1
2
History
Faces of Fluid Mechanics
Archimedes
(C. 287-212 BC)
Navier
(1785-1836)
3
Newton
(1642-1727)
Stokes
(1819-1903)
Leibniz
(1646-1716)
Reynolds
(1842-1912)
Bernoulli
Euler
(1667-1748)
(1707-1783)
Prandtl
Taylor
(1875-1953)
(1886-1975)
Vehicles
Aircraft
High-speed rail
4
Surface ships
Submarines
Environment
Air pollution
5
River hydraulics
Physiology and Medicine
Blood pump
6
Ventricular assist device
Sports & Recreation
Water sports
Auto racing
7
Cycling
Offshore racing
Surfing
Turbofan Engine
8
Flow
9
Course contents
•
•
•
•
•
•
•
•
Concepts and definitions. 1 week
Fluid Static. 2 week
Forces on submerged surfaces and bodies. 2 week
Non-viscous flow. 2 week
Conservation of mass. 1 week
Momentum and energy. 1 week
Bernoulli’s equation. 2 week
Viscous flow ( flow past immersed bodies, pipe flow,
losses in conduit) 2 week
• Laminar and Turbulent flow. 2 week
10
REFERENCES
• Frank M. White ” Fluid Mechanics” 4th Edition,
McGraw-Hill, 2002.
• Munson, Young and Okiish “Fundamentals of Fluid
mechanics“John Wiley & Sons, Inc. 1999.
• Michael J. Moran, Howard N. Shapiro, Bruce R.
Munson, and David P. DeWitt
“Introduction to
Thermal Systems Engineering: Thermodynamics,
Fluid Mechanics, and Heat Transfer” John Wiley &
Sons, Inc. 2003.
• Yunus A. Çengel,
and
Robert
H.
Turner
“FUNDAMENTALS OF THERMAL-FLUID SCIENCES”
McGraw-Hill Higher Education, 2004
11
Introduction
Fluid mechanics is part of applied Mechanics concerned
with the statics and dynamics of liquids and gases.
Fluid Mechanics
Fluid Statics
Fluid Dynamics
Hydro- Dynamics
Hydro-Statics
Aero-Statics
12
Hydraulics
Gas Dynamics
Hydro-Statics:
Deals with the statics of incompressible fluids “ Liquids“.
Aero-Statics:
Deals with the statics of compressible fluids “ Gases “.
Hydro-Dynamics:
Deals with the dynamics of incompressible fluids “ Liquids “.
Gas-Dynamics:
Deals
with the` dynamics of compressible fluids “ Gases “.
13
Fluid
Free Surfaces
Gas
Liquid
No free Surfaces
Easy to compress
14
Difficult to compress
DIMENSIONS AND UNITS
• Any physical quantity can be
characterized by dimensions.
Length = 10 m
The magnitudes assigned to the
dimensions are called units.
Unit
• Primary
or
fundamental Dimension
dimensions, mass m, length L,
time t, and temperature T
• Secondary
dimensions,
or
derived dimensions. velocity V,
energy E, and volume V are
expressed in terms of the primary
dimensions
• The Metric System of units, m, kg, s.
• The English system, which is also known as
the United States Customary System (USCS).
• The SI is being used for scientific and
engineering work in most of the industrialized
nations, including England.
16
Metric
Unit
S.I.
British
Length, L
Mass, m
m
kg
m
kg
Ft
lbm
Density, 
Force, F
kg/m3
Kgf
kg/m3
N
lbm/ft3
lb
Kgf/m3
ata, kgf/m2
kCal
N/m3
bar, Pa
kJ
lb/ft3
psi
BTU
kCal/kg.C
hp
poise, kp.s/m2
kJ/kg.C
kW
N.s/m2
BTU/lb.f
B.hp
lb.S/ft2
C, K
Sec
C, K
Sec
f, R
Sec
Dimension
Specific Weight, 
Pressure, p
Heat, Q
Specific heat, Cp
Power, P
Viscosity, 
Temperature, T
17
Time,

