Download Chapter 2 Fluid Static

Fluid Mechanics for
Mechanical Engineering
Chapter 2: Fluid Static
1
SEQUENCE OF CHAPTER 2
Introduction
Objectives
2.1 Force Equilibrium of a fluid element
2.2 Hydrostatic Pressure Distribution
2.3 Standard Atmosphere
2.4 Pressure Measurement
2.5 Hydrostatic Forces
2.6 Buoyancy and Stability of Floating Bodies
Summary
2
Introduction
• This chapter will begin with basic concepts in fluid
static, which is the hydrostatic pressure distribution.
• It brings the student into the use of this principle in the
measurement of pressure via manometer and in the
determination of hydrostatic forces.
• This chapter also include buoyancy and stability which
is crucial in designing submerged and floating bodies.
3
Objectives
At the end of this chapter, you should be able to :
 understand the concept of hydrostatic pressure distribution,
 use the principle in pressure measurement of using
manometers,
 determine quantitatively hydrostatic forces and centres of
pressure,
 determine quantitatively buoyant forces and centres of
buoyancy,
4
2.1 Force Equilibrium of a Fluid Element
• Fluid static is a term that is referred to the state of a fluid
where its velocity is zero and this condition is also called
hydrostatic.
• So, in fluid static, which is the state of fluid in which the
shear stress is zero throughout the fluid volume.
• In a stationary fluid, the most important variable is pressure.
• For any fluid, the pressure is the same regardless its
direction. As long as there is no shear stress, the pressure is
independent of direction. This statement is known as
Pascal’s law
5
2.1 Force Equilibrium of a Fluid Element
Fluid surfaces
Figure 2.1 Pressure acting uniformly in all directions
Figure 2.2: Direction of fluid pressures on boundaries
6
2.1 Force Equilibrium of a Fluid Element
Pressure is defined as the amount of surface force exerted by
a fluid on any boundary it is in contact with. It can be written
as:
Pr essure 
P
F
A
Force
Area of which the force is applied
(2.1)
Unit: N / m2 or Pascal (Pa).
(Also frequently used is bar, where 1 bar = 105 Pa).
7
2.1 Force Equilibrium of a Fluid Element
• The formulation for pressure gradient can be written as:
p = (g – a)
(2.11)
1. If a = 0, the fluid is stationary.
2. if a ≠ 0, the fluid is in the rigid body motion.
• The formulation for pressure gradient in Eq. (2.11) is derived
under the assumption that there is no shear stress present, or in
other word, there is no viscous effect.
• The rest of this chapter will only concentrate on the first case,
i.e. stationary fluids. The second case is applicable the case for a
fluid in a container on a moving platform, such as in a vehicle, and
is out of scope of this course.
8
2.2 Hydrostatic Pressure Distribution
For a liquid, usually the position is measured as distance from
the free surface, or depth h, which is positive downward as
illustrated in Fig. 2.3. Hence,
p2 – p1 = g (h2 – h1) =  (h2 – h1)
p = g h =  h
9
2.2 Hydrostatic Pressure Distribution
 If the atmospheric pressure p0 is taken as reference and is
calibrated as zero, then p is known as gauge pressure.
Taking the pressure at the surface as atmospheric pressure
p0, i.e., p1=p, p2=p0 when h1=h, h2=0, respectively:
p = p0 + gh = p0 + h
 This equation produces a linear, or uniform, pressure
distribution with depth and is known hydrostatic pressure
distribution.
 The hydrostatic pressure distribution also implies that
pressure is the same for all positions of the same depth.
10
2.2 Hydrostatic Pressure Distribution
• This statement can be explained by using a diagram in Fig. 2.4,
where all points, a, b, c and d, have the same value of pressure,
that is
pa = pb = pc = pd
• However, the pressure at point D is not identical from those at
points, A, B, and C since the fluid is different, i.e.
pA = pB = pB ≠ pD
11
Example 2.1
An underground gasoline tank is accidentally opened during raining causing
the water to seep in and occupying the bottom part of the tank as shown in
Fig. E2.1. If the specific gravity for gasoline 0.68, calculate the gauge
pressure at the interface of the gasoline and water and at the bottom of the
tank. Express the pressure in Pascal and as a pressure head in metres of
water. Use water = 998 kg/m3 and g = 9.81 m/s2.
