8. fm 9 flow in pipes major loses co 3 copy

FLUID MECHANICS – 1FLUID MECHANICS – 1Semester 1 2011 - 2012Semester 1 2011 - 2012

Compiled and modifiedCompiled and modified

byby

Sharma, AdamSharma, Adam

Week – 8, 9 and 10

FLOW IN PIPES- FLOW IN PIPES- CO3CO3

Review

• Conservation of mass (mass balance)

• Continuity Equation

• Bernoulli Equation

• Momentum Equation

3

Objectives• Understand Laminar and Turbulent flow in pipes

• Identify types of flow using Reynolds number

• Explain minor losses and major losses for flow in pipes

• Determine friction factor and major losses using moody chart and simplified Colebrook equation

• Calculate minor losses, major losses, pressure losses and head losses

• Understand equivalent length

• Solve piping system using Bernoulli equation considering all losses.

4

1. Introduction• There are two kind of flow

a. Internal flow ( flow in pipes)

b. External flow ( flow over bodies, drag, lift) – will be covered in fluids mechanics 2 (BMM2543)

• Examples of internal flow

1. Water flow in pipes

2. Blood flow

3. Oil and Gas industry

4. Cooling system of a car (Radiator)

5. Air Conditioning (Chilled water system)

5

examples & pictures

6

7

2. Types of flow• Flow in pipes has three types, Laminar,

Transition, Turbulent

• Laminar flow

―Smooth streamlines (flow)

―Highly ordered motion

―Short in length

― normally appeared in high viscosity flow and small pipe/passage i.e. oil in small pipe

Transition flow

8

2. Types of flow (cont’)• Turbulent flow

―Rough streamlines (flow)

―Highly disordered motion

―Most flow in reality is turbulent

―High momentum, thus high friction

• Transition flow

―flow from laminar does not suddenly change to turbulent, it will enter transition flow first

―Normally ignored in calculation

―fluctuation of laminar and turbulent randomly

Is there other method to distinguish these flow?

9

• Yes, use Reynolds number, Re

pipe ofdiameter Internal

velocitypipe Average

)(kg/m fluid flowing ofdensity

) m.skg( , viscositydynamic

) sm( , viscositykinematic

avg

3

2

D

V

• Re < 2300, Laminar flow

• 2300 < Re < 4000, Transition flow

• Re > 4000, Turbulent flow

Is there other method to distinguish these flow?(cont)

10

• For non-circular pipes, hydraulic diameter, Dh is used for calculating the Reynolds number.

• however, in practical, type of flow depend on smoothness of pipe, vibrations and fluctuation in the flow.

p

AD c

h

4

Perimeter Internal

(internal) Area onalCrosssecti4

Reynolds Number (example)

11

• Q1: Water at 20°C flow with average velocity of 2cm/s inside a circular pipe. Determine flow type if the pipe diameter, a) 2 cm, b) 15 cm, and c) 30 cm

33- 998kg/m m.skg 101.002

C20At 2010) book, (Cengel tableFrom

,

• (c) Turbulent flow59700.001002

0.3)998(0.02)(Re c)

DVave

3980.001002

0.02)998(0.02)(Re a)

DVave • (a) Laminar flow

29880.001002

0.15)998(0.02)(Re b)

DVave • (b) Transition flow

Reynolds Number (example) cont’

12

• Q2: Water at 20°C flow in a circular pipe of 3.5 cm diameter. Determine the range for the average velocity so the flow is always transition flow

33- 998kg/m m.skg 101.002

C20At 2010) book, (Cengel tableFrom

,

m/s 0660.035)0998(

0.0010022300.V min,ave

0.001002

.035)0)(998(2300Re min,aveave

imummin

VDV

m/s 1150.035)0998(

0.0010024000.V max,ave

0.001002

.035)0)(998(4000Remaximum

max,aveave VDV

m/s 0.115 m/s 0660 Range aveV.

Reynolds Number (example) cont’

13

• Q3: Air at 35°C flow inside rectangular pipe of 2cmx5cm. Determine the maximum flow velocity for the pipe before the flow enter transition region

sm 106551 C,35At Air

2010), book, (Cengel tableFrom25- .

m/s31310290

1065512300 5

..

