1
CHAPTER 1 PROPERTIES OF FLUIDS 1.1
Introduction Fluids mechanics deals with study of fluids-liquid and gases. The study can
be behavior of liquid fluids at rest (static) and in motion (dynamic). The study of fluid mechanics is important because our life depend on them. The air we breathe, flight of birds in air, the motion of fish in water, circulation of blood in veins of human body, flow of oil and gas in pipelines, transportation of water in pipe, all follow the principles of fluid mechanics. Engineers have applied these principles in the design of dams, construction of ships, airplanes, turbo-machinery etc. Fluids in motion are potential sources of energy and can be converted into useful work to drive a water turbine or windmill. The principles of fluid mechanics are also applied to fluid power system in which pressured fluid is used to transmit power. Hydraulic drives and controls have become more and important due to automation and mechanization.
Today, a very large part or modern machinery is controlled
completely or partly by fluid power.
Fluid can be defined as substance that has ability to flow. Gases expand whereas liquids do not. Liquid have no shape of it’s own but rather take the shape of the container in which it is placed. That means if liquid or volume less than volume of the container is poured into container the fluids will occupy a volume of the container and will have a free surface. Gases expand and occupy full volume of the container. Gases are compressible which means their volume changes with pressure where as liquids are incompressible. Compressible flows are again divided into subsonic and supersonic depending on gas velocity less or greater than sound velocity. Their application is in jet propulsion system, aircraft and rockets. 1.2
International System of Units (SI) In the text we shall use SI units.
The dimension in any system can be
considered as either primary or secondary dimensions. In the SI units there are 4 primary dimensions.
2 a) Primary Units Dimension Mass Length Time Temperature Electric Current
International Symbol M L T K A
Unit Kilogram (kg) Meter (m) Second (s) Kelvin (K) Ampere (A)
b) Secondary Units Secondary units is a combination of primary units such as Newton (N or kgm/s2), Joule (J or Nm), Watt (W or Nm/s) etc.
1.3
Specific weight and mass density Two important parameters that tend to indicate heaviness of the substance are
specific weight and mass density. The specific weight in the weight of substance per unit volume and is commonly designated by Greek letter ‘gamma’ ( γ ). In equation form,
γ =
Weight W = Volume V
Mass density is the mass per unit volume of the substance. It commonly designated by a Greek letter ‘rho’ ( ρ ). In the equation form,
ρ=
mass m = Volume V
There exist an important relation between specific weight and mass density.
Weight of the substance w = mg
Weight of the substance/unit volume,
γ =
mg = ρg V
3
For ideal gases, the density of gas is depended on the pressure and temperature of the gas. The density can be obtained by the gas equation; PV = mRT
or P = ρRT
where R =
Universal _ gas _ Coefficient ℜ = molar _ gas M
Thus the specific weight is the product of mass density and acceleration due to gravity. In the SI units, γ will be expressed in N/m3 and ρ in kg/m3. The values of specific weight and mass density of water at different temperature are given in Table 1.1 and Table 1.2 gives
Table 1.1 Physical Properties of Water Temperature
Specific
Mass
Dynamic
Kinematic
Surface
Weight
Density
Viscosity
Viscosity
tension
T
γ
ρ
µ
ν
σ
(°C)
(kN/m³)
(Kg/m³)
N-s/m²
m/s²
N/m
0
9.81
1000
1.75 x 10
−3
1.75 x 10
−6
0.0756
30
9.77
996
8.00 x 10
−4
1.02 x 10
−7
0.0712
60
9.65
984
4.60 x 10
−4
4.67 x 10
−7
0.0662
90
9.47
956
3.11 x 10
−4
3.22 x 10
−7
0.0608
the mass density for common fluids. From this table, other fluids can be compared with water in terms of density and specific of weight.
4
Table 1.2 Physical Properties of Common fluids at Standard Atmospheric Pressure Fluids
Specific
Specific
Mass
Dynamic
Kinematic
Gravity
Weight
Density
Viscosity
Viscosity
s
γ
ρ
µ
ν
-
(kN/m³)
(kg/m³)
N-s/m²
m /s
0.0012
11.8
1.20
1.81 x 10
−5
1.51 x 10
−5
Ammonia
0.830
8.31
829
2.20 x 10
4
2.65 x 10
−7
Glycerine
1.263
12.34
1258
950 x 10
−3
7.55 x 10
−4
Kerosene
0.823
8.03
819
1.92 x 10
−3
2.34 x 10
−6
Mercury
13.60
133.1
13570
1.56 x 10
−3
1.14 x 10
−7
Methanol
0.79
7.73
788
5.98 x 10
4
5.58 x 10
−7
SAE 10 Oil
0.87
8.71
869
8.14 x 10
−2
9.36 x 10
−5
SAE 30 Oil
0.89
8.71
888
4.40 x 10
−1
4.95 x 10
−4
Turpentine
0.87
8.51
868
1.38 x 10
−3
1.58 x 10
−6
Water
1.00
9.79
998
1.02 x 10
−3
1.02 x 10
−6
Sea Water
1.03
10.08
1028
1.07 x 10
−3
1.04 x 10
−6
Air
1.4
2
Specific gravity
It is the ratio of specific weight of the substance to the specific weight of water at 4°C. A convenient method to measure sp. gravity is by means of a hydrometer. It is dipped into the liquid and a calibrated scale gives the specific gravity. It should be noted that specific gravity is a dimensionless number and its value for a particular substance is the same regardless of the system of units. It is abbreviated as (s).
Specific _ gravity =
ρf specific _ weight _ of _ fluid = o specific _ weight _ of _ water _ at _ 4 C ρ w
5 The specific gravity can also be expressed as ratio of mass density of the substance to mass density of water at 4°C.
Example 1.1 A tank of glycerol has a mass of 1200kg and volume of 0.95m3. Determine:
(a) Weight of glycerol (b) Density (c) Specific weight (d) Specific gravity
Solution:
(a) From Newton’s Law;
W= mg
Thus, W= 1200 x 9.81 = 11.76kN
(b) From equation (1.2)
ρ=
m 1200 = = 1265kg / m3 V 0.95
(c) From equation (1.3)
γ = ρg = 1263 × 9.81 = 12.38kN / m3
(d) From equation (1.4)
6 s=
1.5
ρ s 1265 = = 1.26 ρW 1000
Viscosity
Fluids offer resistance to shearing force. Viscosity is the property of the fluid that determine amount of this resistance.
