1

REFERENCES

1- Fluid Mechanics, Victor L. Streeter and E. Benjamin Wylie, McGraw Hill book Company.

2- Mechanics of Fluids, Irving H. Shames, McGraw Hill book Company. 3- Fluid Mechanics, F. M. White, McGraw Hill book Company. 4- Introduction to Fluid Mechanics, A. T. McDonald, and R. W. Fox,

John Wiley & Sons book Company. 5- Elementary Fluid Mechanics, John K. Vennard & Robert L. Street,

John Wiley & Sons book Company. FLUID MECHANICS Is a branch of applied mechanics which deals with fluids. Statics MECHANICS Dynamics FLUID MECHANICS ⇔ Statics + Dynamicsof Fluids (Liquids and Gases) Fluids are classified into liquids and gases. The difference between the two is that LIQUIDS OCCUPY A DEFINITE VOLUME, whereas GASES EXPAND TO FILL THE ENTIRE VOLUME OF THE CONTAINER IN WHICH THEY ARE PLACED. FLUID is a substance that deforms continuously when subjected to a shear stress, no matter how small that shear stress may be.

F U

Fixed Plate

y

2

When a force F is applied to the upper plate, it exerts a shear stress F/A on the substance at the interface of the plate and the substance. If the force F causes the upper plate to move with a steady(nonzero) velocity, no matter how small the magnitude of F , one conclude that the substance between the two plates is a fluid. RELATIONSHIP BETWEEN STRESS AND DEFORMATION

ityVisofLawNewtonsdydv

dydv cos'µττ =⇒∝

µ is called viscosity , absolute or dynamic viscosity.

µ is defined as the proportionality factor between dydvandτ

dydv /τ

µ =

Considering two different fluids such as water and tar Applying the same τ for both fluids results in less deformation (dv/dy) for the tar. Hence, µ of the tar is greater than that of the water. So, the viscosity can also be expressed as the fluid characteristic that represent the resistance of the fluid against the shear stress. VISCOSITY IS A FUNCTION OF TEMPERATURE The magnitude of µ depends on:

1- Cohesive force between molecules. 2- Momentum interchange between colliding molecules.

The cohesive force is dominant for Liquids, So, as the temperature of a liquid is raised, the cohesive force between molecules decreases. This results in a decrease in viscosity. The momentum interchange is dominant for gases. So, as the temperature of the gas is raised, providing for greater momentum interchange, the viscosity of the gas increases.

3

RHEOLOGICAL DIAGRAM

Thin Liquids and Gases tend to be Newtonian Fluids. UNITS OF VISCOSITY

sPam

sN

msm

mNdydv

../

// 2

2

====τ

µ

In English units

2

.ft

slb=µ

Kinematic Viscosity

ρµ

ν =

Units s

mmkg

ms

smkg

mkgmsN 2

3

22

3

2

/

.

//.

====ρµ

ν

τ

dydv

Newtonian Fluids

Non-Newtonian Fluids Thixotropic Substance

Ideal Plastic

4

In English units s

ft 2

=ν

P.1.3 A Newtonian liquid flows down an inclined plane in a thin sheet of thickness t: The upper surface is in contact with air, which offers almost no resistance to the flow. Using Newton’s law of viscosity, decide what the value of dydu / , y measured normal to the inclined plane, must be at the upper surface. Would a linear variation Of u with y be expected?

Hence, a linear variation of u with y would not be expected. 1.5 A Newtonian fluid is in the clearance between a shaft and a concentric sleeve. When a force of 500 N is applied to the sleeve parallel to the shaft , the sleeve attains a speed of 1 m/s. If a 1500 N force is applied, what speed will the sleeve attain? The temperature of the sleeve remains constant.

Newtonian Fluid

0=τ

00

0

=⇒≠

==

dydu

dydu

µ

µτ

0≠dydu

5

smVNFsmVNF/?1500

/1500

22

11

=⇒==⇒=

dydV1

1 µτ =

Since the clearance thickness is very small, the velocity distribution can be assumed linear.

smVVV

VFF

VV

AFAFVV

tVt

Vt

Vdy

dV

/3

11500500

//

2

22

1

2

1

2

1

2

1

2

1

2

1

11

11

=⇒

=⇒=⇒=⇒

=⇒

=

=⇒=

ττ

µτ

µτ

1.16 A flywheel weighing 600 N has a radius of gyration of 300 mm. When it is rotating 600 rpm, its speed reduces 1 rpm/s owing to fluid viscosity between sleeve and shaft. The sleeve length is 50 mm; shaft diameter is 20 mm; and radial clearance is 0.05 mm. Determine the fluid viscosity. W = 600 N R = 300 mm=0.3 m

y Concentric Sleeve

Shaft

6

?

/60

21/1

/2.060

01.02600600

2

=

×==

=××

==

µ

πα

ππ

sradsrpm

smrpmV

The system is rotating with 600 rpm. When the power is disconnected, the velocity of the rotating system will be reduced by 1 rpm/s owing to fluid viscosity. In other words the frictional shear force between the rotating shaft and the fluid results in a torque equals to Fr

αµ

α

α

2

2

mRrAtV

mRFrIT

=

=

=

Where t = clearance thickness between shaft and sleeve, A = Shaft-Fluid interface surface area.

sPa.46.16023.0

806.9600)1010()1050()1020(

1005.02.0 2333

3

=

××=××

×××××

×−−−

−

µ

ππ

πµ

50 mm

20 mm V=600rpm 0.05 mm

shaft flywheel

sleeve

7

EXAMPLE -An oil of viscosity µ between the conical body and the container fills the clearance. Find the torque required to be applied on the conical body to produce an angular velocity w

Owing to the fact that the velocity at the surface of the conical body varies with radius of the cone at that point, the torque should be obtained by integration.

w

αw H

Conical Body

Container

Oil Film of thickness h

8

ααµπ

αµπ

αµπ

αµπ

sin2tan

sin24sin2

sin2 444

0

4

0

3

hHw

hRwx

hwdxx

hwT

RR

==

== ∫

w

αw H

Conical Body

Container

Oil Film of thickness t

dx

dy ds

α α

α

sin

sin

dxds

dsdx

=

=

απµ

απµ

πµ

τ

sin2

sin2

2

)(

dxxdsh

wxx

dxxdshVx

xdsdydVx

dAxxdFT

xdFdT

A

A

A

AA

∫

∫

∫

∫∫

×=

×=

×=

==

=

9

αµπ

sin2

4

1

21

tRwT

TTT

=

+=

H

R

w H =4” R = 2” t = 0.01”

25 .105.4

ftslb−×=µ

Find the torque T required to rotate the conical body with angular velocity w

r rdrdA π2=

10

BULK MODULUS OF ELASTICITY K For most purposes a liquid may be considered as incompressible, but for situations involving either sudden or great changes in pressure, its compressibility becomes important.

∀∀−=

/ddPK

The compressibility of a liquid is expressed by its bulk modulus of elasticity K. An increase of the pressure dP results in a liquid volume decrease of ∀d . The ratio

∀∀−

/ddP will be equal to K of that liquid.

Example In order to show the low compressibility of water let’s consider the application of 0.1 MPa (about one atmosphere) to a cubic meter of water:

3102.210)(

9999545.01 9

5)

meeKLnPP

dKdP

KPP

oo

o

P

Po o

=×=∀=∀⇒

∀∀

−=−

∀∀

=−

×

−−−

∀

∀∫ ∫

SURFACE TENSION Surface tension of a liquid is due to the forces of attraction between like molecules called COHESION and those between unlike molecules called ADHESION. In the interior of a liquid, the cohesive forces acting on a molecule are balanced out, since the molecule is surrounded by like molecules. Near a free surface however, since the cohesive force between liquid molecules is greater than that between air molecule and liquid molecules, there is a resultant force on a liquid molecule acting toward the interior of the liquid. This force is balanced out with the surface tension force formed in the membrane type surface. The surface tension force is proportional to the product of a surface tension coefficient σ and the length of the free surface. To illustrate, consider the equilibrium of a water droplet shown in cross-section in the figure:

11

BUBLE

JET

Po Pi

Interface of water and air

RPP

RRPP

oi

oi

σ

πσπ

22)( 2

=−

×=×−

RPP

LRLPP

oi

oi

σ

σ

=−

×=×− 22)(

12

BULB

JET RPP oi

σ=−

BUBLE RPP oi

σ2=−

BULB RPP oi

σ4=−

Po Pi

Two Interfaces of liquid and gas

RPP

RRPP

oi

oi

σ

πσπ

422)( 2

=−

××=×−

13

CAPILARITY When a liquid is in contact with a solid surface, adhesion forces between solid and liquid in addition to cohesive forces within the liquid should be taken into consideration.

Water meniscus

Glass tube

Liquid

Rh

hRR

γθσ

γπθσπcos2cos2 2

=

=

Adhesive Forces > Cohesive Forces

14

Cohesive Forces > Adhesive Forces Adhesive Forces > Cohesive Forces

15

CHAPTER 2 FLUID STATICS STRESS AT A POINT

The stress is the ratio of two vectors. So it is neither vector nor scalar.

kFjFiFF

kAjAiAA

zyx

zyx

∆+∆+∆=∆

∆+∆+∆=∆

x

zxz

x

yxy

x

xxx

AFAFAF

∆∆

=

∆

∆=

∆∆

=

τ

τ

σ

C

A∆ F∆

To obtain the stress at point C, let F∆ be the force vector acting on the surface element A∆ . Then

AFCatStress

A ∆

∆=

→∆lim

0

A∆

y

z

xyτ zA∆

x

xA∆

yA∆

xzτ

xxσ

16

==

zzzyzx

yzyyyx

xzxyxx

ijpoaatStressσττ

τστ

ττσ

τint

The stress is a second degree tensor, since it has two indices. Number of Components = 3degree = 32 = 9 Vector ⇒ Tensor of first degree ⇒ Number of Components = 31=3 Scalar ⇒ Tensor of zero degree ⇒ Number of Components = 30=1 In Fluid Statics there exists only pressure(Normal Stress) and there are no shere stresses. Hence,

==

z

y

x

ij

PP

PpoaatStress

00

00

00

int τ

PRESSURE AT A POINT At a point, a fluid at rest has the same pressure in all directions. To demonstrate this, a tetra-hydral element is considered in a fluid at rest as shown in the figure:

x

xA∆

yA∆

zA∆

xx AP ∆

y

z

zz AP ∆

yy AP ∆

AP∆

17

Direction cosines of PdA are l, m, n, respectively. ∑ =0xF

⇒=∆−∆⇒=∆−∆ 00 xxxxx APAPAlPAP Px = P ∑ = 0zF

⇒=∆−∆⇒=∆−∆ 00 zzzzz APAPAnPAP Pz = P ∑ = 0yF

⇒∆⇒=∆

−−

⇒=∆∆

−∆−∆⇒=∆∆

−∆−∆

zeroapproachesyyPP

yAAPAP

yAAmPAP

y

yyyy

yyy

03

03

03

γ

γγ

Py = P Px = Py = Pz = P BASIC EQUATIONS OF FLUID STATICS PRESSURE VARIATION IN A STATIC FLUID The forces acting on element of fluid at rest consist of surface forces and body forces. Surface forces are the forces that act on the element surfaces and body forces act on the particles of the body. Gravity force is the only body force.

18

amF vv=∑ (1)

kFjFiFF zyx

vvvv∑∑∑∑ ++=

dxdydzxPF

dydzdxxPPPdydzF

x

x

∂∂

−=

∂∂

+−=

∑

∑ )(

Similarly:

dxdydzdydzdxyPF

yγ−

∂∂

−=∑

dzdxdyzPF

y ∂∂

−=∑

Substituting in (1) results in:

)( EquationEulerajP

ajkzPj

yPi

xP

adxdydzkdzdxdyzPjdxdydzdydzdx

yPidxdydz

xP

vvv

vvvvv

vvvv

ργ

ργ

ργ

=−∇−

=−

∂∂

+∂∂

+∂∂

−

=∂∂

−

−

∂∂

−+∂∂

−

This equation is valid for invicid fluids in motion, or fluids so moving that the shear stress is everywhere zero. This equation will be used in the end of this chapter for a fluid which is in container moving with acceleration. For fluid statics 0=av

Hence, =−∇− jPvv

γ 0 ⇒ =−

∂∂

+∂∂

+∂∂

− jkzPj

yPi

xP vvvv

γ 0

Pdydz dydzdxxPP )(

∂∂

+

dx dz

dy

x

y

z

19

⇒ 0,,0 =∂∂

−=∂∂

=∂∂

zP

yP

xP

γ

00 =∂∂

=∂∂

zPand

xP states that two points at the same elevation in the same

continuous mass of fluid at rest have the same pressure. Using Differential Calculus:

dydP

dzzPdy

yPdx

xPdP

γ−=∂∂

+∂∂

+∂∂

=

The last equation holds for both compressible and incompressible fluids. Integrating this equation yields:

P is the absolute pressure and hγ represents the gage pressure. Absolute pressure is expressed as a difference between its value and a complete vacuum. Gage pressure is the difference between its absolute value and the local atmospheric pressure.

