ALL CHAPTER 1-14 COVERED
SOLUTIONS
MANUAL
,TABLE OF CONTENTS
1. Fundamental Concepts
2. Fluid Statics
3. The Principles Governing Fluids in Motion
4. The Momentum Equation
5. Physical Similarity and Dimensional Analysis
6. Laminar Flow Between Solid Boundaries
7. Flow and Losses in Pipes and Fittings
8. Boundary Layers, Wakes and Other Shear Layers
9. The Flow of an Inviscid Fluid
10. Flow with a Free Surface
11. Compressible Flow of Gases
12. Unsteady Flow
13. Fluid Machines
14. Fluid Mechanics in a Changing World
,Chapter 1
1. V1 = T 1p 2
1 Since pV = mRT , V T p
1 2 1
π 3 288.15 1.1 3
= 56.2 m
∴ V1 = (20 m) 233.15 101.3
6
14.2 p 1.4 × 105 N · m−2
̺= = −3
RT 287 J · kg−1 · K−1 = 1.51 kg · m
× 323.15 K
∂p p − p0
14.3 K=̺ Assume K constant. Then ln(̺/0̺ ) =
∂̺ p − p0 3 81.7 K
× 106
− ex
∴ ̺ = ̺0 exp = 1025 kg · m 2.34 × 109
K p
= 1061 kg · m−3
1. µ 2 × 10−5 N · s · m−2
4
−3
̺= = 15 × 10−6 · s−1 = 1.333 kg · m
ν m2
p 1.013 × 105 N · m−2 −1 −1
R= =
̺T 1.333 kg · m−3 × 293.15 K = 259.2 J · kg ·K
8310
∴M= = 32.06
259.2
1.5 µ = ν̺ = 400 × 10−6 m2 · s−1 × 850 kg · m−3 = 0.34 Pa · s
0.12 m · s−1
−1
Velocity gradient = × 10−3 m = 1200 s
0.1
Area = π 0.2 × 1.2 m2 = 0.754 m 2
Force = 0.754 m2 × 0.34 Pa · s × 1200 s−1 = 307.6 N
,4 Solutions manual
1.6 ∂u
Total force on plate Area µ + ∂u
= × ∂y ∂ y side B
side A
0.15 m · 0.15 m · s−1
= (0.25 m)2 × 0.7 Pa · s s−1 +
0.019 m
0.006 m
= 1.439 N
1.7 For annulus, radius r, width δr
Velocity ωr
Force = Area × µ × = 2π rδrµ
Clearance µω c
3
∴ Torque = Force × r = 2π r δ r
c
R
3 µω π R4µω
Total torque = 0 2πr
dr = 2c
c
π(0.1 m)40.14 Pa · s × 2π × 7 rad · s−1
= = 7.44 N · m
2 × 0.00013 m
1.8 2γ 2 × 0.073 N · m−1
p= = = 36.5 Pa
d 0.004 m
1.9 4γ cos 4 × 0.073 N · m−1 × 1
θ
h= =
̺gd 1000 · m−3 × 9.81 · kg−1 × 0.005 m
N
kg
= 0.00595 m = 5.95 mm
1.10 4 × 0.377 N · m−1 × cos 140◦
h=
(13.56 − 1)1000 kg · m−3 × 9.81 N · kg−1 × 0.006 m
= −1.563 mm
1.11 ud̺
4Q ̺ 4 × 0.0025 m3 · s−1 × 900 kg · m−3
Re = = =
µ πdµ π 0.05 m × 0.038 N · s · −2 = 1508
m
2000µ 2000 × 0.038 N · s · m−2
−1
u= = 0.05 × 900 kg · −3 = 1.689 m · s
d̺ m
m
1.12 4Q ̺ 4 × 0.01 m3 · s−1
Re = = × 370 × 10−6 m2 · −1 = 430 ∴ Laminar
πdµ π0.08 m
s
,Chapter 2
p 200 × 103 N · m−2