12
Example 2.1
For gasoline:
g = 0.68(998) = 678.64kg/m3
 At the free surface, take the atmospheric pressure to be zero, or p0 = 0 (gauge
pressure).
p1 = p0 + pgghg = 0 + (678.64)(9.81)(5.5)
= 36616.02 N/m2 = 36.6 kPa
The pressure head in metres of water is:
h1 = p1 – p0 = 36616.02 - 0
pwg
(998)(9.81)
= 3.74 m of water
At the bottom of the tank, the pressure:
p2 = p1 + pgghg = 36616.02 + (998)(9.81)(1)
= 46406.4 N/m2 = 46.6 kPa
And, the pressure head in meters of water is:
h2 = p1 – p0 = 46406.4 - 0
pwg
(998)(9.81)
= 4.74 m of water
13
2.3 Standard Atmosphere
 A pressure is quoted in its gauge value, it usually refers to a
standard atmospheric pressure p0. A standard atmosphere is
an idealised representation of mean conditions in the earth’s
atmosphere.
 Pressure can be read in two different ways; the first is to
quote the value in form of absolute pressure, and the second
to quote relative to the local atmospheric pressure as
reference.
 The relationship between the absolute pressure and the
gauge pressure is illustrated in Figure 2.6.
14
2.3 Standard Atmosphere
 The pressure quoted by the latter approach (relative to the local
atmospheric pressure) is called gauge pressure, which indicates the
‘sensible’ pressure since this is the amount of pressure experienced by our
senses or sensed by many pressure transducers.
 If the gauge pressure is negative, it usually represent suction or partially
vacuum. The condition of absolute vacuum is reached when only the
pressure reduces to absolute zero.
15
2.4 Pressure Measurement
 Based on the principle of hydrostatic pressure distribution, we can develop
an apparatus that can measure pressure through a column of fluid (Fig.
2.7)
16
2.4 Pressure Measurement
 We can calculate the pressure at the bottom surface which has to
withstand the weight of four fluid columns as well as the
atmospheric pressure, or any additional pressure, at the free
surface. Thus, to find p5,
Total fluid columns = (p2 – p1) + (p3 – p2) + (p4 – p3) + (p5 – p4)
p5 – p1 = og (h2 – h1) + wg (h3 – h2) +
gg (h4 – h3) + mg (h5 – h4)
 The p1 can be the atmospheric pressure p0 if the free surface at z1
is exposed to atmosphere. Hence, for this case, if we want the
value in gauge pressure (taking p1=p0=0), the formula for p5
becomes
p5 = og (h2 – h1) + wg (h3 – h2) + gg (h4 – h3) + mg (h5 – h4)
 The apparatus which can measure the atmospheric pressure is
called barometer (Fig 2.8).
17
2.4 Pressure Measurement
 For mercury (or Hg — the chemical symbol for mercury), the height
formed is 760 mm and for water 10.3 m.
patm = 760 mm Hg (abs) = 10.3 m water (abs)
 By comparing point A and point B, the atmospheric pressure in the
SI unit, Pascal,
pB = pA + gh
pacm = pv + gh
= 0.1586 + 13550 (9.807)(0.760)
 101 kPa
18
2.4 Pressure Measurement
 This concept can be extended to general pressure measurement using an
apparatus known as manometer. Several common manometers are given
in Fig. 2.9. The simplest type of manometer is the piezometer tube, which
is also known as ‘open’ manometer as shown in Fig. 2.9(a). For this
apparatus, the pressure in bulb A can be calculated as:
pA = p1 + p0
= 1gh1 + p0
 Here, p0 is the atmospheric
pressure. If a known local
atmospheric pressure value is used
for p0, the reading for pA is in
absolute pressure. If only the
gauge pressure is required, then p0
can be taken as zero.
19
2.4 Pressure Measurement
 Although this apparatus (Piezometer) is simple, it has limitations, i.e.
a) It cannot measure suction pressure which is lower than the
atmospheric pressure,
b) The pressure measured is limited by available column height,
c) It can only deal with liquids, not gases.
