Fluid3a Trawneh PDF

FLUID MECHANICS
Chapter 3a
Fluid Statics
Dr. Ibrahem Altarawneh

1
Fluid Mechanics Chapter 3: Fluid Statics Dr. Ibrahem Altarawneh
Ch3: Fluid statics.
• Determine the variation of pressure

in a fluid at rest
• Calculate pressure using various kinds
of manometers
• Calculate the forces and moments
exerted by a fluid at rest on plane or
curved submerged surfaces
• Analyze the stability of floating and
submerged bodies
• Analyze the rigid-body motion of
fluids in containers during linear
acceleration or rotation
This chapter begins mechanics of fluids in depth by introducing many

concepts related to pressure and by describing how to calculate
forces associated with distributions of pressure. This chapter is
restricted to fluids that are in hydrostatic equilibrium.
Fluid particle
• A fluid particle, is defined as a body of fluid having finite mass and
internal structure but negligible dimensions. Thus, a fluid particle is
very small, but large enough so that the continuum assumption applies.
• The hydrostatic condition means that each fluid particle is in force
equilibrium with the net force due to pressure balancing the weight of
the fluid particle.
The hydrostatic condition.

(a) A fluid particle in a body of fluid.
(b) Forces acting on the fluid particle.
Pressure
• Pressure is defined as the ratio of normal force to area at a point.
• For example, the Fig. shows fluid inside an object such as air inside
a soccer ball.
• The molecules of the fluid interact with the walls to produces a
pressure distribution.
• At each point on the walls, this
pressure distribution creates a
resultant force that acts on an
infinitesimal unit of area A as shown.
Pressure
• Pressure is a scalar quantity; that is, it has magnitude only.
• Pressure is not a force; rather it is a scalar that produces a
resultant force by its action on an area.
• The resultant force is normal to the area and acts in a
direction toward the surface (compressive).
• Some units for pressure give a ratio of force to area.
Newtons per square meter of area, or pascals (Pa), is the SI
unit.
• Other units for pressure give the height of a

column of liquid. For example, gas pressure in a
tank will push on a water column upward about h
meter
• 1.0 atm = 101.3 kPa = 14.70 psi = 33.9 ft-H2O
=760 mm-Hg = 29.92 in-Hg = 2116 psf
Absolute, Gage and Vacuum Pressure

• The pressure in a perfect vacuum is called absolute zero, and the
pressure measured relative to this zero pressure is termed absolute
pressure
• When pressure is measured relative to local atmospheric pressure, the
pressure value is called gage pressure. For example, when a tire gage
gives a value of 300 kPa, this means that the absolute pressure in the
tire is 300 kPa greater than local atmospheric pressure.
• When pressure is less than atmospheric, the pressure can be described
using vacuum pressure, which is the difference between atmospheric
pressure and actual pressure
Pabs  Patm  Pgage
Pabs  Patm  Pvacuum
Pvacuum   Pgage
Absolute, Gage and Vacuum Pressure
Pvacuum   Pgage
Pabs  Patm  Pgage
Pabs  Patm  Pvacuum

Hydraulic Machines
• A hydraulic machine uses components such as pistons, pumps, and
hoses to transmit forces and energy using fluids.
• Hydraulic machines are applied, for
example, to braking systems, forklift
trucks, power steering systems, and
airplane control systems.
• Hydraulic machines provide an
example of Pascal’s law. This law
states that pressure applied to an
enclosed and continuous body of
fluid is transmitted undiminished to
every portion of that fluid and to
the walls of the containing vessel.
Example
Example
Example
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 140 N, the
gage being tested indicates 200 kPa. If
the piston diameter is 30 mm, what
percentage of error exists in the gage?
A Crosby gage tester is applied to calibrate a pressure gage. Indicated
pressure on the gage is p = 200 kPa. W = 140N, D = 0.03m.
Find: Percent error in gage reading.

PLAN
1. Calculate the pressure that the gage should be indicating (true pressure).
2. Compare this true pressure with the actual pressure.
Solution:
Example 2:
If a 200 N force F1 is applied to the piston with the 4 cm diameter, what

is the magnitude of the force F2 that can be resisted by the piston with
the 10 cm diameter? Neglect the weight of the pistons.
Solution:
Solution:
Example 2:
As shown, a mouse can use the mechanical advantage provided by a

hydraulic machine to lift up an elephant.
a) Derive an algebraic equation that gives the mechanical advantge of the
hydraulic machine shown. Assume the pistons are frictionless and
massless.
b) A mouse can have a mass of 25 g and an elephant a mass of 7500 kg.
Determine a value of D1 and D2 so that the mouse can support the
elephant.
Situation:
A hydraulic machine is used to provide a mechanical advantage.

m1 = 0.025 kg, m2 = 7500 kg.
Find:
(a) Derive an algebraic equation for the mechanical advantage.
(b) Calculate D1 and D2 so the mouse can support the elephant.
Assumptions:
• Neglect the mass of the pistons.
• Neglect the friction between the piston and the cylinder wall.
• The pistons are at the same elevation; thus, the pressure acting on the
bottom of each piston is the same.
• A mouse can fit onto a piston of diameter D1 = 70 mm.
PLAN
• Define ”mechanical advantage.”

