Introduction to Fluid Mechanics 7th Edition

Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 1

1.5 M = 5913 kg
1.7 L = 27.25 in. D = 13.75 in.
W2
1.9 y = 2.05 2
gt
1.11 d = 0.109 mm
1.15 y = 0.922 mm
1.17 a) N·m/s, lbf·ft/s b) N/m2, lbf/ft2 c) N/m2, lbf/ft2, d) 1/s, 1/s e) N·m, lbf·ft
f) N·s, lbf·s g) N/m2, lbf/ft2 h) m2/s2·K, ft2/s2·R i) 1/K, 1/R j) N·m·s, lbf·ft·s
1.19 a) 6.89 kPa b) 0.264 gal 47.9 N·s/m2
1.21 a) 0.0472 m3/s b) 0.0189 m3 c) 29.1 m/s d) 2.19 x 104 m2
1.23 101 gpm
1.25 SG = 13.6 v = 7.37 x 10−5 m3/kg γE = 847 lbf/ft3, γM = 144 lbf/ft3
1.27 2.25 kgf/cm2
At ⋅ p0
1.29 c = 0.04 K1/2·s/m m& max = 2.36 (ft2, psi, R)
T0
1.31 CD is dimensionless
1.33 c: N·s/m, lbf·s/ft k: N/m, lbf/ft f: N, lbf
3 2
1.35 H(m) = 0.457 − 3450·(Q(m /s))
1.37 ρ = 0.0765 ± 2.66 x 10−4 lbm/ft3 (± 0.348%)
1.39 ρ = 1130 ± 21.4 kg/m3 SG = 1.13 ± 0.0214
1.41 ρ = 930 ± 27.2 kg/m 3

1.43 t = 1, 5, 5 s Flow rate uncertainty = ± 5.0, 1.0, 1.0%

1.45 μ = 1.01 x 10 N·s/m
−3 2
± 0.609%
1.47 δx = ± 0.158 mm
1.49 H = 57.7 ± 0.548 ft θmin = 31.4o
 Introduction to Fluid Mechanics 7th Edition
Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 2

2.1 1) 1D, Unsteady 2) 1D, Steady 3) 2D, Unsteady 4) 2D, Unsteady

5) 1D, Unsteady 6) 3D, Steady 7) 2D, Unsteady 8) 3D, Steady
c
2.3 Streamlines: y=
x
b
− t
2.5 Streamlines: y=cx a

1
2.7 Streamlines: y=
⎛ b ⎞
2⎜⎜ + c ⎟⎟
⎝ax ⎠
2.9 Streamlines: y = 3x Δt = 0.75 s
2.11 Streamlines: x2 + y2 = c
2.13 Pathlines: y = 2/x Streamlines: y = 2/x
K
2.15 ω=
2πa
( )
1
B t + 12 At 2
2.17 Pathlines: y = y0e , x = x0e
Ct
Streamlines: y=x 1+ 0.5t

b
−
−b t
1
at 2
2.19 Pathlines: y = y0e , x = x0 e 2
Streamlines: y = Cx at

40 ⎛ x ⎞
Pathlines: y = 4t + 1 , x = 3e0.05t y = 1+
2
2.21 Streamlines: ln⎜ ⎟
t ⎝ 3⎠
2.23 Streamlines: y (t0 ) = v0 sin(ωt )(t − t0 ) , x (t0 ) = u0 (t − t0 )
1
(t −τ ) (t −τ )+ 0.1(t 2 −τ 2 ) (1+ 0.2 t )
2.25 Streaklines: y = e , x=e Streamlines: y=x
2
x
2.29 Streamlines: y = +4 (4 m, 8 m) (5 m, 10.25 m)
4
2.31 (2.8 m, 5 m) (3 m, 3 m)
b′T 2
3

2.33 ν= b´ = 4.13 x 10−9 m2/s·K3/2 S´ = 110.4 K

1 + S′
T
2.35 τyx = − 4.56 N/m2
2.39 a = − 0.491 ft/s2
2.41 L = 2.5 ft
2.43 t = 1.93 s
μA
Mgd sin θ ⎛ − t⎞
2.45 V= ⎜1 − e Md ⎟ V = 0.404 m/s μ = 1.08 N·s/m2
μA ⎜⎝ ⎟
⎠
2.47 F = 2.83 N
2.49 F = 0.0254 N
2.51 μ = 8.07 x 10−4 N·s/m2
Mgr 2 a
2.53 μ= μ = 0.0651 N·s/m2
2πVm R 3 H
2.55 t = 4.00 s
A⎛ − t⎞
B
2.57 ⎜
ω = ⎜1 − e ⎟ C ⎟
ωmax = 25.1 rpm t = 0.671 s
B⎝ ⎠
μωr πμωR 4
2.59 τ zθ = T=
h 2h
2.61 Dilatant k = 0.0499 n = 1.21 μ = 0.191 N·s/m2, 0.195 N·s/m2
2.63 Bingham plastic μp = 0.652 N·s/m2
2πμωR 3 H πμωR 4
2.65 T= T=
a 2b
μωR sin θ
2.69 τ= τmax = 79.2 N/m2
a + R (1 − cos θ )
2.75 a=0 b = 2U c = −U
2.77 V = 229 mph
2.79 M = 2.5 xtrans = 0.327 m
2.81 SG = 0.9 γ = 8830 N/m3 Laminar flow
2.83 V = 667 km/hr
 Introduction to Fluid Mechanics 7th Edition
Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 3

3.1 M = 62.4 kg t = 22.3 mm

3.3 z = 9303 ft Δz = 5337 ft
3.5 F = 45.6 N
3.9 Δp = 972 Pa ρ = 991 kg/m3
3.11 D = 0.477 in
3.13 Δρ/ρ0 = 4.34% Δp/p0 = 2.15%
3.15 p = − 0.217 psig
3.17 p = 6.39 kPa (gage) h = 39.3 mm
3.19 p = 128 kPa (gage)
3.21 Δp = 59.5 Pa
3.23 H = 30 mm
3.25 SG = 0.900
3.27 Δp = 1.64 psi
Δp
3.29 L=
[ ]
ρ oil 1 + ( Dd )
2
L = 27.2 mm

