA ≈ 0.0230, ReB ≈ 264000, fB ≈ 0.0226, ReC ≈ 431000, and fC ≈ 0.0222. Repeat the head loss iteration and we converge: VA ≈ 2.06 m/s, VB ≈ 3.29 m/s, VC ≈ 5.35 m/s, Q = (π / 4)(0.08)2(5.35) ≈ 0.0269 m3/s. Ans. 6.117 A blower delivers air at 3000 m3/h to the duct circuit in Fig. P6.117. Each duct is commercial steel and of square cross- section, with side lengths a1 = a3 = 20 cm and a2 = a4 = 12 cm. Assuming sea-level air conditions, estimate the power required if the blower has an efficiency of 75%. Neglect minor losses. Solution: For air take ρ = 1.2 kg/m3 and µ = 1.8E−5 kg/m⋅s. Establish conditions in each duct: Fig. P6.117 = = = = = = − 3 3 1&3 1&32 3000 m 0.833 m /s 1.2(20.8)(0.2) 0.833 ; 20.8 m/s; Re 278,000 3600 s 1.8 5(0.2 m) Q V E = = = = − 3 2&4 2&42 0.833 m /s 1.2(57.8)(0.12) 57.8 m/s; Re 463,000 1.8 5(0.12 m) V E For commercial steel (Table 6.1) ε = 0.046 mm. Then we can find the two friction factors: 1&3 1&3 1&3 0.046 0.00023; Re 278000; Moody chart: 0.0166 200 f D ε = = = ≈| 2&4 2&4 1&3 0.046 0.000383; Re 463000; Moody chart: 0.0170 120 f D ε = = = ≈| 2 2 1&3 1&3 80 (1.2)(20.8) Then (0.0166) 1730 2 0.2 2 L V p f Pa D ρ

∆ = = =      2 2 2&4 1&3 60 (1.2)(57.8) and (0.0170) 17050 2 0.12 2 L V p f Pa D ρ

∆ = = =      Chapter 6 • Viscous Flow in Ducts 437 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. For cast iron, ε = 0.26 mm. Then, ε/d1 = 0.00217, ε/d2 = 0.00325, and ε/d3 = 0.0026. The head losses are the same for each pipe, and the flow rates add: 22 2 3 3 3 31 1 1 2 2 2 1 2 32 5 2 5 2 5 1 2 3 88 8 200 ; and 3600f f L Qf L Q f L Q m h Q Q Q sgd gd gdπ π π = = = + + = 1/2 1/2 3 1 1 2 1 3Substitute and combine: [1 0.418( / ) 0.599( / ) ] 0.0556 /Q f f f f m s+ + = We could either go directly to EES or begin by guessing f1 = f2 = f3, which gives Q1 = 0.0275 m3/s, Q2 = 0.0115 m 3/s, and Q3 = 0.0165 m 3/s. This is very close! Further iteration gives 1 1 2 2 3 3Re 298000, 0.0245; Re 177000, 0.0275; Re 208000, 0.0259f f f= = = = = = 1 2 3, , and . (a)Q Q Q Ans= = = 3 3 30.0281 m /s 0.0111 m /s 0.0163 m /s 3 2 f f51.4 m, (998 kg/m )(9.81 m/s )(51.4 m) . (b)h p gh Ansρ= ∆ = = = 503,000 Pa 6.121 Consider the three-reservoir system of Fig. P6.121 with the following data: L1 = 95 m L2 = 125 m L3 = 160 m z1 = 25 m z2 = 115 m z3 = 85 m All pipes are 28-cm-diameter unfinished concrete (ε = 1 mm). Compute the steady flow rate in all pipes for water at 20°C. Fig. P6.121 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. All pipes have ε/d = 1/280 = 0.00357. Let the intersection be “a.” The head loss at “a” is desired: 22 2 3 31 1 2 2 1 a 1 2 a 2 3 a 3 1 2 3 L VL V L V z h f ; z h f ; z h f d 2g d 2g d 2g − = − = − = plus the requirement that Q1 + Q2 + Q3 = 0 or, for same d, V1 + V2 + V3 = 0 We guess ha then iterate each friction factor to find V and Q and then check if
