Slide07 Extra

HEADLOSS

ESTIMATION Mekanika Fluida 1

HST

1

Friction Factor : Major losses

• Laminar flow

• Hagen-Poiseuille

• Turbulent (Smooth, Transition, Rough)

• Colebrook Formula

• Moody diagram

• Swamee-Jain

2

Hagen-Poiseuille

Darcy-Weisbach

Laminar Flow Friction Factor

3

2

32

lhDV

L

f 2

32 LVh

gD

2

f f2

L Vh

D g

2

2

32f

2

LV L V

gD D g

64 64f

ReVD

Slope of ___ on log-log plot

f 4

128 LQh

gD

-1

Turbulent Pipe Flow Head Loss

• ___________ to the length of the pipe

• Proportional to the _______ of the velocity (almost)

• ________ with surface roughness

• Is a function of density and viscosity

• Is __________ of pressure

4

Proportional

Increases

independent 2

f f2

L Vh

D g

square

(used to draw the Moody diagram)

Smooth, Transition, Rough

Turbulent Flow

• Hydraulically smooth

pipe law (von Karman,

1930)

• Rough pipe law (von

Karman, 1930)

• Transition function for

both smooth and

rough pipe laws

(Colebrook)

5

1 Re f2log

2.51f

1 2.512log

3.7f Re f

D

1 3.72log

f

D

2

f2

f

L Vh

D g

Moody Diagram

6

0.01

0.1

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

fric

tio

n fa

cto

r

laminar

0.05

0.04

0.03

0.02

0.015

0.01 0.008 0.006

0.004

0.002

0.001 0.0008

0.0004

0.0002

0.0001

0.00005

smooth

f p

DC

l

D

Swamee-Jain

• 1976

• limitations

• /D < 2 x 10-2

• Re >3 x 103

• less than 3% deviation

from results obtained

with Moody diagram

• easy to program for

computer or calculator

use

7

5/ 2 f

3/ 2 f

1.782.22 log

3.7

ghQ D

L D ghD

L

0.044.75 5.2

21.25 9.4

f f

0.66LQ L

D Qgh gh

2

0.9

0.25f

5.74log

3.7 ReD

no f

Swamee-Jain gets an f

• The challenge that S-J solved was deriving explicit

equations that are independent of the unknown

parameter.

• 3 potential unknowns (flow, head loss, or diameter): 3

equations for f

• that can then be combined with the Darcy Weisbach

equation

8

2

f 2 5

8f

LQh

g D

2

f f2

L Vh

D g

2

0.9

0.25f

5.74log

3.7 ReD

Colebrook Solution for Q

9

1 2.512log

3.7f Re f

D

2

f 2 5

8f

LQh

g D

2

2 5

f

1 1 8

f

LQ

h g D

Re 4Q

D 2 5

f 2

4Re f

8

Q g Dh

D LQ

3

f21Re f

gh D

L

2

1 2.514 log

f 3.7 Re f

D

f

2 5 2

8f

h g

D LQ

Colebrook Solution for Q

10

2

2

2 5 3f f

1 8 2.514 log

3.7 21

LQ

h g D D gh D

L

5/ 2 3f f

2 2.51log

3.7 21

L Q

gh D D gh D

L

5/ 2 f

3

f

log 2.513.7 22

gh LQ D

L D gh D

Swamee D?

11

0.045 1/ 4 5 1/5

2 2 2 21.250.66

Q Q Q QD

g g Q g g

1/ 251/5 1/ 4 1/5

2 2 25/ 40.66

Q Q QD

g g Q g

2

f 2 5

8f

LQh

g D

25

2

8f

QD

g

25

2

64f

8

QD

g

1/51/ 4 1/5

2 25/ 4

2

64f

Q Q

g Q g

1/52

2

64f

8

QD

g

1/51/5

1/ 4 1/52 2 2

5/ 4

8

Q Q QD

g g Q g

1/51/ 4 1/52 2 2

5/ 41f

4 4

Q Q

g Q g

Pipe Roughness

12

pipe material pipe roughness (mm)

glass, drawn brass, copper 0.0015

commercial steel or wrought iron 0.045

asphalted cast iron 0.12

galvanized iron 0.15

cast iron 0.26

concrete 0.18-0.6

rivet steel 0.9-9.0

corrugated metal 45

PVC 0.12

Solution Techniques

13

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

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

find pipe size given (head, type of pipe,L, Q) 0.04

4.75 5.22

1.25 9.4

f f

0.66LQ L

D Qgh gh

2

2 5

8ff

LQh

g D2

0.9

0.25f

5.74log

3.7 ReD

Re 4Q

D

5/ 2 f

3

f

log 2.513.7 22

gh LQ D

L D gh D

Exponential Friction Formulas

• Commonly used in commercial and industrial settings

• Only applicable over _____ __ ____ collected

• Hazen-Williams exponential friction formula

14

f

n

m

RLQh

D=

C = Hazen-Williams coefficient

range of data

Head loss:

Hazen-Williams Coefficient

C Condition

150 PVC

140 Extremely smooth, straight pipes; asbestos cement

130 Very smooth pipes; concrete; new cast iron

120 Wood stave; new welded steel

110 Vitrified clay; new riveted steel

100 Cast iron after years of use

95 Riveted steel after years of use

60-80 Old pipes in bad condition

15

Hazen-Williams vs

Darcy-Weisbach • Both equations are empirical

• Darcy-Weisbach is dimensionally correct, and ________.

