A selection of answers to problems in F. M. White, Fluid Mechanics ...

P4.84 u = C (2σ 2 ln η − η 2 + 1), where C = g a 2 /(4ν),
A selection of answers to problems in
η = y/h, h = h(x), C = h2 dp
2µ dx . P6.110 P = 0.84 kW.
F. M. White, Fluid Mechanics, 6th edition,
McGraw-Hill, 2008 (Ch. 1 − 10)
February 25, 2010
Chapter 1
P1.45 (a) V t = hW sin θ/(µA); W = mg,
(b) V ≃ 15 m/s (µ = 0.29 Pa s).
P1.50 (a) yes, (b) µ ≈ 0.40 Pa s.
P1.54 M ϕ = πµΩR 4 /h.
P1.56 µ = 3M sin θ/(2πΩR 3 ).
P1.58 µ = 0.040 Pa s; the last two points indicate nonlaminar
flow; laminar flow is guaranteed if Re =
ρV 2r 0 /µ ≤ 2100).
P1.88 (a) µ = 0.29 Pa s, (b) ±4.4%.
P1.89 Ω = 600 rpm ⇒ µ = (0.29 ± 0.020) Pa s.
P1.90 δC D /C D = ±5.6%.
z
η = r/a and σ = b/a; Q = C 2πa 2 (−3σ 4 − 1 + 4σ 2 +
4σ 4 ln σ)/4.
P4.86 Q/b = V A = 31 cm 2 /s
(h = 8 mm, Re = ρV 2h/µ = 51).
P4.87 (a) Re = 1190 < 2100, (b) Q = 98.5 m 3 /h,
(c) h = 99 mm.
Chapter 5
P5.6 D ≃ 0.46 kN, linear interpolation using C D .
D ≃ 0.45 kN, linear interpolation using D/(µV d).
P5.24 F/(ρ V 2 L 2 ) = f(α, Re, L/D, Ma),
Re = ρ V L/µ, Ma = V/a.
P5.25 Ex. F ρ −1 Ω −2 D −4 = g(ΩD/V, ρΩD 2 /µ).
P5.51 P ≈ 0.13 MW, f S = (5 ± 1) Hz
(C D ≈ 0.39, St = 0.25 ± 0.05).
P5.62 Q m = 0.102 m 3 /s; P p ≃ 117 kW
FE1.6 (e) ρLV 2 (ρ
/σ (= We, Weber number).
m = 1.204 kg/m 3 , ρ p = 998 kg/m 3 ).
P5.65 (a) u/u ∗ ≃ 8.9(u ∗ y/ν) 0.13 ; u ∗ = √ τ w /ρ,
Chapter 2
(b) τ w = 2.6 Pa.
FE2.2 (b) 129 m.
FE2.3 (c) p = 107 kPa.
FE2.8 (d) F B = 653 N.
Chapter 3
P3.16 Q = 3 U o bδ/8.
P3.21 (a) Q = U o b δ/8, (b) v w /U o = 3/1000.
P3.51 (i) P max = 8 ρ A j Vj 3/27
j/3;
(ii) P max = ρ A j Vj 3/2
j/2.
P3.53 F = πR 2 [ p 1 − p 2 − ρ U 2 o (β 2 − 1) ],
(a) lam. β 2 = 4/3, (b) turb. β 2 ≃ 1.02.
P3.62 F bolts = 3.1 kN.
P3.72 D = 4.3 kN; C D = 2/3.
P3.77 F = 15 kN.
P5.70 F p = 0.68 kN. For the last two points (Π 2 = Re m =
2.7 × 10 5 , 2.9 × 10 5 ), the dimensionless force (drag
coefficient) is a constant, Π 1 = F/(ρ V 2 L 2 ) = 0.966.
For a diamond-shaped body and high enough Re it
is expected to have a constant drag coefficient (the
form drag dominates and separation is fixed to the
sharp corners). Extrapolation using Π 1 = const. for
Re > 2.9 × 10 5 gives F p = 0.68 kN.
P5.76 F tot = 0.48 MN.
Chapter 6
P6.14 (a) Yes it might be, 0.98 m < H < L = 1.00 m ⇒
Re > 2100,
(b) Q H=50 cm = 6.5 dm 3 /h (Re = 1.1 × 10 3 ).
P6.32 (a) upwards, (b) Q = 1.9 m 3 /h (Re = 68).
P6.36 (a) τ w,avg = 1.4 Pa (u ∗ avg = 1.1 m/s),
(b) u ≃ 21.3 m/s.