2.1 h= = · m−3 × 9.81 · kg−1 = 12.82 m
̺g 1590 kg N
2.2 Pressure depends only on depth below free surface.
(a) p = ̺gh = (820 kg · m−3 × 9.81 N · kg−1)(3 − 0.15) m
= 22 930 N · m−2 = 22.93 kPa
(b) p = 820 × 9.81 N · m−3 × (3 + 2) m = 40.2 kPa
(c) p = 820 × 9.81 N · m−3 × {3 + 2 − (1.2 sin 30◦ + 0.6)} m
= 820 × 9.81 × 3.8 N · m−2 = 30.57 kPa
(d) Load = Pressure × Area
= 820 × 9.81 × 3 N · m−2 × (3.5 × 2.5) m2 = 211.2 kN
p ̺watergh water ̺ ater
w
2.3 hair = = = hwater
̺airg ̺airg ̺air
1000 kg · m−3 × 287 J · kg−1 · K−1 × 288.15 K
= 0.075 m
1.013 × 105 N · m−2
= 61.2 m
2.4 pV = constant
3
d 101.3 × Pa + 1000 kg · m−3 × 9.81 N · kg−1 × 9 m
∴
103
=
4 mm 101.3 × 103 Pa
whence d = 4.93 mm
2.5 p = 820 kg · m−3 × 9.81 N · kg−1 × 2 m + (13.56 − 0.82)
× 1000 kg · m−3 × 9.81 N · kg−1 × 0.225 m = 44.2 kPa
,6 Solutions manual
p∗ 0.225 m(13.56 − 0.82)1000 kg · m−3 × 9.81 N · kg−1
= h= 820 · m−3 × 9.81 · kg−1
̺g
= 3.496 m kg N
44 200 N · m−2 = 820 × 9.81 × 2 N · m−2 + x(0.82 − 0.74)1000
× 9.81 N · m−3
whence x = 35.83 m
2.6 New levels
Movement of fluid
A x y B
in
C = 60 mm × 70 mm 2
= (500 mm2 )x
X X 2
= (800 mm )y
Initial ∴ x = 8.4 mm;
y = 5.25 mm
C
surface of
separation
Measuring above XX: Initially 0.8hA = 0.9hB
Later: 800 × 9.81(Old hA − 60 + 8.4)10−3 Pa
= p + 900 × 9.81(Old hB − 60 + 5.25)10−3 Pa
∴ p = 9.81 × 10−3 (−800 × 51.6 + 900 × 65.25)Pa = 171.1 Pa
g/Rλ
λz
2.7 From eqn 2.7 p = p0 1 −
T0
9.81/287×0.0065
0.0065 × 7500
= 101.5 Pa 1 −
288.15
= 38.3 kPa
2.8 p T0 − λz g/ R λ Ttop g / Rλ
T0 =
= Ttop + λz
p0
Ttop Rλ/g
p0
∴z= λ −1
p
268.1 287×0.0065/9.81
5 749
= m −1
0.0065 566
= 2257 m
, Chapter 2 5
2.9 F = (1.2 × 1.8) m2 × 1000 kg · m−3 × 9.81 N · kg−1
× (x + 0.9 sin 30◦) m
(a) 2160 × 9.81 N · m−1 × 0.45 m = 9.54 kN
(b) 2160 × 9.81 N · m−1 × 0.95 m = 20.13 kN
(c) 2160 × 9.81 N · m−1 × 30.45 m = 645 kN
(bd3/12) + bd(2x + 0.9)2
Centre of pressure is at slant depth bd(2x + 0.9)
d2
= + 2x +
12(2x + 0.9)
0.9(metres)
(1.8
m)2
= + 2x + 0.9 m
12(2x + 0.9)
1.82
that is + 0.9 m from upper edge
12(2x + 0.9)
= (a) 1.2 m; (b) 1.042 m; (c) 0.904 m from upper edge
2.10 By symmetry, centre X X
of pressure is on x r
vertical centre-line
2nd moment about
XX
Depth = 1st moment about XX
r 2 2 2 1/2 dx
= 0 x 2( r − x )
