1

ANALYSIS OF PIPE FLOW

9/21/2015

Pipes of different sizes and shapes are used in many industrial applications

(Acknowledgement: Forchase Corporation Ltd, http://forchase.com)

2

To Malaysia

Water pipelines

Water pipelines between Malaysia and Singapore

(Acknowledgement: skyscrapercity.com)

Some examples of piping system in engineering applications

Pipelines in a petrochemical plant

(Acknowledgement: http://www.fullsupply.co.uk/)

3

Pipelines inside Honda Accord engine compartment

(Acknowledgement: Mark Gallagher)

Pipelines used in household appliances

(Acknowledgement:www.hotshowers.com) (Acknowledgement:www.bengreenplumbing.co.uk )

4

A satisfactory analysis of pipe flow depends on the accuracy involved inestimating the dissipation of energy incurred in maintaining the flow. Thisrequires the knowledge of boundary layer theory.

The following analysis relates to homogeneous fluid of constant viscosity anddensity. The results are also applicable to gases provided the densitychanges are small.

Types of Flow:

There are two types of flow in a pipe:

(a) Laminar Flow(b) Turbulent Flow

5

FLOW

This type of flow is characterized by motion of fluids in layers or laminas, parallel to the boundary surface.

(a) Laminar Flow:

Laminar flowFLOW

(b) Turbulent Flow:

Under certain conditions, a laminar flow can become unstable and become turbulent.

Laminar flow Turbulent flow

FLOW

FLOW

The turbulent flow is characterized by RANDOM, IRREGULAR and UNSTEADY movement of fluid particles, making it impossible to predict the motion of a fluid particle with respect to time and space.

6

Vdor

VdRed

The criteria which ascertain the type of flow is REYNOLD NUMBER (Re)

where =density of fluid, =dynamic viscosity, and =kinematic viscosity.

dV

Osborne Reynolds (a British Engineering Professor) was the first to show that the Reynolds number is the criterion for determining whether the flow is LAMINAR or TURBULENT in a circular pipe.

Red < 2300 (LAMINAR FLOW)

Critical Reynolds number is very sensitive to the initial disturbances in the fluid at the entrance.

By “quieting” the flow, it is possible to extend Red >50,000

The upper limit of critical Reynolds number depends on

(a) initial disturbance of approach flow,(b) shape of pipe entrance and(c) roughness of pipe.

7

FLOW NEAR ENTRANCE OF PIPE

Laminar Boundary Layer Turbulent Boundary Layer

Transitional Boundary Layer

dFlow

FLOW NEAR ENTRANCE OF PIPE

Laminar Flow

Very low velocity

Laminar B/L

Turbulent Flow

Very high velocity

Laminar B/L

Turbulent B/L

8

ENTRANCE LENGTH

Entrance length is defined as the distance from the entrance of the pipe that the flow needs to travel before the flow is fully developed (i.e. the velocity profile does not change with distance).

de Re

Vd

d

L

A B C

Velocity ProfileFully DevelopedProfile

Laminar Boundary Layer

Entrance Length (Le)

Laminar Flow

d

Figure P5(a)

The entrance length (Le) is a function of Reynolds number , i.e.

A B C

Velocity ProfileFully DevelopedProfile

Laminar Boundary Layer

Entrance Length (Le)

Laminar Flow

d

de Re06.0

d

L (LAMINAR)

The accepted correlation is

LAMINAR FLOW:

The maximum laminar entrance length at Red,critical = 2300 is Le=138d, which is the longest development length possible.

9

TURBULENT FLOW:

Laminar Boundary Layer

Turbulent Boundary Layer

Entrance Length (Le)

Fully DevelopedProfile

(a) Turbulent Flow

A B C

In turbulent flow, the boundary layers grow faster, and Le is relatively short. Based on the approximation

61

de Re4.4

d

L (TURBULENT)

The entrance length at various Reynolds number can be calculated as shown in the table below

Red Le/d

4000 18

104 20

105 30

106 44

107 65

EXAMPLE:

SAE 10 oil at 20oC flows through a 3-cm diameter tube. Estimate the entrance length in cm if the volume flow rate is (a) 0.001 m3/s and (b) 0.03 m3/s. The density () and dynamic viscosity () of SAE 10 oil are 870 kg/m3 and 0.104 kg/m.s, respectively.

d

Q4VdRe

Solution:Before we can determine the entrance length, we need to determine whether the flow is laminar or turbulent.

V4

dQ

2Volume flow rate

2d

Q4

V=

(a) Q=0.001m3/s

355)03.0)(104.0(

001.0)870(4Re

Flow is laminar since Re<2300.

