, Contents
Preface iv
Chapter 1 Basic Considerations 1
Chapter 2 Fluid Statics 15
Chapter 3 Introduction to Fluids in Motion 29
Chapter 4 The Integral Forms of the Fundamental Laws 37
Chapter 5 The Differential Forms of the Fundamental Laws 59
Chapter 6 Dimensional Analysis and Similitude 77
Chapter 7 Internal Flows 87
Chapter 8 External Flows 109
Chapter 9 Compressible Flow 131
Chapter 10 Flow in Open Channels 141
Chapter 11 Flows in Piping Systems 161
Chapter 12 Turbomachinery 175
Chapter 13 Measurements in Fluid Mechanics 189
, Chapter 1 / Basic Considerations
CHAPTER 1
Basic Considerations
FE-type Exam Review Problems: Problems 1.1 to 1.13
1.1 (C) m = F/a or kg = N/m/s2 = N·s2/m
1.2 (B) [μ] = [τ/(du/dy)] = (F/L2)/(L/T)/L = F.T/L2
1.3 (A) 2.36 ×10−8 Pa = 23.6 ×10−9 Pa = 23.6 nPa
The mass is the same on earth and the moon, so we calculate the mass using the
weight given on earth as:
1.4 (C) m = W/g = 250 N/9.81 m/s2 = 25.484 kg
Hence, the weight on the moon is:
W = mg = 25.484 × 1.6 = 40.77 N
The shear stress is due to the component of the force acting tangential to the
area:
1.5 (C) Fshear = F sin θ = 4200sin 30 = 2100 N
F 2100 N
τ = shear = −4
= 84 ×103 Pa or 84 kPa
A 250 ×10 m 2
1.6 (B) – 53.6°C
Using Eqn. (1.5.3):
1.7 (D) (T − 4) 2 (80 − 4) 2
ρ water = 1000 − = 1000 − = 968 kg/m3
180 180
du
The shear stress is given by: τ = μ
dr
1.8 (A) We determine du/dr from the given expression for u as:
10 (1 − 2500r2 ) ⎤⎦ = −50, 000r
du d ⎡
=
dr dr ⎣
At the wall r = 2 cm = 0.02 m. Substituting r in the above equation we get:
1
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,Chapter 1 / Basic Considerations
du
= 50, 000r = 1000 1/s
dr
The density of water at 20°C is 10−3 N ⋅ s/m2
Now substitute in the equation for shear stress to get
du
τ =μ = 10−3 N ⋅ s/m 2 × 1000 1/s = 1 N/m 2 = 1 Pa
dr
Using Eqn. (1.5.16), β = 0 (for clean glass tube), and σ = 0.0736 N/m for water
(Table B.1 in Appendix B) we write:
1.9 (D) 4σ cosβ 4 × 0.0736 N/m ×1
h= = = 3 m or 300 cm
ρ gD 1000 kg/m3 × 9.81 m/s2 ×10 ×10−6 m
where we used N = kg×m/s2
1.10 (C) Density
Assume propane (C3H8) behaves as an ideal gas. First, determine the gas
R 8.314 kJ/kmol ⋅ K
constant for propane using R = u = = 0.1885 kJ/kg
M 44.1 kg/kmol
1.11 (C)
pV 800 kN/m 2 × 4 m3
then, m = = = 59.99 kg ≈ 60 kg
RT 0.1885 kJ/(kg ⋅ K) × (10 + 273) K
Consider water and ice as the system. Hence, the change in energy for the
system is zero. That is, the change in energy for water should be equal to the
change in energy for the ice. So, we write ΔEice = ΔEwater
mice × 320 kJ/kg = mwater × cwater ΔT
The mass of ice is calculate using
mice = ρ V = 5 cubes × (1000 kg/m3 ) × ( 40 × 10−6 m3 /cube ) = 0.2 kg
1.12 (B) Where we assumed the density of ice to be equal to that of water, namely 1000
kg/m3. Ice is actually slightly lighter than water, but it is not necessary for
such accuracy in this problem.
