A STUDENT’S SOLUTIONS MANUAL TO ACCOMPANY
MECHANICS of FLUIDS,
4TH EDITION, SI
MERLE C. POTTER
DAVID C. WIGGERT
BASSEM H. RAMADAN
STUDENT'S SOLUTIONS MANUAL
TO ACCOMPANY
MECHANICS of FLUIDS
FOURTH EDITION, SI
MERLE C. POTTER
Michigan State University
DAVID C. WIGGERT
Michigan State University
BASSEM RAMADAN
Kettering University
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
Preface for the Student
This manual provides the solutions to the problems whose answers are provided at the end of our
book, MECHANICS OF FLUIDS, SI. In many cases, the solutions are not as detailed as the
examples in the book; they are intended to provide the primary steps in each solution so you, the
student, are able to quickly review how a problem is solved. The discussion of a subtle point,
should one exist in a particular problem, is left as a task for the instructor. In general, some
knowledge of a problem may be needed to fully understand all of the steps presented. This
manual is not intended to be a self-paced workbook; your instructor is critically needed to provide
explanations, discussions, and illustrations of the myriad of phenomena encountered in the study
of fluids, but it should give you considerable help in working through a wide variety of problems.
The degree of difficulty and length of solution for each problem varies considerably. Some are
relatively easy and others quite difficult. Typically, the easier problems are the first problems for
a particular section.
We continue to include a number of multiple-choice problems in the earlier chapters, similar to
those encountered on the Fundamentals of Engineering Exam (the old EIT Exam) and the
GRE/Engineering Exam. These problems will provide a review for the Fluid Mechanics part of
those exams. They are all four-part, multiple-choice problems and are located at the beginning of
the appropriate chapters.
The examples and problems have been carefully solved with the hope that errors have not been
introduced. Even though extreme care is taken and problems are reworked, errors creep in. We
would appreciate knowing about any errors that you may find so they can be eliminated in future
printings. Please send any corrections or comments to MerleCP@att.net.We have class tested
most of the chapters with good response from our students, but we are sure that there are
improvements to be made.
East Lansing, Michigan
Flint, Michigan
Merle C. Potter
David C. Wiggert
Bassem Ramadan
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 30D = 2100 N
F
2100 N
τ = shear =
= 84 ×103 Pa or 84 kPa
2
−4
A
250 ×10 m
1.6 (B)
– 53.6°C
Using Eqn. (1.5.3):
1.7 (D)
ρ water = 1000 −
(T − 4) 2
(80 − 4) 2
= 1000 −
= 968 kg/m3
180
180
The shear stress is given by: τ = μ
1.8 (A)
du
dr
We determine du/dr from the given expression for u as:
du d ⎡
10 (1 − 2500r2 ) ⎤⎦ = −50, 000r
=
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
τ =μ
du
= 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)
h=
4σ cosβ
4 × 0.0736 N/m ×1
=
= 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)
1.11 (C)
Density
Assume propane (C3H8) behaves as an ideal gas. First, determine the gas
R 8.314 kJ/kmol ⋅ K
= 0.1885 kJ/kg
constant for propane using R = u =
M
44.1 kg/kmol
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 = 7.66D C
2
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Chapter 1 / Basic Considerations
1.13 (D)
Since a dog’s whistle produces sound waves at a high frequency, the speed of
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
a) density =
1.16
M FT 2 /L
=
= FT 2 /L4
3
3
L
L
c) power = F × velocity = F × L/T = FL/T
M/T FT 2 /L
= 2
= FT/L3
A
LT
e) mass flux =
1.18
∴ [C] = N/kg = (kg⋅ m/s2)/kg = m/s2
b) N = [C] kg
m
m
+ c + km = f . Since all terms must have the same dimensions (units)
2
s
s
we require:
kg
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
a) 20 cm/hr = 20
1.24
c) 6.7 × 108 Pa
e) 5.2 × 10−2 m2
cm
m
hr
×
×
= 5.556 × 10−5 m/s
hr 100 cm 3600 s
c) 500 hp = 500 hp ×
745.7 W
= 37, 285 W
hp
2
⎛ 100 cm ⎞
10
2
×
e) 2000 kN/cm = 2000
⎟ = 2 × 10 N/m
2 ⎜
cm ⎝ m ⎠
2
1.26
kN
The mass is the same on the earth and the moon, so we calculate the mass, then
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:
1.28
b) At an elevation of 1000 m the atmospheric pressure is 89.85 kPa. Hence, the
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
1.30
From Table B.3, at 4000 m: p = 61.6 kPa. The percent error is
% error =
62.8 − 61.6
× 100 = 1.95 %
61.6
Using Table B.3 and linear interpolation we write:
1.32
10,600 − 10,000
(223.6 − 216.7) = 221.32 K
12,000 − 10,000
5
or (221.32 − 273.15) = −51.83°C
9
T = 223.48 +
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
% error for ρ =
976 − 978
× 100 = −0.20%
978
% error for γ =
9560 − 9592
× 100 = −0.33%
9592
4
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Chapter 1 / Basic Considerations
1.38
b) m =
γV
g
=
12,400 N/m3 × 500 × 10−6 m3
9.77 m/s 2
= 0.635 kg
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 )
γ=
1.40
W mg
=
= ρ g = 2.915 kg/m3 × 9.81 m/s2 = 28.6 kg/m2 ⋅ s2 = 28.6 N/m3
V
V
From Fig. B.1 at 90°C, μ ≅ 2 ×10−5 N ⋅ s/m2 , so that the kinematic viscosity is
ν=
μ 2 ×10−5 N ⋅ s/m2
=
= 6.861× 10−6 m2 /s
2.915 kg/m3
ρ
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
distribution, u = 120(0.05 y − y2 ) we get ⇒
du
= 120(0.05 − 2 y )
dy
From Table B.1 at 10°C for water, μ = 1.308 ×10−3 N ⋅ s/m2
1.42
So, at the lower plate where y = 0, we have
(
)
du
= 120(0.05 − 0) = 6 s−1 ⇒ τ = 1.308 × 10−3 × 6 = 7.848 ×10−3 N/m2
dy
At the upper plate where y = 0.05 m,
du
dy
= 120(0.05 − 2 × 0.05) = 6 s−1 ⇒ τ = 7.848 × 10−3 N/m2
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
u = 16(1 − r2 ro2 ) ⇒ we get
At the centerline, r = 0, so
1.44
At r = 0.25 cm, ⇒
du
du
= 16(−2 r ro2 ). Hence,
= 32 r ro2
dr
dr
du
= 0, and hence τ = 0.
dr
du
0.25 100
= 32 r ro2 = 32
= 3200 s−1 , ⇒
2
dr
( 0.5 100 )
τ = 3200 × μ = 3200 (1 × 10−3 ) = 3.2 N/m2
At the wall, r = 0.5 cm, ⇒
du
0.5 100
= 32 r ro2 = 32
= 6400 s−1 , ⇒
2
dr
( 0.5 100 )
τ = 6400 × μ = 6400 (1 × 10−3 ) = 6.4 N/m2
Use Eq.1.5.8 to calculate the torque, T =
2π R3ω Lμ
h
where h = (2.6 − 2.54) 2 = 0.03 cm = 0.03 × 10−2 m
The angular velocity ω =
1.46
2π
× 2000 rpm = 209.4 rad/s
60
The viscosity of SAE-30 oil at 21°C is μ = 0.2884 Ns/m2 (Figure B.2)
T=
2π × (1.27 ×10−2 m)3 × 209.4 rad/s ×1.2 m × 0.2884 Ns/m2
(0.03 ×10−2 )m
= 3.10 N ⋅ 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
Due to the area element shown, dT = dF × r = τdA × r
τ
dr
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 dr =
R
h
2h
0
R
T =∫
The viscosity of water at 16°C is μ = 1.12 ×10−3 Ns/m2
4
⎛ 0.15 ⎞
π 1.12 × 10 Ns/m ( 41.9 rad/s ) ⎜
⎟
π μω 4
⎝ 2 ⎠ = 1.16 × 10−3 N ⋅ m
R =
⇒T =
2h
2 × 2 × 10−3 m
(
−3
2
)
(
If τ = μ
AeCy
1.50
)
du
= constant, and μ = AeB/T = AeBy/K = AeCy, then
dy
du
du
= constant. ∴
= De−Cy, where D is a constant.
dy
dy
Multiply by dy and integrate to get the velocity profile
y
u = ∫ De−Cy dy = −
0
(
)
D −Cy y
= E eCy − 1
e
0
C
where A, B, C, D, E, and K are constants.
Compressibility
1.54
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
B = 211×107 Pa , and ρ = 999.7 kg/m3 ⇒ c =
211×107 N/m2
= 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
1.56
The speed of sound in water is calculated using:
c=
B
ρ
⇒c=
( 226 ×107 ) / 993 = 1508 m/s
7
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Chapter 1 / Basic Considerations
Surface Tension
2σ
R
For a spherical droplet the pressure is given by p =
Using Table B.1 at 15°C the surface tension σ = 7.41×10−2 N/m
1.58
⇒ p=
2σ 2 × 0.0741 N/m
=
= 29.6 kPa
5 × 10−6 m
R
For bubbles: ⇒ p =
4σ 2 × 0.0741 N/m
=
= 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:
1.60
Δp =
2σ
2 × 0.025 N/m
⇒ pinside − poutside =
= 10 kPa
5 × 10−6 m
R
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:
1.62
4 × ( 0.47 N/m ) cos130D
4σ cosβ
=
h=
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
1.64
⎛ π d2 ⎞
L⎟
The volume of the needle is V = ⎜
⎝ 4 ⎠
σL
⎛ π d2 ⎞
⇒⎜
L ⎟ ρ g = 2σ L
⎝ 4 ⎠
σL
needle
W
8σ
∴d =
πρg
8
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Chapter 1 / Basic Considerations
There is a surface tension force on the outside
and on the inside of the ring. Each surface
tension force = σ × π D.
1.66
Neglecting the weight of the ring, the free-body
diagram of the ring shows that
F = 2σπ D
F
D
Vapor Pressure
To determine the temperature, we can determine the absolute pressure and then
use property tables for water.
The absolute pressure is p = −80 + 92 = 12 kPa.
1.68
From Table B.1, at 50°C water has a vapor pressure of 12.3 kPa;
so T = 50°C is a maximum temperature. The water would “boil” above
this temperature.
1.70
At 40°C the vapor pressure from Table B.1 is 7.38 kPa. This would be the
minimum pressure that could be obtained since the water would vaporize below
this pressure.
The inlet pressure to a pump cannot be less than 0 kPa absolute. Assuming
1.72
atmospheric pressure to be 100 kPa, we have
10,000 kPa + 100 kPa = 600x
∴x = 16.83 km.
Ideal Gas
Assume air is an ideal gas and calculate the density inside and outside the house
using Tin = 15°C, pin = 101.3 kPa, Tout = −25°C, and pout = 85 kPa
p
101.3 kN/m3
ρin =
=
= 1.226 kg/m3
RT 0.287 kJ/kg ⋅ K × (15 + 273 K)
1.74
ρout =
85
= 1.19 kg/m3
0.287 × 248
Yes. The heavier air outside enters at the bottom and the lighter air inside exits
at the top. A circulation is set up and the air moves from the outside in and the
inside out. This is infiltration, also known as the “chimney” effect.
9
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Chapter 1 / Basic Considerations
The weight can be calculated using, W = γ V = ρ g V
1.76
Assume air in the room is at 20°C and 100 kPa and is an ideal gas:
p
100
⇒ρ=
=
= 1.189 kg/m3
RT 0.287 × 293
⇒ W = γ V = ρ g V = 1.189 kg/m3 × 9.81 m/s 2 × (10 × 20 × 4 m3 ) = 9333 N
The pressure holding up the mass is 100 kPa. Hence, using pA = W, we have
100,000 N/m2 ×1 m2 = m × 9.81 m/s2
This gives the mass of the air m = 10,194 kg
Since air is an ideal gas we can write pV = mRT or
1.78
V =
mRT 10,194 kg × 0.287 kJ/kg ⋅ K × (15+273 K )
=
= 8426 m3
p
100 kPa
Assuming a spherical volume ⇒ V = π d 3 6 , which gives
13
⎛ 6 × 8426 m3 ⎞
d =⎜
⎟ = 25.2 m
π
⎝
⎠
10
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Chapter 1 / Basic Considerations
First Law
The fist law of the thermodynamics is applied to the mass:
Q1− 2 − W1− 2 = ΔPE + ΔKE + ΔU
In this case, since there is no change in potential and internal energy, i.e.,
ΔPE = 0, and Δ U = 0, and there is no heat transfer to or from the system,
Q1− 2 = 0. The above equation simplifies to: −W1−2 = ΔKE
2
where W1− 2 = ∫ Fdl
1
a) F = 200 N ⇒ W1−2 = −200 N ×10 m = −2000 N ⋅ m or J
Note that the work is negative in this case since it is done on the system.
Substituting in the first law equation we get:
1.80
2000 N ⋅ m =
1
2 × 2000 N ⋅ m
m (V22 − V12 ) ⇒ V2 = (10 m/s)2 +
= 19.15 m/s
2
15 kg
10
b)
F = 20 s ⇒ W1− 2 = − ∫ 20 sds = − 10 s2
10
0
= −1000 N ⋅ m
0
1000 N ⋅ m =
1
2 × 1000 N ⋅ m
m (V22 − V12 ) ⇒ V2 = (10 m/s)2 +
= 15.28 m/s
2
15 kg
10
c) F = 200 cos (π s 20 ) ⇒ W1− 2 = − ∫ 200 cos (π s 20 ) ds
0
⇒ W1− 2 = −200 × ( 20 π ) sin (10π 20 ) = − 4000 π N ⋅ m
1
2 × 4000 π N ⋅ m
4000 π N ⋅ m = m (V22 − V12 ) ⇒ V2 = (10 m/s)2 +
= 16.42 m/s
2
15 kg
Applying the first law to the system (auto + water) we write:
Q1− 2 − W1− 2 = ΔPE + ΔKE + ΔU
1.82
In this case, the kinetic energy of the automobile is converted into internal energy
in the water. There is no heat transfer, Q = 0, no work W = 0, and no change in
potential energy, ΔPE = 0 . The first law equation reduces to:
1
mV12 = ΔU H2O = [ mcΔT ]H O
2
2
where, mH2O = ρ V = 1000 kg/m3 × 2000 cm3 ×10−6 m3 cm3 = 2 kg
11
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Chapter 1 / Basic Considerations
For water c = 4180 J/kg ⋅ K. Substituting in the above equation
2
1
⎛ 100 ×1000
⎞
m/s ⎟ = 2 kg × 4180 J/kg ⋅ K × ΔT
×1500 kg × ⎜
2
⎝ 3600
⎠
⇒ ΔT = 69.2°C
2
For a closed system the work is W1− 2 = ∫ pdV
1
For air (which is an ideal gas), pV = mRT , and p =
2
2
1
1
W1− 2 = ∫ pdV = ∫
mRT
V
2
mRT
dV
V
dV = mRT ∫
= mRT ln 2
V
V
V1
1
Since T = constant, then for this process p1V1 = p2 V2 or
1.84
V2 p1
=
V1 p2
Substituting in the expression for work we get: W1−2 = mRT
W1−2 = 2 kg × ( 287 J kg ⋅ K ) × 294 K ln
p2
p1
1
= −116,980 J = −116.98 kJ
2
The 1st law states that Q1− 2 − W1− 2 = mΔu = mcv ΔT = 0 since ΔT = 0.
∴ Q = W = −116.98 kJ
2
For a closed system the work is W1− 2 = ∫ pdV
1
If p = constant, then W1−2 = p ( V2 − V1 )
1.86
Since air is an ideal gas, then p1 V1 = mRT1 , and p2 V2 = mRT2
W1− 2 = mR (T2 − T1 ) = mR ( 2T1 − T1 ) = mRT1 = 2 kg × 0.287 kJ/kg ⋅ K × 423 K = 243 kJ
12
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Chapter 1 / Basic Considerations
Isentropic Flow
Since the process is adiabatic, we assume an isentropic process to estimate the
maximum final pressure:
⎛T ⎞
p2 = p1 ⎜ 2 ⎟
⎝ T1 ⎠
1.88
k / k −1
1.4/0.4
⎛ 423 ⎞
= (150 + 100) ⎜
⎟
⎝ 293 ⎠
= 904 kPa abs or 804 kPa gage.
Note: We assumed patm = 100 kPa since it was not given. Also, a measured
pressure is a gage pressure.
Speed of Sound
b) c = kRT = 1.4 × 188.9 × 293 = 266.9 m/s
1.90
d) c = kRT = 1.4 × 4124 × 293 = 1301 m/s
Note: We must use the units on R to be J/kg.K in the above equations.
1.92
b) The sound will travel at the speed of sound
c = kRT = 1.4 × 287 × 293 = 343 m/s
The distance is calculated using L = cΔt = 343× 8.32 = 2854 m
13
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Chapter 1 / Basic Considerations
14
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Chapter 2 / Fluid Statics
CHAPTER 2
Fluid Statics
FE-type Exam Review Problems: Problems 2-1 to 2-9
The pressure can be calculated using: p = γ Hg h were h is the height of mercury.
2.1 (C)
p = γ Hg h = (13.6 × 9810 N/m3 ) × (28.5 × 0.0254) = 96,600 Pa = 96.6 kPa
Since the pressure varies in a vertical direction, then:
2.2 (D)
2.3 (C)
p = p0 − ρ gh = 84, 000 Pa − 1.00 kg/m3 × 9.81 m/s 2 × 4000 m = 44.76 kPa
pw = patm + γ m hm − γ water hw = 0 + 30,000 × 0.3 − 9810 × 0.1 = 8020 Pa = 8.02 kPa
This is the gage pressure since we used patm = 0.
Initially, the pressure in the air is
p Air ,1 = −γ H = −(13.6 × 9810) × 0.16 = −21,350 Pa.
2.4 (A)
After the pressure is increased we have:
p Air ,2 = −21, 350 + 10, 000 = −11, 350 = −13.6 × 9810 H 2 .
∴ H 2 = 0.0851 m = 8.51 cm
The moment of force P with respect to the hinge, must balance the moment of
the hydrostatic force F with respect to the hinge, that is: (2 × 5 ) × P = F × d
3
2.5 (B)
5
F = γ hA = 9.81 kN/m3 × 1 m × (2 × × 3 m 2 )] ∴ F = 98.1 kN
3
The location of F is at
3
3 ( 3.33) 12
I
yp = y +
= 1.67 +
= 2.22 m ⇒ d = 3.33 − 2.22 = 1.11 m
yA
1.67(3.33 × 3)
3.33 × P = 98.1× 1.11 ∴ P = 32.7 kN
The gate opens when the center of pressure is at the hinge:
2.6 (A)
1.2 + h
I
11.2 + h
b(1.2 + h)3 /12
y=
+ 5. y p = y +
=
+
= 5 + 1.2
2
Ay
2
(1.2 + h)b(11.2 + h) / 2
This can be solved by trial-and-error, or we can simply substitute one of the
answers into the equation and check to see if it is correct. This yields
h = 1.08 m.
15
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Chapter 2 / Fluid Statics
2.7 (D)
The hydrostatic force will pass through the center, and so FH will be balanced
by the force in the hinge and the force P will be equal to FV.
∴ P = FV = 9.81× 4 × 1.2 × w + 9.81× (π × 1.22 / 4) × w = 300. ∴ w = 5.16 m.
2.8 (A)
The weight is balanced by the buoyancy force which is given by
FB = γ V where V is the displaced volume of fluid:
900 × 9.81 = 9810 × 0.01 × 15w. ∴ w = 6 m
The pressure on the plug is due to the initial pressure plus the pressure due to
the acceleration of the fluid, that is: p plug = pinitial + γ gasoline ΔZ where
ΔZ = Δx
2.9 (A)
ax
g
p plug = 20, 000 + 6660 × (1.2 ×
5
) = 24, 070 Pa
9.81
Fplug = p plug A = 24,070 × π × 0.022 = 30.25 N
Pressure
Since p = γ h, then h = p/γ
2.12
a) h = 250,000/9810 = 25.5 m
c) h = 250,000/(13.6 × 9810) = 1.874 m
2.14
2.16
This requires that pwater = pHg
⇒
(γ h )water = (γ h )Hg
b) 9810 × h = (13.6 × 9810) × 0.75 ∴h = 10.2 m
Δp = −γΔz ⇒ Δp = – 1.27 × 9.81 (3000) = –37,370 Pa or –37.37 kPa
From the given information the specific gravity is S = 1.0 + z/100 since S(0) = 1
and S(10) = 1.1.
By definition ρ = 1000 S, where ρwater = 1000 kg/m3.
Using dp = −γ dz then, by integration we write:
2.18
10
⎛
z2 2 ⎞
dp
=
1000(1
+
z
/100)
gdz
=
1000
g
z
+
⎜
⎟
∫0
∫0
100 ⎠
⎝
p
⎛
102 ⎞
p = 1000 × 9.81⎜10 +
⎟ = 103,000 Pa or 103 kPa
2 ×100 ⎠
⎝
16
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Chapter 2 / Fluid Statics
Note: we could have used an average S: Savg = 1.05, so that ρ avg = 1050 kg/m3
and so p = γ h ⇒ p = 1050 × 9.81 × 10 = 103,005 Pa
From Eq. (2.4.8): p = patm [(T0 − α z ) / T0 ]g / α R = 100 [(288 − 0.0065 ×
300)/288]9.81/0.0065 × 287 = 96.49 kPa
Assuming constant density, then
2.20
100
⎛
⎞
p = patm − ρ gh = 100 − ⎜
⎟ × 9.81× 300 /1000 = 96.44 kPa
⎝ 0.287 × 288 ⎠
96.44 − 96.49
× 100 = −0.052%
96.49
% error =
Since the error is small, the density variation can be ignored over heights of
300 m or less.
dp
dρ
Eq. 1.5.11 gives B = ρ
ρ gdh =
B
ρ
dρ
dρ
or
ρ
2
But, dp = ρgdh. Therefore
T
=
g
dh
B
Integrate, using ρ0 = 1064 kg/m3, and B = 2.1 × 109 N/m2
ρ
∫
2
2.22
dρ
h
g
dh
=
2
B
ρ
0
∫
This gives ρ =
⎛
⎛1
1 ⎞ ⎜
9.81
∴ −⎜ −
⎟=⎜
⎝ ρ 1064 ⎠ ⎜ 2.1×109 N/m 2
⎝
(
)
⎞
⎟ h = 4.67 ×10−9 h
⎟⎟
⎠
1
9.4 ×10
h
−4
− 4.67 ×10−9 h
h
∫
Now p = ρ gdh =
0
g
∫ 9.4 × 10−4 − 4.67 ×10−9 h dh
0
=
g
−4.67 × 10
−9
ln(9.4 × 10 −4 − 4.67 × 10−9 h)
If we assume ρ = const: p = ρ gh = 1064 × 9.81× h = 10, 438 h
b) For h = 1500 m: paccurate = 15,690 kPa and pestimate = 15,657 kPa.
% error =
2.24
15, 657 − 15, 690
× 100 = −0.21%
15, 690
Use Eq. 2.4.8:
p=
9.81
×287
0.0065
101(288 − 0.0065 z / 288)
17
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Chapter 2 / Fluid Statics
a) for z = 3000
∴p = 69.9 kPa.
c) for z = 9000
∴p = 30.6 kPa.
Manometers
2.26
Referring to Fig. 2.7b, the pressure in the pipe is:
p = γh = (13.6 × 9810) × 0.25 = 33,350 Pa or 33.35 kPa
Referring to Fig. 2.7a, the pressure in the pipe is p = ρgh. If p = 2400 Pa, then
2400 = ρgh = ρ × 9.81 h or ρ =
2.28
a) ρ =
2400
= 680 kg/m3
9.81× 0.36
c) ρ =
2.30
2400
9.81 h
∴The fluid is gasoline
2400
= 999 kg/m3
9.81× 0.245
∴The fluid is water
See Fig. 2.7b: The pressure in the pipe is given by p1 = –γ1h + γ2H
p1= –0.86 × 9810 × 0.125 + 13.6 × 9810 × 0.24 = 30,695 Pa or 30,96 kPa
pwater − poil = γ oil × 0.3 + γ Hg × H − γ water × 0.2
2.32
40,000 – 16,000 = 920 × 9.81 × 0.3 + 13,600 × 9.81 × H −1000 × 9.81 × 0.2
Solving for H we get: H = 0.1743 m or 17.43 cm
pwater = γ Hg ( 0.16 ) − γ water (0.02) − γ Hg (0.04) − γ water (0.02)
2.34
Using γ water = 9.81 kN/m3 and γ Hg = 13.6 × 9.81 kN/m3
pwater = 15.62 kPa
2.36
pwater – 9.81 × 0.12 – 0.68 × 9.81 × 0.1 + 0.86 × 9.81 × 0.1 = poil
With pwater = 15 kPa, poil = 14 kPa
pgage = pair + γ water × 4 where, pair = patm − γ Hg × H
2.38
⇒ pgage = −γ Hg × H + γ water × 4
pgage = −13.6 × 9.81 × 0.16 + 9810 × 4 = 17.89 kPa
Note: we subtracted atmospheric pressure since we need the gage pressure.
18
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Chapter 2 / Fluid Statics
2.40
p + 9810 × 0.05 + 1.59 × 9810 × 0.07 – 0.8 × 9810 × 0.1 = 13.6 × 9810 × 0.05
∴p = 5873 Pa
or
5.87 kPa
The distance the mercury drops on the left equals the distance along the tube
that the mercury rises on the right. This is shown in the sketch.
Oil (S = 0.87)
B
10 cm
Water
Δh
A
9 cm
7 cm
Δh
Mercury
40
2.42
o
From the previous problem we have:
( pB )1 = pA + γ water × 0.07 − γ HG × 0.09sin 40 − γ oil × 0.1 = 2.11 kPa
(1)
For the new condition:
( pB )2 = pA + γ water × ( 0.07 + Δh ) − γ HG × 0.11sin 40 − γ oil × ( 0.1 − Δh sin 40 )
(2)
where Δh in this case is calculated from the new manometer reading as:
Δh + Δh / sin 40 = 11 − 9 cm ⇒
Δh = 0.783 cm
Subtracting Eq.(1) from Eq.(2) yields:
( pB )2 − ( pB )1 = γ water × ( Δh ) − γ HG × 0.02sin 40 − γ oil × ( −Δh sin 40 )
Substituting the value of Δh gives:
( pB )2 = 2.11 + ⎡⎣( 0.00783) − 13.6 × 0.02sin 40 − 0.87 × ( −0.00783sin 40 )⎤⎦ × 9.81
= 0.52 kPa
a) Using Eq. (2.4.16):
2.44
p1 = γ1 ( z2 − z1 ) + γ 2 h + ( γ 3 − γ 2 ) H where h = z5 − z2 =17 − 16 = 1 cm
4000 = 9800(0.16 – 0.22) + 15,600(0.01) + (133,400 − 15,600)H
∴H = 0.0376 m or 3.76 cm
19
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Chapter 2 / Fluid Statics
From No. 2.30: poil = 14.0 kPa
2.46
From No. 2.36: poil = pwater – 9.81 × (0.12+Δz) – 0.68 × 9.81 × (0.1−2Δz) +
0.86 × 9.81× (0.1−Δz)
∴Δz = 0.0451 m or 4.51 cm
Forces on Plane Areas
The hydrostatic force is calculated using: F = γh A where, h = 10 m, and
2.48
A = π R2 = π ( 0.15 m ) hence, F = 9810 ×10 × π ( 0.15) = 6934 N
2
2
For saturated ground, the force on the bottom tending to lift the vault is:
F = pc A = 9800 × 1.5 × (2 × 1) = 29,400 N
2.50
The weight of the vault is approximately: W = ρgV walls
W = 2400 × 9.81 ⎡⎣ 2 ( 2 × 1.5 × 0.1) + 2 ( 2 × 1× 0.1) + 20 ( 0.8 × 1.3 × 0.1) ⎤⎦ = 28,400 N
The vault will tend to rise out of the ground.
b) Since the triangle is horizontal the force is due to the uniform pressure at a
depth of 10 m. That is, F = pA, where p = γ h = 9.81×10 = 98.1 kN/m2
2.52
The area of the triangle is A = bh 2 = 2.828 × 2 / 2 = 2.828 m2
F = 98.1 × 2.828 = 277.4 kN
a) F = γ hA = 9.81× 6 × π 22 = 739.7 kN
yp = y +
I
π × 24 / 4
= 6+
= 6.167 m
Ay
4π × 6
∴(x, y)p = (0, –0.167) m
c) F = 9.81 × (4 + 4/3) × 6 = 313.9 kN
2.54
y p = 5.333 +
4/2.5 =
1.5
x
3 × 4 3 / 36
= 5.50 m
5.333 × 6
∴y = –1.5
y
∴x = 0.9375
∴(x, y)p = (0.9375, –1.5) m
2.56
F = γh A = 9810 × 6 × 20 = 1.777 × 106 N, or 1177 kN
20
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x
Chapter 2 / Fluid Statics
I
4 × 5 3 / 12
yp = y +
= 7.5 +
= 7.778 m
7.5 × 20
Ay
ΣMHinge = 0 ⇒(10 – 7.778) 1177 = 5 P
∴P = 523 kN
The vertical height of water is h = 1.22 − 0.42 = 1.1314 m
The area of the gate can be split into two areas: A = A1 + A2 or
A = 1.2 × 1.1314 + 0.4 × 1.1314 = 1.8102 m2
Use 2 forces: F1 = γ h1A1 = 9810 × 0.5657 × (1.2 ×1.1314) = 7534 N
F2 = γ h2 A2 = 9810 ×
2.60
1.1314
× (0.4 × 1.1314) = 1674 N
3
2
The location of F1 is at y p1 = ( 1.1314 ) = 0.754 m, and F2 is at
3
yp2 = y +
I2
1.1314
0.4 × 1.13143 / 36
= 0.5657 m
=
+
A2 y
3
0.4 × (1.1314 / 2) × (1.1314 / 3)
ΣM hinge = 0 : 7534 ×
1.1314
+ 1674 × (1.1314 − 0.5657 ) − 1.1314 P = 0
3
∴P = 3346 N
The gate is about to open when the center of pressure is at the hinge.
2.62
b) y p = 1.2 + H = (2.0/2 + H ) +
b × 23 /12
(1 + H )2b
∴H = 0.6667 m
A free-body-diagram of the gate and block is
sketched.
T
Sum forces on the block:
ΣFy = 0
2.64
T
0
∴W = T + FB
where FB is the buoyancy force which is given
by
FB = γ ⎡⎣π R (3 − H ) ⎤⎦
2
F stop
FB
yp
FH
W
Take moments about the hinge:
T × 3.5 = FH × (3 − yp )
Rx
Ry
where FH is the hydrostatic force acting on the gate. It is, using
21
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Chapter 2 / Fluid Statics
h = 1.5 m and A = 2 × 3 = 6 m2
FH = γ hA = ( 9.81 kN/m3 )(1.5 m × 6 m2 ) = 88.29 kN
From the given information
( )
2 33 /12
I
yp = y +
= 1.5 +
=2m
1.5 × 6
yA
∴T =
88.29 × ( 3 − 2 )
3.5
= 25.23 kN
FB = W − T = 70 − 25.23 = 44.77 kN.
H =3 m−
44.77 kN
( 9.81 kN/m ) π (1 m )
2
3
∴ γπ R2 ( 3 − H ) = 44.77
= 1.55 m
The dam will topple if there is a net clockwise moment about “O”
The weight of the dam consists of the weight of the rectangular area + a
triangular area, that is: W = W1 + W2 . The force F3 acting on the bottom of the
dam can be divided into two forces:
Fp1 due to the uniform pressure distribution and
Fp2 due to the linear pressure distribution.
b)
2.66
W1 = 2.4 × 9810 × 18.9 × 1.8 = 801 kN
W2 = 2.4 × 9810 × 18.9 × 7.2 / 2 = 1602 kN
W3 = 9810 × (18 × 6.86/2) = 605.7 kN
F1 = 9810 × 9 × 18 = 1589 kN
assume 1 m deep
W3
F1
W
F2 = 9810 × 1.5 × 3 = 44.14 kN
F2
O
F3
Fp1 = 9810 × 3 × 9 = 265 kN
Fp2 = 9810 × 15 × 9 / 2 = 662.2 kN
ΣM O : (1589)(6) + (265)(4.5) +
(662.2)(6) − (801)(0.9) − (44.14)(3/3)
− (1602)(4.2) − (605.7)(6.37) = 3392 kN ⋅ m > 0. ∴will tip
22
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Chapter 2 / Fluid Statics
Forces on Curved Surfaces
Since all infinitesimal pressure forces pass through the center, we can place the
resultant forces at the center. Since the vertical components pass through the
bottom point, they produce no moment about that point. Hence, consider only
horizontal forces:
2.68
( FH ) water = ( γ hA)
(FH )oil = ( γ hA )
oil
water
= 9.81× 2 × (4 ×10) = 784.8 kN
= 0.86 × 9.81×1× 20 = 168.7 kN
ΣM: 2 P = 784.8 × 2 − 168.7 × 2.
∴P = 616.1 kN
A free-body-diagram of the volume of water in the
vicinity of the surface is shown.
F1
Force balances in the horizontal and vertical
directions give:
FH
FH = F2
F2
FV = W + F1
.
W
FV
.
A
xV
B
where FH and FV are the horizontal and vertical components of the force acting
on the water by the surface AB. Hence
FH = F2 = ( 9.81 kN/m3 ) ( 8 + 1)( 2 × 4 ) = 706.3 kN
The line of action of FH is the same as that of F2. Its distance from the surface
is
2.70
( )
4 23 12
I
yp = y +
= 9+
= 9.037 m
9×8
yA
To find FV we find W and F1:
π
⎡
⎤
W = γ V = ( 9.81 kN/m3 ) ⎢2 × 2 − ( 22 ) ⎥ × 4 = 33.7 kN
4
⎣
⎦
F1 = 9.81 kN/m3 ( 8 × 2 × 4 ) = 628 kN
∴ FV = F1 + W = 33.7 + 628 = 662 kN
To find the line of action of FV, we take moments at point A:
FV × xV = F1 × d1 + W × d2
where d1 = 1 m, and d2 =
2R
2× 2
=
= 1.553 m:
3(4 − π ) 3(4 − π )
23
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Chapter 2 / Fluid Statics
∴ xV =
F1 × d1 + W × d2 628 ×1 + 33.7 ×1.553
=
= 1.028 m
662
FV
Finally, the forces FH and FV that act on the surface AB are equal and opposite
to those calculated above. So, on the surface, FH acts to the right and FV acts
downward.
2.72
Place the resultant FH + FV at the center. FV passes through the hinge. The
moment of FH must equal the moment of P with respect to the hinge:
2 × (9.81 × 1 × 10) = 2.8 P
∴P = 70.1 kN
The resultant FH + FV of the unknown liquid acts through the center of the
circular arc. FV passes through the hinge. Thus, we use only ( FH ) . Assume
1 m wide:
FH = γ x hA = γ x ( R 2 )( R ×1) = γ x R2 2
The horizontal force due to the water is Fw = γ w hA = γ w ( R 2 )( R ×1) = γ w R2 2
2.74
The weight of the gate is W = Sγ w V = 0.2γ w (π R2 4) ×1
Summing moments about the hinge:
Fw ( R 3) +W ( 4R 3π ) = FH × R
⎛
R2 ⎞ R ⎛
πR 2 ⎞ 4 R ⎛ R 2 ⎞
⎟⎟ × + ⎜⎜ 0.2 × 9810
⎟×
⎟× R
a) ⎜⎜ 9810 ×
= ⎜γ x
2 ⎠ 3 ⎝
4 ⎟⎠ 3π ⎜⎝
2 ⎟⎠
⎝
∴ γ x = 4580 N/m3
The pressure in the dome is:
a) p = 60,000 – 9810 × 3 – 0.8 × 9810 × 2 = 14,870 Pa or 14.87 kPa
The force is F = pAprojected = (π × 32) × 14.87 = 420.4 kN
2.76
b) From a free-body diagram of the dome filled with oil:
W
Fweld + W = pA
Using the pressure from part (a):
Fweld = 14,870 × π × 32 – (0.8 × 9810) ×
pA
Fweld
1⎛4
⎞
⎜ π × 3 3 ⎟ = –23,400 or –23.4 kN
⎝
⎠
2 3
24
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Chapter 2 / Fluid Statics
Buoyancy
Under static conditions the weight of the barge + load = weight of displaced
water.
2.78
(a) 20,000 + 250,000 = 9810 × 3 (6d + d 2 /2). ∴d2 + 12d – 18.35 = 0
∴d = 1.372 m
2.80
2.82
The weight of the cars will be balanced by the weight of displaced water:
15,000 × 60 = 9810(7.5 × 90 × Δd )
∴Δd = 0.136 m or 13.6 cm
T + FB = W (See Fig. 2.11 c)
T = 40,000 – 1.59 × 9810 × 2 = 8804 N or 8.804 kN
At the limit of lifting: FB = W+pA where p is the pressure
acting on the plug.
(b) Assume h > 4.5 + R and use the above equation with
2.86
R = 0.4 m and h = 4.92
FB = γwV = γw ×3×(π R2 − Asegment ) = 9810 × 3 × 0.2803 = 8249 N
W + pA = 6670 N + 9810 × 4.92 × π ( 0.1) = 8234 N
2
Hence, the plug will lift for h > 4.92 m
(a) When the hydrometer is completely submerged in water:
⎡ π × 0.0152
⎤
π × 0.0052
W = γ w V ⇒ (0.01 + mHg )9.81 = 9810 ⎢
× 0.15 +
× 0.12 ⎥
4
4
⎢⎣
⎥⎦
2.88
∴mHg = 0.01886 kg
When the hydrometer without the stem is submerged in a fluid:
W = γ x V ⇒ (0.01 + 0.0189)9.81 = S x × 9810 ×
π ( 0.015 )
4
2
× 0.15
∴Sx = 1.089
25
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Chapter 2 / Fluid Statics
Stability
With ends horizontal I o = π d 4 64 The displaced volume is V = W γ water ⇒
V = ( γ xπ d 2 h / 4 ) / 9810 = 8.01× 10−5 γ x d 3 since h = d
The depth the cylinder will sink is
depth =
2.90
V 8.01× 10−5 γ x d 3
=
= 10.20 × 10−5 γ x d
A
π d2 / 4
The distance CG is CG =
h
− 10.2 × 10 −5 γ x d / 2 . Then
2
IO
π d 4 / 64
d
− +10.2 ×10−5γ x d / 2 > 0
GM = − CG =
3
−5
V
8.01×10 γ xd 2
This gives (divide by d and multiply by γx): 612.8 – 0.5 γx + 5.1 × 10−5 γ 2x > 0.
Consequently γx > 8368 N/m3
As shown, y =
2.92
G=
or
16 × 9 + 16 × 4
= 6.5 cm above the bottom edge.
16 + 16
4γ × 9.5 + 16γ × 8.5 + 16SAγ × 4
= 6.5 cm
0.5γ × 8 + 2γ × 8 + SAγ ×16
∴130 + 104 SA = 174 + 64 SA
∴ SA = 1.1
The centroid C is 1.5 m below the water surface
2.94
γx < 1436 N/m3
Using Eq. 2.4.47: GM =
∴ CG = 1.5 m
A × 83 / 12
− 1.5 = 1.777 − 1.5 = 0.277 > 0
A ×8×3
∴The barge is stable
Linearly Accelerating Containers
(a) pmax = −1000 × 20 (0 − 4) – 1000 (9.81) (0 − 2) = 99,620 Pa
2.96
(c) pmax = −1000 × 18 (0 – 3.6) – 1000 (9.81 + 18) (0 – 1.8)
= 114,860 Pa or 114.86 kPa
26
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Chapter 2 / Fluid Statics
Use Eq. 2.5.2:
⎛
8a x ⎞
⎟
b) 60,000 = –1000 ax (–8) – 1000 (9.81 + 10) ⎜ − 2.5 +
⎜
⎟
9
.
81
⎝
⎠
2.98
60 = 8 ax + 49.52 – 19.81
8ax
or ax – 1.31 = 1.574
19.81
ax
∴ax = 0.25, 4.8 m/s2
a x2 – 5.1 ax + 1.44 = 0
a) The pressure on the end AB (z is zero at B) is, using Eq. 2.5.2
p(z) = –1000 × 10 (–7.626) – 1000 × 9.81(z) = 76,260 – 9810 z
2.5
∴ FAB =
∫ (76,260 − 9810z)4dz
= 640,000 N or 640 kN
0
2.100
b) The pressure on the bottom BC is
p(x) = –1000 × 10 (x – 7.626) = 76,260 – 10,000 x
7.626
∴ FBC =
∫
(76,260 − 10,000 x)4dx = 1.163 × 106 N or 1163 kN
0
Use Eq. 2.5.2 with position 1 at the open end:
b) pA = –1000 × 10 (0.9 – 0) = –9000 Pa
z
1
A
pB = –1000 × 10 (0.9) – 1000 × 9.81(−0.6)
= –3114 Pa
2.102
pC = –1000 × 9.81 × (–0.6) = 5886 Pa
C
B
x
e) pA = 1000 × 18 ( −1.5 × 0.625) = −16,875 Pa
pB = 1000 × 18 ( −1.5 × 0.625 ) – 1000 × 9.81 ( −0.625) = –10,740 Pa
pC = –1000 × 9.81 ( −0.625) = 6130 Pa
Rotating Containers
Use Eq. 2.6.4 with position 1 at the open end:
2.104
a) p A =
z
A
1
1
× 1000 × 102 (0 – 0.92) = –40,500 Pa
2
ω
pB = –40,500 + 9810 × 0.6 = –34,600 Pa
r
pC = 9810 × 0.6 = 5886 Pa
C
B
27
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Chapter 2 / Fluid Statics
The air volume before and after is equal
1
∴ π r02 h = π × 0.6 2 × 0.2
2
z
2
∴ r02 h = 0.144
r0
h
(a) Using Eq. 2.6.5: r02 × 5 2 / 2 = 9.81 h
∴h = 0.428 m
1
∴pA = × 1000 × 52 × 0.62 – 9810 (–0.372)
2
1
A
r
= 8149 Pa
(c) For ω = 10, part of the bottom is bared
1
2
1
2
π × 0.6 2 × 0.2 = π r02 h − π r12 h1
2.106
z
Using Eq. 2.6.5:
ω 2 r02
= h,
2g
∴ 0.144 =
2g
ω
ω 2 r12
2
2g
h2 −
h 2 − h12 =
2g
ω2
h12
r0
= h1
h
or
A
0.144 × 10 2
2 × 9.81
r
h1 1
Also, h – h1 = 0.8. 1.6 h – 0.64 = 0.7339. ∴h = 0.859 m, r1 = 0.108 m
1
∴pA = × 1000 × 102 (0.62 – 0.1082) = 17,400 Pa
2
p (r ) =
1
ρω 2r 2 − ρ g[0 − (0.8 − h)]
2
p(r ) = 500ω 2r 2 + 9810(0.8 − h)
p(r ) = 500ω (r − r )
2
2.108
2
2
1
a) F = ∫ p2π rdr = 2π
dA = 2πrdr
if h < 0.8
dr
if h > 0.8
0.6
∫ (12,500r
3
+ 3650r )dr = 6670 N
0
(We used h = 0.428 m)
c) F = ∫ p2π rdr = 2π
0.6
∫
(50,000(r 3 − 0.1082 r )dr = 9520 N
−0.108
(We used r1 = 0.108 m)
28
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Chapter 3 / Introduction to Fluids in Motion
CHAPTER 3
Introduction to Fluids in Motion
FE-type Exam Review Problems: Problems 3-1 to 3-9
nˆ ⋅ V = 0
3.1 (D)
(nxˆi + ny ˆj) ⋅ (3ˆi − 4ˆj) = 0
or
3nx − 4ny = 0
Also nx2 + ny2 = 1 since n̂ is a unit vector. A simultaneous solution yields
nx = 4 / 5 and n y = 3 / 5. (Each with a negative sign would also be OK)
a=
3.2 (C)
∂V
∂V
∂V
∂V
+u
+v
+w
= 2xy (2 yˆi ) − y 2 (2xˆi − 2 yˆj) = −16ˆi + 8ˆi + 16ˆj
∂t
∂x
∂y
∂z
∴ a = (−8)2 + 162 = 17.89 m/s
ax =
3.3 (D)
=
∂u
∂u
∂u
∂u
∂u
10
∂ ⎡
+u +v +w
=u
=
10(4 − x) −2 ⎤
2
⎣
⎦
∂t
∂x
∂y
∂z
∂x (4 − x) ∂x
10
10
1
10(−2)(−1)(4 − x)−3 = × 20 × = 6.25 m/s2
2
4
8
(4 − x)
3.4 (C)
The only velocity component is u(x). We have neglected v(x) since it is quite
small. If v(x) were not negligible, the flow would be two-dimensional.
3.5 (B)
V 2 p γ water h 9810 × 0.800
= =
=
ρ
ρ air
2
1.23
3.6 (C)
V12 p V22
+ =
2g γ 2g
V12
+ 0.200 = 0.600
2g
∴V = 113 m/s
∴V = 2 × 9.81× 0.400 = 2.80 m/s
The manometer reading h implies:
3.7 (B)
V12 p1 V22 p2
2
+
=
+
or V22 =
(60 − 10.2)
ρ
ρ
2
2
1.13
∴V2 = 9.39 m/s
The temperature (the viscosity of the water) and the diameter of the pipe are not
needed.
29
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Chapter 3 / Introduction to Fluids in Motion
3.8 (A)
V12 p1 V22 p2
+
=
+
2g γ
2g
γ
3.9 (D)
p1 =
800,000
V22
=
9810
2 × 9.81
∴V2 = 40 m/s
ρ
902
V22 − V12 ) =
302 − 152 ) = 304,400 Pa
(
(
2
2
Flow Fields
a) u =
3.14
dx
= 2t + 2
dt
v=
dy
= 2t
dt
y
x = t 2 + 2t + c1
y = t 2 + c2
(27, 21)
= y+2 y
∴parabola
x
Lagrangian: Several college students would be hired to ride bikes around the
various roads, making notes of quantities of interest.
Eulerian:
a) cos α =
Several college students would be positioned at each intersection
and quantities would be recorded as a function of time.