From equation P1,

(b) Q=0.03 m3/s

10650)03.0)(104.0(

03.0)870(4Re

Flow is turbulent since Re>2300.

From equation P2,

cm64)3)(355(06.0Re06.0L de

cm62)3()10650(4.4Re4.4L 61

61

de

10

FLOW IN A CIRCULAR PIPE

By applying energy equation between stations 1 and 2, we get

f2

222

1

211 hz

g2

V

g

pz

g2

V

g

p

where hf = friction head loss, V = averaged velocity

(P3)

gsin

L =x2-x1

w

wV1

V2

z1

z2= z1+z

1

g

p1

p2= p1+pr=R

Arbitrary reference datum

2

gR2L

d

Control volume

In general, the energy equation in a pipe is given by

where 1 and 2 are correction factorsf2

2

22

21

2

11

1 hzg2

V

g

pz

g2

V

g

p

NOTE:

g

pzhf

(P3a)

g

pzhf

(P3a)

g

pzhf

(P3a)

For LAMINAR pipe flow

1 = 2 = 2

For TURBULENT pipe flow

1 = 2 1

Using Averaged Velocity

1 = 2 = 1

11

0VVmLR2sinLRgRpRp 12w22

22

1

0LR2sinLRgRp w22

zzzsinLBut 21

From the last slide, the energy loss (or head loss) is given by

-

g

pzhf (P3a)

Apply momentum equation to the control volume, we get

LR2zRgRp w22

)4P(gR

L2h

g

pzor w

f

Therefore, the above equation becomes

gsin

L =x2-x1

w

wV1

V2

z1z2= z1+z

1

gp1

p2= p1+p

r=R

Arbitrary reference datum 2

gR2L

)5P(V

8f

2w

Define the dimensionless parameter f (the Darcy Friction Factor)

By combining P4 and P5, the desired equation for pipe head loss

)4P(gR

L2h

g

pzor w

f

NOTE: Darcy-Weisbach Equation is valid for duct flows of any cross-section and both laminar and turbulent flow.

V = Average velocity

L

d

(Darcy-Weisbach Equation)

g2

V

d

Lfh

2

f (P6)

12

Rp

p +dp

dx

x

gr2dx

rdz

v1

v2

dx

-dz

Laminar Flow in a Circular Pipe

The laminar flow of an incompressible fluid under steady conditions may be completely analysed using Newton’s second law in the direction of motion..

2112222 vvcesin0)vv(mrdx2sindxrgr)dpp(rp

rdx2sindxrgrdp 22

rdx2dzrgrdp 22

gzpdx

d

2

r (P6a)

gzpdx

d

2

r

dr

du

gzpdx

d

2

r

dr

du

)7P(Cgzpdx

d

4

ru 1

2

Integrating gives

)8P(gzpdx

d

4

RC

2

1

Boundary condition: at r=R, u=0.

Note that velocity gradient, du/dr, is negative i.e. as r increases u decreases,

dr

du

The shear stress is related to the velocity gradient by

(P6b)

Substituting (P6b) into (P6a) gives gzpdx

d

2

r (P6a)

Substitute into equation (P7) gives

C1 is a constant that depends on the boundary condition.

13

After substituting (P8) into equation (P7), the velocity distribution is given by

which is a parabolic distribution. Equation (P9) iscalled the Hagen-Poiseuille Flow.

FLOW

uR

r

22 rRgzpdx

d

4

1u

(P9)

Note: Equation (P9) applies to Laminar Flow Only

gzpdx

d

4

ru

2

gzpdx

d

4

R 2

22 rRgzpdx

d

4

1u

)10P(gzpdx

d

4

Ru

2

max

Maximum velocity occurs at the centre of the pipe, i.e. r=0

FLOW

uRr umax

To find average volume flow rate (Q) [m3/s],

R

0

rdru.2Q

R

0

22 rdr2rRgzpdx

d

4

1Q

)11P(gzpdx

d

8

R 4

Therefore, dQ = u.2rdr

r

dr

Figure P8

R

Shaded area = (r+dr)2-(r)2

2rdr (since dr is small)

We consider flow through a GREEN circular ring

22 rRgzpdx

d

4

1u

From laminar velocity profile (P9)

14

Comparing equation (P10) and (P11), it can be seen that

gzpdx

d

4

RR

2

1gzp

dx

d

8

RQ

22

4

umax

max2uR

2

1Q

)a11P(2

u

A

Q)velocityaverage(VBut max

Similarly, shear stress (), friction factor (f) and head loss (h) can be determined as follows

)b11P(R

u2gzp

dx

d

4

R

R

2gzp

dx

d

2

R

dr

du max2

Rrw

)12P(Re

64

V

8f

d2

warminla

)13P(gd

LQ128

g2

V

d

Lfh 4

2

arminlaarminla,f

22 rRgzpdx

d

4

1u

(P9)

Pressure Drop for a Laminar Flow in a Horizontal Pipe

gzpdx

d

8

RQ

4

)14P(dx

dzg

dx

dp

8

R4

From equation (P11),

Assuming that the length of the pipe is L; p=change in pressure over the length of the pipe, and z=change of elevation.