• we consider force to be a secondary dimension
whose unit is derived from Newton’s second
law, that is,
18
Example
• A commonly used equation for determining the volume rate of
flow Q, of a liquid through an orifice located in the side of a
tank is
• Q = 0.61 A 2gh
• Where A is the area of the orifice, g is the acceleration of
gravity and h is the height of the liquid above the orifice.
Investigate the dimensional homogeneity of this formula.
Solution
•
•
•
•
•
19
Left hand side
Q = volume/ time = L3. t-1
Right hand side
0.61 A 2gh = L2 ( L /t2 L)1/2= L3.t-1
The formula is homogenous
Properties of Fluids
• Density
• The density of a substance is that quantity of matter contained in
unit volume of the substance.
• Mass Density
• Mass density ρ is defined as the mass of the substance per unit
volume.
mass

Volume
3
• Units: kg/m
• Dimensions: M L-3
• Typical values at p = 1.013 x 105 N/m2, T = 288.15 K:
• Water, 1000 kg/m3;
• Air, 1.23 kg/m3.
Specific Volume:
20
volume
1
v

mass

• Specific Weight γ
• Specific weight γ is defined as the weight per unit volume.
Weight mass g N

 3
volume Volume m
• Weigh per unit volume = Mass per unit volume x g
γ = ρ.g
•
•
•
•
•
21
Units: N/m3
Dimensions: M L-2-2
Typical values at p = 1.013 x 105 N/m2, T = 288.15 K:
Water, 9.81 x 103 N/m3;
Air, 12.07 N/m3.
Specific gravity
Specific gravity is defined as the ratio between
density and water density.
S .G 
Example
the substance
 subs tan ce
 water
If the specific gravity of an oil is 0.8, find its specific
weight in kg/m3.
Solution:
22
N/m3
γ
Viscosity:
μ
It is the property of a fluid, due to cohesion and
interaction between molecules, which offers resistance
to shear deformation.
Shear Stress
y
u
(
)B
y
B
23
x
Newtonian Fluid
u
A
b
y
No slip condition
u
u
τ xy 
y
b’ c
Fluid
a
d
F
c’
u
F  A
y
u
 F  μ A
y
u
 τ  μ
y
Where
24
μ:
Is the coefficient of viscosity or
the dynamic viscosity.
Dimension of dynamic viscosity
  ML1t 1
Units of dynamic viscosity (Poise)
Poise 
dyne.s
N

0
.
1
Pa
.
S

0
.
1
s
2
2
cm
m
Kinematics viscosity 



Dimension of Kinematic viscosity L2.t-1
Units of kinematic viscosity (Stoke)
2
cm 2
4 m
1stoke 1
10
,
s
s
25
ft 2
 930 stoke
s
Example:
A 6 Cm. diameter shaft runs in 12 Cm long
bearing, if an oil viscosity 0.5 poise and 0.02
thick fills the space between the shaft and the
bearing. Find the power lost due to friction when
the shaft rotates at 200 r.p.m.
L = 12 Cm.
d = 6 Cm
D = 6.02
26
Solution:
A  πdL  π 6 12  226.19
(6.02  6.0)
y
 0.01
2
Cm 2
Cm
2 πN d 2 π 200 6
u  ω r 
 
  62.832
60 2
60
2
Cm/sec
u
62.832
 F  μ A  0.5  226.16 
 710598.5
y
0.01
dyne
F u
710598.5  62.832
 Power 