Similarly, the mass of water is calculated using
mwater = ρ V = (1000 kg/m3 ) × 2 liters (10−3 m3 /liter ) = 2 kg
Solving for the temperature change for water we get
0.2 kg × 320 kJ/kg = 2 kg × 4.18 kJ/kg ⋅ K × ΔT ⇒ ΔT = 7.66 C
2
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, Chapter 1 / Basic Considerations
Since a dog’s whistle produces sound waves at a high frequency, the speed of
1.13 (D) sound is c = RT = 287 J/kg ⋅ K × 323 K = 304 m/s
where we used J/kg = m2/s2.
Dimensions, Units, and Physical Quantities
M FT 2 /L
a) density = 3
= 3
= FT 2 /L4
L L
1.16 c) power = F × velocity = F × L/T = FL/T
M/T FT 2 /L
e) mass flux = = 2 = FT/L3
A LT
1.18 b) N = [C] kg ∴ [C] = N/kg = (kg⋅ m/s2)/kg = m/s2
m m
kg 2
+ c + km = f . Since all terms must have the same dimensions (units)
s s
we require:
1.20
[c] = kg/s, [k] = kg/s2 = N ⋅ s 2 / m ⋅ s 2 = N / m, [f] = kg ⋅ m/s2 = N
Note: we could express the units on c as [c] = kg/s = N ⋅ s2 /m ⋅ s = N ⋅ s/m
1.22 a) 1.25 × 108 N c) 6.7 × 108 Pa e) 5.2 × 10−2 m2
cm m hr
a) 20 cm/hr = 20 × × = 5.556 × 10−5 m/s
hr 100 cm 3600 s
745.7 W
1.24 c) 500 hp = 500 hp × = 37, 285 W
hp
2
⎛ 100 cm ⎞ kN
e) 2000 kN/cm = 2000 ×
2 ⎜
2
⎟ = 2 × 10 N/m
10 2
cm ⎝ m ⎠
The mass is the same on the earth and the moon, so we calculate the mass, then
1.26 calculate the weight on the moon:
m = 27 kg ∴ Wmoon = (27 kg) × (1.63) = 44.01 N
3
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,Chapter 1 / Basic Considerations
Pressure and Temperature
Use the values from Table B.3 in the Appendix:
b) At an elevation of 1000 m the atmospheric pressure is 89.85 kPa. Hence, the
1.28 absolute pressure is 52.3 + 89.85 = 142.2 kPa
d) At an elevation of 10,000 m the atmospheric pressure is 26.49 kPa, and the
absolute pressure is 52.3 + 26.49 = 78.8 kPa
p = po e−gz/RT = 101 e−(9.81 × 4000)/[(287) × (15 + 273)] = 62.8 kPa
From Table B.3, at 4000 m: p = 61.6 kPa. The percent error is
1.30
62.8 − 61.6
% error = × 100 = 1.95 %
61.6
Using Table B.3 and linear interpolation we write:
10,600 − 10,000
T = 223.48 + (223.6 − 216.7) = 221.32 K
1.32 12,000 − 10,000
5
or (221.32 − 273.15) = −51.83°C
9
The normal force due pressure is: Fn = (120, 000 N/m 2 ) × 0.2 × 10−4 m 2 = 2.4 N
The tangential force due to shear stress is: Ft = 20 N/m 2 × 0.2 ×10−4 m 2 = 0.0004 N
1.34 The total force is F = Fn2 + Ft2 = 2.40 N
⎛ 0.0004 ⎞
The angle with respect to the normal direction is θ = tan−1 ⎜ ⎟ = 0.0095°
⎝ 2.4 ⎠
