EIT Review (1)

7/28/2019 EIT Review (1)

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Monroe L. Weber-Shirk School of Civil and

Environmental Engineering

Fluid Mechanics

EIT Review

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7/28/2019 EIT Review (1)


Shear Stress

change in velocity with respect to distance

A

F

2m

N

dy

du

Tangential force per unit area

rate of shear

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P1 = 0 h1

?

h2

Manometers for High Pressures

Find the gage pressure in the center

of the sphere. The sphere contains

fluid with g1 and the manometer contains fluid with g2.

What do you know? _____

Use statics to find other pressures.

1

2

3

=P3

g1

g2

For small h1 use fluid with high density. Mercury!

+ h1g2

- h2g1P1

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Differential Manometers

h1

h3

Mercury

Find the drop in pressure

between point 1 and point

2.

p1 p2 Water

h2

orifice

= p2

p1 - p2 = (h3-h1)gw + h2gHg

p1 - p2 = h2(gHg - gw)

p1

+ h1gw

- h2gHg

- h3gw

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Forces on Plane Areas: Inclined

Surfaces

q

A’

B’

O

O

x

y

c y

c x

R x

R y

Ah F c R g ch

Free surface

centroid

center of pressure

The origin of the y

axis is on the free

surface

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Statics

Fundamental Equations

Sum of the forces = 0

Sum of the moments = 0

A p F c pc is the pressure at the __________________ centroid of the area

y A y

I

A y

A y I y x x

p

2

Line of action is below the centroid

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Properties of Areas

yc

b

a I xc

yc

b

a I xc

A ab=2

c

a y =

3

12 xc

ba I =

2

ab A =

3

c

b d x

+=

3

36 xc

ba I =

2 A R=

4

4 xc

R I

p=

R yc

I xc

0 xyc I =

( )2

272

xyc

ba I b d = -

0 xyc I =

3c

a y =

d

c y R=

7/28/2019 EIT Review (1)


Properties of Areas

3

4 xc

ba I

p= A ab=

4

3c

R y =

a

yc

b

I xc

2

2

R A

p=

4

3c

R y =

4

8 xc

R I

p=

ycR

I xc

0 xyc I =

0 xyc I =

4

16 xc

R I

p=

2

4

R A

p=R

yc

c y a=

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Inclined Surface Summary

The horizontal center of pressure and the

horizontal centroid ________ when the surface

has either a horizontal or vertical axis of symmetry

The center of pressure is always _______ the

centroid

The vertical distance between the centroid and

the center of pressure _________ as the surface

is lowered deeper into the liquid

What do you do if there isn’t a free surface?

y A y

I y x

p

A y

I x x

xy

p coincide

below

decreases

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An elliptical gate covers the end of a pipe 4 m in diameter. If the

gate is hinged at the top, what normal force F applied at the

bottom of the gate is required to open the gate when water is 8 m

deep above the top of the pipe and the pipe is open to theatmosphere on the other side? Neglect the weight of the gate.

hingewater

F

8 m

4 m

Solution Scheme

Magnitude of the force

applied by the water

Example using Moments

Location of the resultant force

Find F using moments about hinge

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Magnitude of the Force

A p F cr

ab A

abh F r g

m2m2.5πm10m

N 9800

3

r F

b = 2 m

a = 2.5 m

pc = ___

F r = ________

h = _____

hingewater

F

8 m

4 m

Fr

hg

10 m Depth to the centroid

1.54 MN

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Location of Resultant Force

4

3ba I x

h y

y A y

I y x

p

ab y

ba y y p

4

3

y

a y y p

4

2

m12.54

m2.52

y y p

ab A

_______ y y p

________ y

hingewater

F

8 m

4 m

Fr

12.5 mSlant distanceto surface

0.125 m __

p x x b = 2 m

a = 2.5 m

cp

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Force Required to Open Gate

How do we find the

required force?

0hinge M

F = ______ b = 2 m

2.5 ml cp=2.625 m

m5

m2.625 N10x1.54 6

F

tot

cpr

l

l F

F

l tot

hingewater

F

8 m

4 m

Fr

Moments about the hinge

=Fl tot - F r l cp

809 kN

cp

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Example: Forces on Curved

Surfaces

Find the resultant force (magnitude and location)

on a 1 m wide section of the circular arc.

FV =

FH = A p

water 2 m

2 m

3 m W1

W2

W1 + W2

= (3 m)(2 m)(1 m)g + p/4(2 m)2(1 m)g

= 58.9 kN + 30.8 kN

= 89.7 kN

= g(4 m)(2 m)(1 m)

= 78.5 kN y

x

7/28/2019 EIT Review (1)



Surfaces

The vertical component line of action goes through

the centroid of the volume of water above the surface.

21V W3

)m2(4W)m1(Fx

water 2 m

2 m

3 m

A

W1

W2

kN89.7

kN30.83

)m2(4kN58.9)m1(

x

Take moments about a vertical

axis through A.

= 0.948 m (measured from A) with magnitude of 89.7 kN

7/28/2019 EIT Review (1)



Surfaces

water 2 m

2 m

3 m

A

W1

W2

The location of the line of action of the horizontal

component is given by

y A y

I

yx

p

12

3bh I x

b

h

x I

y

m4.083m4

m1m2m4

m0.667 4

p y

y

x

(1 m)(2 m)3/12 = 0.667 m4

4 m

7/28/2019 EIT Review (1)



Surfaces

78.5 kN

89.7 kN

4.083 m 0 . 9

4 8 m

119.2 kN

horizontal

vertical

resultant

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C

(78.5kN)(1.083m) - (89.7kN)(0.948m) = ___ 0

0.948 m

1.083 m

89.7kN

78.5kN

Cylindrical Surface Force Check

All pressure forces passthrough point C.

