Fox 7th Answers Selected Problems

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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. 
1.9 2
2
05.2
tg
Wy = 
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 
1.29 c = 0.04 K1/2·s/m 
0
0
max 36.2 T
pAm t ⋅=& (ft2, psi, R) 
1.31 CD is dimensionless 
1.33 c: N·s/m, lbf·s/ft k: N/m, lbf/ft f: N, lbf 
1.35 H(m) = 0.457 − 3450·(Q(m3/s))2 
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/m3 
1.43 t = 1, 5, 5 s Flow rate uncertainty = ± 5.0, 1.0, 1.0% 
1.45 μ = 1.01 x 10−3 N·s/m2 ± 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 
2.3 Streamlines: 
x
cy = 
2.5 Streamlines: 
t
a
b
xcy
−= 
2.7 Streamlines: 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
=
c
xa
b
y
2
1 
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 
2.15 
a
K
πω 2= 
2.17 Pathlines: tCeyy 0= , ( )2210 AttBexx += Streamlines: txy 5.01
1
+= 
2.19 Pathlines: tbeyy −= 0 , 
2
2
1
0
atexx = Streamlines: ta
b
Cxy
−= 
2.21 Pathlines: 14 += ty , 205.03 tex = Streamlines: ⎟⎠
⎞⎜⎝
⎛+=
3
ln401 x
t
y 
2.23 Streamlines: ( ) ( )( )000 sin tttvty −= ω , ( ) ( )000 ttutx −= 
2.25 Streaklines: ( )τ−= tey , ( ) ( )221.0 ττ −+−= ttex Streamlines: ( )txy 2.01
1
+= 
2.29 Streamlines: 4
4
2
+= xy (4 m, 8 m) (5 m, 10.25 m) 
2.31 (2.8 m, 5 m) (3 m, 3 m) 
2.33 
T
S
Tb
′+
′=
1
2
3
ν b´ = 4.13 x 10−9 m2/s·K3/2 S´ = 110.4 K 
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 
2.45 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= − tMd
A
e
A
MgdV
μ
μ
θ 1sin V = 0.404 m/s μ = 1.08 N·s/m2 
2.47 F = 2.83 N 
2.49 F = 0.0254 N 
2.51 μ = 8.07 x 10−4 N·s/m2 
2.53 
HRV
aMgr
m
3
2
2πμ = μ = 0.0651 N·s/m
2 
2.55 t = 4.00 s 
2.57 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= − tC
B
e
B
A 1ω ωmax = 25.1 rpm t = 0.671 s 
2.59 
h
r
z
μωτ θ = h
RT
2
4πμω= 
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 
2.65 
a
HRT
32πμω= 
b
RT
2
4πμω= 
2.69 ( )θ
θμωτ
cos1
sin
−+= Ra
R τmax = 79.2 N/m2 
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 
3.29 ( )[ ]2oil 1 DdpL +Δ= ρ 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 = − ρgwR2π/4 x´ = 4R/3π 
3.75 FV = 1.05 x 106 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 
3.83 ( )
a
dHLM
3
4 2
3−= ρ M = 583 kg 
3.85 ( ) ( ) ( ) ⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛ ++−=
4
2sin
2
sin2 maxmaxmax
θθθρ RRdLRM M = 631 kg 
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 
4.9 
4094.6
240047.3 2
+
+=
h
hyc (yc, h: mm) h = 21.2 mm μs ≥ 0.604 
4.11 Q = − 90 ft3/s ( ) jiAdVV ˆ360ˆ450 ρρρ +−=⋅∫ rrr (slug·ft/s/s; ρ: slug/ft3) 
4.13 sm24 3−=⋅∫ AdV rr ( ) ( ) 24 smˆ60ˆ96ˆ64 kjiAdVV −−=⋅∫ rrr 
4.15 2max21 RuQ π= iRu ˆm.f. 22max31 π= 
4.17 V = 1.56 ft/s Q = 3.82 gpm 
4.19 Qcool = 489 gpm hrlb1045.2 5cool ×=m& hrlb19404moist =m& hrlb14331air =m& 
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/ft3 
4.29 ( )μ
θρ
3
sin 32 hg
w
m =& 
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 
4.51 ( )
Ag
yt
2
tan
5
2 250
2 θπ= 
0
0
5
6
Q
Vt = 
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 slbm2.311 =m& slbm0.322 =m& (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) 
4.71 ( ) ⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛−+−=
22
2 1sin1
4 D
dDVRx θπρ Rx = − 314 N 
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 
4.95 ⎟⎠
⎞⎜⎝
⎛ −= π
πρ 2
8
52Uf 
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 ghVV 220 −= h = 4.28 m 
4.111 V = 175 ft/s F = 2.97 lbf 
4.113 p1 = 68.4 kPa (gage0 F = 209 N 
4.115 gzVV 220 −= 
2
0
0
21
V
gz
AA
−
= 
4.117 
22
0 ⎟⎠
⎞⎜⎝
⎛⎟⎠
⎞⎜⎝
⎛−=
L
x
wh
Qpp ρ 
4.119 ( )tVh
rVV
00
0
2 −= 
4.123 Rx = − 2400 N Ry = 1386 N 
4.125 jVk
Q
k
Q
k
QV ρρρ +⎟⎠
⎞⎜⎝
⎛+−=
2
22
 Vj = 80 m/s 
4.127 F = 3840 lbf 
4.129 F = 4.24 kN t = 4.17 s 
4.131 α= 30o F = 10.3 kN 
4.133 a = 13.5 m/s2 
4.135 t = 0.680 s 
4.137 ⎟⎟⎠
⎞
⎜⎜⎝
⎛
−= VAtM
M
V
U
ρ0
0ln V = 0.61 m/s 
4.139 amax at t = 0 θ = 90o U → V 
4.141 ( )23
2
atV
aMA −= ρ A = 111 mm
2 
4.143 h = 17.9 mm 
4.145 ( )
M
kUAUVa
222 −−= ρ a = 5.99 m/s2 U/Ut = 0.667 
4.147 a = 14.2 m/s2 t = 15.2 m/s 
4.149 ( )
M
AUVa
2+−= ρ 
⎟⎟⎠
⎞
⎜⎜⎝
⎛ +
=
0
1
U
VVA
Mt
ρ
 
4.151 V = 5 m/s xmax = 1.93 m t = 2.51 s 
4.153 a = 2.28 m/s2 
4.155 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+=
0
0 1ln M
tmVUU e
&
 U = 227 m/s 
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 
4.169 
t
M
AU
UU
0
0
0
21 ρ+
= 
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 2rad/s2610=ω& 