V ⋅ ˆi
1+ 2
=
= 0.832
V
32 + 22
V ⋅ nˆ = 0
3.18
(35, 25)
39.8o
∴ x 2 − 2 xy + y 2 = 4 y
3.16
streamlines
t=5s
∴ nx =
c) cos α =
∴α = 33.69D
(3ˆi + 2ˆj) ⋅ (nxˆi + ny ˆj) = 0
2
3
1
, ny = −
or nˆ =
(2ˆi − 3ˆj)
13
13
13
5
V ⋅ ˆi
=
= 0.6202
V
52 + (−8)2
V ⋅ nˆ = 0
3
ny = − nx
3nx + 2ny = 0 ⎫
⎪
2
⎬ ∴
2
2
9
nx + ny = 1 ⎪⎭
nx2 + nx2 = 1
4
(5ˆi − 8ˆj) ⋅ (nx ˆi + ny ˆj) = 0
∴α = −51.67D
8
nx = n y
5nx − 8ny = 0 ⎫
⎪
5
⎬ ∴
2
2
64 2
nx + ny = 1 ⎪⎭
ny + ny2 = 1
25
30
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Chapter 3 / Introduction to Fluids in Motion
∴ny =
5
8
1 ˆ ˆ
, nx =
or nˆ =
(8i + 5j)
89
89
89
∂V
∂V
∂V ∂V
+v
+w
+
= 2 x(2ˆi ) + 2 y (2ˆj) = 4 xˆi + 4 yˆj = 8ˆi − 4ˆj
∂x
∂y
∂z ∂t
3.20
b) u
3.22
The vorticity ω = 2Ω. a) ω = −40ˆi c) ω = 12ˆi − 4kˆ
40 ⎞
40 ⎞ sin θ ⎛ 40 ⎞
⎛
⎛ 80 ⎞
⎛
a) ar = ⎜10 − 2 ⎟ cos θ ⎜ 3 ⎟ cos θ − ⎜10 + 2 ⎟
⎜ 1 − ⎟ ( − sin θ )
r ⎠
r ⎠ r ⎝ r2 ⎠
⎝
⎝r ⎠
⎝
2
1⎛
40 ⎞
− ⎜ 10 + 2 ⎟ sin 2 θ = (10 − 2.5)( −1)1.25( −1) = 9.375 m/s2
r⎝
r ⎠
3.24
40 ⎞
40 ⎞ sin θ ⎛
40 ⎞
⎛
⎛ 80 ⎞
⎛
aθ = ⎜ 10 − 2 ⎟ cos θ ⎜ 3 ⎟ sin θ + ⎜ 10 + 2 ⎟
⎜ 10 + 2 ⎟ cos θ
⎝
⎝r ⎠
⎝
r ⎠
r ⎠ r ⎝
r ⎠
1⎛
1600 ⎞
− ⎜ 100 − 4 ⎟ sin θ cos θ = 0
r⎝
r ⎠
since sin(180°) = 0
aφ = 0
3.26
a=
∂V
∂V
∂V
∂V ∂u ˆ
+u
+v
+w
=
i
∂t
∂x
∂y
∂z ∂t
For steady flow ∂u / ∂t = 0 so that a = 0
3.28
3.30
b) u = 2(1 − 0.52 )(1 − e −t /10 ) = 1.875 m/s at t = ∞
−4
Dρ
∂ρ
∂ρ
∂ρ ∂ρ
=u +v +w +
= 10(−1.23 ×10−4 e −3000×10 )
Dt
∂x
∂y
∂z ∂t
= −9.11× 10 −4 kg/m 3 ⋅ s
3.32
Dρ
∂ρ
=u
= 4 × (0.01) = 0.04 kg/m3⋅s
Dt
∂x
31
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Chapter 3 / Introduction to Fluids in Motion
∂u
⎫
+ V ⋅∇u ⎪
∂t
⎪
∂v
∂V
⎪
+ V ⋅∇v ⎬ ∴ a =
+ (V ⋅∇)V
ay =
∂t
∂
t
⎪
∂w
⎪
+ V ⋅∇w ⎪
az =
∂t
⎭
ax =
3.34
2π
kˆ = 7.272 × 10−5 kˆ rad/s
24 × 60 × 60
V = 5( −0.707ˆi − 0.707kˆ ) = −3.535ˆi − 3.535kˆ m/s
Ω=
A = 2Ω × V + Ω × ( Ω × r )
= 2 × 7.272 ×10−5 kˆ × (−3.535ˆi − 3.535kˆ ) + 7.272 ×10−5 kˆ
× ⎡7.272 ×10−5 kˆ × (6.4 ×106 )(−0.707ˆi + 0.707kˆ ) ⎤
⎣
⎦
3.36
= −52 ×10−5 ˆj + 0.0224ˆi m/s 2
Note: We have neglected the acceleration of the earth relative to the sun since it
is quite small (it is d 2S /dt 2 ). The component ( −51.4 × 10 −5 ˆj) is the Coriolis
acceleration and causes air motions to move c.w. or c.c.w. in the two
hemispheres.
Classification of Fluid Flows
3.38
3.42
a, c, e, f, h
Steady:
a) inviscid
Unsteady: b, d, g
c) inviscid
e) viscous inside the boundary layers and separated regions.
VL
3.46
Re =
3.48
a) Re =
ν
=
VD
ν
0.18 × 0.75
= 9640
1.4 ×10−5
=
1.2 × 0.01
= 795
1.51× 10−5
g) viscous.
∴Turbulent
Always laminar
Assume the flow is parallel to the leaf. Then 3 × 105 = VxT /ν
3.50
∴ xT = 3 × 105ν / V = 3.5 × 105 × 1.4 × 10−4 / 6 = 8.17 m
The flow is expected to be laminar
32
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Chapter 3 / Introduction to Fluids in Motion
Dρ
∂ρ
∂ρ
∂ρ ∂ρ
=u
+v
+w
+
=0
Dt
∂x
∂y
∂z ∂t
3.52
For a steady, plane flow ∂ρ /∂t = 0 and w = 0
∂ρ
∂ρ
+v
=0
∂x
∂y
u
Then
Bernoulli’s Equation
3.54
V2 p
=
2
ρ
Use ρ = 1.1 kg/m3
a) v = 2 p / ρ = 2 × 2000 /1.1 = 60 m/s
3.56
3.58
V2 p
+ =0
2
ρ
∴V =
V 2 p U∞2 p∞
+ =
+
2 ρ
2
ρ
−2 p
ρ
=
2 × 2000
= 57.0 m/s
1.23
b) Let r = rc :
d) Let θ = 90 D :
pT =
ρ
2
U ∞2
3
p 90 = − ρU ∞2
2
V 2 p U ∞2
p
+ =
+ ∞
2 ρ
2
ρ
a) p =
ρ
2
(
)
U ∞2 − u 2 =
ρ⎡
20π ⎞
⎛
⎢102 − ⎜10 +
⎟
2 ⎢⎣
2π x ⎠
⎝
2⎤
⎡ ⎛ 1 ⎞2 ⎤
⎥ = 50 ρ ⎢1 − ⎜ 1 + ⎟ ⎥
⎥⎦
⎢⎣ ⎝ x ⎠ ⎥⎦
⎛2 1 ⎞
= −50 ρ ⎜ + 2 ⎟
⎝x x ⎠
3.60
c) p =
ρ
2
(
)
U ∞2 − u 2 =
ρ⎡
60π ⎞
⎛
⎢302 − ⎜ 30 +
⎟
2 ⎢⎣
2π x ⎠
⎝
2⎤
⎡ ⎛ 1 ⎞2 ⎤
⎥ = 450 ρ ⎢1 − ⎜1 + ⎟ ⎥
⎥⎦
⎢⎣ ⎝ x ⎠ ⎥⎦
⎛2 1 ⎞
= −450 ρ ⎜ + 2 ⎟
⎝x x ⎠
33
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Chapter 3 / Introduction to Fluids in Motion
Assume the velocity in the plenum is zero. Then
3.62
V12 p1 V22 p2
2
+
=
+
or V22 =
(60 − 10.2)
2
2
1.13
ρ
ρ
∴V2 = 9.39 m/s
We found ρ = 113
. kg / m 3 in Table B.2.
Bernoulli from the stream to the pitot probe:
V2
+p
pT = ρ
2
Manometer: pT + γ H − γ Hg H − γ h = p − γ h
3.64
Then, ρ
V2
+ p + γ H − γ Hg H = p
2
a) V 2 =
(13.6 − 1)9800
(2 × 0.04)
1000
b) V 2 =
(13.6 − 1)9800
(2 × 0.1)
1000
∴V 2 =
γ Hg − γ
ρ
(2 H )
∴ V = 3.14 m/s
∴ V = 4.97 m/s
The pressure at 90° from Problem 3.58 is p90 = −3 ρU ∞2 / 2. The pressure at the
3.66
stagnation point is pT = ρU ∞2 / 2. The manometer provides: pT − γH = p90
1
3
× 1.204U ∞2 − 9800 × 0.04 = − × 1.204U ∞2 .
2
2
3.68
Bernoulli:
V22
p
V2 p
+ 2 = 1 + 1
2g
γ
2g γ
Manometer:
p1 + γ z + γ Hg H − γ H − γ z =
∴ U ∞ = 12.76 m/s
V22
γ + p2
2g
Substitute Bernoulli’s into the manometer equation:
V2
p1 + γ Hg − γ H = 1 γ + p1
2g
(
)
b) Use H = 0.05 m:
V12 × 9800
= (13.6 − 1)9800 × 0.05
2 × 9.81
∴ V1 = 3.516 m/s
Substitute into Bernoulli:
p1 =
V22 − V12
202 − 3.5162
γ=
× 9800 = 193,600 Pa
2g
2 × 9.81
34
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Chapter 3 / Introduction to Fluids in Motion
Write Bernoulli’s equation between points 1 and 2 along the center streamline:
p1 +
ρV12
2
+ γ z1 = p2 +
ρV22
2
+ γ z2
Since the flow is horizontal, z1 = z2 and Bernoulli’s equation becomes
3.70
p1 + 1000 ×
0.52
1.1252
= p2 + 1000 ×
2
2
From fluid statics, the pressure at 1 is p1 = γ h = 9810 × 0.25 = 2452 Pa and at 2,
using p2 = γH, Bernoulli’s equation predicts
2452 + 1000 ×
0.52
1.1252
= 9810H + 1000 ×
2
2
∴ H = 0.1982 m or 19.82 cm
Assume incompressible flow (V < 100 m/s) with point 1 outside the wind tunnel
where p1 = 0 and V 1 = 0. Bernoulli’s equation gives
0=
3.72
3.74
V22 p2
+
2
ρ
1
∴ p2 = − ρ V22
2
a) ρ =
p
90
1
=
= 1.239 kg/m3 ∴ p2 = − × 1.239 × 1002 = −6195 Pa
RT 0.287 × 253
2
c) ρ =
p
92
1
=
= 1.094 kg/m3 ∴ p2 = − × 1.094 × 1002 = −5470 Pa
2
RT 0.287 × 293
Bernoulli across nozzle:
V12 p1 V22 p2
+
=
+
2
ρ
2
ρ
Bernoulli to max. height:
p
V12 p1
V22
+ + h1 =
+ 2 + h2
2g γ
2g
γ
∴V2 = 2 p1 / ρ
∴ h2 = p1 / γ
a) V2 = 2 p1 / ρ = 2 × 700,000 /1000 = 37.42 m/s
h2 = p1 / γ = 700,000/9800 = 71.4 m
b) V2 = 2 p1 / ρ = 2 ×1, 400,000 /1000 = 52.92 m/s
h2 = p1 / γ = 1,400,000/9800 = 142.9 m
3.76
V12 p1 V22 p2
+
=
+
2
ρ
2
ρ
p2 = −100,000 Pa, the lowest possible pressure.
35
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Chapter 3 / Introduction to Fluids in Motion
a)
600,000 V22 100,000
=
−
1000
2
1000
∴ V 2 = 37.4 m/s
b)
300,000 V22 100,000
=
−
1000
2
1000
∴ V 2 = 28.3 m/s
b) p1 =
3.78
d) p1 =
ρ
902 2
V22 − V12 ) =
2 − 102 ) = − 43,300 Pa
(
(
2
2
ρ
1.23 2
V22 − V12 ) =
2 − 102 ) = −59.0 Pa
(
(
2
2
Apply Bernoulli’s equation between the exit (point 2) where the radius is R and
a point 1 in between the exit and the center of the tube at a radius r less than R:
3.80
V12 p1 V22 p2
+
=
+
2
ρ
2
ρ
∴ p1 = ρ
V22 − V12
2
Since V 2 < V 1 , we see that p1 is negative (a vacuum) so that the envelope
would tend to rise due to the negative pressure over most of its area (except for a
small area near the end of the tube).
3.82
3.84
A burr downstream of the opening will create a region that
acts similar to a stagnation region thereby creating a high
pressure since the velocity will be relatively low in that
region.
stagnation
region
The higher pressure at B will force the fluid toward the lower
pressure at A, especially in the wall region of slow moving
fluid, thereby causing a secondary flow normal to the pipe’s
axis. This results in a relatively high loss for an elbow.
36
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Chapter 4 / The Integral Forms of the Fundamental Laws
CHAPTER 4
The Integral Forms of the
Fundamental Laws
FE-type Exam Review Problems: Problems 4-1 to 4-15
4.1 (B)
4.2 (D)
m = ρ AV =
p
200
AV =
π × 0.042 × 70 = 0.837 kg/s
RT
0.287 × 293
Refer to the circle of Problem 4.27:
4.3 (A)
4.4 (D)
Q = AV = (π × 0.42 ×
75.7 × 2
− 0.10 × 0.40 × sin 75.5 ) × 3 = 0.516 m3/s
360
WP V22 − V12 p2 − p1
=
+
γQ
2g
γ
∴WP = 40 kW
4.5 (A)
0=
V22 − V12
2g
+
p2 − p1
γ
WP
1200 − 200
=
γ × 0.040
γ
and energy required =
−1202
p
0=
+ 2 . ∴ p2 = 7, 200,000 Pa
2 × 9.8 9810
Manometer: γ H + p1 = ρ g
4.6 (C)
Energy: K
V22
+ p2
2g
7.962 100,000
=
2 × 9.81
9810
or 9810 × 0.02 + p1 = ρ g
4.7 (B)
K
V 2 Δp
=
2g γ
7.962 100,000
=
2 × 9.81
9810
V=
V22
2g
∴ K = 3.15
Combine the equations: 9810 × 0.02 = 1.2 ×
hL = K
40
= 47.1 kW
0.85
V12
2
∴V1 = 18.1 m/s
Q
0.040
=
= 7.96 m/s
A π × 0.042
∴ K = 3.15
37
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 4 / The Integral Forms of the Fundamental Laws
4.8 (C)
WP V22 − V12 Δp
=
+
γQ
2g
γ
WP = QΔp = 0.040 × 400 = 16 kW
36.0 + 15 =
4.9 (D)
η
=
4.582
p
7.162
+ B + 3.2
2 × 9.81 9810
2 × 9.81
16
= 18.0 kW
0.89
∴ pB = 416,000 Pa
In the above energy equation we used
hL = K
V=
4.10 (A)
WP
V2
Q
0.2
with V = =
= 4.42 m/s
2g
A π × 0.22
Q
0.1
=
= 19.89 m/s
A π × 0.042
HP =
Energy—surface to entrance:
∴ HP =
V22 p2
V2
+
+ z2 + K 2
2g γ
2g
19.892 180,000
19.892
+
+ 50 + 5.6
= 201.4 m
2 × 9.81
9810
2 × 9.81
∴WP = γ QH P /η P = 9810 × 0.1× 201.4 / 0.75 = 263,000 W
4.11 (A)
After the pressure is found, that pressure is multiplied by the area of the
window. The pressure is relatively constant over the area.
V12 p1 V22 p2
.
+
=
+
2g γ
2g
γ
4.12 (C)
p1 = 9810 ×
(6.252 − 1) × 12.732
= 3, 085, 000 Pa
2 × 9.81
p1 A1 − F = ρ Q(V2 − V1 ). 3, 085, 000 × π × 0.052 − F = 1000 × 0.1× 12.73(6.25 − 1)
∴ F = 17,500 N
4.13 (D)
4.14 (A)
4.15 (A)
− Fx = m(V2 x − V1x ) = 1000 × 0.01 × 0.2 × 50(50cos 60 − 50) = −2500 N
− Fx = m (Vr 2 x − Vr1x ) = 1000 × π × 0.02 2 × 60 × (40 cos 45 − 40) = 884 N
Power = Fx × VB = 884 × 20 = 17, 700 W
Let the vehicle move to the right. The scoop then diverts the water to the right.
Then, F = m(V2 x − V1x ) = 1000 × 0.05 × 2 × 60 × [60 − (−60)] = 720,000 N
38
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Chapter 4 / The Integral Forms of the Fundamental Laws
Basic Laws
4.16
b) The energy transferred to or from the system must be zero: Q − W = 0
4.20
b) The conservation of mass.
d) The energy equation.
System-to-Control-Volume Transformation
nˆ 1 = −
4.24
1 ˆ 1 ˆ
i−
j = −0.707(ˆi + ˆj)
2
2
nˆ 2 = 0.866ˆi − 0.5ˆj .
nˆ 3 = −ˆj
V1n = V1 ⋅ nˆ1 = 10ˆi ⋅ ⎡⎣−0.707(ˆi + ˆj)⎤⎦ = −7.07 m/s
V2n = V2 ⋅ nˆ 2 = 10ˆi ⋅ (0.866ˆi − 0.5ˆj) = 8.66 m/s
V = V ⋅ nˆ = 10ˆi ⋅ (−ˆj) = 0
3n
4.26
3
2
(B ⋅ nˆ ) A = 15(0.5ˆi + 0.866ˆj) ⋅ ˆj (10 × 12) = 15 × 0.866 × 120 = 1559 cm3
Volume = 15 sin 60 ×10 ×12 = 1559 cm3
Conservation of Mass
Use Eq. 4.4.2 with m V representing the mass in the volume:
0=
dmV
dmV
+ ρ A2V2 − ρ A1V1
+ ∫ ρ nˆ ⋅ VdA =
dt
dt c.s.
4.32
=
Finally
dmV
= m − ρQ
dt
π×
A1V1 = A2V2
4.34
dmV
+ ρQ − m
dt
m = ρ AV = 1000π
Q = AV = π
32
10
4
32
104
32
104
× 18 = π ×
62
10 4
V2
∴V2 = 4.5 m/s
×18 = 51 kg/s
× 18 = 0.051 m3 /s
39
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Chapter 4 / The Integral Forms of the Fundamental Laws
ρ1A1V1 = ρ2 A2V2
4.38
p1
500
=
= 4.433 kg/m3
RT 0.287 × 393
1246
ρ2 =
= 8.317 kg/m3
0.287 × 522
ρ1 =
4.433 π × 0.052 × 600 = 8.317 π × 0.052 V2
∴V2 = 319.8 m/s
m = ρ 1 A1V1 = 20.89 kg/s
Q 1 = A1V1 = 4.712 m3 /s
b) (2 × 1.5 + 1.5 × 1.5) 3 = π
d 22 2
×
4 2
4.40
Q2 = 2.512 m3/s
∴d2 = 4.478 m
a) Since the area is rectangular, V = 5 m/s.
m = ρ AV = 1000 × 0.08 × 0.8 × 5 = 320 kg/s
4.42
c) V × 0.08 = 10 × 0.04 + 5 × 0.02 + 5 × 0.02
m = ρ AV = 1000 × 0.08 × 0.8 × 7.5 = 480 kg/s
4.44
m
ρ
= 0.32 m3/s
∴V = 7.5 m/s
Q=
m
ρ
= 0.48 m3 /s
If dm/dt = 0, then ρ 1 A 1V 1 = ρ 2 A 2 V 2 + ρ 3 A 3 V 3 . In terms of m 2 and Q 3 this
becomes, letting ρ1 = ρ2 = ρ3
1000 × π × 0.02 2 × 12 = m2 + 1000 × 0.01
min = mout + m
4.46
Q=
∴ m2 = 5.08 kg/s
⎡
0.1
⎤
⎢⎣
0
⎥⎦
ρ × 0.2 × 2 × 10 = ⎢ ρ ∫ 10(20 y − 100 y 2 )2dy + ρ × 0.1 × 2 × 10 ⎥ + m
Note: We see that at y = 0.1 m the velocity u(0.1) = 10 m/s. Thus, we integrate to y =
0.1, and between y = 0.1 and 0.2 the velocity u = 10:
⎡4
⎤
4ρ = ⎢ ρ + 2ρ ⎥ + m
⎣3
⎦
m = ∫ ρ VdA =
∴ m = 0.6667 ρ = 0.82 kg/s
0 .1
∫ 1165(1 − 1.42 y )(y − 5 y
2
)12 × 1.5dy
0
0 .1
4.48
= 20,970 ∫ (y − 6.42 y 2 + 7.1 y 3 ) dy = 63.7 kg/s
0
2
2
V = umax = × 0.6 = 0.4 m/s (See Prob. 4.43b)
3
3
40
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Chapter 4 / The Integral Forms of the Fundamental Laws
ρ=
1165 + 1000
= 1082 kg/m3
2
∴ρVA = 1082 × 0.4 × (1.5 × 0.1) = 64.9 kg/s
Thus, ρV A ≠ m since ρ = ρ(y) and V = V(y) so that ρV ≠ ρV
4.50
m3 of H 2O
4
m3 of air
= 1.5 × (1.5 h)
× 9000 × 5
2000 × π × 0.00153 3
3
s
m of air
∴h = 0.565 m
4.52
min = m2 + m3 V1 = 20 m/s (see Prob. 4.43c)
20 × 1000π × 0.022 = 10 + 1000π × 0.022 × V3
∴ V3 = 12.04 m/s
The control surface is close to the
interface at the instant shown.
Ve
∴Vi = interface velocity
4.54
n̂
n̂
Vi
ρe AeVe = ρi AV
i i
1.5 × π × 0.152 × 300 =
8000
π × 12 × Vi
0.287 × 673
∴Vi = 0.244 m/s
For an incompressible flow (low speed air flow)
4.56
∫ udA = A V
2
0.2
∫ 20 y
1/5
2
A1
× 0.8dy = π × 0.152 V2
0
5
20 × 0.8 0.26/5 = π × 0.152 V2
6
∴ V2 = 27.3 m/s
Draw a control volume around the entire set-up:
dmtissue
+ ρV2 A2 − ρV1 A1
dt
⎛ d2 − d2 ⎞ 2
= m tissue + ρπ ⎜⎜ 2
⎟⎟ h2 − ρπ ( h1 tan φ ) h1
4
⎝
⎠
0=
4.58
or
mtissue
⎡ d 2 − d 22
⎤
= ρπ ⎢
h2 + h12 h1 tan 2φ ⎥
4
⎢⎣
⎥⎦
41
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Chapter 4 / The Integral Forms of the Fundamental Laws
dm
+ ρ A2V2 − ρ AV
1 1
dt
0=
4.60
= m + 1000(π × 0.0032 × 0.02 − 10 × 10−6 / 60)
m = 3.99 ×10−4 kg/s
0=
4.62
dm
+ ρ 3 Q 3 − ρ 1 A 1V 1 − m 2 where m = ρ Ah
dt
a) 0 = 1000π × 0.62 h + 1000 × 0.6 / 60 − 1000π × 0.022 × 10 − 10
∴ h = 0.0111 m/s
or
11.1 mm/s
Choose the control volume to consist of the air volume inside the tank. The
conservation of mass equation is
0=
d
∫ ρ dV + CS∫ ρ V ⋅ n dA
dt CV
Since the volume of the tank is constant, and for no flow into the tank, the
equation is
0=V
4.64
dρ
+ ρeVe Ae
dt
Assuming air behaves as an ideal gas, ρ =
p
. At the instant of interest,
RT
dρ
1 dp
. Substituting in the conservation of mass equation, we get
=
dt RT dt
(
)
1.8 kg/m3 ( 200 m/s ) π ( 0.015 m ) ⎡
⎤
ρeVe Ae
dp
kJ
( RT ) = −
0.287
=−
× 298 K ⎥
⎢
3
dt
V
1.5 m
kg ⋅ K
⎣
⎦
∴
2
dp
= −14.5 kPa/s
dt
Energy Equation
W = Tω + pAV + μ
4.66
du
Abelt
dy
= 20 × 500 × 2π /60 + 400 × 0.4 × 0.5 × 10 + 1.81× 10−5 × 100 × 0.5 × 0.8
= 1047 + 800 + 0.000724 = 1847 W
42
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Chapter 4 / The Integral Forms of the Fundamental Laws
80% of the power is used to increase the pressure while 20% increases the
internal energy (Q = 0 because of the insulation). Hence
4.68
mΔu = 0.2 W
1000 × 0.02 × 4.18ΔT = 0.2 × 500
4.70
−
WT
= −40 × 0.89
mg
∴ΔT = 0.836°C
b) WT = 40 × 0.89 × (90,000/60) × 9.81 = 523,900 W
V12 p1
V2 p
+ + z1 = 2 + 2 + z2
2g γ
2g γ
42
42
+2=
+ h2
2 × 9.81
19.62h22
4.72
2.82 =
0.82
+ h2
h22
Continuity: 1 × 4 = h2 V2
This can be solved by trial-and-error:
h2 = 2.7 m : 2.82 ? 2.81
h2 = 2.72 m : 2.82 ? 2.83
∵h2 = 2.71 m
h2 = 0.6 m : 2.82 ? 2.87
h2 = 0.62 : 2.82 ? 2.75
∵h2 = 0.61 m
Manometer: Position the datum at the top of the right mercury level.
9810 × 0.4 + 9810 z2 + p2 +
4.74
V22
× 1000 = (9810 × 13.6) × 0.4 + 9810 × 2 + p1
2
Divide by γ = 9810: 0.4 + z 2 +
Energy:
p2
γ
+
V22
p
= 13.6 × 0.4 + 2 + 1
2g
γ
V12 p1
V2 p
+
+ z1 = 2 + 2 + z2
2g γ
2g γ
Subtract (1) from (2): With z1 = 2 m
(2)
V12
= 12.6 × 0.4
2g
Q = 7 × 10 −3 = π × (0.025) 2V1
4.76
(1)
∴V1 = 9.94 m/s
∴V1 = 3.75 m/s
Continuity:
π × (0.025) 2V1 = π × (0.0375) 2V2
Energy:
V12 p1 V22 p2
V2
+
=
+
+ 0.37 1
2g γ
2g γ
2g
∴V2 = 1.58 m/s
43
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Chapter 4 / The Integral Forms of the Fundamental Laws
⎡
3.57 2 1.582 ⎤
∴ p2 = 414 × 103 + 1000 ⎢ 0.63
−
⎥ = 414.3 kPa
19.62 19.62 ⎥⎦
⎢⎣
V1 = Q/A1 =
4.78
Energy:
0.08
π × 0.032
∴V2 = 9V1 = 254.6 m/s
= 28.29 m/s
V12 p1 V22 p2
V2
+ =
+ + 0.2 1
2g γ 2g γ
2g
⎡ 254.6 2
28.29 2 ⎤
6
∴ p1 = 9810 ⎢
− 0.8
⎥ = 32.1× 10 Pa
2
×
9
.
81
2
×
9
.
81
⎣
⎦
a) Energy:
V02 p0
V2 p
+
+ z0 = 2 + 2 + z2
2g γ
2g γ
∴ V 2 = 2 gz 0 = 2 × 9.81 × 2.4 = 6.862 m/s
Q = AV = 0.8 × 1 × 6.862 = 5.49 m3 /s
4.80
For the second geometry the pressure on the surface is zero but it increases with
depth. The elevation of the surface is 0.8 m:
V22
∴ z0 =
+h
2g
∴ V2 = 2 g( z 0 − h) = 2 × 9.81 × 2 = 6.264 m/s
∴Q = 0.8 × 6.264 = 5.01 m3/s
Note: z0 is measured from the channel bottom in the 2nd geometry.
∴z0 = H + h
4.82
V02 p0
V22 p2
+
+ z0 =
+
+ z2
2g γ
2g γ
80,000
V22
+4=
9810
2 × 9.81
∴V2 = 19.04 m/s
b) Q = A2V2 = π × 0.092 × 19.04 = 0.485 m3/s
Manometer: γ H + γ z + p1 = 13.6γ H + γ z + p2
Energy:
p1
γ
+
Combine energy and manometer:
V2 =
p1
γ
= 12.6 H +
p2
γ
V 12 p 2 V 22
=
+
.
2g
γ
2g
4.84
Continuity:
∴
12.6 H =
V22 − V12
2g
⎛ d4 ⎞
∴ V12 = 12.6 H × 2 g ⎜ 14 − 1⎟
⎜d
⎟
⎝ 2
⎠
d12
V1
d 22
44
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Chapter 4 / The Integral Forms of the Fundamental Laws
1/2
d 2 π ⎛ 12.6 H × 2 g ⎞
∴ Q = V1π 1 = ⎜ 4 4
⎟
4 4 ⎜⎝ d1 / d2 − 1 ⎟⎠
a) Energy from surface to outlet:
4.86
Energy from constriction to outlet:
Continuity:
V1 = 4V2
d12
⎛ H
= 12.35d12d22 ⎜ 4
⎜ d − d4
⎝ 1
2
V22
=H
2g
∴ V22 = 2 gH
V12 p2 V22
+
=
+
γ 2g γ 2g
p1
With p1 = pv = 2450 Pa and p2 = 100,000 Pa
2450
16
100,000
1
+
× 2 gH =
+
× 2 gH
9810 2 × 9.81
9810
2 × 9.81
Energy: surface to surface: z0 = z2 + hL
4.88
Continuity:
V1 = 4V2
∴ V 12 = 160 g
30 =
Energy: surface to constriction:
∴z1 = −40.4 m
1/2
⎞
⎟⎟
⎠
∴H = 0.663 m
∴30 = 20 + 2
V22
2g
∴ V 22 = 10g
160 g (−94,000)
+
+ z1
2g
9810
∴H = 40.4 + 20 = 60.4 m
Velocity at exit = Ve Velocity in constriction = V1 Velocity in pipe = V2
Ve2
Energy: surface to exit:
=H
2g
4.90
Continuity across nozzle: V2 =
∴Ve2 = 2 gH
D2
Ve Also, V1 = 4V2
d2
Energy: surface to constriction: H =
a) 5 =
V12 pv
+
2g γ
⎞ −97,550
1 ⎛
D4
16
×
×
2
g
×
5
+
∴ D = 0.131 m
⎜⎜
⎟
⎟
2g ⎝
9810
0.24
⎠
m = pAV = 1000 × π × (0.025) 2 × 36 = 70.65 kg/s
4.92
⎡ 92 − 362 830 ×1000 ⎤
+
WP = 70.65 × 9.81 ⎢
⎥ / 0.85 =18,500 W or 18.5 kW = 24.8 hp
9810 ⎦
⎣ 2 × 9.81
45
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Chapter 4 / The Integral Forms of the Fundamental Laws
⎡ 0 − 10.2 2 −600,000 ⎤
−WT = 2 × 1000 × 9.81 ⎢
+
⎥ × 0.87
9810 ⎦⎥
⎣⎢ 2 × 9.81
4.94
∴WT =
1.304 × 106 W
We used V2 =
Q
2
=
= 10.2 m/s
A2 π × 0.252
∴ ηT = 0.924
⎡ V 2 − V12 p2 p1
⎤
c
a) Q − WS = mg ⎢ 2
+
− + z2 − z1 + v (T2 − T1 ) ⎥
γ 2 γ1
g
⎢⎣ 2 g
⎥⎦
The above is Eq. 4.5.17 with Eq. 4.5.18 and Eq. 1.7.13
4.96
γ1 =
p1g
85 × 9.81
=
= 9.92 N/m3
RT1 0.287 × 293
γ2 =
600 × 9.81 20,500
=
T2
0.287 T2
⎡ 2002
⎤
600,000T2 85,000 716.5
(T2 − 293) ⎥
∴ −(−1,500,000) = 5 × 9.81 ⎢
+
−
+
20,500
9.92
9.81
⎣ 2 × 9.81
⎦
∴T2 = 572 K or 299°C
Be careful of units. p2 = 600,000 Pa
cv = 716.5 J/kg ⋅ K
⎡V 2
V2⎤
Energy: surface to exit: −WTηT = mg ⎢ 2 − 20 + 4.5 2 ⎥
2 g ⎥⎦
⎢⎣ 2 g
4.98
V2 =
15
= 13.26 m/s
π × 0.6 2
mg = Qγ = 15 × 9810 = 147,150 N/s
⎡ 13.262
13.262 ⎤
−WT × 0.8 = 147,150 ⎢
− 20 + 4.5
⎥
2 × 9.81 ⎥⎦
⎢⎣ 2 × 9.81
4.100
∴WT = 5390 kW
Choose a control volume that consists of the entire system and apply
conservation of energy:
V12
p2 V22
+ z = HT + +
+z +h
HP + +
γ 2g 1
γ 2g 2 L
p1
46
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Chapter 4 / The Integral Forms of the Fundamental Laws
Carburetor
0.5 m
Section (1)
Section (2)
Pump
We recognize that HT = 0, V1 is negligible and hL = 210 V 2/2g where V = Q/A
and V2 = Q / A2 . Rearranging we get:
HP =
p2 − p1 V22
+ + z2 − z1 + hL
γ
2g
(
)
6.3 ×10−6 m3 /s
(0.321) 2
Q
=
0.321
m/s
⇒
=
210
= 1.1 m
V= =
h
L
2 × 9.81
A π 2.5 ×10−3 m 2
V2 =
(
)
Q 6.3 × 10−6 m3 /s
=
= 12.53 m/s
A2 π 4 × 10−4 m 2
(
)
Substituting the given values we get:
95 − 100 ) kN/m2
(
(12.53 m/s )
=
+ 0.5 m +
2
HP
(
6.660 kN/m3
2 9.81 m/s2
)
+ 1.1 m = 8.85 m
The power input to the pump is:
(
)(
)
WP = γ QH P /ηP = 6660 N/m3 6.3 ×10−6 m3/s ( 8.85 m ) / 0.75 = 0.5 W
Energy: across the nozzle:
∴
4.102
p1 V12 p2 V22
+
=
+
γ 2g γ 2g
400,000
V12
6.252V12
+
=
9810
2 × 9.81 2 × 9.81
V2 =
52
V1 = 6.25V1
22
∴V1 = 4.58 m/s , VA = 7.16 m/s
V2 = 28.6 m/s
Energy: surface to exit:
HP +15 =
28.62
4.582
7.162
+1.5
+ 3.2
2× 9.81
2× 9.81
2× 9.81
∴ HP = 36.8 m
47
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Chapter 4 / The Integral Forms of the Fundamental Laws
∴WP =
γ QH P 9810 × (π × 0.012 ) × 28.6 × 36.8
=
= 3820 W
ηP
0.85
Energy: surface to “A”:
15 =
7.162
p
7.162
+ A + 3.2
2 × 9.81 9810
2 × 9.81
∴ p A = 39,400 Pa
Energy: surface to “B”:
4.582
pB
7.162
36.0 + 15 =
+
+ 3.2
2 × 9.81 9810
2 × 9.81
∴ pB = 416,000 Pa
Depth on raised section = y 2 . Continuity: 3 × 3 = V2 y2
32
V2
+ 3 = 2 + (0.4 + y2 )
2g
2g
Energy (see Eq. 4.5.21):
92
∴ 3.059 =
+ y2
2 g y22
4.104
Trial-and-error:
y23 − 3.059 y22 + 4.128 = 0
or
y2 = 2.0 :
y2 = 1.8 :
− 0.11 ? 0 ⎫
⎬
+ 0.05 ? 0 ⎭
y2 = 2.1:
y2 = 2.3 :
− 0.1 ? 0 ⎫
⎬
+ 0.1 ? 0⎭
∴ y2 = 1.85 m
∴ y2 = 2.22 m
The depth that actually occurs depends on the downstream conditions. We
cannot select a “correct” answer between the two.
The average velocity at section 2 is also 8 m/s. The kinetic-energy-correction
factor for a parabola is 2 (see Example 4.9). The energy equation is:
V12 p1
V22 p2
+
= α2
+
+ hL
2g γ
2g γ
4.106
82
150,000
82
110,000
+
=2
+
+ hL
2 × 9.81
9810
2 × 9.81
9810
a) V = 1 ∫ VdA =
A
4.108
α=
=
1
AV 3
∫V
π × 0.012
3
dA =
⎛
r2 ⎞
20
10 ⎜ 1 −
2π rdr =
⎜ 0.012 ⎟⎟
0.012
⎝
⎠
0.01
1
∫
0
π × 0.012 × 53
∫
0
3⎛
⎡ 0.012
0.014 ⎤
−
⎢
⎥ = 5 m/s
4 × 0.012 ⎦⎥
⎣⎢ 2
3
r2 ⎞
10 ⎜1 −
2π rdr
⎜ 0.012 ⎟⎟
⎝
⎠
0.01
1
∴ hL = 0.815 m
2000 ⎛ 0.012 3 × 0.014 3 × 0.016
0.018 ⎞
−
+
−
⎜
⎟ = 2.00
0.012 × 5 3 ⎝ 2
4 × 0.012 6 × 0.014 8 × 0.016 ⎠
48
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Chapter 4 / The Integral Forms of the Fundamental Laws
4.110
⎛ V22 − V12
⎞
Engine power = FD × V∞ + m ⎜
+ u2 − u1 ⎟
⎜
⎟
2
⎝
⎠
⎡ V 2 − V12
⎤
+ cv (T2 − T1 ) ⎥
m f q f = FD V∞ + m ⎢ 2
2
⎣⎢
⎦⎥
0 = α2
4.112
V22 p2
V2 p
ν LV
+ + z2 − 1 − 1 − z1 + 32
2g γ
2g γ
gD2
V2
10−6 × 180V
0=2
− 0.35 + 32 ×
2 × 9.81
9.81× 0.022
V 2 + 14.4V − 3.434 = 0
∴ V = 0.235 m/s
and
Q = 7.37 × 10−5 m3/s
Energy from surface to surface:
HP =
4.114
V22 p2
V2 p
V2
+
+ z2 − 1 − 1 − z1 + K
2g γ
2g γ
2g
a) H P = 40 + 5
Q2
= 40 + 50.7 Q 2
π × 0.04 2 × 2 × 9.81
Try Q = 0.25 :
H P = 43.2 (energy)
H P = 58 (curve)
Try Q = 0.30 :
H P = 44.6 (energy)
H P = 48 (curve)
Solution: Q = 0.32 m3 /s
Momentum Equation
V12 p1 V22 p2
+
=
+
2g γ
2g γ
4.116
b)
V2 =
V12
400,000 16 V12
+
=
2 × 9.81 9810
2 × 9.81
d2
V1 = 4 V1
(d / 2)2
∴ V1 = 7.303 m/s
400,000π × 0.032 − F = 1000π × 0.032 × 7.303(4 × 7.303 − 7.303) ∴ F = 679 N
49
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Chapter 4 / The Integral Forms of the Fundamental Laws
V12 p1 V22 p2
+
=
+
2g γ
2g γ
V0π × 0.012 = Ve × 0.006 × 0.15
∴ Ve = 11.1 m/s
ΣFx = m(V2 x − V1x )
a) V2 =
102
V1 = 1.562 V1
82
V12
400,000 2.441 V12
+
=
2 × 9.81 9810
2 × 9.81
∴ V1 = 23.56 m/s
4.118
∴ p1 A1 − F = m(V2 − V1)
400,000π × 0.052 − F = 1000π × 0.052 × 23.56(0.562 × 23.56)
c) V2 =
102
V12 400,000 39.06 V12
6
25
V
=
.
V
+
=
1
1
2g
9810
2g
42
∴ V1 = 4.585 m/s
400,000π × 0.052 − F = 1000π × 0.052 × 4.585(5.25 × 4.585)
V12 p1 V22 p2
+
=
+
2g γ
2g γ
V2 = 4 V1
b) V12 =
4.120
∴ F = 692 N
∴ F = 2275 N
15 V12 p1
∴
=
2g
γ
2 × 9.81
× 400,000 = 53.33
15 × 9810
∴V1 = 7.30 m/s, V2 = 29.2 m/s
p1 A1 − Fx = m(V2 x − V1x )
∴ Fx = 400,000π × 0.04 2 + 1000π × 0.04 2 × 7.32 = 2280 N
Fy = m(V 2 y − V1y ) = 1000π × 0.042 × 7.3 × ( 29.2 ) = 1071 N
π × 0.0252 × 4 = π (0.0252 − 0.022 )V2
AV
1 1 = A2V2
∴ V2 = 11.11 m/s
4.122
p1
γ
+
V12 p2 V22
=
+
2g γ
2g
⎛ 11.112 − 42 ⎞
p1 = 9810 ⎜
= 53, 700 Pa
⎜ 2 × 9.81 ⎟⎟
⎝
⎠
p1A1
F
p1 A1 − F = m(V2 − V1 )
∴ F = 53,700π × 0.0252 − 1000π × 0.0252 × 4(11.11 − 4)
= 49.6 N
4.124
Continuity:
6 V1 = 0.2 V2
∴ V2 = 30 V1
Energy (along bottom streamline):
50
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Chapter 4 / The Integral Forms of the Fundamental Laws
V12 p1
V2 p
+ + z1 = 2 + 2 + z 2
2g γ
2g γ
V22/900
2 × 9.81
+6=
V22
F
F1
2 × 9.81
+ 0.2
F2
∴ V2 = 10.67, V1 = 0.36 m/s
Momentum: F1 − F2 − F = m(V 2 − V1 )
9810 × 3(6 × 4) − 9810 × 0.1(0.2 × 4) − F = 1000 × (0.2 × 4) × 10.67(10.67 − 0.36)
∴ F = 618,000 N
Continuity:
(F acts to the right on the gate)
V2 y2 = V1y1 = 4V2 y1
∴ y2 = 4 y1
1/2 ⎤
Use the result of Example 4.12:
4.126
⎛
⎞
1⎡
8
y2 = ⎢ − y1 + ⎜ y12 + y1V12 ⎟
2⎢
g
⎝
⎠
⎣
⎥
⎥
⎦
a) y2 = 4 × 0.8 = 3.2 m
1⎡
8
⎛
⎞
3.2 = ⎢ −0.8 + ⎜ 0.82 +
× 0.8 × V12 ⎟
2 ⎣⎢
9.81
⎝
⎠
1/ 2 ⎤
∴ V1 = 8.86 m/s
⎥
⎦⎥
Refer to Example 4.12:
γ
4.128
⎛
y1
6⎞
y1w − γ × 1 × 2w = ρ × 6w ⎜ 3 − ⎟
2
y1 ⎠
⎝
⎛ y −2⎞
γ
∴ ( y12 − 4) = 18 ρ ⎜ 1
⎟
2
⎝ y1 ⎠
or
(y1 + 2)y1 =
V1A1 = 2V2 A2
4.130
(V1y1 = 2 × 3)
36
= 3.67 ∴ y1 = 0.77 m, V1 = 7.8 m/s
9.81
V2 = 15
π × 0.052
= 30 m/s
2π × 0.0252
p1
γ
+
V12 p2 V22
=
+
2g
γ
2g
302 − 152
∴ p1 = 9810
= 337,500 Pa
2 × 9.81
ΣFx = m ( V2 x − V1x )
p1A1 − F = m(−V1)
∴ F = p1A1 + mV1
= 337,500π × 0.052 + 1000π × 0.052 × 152 = 4420 N
51
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Chapter 4 / The Integral Forms of the Fundamental Laws
a) ∑ Fx = m(V2x − V1x ),
V1 =
4.132
− F = −mV1
m
300
=
= 38.2 m/s
ρ A1 1000 × π × 0.052
∴ F = 300 × 38.2 = 11, 460 N
V2
F
V1
b) − F = mr (V1 − VB )( cos α − 1)
∴ F = 300 ×
28.2
(38.2 − 10) = 6250 N
38.2
b) − F = mr (V1 − VB )( cos α − 1)
−700 = 1000π × 0.042 (V1 − 8)2 (0.866 − 1)
V1 = 40.24 m/s
4.134
2
∴ m = ρ AV
1 1 = 1000π × 0.04 × 40.24 = 202 kg/s
VB = Rω = 0.5 × 3.0 = 15 m/s
4.136
−Rx = m(V1 − VB )(cosα − 1) = 1000π × 0.0252 × 40 × 25(0.5 − 1) ∴Rx = 982 N
∴W = 10RxVB = 10 × 982 × 15 = 147,300 W
−Fx = m(V1 − VB )(cos120° − 1) = 4π × 0.022 × (400 − 180)2 (−0.5 − 1)
4.138
∴Rx = 365 N
VB = 1.2 × 150 = 180 m/s
W = 15 × 365 × 180 = 986,000 W
The y-component force does no work.
b)
100sin 30 = Vr1 sin α1 ⎫⎪
⎬ ∴α1 = 47 , Vr1 = Vr 2 = 68.35 m/s
100 cos 30 − 40 = Vr1 cos α1 ⎪⎭
⎫⎪
⎬ V2 = 38.9 m/s, α 2 = 29.5
V2 cos 60 = 68.35cos α 2 − 40⎪⎭
V2 sin 60 = 68.35sin α 2
4.140
− Rx = m(V2 x − V1x ) = 1000π × 0.0152 × 100(−38.9cos 60 − 100cos 30 )
∴ Rx = 7500 N
∴ W = 12VB Rx = 12 × 40 × 7500 = 3.60 × 106 W
4.142
To find F, sum forces normal to the plate: ΣFn = m ⎡ (Vout )n − V1n ⎤
⎣
⎦
52
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Chapter 4 / The Integral Forms of the Fundamental Laws
a) ∴ F = 1000 × 0.02 × 0.4 × 40 ⎡ −( − 40 sin 60 ) ⎤ = 11, 080 N
⎣
⎦
(We have neglected friction)
ΣFt = 0 = m2V2 + m3 (−V3 ) − m1 × 40sin 30
Bernoulli: V1 = V2 = V3
0 = m2 − m3 − 0.5m1 ⎫ ∴ m2 = 0.75m1 = 0.75 × 320 = 240 kg/s
⎬
m3 = 80 kg/s
Continuity: m1 = m2 + m3 ⎭
∴
F = mr (V1r )n = 1000 × 0.02 × 0.4(40 − VB )2 sin 60
4.144
Fx = 8(40 − VB2 ) sin 2 60
W = VB Fx = 8VB (40 − VB ) 2 × 0.75 = 6(1600VB − 80VB2 + VB3 )
dW
= 6(1600 − 160VB + 3VB2 ) = 0
dVB
∴ VB = 13.33 m/s
F = mr (V1 − VB )(cos α − 1) = 90 × 0.8 × 2.5 × 13.89 × (−13.89)(−1) = 34, 700 N
4.146
50 × 1000
⎛
⎞
= 13.89 m/s ⎟
⎜ VB =
3600
⎝
⎠
∴ W = 34, 700 × 13.89 = 482,000 W
or
647 hp
To solve this problem, choose a control volume attached to the reverse thruster
vanes, as shown below. The momentum equation is applied to a free body
diagram.
Momentum:
− Rx = 0.5m [ (Vr 2 ) x + (Vr 3 ) x ] − m × ( Vr1 ) x
Assume the pressure in the gases
equals the atmospheric pressure
and that Vr1 = Vr2 = Vr3. Hence
Vr2
(Vr1 ) x = Vr1 = 800 m/s , and
4.148
(Vr 2 )x = (Vr3 )x = −Vr1 × sin α
where α = 20° . Then, momentum is
Rx
Vr1
− Rx = 0.5m [ −2Vr1 sin α ] − m × Vr1
= − mVr1 ( sin α + 1)
Momentum:
− Rx = 0.5m [ −2Vr1 sin α ] − m × Vr1
= −mVr1 ( sin α + 1)
Vr3
The mass flow rate of the exhaust gases is
m = mair + mfuel = mair (1 + 1 40 ) = 100 ×1.025 = 102.5 kg/s
53
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Chapter 4 / The Integral Forms of the Fundamental Laws
Substituting the given values we calculate the reverse thrust:
(
)
Rx = (102.5 kg/s )( 800 m/s ) 1 + sin 20 = 110 kN
Note that the thrust acting on the engine is in the opposite direction to Rx, and
hence it is referred to as a reverse thrust; its purpose is to decelerate the airplane.
For this steady-state flow, we fix the boat and move the upstream air. This
provides us with the steady-state flow of Fig. 4.17. This is the same as observing
the flow while standing on the boat.