Density and Specific Weight
Using Eq. 1.5.3 we have
ρ = 1000 − (T − 4)2/180 = 1000 − (70 − 4)2/180 = 976 kg/m3
γ = 9800 − (T − 4)2/18 = 9800 − (70 − 4)2/180 = 9560 N/m3
Using Table B.1 the density and specific weight at 70°C are
1.36 ρ = 977.8 kg/m3
γ = 977.8 × 9.81 = 9592.2 N/m3
976 − 978
% error for ρ = × 100 = −0.20%
978
9560 − 9592
% error for γ = × 100 = −0.33%
9592
4
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, Chapter 1 / Basic Considerations
γV 12,400 N/m3 × 500 × 10−6 m3
1.38 b) m = = = 0.635 kg
g 9.77 m/s 2
Viscosity
Assume carbon dioxide is an ideal gas at the given conditions, then
p 200 kN/m3
ρ= = = 2.915 kg/m3
RT ( 0.189 kJ/kg ⋅ K )( 90 + 273 K )
W mg
γ= = = ρ g = 2.915 kg/m3 × 9.81 m/s2 = 28.6 kg/m2 ⋅ s2 = 28.6 N/m3
V V
1.40
From Fig. B.1 at 90°C, μ ≅ 2 ×10−5 N ⋅ s/m2 , so that the kinematic viscosity is
μ 2 ×10−5 N ⋅ s/m2
ν= = = 6.861× 10−6 m2 /s
ρ 2.915 kg/m3
The kinematic viscosity cannot be read from Fig. B.2 since the pressure is not at
100 kPa.
The shear stress can be calculated using τ = μ du /dy . From the given velocity
du
distribution, u = 120(0.05 y − y2 ) we get ⇒ = 120(0.05 − 2 y )
dy
From Table B.1 at 10°C for water, μ = 1.308 ×10−3 N ⋅ s/m2
So, at the lower plate where y = 0, we have
1.42
du
dy
(
= 120(0.05 − 0) = 6 s−1 ⇒ τ = 1.308 ×10−3 × 6 = 7.848 ×10−3 N/m2 )
At the upper plate where y = 0.05 m,
du
= 120(0.05 − 2 × 0.05) = 6 s−1 ⇒ τ = 7.848 × 10−3 N/m2
dy y = 0.05
5
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,Chapter 1 / Basic Considerations
The shear stress can be calculated using τ = μ du /dr . From the given velocity
du du
u = 16(1 − r2 ro2 ) ⇒ we get = 16(−2 r ro2 ). Hence, = 32 r ro2
dr dr
du
At the centerline, r = 0, so = 0, and hence τ = 0.
dr
du 0.25 100
At r = 0.25 cm, ⇒ = 32 r ro2 = 32 = 3200 s−1 , ⇒
1.44
( 0.5 100 )
2
dr
τ = 3200 × μ = 3200 (1 × 10−3 ) = 3.2 N/m2
du 0.5 100
At the wall, r = 0.5 cm, ⇒ = 32 r ro2 = 32 = 6400 s−1 , ⇒
( 0.5 100 )
2
dr
τ = 6400 × μ = 6400 (1 × 10−3 ) = 6.4 N/m2
2π R3ω Lμ
Use Eq.1.5.8 to calculate the torque, T =
h
where h = (2.6 − 2.54) 2 = 0.03 cm = 0.03 × 10−2 m
2π
The angular velocity ω = × 2000 rpm = 209.4 rad/s
1.46 60
The viscosity of SAE-30 oil at 21°C is μ = 0.2884 Ns/m2 (Figure B.2)
2π × (1.27 ×10−2 m)3 × 209.4 rad/s ×1.2 m × 0.2884 Ns/m2
T= = 3.10 N ⋅ m
(0.03 ×10−2 )m
power = Tω = 3.1× 209.4 = 650 W = 0.65 kW
Assume a linear velocity in the fluid between the rotating disk and solid surface.