 0.00446 kW
5
2
3
Const .
10 10 10
27
σ
Surface Tension:
σ
σ
d
θ
h
θ
h
Water
Mercury
θ : is the angle of contact between liquid and solid
Upward pull due to surface tension equal σ π d cos θ
h : is the height of liquid is raised
28
Weight
of liquid raised = γ A h = γ π d2 h/4
σ π d cos θ = γ π d2 h/4
4 σ cosθ
h 
γd
For water in tube 6 mm in diameter h will be 4.5 mm
For mercury the corresponding figure is – 1.5 mm.
7- Bulk Modulus of Elasticity:
29
K
P2  P1
K
V2  V1
V1
Fluid Statics
That means: Fluid at rest and free of shearing stresses.
Pressure Intensity:
P1
P1
A
P3Sinθ
θ
P2
C
P3Cos θ
A
B
θ
B
C
Force on AB = P1 x AB x S
Force on BC = P2 x BC x S
30
Force on AC = P3 x AC x S
P2
P1
Fluid at rest
Appling the forces balance :
P
1
A
P3Sinθ
θ
x AB x S = P3Cos θ x AC x S
 AB = AC x Cos θ
P
1
P3Cos θ
= P3
31
P2 = P3

P1 = P2 = P3
P2
C
Appling the forces balance (Horizontally) :

B
Pressure Head:
h
in m
P  ρgh  γh
h
Fluid
(γ)
P
Variation of pressure along horizontal line:
  Fx  0
P1a  P2a
32
 P1  P2
P1
a
P2
Variation of pressure along vertical line:
P1
 W  γ a h
  Fh  0
P1 a  γ a h  P2  a
 P2  P1  γh
W
h
P2
a
33
Absolute and Gauge Pressure:
kg f
Pressure 2
m
1 bar = 105
or
N/m2
Patm = atmospheric pressure
Patm = 1.013 bar
= 760
mm.hg
= 1.033 kgf /Cm2
34
Force
Pressure
Area
N
Pressure 2
m
Pabs = Patm + Pgauge
35
Vacuum or Negative Pressure:
If the pressure is blew the atmospheric
negative or vacuum
pressure it is called
Vapor Pressure:
The vapor pressure is the absolute pressure below which the
liquid turns into vapor.
36
Example
Convert a pressure of P = 5 bar into meters of :
i- Water.
ii- Mercury
iii- Oil of specific gravity of S = 0.9
P  γ  h
P
P
h  
γ g ρ
i- For water:
h water
5  105