W = FV1
4.150
20,000 = F ×
50 × 1000
3600
∴ F = 1440 N
V2 + 13.89
(V2 − 13.89)
2
30.6 + 13.89
∴ Q = A3V3 = π × 12 ×
= 69.9 m3 /s
2
F = m(V2 − V1 )
ηp =
1440 = 1.23π × 12 ×
V1 13.89
=
= 0.625
V3 22.24
or
V2 = 72 ×
∴ V2 = 30.6 m/s
62.5%
Fix the reference frame to the boat so that V1 = 36 ×
4.152
(V1 = 13.89 m/s)
5
= 10 m/s
18
5
= 20 m/s
18
∴ F = m(V2 − V1 ) = 1000π × ( 0.25 ) ×
2
10 + 20
(20 − 10) = 29.44 kN
2
W = F × V1 = (24 kN) × 10 m/s = 294.4 kW or 395 hp
m = 1000 × π × ( 0.25) ×
2
10 + 20
= 2944 kg/s
2
0.2 = V1A1 = V1 × .2 × 1.0 ∴ V1 = 1 m/s ∴ V1 max = 2 m/s
4.154
0.1
0.1
0
0
∴ V1 ( y ) = 20(0.1 − y )
flux in = 2 ∫ ρ V 2dy = 2 ∫ 1000 × 202 (0.1 − y) 2 dy = 800,000
The slope at section 1 is −20
Continuity:
A1V1 = A2V2
0.13
= 267 N
3
∴ V2 ( y) = −20 y + A
∴ V2 = 2V1 = 2 m/s
54
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Chapter 4 / The Integral Forms of the Fundamental Laws
V2 (0) = A ⎫
⎬ ∴ V2 = A − 1/ 2
V2 (0.05) = A − 1⎭
2 = A−
0.05
flux out = 2
∫
0
1
∴ A = 2.5 ∴V2 (y) = 2.5 − 20y
2
0.05
⎡ ( y − 0.125)3 ⎤
1000(2.5 − 20 y ) dy = 800,000 ⎢
⎥
3
⎣⎢
⎦⎥ 0
2
=
800,000
[0.00153]
3
= 408.3 N ∴change = 408 − 267 = 141 N
From the c.v. shown: ( p1 − p2 )π r02 = τ w 2π r0 L
∴τ w =
4.156
∴
Δp ro
du
=μ
2L
dr w
τw2πroL
du
210 × 0.019
=
dr w 2 × 9 × 0.001
p1A1
p2A2
= 220 m/s/m
mtop = ρ A1V1 − ρ ∫ V2 ( y )dA
4.158
2
⎡
⎤
= 1.23 ⎢ 2 ×10 × 32 − ∫ (28 + y 2 )10dy ⎥ = 65.6 kg/s
⎢⎣
⎥⎦
0
2
F
− = ∫ ρ V 2 dA + mtop V1 − m1V1 = 1.23∫ (28 + y 2 ) 210dy + 65.6 × 32 − 1.23 × 20 × 322
2
0
∴ F = 3780 N
Momentum and Energy
a) Energy:
4.160
V12
V2
+ z1 = 2 + z2 + hL
2g
2g
See Problem 4.125(a)
82
1.9122
+ 0 .6 =
+ 2.51 + hL
2 × 9.81
2 × 9.81
∴ hL = 1.166 m
∴ losses = γ A1V1hL = 9810 × (0.6 × 1) × 8 × 1.166 = 54,900 W/m of width
55
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Chapter 4 / The Integral Forms of the Fundamental Laws
Moment of Momentum
Ve =
m
4
=
= 19.89 m/s
ρ Ae 1000 × 4 × π × 0.0042
MI =
∫ r × (2Ω × V) ρ d V
Velocity in arm = V
0.3
= 4 ∫ rˆi × (−2Ωkˆ × V ˆi ) ρ Adr
c.v.
0
= 8 ρ AV Ωkˆ
0.3
∫ rdr = −0.36ρ AV Ωkˆ
0
∑M = 0
4.164
and
d
∫ (r × V ) ρ d V = 0
dt c.v.
∫ (r × V)V ⋅ nˆ ρ dA = 0.3ˆi × ( 0.707Ve ˆj + 0.707Vekˆ ) Ve ρ Ae
c.s.
The z-component of
∫ r × V(V ⋅ nˆ ) ρ dA = 0.3 × 0.707Ve Ae ρ
2
c.s.
Finally −( M I ) z = 0.36 ρ AV Ω = 4 × 0.3 × 0.707Ve2 Ae ρ
0.36Ω = 4 × 0.3 × 0.707 ×19.89
∴Ω = 46.9 rad/s
m = 10 = ρ AV = 1000π × 0.012 V0
Continuity:
Using AV = Ae Ve
∴ V0 = 31.8 m/s
V0π × 0.012 = V π × 0.012 + Ve × 0.006(r − 0.05)
V0π × 0.012 = Ve × 0.006 × 0.15
∴ Ve = 11.1 m/s
∴ V = V0 − 19.1(r − 0.05)Ve = 42.4 − 212r
0.05
4.166
MI =
∫
2rˆi × (+2Ωkˆ × V0ˆi ) ρ Adr +
0
= 4ΩV0 ρ Akˆ
0.2
∫
2rˆi × [+2Ωkˆ × (42.4 − 212r )ˆi ]ρ Adr
0.05
0.05
∫
rdr + 4Ωρ Akˆ
0
0.2
∫
(42.4r − 212r 2 )dr
0.05
212
⎡ 42.4
⎤
= ⎢
(0.22 − 0.052 ) −
(0.23 − 0.053 ) ⎥ kˆ
3
⎣ 2
⎦
= (0.05Ω + 0.3Ω )kˆ = 0.35Ωkˆ
56
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Chapter 4 / The Integral Forms of the Fundamental Laws
0.2
∫
rˆi × (−Ve ˆj)Ve ρ × 0.006dr = −11.12 × 1000 × 0.006
0.05
0.2
∫
rdr kˆ = −13.86 kˆ
0.05
∴Ω = 39.6 rad/s
∴ −0.35Ω = −13.86
0.0082
V=
× 19.89 = 3.18 m/s
0.022
See Problem 4.165. Ve = 19.89 m/s
0.3
⎡
⎛ dΩ ˆ ⎞ ˆ ⎤
M I = 4 ∫ rˆi × ⎢ (−2Ωkˆ × V ˆi ) + ⎜ −
k ⎟ × ri ⎥ρ Adr A = π × 0.012 , Ae = π × 0.0042
⎝ dt ⎠
⎣
⎦
0
= −8 ρ AV Ωkˆ
0.3
∫
rdr − 4 ρ A
0
dΩ ˆ
k
dt
0.3
∫r
2
dr
0
dΩ ˆ
k
= −360 AV Ωkˆ − 36 A
dt
4.168
∫ (r × V)z (V ⋅ nˆ ) ρ dA = 212Ve Aekˆ
2
c.s.
Thus, 360 AV Ω + 36 A
dΩ
dΩ
= 212Ve2 Ae or
+ 31.8Ω = 373
dt
dt
The solution is Ω = Ce −31.8t + 11.73. The initial condition is Ω( 0 ) = 0
∴ C = −1173
. .
Finally
Ω = 11.73(1 − e −31.8t ) rad/s
57
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Chapter 4 / The Integral Forms of the Fundamental Laws
58
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Chapter 5 / The Differential Forms of the Fundamental Laws
CHAPTER 5
The Differential Forms of the
Fundamental Laws
Differential Continuity Equation
in − m
out =
m
∂melement
. This is expressed as
∂t
⎡
⎣
ρ vr (rdθ dz) − ⎢ ρ vr +
∂
∂
( ρvr ) dr ⎤⎥ (r + dr)dθ dz + ρ vθ drdz − ⎡⎢ ρ vθ + ( ρ vθ )dθ ⎤⎥ drdz
∂r
∂θ
⎦
⎣
⎦
dr ⎞
∂
dr ⎞
∂ ⎡ ⎛
dr ⎞
⎤
⎛
⎡
⎤⎛
+ ρ vz ⎜ r + ⎟ dθ dr − ⎢ ρ vz + ( ρ vz )dz ⎥ ⎜ r + ⎟ dθ dr = ⎢ ρ ⎜ r + ⎟ dθ drdz ⎥
2⎠
∂z
2⎠
∂t ⎣ ⎝
2⎠
⎝
⎣
⎦⎝
⎦
5.2
Subtract terms and divide by rdθdrdz :
−
ρ vr
r
−
r + dr ∂
1 ∂
∂
r + dr / 2 ∂ r + dr / 2
( ρ vr ) −
( ρ vθ ) − ( ρ vz )
= ρ
r ∂r
r ∂θ
∂z
r
∂t
r
Since dr is an infinitesimal, (r + dr )/r = 1 and (r + dr /2)/r = 1. Hence
∂ρ ∂
1 ∂
∂
1
+ ( ρ vr ) +
( ρ vθ ) + ( ρ vz ) + ρ vr = 0
∂t ∂r
r ∂θ
∂z
r
This can be put in various forms. If ρ = const, it divides out
For a steady flow
∂ρ
= 0. Then, with v = w = 0, Eq. 5.2.2 yields
∂t
∂
( ρ u) = 0
∂x
5.4
or
ρ
du
dρ
+u
=0
dx
dx
Partial derivatives are not used since there is only one independent variable.
∂
∂ρ
= 0,
≠ 0. Since water can be considered to be incompressible, we
∂t
∂z
∂ρ
∂ρ
demand that D ρ /Dt = 0. Equation (5.2.8) then provides u
+w
= 0,
∂x
∂z
assuming the x-direction to be in the direction of flow. There is no variation
with y. Also, we demand that ∇ ⋅ V = 0, or
Given:
5.6
59
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Chapter 5 / The Differential Forms of the Fundamental Laws
∂u ∂w
+
=0
∂x ∂z
We can use the ideal gas law, ρ =
5.7
p
. Then, the continuity equation (5.2.7)
RT
Dρ
= − ρ∇ ⋅ V becomes, assuming RT to be constant
Dt
p
1 Dp
=−
∇ ⋅ V or
RT Dt
RT
1 Dp
= −∇ ⋅ V
p Dt
a) Use cylindrical co-ordinates with vθ = vz = 0 :
1 ∂
( rv r ) = 0
r ∂r
Integrate:
rvr = C
5.8
∴ vr =
C
r
b) Use spherical coordinates with vθ = vφ = 0 :
1 ∂ 2
(r vr ) = 0
r 2 ∂r
Integrate:
r 2vr = C
∴ vr =
C
r2
(a) Since the flow is steady and incompressible then VA = constant, where the
constant is determined by using the conditions at the inlet that is,
( VA )inlet = 40 ×1 = 40 m3 /s. And, since the flow is inviscid, the velocity is
5.9
uniform in the channel, so u = V . Hence, at any x position within the channel
the velocity u can be calculated using u = V = 40/A. Since the flow area is not
constant it is given by A = 2hw, where the vertical distance h is a function of x
and can be determined as, h = 0.15x + 0.5H . Substituting, we obtain the
following expression for the velocity:
u( x) =
40
20
=
m/s
2(0.15x + 0.5) 0.15x + 0.5
(b) To determine the acceleration in the x-direction, we use (see Eq. 3.2.9)
du
where
ax = u
dx
60
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Chapter 5 / The Differential Forms of the Fundamental Laws
du
3
=−
dx
(0.15x + 0.5) 2
Hence, the expression for acceleration is
ax =
20
−3
−60
×
=
m/s 2
2
3
0.15x + 0.5 (0.15x + 0.5)
(0.15x + 0.5)
Note that the minus sign indicates deceleration of the fluid in the x-direction.
∂u ∂v
⎛ ∂u ⎞
+
= 0, we write v = − ⎜ ⎟dy + C.
∂x ∂y
⎝ ∂x ⎠
∂u
3
∂v
With the result from Problem 5.9:
=−
= − , we integrate to
2
∂x
∂y
(0.15x + 0.5)
find
∫
(a) Using the continuity equation
v( x, y ) =
3y
(0.15x + 0.5)2
and since v = 0 at y = 0, then C = 0.
5.10
(b) To determine the acceleration in the y-direction, we use (see Eq. 3.2.9)
ay = u
∂v
∂v
+v
∂x
∂y
From part (a) we have
0.9 y
∂v
=−
∂x
(0.15x + 0.5)3
Substituting in the expression for acceleration we get
−0.9 y
3y
−9 y
20
3
ay =
×
+
×
=
3
2
2
0.15x + 0.5 (0.15x + 0.5)
(0.15x + 0.5) (0.15x + 0.5)
(0.15x + 0.5) 4
5.11
⎛ ∂u ∂v ⎞
Dρ
= − ρ∇ ⋅ V = − ρ ⎜ + ⎟ = −2.3(200 × 1 + 400 × 1) = −1380 kg/m3 ⋅ s
Dt
⎝ ∂x ∂y ⎠
5.12
In a plane flow, u = u( x, y) and v = v( x, y). Continuity demands that
∂u
∂u ∂v
∂v
= 0 and hence
+
= 0. If u = const, then
= 0. Thus, we also
∂x
∂x ∂y
∂y
have v = const and D ρ /Dt = 0.
5.13
If u = C1 and v = C2 , the continuity equation provides, for an incompressible
flow
61
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Chapter 5 / The Differential Forms of the Fundamental Laws
∂u ∂v ∂w
+ +
=0
∂x ∂y ∂z
∴
∂w
= 0 so w = C3
∂z
The z-component of velocity w is also constant. We also have
Dρ
∂ρ
∂ρ
∂ρ
∂ρ
=0=
+u
+v
+w
Dt
∂t
∂x
∂y
∂z
The density may vary with x, y, z and t. It is not, necessarily, constant.
5.14
∂u ∂v
+
=0
∂x ∂y
∴ A+
∂v
=0
∂y
∴ v = − Ay
But, v( x, 0) = 0 = f ( x)
∂u ∂v
+
=0
∂x ∂y
5.15
∴ v(x, y) =
∴v =
∴
5y 2 − 5x2
∫ (x
2
( x 2 + y 2 )5 − 5 x(2 x)
5x 2 − 5y 2
∂v
∂u
=−
=−
=
−
∂y
∂x
(x 2 + y 2 )2
(x 2 + y 2 )2
dy + f (x) =
5y
x + y2
2
+ f (x)
f (x) = 0
5y
x + y2
2
From Table 5.1:
5.16
+y )
2 2
∴ v(x, y) = − Ay + f (x)
1 ∂
1 ∂vθ
1⎛
0.4 ⎞
(rvr ) = −
= − ⎜ 10 + 2 ⎟ sin θ
r ∂r
r ∂θ
r⎝
r ⎠
0.4 ⎞
0.4 ⎞
⎛
⎛
∴ rvr = ⎜ 10 + 2 ⎟ sin θ dr + f (θ ) = ⎜10r −
⎟ sin θ + f (θ )
r ⎠
r ⎠
⎝
⎝
∫
0.4 ⎞
⎛
0.2vr (0.2,θ ) = ⎜10 × 0.2 −
⎟ sin θ + f (θ ) = 0
0.2 ⎠
⎝
∴ f (θ ) = 0
0.4 ⎞
⎛
∴ vr = ⎜10 − 2 ⎟ sin θ
r ⎠
⎝
From Table 5.1:
5.17
1 ∂
1 ∂ vθ −20 ⎛
1 ⎞
=
( rvr ) = −
⎜1 + 2 ⎟ cosθ
r ∂r
r ∂θ
r ⎝ r ⎠
1⎞
⎛
⎛ 1⎞
∴ rvr = −20 ⎜1 + 2 ⎟ cos θ dr + f (θ ) = −20 ⎜ r − ⎟ cos θ + f (θ )
⎝ r ⎠
⎝ r⎠
∫
vr (1,θ ) = −20(1 − 1) cos θ + f (θ ) = 0
∴ f (θ ) = 0
1⎞
⎛
∴ vr = −20 ⎜1 − 2 ⎟ cos θ
⎝ r ⎠
62
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Chapter 5 / The Differential Forms of the Fundamental Laws
From Table 5.1, spherical coordinates:
∴
5.18
1 ∂ 2
1 ∂
( r vr ) = −
( vθ sin θ )
2 ∂r
r sin θ ∂θ
r
1 ∂ 2
1 ⎛
40 ⎞
(
r
v
)
=
+
10
⎜
⎟ 2sinθ cosθ
r
r sinθ ⎝
r2 ∂ r
r3 ⎠
40 ⎞
80 ⎞
⎛
⎛
∴ r 2 vr = r ⎜ 10 + 3 ⎟ 2 cos θ dr + f (θ ) = ⎜ 10r 2 − ⎟ cos θ + f (θ )
r ⎠
r ⎠
⎝
⎝
∫
80 ⎞
⎛
4vr (2,θ ) = ⎜10 × 22 − ⎟ cos θ + f (θ ) = 0
2 ⎠
⎝
∴ f (θ ) = 0
80 ⎞
⎛
∴ vr = ⎜10 − 3 ⎟ cos θ
r ⎠
⎝
Continuity:
5.19
ρ=
∂
( ρu) = 0
∂x
∴ρ
du
dρ
+u
=0
dx
dx
p
124 × 103
=
= 1.554 kg/s
RT 287 × 278
∴
du 159 − 137
=
= 220 m/s/m
dx
0.1
dρ
ρ du
1.554
=−
=−
× 220 = −2.32 kg/m4
dx
u dx
147
∂u ∂v
+
=0
∂x ∂y
∂ ⎡
−20( 1 − e − x )⎤ = −20e − x
⎦
∂x⎣
Hence, in the vicinity of the x-axis:
∂v
= 20e − x
∂y
5.20
But v = 0 if y = 0
and
v = 20 ye − x + C
∴C = 0
v = 20 ye − x = 20(0.2)e −2 = 0.541 m/s
From Table 5.1,
1 ∂
∂v
(rvr ) + z = 0
r ∂r
∂z
∂ ⎡
−20(1 − e − z ) ⎤ = −20e − z
⎦
∂z ⎣
Hence, in the vicinity of the z-axis:
1 ∂
r2
(rvr ) = 20e − z and rvr = 20e − z + C
r ∂r
2
5.21
But vr = 0 if r = 0
∴C = 0
vr = 10re − z = 10(0.2)e −2 = 0.271 m/s
63
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Chapter 5 / The Differential Forms of the Fundamental Laws
The velocity is zero at the stagnation point. Hence, 0 = 10 −
The continuity equation for this plane flow is
5.22
we see that
40
R2
∴R = 2 m
∂u
∂u ∂v
= 80 x −3 ,
+
= 0. Using
∂x
∂x ∂y
∂v
= −80 x −3 near the x-axis. Consequently, for small Δy ,
∂y
Δv = −80 x −3Δy so that v = −80(−3)−3 (0.1) = 0.296 m/s
The velocity is zero at the stagnation point. Hence
0 = (40/R2 ) − 10 ∴ R = 2 m
Use continuity from Table 5.1:
1 ∂ 2
1 ∂
20
(40 − 10r 2 ) = −
r vr = 2
2 ∂r
r
r
r ∂r
(
)
Near the negative x-axis continuity provides us with
5.23
1 ∂
20
( vθ sinθ ) =
r sinθ ∂θ
r
Integrate, letting θ = 0 from the y-axis:
vθ sin θ = −20 cos θ + C
Since vθ = 0 when θ = 90°, C = 0. Then, with α = tan −1
vθ = −20
Continuity:
5.24
cos θ
cos88.091
0.0333
= −20
= −20
= 0.667 m/s
sin θ
sin 88.091
0.999
∂u ∂v
+
=0
∂x ∂y
∴
∴Δv = v − 0 = −220Δy
b) ax = u
0. 1
= 1.909°
3
Δv
Δu
13.5 − 11.3
m/s
=−
=−
= −220
Δy
Δx
2 × 0.005
m
∴ v = −220 × 0.004 = −0.88 m/s
∂u
= 12.6 × (+220) = 2772 m/s 2
∂x
Differential Momentum Equation
ΣFy = may . For the fluid particle occupying the volume of Fig. 5.3:
5.25
∂τ yy dy ⎞
∂τ zy dz ⎞
∂τ xy dx ⎞
⎛
⎛
⎛
⎜⎜τ yy +
⎟⎟ dxdz + ⎜⎜ τ zy +
⎟⎟ dxdy + ⎜⎜ τ xy +
⎟ dydz
∂y 2 ⎠
∂z 2 ⎠
∂x 2 ⎟⎠
⎝
⎝
⎝
64
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Chapter 5 / The Differential Forms of the Fundamental Laws
∂τ yy dy ⎞
∂τ zy dz ⎞
∂τ xy dx ⎞
⎛
⎛
⎛
− ⎜⎜ τ yy −
⎟⎟ dxdz − ⎜⎜ τ zy −
⎟⎟ dxdy − ⎜⎜ τ xy −
⎟⎟ dydz
2
2
2
∂
∂
∂
y
z
x
⎝
⎠
⎝
⎠
⎝
⎠
+ ρ g y dx dy dz = ρ dx dy dz
Dv
Dt
Dividing by dx dy dz , and adding and subtracting terms:
∂τ xy
∂x
+
∂τ yy
∂y
+
∂τ zy
∂z
+ ρ gy = ρ
Dv
Dt
Check continuity:
∂ u ∂ v ∂ w ( x 2 + y 2 )10 − 10 x(2x) ( x 2 + y 2 )10 − 10 y(2 y)
+
+
=
+
=0
∂x ∂y ∂z
( x2 + y 2 )2
( x2 + y 2 )2
Thus, it is a possible flow. For a frictionless flow, Euler’s Eqs. 5.3.7, give with
g x = g y = 0:
ρu
5.26
∴
∂p
∂u
∂u
+ ρv
=−
∂x
∂y
∂x
∂p
10 y
−20 xy
100( x 2 + y 2 ) y
10 x 10 y 2 − 10 x 2
= −ρ 2
−
ρ
=
ρ
∂x
x + y 2 (x2 + y 2 )2
x2 + y 2 (x2 + y 2 )2
( x 2 + y 2 )3
∂p
∂v
∂v
ρu + ρ v = −
∂x
∂y
∂y
∂p
−20 xy
10 y 10 x 2 − 10 y 2
100( x 2 + y 2 ) y
10 x
∴ = −ρ 2
−ρ 2
=ρ
∂y
x + y 2 ( x2 + y 2 )2
x + y 2 (x2 + y 2 )2
( x 2 + y 2 )3
∴∇p =
∂p ˆ ∂p ˆ
100 y ρ ˆ
100 x ρ ˆ
100 ρ
j= 2
i
j
i+
+
=
( xˆi + yˆj)
2
2
2
2
2
2
2
2
∂x ∂y
(x + y )
(x + y )
(x + y )
Check continuity (cylindrical coord. from Table 5.1):
5.27
1 ∂
1 ∂v θ 10 ⎛
1⎞
1⎞
−10 ⎛
( rv r ) +
=
⎜ 1 + 2 ⎟ cos θ +
⎜ 1 + 2 ⎟ cos θ = 0. ∴It is a
r ∂r
r ∂θ
r ⎝
r ⎠
r ⎝
r ⎠
possible flow. For Euler’s Eqs. (let v = 0 in the momentum eqns of Table 5.1) in
cylindrical coord:
2
∂p
v2
∂v
v ∂v 100 ρ ⎛
1⎞
1⎞
⎛
⎛ 20 ⎞
1 + 2 ⎟ sin 2 θ − 10 ρ ⎜1 − 2 ⎟ cos 2 θ ⎜ 3 ⎟
= ρ θ − ρ vr r − ρ θ r =
⎜
∂r
r
∂r
r ∂θ
r ⎝ r ⎠
⎝ r ⎠
⎝r ⎠
10ρ ⎛
1⎞ 2 ⎛
10 ⎞
−
⎜1 + 2 ⎟ sin θ ⎜10 − 2 ⎟
r ⎝ r ⎠
r ⎠
⎝
65
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Chapter 5 / The Differential Forms of the Fundamental Laws
vv
∂v
v ∂v
1 ∂p
100 ρ ⎛
1 ⎞
1 − 4 ⎟ sin θ cos θ
= − ρ r θ − ρ vr θ − ρ θ θ =
⎜
r ∂θ
r
r ∂θ
r ⎝ r ⎠
∂r
2
1 ⎞
1 ⎞
⎛
⎛ 20 ⎞ 100 ρ ⎛
−10 ρ ⎜1 − 2 ⎟ cos θ sin θ ⎜ 3 ⎟ −
⎜1 + 2 ⎟ sin θ cos θ
r ⎝ r ⎠
⎝ r ⎠
⎝r ⎠
∴∇p =
∂p ˆ 1 ∂p ˆ
200 ρ ⎡ 1
200 ρ
⎤
ir +
iθ = 3 ⎢ 2 − cos 2θ ⎥ ˆir − 3 sin 2θ ˆiθ
∂r
r ∂θ
r ⎣r
r
⎦
Follow the steps of Problem 5.27. The components of the pressure gradient are
vθ2 + vφ2
∂p
∂v
v ∂v
=ρ
− ρ vr r − ρ θ r
∂r
r
∂r
r ∂θ
5.28
vv
v ∂v
∂v
1 ∂p
= − ρ r θ − ρ vr θ − ρ θ θ
r ∂θ
r
r ∂θ
∂r
⎛ 2μ
⎞
∴p = p −⎜
+λ⎟ ∇⋅V
⎝ 3
⎠
⎛ 2μ
⎞
∴p − p = −⎜
+λ⎟ ∇⋅V
⎝ 3
⎠
∂ sˆ Δsˆ
Δα nˆ
nˆ
≅
=−
=−
∂ s Δs
RΔα
R
∂ sˆ Δsˆ nˆ Δθ
∂θ
≅
=
= nˆ
∂t Δt
Δt
∂t
5.29
∴
DV ⎛ ∂V
∂V ⎞ ⎛ ∂θ V 2 ⎞
=⎜
+V
−
⎟ nˆ
⎟ sˆ + ⎜ V
Dt ⎝ ∂t
∂s ⎠ ⎜⎝ ∂t
R ⎟⎠
⎛ V2 ⎞
∂V
For steady flow, the normal acc. is ⎜ −
, the tangential acc. is V
⎟
⎜ R ⎟
∂s
⎝
⎠
For a rotating reference frame (see Eq. 3.2.15), we must add the terms due to Ω.
5.30
Thus, Euler’s equation becomes
dΩ ⎞
⎛ DV
+ 2Ω × V + Ω × (Ω × r ) +
× r ⎟ = −∇p − ρ g
dt
⎝ Dt
⎠
ρ⎜
5.31
τ xx = − p + 2μ
∂u
+ λ ∇ ⋅ V = −210 kPa
∂x
τ yy = τ zz = − p = −210 kPa
66
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Chapter 5 / The Differential Forms of the Fundamental Laws
⎛ ∂u
τ xy = μ ⎜
⎝ ∂y
+
∂v
∂x
⎞
−4
−3
⎟ = 5 ×10 [32 − 32 ×192 × 0.0025] = 8.32 × 10 Pa
⎠
τ xy
τ xz = τ yz = 0
τ xx
=
8.32 × 10−3
210 × 10
3
16 y
16 y 2
∂v
∂u
=−
=
−
∂y
∂x C x9 / 5 C 2 x13/ 5
v( x, o) = 0
∴ f ( x) = 0
= 3.96 × 10−8
∴ v( x, y) =
8 = C 10004 / 5
8y 2
Cx9 / 5
−
16 y3
3C 2 x13/ 5
+ f ( x)
∴C = 0.0318
∴ u( x, y ) = 629 yx −4 / 5 − 9890 y 2 x −8 / 5
v( x, y) = 252 y 2 x−9 / 5 − 5270 y3x−13/ 5
5.32
τ xx = − p + 2μ
∂u
= −100 + 0 = −100 kPa
∂x
τ yy = τ zz = − p = −100 kPa
⎛ ∂u ∂v ⎞
+ ⎟ = 2 ×10−5 ⎡629 ×1000−4/5 ⎤ = 5.01×10−5 Pa
⎣
⎦
∂
∂x ⎠
y
⎝
τ xy = μ ⎜
τ xz = τ yz = 0
Du ∂u ⎛ ∂
∂
∂ ⎞
=
+ ⎜ u + v + w ⎟ u = ( V ⋅ ∇ )u
∂y
∂z ⎠
Dt
∂t ⎝ ∂x
5.33
Dv ∂v ⎛ ∂
∂
∂ ⎞
=
+ ⎜ u + v + w ⎟ v = (V ⋅ ∇ )v
Dt ∂t ⎝ ∂x
∂y
∂z ⎠
Dw ∂w ⎛ ∂
∂
∂ ⎞
=
+ ⎜ u + v + w ⎟ w = (V ⋅ ∇ )w
Dt
∂t ⎝ ∂x
∂y
∂z ⎠
∴
5.34
DV Du ˆ Dv ˆ Dw ˆ
i+
j+
k = V ⋅ ∇ (uˆi + vˆj + wkˆ ) = (V ⋅ ∇ )V
=
Dt Dt
Dt
Dt
Follow the steps that lead to Eq. 5.3.17 and add the term due to compressible
effects:
ρ
DV
μ ∂
μ ∂
μ ∂
= −∇p + ρ g + μ∇ 2 V +
∇ ⋅ V ˆi +
∇ ⋅ V ˆj +
∇ ⋅ V kˆ
Dt
3 ∂x
3 ∂y
3 ∂z
67
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Chapter 5 / The Differential Forms of the Fundamental Laws
= −∇p + ρ g + μ∇ 2 V +
∴ρ
μ⎛ ∂ ˆ ∂ ˆ ∂ ˆ⎞
j+ k ⎟∇ ⋅ V
⎜ i+
3 ⎝ ∂x
∂y
DV
μ
= −∇p + ρ g + μ∇ 2 V + ∇(∇ ⋅ V )
Dt
3
If u = u ( y ), then continuity demands that
lower plate), v = 0
∴ρ
5.35
∂z ⎠
∂v
= 0 ∴ v = C. But, at y = 0 (the
∂y
∴ C = 0, and v( x, y) = 0
⎛ ∂ 2u ∂ 2u ∂ 2u ⎞
⎛ ∂u
∂p
∂u
∂u
∂u ⎞
Du
= ρ ⎜ + u + v + w ⎟ = − + ρ gx + μ ⎜ 2 + 2 + 2 ⎟
⎜
Dt
∂x
∂y
∂z ⎠
∂x
∂y
∂z ⎠⎟
⎝ ∂t
⎝ ∂x
∴0 = −
∂p
∂ 2u
+μ 2
∂x
ay
ρ
∂p
Dv
=0=−
Dt
∂y
ρ
∂p
Dw
= 0 = − + ρ (− g )
Dt
∂z
∴
∂p
= −ρ g
∂z
The x-component Navier-Stokes equation can be written as
⎛ ∂ 2u ∂ 2u ∂ 2u ⎞
⎛ ∂u
∂p
∂u
∂u
∂u ⎞
+ u + v + w ⎟ = − + ρ gx + μ ⎜ 2 + 2 + 2 ⎟
⎜ ∂x
∂x
∂y
∂z ⎠
∂x
∂y
∂z ⎟⎠
⎝ ∂t
⎝
ρ⎜
Based on the given conditions the following assumptions can be made:
5.36
One-dimensional ( v = w = 0 )
⎛ ∂u
⎞
Steady state ⎜
= 0⎟
⎝ ∂t
⎠
Incompressible ( ρ = constant )
⎛ dp
⎞
Zero pressure-gradient ⎜
= 0⎟
⎝ dx
⎠
⎛ ∂u
⎞
Fully-developed flow ⎜
= 0⎟
⎝ ∂x
⎠
⎛ ∂u
⎞
A wide channel ⎜
= 0⎟
∂
z
⎝
⎠
The Navier-Stokes equation takes the simplified form μ
∂ 2u
∂ 2u
= 0 or
= 0.
∂y 2
∂y 2
Integrating twice yields, u( y ) = ay + b . To determine a and b we apply the
following boundary conditions: u = V1 at y = 0, and u = −V2 at y = h. This gives
b = V1 and a = −(V1 + V2 ) / h. The velocity distribution between the plates is
68
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 5 / The Differential Forms of the Fundamental Laws
then
⎛ V + V2 ⎞
u( y ) = − ⎜ 1
⎟ y + V1
⎝ h ⎠
Using the x-component Navier-Stokes equation with x being vertical and the
following assumptions:
⎛ ∂u
⎞
Steady state ⎜
= 0⎟
⎝ ∂t
⎠
One-dimensional ( v = w = 0 )
Incompressible
⎛ ∂u
⎞
Fully-developed flow ⎜
= 0⎟
⎝ ∂x
⎠
( ρ = constant )
The x-component Navier-Stokes equation reduces to
0=−
5.37
∂p
∂2 u
− ρg + μ 2
∂x
∂y
Integrate the above differential equation twice (see Problem 5.36):
u( y ) =
1 ⎛ dp
⎞
+ ρ g ⎟ y 2 + ay + b
⎜
2 μ ⎝ dx
⎠
Applying the no-slip boundary condition at both plates (see Problem 5.36) we
get
u( y ) =
(
1 ⎛ dp
⎞
+ ρ g ⎟ y 2 − hy
⎜
2 μ ⎝ dx
⎠
)
Assumptions: One-dimensional ( vr = vθ = 0 )
⎛ ∂v
⎞
Steady state ⎜ z = 0 ⎟
⎝ ∂t
⎠
Incompressible ( ρ = constant )
Horizontal ( g z = 0)
⎛ ∂v
⎞
Fully-developed flow ⎜ z = 0 ⎟
⎝ ∂z
⎠
Dvr
1 ∂p
=0=−
ρ ∂r
Dt
5.38
ρ
Dvθ
1 ∂p
=0=−
ρ r ∂θ
Dt
⎛ ∂ 2 v 1 ∂v
⎛ ∂v
∂p
Dvz
∂vz vθ ∂vz
∂v ⎞
∂ 2 vz
1 ∂ 2 vz
z
+ 2
= ρ ⎜ z + vr
+
+ vz z ⎟ = − + μ ⎜ 2z +
+
2
⎜
⎜ ∂r
∂z
Dt
∂r
r ∂θ
∂z ⎠⎟
r ∂r r ∂θ
∂z 2
⎝ ∂t
⎝
∴0 = −
⎛ ∂2v 1 ∂vz ⎞
∂p
+ μ ⎜ 2z +
⎟
⎜ ∂r
r ∂r ⎟⎠
∂z
⎝
69
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
⎞
⎟
⎟
⎠
Chapter 5 / The Differential Forms of the Fundamental Laws
Assumptions:
One-dimensional flow ( vθ = vr = 0 )
⎛ ∂v
⎞
Steady state ⎜ z = 0 ⎟
⎝ ∂t
⎠
Incompressible ( ρ = constant )
Horizontal ( g z = 0)
⎛ ∂v
⎞
Fully-developed flow ⎜ z = 0 ⎟
⎝ ∂z
⎠
The Navier-Stokes equation in cylindrical form provides the following equation:
0=−
⎡ ∂ 2 v 1 ∂vz ⎤
∂p
+ μ ⎢ 2z +
⎥
∂z
r ∂r ⎥⎦
⎢⎣ ∂r
Rearrange the above equation and integrate:
0=−
∂p μ ∂ ⎛ ∂vz ⎞
+
⎜r
⎟
∂z r ∂r ⎝ ∂r ⎠
Integrating again yields: vz (r ) =
5.39
∂vz 1 ∂p ⎛ r ⎞ C1
=
⎜ ⎟+
∂r μ ∂z ⎝ 2 ⎠ r
1 ∂p ⎛ r 2 ⎞
⎜ ⎟ + C ln r + C2
μ ∂z ⎜⎝ 4 ⎟⎠ 1
C1 and C2 are determined using the boundary conditions: vz = 0, at r = ro , and
vz = Vc at r = ri . Hence
Vc =
1 ∂p ⎛ ri2 ⎞
1 ∂p ⎛ ro2 ⎞
+
+
C
=
ln
and
0
C
r
⎜ ⎟
⎜ ⎟ + C ln r + C2
2
μ ∂z ⎜⎝ 4 ⎟⎠ 1 i
μ ∂z ⎜⎝ 4 ⎟⎠ 1 o
Subtracting the second equation from the first yields
C1 =
Vc −
1 ∂p 2 2
ri − ro
4 μ ∂z
ln ( ri ro )
(
)
The drag force on the inner cylinder is zero when the shear stress τ rz on the
⎡ ∂v ∂v ⎤
inner cylinder is zero, i.e., τ rz = μ ⎢ r + z ⎥
= 0 . Since vr = 0 , then
∂r ⎦ r =ri
⎣ ∂z
⎡ ∂v ⎤
τ rz = μ ⎢ z ⎥ = 0 . From the above expression for vz we find
⎣ ∂r ⎦ r =r
i
∂vz
∂r
=
r = ri
C
1 ∂p
1 ∂p 2
ri + 1 = 0. Then C1 = −
ri . Combining with the above
2μ ∂z
ri
2μ ∂z
expressions for C1 we solve for Vc . The result is:
70
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Chapter 5 / The Differential Forms of the Fundamental Laws
Vc =
Continuity:
5.40
−
(
)
1 ∂p ⎡ 2 2
ri − ro − 2ri2 ln ( ri ro ) ⎤
⎣
⎦
4 μ ∂z
1 ∂ 2
(r vr ) = 0
r 2 ∂r
∴ r 2vr = C
∂p
vθ2
⎛ 2v
⎞
ρ = − + μ ⎜ − 2θ cot θ ⎟
r
∂r
⎝ r
⎠
At r = r1, vr = 0
0=−
0=−
∴C = 0
∂p
⎡ 1 ∂ ⎛ 2 ∂vθ
+μ⎢
⎜ r ∂r
∂θ
⎣ r ∂r ⎝
vθ ⎤
⎞
⎟−
⎥
⎠ r sin 2 θ ⎦
1 ∂p
r sin θ ∂φ
Assumptions:
One-dimensional flow ( vz = vr = 0 )
⎛ ∂v
⎞
Steady state ⎜ z = 0 ⎟
⎝ ∂t
⎠
Incompressible ( ρ = constant )
Vertical ( gr = gθ = 0)
⎛ ∂v
⎞
Developed flow ⎜ r = 0 ⎟
⎝ ∂θ
⎠
The simplified differential equation from Table 5.1 is
which can be re-written as
5.41
Integrating we get:
∂ 2 vθ
∂r
2
+
∂ ⎛ vθ
∂r ⎜⎝ r
∂ 2 vθ
∂r 2
+
1 ∂vθ vθ
−
=0
r ∂r r 2
⎞
⎟ = 0.
⎠
∂vθ vθ
+
= C1
r
∂r
The above equation can be re-written as
Integrating again yields rvθ = C1
1 ∂
( rvθ ) = C1
r ∂r
r2
r C
+ C2 ⇒ vθ = C1 + 2
2
2 r
Apply the boundary conditions vθ = 0 at r = ri , and vθ = roω at r = ro . We have
0 = C1
ri C2
+
2 ri
Solving for C1 and C2 yields C1 = −2
roω = C1
and
ω ro2
and C2 =
2
ri2 − ro
ro C2
+
2 ro
ω ri2ro2
ri2 − ro2
71
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Chapter 5 / The Differential Forms of the Fundamental Laws
⎛ ω r 2 ⎞ ⎛ ω r 2r 2 ⎞ 1
vθ = ⎜ − 2 o 2 ⎟ r + ⎜ 2 i o2 ⎟
⎜ r −r ⎟ ⎜r −r ⎟r
o ⎠
o ⎠
⎝ i
⎝ i
Finally
For an incompressible flow ∇ ⋅ V = 0. Substitute Eqs. 5.3.10 into Eq. 5.3.2 and
5.3.3:
ρ
ρ
∂u ⎞ ∂ ⎛ ∂u ∂v ⎞ ∂ ⎛ ∂u ∂w ⎞
Du ∂ ⎛
= ⎜ − p + 2μ ⎟ + μ ⎜ + ⎟ + μ ⎜ +
⎟ + ρ gx
∂x ⎠ ∂y ⎝ ∂y ∂x ⎠ ∂z ⎝ ∂z ∂x ⎠
Dt ∂x ⎝
=−
∂p
∂ 2u
∂ 2u
∂ 2u
∂ ⎛ ∂u ∂v ∂w ⎞
+μ 2 +μ 2 +μ 2 +μ ⎜ +
+
⎟ + ρ gx
∂x
∂x ⎝ ∂x ∂y ∂z ⎠
∂x
∂y
∂z
∴ρ
⎛ ∂ 2u ∂ 2u ∂ 2u ⎞
∂p
Du
= − + μ ⎜ 2 + 2 + 2 ⎟ + ρ gx
∂x
∂z ⎠
Dt
⎝ ∂x ∂y
Dv ∂ ⎛ ∂u ∂v ⎞ ∂ ⎛
∂v ⎞ ∂ ⎛ ∂v ∂w ⎞
=
μ ⎜ + ⎟ + ⎜ − p + 2μ ⎟ + μ ⎜ +
⎟ + ρ gy
Dt ∂x ⎝ ∂y ∂x ⎠ ∂y ⎝
∂y ⎠ ∂z ⎝ ∂z ∂y ⎠
∂p
∂ 2v
∂2v
∂ 2v
∂ ⎛ ∂u ∂v ∂w ⎞
=− +μ 2 +μ 2 +μ 2 +μ ⎜ +
+
⎟ + ρ gy
∂y
∂y ⎝ ∂x ∂y ∂z ⎠
∂x
∂y
∂z
5.42
∴ρ
ρ
⎛ ∂2v ∂2v ∂2v ⎞
∂p
Dv
= − + μ ⎜ 2 + 2 + 2 ⎟ + ρ gy
⎜ ∂x ∂y ∂z ⎟
Dt
∂y
⎝
⎠
Dw ∂ ⎛ ∂u ∂w ⎞ ∂ ⎛ ∂v ∂w ⎞ ∂ ⎛
∂w ⎞
μ⎜ +
=
⎟ + ⎜ − p + 2μ
⎟+ μ⎜ +
⎟ + ρ gz
Dt ∂x ⎝ ∂z ∂x ⎠ ∂y ⎝ ∂z ∂y ⎠ ∂z ⎝
∂z ⎠
=−
∴ρ
∂p
∂2w
∂ 2w
∂ 2w
∂ ⎛ ∂u ∂v ∂w ⎞
+μ 2 +μ 2 +μ 2 +μ ⎜ + +
⎟ + ρ gz
∂z
∂z ⎝ ∂x ∂y ∂z ⎠
∂x
∂y
∂z
⎛ ∂ 2w ∂ 2w ∂ 2w ⎞
∂p
Dw
= − + μ ⎜ 2 + 2 + 2 ⎟ + ρ gz
⎜ ∂x
Dt
∂z
∂y
∂z ⎟⎠
⎝
If we substitute the constitutive equations (5.3.10) into Eqs. 5.3.2 and 5.3.3.,
with μ = μ( x , y , z ) we arrive at
5.43
ρ
⎛ ∂ 2u ∂ 2u ∂ 2u ⎞
∂p
Du
∂μ ∂u ∂μ ⎛ ∂u ∂v ⎞ ∂μ ⎛ ∂u ∂w ⎞
= − + ρ gx + μ ⎜ 2 + 2 + 2 ⎟ + 2
+
⎜ + ⎟+
⎜ +
⎟
⎜
⎟
Dt
∂x
∂x ∂x ∂y ⎝ ∂y ∂x ⎠ ∂z ⎝ ∂z ∂x ⎠
∂y
∂z ⎠
⎝ ∂x
72
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Chapter 5 / The Differential Forms of the Fundamental Laws
If plane flow is only parallel to the plate, v = w = 0. Continuity then demands
that ∂u/∂x = 0. The first equation of (5.3.14) simplifies to
⎛ ∂u
ρ⎜
5.44
⎝ ∂t
+u
⎛ ∂ 2u ∂ 2 u ∂ 2u ⎞
∂p
∂u ⎞
∂u
∂u
⎜
⎟
=
−
+v
+
ρ
g
μ
+w
+
+
+
⎟
x
⎜ ∂x 2 ∂y 2 ∂z 2 ⎟
∂z ⎠
∂x
∂x
∂y
⎝
⎠
ρ
∂u
∂ 2u
=μ 2
∂t
∂y
We assumed g to be in the y-direction, and since no forcing occurs other than
due to the motion of the plate, we let ∂p /∂x = 0.
From Eqs. 5.3.10, −
5.45
τ xx + τ yy + τ zz
⎛ 2μ
⎞
∴p = p −⎜
+λ⎟ ∇⋅V
⎝ 3
⎠
3
= p−
2μ ⎛ ∂u ∂v ∂w ⎞
⎜ + +
⎟− λ ∇⋅V
3 ⎝ ∂x ∂y ∂z ⎠
⎛ 2μ
⎞
∴p − p = −⎜
+λ⎟ ∇⋅V
⎝ 3
⎠
Vorticity
⎛ ∂
∂
∂ ⎞
(V ⋅ ∇ )V = ⎜ u + v + w ⎟ (uˆi + vˆj + wkˆ )
∂y
∂z ⎠
⎝ ∂x
⎡ ∂ ⎛ ∂w
∂w
∂w ⎞ ∂ ⎛ ∂v
∂v
∂v ⎞ ⎤
∇ × (V ⋅ ∇ )V = ⎢ ⎜ u
+v
+w
⎟ − ⎜ u + v + w ⎟ ⎥ ˆi
∂y
∂z ⎠ ∂z ⎝ ∂x
∂y
∂z ⎠ ⎦⎥
⎣⎢ ∂y ⎝ ∂x
⎡ ∂ ⎛ ∂u
∂u
∂u ⎞ ∂ ⎛ ∂w
∂w
∂w ⎞ ⎤ ˆ
+ ⎢ ⎜u + v + w ⎟ − ⎜u
+v
+w
⎟⎥ j
∂y
∂z ⎠ ∂x ⎝ ∂x
∂y
∂z ⎠ ⎦⎥
⎣⎢ ∂z ⎝ ∂x
5.46
⎡ ∂ ⎛ ∂v
∂v
∂v ⎞ ∂ ⎛ ∂u
∂u
∂u ⎞ ⎤
+ ⎢ ⎜ u + v + w ⎟ − ⎜ u + v + w ⎟ ⎥ kˆ
∂y
∂z ⎠ ∂y ⎝ ∂x
∂y
∂z ⎠ ⎥⎦
⎢⎣ ∂x ⎝ ∂x
Use the definition of vorticity: ω = (
∂w ∂v ˆ ∂u ∂w ˆ ∂v ∂u ˆ
) j + ( − )k :
− )i + ( −
∂y ∂z
∂z ∂x
∂x ∂y
⎡ ∂w ∂v ∂
∂u ∂w ∂
∂v ∂x ∂ ⎤
− ) +( −
) + ( − ) ⎥ (uˆi + vˆj + wkˆ )
(ω ⋅ ∇ ) V = ⎢ (
⎣ ∂y ∂z ∂x ∂z ∂x ∂y ∂x ∂y ∂z ⎦
⎡ ∂
∂
∂ ⎤ ⎡ ∂w ∂v ˆ ∂u ∂w ˆ ∂v ∂u ˆ ⎤
) j + ( − )k ⎥
− )i + ( −
( V ⋅ ∇ )ω = ⎢ u + v + w ⎥ ⎢ (
∂y
∂z ⎦ ⎣ ∂y ∂z
∂z ∂x
∂x ∂y ⎦
⎣ ∂x
73
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Chapter 5 / The Differential Forms of the Fundamental Laws
Expand the above, collect like terms, and compare coefficients of ˆi , ˆj, and kˆ .