CyP +−= γ

•

h Depth of point from fluid free surface

y Choosing origin for the coordinates is arbitrary. For simplicity, the origin has been chosen at the free surface. y=0 , P = Patm ⇒ C= Patm y=-h ⇒

.atmPhP += γ

20

EXAMPLE Find the pressure at depth 6 m of water. Then calculate the absolute pressure. The barometer reads 760 mmHg. SHg = 13.57

.24.15996710113158826.24.10113176.0806.9100057.13

).(588266806.91000

.

.

PaPpPPahgP

gagePahp

atm

HgHgatm

=+=+=

=×××===××==

ργ

h=760mm

vacuum

21

PRESSURE VARIATION IN A COMPRESSIBLE FLUID When the fluid is a perfect gas at rest at constant temperature (isothermal conditions), the following can be introduced:

RTP

PPP

PRTP

oo

o

o

oo

ρ

ρρ

ρρ

ρ

=

=⇒=⇒

=

(1)

gdydydP ργ −=−= (2)

Substituting for ρ from Eq.(1) in Eq.(2) results in:

dyP

gP

dPPgdyP

dPy

y o

oP

Po

o

oo

∫∫ −=⇒−=ρρ

[ ] [ ]yy

o

oPP oo

yP

gLnP ρ−=

)(

)(

oo

o yyP

g

o

oo

o

o

ePP

yyP

gPPLn

−−

=

−−=

ρ

ρ

The atmosphere frequently is assumed to have a constant temperature gradient expressed by:

yTT o β+= For standard atmosphere, mCo /00651.0−=β

)( yTRP o βρ += EXAMPLE Assuming isothermal conditions to prevail in the atmosphere, compute the pressure and density at 2000 m elevation if 35 /24.1.10 mkgPaP oo == ρ at sea level.

P,y

Po,yo

datum

22

.).(4.78.7841210)2000(

10806.924.1

5

)(

5 absPaPaeP

ePP

o

oo

o yyP

g

o

===

=×

−

−−ρ

35 /972.078412

1024.1 mkgP

Po

o =×==ρ

ρ

EXAMPLE Find the gage pressure at depth 1500 m of an ocean, assume: a) Incompressible fluid, 3/10050 mN=γ b) Compressible fluid, .1007.2 9 PaK ×=

23

EXAMPLE Cross sections at A and B are equal to 0.004 m2 and 0.4 m2 respectively. WB=40 kN., S = 0.75, Find F. Neglect cylinder A weight A general procedure should be followed in working all manometer’s problems (U type problems):

1- Start at one end and write the pressure there in an appropriate unit. 2- Add to this the change in pressure from one meniscus to the next (plus

if the next meniscus is lower, minus if higher). 3- Continue until the other end of the gage and equate the expression to the

pressure at that point.

F

5 m B NF

FAreaWh

AreaF

B

Bf

A

2534.0

40000

5980675.0004.0

=

=××+

=+γ

A

24

EXAMPLE

A

h1=0.6m

h2=0.9 m water

Mercury

SHg=13.57 Find the pressure at A

.11387709.0980657.136.09806

021

PaPP

hhP

A

A

HgwA

==××−×+

=−+ γγ

25

EXAMPLE

.37.246722.0980612.0980657.131.098068.0075.0980657.1325.09806

65421

PaPPPP

PdddddP

BA

BA

BwHgoilHgwA

=−

=×−××−××+××−×−

=−−+−+ γγγγγ

ASSIGNMENTS Assignment #2 2.15, 2.16, 2.21 Assignment #3 2.28, 2.32, 2.35, 2.55, 2.63 Assignment #4 2.69, 2.78, 2.82, 2.85, 2.90 Assignment #5 2.91, 2.123, 2.133

A

B

water

Mercury

Oil water

Mercury

d1 d2 d3

d4 d5

d6 SHg =13.57 Soil = 0.8 d1 = 0.25 m d2 = 0.075 m d3 = 0.1 m d4 = 0.1 m d5 = 0.12 m d6 = 0.2 m PA – PB = ?

26

FORCES ON PLANE AREAS HORIZONTAL SURFACES

To find the line of action of the resultant:

areatheofcentroidtheofxXA

xdAX

xPdAPAXxdFXF

GA

P

AP

AP

==

=⇒=

∫

∫∫ )()(

Similarly:

areatheofcentroidtheofzZA

zdAZ

zPdAPAZzdFZF

GA

P

AP

AP

==

=⇒=

∫

∫∫ )()(

Hence, for a horizontal area subjected to static fluid pressure, the resultant passes through the centroid of the area.

z

z

F = PA

⊗

zP

xP

APdAPPdAF

PdAdF

A A∫ ∫ ===

=

x

dF

x

y

27

INCLINED SURFACES

In the figure a plane surface is indicated by trace A’B’. It is inclined oθ from the horizontal plane. The intersection of the plane of the area and the free surface is taken as the x-axis. y-axis is taken in the plane of the inclined area with origin O, as shown in the free surface. The xy plane portrays the arbitrary inclined area. The magnitude, direction, and line of action of the resultant force due to the liquid, acting on one side of the area, are sought.

y

x

y

y

x

h h

x

y

O

O

Gx

Gy

28

For an element with area dA

dAyhdAPdAdF θγγ sin=== All force components (dFs) are parallel. Therefore, the resultant force will be equal to the algebraic summation of these components, i. e., the resultant F equals to the integral of dF:

APF

AhAyydAdAydFF

G

GAAA

=

===== ∫∫∫ γθγθγθγ sinsinsin

In other words, the magnitude of force exerted on one side of a plane area submerged in a liquid is the product of the area and the pressure at its centroid. As all force elements are normal to the surface, the line of action of the resultant also is normal to the surface. Any surface may be rotated about any axis through its centroid without changing the magnitude of the resultant if the total area remains submerged in the static liquid. CENTER OF PRESSURE The line of action of the resultant force has its piercing point in the surface at a point called the pressure center, with coordinates (xP, yP). Unlike that of the horizontal surface, the center of pressure of an inclined surface is not at the centroid. To find the pressure center, the moments of the resultant FyFx PP , are equated to the moments of the distributed forces about the y and x axes, respectively; thus:

29

AyxIyy

AyyxIxx

yAy

xIAy

AyxIAy

IxAy

dAyy

xAyyxI

AyAyxyxI

AyI

xAy

dAyxx

dAyyAyyyPdAFy

dAyxAyxxPdAFx

GP

GP

GGx

PA

P

GGGxy

PA

P

AGP

AP

AGP

AP

+=

+=

+=+

==⇒=

+=+

==⇒=

=⇒=

=⇒=

∫

∫

∫∫

∫∫

22

.

.

sinsin

sinsin

θγθγ

θγθγ

When either of the centroidal axes, xx = or yy = is an axis of symmetry for the surface, yxI vanishes and the pressure center lies on xx = . Since yxI may be either positive or negative, the pressure center may lie on either side of the line xx = .

tyeccentriciAy

xIyyAy

xIyy GPGP ==−+=

Py is always below the centroid of the surface.

30

EXAMPLE The triangular gate CDE in the figure is hinged along CD and is opened by a normal force P applied at E. It holds oil, relative density 0.8, above it and is open to the atmosphere on its lower side. Neglecting the weight of the gate, find a) the magnitude of force exerted on the gate. b) the location of pressure center.

31

N

AhAPF G

8.47068

)2

23(30sin498068.0

=

××××=== γ

mxx GP 667.03/2 ===

mAy

xIyy GP

09375.4)2/23(4

125.122

4

3

=××

××

+=+=

D

⊗ G

O

D

⊕

y y

C

O

TOP VIEW

x C

3 m

2m

Center of Pressure

30o

E

SIDE VIEW

x

2.5m

32

EXAMPLE Considering 3ft length of the tank perpendicular to the paper, Calculate the water force and the location of this force on the wall AD.

A

D 30o

Water 6ft

6ft

33

ADedgethefromftx

lb

AhAPF

P

G

5.1

8424

)63()2

30sin66(4.62

=

=

××+×=== γ

ft

AyxIyyG

GP

2.15

)63)(330sin

6(

1263

330sin

6

3

=

×+

×

++=+=

A

B

C

D

O

y

30o

⊗

x

y

Water 6ft

6ft

3ft

x

34

THE PRESSURE PRISM

olumeVismdFForceResultant

dPdAdF

∫ ∀=∀=

∀==

Pr

∀

∀=

∀

∀=

∫∫∀∀

ydy

xdx PP

This indicates that the resultant passes through the centroid of the pressure prism volume.

P=γh

35

EXAMPLE

lb

ftft

ftftPPF

8424

362

)30sin66(4.6264.62

362

21

=

××+×+×

=

××+

=∀=

The influence point of the resultant pressure force is the centroid of the pressure prism volume.

ftPPPP

y

ftx

8.269

629362

36

5.1

12

12 =

×+×

××+×=

+

+=

=

γγγγ

B

C

D

O

y

30o

x

y

Water 6ft

6ft

3ft

x

11 hp γ=

22 hp γ=

36

EXAMPLE

The width of the bate AB (Normal to the gate is 2 m). The gate weight is 15 kN. The gate is hinged at B. Find h so that the gate will start to open. Integration Method

dlyydAPPdAPdAPdF 2)()( 212121 ×−=−=−= γγ y1 and y2 are the depths of the element at left and right of the gate, respectively.

o 1.5 m

2 m W

A

B

Water Water

1.5 m

h

37

25.10

25.2)5.1(2

2)5.1(2

)5.1(2)()(

)5.1(2)(22)(

25.2

0

2

5.2

0

2121

×=

−−=

−=

−==⇒=

−=−=×−=

∫∫

WhlhM

dllhldFMldFdM

dlhyydlyydF

γγ

γ

γγγγ

mh

h

683.125.1150

25.2)5.1(806.92

2

=

×=

−−××

Second Method

122575)25.2()15.1()1(49030)25.2()1(

222

111

=××+==+=××+==

γγ

APFhhAPF

G

G

)25.2)(25.125.1(52.025.125.1

)25.2(25.2

45

12/5.2225.2

45 3

1

111 ×+

++=×

+

×+

+=+=

hh

hh

AyxIyy

GGP

)25.2)(25.125.1(52.0

×+h is e1

16667.0125.3)25.2(

25.25.1

45

12/5.2225.25.1

45 3

2

222 +=

×

+×

×+

+×=+=

AyxIyy

GGP

o 1.5 m

2 m W

A

B

Water Water

1.5 m

h

P1 P2

y1

y2

l

y

38

0.16667 is e2

Hence, )25.2)(25.125.1(52.0

25.2

25.2

11 ×+−=−=

hel

ml 08333.116667.0

25.2

2 =−=

∑ =−×−⇒= 025.10 2211 lFWlFM B

mh

hh

hh

683.132

05.53405.205.2

005.505.23

00833.112257525.115000

)25.2)(25.125.1(52.0

25.2)1(49030

2

2

=×

××+±=

=−−

=×−×−

×+

−×+

It has to remember that this example can also be solved as a curved surface.

39

CURVED SURFACES When the elemental forces PdA vary in direction, as in the case of a curved surface, they must be added as vector quantities; i.e., their components in three mutually perpendicular directions are added as scalors, and then three components are added victorially.

xxx PdAPdAdFPdAdF

===

θcos

P is the pressure on dA

∫ =⇒=xA

xGxxxxx APFequalaredAanddAonPPdAF ,

Similarly, it can be proved that:

zGzz APF = But,

z

x

y

PdA

dAy

dAx

dAz h

⊗ ⊗ Gx

Gz

40

yGyy

Ayyy

APF

equalnotaredAanddAonPessuresPdAFy

≠

⇒= ∫ Pr,

∫ ∫∫ ∀===

yy Ay

Ayy dhAPdAF γγ

∀d is the volume of the small prism of height h and base dAy or the volume

of liquid vertically above the areaelement. Hence, ∀=∀= ∫ γγ dFy The vertical component of pressure force on a curved surface is equal to the weight of liquid vertically above the curved surface and extending up to the free surface. The horizontal component of pressure force exerted on a projection of the curved surface. The vertical plane of projection is normal to the direction of the component. EXAMPLE The equation of an ellipsoid of revolution submerged in water is

1944

222

=++zyx . The center of the body is located 2m below the free surface.

Find the horizontal force components acting on the curved surface that is located in the first octant. Consider the x-z plane to be horizontal and y to be positive upward.