4.185 ( )23 22 3 VARQRVTAR ωρρρω −−=& 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 2rad/s161.0=ω& 
4.195 αρω 22 sinQLT = αωρ 2332 sinALT &=Δ 
4.197 α = 0o α ≈ 42o 
4.199 dT/dt = − 0.064 K/s (oC/s) 
4.201 Δp = 75.4 kPa 
4.203 kW0.96−=sW& 
4.205 kW41.3, −=actualsW& 
4.207 J/kg88.1−=Δ
m
mef
& Δt = 4.49 x 10
−4 K (oC) 
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 
5.11 %)167.0(00167.0
max
=⎟⎠
⎞
U
v 
5.13 ⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛−⎟⎠
⎞⎜⎝
⎛=⎟⎠
⎞ 42
max 2
1
8
3
δδ
δ yy
xU
v 1=δ
y ⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛−⎟⎠
⎞⎜⎝
⎛=⎟⎠
⎞ 42
max 2
1
8
3
δδ
δ yy
xU
v 
5.15 v = − 2Axy3/3 +f(x) ψ = xy3/2 
5.19 ( )rf
r
V +Λ−= 2sinθθ 
5.23 
h
Uy
2
2
=ψ 
2
hy = 
5.25 θθπθ iUir
qUV r ˆsinˆ2
cos +⎟⎠
⎞⎜⎝
⎛ +−=r ( ) ⎟⎠
⎞⎜⎝
⎛= 0,
2
,
U
qr πθ 
5.27 2D Incompressible ψ = zy3 − z3y 
5.29 
h
Uy
2
2
=ψ hhalf = 1.06 m 
5.31 
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛−⎟⎠
⎞⎜⎝
⎛=
32
3
1
δδδψ
yyU 442.0=δ
y (1/4 flow) 652.0=δ
y (1/2 flow) 
5.33 
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛−⎟⎠
⎞⎜⎝
⎛=
42
8
1
4
3
δδδψ
yyU 465.0=δ
y (1/4 flow) 671.0=δ
y (1/2 flow) 
5.35 ψ = − Cln(r) + C1 Q/b = 0.0912 m3/s/m 
5.37 2D Incompressible 2m/sˆ
3
16ˆ
3
32ˆ
3
16 kjia ++=r 
5.39 ( ) 242 m/sˆ10ˆ1086.2 jia −− +−=r dy/dx = 0.0025 
5.41 Incompressible ( )222
2
yx
xax +
Λ−= ( )222
2
yx
yay +
Λ−= 3100ra −= 
5.43 ⎟⎠
⎞⎜⎝
⎛ −−=
L
x
L
Uax 2
1
2
2
 
5.45 3
2 1
2 rh
Qa ⎟⎠
⎞⎜⎝
⎛−= π (Radial) 
5.47 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= −− a
Ut
a
Ut
ee
a
UA
Dt
Dc 2
2
1 x = 2.77 m 
s
ppm1025.1 5
max
−×=
Dt
Dc 
5.49 ∂T/∂x = − 0.0873 oF/mile 
5.53 ( )jyixAa ˆˆ2 +=r jia ˆ2ˆ
2
1 +=r jia ˆˆ +=r jia ˆ
2
1ˆ2 +=r 
5.55 xy = 8 jiV ˆ24ˆ12 −=r 
 jiV ˆ12ˆ6 ππ −=r (Local) jiV ˆ144ˆ72 +=r (Convective) jiV ˆ106ˆ8.9 +=r (Total) 
5.57 ⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛−⎟⎠
⎞⎜⎝
⎛+⎟⎠
⎞⎜⎝
⎛−=
2222
3
1
3
4
δδδ
yyy
x
Uax ax(max) = − 5.22 m/s
2 
5.59 ⎟⎠
⎞⎜⎝
⎛ −=
h
yvv 10 jh
y
h
vi
h
xva ˆ1ˆ
2
0
2
2
0 ⎟⎠
⎞⎜⎝
⎛ −+=r 
5.61 
322
12 ⎟⎠
⎞⎜⎝
⎛
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛−=
r
R
r
R
R
Uar aθ = 0 r = 1.29R (max a) 
 θ22
2
sin4
R
Uar −= θθθ cossin4 2
2
R
Ua = θ = ±π/2 (max a) 
 
5.63 ( )[ ] ( ) ( )22 m/scos4.2sin22032 ttax ωω ++= at x = L 
5.67 Not incompressible Not irrotational 