The velocity of the fluid at the rotating disk is V = rω , and at the solid surface
du rω
V = 0. So, = , where h is the spacing between the disk and solid surface,
dy h
and ω = 2π × 400 60 = 41.9 rad/s . The torque needed to rotate the disk is
1.48 T = shear force × moment arm τ dr
Due to the area element shown, dT = dF × r = τdA × r r
where τ = shear stress in the fluid at the rotating disk and
du rω
dA = 2π rdr ⇒ dT = μ × 2π rdr × r = μ × 2π r2 dr
dy h
6
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, Chapter 1 / Basic Considerations
2πμω 3 πμω 4
R
T =∫ r dr = R
0
h 2h
The viscosity of water at 16°C is μ = 1.12 ×10−3 Ns/m2
4
⎛ 0.15 ⎞
π μω 4
(
π 1.12 × 10 Ns/m ( 41.9 rad/s ) ⎜ −3
⎟
2
)
⎝ 2 ⎠ = 1.16 × 10−3 N ⋅ m
⇒T = R =
2h 2 × 2 × 10−3 m ( )
du
If τ = μ = constant, and μ = AeB/T = AeBy/K = AeCy, then
dy
du du
AeCy = constant. ∴ = De−Cy, where D is a constant.
dy dy
1.50
Multiply by dy and integrate to get the velocity profile
y
u = ∫ De−Cy dy = −
D −Cy y
C
e
0
= E eCy − 1 ( )
0
where A, B, C, D, E, and K are constants.
Compressibility
The sound will travel across the lake at the speed of sound in water. The speed of
B
sound in water is calculated using c = , where B is the bulk modulus of
ρ
elasticity. Assuming (T = 10°C) and using Table B.1 we find
1.54
211×107 N/m2
B = 211×107 Pa , and ρ = 999.7 kg/m3 ⇒ c = = 1453 m/s
999.7 kg/m3
The distance across the lake is, L = cΔt = 1453 × 0.62 = 901 m
b) Using Table B.2 and T = 37°C we find B = 226 ×107 N/m2 , ρ = 993 kg/m3
The speed of sound in water is calculated using:
1.56
c=
B
ρ
⇒c= ( 226 ×107 ) / 993 = 1508 m/s
7
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, Chapter 1 / Basic Considerations
Surface Tension
2σ
For a spherical droplet the pressure is given by p =
R
Using Table B.1 at 15°C the surface tension σ = 7.41×10−2 N/m
1.58 2σ 2 × 0.0741 N/m
⇒ p= = = 29.6 kPa
R 5 × 10−6 m
4σ 2 × 0.0741 N/m
For bubbles: ⇒ p = = = 59.3 kPa
R 5 × 10−6 m
For a spherical droplet the net force due to the pressure difference Δp between
the inside and outside of the droplet is balanced by the surface tension force,
which is expressed as:
2σ 2 × 0.025 N/m
1.60 Δp = ⇒ pinside − poutside = = 10 kPa
R 5 × 10−6 m
Hence, pinside = poutside + 10 kPa = 8000 kPa + 10 kPa = 8010 kPa
In order to achieve this high pressure in the droplet, diesel fuel is usually pumped
to a pressure of about 2000 bar before it is injected into the engine.
See Example 1.4:
4σ cosβ 4 × ( 0.47 N/m ) cos130
h= =
1.62
ρ gD ( )
13.6 × 1000 kg/m 3 × 9.81 m/s 2 × 2 × 10−2 m ( )
= −4.53 × 10−4 m = −0.453 mm
Note that the minus sign indicates a capillary drop rather than a capillary rise in
the tube.
Draw a free-body diagram of the floating needle as shown in the figure.