Consider a fluid in between two parallel plates Figure 1.1 where the upper plate is moving with velocity V and lower plate is stationery. The distance between the plates is y. The layer in contact with the upper plate is moving with velocity V where as the layer is contact with lower plate which is fixed will have zero velocity. The deformation of the fluid under the action of shear stress is assumed proportional to the rate of change of velocity, may be expressed in the equation form, Upper plate (moving) Velocity = V
v + dv dy y
v
The rate of change of velocity = [(v+dv)-v] / dy = dv/dy
Lower plate (fixed) Velocity = 0
Figure 1.1 The displaced fluid due to shear stress acted and the shear stress is assumed proportional with the velocity gradient. It can be illustrated in the equation;
7
τ =µ
∂v ∂y
(1.5)
Where, τ= shear stress μ= proportional coefficient ∂v = velocity gradient ∂y
The shear force, F D acting on the lower plate surface is given by;
FD = τ × A
Where A= surface area of the lower plate The unit of dynamic viscosity, µ is SI unit is (N-sm-2 or Pas). Kinematics viscosity which is usually, denoted by the Greek letter ‘nu’ (v) is determined by dividing dynamic viscosity (µ) by mass density of the fluid ( ρ ). In the equation form;
ν=
µ ρ
(1.6)
When the fluid is at rest the velocity gradient dv/dy is zero and therefore no shearing force exists. The viscosity varies with temperature therefore values of µ for given fluid are usually tabulated at various temperatures. There are experimental methods to calculate viscosity. One such experimental method is Falling Sphere Viscometer. In this method a sphere of known diameter is dropped into a liquid. By determining the time required for the sphere to fall through a certain distance, its terminal velocity (v) can be calculated. The stokes equation can be written as
v=
gd 2 σ − 1 18ν ρ
8
Where
d = diameter of sphere σ = sphere density ρ = fluid density ν = kinematics density
A number of viscometers are available in the market. These viscometers are electronic devices with digital panel and measured viscosity most of the liquids such paint, lubrication oil, polymer compound, chemical compositions etc. Fluids obeying Newton’s law of viscosity (equation 1.6) and for which µ has a constant value are known as Newton’s fluids. Most common fluids such as air, water and oil come under this category for which shear stress in proportional to velocity gradient. The fluids that do not obey Newton’s law of viscosity are known as nonNewtonian fluids such as human blood, lubrication oils, molten rubber and sewage sludge etc. A general relation between shear stress and velocity gradient for nonNewton’s fluids may be written as; dv τ = A + B dy Where A and B are constants.
n
(1.7)
Based on the value of power index ‘n’ non –
Newtonian fluids are classified as;
Pseudoplastic (such as milk, cement, clay)
n<1
Bingham-plastic (such as sewage sludge, toothpaste)
n=1
Dilatent (such as lubrication oil, butter, printing ink)
n>1
A Newtonian fluid is a special case of non–Newtonian fluid for which A = 0 and power index n = 1.
The dynamic viscosity of various fluids at various temperatures is shown in Figure 1.2.
9 0.5 0.4 0.3 0.2
Castor Oil SAE10 oil Glyserin
Absolute viscosity µ, N . s/m2
0.1
SAE 30 oil
0.06 0.04 0.03 0.02
Crude oil (SG 0.86)
0.01 6 4 3 2
Kerosene Aniline
1 x 10 -3
Mercury
Carbon tetrachloride
6 4 3 2
Ethyl alcohol Water Gasoline (SG 0.68)
Benzene
1 x 10-4 6 4 3 2
Helium Carbon Dioxide
Air
1 x 10-5
Hydrogen 5
-20
0
20
40 60 80 Temperature, oC
100
120
Figure 1.2 Dynamic viscosities versus Temperature Example 1.2 The viscosity of a fluid is to be measured by a viscometer constructed of two 75cm long concentric cylinders as shown in Figure E1.2. The outer radius of the inner cylinder is 15cm, and the gap between the two cylinders is 0.12cm. The inner cylinder is rotated at 200rpm, and the torque is measured to be 0.8Nm. Determine the viscosity of the fluid. u=0 t
R
ω
Figure E1.2
10
Solution:
The shear force, FD = τ × A = µ
∂v × 2πRL ∂y
Where shear force can be calculated by
FD =
0.8 Torque = = 5.33 N 0.15 R
V= 2πRf = 2(3.142)(0.15)(200)/60 = 3.14 m/s ∂v V − u 3.14 − 0 = = = 2616.67 m / s ∂y t 0.0012
Thus, the dynamic viscosity:
µ=
1.6
5.33 = 0.0029 Ns / m 2 2π (0.15)(0.75)(2616.67)
Compressibility and Bulk Modulus
Consider a mass of fluid m whose initial pressure and volume is P and V respectively. Let the fluid be compressed by application of force such that final pressure is P+dP and volume reduced to V–dV. Hence, change in pressure is dp and change in volume is –dV. Volumetric strain in defined as change in volume divided by original volume and is –dV/V. The bulk modulus denoted by k and is defined as change in pressure to volumetric strain;
k= Changes in pressure/Volume strain or
11 k = −V
dP dV
(1.8)
Let mass of fluid is m: m = ρV
(1.9)
After differentiation of equation (1.9):
ρdV + Vdρ = 0
−
V ρ = dV dρ
(1.10)
Substitute –V/dV into equation (1.8), thus;
k=ρ
dP dρ
(1.11)
From equation (1.11) the value of k is dependent on the relationship between pressure and density.
For liquids, changes of density with pressure are small and Bulk
modulus k is high. These liquids can be considered incompressible. However, for gases the compressibility is so large that value of k is not a constant but proportional to pressure. For gases relation between pressure and mass density in obtained from characteristic equation of a gas and particular relation between pressure and density is established depending on type of compression process.
(i)
For an isothermal process where the temperature is maintained constant the characteristic equation is written as
dP P P = = const. = const. ρ dρ ρ
dP P = = const. dρ ρ
(1.12)
(1.13)
12
Substitute dP/dρ into equation (1.6) gives k= P
(ii)
(1.14)
For an adiabatic process where no heat is allowed to enter or leave during compression the relation between pressure and density is given by P
ργ
= const.
(1.15)
After differentiation will give;
dP P = γ dρ ρ
(1.16)
where γ= ratio of specific heats at constant pressure and at constant volume or γ= C P /C V Again substitute dP/dρ into equation (1.11) will give; k= γP
(1.17)
The ratio of adiabatic bulk modulus is equal to the ratio of specific heat of fluid as constant pressure to that at constant volume. For liquids is almost equal to one, but for gases the difference is large for example for air = 1.4. 1.7
Mach no. and Compressibility
Mach. No. is defined as ratio of velocity of flow (v) to local velocity of sound (a) and is a measure of compressibility effects. M=
v a
(1.18)
13
The velocity of propagation of sound waves in a fluid, flow is expressed as
a=
dP dρ
or
a2 =
dP dρ
(1.19)
Substituting the value of dP/dρ in equation (1.11) we get k = ρa 2
(1.20)
Substituting value of a in equation (1.12) we get
M=
v2ρ k
(1.21)
For liquids the bulk modulus k is large and velocities small and, hence, Mach. No. is negligible or effect of compressibility is neglected. Gas velocities are high bulk modulus is low, and hence, Mach. no. is high and compressibility cannot be neglected. Gases can only be treated as incompressible if pressure changes are small and Mach. no. is less than 0.3.
1.8
Surface tension
The molecules of the liquid are attracted by the molecules of the same liquid by a force known as ‘Cohesion’. This force keeps the molecules bonded together.
The force of attraction between molecules of two different liquids that do not mix each other or between liquids molecules and solid boundary containing the liquid or between molecules or liquid on side and molecules of air (or gas) on the other side is known as ‘adhesion’.
14 Vapor Liquid B
A
Figure 1.3
Figure 1.3 is shown a molecule of liquid at the surface is acted on by imbalance cohesion and adhesive forces giving rise to surface tension.
It is
commonly denoted by Greek letter, ‘sigma’ (σ) and is defined as force per unit length of the surface. In the equation, it can be written as
σ=
F L
(1.22)
The units of σ in SI units will be N/m. In many engineering problems surface tension forces are very small compared with other forces acting on the fluid and may therefore be neglected. However, surface tension can cause serious errors in capillary effects particularly in manometer.