Q = 0. 2 1 a 1 1 1 95 V m h 75 m: 25 75 ( )50 f , solve f 0.02754, V 10.25 0.28 2(9.81) s
= − = − = ≈ ≈ −   438 Solutions Manual • Fluid Mechanics, Fifth Edition 2 2 2 2 2Similarly, 115 75 f (125/0.28) V /2(9.81) gives f 0.02755. V 7.99
− = ≈ ≈ +  2 3 3 3 3 and 85 75 f (160/0.28) V /2(9.81) m gives f 0.02762, V 3.53 , V s
− =   ≈ ≈ +  = +1.27 Repeating for ha = 80 m gives V1 = −10.75, V2 = +7.47, V3 = +2.49 m/s,
V = −0.79. Interpolate to ha ≈ 78 m, gives V1 = −10.55 m/s, V2 = +7.68 m/s, V3 = +2.95 m/s, or: Q1 = −0.65 m 3/s, Q2 = +0.47 m 3/s, Q3 = +0.18 m 3/s. Ans. 6.122 Modify Prob. 6.121 by reducing the diameter to 15 cm, with ε = 1 mm. Compute the flow rate in each pipe. They all reduce, compared to Prob. 6.121, by a factor of about 5.2. Can you explain this? Solution: The roughness ratio increases to ε/d = 1/150 = 0.00667, and all L/d’s increase. Guess ha = 75 m: converges to f1 = 0.0333, f2 = 0.0333, f4 = 0.0334 and V1 ≈ −6.82 m/s, V2 ≈ +5.32 m/s, V3 ≈ +2.34 m/s,
V ≈ +0.85 We finally obtain ha ≈ 78.2 m, giving V1 = −7.04 m/s, V2 = +5.10 m/s, V3 = +1.94 m/s, or: Q1 = −0.124 m 3/s, Q2 = +0.090 m 3/s, Q3 = +0.034 m 3/s. Ans. 6.123 Modify Prob. 6.121 on the previous page as follows. Let z3 be unknown and find its value such that the flow rate in pipe 3 is 0.2 m3/s toward the junction. (This problem is best suited for computer iteration.) Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. All pipes have ε/d = 1/280 = 0.00357. Let the intersection be “a.” The head loss at “a” is desired for each in order to check the flow rate in pipe 3. In Prob. 6.121, with z3 = 85 m, we found Q3 to be 0.18 m 3/s toward the junction, pretty close. We repeat the procedure with a few new values of z3, closing to
Q = 0 each time: 3 3 a 1 2 3 3 Guess z 85 m: h 78.19 m, Q 0.6508, Q 0.4718, Q 0.1790 m /s 90 m: 80.65 m, 0.6657, 0.6657, 0.2099 m /s = = = − = + = + − + + a 1 2 Interpolate: h 79.89, Q 0.6611, Q 0.4608, , .Ans ≈ ≈ − ≈ + 33 3Q 0.200 m /s z 88.4 m≈ + ≈ Chapter 6 • Viscous Flow in Ducts 439 6.124 The three-reservoir system in Fig. P6.124 delivers water at 20°C. The system data are as follows: D1 = 8 in D2 = 6 in D3 = 9 in L1 = 1800 ft L2 = 1200 ft L3 = 1600 ft All pipes are galvanized iron. Compute the flow rate in all pipes. Fig. P6.124 Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. For galvanized iron, take ε = 0.0005 ft. Then the roughness ratios are 1 2 3/ 0.00075 / 0.0010 / 0.00667d d dε ε ε= = = Let the intersection be “a.” The head loss at “a” is desired: 22 2 3 3 31 1 1 2 2 2 1 a 2 a 3 a 1 2 3 1 2 3 f L Vf L V f L V z h ; z h ; z h ; plus Q Q Q 0 d 2g d 2g d 2g − = − = − = + + = We guess ha then iterate each friction factor to find V and Q and then check if