Studying the vorticity components of Eq. 3.2.21, we see that ωz = −∂u/∂y is the
only vorticity component of interest. The third equation of Eq. 5.3.24 then
simplifies to
Dωz
= ν ∇ 2ωz
Dt
5.47
=ν
∂ 2ωz
∂y 2
since changes normal to the plate are much larger than changes along the plate,
i.e.,
∂ωz
∂ω
>> z
∂y
∂x
If viscous effects are negligible, as they are in a short section, Eq. 5.3.25 reduces
to
Dω z
=0
Dt
that is, there is no change in vorticity (along a streamline) between sections 1
and 2. Since (see Eq. 3.2.21), at section 1
ωz =
∂v ∂u
−
= −10
∂x ∂y
∂u
= 10. This
∂y
means the velocity profile at section 2 is a straight line with the same slope of
the profile at section 1. Since we are neglecting viscosity, the flow can slip at
the wall with a slip velocity u0 ; hence, the velocity distribution at section 2 is
u 2 ( y ) = u 0 + 10 y . Continuity then allows us to calculate the profile:
we conclude that, for the lower half of the flow at section 2,
5.48
V1 A1 = V2 A2
1
(10 × 0.04)(0.04w) = (u0 + 10 × 0.02 / 2)(0.02w)
2
∴ u0 = 0.3 m/s
Finally
u2 ( y) = 0.3 + 10y
5.49
No. The first of Eqs. 5.3.24 shows that
Dωx
∂u
∂u
∂u
= ωx
+ ωy
+ ωz
∂x
∂y
∂z
Dt
74
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Chapter 5 / The Differential Forms of the Fundamental Laws
neglecting viscous effects, so that ω y , which is nonzero near the snow surface,
creates ω x through the term ωy ∂u/∂y, since there would be a nonzero ∂u/∂y
near the tree.
Differential Energy Equation
∫
K∇T ⋅ nˆ dA =
c.s.
c.v.
∫
∫
∇ ⋅ (K∇T )d V =
c.v.
5.50
∴
⎞
∂ ⎛V2
+ gz + u ⎟ρ d V +
⎜⎜
⎟
∂t ⎝ 2
⎠
∫
c.v.
⎛V2
p⎞
+ gz + u + ⎟ ρ V ⋅ nˆ dA
⎜⎜
ρ ⎟⎠
2
c.s. ⎝
∫
⎞
∂ ⎛ V2
+ ρ gz + ρ u ⎟d V +
⎜⎜ ρ
⎟
∂t ⎝ 2
⎠
⎛V2
p⎞
∇ ⋅ ρV ⎜
+ gz + u + ⎟ d V
⎜ 2
ρ ⎟⎠
⎝
c.v.
∫
⎡
⎞
⎛V2
p ⎞⎤
∂ ⎛ V2
2
+ ρ gz + ρ u ⎟ + ∇ ⋅ ρ V ⎜
+ gz + u + ⎟ ⎥ d V = 0
⎢ −K∇ T + ⎜⎜ ρ
⎟
⎜ 2
∂t ⎝ 2
ρ ⎟⎠ ⎥⎦
⎢
⎠
⎝
c.v. ⎣
∫
⎛V2
⎞
p ⎞ V 2 ⎛ ∂ρ
⎡ ∂V
∇p
⎤
∂ V2
+ ∇ ⋅ ρV ⎜
+ gz + ⎟ =
+ ∇ ⋅ ρV ⎟ + ρV ⋅ ⎢
+ V ⋅ ∇V +
+ g∇z ⎥ = 0
ρ
⎜
⎜
⎟
t
2
2
2
ρ
∂t
∂
ρ
∂
t
⎣
⎦
⎝
⎠
⎝
⎠
continuity
momentum
∴−K∇ 2T +
∂
ρ u + ρV ⋅ ∇u = 0
∂t
ρ
or
Du
= K∇ 2T
Dt
Divide each side by dx dy dz and observe that
5.51
∂T
∂x
−
x + dx
dx
∂T
∂x
x
=
∂ T
∂T
∂y
2
∂x 2
,
−
y + dy
∂T
∂y
y
dy
=
∂ T
2
∂x 2
,
∂T
∂z
−
z + dz
∂T
∂z
z
dz
=
∂ 2T
∂z 2
Eq. 5.4.5 follows.
ρ
5.52
D ( h − p /ρ )
Du
Dh Dp p D ρ
Dh Dp p
=ρ
=ρ
−
+
=ρ
−
+ [ − ρ∇ ⋅ V ]
Dt
Dt
Dt Dt ρ Dt
Dt Dt ρ
where we used the continuity equation: D ρ /Dt = − ρ∇ ⋅ V. Then Eq. 5.4. 9
becomes
ρ
Dh Dp p
−
+ [ − ρ∇ ⋅ V ] = K∇ 2T − p∇ ⋅ V
Dt Dt ρ
which is simplified to
ρ
Dp
Dh
= K∇ 2T +
Dt
Dt
75
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Chapter 5 / The Differential Forms of the Fundamental Laws
See Eq. 5.4.9: u = cT
5.53
⎛ ∂T
∂T
∂T
∂T ⎞
2
∴ ρc ⎜
+u
+v
+w
⎟ = K∇ T
∂x
∂y
∂z ⎠
⎝ ∂t
Neglect terms with velocity:
ρc
∂T
= K∇ 2T
∂t
The dissipation function Φ involves viscous effects. For flows with extremely
large velocity gradients, it becomes quite large. Then
ρ cp
5.54
and
DT
is large. This leads to very high temperatures on reentry vehicles.
Dt
u = 10(1 − 10, 000 r 2 )
5.55
DT
=Φ
Dt
∴
∂u
= −2r × 105
∂r
(r takes the place of y )
⎡ 1 ⎛ ∂u ⎞ 2 ⎤
From Eq. 5.4.17, Φ = 2 μ ⎢ ⎜ ⎟ ⎥ = μ × 4r 2 × 1010
⎢ 2 ⎝ ∂y ⎠ ⎥
⎣
⎦
Φ = 1.8 × 10−5 × 4 × 0.012 × 1010 = 72 N/m2 ⋅ s
At the wall where r = 0.01 m,
At the centerline:
∂u
= 0 so Φ = 0
∂r
At a point half-way: Φ = 1.8 × 10−5 × 4 × 0.0052 × 1010 = 18 N/m2 ⋅ s
(a) Momentum:
5.56
∂u
∂ 2u
=ν 2
∂t
∂y
⎛ ∂u ⎞
∂T
∂ 2T
Energy: ρ c
= K 2 +μ⎜ ⎟
∂t
∂y
⎝ ∂y ⎠
2
∂u
∂ 2u ∂μ ∂u
(b) Momentum: ρ
=μ 2 +
∂t
∂y ∂y
∂y
⎛ ∂u ⎞
∂T
∂ 2T
Energy: ρ c
= K 2 +μ⎜ ⎟
∂t
∂y
⎝ ∂y ⎠
2
76
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Chapter 6 / Dimensional Analysis and Similitude
CHAPTER 6
Dimensional Analysis
and Similitude
FE-type Exam Review Problems: Problems 6-1 to 6-6
The dimensions on the variables are as follows:
L L ML2
ML / T 2
M
L
[W ] = [ F × V ] = M 2 × = 3 , [ d ] = L, [ Δp ] =
=
, [V ] =
2
2
T
T
T
T
L
LT
6.1 (A)
First, eliminate T by dividing W by Δp. That leaves T in the denominator so
divide by V leaving L2 in the numerator. Then divide by d 2 . That provides
π=
W
ΔpVd 2
V = f ( d , l , g , ω , μ ). The units on the variables on the rhs are as follows:
[d ] = L, [l ] = L, [ g ] =
6.2 (A)
L
T2
, [ω ] = T −1 , [ μ ] =
ML
T
Because mass M occurs in only one term, it cannot enter the relationship.
6.3 (A)
Re m = Re p
6.4 (A)
Rem = Re p
Vm Lm
νm
Vm Lm
νm
=
=
Vp Lp
∴Vm = V p
νp
Vp Lp
νp
∴Vm = V p
6.5 (C)
Frm = Fr p
Lm ν p
∴Vm = V p
From Froude’s number Vm = V p
6.6 (A)
Fm*
=
Fp*
or
Fm
ρmVm2lm2
=
Fp
ρ pV p2l 2p
Lm
Lp ν m
2
Vp
Vm2
=
lm g m l p g p
Lp
= 12 × 9 = 108 m/s
= 4 × 10
1.51×10−5
= 461 m/s
1.31× 10−6
lm
1
= 2 × = 0.5 m/s
lp
4
lm
. From the dimensionless force we have:
lp
∴ Fp = Fm
V p2 l 2p
Vm2 lm2
= 10 × 25 × 252 = 156,000 N
77
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6 / Dimensional Analysis and Similitude
Introduction
a) [ m ] = kg/s = N ⋅ s2 /m ⋅ s = N ⋅ s/m
6.8
∴
c) [ ρ ] = kg/m = N ⋅ s /m ⋅ m = N ⋅ s /m
3
2
e) [W ] = N ⋅ m
3
2
FT
L
4
FT 2
∴ 4
L
∴ FL
Dimensional Analysis
[V ] =
V = f (A, ρ , μ )
6.10
L
M
M
, [ A ] = L, [ ρ ] = 3 , [ μ ] =
T
LT
L
∴ π1 = f1 (π 20 ) = Const.
[V ] =
V = f ( H , g , m)
6.12
ρV A
μ
∴There is one π − term: π1 =
gHm0
∴ π1 =
V2
FD = f (d, A, V , μ , ρ )
∴ρ
L
,
T
VA
μ
[g] =
∴ π1 = C
[ FD ] =
=C
or Re = Const.
L
,
T2
[ m] = M , [ H ] = L
∴V = gH / C
ML
L
M
M
d
L
V
,
=
,
=
,
μ
=
,
ρ
=
[
]
[
]
[
]
[
]
T
LT
T2
L3
π1 = FD A a1V b1 ρ c1 , π 2 = dV b2 ρ c2 A a2 , π 3 = μ A a3V b3 ρ c3
∴ π1 =
6.14
∴
FD
d
μ
, π2 = , π3 =
2
A
ρV A
V ρA
2
⎛d μ ⎞
FD
= f1 ⎜ ,
⎟
2 2
ρA V
⎝ A ρ AV ⎠
We could write
⎛ 1 π ⎞
⎛A μ ⎞
π1
FD
= f 2 ⎜ , 3 ⎟ or
= f2 ⎜ ,
⎟ . This is
2 2
2
ρd V
⎝ d ρdV ⎠
⎝π2 π2 ⎠
π2
equivalent to the above. Either functional form must be determined by
experimentation.
6.16
h = f (σ , d, γ , β , g). [ h ] = L, [σ ] =
M
M
L
, d = L, [γ ] = 2 2 , [ β ] = 1, [ g ] = 2 Se
2 [ ]
T
LT
T
78
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Chapter 6 / Dimensional Analysis and Similitude
lect d , γ , g as repeating variables:
π 1 = hd a1 γ b1 g c1 , π 2 = σ d a2 γ b2 g c2 , π 3 = β
∴π1 = h/d, π 2 = σ /γ d2 , π 3 = β
∴ h /d = f1 (σ /γ d 2 , β ). Note: gravity does not enter the answer.
σ = f ( M , y, I )
[σ ] =
6.18
M
ML2
,
M
=
, [ y ] = L, [ I ] = L4
2 [ ]
2
LT
T
Given that b = −1 , π1 =
V = f ( H , g, ρ )
6.20
[V ] =
∴π1 = VH a g b ρ c = V
σI
yM
= Const.
∴σ = C
My
I
L
L
M
, [ H ] = L, [ g ] = 2 , [ ρ ] = 3
T
T
L
ρ0
g H
= Const.
∴ V = Const.
gH
Density does not enter the expression.
Δp = f (V , d ,ν , L, ε , ρ )
[ Δp ] =
M
L
L2
M
V
d
L
,
=
,
=
,
ν
=
, [ L ] = L, [ e ] = L, [ ρ ] = 3
[ ]
2 [ ] T [ ]
T
LT
L
Repeating variables: V , d , ρ
6.22
π1 = ΔpV a1 db1 ρ c1 , π 2 = ν V a2 db2 ρ c2 , π 3 = L V a3 db3 ρ c3 , π 4 = e V a4 db4 ρ c4
By inspection: ∴ π1 =
Δp
ρV
2
π1 = f1(π 2 , π 3 , π 4 )
Q = f ( R, A, e, S , g ). [Q ] =
6.24
, π2 =
ν
Vd
, π3 =
L
e
, π4 =
d
d
∴Δp/ρ V 2 = f1(ν /Vd, L/d, e/d)
L3
L
, [ R ] = L, [ A] = L2 , [ e ] = L, [ s ] = 1, [ g ] = 2
T
T
There are only two basic dimensions. Choose two repeating variables, R and g.
Then
π 1 = QR a1 g b1 , π 2 = AR a2 g b2 , π 3 = eR a3 g b3 , π 4 = sR a4 g b4
∴ π1 =
Q
A
e
,
π
=
,
π
=
, π4 = s
2
3
R
g R5/ 2
R2
79
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Chapter 6 / Dimensional Analysis and Similitude
∴π1 = f1 (π 2 , π 3 , π 4 ) ∴ Q/ gR5 = f1 ( A/R2 , e/R, s)
FD = f (V , μ , ρ , e, I , d) Repeating variables: V , ρ , d
[ FD ] =
6.26
ML
L
M
M
, [V ] = , [ μ ] =
, [ ρ ] = 3 , [ e ] = L, [ I ] = 1, [ d ] = L
2
T
LT
T
L
π 1 = FDV a1 ρ b1 d c1 , π 2 = μV a2 ρ b2 d c2 , π 3 = e V a3 ρ b3 d c3 , π 4 = I V a4 ρ b4 d c4
∴ π1 =
FD
μ
e
, π2 =
, π3 = , π 4 = I
2 2
V ρd
d
ρV d
∴ FD /ρ V 2d 2 = f1 ( μ /ρ Vd, e /d, I )
T = f ( ρ , μ , V , D) = ρ μ V D
a b
6.28
c
d
ML
or
T2
a
b
c
⎡M ⎤ ⎡ M ⎤ ⎡ L⎤
=⎢ 3⎥ ⎢
[ L]d
⎥
⎢
⎥
⎣ L ⎦ ⎣ LT ⎦ ⎣ T ⎦
M : 1 = a + b ⇒ a = 1− b
T : − 2 = −b − c ⇒ c = 2 − b
L:
1 = −3a − b + c + d ⇒ 1 = −3(1 − b) − b + (2 − b) + d
⎛ μ ⎞
1−b
2−b
2−b
∴ d = 2 − b. T = ρ ( ) μ bV ( ) D( ) = ρ V 2 D 2 ⎜
⎟
⎝ ρ VD ⎠
b
and
T
ρ V 2 D2
= φ [ Re]
FD = f (V , μ , ρ, d , e, r, c ). Repeating variables: V , ρ , d
[ FD ] =
ML
L
M
M
1
=
μ
=
ρ
=
=
=
=
=
,
[
V
]
,
[
]
,
[
]
,
[
d
]
L
,
[
e
]
L
,
[
r
]
L
,
[
c
]
T
LT
T2
L3
L2
π 1 = FDV a ρ b d c , π 2 = μV a ρ b d c , π 3 = eV a ρ b d c ,
1
1
1
2
2
2
3
3
3
π 4 = rV a ρ b d c , π 5 = cV a ρ b d c
4
6.30
∴ π1 =
∴
4
4
5
5
5
FD
μ
e
r
, π2 =
, π 3 = , π 4 = , π 5 = cd 2
2 2
ρVd
d
d
ρV d
⎛ μ e r
⎞
FD
= f1 ⎜
, , , cd 2 ⎟
2 2
ρV d
⎝ ρVd d d
⎠
FL = f (V , c , ρ , A c , t , α ). Repeating variables: V , ρ , A c
6.32
[ FL ] =
ML
L
L
M
, [V ] = , [c ] = , [ ρ ] = 3 , [A c ] = L, [t ] = L, [α ] = 1
2
T
T
T
L
π1 = FLV a1 ρ b1 A c c1 , π 2 = cV a2 ρ b2 A c c2 , π 3 = tV a3 ρ b3 A c c3 , π 4 = αV a4 ρ b4 A c c4
80
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Chapter 6 / Dimensional Analysis and Similitude
∴π1 =
∴
FL
c
t
,
π
=
,
π
=
, π4 = α
2
3
Ac
V
ρV 2A c2
⎛c t
⎞
FL
= f1 ⎜ , , α ⎟
2 2
ρV A c
⎝ V Ac ⎠
FD = f (V , ρ , μ , d , L , ρ c , ω ) where d is the cable diameter, L the cable length,
ρ c the cable density, and ω the vibration frequency. Repeating variables:
V , d , ρ. The π − terms are
π1 =
6.34
FD
ρV d
2 2
, π2 =
Vd ρ
μ
, π3 =
d
V
ρ
, π4 =
, π5 =
L
ρc
ωd
We then have
⎛ Vdρ d ρ
FD
V ⎞
= f1 ⎜
, ,
,
⎟
2 2
L ρ c ωd ⎠
ρV d
⎝ μ
T = g( f , ω , d , H , A, N , h, ρ ). Repeating variables: ω , d , ρ
[T ] =
6.36
π1 =
ML2
1
1
M
, [ f ] = , [ω ] = , [ d ] = L, [ H ] = L, [A] = L, [ N ] = 1, [ h] = L, [ ρ ] = 3
2
T
T
T
L
T
ρω d
2 5
∴
, π2 =
f
ω
, π3 =
H
h
A
, π4 = , π5 = N, π6 =
d
d
d
h⎞
⎛ f H A
= g1 ⎜ ,
, , N, ⎟
d⎠
ρω d
⎝ω d d
T
2 5
d = f (V, Vj , D, σ , ρ, μ, ρ a ). Repeating variables: V j , D, ρ
[ d ] = L, [V ] =
6.38
π1 =
L
L
M
M
M
M
, [V j ] = , [ D ] = L, [σ ] = 2 , [ ρ ] = 3 , [ μ ] =
, [ρa ] = 3
T
T
LT
T
L
L
d
V
σ
μ
ρ
, π 2 = , π3 =
, π4 =
, π5 = a
2
ρV j D
ρ
D
Vj
ρV j D
∴
⎛V
d
σ
μ
ρa ⎞
⎟
= f1 ⎜ ,
,
,
⎜ V j ρV j2 D ρV j D ρ ⎟
D
⎝
⎠
μ = f ( D, H , A, g, ρ, V ) . D = tube dia., H = head above outlet, A = tube
length.
6.40
Repeating variables: D, V , ρ
π1 =
μ
ρVD
, π2 =
H
A
gD
, π3 = , π 4 = 2
D
D
V
81
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Chapter 6 / Dimensional Analysis and Similitude
∴
μ
⎛ H A gD ⎞
= f1 ⎜ ,
,
⎟
ρVD
⎝ D D V2 ⎠
y 2 = f (V1 , y1 , ρ, g). Neglect viscous wall shear.
[ y2 ] = L, [V1] =
L
M
L
, [ y1] = L, [ ρ ] = 3 , [ g ] = 2
T
L
T
Repeating variables: V1, y1, ρ
6.42
π1 =
∴
y2
gy
, π 2 = 21
y1
V1
( ρ does not enter the problem)
⎛ gy ⎞
y2
= f ⎜ 21 ⎟
⎜V ⎟
y1
⎝ 1 ⎠
Similitude
Qm Vm A 2m Δpm ρ m Vm2 ( Fp ) m ρ m Vm2 A 2m
=
,
=
,
=
ρ p Vp2 ( Fp ) p ρ p Vp2 A 2p
Qp
Vp A 2p Δp p
6.44
Q m ρ m Vm3 A 2m
=
Q p ρ p Vp3 A 2p
τ m ρ m Vm2 Tm ρ m Vm2 A 3m
,
,
=
=
τ p ρ p Vp2 Tp ρ p Vp2 A 3p
(Q has same dimensions as W )
a) Rem = Re p
6.46
Vmdm
νm
=
Vpd p
∴
νp
Vm d p
=
=5
V p dm
m m ρ mA 2mVm 1
1
=
= 2 × 5 ∴ m m = m p = 800 / 5 = 160 kg/s
2
m p ρ p A pV p 5
5
Δpm ρ mVm2
=
= 52 ∴Δpm = 25Δp p = 25 × 600 = 15,000 kPa
2
Δp p ρ pV p
Rem = Re p
6.48
VmA m
νm
=
V pA p
νp
∴
Vm A p ν m
ν
=
= 10 assuming m = 1
Vp A m ν p
νp
∴Vm = 10V p = 1000 km/hr
This velocity is much too high for a model test; it is in the compressibility
region. Thus, small-scale models of autos are not used. Full-scale wind tunnels
are common.
82
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Chapter 6 / Dimensional Analysis and Similitude
Properties of the atmosphere at 8 km altitude: T = −37°C +273 = 236 K and
pressure = 35.7 kPa, density = 0.526 kg/m3, and viscosity = 1.527 × 10−5 N·s/m2
Properties of air at standard atmosphere: T = 20°C, p = 101.325 kPa, density =
1.204 kg/m3, dynamic viscosity = 1.82 × 10−5 N·s/m2
⎛ ρVD ⎞ ⎛ ρVD ⎞
Use the Reynolds number to achieve dynamic similarity, ⎜
⎟ =⎜
⎟
⎝ μ ⎠m ⎝ μ ⎠ p
6.50
⎛ ρD ⎞ ⎛ μ ⎞
Then Vm = Vp ⎜
⎟ ⎜
⎟
⎝ μ ⎠ p ⎝ ρ D ⎠m
−5
⎛ ρ p μ m ⎞ ⎛ Dp ⎞
⎛ 0.526 ⎞ ⎛ 1.82 ×10 ⎞ ⎛ 10 ⎞
=
= 1041 km/hr
and Vm = Vp ⎜
200
⎟ ⎜⎜
⎜
⎟
⎜
⎟⎜
−5 ⎟
⎟ ⎝⎜ 1 ⎠⎟
⎜ ρ m μ p ⎟ ⎝ Dm ⎟⎠
1.204
⎝
⎠
×
1.527
10
⎝
⎠
⎝
⎠
⎛ T
⎞
⎛ T
⎞
To calculate the thrust apply: ⎜
=
⎟
⎜
2 2
2 2⎟
⎝ ρ V D ⎠ p ⎝ ρ V D ⎠m
⎛ ρ p Vp2 Dp2 ⎞
⎛ 0.526 ×10412 ×102 ⎞
⎟
Then Tp = Tm ⎜
10
=
= 1184 N
⎜
2
2 ⎟
⎜ ρmVm2 Dm2 ⎟
1.204
200
1
×
×
⎝
⎠
⎝
⎠
Rem = Re p
Vmdm
νm
=
Vpd p
νp
10−6
∴ dm = d p
= 0.75 × 1×
Vm ν p
5 × 10−3
Vp ν m
= 0.0015 m = 1.5 mm
6.52
Δpm ρmVm2
1000
=
=
× 12 = 1.11
2
Δp p ρ pV p 1000 × 0.9
Find ν oil using Fig. B.2. Then
Re m = Re p ∴
6.54
VmA m
νm
=
V pA p
νp
Vm A p ν m
ν
1
=
= 30 m =
Vp A m ν p
νp
30
2
Frm = Frp ∴
∴
νm
1
=
ν p 164
∴
A
Vm
= m
Ap
Vp
2
Frm = Frp
Vp
Vm2
=
A m gm A p g p
Vp
1
Vm2
V
=
∴ m =
30
Vp
A m gm A p g p
∴ν m = 6.1× 10−9 m 2 /s Impossible!
6.56
a)
Qm Vm A 2m
=
Qp
V p A 2p
∴ Qm = Q p
Vm A 2m
1
1
= 2×
× 2 = 0.00632 m 3 /s
2
Vp A p
10 10
83
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Chapter 6 / Dimensional Analysis and Similitude
Fm ρmVm2A2m
b)
=
Fp ρ pV p2A2p
∴ Fp = Fm
V p2 A2p
Vm2
A2m
= 12 ×10 × 102 = 12,000 N
Neglect viscous effects, and account for wave (gravity) effects.
2
Frm = Frp
6.58
∴
Vp
Vm2
=
A m gm A p g p
∴ ωm = ω p
Tm ρmVm2A3m
=
Tp ρ pV p2A3p
∴
Vm
A
= m
Vp
Ap
ωm Vm / A m
=
ω p Vp / A p
Vm A p
1
= 600 ×
× 10 = 1897 rpm
Vp A m
10
∴Tp = Tm
V p2 A3p
Vm2
A3m
= 1.2 × 10 × 103 = 120,000 N ⋅ m
Check the Reynolds number:
Re p = V pdp /ν p = 15 × 2/10 −6 = 30 × 106
This is a high-Reynolds-number flow
Re m =
6.60
2 × 2/30
10
−6
= 1.33 × 105
This may be sufficiently large for similarity. If so
Wm ρmVm3A 2m 23
1
=
= 3 × 2 = 2.63 × 10−6
3
2
W p ρ pV p A p 15 30
∴W p = (2 × 2.15) / 2.63 × 10−6 = 1633 kW
Re p =
20 × 10
= 13.3 × 10 6 . This is a high-Reynolds-number flow.
−5
1.5 × 10
For the wind tunnel, let Rem = 105 = Vm × 0.4/1.5 × 10−5
For the water channel, let Rem = 105 = Vm × 0.1/1× 10−6
6.62
∴ Vm ≥ 3.75 m/s
∴ Vm ≥ 1.0 m/s
Either could be selected. The more convenient facility would be chosen.
Fm1
Fm2
=
ρ m1 Vm21 A 2m1
ρ m2 Vm22 A 2m2
3.2
=
Fm2
W
ρ mVm3 A 2m 153 × 0.42
m
= ρ V 3A 2 = 203 × 102
W
p
p p p
∴ Fm2
1000 2.42 0.12
= 3.2
×
= 4.16 N
1.23 152 0.42
203 102
∴ Wp = (15 × 3.2) 3 ×
= 71,100 W or 95 hp
15 0.42
84
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Chapter 6 / Dimensional Analysis and Similitude
Mach No. is the significant parameter: M m = M p
6.64
b) V p = Vm
cp
cm
Fp = Fm
= Vm
Tp
Tm
ρ mVp2A 2p
ρ mVm2 A 2m
= 10 × 0.601 ×
Vp2
Vm2
=
A m gm A p g p
a) Frm = Frp
255.7
= 186 m/s
296
= 200
∴
1862
× 202 = 2080 N
2
200
Vm
A
= m
Vp
Ap
ωm Vm A p
1
=
=
× 10
ω p Vp A m
10
6.66
VmA m
b) Rem = Re p
=
νm
V pA p
νp
∴
∴ωm = 2000 ×
Vm A p
=
= 10
Vp A m
ωm Vm A p
1
=
= 10 × = 1
ω p Vp A m
10
Rem = Re p
∴
VmA m
νm
=
VpA p
∴ωm = 2000 rpm
∴ Vm = Vp
νp
10
= 6320 rpm
10
Ap
Am
= 10 × 10 = 100 m/s
This is too large for a water channel. Undoubtedly this is a high-Re flow. Select
a speed of 5 m/s. For this speed
6.68
Rem =
5 × 0.1
= 5 × 105
−6
1× 10
where we used A m = 0.1 (A p = 1 m, i.e., the dia. of the porpoise)
ωm = ωp (Vm/Vp )(A p/A m ) = 1× (5/10) ×10 = 5 motions/s
Normalized Differential Equations
6.70
ρ* =
y
ρ *
u
v
x
, t = tf , u* = , v* = , x* = , y* = . Substitute in:
A
A
ρ0
V
V
f ρ0
∂ρ *
V ∂( ρ *u* )
V ∂(ρ *v* )
ρ
ρ
+
+
=0
0
0
A ∂x*
A ∂y*
∂t*
Divide by ρ0 V /A :
85
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Chapter 6 / Dimensional Analysis and Similitude
∴
f A ∂ρ * ∂
∂
+ * ( ρ *u* ) + * ( ρ *v* ) = 0
*
V ∂t
∂x
∂y
V* =
Parameter =
fA
V
V * tU
p
h
, t =
, ∇* = A ∇ , p * =
, h* = . Euler’s equation is then
2
U
A
A
ρU
U 2 DV*
U2 * *
A
ρ
= −ρ
∇ p − ρ g ∇*h*
*
A Dt
A
A
6.72
Divide by ρU 2 /A :
DV*
gA
= −∇* p* − 2 ∇*h*
*
Dt
U
Parameter =
gA
U2
The only velocity component is u. Continuity then requires that ∂u/∂x = 0
(replace z with x and vz with u in the equations written using cylindrical
coordinates). The x-component Navier-Stokes equation is
⎛ ∂ 2u 1 ∂u 1 ∂ 2u ∂ 2u
∂u
∂u vθ ∂u
∂u
1 ∂p
=−
+ gx +ν ⎜ 2 +
+ vr
+
+u
+
+
⎜ ∂r
ρ ∂x
∂t
∂r
r ∂θ
∂x
r ∂r r 2 ∂θ 2 ∂x 2
⎝
This simplifies to
⎛ ∂ 2u 1 ∂u ⎞
1 ∂p
∂u
=−
+ν ⎜ 2 +
⎟
⎜ ∂r
∂t
ρ ∂x
r ∂r ⎟⎠
⎝
6.74
b) Let u* = u/V , x* = x /d, t* = tν /d 2 , p* = p /ρ V 2 and r* = r /d :
ρ V 2 ∂p* ν V ⎛ ∂ 2u* 1 ∂u* ⎞
+
+
⎜
⎟
ρ d ∂x* d2 ⎜⎝ ∂r*2 r* ∂r* ⎟⎠
d2 ∂t*
The normalized equation is
ν V ∂u*
∂u*
∂t
u* =
6.76
u
,
U
*
=−
= − Re
v* =
∂p*
∂x
*
v
,
U
+
∂ 2u*
∂r
T* =
*2
+
1 ∂u*
r ∂r
T
,
T0
*
*
x
x* = ,
A
where Re =
y* =
y
,
A
Vd
ν
∇*2 = A 2∇ 2
⎡UT0 ∂T * UT0 ∂T * ⎤ K
*2 *
ρc p ⎢
+
⎥ = 2 T0∇ T
*
*
A ∂y ⎥⎦ A
⎢⎣ A ∂x
Divide by ρ c pUT0 /A :
∂T * ∂T *
K
+ * =
∇*2T *
*
c
U
ρ
A
∂x
∂y
p
Parameter =
K μ
1 1
=
μ c p ρU A Pr Re
86
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
⎞
⎟
⎟
⎠
Chapter 7 / Internal Flows
CHAPTER 7
Internal Flows
FE-type Exam Review Problems: Problems 7-1 to 7-12
7.1 (D)
7.2 (A)
7.3 (D)
The flow in a pipe may be laminar at any Reynolds number between 2000 and
perhaps 40,000 depending on the character of the flow.
7.4 (D)
The friction factor f depends on the velocity (the Reynolds number).
7.5 (A)
L V2
Δp = f
D 2g
7.6 (B)
e 0.15
=
= 0.0075 ∴ Moody's diagram gives, assuming Re > 105
D 20
15 V 2
f = 0.034 Then 60,000 = 0.034 ×
0.02 2 × 9.81
∴V = 6.79 m/s and
Q = AV = π × 0.012 × 6.79 = 0.00214 m3 /s
Check the Reynolds number: Re =
7.7 (D)
6.79 × 0.02
10
−6
e 0.26
=
= 0.00325
D
80
Assume Re > 3 × 105
hL
1 V2
= sin θ = f
L
D 2g
sin 30D = 0.026
Check Re: Re =
VD
ν
=
5.49 × 0.08
10
−6
= 1.36 ×105 ∴ OK
Then f = 0.026
1
V2
0.08 2 × 9.81
= 4.39 ×105
∴V = 5.49 m/s
∴ OK
e 0.26
4 × 0.06
=
= 0.0043 Re =
= 3 × 10 4 ∴ f = 0.033 from Moody's diagram
−6
D
60
8 × 10
7.8 (B)
20
42
L V2
Δp = −γ f
+ γ Δh = −9810 × 0.033 ×
+ 9810 × 20 = 108,000 Pa
0.06 2 × 9.81
D 2g
7.9 (A)
R=
A 4× 4
Q
0.02
=
= 1 cm V = =
=12.5 m/s
P 4× 4
A 0.04 × 0.04
87
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Chapter 7 / Internal Flows
Re =
4RV
ν
4 × 0.01×12.5
=
10−6
e
0.046
=
= 0.00115 ∴ f = 0.021
4 R 4 ×10
= 5 ×105
L V2
40
12.52
Δp = f
= 0.021×
×
= 167 Pa
4R 2 g
4 × 0.01 2 × 9.81
7.10 (B)
Viscous effects (losses) are important.
7.11 (A)
A negative pressure must not exist anywhere in a water system for a community
since a leak would suck into the system possible impurities.
1
1
0.8 × 2.4 × 0.482 / 3 × 0.0021/ 2 = 4.39 m3 /s
AR 2 / 3 S 1/ 2 =
0.012
n
Q=
7.12 (C)
where we have used R =
A
Pwetted
=
0.8 × 2.4
= 0.48 m
0.8 + 0.8 + 2.4
Laminar or Turbulent Flow
7.14
Re =
Vh
7.16
Re =
Vh
ν
ν
=
=
Vh
1×10
b) 1500 =
−6
(1/ 2) × 1.4
10 −6
V ×1
∴V = 0.0015 m/s
1×10−6
∴Very turbulent
= 700,000
Entrance and Developed Flow
LE
= 0.065 Re
D
7.18
a) LE = 0.065 ×
c) LE = 0.065 ×
V=
7.20
0.025
π × 0.03
2
Re =
VD
V=
ν
0.1592 × 0.04
1.31×10−6
0.1592 × 0.04
0.661×10−6
= 8.84 Re =
π × 0.022
= 0.1592 m/s
× 0.04 = 12.6 m
× 0.04 = 25.0 m
8.84 × 0.06
1.007 ×10
∴ LE = 120 × 0.06 = 7.2 m
7.22
0.0002
−6
= 5.3 ×105 ∴Turbulent
∴Developed
LE = 0.04 Re × h = 0.04 × 7700 × 0.012 = 3.7 m
88
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Chapter 7 / Internal Flows
( LE )min = 0.04 ×1500 × 0.012 = 0.72 m
Re =
7.24
VD 0.2× 0.04
=
= 8000
ν
10−6
a) If laminar, LE = 0.065 × Re × D = 0.065 × 8000 × 0.04 = 20.8 m
Li = LE /4 = 20.8/4 = 5.2 m
Laminar Flow in a Pipe
7.28
In a developed flow, dh/dx = slope of the pipe and p is a linear function of x so
that dp/dx = const. Therefore, d(p + γh)/dx = const and it can be moved outside
the integral. Then
r
d( p + γ h) 0 2 2
1 d( p + γ h) ⎛ r04 r02 2 ⎞ r02 d( p + γ h)
(
r
r
)
rdr
−
=
⎜⎜ − × r0 ⎟⎟ =
0
∫
2
dx
dx
dx
4 μ r02
2
μ
r
0
⎝ 4 2
⎠ 8μ
0
2
1500 =
VD
ν
Eq. 7.3.14:
7.30
∴α =
8ν V
r02g
=
=
V × 0.01
∴ V = 0.0992 m/s
6.61×10−7
Δp
Δh 8μV
+γ
= 2
L
L
r0
8× 6.61×10−7 × 0.0992
0.005 × 9.81
2
Δh 8μV
∴
=
L r02 ρ g
α
Δh
L
α = Δh/ L
= 0.00214 rad or 0.123D
Q = AV = π × 0.0052 × 0.0992 = 7.79 ×10−6 m3 /s
0.01
Eq. 7.3.14:
Q=
∫
0
9810(0.00015)
2 ×10
−3
(0.02 y − y 2 )100dy
= 73,600 × (0.01× 0.012 −
7.32
0.013
) = 0.049 m3/s
3
For a vertical pipe Δh = L. Thus
ρ gr02 gr02 9.81× .012 1.226 × 10−4
V=
=
=
=
8μ
8ν
8ν
ν
b) V =
1.226 × 10−4
= 0.33 m/s ∴ Q = 1.04 × 10−4 m3 /s Re = 17.8 ∴ laminar
0.34/917
89
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Chapter 7 / Internal Flows
Neglect the effects of the entrance region and assume developed flow for the
whole length. Also, assume p inlet = γh (neglect V 2 /2 g compared to 4 m).
7.34
∴Δp = γ h − 0
∴ ρ gh =
8μ VL
∴V =
r02
ghr02 9.81× 4 × 0.00252
=
= 0.766 m/s
8ν L
8 ×1×10−6 × 40
∴ Q = AV = π × 0.00252 × 0.766 = 1.5 × 10−5 m3s
LE = 0.065 ×
Re = 2000 =
7.36
Δp =
VD
ν
Δp =
=
8μ VL
Re = 40,000 =
7.38
0.766 × 0.005
× 0.005 = 1.2 m
1× 10−6
r02
VD
ν
V × 0.02
1.5 ×10−5
=
=
∴ V = 1.5 m/s
8 ×1.9 × 10−5 × 1.5 × 10
(0.01)2
V × 0.1
= 23 Pa
∴ V = 6.04 m/s ∴ Vmax = 2V = 12.1 m/s
1.51×10−5
8μ VL 8 ×1.81×10−5 × 6.04 ×10
=
= 3.5 Pa LE = 0.065 × 40,000 × 0.1 = 260 m
r02
0.052
V =−
R2 Δp + γΔh
R2 γ (− L) γ R2 9800 × 0.0012
=−
=
=
= 1.225 m/s
8μ
8μ L
8μ
L
8 ×10−3
Q = AV = π × 0.0012 ×1.225 = 3.85×10−6 m3/s
7.40
Re =
VD
ν
=
122.5 × 00.02
= 2450
10 − 6
It probably is not a laminar flow. Since Re > 2000 it would most likely be
turbulent. If the pipe were smooth, disturbance and vibration free, with a wellrounded entrance it could be laminar.
⎛ r2 ⎞
a) Combine Eq. 7.3.12 and 7.3.15: u = umax ⎜ 1 − 2 ⎟
⎝ r0 ⎠
7.42
⎛ r2 ⎞
u(r ) = V = 2V ⎜ 1 − 2 ⎟
⎜ r ⎟
0 ⎠
⎝
∴r2 =
r02
and r = 0.707 r0
2
b) From Eq. 7.3.17: τ = Cr where C is a constant. Then τ w = Cr0
If τ = τ w /2, then τ w /2 = Cr and r = τ w /2C = r0 /2
90
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
Re = 20,000 =
7.44
hL =
Δp
γ
=
VD
ν
8μVL
γ r02
=
=
V × 0.05
1.1×10−6
∴ V = 0.44 m/s
8 ×1.1×10−3 × 0.44 ×10
9810 × (0.025)2
= 0.0063 m
r0 Δp (0.025) × 0.0063 × 9810
=
= 0.08 Pa
2L
2 × 10
LE = 0.065 × 20,000 × 0.05 = 65 m
τ0 =
Q=−
See Example 7.2:
7.46
π (−Δp) ⎡ 4 4 (r22 − r12 ) 2 ⎤
⎢ r2 − r1 −
⎥
8μ L ⎢⎣
ln(r2 / r1 ) ⎥⎦
∴Q =
2
2 2
⎡
−4
4
4 (0.03 − 0.02 ) ⎤
3
0.03
0.02
−
−
⎢
⎥ = 7.25 × 10 m /s
−5
ln(0.03/0.02) ⎦⎥
8 × 1.81× 10 × 10 ⎣⎢
∴V =
7.25 ×10−4
Q
=
= 0.462 m/s
A π (0.032 − 0.022 )
π × 10
τ r1 = μ
Δp ⎡
r22 − r12 1 ⎤
∂u
=−
⎢ 2r1 −
⎥
∂r r =r1
4 L ⎢⎣
ln(r2 / r1 ) r1 ⎦⎥
10 ⎡
0.032 − 0.022
1 ⎤
=−
×
⎢ 2 × 0.02 −
⎥ = 0.0054 Pa
4 × 10 ⎣⎢
ln(0.03/0.02) 0.02 ⎦⎥
From Example 7.2 u(r ) =
1 dp ⎡ 2 2 r22 − r12
r⎤
ln ⎥
⎢ r − r2 +
4 μ dx ⎣⎢
ln(r1 / r2 ) r2 ⎦⎥
As r1 → 0, ln(r1 / r2 ) → −∞ so that
7.48
(
)
r22 − r12
ln(r / r2 ) → 0
ln(r1 / r2 )
Thus, u(r ) =
1 dp 2
r − r02
4 μ dx
As r1 → r2 ,
r22 − r12
0
= . ∴Differentiate w.r.t r1:
ln(r1 / r2 ) 0
Also, ln
w here r2 = r0 . See Eq. 7.3.11
−2r1
= −2r12
1/ r1
⎛
y⎞
y
r
= ln ⎜ 1 − ⎟ ≅ − , where y = r2 − r
r2
r2
⎝ r2 ⎠
91
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Chapter 7 / Internal Flows
∴ u(r ) =
1 dp ⎛ 2 2 2r12
⎜ r − r2 +
4μ dx ⎜⎝
r2
⎞ 1 dp ⎡ 2
2r12
2
−
+
y⎟ =
y
r
y
2
⎟ 4μ dx ⎢⎢
r2
⎠
⎣
⎤ 1 dp 2
( y − ay )
y⎥ =
⎥⎦ 4μ dx
Laminar Flow Between Parallel Plates
There is no pressure gradient ∴Eq. 7.4.13 gives u =
V
y
a
The friction balances the weight component.
7.50
τ A = W sin θ
a) μ
∂u
V
0.2
τ =μ =μ =μ
0.0004
a
∂y
0.2
× 1× 1 = 40 sin 20D
0.0004
dh
= sin θ
dx
Q=−
a /2
∫
0
=
γ sin θ 2
( y − ay)
2μ
y
a = 0.012 m
γ sin θ 2
50γ sin θ ⎛ a3 a3 ⎞ 25γ a3
( y − ay)50dy = −
sin θ
⎜ − ⎟=
2μ
2μ ⎜⎝ 24 8 ⎟⎠ 12μ
25
12 × 10
∴V =
7.52
Hence, u( y) = −
W
∴ μ = 0.0274 N ⋅ s/m 2
The depth of water is a/2 with the maximum velocity at the
surface:
−
τA
−3
× 9810 × sin 20D × 0.0123 = 12.1 m 3 /s
12.1
= 40.3 m/s
0.006 × 50
∴ Re =
Va/2
ν
=
40.3 × 0.006
10−6
= 241,000
The assumption of laminar flow was not a good one, but we shall stay with it to
answer the remaining parts.
umax
9810 × sin 20D ⎛ 0.0122 ⎞
=
⎜
⎟ = 60.4 m/s
2 × 10−3 ⎜⎝ 4 ⎟⎠
τ0 = μ
∂u
γ sinθ
9810sin 20D
=−
(−a) =
× 0.012 = 20.1 Pa
∂y y=0
2
2
Obviously the flow would be turbulent and the above analysis would have to be
modified substantially for an actual flow.
92
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Chapter 7 / Internal Flows
Δp =
Eq. 7.4.17:
12 μ VL
a2
∴V =
50 × 0.022
12 × 1.81×10−5 × 60
= 1.53 m/s
∴ Q = AV = 0.02 × 0.9 ×1.53 = 0.028 m3/s
7.54
This is maximum since laminar flow is assumed. Check the Reynolds number:
Re =
Va
ν
=
1.53 × 0.02
1.51× 10−5
= 2030
This is marginally high. Care should be taken to eliminate vibrations,
disturbances, or rough walls.
7.56
Assume laminar:
−3
12 μ VL
6 12 × 9.4 × 10 V × 0.1
Δp =
∴
4.2
×
10
=
∴ V = 93 m/s
(0.5 × 10−3 ) 2
a2
∴Q = AV = (0.0005 × 0.1) × 93 = 4.6 ×10−3 m3/s = 4.6 L/s
u=
7.58
1 dp 2
U
du 1 dp
U
( y − ay ) + y τ = μ
=
(2 y − a) + μ
2μ dx
a
dy 2 dx
a
a) τ y =0.006 =
0.006
c) Q =
∫
0
1
U
(−20)(0.006) + 1.95 × 10−5
=0
2
0.006
∴U = 18.5 m/s
1
U
⎡
(−20)( y 2 − 0.006 y ) +
⎢
−5
0.006
⎣ 2 × 1.95 × 10
⎤
y ⎥dy ∴U = −6.15 m/s
⎦
a) The solution for laminar flow between two parallel plates is given in
Eq. (7.4.13) as
u( y) =
(
)
1 d
U
( p + γ h ) y2 − ay + y
2 μ dx
a
For a horizontal flow dh/dx = 0, and hence the velocity profile is given by
u( y) =
7.60
(
)
1 dp 2
U
y − ay + y
2 μ dx
a
b) To determine the pressure gradient we set u = 0 at y = a/2, that is
0=
1 dp ⎛ a2 a2 ⎞ U
⎜ − ⎟+
2 μ dx ⎝ 4 2 ⎠ a
⎛a⎞
⎜ ⎟
⎝2⎠
Simplifying and solving for the pressure gradient we get
(
dp
= 4 μU a2 = 4 0.4 N ⋅ s/m2
dx
) ( 2 m/s ) ( 0.01m )
2
= 32 kPa/m
93
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Chapter 7 / Internal Flows
Note that since the pressure gradient is positive in the flow direction, it is
considered to be an adverse pressure gradient.
U
y
a
vθ =
7.62
∴τ = μ
dvθ
0.2 × 30
= 0.1×
= 750 Pa
dy
0.0008
T = F × R = τπ D × L × R = 750π × 0.4 × 0.8 × 0.2 = 151 N ⋅ m
Neglect the shear on the cylinder bottom; assume a linear velocity profile:
vθ =
7.64
U
0.1× 30
y=
y = 3000 y
a
0.001
∴τ = μ
dvθ
= 0.42 × 3000 = 1260 Pa
dy
∴T = F × R = π DL ×τ × R = π × 0.2 × 0.1×1260 × 0.1 = 7.9 N ⋅ m
Assume that all losses occur in the 8-m-long channel. The velocity through the
straws and screens is so low that the associated losses will be neglected. Assume
a developed flow in the channel:
V=
7.66
Re×ν 7000 ×1.5 × 10−5
=
= 8.75 m/s
0.012
h
Δp =
12 × 1.8 × 10 −5 × 8.75 × 8
= 105 Pa
0.012 2
Energy:
Δp
ΔpAV
= Δp m
=
W
( ρ AV ) =
fan
ρη
=
ρη
η
105 × 1.2 × 0.012 × 8.75
= 18.9 W
0.7
Laminar Flow Between Rotating Cylinders
Use Eq. 7.5.19:
T=
4πμ r12r22 Lω1
r22 − r12
=
4π × 0.035 × 0.022 × 0.032 × 0.4 × (3000 × 2π /60)
∴ T = 0.040 N ⋅ m
7.68
Re =
0.032 − 0.022
= Tω = 0.04 × (3000 × 2π /60) = 12.6 W
∴W
ω r1 (r2 − r1 ) ρ (3000 × 2π /60) × 0.02 × (0.01) × 917
=
= 1650
μ
0.035
∴Eq. 7.5.15 is OK
94
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Chapter 7 / Internal Flows
r22ω2 ⎛
r12 ⎞
d ⎛ vθ ⎞ 2 μ r12ω2
With ω1 = 0, Eq. 7.5.15 is vθ = 2 2 ⎜ r − ⎟ τ 2 = μ r ⎜ ⎟ = 2 2
r2 − r1 ⎝
r ⎠
dr ⎝ r ⎠ r2 − r1
7.70
2μr12ω2
4πμr12r22 Lω2
∴T2 = τ 2 A2r2 = 2 2 2π r2 Lr2 =
r2 − r1
r22 − r12
Turbulent Flow
Let u = u + u′, v = v + v′, w = w + w′. The continuity equation becomes
∂
∂
∂
∂u ∂v ∂w ∂u′ ∂v′ ∂w′
(u + u′) + ( v + v′) + (w + w′) =
+
+
+
+
+
∂x
∂y
∂z
∂x ∂y ∂z ∂x ∂y ∂z
Now, time-average the above equation recognizing that
7.72
∂u ∂u
=
and
∂x ∂x
∂u′ ∂
∂u ∂v ∂w
= u′ = 0. Then,
+
+
= 0. Substitute this back into the continuity
∂x ∂x
∂x ∂y ∂z
equation, so that
∂u′ ∂v′ ∂w′
+
+
=0
∂x ∂y ∂z
Use the fact that
7.74
∂
∂
u′v′ = u′v′. See Eq. 7.6.2. This is equivalent to
∂y
∂y
T
⎞ 1T ∂
∂ ⎛1
⎜ ∫ u′v′dt ⎟ = ∫ (u′v′)dt
⎟ T ∂y
∂y ⎜⎝ T 0
0
⎠
which is obviously correct. Also
⎛ ∂u′ ∂v′ ∂w′
∂
∂
∂
u′u′ + u′v′ + u′w′ = u′ ⎜
+
+
⎜ ∂x ∂y ∂z
∂x
∂y
∂z
⎝
⎞
∂u′
∂u′
∂u′
+ v′
+ w′
⎟ + u′
⎟
∂x
∂y
∂z
⎠
Time average both sides and obtain the result.
