Unformatted text preview: 438 Solutions Manual • Fluid Mechanics, Fifth Edition Mercury Glycerin 13,550 1260 1.56E3 1.49 0.046 470. Clearly there are vast differences between fluid properties and their effects on flows. 6.7 Cola, approximated as pure water at 20°C, is to fill an 8oz container (1 U.S. gal = 128 fl oz) through a 5mmdiameter tube. Estimate the minimum filling time if the tube flow is to remain laminar. For what cola (water) temperature would this minimum time be 1 min? Solution: For cola “water”, take ρ = 998 kg/m3 and μ = 0.001 kg/m⋅s. Convert 8 fluid ounces = (8/128)(231 in3) ≈ 2.37E−4 m3. Then, if we assume transition at Re = 2300, Re crit = 2300 = ρ VD 4 ρ Q 2300π (0.001)(0.005) m3 , or: Q crit = = ≈ 9.05E−6 4(998) s μ πμ D Then Δtfill = υ/Q = 2.37E−4/9.05E−6 ≈ 26 s Ans. (a) (b) We fill in exactly one minute if Qcrit = 2.37E−4/60 = 3.94E−6 m3/s. Then
Q crit = 3.94E−6 m 3 2300πν D = s 4 if ν water ≈ 4.36E−7 m 2 /s From Table A1, this kinematic viscosity occurs at T ≈ 66°C Ans. (b) 6.8 When water at 20°C (ρ = 998 kg/m3, μ = 0.001 kg/m⋅s) flows through an 8cmdiameter pipe, the wall shear stress is 72 Pa. What is the axial pressure gradient (∂ p/∂ x) if the pipe is (a) horizontal; and (b) vertical with the flow up? Solution: Equation (6.9b) applies in both cases, noting that τw is negative: (a) Horizontal: dp 2τ w 2(−72 Pa ) Pa = = = −3600 dx R m 0.04 m Ans. (a) (b) Vertical, up: dp 2τ w dz 1 Pa = −3600 − 998(9.81) = −13, 400 = − ρg dx R dx m Ans. (b) Chapter 6 • Viscous Flow in Ducts 439 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 1m intervals along the pipe, as follows: x, m: p, kPa: 0 304 1 273 2 255 3 240 4 226 5 213 6 200 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 Δz = 0, is defined by Eq. (6.8) of the text:
hf = Δp 304,000 − 200,000 Pa = = 11.2 m ρ g (950 kg/m 3 )(9.81 m/s 2 ) Ans. (a) (b) The wall shear stress in the fullydeveloped region is defined by Eq. (6.9b): Δp  fully developed = 13000 Pa = 4τ w = 4τ w , solve for τ w = 163 Pa ΔL 1m 0.05 m d
(c) The overall friction factor is defined by Eq. (6.10) of the text:
foverall = h f , overall
2 d 2g ⎛ 0.05 m ⎞ 2(9.81 m/s ) = (11.2 m) ⎜ = 0.0182 ⎟ 2 L V2 ⎝ 6 m ⎠ (10 m/s) Ans. (b) Ans. (c) NOTE: The fullydeveloped friction factor is only 0.0137. 6.10 Water at 20°C (ρ = 998 kg/m3) flows through an inclined 8cmdiameter 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?