• Derive an equation for the pressure acting on piston 1.
• Derive an equation for the pressure acting on piston 2.
• Derive an equation for mechanical advantage by combining steps 2 and
3.
• Calculate D2 by using the result of step 4.
SOLUTION
Mechanical advantage
= weight lifted by the mouse/weight of the mouse = W_Eleph/W_mouse
= W2/W1
Equilibrium (piston 1):
Equilibrium (piston 2):
Combine
Solve
Calculate
Pressure Variation with Elevation

• Hydrostatic Equation
• The hydrostatic differential equation is derived by
applying force equilibrium to a static body of
fluid. dP   ;   g
dz
• The equation means that changes in pressure
correspond to changes in elevation.
• If one travels upward in the fluid (positive z direction), the pressure
decreases;
• If one goes downward (negative z), the pressure increases;
• If one moves along a horizontal plane, the pressure remains
constant.
• These pressure variations are exactly what a diver experiences when
ascending or descending in a lake or pool.
• The hydrostatic equation is used to predict pressure variation in a fluid

with constant density. This equation is derived by assuming that specific
weight is constant and then integrating dP   ;   g
dz
• The integration yields: P  z  Pz  const .
where z is the elevation (m), which is the height above a fixed

reference point called a datum, Pz is the piezometric pressure (Pa),
γ is the specific weight (N/m3).
• Dividing by specific weight gives:

P  z  Pz  h  const.
 
where h is the piezometric head
• Since h is constant P1 P2
 z1   z2
 
where the subscripts 1 and 2 identify any two points in a static fluid of
constant density. Multiplying by the specific weight, gives
P1  z1  P2  z 2
P  P2  P1; z  z 2  z1  h
P  z  P   gh
where h is the distance below the free liquid surface at
which the pressure of air and vapor on the surface is arbitrarily
taken as zero
• The hydrostatic equation is valid for any constant density fluid in
hydrostatic equilibrium.

Problem 3.11: For the closed tank with Bourdon-tube gages tapped into it,
what is the specific gravity of the oil and the pressure reading on gage C?
Solution:
Solution:
• Pressures at points A, B, C, D, E, F, and G are the same since they are at the
same depth, and they are interconnected by the same static fluid.
• Pressures at points H and I are not the same since these two points cannot be
interconnected by the same fluid
Pressure Measurements
• Barometer.
• An instrument that is used to measure
atmospheric pressure is called a barometer.
• A mercury barometer is made by inverting

a mercury-filled tube in a container of
mercury as shown.
• Atmospheric pressure will push the

mercury up the tube to a height h. It is
calculated using
Patm  h Hg   Hg gh
Example
Example
• Bourdon-tube gage
Bourdon-tube gage measures pressure by sensing the
deflection of a coiled tube. The tube has an elliptical
cross section and is bent into a circular arc. When
atmospheric pressure prevails in the gage, the tube is
undeflected, and the gage pointer is calibrated to read
zero pressure. When the pressure is applied to the
gage, the curved tube tends to straighten, thereby
actuating the pointer to read a positive gage pressure.
Bourdon-tube gage is common because it is low cost,
reliable, easy to install, and available in many different
pressure ranges. But, the dynamic pressures are
difficult to read accurately, and the gage can be
damaged by excessive pressure pulsations.
• Bizometer
• A piezometer is a vertical transparent tube

in which a liquid rises in response to a
positive gage pressure.
• In the case of shown piezometer that is
attached to a pipe, the gage pressure at the
center of the pipe is:
P  h  gh
Piezometer advantages: simplicity, direct measurement, and accuracy.

Piezometer disadvantages: cannot be used for measuring gas pressure, and is
limited to low pressure
• Manometer
• A manometer, often shaped like the letter “U”,
is a device for measuring pressure by raising or
lowering a column of liquid.
• The general equation for the pressure
difference measured by the manometer is:
P2  P1    i hi    i hi
down up
• Differential Manometer
• Used to measure the pressure
difference between two points.
Consider an arrangement as shown
in the Figure. A U-tube partially
filled with a heavier liquid, mercury
in most cases, connected to a pipe
across a restriction in the pipe.
Density of the fluid in the pipe is P1  h1 L  P2  h2 L  h Hg
ρL and the density of the heavy
P1  h1 L g  P2  h2  L g  h Hg g
liquid is ρHg. Pressure at two
tapings to which the manometer   Hg 
P1  P2  h L g   1
arms are connected are P1 and P2  L 
• Inclined Tube Manometer
• To measure small pressure
differences the sensitivity of
the manometer has to be
increased. This is achieved by
inclining one arm of the U-
tube
P1  h1 ρ1 g  P2  h2 ρ2 g  l sinθ ρ3 g
P1  P2  h1 ρ1  h2 ρ2 g  l sinθ ρ3 g

Fluid3a Trawneh PDF

Uploaded by

Copyright:

Available Formats

Fluid3a Trawneh PDF

Uploaded by

Document Information

Original Description:

Original Title

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

Fluid3a Trawneh PDF

Uploaded by

Copyright:

Available Formats

FLUID MECHANICS

Dr. Ibrahem Altarawneh

Ch3: Fluid statics.

• Determine the variation of pressure

This chapter begins mechanics of fluids in depth by introducing many

The hydrostatic condition.

• Other units for pressure give the height of a

Absolute, Gage and Vacuum Pressure

Absolute, Gage and Vacuum Pressure

Pabs  Patm  Pgage

Pabs  Patm  Pvacuum

Find: Percent error in gage reading.

If a 200 N force F1 is applied to the piston with the 4 cm diameter, what

As shown, a mouse can use the mechanical advantage provided by a

A hydraulic machine is used to provide a mechanical advantage.

• Define ”mechanical advantage.”

Equilibrium (piston 2):

Pressure Variation with Elevation

Pressure Variation with Elevation

• The hydrostatic equation is used to predict pressure variation in a fluid

where z is the elevation (m), which is the height above a fixed

• Dividing by specific weight gives:

Pressure Variation with Elevation

Pressure Variation with Elevation

Pressure Variation with Elevation

• A mercury barometer is made by inverting

• Atmospheric pressure will push the

• A piezometer is a vertical transparent tube

Piezometer advantages: simplicity, direct measurement, and accuracy.

You might also like