 50
1000 9.81
meter
 γ1  h1  γ 2  h 2
 γ water  S1  h1  γ water  S2  h 2
37
 S1  h1  S2  h 2
Measurements of pressure:
The pressure is measured by a devices is called Manometers.
Manometers can be classified into :
Vacuum
region
a- Simple manometers:
1 – Barometer:
h
The barometer is a device for
measuring the local atmospheric
pressure.
Mercury
The Standard atmospheric pressure is 10.3 meter of water or
0.758 meter of mercury.
38
Piezometer:
h
Fluid ( water or oil )
The Piezometer tube is used to measure the static pressure head of
the flowing liquid at any section of a pipe.
Note that:
The Piezometer tube would not work for negative pressure.
It is not suitable for measuring high pressure.
It is not suitable for measuring for gas pressure.
39
Differential manometers:
Pgas
First for positive pressure:
PL = PR
x
y
PL = Pgas + gas . x
PR = 1 . y
Pgas = 1.y - g . x
g . x ≃ 0
∴ Pgas = 1.y
40
L
S1
R
Second for negative pressure:
∵ PL = PR
Pgas
PL = Pgas + 1 . y
y
PR = 0
∴ 0 = Pgas + 1 . y
∴ Pgas = - 1.y
41
L
S1
R
Third for liquids:
PLiq
In this case : S2 > S1
x
S1,γ1
y
∵ PL = PR
L
PL = P + 1 . x
S2
PR = 2 . y
∴ P = 2 . y - 1 . x
42
R
The U-tube manometer can used to measure the difference in pressure
In this case : S2 > S1
S1
∵ PL = PR
1
x
2
PL = P1 + 1 . (x + y)
y
PR = P2 + 1 . x + 2. y
L
S2
 P1 + 1(x + y) = P2 + (1x + 2.y)
P1 – P2 = y (2 – 1)
43
P1 – P2 = water.y (S2 – S1)
R
In this case :
S1 > S2
S2
L
R
∵ PL = PR
y
∴ PL = P1 ‫ ــ‬1(x + y)
∴ PR = P2 ‫( ــ‬1x + 2.y)
 P1 ‫ ــ‬1(x + y) = P2 ‫( ــ‬1x + 2.y)
P1 ‫ ــ‬P2 = y (1 ‫ ــ‬2)
 P1 ‫ ــ‬P2 = water.y (S1 ‫ ــ‬S2)
44
x
S1
1
2
Inclined manometer:
P1
P1 = . h
θ
θ
Bourdon pressure tube:
A
A
Section at A-A
45
ℓ
h
Transmission of pressure:
F
∵ P1 = P2
a
F W
A
46
A
a
F
P1 
a
W
P2 
A
W
P1
Oil
P2
Hydraulic press:
F
a
∵ P1 = P2
f
W
 γh
 P2
a
A
47
P1
W
A
h
P2
Hydraulic Jack:
F
L
F1
F. L = F1. L1
a
P1
Oil
48
W
A
L1
P2
Hydrostatic of pressure
Introduction:
When a surface is submerged in a fluid, forces develop on the
surface due to the fluid.
The determination of these forces is important in the design of
storage tanks, ships, dams and other hydraulic structure. For
fluids at rest we know that the force must be perpendicular to the
surface since there are no shearing stresses present.
49
Hydrostatic Forces on Horizontal
and Vertical Surfaces
50
1
F  h. A
2
F  h.A
Fy p   dF y    . y sin  .dA y
F   h.dA
A
A
F    . y. sin  dA
A
A
  sin   y 2 dA
A
  sin  I x
F   sin   y.dA
θ
a
 y dA
yc 
 y dA
A
F   yc sin  A
  hc A
51
o
h
hc
First Moment of Area
hp
F
y
yc yp
cp
dA
x
I x   y dA
2
Second moment of Area
I x  I c  y c2 A
yp 
 sin  ( I c  y A)
2
c
F
2
 sin  ( I c  yc A)