r 2 2 1/2dx
0 x 2(r − x )
= 2
π/2(r cos θ) 2r sin θ(− r sin θdθ)
0
π/2 r cos θ2r sin θ(− r sin θdθ)
0
π/2 2
r 2θ sin θdθ
= 0 cos
π/2 2
0 sin θ cos θdθ
π 1 2
r
0 8 sin 2θd(2θ)
= 3
π/2
1 sin θ
3 0
r/8 [2θ/2 − (1/4) sin 4θ]2θ=π
0
= 1/3
3 π 3π d
= r =
8 2 32
,6 Solutions manual
2.11 X X
60 60 Full depth = (2.5 m) sin 60◦
x
Breadth of strip
(2.5 m) sin 60◦ − x
= 2.5 m (2.5 m) sin 60◦
= 2.5 m − x cosec 60◦
∴ Second moment of area about XX
(2.5 m) sin 60◦ 2.5
= (2.5 m − x cosec 60◦)x2 dx = 4 sin360◦ m4
0
Depth 12
First moment = Area
3
×
2.5 sin 60◦ 2.53 2 3
1
3
sin 60◦ m
= 2.5 × 2.5 sin 60◦ × m = 6
2 3
2.5 Dept
∴ Depth of C.P. = sin 60◦ m =
2 h2
∴ Thrust is equally divided between XX and bottom.
Thrust = Area × Pressure at centroid
1 2 2.5 sin 60◦
sin 60◦ × 1000 × 9.81 × N = 19 160 N
= 2.5 3
2
∴ Load at bottom = 9580 N; at each upper corner 4790 N
2.12 Let shaft be at depth h below free surface. Then force on disc
= πR2̺ gh.
By parallel axes theorem, 2nd moment of area about free
surface = π R4/4 + π R 2h 2.
1st moment of area about free surface = π R2 h
R2 2
∴ Depth of C.P. = + h below free surface, that is, R /4h
4h
below shaft
∴ Turning moment on shaft
R2 πR4̺g
= π R2
̺ g h × 4h = [independent of h]
4
π(0.6 m)41000 kg · m−3 × 9.81 N · kg−1
= = 999 N · m
4
, Chapter 2 7
2.1
3 0.5 m Force on plate
1.5 m
= 1150 kg · m−3
× 9.81 N · kg−1
√
C × 1.5 m( 2 m)2
2m
= 33.84 kN
l Al 2
2
(Ak )c, ⊥ plate =
6
Al 2
∴ (Ak2)c, diagonal =
12
since diagonals are perpendicular
∴ Depth of C.P. below free surface
(Al2/12) + Ay2 l2 (√ 2)2
m
= = y+ = 1.5 + × 1.5
Ay 12y 12
= 1.611 m, that is, 1.111 m from top of aperture
√
∴ Total moment about hinge = 33.84 kN × 1.111/ 2 m
= 26.59 kN · m
2.14 Width of gates = (3 m) sec 30◦ = 3.464 m
Thrust on ‘deep’ side of gate
= (1000 × 9.81 × 4.5)(9 × 3.464)N = 1.376 MN
Trust on ‘shallow’ side of gate
= (1000 × 9.81 × 1.35)(2.7 × 3.464)N
= 0.124 MN
Net thrust = (1.376 − 0.124) MN = 1.252 MN
1.252 MN
∴ Force between gates = 1.252 MN
= 2 sin 30◦
Resultant force F acts at height y given by
h1 h2 2 2
F1 − F2 = Fy, since F 1, F 2 act at h 1, h2 below free surfaces
3 3 × 9/3 − 0.124 × 2.7/3
1.376 3 3
∴ y= m = 3.208 m
1.252
Total hinge reaction R also acts at this height.
