Fluid Mechanics
FLUID STATICS
Pressure Variation with Elevation
F  0
l
dp
 
dz
Variation in pressure with elevation.
Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Variation in pressure with elevation.
• Hydrostatic equation
dp
 
dz
From a vertical datum,
pressure decreases as
z-datum elevation increases.
Pressure Variation in Uniform Density
Assuming that the density (ρ) and specific weight () of a fluid are
uniform through the fluid
dp
Integrate
  to get
dz
p  z  pz
pz : Piezometric pressure
p

  z   constant


Hydrostatic Equation
Piezometric head
 p1
  p2

  z1     z2 

 

Pressure Variation in Uniform Density
 p1
  p2

  z1     z2 

 

Hydrostatic Equation applies only in a fluid with a
constant specific weight.
It applies to two point in the same fluid but not across
an interface of two fluids having different specific
weight.
Example:
What is the water
pressure at a depth
of 35 ft in the tank
shown?
Specific Weight
γ = 62.4 lb/ft3
Example 3.3 (p. 37)
Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Solution:
 p1
  p2

  z1     z2 

 

0 + 250
=
(P2/) + 215
35 ft
=
(P2/)
P2
=
35 * 62.4
P2
=
2180 psfg = 15.2 psig
Psig = pound force per square inch gage
Example:
Oil with a specific gravity of 0.80 forms a layer 0.90m deep in an open
tank that is otherwise filled with water. The total depth of water and oil is 3
m. What is the gage pressure at the bottom of the tank. γ = 9810 N/m3
p2 = 0.90 x (0.8 x 9810) = 7.06 kPa
p3 = 7.06 + 2.1 x 9,810 = 27.7 kPa
Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Hydrostatic Equation
 p1
  p2

  z1     z2 

 

P  h
Change in pressure between two points depends on the
specific weight of the fluid, and the vertical distance
between the two points.
Pressure Variation with Elevation
p0
p0
p0
p
h1
h

p0
1
p1=p0+1h1
p=h
p=p0+h
h2
2
p2=p1+2h2
h
h
p
Pressure Measurements
Barometer → to measure atmospheric pressure
Absolute pressure
pabsolute  patm   h
Gage pressure
ph
Many pressure-measuring device measure not absolute
pressure but only differences in pressure.
Manometer
• A pressure measuring method that utilizes the change in pressure with
elevation to evaluate pressure.
Piezometer (manometer )
attached to a pipe
P  h
- Accurate & simple .
- The problem is that a piezometer is
impractical for high pressure and useless
for gasses.
If h is 10 m, what is
the pressure at the
centre of the pipe?
U-tube Manometer
General manometer equation
p2  p1    i hi    i hi
down
1: Initial point index
2: Final point index
Problem: water in pipe, mercury manometer liquid ( mercury specific weight =133
kN/m3)
∆h = 60 cm l =180 cm
Find the pressure at the centre of the pipe ?
Ans: 62.1 kPa
up
Example 3.7: Manometer Analysis
Question: Pressure of the air?
Given: l1 = 40 cm l2 = 100 cm l3 = 80 cm
slide 14
Differential Manometers
To measure the pressure difference btw two points in a pipe
Here, the pressure difference
between 1 and 2 is:
P  ( m   f )h
(this is for a horizontal pipe… z1 = z2)
γm : the specific weight of the manometer liquid,
γf : the specific weight of the fluid,
Δh : the deflection of this liquid.
Differential Manometers
• Example 3.8:
• Specific gravity of
manometer fluid is 3.
Δh = 5 cm
Δz = 1 m
y = 2 cm
What is the pressure difference?
What is the change in piezometric pressure?
Solution:
Piezometric pressure
p  z  pz
Piezometric difference
P  ( m   f )h
Applying the manometer equation between points 1 and 2:
P2  P1   w (y  h)   mh   w (y  z2  z1 )
( P2   w z2 )  ( P1   w z1 )  h( w   m )
Change in piezometric pressure:
Pz 2  Pz1  h( w  ym )
slide 18
FLUID STATICS:
Hydrostatic Force on Plane Surfaces
Distribution of hydrostatic pressure on a plane surface
Pressure on the differential
area can be computed if the y
distance to the point is known
dF = p dA = ( y sin) dA
Integrating the differential
force over the entire area A
Hydrostatic Force
_
_
_
F  y SinA  p A
Pressure at the centroid
F  Sin  y dA Sin y A
A
Integral is the first moment of the area
Hydrostatic Force
Hydrostatic Force Terms
• Δh: Vertical distance from centroid to the water surface
(This distance determines the pressure at the centroid)
• y: Inclined distance from water surface to the centroid
• ycp: Inclined distance from water surface to centre of pressure
• P¯: the pressure at the centroid
Vertical Location of Line of Action of resultant Hydrostatic Force
- Considering moments of the pressure about the horizontal axis 0-0:
_
_
ycp  y 
I
slide 21
_
yA
•
•
•
•
ycp = (inclined) distance to the centre of pressure
y ¯= (inclined) distance to the centroid
I ¯= area moment of inertia about horizontal axis passing the centriod
A = surface area
Restrictions:
1- One liquid involved
2- Gage pressure is zero at the liquid surface
Lateral Location of Line of Action of resultant Hydrostatic Force
- The same principles as above can be used for the lateral location
- Starts with taking moments about a line normal to line 0-0
Review of Centroid & Area Moment of Inertia
Example: 3.10
• An elliptical gate covers the end of a pipe 4m in diameter. If the gate is hinged at
the top, what normal force F is required to open the gate when water is 8 m deep
above the top of the pipe and the pipe is open to the atmosphere on the other side?
Neglect the weight of gate.
Resultant hydrostatic force:

