W1L2 Fluid Statics

8/12/2019 W1L2 Fluid Statics

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Fluid Statics

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Overview of this lecture

Calculate the forces exerted by a fluid at rest

on plane or curved submerged surfaces

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Introduction

Fluid statics: Deals with problem associated with fluids at rest.

Hydrostatics: When the fluid is a liquid.

Aerostatics: When the fluid is a gas.

In fluid statics, there is no relative motion between adjacentfluid layers, and thus there are no shear stress in the fluid

trying to deform it.

The only stress we deal with in fluid statics is the normal

stress, which is the pressure, and the variation of pressure isdue only to the weight of the fluid.

It only has significance in gravity fields.

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Introduction

The design of many engineering systems such as water dams and

liquid storage tanks requires the determination of the forcesacting on the surfaces using fluid statics.

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Hydrostatic Forces on Submerged

plane Surfaces

A plate, such as a gate valve in a dam, the wall of a liquid storage tank, orthe hull of a ship at rest, is subjected to fluid pressure distributed over its

surface when exposed to a liquid.

On a plane surface, the hydrostatic forces form a system of parallel forces,

and we often need to determine the magnitude of the force and its point

of application, which is called the center of pressure.

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General Concept:Variation of Pressure with Depth

Pressure in a fluid at rest does not change in the

horizontaldirection.

Pressure in a fluid increases linearlywith depth

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General Concept:Variation of Pressure with Depth

PA= Patm+ gh

PA= PB= PC= PD= PE= PF= PG

PH PI 7


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plane Surfaces

Recall: First moment of area

= + =

Po= Local atmospheric pressure = Patmif the liquid is open to the atmosphere

= = + = +

=1

Recall:

= + = + A

= = 8


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plane Surfaces

= =

The magnitude of the resultant force, FRacting on

a plane surface of a completely submerged plate in

a homogeneous (constant density) fluid is equal to

the product of the pressure Pcat the centroid ofthe surface and the areaAof the surface.

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plane SurfacesThe point of intersection of the line of action of the resultant force andthe surface is the center of pressure.

= = +

= +

Recall:

,=

Second moment of area

Parallel axis theorem ,= ,+ where:Ixx,c= the second moment of area about the x axis

passing through the centroid of the area

yc= the distance between two parallel axes

= + ,

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plane Surfaces

Center of pressure:

= + ,+

With atmospheric pressure applied to both

side of the plate

= + ,

The vertical distance of the center of pressure from

the free surface:

= 11


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plane Surfaces

Special case: Submerged Rectangular Plate

= + + 2

= + = + ARecall:

Recall: = + ,

+

= + 2 +

12 + 2 +

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plane Surfaces

= + + 2 = +

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=


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Example 1

A heavy car plunges into a lake during an accident and lands at the bottom of

the lake on its wheel. The door is 1.2 m high and 1 m wide, and the top edgeof the door is 8 m below the free surface of the water. Determine the

hydrostatic force on the door and the location of the pressure center and

discuss if the driver can open the door.

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Example 1 (cont.)Assumption:

1. The bottom surface of the water is horizontal/flat.

2. The passenger cabin is well-sealed so that no water leaks inside.

3. The door can be approximated as a vertical rectangular plate

4. The pressure in the passenger cabin remains at atmospheric value since

there is no water leaking in, and no compression of the air inside. (Patm

cancels out in the calculation since it acts on both sides of the door)5. The weight of the car is larger than buoyant force acting on it.

= = + = +

2

= 1000 9.81 8 + 1.22 = 84.4 /

The resultant hydrostatic force on the door:

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= = 84.4 1 1.2 = .


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Example 1 (cont.)

The pressure center is directly under the midpoint of the door:

= + 2 +

12 + 2= 8 + 1.22 +

1.21 2 8 + 1.22

= .

http://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-

scale.html

This is meant for Education purpose. The demonstrators are well-trained

individuals.

DO NOT try this at home!

Interesting Experiment
http://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.htmlhttp://dsc.discovery.com/videos/mythbusters-underwater-car-escape-small-scale.html


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Curved SurfacesFor submerged curved surface, the determination of the resultant force is more

complicated because it requires the integration of the pressure forces that change

direction along the curve surface.

The easier way: Determine the horizontal and vertical components, FHand FVseparately

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Curved Surfaces

Horizontal force component on curved surface:

Vertical force component on curved surface:

The magnitude of the resultant hydrostatic force acting on the curved surface:

The tangent of angle it makes with the horizontal:

=

= +

=

+

tan =19

d b d


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Curved Surfaces

When a curved surface is above the liquid,

the weight of the liquid and the vertical

component of the hydrostatic force act in the

opposite directions.

The hydrostatic force acting on a circular

surface always passes through the center

of the circle since the pressure forces are

normal to the surface and they all pass

through the center.

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Example 2 (cont.)Assumption:

1. Friction at the hinge is negligible2. Atmospheric pressure acts on both sides of the gate and thus it cancels out.

Horizontal force on vertical surface:

Vertical force on horizontal surface:

= = = = +

2

= 1000 9.81 4.2+ 0.82 0.8 1 = 36.1

= = =

= 1000 9.81 (5 ) 0.8 1 = 39.2

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Example 2 (cont.)

Weight of the fluid block:

= = = 4 1

= 1000 9.81 0.8 2 1

4 (1 ) = 1.3

= = 39.2 1.3 = 37.9 The net upward vertical force:

= += 36.1+ 37.9= . tan = =

37.936.1= 1.05 = .

The resultant force:

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Example 2 (cont.)

(b) When the water level is 5 m high, the gate is about to open and thus the reaction

force at the bottom of the cylinder is zero. Then the forces other than those at thehinge acting on the cylinder are its weight, acting through the center and hydrostatic

force exerted by water. Taking a moment about point A and equating it to zero.

= 0 = = 52.3 (sin46.4) = .

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Example 3 (cont.)

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Example 3 (cont.)

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What You Learned from

FLUID STATICS

Hydrostatic force on submerged surfaces

Center of pressure

Reading Material: Y. A. Cengel, J. M. Cimbala, R. H. Turner, Fundamentals of

Thermal-fluid Sciences, 4thed. Chapter 11.

Next Lecture: Fluid Power System- Hydraulic system

- Pneumatic system

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End of Lecture

Q & A

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W1L2 Fluid Statics

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