L

zg

L

p

8

RQ

4

z1z2

Lp1

p1+p

Datum

Equation (P14) becomes

)15P(R

LQ8p

4

If the pipe is horizontal z=0, therefore

for laminar flow only

Introducing R=d/2

)16P(d

LQ128p 4

15

TURBULENT FLOW IN A CIRCULAR PIPE

Turbulent Flow in a Circular Pipe

d V

Laminar(PARABOLIC)

Turbulent

In turbulent flow, a major part of the mechanical energy in the flow goes into forming and maintaining randomly eddying motion.

Eddying motion dissipates their kinetic energy into heat.

At a given Reynolds number, the drag of the turbulent flow is higher than the drag of a laminar flow.

Turbulent flow is affected by surface roughness, so that increasing roughness increases the drag.

16

TURBULENT FLOW THROUGH SMOOTH PIPE

For turbulent flows, there are no exact solutions available for the velocity profile and the friction factor variation with Reynolds number, and we must always rely on experimental data.

VELOCITY Profile

For turbulent flow near a wall, the boundary layer can be divided into three regions. They are the WALL LAYER, the OVERLAP LAYER and the OUTER LAYER.

Outer layer

Overlap layer

Wall layer (or viscous sub-layer)Figure P10

Wall

Rr

uy

Umax

Turbulent Flow

Wall

uy

Outer layer

Umax

Turbulent Flow

Overlap layer

Wall layer (or viscous sublayer)

Rr

)17P(*yu

offunction*u

u

wwhere y = (R-r), =kinematic viscosity of fluid, and u*= which is called the friction velocity.

FOR THE WALL LAYER: Viscous shear (laminar) dominates. Using dimensional analysis, Prandtl deduced in 1930 that

Experiment indicates that

)17P(*yu

*u

u

*u)rR(

*u

u or

Equation (P17) is called the LAW OF THE WALL.

17

Wall

uy

Outer layer

Umax

Turbulent Flow

Overlap layer

Wall layer (or viscous sublayer)

Rr

FOR THE OUTER LAYER: Turbulent shear dominates. Using dimensional analysis,Von Karman deduced in 1933 that

orR

yfunction

*u

uUmax

)18P(

R

rRfunction

*u

uUmax

Umax-u is called the velocity defect, and equation (P18) is called VELOCITY DEFECT LAW

orB*yu

ln1

*u

u

)19P(B

*u)rR(ln

1

*u

u

FOR THE OVERLAP LAYER: Both viscous and turbulent shear dominate. C. B. Milliken in 1937 showed, using dimensional analysis, that

where and B are constant. Experimental data show that =0.4 and B=5.0. Equation (P19) is called the LOGARITHMIC OVERLAP LAYER

Wall

uy

Outer layer

Umax

Turbulent Flow

Overlap layer

Wall layer (or viscous sublayer)

Rr

18

orB*yu

ln1

*u

u

)19P(B

*u)rR(ln

1

*u

u

=0.4 B=5.0

10o 101 102 103 1040

10

20

30

u/u* Eqn (P17)

5 30

Viscous sublayer Buffer layer Overlap layer

Experimental dataOuter layer

*yu

)a19P(B*Ru

ln1

*u

Umax

(P19) minus (P19a) gives

A good approximation for the OUTER TURBULENT BOUNDARY OF THE PIPE can be obtained from (P19) by setting u=Umax when r=0 because maximum velocity occurs at the centre of the pipe, i.e.

)19P(B*u)rR(

ln1

*u

u

rR

Rln

1

*u

uUmax

19

(b) Relationship between mean velocity (V) and maximum velocity (umax)for a turbulent flow in a pipe

Rr

u

umax

B

*urRn

1

*u

ru

No exact solutions available for the velocity profile of a smooth turbulent flow, however experimental data follow a logarithmic relation.