, 8 Solutions manual
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SOLUTIONS
MANUAL
,TABLE OF CONTENTS
1. Fundamental Concepts
2. Fluid Statics
3. The Principles Governing Fluids in Motion
4. The Momentum Equation
5. Physical Similarity and Dimensional Analysis
6. Laminar Flow Between Solid Boundaries
7. Flow and Losses in Pipes and Fittings
8. Boundary Layers, Wakes and Other Shear Layers
9. The Flow of an Inviscid Fluid
10. Flow with a Free Surface
11. Compressible Flow of Gases
12. Unsteady Flow
13. Fluid Machines
14. Fluid Mechanics in a Changing World
,Chapter 1
1. V1 = T 1p 2
1 Since pV = mRT , V T p
1 2 1
π 3 288.15 1.1 3
= 56.2 m
∴ V1 = (20 m) 233.15 101.3
6
14.2 p 1.4 × 105 N · m−2
̺= = −3
RT 287 J · kg−1 · K−1 = 1.51 kg · m
× 323.15 K
∂p p − p0
14.3 K=̺ Assume K constant. Then ln(̺/0̺ ) =
∂̺ p − p0 3 81.7 K
× 106
− ex
∴ ̺ = ̺0 exp = 1025 kg · m 2.34 × 109
K p
= 1061 kg · m−3
1. µ 2 × 10−5 N · s · m−2
4
−3
̺= = 15 × 10−6 · s−1 = 1.333 kg · m
ν m2
p 1.013 × 105 N · m−2 −1 −1
R= =
̺T 1.333 kg · m−3 × 293.15 K = 259.2 J · kg ·K
8310
∴M= = 32.06
259.2
1.5 µ = ν̺ = 400 × 10−6 m2 · s−1 × 850 kg · m−3 = 0.34 Pa · s
0.12 m · s−1
−1
Velocity gradient = × 10−3 m = 1200 s
0.1
Area = π 0.2 × 1.2 m2 = 0.754 m 2
Force = 0.754 m2 × 0.34 Pa · s × 1200 s−1 = 307.6 N
,4 Solutions manual
1.6 ∂u
Total force on plate Area µ + ∂u
= × ∂y ∂ y side B
side A
0.15 m · 0.15 m · s−1
= (0.25 m)2 × 0.7 Pa · s s−1 +
0.019 m
0.006 m
= 1.439 N
1.7 For annulus, radius r, width δr
Velocity ωr
Force = Area × µ × = 2π rδrµ
Clearance µω c
3
∴ Torque = Force × r = 2π r δ r
c
R
3 µω π R4µω
Total torque = 0 2πr
dr = 2c
c
π(0.1 m)40.14 Pa · s × 2π × 7 rad · s−1
= = 7.44 N · m
2 × 0.00013 m
1.8 2γ 2 × 0.073 N · m−1
p= = = 36.5 Pa
d 0.004 m
1.9 4γ cos 4 × 0.073 N · m−1 × 1
θ
h= =
̺gd 1000 · m−3 × 9.81 · kg−1 × 0.005 m
N
kg
= 0.00595 m = 5.95 mm
1.10 4 × 0.377 N · m−1 × cos 140◦
h=
(13.56 − 1)1000 kg · m−3 × 9.81 N · kg−1 × 0.006 m
= −1.563 mm
1.11 ud̺
4Q ̺ 4 × 0.0025 m3 · s−1 × 900 kg · m−3
Re = = =
µ πdµ π 0.05 m × 0.038 N · s · −2 = 1508
m
2000µ 2000 × 0.038 N · s · m−2
−1
u= = 0.05 × 900 kg · −3 = 1.689 m · s
d̺ m
m
1.12 4Q ̺ 4 × 0.01 m3 · s−1
Re = = × 370 × 10−6 m2 · −1 = 430 ∴ Laminar
πdµ π0.08 m
s
,Chapter 2
p 200 × 103 N · m−2