3.31 h = 1.11 in
3.33 θ = 11.1o S = 5/SG
3.35 patm = 14.4 psi Shorter column at higher temperature
3.37 Δh = 38.1 mm Δh = 67.8 mm
3.39 Δh = 0.389 cm
3.43 Δz = 587 ft Δz = 3062 ft
3.45 p = 57.5 kPa p = 60.2 kPa
3.49 pA = 1.96 kPa pB = 8.64 kPa pC = 21.9 kPa pair = − 11.3 kPa pair = 1.99 kPa
3.51 FA = 79,600 lbf
3.53 W = 68 kN
3.55 FR = 0.407 lbf
3.57 FR = 8.63 MN R = (8.34 MN, 14.4 MN)
3.59 F = 600 lbf
3.63 D = 8.66 ft
3.65 d = 2.66 m
3.67 SG = 0.542
3.69 F = − 137 kN
3.71 FV = 7.62 kN x´FV = 3.76 kN·m FA = − 5.71 kN
3.73 FV = − ρgwR π/4 2
x´ = 4R/3π
6
3.75 FV = 1.05 x 10 N x´ = 1.61 m
3.77 FV = 1.83 x 107 N α = 19.9o
3.79 FV = 416 kN FH = 370 kN α = 48.3o F = 557 kN
3.81 FV = 2.47 kN x´ = 0.645 m FH = 7.35 kN y´ = 0.217 m
4 ρL ( H − d ) 2
3

3.83 M= M = 583 kg
3 a
⎡ ⎛θ sin(2θ max ) ⎞⎤
3.85 M = 2 ρLR ⎢(d − R ) sin(θ max ) + R⎜ max + ⎟⎥ M = 631 kg
⎣ ⎝ 2 4 ⎠⎦
3.87 γ = 8829 N/m3 h = 0.292 m
3.89 VNot submerged/ VSubmerged = 10.5%
3.91 SG = Wa/(Wa − Ww)
3.93 FB = 1.89 x 10−11 lbf V = 1.15 x 10−3 ft/s (0.825 in/min)
3.95 L/VHe = 0.0659 lbf/ft3 L/VH2 = 0.0712 lbf/ft3 L/Vair = 0.0172 lbf/ft3/0.0249 lbf/ft3
3.97 D = 116 m M = 703 kg
3.99 θ = 9.1o (with A = 25 cm2 not A = 20 cm2)
3.101 x = 0.257 m f = 6.1 N
3.103 D = 2.57 ft
3.105 f = 0.288 cycle/s (ω = 1.81 rad/s)
3.107 F = 34.2 lbf
3.113 a = g(h/L)
3.115 ω = 185 rad/s (1764 rpm)
3.117 Δp = ρω2R2/2 ω = 7.16 rad/s
3.119 dy/dx = − 0.25 p = 105 − 1.96x (p: kPa, x: m)
3.121 α = 30o dy/dx = 0.346
3.123 T = 47.6 lbf p = 55.3 lbf/ft2 (gage)
 Introduction to Fluid Mechanics 7th Edition
Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 4

4.1 x = 0.934 m
4.3 x = 747 m t = 23.9 mm
4.5 V0 = 87.5 km/hr
4.7 τ = 1.50 hr
3.47h 2 + 2400
4.9 yc = (yc, h: mm) h = 21.2 mm μs ≥ 0.604
6.94h + 40
4.11 Q = − 90 ft3/s
r r r
( )
ρ ∫ V V ⋅ dA = −450ρiˆ + 360ρˆj (slug·ft/s/s; ρ: slug/ft3)
4.13
r r
∫ V ⋅ dA = −24 m s
3
r r r
( ) (
∫ V V ⋅ dA = 64iˆ − 96 ˆj − 60kˆ m s
4
)
2

4.15 Q = 12 umaxπR 2 m.f. = 13 umax

2
πR 2iˆ
4.17 V = 1.56 ft/s Q = 3.82 gpm
4.19 Qcool = 489 gpm m& cool = 2.45 × 105 lb hr m& moist = 19404 lb hr m& air = 14331 lb hr
4.21 V3 = (4.33 ft/s, − 2.50 ft/s)
4.23 Eight pipes
4.25 1400 units/hr 4220 units/hr Outflow = 9 units/s
4.27 ρ = 10.7 lb/ft 3

m& ρ 2 g sin (θ )h 3
4.29 =
w 3μ
4.31 U = 1.5 m/s
4.33 Q = 1.05 x 10−5 m3/s (10.45 mL/s) Vave = 0.139 m/s umax = 0.213 m/s
4.35 vmin = 5.0 m/s
4.37 ∂Voil/∂t = − 2.43 x 10−2 ft3/s (0.18 gal/s)
4.39 dh/dt = − 8.61 mm/s
4.41 dh/dt = − 0.289 mm/s
4.43 Q = 1.5 x 104 gal/s A = 4.92 x 107 ft2
4.45 t = 22.2 s
4.47 dy/dt = − 9.01 mm/s
4.49 Qcd = 4.50 x 10−3 m3/s Qad = 6.0 x 10−4 m3/s Qbc = 1.65 x 10−3 m3/s
2 π tan 2 (θ ) y05 2 6 V0
4.51 t= t=
5 2g A 5 Q0
4.53 mf = (− 2406, 2113) lbf
4.55 mf2/mf1 = 1.33
4.57 mf = (− 320, 332) N
4.61 T = 3.12 N
4.63 F = 35.7 lbf
4.65 m& 1 = 31.2 lbm s m& 2 = 32.0 lbm s (because of weight plus momentum loss)
4.67 V = 51 m/s V = 18.0 m/s V = 67.1 m/s
4.69 F = 1.81 kN (tension)
2 πD
2 ⎡ ⎛ d ⎞2 ⎤
4.71 R x = − ρV (1 + sin θ )⎢1 − ⎜ ⎟ ⎥ Rx = − 314 N
4 ⎣⎢ ⎝ D ⎠ ⎦⎥
4.73 F = 11.6 kN
4.75 F = (− 714, 498) N
4.77 F = 1.70 lbf
4.79 F = 22.7 kN
4.81 d/D = 0.707 No-dimensional pressure = 0.5
4.83 t = 1.19 mm F = 3.63 kN
4.85 Rx = − 4.68 kN Ry = 1.66 kN
4.87 Rx = − 1040 N Ry = − 667 kN
4.89 F = 2456 N
4.91 Q = 0.141 m3/s Rx = − 1.65 kN Ry = − 1.34 kN
4.93 F = 37.9 N
⎛ 5π 2 ⎞
4.95 f = ρU 2 ⎜ − ⎟
⎝ 8 π⎠
4.97 umax = 60 ft/s Δp = 0.699 lbf/ft2
4.99 D = 0.446 N
4.101 D/w = 0.163 N/m
4.103 h2/h = (1 + sinθ)/2
4.105 h = H/2
4.107 Q = 257 L/min
4.109 V = V02 − 2 gh h = 4.28 m
4.111 V = 175 ft/s F = 2.97 lbf
4.113 p1 = 68.4 kPa (gage0 F = 209 N
A0
4.115 V = V02 − 2 gz A=
2 gz
1− 2
V0
2 2
⎛ Q ⎞ ⎛x⎞
4.117 p = p0 − ρ ⎜ ⎟ ⎜ ⎟
⎝ wh ⎠ ⎝ L ⎠
V0 r
4.119 V =
2(h0 − V0t )
4.123 Rx = − 2400 N Ry = 1386 N
ρQ ⎛ ρQ ⎞ ρQ
2