For a droplet or a half bubble, the surface tension effect can be illustrated by analyzing a free-body diagram as shown in Figure 1.4.
2πRσ
PπR 2
2πRσ
Figure 1.4
The pressure force exerted in the droplet is given by
15
F = PπR 2
The force due to surface tension is F = 2πRσ
The pressure force and tension must be balance each other; PπR 2 = 2πRσ
P=
1.9
2σ R
(1.23)
Capillarity
If a small diameter glass tube is inserted into water through a free surface the water will rise in the tube. This phenomenon is known as capillarity and is caused by cohesive force of the liquid molecules and adhesion of liquid surface to solid glass surface. The rise in level of the capillarity tube will depend on σ and angle of contact, θ as shown in Figure 1.5.
Parallel Plates θ σ A
B dL
A B h F
Tube
Figure 1.5
16 Length of line of contact of the liquid with the tube = πd Vertical component of the surface tension force = (πd).σ.cosθ
π
Weight of column of liquid, W=
4
d 2 hγ
Thus, for the equilibrium of surfaces tension and gravity forces requires as
πdσ cosθ =
h=
4σ cosθ γd
or
π 4
P=
d 2 hγ
4σ cosθ d
(1.24)
Consequently, when one does not wish a meniscus to rise appreciably in a tube, a large value of diameter is chosen. It is believed that trees, even very tall ones, send water to their highest branches by means of capillarity effects. Hence, capillary passages must be extremely fine. In water and certain other liquids that exhibit capillarity the meniscus is ‘concave’. These liquids wet the glass and angle of contact θ is less than 90°. In some other liquids such as mercury the meniscus is convex. The liquids do not wet the solid surface and angle of contact θ is more than 90°. Glass tubes are commonly used in manometer and capillary action is a serious source of error in reading levels in such tubes. They should have as large a diameter as is conveniently possible to minimize errors due to capillarity. 1.10
Vapor Pressure
Cavitation is given to the phenomenon that occurs at the solid boundaries of liquid streams when the pressure of the liquid is reduced to vapor pressure of the liquid at the prevailing temperature. Any attempt to reduce the pressure still further merely causes the liquid to vaporize more quickly and clouds of vapor bubble form. The bubbles of vapor formed in the region of cavitation move downstream to a region
17
of higher pressure where they collapse (see Figure 1.5). It is repeated formation and collapse of vapor bubbles which can have damaging effects upon the walls of the solid surface. The actual time between formation and collapse may not be more than 1/100 of a second, but dynamic force caused by this phenomenon may be very severe. It is only a matter of having enough bubbles formed over a sufficient period of time for the destruction of the metal begins. Cavitation may occur in pumps, turbines, hydrofoils, propellers, and in venture-meters. In the case of turbines, cavitation is most likely to occur on the blade surfaces near the tail race where as for pumps it is most likely to occur at inlet to the impeller. Cavitation can also occur if a liquid contains dissolved air or gases, since solubility of gases in liquid decreases as the pressure is reduced. Gas and air bubbles will be released as vapor bubble with the same damaging effects. Care should be taken to avoid cavitation as far as possible but if this proves impracticable, than the parts likely to be affected by cavitations should be constructed of especially resistant metals such as stainless steel.
Figure 1.5 Cavitation phenomena inside the nozzle
18 Problems
1.
Determine the density of air, hydrogen, and carbon dioxide at an absolute pressure of 300kN/m² and a temperature of 38.8ºC.
2.
Calculate the specific weight and density of air at absolute pressure of 445 kPa and a temperature of 38 ºC.
3.
If the volume of liquid decreases by 0.2% for an increase of pressure from 6867 kN/m² to 15696 kN/m², calculate the bulk modulus of elasticity of the liquid.
4.
A soap bubble 51 mm in diameter has an internal pressure in excess of outside pressure of 2.06 x 10-2 kPa, calculate the tension in the soap film.
5.
If the pressure inside a water droplet is 0.2 kPa in excess of external pressure, and given surface tension of water in contact with air at 20ºC is equal to 0.0736 N/m, determine the diameter of the droplet.
6.
Air is introduced a nozzle into a tank of water to from a stream of bubbles. If the bubbles are intended to have diameter of 2 mm, determine the pressure of air at the nozzle exceed that of surrounding water (given tension of water = 0.0736 N/m).
7.
The air in an automobile tyre is at 2.943 MPa absolute at 26.6ºC. Assuming no change in the volume of air if the temperature rises to rises to 62.2ºC, determine the air pressure.
8.
A gas occupying a volume of 300 liters at a certain temperature and pressure of 0.34 N/mm² is compressed isothermally to 150 liters. Calculate the initial and final bulk modulus of elasticity.
9.
At a certain point in a fluid, the shear stress in 0.22 N/m² and the velocity gradient is 0.167 sec. If the mass density of the fluid is 1293 kg/m3, determine the kinematic viscosity.
19
10.
A liquid flows between two fixed parallel boundaries. The velocity distribution near to lower wall is given in the following table:
y (mm) 1.0 2.0 3.0 4.0 5.0
v (m/s) 1.00 1.99 2.98 3.00 3.00
Determine the maximum and minimum shear stresses (The dynamic viscosity of fluid is 0.05Pas).
11.
Two plates are arranged as in Figure Q11 in the liquid. The top plate is moving with the velocity of 0.5m/s and the middle plate is moving with the velocity of 2m/s in opposite direction. The area of both plates area 0.25m2. Plot the velocity profile on all surfaces and determine the force acting on the middle plate (Take the viscosity of liquid is 0.01Pas). 0.5m/s 3mm 2m/s 3mm
Figure Q11 12. Mercury does not adhere to a glass surface, so when a glass tube immersed in a pool of mercury, the meniscus is depressed, as a shown in Figure Q12. The surface of mercury is 0.514 N/m and the angle contact is 40ºC. Calculate the depression distance in a 1 mm glass tube.
20
d
Mercury
40o
Figure Q12
13.
The vapor pressure of water at 100ºC is 101 kN/m², because water boils under these conditions. The vapor pressure of water decreases approximately linearly with decreasing temperature at a rate of 3.1 kN/m²/ºC. Calculate the boiling temperature of water at an altitude of 3000 m, where the atmospheric pressure is 69 kN/m² absolute.