41

N

APF xGxx

53200

)4

23)(3

242(

−=

×××−−=

−=π

πγ

N

APF zGzz

35400

)42)(

3242(

2

−=

××−−=

−=

ππ

γ

×××−×

×××−=

−×××

×−=

∀−=

8

32234

24

329806

834

4

ππ

ππγ

γ

abcbca

Fy

The influence line of Fy passes through the centroid of the volume ∀ .

2m

2m

3m

Gz Gx

Fz

Fx

42

Liquid pressure force when the liquid is below the curved surface

∀ is the volume above the curved surface and extending up to the free surface or the extension of it. Unlike the previous case yF direction is upward and the influence line passes through the centroid of ∀ . EXAMPLE A cylindrical barrier holds water as shown. The contact between cylinder and wall is smooth. Considering a 1m length of cylinder, determine a) its gravity force and b) the force exerted against the wall.

θ

PdA

dAx

Extension of free surface

h

∀=∀=

==

=

=

=

==

∫

∫∫

∫

γγ

γ

θ

dF

hdAPdAF

PdAdFAPF

PdAF

PdAdFPdAdF

y

yyy

yy

xGxx

Axx

x

x

cos

43

a) In equilibrium: ∑ = 0yF

ABy

BCDy

BCDy

ABy

FFW

FWF

−=

=+−− 0

γπγπ )4(1

4

22 −=××

−=

rrF ABy

A

B

C

D r=2m

44

γπ

γπ

)82(

122

22

+=

××

+= rrF BCD

y

NW 51032.1)43()4()82( ×=+=−−+= γπγπγπ b)

1m

2m

NAPF Gh 19600221 ==××== γγ

⊗ G

45

TENSILE STRESS IN A PIPE AND SPHERICAL SHELL A circular pipe under the action of an internal pressure is in tension around periphery. A section of pipe of unit length is considered. If one-half of the ring is taken as a free body, The horizontal component of force acts through the pressure center of the projected area and is 1×× dDP , in which P is the pressure at the center line.

∑ =⇒= TPdDFx 20

σσ

22 PdeePd =⇒=

EXAMPLE A 100 mm-ID steel pipe has a 6 mm wall thickness. For an allowable tensile stress of 70 MPa., what is the maximum pressure?

PaPPPde 66 102.4

107021.0006.0

2×=⇒

×××

=⇒=σ

46

For a spherical shell

It can be noted that spherical shells with half thickness compared to cylindrical shells can tolerate the same internal pressure.

σ

σππ

4

4

2

Pde

dedP

TPA

=

=×

=

47

EXAMPLE Calculate the force F required to hold the gate in a closed position

θ

θd

r

θcosr

θsinr

F

PdA

∫=×

=×

dMrF

essureinsideofMrF Pr

)1(sin3.1

sin

∫∫

××=×

×=×

θθ

θ

rrdPrF

rPdArF

o

)2(5295.24cos-17650.806.09806398062.1cos6.098069.0

036.06.0cos

θθ

γγγθγ

==××−×+××+

=−++×+

PP

RrP wwwoil

F

S=3.0

o

48

Substituting for P from Eq. (2) in Eq. (1) yields:

[ ]

[ ] [ ]{ }NF

F

dF

ddF

ddrF

drrF

rrdrF

604.22942/2cos62.2647cos0/2cos-17650.86.03.1

2sin62.2647cos-17650.86.03.1

cossin22

5295.24-sin17650.86.03.1

sin5295.24cos-sin17650.83.1

sin)5295.24cos-17650.8(3.1

sin3.1)5295.24cos-17650.8(

2/0

2/

0

2/0

2/

0

2/

0

2/

0

2/

0

2/

0

2

=

−−+××=

−××=

××=

=

×=×

××=×

∫

∫ ∫

∫ ∫

∫

∫

π

ππ

π π

π π

π

θπ

θθθ

θθθθθ

θθθθθ

θθθ

θθθ

BUOYANT FORCE The resultant force exerted on a body by a static fluid in which it is submerged or floating is called the buoyant force. The buoyant force always acts vertically upward. There can be no horizontal component of the resultant because the projection of the submerged body or submerged portion of the floating body on a vertical plane is always zero. The buoyant force on a submerged body is the difference between the vertical component of pressure force on its underside and the vertical component of pressure force on the upperside.

49

∀=∀−∀=∀−∀=

−=

γγ

γγ)( 12

12

)1()2( VVB FFF

∀ is the volume of fluid displaced and γ is the unit gravity force of fluid. Buoyant force BF = Weight of Displaced Fluid The same relationship holds for floating bodies. The buoyant force acts through the centroid of the displaced volume of fluid. The centroid of the displaced volume of fluid is called the center of buoyancy. EXAMPLE A piece of ore having gravity force 1.1 N when submerged in water. What is the volume in cubic centimeters and what is its relative density?

50

75.398060000408.0/5.1/

8.400000408.0

98064.04.01.15.1

33

==∀

=

==∀

∀×∀=⇒=−=

w

w

B

WS

cmm

F

γ

γ

Buoyant force of a body submerged between two fluids

2211 ∀+∀= γγBF The center of buoyancy is located at the centroid of the displaced combined fluid.

2∀

1∀ 1γ

2γ

51

STABILITY OF SUBMERGED AND FLOATING BODIES

A body has stable equilibrium when a small disturbance results in forces tending to return it to its initial position. A body has unstable equilibrium when a small disturbance results in forces tending to increase displacement from its initial position. A body has neutral equilibrium when a small disturbance results in no forces. For floating bodies

Stable Equilibrium Unstable Equilibrium Neutral Equilibrium

52

A light piece of wood with a metal mass at its bottom is stable because a small disturbance (rotation) results in a couple that tends to return it to its initial position.

The metal mass is at the top. A small disturbance results in a couple that tend increase displacement (angular displacement) from its initial position.

⊗ × B

G G ⊗ B ×

⊗ ⊗ × × B

B

G G

53

FLOATING PRISMATIC OBJECTS If G is below B, stable equilibrium will govern. If B is below G, two conditions may occur: - The metacenter M lies above G. This results in stable equilibrium. The intersection of the buoyant force vector and the centerline is called metacenter

- The matacenter M lies below G. This results in unstable equilibrium

× ⊗ ×

⊗

B’ G G

B

M

Metacenter

54

Summary G is below B ⇒ Stable Equilibrium G is above B ⇒ -Metacenter is above G ⇒ Stable Equilibrium -Metacenter is below G ⇒ Unstable Equilibrium - Metacenter is on G ⇒ Neutral Equilibrium

RELATIVE EQUILIBRIUM When a fluid is being accelerated so that no layer moves relative to an adjacent one, i.e., when the fluid moves as if it were a solid, no shear stresses occur and variation in pressure can be determined by writing the equation of motion for an appropriate free body.

× ⊗

× ⊗

B’

G G B

M

55

UNIFORM LINEAR ACCELERATION By selecting a Cartesian coordinate system with y vertical and x such that the acceleration vector a is in the xy plane, the z axis is normal to a and there is no acceleration component in that direction:

ajP vvvργ =−∇−

The pressure gradient P∇

v is then the vector sum of avρ− and j

vγ− as shown

below:

Since P∇

v is in the direction of maximum change in P (the gradient), at right

angles to P∇v

there is no change in pressure. Surfaces of constant pressure, including the free surface must therefore be normal to P∇

v.

)(,

)(

yyx

yx

yx

agayPa

xP

jaiajyPi

xP

jaiajjyPi

xP

+−=−−=∂∂

−=∂∂

+=

+

∂∂

−∂∂

−

+=−

∂∂

+∂∂

−

ρργρ

ρργ

ργ

vvvv

vvvvv

dyagdxa

dyyPdx

xPdP

yx )( +−−=∂∂

+∂∂

=

ρρ

Integrating for an incompressible fluid, yields:

56

)1()( CyagxaP yx ++−−= ρρ Equation of free surface can be obtained by setting P=0:

SurfaceFreeofSlopeag

adxdy

Cyagxa

y

x

yx

+−

=

++−−= )(0 ρρ

In Eq.(1) C can be determined by choosing the origin of coordinates at the intersection of the vertical line passing through the point we want to obtain its pressure and free surface (surface of zero pressure):

hagPCPyx

y )(00,0,0

+==⇒===

ρ

h ×

O x

y

57

EXAMPLE The tank in the figure is filled with oil, relative density 0.8, and accelerated as shown. There is a small opening in the tank at A. Determine the pressure at B and C ; and the acceleration xa required to make the pressure at B zero.

5.0tan

5.00806.9

903.4tan

=⇒

−=+

−=

+−

==

α

θy

x

aga

dxdy

.18.11.11179425.1)0806.9(10008.0)(

425.12/15.02.115.0.35.2.44.2353

3.0806.910008.)(3.09.02.19.08.15.0'

kPaPaPhagP

mhkPaPaP

oghhagPmhmAA

C

CyC

C

B

BByB

B

==

×+××=+=

=++===

×××==+==−=

=×=

ρ

ρρ

For zero pressure at B 667.0

8.12.1tan / ==⇒ α

2

'

/537.6

0806.9667.0tan

sma

aag

adxdy

x

x

y

x

=⇒

+−

=+

−==−=θ

Zero pressure surface

hB

θ α A’

58

EXAMPLE A closed box with horizontal base 6 6× units and a height of 2 units is half filled with liquid. It is given a constant linear acceleration

4/,2/ gaga yx −== . Find the pressure at A.

32tan

32

4/2/tan

=⇒

−=−

−=

+−

=

α

θgg

gag

a

y

x

Fluid Volume = 6)26(

2162

2)3(

×××=××++ xx

ggghagP

x

AyA ρρρ49)5.1

322()

4()(

5.1

=×+×−=+=

=⇒

g/2

A

x

x 3

59

UNIFORM ROTATION ABOUT A VERTICAL AXIS Rotation of a fluid, moving as a solid, about an axis is called forced-vortex motion Every particle of fluid has the same angular velocity. This motion is to be distinguished from free vortex, motion, in which each particle moves in a circular path with a speed varying inversely as the distance from the center. A liquid in a container, when rotated about a vertical axis at a constant angular velocity, moves like a solid after some time interval. No shear stress exist in the liquid, and the only acceleration that occurs is directed radially inward toward the axis of rotation.

60

CrgyP

rdrgdydrrPdy

yPdP

++−=

+−=∂∂

+∂∂

=

2

22

2

ρωρ

ρωρ

Take the origin at the vertex of the paraboloid of revolution, then:

2

00,0,022rgyP

CPyrρω

ρ +−=

=⇒===

The zero pressure surface can be obtained by setting P=0:

gryrgy oo 22

02222 ωρω

ρ =⇒+−=

PdA dAdrrPP )(

∂∂

+ dr r

gyPmaF

rrP

rdAdrdAdrrPPPdA

maF

yy

rr

ρ

ρω

ωρ

−=∂∂

⇒=

=∂∂

−=∂∂

+−

=

∑

∑

2

2 )()(

61

In forced vortex ως 2= , while in free vortex the vorticity is equal to zero.

oy

1y ×

h ghP

yygPgygyP

grggyP

rgyP

o

o

ρ

ρρρ

ωρρ

ρωρ

=

+=+=

+=

+−=

)(

2

2

1

1

22

1

22

r

y

Volume of displaced fluid =Ah =Volume of fluid in the periphery of the paraboloid = 1/2 (Volume of the cylinder ) = 1/2 (AH)

2Hh = h

H

62

EXAMPLE A liquid, relative density 1.2, is rotated at 200 rpm about a vertical axis. At one point in the fluid 1 m from the axis, the pressure is 70 kPa. What is the pressure at a point B, which is 2m higher than A and is 1.5 m from the axis.

EXAMPLE A straight tube 2m long, closed at the button and filled with water, is inclined 30o with the vertical and rotated about a vertical axis through its midpoint 6.73 rad/s. Draw the paraboloid of zero pressure, and determines the pressure at the bottom and midpoint of the tube.

r ×A

×B

2m

1.5m

1m

mymy

y

rgyP

PaP

BA

A

AAA

A

42.18242.1642.16

2/160

220010002.1

806.910002.1700002

70000

22

22

=+=⇒=

×

×

××+

×××−=

+−=

=

π

ρωρ

.44.375424

2/5.160

220010002.142.18806.910002.1 22

PaP

P

B

B

=

×

×

××+×××−=π

y

63

?,? == AB PP

yA

1 m

1 m

64

.1698030cos2806.91000.2830

)]577.030cos1([806.910002

577.0806.92

)30sin1(73.6806.92

73.62

22

22

2222

PahgPPa

rygP

m

y

rgry

CC

BB

A

=××===

−−××−=

+−=

=×

=

×==

ρ

ρωρ

ω

65

FLUID DYNAMICS SYSTEM A system refers to a definite fixed mass of material and distinguishes it from all other matter, called its surroundins.