Q = 0. 2 1 1 a 1 1 f (1800)V Guess h 50 ft: 20 50 ( )30 ft , (8/12)2(32.2) ft solve f 0.0194, V 6.09 s = − = − = = = − 3 2 2 3Similarly, f 0.0204, V 8.11 ft/s and of course V 0. Get Q 0.54 ft /s= ≈ + =
= − Try again with a slightly lower ha to reduce Q1 and increase Q2 and Q3: 3 3 a 1 2 3 3 ft ft h 48 ft: converges to Q 2.05 , Q 1.62 , s s ft Q 0.76 , Q 0.33 s = = − = + = +
= + Interpolate to ah 49.12 ft: / / / .Ans= 3 3 3 1 2 3Q 2.09 ft s, Q 1.61 ft s, Q 0.49 ft s= − = + = + 442 Solutions Manual • Fluid Mechanics, Fifth Edition And there are three independent junctions which have zero net flow rate: AB AC AB BC BD AC BC CDJunction A: Q Q 2.0; B: Q Q Q ; C: Q Q Q+ = = + + = These are five algebraic equations to be solved for the five flow rates. The answers are: AB AC BC CD BD Q , Q , Q , Q , Q (a)Ans.= = = = = 3ft 1.19 0.81 0.99 1.80 0.20 s The pressures follow by starting at A (120 psi) and subtracting off the friction losses: 2 2 B A AB AB B p p gK Q 120 144 62.4(19.12)(1.19) p 15590 psf 144 ρ= − = × − = ÷ = 2 lbf 108 in C DSimilarly, p and p (b)Ans.≈ ≈103 psi 76 psi 6.128 Modify Prob. 6.127 above as follows: Let the inlet flow at A and the exit flow at D be unknown. Let pA − pB = 100 psi. Compute the flow rate in all five pipes. Solution: Our head loss coefficients “K” from above are all the same. Head loss AB is known, plus we have two “loop” equations and two “junction” equations: 2 2A B AB AB AB AB p p 100 144 231 ft K Q 19.12Q , or Q / g 62.4ρ − ×= = = = = 33.47 ft s 2 2 BC ACTwo loops: 231 13.26Q 60.42Q 0+ − = 2 2 2 BC CD BD13.26Q 19.12Q 1933.0Q 0+ − = Two junctions: QAB = 3.47 = QBC + QBD; QAC + QBC = QCD The solutions are in exactly the same ratio as the lower flow rates in Prob. 6.127: AB BC BD CD AC Q , Q , Q , Q , Q .Ans = = = = = 3 3 3 3 3 ft ft ft 3.47 2.90 0.58 s s s ft ft 5.28 2.38 s s Chapter 6 • Viscous Flow in Ducts 443 6.129 In Fig. P6.129 all four horizontal cast-iron pipes are 45 m long and 8 cm in diameter and meet at junction a, delivering water at 20°C. The pressures are known at four points as shown: p1 = 950 kPa p2 = 350 kPa p3 = 675 kPa p4 = 100 kPa Neglecting minor losses, determine the flow rate in each pipe. Fig. P6.129 Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. All pipes are cast iron, with ε/d = 0.26/80 = 0.00325. All pipes have L/d = 45/0.08 = 562.5. One solution method is to guess the junction pressure pa, iterate to calculate the friction factors and flow rates, and check to see if the net junction flow is zero: 2 21 1 1 a fl 1 12 5 1 950000 500000 8f L Q Guess p 500 kPa: h 45.96 m 1.135E6f Q 998(9.81) gdπ −= = = = = 3 1 1 1 1 1 1-new 3 1 1 then guess f 0.02, Q 0.045 m /s, Re 4 Q /( d ) 715000, f 0.0269 converges to f 0.0270, Q 0.0388 m /s ρ πµ≈ = = = ≈ ≈ ≈ 3 2 3 4 m Iterate also to Q 0.0223 (away from ), Q 0.0241, Q 0.0365 s a= − = = − aQ 0.00403, so we have guessed p a little low.