The weight of the needle and the surface tension force must balance:
W = 2σ L or ρ gV = 2σ L
⎛ π d2 ⎞
The volume of the needle is V = ⎜ L⎟
1.64 ⎝ 4 ⎠
σL σL
⎛ π d2 ⎞
⇒⎜ L ⎟ ρ g = 2σ L
⎝ 4 ⎠ needle
W
8σ
∴d =
πρg
8
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Preface iv
Chapter 1 Basic Considerations 1
Chapter 2 Fluid Statics 15
Chapter 3 Introduction to Fluids in Motion 29
Chapter 4 The Integral Forms of the Fundamental Laws 37
Chapter 5 The Differential Forms of the Fundamental Laws 59
Chapter 6 Dimensional Analysis and Similitude 77
Chapter 7 Internal Flows 87
Chapter 8 External Flows 109
Chapter 9 Compressible Flow 131
Chapter 10 Flow in Open Channels 141
Chapter 11 Flows in Piping Systems 161
Chapter 12 Turbomachinery 175
Chapter 13 Measurements in Fluid Mechanics 189
, Chapter 1 / Basic Considerations
CHAPTER 1
Basic Considerations
FE-type Exam Review Problems: Problems 1.1 to 1.13
1.1 (C) m = F/a or kg = N/m/s2 = N·s2/m
1.2 (B) [μ] = [τ/(du/dy)] = (F/L2)/(L/T)/L = F.T/L2
1.3 (A) 2.36 ×10−8 Pa = 23.6 ×10−9 Pa = 23.6 nPa
The mass is the same on earth and the moon, so we calculate the mass using the
weight given on earth as:
1.4 (C) m = W/g = 250 N/9.81 m/s2 = 25.484 kg
Hence, the weight on the moon is:
W = mg = 25.484 × 1.6 = 40.77 N
The shear stress is due to the component of the force acting tangential to the
area:
1.5 (C) Fshear = F sin θ = 4200sin 30 = 2100 N
F 2100 N
τ = shear = −4
= 84 ×103 Pa or 84 kPa
A 250 ×10 m 2
1.6 (B) – 53.6°C
Using Eqn. (1.5.3):
1.7 (D) (T − 4) 2 (80 − 4) 2
ρ water = 1000 − = 1000 − = 968 kg/m3
180 180
du
The shear stress is given by: τ = μ
dr
1.8 (A) We determine du/dr from the given expression for u as:
10 (1 − 2500r2 ) ⎤⎦ = −50, 000r
du d ⎡
=
dr dr ⎣
At the wall r = 2 cm = 0.02 m. Substituting r in the above equation we get:
1
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,Chapter 1 / Basic Considerations
du
= 50, 000r = 1000 1/s
dr
The density of water at 20°C is 10−3 N ⋅ s/m2
Now substitute in the equation for shear stress to get
du
τ =μ = 10−3 N ⋅ s/m 2 × 1000 1/s = 1 N/m 2 = 1 Pa
dr
Using Eqn. (1.5.16), β = 0 (for clean glass tube), and σ = 0.0736 N/m for water
(Table B.1 in Appendix B) we write:
1.9 (D) 4σ cosβ 4 × 0.0736 N/m ×1
h= = = 3 m or 300 cm
ρ gD 1000 kg/m3 × 9.81 m/s2 ×10 ×10−6 m
where we used N = kg×m/s2
1.10 (C) Density
Assume propane (C3H8) behaves as an ideal gas. First, determine the gas
R 8.314 kJ/kmol ⋅ K
constant for propane using R = u = = 0.1885 kJ/kg
M 44.1 kg/kmol
1.11 (C)
pV 800 kN/m 2 × 4 m3
then, m = = = 59.99 kg ≈ 60 kg
RT 0.1885 kJ/(kg ⋅ K) × (10 + 273) K
Consider water and ice as the system. Hence, the change in energy for the
system is zero. That is, the change in energy for water should be equal to the
change in energy for the ice. So, we write ΔEice = ΔEwater
mice × 320 kJ/kg = mwater × cwater ΔT
The mass of ice is calculate using
mice = ρ V = 5 cubes × (1000 kg/m3 ) × ( 40 × 10−6 m3 /cube ) = 0.2 kg
1.12 (B) Where we assumed the density of ice to be equal to that of water, namely 1000
kg/m3. Ice is actually slightly lighter than water, but it is not necessary for
such accuracy in this problem.