SYSTEMS LAGRANGIAN APPROACH The Lagrangian approach is used wherein the basic equations are derived for a given mass. In this approach we follow the given mass (system) With a reference to Fig. 2 to obtain position, velocity, momentum, energy or acceleration of the car, the Lagrangian approach in which the mass is followed is used.

The relative position of various particles comprising the mass of a solid stays in the same relative position during subsequent motion. Therefore, in order to

t1 t2 to

Solid Brick Fluid Flow in apipe

66

obtain position, velocity and acceleration of all particles or points of a solid it is sufficient to determine these unknowns at one point such as the center of gravity of the solid. Solid Mechanics Lagrangian Approach System Let’s try the lagrangian approach in fluid mechanics:

Since the relative position of various particles comprising the mass of the system does not stay in the same relative position during subsequent motion, knowing the situation of one point will not help to determine other other points situations. Unless, infinite number of equations for infinite number of particles is used. This approach is not reasonable. Therefore, another approach called Eulerian approach is adopted for most analyses in fluid dynamics. CONTROL VOLUME A fixed volume in space is called a control volume. In the Eulerian approach a control volume is adopted, and equations are described to express changes in mass, momentum, and energy as the fluid passes through or by the fixed volume or point (very small control volume). The boundary of a control volume is its control surface. The size and shape of the control volume are entitrly arbitrary but frequently they are made to coincide with the solid boundaries.

System System

67

To calculate the point by point variation of the fluid variable over the whole domain, the field approach is used. This approach essentially allows the domain to consist of a large number of control volumes or systems such that when they shrink to zero a series of nonlinear partial differential equations result. These equations are so difficult that a general solution for the three-dimensional distribution of relevant variables for the type of complex geometries encountered in practice has not yet been achieved, not even on the most powerful parallel processing super computers. THE GENERAL CONTROL VOLUME CONSERVATION EQUATION Consider some general flow situation in which the velocity of a fluid is given relative to an xyz coordinate system. At time t consider a certain mass of fluid that is contained within a system, having the dotted line boundaries indicated. Also consider a control volume, fixed relative to the xyz axes, that exactly coincide with the system at time t. At t+Δt the system has moved somewhat, since each mass particle moves at the velocity associated with its location. Let N be the total amount of some property (e.g., mass, energy, or momentum) within the system at time t, and let η be the amount of this property, per unit unit mass, throughout the fluid. The time rate of increase of N for the system is now formulated in terms of the control volume:

68

tN

tN

tN

tN

tN

tNN

tNNNN

tNN

dtdN

ttC

t

ttB

t

VC

ttC

t

ttB

t

tVCttVC

t

tVCttBCVC

t

tstts

tsystem

∆−

∆+

∂∂

=

∆−

∆+

∆−

=

∆−−−

=

∆−

=

∆+

⇒∆

∆+

⇒∆

∆+

⇒∆

∆+

⇒∆

∆+

⇒∆

∆+

⇒∆

∆+

⇒∆

)(lim

)(lim

)(lim

)(lim

)()(lim

)()(lim

)()(lim

00

.

00

...

0

...

0

0

Consider a small element with mass of dm, then

∫ ∫ ∀==⇒=.. ..VC VC

ddmNdmdN ηρηη

∫ ∀

∂∂

=∂

∂

..

.

VC

VC dtt

Nηρ

69

tN ttB

t ∆∆+

⇒∆

)(lim

0= time rate of flow of N out of the control volume

= ∫∫ ∀

=∆

∀ ∆+

⇒∆SBC

ttB

t dtd

t

d

.0

)(lim ηρ

ηρ

∫∫∫

=∀

=∆

∀=

∆

∆+

⇒∆

∆+

⇒∆)(..

00.

)(lim

)(lim

areaoutflowSBCSBC

ttB

t

ttB

tAdV

dtd

t

d

tN rr

ηρηρηρ

Similarly, for inflow to the control volume:

AdV

dAV

dAdtdl

dtd

dldAd

rr.

cos

cos

cos

=

=

=∀

=∀

α

α

α

dl

αcosdA

70

∫∫∫

−=∀

=∆

∀=

∆

∆+

⇒∆

∆+

⇒∆)(inf..

00.

)(lim

)(lim

arealowSCCSCC

ttC

t

ttC

tAdV

dtd

t

d

tN rr

ηρηρηρ

∫∫

∫∫∫

+∀∂∂

=

++∀∂∂

=

SCVCsystem

arealowSCCareaoutflowSBCVCsystem

AdVdtdt

dN

AdVAdVdtdt

dN

...

)(inf.)(...

.

..

rr

rrrr

ηρηρ

ηρηρηρ

The last equation states that the time rate of increase of N within a system is just equal to the time rate of increase of the property N within the control volume (fixed relative to xyz coordinate sytem) plus the the net rate of efflux of N across the control volumeboundaries. CONSERVATION OF MASS

∫∫ +∀∂∂

=

SCVCsystem

AdVdtdt

dN

...

.rr

ηρηρ

N =Property=m , η =Property per unit mass=1 Then:

∫∫ +∀∂∂

=

SCVCsystem

AdVdtdt

dm

...

.rr

ρρ

0.0...

=+∀∂∂

⇒=

∫∫SCVCsystem

AdVdtdt

dm rrρρ (conservation of mass or continuity

equation) The continuity equation states that the time rate of change of mass in the control volume plus the net rate at which mass leaves through the control surface equals zero.

71

Consider the cylindrical tube in the figure, flow enters the tube atstation 1 and exists at station 2. No flow is permitted through the solid surface comprising the tube. The application of the conservation of mass proceeds as follows:

1- The control volume is defined to include all the fluid in the tube out to the solid wall and from station 1 to station 2. If at all possible the inlet and outlet stations should be defined or placed in regions where the stream lines (or tubes) are parallel to the boundary such that the entrance and exit fluid velocities are perpendicular to the respective areas.

2- For steady flow:

0..0.2.

2221.

111.

=+⇒= ∫∫∫SCSCSC

dAVdAVAdVrrrr

ρρρ

3- If the inlet and outlet velocity vectors are at each inlet and outlet,

perpendicular to their respective areas; 4- - 0

2.222

1.111 =+ ∫∫

SCSC

dAVdAV ρρ

5- If average velocities are used; mAVAVAVAV &==⇒=+− 222111222111 0 ρρρρ

72

m& is the mass flow rate in kg/s or slug/s. If the discharge Q (also called the volumetric flow rate, flow or discharge) is defined as VAQ = For incompressible, steady flow:

2211 AVAVQ == Go back to the general equation:

0....

=+∀∂∂

∫∫SCVC

AdVdt

rrρρ

For incompressible fluids:

0..

=∫SC

AdVrr

which states that the net volume efflux is zero. This implies that

the control volume is filled with liquid at all times:

22112211 0 AVAVAVAV =⇒=+− EXAMPLE Apply the steady continuity equation for the figure:

73

0

0.0.

222333111

....

=+−−

=⇒=+∀∂∂

∫∫∫AVAVAV

AdVAdVdt SCSCVC

ρρρ

ρρρrrrr

For incompressible fluid flow ( tcons tan=ρ ):

223311

223311 0AVAVAVAVAVAV

=++=+−−

CONTINUITY DIFFERENTIAL EQUATION

0....

=+∀∂∂

∫∫SCVC

AdVdt

rrρρ Continuity Equation

dxdydzt

dxdydzt

dt VC ∂

∂=

∂∂

=∀∂∂

∫ρ

ρρ )(..

zyxSC

mmmAdV &&&rr

++=∫.

.ρ (Summation of net effluxes in x, y and z directions)

dxdydzxudydzdx

xuuudydzmx ∂

∂=

∂∂

++−=ρρ

ρρ&

Similarly, it can be proved that:

udydzρ dydzdxxuu )(

∂∂

+ρ

ρ

dx dz

dy

x

y

z

74

dzdxdyzwm

dydzdxyvm

z

y

∂∂

=

∂∂

=

ρ

ρ

&

&

Hence;

0).(

0

0

=∇+∂∂

=∂

∂+

∂∂

+∂

∂+

∂∂

=∂

∂+

∂∂

+∂

∂+

∂∂

Vt

Or

zw

yv

xu

t

dzdxdyzwdydzdx

yvdxdydz

xudxdydz

t

rrρ

ρ

ρρρρ

ρρρρ

For steady flow 0).( =∇⇒ V

rrρ

For incompressible fluid flow 0. =∇⇒ Vrr

75

CONTINUITY DIFFERENTIAL EQUATION IN CYLINDRICAL COORDINATES

0....

=+∀∂∂

∫∫SCVC

AdVdt

rrρρ Continuity Equation

zrSC

mmmAdV &&&rr

++=∫ θρ.

. (Summation of net effluxes in r, θ and z directions)

dzrdrdt

dzddrt

dzrdrdt

drdzddrrrdtt

m

θρ

θρθ

ρ

θθρ

∂∂

=

∂∂

+∂∂

=

++

∂∂

=∂∂

2)(

2)(

2

dzdrrdrVdzdrdV

dzddrrVdzdrrd

rVdzdrdVdzrdVdzrdV

dzddrrdrrVVdzrdVm

rr

rrrrr

rrrr

θρ

θρ

θρ

θρ

θρθρθρ

θρ

ρθρ

∂∂

+=

∂∂

+∂

∂+++−=

+

∂∂

++−=

2)(

)(&

rVρ

DrVV r

r ∂∂

+ρ

ρ

θρV θ

ρρ θ

θ ∂∂

+VV

zVρ

zVV z

z ∂∂

+ρ

ρ

76

drdzdV

drdzdVVdrdzVm

θθ

ρ

θθ

ρρρ

θ

θθθθ

∂∂

=

∂

∂++−=&

drdzrdzV

ddrdrrddzzVVddrdrrdV

drddrrrddzzVVdrddrrrdVm

z

zzz

zzzz

θρ

θθ

ρρ

θθρ

θθρρ

θθρ

∂∂

=

+

∂∂

++

+−=

++

∂∂

++

++

−=

2)(

2)(

2)(

2)(

22

&

Hence;

01

0

=∂

∂+

∂∂

+∂

∂++

∂∂

=∂

∂+

∂∂

+∂

∂++

∂∂

zVV

rrV

rV

t

drdzrdzVdrdzdVdzdrrd

rVdzdrdVdzrdrd

tzrr

zrr

ρθ

ρρρρ

θρ

θθ

ρθ

ρθρθ

ρ

θ

θ

FLUID FLOW CLASSIFICATION Flow can be classified in many ways such as laminar versus turbulent; steady versus unsteady; uniform versus non-uniform. Laminar Flow

Laminar Flow

Turbulent Flow

77

In laminar flow, fluid particles move along smooth paths in laminas, or layers, with one layer gliding smoothly over an adjacent layer. Laminar flow is governed by Newton’s law of viscisity. TURBULENT FLOW In turbulent flow the fluid particles (small molar masses) move in very irregular, swirling paths, causing an exchange of momentum from portion of the fluid to another. The turbulent swirls continuously range in size from very small (say a few thousand fluid molecules) to very large (a large swirl in a river or in an atmospheric gust). The turbulent sets up greater shear stresses through out the fluid and causes more irreversibilities or less.

ityviseddythecalledisdydV cos, ηητ =

The eddy viscosity is not a property of the fluid, varies from time to time and point to point. The eddy viscosity is not constant and it is function of flow characteristics (velocity, velocity gradient, fluid density and distance from wall)

kyldydul

=

= 2ρη

STEADY FLOW Steady flow is the flow that does not change with time:

00,0 =∂∂

=∂∂

=∂∂

tPand

tQ

tV

78

UNSTEADY FLOW Unsteady flow is the flow that changes with time:

00,0 ≠∂∂

≠∂∂

≠∂∂

tPand

tQ

tV

Changes with time

Fixed

79

UNIFORM FLOW Uniform flow occurs when at every point, the velocity vector or any of the other fluid variables is identically the same for any given instant:

0=∂∂

tVr

Uniform over the cross section means that the velocity is the same at all points over the cross section: NON-UNIFORM FLOW Non-uniform flow is the flow in which the velocity vector varies from place to place at any instant.