95
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Chapter 7 / Internal Flows
u = Σui /11 = 16.2 m/s
7.76
t
0
u′
−0.1
v = Σvi /11 = −1.6 m/s
0.01
0.02
0.03
9.5
−5.6
1.1
u′2
0.01
90.2
31.4
1.2
v′
3.2
−3.8
−7
5.1
v′ 2
10.2
14.4
49
u′v′
−0.32 −36.1
39.2
u′2 = 51.2 m 2 /s 2
η =−
26
5.6
0.05
0.06
−11
−6
0.9
121
36
0.81
5.7
−4.4
0.2
32.5
19.4
−62.7
26.4
0.07
v′ = v − v
0.08
0.09
0.1
12.4
−9.5
3
5.4
153.8
90.2
9
29.2
8.3
−3.6
−6.6
3.1
0.04
68.9
13.0
43.6
9.6
0.2
102.9
34.2
−19.8
16.7
u′v′ = 9.7 m2 /s2
v ′2 = 26.1 m 2 /s 2
u′v′
−2.52
=−
= 0.016 m2 /s
(23
18.2)
/
0.03
−
du / dy
u′v′
Kuv =
7.78
0.04
u′ = u − u
=
u′2 v′2
−2.52
= −0.125
29 14
A m = η / ∂u / ∂y = 0.016/(23 −18.2) / 0.03 = 0.01 m or 1 cm
b) Re =
7.80
0.2 × 0.2
10
−6
= 40,000,
e
= 0.0013
D
From the Moody diagram, this is in the transition zone where δ ν may be near, in
magnitude, to e. The pipe is rough.
b) With Re = 40,000, e /D = 0.0013 the Moody diagram gives f = 0.026. Then
0.026
uτ = 0.130/1000 = 0.01140 m/s
1000 × 0.2 2 = 0.130 Pa
τ0 =
8
7.82
Eq. 7.6.17:
V=
7.84
r
⎛
⎞
umax = uτ ⎜ 2.44 ln o + 8.5 ⎟
e
⎝
⎠
0.1
⎛
⎞
= 0.0114 ⎜ 2.44 ln
+ 8.5 ⎟ = 0.262 m/s
0.00026
⎝
⎠
Q 70 ×10−3
=
= 6.2 m/s
A π × (0.6)2
Re =
VD
ν
=
6.2 × 0.12
= 7.4 × 105
−6
1.007 ×10
From Table 7.1, n ≅ 8.5. From Eq. 7.6.20
umax =
(n + 1)(2n + 1)
9.5 ×18
×V =
× 6.2 = 7.34 m/s
2
2n
2 × 8.52
96
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
With n = 7, Eq. 7.6.21 gives f =
1
n
2
=
1
= 0.0204
49
y
1
1
∴τ 0 = ρ V 2 f = × 1000 × 102 × 0.0204 = 1020 Pa
8
2
Since τ varies linearly with r and is zero at r = 0
τ = τ0
7.86
τturb
C (r = 0)
τlam
τ
r 1020
r = 20,400 r
=
r0 0.05
τ lam = μ
⎡
∂u
1 1 −6 / 7 ⎤ 10−3 × 12.24 −6 / 7
= 10−3 ⎢umax
= 0.00268 y −6 / 7
y
y
⎥=
1/ 7
1/ 7
∂y
7 r0
⎣
⎦ 7 × 0.05
τturb =τ −τlam = 20,400r − 0.00268y−6/7 where y + r = 0.05
τ lam ( y ) is good away from y = 0 (the wall)
τ lam (0.00625) = 0.21, τ lam (0.003125) = 0.38
τ lam (0.00156) = 0.68, τ lam (0.00078) = 1.24
dp
2τ
2 ×1020
=− 0 =−
= −40,800 Pa/m
0.05
dx
r0
We must find uτ :
V=
Q
1.2
=
= 19.63 m/s
A π × 0.42
Re =
19.63 × 0.8
= 71,400
2.2 × 10−4
e 0.26
=
= 0.000325 ∴ Moody diagram ⇒ f = 0.021
D 800
7.88
τ0 =
1
1
fρV 2 = × 0.021 × 917 × 19.63 2 = 928 Pa
8
8
uτ = τ 0 /ρ = 928/917 = 1.006 m/s
1.006 × 0.4
⎡
⎤
∴ umax = 1.006 ⎢ 2.44 ln
+ 5.7 ⎥ = 24.2 m/s
−4
2.2 × 10
⎣
⎦
a) From a control volume of the 10-m section of pipe
7.90
Δp
π D2
4
= τ 0π DL
∴τ 0 =
0.12 × 5000
= 15 Pa
4 × 10
Assume n = 7. Then Eq. 7.6.21 gives f =
1
1
=
= 0.0204
2
n
49
97
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
From Eq. 7.3.19,
8τ 0
8 × 15
=
= 6.41 or V = 2.53 m/s
ρ f 917 × 0.0204
2.53 × 0.12
= 1381. This suggests laminar flow. Use Eq. 7.3.14:
2.2 × 10 −4
Check: Re =
V=
V2 =
r02 Δp
0.062 × 5000
=
= 1.12 m/s
8μ L 8 × (917 × 2.2 ×10−4 ) ×10
Check: Re =
1.12 × 0.12
= 611
2.2 × 10−4
OK
Finally, Q = AV = π × 0.06 2 × 1.12 = 0.0127 m 3 /s
Turbulent Flow in Pipes and Conduits
a) Re =
VD
ν
=
(0.020 / π × 0.042 ) × 0.08
= 3.18 × 105
−6
10
e
= 0 ∴ f = 0.0143
D
b) Eq. 7.6.26 provides f by trial-and-error. Try f = 0.0143 from Moody’s
diagram:
(
7.92
)
1
= 0.86 ln 3.18 ×105 0.0143 − 0.8 ∴ f = 0.0146
f
Another iteration may be recommended but this is quite close. The value for f is
essentially the same using either method. The equations could be programmed
on a computer.
a) Re =
0.025 × 0.04
64ν
64 × 10−6
=
1000
∴
laminar
∴
f
=
=
= 0.064
VD 0.025 × 0.04
10−6
c) Re =
2.5× 0.04
= 100,000
10−6
7.94
V=
Q 1.65 × 10−3
VD 1.45 × 0.038
=
= 1.45 m/s Re =
=
= 4.8 × 104
2
−6
ν
A π (0.019)
1.14 × 10
hL = f
7.96
e 0.26
=
= 0.0065 ∴ f = 0.034
D 40
L V2
180 1.452
= f
= 508 f
D 2g
0.038 19.62
a)
e 0.00026
=
= 0.0068
D
0.038
c)
e 0.000046
=
= 0.0012
D
0.038
∴ f = 0.035
∴ f = 0.0245
∴ hL = 508 × 0.035 = 17.8 m
∴ hL = 508 × 0.0245 = 12.4 m
98
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
V=
Q
0.08
=
= 4.73 m/s
A π × 0.0752
Δp = γ f
a) Re =
7.98
e 0.15
=
= 0.001
D 150
L V2
10
4.732
= (917 × 9.81) f
×
= 6.84 ×105 f
D 2g
0.15 2 × 9.81
4.73 × 0.15
64
= 322 ∴ laminar and Δp = 6.84 × 105 ×
= 1.36 × 105 Pa
0.0022
Re
Eq. 7.6.29 is not applicable to a laminar flow
c) Re =
4.73 × 0.15
= 16,300
0.000044
∴ f = 0.029
Δp = 6.84 × 105 × 0.029 = 20,000 Pa
0.9
⎧
⎛ 4.4 × 10−5 × 0.15 ⎞ ⎤ ⎫⎪
0.08 × 10 ⎪ ⎡ 0.15
⎢
Δp = 9.8 × 917 ×1.07
+ 4.62 ⎜
⎨ln
⎟ ⎥⎬
0.08
9.8 × 0.155 ⎪ ⎢ 3.7 × 50
⎝
⎠ ⎥⎦ ⎭⎪
⎩ ⎣
= 20,700 Pa
−2
2
V=
7.100
Q 8.3 × 10−3
=
= 2.94 m/s
A π (0.03) 2
L V2
∴ f = 0.0167 ∴Δp − γΔh = γ f
D 2g
∴ Δp = 9810 × 0.0167
V=
7.102
2.94 × 0.06
= 1.3 × 105
1.31× 10−6
e
=0
D
(Recall: Δp = p1 − p 2 , Δh = h2 − h1 )
90 2.942
×
+ 9810 × 90sin 30D = 550 × 103 Pa. = 550 kPa
0.06 19.62
5
Q
=
= 9.95 m/s
A π × 0.42
Re =
9.95 × 0.8
= 8 × 106
−6
10
e 1.6
=
= 0.002 (using an average “e” value) ∴ f = 0.0237
D 800
∴Δp = γ f
L V2
100 9.952
= 9810 × 0.0237 ×
×
= 147,000 Pa
D 2g
0.8 2 × 9.81
Use Eq. 7.6.30: hL =
7.104
Re =
Δp
γ
=
200,000
= 20.4 m ν = 10−6 m2/s
9810
⎡ 9.81× 0.045 × 20.4 ⎤
b) Q = −0.965 ⎢
⎥
100
⎣
⎦
0.5
⎡ 0.046 ⎛ 3.17 ×10−12 × 100 ⎞0.5 ⎤
ln ⎢
+⎜
⎟ ⎥
⎢ 3.7 × 40 ⎝ 9.81× 0.043 × 20.4 ⎠ ⎥
⎣
⎦
= 0.0033 m3 /s
99
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
Use Eq. 7.6.30:
p
200
=
= 3.38 kg/m3
RT 0.189 × 313
b) ρ =
7.106
e
0.046
=
= 0.000104
3.7 D 3.7 ×120
= ρQ
m
ν=
∴ hL =
400
= 12.1 m
9.81× 3.38
μ 1.3 ×10−5
=
= 3.8 ×10−6 m2 /s
ρ
3.38
⎡ 9.81 × 0.125 × 12.1 ⎤
Q = − 0.965 ⎢
⎥
400
⎣
⎦
0.5
0.5
⎡
⎛ 3.17 × 3.82 × 10−12 × 400 ⎞ ⎤
ln ⎢ 0.000104 + ⎜
⎟ ⎥
9.81 × 0.123 × 12.1 ⎠ ⎥
⎢
⎝
⎣
⎦
= 0.0205
= 0.069 kg/s
∴m
5.2 ⎤
2 4.75
⎡
1.25 ⎛ LQ ⎞
9.4 ⎛ L ⎞
⎢
Use Eq. 7.6.31: e = 0 D = 0.66 e ⎜
⎟ +ν Q ⎜
⎟ ⎥
gh
gh
⎢
⎝ L ⎠ ⎥⎦
⎝ L ⎠
⎣
a) hL =
Δp
γ
=
200,000
= 20.4 m
9810
ν = 10−6 m2 /s
5.2
⎡ −6
100
⎞ ⎤
9.4 ⎛
D = 0.66 ⎢10 × 0.002 ⎜
⎟ ⎥
⎝ 9.81× 20.4 ⎠ ⎦⎥
⎣⎢
7.108
c) hL =
Δp
γ
=
200,000
= 25.2 m
7930
0.04
= 0.032 m
ν = 2.1× 10−6 m 2 /s
5.2
⎡
⎛ 100 ⎞ ⎤
D = 0.66 ⎢2.1×10−6 × 0.0029.4 ⎜
⎟ ⎥
⎝ 9.81× 25.2 ⎠ ⎦⎥
⎣⎢
a) V =
0.04
0.4/60
L V2
2
0.00849/
D
,
h
f
=
=
L
D 2g
π D 2 /4
0.04
or 3 = f
= 0.031 m
1200 (0.00849/D 2 ) 2
or
D
2 × 9.8
f = 680 D5
7.110
Guess: f = 0.02. Then the above equation gives D = 0.124 m
Check: V = 0.551 m/s, Re =
0.551× 0.25
e 0.0015
= 1.05 × 105 ,
=
= 1.2 × 10−5
−6
124
D
1.31× 10
∴ f = 0.0175 . Use f = 0.0175 and D = 0.121 m
b) Use Eq. 7.6.31:
100
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Chapter 7 / Internal Flows
9.4
5.2 ⎤
2 4.75
⎡
⎛ 1200 ⎞ ⎥
−5 1.25 ⎛ 1200 × (0.4/60) ⎞
−6 ⎛ 0.4 ⎞
D = 0.66 ⎢ (1.5 × 10 ) ⎜
1.31
10
+
×
×
⎟
⎜
⎟ ⎜
⎟
9.8 × 3
⎢
⎝ 60 ⎠ ⎝ 9.8 × 3 ⎠ ⎥
⎝
⎠
⎣
⎦
0.04
= 0.127 m
Exiting K.E. =
V 2 0.8922
=
= 0.041 m
2 g 2 × 9.8
∴ negligible
Use Eq. 7.6.30:
80
0.02 × 0.04
, ν = 10−6 m 2 /s, use D = 4 R = 4
= 0.027 m
9810
2(0.02 + 0.04)
0.5
⎡ 0.0015 ⎛ 3.17 × 10−12 × 2 × 9810 ⎞0.5 ⎤
⎡ 9.81× 0.0275 80 ⎤
×
+⎜
Q = −0.965 ⎢
⎟ ⎥
⎥ ln ⎢
3
×
2
9810
3.7
27
×
×
9.81
0.027
80
⎢
⎣
⎦
⎝
⎠ ⎥⎦
⎣
hL =
7.112
= 0.000143 m3 /s
b) R =
7.114
A
0.6 × 1.2
=
= 0.3 m, hL = 10,000 × 0.0015 = 15 m (from energy eq.)
Pwet
2.4
⎡ 9.8 × 1.25 × 15 ⎤
Q = −0.965 ⎢
⎥
⎣ 10,000 ⎦
0.5
0.5
⎡ 1.5
⎛ 3.17 × 10 −12 × 10,000 ⎞ ⎤
3
+⎜
ln ⎢
⎟ ⎥ = 0.257 m /s
9.8 × 1.23 × 15
⎢ 3.7 × 1200 ⎝
⎥
⎠ ⎦
⎣
Minor Losses
7.116
V2
Referring to the equation Δp = − ρ
Δn (n is in the direction
R
of the center of curvature), we observe that Δp is negative all
along the line CB from C to B in Fig. P7.115. Hence, the
pressure decreases from C to B with pC > pB. Using Bernoulli’s p B
equation we see that VB > VC. Fluid moves from the high
pressure region at the outside of the bend toward the low
pressure region at the outside of the bend creating a secondary
flow.
a) V1 = 63.7, V2 = 15.9,
7.118
7.120
Energy: 0 =
V=
pC
ρ1 = 1.78 kg/m3
15.92 − 63.7 2 p2 − 50,000
(63.7 − 15.9)2
+
+ 0.4
2 × 9.81
1.78 × 9.81
2 × 9.81
∴ p2 = 52,600 Pa
Q 3.3 ×10−3
L V2
Neglect
pipe
friction,
i.e.,
f
≅0
=
=
1.6
m/s
D 2g
A π (0.025)2
101
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
Energy: 0 =
1.682 − 02
1.682
+ 0 − 1.8 + (K + 0.03)
2 × 9.81
2 × 9.81
∴ K = 11.48
Simple Piping Systems
Assume completely turbulent regime:
e 0.046
=
= 0.0012
D
40
Energy:
7.122
0=
∴ f = 0.0205, Re > 106
2
V2
⎡ L
⎤V
− 40 + ⎢ f + 2Kelbow + Kentrance ⎥
2g
⎣ D
⎦ 2g
2
110
⎡
⎤ V
40 = ⎢1 + 0.0205 ×
+ 2 × 1.0 + 0.5⎥
0.04
⎣
⎦ 2 × 9.81
∴ V = 3.62 m/s Re =
∴ V = 3.50 m/s
(Assume a screwed elbow)
3.62 × 0.04
= 1.4 ×105
−6
10
∴ Try f = 0.022
∴Q = AV = π × 0.022 × 3.5 = 0.0044 m3/s
The EGL and HGL have sudden drops at the elbows, and a gradual slope over
the pipe length.
Assume screwed elbows: 0 =
V2
L ⎞V2
⎛
− 2 + ⎜ Kentrance + 2K elbows + f ⎟
2g
D ⎠ 2g
⎝
b) Assume
7.124
26 ⎞ V 2
⎛
f = 0.018 : 2 = ⎜ 0.8 + 1 + 2 × 0.8 + 0.018
⎟
0.08 ⎠ 2 × 9.81
⎝
∴ Re =
2.06 × 0.08
1.14 × 10
−6
= 1.5 × 105
∴ V = 2.06 m/s
f = 0.0164
∴V = 2.12 Q = π × 0.042 × 2.12 = 0.011 m3/s
Assume constant water level, neglect the velocity head at the pipe exit and let
p exit = 0. Then the energy equation says hL = 0.8 m:
0.8 = f
7.126
V2
V2
10
+ 2 × 1.5
0.025 2 × 9.8
2 × 9.8
Try f = 0.02. Then V = 0.319 m/s Re =
or
0.8 = 400 fV 2 + 0.153V 2
0.313 × 0.025
= 1.1× 104 ∴ f = 0.029
−6
0.73 ×10
Try f = 0.029 Then V = 0.262 m/s
∴ Q = AV = π × 0.01252 × 0.262 = 1.28 × 10−4 m3 /s
102
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
Finally
7.128
Δt =
−
V
10
=
= 78.1 sec or 1.3 min
Q 0.128
V32 p0
V2
1.2 V12
1.2 V22
−
− 1.2 + 0.8 1 + f1
+ f2
2g γ
2g
0.008 2 g
0.005 2 g
Energy:
0=
Continuity:
V3 =
Assume
f1 = f 2 = 0.02 Then energy says V1 = 2.09 m/s
Check:
Re1 =
Finally
V3 = 16V1 = 46.1 m/s
Energy:
0 = −3 + f
Assume turbulent:
Re =
Check:
25
64
V2 , V2 = V1
4
25
2.09 × 0.008
10−6
= 16,700
∴ f = 0.026 Then V1 = 2.88 m/s
300 V 2
0.0094 2 × 9.8
f = 0.04 Then V = 0.215 m/s
0.215 × 0.0094
10−6
= 2020 ∴ marginally laminar
Assume laminar flow with f = 64/ Re. Then
7.130
3=
64 ×10−6
300
V2
×
×
V × 0.0094 0.0094 2 × 9.8
Check:
Re =
0.271× 0.0094
10−6
∴ V = 0.271 m/s
= 2540 ∴turbulent
The flow is neither laminar nor turbulent but may oscillate between the two. The
wall friction is too low in the laminar state so it speeds up and becomes
turbulent. The wall friction is too high in the turbulent state so it slows down
and becomes laminar, etc., etc.
Q
0.01
=
= 7.96 m/s
A π × 0.022
e 0.0015
=
= 3.8 × 10−5
D
40
V=
7.132
Re =
7.96 × 0.04
1.14 × 10
−6
= 2.8 × 105
∴ f = 0.0145
800 ⎞ 7.962
⎛
H P = 80 − 10 + ⎜ 0.5 + 1.0 + 0.0145
= 1010 m.
⎟
0.04 ⎠ 2 × 9.81
⎝
= γ QH P = 9810 × 0.01×1010 = 117,000 W
∴W
P
ηP
0.85
103
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
0=
7.962 1670 − 100,000
L ⎞ 7.962
⎛
+
+ 0 − 10 + ⎜ 0.5 + 0.0145
∴ L = 13.0 m
⎟
2 × 9.81
9810
0.04 ⎠ 2 × 9.81
⎝
V12 690×103 V22 (4V1)2
+
= =
Energy across nozzle (neglect losses):
2g
9810
2g
2g
∴ V1 = 9.6 m/s
Re =
9.6 × 0.05
1.007 × 10
−6
= 4.8 × 105
e 0.000045
=
= 0.0009
D
0.05
∴ f = 0.02
7.134
HP =
(4 × 9.62 )2
360 ⎞ 9.62
⎛
− 18 + ⎜ 0.5 + 0.02
= 677.2 m
⎟
19.62
0.05 ⎠ 19.62
⎝
= γ QH P = 9810 × (π × ( 0.025 ) × 9.6) × 677.2 =166.97 ×103 W or 166.97 kW
∴W
P
ηP
0.75
2
0=
9.62 (2.3 − 100)103
L ⎞ 9.62
⎛
+
− 18 + ⎜ 0.5 + 0.02
∴ L = 11.1 m
⎟
19.62
9810
0.05 ⎠ 19.62
⎝
e
= 0.0013 ∴ f = 0.021 Q = 0.0314V
D
300 ⎞ V 2
⎛
H P = −20 + ⎜ 0.5 + 1.0 + 0.021
= −20 + 1700Q 2
⎟
0.2 ⎠ 2 g
⎝
7.136
Try Q = 0.23 m3 /s: ( H P )E = 70 m
( H P )C = 70 m
= 9810 × 0.23 × 70 = 190,000 W
∴W
P
0.83
EGL
HGL
Re = 1.5 × 106 ∴OK
e 0.046
=
= 0.00029 ∴ f = 0.015 Q = π × 0.082 V = 0.0201V
D 160
50 ⎞
Q2
⎛
H P = 25 + ⎜ 0.5 + 2 × 0.4 + 1 + 0.015
= 25 + 882Q 2
⎟
2
0.16
⎝
⎠ 2 g × 0.0201
Try Q = 0.23 m3/s (H P )E = 72 m ( H P )C. = 70 m
7.138
= 9810 × 0.23 × 72 = 195,000 W
a) ∴ W
P
0.83
b) 0 =
p
11.442
10 ⎞ 11.442
⎛
+ 2 − 8 + ⎜ 0.5 + 0.015
∴ pin = −81,000 Pa
⎟
2 × 9.81 9810
0.16 ⎠ 2 × 9.81
⎝
c) 72 =
p
11.442
10 ⎞ 11.442
⎛
+ 3 − 8 + ⎜ 0.5 + 0.015
∴ pout = 625,000 Pa
⎟
2 × 9.81 9810
0.16 ⎠ 2 × 9.81
⎝
104
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Chapter 7 / Internal Flows
d)
EGL
HGL
e 1.65
=
= 0.0014 ∴ f = 0.021 Q = π × 0.62 V = 1.13V
D 1200
1000 ⎞
Q2
⎛
− HT = −60 + ⎜ 0.8 + 1 + 0.021
∴ HT = 60 − 0.77Q 2
⎟
2
1.2
⎝
⎠ 1.13 × 2 × 9.81
7.140
= γ QH η = 9810 × (60Q − 0.77Q3 ) × 0.9 = 530,000Q − 6800Q3
W
T
T T
= 500,000 Q (power in watts)
But, the characteristic curve is W
T
Hence, 500,000Q = 530,000Q − 6800Q3
∴Q = 2.1 m3/s
= 1.05 MW
∴W
T
Open-Channel Flow
τ w Aw = W sin θ
7.142
sin θ = 0.001
L
τwAw
or τ w × w ×1 = ( w × d ×1) ρ w g sin θ
τ w = ρ w dg sin θ = 1000 ×1.8 × 9.81× 0.001
W
θ
= 17.6 Pa
a) Q =
1.0
1.0
⎛ 2 × 0.6 ⎞
AR2 / 3S1/ 2 =
× (2 × 0.6) ⎜
⎟
n
0.012
⎝ 2 + 2 × 0.6 ⎠
2/3
× 0.0011/ 2 = 1.64 m3/s
b) Planed wood and finished concrete have the same “n” (i.e., roughness)
7.144
∴Use a value for “e” which is relatively small, say e = 0.6 mm. The solution
is not too sensitive to this choice. Then, with R = 0.375 m
e
e
0.6
=
=
= 0.0004
D 4 R 4 × 375
∴ f = 0.016
105
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 7 / Internal Flows
hL = f
L V2
= LS
D 2g
∴V 2 =
2 × 9.81× 0.001× (4 × 0.375)
0.016
∴V = 1.36 m/s
1.36 × (4 × 0.375)
⎛
⎞
∴ Q = 1.36 × 2 × 0.6 = 1.64 m 3 /s ⎜ Check: Re =
= 2.0 × 106 ∴ OK ⎟
−6
10
⎝
⎠
a) R = 1.2 × 0.5 + 0.5 × 0.5 = 0.325 m ∴ Q = 1.0 × 0.85 × 0.3252 / 3 × 0.0011/ 2 = 0.794 m3 /s
1.2 + 2 × 0.5/0.707
V=
7.146
0.016
Q 0.794
=
= 0.934 m/s
A 0.85
b) Use an “e” for rough concrete, e = 3 mm
e
3
∴ =
= 0.0023 ∴ f = 0.025
D 4 × 325
hL = LS = f
L V2
4R 2g
∴V 2 =
4 × 0.325 × 2 × 9.81× 0.001
and V = 1.01 m/s
0.025
∴Q = 1.01× 0.85 = 0.86 m3/s
Assume y > 3. R =
30 + 20( y − 3) 20 y − 30 10 y − 15
=
=
26 + 2( y − 3)
2 y + 20
y + 10
⎛ 10 y − 15 ⎞
1.0
100 =
(20 y − 30) ⎜
⎟
0.022
⎝ y + 10 ⎠
7.148
Or
⎛ 10 y − 15 ⎞
6.96 = (2 y − 3) ⎜
⎟
⎝ y + 10 ⎠
Try y = 5: 6.96 =? 12.3
2/3
× 0.0011/2
2/3
y
y = 4: 6.96 =? 7.36
⎫
y = 3.8: 6.96 =? 6.47 ⎪
⎪
⎬ ∴ y = 3.91 m
?
y = 3.9: 6.96 = 6.91⎪
⎪⎭
106
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Chapter 7 / Internal Flows
2/3
2⎞
⎛π
1 π
2 2 × 0.6
⎟ × 0.0011/2
Try y = 0.6 m : Q =
× × (0.6) ⎜
⎜ 0.6π ⎟
0.013 2
⎝
⎠
y − 2 y − 0.6
= 0.62 m3/s cosα =
=
2
0.6
∴ y > 0.6 m. R =
7.150
A = 2π (0.6)
Q=
A
2π (0.6) × α /180
180 − α
+ ( y − 0.6)0.6 sin α
180
α
0.6
1
AR2 / 3 × 0.0011/ 2 = 0.66 ∴ AR2 / 3 = 0.27
0.013
⎫
Try y = 0.63: α = 87.13D , A = 0.6012 m 2 , R = 0.33 m. 0.287 =? 0.27 ⎪
⎪
⎬ y = 0.612 m
D
2
?
Try y = 0.612: α = 88.85 , A = 0.58 m , R = 0.312 m. 0.267 = 0.27 ⎪
⎪⎭
107
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y
Chapter 7 / Internal Flows
108
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Chapter 8 / External Flows
CHAPTER 8
External Flows
FE-type Exam Review Problems: Problems 8-1 to 8-8
8.1 (C)
8.2 (C)
8.3 (B)
Re =
VD
ν
=
0.8 × 0.008
1.31× 10−6
= 4880
Assume a large Reynolds number so that CD = 0.2. Then
8.4 (B)
2
F=
1
1
⎛ 80 ×1000 ⎞
2
ρV 2 ACD = ×1.23 × ⎜
⎟ × π × 5 × 0.2 = 4770 N
2
2
⎝ 3600 ⎠
Assume a Reynolds number of 105. Then CD = 1.2
8.5 (D)
8.6 (C)
F=
1
1
ρV 2 ACD ∴ 60 = × 1.23 × 402 × 4 × D × 1.2 ∴ D = 0.0041 m
2
2
Re =
VD
Re =
VD
ν
ν
=
=
40 × 0.0041
10
−6
4 × 0.02
1.6 × 10
−5
= 1.64 ×105 ∴ CD = 1.2 The assumption was OK
= 5000 ∴ St = 0.21 =
∴ f = 42 Hz (cycles/second) distance =
8.7 (C)
fD f × 0.02
=
V
4
V
4 m/s
=
= 0.095 m/cycle
f 42 cycles/s
By reducing the separated flow area, the pressure in that area increases thereby
reducing that part of the drag due to pressure.
From Fig. 8.12a, C L = 1.1 C L =
8.8 (B)
∴V 2 =
1
2
FL
ρ V 2cL
2W
2 ×1200 × 9.81
=
= 1088 and V = 33.0 m/s
ρ cLCL 1.23 ×16 ×1.1
109
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Chapter 8 / External Flows
Separated Flows
Re = 5 =
8.10
VD
5 × 1.51× 10−5
= 3.78 × 10−5 m
20
∴D =
ν
inviscid
flow
inviscid
flow
no separation
separated
region
boundary layer
near surface
viscous flow
near sphere
boundary
layer
separated
region
8.12
building
wake
inviscid
flow
Re =
VD
8.14
ν
=
20 × D
1.51×10
−5
= 13.25 ×105 D
b) Re = 13.25 ×105 × 0.06 = 7.9 ×104
∴Separated flow
Ftotal = Fbottom + Ftop = 20,000 × 0.3 × 0.3+10,000 × 0.3 × 0.3 = 2700 N
Flift = 2700 cos 10° = 2659 N
Fdrag = 2700 sin 10° = 469 N
8.16
CL =
CD =
FL
1
2
ρV A
2
FD
1
2
ρV A
2
=
=
2659
1
2
× 1000 × 5 × 0.3 × 0.3
2
469
1
2
× 1000 × 5 × 0.3 × 0.3
2
= 2.36
= 0.417
If C D = 1.0 for a sphere, Re = 100 (see Fig. 8.8).
∴
8.18
V × 0.1
ν
= 100, V =1000ν
b) V = 1000 ×
1.46 ×10−5
= 0.798 m/s
0.015 × 1.22
1
∴ FD = × (0.015 ×1.22) × 0.7982 π × 0.052 × 1.0
2
110
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Chapter 8 / External Flows
= 4.58 ×10−5 N
The velocities associated with the two Reynolds numbers are
V1 =
8.20
Re1ν 3 × 105 ×1.5 × 10−5
=
= 101 m/s
0.0445
D
Re 2 ν 6 × 104 ×1.5 ×10−5
=
= 20 m/s
V2 =
0.0445
D
The drag, between these two velocities, is reduced by a factor of 2.5, i.e.,
(CD )high = 0.5 and (CD )low = 0.2. Thus, between 20 m/s and 100 m/s, the drag is
reduced by a factor of 2.5. This would significantly lengthen the flight of the
ball.
8.22
1
V × 0.2
4.2 = × 1000V 2π × 0.12 C D ∴ V 2C D = 0.267 Re =
= 2 × 105 V
−
6
2
10
Try CD = 0.5 :
a) Re1 =
25 × 0.05
1.08 ×10
−5
∴ V = 0.73 m/s Re = 1.46 ×105 ∴ OK
= 1.2 ×105
Re2 = 1.8 × 105
Re3 = 2.4 × 105
Assume a rough cylinder (the air is highly turbulent).
∴ ( CD )1 = 0.7, ( CD )2 = 0.8, ( CD )3 = 0.9
8.24
1
∴ FD = × 1.45 × 252 (0.05 × 10 × 0.7 + 0.075 × 15 × 0.8 + 0.1× 20 × 0.9) = 1380 N
2
1
M = ×1.45 × 252 (0.05 ×10 × 0.7 × 40 + 0.075 ×15 × 0.8 × 27.5 + 0.1× 20 × 0.9 × 10)
2
= 25,700 N ⋅ m
Since the air cannot flow around the bottom, we imagine the structure to be
mirrored as shown. Then
L / D = 40/5 = 8
8.26
Remin =
VDmin
∴CD = 0.66CD∞
=
30 × 2
1.5 × 10−5
∴ C D = 1.0 × 0.66 = 0.66
ν
= 4 × 106
1
⎛ 2+8
⎞
∴ FD = × 1.22 × 302 × ⎜
× 20 ⎟ × 0.66 = 36,000 N
2
⎝ 2
⎠
111
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Chapter 8 / External Flows
8.28
4
1
4
FB + FD = FW 1.22 × π (0.25)3 + ×1.22V 2π (0.25)2 CD = 9810S π (0.25)3
3
2
3
V × 0.5
Re =
= 3.4 ×104 V
∴1 + 0.153V 2C D = 820S
−5
1.47 ×10
a) S = 0.005
V 2C D = 20.26 Assume atmospheric turbulence, i.e., rough
Try C D = 0.4 : V = 7.1 m/s Re = 2.4 ×105 ∴ C D = 0.3 and V = 8.22 m/s
From Table 8.2 CD = 0.35
8.30
1
FD = ×1.22V 2 × 3.2 × 0.35 = 0.683V 2
2
2
80 ×1000
⎛ 80 ×1000 ⎞
a) FD = 0.683 × ⎜
= 7500 W or 10 hp
⎟ = 337 N ∴ W = 337
3600
⎝ 3600 ⎠
Re =
8.32
VD
ν
a) FD =
=
(40,000/3600)0.6
= 4.42 × 105 ∴ C D = 0.35 from Fig. 8.8.
−5
1.51× 10
1
1
ρ V 2 AC D = ×1.204 × (40, 000/3600) 2 × 0.6 × 6 × 0.35 = 93.6 N
2
2
With the deflector the drag coefficient is 0.76 rather than 0.96. The required
power, directly related to fuel consumed, is reduced by the ratio of 0.76/0.96.
The cost per year without the deflector is
Cost = (200,000/1.2) × 0.25 = $41,667
8.34
With the deflector it is
Cost = 41,667 × 0.76/0.96 = $32,986
The savings is $41,667 − 32,986 = $8,800
8.36
FD =
1
1
ρ V 2 ACD = ×1.22 × (27.8 ×1.6)2 × π × 0.052 ×1.1 = 10.43 N
2
2
= F × V × 2 = 10.43 × (27.8 ×1.6) × 2 = 226 W or 1.24 hp
W
D
The net force acting up is (use absolute pressure)
4
4
120
Fup = π × 0.43 × 1.21 × 9.8 − 0.5 − π × 0.43
× 9.8 = 2.16 N
3
3
2.077 × 293
8.38
From a force triangle (2.16 N up and FD to the right), we see that tan α = Fup /FD
a) FD = 2.16 / tan 80° = 0.381. Assume CD = 0.2:
1
0.381 = × 1.21V 2π × 0.42 × 0.2
2
∴ V = 2.50 m/s
112
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Chapter 8 / External Flows
Check Re: Re =
2.5 × 0.8
= 1.33×105
−5
1.51×10
Too low. Use C D = 0.5:
1
0.381 = × 1.21V 2π × 0.42 × 0.5
2
c) FD = 2.16 / tan 60° = 1.25.
Assume CD = 0.5:
1
1.25 = × 1.21V 2π × 0.42 × 0.5
2
Check Re: Re =
∴ V = 1.58 m/s
∴ V = 2.86 m/s
2.86 × 0.8
= 1.5 ×105
1.51×10−5
∴OK
Power to move the sign:
FDV =
1
ρ V 2 ACD × V
2
1
= × 1.21× 11.112 × 0.72 × 1.1× 11.11 = 657 J/s.
2
8.40
This power comes from the engine:
657 = (12,000 × 1000)m × 0.3
∴ m = 1.825 × 10−4 kg/s
Assuming the density of gas to be 900 kg/m3
1825
.
× 10 − 4 × 10 × 3600 × 6 × 52 ×
8.42
1000
× 0.30 = $683
900
= 40 × 746η = F × V = 1 ρ V 2 AC × V = 1 ρ AC V 3
W
D
D
D
2
2
1
∴ 40 × 746 × 0.9 = × 1.22 × 3 × 0.35V 3 ∴ V = 34.7 m/s or 125 km/hr
2
Vortex Shedding
40 >
8.44
VD
ν
=
1.8D
1.141×10
10,000 <
VD
ν
−6
=
∴ D < 2.5 ×10−5 m
1.8 D
1.141×10−6
∴ D > 0.0063 m or 0.63 m
113
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Chapter 8 / External Flows
St =
8.46
fD 0.002 × 2
=
V
V
Re =
VD
ν
=
Try St = 0.21: V = 0.0191 m/s
V ×2
10−6
Use Fig. 8.9
Re = 38 ×103
∴ OK
Streamlining
Re =
8.48
25 × 0.15
1.5 × 10
−5
1
= 2.5 × 105 FD = × 1.225 × 252 × 1.0 × 0.8 × (1.8 × 0.5 ) = 82.7 N
2
The coefficient 1.0 comes from Fig. 8.8 and 0.8 from Table 8.1. We have
= F × V = 82.7 × 25 = 2067.5 W or
W
D
(C D )streamlined = 0.035
Re =
VD
ν
=
2 × 0.8
10
−6
L
4
=
=5
D 0.8
8.50
∴ FD = 2.9 N
= 1.6 × 106
2.77 kW
= 72.5 W or 0.72 kW. = 0.1 hp
W
∴ C D = 0.45 from Fig. 8.8.
∴ C D = 0.62 × 0.45 = 0.28
Because only one end is free, we double the length.
FD =
1
1
ρ V 2 AC D = ×1000 × 22 × 0.8 × 2 × 0.28 = 900 N
2
2
If streamlined, CD = 0.03 × 0.62 = 0.0186
1
∴ FD = ×1000 × 22 × 0.8 × 2 × 0.0186 = 60 N
2
V = 50 ×1000/3600 = 13.9 m/s
Re =
13.9 × 0.3
= 2.8 × 105 ∴ C D = 0.4
−5
1.5 × 10
We assumed a head diameter of 0.3 m and used the rough sphere curve.
8.52
FD =
1
1
ρ V 2 AC D = × 1.2 × 13.92 (π × 0.32 /4) × 0.4 = 3.3 N
2
2
FD =
1
1
ρ V 2 ACD = ×1.2 ×13.92 (π × 0.32 /4) × 0.035 = 0.29 N
2
2
114
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Chapter 8 / External Flows
Cavitation
CL =
8.54
FL
1
2
ρV A
2
=
200,000
1
× 1000 × 122 × 0.4 × 10
2
CD = 0.0165 =
?
σ crit = 0.75 >
∴α ≅ 3D
= 0.69
FD
1
×1000 ×122 × 0.4 ×10
2
∴ FD = 4800 N
(9810 × 0.4 + 101,000) − 1670
1
×1000 ×122
2
= 1.43
p∞ = 9810 × 5 + 101,000 = 150,000 Pa pv = 1670 Pa Re =
σ=
8.56
150,000 − 1670
1
×1000 × 202
2
∴ FD =
= 0.74
∴ no cavitation
20 × 0.8
10
−6
= 16 × 106
∴ C D = C D (0)(1 + σ ) = 0.3(1 + 0.74) = 0.52
1
1
ρ V 2 AC D = ×1000 × 202 × π × 0.42 × 0.52 = 52,000 N
2
2
Note: We retain 2 sig. figures since C D is known to only 2 sig. figures.
Added Mass
ΣF = ma
8.58
4
400
a) 400 − 9810 × π × 0.23 =
a
3
9.81
∴ a = 1.75 m/s 2
4
4
⎛ 400 1
⎞
b) 400 − 9810 × π × 0.23 = ⎜
+ × 1000 × π × 0.23 ⎟ a
3
3
⎝ 9.81 2
⎠
∴ a = 1.24 m/s 2
Lift and Drag on Airfoils
The total aerodynamic drag consists of both lift and drag that is:
FTotal = FL + FD ⇒ FTotal = FL2 + FD2 = 18 kN ⇒ FL2 + FD2 = (18 kN )
8.60
2
which can be combined with FL = 3FD to yield
9 FD2 + FD2 = 324 ⇒ FD = 324 10 = 5.69 kN
and hence, FL = 3 × 5.69 = 17.1 kN so
115
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Chapter 8 / External Flows
CL =
CL =
8.62
FL
1
2
ρ V 2cL
FL
1
2
ρV A
2
=
=
17.1×103
1 × 1.2 × 61.12 × 1.3 ×10
2
1000 × 9.81
1
× 0.412 × 802 × 15
2
= 0.496
= 0.587
∴α = 3.2D C D = 0.0065
= F V = ⎛ 1 × 0.412 × 802 × 15 × 0.0065 ⎞ × 80 = 10,300 W or 13.8 hp
W
D
⎜
⎟
⎝2
⎠
8.64
C L = 1.22 =
8.66
C L = 1.22 =
1500 × 9.81 + 3000
1
×1.007 × V 2 × 20
2
1500 × 9.81 + 9000
1
× 1.22 × V 2 × 20
2
a) C L = 1.72 =
8.68
∴ V = 38.0 m/s
∴ V = 39.9 m/s
250,000 × 9.81
∴ V = 75.2 m/s
1
×1.05 × V 2 × 60 × 8
2
% change =
75.2 − 69.8
× 100 = 7.77% increase
69.8
Vorticity, Velocity Potential, and Stream Function
∇p
⎡∂ V
⎤
∇× ⎢
+ (V ⋅∇)V +
− ν∇ 2 V ⎥ = 0
ρ
⎣ ∂t
⎦
∇×
∂V ∂
∂ω
= (∇ × V ) =
∂t ∂t
∂t
∇×
∇p
ρ
=
1
ρ
∇ × ∇p = 0
∇ × (∇ 2 V ) = ∇ 2 (∇ × V ) = ∇ 2ω (we have interchanged derivatives)
8.70
⎡1
⎤ 1
∇ × [ (V ⋅∇ )V ] = ∇ × ⎢ ∇V 2 − V × (∇ × V ) ⎥ = (∇ × ∇V 2 ) − ∇ × (V × ω)
⎣2
⎦ 2
= V (∇ ⋅ ω) − ω (∇ ⋅ V ) + (V ⋅∇)ω − (ω ⋅∇)V
= ( V ⋅ ∇ )ω − (ω ⋅ ∇ ) V
since ∇ ⋅ ω = ∇ ⋅ (∇ × V ) = 0 and ∇ ⋅ V = 0
There results:
∂ω
+ (V ⋅∇)ω − (ω ⋅∇)V −ν∇ 2ω = 0
∂t
This is written as
Dω
= (ω ⋅∇)V + ν∇ 2ω
Dt
116
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Chapter 8 / External Flows
x-comp:
8.72
y-comp:
z-comp:
∂ωx
∂ω
∂ω
∂ω
∂u
∂u
∂u
+ u x + v x + w x = ωx
+ ωy
+ ωz
+ ν∇ 2ωx
∂t
∂x
∂y
∂z
∂x
∂y
∂z
∂ω y
∂t
+u
∂ω y
∂x
+v
∂ω y
∂y
+w
∂ω y
∂z
= ωx
∂v
∂v
∂v
+ ωy
+ ωz
+ ν∇ 2ω y
∂x
∂y
∂z
∂ωz
∂ω
∂ω
∂ω
∂w
∂w
∂w
+ u z + v z + w z = ωx
+ ωy
+ ωz
+ ν∇ 2ωz
∂t
∂x
∂y
∂z
∂x
∂y
∂z
⎛ ∂w ∂v ⎞ ˆ ⎛ ∂u ∂w ⎞ ˆ ⎛ ∂v ∂u ⎞ ˆ
a) ∇× V = ⎜
− ⎟i +⎜ −
⎟ j+ ⎜ − ⎟k = 0
⎝ ∂y ∂z ⎠ ⎝ ∂z ∂x ⎠ ⎝ ∂x ∂y ⎠
∴irrotational
∂φ
= 10x ∴φ = 5x2 + f ( y)
∂x
∂φ ∂f
= = 20y
∂y ∂y
∴ f = 10y2 + C Let C = 0
∴ φ = 5 x 2 + 10 y 2
8.74
⎛ − y 1 ( x 2 + y 2 ) −1/ 2 2 x − x 1 ( x2 + y 2 )−1/ 2 2 y ⎞
2
2
ˆ
ˆ
⎟ kˆ = 0
c) ∇ × V = 0i + 0 j + ⎜
−
2
2
2
2
⎜
⎟
x +y
x +y
⎝
⎠
∴irrotational
∂φ
x
=
∂x
x2 + y 2
∴φ = x 2 + y 2 + f ( y )
∂f
y
∂φ 1 2
= ( x + y 2 )−1/ 2 2 y +
=
∂y 2
∂y
x2 + y 2
∴
∂f
= 0 ∴ f = C. Let C = 0
∂y
∴φ = x 2 + y 2
u=
∂ψ
= 100 ∴ψ = 100 y + f ( x)
∂y
v=−
df
∂ψ
=−
= 50
∂x
dx
∴ f = −50 x + C
∴ψ ( x, y ) = 100 y − 50 x
8.76
u=
(We usually let C = 0)
∂φ
= 100 ∴φ = 100x + f ( y)
∂x
v=
∂φ df
=
= 50
∂y dy
∴ f = 50 y + C
∴φ ( x, y ) = 100 x + 50 y
117
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Chapter 8 / External Flows
8.78
u=
2y
∂ψ
∂φ
= 20 2
=
2
∂y
∂x
x +y
v=
∂f
∂f
∂φ
40 / x
40 x
2x
=−
+
=− 2
+
= −20 2
∴ f = C. Let C = 0
2
2
2
∂y
∂y
∂y
x +y
x + y2
1+ y / x
φ = −40 tan −1
∴φ = −40 tan −1
y
+ f ( y)
x
y
x
b) Polar coord: φ = 10r cos θ + 5ln r 2
∂φ
10r 1 ∂ψ
= 10 cos θ + 2 =
∂r
r ∂θ
r
(See Eq. 8.5.14.)