2.1 h= = · m−3 × 9.81 · kg−1 = 12.82 m
̺g 1590 kg N
2.2 Pressure depends only on depth below free surface.
(a) p = ̺gh = (820 kg · m−3 × 9.81 N · kg−1)(3 − 0.15) m
= 22 930 N · m−2 = 22.93 kPa
(b) p = 820 × 9.81 N · m−3 × (3 + 2) m = 40.2 kPa
(c) p = 820 × 9.81 N · m−3 × {3 + 2 − (1.2 sin 30◦ + 0.6)} m
= 820 × 9.81 × 3.8 N · m−2 = 30.57 kPa
(d) Load = Pressure × Area
= 820 × 9.81 × 3 N · m−2 × (3.5 × 2.5) m2 = 211.2 kN
p ̺watergh water ̺ ater
w
2.3 hair = = = hwater
̺airg ̺airg ̺air
1000 kg · m−3 × 287 J · kg−1 · K−1 × 288.15 K
= 0.075 m
1.013 × 105 N · m−2
= 61.2 m
2.4 pV = constant
3
d 101.3 × Pa + 1000 kg · m−3 × 9.81 N · kg−1 × 9 m
∴
103
=
4 mm 101.3 × 103 Pa
whence d = 4.93 mm
2.5 p = 820 kg · m−3 × 9.81 N · kg−1 × 2 m + (13.56 − 0.82)
× 1000 kg · m−3 × 9.81 N · kg−1 × 0.225 m = 44.2 kPa
,6 Solutions manual
p∗ 0.225 m(13.56 − 0.82)1000 kg · m−3 × 9.81 N · kg−1
= h= 820 · m−3 × 9.81 · kg−1
̺g
= 3.496 m kg N
44 200 N · m−2 = 820 × 9.81 × 2 N · m−2 + x(0.82 − 0.74)1000
× 9.81 N · m−3
whence x = 35.83 m
2.6 New levels
Movement of fluid
A x y B
in
C = 60 mm × 70 mm 2
= (500 mm2 )x
X X 2
= (800 mm )y
Initial ∴ x = 8.4 mm;
y = 5.25 mm
C
surface of
separation
Measuring above XX: Initially 0.8hA = 0.9hB
Later: 800 × 9.81(Old hA − 60 + 8.4)10−3 Pa
= p + 900 × 9.81(Old hB − 60 + 5.25)10−3 Pa
∴ p = 9.81 × 10−3 (−800 × 51.6 + 900 × 65.25)Pa = 171.1 Pa
g/Rλ
λz
2.7 From eqn 2.7 p = p0 1 −
T0
9.81/287×0.0065
0.0065 × 7500
= 101.5 Pa 1 −
288.15
= 38.3 kPa
2.8 p T0 − λz g/ R λ Ttop g / Rλ
T0 =
= Ttop + λz
p0
Ttop Rλ/g
p0
∴z= λ −1
p
268.1 287×0.0065/9.81
5 749
= m −1
0.0065 566
= 2257 m
, Chapter 2 5
2.9 F = (1.2 × 1.8) m2 × 1000 kg · m−3 × 9.81 N · kg−1
× (x + 0.9 sin 30◦) m
(a) 2160 × 9.81 N · m−1 × 0.45 m = 9.54 kN
(b) 2160 × 9.81 N · m−1 × 0.95 m = 20.13 kN
(c) 2160 × 9.81 N · m−1 × 30.45 m = 645 kN
(bd3/12) + bd(2x + 0.9)2
Centre of pressure is at slant depth bd(2x + 0.9)
d2
= + 2x +
12(2x + 0.9)
0.9(metres)
(1.8
m)2
= + 2x + 0.9 m
12(2x + 0.9)
1.82
that is + 0.9 m from upper edge
12(2x + 0.9)
= (a) 1.2 m; (b) 1.042 m; (c) 0.904 m from upper edge
2.10 By symmetry, centre X X
of pressure is on x r
vertical centre-line
2nd moment about
XX
Depth = 1st moment about XX
r 2 2 2 1/2 dx
= 0 x 2( r − x )
r 2 2 1/2dx
0 x 2(r − x )
= 2