4.125 V =− + ⎜ ⎟ + Vj Vj = 80 m/s
2k ⎝ 2k ⎠ k
4.127 F = 3840 lbf
4.129 F = 4.24 kN t = 4.17 s
4.131 α = 30 o
F = 10.3 kN
4.133 a = 13.5 m/s2
4.135 t = 0.680 s
U ⎛ M0 ⎞
4.137 = ln⎜⎜ ⎟⎟ V = 0.61 m/s
V ⎝ M 0 − ρVAt ⎠
4.139 amax at t = 0 θ = 90o U→V
2aM
4.141 A = A = 111 mm2
3ρ (V − at )
2

4.143 h = 17.9 mm
2 ρ (V − U ) A − kU 2
2
4.145 a = a = 5.99 m/s2 U/Ut = 0.667
M
4.147 a = 14.2 m/s2 t = 15.2 m/s
 ρ (V + U )2 A M
4.149 a = − t=
M ⎛ V ⎞
ρVA⎜⎜1 + ⎟⎟
⎝ U0 ⎠
4.151 V = 5 m/s xmax = 1.93 m t = 2.51 s
4.153 a = 2.28 m/s2
⎛ m& t ⎞
4.155 U = U 0 + Ve ln⎜⎜1 − ⎟ U = 227 m/s
⎝ M 0 ⎟⎠
4.157 Vmax = 834 m/s amax = 96.7 m/s2
4.159 mf = 82.7 kg
4.161 a = 83.3 m/s2 U = 719 m/s
4.163 V = 3860 ft/s Y = 33,500 ft
4.165 V = 641 m/s
4.167 θ = 19o
U0
4.169 U =
2 ρU 0 A
1+ t
M0
4.171 Vmax = 138 m/s ymax = 1085 m
4.175 h = 10.7 m
4.181 M = − 192 N·m
4.183 T = 0.193 N·m ω& = 2610 rad/s 2
4.185 ω& =
3
2 ρAR 3
(T − ρQRV − 2ωρVAR 2 ) ωmax = − 20.2 rad/s (− 193 rpm)

4.187 T = 30 N·m ω = 1434 rpm

4.189 ω = 78.3sin(θ) rad/s ω = 39.1 rad/s
4.191 ω& = 0.161 rad/s 2

4.195 T = ρωQL2 sin 2 α ΔT = 23 ρω& AL3 sin 2 α

4.197 α = 0o α ≈ 42o
o
4.199 dT/dt = − 0.064 K/s ( C/s)
4.201 Δp = 75.4 kPa
4.203 W& s = −96.0 kW
4.205 W& = −3.41 kW
s , actual

Δmef
4.207 = −1.88 J/kg Δt = 4.49 x 10−4 K (oC)
m&
 Introduction to Fluid Mechanics 7th Edition
Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 5

5.1 a) ρ ≠ const b) Possible c) Possible d) Possible

5.3 A + E + J = 0 (Others arbitrary)
5.5 v = − 3x2y + y3 + f(x)
5.7 u = x4/2 − 3x2y3
v⎞
5.11 ⎟ = 0.00167 (0.167%)
U ⎠ max
3 δ ⎡⎛ y ⎞ 1 ⎛ y ⎞ ⎤ 3 δ ⎡⎛ y ⎞ 1 ⎛ y ⎞ ⎤
2 4 2 4
v⎞ y v⎞
5.13 ⎟ = ⎢⎜ ⎟ − ⎜ ⎟ ⎥ =1 ⎟ = ⎢⎜ ⎟ − ⎜ ⎟ ⎥
U ⎠ max 8 x ⎣⎢⎝ δ ⎠ 2 ⎝ δ ⎠ ⎦⎥ δ U ⎠ max 8 x ⎣⎢⎝ δ ⎠ 2 ⎝ δ ⎠ ⎦⎥
5.15 v = − 2Axy3/3 +f(x) ψ = xy3/2
Λ sin θ
5.19 Vθ = − + f (r )
r2
Uy 2 h
5.23 ψ= y=
2h 2
r ⎛
5.25 V = ⎜ − U cosθ +
q ⎞ˆ
⎟ir + U sin θiˆθ (r, θ ) = ⎛⎜
q ⎞
,0 ⎟
⎝ 2π r ⎠ ⎝ 2πU ⎠
5.27 2D Incompressible ψ = zy3 − z3y
Uy 2
5.29 ψ= hhalf = 1.06 m
2h
⎡⎛ y ⎞ 2 1 ⎛ y ⎞ 3 ⎤ y y
5.31 ψ = Uδ ⎢⎜ ⎟ − ⎜ ⎟ ⎥ = 0.442 (1/4 flow) = 0.652 (1/2 flow)
⎣⎢⎝ δ ⎠ 3 ⎝ δ ⎠ ⎦⎥ δ δ
⎡ 3 ⎛ y ⎞2 1 ⎛ y ⎞4 ⎤ y y
5.33 ψ = Uδ ⎢ ⎜ ⎟ − ⎜ ⎟ ⎥ = 0.465 (1/4 flow) = 0.671 (1/2 flow)
⎢⎣ 4 ⎝ δ ⎠ 8 ⎝ δ ⎠ ⎥⎦ δ δ
5.35 ψ = − Cln(r) + C1 Q/b = 0.0912 m3/s/m
r 16 32 ˆ 16 ˆ
5.37 2D Incompressible a = iˆ + j + k m/s2
3 3 3
5.39
r
(
a = −2.86 10−2 iˆ + 10−4 ˆj ) m/s2 dy/dx = 0.0025
Λ2 x Λ2 y 100
5.41 Incompressible ax = − ay = − a=−
(x 2
+y )
2 2
(x 2
+y )
2 2 r3
U ⎛
2
x ⎞
5.43 ax = − ⎜1 − ⎟
2L ⎝ 2L ⎠
2
⎛ Q ⎞ 1
5.45 a = −⎜ ⎟ 3 (Radial)
⎝ 2πh ⎠ r
Dc UA ⎛ 1 − 2 a ⎞
Ut Ut
− Dc ppm
5.47 = ⎜ e −e a ⎟ x = 2.77 m = 1.25 × 10− 5
Dt a ⎜⎝ 2 ⎟
⎠ Dt max s
5.49 ∂T/∂x = − 0.0873 oF/mile
5.53
r
(
a = A2 xiˆ + yˆj ) r 1
a = iˆ + 2 ˆj
2
r
a = iˆ + ˆj
r 1
a = 2iˆ + ˆj
2
r
5.55 xy = 8 V = 12iˆ − 24 ˆj
r r r
V = 6πiˆ − 12πˆj (Local) V = 72iˆ + 144 ˆj (Convective) V = 9.8iˆ + 106 ˆj (Total)
U2 ⎡ ⎛ y ⎞2 4 ⎛ y ⎞2 1 ⎛ y ⎞2 ⎤
5.57 ax = ⎢− ⎜ ⎟ + ⎜ ⎟ − ⎜ ⎟ ⎥ ax(max) = − 5.22 m/s2
x ⎢⎣ ⎝ δ ⎠ 3 ⎝ δ ⎠ 3 ⎝ δ ⎠ ⎥⎦
⎛ y⎞ r v2 x v2 ⎛ y ⎞
5.59 v = v0 ⎜ 1 − ⎟ a = 02 iˆ + 0 ⎜ − 1⎟ ˆj
⎝ h⎠ h h ⎝h ⎠
2U 2 ⎡ ⎛ R ⎞ ⎤⎛ R ⎞
2 3