∴ψ = 10r sin θ + 10θ + f (r )
1 ∂φ
∂ψ
df
= −10sin θ = −
= −10sin θ −
∴ f = C ∴ψ = 10r sin θ + 10θ
∂r
r ∂θ
dr
∴ψ ( x, y ) = 10 y + 10 tan −1
c) v =
8.80
u=
10y
∂φ
= 2
∂y x + y 2
∂φ
10 x
= 10 + 2
∂x
x + y2
Bernoulli:
y
x
Along x-axis (y = 0) v = 0
Along x-axis u = 10 +
V2 p
V2 p
+ + gz = ∞ + ∞ + gz∞
2 ρ
2
ρ
(10 + 10/x)2 p 102 100,000
+ =
+
ρ
ρ
2
2
e)
(assume z = z∞ )
⎛2 1 ⎞
∴ p = 100 − 50 ⎜ + 2 ⎟ kPa
⎝x x ⎠
ay = v ∂v/∂y + u ∂v/∂x = 0 on x-axis
(
10
x
ax = u∂u/∂x + v ∂u/∂y = (10 + 10/x ) −10/x2
)
∴ ax ( −2, 0) = (10 − 5) ( −10/4 ) = −12.5 m/s 2
118
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Chapter 8 / External Flows
Superposition of Simple Flows
ψ = 9y +
a) vr =
0.45π
θ = 9r sin θ + 0.225θ
2π
1 ∂ψ
0.225
= 9cosθ +
=0
r ∂θ
r
At θ = π ,
8.82
Stag. pt:
0.225
= 9 ∴ rs = 0.025 m
rs
(−0.025, 0)
c) q = U × H = Δψ
∴9H = 0.225π
Thickness = 2 H =
π
20
d) vr (1, π ) = 9 cos π +
φ=
∴H =
π
40
m or 0.157 m
0.225
= −8.25 ∴ u = 8.25 m/s
3
1/2 −2π
1/2
2π ⎡
ln ( x + 1) 2 + y 2 ⎤ +
ln ⎡ ( x − 1) 2 + y 2 ⎤ + 2 x
⎦
⎣
⎦
2π ⎣
2π
1
1
= ln ⎡(x +1)2 + y2 ⎤ − ln ⎡(x −1)2 + y2 ⎤ + 2x
⎣
⎦
⎦
2
2 ⎣
1
1
2( x + 1)
2( x − 1)
∂φ
2
2
=
−
+2
u=
∂x ( x + 1) 2 + y 2 ( x − 1) 2 + y3
8.84
Along the x-axis (y = 0), v = 0 and u =
Set u = 0 :
Stag. pts.:
( x − 1) + y
2
2
−
y
( x − 1) 2 + y 2
1
1
−
+2
x +1 x −1
( 2, 0), (− 2, 0)
u(0, 4) =
φ=
y
1
1
−
= 2, or x 2 = 2 ∴ x = ± 2
x −1 x +1
u(4, 0) =
8.86
v=
1
1
−
+ 2 = 1.867 m/s
−4 + 1 −4 − 1
1
1 + 42
−
−1
1 + 42
+ 2 = 2.118 m/s
v(−4, 0) = 0
v(0, 4) =
4
1 + 42
−
4
1 + 42
=0
1/ 2 2π
1/ 2
2π ⎡
ln ( y − 1) 2 + x 2 ⎤ +
ln ⎡ ( y + 1) 2 + x 2 ⎤ + U∞ x
⎦
⎦
2π ⎣
2π ⎣
119
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Chapter 8 / External Flows
1
1
= ln ⎡( y − 1)2 + x2 ⎤ + ln ⎡( y + 1)2 + x2 ⎤ + U∞ x
⎦ 2 ⎣
⎦
2 ⎣
y
a) Stag. pts. May occur on x-axis, y = 0
∂φ
u=
∂x
=
y =0
x
1 + x2
∴ x 2 + 0.2 x + 1 = 0
+
x
x
1 + x2
+ 10
∴no stagnation points exist on the x-axis
(They do exist away from the x-axis)
h
1
Along the y-axis: u( y) = 10 q = ∫ udy = (2π ) = π m2 /s
2
0
h
∴π = ∫ 10dy = 10 h
∴ h = 0.314 m
0
ψ=
4π
20π
ln r = 2θ + 10 ln r at ( x, y) = (0,1) which is (r,θ ) = (1, π /2).
θ+
2π
2π
b) vr =
2
10
and vθ = − . From Table 5.1 (use the l.h.s. of momentum):
r
r
Dvr vθ2
∂vr vθ2 2 ⎛ 2 ⎞ 100
ar =
+
= vr
−
= ⎜ − 2 ⎟ − 3 = −104 m/s 2
Dt
r
∂r
r
r⎝ r ⎠ r
aθ =
8.88
Dvθ vr vθ
∂v
vv
2 ⎛ 10 ⎞ 2(−10)
+
= vr θ + r θ = ⎜ 2 ⎟ +
=0
Dt
r
∂r
r
r⎝r ⎠
r3
∴ a(0,1) = −104i r or (ax , ay ) = (0, −104) m/s2
c) vr (14.14, π /4) =
2
= 0.1414,
14.14
vr (0.1, π /2) =
2
= 20,
0.1
vθ (14.14, π /4) = −
vθ (0.1, π /2) =
10
= −0.707 m/s
14.14
−10
= −100 m/s
−0.1
20,000 0.14142 + 0.707 2
p 202 + 1002
Bernoulli:
+
=
+
∴ p = 13,760 Pa
1.2
2
1.2
2
We used ρair = 1.2 kg/m3 at standard conditions.
8.90
b) vθ = −
∂ψ
μ sin θ ⎛
4⎞
= −U∞ sin θ −
= ⎜ −4 − 2 ⎟ sin θ = −8sin θ
2
∂r
rc
⎝
1 ⎠
120
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Chapter 8 / External Flows
c) pc = p∞ + ρ
v2
V∞2
42
82 sin 2 θ
− ρ θ = 50,000 + 1000 × − 1000
2
2
2
2
−x
∴ pc = 58 − 32sin 2 θ kPa
π /2
d) Drag = 2
∫
(58 − 32sin 2 α ) cos α ×1×1dα − 26 × 2 ×1
x = −1
0
−vr
⎡
⎛ 1 ⎞⎤
= 2 ⎢58 − 32 ⎜ ⎟⎥ − 52 = 42.7 kN
⎝ 3 ⎠⎦
⎣
vθ = −2 × 20sin θ −
Γ
2π × 0.4
0 = −2 × 20sin 270D −
Γ = 2π rc2ω
∴ω =
8.92
For one stag. pt.: vθ = 0 at θ = 270D :
Γ
2π × 0.4
Γ
∴Γ = 2 × 20 × 2π × 0.4 = 100.5 m 2 /s
100.5
=
2π rc2
(See the figure in Problem 8.89c)
2π × 0.42
= 100 rad/s (See Example 8.12)
Min. pressure occurs where vθ is max, i.e., θ = π / 2. There
vθ = −2 × 20 ×1 −
∴ pmin = p∞ +
8.94
100.5
= 80 m/s
2π × 0.4
v2
V∞2
202
802
ρ − θ ρ = 0+
× 1.22 −
× 1.22 = −3660 Pa
2
2
2
2
At 9000 m, ρ = 0.47 kg/m3
Lift = ρU ∞ ΓL = 0.47 × (105) × (1400) × 18 = 1244 × 103 N = 1244 kN
Place four sources as shown. Then, with q = 2π for each:
u( x, y ) =
8.96
+
v ( x, y ) =
x−2
( x − 2) + ( y − 2)
x+2
2
2
( x + 2) 2 + ( y + 2) 2
y−2
( x − 2) + ( y − 2)
2
2
+
x+2
+
+
y
( x + 2) 2 + ( y − 2) 2
x−2
x
( x − 2) 2 + ( y + 2) 2
y+2
( x − 2) + ( y + 2)
2
2
+
y−2
( x + 2) + ( y − 2)
2
2
+
y+2
( x + 2) 2 + ( y + 2) 2
∴v(4, 3) = (0.729, 0.481) m/s
121
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Chapter 8 / External Flows
Boundary Layers
Recrit =
8.98
b) ν =
8.100
U ∞ xT
∴ xT =
ν
6 × 105ν
= 2700ν
222.2
μ
= 3.22 ×10−5 m2 /s ∴ xT = 2700 × 3.22 ×10−5 = 8.7 ×10−2 m or 8.7 cm
ρ
a) Use Recrit = 3 × 105 =
10 xT
∴ xT = 0.03 m or 3 cm
c) Use Recrit = 3 × 105 =
10 xT
∴ xT = 0.03 m or 3 cm
e) Re = 6 × 104 =
10−6
10−6
10 xgrowth
∴ xgrowth = 0.006 m
10−6
or
6 mm
The x-coordinate is measured along the cylinder surface as shown in Fig. 8.19.
The pressure distribution (see solution 8.89) on the surface is
p = p 0 − 2 ρU ∞2 sin 2 α where rc α = x (α is zero at the stagnation point).
Then
8.102
p( x) = 20,000 − 2 ×1000 ×102 sin 2 ( x/2)
= 20 − 200 sin 2 ( x /2) kPa
The velocity U(x) at the edge of the b.l. is U(x) on the cylinder wall:
vθ ( r = 2) = −10 sin θ − 10 sin θ = −20 sin(π − α ) = 20 sin α
∴ U ( x) = 20 sin( x /2)
The height h above the plate is
h( x) = mx + 0.4
0.1 = m × 2 + 0.4
∴ m = −0.15
∴h(x) = 0.4 − 0.15x Continuity: 6× 0.4 = U(x)h ∴U(x) =
8.104
U ( x) =
Euler’s Eqn:
2.4
or
0.4 − 0.15x
16
2.67 − x
ρu
∂p
∂u
=−
∂x
∂x
∴
dp
16
16
256
=ρ
×
=
2
dx
2.67 − x (2.67 − x)
(2.67 − x)3
122
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Chapter 8 / External Flows
Von Kármán Integral Equation
δ
τ 0 = −δ
= −δ
δ
dp
d
d
+ U ( x ) ∫ ρudy − ∫ ρu 2 dy
dx
dx 0
dx 0
δ
⎞ d δ
dp d δ
dU ⎛
uUdy
udy
+
ρ
−
ρ
ρu 2 dy
⎜
⎟−
∫
∫
∫
dx dx 0
dx ⎝ 0
⎠ dx 0
8.106
where we have used g
δ
⎞
df dfg
dg ⎛
=
−f
⎜⎜ Here U = g , f = ∫ ρudy ⎟⎟
dx dx
dx ⎝
0
⎠
δ
δ
dp d
dU
∴τ 0 = −δ
+ ∫ ρu(U − u)dy − ρ
udy
dx dx 0
dx ∫0
( ρ = const.)
δ
dU
d
a) If dp/dx = 0 then
= 0 and τ 0 = ρ ∫ u(U∞ − u)dy
dx
dx 0
τ0 = ρ
δ
δ
πy⎛
πy⎞
πy y⎤
d
d ⎡ 2δ
d ⎛ 2δ δ ⎞
U∞2 sin
1 − sin
dy = ρU ∞2
cos
−
−
= ρU∞2
− ⎟
⎜
⎜
⎟
∫
⎢
dx 0
2δ ⎝
2δ ⎠
dx ⎣ π
2δ 2 ⎦⎥ 0
dx ⎝ π 2 ⎠
Also, we have τ 0 = μ
∴ μU ∞
8.108
b) τ 0 = μU∞
∂u
π
= μU∞
cos 0
∂y y =0
2δ
π
dδ
= 0.137 ρU ∞2
2δ
dx
∫
∴ v = U∞
0
8.110
U∞
×
νx
2
4.79
⎝
∴ δ = 4.79
νx
U∞
⎛ ayx −3/ 2 ⎞
⎞
ay
∂
∂v
⎛ ay ⎞
=
=
−
=−
U
U
sin
⎟⎟ cos
⎟⎟
∞
∞⎜
⎜
⎟
⎜
∂
∂
x
y
2
x
⎝ x⎠
⎠
⎝
⎠
⎛
U∞ ⎞
U∞3
cos ⎜⎜ 0.328y
⎟⎟ dy = 0.0316
ν
νx ⎠
ν
⎝
0.164 y U∞
x
U∞
dx
U∞
U∞
= 0.328μU∞
2 4.79 ν x
νx
∂x
δ
ν
π 1
⎛
c) ∂u = U∞ ∂ sin ⎜ π y
⎜
∂x
∴ δ dδ = 11.5
3/ 2
δ
⎡⎛
∫ y cos ⎢⎣⎢⎜⎜⎝ 0.189
0
U∞ ⎞ ⎤
⎟ y ⎥ dy
ν ⎟⎠ ⎦⎥
δ /2
⎧δ /6
y⎞
d ⎪
⎛ y 1 ⎞⎛ y 1 ⎞
2 y⎛
τ 0 = ⎨ ρ 3U∞ ⎜1 − 3 ⎟ dy + ρU∞2 ⎜ + ⎟⎜1 − − ⎟ dy
δ⎝
δ⎠
dx ⎪
⎝ δ 3 ⎠⎝ δ 3 ⎠
δ /6
⎩0
∫
∫
123
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8 / External Flows
δ
+
∫
δ /2
=
3U
d
ρU∞2 (0.1358δ ) = μ ∞
dx
δ
∴δ
⎫
y 2 ⎞⎛
y 2⎞ ⎪
+ ⎟ ⎜1 −
− ⎟ dy ⎬
⎝ 3δ 3 ⎠ ⎝ 3δ 3 ⎠ ⎪
⎭
⎛
ρU∞2 ⎜
dδ
μ
= 22.08
dx
ρU∞
Thus
⎛ 6.65 ν ⎞
2
−1/ 2
, τ 0 (x) = 0.1358ρU∞2 ⎜⎜
⎟⎟ = 0.451ρU∞ Rex
U∞
2
U
x
∞ ⎠
⎝
νx
δ (x) = 6.65
% error for δ =
6.65 − 5
×100 = 33%
5
% error for τ 0 =
0.451 − 0.332
× 100 = 36%
0.332
The given velocity profile is that used in Example 8.13. There we found
δ = 5.48 ν x/U∞ = 5.48 10−6 x/10 = 0.00173 x = 0.00173 3 = 0.003 m
Assume the streamline is outside the b.l. Continuity is then
0.003
10 × 0.02 =
∫
0
8.112
⎛ 2y
y2 ⎞
10 ⎜
−
dy + (h − 0.003)10
⎜ 0.003 0.0032 ⎟⎟
⎝
⎠
= 0.02 + 10h − 0.03
1
δd =
10
0.003
∫
0
∴ h = 0.021 m or 2.1 cm
⎛
20 y
10 y 2 ⎞
1
+
⎜⎜10 −
⎟⎟ dy = ( 0.03 − 0.03 + 0.01) = 0.001 m
2
0.003 0.003 ⎠
10
⎝
h − 2 = 2.1 − 2 = 0.1 cm or 0.001 m
The streamline moves away from the wall a distance δ d
δ
3⎞
3⎞
⎛
⎛
a) u = U ∞ ⎜ 3 y − 1 y ⎟ δ d = 1 U ∞ ⎜ 1 − 3 y + 1 y ⎟ dy = δ − 3 δ + 1 δ = 0.375δ
⎜ 2 δ 2 δ3 ⎟
⎜ 2 δ 2 δ3 ⎟
4
8
U∞
⎝
⎠
⎝
⎠
∫
0
8.114
From Eq. 8.6.16, δ d = 0.375 × 4.65
θ=
νx
U∞
= 1.74
νx
U∞
% error = 1.2%
δ
3 ⎞⎛
⎛
3 y 1 y3 ⎞
2 3 y 1 y
U
−
1
−
+
dy = 0.139δ
⎜
⎟⎜
∞⎜
3 ⎟⎜
3⎟
⎟
2
δ
2
2
δ
2
U∞2
δ
δ
⎝
⎠⎝
⎠
0
1
∫
124
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8 / External Flows
∴θ = 0.139 × 4.65
a) δ = 4.65
b)
νx
U∞
νx
U∞
= 0.648
νx
% error =
U∞
1/ 2
⎛ 1.5 ×10−5 × 6 ⎞
= 4.65 ⎜
⎟⎟
⎜
4
⎝
⎠
τ 0 = 0.323ρU∞2
0.648 − 0.644
× 100 = 0.62%
0.644
ν
xU∞
= 0.0221 m
2 ⎛ 1.5 × 10
= 0.323 × 1.22 × 4 ⎜
⎜
⎝
6× 4
−5 ⎞1/ 2
⎟⎟
⎠
= 0.00498 Pa
1/ 2
−5 ⎞
⎛
c) Drag = 1 ρU∞2 Lw ×1.29 ν = 1 ×1.22 × 42 × 6 × 5 ×1.29 ⎜ 1.5 ×10 ⎟
⎜ 6× 4 ⎟
2
LU∞ 2
⎝
⎠
d)
8.116
= 0.299 N
⎡ 3y 3 y3 ⎤ dδ
∂u
= U∞ ⎢ − 2 +
⎥
∂x
2 δ 4 ⎥⎦ dx
⎢⎣ 2δ
⎡
⎤ dδ
42 y 3
3× 4
3
= 4 ⎢−
y
+
×
⎥
2
−5
2 4.654 × (1.5 ×10−5 × 3)2 ⎦⎥ dx
⎣⎢ 2 × 4.65 ×1.5 ×10 × 3
∴
∂u
4.65 1.5 × 10−5 1
= 4 ⎡ −6166 y + 2.53 × 107 y3 ⎤
= −64.1y + 2.63 × 105 y3
⎣
⎦
∂x
2
4
3
δ
∫
∴v = −
0
∂u
64.1
2.63 × 105
× 0.01562 −
× 0.01564 = 0.00391 m/s
dy =
∂x
2
4
where δ x =3 = 4.65
1.5 ×10−5 × 3
= 0.01560 m
4
Laminar and Turbulent Boundary Layers
8.118
−5 ⎞
⎛
a) δ = 0.38 × 6 ⎜ 1.5 × 10 ⎟
⎜ 20 × 6 ⎟
⎝
⎠
0.2
⎛ 1.5 × 10−5 ⎞
1
= 0.0949 m τ 0 = × 1.22 × 202 × 0.059 ⎜
⎜ 20 × 6 ⎟⎟
2
⎝
⎠
= 0.6 Pa
8.120
0.2
⎡
⎛ 1.47 ×10−5 ⎞
⎛ 1.47 × 10−5 ⎞ ⎤
1
2
⎢
⎥
a) Drag = ×1.225 × 6 × (3.6 × 15) 0.074 ⎜
− 1060 ⎜
⎟
⎜ 6 × 3.6 ⎟
⎜ 6 × 3.6 ⎟⎟ ⎥
⎢
2
⎝
⎠
⎝
⎠⎦
⎣
= 1191(4.32 ×10−3 − 0.72 ×10−3 ) = 4.3 N
125
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0.2
Chapter 8 / External Flows
⎛ 1.5 ×10−5 ⎞
60 × 1000
a) U∞ =
= 16.67 m/s δ = 0.38 × 100,000 ⎜
⎜ 16.67 × 105 ⎟⎟
3600
⎝
⎠
0.2
= 235 m
0.2
⎡
⎛ 1.5 ×10−5 ⎞ ⎤
1
1
2
2 ⎢
⎥ = 0.0618 Pa
τ 0 = ρU∞ c f = ×1.22 ×16.67 × 0.059 ⎜⎜
5⎟
⎟ ⎥
⎢
2
2
16.67
10
×
⎝
⎠ ⎦
⎣
8.122
b) τ 0 =
∴ uτ =
1
1
0.455
ρU∞2 c f = ×1.22 ×16.672
= 0.151 Pa
2
2
2
5
⎡ ⎛
16.67 ×10 ⎞ ⎤
⎢ln ⎜⎜ 0.06
⎟⎥
1.5 ×10−5 ⎠⎟ ⎥⎦
⎢⎣ ⎝
0.151
16.67
0.351δ
= 0.351 m/s ∴
= 2.44 ln
+ 7.4 ∴ δ = 585 m
1.22
0.351
1.5 ×10−5
Both (a) and (b) are in error; however, (b) is more accurate.
a) Use Eq. 8.6.40: c f =
b) τ 0 =
8.124
uτ =
c) δν =
d)
0.455
⎡ ⎛
90 × 6 ⎞ ⎤
⎟⎥
⎢ ln ⎜ 0.06
1.47 ×10−5 ⎠ ⎦
⎣ ⎝
2
= 0.00213
1
1
ρU ∞2 c f = × 1.225 × 902 × 0.00213 = 10.52 Pa
2
2
10.52
= 2.93 m/s
1.225
5ν 5 ×1.47 × 10−5
=
= 2.51× 10−5 m
uτ
2.93
90
2.93δ
= 2.44 ln
+ 7.4
2.93
1.47 ×10−5
∴δ = 0.071 m
Assume flat plates with dp/dx = 0. C f =
8.126
0.523
⎡ ⎛
10 ×100 ⎞ ⎤
⎟⎥
⎢ln ⎜ 0.06
10−6 ⎠ ⎦
⎣ ⎝
2
= 0.00163
1
∴Drag = 2 × × 1000 × 102 × 10 × 100 × 0.00163 = 163,000 N
2
To find δ max we need uτ :
126
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Chapter 8 / External Flows
1
2
τ 0 = × 1000 × 102
0.455
⎡ ⎛
10 × 100 ⎞ ⎤
⎟⎥
⎢ ln ⎜ 0.06
10−6 ⎠ ⎦
⎣ ⎝
2
= 70.9 Pa ∴ uτ =
10
0.266δ
= 2.44 ln
+ 7.4
0.266
10−6
70.9
= 0.266 m/s
1000
∴δ max = 0.89 m
Laminar Boundary-Layer Equations
∂ψ ∂u ∂ 2ψ
∂ψ
u=
,
=
, v=−
,
∂y
∂x ∂x∂y
∂x
8.128
∂u ∂ 2ψ
=
,
∂y ∂y 2
∂ 2u
∂y 2
=
∂ 3ψ
∂y 3
Substitute into Eq. 8.6.45 (with dp /dx = 0) :
∂ψ ∂ 2ψ ∂ψ ∂ 2ψ
∂3ψ
−
=ν 3
∂y ∂x∂y ∂x ∂y 2
∂y
u=
U∞
dF ∂η
∂ψ
= U∞ν x
= U∞ν x F '(η )
= U∞ F '(η )
dη ∂y
νx
∂y
We used Eq. 8.6.50 and Eqs. 8.6.48.
8.130
v=−
∂ψ
∂
=−
∂x
∂x
(
1 U∞ν
∂F ∂η
F − U∞ν x
2
x
∂η ∂x
=−
U∞ ⎛ 1 −3/2 ⎞
1 U∞ν
F − U∞ν x F ' y
− x
⎟
2
x
ν ⎜⎝ 2
⎠
=−
y U ∞ U ∞ν
1 U ∞ν
1 U ∞ν
F+
F'=
(η F '− F )
2
x
2 νx
x
2
x
a) τ 0 = 0.332 × 1.22 × 52
b) δ = 5
8.132
c) vmax =
δ
)
U∞ν x F = −
1.5 × 10−5
= 0.0124 Pa
2×5
1.5 × 10−5 × 2
= 0.0122 m
5
1.5 ×10−5 × 5
⎤
η
(
F
'
−
F
)
=
× 0.8605 = 0.00527 m/s
⎥⎦
x ⎢⎣ 2
2
max
ν U∞ ⎡ 1
δ
δ
dF
dF
vx
d) Q = ∫ u(1× dy) = ∫ U∞
dy = ∫ U∞
dη
dη
dη
U∞
0
0
0
127
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Chapter 8 / External Flows
= U∞
νx
U∞
[ F (δ ) − F (0)] = 5
1.5 ×10−5 × 2
× 3.28 = 0.04 m3 /s/m
5
At x = 2 m, Re = 5 × 2/10−6 = 107 ∴Assume turbulent from the leading edge.
1
0.455
1
0.455
ρU∞2
= ×1000 × 52
= 32.1 Pa
2
2
2
⎡(ln 0.06 ×107 ) ⎤
[ln(0.06 Re x )] 2
⎣
⎦
a) τ 0 =
b) uτ =
8.134
τ0
32.1
=
= 0.1792 m/s
ρ
1000
5
0.1792δ
= 2.44 ln
+ 7.4
0.1792
10−6
∴δ = 0.0248 m
c) Use the 1/7 the power-law equation:
0.0248
Q=
∫
0
1/7
⎛ y ⎞
5⎜
⎟
⎝ 0.0248 ⎠
dy = 0.109 m3 /s/m
From Table 8.5 we interpolate for F' = 0.5 to be
η=
0.5 − 0.3298
( 2 − 1) + 1 = 1.57
0.6298 − 0.3298
1.57 =
8.136
v=
U∞
5
=y
νx
1.5 ×10−5 × 2
νU ∞ ⎛ 1 ⎞
⎜ ⎟ (ηF '− F )
x ⎝ 2⎠
=
∴ y = 0.00385 m or 3.85 mm
1.5 ×10−5 × 5
(0.207) = 0.00127 m/s
2
−5
⎛ ∂u ∂v ⎞
ν
2 1.5 × 10
2
= 0.291(1.2)5
= 0.011 Pa
τ = μ⎜ +
⎟ = F " ρU ∞
xU∞
2×5
⎝ ∂y ∂x ⎠
⎛ 3 y 1 y3 ⎞
u = U∞ ⎜
−
⎜ 2 δ 2 δ 3 ⎟⎟
⎝
⎠
8.138
y
U
cubic
For the Blasius profile: see Table 8.5. (This is
Blasius
only a sketch. The student is encouraged to draw
the profiles to scale.)
8.140
A:
∂p
< 0. (favorable)
∂x
128
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Chapter 8 / External Flows
B:
∂p
≅0
∂x
y
C
∂p
C:
> 0 (unfavorable)
∂x
D:
∂p
>0
∂x
E:
∂p
<0
∂x
D
δD δ
C
E
δΒ δ δ
E Α
A
129
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B
Chapter 8 / External Flows
130
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Chapter 9 / Compressible Flow
CHAPTER 9
Compressible Flow
Introduction
c p = cv + R
c p = kcv
∴ cp =
9.2
∴ cp =
cp
⎛ 1⎞
+ R or c p ⎜1 − ⎟ = R
k
⎝ k⎠
Rk
k −1
Speed of Sound
Substitute Eq. 4.5.18 into Eq. 4.5.17 and neglect potential energy change:
Q − W
V 2 − V12 p2 p1
S
= 2
+
− + u − u
m
2
ρ 2 ρ1 2 1
~ + pv = u
~ + p / ρ. Therefore
Enthalpy is defined in Thermodynamics as h = u
Q − W
V 2 − V12
S
= 2
+ h2 − h1
2
m
9.4
Assume the fluid is an ideal gas with constant specific heat so that Δh = c p ΔT .
Then
2
2
Q − W
S = V2 − V1 + c (T − T )
1
p 2
m
2
Next, let c p = cv + R and k = c p /c v so that c p / R = k /( k − 1). Then, with the
ideal gas law p = ρ RT , the first law takes the form
Q − W
V22 − V12
k ⎛ p2 p1 ⎞
S
=
+
− ⎟
⎜
m
k − 1 ⎝ ρ 2 ρ1 ⎠
2
The speed of sound is given by
c = dp/dρ
9.6
For an isothermal process TR = p /ρ = K, where K is a constant. This can be
differentiated: dp = Kdρ = RTdρ. Hence, the speed of sound is c = RT .
131
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Chapter 9 / Compressible Flow
For water
Bulk modulus = ρ
dp
= 2110 × 106 Pa
dρ
9.8
Since ρ = 1000 kg/m 3 , we see that
c=
dp
2110 ×106
=
= 1453 m/s
1000
dρ
Since c = 1450 m/s for the small wave, the time increment is
9.10
9.12
Δt =
d
10
=
= 0.0069 s
c 1450
c = kRT = 1.4 × 287 × 263 = 325 m/s ∴ d = ct = 256 × 1.21 = 393 m
c = 1.4 × 287 × 263 = 256 m/s
sin α =
sin α = 0.256 ∴ tan α = 0.2648 =
9.14
Δt =
1
c
=
M V
1000
∴ L = 3776 m
L
3776
= 3.776 s
1000
V
1000 m
L
Eq. 9.2.4:
ΔV = −
Δp
Δp
0.3
=−
=−
= −0.35 m/s
ρc
ρ kRT
1.225 1.4 × 287 × 288
Energy Eq:
9.16
V2
(V + ΔV )2
+ c pT =
+ c p (T + ΔT )
2
2
∴ΔT =
(ΔV )2
∴ 0 = V ΔV +
+ c p ΔT
2
−cΔV − 1.4 × 287 × 288 m/s (−0.35 m/s)
=
= 0.012 K
cp
1005 (N-m)/(kg ⋅ K)
Use N = kg ⋅ m ⋅ s −2 (Units can be a pain!)
132
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Chapter 9 / Compressible Flow
Isentropic Flow
a) ps = patm + 10 = 69.9 + 10 = 79.9 kPa abs
V
p1 = 69.9 kPa abs
1
s
Vs=0
From 1 → s :
9.18
V12 p1 ps
+ =
2 ρ1 ρs
1/ k
⎛p ⎞
ρs = ρ1 ⎜ s ⎟
⎝ p1 ⎠
V12 69,900 79,900
∴
+
=
2
0.906
0.997
Is pr < 0.5283p0 ?
1/1.4
⎛ 79.9 ⎞
= 0.906 ⎜
⎟
⎝ 69.9 ⎠
∴ V1 = 77.3 m/s
0.5283 × 200 = 105.7 kPa
a) pr < 0.5283p0 ∴ choked flow ∴ Me = 1
1000 × 298 =
9.20
ρe =
= 0.997 kg/m3
∴ Ve2 = kRTe
pe = 105.7 kPa
1.4 × 287Te
+ 1000Te . ∴ Te = 248.1 K, Ve = 315.8 m/s.
2
105.7
= 1.484 × π × 0.012 × 315.8 = 0.1473 kg/s
= 1.484 kg/m3 ∴ m
0.287 × 248.1
b) pr > 0.5283p0
ρ0 =
∴ M e < 1 1000 × 298 =
Ve2 1.4 130,000
+
×
2 0.4
ρe
1.4
130 ⎛ ρe ⎞
=
200 ⎜⎝ 2.338 ⎟⎠
200
= 2.338 ∴ ρ e = 1.7187 kg/m3 ∴ Ve = 257.9 m/s
0.287 × 298
= 1.7187 × π × 0.012 × 257.9 = 0.1393 kg/s
∴m
a) pr < 0.5283 p0 ∴ Me = 1 ∴ pe = 0.5283 × 200 = 105.7 kPa Te = 0.8333 × 298 = 248.3 K
ρe =
105.7
= 1.483 kg/m3
0.287 × 248.3
Ve = 1.4 × 287 × 248.3 = 315.9 m/s
= 1.483 × π × 0.012 × 315.9 = 0.1472 kg/s
∴m
9.22
b) pr > 0.5283 p0 ∴ pe = 130 kPa,
ρe =
pe
= 0.65 ∴ M e = 0.81, Te = 0.884T0
p0
130
= 1.719 kg/m3 , Ve = 0.81 1.4 × 287 × 263.4 = 263.5 m/s
0.287 × 263.4
= 1.719 × π × 0.012 × 263.5 = 0.1423 kg/s
∴m
133
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Chapter 9 / Compressible Flow
pe = 0.5283 × 400 = 211.3 kPa abs
9.24
Te = 0.8333 × 303 = 252.5 K
=
Ve = 1.4 × 287 × 252.5 = 318.5 m/s ∴ m
211.3
π × 0.052 × 318.5 = 7.29 kg/s
0.287 × 252.5
pe = 0.5283 p0 = 101.3 kPa ∴ p0 = 192 kPa Te = 0.8333 × 280 = 233.3 K
Ve = 1.4 × 287 × 233.3 = 306.2 m/s
9.26
= [101.3 ×103 /(287 × 233.3)] × π × (0.3)2 × 306.2 = 1.31 kg/s
m
p0 = 2 ×192
pe = 0.5283 p0 = 203 kPa,
Te = 233.3 K, Ve = 306.2 m/s
= [203×103 /(287 × 233.3)]× π × (0.03)2 × 306.2 = 2.62 kg/s
m
1.667 / 0.667
5193 × 300 =
1.667 × 2077 Te
⎛ 225 ⎞
+ 5193 Te ∴ Te = 225 K ∴ pe = 200 ⎜
⎟
2
⎝ 300 ⎠
= 97.45 kPa abs
Next, Tt = 225 K, pt = 97.45 kPa; ∴ Vt = 1.667 × 2077 × 225 = 882.6 m/s
ρt =
9.28
97.45
= 0.2085 kg/m3
2.077 × 225
V 2 1.667 pe
5193 × 300 = e +
2 0.667 ρ e
=
0.2085×π ×0 .032 ×882.6 = ρ eπ × 0.0752 Ve
1.667
ρe
⎛
⎞
pe = 200 ⎜
⎟
⎝ 200 / 2.077 × 300 ⎠
= 1330 ρ e1.667 kPa
Ve2
+ 3324 × 103 × 9.54Ve−0.667
2
or 3.116 ×106 = Ve2 + 63 420 ×103 Ve−0.667 Trial-and-error: Ve = 91.8 m/s
∴ ρe = 0.3203 kg/m3 and pe = 199.4 kPa abs
We need to determine the Mach number at the exit. Since the M = 1 at the
throat, then A* = Athroat = 9.7 cm 2 . Hence, the area ratio at the exit is
9.30
Ae A* = 13 9.7 = 1.34 . Using the air tables, we find two possible solutions, one
for subsonic flow, and the other for supersonic flow in the diverging section of
the nozzle. At the exit:
Subsonic Flow: ⇒ M e = 0.5, Te T0 = 0.9524, and pe p0 = 0.8430
Hence, Ve = Me ce = Me kRTe = 0.5 1.4 × 287 ( 0.9524 × 295) = 168 m/s
Supersonic Flow: ⇒ M e = 1.76, Te T0 = 0.6175, and pe p0 = 0.1850.
134
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Chapter 9 / Compressible Flow
Hence, Ve = M e ce = M e kRTe = 1.76 1.4 × 287 ( 0.6175 × 295 ) = 476 m/s
ρ1 = p1 / RT1 = (310 + 101.3)103 /(287 × 289) = 4.96 kg/m3
1/1.4
⎛ 351.3 ⎞
ρ 2 = 4.96 ⎜
⎟
⎝ 411.3 ⎠
9.32
= 4.43 kg/m3
V1 × 4.96 × (0.1)2 = V2 × 4.43 × (0.05)2
∴ V2 = 4.48 V1
V12 1.4 411.3 × 103 4.482 V12 1.4 351.3 × 103
+
×
=
+
×
2 0.4
4.96
2
0.4
4.43
∴ V1 = 36.5 m/s
= 4.96π × (0.05) 2 × 36.5 = 1.42 kg/s
∴m
Vt2 = kRT 1000 × 293 =
1.4 / 0.4
⎛ 244 ⎞
∴ pt = 500 ⎜
⎟
⎝ 293 ⎠
9.34
1000 × 293 =
pe
ρ e1.4
=
1.4 × 287 Tt
+ 1000 Tt ∴ Tt = 244.0 K Vt = 313.1 m/s
2
= 263.5 kPa abs ∴ ρt =
Ve2 1.4 pe
+
2
0.4 ρ e
263.5
= 3.763 kg/m3
0.287 × 244
3.763 × π × 0.0252 × 313.1 = ρ eπ × 0.0752Ve
263,500
3.7631.4
∴ 293,000 =
Ve2
+ 1.014 ×106 Ve−0.4 . Trial-and-error: Ve = 22.2 m/s, 659 m/s
2
∴ ρe = 5.897, 0.1987 kg/m3
∴ pe = 494.2 kPa, 4.29 kPa abs
M t = 1 ∴ pt = 0.5283 × 830 = 438.5 kPa, Tt = 0.8333 × 289 = 241 K
∴ ρt = 6.34 kg/m 3
= 14.6 = 6.34
m
9.36
π dt2
4
1.4 × 287 × 241 ∴ dt = 0.097 m
p 103
=
= 0.125 ∴ M e = 2.014, Te = 0.552 × 289 = 160 K, Ve = 2.014 1.4 × 287 × 160
p0 830
= 253 m/s
A
A*
= 1.708 ∴
π de2
4
= 1.708
π × 0.0972
4
∴ de = 0.127 m
135
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Chapter 9 / Compressible Flow
Using compressible flow tables for air, we determine the pressure ratio and
temperature ratio for M = 2.8 to be:
p
T
= 0.03685, and = 0.3894
p0
T0
9.38
∴ p = 0.03685 × p0 = 129 kPa abs
and T = 0.3894 × T0 = 125 K
∴ V = Mc = 2.8 kRT = 2.8 1.4 × 287 × 320 = 1004 m/s
Let Mt = 1. Neglect viscous effects. M1 =
9.40
∴
A
A*
= 1.5007 ∴ At =
150
= 0.430
1.4 × 287 × 303
A1
π × 0.052 π dt2
=
=
∴ dt = 0.0816 m or 8.16 cm
1.5007
1.5007
4
Isentropic flow. Since k = 1.4 for nitrogen,
the isentropic flow table may be used.
At M = 3,
A
A*
i
= 4.235
M>1
Vi = 3 1.4 × 297 × 373 = 1181 m/s
9.42
∴ Ai =
t
ρi =
Ve ~= 0
M<1
e
Mt = 1
100
= 0.9027 kg/m3
0.297 × 373
m
10
0.00938
=
= 0.00938 m2 ∴ At =
= 0.00221 m2
ρi Vi 0.9027 ×1181
4.235
At M = 3, T = 0.3571 T0 , p = 0.02722 p0
∴T0 = Te =
373
100
= 1044 K or 772DC p0 =
= pe = 3670 kPa abs
0.3571
0.02722
Assume pe = 101 kPa. Then ρ e =
9.44
= ρ AV 2
F = mV
Mt = 1
9.46
101
= 0.4198 kg/m3
0.189 × 1273
80,000 × 9.81
= 0.4198π × 0.252 V 2
6
∴ V = 1260 m/s
Ae
= 4; ∴ M e = 2.94, pe = 0.02980 p0
A*
Te = 0.3665 T0 = 0.3665 × 300 = 110.0 K,
pe = 100 = 0.0298 p0 ∴ p0 = 3356 kPa abs
F
p0A0
∴Ve = 2.94 1.4 × 287 ×109.95 = 618 m/s
136
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Ve
Chapter 9 / Compressible Flow
∴ FB =
−100
π × 0.052 × 6182 + 3,356,000π × 0.22 = 412,000 N
0.287 ×109.95
Normal Shock
a) 0.9850 ×1000 = ρ 2V2 80,000 − p2 = 0.985 × 1000(V2 − 1000)
⎞
V22 − 10002 1.4 ⎛ p2
80
⎛
⎞
+
− 287 × 283 ⎟ = 0 ⎜ ρ1 =
= 0.9850 kg/m3 ⎟
⎜
2
0.4 ⎝ ρ 2
0.287 × 283
⎝
⎠
⎠
V22 10002 1.4 V2
−
+
×
(−985V2 + 1,065,000) − 284,300 = 0
2
2
0.4 985
9.48
∴ 3V22 − 3784V2 + 784,300 = 0 ∴ V2 = 261 m/s
ρ 2 = 3.774 kg/m3
Substitute in and find p2 = 808 kPa abs
M1 =
1000
808
= 2.966 T2 =
= 746 K or 473DC
0.287 × 3.774
1.4 × 287 × 283
M2 =
261
= 0.477
1.4 × 287 × 746
ρ 2 p2 T1 2 k M12 − k + 1
=
=
ρ1 p1 T2
k +1
⎡
( k + 1) 2 M12
( k + 1)M12
=
k −1 2 ⎤
2
2 + ( k − 1)M12
⎢⎣1 + 2 M1 ⎥⎦ [4 k M1 − 2 k + 2]
M12 =
9.50
k + 1 p2 k − 1
(This is Eq. 9.4.12). Substitute into above:
+
2k p1 2k
⎡
⎤
⎡
⎤
p
p
( k + 1) ⎢ ( k + 1) 2 + ( k − 1) ⎥
( k + 1) ⎢ ( k + 1) 2 + k − 1⎥
p1
p1
ρ2
⎣
⎦
⎣
⎦
=
=
⎧
⎫ ( k + 1) 2 + ( k − 1)( k + 1) p 2
ρ1
p2
+ ( k − 1) ⎬
4 k + ( k − 1) ⎨ ( k + 1)
p1
p1
⎩
⎭
=
k − 1 + ( k + 1) p2 / p1
k + 1 + ( k − 1) p2 / p1
p2
ρ
k +1
>> 1, 2 =
p1
ρ1 k − 1
For a strong schock in which
If M 2 = 0.5, then M1 = 2.645 ∴ V1 = 2.645 1.4 × 287 × 293 = 908 m/s
9.52
p2 = 8.00 × 200 = 1600 kPa abs
ρ2 =
1600
= 8.33 kg/m3
0.287 × (2.285 × 293)
137
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Chapter 9 / Compressible Flow
p1 = 0.2615 × 101 = 26.4 kPa
9.54
T1 = 223.3 K
M1 = 1000 / 1.4 × 287 × 223.3 = 3.34
∴ M 2 = 0.4578 p2 = 12.85 × 26.4 = 339 kPa T2 = 3.101× 223.3 = 692.5 K
For isentropic flow from 2 → 0: For M = 0.458, p = 0.866p0 and
T = 0.960 T0
A
*
A
T0 = 692.5 / 0.96 = 721 K or 448D C
∴ p0 = 339 / 0.866 = 391 kPa abs
= 4 ∴ M e = 0.147 pe = 0.985 p0
∴ p0 =
101
= 102.5 kPa abs
0.985
M t = 1 pt = 0.5283 ×102.5 = 54.15 kPa Tt = 0.8333 × 298 = 248.3 K
9.56
∴ ρt =
54.15
= 0.7599 kg/m3
0.287 × 248.3
Vt = 1.4 × 287 × 248.3 = 315.9 m/s
= 0.7599 × π × 0.0252 × 315.9 = 0.471 kg/s. If throat area is reduced, M t
∴m
= 0.7599 × π × 0.022 × 315.9 = 0.302 kg/s
remains at 1, ρt = 0.7599 kg/m3 and m
pe = 100.3 kPa = p2
∴ p1 =
A
A*
= 4 ∴ M1 = 2.94, and p2 / p1 = 9.918
100.3
10.11
= 10.11 kPa At M1 = 2.94, p / p0 = 0.0298 ∴ p0 =
= 339.3 kPa
9.918
0.0298
M t = 1, pt = 0.5283 × 393.3 = 179.2 kPa Tt = 0.8333 × 290 = 242 K
9.58
∴Vt = 1.4 × 287 × 242 = 312 m/s
M1 = 2.94, p1 = 10.11 kPa T1 = 0.3665 × 290 = 106.3 K
∴V1 = 2.94 1.4 × 287 × 106.3 = 607.6 m/s
M 2 = 0.4788, pe = 100.3 kPa
Te = T2 = 2.609 ×106.3 = 277.3
∴ V2 = 0.4788 1.4 × 287 × 277.3 = 160 m/s
Vapor Flow
⎛ 655 ⎞
pt = 0.546 p0 = 0.546 ×1200 = 655 kPa Tt = 673 ⎜
⎟
⎝ 1200 ⎠
9.60
∴ ρt =
655
= 2.42 kg/m3
0.462 × 585
= ρt At Vt ∴ 4 = 2.42 ×
m
π × dt2
4
0.3/1.3
= 585 K
Vt = 1.3 × 462 × 585 = 593 m/s (M t = 1)
× 593 ∴ dt = 0.060 m or 6 cm
138
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Chapter 9 / Compressible Flow
⎛ 101 ⎞
Te = 673 ⎜
⎟
⎝ 1200 ⎠
0.3/1.3
= 380.2 K ∴ ρe =
101
= 0.575 kg/m3
0.462 × 380.2
Ve2
+ 1872 × 380.2 = 1872 × 673 (Energy from 0 → e) (c p = 1872 J/kg ⋅ K)
2
∴ Ve = 1050 m/s ∴ 4 = 0.575(π de2 /4) × 1050 ∴ de = 0.092 m or 9.2 cm
⎛ 546 ⎞
M e = 1, pe = 0.546 ×1000 = 546 kPa Te = 643 ⎜
⎟
⎝ 1000 ⎠
9.62
∴ ρe =
0.3/1.3
= 559 K
546 ×103
= 2.11 kg/m3 Ve = 1.3 × 462 × 559 = 579 m/s
462 × 559
3.65 = 2.11
π de2
4
× 579 ∴ de = 0.0617 m or 6.17 cm
Oblique Shock Wave
M1 =
800
= 2.29
1.4 × 287 × 303
V2
V1
From Fig. 9.15, β = 46°, 79°
a) β = 46° ∴ M1n = 2.29 sin 46° = 1.65
∴ M 2n = 0.654 = M 2 sin(46° − 20°) ∴ M 2 = 1.49
9.64
p2 = 3.01× 40 = 120.4 kPa abs T2 = 1.423 × 303 = 431 K
V2 = 1.4 × 287 × 431 × 1.49 = 620 m/s
a detached
shock
c)
V1
= 35o
M1n = 3.5sin 35° = 2.01 ∴ M 2n = 0.576 T2 = 1.696 × 303 = 514 K
M 2 = 0.576/ sin(35° − 20°) = 2.26
9.66
θ1 = 20° = θ 2
∴ β 2 = 47°
M2n = 2.26sin 47° = 1.65 ∴ M3n = 0.654 = M3 sin(47° − 20°) ∴ M3 = 1.44
T3 = 1.423 × 514 = 731 K
V3 = M3 kRT3 = 1.44 1.4 × 287 × 731 = 780 m/s
139
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= 20o
Chapter 9 / Compressible Flow
M1 = 3, θ = 10° ∴ β1 = 28° M1n = 3sin 28° = 1.41 ∴ M 2n = 0.736
∴ p2 = 2.153 × 40 = 86.1 kPa abs
9.68
M2 =
0.736
= 2.38
sin(28° − 10°)
∴ p3 = 6.442 × 86.1 = 555 kPa abs
( p3 )normal = 10.33 × 40 = 413 kPa abs
Expansion Waves
θ1 = 26.4° For M = 4, θ = 65.8° (See Fig. 9.18.)
∴θ = 65.8 − 26.4 = 39.4°
9.70
T2 = T1
T0 T2
1
= 273
× 0.2381 = 117 K
T1 T0
0.5556
T2 = −156°C
∴ V2 = 4 1.4 × 287 × 117 = 867 m/s
a) θ1 = 39.1° θ 2 = 39.1 + 5 = 44.1° ∴ M u = 2.72
p2u = (20/0.0585) × 0.04165
= 14.24 kPa abs
9.72
For θ = 5° and M = 2.5, β = 27° M1n = 2.5sin 27° = 1.13 ∴ M 2n = 0.889
∴ p2A = 1.32 × 20 = 26.4 kPa abs
MA = M2 = 0.889/ sin(27° − 5°) = 2.37
If θ = 5° with M1 = 4, then Fig. 9.15 → β =18°
M1n = 4sin18° = 1.24 ∴ M 2n = 0.818
M1
M2u
∴ p2A = 1.627 × 20 = 32.5 kPa
∴ M 2A
shock
0.818
=
= 3.64
sin(18° − 5°)
At M1 = 4, θ1 = 65.8° At 75.8°, M2u = 4.88 p2u = p1
shock
M2l
p0 p2
0.002177
= 20
0.006586
p p0
= 6.61 kPa
CL =
CD =
Lift
32.5 A cos 5° − 20 × A/2 − 6.61× ( A/2) × cos10°
=
= 0.0854
1
1
2
2
ρ V1 A
×1.4 × 4 × 20 A
2
2
Drag
1
ρV12 A
2
=
32.5A sin5° − 6.61× ( A/2) × sin10°
= 0.010
1
2
×1.4 × 4 × 20A
2
140
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Chapter 10 / Flow in Open Channels
CHAPTER 10
Flow in Open Channels
Introduction
α = cos −1 (1 − 2 × 0.3) = 1.159 rad or 66.4°
Q2B
gA 3
=
10.2
=
Q 2d sin α
g ⎡ d 2/ 4 (α − sin α cos α ) ⎤
⎣
⎦
3
42 sin(1.159)
=1
9.81× (d5/ 64) × 1.159 − sin1.159cos1.159)3
This equation reduces to:
192.1 = d 5
∴ d = 2.86 m
Uniform Flow
10.4
141
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Chapter 10 / Flow in Open Channels
c1 A 5/ 3
S 0 . Substitute in
n P 2/ 3
appropriate expressions for A and P, and solve for y 0 by trial and error.