The pressure forceapplies no moment about point C.

The resultant must pass

through point C.

7/28/2019 EIT Review (1)


Curved Surface Trick

Find force F required to open

the gate.

The pressure forces and force F pass through O. Thus the hinge

force must pass through O!

All the horizontal force iscarried by the hinge

Hinge carries only horizontal

forces! (F = ________)

water 2 m

3 m

A

W1

W2 F

O

W1 + W2 11.23

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Dimensionless parameters

Reynolds Number

Froude Number

Weber Number

Mach Number

Pressure Coefficient

(the dependent variable that we measure experimentally)

Vl R

gl

V

F

2

2C

V

p p

l V W

2

cV M

AV d

2

Drag2C

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Model Studies and Similitude:

Scaling Requirements

dynamic similitude

geometric similitude

all linear dimensions must be scaled identically

roughness must scale

kinematic similitude

constant ratio of dynamic pressures at corresponding

points streamlines must be geometrically similar

_______, __________, _________, and _________

numbers must be the same

Mach Reynolds Froude Weber

C f p M,R,F,W,geometry

7/28/2019 EIT Review (1)


Froude similarity

Froude number the same in model and

prototype

________________________ define length ratio (usually larger than 1)

velocity ratio

time ratio

discharge ratio

force ratio

gl

V F

pm FF

p p

2

p

mm

2

m

Lg

V

Lg

V

p

2

p

m

2

m

LV

LV

m

p

r L

LL r r LV

r

r

r r LV

Lt

2/5

r r r LLL r r r r L AV Q

3 3r r r r r r r 2

r

LF M a L L

t

r = = =

difficult to change g

11.33

7/28/2019 EIT Review (1)


Control Volume Equations

Mass

Linear Momentum

Moment of Momentum

Energy

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Conservation of Mass

cvcs

d t

d Av

021

222111 cscs

d d AvAv

0222111 AV AV

m AV AV 222111

1

2

Q AV AV 2211

v1

A1

V = spatial average of v

If mass in cv

is constant

[M/t]

If density is constant [L3/t]

Area vector is normal to surface and pointed out of cv

7/28/2019 EIT Review (1)


Conservation of Momentum

F M M 1 2

( ) ( )1 1 1 1 1 1V A Qr r = - = -M V V

M V V2 2 2 2 2 2 V A Q

F V V Q Q1 2

F V V Q 2 1 ss p p FFFWF 21

7/28/2019 EIT Review (1)


Energy Equation

l t p h H g

V z

p H

g

V z

p

22

2

222

2

2

2

111

1

1

g

g

g

V K hl

2

2

g

V

D

L f h f

2

2

R

64f

laminar turbulent

Moody Diagram

7/28/2019 EIT Review (1)


z

Example HGL and EGL

z = 0

pump

energy grade line

hydraulic grade line

velocity head

pressure head

elevation

datum

2g

V 2

g

p

2 2

2 2in in out out

in in P out out T L p V p V z h z h h g g

a ag g

+ + + = + + + +

h i i h

7/28/2019 EIT Review (1)


Smooth, Transition, Rough

Turbulent Flow

Hydraulically smooth

pipe law (von Karman,

1930) Rough pipe law (von

Karman, 1930)

Transition function for

both smooth and rough

pipe laws (Colebrook)

51.2

Relog2

1 f

f

D

f

7.3log2

1

g

V

D

L f h f

2

2

(used to draw the Moody diagram)

f

D

f Re

51.2

7.3log2

1

7/28/2019 EIT Review (1)


Moody Diagram

0.01

0.10

1E+03 1E+04 1E+05 1E+06 1E+07 1E+08R

f r i c t i o n f a c t o r

laminar

0.05

0.04

0.03

0.020.015

0.010.0080.006

0.004

0.002

0.0010.0008

0.0004

0.0002

0.0001

0.00005

smooth

l

DC pf

D

0.02

0.03

0.040.05

0.06

0.08

7/28/2019 EIT Review (1)


find head loss given (D, type of pipe, Q)

find flow rate given (head, D, L, type of pipe)

find pipe size given (head, type of pipe,L, Q)

Solution Techniques

Q D gh

L D

D

gh

L

f

f

H

GGG

2 2237

1785 2

2 3

. log.

./

/

DLQ

ghQ

L

gh f f

HG KJ

HG KJ

N

MM

Q

PP

0 66 1 252

4 75

9 4

5 2 0 04

. .

.

.

. .

h f g

LQ

D f

82

2

5

f

D

FH

IK

L

NM

0 25

37

5740 9

2

.

log

.

.

Re

.

Re 4Q

D

7/28/2019 EIT Review (1)


Power and Efficiencies

Electrical power

Shaft power

Impeller power

Fluid power

electric P

water P

shaft P

impeller P

IE

Tw

Tw

gQH p

Motor losses

bearing losses

pump losses

7/28/2019 EIT Review (1)


Manning Formula

1/2

o

2/3

h SR 1

nV

The Manning n is a function of the boundary roughness as well

as other geometric parameters in some unknown way...

RA

P h

A bh P b h 2

Rbh

b h

h

2

Hydraulic radius for wide channels

7/28/2019 EIT Review (1)


Drag Coefficient on a Sphere

0.1

1

10

100

1000

0.1 1 10 102 103 104 105 106 107

Reynolds Number

D r a g C o e f f i c

i e n t

Stokes Law

24DC =

2U

ACF dd

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EIT Review (1)