= + Trying pa = 530 kPa gives
Q = −0.00296, hence iterate to pa ≈ 517 kPa: 1 2 3 4 Q (toward a), Q , Q , Q Ans. = = = = + − + − 3 3 3 3 m m 0.0380 0.0236 s s m m 0.0229 0.0373 s s 6.130 In Fig. P6.130 lengths AB and BD are 2000 and 1500 ft, respectively. The friction factor is 0.022 everywhere, and pA = 90 lbf/in2 gage. All pipes have a diameter of 6 in. For water at 20°C, determine the flow rate in all pipes and the pressures at points B, C, and D. Fig. P6.130 444 Solutions Manual • Fluid Mechanics, Fifth Edition Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. Each pipe has a head loss which is known except for the square of the flow rate: 22 2AC f AC AC AC2 5 2 5 8(0.022)(1500)Q8fLQ Pipe AC: h K Q , where K gd (32.2)(6/12)π π = = = ≈ 26.58 Similarly, KAB = KCD = 35.44, KBD = 26.58, and KBC = 44.30. The solution is similar to Prob. 6.127, except that (1) the K’s are different; and (2) junctions B and C have additional flow leaving the network. The basic flow relations are: 2 2 2 AB BC ACLoop ABC: 35.44Q 44.3Q 26.58Q 0+ − = 2 2 2 BC CD BDLoop BCD: 44.3Q 35.44Q 26.58Q 0+ − = AB AC AB BC BD AC BC CD Junctions A,B,C: Q Q 2.0; Q Q Q 1.0; Q Q Q 0.5 + = = + + + = + In this era of PC “equation solvers” such as MathCAD, etc., it is probably not necessary to dwell upon any solution methods. For hand work, one might guess QAB, then the other four are obtained in sequence from the above relations, plus a check on the original guess for QAB. The assumed arrows are shown above. It turns out that we have guessed the direction incorrectly on QBC above, but the others are OK. The final results are: AB ACQ / (toward B); Q / (toward C)= = 3 30.949 ft s 1.051 ft s BC CD BDQ (toward B); Q (toward D); Q (to D) (a)Ans.= = =0.239 0.312 0.188 The pressures start at A, from which we subtract the friction losses in each pipe: 2 2 B A AB ABp p gK Q 90 144 62.4(35.44)(0.949) 10969 psf 144ρ= − = × − = ÷ = 76 psi C DSimilarly, we obtain p 11127 psf ; p 10911 psf . (b)Ans= = = ≈77 psi 76 psi 6.131 A water-tunnel test section has a 1-m diameter and flow properties V = 20 m/s, p = 100 kPa, and T = 20°C. The boundary-layer blockage at the end of the section is 9 percent. If a conical diffuser is to be added at the end of the section to achieve maximum pressure recovery, what should its angle, length, exit diameter, and exit pressure be? Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The Reynolds number is very high, Re = ρVd/µ = (998)(20)(1)/(0.001) ≈ 2.0Ε7; much higher than the diffuser data in Fig. 6.28b (Re ≈ 1.2E5). But what can we do (?) Let’s use it anyway: t p,maxB 0.09, read C 0.71 at L/d 25, 2 4 , AR 8:θ= ≈ ≈ ≈ ° ≈ Chapter 6 • Viscous Flow in Ducts 447 Solution: For air at 20°C and 1 atm, take ρ = 1.2 kg/m3 and µ = 1.8E−5 kg/m⋅s. For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The manometer reads o water airp p ( )gh (998 1.2)(9.81)(0.040) 391 Paρ ρ− = − = − ≈ 1/2 1/2 CLTherefore V [2 p/ ] [2(391)/1.2] / (a)Ans.