Answers to selected problems in P = 0 F. M. White, Fluid ... · Answers to selected problems in F....
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Answers to selected problems in F. M. White, Fluid Mechanics, 7th Edition in SI Units, McGraw-Hill, 2011 (Ch. 1 − 10) February 27, 2013 P1.50 (a) V t = hW sin θ/(µA); W = mg, (b) V ≃ 15 m/s(µ =0.29 Pa s). P1.55 (a) yes, (b) µ ≈ 0.40 Pa s. P1.59 M φ = πµΩR 4 /h. P1.61 µ =3M sin θ/(2πΩR 3 ). P1.65 µ =0.040 Pa s; last two points are not for laminar flow, which requires (Re = ρV 2r 0 /µ ≤ 2100; laminar for point no. 3 means that ρ ≤ 878 kg/m 3 ; candidate: SAE 10W oil at around 40 ◦ C). P1.98 (a) µ =0.29 Pa s, (b) ±4.4%. P1.99 Ω = 600 rpm ⇒ µ = (0.29 ± 0.020) Pa s. P1.100 δC D /C D = ±5.6%. FE1.6 (e) ρLV 2 /σ (= We, Weber number). FE2.2 (b) 129 m. FE2.3 (c) p = 107 kPa. FE2.8 (d) F B = 653 N. P3.18 Q =3 U o bδ/8. P3.29 (a) ρ 02 /ρ 01 = exp(−αA √ RT 0 t/V 0 ), (b) t =0.42V 0 /(A √ RT 0 ). P3.60 (a) F x = ρA j (V j − V c ) 2 (1 − cos θ), (b) P = V c F x , (c) V c = 0, (d) V c = V j /3. P3.67 F bolts =3.1 kN. P3.77 D =4.3 kN; C D =2/3. P3.82 F = 15 kN P3.107 V 2 = V 1 h 1 /h 2 ; h 2 /h 1 =[ −1 + (1 + 8Fr 2 1 ) 1/2 ]/2, Fr = V/ √ gh P3.121 (a) V 1 =0.90 m/s; V 2 = 10 m/s, (b) h =0.40 m. P3.133 (a) Q =5.6 dm 3 /s, (b) Q =2.5 dm 3 /s. P3.177 ˙ W = 33.7 kW. P3.182 (a) P turbine =0.30 MW, (b) P pump =0.40 MW. P4.94 Combined Couette-Poiseuille flow; u(0) = U , u(h)=0 ⇒ u = U (1 − η) − Cη (1 − η), η = y/h, h = h(x), C = h 2 2µ dp dx . P4.95 u z = C (2σ 2 ln η − η 2 + 1), d¨ ar C = ga 2 /(4ν ), η = r/a och σ = b/a; Q = C 2πa 2 (−3σ 4 − 1+4σ 2 + 4σ 4 ln σ)/4. P4.97 Q/b = VA = 31 cm 2 /s (h = 8 mm, Re = ρV 2h/µ = 51). P4.98 (a) Re = 1190 < 2100, (b) Q = 98.5m 3 /h, (c) h = 99 mm. P5.5 (a) F =0.94 kN, (b) P = 27 kW. P5.54 P ≈ 95 kW, f S = (5 ± 1) Hz (C D ≈ 0.30, St = 0.24 ± 0.04). P5.64 (a) ∆p = 34 kPa, (b) P =0.41 kW. P5.66 (a) u/u ∗ ≃ 8.9(u ∗ y/ν ) 0.13 ; u ∗ = √ τ w /ρ, (b) τ w =2.6 Pa. P5.69 F p =0.68 kN. For the last two points (Π 2 = Re m = 2.6 × 10 5 , 2.8 × 10 5 ), the non-dimensional force (force coefficient) is Π 1 = F/(ρV 2 L 2 )=0.975 (±0.001); prototype: Re p =8.3 × 10 5 . The expected scenario for all bodies is that the drag coefficient becomes a constant at high enough Re; extrapolation with Π 1 = 0.975 for Re > 2.8 × 10 5 gives F p =0.68 kN. P5.85 F tot ≃ 0.46 MN. P6.36 (a) upwards, (b) Q =1.9m 3 /h (Re = 68). P6.58 p 1 − p atm =2.4 MPa (f ≃ 0.0136). P6.73 (a) ∆p = 56 kPa, (b) Q = 85 m 3 /h, (c) u =3.5m/s(C =1.1). P6.87 Q =0.86 m 3 /s(f ≃ 0.0191), with the assumption of a free outlet, pumping from a large basin and without elevation change (∆p f +ρV 2 /2= ρgh P ); ∆p f = ρgh P ⇒ Q =0.90 m 3 /s. P6.95 Q = 20 m 3 /h(f ≃ 0.0263, D eff = 20.3 mm). P6.101 (a) Q =0.43 m 3 /s(D eff = 337 mm), (b) p − p a = −6.3 Pa. P6.113 Q = 25 dm 3 /s(f a ≃ 0.0389, f b ≃ 0.0321); ϵ = 0.046 mm ⇒ Q = 31 dm 3 /s(f a ≃ 0.0240, f b ≃ 0.0222). P6.118 (a) Q =7.7m 3 /h (0.3% more), (b) Q = 38 m 3 /h (5 times more). P6.121 P =0.84 kW. P6.126 Q 1 =0.023 m 3 /s; Q 2 =0.013 m 3 /s; ∆p =2.2 MPa. P6.152 u CL ≃ 47 m/s(u CL /V =1+1.3 √ f , f ≃ 0.0162). P6.161 (a) ϵ =0 ⇒ Q = 21 m 3 /h=5.7×10 −3 m 3 /s, ( ∑ K = 12.9), (b) D : 1 2 D ⇒ ∆p = 76 kPa (C d =0.609). P7.28 F a =2.83F 1 ; F b =2.0F 1 . P7.46 (a) U ≈ 34 m/s, (b) δ ≈ 36 mm, (c) u ≈ 26 m/s. P7.51 x sep /L =0.158. P7.53 θ sep =2.3 ◦ . P7.63 (a) V 3 + (2V w + V α )V 2 + V 2 w V − (V 3 nw + V α V 2 nw ) = 0, V α =2C RR /(ρC D A), (b) V ≈ 7.4m/s, (c) D ∝ V 2 rel . P7.84 θ ≃ 72 ◦ . P7.97 Maximal cone angle θ ≃ 60 ◦ (C D,rod ≃ 1.2). P7.127 (a) V stall = 15 m/s(z = 1200 m), (b) θ min =1.6 ◦ (C D,∞ =0.007, C D,fuselage = 0), (c) gliding dis- tance 44 km. C D,fuselage =0.002 ⇒ (b) θ min =2.0 ◦ (C D,∞ =0.009), and (c) approx. 34 km. P7.128 (a) V min =6.7m/s, (b) V max = 15.0m/s (130 hp = 96.9 kW ⇒ V max = 13.5m/s). P8.12 a = 1.27 cm; cavitation appears when U ∞ > 18.7m/s. 1
Transcript of Answers to selected problems in P = 0 F. M. White, Fluid ... · Answers to selected problems in F....
Answers to selected problems inF. M. White, Fluid Mechanics, 7th Edition in SI
Units, McGraw-Hill, 2011 (Ch. 1− 10)February 27, 2013
P1.50 (a) Vt = hW sin θ/(µA); W = mg,(b) V ≃ 15 m/s (µ = 0.29 Pa s).
P1.55 (a) yes, (b) µ ≈ 0.40 Pa s.
P1.59 Mφ = πµΩR4/h.
P1.61 µ = 3M sin θ/(2πΩR3).
P1.65 µ = 0.040 Pa s; last two points are not for laminarflow, which requires (Re = ρV 2r0/µ ≤ 2100; laminarfor point no. 3 means that ρ ≤ 878 kg/m3; candidate:SAE 10W oil at around 40C).
P1.98 (a) µ = 0.29 Pa s, (b) ±4.4%.
P1.99 Ω = 600 rpm ⇒ µ = (0.29± 0.020) Pa s.
P1.100 δCD/CD = ±5.6%.
FE1.6 (e) ρLV 2/σ (= We, Weber number).
FE2.2 (b) 129 m.
FE2.3 (c) p = 107 kPa.
FE2.8 (d) FB = 653 N.
P3.18 Q = 3Uobδ/8.
P3.29 (a) ρ02/ρ01 = exp(−αA√RT0t/V0),
(b) t = 0.42V0/(A√RT0).
P3.60 (a) Fx = ρAj(Vj − Vc)2(1− cos θ),
(b) P = VcFx, (c) Vc = 0,(d) Vc = Vj/3.
P3.67 Fbolts = 3.1 kN.
P3.77 D = 4.3 kN; CD = 2/3.
P3.82 F = 15 kN
P3.107 V2 = V1 h1/h2;h2/h1 = [−1 + (1 + 8Fr21)
1/2 ]/2, Fr = V/√gh
P3.121 (a) V1 = 0.90 m/s; V2 = 10 m/s, (b) h = 0.40 m.