Use Chezy-Manning equation in the form Q =
(a)
35 =
1 2
⎡
⎤
⎢⎣ 4.5 y 0 + 2 y 0 (2.5 + 3.5) ⎥⎦
5/3
0.015 ⎡ 4.5 + y 0 ( 1 + 2.52 + 1 + 3.52 ) ⎤
⎣⎢
⎦⎥
0.00035
2/3
(4.5 y0 + 3y02 )5/3
This reduces to 28.06 =
(4.5 + 6.33y0 ) 2/3
Solving, y0 = 2.15 m, A = 4.5 × 2.15 + 3 × 2.152 = 23.5 m 2 ,
10.6
P = 4.5 + 6.33 × 2.15 = 18.1 m
(c)
⎡1.82 / 4 (α − sin α cos α ) ⎤
⎦
3.3 = ⎣
2/3
0.012(1.8α )
This reduces to 2.63 =
0.001
(α − sin α cos α )5 / 3
α 2/3
Solving, α = 1.94 rad, y0 =
A=
5/3
1.8
(1 − cos1.94) = 1.255 m
2
1.82
(1.94 − sin1.94 cos1.94) = 1.845 m2
4
P = 1.8 × 1.94 = 3.5 m
b = 0, m1 = 8, m2 = 0
1
1
8
A = by + y 2 (m1 + m2 ) = m1 y 2 = y 2 = 4 y 2
2
2
2
P = b+ y
10.8
Q = AR
(
2/3
) (
1 + m12 + 1 + m22 = y
2
⎞
S0
2 ⎛ 4y
= 4y ⎜
⎟
n
⎝ 9.062 y ⎠
2/3
) (
1 + m12 + 1 = y
0.0005
= 3.456 y8/3
0.015
(a)
y = 0.12 m, Q = 3.956(0.12)8 / 3 = 0.0121 m 3 /s
(b)
⎛ 0.08 ⎞
Q = 0.08 m /s, y = ⎜
⎟
⎝ 5.456 ⎠
3
)
65 + 1 = 9.062 y
3/8
= 0.244 m
142
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Chapter 10 / Flow in Open Channels
Energy Concepts
q = 2gy 2 (E − y ) , E = constant,
dq 1 4 gy ( E − y ) − 2 gy 2 2 gy ( E − y ) − gy 2
=
=
dy 2
q
2 gy 2 ( E − y )
10.10
Setting dq / dy = 0 and noting that q ≠ 0, so that the numerator is zero, one can
solve for y at the condition q = qmax (note: y = 0 is a trivial solution):
gy[ 2(E − y ) − y ] = 0,
2E − 3y = 0,
∴ y = 2 E/3 = yc
q = V1y1 = 3 × 2.5 = 7.5 m 2 /s
E1 = y1 +
q2
2 gy12
= 2.5 +
32
= 2.96 m
2 × 9.81
yc = 3 q 2/ g = 7.52/ 9.81 = 1.79 m, Ec =
3
3
yc = × 1.79 = 2.68 m
2
2
(a) E2 = E1 − h = 2.96 − 0.2 = 2.76 m, E2 > Ec ∴ y2 is subcritical
10.12
q2
7.5 2
2.87
∴ E2 = y 2 +
, or 2.76 = y 2 +
= y2 + 2
2
2
2gy 2
2 × 9.81y 2
y2
Solving, y2 = 2.13 m, and y2 + h − y1 = 2.13 + 0.2 − 2.5 = −0.17 m
(c) Set E2 = Ec = 2.68 m and ∴ maximum h is
hmax = E1 − E2 = E1 − Ec = 2.96 − 2.68 = 0.28 m
∴ yc + hmax − y1 = 1.79 + 0.28 − 2.5 = −0.43 m
(a) q1 = V1y1 = 3 × 3 = 9 m 2 /s, Q = b1q1 = 3 × 9 = 27 m 3 /s
E1 = y1 +
10.14
q12
2 gy12
= 3+
32
3
= 3.46 m, yc1 = 92 /9.81 = 2.02 m
2 × 9.81
3
Ec1 = × 2.02 = 3.03 m
2
Without change in width at loc. 2, E2 = E1 − h = 3.46 − 0.7 = 2.76 m
Since E2 < Ec1 , width must change to prevent choking. Set Ec2 = 2.76 m
2
∴ yc2 = × 2.76 = 1.84 m, q2 = gyc32 = 9.81×1.843 = 7.81 m2 /s
3
∴ b2 = Q / q2 = 27 / 7.81 = 3.46 m
143
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Chapter 10 / Flow in Open Channels
q2
5.52
E1 = y1 +
= 2.15 +
= 2.484 m
2 gy12
2 × 9.81× 2.152
Fr1 =
Ec =
q
gy13
=
5.5
9.81× 2.153
= 0.557 yc = 3
q 2 3 5.52
=
= 1.456 ≈ 1.46 m
9.81
g
3
3
yc = × 1.456 = 2.183 ≈ 2.18 m
2
2
(a) The maximum height of the raised bottom at location 2 will be one for
which the energy is a minimum:
E1 = Ec + h , 2.484 = 2.183 + h , ∴ h = 2.484 − 2.183 = 0.30 m
10.16
(b)
(c)
(d) Since Fr1 < 1, if h > 0.30 m, subcritical nonuniform flow will occur upstream
of the transition.
At the canal entrance, the condition of critical flow is Fr 2 = Q 2 B / gA 3 = 1, to be
solved for the unknown width b.
10.18
(a) Rectangular channel: B = b, A = by, y = yc
∴
10.20
Q 2b
3 3
gb y
=
Q2
2 3
gb y
=1 ∴b =
Q
gy 3
=
18
= 5.75 m
9.81 × 1
Objective is to determine the zero of the function f ( y ) = Q 2 B( y ) / g[ A ( y )]3 − 1,
in which eqns. for B(y) and A(y) representing either a circular or trapezoidal
section area can be substituted. The false position algorithm is explained in
Example 10.10; this was used to determine the roots. The solutions are:
(a) yc = 1.00 m, (c) yc = 1.81 m
144
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Chapter 10 / Flow in Open Channels
In the lower section ( y ≤ 1 m), the area is
y
y
A = ∫ bdη = ∫ 3 η dη = 2 y 3/ 2
0
Assume y < 1 m; then Fr 2 =
0
If yc = 1 m, then
Q2B
gA3
=
Q23 y
g 8 y 9/ 2
=
3Q 2
8 gy 4
3Q 2
= 1, or Q = 8 × 9.81/ 3 = 5.1 m 3 /s
4
8 × 9.81× 1
∴ when Q > 5.1 m 3 /s, critical depth will be > 1 m. Cross-sectional area including
10.22
upper region (y > 1 m) is
A = 2( 1) 3 /2 + 23( y c − 1) = 23 y c − 21 , and B = 23 m
(a)
Q 2B
55 2 × 23
=
= 1, ( 23y c − 21) 3 = 7092,
3
3
gA
9.81( 23y c − 21)
(7092)1/ 3 + 21
∴ yc =
= 1.75 m
23
(b)
Q2 B
3 × 3.52
=
= 1, yc4 = 0.468, ∴ yc = 0.83 m
gA3 8 × 9.81yc4
22.5
1.52
2
3
= 1.5 m /s, yc =
(a) q =
= 0.612 m
15
9.81
3
1.52
Ec = × 0.612 = 0.919 m, E0 = 1.2 +
= 1.28 m = E3
2
2 × 9.81× 1.2 2
E2 = E0 − h = 1.28 − 0.2 = 1.08 m > E3 ∴ no choking.
Since losses are neglected throughout the transition, y1 = y0, and y2 can be
computed by writing the energy equation between locations 1 and 2:
10.26
1.28 = y2 +
1.52
2 × 9.81× 1.2
2
+ h = y2 +
0.115
y22
+ 0.2 , ∴ y2 ≈ 0.95 m
(b)
145
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Chapter 10 / Flow in Open Channels
(a) B′ = 1 m, H′ = 0.36 m
10.28
Q = 0.11BH (1.522 B)
(b) H′ (m)
Q m3/s
0.026
⇒B=
B′
H′
= 3.281 and H =
= 1.18
0.3048
0.3048
= 0.11× (3.281)(1.18)(1.522×3.218)
0.026
= 0.42 m3 /s
0.15
0.2
0.25
0.30
0.35
0.4
0.45
0.17
0.23
0.29
0.35
0.41
0.47
0.53
⎛2 2 ⎞
⎛2 2 ⎞
Q1 = b ⎜⎜
g ⎟⎟ ( y1 − h)3 / 2 , Q2 = b ⎜⎜
g ⎟⎟ ( y2 − h)3 / 2
⎝3 3 ⎠
⎝3 3 ⎠
given Q1 , y1 , Q2 , y2 , solve for b and h
y1 − h ⎛ Q1 ⎞
=⎜
⎟
y2 − h ⎝ Q2 ⎠
10.30
(a)
h=
2/3
or h =
y1 − y2 (Q1 / Q2 ) 2 / 3
1 − (Q1 / Q2 ) 2 / 3
1.05 − 1.75(0.15 / 30) 2 / 3
= 1.03 m
1 − (0.15 / 30) 2 / 3
∴b =
Q2
30
=
= 28.8 m
⎞
2⎛ 2 ⎞
2⎛ 2
3/ 2
3/ 2
g ⎟ ( y 2 − h)
× 9.81 ⎟ (1.75 − 1.03)
⎜
⎜
3⎝ 3 ⎠
3⎝ 3
⎠
Momentum Concepts
⎛ y2 q2 ⎞
M = b⎜
+ ⎟
⎝ 2 gy ⎠
2
q2
M
1⎛ y ⎞
= ⎜ ⎟ +
by c2 2 ⎝ y c ⎠
gy c2 y
10.32
2
q2
1⎛ y ⎞
= ⎜ ⎟ + 3
2 ⎝ yc ⎠
gyc ( y / yc )
But q 2 /gyc3 = 1
2
M
1⎛ y ⎞
1
∴ 2 = ⎜ ⎟ +
(y / yc )
by c 2 ⎝ y c ⎠
146
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Chapter 10 / Flow in Open Channels
(a) Conditions at location 1:
E1 = y1 +
q12
2 gy12
= 1.8 +
d/2
1.5 2
2 × 9.81 × 1.8 2
= 1.835 m
q1
b2
b1
q2
d/2
3
yc = 3 q12 /g = 1.52 /9.81 = 0.612 m.
The smallest constriction at location 2 is one that establishes critical flow, i.e.,
where minimum energy exists:
∴ E 2 = E c , or since E 1 = E 2 ,
2
2
2
yc2 = Ec2 = E1 = × 1.835 = 1.22 m
3
3
3
∴ q2 = g ( yc2 )3 = 9.81×1.223 = 4.22 m 2 /s
10.34
b2 =
q1b1 1.5 × 6
=
= 2.13 m
4.22
q2
Hence the maximum diameter is d = b 1 − b 2 = 6 − 2.13 = 3.87 m
The momentum eqn. cannot be used to determine the drag since no information
is given with regard to location downstream of cofferdam. But we do have
available the drag relation, which will provide the required drag force:
Frontal area:
A = y1d = 1.8 × 3.87 = 6.97 m 2
Approach velocity: V1 = q1 / y1 = 1.5 /1.8 = 0.833 m/s
∴ F = C D A ρ V12 / 2 = 0.15 × 6.97 × 1000 × 0.8332 / 2 = 363 N , a rather
insignificant drag force!
Asill =
=
1
wh
2
1
× 2 × 3 × 0.3
2
= 0.27 m 2
10.36
y=
1
y
3
A = my 2 = 3 y 2
147
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Chapter 10 / Flow in Open Channels
M1 = A1y1 +
Q2
Q2
Q2
1
= 3 × 0.52 × × 0.5 +
=
+
0.125
,
gA1
3
7.358
9.81× 3 × 0.52
Q2
1
Q2
Q2
2
M 2 = A2 y 2 +
= 3 × 1.8 × × 1.8 +
= 5.832 +
,
gA2
3
95.35
9.81 × 3 × 1.8 2
F
γ
=
CD AsillQ2
2 gA12
=
0.4 × 0.27Q2
2 × 9.81× (3 × 0.52 )2
=
Q2
102.2
Substitute into momentum eqn: M1 = M 2 + F /γ
∴ 0.125 +
Q2
Q2
Q2
= 5.832 +
+
7.358
95.35 102.2
1
1 ⎞
⎛ 1
−
−
Q2 = (5.832 − 0.125) ⎜
⎟
⎝ 7.58 95.35 102.2 ⎠
−1
= 49.37
∴ Q = 7.03 m3 /s
10.38
Given condition:
V2 y2 = 0.4V1y1
Continuity eqn:
y1 (V1 + w) = y2 (V2 + w)
Momentum eqn:
y2 1 ⎡
(V + w ) 2 ⎤
= ⎢ 1+ 8 1
− 1⎥
y1 2 ⎢
gy1
⎥⎦
⎣
Unknowns are V2 , y2 and w. Eliminate V2 and w, and solve for y2 .
(a) V2 y2 = 0.4 × 1 × 1.5 = 0.6
1.5( 1 + w ) = y 2V2 + y 2 w = 0.6 + y 2 w, or w = 0.9 /( y2 − 1.5)
2
⎡
⎤
⎛
y2 1 ⎢
0.9 ⎞
⎥
∴
=
1 + 8 ⎜1 +
⎟ /(9.81 × 1.5) − 1⎥ or
⎢
1.5 2
⎝ y2 − 1.5 ⎠
⎣⎢
⎦⎥
2
⎛
0.9 ⎞
1 − 1 + 0.5437 ⎜1 +
⎟ + 1.333y2 = 0
⎝ y2 − 1.5 ⎠
148
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Chapter 10 / Flow in Open Channels
∴Solving, y2 = 1.77 m
∴ V2 =
∴w =
0.6 0.6
=
= 0.339 m/s
y 2 1.77
0.9
0.9
=
= 3.33 m/s
y2 − 1.5 1.77 − 1.5
(a) To compute the discharge, write the energy equation between locations
1 and 2, and solve for q:
2 g ( y2 − y1 )
q=
(
y1−2
− y2−2
)
=
2 × 9.81(0.10 − 2.5)
( 2.5
−2
− 0.10
−2
)
= 0.687 m 2 /s
∴ Q = bq = 5 × 0.687 = 3.43 m 3 /s
(b) To compute depth downstream, first compute the Froude number at 2:
10.40
Fr2 =
∴ y3 =
q
gy
y2
2
3
2
(
=
0.687
9.81 × 0.10 3
)
1 + 8 Fr22 − 1 =
= 6.934
0.10
2
(
)
1 + 8 × 6.934 2 − 1 = 0.932 m
(c) To calculate power lost, first compute the head loss in the jump:
( y 3 − y 2 )3
(0.932 − 0.10 ) 3
hj =
=
= 1.544 m
4y 3 y 2
4 × 0.932 × 0.10
= γ Qh = 9810 × 3. 43 × 1.544 = 5.20 × 10 4 watt, or 52.0 kW
∴W
j
Use Eqn. 10.5.16 with F = 0:
M1 =
=
10.42
y 12
Q2
( 2my 1 + 3b) +
6
g(by 1 + my 12 )
1.12
602
(2 × 3 × 1.1 + 3 × 5) +
= 44.55 m3
2
6
9.81(5 × 1.1 + 3 × 1.1 )
∴ M 2 = M1 or
y22
Q2
(2my2 + 3b) +
= 44.55
6
g (by2 + my22 )
y22
602
(2 × 3y2 + 3 × 5) +
= 44.55
6
9.81(5y2 + 3y22 )
149
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Chapter 10 / Flow in Open Channels
y 23 + 2.5 y 22 +
367
= 44.55
5 y 2 + 3 y 22
∴Solving, y2 = 2.55 m
Energy loss across jump is h j :
h j = E1 − E2 = y1 +
Q2
Q2
y
−
−
2
2 gA12
2 gA22
A1 = 5 × 1.1 + 3 × 1.12 = 9.13 m 2 , A2 = 5 × 2.55 + 3 × 2.552 = 32.26 m 2 ,
∴ h j = 1.1 +
60 2
60 2
2.55
−
+
= 0.575 m
2 × 9.81 × 9.132
2 × 9.81 × 32.26 2
:
∴Power dissipated is W
j
= γ Qh = 9810 × 60 × 0.575 = 3.38 ×105 W
W
j
j
First find Q, then compute y 1 :
V2 = Fr2 gy2 = 0.4 9.81× 1.5 = 1.534 m/s
∴ Q = V 2 A2 = 1.534 × 5 × 1.5 = 11.51 m 3 /s
q = Q / b = 11.51/ 5 = 2.30 m 2 /s
yc = 3 q 2 / g = 3 2.302 / 9.81 = 0.814 m
To compute y2 , use momentum eqn, M1 − F / γ = M2 or
10.44
⎛ y12 q 2 ⎞ C D Aρ V12 / 2 ⎛ y22 q 2 ⎞
=⎜ +
⎜⎜ +
⎟⎟ −
⎜ 2 gy ⎟⎟
bγ
2 ⎠
⎝ 2 gy1 ⎠
⎝
Substitute in known data, plus the relation V1 = q / y 1 :
y12 2.302 0.35 × 5 × 0.17 × 1000 × 2.302 1.52
2.302
+
−
=
+
2 9.81y1
2
5 × 9810 y12
9.81 × 1.52
which reduces to y12 +
1.078 0.0321
−
= 2.97
y1
y12
∴Solving, y1 = 0.346 m
150
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Chapter 10 / Flow in Open Channels
Nonuniform Gradually Varied Flow
3
yc = 3 q 2 /g = 0.1942 /9.81 = 0.157 m
(a) q = Q / b = 0.35 /1.8 = 0.194 m 2 /s
Use Chezy-Manning eqn. to compute y 0 :
(1.8 y0 )5/3
0.35 × 0.012
Qn
=
= 0.0329 =
(1.8 + 2 y0 ) 2/3
c1 S0 1× 0.0163
∴Solving, y0 = 0.094 m
Ec + h =
3
3
yc + h = × 0.157 + 0.1 = 0.335 m
2
2
E0 = y0 +
q2
2 gy02
10.46
0.1942
= 0.311 m
2 × 9.81× 0.0942
= 0.094 +
Since E 0 < E c + h , normal conditions cannot exist at
loc. 1, and choking will occur at loc. 2. Compute
alternate depths at locs. 1 and 3: E 1 = E 3 = E c + h.
y+
0.1942
2 × 9.81y
2
= y+
0.00192
y2
= 0.335
∴Solving, y 1 = 0.316 m, y 3 = 0.088 m. (Note: loc. 2 is a critical control, with
subcritical flow upstream and supercritical flow downstream.) A jump is located
upstream of loc. 1. Find the depth conjugate to y0:
Fr03 = q2/gyo3 = 0.1942 /(9.81× 0.0943 ) = 4.62
∴ycj =
y0
2
( 1+ 8× 4.62 −1) = 0.243 m
( 1+ 8Fr −1) = 0.094
2
2
0
3
yc = 3 q 2 /g = 8.252 /9.81 = 1.91 m
q = Q / b = 33 / 4 = 8.25 m 2 /s
Use Chezy-Manning eqn. to compute y 0 :
10.48
(4 y0 )5 / 3
Qn
33 × 0.012
=
= 13.43 =
c1 S0 1 × 0.00087
(4 + 2 y0 ) 2 / 3
∴Solving, y0 = 2.98 m
Since y 0 > y c , mild slope conditions prevail. With yentrance < yc , an M3 profile
exists downstream of the entrance. For free outfall conditions, yexit ≅ yc , and an
M 2 profile exists upstream of the exit. The M 3 and M 2 profiles are separated by
a hydraulic jump located approximately 260 m downstream of the entrance
(determined by numerical analysis).
151
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Chapter 10 / Flow in Open Channels
A spreadsheet solution is shown below. An M3 profile is situated upstream, with a
hydraulic jump at approximately 150 m followed by an M1 profile.
Q=
20
n=
0.014
S0 =
0.001
L=
300 g =
9.81
b=
0
m1 =
3.5
yu =
0.50
yd =
2.50 c1 =
1.00
m2 =
2.5
Depth Residual
10.50
Station
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
yc =
1.554
y0 =
1.809 -1.8E-09
y
0.500
0.600
0.700
0.800
0.900
1.000
1.100
1.200
1.300
1.400
2.500
2.450
2.400
2.350
2.300
2.250
2.200
A
0.750
1.080
1.470
1.920
2.430
3.000
3.630
4.320
5.070
5.880
18.750
18.008
17.280
16.568
15.870
15.188
14.520
0.00
V
E
ym
S(ym)
ycj
FM Residual
Δx x
26.667 36.744
0 3.769 54.49 2.77E-04
18.519 18.079 0.550 5.720E-01 33 33 3.324 37.97 -5.67E-05
13.605 10.135 0.650 2.347E-01 34 67 2.837 28.08 3.64E-06
10.417 6.330 0.750 1.094E-01 35 102 2.578 21.75 -5.48E-04
8.230 4.353 0.850 5.612E-02 36 138 2.365 17.51 -1.19E-04
6.667 3.265 0.950 3.101E-02 36 174 2.184 14.59 -2.52E-05
5.510 2.647 1.050 1.818E-02 36 210 2.028 12.56 -4.53E-06
4.630 2.292 1.150 1.119E-02 35 245 1.893 11.17 -1.48E-04
3.945 2.093 1.250 7.175E-03 32 277 1.778 10.24 -6.60E-05
3.401 1.990 1.350 4.760E-03 28 304 1.684 9.68 -3.90E-05
1.067 2.558
300
1.111 2.513 2.475 1.878E-04 -56 244
1.157 2.468 2.425 2.094E-04 -56 188
1.207 2.424 2.375 2.340E-04 -57 131
1.260 2.381 2.325 2.621E-04 -59 72
1.317 2.338 2.275 2.943E-04 -60 12
1.377 2.297 2.225 3.313E-04 -62 -51
q = Q /b = 15/4 = 3.75 m 2 /s
3
yc = 3 q 2 /g = 3.752 /9.81 = 1.13 m
y01 = 0.93 m, yc > y01
∴upstream reach has a steep slope.
y02 = 1.42 m, yc < y02
10.52
∴downstream reach has a mild slope.
Compute depth conjugate to y 0 1 :
Fr012 = q 2 /( gy013 ) = 3.752 /(9.81 × 0.933 ) = 1.782
∴ ycj =
y01 ⎡
0.93 ⎡
1 + 8Fr012 −1⎤ =
1 + 8 ×1.782 −1⎤⎦ = 1.35 m
⎢
⎥
⎦
2 ⎣
2 ⎣
152
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Chapter 10 / Flow in Open Channels
∴Hydraulic jump occurs upstream, followed by an S 1 curve up to the transition.
10.54
q 2 3 5.752
=
= 1.50 m. Since y01 < yc and y02 > yc, there will be a
g
9.81
hydraulic jump between A and C.
(a) yc = 3
(a) Compute y c using Fr 2 = 1:
Q2B
gA 3
=
Q 2 d sin α
g ⎡ (d 2 /4)(α − sin α cos α ) ⎤
⎣
⎦
= 0.4175
3
=
2.52 × 2.5sin α
9.81 ⎡ (2.52 /4)(α − sin α cos α ) ⎤
⎣
⎦
3
sin α
=1
(α − sin α cos α )3
∴Solving
d
2
α = 1.115 rad and ∴ yc = (1 − cosα ) =
2.5
(1 − cos1.115) = 0.70 m
2
Compute y 0 using
10.56
⎡ (d 2 /4)(α − sin α cos α ) ⎤
Qn
⎦
=⎣
c1 S0
(α d ) 2/3
which reduces to 1.039 =
5/3
2
⎡
⎤
2.5 × 0.015 ⎣(2.5 /4)(α − sin α cos α ) ⎦
=
,
1× 0.001
(α × 2.5)2/3
5/3
(α − sin α cos α )5 / 3
α 2/3
∴Solving, α = 1.361 rad, and
d
2.5
∴ y0 = (1 − cosα ) =
(1 − cos1.361) = 0.99 m
2
2
Since y 0 > y c , a mild slope condition exists. The water surface consists of an
M 3 curve beginning at the inlet, followed by a hydraulic jump to an M 2 curve,
which terminates at critical depth at the outlet. The numerically-predicted water
surface and energy grade line are provided in the following table.
153
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Chapter 10 / Flow in Open Channels
x
(m)
0
7
14
20*
20*
230
350
(a) q =
E
(m)
1.64
1.44
1.29
1.18*
1.07*
1.065
1.05
30
= 6 m2 /s
5
62
E1 = 2 +
2 × 9.81× 22
E2 = 1.8 +
10.58
y
(m)
0.40
0.43
0.46
0.49*
0.97*
0.96
0.93
x
y
E
(m)
(m)
(m)
406
0.90 1.03
442
0.87 1.01
465
0.84 1.00
480
0.82 0.98
490
0.79 0.97
500
0.70 0.95
*
hydraulic jump
yc = 3
62
=1.54 m
9.81
= 2.459 m
62
2 × 9.81× 1.82
= 2.366 m
ym = 1.9 m, Am = 5 × 1.9 = 9.5 m 2 , Pm = 5 + 2 × 1.9 = 8.8 m, Rm = 9.5 / 8.8 = 1.08 m
Q 2 n2
1
⎛ 30 × 0.012 ⎞
=⎜
= 0.0036
⎟
2 4/3
4/3
Am Rm
9.5
⎝
⎠ 1.08
2
S( y m ) =
∴Δx =
E2 − E1
2.366 − 2.459
=
= 20.2 m
S0 − S( ym )1 −0.001 − 0.0036
(b) An A2 profile is contained within the reach (which has an adverse slope with
y2 < y1)
Evaluate Q using Δx =
10.60
E2 − E1
S0 − S( ym )
E1 = y1 +
Q2
Q2
Q2
=
1.05
+
=
1.05
+
135.2
2 gA12
2 × 9.81 × (2.5 × 1.05) 2
E2 = y2 +
Q2
Q2
Q2
=
1.2
+
=
1.2
+
176.6
2 gA22
2 × 9.81× (2.5 × 1.2)2
ym =
1
1
( y1 + y2 ) = (1.05 + 1.2) = 1.125 m
2
2
2
⎛ Qn ⎞ (b + 2 ym ) 4 / 3
S ( ym ) = ⎜
⎟
10 / 3
⎝ c1 ⎠ (bym )
154
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Chapter 10 / Flow in Open Channels
=
Q2 × 0.0132 × (2.5 + 2 ×1.125)4/ 3
= 4.296 × 10−5 Q 2
12 × (2.5 ×1.125)10/ 3
∴Δ x[S0 − S( ym )] = E2 − E1
50(0.005 − 4.296 × 10−5 Q 2 ) = 1.2 +
Q2
Q2
− 1.05 −
176.6
135.2
which reduces to 4.141 × 10 −4 Q 2 = 0.10 ∴ Solving, Q = 15.5 m3 /s
⎛Q⎞
yc = 3 ⎜ ⎟
⎝b⎠
2
⎛ 15.5 ⎞
g =3⎜
⎟
⎝ 2.5 ⎠
2
9.81 = 1.58 m
∴The profile is an S3 curve
Write energy eqn. across the gate to compute the discharge:
E1 = E2
or
y1 +
q2
2 gy12
= y2 +
q2
2 gy22
⎛ 1
1⎞
∴ q = 2g( y 1 − y 2 )⎜ 2 − 2 ⎟ − 1
⎝ y2 y1 ⎠
1 ⎞
⎛ 1
= 2 × 9.81× (1.85 − 0.35) ⎜
−
− 1 = 1.93 m2 /s
2
2⎟
⎝ 0.35 1.85 ⎠
∴Q = bq = 4 ×1.93 = 7.72 m3/s
yc = 3 q 2 / g = 1.932 / 9.81 = 0.72 m
10.62
Compute y 0 using Chezy-Manning relation:
(4 y0 )5/ 3
Qn
7.72 × 0.014
=
= 3.821 =
(4 + 2 y0 )2 / 3
c1 S0 1× 0.0008
∴ Solving, y 0 = 1.17 m. Since y 0 > y c , mild channel conditions exist.
Upstream of the gate there will be an M 1 profile, with an M 3 downstream of the
gate, terminating in a hydraulic jump to normal flow conditions.
Compute the depth upstream of the jump, conjugate to y 0 :
Fr02 = q 2 /gy03 = 1.932 /(9.81× 1.173 ) = 0.237
∴ ycj =
y0
2
( 1 + 8Fr − 1) = 1.172 ( 1 + 8 × 0.237 − 1) = 0.41 m
2
0
155
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Chapter 10 / Flow in Open Channels
Use the step method to compute water surface and energy grade line:
q2
1.93 2
0.190
E=y+
=
y
+
=y+
2
2
y2
2gy
2 × 9.81y
S( y ) =
Q 2 n 2 ( 4 + 2y ) 4 /3 7.72 2 × 0.014 2 ( 4 + 2y ) 4 /3
=
4 10 /3 y 10 /3
( 4 y ) 10 /3
= 1.15 × 10 −4 ( 4 + 2y ) 4 /3 y −10/3
M 1 curve upstream of gate
y
E
ym
(m)
1.85
(m)
1.906
(m)
1.7
1.5
1.3
1.2
S( y m )
Δx
x
(m)
(m)
0
1.775
2.514×10−4
−255
1.6
3.336×10−4
−390
1.4
4.824×10−4
−542
1.25
6.627×10−4
−583
1.766
1.584
1.412
1.332
−255
−645
−1187
−1770
M 3 curve downstream of gate
y
E
ym
(m)
0.35
(m)
1.901
(m)
0.37
1.758
0.39
0.41
S( y m )
Δx
x
(m)
(m)
0
0.36
0.02741
5
0.38
0.02315
5
0.40
0.01973
5
5
1.639
10
1.540
15
Hydraulic jump occurs ~5 m from the gate.
(a)
10.64
156
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Chapter 10 / Flow in Open Channels
Compute normal depth using Chezy-Manning eqn:
(3.66 y02 )5/3
Qn
15.38 × 0.017
=
= 5.476 =
(3.66 + 2 y02 ) 2/3
c1 S0 1× 0.00228
∴ Solving, y02 = 1.65 m
Compute critical depth:
q = Q /b = 15.38/3.66 = 4.20 m 2 /s
3
∴ yc = 3 q 2 /g = 4.202 /9.81 = 1.22 m
∴Upstream of slope change the channel is steep, and downstream, the channel is
mild. The hydraulic jump will terminate at normal depth.
(a) Find depth conjugate to y 02 :
Fr022 = q 2 /gy023 = 4.20 2 /(9.81 × 1.653 ) = 0.400
1.65
1 + 8Fr − 1) =
( 1 + 8 × 0.4 − 1) = 0.87 m
(
2
2
y0 2
∴ ycj =
10.66
2
02
(b) Step method:
E= y+
q2
2 gy 2
= y+
4.20 2
0.900
= y+ 2
2
2 × 9.81y
y
Q 2 n 2 (b + 2y ) 4 /3 15.38 2 × 0.017 2 ( 3.66 + 2y ) 4 /3
S( y ) =
=
(by ) 10 /3
( 3.66) 10 /3 y 10 /3
= 9.052 ×10−4 (3.66 + 2y)4/3 y−10/3
E
ym
(m)
0.6
(m)
3.100
(m)
0.7
2.537
y
0.8
0.87
S( y m ) Δx
x
(m)
0.65
0.03216
18.8
0.75
0.02104
17.6
0.835
0.01536
11.2
(m)
0
18.8
2.206
36.4
2.059
47.6 ≅ L j
(c) An M 3 curve exists downstream of the change in channel slope, terminating
in a hydraulic jump, approximately 50 m from the slope change.
157
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Chapter 10 / Flow in Open Channels
This is a design analysis problem. Some extensive calculations are required.
(a) Find the depth of flow immediately upstream of the jump (call it location 1):
V2 =
Fr2 =
Q
19
19
=
= 1.267 m/s, q =
=0.95 m 2 /s.
A2 0.75 × 20
20
1.267
= 0.467,
9.81× 0.75
y1 =
0.75
2
( 1 + 8 × 0.467 − 1) = 0.246 m
2
Compute the loss across the jump, and subsequently the dissipated power:
0.952
0.952
−
0.75
−
= 3.87 m
2 × 9.81× 0.102
2 × 9.81× 0.752
h j = E1 − E2 = 0.10 +
= γ Qh = 9810 × 19 × 3.87 = 7.21× 105 watt, or 721 kW
∴W
j
10.68
(b) The length of the apron is the distance from the toe to downstream of the
jump. First, compute the distance from ytoe to y1 using a single increment of
length:
Etoe = 0.10 +
ym =
0.95 2
2 × 9.81 × 0.10 2
= 4.700 m, E1 = 0.246 +
0.95 2
2 × 9.81 × 0.246 2
= 1.006 m,
1
(0.10 + 0.246) = 0.173 m, A m = 20 × 0.173 = 3.46 m 2 ,
2
⎛ 19 × 0.014 ⎞
Pm = 20 + 2 × 0.173 = 20.35 m, Sm = ⎜
⎟
⎝ 3.46 ⎠
∴ x1 − xtoe =
2
1
( 3.46 / 20.35 )4/3
= 0.06275,
E1 − Etoe
1.006 − 4.700
=
= 59.3 m
S0 − Sm 0.0005 − 0.06275
The length of the jump is six times the downstream depth, or 6 × 0.75 = 4.5 m.
Hence the required length of the apron is L = 59.3 + 4.5 = 63.8 m, or
approximately 65 m. Normal design would require additional length as a safety
factor.
Use the varied flow function to compute the water surface profile.
⎛ q 2 n2 ⎞
y0 = ⎜
⎜ S ⎟⎟
⎝ 0 ⎠
10.70
3/10
1/ 3
⎛ q2 ⎞
yc = ⎜ ⎟
⎜ g⎟
⎝ ⎠
⎛ 1.52 × 0.0152 ⎞
=⎜
= 1.627 m
⎜ 0.0001 ⎟⎟
⎝
⎠
⎛ 1.52 ⎞
=⎜
= 0.612 m
⎜ 9.81 ⎟⎟
⎝
⎠
Therefore the estuary has a mild slope and an M1 curve exists upstream of the
outlet. For a wide rectangular channel (Ex. 10.16), J = 2.5, N = 3.33, M = 3, and
N/J = 1.33. Use Eq. 10.7.13 to compute x:
158
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Chapter 10 / Flow in Open Channels
3
⎤
1.627 ⎡
⎛ 0.612 ⎞ 2.5
x=
F (v, 2.5) ⎥
⎢u − F (u,3.33) + ⎜
⎟
0.0001 ⎣⎢
⎝ 1.627 ⎠ 3.33
⎦⎥
= 16, 270 [u − F (u,3.33) − 0.04 F (v, 2.5) ]
The results of the calculations are as follows:
y (m)
7
6
5
4
2
u
4.30
3.67
3.07
2.46
1.23
F(u,3.33)
0.015
0.021
0.032
0.052
0.348
v
6.96
5.67
4.45
3.31
1.32
F(v,2.5)
0.038
0.053
0.074
0.117
0.572
x (m)
69,800
59,700
49,500
39,500
14,700
x – 69,800 (m)
0
−10,100
−20,300
−30,300
−55,100
This is a design analysis problem, and requires extensive calculations that would
best be performed on a spreadsheet. Let location A be 400 m upstream of location
B. Conditions at B are known. The energy equation between A and B is predicted
using total energy H = y + z + α V 2 / 2 g :
L
H*A ≈ HB + hL ≈ HB + (SA + SB )
2
In the equation, HB and SB are known, and HA and SA are unknown. A trial and
error solution is required. With the table of given data, location A is associated
with x = 0, and location B with x = 400 m. The following parameters are
computed at location B, where yB = 3.0 m:
Location
10.72
Side channel
Main channel
AB (m2)
60
273
PB
(m)
149
97
KB (Eq. 10.7.8)
RB
(m)
0.40
2.83
11
18,210
Now αB, VB, HB, and SB can be computed:
(60 + 273)2 ⎛ 113 18,2103 ⎞
+
⎜
⎟ = 1.48
(11 + 18,210)3 ⎝ 602
2732 ⎠
Q
280
VB =
=
= 0.84 m/s
AB 60 + 273
αB =
SB =
280 2
= 0.000236
(11 + 18, 210)3
H B = 3.0 + 15.1 + 1.48 ×
(Eq. 10.7.9)
(Eq. 10.7.7)
0.84 2
= 18.153 m
2 × 9.81
159
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Chapter 10 / Flow in Open Channels
With L = 400 m, the energy equation between A and B is predicted:
H *A = 18.153 +
400
(SA + 0.000236) = 18.20 + 200SA
2
The total energy at location A is
H A = yA + zA + α A
VA2
V2
= y A + 15.0 + α A A
2g
2g
The trial and error solution proceeds by assuming values of yA, computing the
corresponding αA, VA, SA, HA, and H *A until the latter two are in close agreement.
For yA = 3.2 m, the hydraulic parameters at location A are provided in this table:
AA (m2)
172
342
Location
Side channel
Main channel
PA (m)
116
112
RA (m)
1.48
3.05
KA (Eq. 10.7.8)
4470
24,000
The corresponding values of HA and H A* are
αA =
⎛ 44703 24, 0003 ⎞
+
⎜
⎟ = 1.39
(4470 + 24, 000)3 ⎜⎝ 1722
3422 ⎟⎠
VA =
Q
280
=
= 0.54 m/s
AA 172 + 342
SA =
(172 + 342) 2
2802
(4470 + 24, 000)
3
= 9.67 × 10−5
H A = 3.2 + 15.0 + 1.39 ×
(Eq. 10.7.9)
(Eq. 10.7.7)
0.542
= 18.22 m
2 × 9.81
H *A = 18.200 + 200 × 0.0000967 = 18.22 m
Hence the depth at the upstream location is yA = 3.2 m
160
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Chapter 11 / Flows in Piping Systems
CHAPTER 11
Flows in Piping Systems
Steady Flows
W f
Q
Substitute V = and H P =
into the energy equation:
A
γQ
p1
γ
+
W f
p2
Q2
Q2
+
=
+
+ hL
γ 2 gA22
2 gA12 γ Q
(a) Note that the kinetic energy terms can be neglected:
11.2
π
π
A1 = × 0.052 = 0.00196 m2 , A2 = × 0.082 = 0.00503 m2 ,
4
4
Q=
0.095
= 0.00158 m3 /s
60
W f
350 × 103
0 .0 0 1 5 8 2
760 × 103
0 .0 0 1 5 8 2
+
+
=
+
+ 6 .6
9810
9 8 1 0 × 0 .0 0 1 5 8
9810
2 × 9 .8 1 × 0 .0 0 1 9 6 2
2 × 9 .8 1 × 0 .0 0 5 0 3 2
= 750 watt
∴W
f
= 77.5 + 0.005 + 6.6
35.7 + 0.03 + 0.0645W
f
Write energy eqn. between locs. 1 and 2:
H P + z1 = z2 +
f
( L + Le )
D
Q2
⎛π
⎞
2g ⎜ D2 ⎟
4
⎝
⎠
2
1/ 5
⎡
⎤
Q2 ( L + Le ) f
Solving for D 1 , D = ⎢
⎥
2
⎣ 2g(π / 4) ( HP − (z2 − z1 )) ⎦
11.4
Substitute the relations
⎡ ⎛
e
⎞⎤
f = 1.325 ⎢ ln ⎜ 0.27 + 5.74 Re −0.9 ⎟ ⎥
D
⎠⎦
⎣ ⎝
−2
and Re =
4Q
ΣkD
, Le =
,
f
π Dν
and solve for D by successive substitution.
(a) Compute H P from pump data using linear interpolation:
H P = 136 − (136 − 127)
(0.757 − 0.95)
= 129.1 m
(0.757 − 1.01)
161
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Chapter 11 / Flows in Piping Systems
Q = 0.95 m3 /s
1/ 5
⎡
⎤
0.952 (500 + Le ) f
∴D = ⎢
⎥
2
⎣ 2 × 9.81(π / 4) (129.1 − 36) ⎦
Re =
D (m) e/D
0.3 0.003
0.4 0.0023
0.39 0.0023
0.39
= 0.24[(500 + Le ) f ]1/ 5
1.401×106
D
Le =
f
4.67 × 106
3.5 × 106
3.6 × 106
0.026
0.024
0.024
2.50
(m)
f
29
41.6
40.6
∴ D = 1.39 m.
(a) Energy eqn. from A to B:
⎛ ΣK1 ΣK2 ⎞ 2
⎛p
⎞ ⎛p
⎞
1.85
+
Q
⎜ + z ⎟ − ⎜ + z ⎟ = ( R1 + R2 + R3 )Q + ⎜⎜
2
2⎟
⎟
⎝γ
⎠A ⎝ γ
⎠B
⎝ 2 gA1 2 gA2 ⎠
Use Hazen-Williams resistance formula for R:
R=
10.59 L
C1.85D4.87
∴ R1 =
ΣK1
2 gA12
11.6
ΣK 2
2 gA22
10.59 × 200
= 1071
1001.85 × 0.24.87
=
10.59 ×150
2
= 193.4,
= 103.3, R2 =
2
4
1201.85 × 0.254.87
2 × 9.81 × (π / 4) × (0.2)
=
3
= 63.5,
2 × 9.81× (π / 4)2 × (0.3)4
R3 =
10.59 × 300
= 271.1
901.85 × 0.34.87
Substitute known data into energy eqn:
250 − 107 = (1071 + 193 + 271)Q1.85 + (103.3 + 63.5)Q 2
143 = 1535Q1.85 + 167Q2
F (Q) = 1535Q1.85 + 167Q2 − 143,
Iter.
1
2
3
Q(m3 /s)
0.277*
0.265
0.265
F ′(Q) = 2840Q0.85 + 334Q
F′
F
12.60
0.284
Δ Q = − F /F ′
1046
1007
− 0.01205
− 0.00028
162
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Chapter 11 / Flows in Piping Systems
∴Q = 0.265 m3 /s
*
1/1.85
⎛ 143 ⎞
Q≅⎜
⎟
⎝ 1535 ⎠
(1st iteration)
Write the energy equation from the liquid surface (location 1) to the valve exit
(location 2):
p1
γ
+ z1 +
W f
2
L
⎛
⎞ Q
= ⎜ ΣK + f
+ 1⎟
+ z2
D
γQ ⎝
⎠ 2 gA 2
(a) Rearrange and substitute in known data:
P1
11.8
γ
+ z1 − z2 +
1
W
fL ⎞ Q 2
⎛
f
− ⎜ ΣK + + 1⎟
=0
D ⎠ 2 gA2
γ Q ⎝
110 ×103
1×104 1
+ 24 − 18 +
0.68 × 9810
0.68 × 9810 Q
(0.5 + 3 × 0.26 + 2 + 0.015 × 450 / 0.30 + 1) 2
Q =0
−
2 × 9.81× (0.7854 × 0.302 )2
1.50
22.4 +
− 273.2Q 2 = 0
Q
By trial and error, Q ≈ 0.32 m3/s
Initially work between locs. B and C:
11.10
∴W =
Le =
8 f ( L + Le )
DΣk
, R=
f
gπ 2 D5
Le1 =
1.2 × 2
8 × 0.015 × 260
= 160, R1 =
= 0.1295,
0.015
9.81× π 2 × 1.25
Le 2 =
1× 3
8 × 0.02 × 1150
= 150, R2 =
= 1.900,
0.02
9.81× π 2 × 15
Le 3 =
0.5 × 2
8 × 0.018 × 1556
= 56, R3 =
= 74.05,
0.018
9.81× π 2 × 0.55
Le 4 =
0.75 × 4
8 × 0.021 × 943
= 143, R4 =
= 6.895
0.021
9.81 × π 2 × 0.755
Q2
⎛ 4
⎞
⎜Σ 1 ⎟
⎜ i =2 Ri ⎟
⎝
⎠
2
=
32
1
1 ⎞
⎛ 1
+
+
⎜
⎟
74.05
6.895 ⎠
⎝ 1.9
2
= 6.022 m,
163
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Chapter 11 / Flows in Piping Systems
∴Q2 = W / R2 = 6.022 /1.9 = 1.780 m3 /s,
Q3 = W / R3 = 6.022 / 74.05 = 0.285 m3 /s,
Q4 = W / R4 = 6.022 / 6.895 = 0.935 m3 /s.
Check continuity: Q2 + Q3 + Q4 = 1.78 + 0.285 + 0.935 = 3 = Q1, ∴ OK
, write energy equation from loc. A to B:
To find W
P
⎛p
⎞
H P = R1Q12 + ⎜ + z ⎟ = R1Q12 + W + 20
⎝γ
⎠B
∴ H P = 0.1295 × 32 + 6.022 + 20 = 27.2 m
= 1.07 MW
= γ Q1HP = 9810 × 3 × 27.2 = 1.07 ×106 W, or W
∴W
P
P
0.75
η
Use SI data from Pbm 11.11 (a):
ΣQ = Q1 − Q2 − Q3 = 0, or
w( H ) =
30 − H
H − 20
H − 18
−
−
142.1
232.4
1773
Use false position method of solution: Hr =
11.12
Iter
1
2
3
HA
25
24.57
24.46
Hu
20
20
20
w( H A )
−0.02193
−0.00562
−0.00144
H A w ( Hu ) − Hu w ( H A )
w ( Hu ) − w( H A )
Sign
w(H A )w( H r ) ε
0.2317 24.57 −0.00562
+
⎯
0.2317 24.46 −0.00144
+
0.0045
0.2317 24.43 −0.00030
+
0.0012
w( H u )
Hr
w( H r )
∴ Q1 = (30 − 24.43) /142.1 = 0.198 m3/s
Q2 = (24.43 − 20) / 232.4 = 0.138 m3/s
Q3 = (24.43 − 18) /1773 = 0.060 m3/s
Continuity at junction: Q1 − (Q 2 + Q3 + Q 4 ) = 0.
11.14
Energy eqn. across pipe 1:
p0
γ
= R1Q12 + H , or Q1 = ( p0 /γ − H ) / R1
in which H = piezometric head at junction. For each branch (i= 2, 3, 4):
164
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Chapter 11 / Flows in Piping Systems
H = RiQ12 or Qi = H / Ri
1/ 2
Substitute into continuity eqn:
⎡ p0 / γ − H ⎤
⎢
⎥
⎣ R1 ⎦
1/ 2
⎡H⎤
= ∑⎢ ⎥
i = 2 ⎣ Ri ⎦
4
4
= H∑
i=2
1
Ri
∴Solving for H,
H=
=
p 0γ
⎡
1 + R1 ⎢
⎢⎣
Σ
1 ⎤
⎥
R i ⎥⎦
2
300 × 103 / 9810
⎡
⎤
1
1
1
1 + 1.6 × 10 ⎢
+
+
⎥
5
1× 106
1.8 × 106 ⎥⎦
⎣⎢ 5.3 × 10
2
4
=
30.58
= 26.46 m
1.556
∴Q1 = (30.58 − 26.46) /1.6 × 104 = 0.01605 m3/s
Q2 = 26.46 / 5.3 × 105 = 0.00707 m3/s
Q3 = 26.46 /1×106 = 0.00514 m3/s
Q4 = 26.46 /1.8 ×106 = 0.00383 m3/s
(c) Substitute known data into energy equation, and solve for Q 1 :
−2 ⎫
⎧
⎡
1 ⎤ ⎪ 2
⎪
a0 = ⎨a1 + R1 + ⎢ ∑
⎥ ⎬ Q1
R
⎢
⎪
i ⎥
⎣
⎦ ⎪⎭
⎩
11.16
−2
⎧⎪
⎡
⎤ ⎫⎪ 2
1
1
1
4
2
∴ 45 = ⎨10 + 34,650 + ⎢
+
+
⎥ ⎬ Q1 = 56, 410Q1
82,500
127,900
115,500
⎣
⎦ ⎪⎭
⎪⎩
∴ Q1 = 45 / 56,410 = 0.0282 m3/s
∴ H = 45 − (104 + 34,650) × 0.02822 = 9.49 m
Q2 = 9.49 / 82,500 = 0.0107 m3/s
Q3 = 9.49 /127,900 = 0.0086 m3/s
Q4 = 9.49 /115,500 = 0.0091 m3/s
165
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Chapter 11 / Flows in Piping Systems
(a) For each pipe compute equivalent length and resistance coefficient:
Pipe 1: Le = DΣK / f = 0.05 × 3/ 0.02 = 7.5
R1 =
8 f ( L + Le )
8 × 0.02 × 37.5
=
= 1.983 × 105
2 5
2
5
gπ D
9.81 × π × 0.05
Pipe 2: Le = 0.075 × 5 / 0.025 = 15, R2 =
Pipe 3: Le = 0.06 × 1/ 0.022 = 2.7 ,
11.18
W=
Q2
⎛ 3
⎞
⎜ Σ( Ri ) −1/2 ⎟
⎜ i =1
⎟
⎝
⎠
2
=
8 × 0.025 × 55
= 4.788 × 10 4
9.81 × π 2 × 0.0755
R3 =
8 × 0.022 × 62.7
= 1.466 × 105
2
5
9.81× π × 0.06
(0.6 / 60) 2
= 1.125 m
[(1.983 × 105 ) −1/ 2 + (4.788 × 104 ) −1/ 2 + (1.466 × 105 )−1/ 2 ]2
∴ Q1 = W / R1 = 1.125 /1.983 ×105 = 0.00238 m3 /s, or 143 L/min
Q2 = W / R2 = 1.125 / 4.788 ×104 = 0.00485 m3 /s, or 290 L/min
Q3 = W / R3 = 1.125 /1.466 ×105 = 0.00277 m3 /s, or 166 L/min
Write energy eqn. from junction at A to suction side of pump at B:
⎛
ΣK ⎞⎟ Q2 + pB + z
H PA = ⎜ R1 +
B
⎜
γ
2 gA12 ⎟⎠
⎝
where H PA = pump head and pB = pressure. The only unknown in the relation
is Q, since
H PA
11.20
η
W
PA
1×106 × 0.76
95.6
=
=
=
γQ
Q
0.81× 9810 × Q
Evaluate the constants:
R1 =
8 f1L1
8 × 0.23 × 5000
=
= 40.0
2 5
gπ D1 9.81× π 2 × 0.755
pB
γ
+ zB =
ΣK =
2
= 0.52
2 gA12 2 × 9.81× (π / 4)2 × 0.754
150 × 103
+ 27 = 45.9 m.
0.81× 9810
Substituting into the energy eqn.,
95.6
= 40.5Q2 + 45.9
Q
Solve using Newton’s method: Q = 1.053 m3/s and HPA = 95.6/1.053 = 90.8 m
166
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Chapter 11 / Flows in Piping Systems
The second reach is now analyzed. Write energy eqn. from suction side of pump
at B to the downstream reservoir:
⎛
ΣK ⎟⎞ Q2 + 50
+ z B + H PB = ⎜ R2 +
⎜
γ
2 gA22 ⎟⎠
⎝
pB
Evaluate the constants:
R2 =
8 f2 L2 8 × 0.023 × 7500
=
= 60.1
gπ 2D25 9.81× π 2 × 0.755
ΣK =
10
= 2.61
2 gA12 2 × 9.81 × (π / 4) 2 × 0.75 4
∴ 45.9 + H PB = (60.1 + 2.6) ×1.0532 + 50 or H PB = 73.6 m
∴W PB =
γ QH PB 0.81× 9810 ×1.053 × 73.6
=
= 8.10 ×105 W or
0.76
η
= 810 kW
W
PB
Compute resistance coefficients:
R1 =
10.59 L
10.59 × 200
=
= 7.6
1.85 4.87
C D
1301.85 × 0.5 4.87
R2 =
10.59× 600
= 274.6
1301.85 × 0.34.87
R3 =
10.59×1500
= 686.5
1301.85 × 0.34.87
R4 =
10.59×1500
=169.1
1301.85 ×0.44.87
(a) Let H J = piezometric head at junction J. Assume Q1 = 2 m3 /s (out of J).
Then H J = 30 + R1Q12 = 30 + 7.6 × 22 = 60.4 m
Q2 = (250 − 60.4) / 274.6 = 0.831 m3/s (into J )
11.22
Q3 = (300 − 60.4) / 686.5 = 0.591 m3/s (into J )
Q4 = (200 − 60.4) /169.1 = 0.909 m3/s (into J )
∴ ΔQ = −2 + 0.831 + 0.591 + 0.909 = +0.331
Assume Q1 = 2.5 m3/s. Then
H J = 30 + 7.6 × 2.52 = 77.5 m
Q2 = (250 − 77.5) / 274.6 = 0.793 m3/s
Q3 = (300 − 77.5) / 686.5 = 0.569 m3/s
Q4 = (200 − 77.5) /169.1 = 0.851 m3/s
∴ΔQ = −2.5 + 0.793 + 0.569 + 0.851 = −0.287
Linear interpolation to find the next estimate of Q 1 yields
167
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Chapter 11 / Flows in Piping Systems
−0.287 − 0.331 −0.287
=
2.5 − 2
2.5 − Q1
∴Q1 = 2.27 m3 / s
Then H J = 30 + 7.6 × 2.27 2 = 69.2 m
Q2 = (250 − 69.2)/ 274.6 = 0.811 m3 /s
Q3 = (300 − 69.2)/ 686.5 = 0.580 m3 /s
Q4 = (200 − 69.2) /169.1 = 0.880 m3 / s
∴ ΔQ = −2.27 + 0.811 + 0.580 + 0.880 = −0.001 ∴ OK
Compute the R-values for each pipe:
Pipe 1: Le =
1 × 0.1
8 × 0.04 × 5.5
= 2.5 m, R1 =
= 1819.6
0.04
9.87 × π 2 × 0.15
Pipe 2: Le =
2 × 0.1
8 × 0.04 × 155
= 5 m, R2 =
= 5.31 × 104
2
5
0.04
9.81× π × 0.1
Pipe 3: Le =
2 × 0.1
8 × 0.04 × 605
= 5 m, R3 =
= 2 × 105
2
5
0.04
9.81 × π × 0.1
Pipe 4: Le =
4 × 0.1
8 × 0.04 ×1,817
= 10 m, R4 =
= 77,750
0.04
9.81× π 2 × 0.15
(a) Compute the discharge and head loss in pipe 2:
11.24
Q2 = 5.2 ×10−3 m3/s
∴(hL )2 = R2Q22 = 5.31×104 × (5.2 ×10−3 )2 = 1.387 m
(c) Q2 = 11.6 ×10−3 m3/s = 696 L/min, and from the pump curve,
HP = 136 m
Write the energy equation from A to C:
H C = H P − ( R1 + R2 )Q22 = 136 − 53,120 × (11.6 × 10−3 )2 = 129 m
The energy equation from C to D is HC = zD + R4Q42 , where HC is the hydraulic
grade line at location C. Therefore the discharge in pipe 4 is
Q4 =
H C − zD
129 − 127
=
= 5.07 L/ s, or 300 L/min
77,750
R4
168
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Chapter 11 / Flows in Piping Systems
(a) From the pump curve, for HP = 138 m, Q = 0.01 m3/s. Write the energy
equation from A to C and evaluate the hydraulic grade line
HC = H P − (hL )1 − (hL )2 = 138 − (53,120) × (0.01)2 = 132.7 m
11.26
Thus the discharge in pipe 3, between location C and location B is
Q3 =
ΔQI =
HC − HB
132.7 −130
=
= 3.67 ×10−3 m3/s or 220 L/min
5
R3
2 ×10
−(±W1 ± W2 ) − 10
−(±W2 ± W3 ) − 15
, ΔQII =
G1 + G2
G2 + G3
± Wi = RQi Qi , Gi = 2 R Qi
Loop Pipe
I
Q
1
−3.5
2
0
W
G
−24.50
0
0
Σ 24.50
ΔQI =
11.28
II
+24.50 − 10
= +1.04
14.00
2
0
3
− 3.5
ΔQII =
ΔQI =
W
G
Q
−2.46
−12.10
9.84
−2.30
+0.36
+0.26
1.44
+0.43
Σ −11.84
11.28
14.00
+11.84 − 10
= +0.16
11.28
− 0.36
− 0.26
1.44
− 0.43
− 24.5
14.00 − 2.82
− 15.90
11.28
+ 2.73
−24.50
14.00
Σ −16.16
12.72
0
Σ
14.00
Q
+24.50 − 15
= +0.68
14.00
0
ΔQII =
+16.16 − 15
= + 0.09
12.72
HJunction = 50 − R1Q12 = 50 − 2 × 2.32 = 39.4
∴Q1 = 2.30 (into J ),
Q2 = 0.43 (into J ),
Q3 = 2.73 (out of J )
169
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Chapter 11 / Flows in Piping Systems
First, continuity is satisfied as shown in the figure:
(a) The hydraulic grade lines at the nodes (i.e., the nodal piezometric heads can
now be computed:
11.30
H A = 100 − R1Q12 = 100 − 20 × 0.452 = 95.95 m
H B = H A − R2Q22 = 95.95 − 51× 0.282 = 91.95 m
HC = H A − R3Q32 = 95.95 − 280 × 0.172 = 87.86 m
HD = HB − R5Q52 = 91.95 − 310 × 0.182 = 81.91 m
(b)
p A = γ ( H A − z A ) = 9810(95.95 − 10) = 8.43 × 105 Pa
p B = γ ( H B − z B ) = 9810(91.95 − 20) = 7.06 × 105 Pa
pC = γ ( HC − zC ) = 9810(87.86 − 5) = 8.13 ×105 Pa
pD = γ ( H D − z D ) = 9810(81.91 − 0) = 8.04 × 105 Pa
Assume flow directions as shown.
Then ΔQ =
−( R1Q12 + R2Q22 − R3Q32 )
2( R1Q1 + R2Q2 + R3Q3 )
Initial flow assumption:
Q1 = 25, Q 2 = 10, Q3 = 25
11.32
1st iteration: ΔQ =
−(3 × 252 + 5 × 102 − 2 × 252 )
= −3.21
2(3 × 25 + 5 × 10 + 2 × 25)
∴ Q1 = 25 − 3.21 = 21.79, Q2 = 10 − 3.21 = 6.79 , Q3 = 25 + 3.21 = 28.21
2nd iteration: ΔQ =
−(3 × 21.792 + 5 × 6.792 − 2 × 28.212 )
= −0.20
2(3 × 21.79 + 5 × 6.79 + 2 × 28.21)
∴ Q1 = 21.79 − 0.20 = 21.59 , Q2 = 6.79 − 0.20 = 6.59 ,
Q3 = 28.21 + 0.20 = 28.41
170
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Chapter 11 / Flows in Piping Systems
3rd iteration: ΔQ =
−(3 × 21.592 + 5 × 6.592 − 2 × 28.412 )
= − 0.0041
2(3 × 21.59 + 5 × 6.59 + 2 × 28.41)
∴ Q1 ≅ 21.6, Q2 ≅ 6.6, Q3 ≅ 28.4. Units are L/s
A Mathcad solution is provided below. The solution converges after 3 iterations.
ORIGIN := 1
Given:
L :=
⎛ 200 ⎞
⎜ 300 ⎟ ⋅ m
⎜
⎟
⎝ 120 ⎠
D :=
ΔZ := 50⋅ m
⎛ 1500 ⎞
⎜ 1000 ⎟ ⋅ mm
⎜
⎟
⎝ 1200 ⎠
e := 1⋅ mm
f := 1.325⋅ ⎜⎛ ln ⎛⎜ 0.27⋅
11.34
⎝ ⎝
γ := 9810⋅
newton
3
m
W P := 1920⋅ kW
⎞⎞
D ⎟⎟
i ⎠⎠
e
−2
η := 0.82
⎛ 0.018 ⎞
f = ⎜ 0.02 ⎟
⎜
⎟
⎝ 0.019 ⎠
D ⋅ K ⎞⎤
⎡⎢ ⎛⎜
i i ⎟⎥
8⋅ f ⋅ L +
⎢ i⎜ i
f ⎟⎥
i ⎠⎦
⎣ ⎝
R :=
i
2
⎛ 2 ⎞
⎜ 0 ⎟
⎜ ⎟
⎝ 10 ⎠
i := 1 .. 3
Resistance coefficients:
i
K :=
R =
( i)5
g⋅π ⋅ D
⎛ 0.071 ⎞ 2
⎜ 0.487 ⎟ s
⎜
⎟ 5
⎝ 0.473 ⎠ m
3
Q := 2⋅
Initial flow estimate (note that the discharge is common to all three lines):
1
Hardy Cross iteration:
(
3)
− R + R + R ⋅ Q⋅ Q +
ΔQ( Q) :=
1
2
(
W P⋅ η
3)
2⋅ R + R + R ⋅ Q +
N := 4
Solution:
j := 1 .. N
1
Q
j+ 1
2
γ⋅Q
− ΔZ
W P⋅ η
γ⋅Q
2
( j)
:= Q + ΔQ Q
j
( j) =
ΔQ Q
⎛ 2 ⎞
⎜ 2.59 ⎟
⎜
⎟ m3
Q = ⎜ 2.762 ⎟
⎜ 2.771 ⎟ s
⎜
⎟
⎝ 2.771 ⎠
0.59
Residual:
0.172
8.456·10 -3
3
m
s
1.762·10 -5
171
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m
s
Chapter 11 / Flows in Piping Systems
The solution was obtained using EPANET, Version 2.0.
Link - Node Table:
----------------------------------------------------------------Link
Start
End
Length Diameter
ID
Node
Node
m
mm
---------------------------------------------------------------1
1
5
200
100
2
5
3
150
50
3
5
6
500
100
4
6
4
35
50
5
6
2
120
100
11.36
Node Results:
---------------------------------------------------------------Node
Demand
Head Pressure
Quality
ID
LPS
m
m
---------------------------------------------------------------5
0.00
121.72
121.72
0.00
6
0.00
116.23
116.23
0.00
1
-9.28
125.00
0.00
0.00 Reservoir
2
7.15
115.00
0.00
0.00 Reservoir
3
1.58
118.00
0.00
0.00 Reservoir
4
0.55
116.00
0.00
0.00 Reservoir
Link Results:
---------------------------------------------------------------Link
Flow Velocity Headloss
Status
ID
LPS
m/s
m/km
---------------------------------------------------------------1
9.28
1.18
16.40
Open
2
1.58
0.81
24.80
Open
3
7.69
0.98
10.98
Open
4
0.55
0.28
6.55
Open
5
7.15
0.91
10.24
Open
(c) ΔQ =
11.38
(d) R1 =
−(R1 + R2 + R3) ) Q Q + (a0 + a1Q + a2Q2 + a3Q3 ) + zA − zB
2(R1 + R2 + R3) ) Q − (a1 + 2a2Q + 3a3Q2 )
8 f L1
8 f L2
8 f L3
= 0.00516, R2 =
= 2.538, R3 =
= 0.0103
2
5
2
2 5
gπ D1
gπ N D2
gπ 2 D35
Using the equation in part (c), Q = 2.061 m3/s after 5 iterations, beginning with
an initial estimate Q = 3 m3/s.
172
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Chapter 11 / Flows in Piping Systems
Iteration
Q
ΔQ
1
3
− 0.611
2
2.389
− 0.277
3
2.113
− 0.050
4
2.062
− 0.0014
5
2.061
− 5.1 × 10−13
Unsteady Flow
Vss =
11.44
2 g ( H1 − H3 )
2 × 9.81× 3
=
= 0.307 m/s
fL / D + K
0.025 × 2500 / 0.1 + 0.15
V = 0.99Vss = 0.99 × 0.307 = 0.304 m/s
⎡ (V ss + V
V ss L
ln ⎢
2 g(H1 − H 3 )
⎣ (V ss − V
⎡ ( 0 .3 0 7 +
0 .3 0 7 × 2 5 0 0
=
ln ⎢
2 × 9 .8 1 × 3
⎣ ( 0 .3 0 7 −
t =
)( V s s − V 0 ) ⎤
⎥
)( V s s + V 0 ) ⎦
0 .3 0 4 )( 0 .3 0 7 − 0 ) ⎤
= 6 9 .0 s
0 .3 0 4 )( 0 .3 0 7 + 0 ) ⎥⎦
First compute the initial velocity V0, with H1-H3 = 8 m, and K0 = 275:
V0 =
2 × 9.81 × 8
= 0.552 m/s
0.015 × 800
+ 275
0.05
The final steady-state velocity is, with Kss = 5:
VSS =
11.46
2 × 9.81× 8
= 0.800 m/s
0.015 × 800
+5
0.05
The steady-state discharge is Q = 0.7854 × 0.052 × 0.800 = 0.00157 m3 / s , and the
time to reach 95% of that value is
V = 0.95Vss = 0.95 × 0.800 = 0.760 m/s
∴t =
0.800 × 800 ⎡ (0.800 + 0.760)(0.800 − 0.552) ⎤
ln
= 8.03 s
2 × 9.81× 8 ⎢⎣ (0.800 − 0.760)(0.800 + 0.552) ⎥⎦
173
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Chapter 11 / Flows in Piping Systems
174
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Chapter 12 / Turbomachinery
CHAPTER 12
Turbomachinery
Elementary Theory
ω = 800 × π / 30 = 83.8 rad/s
u1 = ωr1 = 83.8 × 0.04 = 3.35 m/s
u 2 = ω r2 = 83.8 × 0.125 = 10.48 m/s
Q = 2π r1b1Vn1 , but Vn1 = u1 since β1 = 45D
12.2
∴Q = 2π × 0.04 × 0.05 × 3.35 = 0.0421 m3 / s
Vn2 =
Q
0.0421
=
= 2.14 m/s
2π r2b2 2π × 0.125 × 0.025
Vt2 = u2 −
Vt1 = 0
Vn2
tan β 2
= 10.48 −
2.14
= 6.77 m/s
tan 30D
(α1 = 90D under ideal conditions)
(
)
∴T = ρQ r2Vt2 − rV
1 t1 = 1000 × 0.0421(0.125 × 6.77 − 0) = 35.6 N ⋅ m
W P = ωT = 83.8 × 35.6 = 2980 W
H t = ωT / γ Q = 2980 /(9810 × 0.0421) = 7.22 m
Q = 7.6 ×10−3 m3/s
12.4
ω = 2000 ×
π
30
= 209 rad/s
175
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Chapter 12 / Turbomachinery
Q
7.6 ×10−3
=
= 1.94 m/s
2π r2b2 2π × 0.0625 × 0.01
Vn2 =
u2 = ωr2 = 209 × 0.0625 = 13.06 m/s
∴ Ht =
13.06
u2
(13.06 − 1.94cot 60°) = 15.9 m
(u2 − Vn2 cot β 2 ) =
9.81
g
= γ QH = 9810 × 0.8 × 7.6 ×10−3 ×15.9 = 948 W
W
t
= 948/ 746 = 1.21 hp
W
Compute loss in suction pipe:
2
⎛ L
⎞ Q
hL = ⎜ f + Σ K ⎟
⎝ D
⎠ 2 gA 2
11
⎛
⎞
= ⎜ 0.015 ×
+ 2 × 0.19 + 0.8⎟
⎝
⎠
0.1
12.6
0.05 2
2
⎛ π⎞
2 × 9.81 × ⎜ ⎟ × 0.1 4
⎝ 4⎠
= 5.85 m
Water at 20°C: γ = 9792 N/m3 , pv = 2340 Pa. Substitute known data into NPSH
relation, solving for Δz:
∴Δ z =
patm − pv
γ
101×103 − 2340
− 5.85 − 3 = 1.23 m
− hL − NPSH =
9792
Dimensional Analysis and Similitude
Q1 =
N
ω1
970
Q = 1Q =
× 0.8 = 0.65 m 3 / s
ω 2 2 N 2 2 1200
∴from Fig. 12.9, H1 ≅ 11.5 m and W 1 ≅ 91 kW.
2
⎛ω ⎞
⎛ 1200 ⎞
∴ H 2 = ⎜ 2 ⎟ H1 = ⎜
⎟ ×11.5 = 17.6 m
⎝ 970 ⎠
⎝ ω1 ⎠
12.8
2
3
⎛ 1200 ⎞
= ⎛ ω2 ⎞ W
∴W
⎜
⎟ 1=⎜
2
⎟ × 91 = 172 kW
⎝ 970 ⎠
⎝ ω1 ⎠
12.10
CQ =
3
Q
Q / 3600
=
= 1.061× 10 −4 Q
3
ω D 304 × 0.2053
(Q in m 3 /h)
176
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Chapter 12 / Turbomachinery
CH =
gH
9.81H
=
= 2.526 × 10−3 H (H in m)
2
2
2
2
ω D 304 × 0.205
Tabulate CQ and CH using selected values of Q and H from Fig. 12.6:
CQ×10 −3
0
0.53
1.06
1.59
2.12
2.65
3.18
Q (m3/h)
0
50
100
150
200
250
300
H (m)
54
53
52
50
47
41
33
CH ×10−1
1.36
1.34
1.31
1.26
1.19
1.04
0.83
The dimensionless curve shown in Fig. 12.12 is for the 240-mm impeller. Since
the impellers are not the same (240 mm versus 205 mm), dynamic similitude
does not exist, and thus the curves are not the same.
Compute the specific speed: Ω P =
12.12
ω Q
( gH P )3 / 4
=
1800 ×
π
× 0.15
30
= 1.30 ,
(9.81 × 22)3 / 4
hence use a mixed flow pump. As an alternate, since Ω P is close to unity, a
radial flow pump could be employed.
Fig. 12.13: At η = 0.75 (best eff. ), CQ ≅ 0.048 ,
C H ≅ 0.018, C W ≅ 0.0011, C NPSH ≅ 0.023.
ω = 750 ×
π
30
= 78.5 rad/s
1/3
(a)
12.14
⎛ 1240 ×10−3 ⎞
D = ⎜⎜
⎟⎟
⎝ 0.049 × 78.5 ⎠
H=
= 0.685 m
0.018 × 78.52 × 0.6852
= 5.3 m
32.2
H NPSH =
0.023 × 78.52 × 0.6852
= 6.78 m
9.81
= 0.0011×1000 × 78.53 × 0.6855 = 80,250 W = 80.25 kW
W
or
12.16
= 82.25/0.746 = 107.6 hp
W
ω = 600 ×
π
30
= 62.8 rad /s,
Q = 22.7 / 60 = 0.378 m 3 / s,
177
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Chapter 12 / Turbomachinery
ΩP =
ω Q
( gH P )
=
3/ 4
2
62.8 0.378
= 0.751
(9.81×19.5)3/ 4
2
H 2 ⎛ ω 2 ⎞ ⎛ D2 ⎞
=⎜
⎟ ⎜
⎟ =2
H1 ⎝ ω1 ⎠ ⎝ D1 ⎠
2
3
Q2 ⎛ ω 2 ⎞⎛ D2 ⎞
=⎜
⎟⎜
⎟ =2
Q1 ⎝ ω1 ⎠⎝ D1 ⎠
2
⎛ ω ⎞ ⎛ D ⎞ ⎛ ω ⎞⎛ D ⎞
∴ ⎜ 2 ⎟ ⎜ 2 ⎟ = ⎜ 2 ⎟⎜ 2 ⎟
⎝ ω1 ⎠ ⎝ D1 ⎠ ⎝ ω1 ⎠⎝ D1 ⎠
12.18
∴Use a radial flow pump.
3
or
ω 2 D2
=
ω1 D1
4
⎛ω ⎞
∴ ⎜ 2 ⎟ = 2, ω2 = 4 2 ω1 = 1.19 ω1 and D2 = 4 2 D1 = 1.19 D1
⎝ ω1 ⎠
Assume a pump speed N = 2000 rpm or ω = 2000 ×
π
30
= 209 rad/s
/(γ Q) = 200 ×103 /(8830 × 0.66) = 34.3 m
HP = W
f
∴Ω P =
ω Q
( gH P )
3/ 4
=
209 0.66
= 2.16
(9.81× 34.3)3/ 4
The specific speed suggests a mixed-flow pump. However, if N = 1000 rpm, a
radial-flow pump may be appropriate. Consider both possibilities.
Mixed flow: from Fig. 12.14, at best η :
CW ≅ 0.0117, CQ ≅ 0.148, CH ≅ 0.067
Use
12.20
CQ =
gH
Q
and CH = 2 P2
3
ωD
ωD
Combining and solving for D and ω
Q / CQ
D=
gHP / CH
ρ=
Q
CQD3
0.66 / 0.148
= 0.251 m
9.81× 34.3/ 0.067
∴D =
ω=
and ω =
0.66
= 282 rad/s or N = 282 × 30 / π = 2674 rpm
0.148 × 0.2513
γ
g
=
8830
= 900 kg/m3
9.81
178
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Chapter 12 / Turbomachinery
= C ρω 3D5 = 0.0117 × 900 × 2823 × 0.2515
∴W
P
W
= 2.35 × 105 W or 235 kW
Radial flow: from Fig. 12.12, at best η :
CW ≅ 0.027, CQ ≅ 0.0165, C H ≅ 0.125
∴D =
ω=
0.66 / 0.0165
= 0.878 m
9.81× 34.3/ 0.125
0.66
59.1× 30
= 59.1 rad/s or N =
= 564 rpm
3
π
0.0165 × 0.878
= 0.0027 × 900 × 59.13 × 0.8785 = 2.62 × 105 W
W
P
or 262 kW
Hence, a mixed-flow pump is preferred.
Use of Turbopumps
12.22
The intersection of the system demand curve with the head-discharge curve
yields Q = 2.75 m3 / min, H P = 12.6 m, W P = 7.2 kW.
N 1 = 1350 rpm, Q 1 = 2.75 m 3 / min, H 1 = 12.6 m,
12.24
W 1 = 7.2 kW, N 2 = 1200 rpm, D 2 = D1
179
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Chapter 12 / Turbomachinery
3
⎛ N ⎞⎛ D ⎞
⎛ 1200 ⎞
3
∴Q2 = Q1 ⎜ 2 ⎟⎜ 2 ⎟ = 2.75 ⎜
⎟ = 2.44 m /min
1350
N
D
⎝
⎠
⎝ 1 ⎠⎝ 1 ⎠
2
2
3
5
⎛N ⎞ ⎛D ⎞
⎛ 1200 ⎞
H 2 = H1 ⎜ 2 ⎟ ⎜ 2 ⎟ = 12.6 ⎜
⎟ = 9.96 m
⎝ 1350 ⎠
⎝ N1 ⎠ ⎝ D1 ⎠
2
⎛N ⎞ ⎛D ⎞
⎛ 1200 ⎞
W 2 = W 1 ⎜ 2 ⎟ ⎜ 2 ⎟ = 7.2 ⎜
⎟ = 5.06 kW
⎝ 1350 ⎠
⎝ N1 ⎠ ⎝ D1 ⎠
3
Efficiency will remain approximately the same.
Compute system demand:
2
2
14
⎛ L
⎞V
⎛
⎞ 3
= ⎜ 0.01 ×
+ 4 × 0.1⎟
= 0.40 m
H P = ⎜ f + ΣK ⎟
⎝ D
⎠ 2g ⎝
0.3
⎠ 2 × 9.81
12.26
Q = VA = 3 ×
∴Ω P =
π
4
× 0.32 = 0.212 m2
ω Q
( gH P )
3/ 4
=
31.4 0.212
= 5.19
(9.81 × 0.4)3/ 4
ω = 300 × π / 30 = 31.4 rad/s
∴Axial pump is appropriate.
(a)
12.28
The intersection of the pump curve with the demand curve yields H P ≅ 64 m
and Q ≅ 280 m3 / h. ∴ From Fig. 12.6, W P ≅ 64 kW and NPSH ≅ 8.3 m
180
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Chapter 12 / Turbomachinery
Use energy eqn. to establish system demand:
e⎤
⎡
f ≅ 1.325 ⎢ln 0.27 ⎥
D⎦
⎣
H P = Δz + f
−2
0.000255 ⎤
⎡
= 1.325 ⎢ln 0.27 ×
0.45 ⎥⎦
⎣
−2
= 0.017
0.017 × 5000
L Q2
= 192 +
Q2 = 192 + 381Q2
2
5
2
D 2 gA
2 × 9.81× (π / 4) × (0.45)
From Fig. 12.6, at best ηP , Q ≅
12.30
240 m3
= 0.067 m3/s, and HP ≅ 65 m
3600 s
Assume three pumps in series, so that HP = 3× 65 = 195 m. Then the demand
head is
H P = 192 + 381× (0.067) 2 = 193.7 m
Hence three pumps in series are appropriate. The required power is
WP = γ QHP/ηP = 9810× 0.86× 0.067 ×195/0.75 = 146,965 W
W P = 146,965/746 = 197 hp
or
(a) For water at 80°C, pv = 46.4 × 103 Pa, and ρ = 9553 kg/m3. Write the energy
equation from the inlet (section i) to the location of cavitation in the pump:
pi
γ
12.32
+
Vi2 pv
=
+ NPSH
2g γ
∴ NPSH =
pi − pv
γ
Vi2 (83 − 46.4) ×103
62
+
=
+
= 5.67 m
2g
9533
2 × 9.81
(b) NPSH1 = 5.67 m, N1 = 2400 rpm, N2 = 1000 rpm, D2/D1 = 4
2
2
⎛N ⎞ ⎛D ⎞
⎛ 1000 ⎞
∴ NPSH 2 = NPSH1 ⎜ 2 ⎟ ⎜ 2 ⎟ = 5.67 ⎜
⎟
⎝ 2400 ⎠
⎝ N1 ⎠ ⎝ D1 ⎠
2
( 4 )2 = 15.8 m
Given:
12.34
L = 200 m , D = 0.05 m , Δz = z 2 − z1 = −3 m , V = 3 m/s , ν = 6 × 10 −7 m 2 /s ,
ρ = 1593 kg/m 3 , pv = 86.2 × 103 Pa , pa = 101 × 103 Pa
Compute the pump head:
181
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Chapter 12 / Turbomachinery
Re =
3 × 0.05
6 × 10
−7
= 2.5 × 105
(
)
f = 1.325 ⎡ ln 5.74Re-0.9 ⎤
⎣
⎦
−2
= 1.325 ⎡ ln 5.74 × (2.5 × 105 ) −0.9 ⎤
⎣
⎦
−2
= 0.015
2
fL ⎞ V 2
⎛
⎛ 0.015 × 200 ⎞ 3
= −3 + ⎜ 1 +
= 25.0 m
H P = Δz + ⎜ 1 + ⎟
⎟
0.05
D ⎠ 2g
⎝
⎠ 2 × 9.81
⎝
(a) Choose a radial-flow pump. Use Fig. P12.35 to select the size and speed:
C H = 0.124, CQ = 0.0165, η = 0.75
Q = 3×
∴D =
∴ω =
π
4
4
× 0.052 = 0.00589 m 2
CHQ 2
CQ2 gH P
Q
CQ D
3
=
=
4
0.124 × 0.00589 2
0.01652 × 9.81 × 25.0
0.00589
0.0165 × 0.090
3
= 0.090 m
= 490 rad/s, or N = 490 ×
30
π
= 4680 rpm
(b) Available net positive suction head:
NPSH =
pa − pv
101×103 − 86.2 ×103
− Δz =
+ 3 = 3.95 m
1593 × 9.81
ρg
(c)
182
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Chapter 12 / Turbomachinery
Turbines
ω = 120 × π / 30 = 12.6 rad/s
α1 = cot −1 (2π r12b1ω / Q + cot β1 )
= cot −1 (2π × 4.52 × 0.85 × 12.6 /150 + cot 75D ) = 6.1D
Vt1 = u1 + Vn1 cot β1 = ω r1 +
= 12.6 × 4.5 +
12.36
Q
cot β1
2π r1b1
150
cot 75D = 58.37 m/s
2π × 4.5 × 0.85
Vt2 = u2 + Vn2 cot β 2 = ω r2 +
= 12.6 × 2.5 +
Q
cot β 2
2π r2b2
150
cot100D = 29.52 m/s
2π × 2.5 × 0.85
∴ T = ρQ(rV
1 t1 − r2Vt2 )
= 1000 ×150(4.5 × 58.37 − 2.5 × 29.52) = 2.83 × 107 N ⋅ m
= ωT = 12.6 × 2.83 × 107 = 3.57 × 108 W or 357 MW
W
T
=W
, hence
Under ideal conditions ηT = 1, and W
T
f
HT = W T / γ Q = 3.57 × 108 /(9810 × 150) = 243 m
N 2 = 240 rpm , W 2 = 2200 kW , D 2 = 0.9 m, W 1 = 9kW , H1 = 7.5 m
3
5
2
2
W 2 ⎛ N 2 ⎞ ⎛ D2 ⎞
H 2 ⎛ N 2 ⎞ ⎛ D2 ⎞
From the similarity rules = ⎜
=⎜
⎟ ⎜
⎟ and
⎟ ⎜
⎟
H1 ⎝ N1 ⎠ ⎝ D1 ⎠
W1 ⎝ N1 ⎠ ⎝ D1 ⎠
12.38
Substitute second eqn. into the first to eliminate ( N 2 / N 1 ), and solve for D 1 :
1/ 2
3/ 4
⎞1/ 2 ⎛ H ⎞3/ 4
⎛W
⎛ 9 ⎞ ⎛ 45 ⎞
1
2
∴ D1 = D2 ⎜ ⎟ ⎜
⎟ = 0.9 ⎜
⎟ ⎜
⎟ = 0.221 m
⎝ 2200 ⎠ ⎝ 7.5 ⎠
⎝ W2 ⎠ ⎝ H1 ⎠
1/ 2
⎛H ⎞
N 2 = N1 ⎜ 1 ⎟
⎝ H2 ⎠
⎛ D2 ⎞
⎛ 7.5 ⎞
⎜
⎟ = 240 ⎜
⎟
D
⎝ 45 ⎠
⎝ 1⎠
1/ 2
⎛ 0.9 ⎞
⎜
⎟ = 399 rpm
⎝ 0.221 ⎠
183
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Chapter 12 / Turbomachinery
Write energy eqn. from upper reservoir (loc. 1) to surge tank (loc. 2) and solve
for Q:
1/ 2
⎡
⎤
D
Q = ⎢ 2 gA2
( z1 − z2 ) ⎥
fL
⎣
⎦
1/ 2
⎡ 2 × 9.81 × (π / 4) 2 × 0.855
⎤
=⎢
(650 − 648.5) ⎥
0.025 × 2000
⎣
⎦
= 0.401 m 3 / s
Apply energy eqn. from loc. 2 to lower reservoir (loc. 3) and determine H T :
2
⎛ L
⎞ Q
HT = z2 − z3 − ⎜ f + K v ⎟
⎝ D
⎠ 2 gA2
12.40
⎛ 0.02 × 100 ⎞
= 648.5 − 595 − ⎜
+ 1⎟
⎝ 0.85
⎠
0.4012
⎛π ⎞
2 × 9.81× ⎜ ⎟ × 0.855
⎝4⎠
2
= 53.4 m
∴ W T = γ QHTηT = 9810 × 0.401 × 53.4 × 0.9 = 1.89 × 105 W or 189 kW
∴ From Fig. 12.32, use a Francis turbine.
A representative value of the specific speed is 2 (Fig. 12.20):
ΩT ( gHT )5 / 4
2(9.81× 53.3)5 / 4
∴ω =
=
= 306 rad/s
/ ρ )1/ 2
(W
(2.67 × 105 /1000)1/ 2
T
or
N = 306 × 30 / π = 2920 rpm
ω = 200 × π / 30 = 20.94 rad/s and from Fig. P12.42 at best efficiency
(ηT = 0.8), φ ≈ 0.42, cv = 0.94
V1 = cv 2 gHT = 0.94 2 × 9.81× 120 = 45.6 m/s
∴Q =
12.42
WT
4.5 × 10 6
=
= 4.78 m 3 / s
γ H T η T 9810 × 120 × 0.8
This is the discharge from all of the jets.
Determine the wheel radius r : r =
φ 2 gHT 0.42 2 × 9.81× 120
=
= 0.973 m
20.94
ω
Hence, the diameter of the wheel is 2r = 2 × 0.973 = 1.95 m
Compute diameter of one jet: D j = 2r / 8 = 1.95 / 8 = 0.244 m, or 244 mm
Let N j = no. of jets. Then each jet has a discharge of Q / N j and an area
184
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Chapter 12 / Turbomachinery
Q / Nj
V1
Nj =
=
π
4
D2j ∴Solving for N j :
Q
1
4.78
=
= 2.24
V1 ⎛ π 2 ⎞
⎛π ⎞
2
D
45.6
0.244
×
×
⎜
⎜ ⎟
j ⎟
⎝4
⎠
⎝4⎠
∴Use three jets
ΩT =
ω (W T / ρ )1/ 2
( gHT )5 / 4
=
20.94(4.5 × 106 /1000)1/ 2
= 0.204
(9.81 × 120)5 / 4
Q = 2082 / 6 = 347 m3 / sec (one unit)
HT =
WT
427,300 × 746
=
= 110.1 m
γ QηT 9810 × 347 × 0.85
Write energy eqn. from upper reservoir (loc. 1) to lake (loc. 2):
2
⎛ L
⎞ Q
+ HT + z2
z1 = ⎜ f + ΣK ⎟
2
⎝ D
⎠ 2gA
12.44
⎛ 0.01 × 390
⎞
∴ 309 = ⎜
+ 0.5 ⎟
D
⎝
⎠
which reduces to 2.4 −
347 2
⎛π ⎞
2 × 9.81 × ⎜ ⎟ D 4
⎝4⎠
2
+ 110.1 + 196.5
38,840 4980
− 4 =0
D5
D
∴ Solving, D ≅ 8 m
∴From Fig. 12.32, a Francis or pump/turbine unit is indicated.
(a) Let H be the total head and Q the discharge delivered to the turbine; then
HT = 0.95H = 0.95 × 305 = 289.8 m
and
12.46
Q=
W
10.4 × 10 6
T
=
= 4.30 m 3 / s.
γHT ηT 9810 × 289.8 × 0.85
Write energy eqn. from reservoir to turbine outlet:
2
⎛ L
⎞ Q
H = HT +⎜ f +∑K⎟
2
⎝ D
⎠ 2gA
4.32
⎛ 0.02 × 3000
⎞
∴ 305 = 289.8 + ⎜
+ 2⎟
D
⎝
⎠ 2 × 9.81× (π / 4)2 D 4
185
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Chapter 12 / Turbomachinery
which reduces to 15.2 −
91.67 3.06
− 4 =0
D5
D
∴Solving D = 1.45 m
(b) Compute jet velocity: V1 = cv 2 gHT = 0.98 2 × 9.81× 289.8 = 73.9 m/s
The flow through one nozzle is Q/4 and the jet area is πD 2j / 4. Hence
π D 2j
4
=
Q / 4 4.3/ 4
=
= 0.0146 m 2
V1
73.9
4 × 0.0146
∴ Dj =
π
= 0.136 m
2
12.48
⎡ 4.15 × (9.81 × 3.7)5 / 4 ⎤
6
WT = 1000 ⎢
⎥ = 4.99 × 10 W (one unit).
×
50
/
30
π
⎣
⎦
Total power developed is 9.21 × 10 6 W. Hence, required number of units is
9.21/ 4.99 = 1.8 ∴ Use two turbines.
(a)
H T = z1 − z 2 −
fL Q 2
D 2 gA 2
0.015 × 350 × 0.252
= 915 − 892 −
= 11.8 m
0.3 × 2 × 9.81× (0.7854 × 0.32 ) 2
= γ QH η = 9810 × 0.25 ×11.8 × 0.85 = 2.46 ×104 W or 24.6 kW
W
T
T T
(b) Compute the specific speed of the turbine:
12.50
ω=N
π
30
∴ΩT =
= 1200 ×
/ρ
ω W
T
( gHT )
5/4
π
30
=
= 126 rad/s
126 2.46 × 104 /1000
( 9.81×11.8)
5/4
= 1.65
Hence, from Fig. 12.20, a Francis turbine is appropriate.
(c) From Fig. 12.24, the turbine with ΩT = 1.063 is chosen:
CH = 0.23, CQ = 0.13, and ηT = 0.91.
186
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Chapter 12 / Turbomachinery
D=
ω=
4
CHQ2
=
CQ2 gHT
4
0.23 × 0.252
= 0.293 m or approximately 0.30 m
0.132 × 9.81 × 11.8
Q
0.25
30
=
= 71.2 rad/s or N = 71.2 ×
= 680 rpm
3
3
π
CQ D
0.13 × 0.30
W T = γ QHTηT = 9810 × 0.25 × 11.8 × 0.91 = 2.63 × 10 4 W, or 26.3 kW
Calculate a new specific speed based on the final design data:
ΩT =
71.2 × 2.63 ×104 /1000
(9.81×11.8)5/4
= 0.96
An acceptable value according to Fig. 12.20.
187
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Chapter 12 / Turbomachinery
188
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Chapter 13 / Measurements in Fluid Mechanics
CHAPTER 13
Measurements in Fluid Mechanics
Introduction
a)
γ Hg h
V2
V2
h
×ρ = p =
∴
× 1.22 = ( 9810 × 13.6 )
2
100
2
100
∴V = 46.8 h and C = 46.8
13.2
b)
V2
h
× (0.8217 ×1.22) = (9810 × 13.6)
2
100
Q=
∴ V = 51.6 h and
C = 51.6
V 1.9 ×10−3
=
= 3.2 ×10−6 m3/s
Δt
10 × 60
= ρQ = 997 × 3.2 ×10−6 = 3.19 ×10−3 kg/s
m
13.4
Q
3.2 ×10−6
0.65 × 0.0025
V= =
= 0.65 m/s ∴ Re =
= 1805
2
A (π × (0.0025) /4)
9 ×10−7
The flow is laminar.
Q = ⎡π ×12 ×10 + π (22 − 12 ) × 9.9 + π (32 − 22 ) × 9.5 + π (42 − 32 ) × 8.65 +
⎣
π (4.52 − 42 ) × 6.65 + π (52 − 4.52 ) × 2.6⎤⎦ ×10−4 = 0.0592 m3/s
13.6
V=
Q
0.0592
=
= 7.54 m/s
A π × 0.052
See the sketch above:
13.8
p1 − p2
= H. ∴ p1 − p2 = 9810(13.6 − 1) × 0.12 = 14,830 Pa.
γ Hg − γ
∴ h1 − h2 =
p1 − p2 14,830
=
= 1.512 m
9810
9810
D0 15
=
= 0.625
D 24
189
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Chapter 13 / Measurements in Fluid Mechanics
a) Assume:
Re = 105 ∴ K = 0.68 ∴Q = 0.68π × 0.0752 2 × 9.81×1.512 = 0.0654 m3/s
Check: V =
Q
= 1.45 m/s
A
Re =
1.45 × 0.24
10
−6
= 3.5 × 105
∴ K = 0.67 and Q = 0.064 m3/s
Q = KA
RΔp
6.5
where K = 1 −
Dρ
Re
Re =
VD
ν
=
Q4
π Dν
a) Assume
13.10
Re = 105 Then K = 0.98 Q = 0.98π × 0.052
Check V =
0.2 × 80,000
= 0.0974 m3/s
0.1×1000
0.0974
12.4 × 0.1
Q
=
=
∴
=
12.4
m/s
Re
=1.2 ×106 ∴ K = 0.99
−
2
6
A π × 0.05
10
∴OK
A0 =
K=
π ⎛ 33.3 ⎞
2
−4 2
×⎜
⎟ = 8.709 × 10 m
4 ⎝ 1000 ⎠
A0
Q
2 g (S − 1)Δh
13.12
=
8.709 × 10
−4
β = 33.3/ 54 = 0.617
(Δh in meters of mercury)
Q
Q
= 73.03
= K(Q , Δh)
Δh
2 × 9.81 × ( 13.6 − 1)Δh
δK = [ K(Q , Δh) − K(Q + δQ , Δh)] 2 + [ K(Q , Δh) − K(Q , Δh + δ ( Δh))] 2
190
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Chapter 13 / Measurements in Fluid Mechanics
No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
K (Q + δ Q, Δh)
1.147
1.093
1.082
1.095
1.100
1.112
1.096
1.084
1.063
1.083
1.099
1.085
1.103
1.109
1.050
K (Q, Δh)
1.076
1.024
1.017
1.026
1.031
1.041
1.027
1.025
1.005
1.025
1.041
1.027
1.043
1.040
0.9935
K (Q, Δh + δ (Δh))
1.002
0.9805
0.9862
1.001
1.012
1.025
1.012
1.013
0.9938
1.015
1.032
1.018
1.036
1.033
0.9864
δK
0.1026
0.0816
0.0719
0.0734
0.0716
0.0728
0.0706
0.0602
0.0591
0.0589
0.0587
0.0587
0.0604
0.0694
0.0569
Compare the above with Fig. 13.10, curve labeled “Venturi meters and nozzles”
β = 0.6.
13.14
i
(S − 1)Δh
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
(m water)
0.164
0.277
0.403
0.504
0.680
0.781
0.882
1.033
1.147
1.260
1.424
1.550
1.688
1.764
1.764
x i( 1)
y i( 2 )
-1.8079
-1.2837
-0.9088
-0.6852
-0.3857
-0.2472
-0.1256
0.03247
0.1371
0.2311
0.3535
0.4383
0.5235
0.5676
0.5676
Σ -2.5929
-6.3890
-6.1754
-5.9955
-5.8746
-5.7199
-5.6408
-5.5940
-5.5165
-5.4846
-5.4171
-5.3412
-5.3124
-5.2533
-5.2344
-5.2805
-84.2292
xi y i
11.5507
7.9274
5.4487
4.0253
2.2062
1.3944
0.7026
-0.1791
-0.7519
-1.2519
-1.8881
-2.3284
-2.7501
-2.9710
-2.9972
+18.1376
x i2
3.2685
1.6479
0.8259
0.4695
0.1488
0.06111
0.01578
0.001054
0.01880
0.05341
0.1250
0.1921
0.2741
0.3222
0.3222
+7.7464
191
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Chapter 13 / Measurements in Fluid Mechanics
(1)
xi = ln[(S − 1)Δh] S = 13.6 (Δh in meters mercury)
∴m =
b=
(2)
yi = ln Q (Q in m3 /s)
18.1376 − ( −2.5929)( −84.2292) / 15
= 0.4902 ,
7.7464 − ( 2.5929) 2 / 15
−84.2292 − 0.4902( −2.5929)
= −5.5305
15
∴ C = exp(−5.5305) = 0.00396 .
From Problem 13.12, A0 = 8.709 ×10−4 m2 and Kavg = 1.029 . Hence in
Eq. 13.3.8
Kavg A 0 2 g = 1.029 × (8.709 ×10−4 ) 2 × 9.81 = 0.00397, and the exponent is
0.500.
192
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