5.69 Γ = 0 
5.71 Incompressible Irrotational 
5.73 Incompressible Not irrotational 
5.75 jxiyV ˆ2ˆ2 −−=r 
5.77 rad/s)(kˆ−=ωr 
5.79 
h
u
2
−=ω ω = − 0.5 s−1 
5.81 ⎟⎠
⎞⎜⎝
⎛ −−=Γ
b
h
b
hUL 1 Γ = − UL/4 (h = b/2) Γ = 0 (h = b) 
5.83 2
max2
b
yU−=γ k
b
yU ˆ2
2
max=ζr 
5.85 
2
max
2 ⎟⎠
⎞⎜⎝
⎛−= δ
πμU
Vd
dF 3
max
m
kN85.1−=
Vd
dF 
 
Introduction to Fluid Mechanics 7th Edition 
Fox, Pritchard, & McDonald 
Answers to Selected Problems, Chapter 6 
 
6.1 2ft/sˆ7ˆ9 jia +=r psi/ftˆ544.0ˆ125.0 jip −−=∇ 
6.3 ( ) 2local ft/sˆˆ10 jia +=r ( ) ( ) jBtAyAiBtAxAa ˆˆconv +−−+=r 
 2conv ft/sˆ200ˆ300 jia −=r psi/ftˆ43.0ˆ56.2ˆ17.4 kjip −+−=∇ 
6.5 2ft/sˆ7ˆ1 jia +=r psi/ftˆ544.0ˆ0139.0 jip −−=∇ 
6.7 v = − Ay 2m/sˆ4ˆ8 jia +=r 
 Pa/mˆ7.14ˆ6ˆ12 kjip −−−=∇ p(x) = 190 − 3x2/1000 (p in Pa, x in m) 
6.9 Incompressible Stagnation point: (2.5, 1.5) 
 ( ) ( )[ ]kgjyixp ˆˆ64ˆ104 +−+−−=∇ ρ Δp = 9.6 Pa 
6.11 ⎟⎠
⎞⎜⎝
⎛ −−=
L
x
L
Uax 2
1
2
2
 ⎟⎠
⎞⎜⎝
⎛ −=
L
x
L
U
dx
dp
2
1
2
2
ρ pout = 43.3 kPa (gage) 
6.13 2m/sˆ0507.0ˆ101.0 θeea r +−=r 2m/sˆ0507.0ˆ101.0 θeea r +−=r 
 2m/sˆ00633.0ˆ0127.0 θeea r +−=r Pa/mˆ5.50ˆ101 θeep r −=∇ 
 Pa/mˆ5.50ˆ101 θeep r −=∇ Pa/mˆ33.6ˆ7.12 θeep r −=∇ 
6.15 ( )( )3
222
xAxALA
AAuLA
x
p
iei
ieii
−+
−−=∂
∂ ρ ( )
⎪⎪⎭
⎪⎪⎬
⎫
⎪⎪⎩
⎪⎪⎨
⎧
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
−−
−+=
2
2
1
2 x
L
AAA
Aupp
ei
i
ii
i
ρ 
6.17 ( )( ) 5
2
1
2
⎥⎦
⎤⎢⎣
⎡ −+
−−=
x
LD
DDLD
DDVa
i
io
i
ioi
x MPa/m10
max
−=∂
∂
x
p L ≥ 1 m 
6.19 Fz = − 1.56 N (Acts downwards) 
6.21 2
22
b
xVax = 2
22
b
xV
x
p ρ−=∂
∂ 
⎥⎥⎦
⎤
⎢⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛−+=
2
2
22
atm 1 L
x
b
LVpp ρ 2
32
3
4
b
WLVFy
ρ= 
6.23 2
2
h
xqax = 2
2
h
xq
x
p ρ−=∂
∂ 
6.25 θ
ρ ee
r
p r ˆ0ˆ
2
5
2
+Λ=∇ 
6.27 Δp = − 30.6 Pa 
6.31 B = − 0.6 m−2·s−1 2m/sˆ3ˆ6 jia +=r an = 6.45 m/s2 
6.33 2ft/sˆ2ˆ4 jia +=r R = 5.84 ft 
6.35 2m/sˆˆ5.0 jia +=r 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 
6.53 ( )
2
1
1
1 21
1
V
zzgAA −+
= 
6.55 V = 262 m/s 
6.57 Q = 304 gpm (0.676 ft3/s) 
6.59 ( )θρ 22 sin41
2
1 −+= ∞ Upp θ = 30o, 150o, 210o, 330o 
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 
6.79 ρpAm 2=& 
dt
Vd
dt
dM air
wρ−= 
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧
⎥⎦
⎤⎢⎣
⎡ +−=
588.0
0
0
0
0
270.11