B B Solution: Guess that the flow is from A to B and write the steady flow energy equation:
2 pA VA p V2 186000 260000 + + z A = B + B + zB + h f , or: + 24.5 = + 9.1 + h f , ρg 2 g ρg 2 g 9790 9790 or: 43.50 = 35.66 + h f , solve: h f = +7.84 m Yes, flow is from A to B. Ans. (a, b) 442 Solutions Manual • Fluid Mechanics, Fifth Edition 6.14 Water at 20°C is to be siphoned through a tube 1 m long and 2 mm in diameter, as in Fig. P6.14. Is there any height H for which the flow might not be laminar? What is the flow rate if H = 50 cm? Neglect the tube curvature. Fig. P6.14 Solution: For water at 20°C, take ρ = 998 kg/m3 and μ = 0.001 kg/m⋅s. Write the steady flow energy equation between points 1 and 2 above:
patm 02 p V2 V2 32 μ L V + + z1 = atm + tube + z 2 + h f , or: H − = hf = 2g 2g ρg 2g ρg ρgd 2 Enter data in Eq. (1): 0.5 − V2 32(0.001)(1.0)V m = , solve V ≈ 0.590 2 2(9.81) (998)(9.81)(0.002) s (1) Equation (1) is quadratic in V and has only one positive root. The siphon flow rate is
m3 m3 Q H=50 cm = (0.002) (0.590) = 1.85E−6 ≈ 0.0067 if H = 50 cm Ans. 4 s h Check Re = (998)(0.590)(0.002) /(0.001) ≈ 1180 (OK, laminar flow)
2 π It is possible to approach Re ≈ 2000 (possible transition to turbulent flow) for H < 1 m, for the case of the siphon bent over nearly vertical. We obtain Re = 2000 at H ≈ 0.87 m. 6.15 Professor Gordon Holloway and his students at the University of New Brunswick went to a fastfood emporium and tried to drink chocolate shakes (ρ ≈ 1200 kg/m3, μ ≈ 6 kg/m⋅s) through fat straws 8 mm in diameter and 30 cm long. (a) Verify that their human lungs, which can develop approximately 3000 Pa of vacuum pressure, would be unable to drink the milkshake through the vertical straw. (b) A student cut 15 cm from his straw and proceeded to drink happily. What rate of milkshake flow was produced by this strategy? Solution: (a) Assume the straw is barely inserted into the milkshake. Then the energy equation predicts p1 V1 p2 V22 = +z = = + z + hf ρg 2 g 1 ρg 2 g 2 = 0+0+0 = V2 (−3000 Pa) + tube + 0.3 m + h f (1200 kg/m3 )(9.81 m/s 2 ) 2 g
V2 tube < 0 which is impossible 2g
2 Solve for h f = 0.255 m − 0.3 m − Ans. (a) 444 Solutions Manual • Fluid Mechanics, Fifth Edition Solution: (a) Assume no pressure drop and neglect velocity heads. The energy equation reduces to: p1 V12 p V2 + + z1 = 0 + 0 + ( L + l ) = 2 + 2 + z2 + h f = 0 + 0 + 0 + h f , or: h f ≈ L + l ρ g 2g ρ g 2g For laminar flow, h f = 128μ LQ πρgd 4 and, for uniform draining, Q = υ
Δt Solve for Δt = 128 μ Lυ πρ gd 4 ( L + l ) Ans. (a) (b) Apply to Δt = 6 s. For water, take ρ = 998 kg/m3 and μ = 0.001 kg/m⋅s. Formula (a) predicts: Δt = 6 s = 128(0.001 kg/m⋅s)(0.12 m)(8E−6 m3 ) , π (998 kg/m3 )(9.81 m/s2 )d 4 (0.12 + 0.02 m) Ans. (b) Solve for 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 30cmlong 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:
V= Q 1.9 E−6 m 3 /s m = = 0.605 2 A (π /4)(0.002 m) s Evaluate ρ liquid = 0.95(998) = 948 kg/m3. Write the energy equation between the top surface and the tube exit:
2 pa Vtop p V2 = + ztop = a + + 0 + hf , ρg 2 g ρg 2 g V 2 32μ LV (0.605)2 32μ (0.3)(0.605) = + + or: 0.42 = 2 2g 2(9.81) 948(9.81)(0.002)2 ρgd Chapter 6 • Viscous Flow in Ducts 445 Note that “L” in this expression is the tube length only (L = 30 cm). Solve for μ = 0.00257 kg (laminar flow) Ans. m⋅ s ρVd 948(0.605)(0.002) Red = = = 446 (laminar ) μ 0.00257 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 steadyflow energy:
patm 02 p V2 + + z1 = atm + 2 + z 2 + h f , ρg 2g ρg 2g