.V max,ave

5-101.655

.029)0(2300Re

max,aveave

imummax

VDV

m 02902(0.05)2(0.02)

05)4(0.02)(0.4.

p

AD c

h

4. Entrance Region

14

• Between the entrance and fully developed flow

• Uniform velocity profile at entrance

• because of no slip boundary condition, friction at the wall reduce the velocity of flow near the wall

• to conserve mass, velocity at the center increase (compensate velocity decreased near the wall)

4. Entrance Region (cont’)

15

• As the fluid move deeper in the pipe, the velocity near the wall decreased further and velocity at center increase (developing velocity profile)

• Both (up & down @ left & right) velocity profile increase till it merge with the other side

• fully developed flow is when the velocity profile stop to develop as it flows deeper inside the pipe

4. Entry Region (cont’)

16

• hydrodynamic entry length, Lh is evaluated from pipe entrance to where wall shear stress achieve 2% of fully developed value or approximately

D.L arminla,h Re050 shorter length for turbulent pipe flow

DL turbulent,h 10

4. Entry Region (example)

17

• Q4: Determine the hydrodynamic entry region for Q1 (a & c).

D.L arminla,h Re050

• (c) Turbulent flow59700.001002

0.3)998(0.02)(Re c)

DVave

3980.001002

0.02)998(0.02)(Re a)

DVave • (a) Laminar flow

m 3980020398050Re050 ...D.L arminla,h

m 200201010 ..DL turbulent,h

DL turbulent,h 10

4.1. Turbulence velocity profile VS Laminar velocity profile (fully developed)

18

• Velocity profile based on analysis

• Consist of 1 layer

• Small velocity gradient

• The average velocity in fully developed laminar pipe flow is ½ of the maximum velocity

• Velocity profile is based on analysis and empirical

• Consist of 4 layer

• High velocity gradient

avemax Vu 2

5. Losses in piping system

19

• There are two type of losses which is major losses and minor losses. Losses are mainly due to friction and obstruction

• Total losses, hL is major losses + minor losses

5.1 Major Losses (pg345)• Major losses, hL, major , is also known pressure

losses or head losses

• The major losses is solely depend on the pipe, nothing else

5.1 Major Losses (cont’)

20

• Pressure drop across pipe can be formulated as

• f, Darcy friction factor. There is another friction factor, Fanning friction factor, but will not be covered

• for fully developed laminar flow, friction factor is obtained from combining Darcy equation with Pressure drop.

L

PP

dx

dP 21

2221

328

D

LV

R

LVPPP aveave

2 Rough)or (Smooth loss Pressure

2ave

L

V

D

LfP

g

V

D

Lf

g

Ph aveL

major,L 2 loss, headmajor

2


21

2

32

2

2aveave

L

V

D

Lf

D

LVPP

2

32 aveV

fD

D

Vf ave 32

2

232

DVf ave

64Re f

flow)laminar developedfully for factor (friction Re

64 f

• from the eq. ,friction factor in Laminar flow is independent of roughness

• The head loss represents the additional height that the fluid needs to be raised by a pump in order to overcome the frictional losses in the pipe


22

• For fully developed turbulent flow pressure losses or head losses, the equation is more complex

• Unlike laminar flow, friction factor, f , for turbulent flow is depend on internal pipe roughness and Reynolds number base on Colebrook equation.

• Whereas, ε is roughness of pipe and D is pipe diameter.

• However this equation is implicit and cannot be solved directly. Using iteration (numerical method) is possible but the method is tedious

flow) (turbulent Re

512

73 log 2.0

1

f

.

.

D-

f


23

• Hence, there are two option, (a), Moody chart, (b) Modified Colebrook equation

• In Moody Chart, friction factor can be obtain by knowing the roughness ratio, ε/D and Reynolds number

ε/D=0.02

Re=100,000

f =0.025


24

• Haaland modified Colebrook Equation (1983).

• Easier to calculate

• The formula has about 2% error compare to original equation

• Formula for major head loss for turbulent is the same

flow)ent for turbul (Modified 73Re

96 log 81

1

111

.

.

D..-

f

d)(simplifie 73Re

96 log 81

-2111

.

.

D..-f

g

V

D

Lfh ave

major,L 2

2


25

• In real pipe application (turbulent flow), friction is unwanted because the rougher the surface, the higher the friction

• Old piping system such as Cast Iron, GI, the performance deteriorate through time as corrosion reduce smoothness and size of internal pipe.