-oooOOOooo-
21
CHAPTER 2 HYDROSTATIC PRESSURE
2.1
Introduction A fluid at rest is characterized by absence of relative motion between adjacent
fluid layers. Under such a condition, the velocity gradient is zero and there is no shear stress, therefore, viscosity of fluid has no effect on fluids at rest. But fluids at rest do exert forces on the solid boundary. Knowledge of force variation or more appropriately pressure variations in a static fluid is important to an engineer. There are so many practical examples of fluids at rest such as water retained by a dam, an overhead tank supplying water to the public, gas or fuel in a tank truck. The object of this chapter is to measure pressure variations in a static fluid to discuss pressure and pressure measurement and laws of fluid pressure. 2.2
Hydrostatic Pressure
In the gravity environment, the static pressure in the fluid is proportionally increased linearly with the depth and always acted perpendicular or normal to the surface as shown in Figure 2.1.
h1 depth
h2
1 Δh 2
Figure 2.1 Static pressure increased linearly with the depth The hydrostatic pressure at every location with the depth, h from the free surface is given as
22
P = Patm + ρgh
(2.1)
by neglecting P atm equation 2.1 becomes P = ρgh
(2.2)
and this is called gage pressure that always use in the calculation of static pressure. As a example, for point 1 and 2 in Figure 1 can be written as:
P1 = ρgh1
(2.3)
P2 = ρgh2
(2.4)
Thus, the pressure different between points 1 and 2 can be written as:
∆P = P2 − P1 = ρg∆h
(2.5)
where Δh=h 2 -h 1 The fluid pressure at rest is constant along the horizontal line. In other words, the pressures for all points with similar depth have same magnitude, and it is independent of the shape or cross section area of the fluid container (see Figure 2.2). This may be stated as the equal level-equal pressure principle that forms the basis for many pressure-measuring devices such as barometer and manometer. It also contributes to the operation of a hydraulic jack that a small input force creates a larger output force.
F A
B
C
D
E
Figure 2.2 Pressures are equal for all points along horizontal plane
23
Example 2.1 A tank is connected to a vertical tube is filled with water (γ= 9810N/m2). Determine: (a) Absolute pressure at levels A, B, C, D, E, and F (b) Gage pressure at levels A, B, C, D, E, and F Patm 0.7
A
Water
B 1.8
E 0.9
0.4
C
D
F
Figure E2.1
Solution:
(a) Absolute pressure
From equation 2.1, it can be written as Point A: P A = P atm + γh A , where h A = 0 = 101.3(103) + 9810(0) = 101.3kPa Point B: P B = P atm + γh B , where h B = 0.7m = 101.3(103) + 9810(0.7) = 108.2kPa Point C: P C = P atm + γh C , where h C = 2.5m = 101.3(103) + 9810(2.5) = 125.8kPa Point D: P D = P atm + γh D , where h D = 3.4m
24 = 101.3(103) + 9810(3.4) = 134.7kPa
Point E: P E = P C
since h E = h C
Thus, P E = 125.8kPa Point F: P F = P atm + γh F ,
where h F = 3.8m
= 101.3(103) + 9810(3.8) = 138.6kPa
(b) Gage pressure P gA = P A - P atm = 0kPa P gB = P B - P atm = 6.87kPa P gC = P C - P atm = 24.5kPa P gD = P D - P atm = 33.4kPa P gE = P E - P atm = 24.5kPa P gF = P F - P atm = 37.3kPa Now consider a simple hydraulic jack in Figure 2.3: F1
A1
F2
2
1
Hydraulic fluid, γ
Figure 2.3 Hydraulic Jack
Plunger or piston
A2
25
Using equilibrium principle, the pressure exerted by the fluid on a plunger or a piston is written as P=
P1 =
From which it follows that
F A
F1 A1
(2.6)
P2 =
and
F2 A2
Since the pressure at points 1 and 2 is equal (i.e. same level), we have
F1 F2 = A1 A2
(2.7)
which can be used to solve for F 2 if F 1 , A 1 and A 2 are given, or vice versa.
Example 2.2: For a hydraulic jack as shown in Figure E2.2, determine the weight that could be lifted if the 400N input force F 1 is applied to the plunger. The diameters of the plunger and the piston are10mm and 75mm, respectively F1
A1
F2
2
1
Hydraulic fluid, γ
Figure E2.2
Plunger or piston
A2
26 Solution:
From equation 2.7 above;
F2 = F1
where
A2 A1
A1 = π (0.01) 2 / 4 = 7.855 × 10 −5 m2 A2 = π (0.075) 2 / 4 = 4.42 × 10 −5 m2
Therefore, F 2 = 22.5kN
2.2
Pressure and head
There are several types of pressure-measuring devices available. Devices such as a barometer, Bourdon gage, and manometer are among commonly used instruments to measure, either a gage, vacuum or absolute pressure. A pressure gage measures a pressure relative to the local atmospheric pressure when the pressure is above the atmospheric pressure whereas a vacuum gage is used when the pressure is below the atmospheric pressure.
Pgage
P absolute Pvacuum Patm
P absolute P absolute = 0
Figure 2.4 Type of pressure
27 2.3
Mercury Barometer A typical mercury barometer is shown in Figure 2.5 and used to measure
atmospheric pressure. It consists of a vertical closed glass tube with a column of mercury inside. The barometer is constructed so as to avoid having any trapped air at the end of the tube. It can be assumed that the space between the mercury and the end of the tube contains a vacuum with zero pressure. At the bottom of the column, mercury is contained in a small reservoir in a small reservoir. The pressure acting on the surface of the reservoir is atmospheric pressure P atm . Thus, the atmospheric pressure can be calculated as P atm = ρgh
(2.8)
P=0
P atm h
Figure 2.5
2.4
The Bourdon Gauge This is most common type of pressure gauge which is compact, reasonably
robust and simple to use. A curved tube of elliptical cross-section is closed at one end is free to move, but the other end-through which the fluid enters is rigidly fixed to the frame as shown in Figure 2.6.
28 X
Section at X-X
Pointer X
Flattened phosphor-bronze tube
Figure 2.6
When the pressure inside the tube exceeds outside pressure (usually atmospheric), the cross-section tends to become circular, thus causing the tube uncurve slightly. The movement of the free end of the tube is transmitted by a suitable mechanical linkage to a pointer moving over a scale. Zero reading is of course obtained when the pressure inside the tube equals the local atmospheric pressure. By using tubes of appropriate stiffness, gauges for a wide range of pressure may be made.
2.5
Manometric Pressure The barometer analysis shows that vertical columns of liquid can be used to
measure pressure. The science of this measurement is called manometer. There are different types of manometers with varying degrees of sensitivity which embody the principle already derived and used for pressure measurement.
2.5.1
Simple U-tube Manometer
29 Patm
D
h2
P A
h1 B
ρ1
C
ρ2 Figure 2.7 Consider the U-tube manometer connected via a small hole to a pipe (Figure 2.7) carrying a fluid of density ρ 1 at pressure P A (which is to be measured). Let the open end of the U-tube be subjected to atmospheric pressure, P atm . At the common surface B-C with the configuration as shown in the diagram we have : P A + ρ 1 gh 1 = P B = P C = P D + ρ 2 gh 2 , or P A + ρ 1 gh 1 = P D + ρ 2 gh 2
(2.9)
Now P A is the pressure to be measured (P) and P D = P atm . Thus
P – P atm = (ρ 2 h 2 - ρ 1 h 1 )g
2.5.2
Differential Manometer
(2.10)
This is used to measure the pressure differential between two fluid reservoirs as shown in Figure 2.8. P A + ρ 1 gh 1 = P D + ρ 3 gh 3 + ρ 2 gh 2 ,
Or differential pressure is given by
30
P A – P D = (ρ 3 h 3 + ρ 2 h 2 - ρ 1 h 1 )g (2.11)
D ρ 3
h3 C
h2
A h1
ρ2
B
ρ1
Figure 2.8 2.5.3
Inverted U-tube Manometer
Another type of differential manometer as shown in Figure 2.9 P B + ρ 1 gh 1 = P A P B + ρ 2 gh 2 = P C P C + ρ 3 gh 3 = P D
or
P D – ρ 3 gh 3 – ρ 2 gh 2 + ρ 1 gh 1 = P A
(2.12)
or
P A – P D = (ρ 1 h 1 – ρ 2 h 2 – ρ 3 h 3 )g
(2.13)
31
ρ2 h2
B
ρ1
C h3
h1 D
ρ 3 A
Figure 2.9
2.5.4
An Inclined Manometer
An inclined manometer as shown in Figure 4 is used to achieve a greater accuracy and sensitivity in pressure measurement. This is because the slight pressure change could cause a noticeable change in L.