V
Atp
V
VVM
w
t
ww ρρ 
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 ( ) tBxyxyA ⎥⎦⎤⎢⎣⎡ +−= 222φ 
6.97 ⎥⎦
⎤⎢⎣
⎡ ⎟⎠
⎞⎜⎝
⎛ ++⎟⎠
⎞⎜⎝
⎛ −= −−
x
hy
x
hyq 11 tantan
2πψ ( )[ ] ( )[ ]{ }2222ln2 hyxhyxq ++−+−= πφ 
6.99 NOTE: Error – function is ψ = Ax2y − By3 32
3
3 xABxy −=φ 
6.101 ( ) AxyyxB 2
2
22 −−=ψ 
6.105 ( ) jByiBxAV ˆ2ˆ2 ++−=r ψ = − (Ay + 2Bxy) Δp = 12 kPa 
6.107 V = x2 + y2 32
3
3 xByAx −=ψ 
6.109 Incompressible Irrotational Stagnation point: (− 2, 4/3) 
 ( ) CyBxxyA −−−= 22
2
φ Δp = 55.8 kPa 
6.113 ( ) θθθπψ sin2 21 Ur
q +−= θπφ cosln2 1
2 Ur
r
rq −⎟⎟⎠
⎞
⎜⎜⎝
⎛= 
 j
rr
qiU
rr
qV ˆsinsin
2
ˆcoscos
2 2
2
1
1
2
2
1
1 ⎟⎟⎠
⎞
⎜⎜⎝
⎛ −+⎥⎦
⎤⎢⎣
⎡ +⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= θθπ
θθ
π
r
 
 Stagnations points: θ = 0, π r = 0.367 m ψstag = 0 
6.115 rKUr ln
2
sin πθψ −= θπθφ 2cos
KUr −−= 
 θθπθ eUr
KeUVr ˆsin2
ˆcos ⎟⎠
⎞⎜⎝
⎛ −+=r 
 Stagnations point: θ = π/2 
U
Kr π2= 
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 
 
7.1 
gL
V 20 
7.3 22ωρL
E 
7.5 ⎟⎠
⎞⎜⎝
⎛=
Re
1
0LV
ν 
7.7 
D
L
VDV
p ,,2
ν
ρ
Δ 
7.9 VDF μ∝ 
7.11 ⎟⎟⎠
⎞⎜⎜⎝
⎛=Δ
D
d
VD
f
V
p ,2 ρ
μ
ρ 
7.13 ⎟⎟⎠
⎞⎜⎜⎝
⎛=
UL
f
U
w
ρ
μ
ρ
τ
2 
7.15 33 , pgpg
W
ρ
σ
ρ 
7.17 ⎟⎠
⎞⎜⎝
⎛=
D
fgDV λ 
7.19 ⎟⎠
⎞⎜⎝
⎛=
h
bfghhQ 2 
7.21 ⎟⎠
⎞⎜⎝
⎛=
D
c
D
Lf
D
W ,2ωμ 
7.23 
D
d
D
d
D
d
D
p
D
oi ,,,, 2235 ωρωρ
ΔP 
7.25 Four parameters 21231 gdρ
μ=Π 
7.27 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
gh
VVhf
Vh
Q 2
2 ,μ
ρ 
7.29 2,, DVVDD
d
ρ
σ
ρ
μ 
7.31 
d
D
d
h
VddV
W ,,,22 ρ
μ
ρ 
7.33 
pDpDD
d
ΔΔ
σ
ρ
μ ,, 2
2
 
7.35 
T
I
T
D
D
L
D
23
,,, ωμωδ 
7.37 2
3
222 ,,,,, ωρ
ρ
ωρ
μ
ωρ D
gNcD
DD
p pΔ 
7.39 ⎟⎟⎠
⎞⎜⎜⎝
⎛=
D
l
D
c
p
f
Dp
,,3
ρω
ω
P 
7.41 Four primary dimensions ⎟⎟⎠
⎞
⎜⎜⎝
⎛ Θ=
VLV
c
fLVQ p ρ
μρ ,223& 
7.43 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
LVcL
k
c
cf
V
Lc
dt
dT
pp
p
ρ
μ
ρ ,, 23 
7.45 
U
u=Π1 δ
y=Π2 ( )U
dydU δ=Π3 Uδ
ν=Π4 
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 
7.61 ⎟⎟⎠
⎞⎜⎜⎝
⎛= μ
ρVdF
V
fd 
2
1
2
1 =
V
V 
4
1
2
1 =
f
f 
7.63 Vm = 0.618 m/s – 1.03 m/s 
7.65 ⎟⎟⎠
⎞⎜⎜⎝
⎛= 21212 VAfAV
FD
ρ
μ
ρ kN46.2=pDF P =55.1 kW (73.9 hp) 
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 kN89.3=
pD
F Vm = 250 m/s (model is impractical, compressible flow) 
7.73 Model = 
50