0≠∂∂

tVr

1Vr

2Vr

Velocity is the same over cross section

1Vr

2Vr

80

EXAMPLES -Liquid flow through a long pipe at a constant rate is steady uniform flow. - Liquid flow through a long pipe at a decreasing rate is unsteady uniform flow. -Flow through an expanding tube at a constant rate is steady non-uniform flow. -Flow through an expanding tube at an increasing rate is unsteady non-uniform flow. STREAM LINE A stream line is a continuous line drawn through the fluid so that it has the direction of the velocity vector at every point. There can be no flow a cross a stream line. If a particle of a fluid which moves along a stream line has a displacement dS , its displacement components in xyz directions will be: dx, dy and dz:

dx

dy

dz

dS

A stream line

81

wdzdt

dtdzw

wdz

vdy

udx

vdydt

dtdyv

udxdt

dtdxu

dtdSV

=⇒=

==⇒=⇒=

=⇒=

=r

EXAMPLE A three-dimensional velocity distribution is given by u=-x , v=2y , w=5-z Find the equation of the stream line through (2,1,1).

yx

Cyx

Cyx

CyLnx

LnCLnyLnxLny

dyx

dxv

dyudx

22/11,2

1)()1()()(21)(

22/12/1

=

=⇒==

=⇒=⇒+=−⇒=−

⇒=

82

252

2521,2

51)

5()1()()5()(5

1

111

zy

x

zxCzx

zC

xzCLn

xLnCLnzLnxLnz

xdx

wdz

udx

−==∴

−=⇒=⇒==

−=⇒

−=⇒+−=−⇒−=

−⇒=

EULER EQUATION ALONG A STREAM LINE A prismatic control volume of very small size (a bar element) with cross sectional area dA and length of dS , is selected along a stream line as shown in the figure:

Assuming that the viscosity is zero, i.e., the flow is frictionless, the only forces acting on the element in the s-direction are end forces and the weight. Other pressure forces are acting perpendicular to s.

ss maF =∑

Z

83

sZ

agsP

dAdsdAdsagdAdsdsdAsP

dAdsagdAdsdAdssPPPdA

s

s

s

∂∂

=

=++∂∂

÷

=−

∂∂

−

=−∂∂

+−

θ

θρ

ρρθρ

ρθρ

cos

)1(0cos1

cos

cos)(

V

sV

tV

ts

sV

tV

dttsdVas ∂

∂+

∂∂

=∂∂

∂∂

+∂∂

==),(

Substiyuting in equation (1) yields:

01=

∂∂

+∂∂

+∂∂

+∂∂

sVV

tV

sZg

sP

ρ Euler Equation for Unsteady Flow

For steady flow:

01=

∂∂

++∂∂

+∂∂

sVV

sZg

sP

ρ

Now s is the only independent variable. Therefore, total differentials can replace the partials:

0

01

=++

×

=+++

VdVdPgdZ

dsdsdVV

dsdZg

dsdP

ρ

ρ

The last equation is The Euler equation for steady flow. Assumptions used in the derivation

1- The equation holds along a stream line. 2- Frictionless flow ( 0=µ ) 3- Steady flow

84

THE BERNOULLI EQUATION Integrating of Euler equation for constant density yields the Bernoulli equation:

tconsVPgZ tan2

2

=++ρ

The last equation represents energy per unit mass. Dividing the equation by g yields:

Cg

VPZ =++2

2

γ

This can be interpreted as energy per unit weight in Newton.meter per Newton or Pound.feet per pound. This equation is derived by applying the momentum equation on a small element.

⊗ 2

⊗ 1 g

VPZg

VPZ

EE

22

222

2

211

1

21

++=++

=

γγ

85

INTERPRETATION OF BERNOULLI EQUATION’S TERMS

EnergyKineticEnergyPotential

mVmgHEmg

gVHmgEmg

gVH

gVhZE

gVPZE

+=

+=

+=

+=++=

++=

2

2

22

2

21

2

22

2γ

γPZ + = Potential Energy per unit weight of fluid

gV2

2

= Kinetic Energy per unit weight of fluid

h= H

86

Units⇒

ftlb

ftlb

or

mNmN

Newtonjoul

fluidofWeightEnergy

==

===

.

.

EXAMPLE Water is flowing in an open channel as shown in the figure at a depth of 2 m and a velocity of 3 m/s . It then flows down a chute into another channel where the depth is 1 m and the velocity is 10 m/s. Assuming frictionless flow, determine the elevation of the channel floors. The velocities are assumed to be uniform over the cross sections, and the pressures hydrostatic.

The points 1 and 2 may be selected on the free surface, as shown, or they could be selected at other depths. If the difference in elevation of floors is y, Bernouli’s equation is:

gVPZ

gVPZ

EE

22

222

2

211

1

21

++=++

=

γγ

87

my

y

64.3

806.921010

806.9232

22

=

×++=

×++

EXAMPLE a) Determine the velocity of efflux from the nozzle in the wall of the reservoir of the figure b) find the discharge through the nozzle.

The jet issues as a cylinder with atmospheric pressure uniformly distributed a cross the outlet and around its periphery. The energy equation without losses

88

is applied between a point on the water surface and a point downstream from the nozzle:

gVPZ

gVPZ

EE

22

222

2

211

1

21

++=++

=

γγ

gV2

21 is negligible compared to

gV2

22

sLsmAVQ

smghVg

Vh

/70/07.041.086.8

/86.84806.9222

0000

32

22

2

22

==×==

=××==⇒++=++∴

π

If losses are to be considered, then :

ghCV 2= C is called the orifice coefficient.

1<C EXAMPLE Find the discharge of the previous example by integration assuming non-uniform outflow.

4 m h

dh x .

sm

dhhghQ

dhhRgh

xdhghVdAdQ

hRx

hxR

/0695.0

)4(05.022

)4(22

22

)4(

)4(

3

05.4

95.3

22

22

22

222

=

−−××=

−−××=

×==

−−=

−+=

∫

89

EXAMPLE Calculate the time required to lower the water surface in the tank from elevation 4 m to 2 m. The tank cross sectional area is 50 m2.

y

dy 1

2

Let’s assume that the water surface level at time t is y. Then:

gyVEE 2221 =⇒= Continuity 2211 AVAV =⇒

[ ]4

22

2

42

2

21

2)05.0(50

22)05.0(50

2)05.0(50

)05.0(2

yg

ydy

gt

ygdydt

gyAdtdy

××−

=

××−

=

××−=

××=−

∫

π

π

π

π

90

EXAMPLE

A hydroelectric plant as shown in the figure has a difference in elevation from headwater to tailwater of H=50 m and a flow Q=5 m3/s of water through the turbine. The turbine shaft rotates at 180 rpm and the torque in the shaft is measured to be mNT .1016.1 5×= . The output of the generator is 2100 kW. Determine:

a) the reversible power for the system b) losses in the system c) losses and efficiency in the turbine d) losses and efficiency in the generator

Solution a) mHEHE TT 5021 =⇒=− For perfect conversion, the reversible power is :

kWNmNQHT 5.2451/.24515005059806 ==××=γ b) Lost power in the system is the difference between the power into and out of the system, or:

91

2451.5-2100=351.5 kW c) The rate of work by the turbine is the product of the shaft torque and the rotational speed:

kWT 5.218660

21801016.1 5 =×

×=π

ω

Lost power through the turbine = 2451.5-2186.5 = 265 kW Head loss through the turbine = lost power per unit fluid weight m

QPower 4.5

5806.9265

=×

==γ

The generator power loss = 2186.5-2100 = 86.5 kW The generator head loss = m

QPower 76.1

5806.95.86

=×

==γ

Efficiency of the turbine = %19.89100

504.550

=×−

Efficiency of the generator = %05.96100

4.55076.14.550

=×−

−−

KINETIC ENERGY CORRECTION FACTOR α Often the flow entering or leaving a port is not strictly one-dimensional. In particular, the velocity may vary over the cross section. In this case the kinetic energy term for a given port should be modified by a correction factor α : Kinetic energy per unit time by integration = α × Kinetic energy per unit time using average velocity

dAVv

A

gVVA

gvdAv

A

A3

22

1

22

∫

∫

=

=

α

αγγ

Find α for the velocity distribution of flow in a pipe, given below:

92

−= 2

2

max 1RrVv Ans. 2=α

EXAMPLE In the figure, a pump with a water horsepower (WHP) rating of 10 hp draws water from the reservoir as indicated and delivers water to an outlet 15 ft higher than the reservoir surface for crop irrigation. What is the outlet discharge? Total system losses from the pump to the outlet are parametrized

as g

V2

82

, but there are no losses from the reservoir inlet to the pump. The

delivery pipe diameter is 4.67 in.

93

sftVAQsftVg

VVg

V

gV

QgV

gVPZHh

gVPZ

EHhE

Pf

Pf

/83.1/4.1524/)12/67.4(4.62

550102

8

201555010

28000

22

3

2

2

2

22

222

221

211

1

2211

==⇒=

=××

×+−

++=×

+−++

++=+−++

=+−

−

−

π

γ

γγ

EXAMPLE A venturimeter, consisting of a converging portion followed by a throat portion of constant diameter and then a gradually diverging portion, is used to determine rate of flow in a pipe. The diameter at section 1 is 0.15 m and at section 2 it is 0.1 m. Find the discharge through the pipe when

.9.0.,200021 ==− SPaPP

94

21

12

2

2

2

1

22

2

21

1

2211

)1(25.24

1.04

15.044

EE

VVVV

DVDV

AVAVQ

=

=⇒×

×=×

×

×=×

==

ππ

ππ

sLsmAVQ

smVVV

VV

gVVPP

gVPZ

gVPZ

/4.58/0584.0415.031.3

/31.3806.92

)25.2(98069.0

20000)2(),1(

)2(806.9298069.0

20000

20

22

32

11

1

21

21

21

22

21

2221

222

2

211

1

==××==

=⇒×

−=

×⇒

×−

=×

−=

−+

++=++

π

γ

γγ

Assignments Chapter1 1.18, 1.20 plus class assignment Chapter 2 2.3, 2.5, 2.15, 2.17, 2.35, 2.50, 2.51, 2.52, 2.69, 2.78, 2.82, 2.91, 2.94, 2.100, 2.124, 2.133 Chapter 3 3.19, 3.29, 3 .34, 3.36, 3.46, 3.61, 3.62, 3.73 Chapter 4 4.13, 4.16, 4.19

95

THE CONTROL VOLUME LINEAR-MOMENTUM EQUATION

∫∫

∫∫

+∀∂∂

=

====

+∀∂∂

=

SCVCsystem

SCVCsystem

AdVVdVtdt

Vdm

VmVm

mNVmN

AdVdtdt

dN

...

...

.

,

.

rrrrr

rr

r

rr

ρρ

η

ηρηρ

∫∫∑ +∀∂∂

=SCVC

AdVVdVt

F...

.rrrrr

ρρ

∫∫∑

∫∫∑

+∀∂∂

=

+∀∂∂

=

SCy

VCyy

SCx

VCxx

AdVVdVt

F

AdVVdVt

F

...

...

.

.

rr

rr

ρρ

ρρ

96

EXAMPLE

Water flows in the pipe D1=25 cm D2=15 cm Q=50 L/s P1=8.5 kPa P2=5.83 kPa W=2.0 N (total fluid weight in the pipe) Find the horizontal and vertical forces required to hold the pipe in place.

∑∑∫∫∫∑

∑∑∫∫∫∑

===+∀∂∂

=

===+∀∂∂

=

QVAdVVAdVVAdVVdVt

F

QVAdVVAdVVAdVVdVt

F

yySC

ySC

yVC

yy

xxSC

xSC

xVC

xx

ρρρρρ

ρρρρρ

rrrrrr

rrrrrr

...

...

....

....

∑∑ = QVF xx ρ

NR

R

QVQVRAPAP

x

x

x

4.801

)05.0(30cos4/15.0

050.01000

)05.0(45cos4/25.0

050.0100030cos415.0583045cos

425.08500

)(cos)(coscoscos

2

2

22

2211222111

=

+××

×

+−××

×=+××−××

++−=+−

π

πππ

θρθρθθ

Rx

Ry

97

∑∑ = QVF yy ρ

NR

R

QVQVWRAPAP

y

y

y

2.963

)05.0(30sin4/15.0

050.01000

)05.0(45sin4/25.0

050.01000230sin415.0583045sin

425.08500

)(sin)(sinsinsin

2

2

22

2211222111

=

+××

×

+−××

×=−+××−××

++−=−+−

π

πππ

θρθρθθ

EXAMPLE Find the reaction force exerted on a fixed vane when a jet discharging 60 L/s of water at 50 m/s is deflected through 135o. The frictional resistance between jet and vane is neglected. The velocity is assumed to be uniform throughout the jet upstream and downstream from the vane, that is, the control volume. When the small change in elevation between ends, if any, is neglected, application of the energy equation shows that the magnitude of the velocity is unchanged for fixed vanes.

Ry

Rx

uV −0

uV −0

98

kNN

QVQVRQVF

x

xx

121.55121)06.0(135cos501000)06.0(501000

)(135cos)( 00

==+×××+−××=

+×+−=−

= ∑∑ρρ

ρ

kNN

QVRQVF

y

yy

121.22121)06.0(135sin501000

)(135sin0

==+×××=

+×=

= ∑∑ρ

ρ

EXAMPLE Fluid issues from a long slot and strikes against a smooth inclined flat plate. Determine the division of flow and the force exerted on the plate, neglecting losses due to impact.

As there are no changes in the elevation or pressure before and after impact, the magnitude of the velocity leaving is the same as the initial speed of the jet. The plate surface is assumed to be smooth. Therefore, there is no shear force exerted on the surface.