ρ= ∆ = ≈ 25.5 m s We can estimate the friction factor and then compute average velocity from Eq. (6.43): avg CL d m Vd 1.2(21.7)(0.08) Guess V 0.85V 21.7 , then Re 115,700 s 1.8E 5 ρ µ ≈ ≈ = = ≈ − smooth better 25.5 m Then f 0.0175, V 21.69 (converged) s[1 1.33 ≈ = ≈ + √0.0175] 2Thus the volume flow is Q ( /4)(0.08) (21.69) / . . (b)Ansπ= ≈ 30.109 m s 2 2 w f 0.0175 Finally, V (1.2)(21.69) (c) 8 8 Ans.τ ρ= = ≈ 1.23 Pa 6.137 For the 20°C water flow of Fig. P6.137, use the pitot-static arrange- ment to estimate (a) the centerline velocity and (b) the volume flow in the 5-in- diameter smooth pipe. (c) What error in flow rate is caused by neglecting the 1-ft elevation difference? Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. For the manometer reading h = 2 inches, oB A merc waterp p (SG 1)( g) hρ− = − Fig. P6.137 waterg(1 ft) but from the energy equation,ρ+ A B water f-AB water oB B mano f-ABp p gh g(1 ft) Therefore p p (SG 1) gh ghρ ρ ρ ρ− = − − = − + 2 f-ABwhere friction loss h f( L/d)(V /2g)≈ ∆ Thus the pitot tube reading equals the manometer reading (of about 130 psf) plus the friction loss between A and B (which is only about 3 psf), so there is only a small error: 1/2 1/2 CL 2 p 2(130.6) (SG 1) gh (13.56 1)(62.4)(2/12) 130.6 psf, V 1.94 ρ ρ
∆
− = − ≈ ≈ =      448 Solutions Manual • Fluid Mechanics, Fifth Edition CL avg CL ft ft 1.94(9.9)(5/12) or V 11.6 , so V 0.85V 9.9 , Re 381500, s s 2.09E 5 ≈ ≈ ≈ = ≈ − 2 2 smooth frictionso f 0.0138, or p f(L/d) V /2 3.2 lbf/ftρ≈ ∆ = ≈ If we now correct the pitot tube reading to pitotp 130.6 3.2 133.8 psf,∆ ≈ + = we may iterate and converge rapidly to the final estimate: CL avgf 0.01375, V ; Q ; V (a, b)Ans.≈ ≈ ≈ ≈ 3ft ft ft 11.75 1.39 10.17 s s s The error compared to our earlier estimate V 9.91 ft/s is about (c)Ans. ≈ 2.6% 6.138 An engineer who took college fluid mechanics on a pass-fail basis has placed the static pressure hole far upstream of the stagnation probe, as in Fig. P6.138, thus contaminating the pitot measurement ridiculously with pipe friction losses. If the pipe flow is air at 20°C and 1 atm and the manometer fluid is Meriam red oil (SG = 0.827), estimate the air centerline velocity for the given manometer reading of 16 cm. Assume a smooth-walled tube. Fig. P6.138 Solution: For air at 20°C and 1 atm, take ρ = 1.2 kg/m3 and µ = 1.8E−5 kg/m⋅s. Because of the high friction loss over 10 meters of length, the manometer actually shows poB less than pA, which is a bit weird but correct: A oB mano airp p ( )gh [0.827(998) 1.2](9.81)(0.16) 1294 Paρ ρ− = − = − ≈ 2 2 2 A B f oB B CL L V fL V Meanwhile, p p gh f , or p p 1294 V d 2 d 2 2 ρ ρ ρρ− = = − = − = 2 2 CL 10 1.2 1.2 V Guess f 0.02, V 0.85V , whence 0.02 V 1294 0.06 2 2 0.85


≈ ≈ − =           d better m 1.2(33.3)(0.06) Solve for V 33.3 , Re 133000, f 0.0170, s 1.8E 5 ≈ = ≈ ≈ − CL CLV V /[1 1.33 f] 0.852V , repeat to convergence≈ + √ ≈ CLFinally converges, f 0.0164, V 39.87 m/s, V V/0.8546 . Ans.