Similarly, the mass of water is calculated using
mwater = ρ V = (1000 kg/m3 ) × 2 liters (10−3 m3 /liter ) = 2 kg
Solving for the temperature change for water we get
0.2 kg × 320 kJ/kg = 2 kg × 4.18 kJ/kg ⋅ K × ΔT ⇒ ΔT = 7.66 C
2
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, Chapter 1 / Basic Considerations
Since a dog’s whistle produces sound waves at a high frequency, the speed of
1.13 (D) sound is c = RT = 287 J/kg ⋅ K × 323 K = 304 m/s
where we used J/kg = m2/s2.
Dimensions, Units, and Physical Quantities
M FT 2 /L
a) density = 3
= 3
= FT 2 /L4
L L
1.16 c) power = F × velocity = F × L/T = FL/T
M/T FT 2 /L
e) mass flux = = 2 = FT/L3
A LT
1.18 b) N = [C] kg ∴ [C] = N/kg = (kg⋅ m/s2)/kg = m/s2
m m
kg 2
+ c + km = f . Since all terms must have the same dimensions (units)
s s
we require:
1.20
[c] = kg/s, [k] = kg/s2 = N ⋅ s 2 / m ⋅ s 2 = N / m, [f] = kg ⋅ m/s2 = N
Note: we could express the units on c as [c] = kg/s = N ⋅ s2 /m ⋅ s = N ⋅ s/m
1.22 a) 1.25 × 108 N c) 6.7 × 108 Pa e) 5.2 × 10−2 m2
cm m hr
a) 20 cm/hr = 20 × × = 5.556 × 10−5 m/s
hr 100 cm 3600 s
745.7 W
1.24 c) 500 hp = 500 hp × = 37, 285 W
hp
2
⎛ 100 cm ⎞ kN
e) 2000 kN/cm = 2000 ×
2 ⎜
2
⎟ = 2 × 10 N/m
10 2
cm ⎝ m ⎠
The mass is the same on the earth and the moon, so we calculate the mass, then
1.26 calculate the weight on the moon:
m = 27 kg ∴ Wmoon = (27 kg) × (1.63) = 44.01 N
3
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,Chapter 1 / Basic Considerations
Pressure and Temperature
Use the values from Table B.3 in the Appendix:
b) At an elevation of 1000 m the atmospheric pressure is 89.85 kPa. Hence, the
1.28 absolute pressure is 52.3 + 89.85 = 142.2 kPa
d) At an elevation of 10,000 m the atmospheric pressure is 26.49 kPa, and the
absolute pressure is 52.3 + 26.49 = 78.8 kPa
p = po e−gz/RT = 101 e−(9.81 × 4000)/[(287) × (15 + 273)] = 62.8 kPa
From Table B.3, at 4000 m: p = 61.6 kPa. The percent error is
1.30
62.8 − 61.6
% error = × 100 = 1.95 %
61.6
Using Table B.3 and linear interpolation we write:
10,600 − 10,000
T = 223.48 + (223.6 − 216.7) = 221.32 K
1.32 12,000 − 10,000
5
or (221.32 − 273.15) = −51.83°C
9
The normal force due pressure is: Fn = (120, 000 N/m 2 ) × 0.2 × 10−4 m 2 = 2.4 N
The tangential force due to shear stress is: Ft = 20 N/m 2 × 0.2 ×10−4 m 2 = 0.0004 N
1.34 The total force is F = Fn2 + Ft2 = 2.40 N
⎛ 0.0004 ⎞
The angle with respect to the normal direction is θ = tan−1 ⎜ ⎟ = 0.0095°
⎝ 2.4 ⎠
