fox 7th - answers selected problems

23
Introduction to Fluid Mechanics 7 th 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 t g W y = 1.11 d = 0.109 mm 1.15 y = 0.922 mm 1.17 a) N·m/s, lbf·ft/s b) N/m 2 , lbf/ft 2 c) N/m 2 , lbf/ft 2 , d) 1/s, 1/s e) N·m, lbf·ft f) N·s, lbf·s g) N/m 2 , lbf/ft 2 h) m 2 /s 2 ·K, ft 2 /s 2 ·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/m 2 1.21 a) 0.0472 m 3 /s b) 0.0189 m 3 c) 29.1 m/s d) 2.19 x 10 4 m 2 1.23 101 gpm 1.25 SG = 13.6 v = 7.37 x 10 5 m 3 /kg γ E = 847 lbf/ft 3 , γ M = 144 lbf/ft 3 1.27 2.25 kgf/cm 2 1.29 c = 0.04 K 1/2 ·s/m 0 0 max 36 . 2 T p A m t = & (ft 2 , psi, R) 1.31 C D 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(m 3 /s)) 2 1.37 ρ = 0.0765 ± 2.66 x 10 4 lbm/ft 3 (± 0.348%) 1.39 ρ = 1130 ± 21.4 kg/m 3 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 3 N·s/m 2 ± 0.609% 1.47 δx = ± 0.158 mm 1.49 H = 57.7 ± 0.548 ft θ min = 31.4 o

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Page 1: Fox 7th - Answers Selected Problems

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

05.2tg

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

0max 36.2

TpAm 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

Page 2: Fox 7th - Answers Selected Problems

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

ab

xcy−

=

2.7 Streamlines:

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

cxa

by

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

ω2

=

2.17 Pathlines: tCeyy 0= , ( )221

0AttBexx += Streamlines: txy 5.01

1+=

2.19 Pathlines: tbeyy −= 0 , 2

21

0atexx = Streamlines: ta

b

Cxy−

=

2.21 Pathlines: 14 += ty , 205.03 tex = Streamlines: ⎟

⎠⎞

⎜⎝⎛+=

3ln401 x

ty

2.23 Streamlines: ( ) ( )( )000 sin tttvty −= ω , ( ) ( )000 ttutx −=

2.25 Streaklines: ( )τ−= tey , ( ) ( )221.0 ττ −+−= ttex Streamlines: ( )txy 2.011

+=

2.29 Streamlines: 44

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

STb′+

′=

1

23

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

eA

MgdVμ

μθ 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

aMgrm

3

2

2πμ = μ = 0.0651 N·s/m2

Page 3: Fox 7th - Answers Selected Problems

2.55 t = 4.00 s

2.57 ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

− tCB

eBA 1ω ωmax = 25.1 rpm t = 0.671 s

2.59 h

rz

μωτ θ = hRT

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

HRT32πμω

= bRT

2

4πμω=

2.69 ( )θθμωτ

cos1sin−+

=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

Page 4: Fox 7th - Answers Selected Problems

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 D

d

pL+Δ

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

dHLM3

4 23

−=

ρ M = 583 kg

3.85 ( ) ( ) ( )⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ ++−=

42sin

2sin2 maxmax

maxθθθρ RRdLRM M = 631 kg

Page 5: Fox 7th - Answers Selected Problems

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)

Page 6: Fox 7th - Answers Selected Problems

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−=⋅∫ AdVrr

( ) ( ) 24 smˆ60ˆ96ˆ64 kjiAdVV −−=⋅∫rrr

4.15 2max2

1 RuQ π= iRu ˆm.f. 22max3

1 π= 4.17 V = 1.56 ft/s Q = 3.82 gpm 4.19 Qcool = 489 gpm hrlb1045.2 5

cool ×=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 ( )μθρ

3sin 32 hg

wm=

&

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

yt2

tan52 25

02 θπ

= 0

0

56

QVt =

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 ( )⎥⎥⎦

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−+−=

222 1sin1

4 DdDVRx θπρ Rx = − 314 N

Page 7: Fox 7th - Answers Selected Problems

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 ⎟⎠⎞

⎜⎝⎛ −=

ππρ 28

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 22

0 −= 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 −=

20

0

21Vgz

AA−

=

4.117 22

0 ⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛−=

Lx

whQpp ρ

4.119 ( )tVhrVV

00

0

2 −=

4.123 Rx = − 2400 N Ry = 1386 N

4.125 jVkQ

kQ

kQV 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

MVU

ρ0

0ln V = 0.61 m/s

4.139 amax at t = 0 θ = 90o U → V

4.141 ( )232

atVaMA−

A = 111 mm2

4.143 h = 17.9 mm

4.145 ( )M

kUAUVa222 −−

=ρ a = 5.99 m/s2 U/Ut = 0.667

4.147 a = 14.2 m/s2 t = 15.2 m/s

Page 8: Fox 7th - Answers Selected Problems

4.149 ( )M

AUVa2+

−=ρ

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

0

1UVVA

Mtρ

4.151 V = 5 m/s xmax = 1.93 m t = 2.51 s 4.153 a = 2.28 m/s2

4.155 ⎟⎟⎠

⎞⎜⎜⎝

⎛−+=

00 1ln

MtmVUU 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

MAU

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 2

23 VARQRVTAR

ωρρρ

ω −−=& 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 23

32 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−=Δmmef&

Δt = 4.49 x 10−4 K (oC)