5.61 ar = −
⎢1 ⎜ ⎟ ⎥⎜ ⎟ aθ = 0 r = 1.29R (max a)
R ⎣⎢ ⎝ r ⎠ ⎦⎥⎝ r ⎠
4U 2 2 4U 2
ar = − sin θ aθ = sin θ cos θ θ = ±π/2 (max a)
R2 R2

5.63 a x = 32[20 + 2 sin(ωt )] + 2.4 cos(ωt )

2
(m/s ) at x = L
2

5.67 Not incompressible Not irrotational

5.69 Γ=0
5.71 Incompressible Irrotational
5.73 Incompressible Not irrotational
r
5.75 V = −2 yiˆ − 2 xˆj
r
5.77 ω = −k̂ ( rad/s)
u
5.79 ω=− ω = − 0.5 s−1
2h
h⎛ h⎞
5.81 Γ = −UL ⎜1 − ⎟ Γ = − UL/4 (h = b/2) Γ = 0 (h = b)
b⎝ b⎠
2 yU max r 2 yU max ˆ
5.83 γ =− ζ = k
b2 b2
⎛π ⎞
2
dFmax dFmax kN
5.85 = − μU ⎜ ⎟ = −1.85 3
dV ⎝ 2δ ⎠ dV m
 Introduction to Fluid Mechanics 7th Edition
Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 6

r
6.1 a = 9iˆ + 7 ˆj ft/s2 ∇p = −0.125iˆ − 0.544 ˆj psi/ft
6.3
r
( )
alocal = 10 iˆ + ˆj ft/s2
r
aconv = A( Ax + Bt )iˆ − A(− Ay + Bt ) ˆj
r
aconv = 300iˆ − 200 ˆj ft/s2 ∇p = −4.17iˆ + 2.56 ˆj − 0.43kˆ psi/ft
r
6.5 a = 1iˆ + 7 ˆj ft/s2 ∇p = −0.0139iˆ − 0.544 ˆj psi/ft
r
6.7 v = − Ay a = 8iˆ + 4 ˆj m/s2
∇p = −12iˆ − 6 ˆj − 14.7kˆ Pa/m p(x) = 190 − 3x2/1000 (p in Pa, x in m)
6.9 Incompressible Stagnation point: (2.5, 1.5)
[
∇p = − ρ (4 x − 10)iˆ + (4 y − 6) ˆj + gkˆ ] Δp = 9.6 Pa
U2 ⎛ x ⎞ dp U2 ⎛ x ⎞
6.11 ax = − ⎜1 − ⎟ =ρ ⎜1 − ⎟ pout = 43.3 kPa (gage)
2L ⎝ 2L ⎠ dx 2L ⎝ 2L ⎠
r r
6.13 a = −0.101eˆr + 0.0507eˆθ m/s2 a = −0.101eˆr + 0.0507eˆθ m/s2
r
a = −0.0127eˆr + 0.00633eˆθ m/s2 ∇p = 101eˆr − 50.5eˆθ Pa/m
∇p = 101eˆr − 50.5eˆθ Pa/m ∇p = 12.7eˆr − 6.33eˆθ Pa/m
⎧ ⎡ ⎤ ⎫
2

∂p ρAi2 L2ui2 ( Ae − Ai ) ρui2 ⎪⎪ ⎢ Ai ⎥ ⎪⎪

=− p = pi + ⎨1 − ⎢ ⎬
2 ⎪ ⎢ A − ( Ai − Ae ) x ⎥⎥ ⎪
6.15
∂x ( Ai L + Ae x − Ai x )3
⎩⎪ ⎣ ⎦ ⎭⎪
i
L
2Vi 2 (Do − Di ) ∂p
6.17 ax = − = −10 MPa/m L≥1m
⎡ (D − Di ) ⎤ ∂x max
5

Di L ⎢1 + o x⎥
⎣ Di L ⎦
6.19 Fz = − 1.56 N (Acts downwards)
2V 2 x ∂p 2 ρV 2 x
6.21 ax = 2 =−
b ∂x b2
ρV 2 L2 ⎡ ⎛ x ⎞ ⎤ 4 ρV 2 L3W
2

p = patm + ⎢1 − ⎜ ⎟ ⎥ Fy =
b 2 ⎢⎣ ⎝ L ⎠ ⎥⎦ 3b 2
q2 x ∂p ρq 2 x
6.23 ax = 2 =− 2
h ∂x h
2Λ2 ρ
6.25 ∇p = 5 eˆr + 0eˆθ
r
6.27 Δp = − 30.6 Pa
r
6.31 B = − 0.6 m−2·s−1 a = 6iˆ + 3 ˆj m/s2 an = 6.45 m/s2
r
6.33 a = 4iˆ + 2 ˆj ft/s 2 R = 5.84 ft
r
6.35 a = 0.5iˆ + ˆj m/s2 R = 5.84 ft
6.37 Δh = 1.37 in
6.39 F = 0.379 lbf F = 1.52 lbf
6.41 h = 628 mm
6.47 p2 = 291 kPa (gage)
6.49 p = 9.53 psig
6.51 h = 17.0 ft
1
A = A1
2 g (z1 − z )
6.53
1+
V12
6.55 V = 262 m/s
6.57 Q = 304 gpm (0.676 ft3/s)
p = p∞ + ρU 2 (1 − 4 sin 2 θ )
1
6.59 θ = 30o, 150o, 210o, 330o
2
6.61 F = − 278 N/m
6.63 Q = 2.55 x 10−3 m3/s
6.65 p1 = 7.11 psig Kx = 12.9 lbf
6.67 p2 = 13.2 kPa (gage) (98.9 mm Hg) p3 = 706 Pa (gage) (5.29 mm Hg)
Rx = 0.375 N Ry = 0.533 N
6.69 p1 = 1.35 psig p0 = 1.79 psig
6.73 Δh = 6.60 in F = 0.105 lbf F = 18.5 lbf
6.75 F = 83.3 kN
6.77 p1 = 11.4 psig F = 14.1 lbf
dM dV
6.79 m& = A 2 pρ = − ρ w air
dt dt
⎧⎪ V ⎡ 2 p0 At ⎤ ⎫⎪
0.588

