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Chapter 2
Pressure & Fluid Statics
STATIC FLUIDS
Static fluids
Objectives &
Learning outcomes
PressurePressure
Measurement
Pressure
Pascal’s Law
Variation of Pressure
in a static fluids
Pressure & Head
Barometer
Piezometer
Manometer
(u-tube)
OBJECTIVES- Introduce the concept of pressure
- Prove it has a unique value at any particular elevation
- Show it varies with depth according to the hydrostatic equation
can be expressed in terms of head of fluid
LEARNING OUTCOMES- Understand the concept of pressure
- Able to derive an equation for variation of pressure
- Solve the simple problems related to hydrostatic
characteristics such as pressure, pressure distributions,
pressure measurements, hydrostatic force on a plane
and curved surface immersed in a liquid
STATIC OF FLUID SYSTEMS
Static of fluid = Fluid in a rest condition
• No shearing occurs
• In equilibrium conditions
• Moment at any point is zero
• All forces exerted between the fluid and solid boundary must
act at right angle to the boundary
PRESSURE• Pressure is defined as a normal force exerted by a fluid per unit
area.
• Pressure only deal with fluid and gas. The counterpart of pressure
in solids is normal stress.
PressureappliedisforcethewhichoverArea
Force
A
Fp
• Units: Newton's per square meter (N/m2) which call a Pascal (Pa).
1 N/m2 = 1 Pa
• Other units commonly used in practice are bar and atm.
1 bar = 105 Pa
1 atm = 101.325 kPa
*English system: 1 atm = 14.696 psi (pound force per square inch)
PASCAL’S LAW (pressure at a point)
Px
Ps
Py
δy
δs
δz
δx
A
F
D
B
C
E
θ
• Consider the equilibrium of a small fluid element in the form of
triangular prism below.
• By assuming a fluid is at rest, px will act at right angles to the
plane ABEF, py at right angles to CDEF and ps at right angles to
ABCD.
δyδzxp
ABFEAreaxpxptodueForce
δyδzsp
δs
δyδsδzsp
θsinABCDAreasp
:sptodueforceofComponent
zpxp
0δyδzspδyδzxp
• Considering the force in x-direction:
• The element will be in equilibrium if:
(2.1)
δxδzyp
CDEFAreaypyptodueForce
δxδzsp
δs
δxδsδzsp
θcosABCDAreasp
:sptodueforceofComponent
zp
yp
0δxδyδz2
1ρgδxδz
spδxδz
xp
• Considering the force in y-direction:
• The element will be in equilibrium if:
δxδyδz2
1ρg -
VolumeeightSpecific welement of Weight
* Since δx, δy and δz are all very small quantities, δxδyδz is negligible in comparison with the other two terms.
(2.2)
• Thus from equation 2.1 and 2.2 :
• From the equation 2.3 we can conclude that:
“ Pressure at any point is the same in all directions. This known as
Pascal’s Law (Blaise Pascal-France Philosophy, 1632-1662), and
applies to a fluid at rest”
zp
yp
xp (2.3)
Atmospheric pressure, Gauge pressure and Absolute pressure.
• Absolute pressure - The actual pressure at a given position. It is measured
relative to absolute vacuum (i.e., absolute zero pressure).
• Gauge pressure - Difference between the absolute pressure and the local
atmospheric pressure. Most pressure-measuring devices are calibrated to read
zero in the atmosphere, and so they indicate the gauge pressure.
• Vacuum pressures and are measured by vacuum gages that indicate the
difference between the atmospheric pressure and the absolute pressure.
Absolute, gage, and vacuum pressures are all positive quantities and are
related to each other by:
atmP
absP
gaugeP
absP
atmP
vacP
(2.4)
(2.5)
Patm
Pvac
Pabs
Vacuum
Absolute Pabs=0
Patm
Pgauge
Pabs
Patm
Vacuum
Absolute
Illustrated of atmospheric pressure, gauge pressure and absolute pressure.
VARIATION OF PRESSURE IN A STATIC FLUIDS
z
δs
area, A
p
area, A
p+δp
θ
mg
z+δz
2
1
• The axis of the prism is inclined at right angle θ to the vertical, the
height of 1 above horizontal datum is z and that of 2 is z+δz.
• The force acting on the element are:
i) F1= pA → acting at right angles to the end face at 1 along the axis of
prism.
ii) F2= (p+δp).A → acting at 2 along the axis in the opposite directions.
iii) FB = mg → weight of the element due to gravity
• For equilibrium; ΣF = 0
ie: F1+F2+FB = 0
cos
0
sgA
mgA
pA-
pp-pA
cosgds
dp
gdz
dp
• In differential form:
• Taking the (x,y) plane as horizontal:
• If 1-2 parallel with z-axis, i.e θ=00, ds=dz
0dz
dp
2
1
12
z
z
dzgppordzgp
• If 1-2 perpendicular to z-axis, i.e θ=900, ds=dz
• The actual pressure variation with elevation is found by integrating the
above equation.