Page 9: Fox 7th - Answers Selected Problems

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(00167.0max

=⎟⎠⎞

Uv

5.13 ⎥⎥⎦

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

42

max 21

83

δδδ yyxU

v 1=δy

⎥⎥⎦

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

42

max 21

83

δδδ yyxU

v

5.15 v = − 2Axy3/3 +f(x) ψ = xy3/2

5.19 ( )rfr

V +Λ

−= 2sinθ

θ

5.23 h

Uy2

2

=ψ 2

hy =

5.25 θθπθ iUi

rqUV r

ˆsinˆ2

cos +⎟⎠⎞

⎜⎝⎛ +−=

r ( ) ⎟

⎠⎞

⎜⎝⎛= 0,

2,

Uqrπ

θ

5.27 2D Incompressible ψ = zy3 − z3y

5.29 h

Uy2

2

=ψ hhalf = 1.06 m

5.31 ⎥⎥⎦

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

32

31

δδδψ yyU 442.0=

δy (1/4 flow) 652.0=

δy (1/2 flow)

5.33 ⎥⎥⎦

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

42

81

43

δδδψ yyU 465.0=

δy (1/4 flow) 671.0=

δy (1/2 flow)

5.35 ψ = − Cln(r) + C1 Q/b = 0.0912 m3/s/m

5.37 2D Incompressible 2m/sˆ3

16ˆ3

32ˆ3

16 kjia ++=r

5.39 ( ) 242 m/sˆ10ˆ1086.2 jia −− +−=r dy/dx = 0.0025

5.41 Incompressible ( )222

2

yxxax

+

Λ−= ( )222

2

yxyay

+

Λ−= 3

100r

a −=

5.43 ⎟⎠⎞

⎜⎝⎛ −−=

Lx

LUax 2

12

2

5.45 3

2 12 rhQa ⎟

⎠⎞

⎜⎝⎛−=π

(Radial)

5.47 ⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−−a

Uta

Ut

eea

UADtDc 2

21 x = 2.77 m

sppm1025.1 5

max

−×=DtDc

5.49 ∂T/∂x = − 0.0873 oF/mile

Page 10: Fox 7th - Answers Selected Problems

5.53 ( )jyixAa ˆˆ2 +=r jia ˆ2ˆ

21

+=r jia ˆˆ +=

r jia ˆ21ˆ2 +=

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

31

34

δδδyyy

xUax ax(max) = − 5.22 m/s2

5.59 ⎟⎠⎞

⎜⎝⎛ −=

hyvv 10 j

hy

hvi

hxva ˆ1ˆ

20

2

20 ⎟

⎠⎞

⎜⎝⎛ −+=

r

5.61 322

12⎟⎠⎞

⎜⎝⎛

⎥⎥⎦

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

rR

rR

RUar aθ = 0 r = 1.29R (max a)

θ22

2

sin4RUar −= θθθ cossin4

2

2

RUa = 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)(k̂−=ωr

5.79 hu

2−=ω ω = − 0.5 s−1

5.81 ⎟⎠⎞

⎜⎝⎛ −−=Γ

bh

bhUL 1 Γ = − UL/4 (h = b/2) Γ = 0 (h = b)

5.83 2max2

byU

−=γ kb

yU ˆ22

max=ζr

5.85 2

max

2 ⎟⎠⎞

⎜⎝⎛−=δπμU

VddF 3

max

mkN85.1−=

VddF

Page 11: Fox 7th - Answers Selected Problems

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 ˆˆ

conv +−−+=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 ⎟⎠⎞

⎜⎝⎛ −−=

Lx

LUax 2

12

2

⎟⎠⎞

⎜⎝⎛ −=

Lx

LU

dxdp

21

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

xAxALAAAuLA

xp

iei

ieii

−+−

−=∂∂ ρ ( )