)19Peqnfrom(B

*urRn

1

*u

ru

where =0.4 and B=5.0.

ru means that u is a function of r.NOTE:

)20P(

R

dr.r2.ruV

2

R

0

A

QV

The mean (or average velocity) V in the pipe is given by

)21P(rdr2B

*urRn

1*u

R

1V

R

02

B

*urRn

1*uru Substituting from the above into eqn (P20a),

we get

Mean (or average velocity) V

r

dr

Figure P8

R

Shaded area = (r+dr)2-(r)2

2rdr (since dr is small)

where Q= volume flow rate and A is the pipe cross-sectional area

20

b

ax

2

x

2

1bxan

b

ax

2

1xdxbxan

2

2

22

To solve the above integral (P21), we can refer to the Handbook of Mathematical Formulas and Integrals by Jeffrey Alan ( 2nd edition, Academic Press, pg 176). Relevance formulation is reproduced here

3B2

*Run

2*u

2

1V

Using the above formulation, equation P21 can be simplified to give

Substituting =0.40 and B=5.0 into the above equation, we obtain

)22P(25.1*Ru

n5.2*u

V

Since the maximum velocity (umax) occurs at the location where r=0. (P20) becomes

)23P(0.5*Ru

n5.2*u

umax

25.10.5*u

Vumax

)24P(V

*u75.31

V

umax

f

8

*u

V

f33.118

f75.31

V

umax

)25P(f33.11

1

u

V

max

Subtract (P22) from (P23), we get

Substitute the above into (P24) gives

w*u 2w

V

8f

But , and simple manipulation givesand

)22P(25.1*Ru

n5.2*u

V

)23P(0.5*Ru

n5.2*u

umax

21

(c) Friction Factor

)26P(8.0fRelog0.2f

1d

The relationship between friction factor and Reynolds number for a smooth turbulent pipe flow is given semi-empirical as

Which is known as PRANDTL’S UNIVERSAL LAW OF FRICTION for smooth pipe (for derivation, see Shames)

Some numerical values of equation P26 is listed below

Red 4000 104 105 106 107 108

f 0.0399 0.0309 0.0180 0.0116 0.0081 0.0059

Note that f drops by only a factor of 5 over a 10,000-fold increase in Reynolds number.

Graphically, equation (P26) can be plotted together with equation (P12) for smooth pipe laminar flow as shown in the graph below

)12P(f arminlaRe

64

d

103 104 105 106 107 108

0.01

0.02

0.03

0.04

0.05

0.06

0.070.080.090.10

f

Reynolds number (Red)

Smooth pipe: Laminar flow (eqn P12)

dRe

64f

Smooth pipe: Turbulent flow (eqn 26)

8.0fRelog0.2f

1d

)27P(10Re4000Re316.0f 5d

41

d

There are many alternatives of finding f, one of which is by Blasius is

Equation P26 is quite cumbersome to solve if Red is known and f is required.

22

Pressure Drop for a Turbulent Flow in a Smooth Horizontal Pipe

Frictional loss in the pipe (laminar or turbulent) is given by

g2

V

d

Lfz

g

ph

2

f

Eqn (3a) Eqn (6)

g2

V

d

Lf

g

ph

2

f

If the pipe is horizontal z=0

Substituting f from equation (P27) into the above equation we get

g2

V

d

L

Vd316.0

g2

V

d

Lf

g

ph

24

12

f

Eqn (27)

4

7

4

54

1

4

3

VdL158.0p

75.125.141

43

VdL158.0p

Vd4

1QgIntroducin 2

)27P(Re316.0f 41

d

flowturbulentforQdL241.0p 75.175.441

43

(P28)

Compare with pressure drop in laminar flow for a horizontal pipe

Note:

(a) For a given Q, the pressure drop in turbulent flow decrease more quickly than in laminar flow.

(b) The quickest way to reduce the required pumping pressure is to increase the diameter of the pipe. Doubling the pipe reduces p by a factor of about 27.

ord

LQ128p

4

)29P(QLd128

p4

For LAMINAR flow

(P28)QdL241.0p 75.175.44

1

4

3 For TURBULENT flow

23

Pipe 1

Pipe 2

Q

Q

d

2d

L

L

Pressure Drop (P1)

27

PP 1

2

TURBULENT FLOW

Better to transport water from Malaysia to Singapore using a big pipe, but the cost of the pipe will go up!

TURBULENT FLOW through A ROUGH PIPE

24

FLOW THROUGH ROUGH PIPES

•Most of the pipes used in engineering applications such as cement pipes, cast iron pipes etc cannot be regarded as hydraulically smooth especially at high Reynolds numbers.

•In fact, they are actually behaving as hydraulically rough.

•The resistance to fluid flow offered by rough boundaries is larger that for smooth boundaries on account of formation of eddies behind rough protrusions.

•For LAMINAR FLOW, all rough pipes irrespective of their roughness size and pattern offer the same resistance as that offered by a smooth pipe under similar conditions of flow.