M w = ρ wV0 ⎨ − ⎢1 + 1.70
t
⎥ ⎬
⎪⎩V0 ⎣ ρ w V0 ⎦ ⎪
⎭
6.83 Cc = ½
6.87 ax = 10.5 ft/s2
6.89 dQ/dt = 0.516 m3/s/s
6.91 Dj/D1 = 0.32
6.93 Bernoulli can be applied
6.95 Incompressible Unsteady Irrotational φ=⎢ (
⎡A 2
y − x 2 ) + Bxy ⎥t
⎤
⎣2 ⎦

6.97 ψ=
q
2π
⎡ −1 ⎛ y − h ⎞ −1 ⎛ y + h ⎞ ⎤
⎢ tan ⎜ x ⎟ + tan ⎜ x ⎟⎥ φ=−
q
2π
{[ 2
][
ln x 2 + ( y − h ) x 2 + ( y + h )
2
]}
⎣ ⎝ ⎠ ⎝ ⎠⎦
A
6.99 NOTE: Error – function is ψ = Ax2y − By3 φ = 3Bxy 2 − x 3
3
6.101 ψ = (x 2 − y 2 ) − 2 Axy
B
2
r
6.105 V = −( A + 2 Bx )iˆ + 2 Byˆj ψ = − (Ay + 2Bxy) Δp = 12 kPa
B 3
6.107 V = x2 + y2 ψ = 3 Ax 2 y − x
3
6.109 Incompressible Irrotational Stagnation point: (− 2, 4/3)
φ = ( y − x 2 ) − Bx − Cy
A 2
Δp = 55.8 kPa
2
q ⎛ r2 ⎞
6.113 ψ =
q
(θ1 − θ 2 ) + Ur sin θ φ= ln⎜ ⎟ − Ur cos θ
2π 2π ⎜⎝ r1 ⎟⎠
 r ⎡ q ⎛ cos θ1 cos θ 2 ⎞ ⎤ q ⎛ sin θ1 sin θ 2 ⎞ ˆ
V = ⎢ ⎜⎜ − ⎟⎟ + U ⎥iˆ + ⎜⎜ − ⎟j
⎣ 2π ⎝ r1 r2 ⎠ ⎦ 2π ⎝ r1 r2 ⎟⎠
Stagnations points: θ = 0, π r = 0.367 m ψstag = 0
K K
6.115 ψ = Ur sin θ − ln r φ = −Ur cos θ − θ
2π 2π
r ⎛ K ⎞
V = U cos θeˆr + ⎜ − U sin θ ⎟eˆθ
⎝ 2πr ⎠
K
Stagnations point: θ = π/2 r=
2πU
6.117 Stagnations points: θ = 63o, 297o r = 1.82 m Δp = 317 Pa
 Introduction to Fluid Mechanics 7th Edition
Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 7

V02
7.1
gL
E
7.3
ρL2ω 2
ν ⎛ 1 ⎞
7.5 ⎜= ⎟
V0 L ⎝ Re ⎠
Δp ν L
7.7 , ,
ρV 2 DV D
7.9 F ∝ μVD
Δp ⎛ μ d⎞
7.11 = f ⎜⎜ , ⎟⎟
ρV 2
⎝ ρVD D ⎠
τw ⎛ μ ⎞
7.13 = f ⎜⎜ ⎟⎟
ρU 2
⎝ ρUL ⎠
W σ
7.15 ,
gρp gρp 3
3

⎛λ⎞
7.17 V = gD f ⎜ ⎟
⎝D⎠
⎛b⎞
7.19 Q = h 2 gh f ⎜ ⎟
⎝h⎠
W ⎛L c⎞
7.21 = f⎜ , ⎟
D ωμ
2
⎝D D⎠
P Δp d d d
7.23 , , , i, o
ρ D ω ρD ω D D D
5 3 2 2

μ
7.25 Four parameters Π1 =
ρd g 1 2
32

Q ⎛ ρVh V 2 ⎞
7.27 = f ⎜⎜ , ⎟⎟
Vh 2 ⎝ μ gh ⎠
d μ σ
7.29 , ,
D ρVD ρDV 2
W μ h D
7.31 , , ,
ρV d ρVd d d
2 2

d μ2 σ
7.33 , ,
D ρΔpD DΔp
2

δ L μωD 3 Iω 2
7.35 , , ,
D D T T
 Δp μ ρ g
7.37 , , cD 3 , N , p ,
ρD ω ρD ω
2 2 2
ρ Dω 2
P ⎛ ρω c l ⎞
7.39 = f ⎜⎜ , , ⎟⎟
pωD 3
⎝ p D D⎠
⎛c Θ μ ⎞
7.41 Four primary dimensions Q& = ρV 3 L2 f ⎜⎜ p 2 , ⎟⎟
⎝ V ρVL ⎠
dT Lc p ⎛ c k μ ⎞⎟
7.43 = f⎜ , 2 ,
dt V 3 ⎜ c ρL c ρLV ⎟
⎝ p p ⎠
7.45 Π1 =
u
Π2 =
y
Π3 =
(dU dy )δ Π4 =
ν
U δ U δU
7.47 Vw = 6.90 m/s Fair = 522 N
7.49 Vair > Vwater Vair = 15.1· Vwater
7.51 ωm = 395 rpm ωm = 12500 rpm Froude number modeling is most likely
7.53 Vm = 40.3 m/s Vp = 40.3 m/s
7.55 Vm = 5.07 m/s Fm/Fp = 3.77
7.57 Qm = 0.125 m3/s Pp = 127 kW
7.59 pm = 2.96 psia
fd ⎛ ρVd ⎞ V1 1 f1 1
7.61 = F ⎜⎜ ⎟⎟ = =
V ⎝ μ ⎠ V2 2 f2 4
7.63 Vm = 0.618 m/s – 1.03 m/s
FD ⎛ μ ⎞
7.65 = f ⎜⎜ ⎟
12 ⎟
FD p = 2.46 kN P =55.1 kW (73.9 hp)
ρV A2 12
⎝ ρVA ⎠
7.67 τp = 1070 hr (~ 45 days)
7.69 Vm = 1.88 m/s Vp = 7.36 m/s Fp/Fm = 1.13 (submerged), = 2.77 x 104 (surface)
7.71 CD = 1.028 FDp = 3.89 kN Vm = 250 m/s (model is impractical, compressible flow)
1
7.73 Model = x Prototype Adequate Reynolds number not achievable
50
7.77 DTotal = 1305 N DTotal = 2316 N (Wave drag negligible)
7.79 hm = 13.8 J/kg Qm = 0.166 m3/s Dm = 0.120 m
Ft ⎛ V g ⎞ T ⎛ V g ⎞ P ⎛ V g ⎞
7.81 = f1 ⎜ , 2 ⎟ = f2 ⎜ , 2 ⎟ = f3 ⎜ , 2 ⎟
ρω D2 4
⎝ ωD ω D ⎠ ρω D
2 5
⎝ ωD ω D ⎠ ρω D
3 5
⎝ ωD ω D ⎠
7.83 K.E. ratio = 7.22
7.85 FB ≈ 0.273 N (to right) CD m = 0.443 FD p = 1.64 kN
 Introduction to Fluid Mechanics 7th Edition
Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 8