π/2(r cos θ) 2r sin θ(− r sin θdθ)
0
π/2 r cos θ2r sin θ(− r sin θdθ)
0
π/2 2
r 2θ sin θdθ
= 0 cos
π/2 2
0 sin θ cos θdθ
π 1 2
r
0 8 sin 2θd(2θ)
= 3
π/2
1 sin θ
3 0
r/8 [2θ/2 − (1/4) sin 4θ]2θ=π
0
= 1/3
3 π 3π d
= r =
8 2 32
,6 Solutions manual
2.11 X X
60 60 Full depth = (2.5 m) sin 60◦
x
Breadth of strip
(2.5 m) sin 60◦ − x
= 2.5 m (2.5 m) sin 60◦
= 2.5 m − x cosec 60◦
∴ Second moment of area about XX
(2.5 m) sin 60◦ 2.5
= (2.5 m − x cosec 60◦)x2 dx = 4 sin360◦ m4
0
Depth 12
First moment = Area
3
×
2.5 sin 60◦ 2.53 2 3
1
3
sin 60◦ m
= 2.5 × 2.5 sin 60◦ × m = 6
2 3
2.5 Dept
∴ Depth of C.P. = sin 60◦ m =
2 h2
∴ Thrust is equally divided between XX and bottom.
Thrust = Area × Pressure at centroid
1 2 2.5 sin 60◦
sin 60◦ × 1000 × 9.81 × N = 19 160 N
= 2.5 3
2
∴ Load at bottom = 9580 N; at each upper corner 4790 N
2.12 Let shaft be at depth h below free surface. Then force on disc
= πR2̺ gh.
By parallel axes theorem, 2nd moment of area about free
surface = π R4/4 + π R 2h 2.
1st moment of area about free surface = π R2 h
R2 2
∴ Depth of C.P. = + h below free surface, that is, R /4h
4h
below shaft
∴ Turning moment on shaft
R2 πR4̺g
= π R2
̺ g h × 4h = [independent of h]
4
π(0.6 m)41000 kg · m−3 × 9.81 N · kg−1
= = 999 N · m
4
, Chapter 2 7
2.1
3 0.5 m Force on plate
1.5 m
= 1150 kg · m−3
× 9.81 N · kg−1
√
C × 1.5 m( 2 m)2
2m
= 33.84 kN
l Al 2
2
(Ak )c, ⊥ plate =
6
Al 2
∴ (Ak2)c, diagonal =
12
since diagonals are perpendicular
∴ Depth of C.P. below free surface
(Al2/12) + Ay2 l2 (√ 2)2
m
= = y+ = 1.5 + × 1.5
Ay 12y 12
= 1.611 m, that is, 1.111 m from top of aperture
√
∴ Total moment about hinge = 33.84 kN × 1.111/ 2 m
= 26.59 kN · m
2.14 Width of gates = (3 m) sec 30◦ = 3.464 m
Thrust on ‘deep’ side of gate
= (1000 × 9.81 × 4.5)(9 × 3.464)N = 1.376 MN
Trust on ‘shallow’ side of gate
= (1000 × 9.81 × 1.35)(2.7 × 3.464)N
= 0.124 MN
Net thrust = (1.376 − 0.124) MN = 1.252 MN
1.252 MN
∴ Force between gates = 1.252 MN
= 2 sin 30◦
Resultant force F acts at height y given by
h1 h2 2 2
F1 − F2 = Fy, since F 1, F 2 act at h 1, h2 below free surfaces
3 3 × 9/3 − 0.124 × 2.7/3
1.376 3 3
∴ y= m = 3.208 m
1.252
Total hinge reaction R also acts at this height.
, 8 Solutions manual
THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