•In fact, there is no surface which may be regarded as perfectly smooth. The term “smooth” and “rough” are relative.

•In turbulent flow, there is a thin layer very close to the boundary in which the flow exhibits characteristics of laminar flow. This layer is call “laminar sublayer”

Limit of laminarsublayer

Turbulent flow

Turbulent boundary layer

Limit of laminar sublayer Roughness elements completely

submerged in the laminar sublayer

Figure P13

This occurs when the roughness elements are submerged in the laminar sublayer, and therefore have no effects on friction

)30P(5*u

Criteria for Hydraulically Smooth Walls:

w*urecall that and = kinematic viscosity of fluid

(a)Hydraulically Smooth Walls

25

Limit of laminar sublayerRoughness elements protrude into the main flow

Figure P14

(b) Hydraulically Rough Walls

This occurs when the roughness elements protrude into the main flow causing it to break up into vortices or eddies.

)31P(70*u

Criteria for Hydraulically Rough Walls:

Limit of laminar sublayer

Some roughness elements are submerged and some are protruded into the main flow

Figure P15

(c) Transitional Roughness:

The regime lies between hydraulically smooth wall and hydraulically rough walls.

)32P(70*u

5

Criteria for Transitional Roughness:

26

EXAMPLE:

Water at 20 oC flows in a 10 cm diameter pipe at an average velocity of 1.6 m/s. If the roughnesselements are 0.046 mm high and the friction factor is f = 0.0204, would the wall be considered roughor smooth? Assume the kinematic viscosity of water () 10-6 m2/s.

5*u

If

70*u

If

Solution:

then the wall is hydraulically smooth.

then the wall would be rough.

To find if the wall is smooth, we need to make use of the flow criteria:

Pa53.60204.06.110008

1fV

8

1 22

s/m0808.01000

53.6*u w

7.3

10

0808.010x046.0*u6

3

The friction velocity is

Therefore the wall is hydraulically smooth, even though it is physically rough.

Therefore, the viscous wall layer thickness is

Velocity Profile in a Rough Pipe

27

Limit of laminar sublayer

Roughness elements completely submerged in the laminar sublayer

(a) Velocity Profile in a Rough Pipe:

The velocity profiles in a rough pipe depend on whether the pipe is hydraulically smoothor completely rough or Transitional rough.

(a) hydraulically Smooth

For hydraulically smooth pipe, the roughness elements are submerged inside laminar boundary layer.

)33P(0.5*yu

ln4.0

1

*u

u

We may use the velocity profile for a smooth pipe, i.e.

Limit of laminar sublayer

Roughness elements protrude through laminar sublayer

Figure P16

(b) Completely Rough

The velocity profile for a completely rough pipe is

5.8y

ln4.0

1

*u

u

28

)35P(By

ln4.0

1

*u

u

*u

n

Figure P18 Acknowledgment: F.M. White: Fluid Mechanics, 3rd Edition

(c) Transitional Roughness

where B has been developed from experimental data of Nikuradse

The velocity profile is also given by

NIKURADSE’S EXPERIMENTS ON ARTIFICIALLY ROUGH PIPES

Irregular roughness

Commercial pipe

Naturally rough surfaces of commercial pipes used in engineering practices are generallycomplex and irregular as shown above.

Because of this, most of the advance in understanding the parameters of pipe resistancehave been developed around experiments on artificially roughened pipes.

An German engineer, J. Nikuradse, conducted a series of systematic experiments on pipes roughened artificially by glueing uniform sand grains as closely spaced as possible on the inner side of the pipe wall.

29

NIKURADSE’S EXPERIMENTS ON ARTIFICIALLY ROUGH PIPES:

Sand grains of different UNIFORM sizes

Pipe

f

Re=Ud/

L

d

2V

g2

L

d

g

pf

V

x

x

p

g2

V

d

Lf

g

p 2

(P27a)

xxx

x

x

x

x

x x

STANTON orMOODY DIAGRAM.

xx

x

x

x x

V

30

A

B

COMPLETELY TURBULENT REGIME

C

D

LAMINARFLOW

F

E

SMOOTH PIPES

8.0fRelog0.2f

1d

STANTON or MOODY DIAGRAM

STANTON or MOODY DIAGRAM

For all the curves on the right hand side of AB (red curve), the f versus Reynolds number relationship becomes horizontal indicating that friction factor is independent of the Reynolds number. This region is identified as a fully rough flow, and are described by

7.3

dlog0.2

f

1(P36) (see page 163 of your notes)

A

B

COMPLETELY TURBULENT REGIME

8.0fRelog0.2f

1d

Smooth pipe

(P26)

31

Between lines EF and AB, the friction factor is dependent on both Reynolds number as well as the relative roughness.