8.1 Q = 0.146 ft3/s L = 12.5 – 20 ft (turbulent) L = 69.0 ft (laminar)

8.3 Smallest turbulent first Qlarge = 0.0244 ft3/s Non are fully developed
Qmid = 0.0122 ft3/s Smallest fully developed
Qsmall = 0.00610 ft3/s Smallest fully developed
8.7 V umax = 2 3
8.9 τyx = – 1.88 Pa Q/b = – 5.63 x 10– 6 m2/s
8.11 Q = 1.25 x 10– 5 ft3/s (0.0216 in3/s)
8.13 Q = 3.97 x 10– 9 m3/s (3.97 x 10– 6 L/s)
π ΔpD 3
8.15 M = 4.32 kg Q= a a = 1.28 x 10– 5 m (12.8 μm)
12 μL
Q dp 6μQ 6μQ ⎛ r ⎞
8.17 V = =− p = patm − ln⎜ ⎟ (p = p0, r < R0)
2πrh dr πrh 3
πrh 3 ⎝ R ⎠
8.19 n = 1.48 (dilatant)
8.21 ∂p/∂x = – 92.6 Pa/m
8.23 uinterface = 15 ft/s
8.25 ∂p/∂x = – 2Uμ/a2 ∂p/∂x = 2Uμ/a2
8.27 ν = 1.00 x 10– 4 m2/s
8.29 τ = ρgsin(θ)(h – y) Q/w = 217 mm3/s/mm Re = 0.163
8.31 Q/w = 0.0208 ft3/s/ft τ = 1.58 x 10– 6 psi ∂p/∂x = 7.58 x 10– 4 psi/ft
8.33 ∂p/∂x = 34.4 Pa/m
ρg ⎛ y 2 ⎞
8.35 B.C.: y = 0, u = U0; y = h, τ = 0 u= ⎜⎜ − hy ⎟⎟ + U 0
μ ⎝ 2 ⎠
dV πwL
8.37 =− V t = 1.06 s
dt mh
⎛ 2Q ⎞
⎜1 − ⎟
6 μLRω ⎛ 2Q ⎞ μLb(Rω ) ⎛
2
6Q ⎞ 6Q ⎝ abRω ⎠
8.39 Δp = ⎜1 − ⎟ P= ⎜4 − ⎟ η=
a 2
⎝ abRω ⎠ a ⎝ abRω ⎠ abRω ⎛ 6Q ⎞
⎜4 − ⎟
⎝ abRω ⎠
πμω 2 D 3 L πDa 3Δp 2
8.41 Pv = Pp = Pv = 3Pp
4a 12 μL
8.45 r = 0.707R
8.47 Q = 1.43 x 10– 3 in3/s (0.0857 in3/min)
c μV0 V0 ln ro
8.49 τ = c1/r u = 1 ln r + c2 c1 = c2 = −
μ ln(ri ro ) ln(ri ro )
⎡ 1 (1 − k 2 )⎤
12

8.51 r = R⎢ ⎥
⎣ 2 ln (1 k ) ⎦
8.53 % change = – 100/(1 + lnk)
8.55 τw = – 131 Pa
8.57 τw = – 0.195 lbf/ft2 τw = – 1.35 x 10– 3 psi
8.59 Q = 4.52 x 10– 7 m3/s Δp = 235 kPa τw = 294 Pa
8.61 n = 6.21 n = 8.55
8.63 βlam = 4/3 βturb = 1.02
8.65 α=2
8.67 HlT = 1.33 m hlT = 13.0 J/kg
8.69 V1 = 3.70 m/s
8.71 Q = 411 gpm
8.73 V1 = 2 m s
8.75 hlT = 913 ft2/s2 (HlT = 28.4 ft)
du
8.77 = 963 s −1 τw = 3.58 x 10– 4 lbf/ft2 τw = 4.13 x 10– 4 lbf/ft2
dy
8.79 f = 0.0390 Re = 3183 Turbulent
8.81 Maximum = 2.12% at Re = 10000 and e/D = 0.01
8.85 p2 = 177 kPa p2 = 175 kPa
8.87 Q = 0.0406 ft3/s (2.44 ft3/min, 18.2 gpm)
8.91 K = 9.38 x 10– 4
8.93 Q = 12.7 gpm Q = 11.6 gpm (ΔQ = – 1.1 gpm) Q = 13.7 gpm (ΔQ = 1.0 gpm)
8.95 Δp = 23.7 psi K = 0.293
2
hl m = (1 − AR ) 1
2V
8.97
2
2Δp
V1 = Inviscid assumption: Lower indicated flow/larger Δp
ρ (1 − AR 2 − K )
8.99