P1
Diameter, D P2 L h
AIR y Fluid density Diameter of tube, d Figure 2.10 Inclined Manometer
Applying the equal level-equal pressure principle, we have
P 1 = P Gas But P 1 = P 2 + γh and h= L sinθ. The above equation then becomes,
Zero level
32
P1 = P2 + γ (L sin θ )
(2.14)
Example 2.3 U-tube manometer containing a mercury (sg= 13.6) as a working fluid is connected to a tank that contains air as shown in Figure E2.3. The other end of the manometer is exposed to the atmosphere. Determine the pressure in the tank if h= 0.4m. Patm Pair
h air, γair A
B Mercury, sg = 13.6
Figure E2.3 Solution:
Since the density of air is neglected, the pressure at point A will correspond to the pressure of air in the tank.
P A = P Air
(1)
Applying equal level-equal pressure principle, we have
PA = PB But
(2)
33 P B = P atm + γ Hg h
(3)
Where γ Hg = sg (γ Hg at 4deg. ) = 13.6 (9810) = 133416N/m2 Substitute (3) and (2) into (1) will give
Pudara = Patm + γ Hg h = 100(103 ) + 13341(0.4) ∴ Pudara = 153.4kPa Example 2.4 The manometer is used to measure the pressure in the pipe and is connected to the pressurized tank containing gas and water as shown Figure E2.4. Pressure gage attached to the tank reads 80 kPa, determine the gage pressure in the pipe. (Given h 1 = 40cm, h 2 = 84cm, h 3 = 57cm, h 4 = 45cm)
Gas
h1 h2 h3 h4
h1
= = = =
40 cm 84 cm 57 cm 45 cm
h2
Water h3 Oil , sg = 0.8 Mercury , sg = 13 .6
h4
A
Figure E2.4 From equal level-equal pressure, PB = PC = PD But, PD = PE + γ Hg h3 Where PE = PF + γ w (h2 )
And
PF = Pgas = 80kPa
34
And also PA = PB + γ oil h4
Substitute (2), (3) and (4) into (1); PA = Pgas + γ water (h2 ) + γ Hg h3 + γ oil h4
P A = 80(103) + 9810(0.84) +13.6(9810)(0.57)+ 0.8(9810)(0.45) = 167kPa. Example 2.5 Figure E2.5 show a three-fluid manometer containing oil (sg= 0.82), mercury (sg= 13.6) and water used to measure large pressure differences. Determine the pressure difference between A and B. Given h 1 = 45cm, h 2 = 70cm, h 3 = 25cm and h 4 = 20cm. Neglect the air density. Oil, sg = 0.82
Water h4 h3
B
h2 h1 Gasoline, Sg = 0.7
Mercury, Sg = 13.6
A
Figure E2.5 Solution: PA = PC = PD
but
PE = PD + γ W h1 = PA + γ W h1
Gas
35 and
PF = PE = PA + γ W h1
Also
PF = PG + γ Hg h2
but
Also
PH = PG PB = PH + γ gasolin h3 + γ oil h4
Substituting 1, 2, 3, 4, 5, 6 and 7 give PA − PB = γ Hg h2 −γ Wh − γ gasolin h3 − γ oil h4
or
PA − PB = γ W ( sg Hg h2 − h − sg gasolin h3 − sg oil h4 )
= (9810)[(13.6)(0.7)-(0.45)-0.7(0.25)-0.82(0.2)] = 85.7kPa.
Problems
1
The Crosby gage tester shown in the figure is used to calibrate or to test pressure gages. When the weights and the piston together weigh 89.0 N, the gage being tested indicates 179kPa. If the piston diameter is 20mm, what is the percentage error exists in the gage? Weight
Piston Air
Oil
Figure Q1 2.
Two hemispheric shells are perfectly sealed together and the internal pressure is reduced to 10% of atmospheric pressure. The inner radius is 15 cm, and the
36 outer radius is 15.5 cm. The seal is located half way between the inner and other radius. If the atmospheric pressure is 100kPa, what force is required to pull the shells apart?
3.
If exactly 20 bolts of 2.5-cm diameter are needed to hold the air chamber together at A-A as a result of the high pressure within, how many bolts will be needed at B-B? Here D = 50 cm and d = 25 cm.
Figure Q3 4.
The reservoir shown in the figure contains two immiscible liquids of specific weights γ A and γ B , respectively, one above the other where γ A > γ B . Which graph depicts the correct distribution of gage pressure along a vertical line through the liquids?
Liquid B
Liquid A
p
p
p
p
(a)
(b)
(c)
(d)
Figure Q4
5.
This manometer contains water at room temperature. The glass tube on the left has an inside diameter of 1 mm (d = 1.0 mm). The glass tube on the right is three times as large. For these conditions, the water surface level in the left tube will be a) higher than the water surface level in the right tube, b) equal to
37 the water surface level in the right tube, c) less than the water surface level in the right tube. State your main reason or assumption for making your choice. d
3d
Figure Q5 6. A tank is fitted with a manometer on the side, as shown. The liquid in the bottom of the tank and in the manometer has a specific gravity (s) of 3.0, the depth of this bottom liquid is 20 cm. A 10-cm layer of water lies on top bottom liquid. Find the position of the liquid surface in the manometer. 7.
10 cm
Water
∆h = ?
Liquid 20 cm
Figure Q6 7.
Determine the gage pressure in pipe A. Oil (SG = 0.90)
0.5 m
1.3 m 1.0 m A
Water
Mercury
38 Figure Q7 8.
Considering the effects of surface tension, estimate the gage pressure at the
center of pipe A. Glass tube ( 1mm ID. 4 mm OD) Water level in tube
10 cm
A
Figure Q8 9.
What is the pressure at the center of pipe B? B
γ = 10 kN/m3 γ = 20 kN/m3
50 cm
4 3 50 cm 10 cm
Figure Q9
39
10.
Find the pressure at the center of pipe A. Oil (SG=0.8)
Water
Water
150 cm 90 cm
Mercury (SG =13.6)
30 cm 30 cm
A
Figure Q10
11. The top of an invented U-tube manometer is filled with an oil of specific gravity of 1.01. Determine the pressure difference in Pa between two points A and B at the same level at the base of the legs when the difference in water level h is 75mm.
40
CHAPTER 3 FLUID STATICS
3.1 Introduction In the previous chapter it was noted that the hydrostatic pressure parts of fluid static. In this chapter we shall develop equations to calculate the magnitude and location of forces acting on submerged surfaces. We shall also examine problems involving ability of floating bodies. Such analysis of fluid helps in the design of dams, gates, ships and submarines. In this chapter, the submerged surfaces are divided into the following types; a) straight horizontal and vertical surfaces, b) straight inclined surfaces, and c) curved surfaces.