• New piping (HDPE), Anti-corrosion metal pipes, smoothness maintain, thus performance is maintained

5.1 Major Losses (example)

26

• Q5: Water at 40°C ( = 992.1 kg/m3 and µ = 0.653×10-3 kg/m.s) is flowing steadily in a 5cm-diameter horizontal pipe made of stainless steel at a rate of 300 l/min. Determine the pressure drop, the head loss, and the required pumping power input for flow over a 50m-long section of the pipe.

m/s5524(0.05)

(1000)/60300

2avgavg ./

/

A

QVVAQ

cc

Turbulent ,1937100.000653

.05)0)(992.1(2.55Re

DVavg

0.00004mm50mm0020 ratio, Roughness /.D0.016 101.94 Re & 0.00004 Chart,Moody from 5 f,D

0160 73

000040

193710

96 log81

-2111

..

...f

.

Pa 516092

(2.55)1992

050

500160

2

22

.

..

V

D

LfP ave

L

m 35(9.81)1992

51609.

.g

Ph L

L

Watt25851609601000300 //PQW L

5.1 Major Losses (example cont’)

27

• Q6: Air at 40°C (ρ = 1.127 kg/m3, ν = 1.702×10-6 m2/s) is flowing steadily in a 50cm-diameter horizontal pipe made of plastic at a rate of 30 l/min. Determine the head loss, and the required pumping power input for flow over a 150m-long section of the pipe.

m/s105524(0.5)

(1000)/6030 3

2avgavg .

/

/

A

QVVAQ

cc

Laminar ,74820.00000170

.5)0)(10552(Re

3

.DVavg

08560748

64

Re

64 .f

Pa 1041981)(1.127)(9.10518 56 ..ghP LL

m 10518(2)(9.81)50

)10552(15008560

26

232

..

..

g

V

D

Lfh ave

L

Watt107141041960100030 85 ..//PQW L

28

5.2 Minor Losses• In piping system there are various fittings,

valves, bends, elbows, tees, inlets, exits, enlargements, and contractions in addition to the pipes.

• These component Interrupt the smoothness of flow and cause additional losses

• Usually, these additional losses is called minor losses and smaller than pipe losses .

• Rarely, minor losses will be greater than major losses especially when the component installed frequently along the pipe system (between short distance)

29

5.2 Minor Losses (cont’)

30


• usually expressed in terms of the loss coefficient, KL

• KL , is provided by manufacturer and the value varies for different components.

• Consider a component, valve, the pressure losses is losses due to valve minus losses by imaginary pipe section without valves

31


2

2avg

LL

VKP

• Independent of Re, Reynolds number

• Usually expressed in term of head losses, hL

• Also, In industrial application, the manufacturing data is expressed in terms of equivalent length, Lequiv.

• for this method, simply add Lequiv to the total length of pipe for the total head losses calculation.

• However, the former expression will be used thoroughly

components todue lossminor 2

2

g

VKh avg

LL

32


Lequivavgequivavg

LL Kf

DL

g

V

D

Lf

g

VKh

22

22

• Component with sharp edge such as sharp edge exit has higher loss coefficient compare to well rounded

• Sharp edge introduce recirculating flow due to fluid unable to make sharp 90° turn especially at high speed

• Same for sudden expansion/contraction

33


34


• These loss coefficient depends on the manufacturer data

35


36


37


38

Globe valve

Angle valve

Ball valve

Swing checkvalve

• Total head losses, hL, total

• Calculation of Minor losses is straight forward. If the piping system consist same components such as bends, simply multiply the loss coefficient with the number of the same bends.

g

VK

D

Lfhhh avg

LMinor,LMajor,Ltotal,L 2

2

39


diameter pipedifferent for different is

diameter) pipe one than more with system pipe(for

2

avg

2avg,

V

g

VK

D

Lfh i

Li

iitotal,L i

• Normally, as an engineer and consultant, piping system for example water storage, sprinkler system, hose reel system is the main concern.

• To solve piping system, extended Bernoulli equation is required which the total losses is placed at the right hand side (at point 2)

• Piping system usually constructed to deliver fluid at higher level or to create a pressurized system

total,Lhzg

V

g

Pz

g

V

g

P 2

222

1

211

2

)(

2

)(

40

6 Piping system

• Two principles in analyzing piping system which are

a) Conservation of mass throughout the system must be satisfied

b) Pressure drop (and thus head loss) between two junctions must be the same for all paths between the two junctions

41

6 Piping system (cont’)

1

2

point 1, Just above water level, P1=?, V1 = ?, Z1 = ?

point 2, Just above water level, P2=?, V2 = ?, Z2 = ?

h

• Q7:A piping system delivering water at 25°C from tank 1 to tank 2. The system consist two 45º, a sharp entrance and a sharp exit. The diameter of the stainless steel pipe is 2cm and length of 55 m. Determine h so that the flowrate is 83.3 L/min.

42

6 Piping system (example)

point 1, Just above water level, P1=0, V1 = 0, Z1 = 0

point 2, Just above water level, P2=0, V2 = 0, Z2 = h

2

1

h=?m45°

45°

Tank 1

Tank 2

43


OH

22avg

major 2m 2949

(9.81)2

(4.42)

020

550180

2.

..

g

V

D

Lfh ,L

m/s424(0.01)

(60000)383

2avgavg ./.

A

QVVAQ

cc

Turbulent ,98916100.891

.02)0997(4.42)(Re

3-

DVavg

000100.02102 , ratio Roughness

m102mm0020 , Roughness steel, Stainless6

6

.D/

.

0180 73

00010

98916

96 log81

-2111

..

...f

.

44


total,Lhh 00000

total,Lhzg

V

g

Pz

g

V

g

P 2

222

1

211

2

)(

2

)(

0 0 0 0 Reference point

m 5851.h

OH

22avg

minor 2m 292

(9.81)2

(4.42)140250

2...

g

VKh L,L

m58512922949 ...hhh Minor,LMajor,Ltotal,L

• Q8:A piping system delivering water at 25°C from pressurised tank 1 to tank 2. Initial plastic pipe diameter is 2 cm has a sharp entrance(inlet), fully open globe valve and two 90º smooth flanged bend, and a sudden expansion at the 1/3 of the pipe length. After expansion, Galvanized Iron (GI) pipe diameter is installed with 8 cm diameter and along the GI pipe, there are two 90º miter bend, a fully open angle valve and a sharp exit. The total pipe length is 75 m. Determine gauge pressure at tank 1 required to deliver water at 226 liter/hour.

45


point 1, Just above water level, P1≠0, V1 = 0, Z1 = 0

point 2, Just above water level, P2=0, V2 = 0, Z2 = h

1

h=25m

2

Tank 2

Tank 1

GI pipe

Plastic pipe

46

6 Piping system (example)m/s20

(0.01)

000)1(3600226

21

avg1avg11 ./

A

QVVAQ

cc

Turbulent ,4475100.891

.02)0997(0.2)(Re

3-

avg11

DV

pipessmooth 0 , Roughness pipe, Plastic ,

factor)friction (turbulent 0390 73

0

4475

96 log81

-2111

1 ..

..f

.

m/s01250(0.04)

(3600000)226

22

avg2avg22 ./

A

QVVAQ

cc

Laminar ,1119100.891

.08)0)(997(0.0125Re

3-

avg22

DV

factor)friction (laminar 0570 1119

64

Re

64 2 .f

47


g

VKKKK

D

Lfh ,L,L,L,Lplastictotal,L 2

22

avg1expansionsudden bend flangesmooth g.valveinlet

1

11

880080

02011

2

2

22

22

21

expansionsudden ..

.

D

DK ,L

m1240(9.81)2

20880(0.3)21050

020

3750390

2

..

...

/.h plastictotal,L

m000350(9.81)2

0125015(1.1)2

080

375750570

2

I ..

.

/.h Gtotal,L

g

VKKK

D

Lfh ,L,L,LGtotal,L 2

22

avg2exit sharp valveanglebendmiter

2

22I

m1243500003501240I plastic ...hhh Gtotal,Ltotal,Ltotal,L

48


1243502500001 .g

P

total,Lhzg

V

g

Pz

g

V

g

P 2

222

1

211

2

)(

2

)(

0 0 0Reference point

small very is losses as differenceelevation the

overcome tois 1at tank pressure theofMost

Pa24573081999712435251 ..P

49

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8. fm 9 flow in pipes major loses co 3 copy