⎪⎪⎭

⎪⎪⎬

⎪⎪⎩

⎪⎪⎨

⎥⎥⎥

⎢⎢⎢

−−

−+=

2

2

12 x

LAAA

Auppei

i

iii

ρ

6.17 ( )( ) 5

2

1

2

⎥⎦

⎤⎢⎣

⎡ −+

−−=

xLDDDLD

DDVa

i

ioi

ioix MPa/m10

max

−=∂∂xp L ≥ 1 m

6.19 Fz = − 1.56 N (Acts downwards)

6.21 2

22b

xVax = 2

22b

xVxp ρ

−=∂∂

⎥⎥⎦

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−+=

2

2

22

atm 1Lx

bLVpp ρ 2

32

34

bWLVFy

ρ=

6.23 2

2

hxqax = 2

2

hxq

xp ρ

−=∂∂

6.25 θρ ee

rp r ˆ0ˆ2

5

2

=∇

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

Page 12: Fox 7th - Answers Selected Problems

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

11 21

1

VzzgAA

−+

=

6.55 V = 262 m/s 6.57 Q = 304 gpm (0.676 ft3/s)

6.59 ( )θρ 22 sin4121

−+= ∞ 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=& dtVd

dtdM air

wρ−=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡+−=

588.0

0

0

00

270.11VAtp

VVVM

w

tww ρ

ρ

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⎥⎦⎤

⎢⎣⎡ +−= 22

6.97 ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +

+⎟⎠⎞

⎜⎝⎛ −

= −−

xhy

xhyq 11 tantan

2πψ ( )[ ] ( )[ ]{ }2222ln

2hyxhyxq

++−+−=π

φ

6.99 NOTE: Error – function is ψ = Ax2y − By3 32

33 xABxy −=φ

6.101 ( ) AxyyxB 22

22 −−=ψ

6.105 ( ) jByiBxAV ˆ2ˆ2 ++−=r

ψ = − (Ay + 2Bxy) Δp = 12 kPa

6.107 V = x2 + y2 32

33 xByAx −=ψ

6.109 Incompressible Irrotational Stagnation point: (− 2, 4/3)

( ) CyBxxyA−−−= 22

2φ Δp = 55.8 kPa

6.113 ( ) θθθπ

ψ sin2 21 Urq

+−= θπ

φ cosln2 1

2 Urrrq

−⎟⎟⎠

⎞⎜⎜⎝

⎛=

Page 13: Fox 7th - Answers Selected Problems

jrr

qiUrr

qV ˆsinsin2

ˆcoscos2 2

2

1

1

2

2

1

1⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎥

⎤⎢⎣

⎡+⎟⎟

⎞⎜⎜⎝

⎛−=

θθπ

θθπ

r

Stagnations points: θ = 0, π r = 0.367 m ψstag = 0

6.115 rKUr ln2

sinπ

θψ −= θπ

θφ2

cos KUr −−=

θθπ

θ eUr

KeUV r ˆsin2

ˆcos ⎟⎠⎞

⎜⎝⎛ −+=

r

Stagnations point: θ = π/2 U

Krπ2

=

6.117 Stagnations points: θ = 63o, 297o r = 1.82 m Δp = 317 Pa

Page 14: Fox 7th - Answers Selected Problems

Introduction to Fluid Mechanics 7th Edition

Fox, Pritchard, & McDonald

Answers to Selected Problems, Chapter 7

7.1 gLV 2

0

7.3 22ωρLE

7.5 ⎟⎠⎞

⎜⎝⎛=

Re1

0LVν

7.7 DL

VDVp ,,2

νρΔ

7.9 VDF μ∝

7.11 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

ΔDd

VDf

Vp ,2 ρ

μρ

7.13 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

ULf

Uw

ρμ

ρτ

2

7.15 33 ,pgpg

Wρσ

ρ

7.17 ⎟⎠⎞

⎜⎝⎛=

DfgDV λ

7.19 ⎟⎠⎞

⎜⎝⎛=

hbfghhQ 2

7.21 ⎟⎠⎞

⎜⎝⎛=

Dc

DLf

DW ,2ωμ

7.23 Dd

Dd

Dd

Dp

Doi ,,,, 2235 ωρωρ

ΔP

7.25 Four parameters 21231 gdρμ

7.27 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

ghVVhf

VhQ 2

2 ,μρ

7.29 2,,DVVDD

dρσ

ρμ

7.31 dD

dh

VddVW ,,,22 ρ

μρ

7.33 pDpDD

dΔΔσ

ρμ ,, 2

2

7.35 T

ITD

DL

D

23

,,, ωμωδ

Page 15: Fox 7th - Answers Selected Problems

7.37 23

222 ,,,,,ωρ

ρωρ

μωρ D

gNcDDD

p pΔ

7.39 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

Dl

Dc

pf

Dp,,3

ρωωP

7.41 Four primary dimensions ⎟⎟⎠

⎞⎜⎜⎝

⎛ Θ=

VLVc

fLVQ p

ρμρ ,2

23&

7.43 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

LVcLk

ccf

VLc

dtdT

pp

p

ρμ

ρ,, 23

7.45 Uu

=Π1 δy

=Π2 ( )U

dydU δ=Π3

Uδν

=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 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