 The restriction possessed by the piezometer tube can be overcome by the
U-tube manometer, as shown in Fig. 2.9(b). The U-tube manometer is
also an open manometer and the pressure pA can be calculated as
followed:
p2 = p3
pA + 1gh1 = 2gh2 + p0
 pA = 2gh2 - 1gh1 + p0
20
2.4 Pressure Measurement
 If fluid 1 is gas, further simplification can be made since it can be
assumed that 1   2, thus the term 1gh1 is relatively very small
compared to 2gh2 and can be omitted with negligible error. Hence, the
gas pressure is:
pA  p2 = 2gh2 - p0
 There is also a ‘closed’ type of manometer as shown in Fig. 2.9(c), which
can measure pressure difference between two points, A and B. This
apparatus is known as the differential U-tube manometer. For this case,
the formula for pressure difference can be derived as followed:
p2 = p3
pA + 1gh1 = pB + 3gh3 + 2gh2
 pA - pB = 3gh3 + 2gh2 - 1gh1
21
2.5 Hydrostatic Forces
 If a solid plate is immersed into the fluid, the pressure is also
acted upon the surface of the solid.
 This pressure acts on the submerged area thus generating a
kind of resultant force known as hydrostatic force.
 Hence, the hydrostatic force is an integration of fluid
pressure on an area.
 Similar to pressure, the direction in which the force is acting
is always perpendicular to the surface.
 To derive the hydrostatic force for a planar surface, consider
the solid plate shown in Fig. 2.10. For a small elementary
area dA, the force magnitude is:
22
2.5 Hydrostatic Forces
 If a solid plate is immersed into the fluid, the pressure is also acted upon
the surface of the solid.
dF = pdA
 Thus for the entire area A, the total magnitude of resultant force is
FA = A ( p0 + gh) dA
 Here, the specific gravity the fluid
g can be taken as constant, which
is valid for liquid, then the
resultant force becomes
FR = p0 A + g Ah dA
23
2.5 Hydrostatic Forces
 From Fig.2.9, a trigonometry relation can be used to represent h, i.e.,
h = y sin . Knowing that the angle  is constant for a planar surface, then,
the above expression can be written as
FR = p0 A + g sin A y dA
 If C is the centroid for the area A, by using the centroidal relationship, i.e.
A y dA = yC A and hc = yc sin , then
FR = p0 A + gyC A sin = p0 A + ghC A
 For isotropic materials where mass is uniformly distributed, the centroid C
is identical to the centre of gravity CG. Hence, the resultant hydrostatic
force FR is a product of pressure p at C and the surface area A, i.e.
FR = ( p0 A + ghC )A = pC A
 In many applications, hydrostatic forces acting on any surfaces such as
walls of a tank are balanced by opposite forces generated by the
atmospheric pressure p0 time the same surface area A.
24
2.5 Hydrostatic Forces
 Therefore, based on this reason, we can omit the term p0A in Eq. (2.20),
and thus it can be reduced to
FR = ghC A
 The horizontal depth of the center of pressure, yR, (along y axis) is given
by:
yR = 1xx + yC
yC A
 The x coordinate for this point can be derived using moment equilibrium
of the force FR about the y-axis at centroid C, and the formula can be
written as followed,
xR = 1xy + xC
yC A
where Ixy is known as the product of inertia. For your convenience,
Table 2.2 shows the formula for centroids and the moments of inertia, Ixx,
Iyy and Ixy, for typical shapes.
25
26
2.5 Hydrostatic Forces
 The hydrostatic force is the resultant of a linear distributed force formed
by the liquid pressure acting perpendicular to the surface.
 In the case where the surface is a wall of a liquid tank, the pressure
distribution is as illustrated in Fig. 2.11.
 Here, the application of Eq. (2.21) leads to the volume of the prism
known as hydrostatic prism, which is generated by the linear distributed
pressure, i.e.
 This prism shape can represent
the hydrostatic force for a
partially immersed surface. As
shown in Fig. 2.11, the centre of
pressure CP is actually the
centroid of the prism.
FR = volume of prism =
½(gh)(bh) = g • ( ½ h) A
27
2.5 Hydrostatic Forces
 For a completely immersed surface, the hydrostatic prism becomes as
shown in Fig. 2.12(a) for vertical surface and Fig. 2.12(b) for inclined
surface, where the cross section of the prism is a trapezium.
 For both cases, the hydrostatic force calculated using Eq. (2.21) gives the
volume of the trapezoidal prism, i.e.
 All the formula in this
section are derived for the
planar surface with a fixed
angle. However,
hydrostatic forces for
curved surfaces, is out of
the scope of this course.