R

99

)2(

)1(cos0))(()()(cos0

210

210

201000

QQQContinuity

QQQQVQVQV

QVF xx

+=⇒

−+−=

+−+++−=

= ∑∑

θ

ρρθρ

ρ

Equations (1) and (2) yields: )cos1(2

01 θ+=

QQ , )cos1(2

02 θ−=

QQ

The force R exerted on the plate must be normal to it.

00

00

sin

))(sin(

QVQVR

QVF yy

θρ

θρ

ρ

=

−−=

= ∑∑

100

101

MOVING VANES To analyse the problem, it is reduced to steady state by superposition of vane velocity u to the left. The control volume relationship has been proved with respect to fixed coordinates. Therefore, in order to be able to use the same relationship, the coordinate system has to be moved with the vane velocity u to the right. By this way the control volume will have a stationary position with respect to the coordinate system. EXAMPLE The vane in the figure is moving with a velocity u= 30 ft/s. Determine the force components due to the water jet and the rate of work done on the vane.

Vo Vo-u

Vo-u Vo

θ

u

θ

Rx

Ry

x

y

z x

y

z

u

102

sftu

sftVftA

/3060

/10004.0

0

0

20

==

==

θ

lbR

AuV

AuVuVAuVuVRQVF

x

x

xx

569

)5.01(04.0)30100(17.324.62)cos1()(

])(][cos)([])()[(

20

20

000000

=⇒

+××−−=+−−=

−+−−+−−−=−

= ∑∑

θρ

θρρ

ρ

lb

AuV

AuVuVRQVF

y

yy

329

60sin04.0)30100(17.324.62sin)(

])][(sin)[(

20

20

000

=

×−=−=

−−=

= ∑∑

θρ

θρ

ρ

The power exerted on the vane is sftlbuRx /.1707056930 =×=

103

THE MOMEN OF MOMENTUM EQUATION The general unsteady linear momentum equation applied to a control volume:

∫∫∑ +∀∂∂

=SCVC

AdVVdVt

F...

.rrrrr

ρρ

Taking the cross, or vector, product of the equation and the position vector r of a point on the line of action of the vector from any point O, yields:

∫∫∑ ×+∀×∂∂

=×SCVC

AdVVrdVrt

Fr...

.rrrvrvrv ρρ

Moment of the resultant force = Summation of moments of components; Hence:

∫∫∑ ×+∀×∂∂

=×SCVC

AdVVrdVrt

Fr...

.rrrvrvrv ρρ

For steady flow and uniform average flow on the cross section:

∑∑∑∑∑

×=

×=×=×

QVrM

QVrAVVrFrrv

rvrrrvrv

ρ

ρρ ).(

104

105

lbftT

T

AVhVAVhVhAPhAPTQVrM

A

A

A

.455

)4

)12/3(40)(12/1040(94.1)4

)12/3(40)(12/240(94.1

1210

4)12/3(14480

122

4)12/3(144100

))(())((

22

2222221111222111

=⇒

×+××+×−××

=××××−××××+

++−=−+

×= ∑∑

ππ

ρρ

ρrv

3.79 In Fig. 3.66, a jet, 3/2 ftslug=ρ , is deflected by a vane through 1800. Assume that the cart is frictionless and free to move in a horizontal direction. The cart weighs 200 lb. Determine the velocity and the distance traveled by the cart 10 s after the jet is directed against the vane A0 = 0.02 ft2 and V0 = 100 ft/s.

V0

106

2100

010001010

)(2

])()][([])()[(

VVAFAVVuVAVVVVF

QVF

x

x

xx

−=⇒

−+−−+−−−=−

= ∑∑

ρ

ρρ

ρ

Fx is variable. It varies with respect to V1. For the cart:

21

1

121

2100

)100(71.77

17.32200)100(02.022

)(2

VdVdt

dtdVV

dtdVmVVAmaF xx

−=

=−××

=−⇒= ρ

ftxsttLntxCxt

CtLntxtdt

dxV

sftVstt

VV

t

CVtCV

t

796103.19)7771.0(71.771003.190,0

)7771.0(71.771007771.071.77100

/79.92107771.071.771007771.0

10071.77

7771.00,0,100

71.77

2

21

1

11

1111

=⇒=−+−=⇒−=⇒==

++−=⇒+

−==

=⇒=+

−=⇒−−

=

−=⇒==+−

=

V0-V1

V0-V1

Fx

107

DIMENSIONAL ANALYSIS AND DYNAMIC SIMILITUDE Solving practical design problems in fluid mechanics usually requires both theoretical developments and experimental results. By grouping significant quantities into dimensionless parameters, it is possible to reduce the number of variables and make this compact result (equations or data plots) applicable to all similar situations. Dimensions Quantity Symbol Dimension Length l L Time t T Mass m M Force F F or MLT-2 Velocity V LT-1 Acceleration a LT-2 Area A L2 Discharge Q L3/s Pressure P FL-2 or ML-1T-2 Density ρ ML-3 Specific weight γ FL-3 or ML-2T-2 Dynamic viscosity µ ML-1T-1 Kinematic viscosity ν L2T-1 EXAMPLE Obtain an equation for distance traveled by a body in a free fall after time t. Assume distance =S Gravity acceleration = g

),( tgfS = According to dimensional homogeneity:

batKgS = Where K is a dimensionless coefficient.

108

202,1)( 22

=⇒=+−=⇒=⇒= +−−

bbaaTLLTLTL baaba

2KgtS =∴

K can be determined by experiment. EXAMPLE Determine the power P of a pump as a function of discharge Q, total dynamic head H and fluid specific weight γ .

HQKPccbaab

TLFLFLTLFLT

HKQP

acbab

cba

cba

γ

γ

==⇒=+−==⇒

=

=

=

−+−

−−−

1133,1,1

)()(33

3131

In S.I. units K=1 which can be determined by experiment. EXAMPLE Find a relationship for the drag force FD exerted by a fluid on a sphere in terms of flow velocity V, sphere diameter D, fluid viscosity µ , and fluid density ρ . Method 1 Obtain the relationship for FD versus D by keeping other variables as constants. 10 points may be appropriate for obtaining a curve. These curves should be plotted for various cases including combinations of 10 µ , 10 V and 10 ρ . Similar curves for FD versus other variables can be obtained. Hence, number of required experiments will be: 10×10×10×10=104 and number of

curves should be 103. If it is assumed to spend half an hour for every experiment and work 8 hours daily, then it takes 2.5 years to prepare the data of the mentioned 1000 curves.

109

Method 2

dcbaD VDF µρ=

To obtain the unknowns a, b, c, and d four equations are required, while only three equations can be written. Method 3 Grouping significant variables into dimensionless parameters by using the pai theorem (Buckingham П theorem) as explained in the following:

1- Select repeating variables. Repeating variables should have the following characteristics:

• Number of them should be equal to number of dimensions. • They should be significant. • They should not be dependant variable. • They should not have similar dimensions. • They should altogether include all dimensions in the problem.

⇒ According to the above-mentioned characteristics DandV ,,ρ are selected as repeating variables.

2- µρπ 1111

zyx DV= In which 1π is dimensionless.

µρ

πρ

µπ

VDOrVD

zyxTLMTMLLLTMLTLM yzyxxzyx

==

−=−=−=⇒==∴ −−−++−+−−−−

11

1111113111111113000

11,11,11)()()(

A dimensionless parameter can be inverted if desired. Similarly:

Dzyx FDV 111

1 ρπ =

222

221222312222123000

22,22,12)()()(

DVF

zyxTLMMLTLLTMLTLM

D

yzyxxzyx

ρπ =

−=−=−=⇒==∴ −−+++−+−−−

110

2/

2/4

(Re)8(Re)

(Re))(),(

0),(

2

22

1

221

112222

21

AVCF

VDfF

DVfF

fVDfDV

FDV

FVDf

f

DD

D

D

DD

ρ

ρππ

ρ

µρ

ρρµρ

ππ

=

=

=

==⇒

=

Where CD is a function of Re..

EXAMPLE The losses hf in turbulent flow through a smooth pipe depend on pipe length L, flow velocity V, diameter D, dynamic viscosity µ , g, and density ρ . Use dimensional analysis to determine the general form of the equation.

0),,,,,/(0),,,,,,(

=

=

gDVLhFgDVLhF

f

f

µρ

µρ

hf/L is a dimensionless parameter, say Lh f /1 =π . Select repeating variables among other variables. The repeating variables will be similar to those of the previous example. Repeating variables are DV ,,ρ

µρ

πµρπVDDV zyx =⇒=∴ 2

2222

gDV zyx 333

3 ρπ =

gDV

VgD

zyxTLMLTLLTMLTLM yzyxxzyx

2

323

23133333233133000

13,23,03)()()(

=⇒=

=−==⇒==∴ −−+++−−−−

ππ

111

gV

DLfh

gV

DLfh

gDVf

Lh

gDVVDf

Lh

gDVVD

Lh

F

f

f

f

f

f

2

2(Re)2

(Re

,

0,,

2

2

2

2

2

2

1

2

=

=

=

=

=

∴

µρ

µρ

The last equation is called the Darcy-Weisbach equation. For rough pipes with absolute roughness ε , another dimensionless parameter

Dε

π =4 will be added . Hence, the equation will be as follows:

gV

DL

Dfh f 2

Re,2

=

ε

Second Method -Select Repeating Variables:

LDLTVML === −− ,, 13ρ -Obtain dimensions in terms of repeating variables:

33,, DLM

VD

VLTDL ρρ =====

-Obtain non-repeating variables in terms of repeating variables:

µρ

πρρµVDVDVDDDTML =⇒=== −−−−

111311 )/)()((

gDVDVVDDLTg

2

2222 /)/( =⇒=== −− π

112

DlDLl =⇒== 3π

Dh

DLh ff =⇒== 4π

DDL ε

πε =⇒== 5

lh

DlDh ff ===

//

3

46 π

ππ

gV

Dl

Dfh

gV

Dl

Dfh

gDV

DVDf

lh

Dlh

gDVVDF

F

f

f

f

f

2)(Re,

2)(Re,2

),(

0),,,(

0),,,(

2

2

1

2

1

2

5621

ε

ε

εµ

ρ

εµ

ρ

ππππ

=

=⇒

=

=

=

DIMENSIONLESS PARAMETRS Reynold’s Number

ForceViscousForceInertial

Ama

LLV

TLL

LLV

LTL

LLV

LVVLVLVDVD

=====×==τµ

ρ

µ

ρ

µ

ρµ

ρµ

ρ2

23

2

22

2

2

22

Re

Froude Number

ForceGravityForceInertial

gLLV

LL

gLVFr

gLVFr ==×=⇒= 3

22

2

222

ρρ

ρρ

Mach Number

ForceElasticForceInertial

CLLV

LL

CVMa

CVMa ==×=⇒= 22

22

2

2

2

22

ρρ

ρρ

113

MODEL STUDIES AND SIMILITUDE Model studies of proposed hydraulic structures and machines are frequently undertaken as an aid to the design. Geometric Similitude The ratios of all corresponding dimensions must be the same.

rprototype

el LLL

=mod =constant

Dynamic Similitude The ratios of the various types of forces must be the same at corresponding points.