 yc sin  A
52
Ic
 yc 
yc A
Figure
Rectangular
h
G
G
d/2
Area
A
2nd moment of
area Ic
b.h
b h 3
12
b h
2
b h 3
36
b
Triangle
h
h/3
b
Circle
Semi-circle
53
G
4R/3π
G
π R2
π R4
4
R
G
π R2
2
0.1104 R 4
R
G
Pressure on Curved Surface:
Problems involving the resultant pressure of a liquid on a curved
surface can usually be solved most conveniently by considering
the horizontal and vertical components of the resultant pressure
separately.`
D
E
C
P
A
Liquid
G
H
V
(a)
D
E
B
B
Liquid
P
G
R
Liquid
O
R
A
(b)
R
(c)
H and V are the horizontal and vertical components of the
resultant pressure R of the liquid on the surface.
54
D
E
C
P
A
Liquid
G
H
V
B
R
(a)
Figure (a), if ACE is vertical plane through A the section of fluid is
in equilibrium horizontally under the action of H and the resultant
pressure P of the liquid on AC, therefore H = P.
But AC is the projection of AB on vertical plane, therefore
Horizontal component H = resultant pressure on the projection of
B on a vertical plane.
55
D
E
B
Liquid
P
G
R
A
(b)
In Figure (b), if the surface AB is removed and the space ABDE
is filled with liquid, this liquid would be in equilibrium under its
weight and the vertical force V on the boundary AB, therefore;
Vertical component V = weight of the fluid which would lie
vertically above AB if AB were removed and V will act upwards
through the centre of gravity G of the imaginary fluid.
56
Example
The 4 –m diameter circular gate is located in the
inclined wall (θ=60o) of a large reservoir containing
water ( =9.8 kN/m3). The gate is mounted on a shaft
along its horizontal diameter for a water depth of 10
m above the shaft, determine:
A- the magnitude and location of the resultant force
exerted on the gate by the water, and
B- the moment that would have to be applied to the
shaft to open the gate.
57
Solution
•
•
•
•
•
•
•
•
•
•
•
58
F= hc A
= 9800x10x(πx42/4)
=1.23 MN
yp = yc + I.c/(yc. A)
Ic = πxR4/4
= πx24/4=4 π
yp = (10/sin60)+4
π/((10/sin60)x πx42/4)=11.6 m
yp –yc= 0.0866 m
∑Mc = 0
M = F(yp-yc)=123x1000x0.0866
=1.07x105 N.m
60
F
h= 10 m
O
shaft
4m
Example
• A rectangular gate in the vertical side of a
reservoir can turn freely about its upper
edge which is horizontal and is fastened at
its two lower corners. The gate is 1.0 m
wide and 2.0 m high and its upper edge is
2.0 m below the water level. Determine the
reactions at the lower corners assuming
them to be equal.
59
Solution
•
•
•
•
•
•
•
FH=γ hc A
= 9810*3*(1*2)= 58750 N
hp =hc + Ic/ (hc*A)
I c=1*23/12=0.67
hp =3+0.75/[3x(1x2)]=3.11 m
FH* (3.11-2)=2R*2.0
R=58750*1.11/4=16340 N
60
2m
hb
2m
hc = 3 m
1m
2m
FH
R
Determine the resultant force due to water acting on the 1 m by 2 m
rectangular area AB and on 1.25 m by 2 m triangle area CD shown
in Figure below. The apex of triangle is at C.
1- For rectangular area AB
 Let the Resultant Force P
A
F  9810  (1.22  1)  (1 2)
 43560 N
2.0 m
B
1m
The resultant force Acts on the center of pressure
61
1.0 m
C
2m
 F  ρ g h C.G  A
1.22 m
45o
D
2- For triangle area AB
 F  ρ g h C.G  A
2
1
F  9810  (1   0.707  2)  ( 1.25  2)  23.8 103 N
3
2
The resultant force Acts on the center of pressure
62
Determine and locate the components of force due to the water
acting on the curved area AB in Figure per meter of its length.
1- For the horizontal component PH
PH : is the force on vertical projection CB.
A
 PH  ρ g h C.G  A CB
C
y
 PH  9810 (1)  (2 1) 19620 N
x
PH
B
2- For the Vertical component PV
PV : is the weight of water above area AB.
PV
π 22
 PV  ρ gV  9810  (
1)  30820 N
4
 R  PH2  PV2  (19620) 2  (30820) 2
63
 29275067368000 
Hinge
N
Fluids at rest Floating bodies
(buoyance and stability)
As pressure increases with depth, the fluid exerts a resultant
upward (buoyancy) force on a body wholly or partially immersed
in a fluid.
If the buoyancy force equals the weight of the body then it will