Colebrook in 1939 cleverly combined the smooth wall (equation P26) and fully rough flow (equation P36) relations into an interpolated formula.

A

B

COMPLETELY TURBULENT REGIME

E

F

7.3

dεlog0.2

f

1

(P36)

8.0fRelog0.2f

1d

Smooth pipe

(P26)

fRe

51.2

7.3

dεlog0.2

f

1

d

(P37)

But equation (P37) is difficult to use

(see page 163 of your notes)

An alternative formula given by Haaland is given by

11.1

d 7.3

d

Re

9.6log8.1

f

1

which varies less than 2% from equation

(P38)

fRe

51.2

7.3

dεlog0.2

f

1

d

(P37)

A

B

COMPLETELY TURBULENT REGIME

E

F

7.3

dεlog0.2

f

1

(P36)

8.0fRelog0.2f

1d

Smooth pipe

(P26)

32

0.0128

0.0185

0.0125

The Moody Diagram is accurate to 15% for design calculations over the fullrange shown in the figure. The shaded area in the Moody diagram indicates the rangewhere transition from laminar to turbulent flow occurs. There are no reliable frictionfactors in this range, 2000<Red<4000.

33

AVERAGE ROUGHNESS OF COMMERCIAL PIPES

Material (new) (mm)

Riveted steel 0.9-9.0

Concrete 0.3-3.0

Cast iron 0.26

Galvanised iron 0.15

Asphalted cast iron 0.12

Commercial steel or wrought iron 0.046

Drawn tubing 0.0015

Glass “Smooth”

From: F.M. White, Fluid Mechanics, 3rd Edition

Friction in Non-circular Pipes

Most of the pipes or conduits used in engineering applications are circular in cross-section.

On some occasions, we also use rectangular ducts and cross sections of other geometry.

We can modify many of the equations that we have derived earlier for circular cross-sections to noncircular sections by using the concept of HYDRAULIC DIAMETER.

Circular pipes

(Acknowledgement: Rigidtools.org)

Non-circular pipes

(Acknowledgement: itctubeco.com)

34

)39P(perimeter

areationalseccrossx4DH

Examples:

H2)W(2

WH4

WH2

WH4DH

W

H

(b) if W>>H (i.e Elongated rectangular section)

WH2

WH4DH

W

H

(a) Rectangular Cross-section:

Hydraulic Diameter

HHH2

HH4DH

g2

V

D

LfLossFrictionalh

2

HL

)12equationfrom(DV

64

Re

64fBut

Hdarminla

H

H

(c) If W=H (i.e. Square)

Therefore, using hydraulic diameter (DH),

35

HYDRAULIC DIAMETER APPROACH gives

A reasonable accurate result for turbulent flow.

But not so accurate for laminar flow.

WHY??

In laminar flow viscous action causes friction phenomenon to occur throughout the fluid.

In turbulent flow, most of the action occurs in the region close to the wall, i.e. it depends onthe wetted perimeter.

d V

Laminar(PARABOLIC)

Turbulent

Minor Losses in Pipes

36

Bends

Valve

Sudden enlargement

Sudden contraction

FittingsInlet

Outlet

The minor losses are those caused by change in pipe cross-section, presence of bends, valves, elbows, enlargements, contractions, inlets, outlets and fittings of all kinds.

A minor loss is expressed in terms of loss coefficient K, defined by

g2

VKh

2

L

It is often the practice to express a loss coefficient as an equivalent length (Leq ) of the pipe.

g2

V

D

Lf

g2

VK

2eq

2

f

DKLTherefore eq

LOSS IN SUDDEN ENLARGEMENT

If the head loss due to sudden enlargement is hL, the energy equation between sections (1) and (2) gives

L2

222

1

211 hz

g2

v

g

pz

g2

v

g

p

Applying momentum equation to the controlvolume ABCDEF, we get

12221211 vvQAp)AA('pAp

22112 AppvvQ

122

21 vvA

Q)pp(

122 vvv

since p’ p1

g2

vv

g

pph

22

2121

L

g2

vv

g

vvvh

22

21122

L

(1) (2)p’

p1A1v1z1

p2A2v2z1

A

B

C D

EF

A1v1=A2v2

g2

vv 221

37

2

2

121

L A

A1

g2

vh

The above equation was first obtained by J.C. Borda (1733-99)and L. Carnot (1753-1823) and is frequently known as theBorda-Carnot head loss.