8.101 Q = 0.345 L/min d = 3.65 m

8.103 d = 6.13 m (or 6.16 m if α = 2, laminar)
8.105 Analogy fails at Q = 7.34 x 10– 7 m3/s
8.107 118800%! (A huge increase because V ~ 1/d2, and Δp ~ V2)
8.111 (a) Δp = 25.2 kPa (b) Δp = 32.8 kPa (c) Δp = 43.3 kPa ((a) is best)
8.113 VB = 4.04 m/s LA = 12.8 m (Not feasible!) Δp = 29.9 kPa
8.115 Δp/L = 7.51 x 10– 3 lbf/ft2/ft (round) Δp/L = 8.68 x 10– 3 lbf/ft2/ft (1:1) (+15.6%)
Δp/L = 9.32 x 10 lbf/ft /ft (2:1) (+24.1%)
–3 2
Δp/L = 0.010 lbf/ft2/ft (3:1) (+33.2%)
8.117 p1 = 179 psig
8.121 L = 26.5 m
8.123 Friction ≈ 77%, Gravity ≈ 23% (Turbulent)
8.125 Q = 0.0395 m3/s
8.127 V1 = 0.0423 m/s (down) (Falls at 42.3 mm/s)
8.129 Rate of downpour = 0.418 cm/min
8.135 Q = 6.68 x 10– 3 m3/s pmin = – 35.5 kPa (gage)
8.137 Q = 5.30 x 10– 4 m3/s Q = 5.35 x 10– 4 m3/s (diffuser)
8.139 L = 0.296 m
8.141 Your boss was wrong (which is s-w-e-e-e-e-e-t!)
8.143 D = 5.0 – 5.1 cm (corresponds to standard 2 in. pipe)
8.145 D = 6 in. (nominal)
8.149 V = 6.46 m s pF = 705 kPa (gage) P = 832 kW τw = 88.6 Pa
3
8.151 dQ/dt = – 0.524 m /s/min
8.153 P = 8.13 hp
8.155 Δp = 53.2 psi
8.157 D = 48 mm Δp = 3840 kPa Ppump = 24.3 kW (32.6 hp)
8.159 Q = 5.58 x 10– 3 m3/s (0.335 m3/min) V = 37.9 m/s P = 8.77 kW
8.161 Cost = $4980/year
8.163 Q = 0.0419 m3/s Δp = 487 kPa P = 29.1 kW
8.165 Q = 2.51 m3/s
8.167 Q0 = 0.00928 m3/s Q1 = 0.00306 m3/s Q2 = Q3 = 0.00311 m3/s Q4 = 0.00623 m3/s
Q0 = 0.00862 m3/s Q1 = 0.0 m3/s Q2 = Q3 = 0.00431 m3/s Q4 = 0.00862 m3/s
8.169 Δp = 22.2 psi Q22 ≈ 5.2 gpm Q24 ≈ 24.8 gpm
8.171 Δp = 25.8 kPa
8.173 Q = 1.49 ft3/s
8.175 Q = 0.00611 m3/s
8.177 Δt = 40.8 mm m& min = 0.0220 kg/s
8.179 Q = 1.37 ft3/s
8.183 Red = 1800 f = 0.0356 p2 = – 290 Pa (gage) (29.6 mm Hg)
8.185 Kc = $9890/in2 13
Kp = 1.81 x 10 $·in 5
 Introduction to Fluid Mechanics 7th Edition
Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 9

9.1 xp = 18.6 cm xm = 10.3 mm

9.3 xp = 10.4 cm xp = 7.47 mm
9.5 ReD = 1 (reasonable) ReD = 2.5 x 105 (not reasonable) Use water and D = 10 cm
9.7 L increases with elevation
π
9.9 A=U B= C=0
2δ
δ* θ
9.11 = 0.375 = 0.139
δ δ
δ* θ
9.13 = 0.396 = 0.152
δ δ
θ θ θ
9.15 Linear: = 0.167 Sinusoidal: = 0.137 Parabolic: = 0.133
δ δ δ
δ* θ δ* θ
9.17 Power: = 0.125 , = 0.0972 Parabolic: = 0.333 , = 0.133
δ δ δ δ
9.19 m& ab = 50.4 kg s FD = 50.4 N
9.21 dexit = 3.13 mm ΔU = 3.91%
9.23 U2 = 13.8 m/s Δp = 20.6 Pa
9.25 Δp = – 1.16 lbf/ft2
9.27 U2 = 24.6 m/s p2 = – 44.5 mm H2O
9.29 δ 2* = 2.54 mm Δp = – 107 Pa FD = 2.00 N
ρU 2 ρU 2bL
9.35 y = 0.121 in dy/dx = 0.00326 τ w = 0.3321 FD = 0.6642
Re x Re L
θL = 0.0454 in
9.39 θL = 0.0454 in FD = 0.850 N
9.41 FD = 26.3 N FD = 45.5 N
9.43 FD = 8.40 x 10–4 N (or FD = 1.68 x 10–3 N for two sides) (Higher than Problem 9.42)
9.45 FD = 3.45 x 10–3 N (or FD = 6.90 x 10–3 N for two sides) (Higher than Problem 9.44)
9.51 FD = 0.557 N
9.53 U = 1.81, 2.42, 3.63, and 7.25 m/s
ρU 2 ρU 2bL
9.55 τ w = 0.0297 1 FD = 0.0360 1
FD = 2.34 N
Re x 5 Re L5
9.57 FD = 4.57 x 10–3 N (or FD = 9.14 x 10–3 N for two sides)
9.59 FD = 55.8 N (or FD = 112 N for two sides)
δ 0.353 0.0612
9.61 = 1
cf = 1
FD = 2.41 N
x Re 5 Re 5
x x

9.63 δ L = 31.3 mm τ w = 0.798 Pa

L
FD = 0.700 N
9.65 w2 = 80.3 mm
9.67 Δp = 6.16 Pa L = 0.233 m
9.69 Petroleum used ≈ 0.089% (about 15% of pipeline use)
9.71 m& lam = 0.525ρU 2δ m& turb = 0.777 ρU 2δ
9.73 a=0 b=0 c=3 d=–2 H = 3.89
9.75 U2 = 2.50 m/s Δp = 0.00370 in H2O
4
⎡ 1
⎤
U ⎡ cf ⎤ U ⎢ ⎛ ν 4⎞ x ⎥ ( τ ≠ constant)
9.77 = ⎢1 + x ⎥ (constant τ w ) = 1 + 0.00583⎜⎜ ⎟⎟
⎣ θ (H + 2 ) ⎦ ⎢ U1δ ⎠ θ (H + 2 )⎥
w
U1 U1 ⎝
⎢⎣ ⎦⎥
9.79 FD = 5.58 N (11.2 N for both sides) One system: FD = 4.23 N (8.46 N for both sides)
9.81 FD = 1500 lbf P = 2000 hp
9.85 xlam/L = 0.0352% FD = 5.49 x 105 N P = 7.63 MW
4
9.87 FD = 3.02 x 10 N Savings = FD = 7.94 x 104 kg/yr
9.91 FD = 3610 lbf P = 77.6 hp
9.93 di = 96.5 mm
9.95 t = 9.29 s, x = 477 m (t = 7.93 s, x = 407 m for three parachutes) “g” = – 3.66
9.97 B is 20.8% better than A (H > D)
9.99 C D = 0.299
9.101 V = 24.7/35.8 km/hr New tires: V = 26.8/32.6/39.1 km/hr Plus fairing: V = 29.8/35.7/42.1 km/hr
9.105 M = 0.0451 kg
⎡ 2mg sin θ ⎤
9.107 V = ⎢ ⎥ t = 1.30 mm
⎣ C D Aρ cos θ ⎦
2