≈ ≈ = ≈ 46.65 m/s Chapter 6 • Viscous Flow in Ducts 449 6.139 Professor Walter Tunnel must measure velocity in a water tunnel. Due to budgetary restrictions, he cannot afford a pitot-static tube, so he inserts a total-head probe and a static-head probe, as shown, both in the mainstream away from the wall boundary layers. The two probes are connected to a manometer. (a) Write an expression for tunnel velocity V in terms of the parameters in the figure. (b) Is it critical that h1 be measured accurately? (c) How does part (a) differ from a pitot-static tube formula? Fig. P6.139 Solution: Write Bernoulli from total-head inlet (1) to static-head inlet (2): 2 1 1 2 2( g ) g g , Solve 2 w o s w o w s w w p p h p z p V z V ρ ρρ ρ ρ − + + = + + = Combine this with hydrostatics through the manometer: 2 3 1 2 3 2, cancel outs w m o w w w wp gh gh p gh gh gh ghρ ρ ρ ρ ρ ρ+ + = + + + 1 3: ( )o s w m wor p p gh ghρ ρ ρ− + = − Introduce this into the expression for V above, for the final result: 32( ) . (a)m wtunnel w gh V Ans ρ ρ ρ −= This is exactly the same as a pitot-static tube—h1 is not important. Ans. (b, c) 6.140 Kerosene at 20°C flows at 18 m3/h in a 5-cm-diameter pipe. If a 2-cm-diameter thin-plate orifice with corner taps is installed, what will the measured pressure drop be, in Pa? Solution: For kerosene at 20°C, take ρ = 804 kg/m3 and µ = 1.92E−3 kg/m⋅s. The orifice beta ratio is β = 2/5 = 0.4. The pipe velocity and Reynolds number are: 2 Q 18/3600 m 804(2.55)(0.05) V 2.55 , Re 53300 A s 1.92E 3( /4)(0.05)π = = = = = − From Eqs. (6.112) and (6.113a) [corner taps], estimate Cd ≈ 0.6030. Then the orifice 452 Solutions Manual • Fluid Mechanics, Fifth Edition (b) From Fig. 6.44, the non-recoverable head loss coefficient is K ≈ 1.8, based on Vt: t 2 t 2 2 loss t Q 0.0150 m V 7.66 , A s(0.025) 998 p K V 1.8 (7.66) (b) 2 2 Ans. π ρ = = ≈
∆ = = ≈   53000 Pa 6.144 Accurate solution of Prob. 6.143, using Fig. 6.41, requires iteration because both the ordinate and the abscissa of this figure contain the unknown flow rate Q. In the spirit of Example 5.8, rescale the variables and construct a new plot in which Q may be read directly from the ordinate. Solve Prob. 6.143 with your new chart. Solution: Figure 6.41 has Cd versus ReD, both of which contain Q: 1/2 1 d D d D4 1/2 4 t Q 4 Q d 2 p C ; Re , then C Re DA [2 p/ (1 )] (1 ) ρ βρζ πµ µρ β ρ β −
∆= = = =  ∆ − −  The quantity ζ is independent of Q—sort of a Q-less Reynolds number. If we plot Cd versus ζ, we should solve the problem of finding an unknown Q when ∆p is known. The plot is shown below. For the data of Prob. 6.143, we compute 1/2 1/2 4 4 2 p 2(75000) m 0.5(998)(0.05)(12.7) 12.7 , 316000 s 0.001(1 ) 998(1 .5 ) ζ ρ β

∆ = = = ≈   − −   From the figure below, read Cd ≈ 0.605 (!) hence Q = CdAt[2∆p/ρ(1 − β 4)]1/2 = 54 m3/h. Fig. P6.144 Chapter 6 • Viscous Flow in Ducts 453 6.145 The 1-m-diameter tank in Fig. P6.145 is initially filled with gasoline at 20°C. There is a 2-cm-diameter orifice in the bottom. If the orifice is suddenly opened, estimate the time for the fluid level h(t) to drop from 2.0 to 1.6 meters. Solution: For gasoline at 20°C, take ρ = 680 kg/m3 and µ = 2.92E−4 kg/m⋅s. The Fig. P6.145 orifice simulates “corner taps” with β ≈ 0, so, from Eq. (6.112), Cd ≈ 0.596. From the energy equation, the pressure drop across the orifice is ∆p = ρgh(t), or 2 d t 4 2 gh Q C A 0.596 (0.02) 2(9.81)h 0.000829 h 4(1 ) ρ π ρ β