2

1

222

L 1A

A

g2

vh

(P42)

g2

vvh

221

L

(1) (2)p’

p1A1v1z1

p2A2v2z1

A

B

C D

EF

A1v1=A2v2

21

21 v

A

Av

A1v1=A2v2

12

12 v

A

Av

A1 A2

A2>> A1

A1 A2

2

2

121

L A

A1

g2

vh

g2

vh

21

L 0A

A

2

1

Exit Loss

A1 A2

Large tank

Equivalent

)lossexit(g2

vh

21

L

38

Vena Contracta(Area=Ac)

d2d1

Area (A1)(Section 1)

Area (A2)(Section 2)

CD

V1 V2

Figure P23

LOSS IN SUDDEN CONTRACTION

Although a SUDDEN CONTRACTION is geometrically the reverse of a SUDDEN ENLARGEMENT, it is not possible to apply the momentum equation to a control volume between sections (1) and (2).

Between the vena contracta and the downstreamsection (2) the flow pattern is similar to that after anabrupt enlargement, and the loss of head is assumedto be given by equation (P42)

2

c

22

2

c

222

L 1C

1

g2

V1

A

A

g2

Vh

Where Ac represents the cross-sectional area of vena contracta,and the coefficient of contractionCc=Ac/A2.

g2

VKh

22

scL

In general, the loss in a Sudden Contraction can be expressed as

Because of the complexity of the flow, the loss coefficient Ksc is obtained experimentally. Representative values are shown below.

d2/d1 0 0.2 0.4 0.6 0.8 1.0

Ksc 0.5 0.45 0.38 0.28 0.14 0

Note:

(a) When d2/d1=1.0, there is no sudden contraction and the pipe is a normal straight pipe, and Ksc=0.

(b) As A1 , d2/d10, the value of KSC =0.5 d2d1

g2

VKh

22

scL

39

ENTRANCE LOSS

V

K=0.5

Square-edge(flush)

V

Protrusions

(a) (b)

Figure P24

Figure P25Acknowledgment: F.M. White: Fluid Mechanics, 3rd Edition

t/d=0.02

0.04

Rounded Edge

V

Bell-mouthed

r

Figure P27

Figure P26

Acknowledgement: F.M. WhiteFluid Mechanics, 3rd Edition

40

GRADUAL EXPANSION

V1 V2

DIFFUSER

A1A2

g2

V

A

A1K

g2

VVKh

21

2

2

1L

221

LL

Energy loss can be considerably reduced if the pipe transition is more gradual.

Such a transition device is called a DIFFUSER.

Figure P29

KL

Acknowledgement: B.S. MasseyMechanics of Fluids, 3rd Edition

6o

A2/A1=4

A2/A1=2.25

9

LOSSES IN BENDS

A

BD

C

Flow separation

Flow separation

Secondary flow

Figure P30

Losses through a pipe bend = Loss due to flow separation + Loss due to friction on the wall + lossdue to secondary flow.

The loss is expressed as )46P(g2

VK

2

K depends on the total angle of the bend, surface roughness () and on the relative radius of curvature R/d, where R is the radius of curvature of the pipe centreline and d is the pipe diameter.

41

Note: (a) K is inclusive of frictional losses.(b) When R/d=0 , then K 1.1

Nominal Loss Coefficients ( K) for some pipe fittings (turbulent Flow) are given in pages 182 to 184 of your lecture notes

42

Nominal Loss Coefficients ( K) (turbulent Flow)

Type of fitting K

1.1

0.2

=30o 0.02

=70o 0.07

2

2

1

A

A1

A2A1

Sudden enlargement

90o mitre bend (without vanes)

90o mitre bend (with vanes)

General contraction

MULTIPLE PIPE SYSTEM

43

Frequently, problems of dividing pipelines are encountered in engineering practice. These problems include looping pipes (pipe connected in parallel), branching pipes and pipe network

(a) Pipes connected in Series(b) Pipes connected in Parallel(c) Branched Pipes

Q1 Q2Q3

1 23

Figure P32

(a) PIPES IN SERIES

If one or more pipes are connected in series, conservation of mass gives

233

222

211 dVdVdV

Q = Q1 = Q2 = Q3 (P47)Or

The total head loss is the sum of the total losses in each of the individual pipes and fittings

HL = HL1 + HL2 + HL3 (P48)

In terms of friction and minor losses in each pipe, equation (P48) may be re-written as

ii,2

2

2222

ii,1

1

112

1L K

d

Lf

g2

VK

d

Lf

g2

VH

ii,3

3

3323 K

d

Lf

g2

V

Friction loss Minor loss

44

PIPES IN PARALLEL:

Q1

Q2

Q3

Q Q

1

2

3

(A) (B)

Figure P33

Q2

Q3

Q Q3

Q1

1

(A) (B)2

When two or more pipes are connected so that the flow divides and subsequently comes together again.