P3.130 Ẇ = 33.7 kW.
P6.52 p 1 − p atm = 2.4 MPa (f ≃ 0.0136).
P3.131 (a) V 2 = 28 m/s, (b) Q = 0.055 m 3 /s.
P3.151 (a) V 1 = 0.90 m/s; V 2 = 10 m/s, (b) h = 0.40 m.
P3.163 (a) Q = 5.6 dm 3 /s, (b) Q = 2.5 dm 3 /s.
P3.173 (a) p 2 = 82 kPa, (b) p 3 = 164 kPa,
(c) R x = −1.2 kN (to left); R y = 22 N (upwards).
FE3.3 (b) F = 36 N.
FE3.4 (c) V = 2.95 m/s (assuming negligible vertical distance
P6.66 (a) ∆p = 56 kPa, (b) Q = 85 m 3 /h,
(c) u = 3.5 m/s (C = 1.1).
P6.76 Q = 15 m 3 /h (f 1 = 0.0214, f 2 = 0.0220).
P6.80 Q = 0.86 m 3 /s (f ≃ 0.0191), assuming a free outlet,
pumping from large reservoir and neglecting any
height difference between reservoir level and outlet.
(∆p f + ρV 2 /2 = ρgh P ); with ∆p f = ρgh P the answer
is Q = 0.90 m 3 /s.
between the level of manometer liquid against
air and the cylinder center).
P6.90 Q = 0.52 m 3 /s
(V 1 ≪ V 2 = V ; V 1 = V 2 ⇒ Q = 0.56 m 3 /s).
FE3.7 (a) Q = 188 gal/min = 11.9 dm 3 /s.
Chapter 4
P4.83 Combined Couette-Poiseuille flow;
u(0) = U, u(h) = 0 ⇒ u = U(1 − η) − C η (1 − η),
P6.92 (a) Q = 0.43 m 3 /s (D eff = 337 mm),
(b) p − p a = −6.3 Pa.
P6.107 (a) Q ≈ 7.7 m 3 /h (K 2 ≈ 80); 0.3% more,
(b) Q = 38 m 3 /h; about 5 times more.
1

P6.116 Q = 97 m 3 /h (0.0269 m 3 /s); (V 100 = 3.3 m/s,
V 250 = 2.1 m/s, V 150 = 5.4 m/s).
P6.138 u CL ≃ 47 m/s (V/u CL ≃ 0.848, f ≃ 0.0162).
P6.146 (a) ɛ = 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).
C6.5 (a) b = 20 mm, (b) (h f /L) round = 0.078;
(h f /L) annul,Dh = 0.32; (h f /L) annul,Deff = 0.36,
Chapter 7
P7.26 F a = 2.83F 1 ; F b = 2.0F 1 .
P7.43 (a) U ≈ 34 m/s, (b) δ ≈ 36 mm,
(c) u ≈ 26 m/s.
P7.47 x sep /L = 0.268; Note: (1 − η 2 ) 5 = 1 − 5η 2 + 10η 4 −
10η 6 + 5η 8 − η 10 .
P7.48 x sep /L = 0.158.
P7.50 θ sep = 2.3 ◦ .
P7.59 (a) V 3 + (2V w + V α )V 2 + V 2 wV − (V 3 nw + V α V 2 nw) = 0,
V α = 2C RR /(ρC D A), (b) V ≈ 7.4 m/s, (c) D ∝ V 2
rel .
P7.80 θ ≃ 72 ◦ .
P7.92 d ≈ 3.9 m.
P7.93 Maximal cone angle: θ ≃ 60 ◦ (C D,rod ≃ 1.2).
P7.121 (a) V stall = 15 m/s (z = 1200 m), (b) θ min = 1.6 ◦
(C D∞ = 0.007), (c) gliding distance 44 km.
P7.122 (a) V min = 6.7 m/s, (b) V max = 15.0 m/s (130 hp =
96.9 kW ⇒ V max = 13.5 m/s).
FE7.3 (a) V f = 2.3 m/s.
FE7.5 (c) τ w = 0.016 Pa (NOTE!).
Chapter 8
P8.10 a = 12.7 mm; cavitation appears if U ∞ > 20 m/s.
P8.14 The outer flow is irrotational (line vortex): p = p ∞ −
ρ ω 2 R 4 /(2r 2 ); inner flow is rotational (rotation like a
solid body): p = p ∞ − ρ ω 2 R 2 + ρ ω 2 r 2 /2; minimum
pressure ar vortex center, p(0) = p ∞ − ρ ω 2 R 2 .