1 x Prototype Adequate Reynolds number not achievable 
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 
7.81 ⎟⎠
⎞⎜⎝
⎛=
D
g
D
Vf
D
Ft
2142 ,ωωρω ⎟⎠
⎞⎜⎝
⎛=
D
g
D
Vf
D
T
2252 ,ωωρω ⎟⎠
⎞⎜⎝
⎛=
D
g
D
Vf
D 2353
,ωωρω
P 
7.83 K.E. ratio = 7.22 
7.85 FB ≈ 0.273 N (to right) 443.0mD =C kN64.1pD =F 
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 32max =uV 
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) 
8.15 M = 4.32 kg 3
12
a
L
pDQ μ
π Δ= a = 1.28 x 10– 5 m (12.8 μm) 
8.17 
rh
QV π2= 3
6
rh
Q
dr
dp
π
μ−= ⎟⎠
⎞⎜⎝
⎛−=
R
r
rh
Qpp ln6 3atm π
μ (p = p0, r < R0) 
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 
8.35 B.C.: y = 0, u = U0; y = h, τ = 0 0
2
2
Uhyygu +⎟⎟⎠
⎞
⎜⎜⎝
⎛ −= μ
ρ 
8.37 V
mh
wL
dt
dV π−= t = 1.06 s 
8.39 ⎟⎠
⎞⎜⎝
⎛ −=Δ ω
ωμ
abR
Q
a
LRp 216 2 
( ) ⎟⎠
⎞⎜⎝
⎛ −= ω
ωμ
abR
Q
a
RLb 64
2
P 
⎟⎠
⎞⎜⎝
⎛ −
⎟⎠
⎞⎜⎝
⎛ −
=
ω
ω
ωη
abR
Q
abR
Q
abR
Q
64
21
6 
8.41 
a
LD
4
32
v
πμω=P 
L
pDa
μ
π
12
23
p
Δ=P pv 3PP = 
8.45 r = 0.707R 
8.47 Q = 1.43 x 10– 3 in3/s (0.0857 in3/min) 
8.49 τ = c1/r 21 ln crcu += μ ( )oi rr
Vc
ln
0
1
μ= ( )oi
o
rr
rVc
ln
ln0
2 −= 
8.51 ( )( )
212
1ln
1
2
1 ⎥⎦
⎤⎢⎣
⎡ −=
k
kRr 
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 sm21 =V 
8.75 hlT = 913 ft2/s2 (HlT = 28.4 ft) 
8.77 1s963 −=
dy
ud τw = 3.58 x 10– 4 lbf/ft2 τw = 4.13 x 10– 4 lbf/ft2 
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 
8.97 ( )
2
1
2
12 VARh
ml
−= 
8.99 ( )KARpV −− Δ= 21 1 2ρ Inviscid assumption: Lower indicated flow/larger Δp 
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– 3 lbf/ft2/ft (2:1) (+24.1%) 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 sm46.6=V pF = 705 kPa (gage) P = 832 kW τw = 88.6 Pa 
8.151 dQ/dt = – 0.524 m3/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 kg/s0220.0min =m& 
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 Kp = 1.81 x 1013 $·in5 
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 
9.9 A = U δ
π
2
=B C = 0 
9.11 375.0* =δ
δ 139.0=δ
θ 
9.13 396.0* =δ
δ 152.0=δ
θ 
9.15 Linear: 167.0=δ
θ Sinusoidal: 137.0=δ
θ Parabolic: 133.0=δ
θ 
9.17 Power: 125.0* =δ
δ , 0972.0=δ
θ Parabolic: 333.0* =δ
δ , 133.0=δ
θ 
9.19 skg4.50=abm& 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 *2δ = 2.54 mm Δp = – 107 Pa FD = 2.00 N 