Continuity dictates that

Q = Q1 + Q2 + Q3 (P49)

At any point in the pipe, there can be only one value of total head (energy).

In other words, all fluid passing point (A) has the same total head

A

2AA zg2

V

g

P

Similarly, at point (B),

B

2BB zg2

V

g

P

LB

2BB

A

2AA Hz

g2

V

g

Pz

g2

V

g

P

The steady-energy equation may be written

Therefore, all the fluid in EACH PIPE suffer the same loss of head HL

HL1 = HL2 = HL3 (P50)

BRANCHED PIPES

Another example of practical importance involving a pipe system is when a number of pipes meeting at a junction as shown above.

The basic principles must be satisfied:

(a) Continuity: At a given junction, Mass flow rate towards the junction= Mass flow rate away from the junction.

(b) There can be only one energy level (Head) at a given point.

(c) The friction equation must be satisfied for each pipe.

Q1

Q3

z1

z2

z3

Arbitrary datum

Tank 1Tank 2

Tank 3

12

3

HJ

JQ2

45

The energy (or head) at:

(a) tank 1=H1, (b) tank 2 =H2,(c) tank 3=H3,

(d) location J = HJ

H1 - HJ = HL1 (head loss in pipe 1)H2 - HJ = HL2 (head loss in pipe 2)HJ –H3 = HL3 (head loss in pipe 3)Q3 =Q1 + Q2

Assuming that MINOR LOSSES are negligible, it can be shown that

g2

V

d

Lf

g2

V

d

Lf 23

3

3322

2

22H2-H3 = HL2 + HL3 =

g2

V

d

Lf

g2

V

d

Lf 23

3

332

1

1

11H1-H3 = HL1 + HL3 =

Under steady condition

Q1

Q3

z1

z2

z3

Arbitrary datum

Tank 1Tank 2

Tank 3

12

3

zJ

JQ2

1

2

111 z

g2

V

g

PH

2

2

222 z

g2

V

g

PH

3

2

333 z

g2

V

g

PH

J

2JJ

J zg2

V

g

PH

In general,

If P1 = P2 = P3 = Patmosphere

Assuming that tank 1, tank 2 and tank 3 are so large that

V1=V2=V3 0

If gauge pressure is used throughout, it can be shown that

11 zH

22 zH

33 zH

J

2

JJg

J zg2

V

g

PH

Q1

Q3

z1

z2

z3

Arbitrary datum

Tank 1Tank 2

Tank 3

1 2

3

HJ

JQ2

P1 P2

P3

46

EXAMPLE 4 (page 196 of your lecture note )

Water flows from reservoir A through a 100 m long pipe of diameter 120 mm to a branch point D where it is diverted to reservoirs B and C in separate pipes as shown in the figure below. Assuming that f =0.02 for all the pipes and neglecting all losses other than those due to friction, determine the elevation of the reservoir B. The flow from the reservoir A is 0.02 m3/s.

Arbitrary datum

QA

QB

QC

zA=50m

zB=?m

Tank A

Tank C

zC=25m

LA= 100mDA= 120mm

LB= 60mDB= 75mm

Tank B

Lc= 40mDc= 60mm

D

The velocity of water between A and D is

s/m768.1)12.0(

4

10x20

A

QV

2

3

The head loss in friction between A and D is

m655.212.0x81.9x2

768.1x100x02.0

g2

V

D

fLh

22A

A

AfA

Arbitrary datum

QA

QB

QC

zA=50m

zB=?m

Tank A

Tank C

zC=25m

LA= 100mDA= 120mm

LB= 60mDB= 75mm

Tank B

Lc= 40mDc= 60mm

D

365.47655.250hzH fAAD

The head loss in friction between D and C is

365.2225365.47zH CD

.

06.0x81.9x2

xV40x02.0365.22But

2C

Therefore, VC= 5.737 m/s

Flow to reservoir C

s/m01622.0737.5x)06.0(4

32

47

Arbitrary datum

QA

QB

QC

zA=50m

zB=?m

Tank A

Tank C

zC=25m

LA= 100mDA= 120mm

LB= 60mDB= 75mm

Tank B

Lc= 40mDc= 60mm

D

s/m596.0075.0x81.9x2

855.0x60x02.0 2

m77.46596.0365.47hHz fBDB

Head loss in friction between D and B =

s/m855.0)075.0(

4

10x78.3V,Therefore

2

3

B

From continuity equation,

)01622.002.0(QQQ CAB

s/m10x78.3 33

THE END