Lecture #7
• Recap lecture #6
- Developed equation to describe pressure in the atmosphere
g
T   z   R

p  p a  a
 Ta 
- Applied hydrostatics principles to problems
z
dp
   g  
dz
x
y
Problem Set # 2, due on Friday, September 10th: 2.10, 2.20, 2.24, 2.28, 2.36
• Focus of today’s lecture
- Verify problem solution you were expected to compete from last lecture
- Examine hydrostatic forces on planar submerged surfaces with the goal
of finding magnitude and line of action of net hydrostatic force
Fluid Mechanics: Mahalingam
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Lecture #7 - continued
PA  PA  known - (1)
PD  PA  (l2  l1 ) w
PE  PA
PF  PE
- (3)
PG  PF   w SG l2
 ( 4)
A
Water
PB  PG   w l1 - (6)
- (5)
PC  PA   w R  (8)
- (2)
C•
=>
l3
E
l2
l1
F
G
B
D
PB  PA   w SG l2   w l1  (7)
To Express Pc in terms of mm’s of Hg
PC (mm' s of Hg) 
PA   w R
 Hg
 (9)
• Substitute numerical values
PB  60  [(9800 )(0.8)(3)  (9800 )(2)] / 1000  103.1 kPa
 [(60)(1000)  (9800)(3)] 
PC (mm' s of Hg)  
 103  229.6 mm  (9)
(13.6)(9800)


Fluid Mechanics: Mahalingam
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Lecture #7 - continued
• Read section 2.5, MYO, 3rd edition on absolute/gage pressure/textbook
convention, concept of pressure head on page 46
• Manometry - Demo
1
A
h1
2
h2
B
Assumptions: Static fluid, gravity is the only
body force, z axis vertical
dp / dz  
PB  PA   1 h1
PB  PAtm   2 h2
 (1)
 (2)
Thus, PA  PAtm   2 h2   1 h1
 (3)
or PA ]gage   2 h2   1 h1
• Read 2.6.3 and 2.7, MYO, 3rd edition for other pressure measuring devices
Fluid Mechanics: Mahalingam
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Lecture #7 - continued
• Hydrostatic forces on submerged surfaces
- Begin by considering forces on planar surfaces
- Basic ideas are the same when considering curved surfaces
• Planar surface analysis
Goal
• To find magnitude of force on a planar surface submerged in a
fluid
• To find location of net force or Center of Pressure
Fluid Mechanics: Mahalingam
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Lecture #7 - continued
h
c
F
z
dF  p dA
  h dA
q
dF dy
h
F    h dA    ( y sin q ) dA
y
x
yR
A
x
A
  sinq  y dA
 (1)
A
y
dA
y
Define  y dA  yc A  (2)
yc
A
Here yc is the y-coordinate of centroid of area
Thus F    Ayc sinq   hc A  (3)
F yR   y dF   sin q  y 2 dA
yR 
A
A yc
A
 A yc sin q  yR   sin q  y 2 dA
A
2
 y dA
A
 ( 4)
Fluid Mechanics: Mahalingam
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Lecture #7 - continued
• Introduce Module 2 briefly (Surface tension module) at the end of lecture
• At the beginning of lecture, point to website for course
http://stripe.colorado.edu/~mcen3021
Fluid Mechanics: Mahalingam
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