P3.133 (a) Q = 5.6 dm3/s, (b) Q = 2.5 dm3/s.
P3.177 W = 33.7 kW.
P3.182 (a) Pturbine = 0.30 MW, (b) Ppump = 0.40 MW.
P4.94 Combined Couette-Poiseuille flow;u(0) = U , u(h) = 0 ⇒ u = U(1− η)− C η (1− η),
η = y/h, h = h(x), C = h2
2µdpdx .
P4.95 uz = C (2σ2 ln η − η2 + 1), dar C = g a2/(4ν), η =r/a och σ = b/a; Q = C 2πa2(−3σ4 − 1 + 4σ2 +4σ4 lnσ)/4.
P4.97 Q/b = V A = 31 cm2/s(h = 8 mm, Re = ρV 2h/µ = 51).
P4.98 (a) Re = 1190 < 2100, (b) Q = 98.5 m3/h,(c) h = 99 mm.
P5.5 (a) F = 0.94 kN, (b) P = 27 kW.
P5.54 P ≈ 95 kW, fS = (5± 1) Hz(CD ≈ 0.30, St = 0.24± 0.04).
P5.64 (a) ∆p = 34 kPa, (b) P = 0.41 kW.
P5.66 (a) u/u∗ ≃ 8.9(u∗y/ν)0.13; u∗ =√τw/ρ,
(b) τw = 2.6 Pa.
P5.69 Fp = 0.68 kN. For the last two points (Π2 = Rem =2.6× 105, 2.8× 105), the non-dimensional force (forcecoefficient) is Π1 = F/(ρ V 2L2) = 0.975 (±0.001);prototype: Rep = 8.3 × 105. The expected scenariofor all bodies is that the drag coefficient becomes aconstant at high enough Re; extrapolation with Π1 =0.975 for Re > 2.8× 105 gives Fp = 0.68 kN.
P5.85 Ftot ≃ 0.46 MN.
P6.36 (a) upwards, (b) Q = 1.9 m3/h (Re = 68).
P6.58 p1 − patm = 2.4 MPa (f ≃ 0.0136).
P6.73 (a) ∆p = 56 kPa, (b) Q = 85 m3/h,(c) u = 3.5 m/s (C = 1.1).
P6.87 Q = 0.86 m3/s (f ≃ 0.0191), with the assumption ofa free outlet, pumping from a large basin and withoutelevation change (∆pf+ρV 2/2 = ρghP); ∆pf = ρghP
⇒ Q = 0.90 m3/s.
P6.95 Q = 20 m3/h (f ≃ 0.0263, Deff = 20.3 mm).
P6.101 (a) Q = 0.43 m3/s (Deff = 337 mm),(b) p− pa = −6.3 Pa.
P6.113 Q = 25 dm3/s (fa ≃ 0.0389, fb ≃ 0.0321); ϵ =0.046 mm ⇒ Q = 31 dm3/s (fa ≃ 0.0240, fb ≃0.0222).
P6.118 (a) Q = 7.7 m3/h (0.3% more),(b) Q = 38 m3/h (5 times more).
P6.121 P = 0.84 kW.
P6.126 Q1 = 0.023 m3/s; Q2 = 0.013 m3/s;∆p = 2.2 MPa.
P6.152 uCL ≃ 47 m/s (uCL/V = 1 + 1.3√f , f ≃ 0.0162).
P6.161 (a) ϵ = 0 ⇒Q = 21 m3/h = 5.7×10−3 m3/s, (∑
K =12.9), (b) D : 1
2D ⇒ ∆p = 76 kPa (Cd = 0.609).
P7.28 Fa = 2.83F1; Fb = 2.0F1.
P7.46 (a) U ≈ 34 m/s, (b) δ ≈ 36 mm,(c) u ≈ 26 m/s.
P7.51 xsep/L = 0.158.
P7.53 θsep = 2.3.
P7.63 (a) V 3 + (2Vw + Vα)V2 + V 2
wV − (V 3nw + VαV
2nw) = 0,
Vα = 2CRR/(ρCDA), (b) V ≈ 7.4 m/s, (c) D ∝ V 2rel.
P7.84 θ ≃ 72.
P7.97 Maximal cone angle θ ≃ 60 (CD,rod ≃ 1.2).
P7.127 (a) Vstall = 15 m/s (z = 1200 m), (b) θmin = 1.6
(CD,∞ = 0.007, CD,fuselage = 0), (c) gliding dis-tance 44 km. CD,fuselage = 0.002 ⇒ (b) θmin = 2.0
(CD,∞ = 0.009), and (c) approx. 34 km.