Fp  p A  hA  (9810x10)ab
(a, b: half of major and minor axes)
Fp = 1.54 MN
_
_
ycp  y 
I
_
yA

1 / 4a 3b

y ab
 0.125 m
y¯(slant distance from surface to centroid): 12.5m
Example cont’d
Moment about the hinge. Moment arm
for the hydrostatic force:
2.5 +0.125 = 2.625m
M hinge  0
(1.541x106 N x 2.625m) - (F x 5 m)  0
F  809 kN
Normal Force required to open gate
Pressure Prism
The volume called the pressure prism, that is a geometric representation
of the hydrostatic force on a plane surface
The resultant force
must pass through the
centroid of the
pressure prism.
Pressure Prism
• An informative and useful graphical interpretation can be made for the force
developed by a fluid acting on a plane area.
• Consider the pressure distribution along a vertical wall of a tank of width b,
which contains a liquid having a specific weight .
• Since the pressure must vary linearly with depth, we can represent the
variation as is shown in Figure below, where the pressure is equal to zero at
the upper surface and reach to maximum at the bottom.
• It is apparent from this diagram that the average pressure occurs at the
depth h/2 and therefore the resultant force acting on the rectangular area
(A = b h) is
FLUID STATICS:
Hydrostatic Forces on Curved surfaces
Hydrostatic Forces on Curved Surfaces
Find the magnitude and line of action of the hydrostatic force acting
on surface AB
Important Questions to Ask
1. What is the shape of
the curve?
2. How deep is the
curved surface?
3. Where does the curve
intersect straight
surfaces?
4. What is the radius of
the curve?
Hydrostatic Forces on Curved Surfaces
A free-body diagram of a suitable volume of fluid can be used to
determine the resultant force acting on a curved surface.
Hydrostatic forces on Curved surfaces.
Find the magnitude and line of action of the hydrostatic force
acting on surface AB
Forces acting on the fluid element
1.
2.
3.
4.
FV : Force on the fluid element due to the
weight of water above CB
FH : Force on the fluid element due to
horizontal hydrostatic forces on AC
W : Weight of the water in fluid element
ABC
F : The force that counters all other forces
- F has a horizontal component: Fx
- F has a vertical component: Fy
Hydrostatic forces on Curved surfaces
Find the magnitude and line of action of the hydrostatic force acting on surface AB
- Given: Surface AB with a width of 1 m
Problem Solving Preparation
1. By inspection, curve is a ¼ circle.
2. The depth to the beginning of the
curve (4 m depth to B)
3. The curve radius (2 m horizontal
curve projection distance = curve
radius)
4. Label relevant points:
• BCDE is water above fluid
element defined by the curve
•
ABC is the fluid element
defined by the curve
Example 3.11: Hydrostatic forces on Curved surfaces
Find Fv, FH, W, Fx, Fy, F, Line of action for FH & Fv
Given: Surface AB goes 1 m into the paper
Fx= FH = (5 x 9810) (2 x 1) = 98.1 kN
Pres. at the cenroid
AC side area
Fy= W + Fv
Fv= 9810 x 4 x 2 x 1 = 78.5 kN
W= γVABC= 9810 (1/4 x  r2) 1 = 30.8 kN
Fy= 78.5 + 30.8 = 109.3 kN
The hydrostatic force acting on AB
is equal and opposite to the force
F shown
The centroid of the quadrant
Location of the resultant force
Slide 33
FLUID STATICS:
Buoyancy
Buoyancy, Flotation & Stability
Archimedes’ Principle
The resultant fluid
force acting on a
body that is completely
Submerged or floating
in a fluid is called the
buoyant force.
Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Buoyancy: Floating Object
Depends on submerged portion of
the volume
VD is the submerged volume
Buoyant force
FB  VD
where g is the specific weight of the
fluid and VD is the volume of the
body
Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Example 3.12: Bouyant force on a metal part
- Wood block (1) has dimensions
of 10‐mm x 50 mm x 50 mm
-Specific gravity of 0.3
- Metal object (2) has volume of
6600 mm3
– Find the tension in the cable
and mass of object 2.
Steps
• Find the buoyant forces.
• Find the weight of the block.
• Perform force balances on both objects.
Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson
Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.
Solution: