chapter 1: water flow in pipes university of palestine engineering hydraulics 2 nd semester...

CHAPTER 1:Water Flow in Pipes

CHAPTER 1:Water Flow in Pipes

University of PalestineEngineering Hydraulics2nd semester 2010-2011

1

2

Description of A Pipe Flow

• Water pipes in our homes and the distribution

system

• Pipes carry hydraulic fluid to various components

of vehicles and machines

• Natural systems of “pipes” that carry blood

throughout our body and air into and out of our

lungs.

3

• Pipe Flow: refers to a full water flow in a closed conduits or circular cross section under a certain pressure gradient.

• The pipe flow at any cross section can be described by:

cross section (A), elevation (h), measured with respect to a

horizontal reference datum. pressure (P), varies from one point to another,

for a given cross section variation is neglected The flow velocity (v), v = Q/A.

Description of A Pipe Flow

4

Pipe flow• The pipe is completely filled with the fluid being transported.

• The main driving force is likely to be a pressure gradient along the pipe.

Open-channel flow• Water flows without completely filling the pipe.

• Gravity alone is the driving force, the water flows down a hill.

Difference between open-channel flow and the pipe flow

5

Steady and Unsteady flow The flow parameters such as velocity (v),

pressure (P) and density (r) of a fluid flow are independent of time in a steady flow. In unsteady flow they are independent. 0

ooo ,z,yxtvFor a steady flow

0ooo ,z,yxtvFor an unsteady flow

If the variations in any fluid’s parameters are small, the average is constant, then the fluid is considered to be steady

Types of Flow

6

Uniform and non-uniform flow

A flow is uniform if the flow characteristics at any given instant remain the same at different points in the direction of flow, otherwise it is termed as non-uniform flow.

0ot

svFor a uniform flow

For a non-uniform flow 0ot

sv

Types of Flow

7

Examples:

The flow through a long uniform pipe diameter at a constant rate is steady uniform flow.

The flow through a long uniform pipe diameter at a varying rate is unsteady uniform flow.

The flow through a diverging pipe diameter at a constant rate is a steady non-uniform flow.

The flow through a diverging pipe diameter at a varying rate is an unsteady non-uniform flow.

Types of Flow

8

Laminar and turbulent flowLaminar flow:

Turbulent flow:

The fluid particles move along smooth well defined path or streamlines that are parallel, thus particles move in laminas or layers, smoothly gliding over each other.

The fluid particles do not move in orderly manner and they occupy different relative positions in successive cross-sections. There is a small fluctuation in magnitude and direction of the velocity of the fluid particles

Transitional flowThe flow occurs between laminar and turbulent flow

Types of Flow

9

Reynolds Experiment Reynolds performed a very carefully prepared

pipe flow experiment.

Types of Flow

10

Increasing flow velocity

11

Reynolds Experiment• Reynold found that transition from laminar to

turbulent flow in a pipe depends not only on the velocity, but only on the pipe diameter and the viscosity of the fluid.

• This relationship between these variables is commonly known as Reynolds number (NR)

ForcesViscous

ForcesInertialVDVDNR

It can be shown that the Reynolds number is a measure of the ratio of the inertial forces to the viscous forces in the flow

FI ma AFV

Types of Flow

12

Reynolds number

VDVD

NR

where V: mean velocity in the pipe [L/T]

D: pipe diameter [L]: density of flowing fluid

[M/L3]: dynamic viscosity [M/LT]: kinematic viscosity [L2/T]

Types of Flow

13

Types of Flow

14

Flow laminar when NR < Critical NR

Flow turbulent when NR > Critical

NR

It has been found by many experiments that for flows in circular pipes, the critical Reynolds number is about 2000

The transition from laminar to turbulent flow does

not always happened at NR = 2000 but varies due to

experiments conditions….….this known as

transitional range

Types of Flow

15

Laminar flows characterized by:

• low velocities•small length scales

•high kinematic

viscosities

•NR < Critical NR

•Viscous forces are

dominant.

Turbulent flows characterized by

•high velocities

• large length scales

• low kinematic

viscosities

•NR > Critical NR

• Inertial forces are

dominant

Laminar Vs. Turbulent flows

Types of Flow

16

Example 1 40 mm diameter circular pipe carries water at

20oC. Calculate the largest flow rate (Q) which laminar flow can be expected.

mD 04.0

CTat o20101 6

sec/1028.6)04.0(4

05.0. 352 mAVQ

sec/05.02000101

)04.0(2000

6mV

VVDNR

Types of Flow

17

Water flow in pipes may contain energy in three basic forms:

1- Kinetic energy.

2- potential energy.

3- pressure energy.

22

22

11

21

22h

P

g

Vh

P

g

V

Bernoulli Equation

Energy per unit weight of waterOR: Energy Head

Energy Head in Pipe Flow

18

Energy head and Head loss in pipe flow


19

11

21

1 2h

P

g

VH

22

22

2 2h

P

g

VH

Kinetic head

Elevation head

Pressure head

Energy head

= + +

LhhP

g

Vh

P

g

V 2

22

21

12

1

22

Notice that:• In reality, certain amount of energy loss (hL) occurs when the water mass flow from one section to another.• The energy relationship between two sections can be written as:


20

Example


ExampleIn the figure shown:Where the discharge through the system is 0.05 m3/s, the total losses through the pipe is 10 v2/2g where v is the velocity of water in 0.15 m diameter pipe, the water in the final outlet exposed to atmosphere.


21

Calculate the required height (h =?)below the tank

mh

h

hzg

V

g

pz

g

V

g

p

smV

smV

L

AQ

AQ

147.2181.9*2

83.21020

81.9*2

366.60)5(00

22

/366.610.0

05.0

/83.215.0

05.0

22

2

222

1

211

24

24


22

Without calculation sketch the (E.G.L) and (H.G.L)


23

24

Basic components of a typical pipe system

25

In General:When a fluid is flowing through a pipe, the fluid

experiences some resistance due to which some of energy (head) of fluid is lost.

Energy Losses(Head losses)

Major Losses Minor losses

loss of head due to pipe friction and to viscous dissipation in flowing water

Loss due to the change of the velocity of the flowing fluid in the magnitude or in direction as it moves through fitting like Valves, Tees, Bends and Reducers.

Calculation of Head (Energy) Losses

Part A:Major Losses

Head Losses in Pipelines


26

• Energy loss through friction in the length of pipeline is commonly termed the major loss hf

• This is the loss of head due to pipe friction and to the viscous dissipation in flowing water.

• Several studies have been found the resistance to flow in a pipe is:

- Independent of pressure under which the water flows- Linearly proportional to the pipe length, L- Inversely proportional to some water power of the pipe

diameter D- Proportional to some power of the mean velocity, V- Related to the roughness of the pipe, if the flow is

turbulent

Losses of Head due to Friction

27

Energy Head & Head loss in pipe flow

Losses of Head due to Friction

28

Several formulas have been developed in

the past. Some of these formulas have

faithfully been used in various hydraulic

engineering practices.

1. Darcy-Weisbach ( f )

2. Hazen-William (CHW)

3. Manning (n)

4. The Chezy Formula

5. The Strickler Formula

Major losses formulas

29

30

The resistance to flow in a pipe is a function of:

• The pipe length, L

• The pipe diameter, D

• The mean velocity, V

• The properties of the fluid .

• The roughness of the pipe, (the flow is turbulent).


Darcy-Weisbach Equation

25

22

8

2

Dg

QLf

g

V

D

LfhL

Where: f is the friction factorL is pipe lengthD is pipe diameterQ is the flow ratehL is the loss due to friction

It is conveniently expressed in terms of velocity (kinetic) head in the pipe

D

VDF

D

VDF

DRFf e

,,,

The friction factor is function of different terms:

Renold number Relative roughness

Re

s

NR

ke


31

32

Friction Factor: (f )

• For Laminar flow: (NR < 2000) [depends only on Reynolds’ number and not on the surface roughness]

RN

64f

• For turbulent flow in smooth pipes (e/D = 0) with 4000 < NR < 105 is

4/1

316.0

RNf


2.51log2

1 fN

fR

510e

7.3log21

RNfor

D

f• Colebrook-White Equation for f

fND

e

f R

51.2

7.3ln86.0

1

For turbulent flow ( NR > 4000 ) with e/D > 0.0, the friction factor can be founded from:• Th.von Karman formulas:

33


There are other Equation such as Karman Equation see Text book 2

Pandtle - Colebrook Equation

510

713log21

eRfor

use

D.

f


There is some difficulty in solving this equationSo, Miller improve an initial value for f , (fo)

2

9.0

74.5

7.3log25.0

R

oND

ef

The value of ffoo can be use directly as ff if: 26

83

101101

101104-

R

D e

N

34

e7.1'

ee 7.108.0 '

RNf

64

pipe wall

e

51.2log2

110

fN

fR

e

pipe wall

transitionallyrough

e

pipe wall

rough

f independent of relative roughness e/D

f independent of NR

f varies with NR and e/D

turbulent flow

NR > 4000

laminar flow

NR < 2000

e08.0'

e

D

f7.3log2

110

fN

De

f R

51.2

7.3log2

110

Colebrook formula

The thickness of the laminar sublayer decrease with an increase in NR

Smooth

Friction Factor f

35

A convenient chart was prepared by Lewis F. Moody and commonly called the Moody diagram of friction Moody diagram of friction factors for pipefactors for pipe flow, There are 4 zones of pipe flow in the chart:

• A laminar flow zone where f is simple linear function of Re

• A critical zone (shaded) where values are uncertain because the flow might be neither laminar nor truly turbulent

• A transition zone where f is a function of both Re and relative roughness

• A zone of fully developed turbulence where the value of f depends solely on the relative roughness and independent of the Reynolds Number

Moody diagram

36

Moody diagram

37

Laminar

Marks Reynolds Number independence

critical

Transition

Moody diagram

38

39

Notes:

• Colebrook formula

is valid for the entire nonlaminar range (4000 < Re < 108) of the Moody chart

12

3 7

2 51

f

e D

f

log

/

.

.

Re

In fact , the Moody chart is a graphical representation of this equation

Moody diagram

Bonus:

Find the theoretical formulation for friction factor for laminar flow.

Re

64f

Moody diagram

40

41

Typical values of the absolute roughness (e) are given

Absolute roughness

Materials Roughness

Absolute roughness

42

Three types of problems for uniform flow in a single pipe:

Type 1:Given the kind and size of pipe and the flow rate head loss ?

Type 2:Given the kind and size of pipe and the head loss

flow rate ?

Type 3:Given the kind of pipe, the head loss and flow rate

size of pipe ?

Problems (head loss)

43

Problems type I (head loss)

44

45

The water flow in Asphalted cast Iron pipe (e = 0.12mm) has a diameter 20cm at 20oC. Is 0.05 m3/s. determine the losses due to friction per 1 km

1.59m/sm0.2π/4

/s0.05mV

22

3

56

26

1015.33148521001.1

2.059.1

0006.0200

12.0

12.0

/sm101.01υ20

VD

N

mm

mm

D

e

mme

CT

R

o

f = 0.018 Moody

m

m/s.

.

m.

m,.

g

V

D

Lfh f

55.11

8192

591

200

00010180

2 2

22


Example 2

The water flow in commercial steel pipe (e = 0.045mm) has a diameter 0.5m at 20oC. Q=0.4 m3/s. determine the losses due to friction per 1 km

sm

A

QV / 037.2

45.0

4.02

013.0

109105.0

045.0

10012.110006.1

037.25.0

10006.15.4220

10497

5.42

10497

53

66

65.1

6

5.1

6

f

D

e

N

T

Moody

R

kmmh f / 5.581.92

037.2

5.0

1000013.0

2


Example 3

46

fRD

k

f e

s 51.2

7.3ln86.0

1

Use other methods to solve f

01334.0

10012.1

74.5

7.3

109log25.0

74.5

7.3log25.0

2

9.06

52

9.0

e

so

R

Dkf

678.866.8

01334.0

51.2

7.3

109ln86.0

01334.0

1 5

eR

1 -Cole brook

kmmh f / 5.581.92

037.2

5.0

100001334.0

2

Example 3-cont.

47

2- Pandtle - Colebrook Equation

solutionlastf

.f

D.

f

01334.0012.0

045.0

500713log2

1

713log21

Example 3-cont.

48

Problems type II (head loss)

49

Method for solution of Type 2 problems

50

Re

lati

ve

ro

ug

hn

es

s e

/D

51

Example 4:

52

Example 4:

54

Cast iron pipe (e = 0.26), length = 2 km, diameter = 0.3m. Determine the max. flow rate Q , If the allowable maximum head loss = 4.6m. T=10oC

10135.0

81.923.0

20006.4

2

2

2

2

fV

Vf

g

V

D

LfhF

00009.01067.8103.0

26.0

210296.21031.1

3.0

1031.15.4210

10497

5.42

10497

53

66

65.1

6

5.1

6

D

e

VV

N

T

R

Type 2:Given the kind and size of pipe and the head loss flow rate ?

Example 5:

61

02.0

1067.8

10668.2

m/s 16.101.0

4

52

1

f

D

e

N

Vf

Moody

Req

eq

021.0

1067.8

10886.1

m/s 82.002.0

4

52

1

f

D

e

N

Vf

Moody

Req

eq

10135.02 f

V

210296.2 6 VNR

Trial 1

Trial 2

V= 0.82 m/s , Q = V*A = 0.058 m3/s

Another solution?

Example 5:cont.

62

Compute the discharge capacity of a 3-m diameter, wood stave pipe in its best condition carrying water at 10oC. It is allowed to have a head loss of 2m/km of pipe length.

hf fL

D

V 2

2g

V 2ghf

L

1/ 2D

f

1/ 2

fV

Vf

12.0

)81.9(23

10002 2

2

Table 3.1 : wood stave pipe: e = 0.18 – 0.9 mm, take e = 0.3 mm

Solution 1:

0001.03

3.0

D

e

At T= 10oC, = 1.31x10-6 m2/sec VVVD

NR .1029.21031.1

3 6

6

Type 2:Given the kind and size of pipe and the head loss flow rate ?

Example 6:

63

• Solve by trial and error:• Iteration 1:• Assume f = 0.02 sec/45.2

02.0

12.02 mVV

66 106.545.2.1029.2 RN

From moody Diagram: 0122.0f

Iteration 2:update f = 0.0122 sec/14.3

0122.0

12.02 mVV

66 102.714.3.1029.2 RN

From moody Diagram: 0122.00121.0 f

0 0.02 2.45 5.6106

1 0.0122 3.14 7.2106

2 0.0121

Iteration f V NR

Convergence

Solution:

/sm 2.2724

3.15.3

m/s 15.3

3

2

2

VAQ

V

Another solution?64

Compute the discharge capacity of a 3-m diameter, wood stave pipe in its best condition carrying water at 10oC. It is allowed to have a head loss of 2m/km of pipe length.

5

6

232/12/3

1062.91000

)3)(81.9(2

1031.1

)3(2

L

ghDfN f

R

Table 3.1 : wood pipe: e = 0.18 – 0.9 mm, take e = 0.3 mm

Solution 2:

0001.03

3.0

D

e

Type 2: Given the kind and size of pipe and the head loss flow rate?

At T= 10oC, = 1.31x10-6 m2/sec

From moody Diagram: 0121.0f

sec/15.32

2

2/12/12

mf

D

L

ghV

g

V

D

Lfh f

f

/sm 2.2724

3.15.3,

3

2

VAQ

Example 6:Another solution?

65

f = 0.0121

66

Problems type III (head loss)

67

Example 7:

68

Example 7:cont.

70

Estimate the size of a uniform, horizontal welded-steel pipe installed to carry 14 ft3/sec of water of 70oF (20oC). The allowable pressure loss is 17 ft/mi of pipe length.

From Table : Steel pipe: ks = 0.046 mm

Darcy-Weisbach:

hL fLD

V 2

2g

Q VA

hL fLD

QA

2

2gf

LD

Q 2

2g42

2D 4 1

D 5

16fLQ 2

2g 2

5/1

2

28

Lhg

fLQD

afff

D 5/15/1

5/1

2

2

33.41781.9

1452808

Let D = 2.5 ft, then V = Q/A = 2.85 ft/sec

Now by knowing the relative roughness and the Reynolds number:

55

10*6.610*08.1

5.2*85.2

VDNR

0012.05.2

003.0

D

e

We get f =0.021

Solution 2:

Example 8:

71

A better estimate of D can be obtained by substituting the latter values into equation a, which gives

ftfD 0.2021.0*33.433.4 5/15/1

A new iteration provide V = 4.46 ft/secNR = 8.3 x 105 e/D = 0.0015f = 0.022, andD = 2.0 ft.More iterations will produce the same results.

Example 8:cont.

72

Example 9:

74

• Hazen-Williams

54.063.085.0 SRCV hHW

tCoefficien iamsHazen Will

4 P wetted

A wetted Radius hydraulic

CL

hS

D

R

HW

f

h

UnitSI

0.71 852.1

87.4852.1 QDC

Lh

HW

f

Sim

pli

fied

UnitsBritishSRCV

mVcmD

hHW54.063.0318.1

sec/0.35

Empirical Formulas 1

75


76

tCoefficien iamsHazen WillHWC


77


78

79

081.0

sec/0.3

V

VCC

mVWhen

oHoH

Where:CH = corrected valueCHo = value from tableVo = velocity at CHo

V = actual velocity


• Manning

tCoefficien M

4 P wetted

A wetted Radius hydraulic

anningnL

hS

D

R

f

h

Sim

pli

fied

UnitSI

0.3133.5

2

D

nQLh f

2/13/21SR

nV h

• This formula has extensively been used for open channel designs. It is also quite commonly used for pipe flows


80

81

• n = Manning coefficient of roughness (See Table)• Rh and S are as defined for Hazen-William

formula.

Vn

R Sh1 2 3 1 2/ /

3/16

223.10

D

QLnh f

2233.1

35.6 VnD

Lh f


82


tCoefficien Manningn


83

84

The Chezy Formula

V C R Sh 1 2 1 2/ /

2

4

C

V

D

Lh f

where C = Chezy coefficient


85

• It can be shown that this formula, for circular pipes, is equivalent to Darcy’s formula with the value for

[f is Darcy Weisbeich coefficient]

• The following formula has been proposed for the value of C:

[n is the Manning coefficient]

Cg

f

8

C S n

S

n

Rh

230 00155 1

1 230 00155

.

(.

)


86

The Strickler Formula:

V k R Sstr h 2 3 1 2/ /

2

33.135.6

strf k

V

D

Lh

where kstr is known as the Strickler coefficient.

Comparing Manning formula and Strickler formula, we can see that

1

nkstr


87

Relations between the coefficients in Chezy, Manning , Darcy , and Strickler formulas.

nkstr

1

6/1hstr RkC

g

Rfn h

8

3/1

Empirical Formulas

New Cast Iron (CHW = 130, n = 0.011) has length = 6 km

and diameter = 30cm. Q= 0.32 m3/s, T=30o. Calculate the head loss due to friction using:

a) Hazen-William Method

b) Manning Method

33332030130

6000710

710

85218748521

8521

8748521

m . .

. h

Q DC

L.h

...f

.

..HW

f

m

.

.. .h

D

nQ L h

.f

.f

47030

32001106000310

3.10

335

2

335

2

Example 10

88

Part B:Minor Losses

Head Losses in Pipelines


89

• Additional losses due to entries and exits, fittings and valves are traditionally referred to as minor losses

2

22

22 gA

Qk

g

Vkh LLm

Minor Losses

90

91

It is due to the change of the velocity of the flowing fluid in the magnitude or in direction [turbulence within bulk flow as it moves through and fitting]

Flow pattern through a valve

Minor Losses

92

• The minor losses occurs du to :• Valves • Tees• Bends• Reducers• Valves• And other appurtenances

• It has the common form

2

22

22 gA

Qk

g

Vkh LLm

can be the dominant cause of head loss in shorter pipelines

“minor” compared to friction losses in long pipelines but,

Minor Losses

Losses due to contraction

g

Vkh cc 2

22

A sudden contractionA sudden contraction in a pipe usually causes a marked drop in pressure in the pipe due to both the increase in velocity and the loss of energy to turbulence.

Along centerline

Along wall

Minor Losses

94

Value of the coefficient Kc for sudden contraction

VV22

Minor Losses

95

96

Head Loss Due to a Sudden Contraction

h KV

gL L 22

2

g

VhL 2

5.02

2

Minor Losses

96

Head losses due to pipe contraction may be greatly reduced by introducing a gradual pipe gradual pipe transition transition known as a confusor confusor

g

V'k'h cc 2

22

'kc

Minor Losses

97

98

Head Loss Due to Gradual Contraction (reducer or nozzle)

g

VVKh LL 2

21

22

100200300400

KL0.20.280.320.35

A different set of data is:

Minor Losses

98

Losses due to Enlargement

g

VVhE 2

)( 221

A sudden EnlargementA sudden Enlargement in a pipe

Minor Losses

99

Head losses due to pipe enlargement may be greatly reduced by introducing a gradual pipe gradual pipe transition transition known as a diffusor diffusor

g

VV'k'h EE 2

22

21

Minor Losses

100

Note that the drop in the energy line is much larger than in the case of a contraction

abrupt expansion

gradual expansion

smaller head loss than in the case of an abrupt expansion

Minor Losses

101

102

Head Loss Due to a Sudden Enlargement

h KV

gL L 12

2

KA

AL

1 1

2

2

h

V V

gL 1 2

2

2

or:

Minor Losses

103

Head Loss Due to Gradual Enlargement (conical diffuser)

g

VVKh LL 2

22

21

100200300400

KL0.390.801.001.06

Minor Losses

104

Gibson tests Minor Losses

Loss due to pipe entranceGeneral formula for head loss at the entrance of a pipe is also expressed in term of velocity head of the pipe

g

VKh entent 2

2

Minor Losses

105

Different pipe inlets

increasing loss coefficient

Minor Losses

106

107

Head Loss at the Entrance of a Pipe (flow leaving a tank)

Reentrant)embeded(

KL = 0.8

Sharpedge

KL = 0.5

Wellrounded

KL = 0.04

SlightlyroundedKL = 0.2

h KV

gL L2

2

Minor Losses

108

Another Typical values for various amount of rounding of the lip

Minor Losses

Loss at pipe exit (discharge head loss)

In this case the entire velocity head of the pipe flow is dissipated and that the discharge loss is

g

Vhexit 2

2

Minor Losses

109

110

Head Loss at the Exit of a Pipe (flow entering a tank)

hV

gL 2

2

the entire kinetic energy of the exiting fluid (velocity V1) is dissipated through viscous effects as the stream of fluid mixes with the fluid in the tank and eventually comes to rest (V2 = 0).

KL = 1.0 KL = 1.0

KL = 1.0 KL = 1.0

Minor Losses

Loss of head in pipe bends

g

Vkh bb 2

2

Minor Losses

111

112

Miter bends

For situations in which space is limited,

Minor Losses

Loss of head through valves

g

VKh vv 2

22

Minor Losses

113

Minor Losses

114

115

The loss coefficient for elbows, bends, and teesMinor Losses

Loss coefficients for pipe components (Table)

Minor Losses

116

Minor loss coefficients (Table)

117

Minor loss calculation using equivalent pipe length

f

DkL l

e

Minor Losses

118

119

• Note that the above values are average typical values, actual values will depend on the make (manufacturer) of the components.

• See:– Catalogs – Hydraulic handbooks !!

Minor Losses

Energy and hydraulic grade lines

Unless local effects are of particular interests the changes in the EGL and HGL are often shown as abrupt changes (even though the loss

occurs over some distance)

Minor Losses

120

Example 11In the figure shown two new cast iron pipes in series, D1 =0.6m , D2 =0.4m length of the two pipes is 300m, level at A =80m , Q = 0.5m3/s (T=10oC).there are a sudden contraction between Pipe 1 and 2, and Sharp entrance at pipe 1.Fine the water level at B

e = 0.26mmv = 1.31×10-

6Q = 0.5 m3/s

Minor Losses

121

exitcentffL

fBA

hhhhhh

hZZ

21

g

Vk

g

Vk

g

Vk

g

V

D

Lf

g

V

D

Lfh exitcentL 22222

22

22

21

22

2

22

21

1

11

01800170

000650000430600

26.0

102211018

sec98340

4

50sec771

604

50

21

11

6222

5111

222

211

.f .f

,.D

, .D

,.υ

DV R , .

υ

DVR

, m/..

π.

A

Q, V m/.

.π

.

A

QV

moodymoody

ee

1 ,27.0 ,5.0 exitcent hhh

Solution

Minor Losses

122

m.g

.

g

..

g

..

g

. .

. .

g

. .

. .h f

36132

983

2

983270

2

77150

2

983

40

3000180

2

771

60

3000170

222

22

ZB = 80 – 13.36 = 66.64 m

g

Vk

g

Vk

g

Vk

g

V

D

Lf

g

V

D

Lfh exitcentL 22222

22

22

21

22

2

22

21

1

11

Minor Losses

123

Example 12A pipe enlarge suddenly from D1=240mm to D2=480mm. the H.G.L rises by 10 cm calculate the flow in the pipe

Minor Losses

124

Solution

125

smAVQsmV

g

V

g

VV

g

V

g

V

VV

VV

AVAV

g

VV

g

V

g

V

zg

pz

g

ph

g

V

g

V

hzg

V

g

pz

g

V

g

p

e

e

/103.048.057.0/57.0

1.02

6

1.02

4

22

16

4

48.024.0

1.0222

22

22

324222

22

2

222

22

2

21

242

241

2211

2

212

22

1

11

22

22

21

2

222

1

211

Solution

Minor Losses

126

http://www.haestad.com/library/books/awdm/online/wwhelp/wwhimpl/java/html/wwhelp.htm

127

chapter 1: water flow in pipes university of palestine engineering hydraulics 2 nd semester...

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