Density and Specific Weight
Using Eq. 1.5.3 we have
ρ = 1000 − (T − 4)2/180 = 1000 − (70 − 4)2/180 = 976 kg/m3
γ = 9800 − (T − 4)2/18 = 9800 − (70 − 4)2/180 = 9560 N/m3
Using Table B.1 the density and specific weight at 70°C are
1.36 ρ = 977.8 kg/m3
γ = 977.8 × 9.81 = 9592.2 N/m3
976 − 978
% error for ρ = × 100 = −0.20%
978
9560 − 9592
% error for γ = × 100 = −0.33%
9592
4
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, Chapter 1 / Basic Considerations
γV 12,400 N/m3 × 500 × 10−6 m3
1.38 b) m = = = 0.635 kg
g 9.77 m/s 2
Viscosity
Assume carbon dioxide is an ideal gas at the given conditions, then
p 200 kN/m3
ρ= = = 2.915 kg/m3
RT ( 0.189 kJ/kg ⋅ K )( 90 + 273 K )
W mg
γ= = = ρ g = 2.915 kg/m3 × 9.81 m/s2 = 28.6 kg/m2 ⋅ s2 = 28.6 N/m3
V V
1.40
From Fig. B.1 at 90°C, μ ≅ 2 ×10−5 N ⋅ s/m2 , so that the kinematic viscosity is
μ 2 ×10−5 N ⋅ s/m2
ν= = = 6.861× 10−6 m2 /s
ρ 2.915 kg/m3
The kinematic viscosity cannot be read from Fig. B.2 since the pressure is not at
100 kPa.
The shear stress can be calculated using τ = μ du /dy . From the given velocity
du
distribution, u = 120(0.05 y − y2 ) we get ⇒ = 120(0.05 − 2 y )
dy
From Table B.1 at 10°C for water, μ = 1.308 ×10−3 N ⋅ s/m2
So, at the lower plate where y = 0, we have
1.42
du
dy
(
= 120(0.05 − 0) = 6 s−1 ⇒ τ = 1.308 ×10−3 × 6 = 7.848 ×10−3 N/m2 )
At the upper plate where y = 0.05 m,
du
= 120(0.05 − 2 × 0.05) = 6 s−1 ⇒ τ = 7.848 × 10−3 N/m2
dy y = 0.05
5
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,Chapter 1 / Basic Considerations
The shear stress can be calculated using τ = μ du /dr . From the given velocity
du du
u = 16(1 − r2 ro2 ) ⇒ we get = 16(−2 r ro2 ). Hence, = 32 r ro2
dr dr
du
At the centerline, r = 0, so = 0, and hence τ = 0.
dr
du 0.25 100
At r = 0.25 cm, ⇒ = 32 r ro2 = 32 = 3200 s−1 , ⇒
1.44
( 0.5 100 )
2
dr
τ = 3200 × μ = 3200 (1 × 10−3 ) = 3.2 N/m2
du 0.5 100
At the wall, r = 0.5 cm, ⇒ = 32 r ro2 = 32 = 6400 s−1 , ⇒
( 0.5 100 )
2
dr
τ = 6400 × μ = 6400 (1 × 10−3 ) = 6.4 N/m2
2π R3ω Lμ
Use Eq.1.5.8 to calculate the torque, T =
h
where h = (2.6 − 2.54) 2 = 0.03 cm = 0.03 × 10−2 m
2π
The angular velocity ω = × 2000 rpm = 209.4 rad/s
1.46 60
The viscosity of SAE-30 oil at 21°C is μ = 0.2884 Ns/m2 (Figure B.2)
2π × (1.27 ×10−2 m)3 × 209.4 rad/s ×1.2 m × 0.2884 Ns/m2
T= = 3.10 N ⋅ m
(0.03 ×10−2 )m
power = Tω = 3.1× 209.4 = 650 W = 0.65 kW
Assume a linear velocity in the fluid between the rotating disk and solid surface.