The analysis of hydrostatic force on submerged surfaces assumes the following conditions. 1. Force is always perpendicular to the surface since there is no shear stress for fluids at rest. 2. Pressure varies linearly with depth for incompressible fluid. 3.
The resultant fluid force passes through the point called the center of
pressure 3.2 Horizontal and vertical surface Any given depth h, the resultant fluid force F R on the horizontal and vertical surfaces may be represented as shown in Figure 3.1.
FR h
FR P=ρgh
Figure 3.1
h
41
Magnitude of the resultant force F R at the bottom of the tank, Figure 3.1 is given by
FR = PA
(3.1)
where A is the area of the surface upon which the pressure is acting. For the vertical Surface. Figure 5.8(b), we have FR =
1 PA 2
(3.2)
3.3 Inclined Surface The surface tilted at an angle θ from the horizontal is shown in Figure 3.2. The pressure variation and hence the resultant hydrostatic force F R , on the surface can be presented as shown in Figure 3.2.
O
θ pc
yc
FR
yp
hc hp
C A
XC XP
P
Figure 3.2
y
42
Magnitude F R can be written as FR = PC A
(3.3)
where P C is a pressure at the centroid surface (point C) and is written as: PC = Po + ρghC
(3.4)
This pressure is equivalent to the average pressure on the surface. we can see that a) the magnitude of the force is independent of the angle θ, b) it is perpendicular to the surface, and c) it passes through the point of application called the center of pressure (point P).
In many cases, the pressure at point O is the atmospheric pressure and may be ignored in the analysis. This simplifies equation (3.4) to PC = ρghC
(3.5)
Noticed that when θ = 90°, the surface becomes vertical and when θ = 0°, the surface becomes horizontal. 3.3.1 Center of Pressure Next we need to determine the line of action of the resultant force F R distance y P and hence its location below the free surface, hp. It is found that T action y P is given by
yP =
IX AyC
(3.6)
43 where I X is the moment of inertia of the area with respect to the x axis. Using parallel axis I X = I X + Ay C2 , equation (3.6) can be rewritten as
y P = yC +
IX AyC
(3.7)
where I X is the moment of inertia of the area with respect to the centroid axis. I X commonly used shapes is given in Figure 3.3. Observe that the center pressure P below the centroid C of the surface since I X / AyC > 0 , Expressing y P = h P /sinθ and yC = h C /sinθ, we obtain from equation 3.7
hP = hC +
I AhC
Where I = I X sin 2 θ . In some textbooks, the location of the centroid of the submerge surface from the fluid free surface, hC is denoted as h and y C as y .
Figure 3.3 shows the centroid y for the simple shape objects and Figure 3.4 shows the second moment of area
I for the simple shapes. X
44
45
Figure 3.3 Note: 1. First Moment of Area It is used to determine the centroid for a complex shape. It can be calculated using the following method:
46
A2 A3 =
-
A1
x a
b
A1 x = A2 a − A3 b Where A 1 , A 2 , and A 3 are the areas and x, a, and b are the centroid for the above shapes respectively. 2. Axial Moment of Inertia of an Area The axial moment of inertia of an area is the summation of the axial moments of inertia of the elements. y
x
dA
y
x
I x = ∫ y 2 dA I y = ∫ x 2 dA
3. Parallel Axis Theorem The parallel axis theorem states that the axial moment of inertia of an area about any axis equals the axial moment of inertia of the area about a parallel axis through the centroid of the area plus the product of the area and the square of the distance between the two parallel axes.
47
y’
y n
x’ m x
0
I x = I x′ + Am 2
I y = I y′ + An 2 3.4 Moment of a Force The resultant force F R and its location yP or h P , are determined, thus the calculation of the moment needed to overcome the resulting moment due to this force about certain point. Consider a force F acting perpendicular to the body at point B as shown in Figure 3.4.
M=Fd F C
B
d
= A
C
Figure 3.4
This force tends to rotate the body about point A in a counter-clockwise direction. The tendency of a force to rotate the body is called the moment of a force about that point. The magnitude of this moment about point A is given by:
M A = Fd
A
48
where d is the perpendicular distance from point A to the point of application of the force. The direction of the moment is indicated using either clockwise or counterclockwise. In this case, M A is in the counter-clockwise direction Figure 3.4. Example 3.1: Consider a rectangular gate AB hinged along A to support the water pressure as shown in Figure E3.1. Determine:
(a) The resultant hydrostatic force exerted on the gate AB (b) The center of pressure (c) The force acting on the stopper at B
The gate width is 3m.
h = 10 m l =6 m Water
γ=9810N/m
A 3
h l B
Figure E3.1 Solution:
(a)
Resultant force
F R = γ h C A (1) where A= 6(3) = 18m2 and h C = 4 + 3 = 7m From equation (1);
49
F R = 9810 (7) (18) = 1.24MN (b)
Location of center of pressure
hP = hC +
I AhC
where I = I X sin 2 θ =
hP = 7 +
(c)
1 (3)(6)3 sin 2 (90°) = 54m4 12
54 = 7.43m (18)(7)
The force acting at the stopper B
Taking moment at A,
FB × l = FR × (hP − 4)
FB =
1.24 × 3.43 = 0.71MN 6
Example 3.2: Determine the friction coefficient required to hold the dam from moving as shown in Figure E1.2. The normal force of dam is 50MN/m.
Water 10m
60°
Figure E1.2
F=µΝ
50
Solution: Hydrostatic force per unit width acting on the dam surface: F = γh A = 9810(5)(10) / sin 60 = 424.8kN / m For force acts along horizontal component: FH = F cos 30 = 424.8 × cos 30 = 367.9kN / m
Therefore, the friction factor required to hold the dam;
µ=
367.9 × 10 −3 = 0.0074 50
3.5 Curved Surface
For the curved surface as shown in Figure 3.5, the pressure variation and hence the resultant hydrostatic force F R on the surface can be represented as shown.
F1 W
FR
D A
A
Fh
G
F2
P
P FR
B
B FV
Figure 3.5 Let
F h = horizontal force exerted on the fluid mass F v = vertical force exerted on the fluid mass F 1 = resultant force of the fluid exerted on the surface AD
51 F 2 = resultant force of the fluid exerted on the surface BD W = the weight of the enclosed volume supported by the curved AB and that W = ρgV where V is volume of the enclosed liquid block, and it passes through the centroid of this volume.
The force balances under static equilibrium and obtain the following equations: Fh = F2 Fv = F1 + W
The magnitude of the total resultant fluid force acting on the curved surface is given by:
FR = Fh2 + Fv2
Example 3.3: A water channel with a 3m long quarter-circular section AB of radius 2.4m is designed as shown in Figure E3.3. Determine the total resultant force exerted by water on section AB. Given γ air = 9.81kN/m3.
WATER
3.6 m
A 2.4 m
B
Figure E3.3 Solution:
Total resultant force is given by
52
FR = Fh2 + Fv2
where
F v = F 1 + W and F h = F 2
For each force components; F1 = γhC A where A= 2.4(3) = 7.2m2 and h C = 3.6m
∴ F1 = 9810(3.6)(7.2) = 254.3kN W = ρgV = γV
1 1 Where V = πr 2l = π ( 2.4) 2 (3) = 13.57m3 4 2 ∴W = 9810(13.57) = 133.2kN
F2 = γhC A Where A= 2.4(3) = 7.2m3 and h C = 3.6 + 1.2 = 4.8m
∴ F2 = 9810( 4.8)(7.2) = 339kN
Therefore,
F v = 254.3 + 133.2 = 387.4kN F h = F 2 = 339kN
And F R = (387.42 + 3392)1/2 = 514.8kN
53
Example 3.4 Determine the force per unit width P, required to hold the gate as shown in Figure E3.4.
