Fluid Mechanics FLUID STATICS. Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson...

Fluid Mechanics

FLUID STATICS

Engineering Fluid Mechanics 8/E by Crowe, Elger, and RobersonCopyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

Variation in pressure with elevation.

Pressure Variation with Elevation

0lF

dz

dp

Variation in pressure with elevation.

• Hydrostatic equation

dz

dp From a vertical datum,pressure decreases asz-datum elevation increases.

Pressure Variation in Uniform Density

Assuming that the density (ρ) and specific weight () of a fluid are uniform through the fluid

Integrate to get dz

dpzpzp

pz : Piezometric pressure

constant

z

p

Piezometric head

2

21

1 zp

zp

Hydrostatic Equation

Pressure Variation in Uniform Density

2

21

1 zp

zp

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 3.3 (p. 37)

Example:

What is the water pressure at a depth of 35 ft in the tank shown?

Specific Weight

γ = 62.4 lb/ft3

Solution:

2

21

1 zp

zp

0 + 250 = (P2/) + 215

35 ft = (P2/)

P2 = 35 * 62.4

P2 = 2180 psfg = 15.2 psig

Psig = pound force per square inch gage

Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson

Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

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

Hydrostatic Equation

hP

2

21

1 zp

zp

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

h

p=h

p

h

pp0

p=p0+h

p0

p0

p1=p0+1h1

p2=p1+2h2

h

h1

h2

1

2

Pressure Measurements

hpp atmabsolute

hp

Absolute pressure

Gage pressure

Many pressure-measuring device measure not absolute pressure but only differences in pressure.

Barometer → to measure atmospheric pressure

Manometer

• A pressure measuring method that utilizes the change in pressure with elevation to evaluate pressure.

Piezometer (manometer ) attached to a pipe

- Accurate & simple .

- The problem is that a piezometer is impractical for high pressure and useless for gasses.

hP

If h is 10 m, what isthe pressure at thecentre of the pipe?

U-tube Manometer

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

iidown

ii hhpp 12

General manometer equation

1: Initial point index

2: Final point index

Example 3.7: Manometer Analysis slide 14

Question: Pressure of the air?

Given: l1 = 40 cm l2 = 100 cm l3 = 80 cm

Differential Manometers

Here, the pressure difference

between 1 and 2 is:

hP fm )(

(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.

To measure the pressure difference btw two points in a pipe

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:

zpzp

hP fm )(

Piezometric pressure

Piezometric difference

)()( 1212 zzyhhyPP wmw

Applying the manometer equation between points 1 and 2:

)()()( 1122 mwww hzPzP

)(12 mwzz yhPP

Change in piezometric pressure:

FLUID STATICS:

Hydrostatic Force on Plane Surfaces

slide 18

Distribution of hydrostatic pressure on a plane surface

AySindAySinFA

_

ApASinyF__

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

Integral is the first moment of the areaPressure at the centroid

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

• 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

Ay

Iyycp _

__

Restrictions:

1- One liquid involved

2- Gage pressure is zero at the liquid surface

- Considering moments of the pressure about the horizontal axis 0-0:

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

slide 21

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.

abxhAApFp )109810(

maby

ba

Ay

Iyycp 125.0

4/1 3

_

__

(a, b: half of major and minor axes)

y¯(slant distance from surface to centroid): 12.5m

Resultant hydrostatic force:

Fp = 1.54 MN

Example cont’d

kN 809F

0m) 5 x (F -m) 2.625 x N (1.541x10

06

hingeM

Moment about the hinge. Moment arm for the

hydrostatic force:

2.5 +0.125 = 2.625m

Normal Force required to open gate

Pressure PrismThe 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

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?

Important Questions to Ask

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.


1. FV : Force on the fluid element due to the weight of water above CB

2. FH : Force on the fluid element due to horizontal hydrostatic forces on AC

3. W : Weight of the water in fluid element ABC

4. F : The force that counters all other forces - F has a horizontal component: Fx

- F has a vertical component: Fy

Forces acting on the fluid element

Hydrostatic forces on Curved surfaces


- Given: Surface AB with a width of 1 m

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

Problem Solving Preparation

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

The hydrostatic force acting on ABis equal and opposite to the forceF shown

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 centroid of the quadrant

Location of the resultant force

Slide 33

FLUID STATICS:

Buoyancy

Buoyancy, Flotation & Stability


Archimedes’ Principle

The resultant fluidforce acting on a

body that is completelySubmerged or floating in a fluid is called the

buoyant force.

Buoyancy: Floating Object

DB VF

where g is the specific weight of the fluid and VD is the volume of the body

Depends on submerged portion of the volume

VD is the submerged volume

Buoyant force


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 cableand mass of object 2.

Steps• Find the buoyant forces.• Find the weight of the block.• Perform force balances on both objects.

Solution:

Fluid Mechanics FLUID STATICS. Engineering Fluid Mechanics 8/E by Crowe, Elger, and Roberson...

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