• Hazen-Williams can be considered valid only over the

range of gathered data.

• Hazen-Williams can’t be extended to other fluids without

further experimentation.

16 1.852

f 4.8704

10.675 SI units

L Qh

D C

2

f 2 5

8f

LQh

g D

preferred

Minor Losses

• Most minor losses can not be obtained analytically, so

they must be measured

• Minor losses are often expressed as a loss coefficient, K,

times the velocity head.

17

2

2l

Vh K

g=

( )geometry,RepC f=

2

2C

V

ghlp

g

Vh pl

2C

2

High Re

Head Loss due to Sudden Expansion:

Conservation of Energy

18

1 2

ltp hHg

Vz

pH

g

Vz

p

22

2

222

2

2

2

111

1

1

lhg

VVpp

2

2

1

2

221

g

VVpphl

2

2

2

2

121

z1 = z2

What is p1 - p2?

Apply in direction of flow

Neglect surface shear

Divide by (A2 )

Head Loss due to Sudden Expansion:

Conservation of Momentum

19

Pressure is applied over all of section 1. Momentum is transferred over area corresponding to upstream pipe diameter. V1 is velocity upstream.

sspp FFFWMM 2121

1 2

xx ppxx FFMM2121

1

2

11 AVM x 2

2

22 AVM x

22212

2

21

2

1 ApApAVAV

g

A

AVV

pp 2

12

1

2

2

21

A1 A2

x

Energy

Head Loss due to

Sudden Expansion

20

g

VVpphl

2

2

2

2

121

g

A

AVV

pp 2

12

1

2

2

21

1

2

2

1

V

V

A

A

g

VV

g

V

VVV

hl2

2

2

2

11

22

1

2

2

g

VVVVhl

2

2 2

121

2

2

g

VVhl

2

2

21

2

2

1

2

1 12

A

A

g

Vhl

2

2

11

A

AK

Momentum

Mass

Contraction

• losses are reduced with a gradual contraction

21

V1 V2

EGL

HGL

vena contracta

g

VKh

cc2

2

2

Expansion!!!

Entrance Losses

• Losses can be reduced

by accelerating the flow

gradually and eliminating

the vena contracta

22

Ke 0.5

Ke 1.0

Ke 0.04

he Ke

V 2

2g

reentrant

Head Loss in Valves

• Function of valve type and valve

position

• The complex flow path through

valves often results in high head

loss

• What is the maximum value that Kv

can have? _____

23

hv Kv

V 2

2g

How can K be greater than 1?

Example

24

D=40 cm L=1000 m

D=20 cm L=500 m

valve 100 m

Find the discharge, Q.

Use S-J on small pipe

ltp hHg

Vz

pH

g

Vz

p

22

2

222

2

2

2

111

1

1

cs1

cs2

2

21002

l

Vm h

g= +

Non-Circular Conduits:

Hydraulic Radius Concept

• A is cross sectional area

• P is wetted perimeter

• Rh is the “Hydraulic Radius” (Area/Perimeter)

• Don’t confuse with radius!

25

2

f2

f

L Vh

D g=

2

f f4 2h

L Vh

R g=

2

44

h

DA D

RP D

p

p= = = 4 hD R=

For a pipe

We can use Moody diagram or Swamee-Jain with D = 4Rh!

GENERAL

CONSIDERATION

HGL-EGL DRAWING

26

EGL & HGL for a Pipe System

• Energy equation

• All terms are in dimension of length (head, or energy per unit weight)

• HGL – Hydraulic Grade Line

• EGL – Energy Grade Line

• EGL=HGL when V=0 (reservoir surface, etc.)

• EGL slopes in the direction of flow

22

22

211

21

122

zp

g

Vhz

p

g

VL

27

zp

HGL

g

VHGLz

p

g

VEGL

22

22


• A pump causes an abrupt rise in EGL (and

HGL) since energy is introduced here

28


• A turbine causes an

abrupt drop in EGL

(and HGL) as energy

is taken out

• Gradual expansion

increases turbine

efficiency

29


• When the flow passage changes diameter,

the velocity changes so that the distance

between the EGL and HGL changes

• When the pressure becomes 0, the HGL

coincides with the system

30


• Abrupt expansion into reservoir causes a

complete loss of kinetic energy there

31


• When HGL falls below the pipe the

pressure is below atmospheric pressure

32

Example (1)

33

Example (2)

34

Example (3)

35

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Slide07 Extra