2m
P 0.5m
2m
Water
Hinge
Figure E3.4 Solution: Hydrostatic force acting on the control volume; FH = γh A = 9810(1)(2) = 19.62kN / m
By taking moment at hinge: P=
2 × 19.62 = 15.7 kN / m 2.5
3.6 Buoyancy, Floatation and Stability When a body is completely submerged or floating in a fluid, the resultant fluid force acting in an upward direction on the body is called the buoyancy force. This force tends to lift the body upward and its existence is due to the fact that (1) the fluid pressure increases with depth, and (2) the pressure force acting from below is larger than the pressure force acting from above.
3.6.1 Buoyancy and Floatation Consider a body submerged completely in a fluid as shown in Figure 5.12. The resultant force on the bottom surface of the body is greater than the resultant force on the top surface of the body.
54
h1
Ftop=γ fh1A
h2 h
Fbottom=γ fh2A
FB
Figure 3.6 Buoyancy force acting on a submerged body The difference between these two forces is the buoyant force which gives the net upward force. This buoyant force will pass through the point called center of buoyancy or the centroid of the displaced volume, C B which happened to be at the same point as the center of gravity of the body, G in the case of a completely submerged body. Writing force balance on the body, we have FB = Fbottom − Ftop = γ f h2 A − γ f h1 A = γ f hA
However, the term hA is basically the volume of the fluid body (or volume of the displaced fluid by the body). Expressing this volume as V = hA, we may write the equation above as FB = γ f V
where V is volume of the displaced fluid. Thus, we conclude that the buoyant force acting on the body is equal to the weight of the fluid displaced by the body and therefore, proportional to the density of the fluid.
In the case of a floating body, Figure 3.7, the weight of the entire body must be equal to the buoyant force, which is the weight of the fluid whose volume is equal to the volume of the submerged portion of the floating body.
55
mg
Vtotal mg
FB
FB
FB=mg
FB
(c)
(b)
(a)
Figure 3.7
Buoyancy force acting on a floating body
Thus, for a floating body in static equilibrium, we may write FB = W
or
γ f Vsub = γ body Vtotal
where Vtotal = total volume of the body or volume of the entire body Vsub = volume of the submerged, portion of the body, which is equal to the volume of the displaced fluid (Figure 3.7(c))
Rewriting the above equation as Vsub γ body ρ body = = Vtotal γf ρf
We observe that the body is completely submerged when the density ratio is equal to 1 that is when the density of the body is equal to the fluid density. We can conclude that a body immersed in a fluid will
(1)
Rise to the surface of the fluid and float when the density of the body
is less than the fluid density,
(2)
Remain at rest at any point in the fluid when its density is equal to the
fluid density, and
Vsub
56 (3)
Sink to the bottom when the density of the body is greater than the
fluid density.
Ρbody < Ρf : floating
Ρbody < Ρf : suspended
Ρbody < Ρf : sinking
Figure 3.8 Situations of a body immersed in a fluid: float, remain at rest, or sink depending on the density of the body relative to the fluid density Example 3.5 A cuboids has the size of b×h×w is floating in the waterγ(
air =
9.81kN/m3) as
shown in Figure E3.5. Determine the portion of body that above the water surface, a if γ Jasad = 3.5kN/m3, b= 5m, h= 8m dan w= 6m.
a
h
w Figure E3.5 Solution: Vsub = ( h − a ) b w Vtotal = h b w
From the relationship, Vsub γ body ρ body = = Vtotal γf ρf
water
57
a = h( 1 −
γ body γf
3500 ) = 81 − 9800
a = 5.14m
Note: Observe that from equation above, the portion above the water surface will be less if the density of the body is higher
Example 3.6 Balloon has a mass of 20kg and the diameter of 10m is filled with the helium gas (Figure E3.6). The density of helium gas is 0.81kg/m3. Determine the acceleration of the balloon after it releases from the ground (Take air density as 1.2kg/m3). Helium
10m
Figure E3.6 Solution: The forces in equilibrium position as in equation below:
FB − mT g = mT a
a=
γ airV − (20 + ρ HeV ) g (20 + ρ HeV )
Thus, the acceleration is
,
4 where V = π × 5 3 = 523.6m 3 3
58 a=
(1.2 × 9.81)(523.6) − (20 + 0.81 × 523.6) × 9.81 = 4.07 m / s 2 (20 + 0.81 × 523.6)
3.6.2 Stability of Floating and Immersed Bodies Stability is an important issue for a floating and immersed body such as in the design of a ship, submarine and barge (Figure 3.9).
Figure 3.9 A barge used for transportation at Kuala Kurau In the vertical direction under static equilibrium, the weight and the buoyant force on a floating or immersed body will balance each other, and such body is said to be vertically stable (vertical stability).
For the rotational stability, the condition depends upon the relative location of the center of gravity G of the body and the center of buoyancy C B , Figure 3.10. A floating or immersed body is stable if the point G is below point C B , Figure 3.9(a). Under this condition, the body will return to its original stable position due to the restoring moment or couple produced by the body.
59
w
FB
FB CB G
G
G CB
FB
w
w
(b) neutral
(a) stable
CB
(c) unstable
w
FB FB
w
(d) Restoring Couple
Figure 3.10
Overturning Moment
Restoring Moment
(e) Overturning Couple
Stability of an immersed body (a) stable with restoring couple as shown in (d), (b) neutral, and (c) unstable with overturning couple as shown in (e).
However, a floating body will still be stable even if G is above C B . See Figure 3.11. This is because the body will still produce the restoring moment since the centroid of the displaced volume is now shifted to the side to point C B . The lines of action of the buoyancy force before and after rotation will meet at the point called the meta-center C M . The distance between G and C M is called the metacentric height h M and is used as a measure of stability for a floating body. The larger it is, the more stable the floating body will be. Note that the floating body is unstable if the point C M is below point G. CM
W
G CB
hM
W
G CB FB
Restoring Couple
Figure 3.11 Stability of an immersed body
To determine whether the floating body is stable or not is given by the following equations:
60
C M − CB =
I V
and for the body is in stable condition when it rotate at certain angle resulting from the reversed moment, it must has the following conditions: hM = C M − CG > 0 hM = C M − C G = 0 hM = C M − CG < 0
Stable Neutral Unstable
Example E3.7: The shifting of a portion of cargo of mass 25000kg, through a distance of 6m at right angles to the vertical plane containing the longitudinal axis of a vessel, causes it to heel through an angle of 5°. The displacement of the vessel is 5000 metric tons and the value of second moment of area is 5840m4. The density of seawater is 1025kg/m3. Determine the metacentric height and the height of the centre of gravity of the vessel above the centre of buoyancy.