.mod constFF

prototype

el =

pviscous

pinertial

p

pm

pviscous

mviscous

pinertial

minertial

pviscous

pinertial

mviscous

minertial

p

m

mviscous

minertial

m

FF

FF

FF

FFFF

FF

=

=⇒===⇒

=

Re

ReRe1ReRe

Re

Similarly, it can be proved that pmpm MaMaFrFr == , and all dimensionless parameters should be equal when a dynamic similitude exists. In order to achieve exact similitude between model and prototype, the scale ratio should be 1:1(Strict fulfillment of these requirements (equality of the dimensionless parameters) is generally impossible to achieve unless the scale ratio is 1:1

114

Therefore, the dominant forces and consequently the corresponding dimensionless number is considered to be equal for model and prototype. Pipe Flow⇒ Rem =Rep Hydraulic structures ⇒ Frm=Frp Very high velocities of compressible flow, i.e., sonic and super-sonic velocities ⇒ Mam=Map

λ==⇒

=⇒=

m

p

m

p

p

p

m

mpm

LL

VV

gLV

gLVFrFr

2222

λ

λλ

=

×=×=⇒==

m

p

p

m

m

p

m

p

p

pp

m

mm

tt

VV

LL

tt

VL

tVLt 1,

( )

2/52

2/533

3

3

//

//

λλλ

λλ

λ

=×==

====

mm

pp

m

p

mp

mp

mm

pp

m

p

AVAV

QQ

ttLL

tLtL

QQ

Force ratio, for example, on gates:

3λγ

γ==

mm

pp

m

p

AhAh

FF

Assignment#5 5.6.1, 5.6.2, 5.2

115

VISCOUS FLOW This chapter deals with real fluids, that is, situations in which irreversibilities (losses) are important, viscosity is the fluid property that causes shear stresses in a moving fluid, viscosity is also one mean by which losses are developed. In turbulent flows random fluid motions, superposed on the average; create apparent shear stresses that are more important than those due to viscous shear. Turbulent flow, however, has very erratic motion of fluid particles, with a violent transverse interchange of momentum. We have used continuity and Euler equations which was integrated to Bernoulli equation to solve problems without viscous effects. In order to solve problems with viscous effects continuity and Navier-Stokes equations should be used. Navier-Stokes Equations

x-component ∑ = xx maF

xViscousx

essurex

Gravityx adFFF ∀=++ ρPr

116

zuw

yuv

xuu

tu

dtdz

zu

dtdy

yu

dtdx

xu

tu

dtduax

∂∂

+∂∂

+∂∂

+∂∂

=

∂∂

+∂∂

+∂∂

+∂∂

==

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

+

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂

+∂∂

=∂

∂+

∂

∂+

∂∂

−

2

2

2

2

2

2

2

2

2

2

2

2

1

1

zu

yu

xu

xp

zuw

yuv

xuu

tu

gzu

yu

xu

xp

zuw

yuv

xuu

tu

zuw

yuv

xuu

tu

zyxpg

x

xzxyx

νρ

νρ

ρττ

ρ

Or

uxp

zuw

yuv

xuu

tu 21

∇+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

νρ

Similarly: y-component

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

+

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

2

2

2

2

2

2

2

2

1

1

zv

yv

xv

yp

zvw

yvv

xvu

tv

gzv

yv

xv

yp

zvw

yvv

xvu

tv

y

νρ

νρ

vyp

zvw

yvv

xvu

tv 21

∇+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

νρ

z-component

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

+

∂∂

+∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

2

2

2

2

2

2

2

2

2

2

2

2

1

1

zw

yw

xw

zp

zww

ywv

xwu

tw

gzw

yw

xw

zp

zww

ywv

xwu

tw

z

νρ

νρ

wzp

zww

ywv

xwu

tw 21

∇+∂∂

−=∂∂

+∂∂

+∂∂

+∂∂

νρ

117

Two Dimensional Navier Stokes equations

y

x

gy

vxv

yp

yvv

xvu

tv

gyu

xu

xp

yuv

xuu

tu

+

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

+

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

2

2

2

2

2

2

2

2

1

1

νρ

νρ

Or

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

2

2

2

2

2

2

2

2

1

1

yv

xv

yp

yvv

xvu

tv

yu

xu

xp

yuv

xuu

tu

νρ

νρ

EXAM PLE Derive an expression for the flow past a fixed cross section of the figure for laminar flow between the two moving plates. Find the shear stress at each plate.

x

y

118

)(0

)0(0

0

)0(0

)(0

)(0

1

2

2

2

2

2

2

directionxtheincomponentnohasggxu

xu

xp

vyuv

continuityxu

FlowSteadytu

gyu

xu

xp

yuv

xuu

tu

x

x

−=

=∂∂

=∂∂

=∂∂

==∂∂

=∂∂

=∂∂

+

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

νρ

12

2

2

2

00 CCyuCdydu

dyud

yu

+=⇒=⇒=⇒=∂∂

Boundary Conditions y=0, u=-V 1CV =−⇒

y=a, u=Ua

VUVCaU +⇒−=⇒

Vya

VUu −+

=

aVU

yu +

=∂∂

= µµτ

EXAM PLE In the figure one plate moves relative to the other as shown; sPa.08.0=µ and

3/850 mkg=ρ . Determine the velocity distribution, the discharge, and the shear stress exerted on the upper plate.

y

119

xp

yu

yu

xp

yu

xu

xp

yuv

xuu

tu

∂∂

=∂∂

∂∂

+∂∂

−=

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

µ

νρ

νρ

1

10

1

2

2

2

2

2

2

2

2

At point 1 pa

hPP264053806.98501400 =×××=

+= γ

At point 2 pahPP

800=+= γ

1m/s

P1=1.4kpa

P2=0.8kpa

3m

3m

x 6 mm

120

212

1

2

2

2

2

3771808.0

603508.0

6035603508.01

/603523

26405800

CyCyu

Cydydu

dyud

yu

mpadxPd

++−=

+−=

−=⇒−×=∂∂

∴

−=−

=

Boundary Conditions

yyuCsmuy

Cuy

646.5937718

646.59/1,006.000,0

21

2

+−=∴

=⇒−===⇒==

smdyyyudyQudydQ /00164.0)646.5937718(1 3006.0

0

2006.0

0

−=+−==⇒×= ∫∫

pay

yu

y 44.31)646.59377182(08.0 006.0 −=+−××=∂∂

= =µτ

EXAMPLE For the fluid flowing downward on the inclined surface, find u , Q and the shear stress at the bed. Fluid viscosity = µ , fluid specific gravity = γ.

θ

b

y

x

121

1

2

2

2

2

2

2

2

2

sin

sinsin0

1

Cygdydu

gdy

udgyu

gyu

xu

xp

yuv

xuu

tu

x

+−

=⇒

−=⇒+

∂∂

=

+

∂∂

+∂∂

+∂∂

−=∂∂

+∂∂

+∂∂

µθρ

µθρ

θν

νρ

Boundary Conditions

22

1

2sin

sin00,0

Cyu

ygdydu

Cuy

+−

=

−=

=⇒==

µθγµ

θρ

2

2 2sin0, bCuby

µθγ

=⇒==

22

2sin

2sin byu

µθγ

µθγ

+−

=

)(2sin 22 ybu −=

µθγ

µθγ

µθγ

3sin)(

2sin 3

0

22

0

bdyybudyQbb

=−== ∫∫

bg

ygygyu

bed θρτ

θρµ

θρµµτ

sin

sinsin

−=

−=−

=∂∂

=

122

LAMINAR FLOW THROUGH CIRCULAR PIPES Cylindrical Coordinates

z-Component

∂∂

+∂∂

+

∂∂

∂∂

+∂∂

−=

∂

∂+

∂∂

+∂

∂+

∂∂

2

2

2

2

2

110zVV

rrVr

rrzP

zVVV

rV

rVV

tV zzzz

zzz

rz

θµ

θρ θ

123

1

2

2

11

10

Cdr

dVrdzPdr

drdVr

drd

dzPdr

drdVr

drd

rdzPd

rVr

rrzP

z

z

z

z

+=

=

=

∂∂

∂∂

+∂∂

−=

µ

µ

µ

µ

At r=0 , Vz =Vmax ,i.e. , 00 1 =⇒= Cdr

dVz

2

2

2

4

22

CdzPdrV

dzPdr

drdV

drdVr

dzPdr

z

zz

+=

=⇒=

µ

µµ

EquationPoiseuilleHagenRr

dzPdRV

dzPdRCC

dzPdRVRr

z

z

−

−−=∴

−=⇒+=⇒==

2

22

2

22

2

14

4400,

µ

µµ

−=⇒−= 2

2

max

2

max 14 R

rVVdzPdRV zµ

Average Velocity V

)1(82

2122

max

02

2

max0 dz

PdRVrdrRrVrdrVdAVV

RR

zz µππ −==×

−=×== ∫∫∫

Darcy-Weisbach Equation

)2()/

2()/

2(

)/()/()/()(

2

2

1

2

12

Lh

L

ZPg

VZPg

V

dzPd

LZPZPZP

dzdZP

dzd

dzPd

fγγγ

γ

γγγγγγ

−=++−++

−=

+−+=+=+=

Substituting for dzPd from equation (2) in equation (1) yields:

124

)3(32888 22

22

DV

RVh

LhR

dzPdRV f

f

γµ

γµ

γµµ

==⇒−×−=−=

It should be noted that in laminar flow Vh f ∝ On the other hand the Darcy-Weisbach equation is as follows:

)4(2

2

gV

DLfh f =

Comparing equations (3) and (4), it can be concluded that :

Re64

=f

EXAMPLE Find the discharge in the pipe shown below.

gV

gV

gVPZE

gV

gV

gVPZE

25.37

280003000000

2

230

280002000005

2222

2222

22211

11

+=++=++=

+=++=++=

γ

γ

E2 >E1 Hence, flow direction should be from 2 to 1.

125

mg

Vg

VEEh f 5.7)2

30()2

5.37(22

12 =+−+=−=

gV

DLfh f 2

5.72

==

Assuming laminar flow (which is the opposite in practice);

gV

V

gV

DL

VDgV

DLfh f

201.010

04.0

01.0806.9

8000645.7

2/64

25.7

2

22

××

××=

===µρ

V= 0.47 m/s 6.9504.0

01.047.0806.9

8000

Re =××

==⇒µ

ρVD

Hence, the assumption of laminar flow is correct. REYNOLD’S EXPERIMENT

Glass tube

dye

126

Laminar flow is defined as flow in which the fluid moves in layers, or laminas, one layer gliding smoothly over an adjacent layer with only a molecular interchange of momentum. Any tendencies toward instability and turbulence are damped out by viscous shear forces that resist relative motion of adjacent fluid layers. Turbulent flow, however, has very erratic motion of fluid particles, with a violent transverse interchange of momentum. The nature of the flow, that is, whether laminar or turbulent, and its relative position along a scale indicating the relative importance of turbulent to laminar tendencies are indicated by the Reynolds number. For small flows, the dye stream moved as a straight line through the tube, showing that the flow was laminar. As the flow rate increased, the Reynolds number increased, since D, ρ and μ were constant and V was directly proportional to the rate of flow. With increasing discharge a condition was reached at which the dye stream wavered and then suddenly, broke up and was diffused or dispersed throughout the tube. The flow had changed to turbulent flow with its violent interchange of momentum that had completely disrupted the orderly movement of laminar flow. By careful manipulation Reynolds was able to obtain a value Re=12000 before turbulence set in. A later investigator, using Reynolds’ original equipment, obtained a value of 40000 by allowing the

127

water to stand in the tank for several days before the experiment and by taking precautions to avoid vibrations of the water or equipment. These numbers, referred to as Reynolds upper critical numbers, (lately, Re=1000000 was obtained before turbulence at Brown university) have no practical significance in that an ordinary pipe installation has irregularities that cause turbulent flow at a much smaller value of the Reynolds number. Starting with turbulent flow in the glass tube, Reynolds found that it always becomes laminar when the velocity is reduced to make Reynolds’ number less than 2000. This is the Reynolds lower critical number for pipe flow and is of practical importance. With the usual piping installation, the flow will change from laminar to turbulent in the range of Reynolds’ numbers from 2000 to 4000. For the purpose of this treatment it is assumed that the change occurs at Re=2000. In laminar flow the losses are directly proportional to the average velocity, while in turbulent flow the losses are proportional to the velocity to a power varying from 1.7 to 2.

Starting with low Reynolds’ number (laminar flow), the upper critical value was 12000, 40000, and 1000000. Starting with high Reynolds’ number (turbulent flow), the lower critical value was always 2000. This can be interpreted due to stability principles. Stable flow is the flow in which the disturbance energy is damped. Laminar flow with Reynolds’ number less than 2000 is stable, i.e., disturbance energy damps.

106 105 12000 10000 5000 2000 1000 100

106 105 12000 10000 5000 2000 1000 100

128

Laminar flow with Reynolds’ number greater than 2000 is unstable, i.e., disturbance energy does not damp and changes the laminar flow to turbulent. At Brown university, disturbance and vibrations were avoided by using more fixed equipments. MOODY DIAGRAM

Laminar Zone (Re =0-2000)

Re64

=f which plots a straight line with slope equals to -1 on a log-log chart.

Head-loss hf is independent of roughness. Critical Zone (Re=2000-4000) The flow may be either laminar or turbulent.

129

Transition Zone The zone in which hf is a function of Re and

Dε . Colebrook formula in this

zone holds:

+−=

fDLn

f Re51.2

7.3/86.01 ε

Complete Turbulent Zone In this zone all curves are almost horizontal which means that hf is a function of

Dε only and is not a function of Re.