P8.15 (a) p min = p ∞ − ρU 2 max = 97.2 kPa,
(b) p(r = R) = p ∞ − ρU 2 max/2 = 99.3 kPa.
P8.23 V = 11.3 m/s, θ = 44.2 ◦ .
P8.50 h = 3a/2, U max = 5U ∞ /4.
P8.74 Without walls: u = v = K/(2a);
with walls: u = 8K/(15a), v = 4K/(15a).
P8.81 (a) V = a12 1/4 /(C D∞ πAR) 1/4 ; a = √ W/(ρA),
(b) V = 176 m/s.
P8.82 (a) V = 4.5 m/s,
(b) C L = 1.12,
(c) P = 0.41 kW (C D = 0.022).
P8.86 (a) b = 26 m, (b) AR = 8.7, (c) D i = 1.6 kN.
P8.87 (a) C = 0.174 m, (b) P = 1.25 kW,
(c) V = 31.3 m/s (α = −1.9 ◦ ).
P8.88 Thrust per engine, F = 12.9 kN (α = 3.18 ◦ ).
P8.100 U ∞ = 14.1 m/s ⇒ p(θ = 90 ◦ ) = p v = 2.34 kPa
(cavitation); p A = 115 kPa.
P8.101 (a) V f = 2.0 m/s, (b) t = 0.64 s (approx. +6%).
Chapter 9
P9.28 (a) ṁ = 0.17 kg/s, (b) Ma e = 0.90.
P9.40 (a) Ma 1 = 2.5, (b) A ∗ = 7.6 cm 2 ,
(c) ṁ = 1.3 kg/s, (d) Ma 2 = 1.5.
P9.55 (a) p 2 = 261 kPa, (b) p 3 = 302 kPa,
(c) A ∗ 2 = 16.5 cm 2 , (d) Ma 3 = 0.34.
P9.60 Ma 1 = 1.92; V 1 = 585 m/s
P9.63 (a) p 3 = 59 kPa, (b) ṁ = 0.024 kg/s.
P9.82 V e ≃ 110 m/s; after shock: Ma 2 = 0.67.
P9.89 (a) ∆p = 125 kPa, (b) ∆p = 88 kPa, (c) L ∗ 1 = 33 m.
P9.90 (a) ṁ = 0.764 kg/s, (b) ṁ = 0.591 kg/s,
(c) ṁ = 0.314 kg/s.
P9.91 (a) Ma 1 = 0.30, (b) p = 1.03 MPa.
P9.99 Adiabatic flow: ṁ = 0.26 kg/s (f = 0.0142).
P9.107 (a) q = 0.58 MJ/kg, (b) Ma 2 = 0.71,
(c) T 2 ≃ 733 K.
P9.108 New Mach number, Ma 1 = 0.20; ṁ new /ṁ old = 0.68.
P9.109 (a) D e = 0.74 m, (b) V e ≃ 508 m/s,
(c) F ≃ 23 kN (V i ≪ V e , p i = p e ).
P9.141 Exact: Ma 2 = 3.274; p 2 = 66.7 kPa;
linear theory: p 2 = 61 kPa.
P9.142 (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.
Chapter 10
P10.17 Q = 5.2 m 3 /s (n = 0.015).
P10.19 y < 1.025 m (n = 0.022).
P10.22 S o = 0.38 m/km ↔ 0.022 ◦ (n = 0.014).
P10.49 Subcritical flow in both cases (y c < y n );
b = 4 ft ⇒ y c = 4.7 ft < y n = 9.3 ft;
b = 8 ft ⇒ y c = 3.0 ft < y n = 4.1 ft.
P10.59 (a) V 2 = 1.13 m/s,
(b) q = 0.566 m 2 /s (Fr 1 = 0.37).
P10.73 q = 0.46 m 2 /s; y 2 = 0.092 m (Fr 2 = 5.6).
P10.84 y 2 = 0.25 m, y 3 = 1.56 m, h f /E 1 = 47%.
P10.95 (a) V 4 = 2.6 m/s, (b) y 4 = 20 cm,
(c) V 1 = 3.1 m/s, (d) y 1 = 16.5 cm.
Christoffer Norberg
tel. +46-46-2228606
Department of Energy Sciences
Lund Institute of Technology, Sweden
Christoffer.Norberg@energy.lth.se
2