2.0 FLUIDS IN RELATIVE
EQUILIBRIUM
2.1 PRESSURE
Pressure is defined as a normal force exerted by a fluid per unit area.
Pressure is defined as force per unit area; it has the unit of Newton per
square meter (N/m2), which is called a Pascal (Pa). That is, 1 Pa =
1 N/m2
EXAMPLE: Absolute Pressure
of a Vacuum Chamber
A vacuum gage connected to a
chamber reads 30kPa at a location where the atmospheric pressure is 100kPa.
Determine the absolute pressure in the chamber.
Analysis: The absolute
pressure is easily determined.
Discussion: Note that the
local value of the atmospheric pressure is used when determining the absolute
pressure.
Pabs = Patm
- Pvac
= 100 - 30
= 70 kPa
2.2
PRESSURE MEASUREMENT
Manometer
P2 = P1
P2 = Patm + pgh
Patm + p1gh1 + p2gh2 + p3gh3 = P1
A relation for the pressure difference P1 – P2
can be obtained by starting at point 1 with P1, moving along the
tube by adding or subtracting the rgh terms until we reach point 2, and setting
the result equal to P2
Assume that, PA = PB, so
P1 + p1g (a + h) = P2
P1 – P2
= (p2- r1) ph
EXAMPLE: Measuring Pressure with a Multifluid Manometer
The water in a tank is pressurized by air, and the pressure is measured
by a multifluid manometer as shown in figure above. The tank is located on a
mountain at an altitude of 1200m where the atmospheric pressure is 85.6 kPa.
Determine the air pressure in the tank if h1 0.1 m, h2
0.2 m, and h3 0.35 m. Take the densities of water, oil, and mercury
to be 1000kg/m3, 850kg/m3, and 13,600kg/m3,
respectively.
PA = PB
P1 + pwater gh2 + poil gh1 – pmercury gh3 = Patm
P1 = Patm
– pwater gh2 – poil gh1 + pmercury gh3
Patm + g (pmercuryh3
– pwater h1 – poilh2)
85.6kPa + (9.81m/s2) [(13600)(0.35m) – (1000)(0.1m) –
(850)(0.2)][
[
=130kPa
2.3 HYDROSTATIC FORCES
2.3.1 ON SUBMERGED PLANE
SURFACES
A plate exposed to a liquid, such as a gate valve in a dam, the wall of
a liquid storage tank, or the hull of a ship at rest, is subjected to fluid
pressure distributed over its surface. On a plane surface, the hydrostatic
forces form a system of parallel forces, and we often need to determine the
magnitude of the force and its point of application, which is called the center
of pressure. When analyzing hydrostatic forces on submerged surfaces, the
atmospheric pressure can be subtracted for simplicity when it acts on both
sides of the structure.
EXAMPLE: Determine the force and center of force on the wall if:
Fluid = gasoline = Sgasoline = 0.68
Total depth = 12 ft
Length of wall = 40 ft
Average pressure = γ (h/2)
= (0.68) (62.4) (12/2) = 254.6
Force on wall, F = pA
= (254.6) (12) (40) = 122204 lb
Center of pressure = 12/3 from bottom
= 4 ft from bottom
Submerged Rectangular Plate
The resultant hydrostatic
force on the upper surface is equal to the average pressure, which is the
pressure at the midpoint of the surface, times the surface area A. That is,
FR = PcA
=
(P0 + pg (s + b/2) sinØ) ab
EXAMPLE: Hydrostatic Force Acting on the Door of a Submerged Car
A heavy car plunges into a lake during an accident and lands at the
Bottom of the lake on its wheels as shown. The door is 1.2 m high and 1 m wide,
and the top edge of the door is 8 m below the free surface of the water.
Determine the hydrostatic force on the door and the location of the pressure
center, and discuss if the driver can open the door.
Analysis: The average pressure on the door is the pressure value at the
centroid (midpoint) of the door and is determined to be. Then the resultant
hydrostatic force on the door becomes
Pave = PC = rghC = rg (s + b/2)
= (1000 kg/m3) (9.81 ms2)
(8+ 1.2/2 m) (1 kN/1000 kgms2)
= 84.4 kNm2
Then the resultant hydrostatic force on the door becomes
FR = PaveA = (84.4 kN/m2) (1 m x1.2 m) = 101.3kN
The pressure center is directly under the midpoint of the door, assume
P0 = 0
Yp = s+ b/ 2 +b2/ (12(s+ b))
= 8 +1.2/2 + 1.22/ (12(8 +
1.2)) = 8.61 m
2.3.2 ON SUBMERGED CURVED SURFACES
·
For
a submerged curved surface, the determination of the resultant hydrostatic
force is more involved since it typically requires the integration of the
pressure forces that change direction along the curved surface.
·
The
pressure at the centroid of the portion of the surface in fluid i and Ai is the
area of the plate in that fluid. The line of action of this equivalent force
can be determined from the requirement that the moment of the equivalent force
about any point is equal to the sum of the moments of the individual forces
about the same point.
Noted: The concept of
the pressure prism in this case is not much help either because of the
complicated shapes involved.
Pc,i = Po + ri ghc,i
Example
:
The height of a
water reservoir is controlled by a cylindrical gate hinged to the reservoir.
The hydrostatic force on the cylinder and the weight of the cylinder per ft.
length are to be determined.
Assume the
density of water, p = 62.4 lbm/ft3
Then the
magnitude and direction of the hydrostatic force acting on the cylindrical
surface become
Noted: When the
water level is 15-ft high, the gate opens and the reaction force at the bottom
of the cylinder becomes zero. Then the forces other than those at the hinge
acting on the cylinder are its weight, acting through the center, and the
hydrostatic force exerted by water. Taking a moment about the point A where the
hinge is and equating it to zero gives,
2.4 BUOYANCY & STABILITY
What
is buoyancy?
The force that tends to lift the body and
is denoted by Fb.
Why
buoyant force occur?
The increase of pressure in a fluid with
depth.
The difference between these two forces is
a net upward force, which is the buoyant force,
Fb = F bottom – F top
= rfg(s +h) A - rfgsA
= rfghA
= rfgV
Example :
A crane
is used to lower weights into the sea (density 1025 kg/m3) for an underwater
construction project as shown. Determine the tension in the rope of the crane
due to a rectangular 0.4-m x 0.4-m x 3-m concrete block (density = 2300 kg/m3)
when it is (a) suspended in the air and (b) completely immersed in water.
Analysis:
(a) Consider the free-body diagram of the concrete
block. The forces acting on the concrete block in air are its weight and the
upward pull action (tension) by the rope. These two forces must balance each
other, and thus the tension in the rope must be equal to the weight of the
block
V
= (0.4 m)(0.4 m)(3 m) =
0.48 m3
FT,air =
W =r concrete gV
(b) When
the block is immersed in water, there is the additional force of buoyancy
acting upward. The force balance in this case gives : Fb = rfgV
Therefore, F T ,water = W – FB = 10.8 -4.8 = 6.0 kN
Stability of Immersed
and Floating Bodies
·
Analogy of
“ball on the floor” can explain the fundamental concepts of stability and
instability.
I.
Stable
since any small disturbance (someone moves the ball to the right or left)
generates a restoring force (due to gravity) that returns it to its initial
position
II.
Neutrally
stable because if someone moves the ball to the right or left, it would stay
put at its new location. It has no tendency to move back to its original
location, nor does it continue to move away.
III.
Unstable
when in a situation which is the ball may be at rest at the moment, but any
disturbance, even an infinitesimal one, causes the ball to roll off the hill—it
does not return to its original position; rather it diverges from it.