P7.128 (a) Vmin = 6.7 m/s, (b) Vmax = 15.0 m/s (130 hp =96.9 kW ⇒ Vmax = 13.5 m/s).
P8.12 a = 1.27 cm; cavitation appears when U∞ >18.7 m/s.
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P8.16 Irrotational outer region (line vortex): p = p∞ −ρω2R4/(2r2); rotational inner region (solid-body ro-tation): p = p∞ − ρω2R2 + ρω2r2/2; lowest pressureat vortex center, p(0) = p∞ − ρω2R2.
P8.17 (a) pmin = 97.2 kPa, (b) p(r = R) = 99.3 kPa.
P8.27 V = 11.3 m/s, θ = 44.2.
P8.56 h = 3a/2, Umax = 5U∞/4.
P8.80 Without walls: u = v = K/(2a);with walls: u = 8K/(15a), v = 4K/(15a).
P8.89 (a) V = 4.5 m/s,(b) CL = 1.1,(c) P = 0.41 kW (CD = 0.022).
P8.93 (a) b = 26 m, (b) AR = 8.7, (c) Di = 1.6 kN.
P8.94 (a) C = 0.22 m, (b) P = 1.1 kW,(c) V = 20 m/s (CL = 0.053, α = −1.7).
P8.107 U∞ = 14.1 m/s ⇒ p(θ = 90) = pv = 2.34 kPa(cavitation); pA = 115 kPa.
P8.108 (a) Vf = 2.0 m/s, (b) t = 0.64 s; ca. +6%.
P9.29 (a) m = 0.17 kg/s, (b) Mae = 0.90.
P9.45 (a) V2 = 806 m/s, (b)Ma2 = 2.40, (c) T2 = 281 K,(d) m = 0.429 kg/s, (e) yes, At = A∗ = 8.32 cm2.
P9.57 (a) p2 = 261 kPa, (b) p3 = 302 kPa,(c) A∗
2 = 16.5 cm2, (d) Ma3 = 0.34.
P9.62 Ma1 = 1.92; V1 ≃ 585 m/s
P9.65 (a) p3 = 59 kPa, (b) m = 0.024 kg/s.
P9.84 Ve ≃ 110 m/s; normal shock.
P9.88 V2 = 107 m/s; p2 = 371 kPa; T2 = 330 K;p02 = 395 kPa; L∗
2 = 9.36 m.
P9.93 (a) Ma1 = 0.30, (b) p = 1.03 MPa.
P9.101 Adiabatic flow: m = 0.26 kg/s (f = 0.0142).
P9.110 (a) q = 0.58 MJ/kg, (b) Ma2 = 0.71,(c) T2 ≃ 733 K.
P9.111 New inlet Mach number, Ma1 = 0.20;mnew/mold = 0.68.
P9.116 (a) V2 ≃ 163 m/s, (b) T2 ≃ 464 K,(c) q = 0.49 MJ/kg.
P9.128 (a) Ma1 = 1.87, (b) p2 = 293 kPa, (c) T2 = 404 K,(d) V2 = 415 m/s.
P9.145 Exact: Ma2 = 3.274; p2 = 66.7 kPa;linear theory: p2 = 61 kPa.
P9.146 (a) Ma = 2.64, p = 60.3 kPa, (b) weak shock:Ma = 2.30, p = 32.5 kPa; strong: Ma = 0.522,p = 74.0 kPa.
P10.19 Q = 5.2 m3/s (n = 0.015).
P10.25 S0 = 0.38 m/km (0.022; n = 0.014).
P10.54 Subcritical flow in both cases (yc < yn);b = 1.2 m ⇒ yc = 1.44 m < yn = 2.90 m;b = 2.4 m ⇒ yc = 0.91 m < yn = 1.26 m.
P10.65 (a) V2 = 1.13 m/s,(b) q = 0.566 m2/s (Fr1 = 0.37).
P10.81 (a) H = 0.24 m, (b) y2 = 0.15 m (Fr2 = 4.8).
P10.93 y2 = 0.24 m, y3 = 1.53 m, hf/E1 = 47%.
P10.104 (a) V4 = 2.6 m/s, (b) y4 = 20 cm,(c) V1 = 3.1 m/s, (d) y1 = 16.5 cm.
Christoffer Norberg, tel. 046-2228606Department of Energy SciencesLund Institute of Technology, [email protected]
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