μρVdF

Vfd

21

2

1 =VV

41

2

1 =ff

7.63 Vm = 0.618 m/s – 1.03 m/s

7.65 ⎟⎟⎠

⎞⎜⎜⎝

⎛= 21212 VA

fAV

FD

ρμ

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=

pDF Vm = 250 m/s (model is impractical, compressible flow)

7.73 Model = 501 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 ⎟⎠⎞

⎜⎝⎛=

Dg

DVf

DFt

2142 ,ωωρω

⎟⎠⎞

⎜⎝⎛=

Dg

DVf

DT

2252 ,ωωρω

⎟⎠⎞

⎜⎝⎛=

Dg

DVf

D 2353 ,ωωρω

P

7.83 K.E. ratio = 7.22 7.85 FB ≈ 0.273 N (to right) 443.0

mD =C kN64.1pD =F

Page 16: Fox 7th - Answers Selected Problems

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

12a

LpDQμ

π Δ= a = 1.28 x 10– 5 m (12.8 μm)

8.17 rh

QVπ2

= 3

6rhQ

drdp

πμ

−= ⎟⎠⎞

⎜⎝⎛−=

Rr

rhQpp ln6

3atm πμ (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

2Uhyygu +⎟⎟

⎞⎜⎜⎝

⎛−=

μρ

8.37 VmhwL

dtdV π

−= t = 1.06 s

8.39 ⎟⎠⎞

⎜⎝⎛ −=Δ

ωωμ

abRQ

aLRp 216

2 ( )⎟⎠⎞

⎜⎝⎛ −=

ωωμ

abRQ

aRLb 64

2

P ⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −

=

ω

ωω

η

abRQ

abRQ

abRQ

64

216

8.41 a

LD4

32

vπμω

=P LpDa

μπ

12

23

=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 rrVc

ln0

= ( )oi

o

rrrVc

lnln0

2 −=

8.51 ( )( )

212

1ln1

21

⎥⎦

⎤⎢⎣

⎡ −=

kkRr

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

Page 17: Fox 7th - Answers Selected Problems

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 −=dyud 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

12

12 VARhml

−=

8.99 ( )KARpV−−

Δ= 21 1

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

Page 18: Fox 7th - Answers Selected Problems

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

Page 19: Fox 7th - Answers Selected Problems

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 δπ2

=B C = 0

9.11 375.0*=

δδ 139.0=

δθ

9.13 396.0*=

δδ 152.0=

δθ

9.15 Linear: 167.0=δθ Sinusoidal: 137.0=

δθ Parabolic: 133.0=

δθ

9.17 Power: 125.0*=

δδ , 0972.0=

δθ Parabolic: 333.0*

=δδ , 133.0=

δθ

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

wURe

3321.02ρτ =

LD

bLUFRe

6642.02ρ

=

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

Re0297.0

x

wUρτ =

51

2

Re0360.0

L

DbLUF 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

1Re

353.0

xx=

δ 5

1Re

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)

Page 20: Fox 7th - Answers Selected Problems

9.71 δρ 2lam 525.0 Um =& δρ 2

turb 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 ( ) ⎥⎦

⎤⎢⎣

⎡+

+= xHc

UU f

21

1 θ (constant wτ ) ( )

4

41

11 200583.01

⎥⎥⎥

⎢⎢⎢

+⎟⎟⎠

⎞⎜⎜⎝

⎛+=

Hx

UUU

θδν ( ≠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 ⎥⎦

⎤⎢⎣

⎡=

θρθ2cos

sin2ACmgV

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

AA

AAA

mgVρ

9.119 ( )221 UVACF DD −= ρ ( ) RUVACT D

2

21

−= ρ ( ) UUVACP D2

21

−= ρ R

V3opt =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 DD2

21

97 ρ= 22

21

167 DHUCM D ρ=

97

uniform

=D

D

FF

87

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

Rwb ACAC

FVV+

−=ρ

2 Vb = 4.56 m/s (16.4 km/hr)

9.147 x ≈ 203 m

Page 21: Fox 7th - Answers Selected Problems

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

Page 22: Fox 7th - Answers Selected Problems

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 51

2

22⎟⎟⎠

⎞⎜⎜⎝

⎛=

gzQyc

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

Page 23: Fox 7th - Answers Selected Problems

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

10T =20T = 344 K

10p = 223 kPa 20p = 145 kPa Δs = 0.124 kJ/kg·K

12.63 10T =

20T = 445 K 10p = 57.5 kPa

20p = 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