It should be noted that the relative roughness curves 001.0<

Dε approach the

smooth pipe curve for decreasing Reynolds’ numbers. This can be explained by the presence of a laminar film at the wall of the pipe that decreases in thickness as the Reynolds’ number increases. For certain ranges of Reynolds’ number in the transition zone the film completely covers small roughness projections, and the pipe has a friction factor the same as that of a smooth pipe. The larger Reynolds’ numbers projections protrude through the laminar film, and each projection causes extra turbulence that increases the head loss. For the zone marked “complete turbulent rough pipes”, the film thickness is negligible compared with the height of roughness projections and each projection contributes fully to the turbulence. Viscosity does not affect the head-loss in this zone, as evidenced by the fact that the friction factor does not change with the Reynolds’ number. In this zone 2Vh f ∝

130

SIMPLE PIPE PROBLEM Type 1 fh⇒ is unknown Type 2 QorV⇒ is unknown Type 3 D⇒ is unknown EXAMPLE TYPE 1 Q=140 L/s

sm /00001.0 2=ν L = 400 m D = 200 mm cast iron pipe mm25.0=⇒ ε

8912700001.0

2.0456.4Re

/456.44/2.0

140.04/ 22

=×

==

=×

==

ν

ππVD

smD

QV

00125.0

20025.0

==mmmm

Dε

⇒ Colebrook Formula

+−=

fDLn

f Re51.2

7.3/86.01 ε

⇒

×+−=

fLn

f 8912751.2

7.300125.086.01 023.0=f

mg

VDLfh f 58.46

806.924/2.0

140.0

2.0400023.0

2

2

22

=×

×××== π

EXAMPLE TYPE 2 T =150C water flow D = 300mm riveted steel pipe

mhmm

f 63

==ε

L = 300 m Q =?

131

termondtheneglectingequationColebrookgumm

mmD

secsin01.0300

3⇒==

ε

04.007.301.086.01

=⇒

+−= fLn

f

Darcy-Wesbach equation

smVV /715.1806.923.0

30004.062

=⇒×

=

From appendix C , sm /1013.1 26−×=ν

01.0,4550001013.1

3.0715.1Re 6 ==×

×==

− DVD εν

Colebrook formula 038.0=f ⇒ smVV /76.1806.923.0

300038.062

=⇒×

=

⇒==××

==−

01.0,4672561013.1

3.076.1Re 6 DVD εν

Colebrook formula 038.0=f

V= 1.76 m/s

Q=VA = sm /1244.043.076.1 3

2

=××π

EXAMPLE TYPE 3 Determine the size of clean wrought iron pipe ( 00015.0=ε ) required to convey 8.93 cfs oil, sft /0001.0 2=ν , for 10000 ft with a head-loss of 75 ft. Try 02.0=f in the middle of the diagram

ftDDD

398.1806.92

4/93.8

1000002.075

2

2

=⇒×

×××= π

00011.0398.1

00015.0,814000001.0

398.14/398.1

93.8

Re2

===×

×==D

VD επν

inftDfformulaColebrook 6.16382.1019.0 ==⇒=⇒

132

NON-CIRCULAR CROSS SECTIONS

gV

DLfh

hf 2

2

= ,

+−=

fDLn

fh

Re51.2

7.3/

86.01 ε , ν

hVD=Re

Where

pA

perimeterwettedareationalcrossDh 4sec4 ==

,

Where EXAMPLE

gV

DLfh

hf 2

2

=

+−=

fDLn

fh

Re51.2

7.3/86.01 ε

νhVD

=Re

pA

perimeterwettedareationalcrossDh 4sec4 ==

133

gV

gV

DLfh

gVPZ

gV

DLf

gVPZ

EhE

h

h

f

200

200

22222

222

2

2211

1

21

++=−++

++=−++

=−

γγ

134

MINOR LOSSES Losses which occur in pipe lines because 0f bends, elbows, joints, valves, etc., are called local or minor losses. In almost all cases the minor loss is determined by experiment. However, one important exception is the head-loss due to a sudden expansion in a pipe line. Sudden Expansion in A pipe line

gV

DD

gVV

gVKhm 2

12

)(2

21

22

2

12

212

−=

−==

135

Continuity:

)1(2211 AVAV = Energy:

)2(22

222

211

gVPhm

gVP

+=−+γγ

Momentum:

)3()( 212212111 QVQVAPAAPAP ρρ +−=−−+ Having known P1, A1, A2, Q , the unknowns V2, P2, hm by solving equations (1), (2) and (3), simultaneously can be obtained:

22

2

1

21

22

2

12

21

1

21

2)(

−=⇒

−=

−=

DDK

gV

DD

gVVhm

For the joint of a pipe and a reservoir , D1/D2 approaches zero and hence, K=1 This means that all kinetic energy will be dissipated. Head-loss due to Gradual Expansion

gVVKhm 2

)( 221 −

=

136

Head loss due to a sudden contraction

The head-loss from section 1 to the vena contrata ( the section of greatest contraction of the jet) is small compared with the loss from section 0 to section 2 , where velocity head being reconverted into pressure head (expansion):

137

gVK

gV

Cg

VCV

h

CV

AAVV

AAVAVAV

gVVh

C

Cm

C

m

2211

2

/

2)(

22

22

2

2

22

2

20

22

0

202200

220

=

−=

−

=

===⇒=

−=

K

K K

K K

K

138

EXAMPLE Find the discharge through the pipe line in the figure for H=10 m and determine the head-loss for Q =60 L/s.

200C

60 m 30 m

10 m

12 m

150mm

139

)1(2

)6803.13(

200

210

29.02

25.0

215.010200

2210

29.02

25.0

22

2

22222

222

2

2222211

1

2211

gVfH

gV

gV

gV

gV

gVfH

gVPZ

gV

gV

gV

gV

DLf

gVPZ

EhE f

+=

++=

+××++−++

++=

+××++−++

=− −

γγ

sLsmVAQsmV

fColebrookD

smVfD

/9.45/0459.0415.06.2/6.2)1(

023.00017.0,3910001001.1

15.063.2Re

/63.2)1(022.00017.0

32

6

==××==⇒=⇒⇒

=⇒⇒==××

=⇒

=⇒⇒=⇒=

−

π

ε

ε

For the second part, with Q = 60 L/s , the solution is straight forward:

mH

fD

smAQV

06.17)1(

023.00017.0,505000Re/4.34/15.0

06.02

=⇒⇒

=⇒==⇒=×

==ε

π

Assignment 6.3.1, 6.3, 6.14, 6.20, 6.21, 6.28, 6.34, 6.78, 6.79, 6.100, 6.110

140

BOUNDARY LAYER Consider flow over a flat plate as shown in the figure:

The flow approaching the plate is of uniform velocity, U0 . From the no-slip condition, we know the velocity at the plate must be zero. The plate exerts a retardation force on the flow; it slows the fluid in the neighborhood of the surface. At , the flow will not be influenced by the presence of the plate. At some intermediate point, the velocity has a value lies between zero and U0.

The flow field can be divided into two general regions. In the region adjacent to the boundary, shear stresses are present; the region is called the boundary layer. Outside the boundary layer the velocity gradient is zero and hence the shear stresses are zero.

U0 y=δ

y=δ

141

is the boundary layer thickness. The plate would influence a greater region of the flow field as we move farther down the plate. DRAG FORCE ON IMMERSED BODIES In the region between A and B, the pressure force behind fluid particles is sufficient to overcome the resisting shear force and the motion in the flow direction is maintained. Now consider an element of fluid inside the boundary layer on the back of the cylinder beyond point B. Since the pressure increases in the direction of flow, the fluid element experiences a net pressure force opposite to its direction of motion. Finally the momentum of the fluid in the boundary layer is insufficient to carry the element further into the region of increasing pressure. The fluid layers adjacent to the solid surface will be brought to rest and the flow will separate from the surface. The point at which this occurs is called the point of separation.

δ

A

Stagnation Point V=0

B Velocity increases Pressure decreases 0>

dxdP

MaximumVelocity

Velocity decreases Pressure increases 0<

dxdP

142

Boundary layer separation results in the formation of a relatively low pressure region behind a body; this region which is deficient in momentum, is called the wake. Thus, for separated flow over a body, there is a net unbalanced of pressure forces in the direction of flow; this results in a pressure drag on the body. The greater the size of the wake behind a body, the greater is the pressure drag.

143

Reducing the size of wake results in reducing the pressure drag. Since a large wake results from boundary layer separation, which in turn is related to the pressure of an adverse pressure gradient (increase of pressure in the direction of flow), reducing the adverse pressure gradient should delay the onset of separation and hence, reduce the drag. STREAMLINING A BODY REDUCES THE ADVERSE PRESSURE GRADIENT. Streamling the body delays the onset of separation; although the surface area of the body and, hence, the total shear force acting on the body is increased, the drag is reduced significantly.

Wake

A

Stagnation Point V=0

B Velocity increases Pressure decreases 0>

dxdP

MaximumVelocity

Velocity decreases Pressure increases

Wake

Boundary Layer

144

COMPRESSIBLE FLOW Flows in which variation in density are not negligible are termed compressible. For many practical cases liquids are incompressible. However, water hammer and cavitation are examples of the importance of compressibility effects in liquid flows. Gas flows also may be considered incompressible provided the flow speeds are small relative to the speed of sound; the ratio of flow speed V, to the local speed of sound, a, in the gas is defined as the Mach number,

145

For values of M <0.3, the maximum density variation is less than 5 percent. Thus gas flows with M<0.3 can be treated as incompressible; a value of M =0.3 in air at standard conditions corresponds to a speed of approximately 100 m/s. Compressibility effects are very important in the design of modern high-speed aircraft and missiles, power plants, fans, and compressors.

RTuPuh +=+= ρ/ h is the enthalpy of the gas and u is the internal energy

vv T

uC

∂∂

= specific heat at constant volume ,

p

p ThC

∂∂

= specific heat at constant pressure

v

p

CC

k = k is the ratio of specific heats

kRTa = a is the sound speed CONTINUITY EQUATION

.consVA =ρ ENERGY EQUATION When there is no heat transfer and no work done by pumps and turbines in the flow of an ideal fluid, the motion is isentropic (Adiabatic[dQ=0] + reversible[dS=0]) and the steady flow energy equation for the streamline becomes:

.2

2

constVPu =++ρ

, .2

2

constVh =+ or .2

2

constVTC p =+

146

EULER EQUATION

0=++ VdVdPgdZρ

For compressible flow, however, the term gdZ is usually dropped (change of elevation is negligible) and the Euler equation written:

0=+VdVdPρ

INTEGRATION OF THE EULER EQUATION When the Euler equation is integrated along the streamline for isentropic flow of a perfect gas, it becomes:

∫=− 1

2

21

22

2

P

P

dPVVρ

Obtaining ρ from kk

PP

1

1

ρρ= and substituting in the above integral,

yields:

−

−=

−−k

k

PP

kkPVV

1

1

2

1

12

12

2 112 ρ

Or obtaining ρ from kk

PP

2

2

ρρ= and substituting in the above

integral, yields:

−

−

=−

−

112

1

2

1

2

22

12

2k

k

PP

kkPVV

ρ

147

EXAMPLE An airplane flies at 644 km/h (179 m/s) through still air at 90 kPa and -20oC. Calculate pressure, temperature, and air density at the stagnation point (On nose of fuselage of the wings).

31

1

111

/24.1)27320(8.286100090

mkg

RTP

=

+−××=×=

ρ

ρρ

smkRTa /3192538.2864.111 =××==

562.0319179

1 ==M

CKT

T

VTCVTC

o

pp

4269

010032

1792531003

22

2

2

2

22

2

21

1

−==

+=+×

+=+

kPaP

P

PP

kkPVV k

k

5.111

901

14.14.1

24.1100090

21790

112

2

4.114.1

22

1

1

2

1

12

12

2

=

−

−×

=−

−

−=

−

−

−

ρ

31

2

222

/45.1)2734(8.28610005.111

mkg

RTP

=

+−××=×=

ρ

ρρ

SUBSONIC AND SUPERSONIC VELOCITIES Taking the differential of the continuity equation .consVA =ρ and dividing by VAρ yields:

148

)1(0=++VdVd

AdA

ρρ

And considering the Euler equation:

22 00

00

aVdVdVdVdaVdVd

ddP

VdVdddPVdVdP

−=⇒=+⇒=+×

=+×⇒=+

ρρ

ρρ

ρρ

ρ

ρρ

ρρ

Substituting the last expression for

ρρd in equation (1) gives:

( )12 −= MVdV

AdA

Subsonic Flow 001,0 2 >⇒<−<

VdVM

AdA Velocity increases

Convergent Nozzle

149

001,0 2 <⇒<−>

VdVM

AdA Velocity decreases

Supersonic Flow 001,0 2 <⇒>−<

VdVM

AdA Velocity decreases

Divergent Nozzle

Convergent Nozzle

150

001,0 2 >⇒>−>

VdVM

AdA Velocity increases

THROAT 0<

AdA 0>

AdA

0=

AdA throat

⇒= 0

AdA Either 12 −M ; 1=M or 1≠M , Velocity

VdV

⇒= 0 is either

minimum or Maximum. Accordingly it can be concluded that if the throat velocity is not sonic it will be a maximum in subsonic flow and a minimum in supersonic flow. For flow at sonic speed (M=1) the rate of change of area must be zero; that is, this might occur at a maximum or minimum cross section; use of the preceding conclusions will show this to be restricted to a minimum section (a throat) only. SONIC VELOCITY (M=1) OCCURS ONLY AT A THROAT.

Divergent Nozzle