Fig. P6.19 where V2 = Q 35/3600 ft = ≈ 7.13 2 A π (0.25 /12) s
Solve h f = z1 − z2 −
2 V2 (7.13)2 = 10 − = 9.21 ft 2g 2(32.2) Assuming laminar pipe flow, use Eq. (6.12) to relate head loss to viscosity:
128ν LQ 128(6)(35/3600)ν μ ft 2 h f = 9.21 ft = , solve ν = ≈ 3.76E−4 = ρ s π gd 4 π (32.2)(0.5/12)4 Ans. Check Re = 4Q/(πν d) = 4(35/3600)/[π (3.76E−4)(0.5/12)] ≈ 790 (OK, laminar) P6.20 The oil tanks in Tinyland are only 160 cm high, and they discharge to the Tinyland oil truck through a smooth tube 4 mm in diameter and 55 cm long. The tube exit is open to the atmosphere and 145 cm below the tank surface. The fluid is medium fuel oil, ρ = 850 kg/m3 and μ = 0.11 kg/ms. Estimate the oil flow rate in cm3/h. Solution: The steady flow energy equation, with 1 at the tank surface and 2 the exit, gives Chapter 6 • Viscous Flow in Ducts 449 6.25 For the configuration shown in Fig. P6.25, the fluid is ethyl alcohol at 20°C, and the tanks are very wide. Find the flow rate which occurs in m3/h. Is the flow laminar? Solution: For ethanol, take ρ = 789 kg/m3 and μ = 0.0012 kg/m⋅s. Write the energy equation from upper free surface (1) to lower free surface (2): Fig. P6.25 p1 V12 p V2 + + z1 = 2 + 2 + z 2 + h f , with p1 = p2 and V1 ≈ V2 ≈ 0 ρg 2g ρg 2g 128μ LQ 128(0.0012)(1.2 m)Q = Then h f = z1 − z 2 = 0.9 m = πρgd 4 π (789)(9.81)(0.002)4
Solve for Q ≈ 1.90E−6 m 3 /s = 0.00684 m 3 /h. Ans. Check the Reynolds number Re = 4ρQ/(πμd) ≈ 795 − OK, laminar flow. ________________________________________________________________________ za = 22 m
P6.26 Two oil tanks are connected by
two 9mlong pipes, as in Fig. P6.26. Pipe 1 is 5 cm in diameter and is 6 m higher than pipe 2. It is found that the flow rate in pipe 2 is twice as large as the flow in pipe 1. (a) What is the diameter
SAE 30W oil at 2 0 °C D2 6m D1 = 5 cm zb =
15 m L=9m Fig. P6.26
of pipe 2? (b) Are both pipe flows laminar? (c) What is the flow rate in pipe 2 (m3/s)? 6.49 The tankpipe 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 V= Q 11/3600 m ρVd 998(4.32)(0.03) = = 4.32 ; Re = = = 129000 2 A (π /4)(0.03) s 0.001 μ
(4.32)2 = 3.05 m 2(9.81) The energy equation yields the value of the head loss: patm V12 p V2 + + z1 = atm + 2 + z 2 + h f ρg 2g ρg 2g But also h f = f or h f = 4 − 2 L V2 ⎛ 5.0 ⎞ (4.32) , or: 3.05 = f ⎜ , solve for f ≈ 0.0192 ⎝ 0.03 ⎟ 2(9.81) ⎠ d 2g With f and Re known, we can find ε/d from the Moody chart or from Eq. (6.48): ⎡ ε /d ⎤ 1 2.51 ε = −2.0 log10 ⎢ + , solve for ≈ 0.000394 1/2 1/2 ⎥ d (0.0192) ⎣ 3.7 129000(0.0192) ⎦
Then ε = 0.000394(0.03) ≈ 1.2E−5 m ≈ 0.012 mm (very smooth) Ans. 6.50 Ethanol at 20°C flows at 125 U.S. gal/min through a horizontal castiron pipe with L = 12 m and d = 5 cm. Neglecting entrance effects, estimate (a) the pressure gradient, dp/dx; (b) the wall shear stress, τw; and (c) the percent reduction in friction factor if the pipe walls are polished to a smooth surface. Solution: For ethanol (Table A3) take ρ = 789 kg/m3 and μ = 0.0012 kg/m⋅s. Convert 125 gal/min to 0.00789 m3/s. Evaluate V = Q/A = 0.00789/[π (0.05)2/4] = 4.02 m/s.
Red = ρVd 789(4.02)(0.05) ε 0.26 mm = = 132,000, = = 0.0052 Then f Moody ≈ 0.0314 μ 0.0012 d 50 mm
(b) τ w = f 0.0314 (789)(4.02)2 = 50 Pa Ans. (b) ρV 2 = 8 8 4τ dp −4(50) Pa =− w = = −4000 Ans. (a) 0.05 dx d m (a) ...
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This note was uploaded on 05/18/2011 for the course MEEG 332 taught by Professor Staff during the Spring '08 term at University of Delaware.
 Spring '08
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