Solution: CM CM
6m 5°
CG
2500g CG
CB
x CB
Force diagram Let the horizontal distance between C G and C B is x, By taking moment at C G : − 2500(9.81)(6) + 5000 × 10 3 × x(9.81) = 0 ; x = 0.03m
where F B = m T× g
61 5000 × 10 3 = 4878m 3 1025
Volume of displacement, V =
CM − CB =
And,
hM =
Thus, metacentric height,
And
the
(C M − C B ) −
vertical
I 5840 = = 1.2m V 4878
0.03 x = = 0.344m sin 5 sin 5
distance
between
CG
and
CB
is
0.03 x = 1.2 − = 0.857 m tan 5 tan 5
3.7 Liquids in Relative Equilibrium If a vessel containing a liquid is at rest or moving with constant velocity the liquid is not affected by the motion of the container, but if the container is given continuously acceleration this will be imparted to the liquid which will take up a new position and come to rest with respect to its container and come to rest relative to the vessel. The liquid is in relative equilibrium and is at rest with respect to its container. There is no relative motion of the particles of the fluid and therefore no shear stress. Fluid pressure is everywhere normal to the surface on which it acts.
3.7.1 Horizontal Acceleration Consider a particle O of mass m on the free surface of the liquid as in Figure 3.11. R θ O
F θ
W
A
a
C
Figure 3.12
B
62
Since the particle is at rest relative to the tank, it will have the same acceleration a, and will be subjected to an accelerating force, F F = ma =
W a g
where W= weight of particles.
The accelerating force, F is the resultant of the weight W of the particle acting vertically downward and the pressure force, R acting normal to the free surface due to surrounding fluid. For equilibrium F= W tanθ, where θ is the angle of the free surface to the horizontal. Thus, tan θ =
a g
and is constant for all points on the surface.
3.7.2 Vertical Acceleration As the acceleration is vertical the free surface will remain horizontal. Consider a vertical prism of height h (Figure 3.13) extending from the free surface to x and let the pressure intensity at x be P.
A
a
B
h
Prism Cross – sectional Area = a x P
Figure 3.13
63
Accelerating force at x, F= Force due to pressure – weight of prism = PA - ρghA By Newton’s second Law: F = mass × acceleration F = ρhA × a
Thus, PA − ρghA = ρhA × a
or a P = ρgh1 + g 3.7.3 Forced Vortex The liquid in the vessel is rotated with the vessel at the same angular velocity, ω. A particle on the free surface will be in equilibrium under the action of its weight W (Figure 3.14), the centrifugal accelerating force, F acting
x
θ Fy A
Axis of rotation
horizontally and the fluid reaction R.
o w
Figure 3.14
D
B
64
For any point at radius x and a height y from the lowest point O, if θ is the angle of inclination of the water surface to the horizontal,
tan θ =
dy F = dx W
For a constant value of ω, F will vary with x, since the centrifugal acceleration is ω2x dan F= (W/g)ω2x.
The surface angle therefore varies and,
tan θ =
dy ω 2 x = dx g
Integrating will give,
x
y=∫ 0
ω2x g
dx =
ω 2 x2 2g
+ malar
If y is measured from AB, y= 0 when x= 0 and
y=
ω 2 x2 2g
The water surface is therefore a parabolic revolution.
Example 3.8 A tank containing water moves horizontally with a constant linear acceleration a of 3m/s2. The tank is 3m long and the depth of water when the tank is at rest is 1.5m. Calculate: (a)
The angle of water surface when the tank is 1.5m
(b)
The maximum pressure intensity on the bottom
65 (c)
The minimum pressure intensity on the bottom
Solution: Given: a= 3m/s2, h= 1.5m
(a)
The angle of water surface to horizontal, θ
θ = tan −1
(b)
a = 17 o g
The depth at A (maximum pressure),
h A = h + h tan θ = 1.5 + 1.5 tan 17 o = 1.96m
(c)
The depth at B (minimum pressure),
hB = h + h tan θ = 1.5 − 1.5 tan 17 o = 1.04m
Problems 1.
Consider the two rectangular gates shown in the figure. They are both the same size, but one (G a t e A) is held in place by a horizontal shaft through its midpoint and the other (G a t e B) is cantilevered to a shaft at its top. Now consider the torque 1' required to hold the gates in places as H is increase. Choose the valid statement(s): a) T A increases with H. b) T B increases with H. c) T A does not change with H. d) T B does not change with H.
Shaft Water H
Gate A Shaft Atmospheric Pressure
Figure Q1 2.
Find the force of the gate on the block.
Water H
Gate B Atmospheric Pressure
66
Water 4 m x 4 m gate
10 m 2m 2m
Pivot
Block
Figure Q2 3.
Neglecting the weight of the gate, determine the force acting on the hinge of the gate. 3m hinge water
9 m x 9 m gate 9m
Atmospheric pressure On this side of gate
Figure Q3 4.
The rectangular gate measures 6m by 4m and is pin-connected at point A. If the surface on which the gate rests at A is frictionless. What is the reaction at A? Neglect the weight of the gate.
3m
Hinge A
Atmospheric Pressure
30o Stop 6m Water
Figure Q4 5.
A 12m x 12m gate is installed at the end of water reservoir, as shown, and is hinged at the top. The gate hinge is G m below the reservoir water surface. The gate is connected to a rectangular tank of water which is 12 m wide (into
67 the paper) and filled with 6 m of water. The weight to the tank is negligible. How long (L) would the tank have to be open the gate?
Figure Q5 6.
The triangular gate ABC is pivoted at the bottom edge AC and closes a triangular opening ABC in the wall of the tank. The opening is 4 m wide (W = 4 m) and 9 m high (H = 9 m). The depth d of water in the tank is 10 m. Determine the hydrostatic force on the gate and the horizontal force P required at B to hold the gate closed. D
E
B P
Water T = 20oC
d
H
A
A,C View D-D
E
C W
View E-E
D
Figure Q6 7. Estimate the depth d needed for the rectangular gate to automatically open if the weight W= 60kN as shown in Figure Q7. The gate is 4m high and 2m wide. Neglect the weight of the gate.
68
W
5m Hinge
4m Water
Atmospheric pressure
d
Gate stop
Figure Q7 8. For the plane rectangular gate (L x W in size), Figure 8(a), what is the magnitude of the reaction at A in terms of γ w and the dimensions L and W? For the cylindrical gate, Figure 9(b), will the magnitude of the reaction of A be greater than, less than, or the same as that for the plane gate? Neglect the weight of the gates.
1/4L
1/4L Hinge
Water
Hinge
Water
B
L
Smooth Boundary
n
Si 45
o
L
45
o
A
B
Rectangular Gate
A
(a)
(b) Figure 8
9. The floating platform shown is supported at each corner by a hollow sealed cylinder 1 m in diameter. The platform itself weighs 30kN in air, and each cylinder weighs 1.0kN per meter of length as in Figure Q9. What total cylinder length L is required for the platform to float 1 m above the water surface? Assume that the specific weight of the water is 10,000 N/m. The platform is square in plan view. 10 m
1m L=? Diameter = 1 m
Figure Q9
Floating Platform Weight = 30kN
69
10. The coffee cup in Figure Q10 is removed from the drag race, placed on a turntable, and rotated about its central axis until a rigid-body mode occurs. Determine:
(a)
the angular velocity which will cause the coffee to just reach the lip of the
(b)
the gauge pressure at point A for this condition
cup (Take the density of coffee as 1100kg/m3)
z 3cm
0 7cm
r ω
3cm
3cm
Figure Q10