28
Example 2.3
 A circular door having a diameter of 4 m is positioned at the inclined wall
as shown in Fig. E2.3(a), which forms part of a large water tank. The
door is mounted on a shaft which acts to close the door by rotating it and
the door is restrained by a stopper. If the depth of the water is 10 m at
the level of the shaft, Calculate:
(a) Magnitude of the hydrostatic
force acting on the door and its
centre of pressure,
(b) The moment required by the
shaft to open the door.
Use  water = 998 kg/m3 and
g = 9.81 m/s2.
29
Example 2.3
(a) The magnitude of the hydrostatic force FR is
FR = ghC A
= (998)(9.81)(10) [ ¼ x(4)2]
= 1.230 x 106 N
= 1.23 MN
For the coordinate system shown in Figure E2.3(b), since circle is a
symmetrical shape, Ixy = 0, then xR = 0. For y coordinate,
yR = 1xx + yC = ¼ R4 + yC
yC A
 y CR2
¼  (2)4
+ 10
 (10/sin 60°)(2)2 sin 60°
= 11.6 m
=
or,
30
Example 2.3
yR = 1xx + yC =
¼  (2)4
= 0.0866 m
yC A
 (10/sin 60°)(2)2
(b) Use moment equilibrium M - 0 about the shaft axis. With reference to
Figure E2.3(b), the moment M required to open the door is:
M = FR ( y R - yC )
= (1.230 x 105) (0.0866)
= 1.065 x 105 N • m
= 107 kM • m
31
2.6 Buoyancy and Stability of Floating
Bodies
 This section will cover the interaction of fluid with the
whole mass in a gravitational field which produce another
form of force known as buoyant force, and the field in Fluid
Mechanics which studies the behaviour of floating bodies is
called buoyancy.
 Buoyant force can be defined as the resultant fluid force
which acts on a fully submerged or floating body.
 To derive the formula for the buoyant force, consider
Fig. 2.13:
32
2.6 Buoyancy and Stability of Floating
Bodies
 From Fig. 2.13, if a rectangular block ABCD is drawn to cover the body, by
using the force equilibrium in the x and y directions, we should obtain:
Fx = 0 : F3 = F4
Fy = 0 : FB = F3 – F1 – W
where W is the weight of
the fluid volume in
ABCD, and FB is the force
exerted by the body to
the fluid.
33
2.6 Buoyancy and Stability of Floating
Bodies
 If the fluid is incompressible, which is a valid assumption for
liquid, by taking A as the area for the top AB plane and bottom CD
plane and V is the volume of the body, the force equilibrium in the
y direction becomes
FR = g (h2 - h1) A - g [(h2 - h1) A – V ]
FB = g V
(2.24)
 From Fig. 2.13, by using the principle of force equilibrium, the
fluid force acting onto the body must be of the same magnitude
with the force exerted by the solid to the fluid FB, but in the
opposite direction. Hence Eq. (2.24) give the formula for the
buoyant force which is always in upward direction, i.e., opposite
to the weight of the body.
34
2.6 Buoyancy and Stability of Floating
Bodies
 The relation also shows that the buoyant force is equal to the weight of
the volume of fluid which is displaced by the body.
 The point of action for this is the centroid of the displaced fluid, which is
also called the centre of buoyancy CB.
 The difference between the centre of buoyancy and the centre of gravity
of the floating body CG may lead to stability issue.
 In general, a body in equilibrium may be in two possible positions, as
shown in Fig. 2.14 for a completely submerged body:
(a) Stable equilibrium — a small displacement causing it to return to
its original position.
(b) Unstable equilibrium — a small displacement causing it to shift
to another position.
 Hence, it can be concluded that for a completely submerged body where
W  FB, the restoring moment produced causes the body to return to its
stable condition, and as a result, the centre of buoyancy CB should be
always lower than the centre of gravity CG.
35
Summary
This chapter has summarized on the aspect below:
 Should be able to understand the principle of hydrostatics,
 The conditions of standard atmosphere, and consequently be
able to distinguish between absolute and gauge pressure
readings.
 Should also be able to apply this knowledge in calculation of
pressure measured using manometers, in evaluating the
hydrostatic force acted onto a planar surface and locating the
corresponding centre of pressure, as well as in evaluating the
buoyant force exerted to a submerged or floating body and
locating the corresponding centre of buoyancy.
 In addition, you should also be able to analyse the stability of a
body immersed in fluid.
 This concludes the scope of this course for fluid static.
36