(2.6)
(2.7)
(2.8)
(2.9)
PRESSURE AND HEAD
- In a fluid of constant density;
gdz
dp
constant ghp
- Integrate immediately to give;
constant gzp
- The pressure ‘p’ at any depth ‘z’ measured downwards from the
free surface so that;
hz i.e,
- Since the pressure at the free surface will normally be
atmospheric pressure, patm ;
atmpghp
(2.10)
(2.11)
p
h
patm
Liquid with
density ρ
PRESSURE AT LAYERED FLUID
The pressure at the bottom of the tank in Fig. 3–13 can be
determined by starting at the free surface where the pressure is
Patm, moving downward until we reach point 1 at the bottom, and
setting the result equal to P1. It gives
Patm + ρ1gh1 +ρ 2gh2+ ρ 3gh3 = P1
PRESSURE MEASUREMENT
• Barometer- Use to measure atmospheric pressure, thus the atmospheric pressure
referred to as the barometric pressure.
- Pressure at point B is equal to the atmospheric pressure.
- Pressure at point C can be taken to be zero (only mercury vapor above the point C and the pressure is very low relative to patm).
ghpatm
h
B
C
hA
mercury
W=ρghA
patm
Evangelista Torricelli (1608-1647)
prove that the atmospheric pressure
can be measured by inverting a
mercury-filled tube into a mercury
container that is open to the
atmosphere.
(2.12)
Example 2
A mountain lake has an average temperature of 10 °C
and a maximum depth of 40 m. For a barometric
pressure of 598 mm Hg, determine the absolute
pressure (in pascals) at the deepest part of the lake.
Example 1
Determine the atmospheric pressure at a location
where the barometric reading is 740 mm Hg and the
gravitational acceleration is g = 9.81 m/s2. Assume
the temperature of mercury to be 10°C, at which its
density is 13,570 kg/m3.
• Piezometer
- The simplest manometer.
- Consist of tube that open at top which is attached to the vessel containing liquid at a pressure (higher than atmospheric pressure) to be measured.
- As the tube is open at atmosphere, the pressure measured is relative to atmospheric so is gauge pressure.
- Pressure at point A
rhgpA
(2.13)
r
h
A
• U-tube Manometer
- U-tube manometer enables to measure pressure for both liquids and gases with the same instrument.
- The U-tube is connected as in figure and filled with a fluid called manometric fluid.
- The fluid whose pressure is being measured should have a mass density less than that of the manometric fluid and the two fluids should not be able to mix readily.
- For the left hand arm:
1ghpp
AB
- For the right hand arm:
2ghpp
mDC
- Pressure are same at same level, i.e:
D
A
fluid density, ρ
B C
h1
h2
manometric fluid
density, ρm
*As we measuring gauge pressure we can subtract patm (pD)
CBpp
- Pressure at point A:
12ghghp
mA (2.14)
Example 3
A manometer is used to measure the pressure in a tank. The
fluid used has a specific gravity 0.85, and the manometer
column height is 55 cm, as shown in Fig. 3–12. If the local
atmospheric pressure is 96 kPa, determine the absolute
pressure within the tank.
Example 4
A closed tank contains compressed air and oil (SGoil = 0.90) as is
shown in Fig. E2.4. A U-tube manometer using mercury (SGoil = 13.6)
is connected to the tank as shown. For column heights h1 = 36 in.,
and h2 = 6 in, and h3 = 9 in, determine the pressure reading (in psi)
of the gage.
• U-tube Manometer (measurement of pressure difference)
A
CB
D
E
h2
h3
h1
ρ1
ρ2
ρ3
- If the U-tube is connected to a pressurized vessel at two points, the pressure difference between these two points can be measured.
- For the manometer that arranged as in figure:
- For the left hand arm:
CBpp
11ghpp
AB
332211ghghpghp
EA
- For the right hand arm:
- Since PB=PC;
3322ghghpp
EC
- Pressure difference between these two
points (point A and point E) in the vessel is:
113322ghghghpp
EA (2.15)
Example 5
The water in a tank is pressurized by air, and the pressure is
measured by a multifluid manometer as shown in Fig. 2.10. The
tank is located on a mountain at an altitude of 1400 m 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 1000 kg/m3,
850 kg/m3, and 13,600 kg/m3, respectively
Example 6
• Choice of Manometer
Care must be taken when attaching the manometer to vessel. No burrs must be present around this joint. Burrs will alter the flow causing local pressure variations to affect the measurement.
Some disadvantages of Manometers:
Slow response-only really useful for very slowly varying pressure.
For U-tube manometer, two measurements must be taken simultaneously to get h value.
For very accurate work the temperature and relationship between temperature and density must be known.
Some advantages of Manometers:
Very simple
No calibration is required.
PRESSURE MEASURING DEVICES
Bourdon pressure gages
Pressure transducers