float.
No horizontal component of the buoyancy force because the
vertical projection of a body is the same in both directions.
64
Fluids at rest Floating bodies
Upward thrust on
M
Water
surface
Q
P
S
N
lower surface PSR
 weight of fluid in volume PSRNM
R
Downward thrust on upper
surfacePQR
 weight of fluid in volume PQRNM
65
M
Water
surface
Q
N
buoyancy, FB :
FB  weight PSRNM  weight PQRNM
R
P
Resultant upward force due to
FB  weight PQRS
S
Buoyancy force  weight of displaced fluid
66
• Buoyancy force acts through the centre of gravity of
the volume of fluid displaced. This is called the
centre of buoyancy. The centre of buoyancy does not
in general correspond to the centre of gravity of the
body.
If the density is constant the
centre of gravity of the displaced
fluid is at the centroid of the
immersed volume.
G
B
67
Fluids at rest
Floating bodies - Archimedes Principle
• The buoyancy force is equal to gV where  is the
density and V the volume of fluid displaced.
• For equilibrium the buoyancy force must equal the
weight of the body:
FB  W
 equilibriu m
FB  W
 body rises (bubbles of air in water)
FB  W
 body sinks (stone in water)
68
Stability and Metacentre:
M
G
θ
B
R=W
W
R
1. If M lies above G a righting moment is produced, GM is
as positive,
equilibrium
is stable.
2. Ifregarded
M lies below
G anand
overturning
moment
is produced, GM
is regarded as negative, and equilibrium is unstable.
3.
69
If M and G coincide the body is in neutral equilibrium.
W
G
W
G M
M
M
W
G
B
B1
B
B1
FB1
FB1
GM
is
+ve positive
Stable
70
B′
B
GM
is
-ve positive
Unstable
FB
GM
is
Neutral
0
Determination of the Metacentric height
W
M
FB = ρ g V
BB1 = BM sin dθ
G
B
dθ
B1
BB1 = BM . dθ
FB1
GM = BM - BG
BM 
Where:
I
V
71
I
 BG
V
Is the Area Moment of Inertia of the top view about the vertical axis
Is the volume of liquid displaced by the body.
Example
A block of wood 0.2 x 0.5 m cross-section and 0.8 m long has a mass
of 64 kg. Can the block float with 0.5 m side vertical?
0.5 m
Solution:
0.5 m
G
0.2 m
0.25 m
0.4 m
Water Level
B
0.2 m
0.8 m
Volume of displaced water
64
V
 0.064 m3
1000
The height of submergence
0.064
h
 0.4 m
0.2  0.8
72
The center of buoyancy is 0.2 m from the base.
The Metacentric height
I
BM   BG
V
0.8  0.23
I
 0.00053 m 4
12
BG = 0.25 – 0.2 = 0.05 m
GM = BM - BG
The Metacentric height
∵ GM < 0
73
0.00053
GM 
 0.05   0.042 m
0.064
The Block is unstable with its 0.5 m side vertical
• Worked Example
• A cube of side 35 mm is floating in water of
density 1000 kg/m3. a) What is the density of
the cube material if the depth of submersion is
27 mm? b) A hemispherical scoop, diameter
15 mm, is removed from the top face (volume
of a sphere = ). What is the new depth of
immersion?
74
74
• Worked Example
a)
27 mm
When the cube is floating in a stable position the buoyancy
force is equal to the weight of the cube:
75
75
• Worked Example
Weight of cube :
W  mg  cubeVcube g  0.035 cube g  4.21x10 4 cube
3
But, the buoyancy force is also equal to the weight of
displaced fluid :
FB   waterVdisplaced g  10000.0352 x 0.027 g  0.324 N
where 0.027 is depth of immersion.
76
76
• Worked Example
Equate :
W  FB
4.21x10 4 cube  0.324
cube  770 kg / m3
77
77
• Worked Example
b)
h
78
78
• Worked Example
Find the new weight of the cube:
W  0.324  12 43 0.00753 cube g  0.317 N
When the depth of immersion is h,
the buoyancy force is :
F B 0.0352 h  water g
Equate :
FB  W
79
12.017h  0.317
h  26.4 mm
79
LIQUID IN RELATIVE EQUILIBRIUM
- If the vessel containing a liquid is at rest or moving with constant
linear velocity, the liquid is not affected by the motion
- But if the container is given continuous acceleration this will be
imparted to the liquid which will take up a new position.
- Fluid pressure is every where normal to the surface on which it acts.
5.2 Horizontal acceleration:
at rest
or