= ≈ ≈  − 2 tank tank d dh dh But also Q ( ) A (1.0 m) dt dt 4 dt πυ= − = − = − Set the Q’s equal, separate the variables, and integrate to find the draining time: finalt1.6 final 2.0 0 dh 2 2 1.6 0.001056 dt, or t 283 s 0.001056h [ ] Ans. −− = = = ≈ √

4.7 min 6.146 A pipe connecting two reservoirs, as in Fig. P6.146, contains a thin-plate orifice. For water flow at 20°C, estimate (a) the volume flow through the pipe and (b) the pressure drop across the orifice plate. Solution: For water at 20°C, take ρ = 998 kg/m3 and µ = 0.001 kg/m⋅s. The energy equation should include the orifice head loss and the entrance and exit losses: Fig. P6.146 2 0.6 entr exit orifice V L z 20 m f K , where K 0.5, K 1.0, K 1.5 (Fig. 6.44) 2g d β=
∆ = = +  ≈ ≈ ≈   2 smooth 2(9.81)(20) 392.4 V ; guess f 0.02,V 3.02 m/s [f(100/0.05) 0.5 1.0 1.5] 2000f 3.0 Iterate to f 0.0162, V 3.33 m/s = = ≈ ≈ + + + + ≈ ≈ The final Re = ρVD/µ ≈ 166000, and Q = (π/4)(0.05)2(3.33) ≈ 0.00653 m3/s Ans. (a) 454 Solutions Manual • Fluid Mechanics, Fifth Edition (b) The pressure drop across the orifice is given by the orifice formula: ReD = 166000, β = 0.6, Cd ≈ 0.609 (Fig. 6.41): 1/2 1/2 2 d t 4 4 2 p 2 p Q 0.00653 C A 0.609 (0.03) , 4(1 ) 998(1 0.6 ) p Ans. π ρ β

∆ ∆ = = =      − − ∆ = 100 kPa 6.147 Air flows through a 6-cm-diameter smooth pipe which has a 2 m-long per- forated section containing 500 holes (diameter 1 mm), as in Fig. P6.147. Pressure outside the pipe is sea-level standard air. If p1 = 105 kPa and Q1 = 110 m 3/h, estimate p2 and Q2, assuming that the holes are approximated by thin-plate orifices. Hint: A momentum control volume may be very useful. Fig. P6.147 Solution: For air at 20°C and 105 kPa, take ρ = 1.25 kg/m3 and µ = 1.8E−5 kg/m⋅s. Use the entrance flow rate to estimate the wall shear stress from the Moody chart: 1 1 1 smooth2 Q 110/3600 m 1.25(10.8)(0.06) V 10.8 , Re 45000, f 0.0214 A s 1.8E 5( /4)(0.06)π = = = = ≈ ≈ − 2 2 wall f 0.0214 then V (1.25)(10.8) 0.390 Pa 8 8 τ ρ= = ≈ Further assume that the pressure does not change too much, so ∆porifice ≈ 105000 − 101350 ≈ 3650 Pa. Then the flow rate from the orifices is, approximately, 1/2 1/2 2 d d t 2(3650) 0, C 0.61: Q 500C A (2 p/ ) 500(0.61) (0.001) 4 1.25 πβ ρ
  ≈ ≈ ≈ ∆ =     3 3 2 110 or: Q m /s, so Q 0.0183 0.01225 m /s 3600 ≈ = − ≈0.0183 Then V2 = Q2/A2 = 0.01225/[(π/4)(0.06)2] ≈ 4.33 m/s. A control volume enclosing the pipe walls and sections (1) and (2) yields the x-momentum equation: 2 2 x 1 2 w 2 2 1 1 2 1F p A p A DL m V m V AV AV , divide by A:τ π ρ ρ
= − − = − = −

Chapter 6 • Viscous Flow in Ducts 457 6.150 Gasoline at 20°C flows at 0.06 m3/s through a 15-cm pipe and is metered by a 9-cm-diameter long-radius flow nozzle (Fig. 6.40a). What is the expected pressure drop across the nozzle? Solution: For gasoline at 20°C, take ρ = 680 kg/m and µ = 2.92E−4 kg/m⋅s. Calculate the pipe velocity and Reynolds number: D2 Q 0.06 m 680(3.40)(0.15) V 3.40 , Re 1.19E6 A s 2.92E 4( /4)(0.15)π = = = = ≈ − The ISO correlation for discharge (Eq. 6.114) is used to estimate the pressure drop: 1/2 1/26 6 d D 10 10 (0.6) C 0.9965 0.00653 0.9965 0.00653 Re 1.19E6 0.9919 β