The velocity of the fluid at the rotating disk is V = rω , and at the solid surface
du rω
V = 0. So, = , where h is the spacing between the disk and solid surface,
dy h
and ω = 2π × 400 60 = 41.9 rad/s . The torque needed to rotate the disk is
1.48 T = shear force × moment arm τ dr
Due to the area element shown, dT = dF × r = τdA × r r
where τ = shear stress in the fluid at the rotating disk and
du rω
dA = 2π rdr ⇒ dT = μ × 2π rdr × r = μ × 2π r2 dr
dy h
6
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, Chapter 1 / Basic Considerations
2πμω 3 πμω 4
R
T =∫ r dr = R
0
h 2h
The viscosity of water at 16°C is μ = 1.12 ×10−3 Ns/m2
4
⎛ 0.15 ⎞
π μω 4
(
π 1.12 × 10 Ns/m ( 41.9 rad/s ) ⎜ −3
⎟
2
)
⎝ 2 ⎠ = 1.16 × 10−3 N ⋅ m
⇒T = R =
2h 2 × 2 × 10−3 m ( )
du
If τ = μ = constant, and μ = AeB/T = AeBy/K = AeCy, then
dy
du du
AeCy = constant. ∴ = De−Cy, where D is a constant.
dy dy
1.50
Multiply by dy and integrate to get the velocity profile
y
u = ∫ De−Cy dy = −
D −Cy y
C
e
0
= E eCy − 1 ( )
0
where A, B, C, D, E, and K are constants.
Compressibility
The sound will travel across the lake at the speed of sound in water. The speed of
B
sound in water is calculated using c = , where B is the bulk modulus of
ρ
elasticity. Assuming (T = 10°C) and using Table B.1 we find
1.54
211×107 N/m2
B = 211×107 Pa , and ρ = 999.7 kg/m3 ⇒ c = = 1453 m/s
999.7 kg/m3
The distance across the lake is, L = cΔt = 1453 × 0.62 = 901 m
b) Using Table B.2 and T = 37°C we find B = 226 ×107 N/m2 , ρ = 993 kg/m3
The speed of sound in water is calculated using:
1.56
c=
B
ρ
⇒c= ( 226 ×107 ) / 993 = 1508 m/s
7
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, Chapter 1 / Basic Considerations
Surface Tension
2σ
For a spherical droplet the pressure is given by p =
R
Using Table B.1 at 15°C the surface tension σ = 7.41×10−2 N/m
1.58 2σ 2 × 0.0741 N/m
⇒ p= = = 29.6 kPa
R 5 × 10−6 m
4σ 2 × 0.0741 N/m
For bubbles: ⇒ p = = = 59.3 kPa
R 5 × 10−6 m
For a spherical droplet the net force due to the pressure difference Δp between
the inside and outside of the droplet is balanced by the surface tension force,
which is expressed as:
2σ 2 × 0.025 N/m
1.60 Δp = ⇒ pinside − poutside = = 10 kPa
R 5 × 10−6 m
Hence, pinside = poutside + 10 kPa = 8000 kPa + 10 kPa = 8010 kPa
In order to achieve this high pressure in the droplet, diesel fuel is usually pumped
to a pressure of about 2000 bar before it is injected into the engine.
See Example 1.4:
4σ cosβ 4 × ( 0.47 N/m ) cos130
h= =
1.62
ρ gD ( )
13.6 × 1000 kg/m 3 × 9.81 m/s 2 × 2 × 10−2 m ( )
= −4.53 × 10−4 m = −0.453 mm
Note that the minus sign indicates a capillary drop rather than a capillary rise in
the tube.
Draw a free-body diagram of the floating needle as shown in the figure.
The weight of the needle and the surface tension force must balance:
W = 2σ L or ρ gV = 2σ L
⎛ π d2 ⎞
The volume of the needle is V = ⎜ L⎟
1.64 ⎝ 4 ⎠
σL σL
⎛ π d2 ⎞
⇒⎜ L ⎟ ρ g = 2σ L
⎝ 4 ⎠ needle
W
8σ
∴d =
πρg
8
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