9.109 t = 2.95 s x = 624 ft

9.111 V = 47.3 mph (1970’s car) V = 59.0 (current car)
9.113 FE = 6.72 mpg ΔQ = 1720 gal/yr (8.78%)
9.115 CD = 1.17
⎡ ⎤
⎢ ⎥
9.117 V= ⎢ 2mg 1 ⎥
⎢ ρA ⎛ ⎞ 2
A ⎛A ⎞ ⎥
⎢ ⎜⎜ 1 ⎟⎟ − 2⎜⎜ 1 ⎟⎟ + 1⎥
⎢⎣ ⎝ A2 ⎠ ⎝ A2 ⎠ ⎥⎦

FD = C D A ρ (V − U ) T = CD A ρ (V − U ) R P = CD A ρ (V − U ) U
1 1 1 V
9.119
2 2 2
ωopt =
2 2 2 3R
9.121 M = 11.0 N·m
9.123 P = 3.00 kW
9.125 V = 23.3 m/s Re = 48,200 FD = 0.111 N
9.127 x = 13.9 m
9.129 CD = 61.9 ρ s = 3720 kg/m3 V = 0.731 m/s
9.131 M = 0.0471 kg
9.133 CL = 1.01 CD = 0.0654
7 1 7 1 FD 7 M 7
9.135 FD = CD ρU 2 DH M = C D ρU 2 DH 2 = =
9 2 16 2 FDuniform 9 M uniform 8
9.137 D = 7.99 mm y = 121 mm
9.139 t = 4.69 s x = 70.9 m
9.141 xmax = 48.7 m (both methods)
9.143 CD = 0.606 V = 37.4 mph
2 FR
9.145 Vb = Vw − Vb = 4.56 m/s (16.4 km/hr)
ρ (C Du Au + C Db Ab )
9.147 x ≈ 203 m
9.151 ΔP = 16.3 kW (94%)
9.157 Vmin = 5.62 m/s (10.9 kt) Pmin = 31.0 kW Vmax = 19.9 m/s (38.7 kt)
9.159 Vmin = 144 m/s R = 431 m
9.161 M = 37.9 kg P = 1.53 kW (or 3.02 kW if treated as two wings)
9.163 T = 17,300 lbf
9.165 FD = 524 lbf P = 209 hp
9.167 θ = 3.42 o
L = 168 km
9.169 For a race car, effective; for a passenger car, not effective
9.175 FL = 0.00291 lbf
9.177 FL = 50.9 kN FD = 18.7 kN F = 54.2 kN P = 5.94 kW
9.179 ω = 14,000 – 17,000 rpm x = 3.90 ft
 Introduction to Fluid Mechanics 7th Edition
Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 11

11.1 Q = 3.18 m3/s

11.3 y = 2.61 ft
11.5 y2 = 0.507 m Fr2 = 2.51
11.7 S0 = 0.00186
11.9 S0 = 0.00160
11.11 Q = 0.194 m3/s
11.13 y = 2.47 ft
11.15 y = 0.775 m
11.19 y = 4.83 ft V = 3.69 ft/s
11.23 y = 7.38 ft
11.25 yc = 0.365 ft, Ec = 0.547 ft yc = 0.759 ft, Ec = 1.14 ft
yc = 1.067 ft, Ec = 1.60 ft yc = 1.46 ft, Ec = 2.19 ft
11.27 yc = 0.637 m
11.31 Δx = 197 ft
11.33 y = 0.645 ft y = 4.30 ft
1
⎛ 2Q 2 ⎞ 5
11.35 yc = ⎜⎜ 2 ⎟⎟
⎝z g⎠
11.37 Q = 3.24 ft3/s
11.39 y2 = 0.610 ft (– 32.2%)
11.41 y2 = 1.31 ft
11.43 Q = 49.5 ft3/s
11.45 Q = 10.6 m3/s yc = 0.894 m Hl = 0.808 m
11.47 y2 = 5.94 ft
11.49 Q = 54.0 ft3/s Hl = 1.62 ft
11.51 y2 = 4.45 m Hl = 9.31 m
3
11.53 Q = 26.6 ft /s
11.55 H = 0.514 m
11.57 Cw = 1.45
 Introduction to Fluid Mechanics 7th Edition
Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 12

12.1 T = const. p decreases ρ decreases (Irreversible adiabatic process)

12.3 Δs > 0 so it is feasible for a real (irreversible) adiabatic process
12.5 T2 = 20oF p2 = 100 kPa
12.7 Δs = – 346 kJ/kg·K (ΔS = – 1729 J/K) Δu = – 143 kJ/kg (ΔU = – 717 kJ)
Δh = – 201 kJ/k (ΔH = – 1004 kJ)
12.9 h = 57.5%
12.11 W = 176 MJ Ws = 228 MJ Ts (max) = 858 K Qs = – 317 MJ
12.13 m& = 36.7 kg/s T2 = 572 K V2 = 4.75 m/s W& = 23 MW
12.15 Δt = 828s (≈ 14 min)
12.17 Δρ = 1.70 x 10–4kg/m3 ΔT = 0.017 K ΔV = 0.049 m/s
12.19 Δt = 198 μs Ev = 12.7 GN/m 2

12.21 x = 19.2 km
12.23 c = 299 m/s V = 987 m/s V/Vbullet = 1.41
12.29 c = 340 m/s (sea level)
12.31 V = 1471 mph α = 31.8o
12.33 V = 642 m/s (2110 ft/s)
12.35 V = 493 m/s Δt = 0.398 s
12.37 V = 515 m/s t = 6.92 s
12.39 Δx ≈ 1043 – 1064 m
12.41 Density change < 1.21%, so incompressible
12.43 M = 0.142 (1%) M = 0.322 (5%) M = 0.464 (10%)
12.45 Δρ/ρ = 48.5% (Not incompressible)
12.47 pdyn = 54.3 kPa p0 = 152 kPa
12.49 p0 = 546 kPa h0 – h = 178 kJ/kg T0 = 466 K
12.51 p0 – p = 8.67 kPa V = 195 m/s V = 205 m/s Error using Bernoulli = 5.13%
12.55 T0 = const (isoenergetic) p0 decreases (irreversible adiabatic)
12.57 V = 890 m/s T0 = 677 K p0 = 212 kPa
12.59 T0 = const = 294 K (20.6o) (isoenergetic) p01 = 1.01 MPa, p0 2 = 189 kPa (irreversible adiabatic)
Δs = 480 J/kg·K Flow accelerates even with friction due to large pressure drop
12.61 T01 = T0 2 = 344 K p01 = 223 kPa p0 2 = 145 kPa Δs = 0.124 kJ/kg·K
12.63 T01 = T0 2 = 445 K p01 = 57.5 kPa p0 2 = 46.7 kPa Δs = 59.6 J/kg·K
12.65 Δp = 48.2 kPa (inside higher)
12.67 T* = 260 K p* = 24.7 MPa V* = 252 m/s
12.69 Tt = 2730 K pt = 25.5 MPa Vt = 1030 m/s