  ≈ − = − ≈    2 4 2 p Then Q 0.06 (0.9919) (0.09) , 4 680(1 0.6 ) Solve Ans. π ∆
= =    − ≈∆p 27000 Pa 458 Solutions Manual • Fluid Mechanics, Fifth Edition 6.151 Ethyl alcohol at 20°C, flowing in a 6-cm-diameter pipe, is metered through a 3-cm-diameter long-radius flow nozzle. If the measured pressure drop is 45 kPa, what is the estimated flow rate in m3/h? Solution: For ethanol at 20°C, take ρ = 789 kg/m3 and µ = 0.0012 kg/m⋅s. Not knowing Re, we estimate Cd ≈ 0.99 and make a first calculation for Q: 2 3 d 4 2(45000) If C 0.99, then Q 0.99 (0.03) 0.00772 m /s 4 789(1 0.5 ) π
≈ ≈ ≈   − D d 4(789)(0.00772) Compute Re 108000, compute C 0.9824 from Eq. 6.114 (0.0012)(0.06)π = ≈ ≈ Therefore a slightly better estimate of flow rate is Q ≈ 0.0766 m3/s. Ans. 6.152 Kerosene at 20°C flows at 20 m3/h in an 8-cm-diameter pipe. The flow is to be metered by an ISA 1932 flow nozzle so that the pressure drop is 7 kPa. What is the proper nozzle diameter? Solution: For kerosene at 20°C, take ρ = 804 kg/m3 and µ =1.92E−3 kg/m⋅s. We cannot calculate the discharge coefficient exactly because we don’t know β, so just estimate Cd: 3 2 d 4 2(7000) 20 m Guess C 0.99, then Q 0.99 (0.08 ) 4 3600 s804(1 ) π β β
≈ ≈ =   − 2 4 1/2 D or: 0.268, solve 0.508, (1 ) 4(804)(20/3600) Re 37000 (1.92E 3)(0.08) β β β π ≈ ≈ − = ≈ − Now compute a better Cd from the ISA nozzle correlation, Eq. (6.115): 1.156 4.1 4.7 d D 10 C 0.99 0.2262 (0.000215 0.001125 0.00249 ) 0.9647 Re β β β
 ≈ − + − + ≈    Iterate once to obtain a better β ≈ 0.515, d = 0.515(8 cm) ≈ 4.12 cm Ans. Chapter 6 • Viscous Flow in Ducts 459 6.153 Two water tanks, each with base area of 1 ft2, are connected by a 0.5-in- diameter long-radius nozzle as in Fig. P6.153. If h = 1 ft as shown for t = 0, estimate the time for h(t) to drop to 0.25 ft. Solution: For water at 20°C, take ρ = 1.94 slug/ft3 and µ = 2.09E−5 slug/ft⋅s. For a long-radius nozzle with β ≈ 0, guess Cd ≈ 0.98 and Kloss ≈ 0.9 from Fig. 6.44. The elevation difference h must balance the head losses in the nozzle and submerged exit: Fig. P6.153 2 2 t t loss nozzle exit t V V z h K (0.9 1.0 ) h, solve V 5.82 h 2g 2(32.2) ∆ =
=
= + = = 2 t tank 1/2 dh dh hence Q V 0.00794 h A 0.5 4 12 dt dt π

= ≈ = − = −       1 2 The boldface factor 1/2 accounts for the fact that, as the left tank falls by dh, the right tank rises by the same amount, hence dh/dt changes twice as fast as for one tank alone. We can separate and integrate and find the time for h to drop from 1 ft to 0.25 ft: ( )finalt1.0 final 0.25 0 2 1 0.25dh 0.0159 dt, or: t 0.0159h Ans. − = = ≈

63 s 6.154 Water at 20°C flows through the orifice in the figure, which is monitored by a mercury manometer. If d = 3 cm, (a) what is h when the flow is 20 m3/h; and (b) what is Q when h = 58 cm? Solution: (a) Evaluate V = Q/A = 2.83 m/s and ReD = ρVD/µ = 141,000, β = 0.6, thus Cd ≈ 0.613. Fig. P6.154 2 2 4 4 20 2 2(13550 998)(9.81) (0.613) (0.03) 3600 4 4(1 ) 998(1 0.6 ) d p h Q C d π π ρ β ∆ −= = = − − where we have introduced the manometer formula ∆p = (ρmercury − ρwater)gh. Solve for: 0.58 m (a)Ans.≈ =h 58 cm