Lecture #7

• Recap lecture #6 - Developed equation to describe pressure in the atmosphere

p



p a

 

T a



T a



z

  

g R

- Applied hydrostatics principles to problems

z dp

  

g dz

  

y x

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 1

Lecture #7 - continued

P A



P A

 known (1)

P E



P A

(3)

P G



P F

 

w l

2 (5)

P D



P A

 (

l

2 

l

1 ) 

w P F



P E

 ( 4 ) (2)

P B



P G

 

w l

1 (6) =>

P B



P A

 

w P C



P A

 

w R

 ( 8 ) A Water

l

3 E

l

2

l

1 D

l

2  

w l

1  ( 7 ) F G B To Express

P c

in terms of mm’s of Hg

P C

( mm' s of Hg) 

P A

  

Hg w R

• Substitute numerical values

P B

 60  [( 9800 )( 0 .

8 )( 3 )  ( 9800 )( 2 )] / 1000  103 .

1 kPa

P C

( mm' s of Hg)    [( 60 )( 1000 ) ( 13 .

6 )(  ( 9800 9800 ) )( 3 )]    10  3  229 .

6 mm Fluid Mechanics: Mahalingam  ( 9 ) 2  ( 9 )

Lecture #7 - continued

• Read section 2.5, MYO, 3rd edition on absolute/gage pressure/textbook convention, concept of pressure head on page 46

h

1 • Manometry - Demo  1 A B

h

2  2 Assumptions: Static fluid, gravity is the only body force,

z

axis vertical

dp

/

dz

  

P B



P A

  1

h

1

P B



P

Atm   2

h

2  ( 1 )  ( 2 ) Thus,

P A

or

P A

] gage 

P

Atm   2

h

2   2

h

2   1

h

1   1

h

1  ( 3 ) • Read 2.6.3 and 2.7, MYO, 3rd edition for other pressure measuring devices Fluid Mechanics: Mahalingam 3

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 4

y F h y h c dF dy

q

Lecture #7 - continued

z y y R dA x x y c dF

  

p dA h dA F



A

 

h dA



A

  (

y

sin q )

dA

  sin q 

A y dA

Define 

A y dA



y c A

 ( 1 )  ( 2 ) Thus

F

  

Ay c

 sin q  

h c A

Here

y c

is the

y

-coordinate of centroid of area  ( 3 )  

F y R A y c

  sin

A

q

y dF



y R

    sin q sin q 

A y

2

dA



A y

2

dA y R

 

A y

2

dA A y c

 ( 4 ) Fluid Mechanics: Mahalingam 5

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 6