moving with constant linear velocity
W
B
A
80
1.5 m
C
1.5 m
W
P  m f 
f
g
R

f
P
For equilibrium P = W . tan θ
1.5 m
f
W  tan θ  P  W 
g
W
B
A
f
tan θ 
g

C
1.5 m
1.5 m
and is constant for all points on the free surface.
Example
If the linear acceleration is given by f = 3 m/sec2, the tank is 3 m
long and the depth of water when the tank is at rest is 1.5 m.
Calculate: a- The angle of the water surface to the horizontal.
b- The maximum pressure intensity on the bottom.
81
c- The minimum pressure intensity on the bottom.
Solution:
(a)
(b)
f
3
tan θ  
 0.306
g 9.81
θ  17 o
Maximum pressure intensity occurs at point A
hA = 1.5 + 1.5 tan  = 1.96 m
 The pressure intensity at A =  h A = 9.81 x 103 x 1.96
= 19.2 KN/m2
(c) The minimum intensity of pressure occurs at point B.
hB = 1.5 – 1. 5 tan  = 1.04
82
The pressure intensity at B =  h B = 9.81 x 103 x 1.04
= 10.2 KN/m2
5.3 Vertical acceleration:
Prism cross-section
area is = a
A
f
B
h
Accelerating force at X = P
force due to pressure weight of prism = P . a -
X
 . h .a
By Newton's 2nd law
h a
f
P = mass x acceleration =
g
h a
Pa  γ h  a 
f
g
83

f
 Pa  γ h  1  
g

C
dy
tan θ 
dx
For a constant value of ω , P will vary
with x, since the centrifugal acceleration
is ω2. x and,
W 2
P  ω  x
g
dy ω 2 x
tan θ 

dx
g
x
P
R

A
O
W
ω2  x
ω2  x 2
y
 dx 
 Constant
g
2g
0
84
D
B
Integrating the
x
Axis of rotation
5.4 Forced vortex:
If y is measured from the AB, i.e,
∴ Constant = 0
y = 0 when x = 0 .
ω2  x 2
 y
2g
Example
A cylindrical tank is spun at 300 r.p.m. with its vertical axis. The tank
is 0.6 m high and 45 Cm. diameter and filled of water before spinning.
Show that the water surface will take the form of paraboloid when the
container is spun and calculate:
(a) The speed at which the water surface will just touch the top
rim and the center bottom of the tank.
(b) The level to which the water will return when the tank stops
spinning and the amount of water lost.
85
Solution:
(a) When the surface just touched the rim of the base: y = 0.6 m and
x = 0.225 m
2g y
 ω
15.25
2
x
N = 145.6
rad . /sec.
r.p.m
(b) In figure the volume of the paraboloid will be half that of the
containing cylinder ABCD . When the water surface touches the rim and
the center of the bottom,
volume of water left in tank = half of the original volume
= ½ π (0.225)2 x 0.6 = 0.0475 m3
Depth when tank ceases to rotate = 0.3
86
∴ Amount of water thrown out = 0.0475
m
m3
The figure shows a semicircular gate of 1.2 m diameter and
length 0.9m in the wall of a tank containing water. If the gate has
a mass of 1000 kg, what force, F, is required to lift the gate?
Resultant force
R = pressure at centroid × area,
2 m
π 42
R  ρg(2  2)
 43104 N
4
1.2 m
Depth to centre of pressure, the point
where the resultant force acts
87
I CG
 y Cp  y CG 
y CG A
π 4 12.566
 y Cp 

 4.25 m
4
π4
4
4
4
2
Moment of R about pivot
MR = 493104 (4.25-4.0) = 123276 N.m
Moment required from F applied at base is MF is equal to MR
MF = M R
F x 2 = 123267
88
F = 61638 N
b) Force Down = weight of water above gate (volume A) +
weight of gate
Force Up = F + Weight of water in volume B
2.6 m
A
0.6 m
89
2.6 m
B
0.6 m

π 0.62 
FD  2.6  0.6 
  0.9 1000 9.81  21087 N
4 


π 0.62 
FV  2.6  0.6 
  0.9 1000 9.81  F 16270
4 

FD  FV
21087  F16270
F  2108716270 4817 N
90
91
92
93