9.35 y = 0.121 in dy/dx = 0.00326 
x
w
U
Re
3321.0
2ρτ= 
L
D
bLUF
Re
6642.0
2ρ= 
 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 
9.55 
5
1
2
Re
0297.0
x
w
Uρτ = 
5
1
2
Re
0360.0
L
D
bLUF ρ= FD = 2.34 N 
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) 
9.61 
5
1
Re
353.0
x
x
=δ 
5
1
Re
0612.0
x
fc = FD = 2.41 N 
9.63 Lδ = 31.3 mm =Lwτ 0.798 Pa 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 δρ 2lam 525.0 Um =& δρ 2turb 777.0 Um =& 
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 
9.77 ( ) ⎥⎦
⎤⎢⎣
⎡
++= xH
c
U
U f
2
1
1 θ (constant wτ ) ( )
4
4
1
11 2
00583.01
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢
⎣
⎡
+⎟⎟⎠
⎞
⎜⎜⎝
⎛+=
H
x
UU
U
θδ
ν ( ≠wτ constant) 
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 
9.87 FD = 3.02 x 104 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 =DC 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 
9.107 ⎥⎦
⎤⎢⎣
⎡= θρ
θ
2cos
sin2
AC
mgV
D
 t = 1.30 mm 
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 
⎥⎥
⎥⎥
⎥
⎦
⎤
⎢⎢
⎢⎢
⎢
⎣
⎡
+⎟⎟⎠
⎞
⎜⎜⎝
⎛−⎟⎟⎠
⎞
⎜⎜⎝
⎛=
12
12
2
1
2
2
1
A
A
A
AA
mgV ρ 
9.119 ( )2
2
1 UVACF DD −= ρ ( ) RUVACT D 22
1 −= ρ ( ) UUVACP D 22
1 −= ρ 
R
V
3opt
=ω 
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 =sρ 3720 kg/m3 V = 0.731 m/s 
9.131 M = 0.0471 kg 
9.133 CL = 1.01 CD = 0.0654 
9.135 DHUCF DD
2
2
1
9
7 ρ= 22
2
1
16
7 DHUCM D ρ= 9
7
uniform
=
D
D
F
F 
8
7
uniform
=
M
M 
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 
9.145 ( )bDbuDu
R
wb ACAC
FVV +−= ρ
2 Vb = 4.56 m/s (16.4 km/hr) 
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 = 3.42o 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 
11.35 
5
1
2
22 ⎟⎟⎠
⎞
⎜⎜⎝
⎛=
gz
Qyc 
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 
11.53 Q = 26.6 ft3/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/m2 
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) 10p = 1.01 MPa, 20p = 189 kPa (irreversible adiabatic) 
 Δs = 480 J/kg·K Flow accelerates even with friction due to large pressure drop 
12.61 
10
T =
20
T = 344 K 
10
p = 223 kPa 
20
p = 145 kPa Δs = 0.124 kJ/kg·K 
12.63 
10
T =
20
T = 445 K 
10
p = 57.5 kPa 
20
p = 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 
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