1

Chapter 6

• Losses due to Fluid Friction

2

Objectives

ä To measure the pressure drop in the straight section of smooth, rough, and packed pipes as a function of flow rate.

ä To correlate this in terms of the friction factor and Reynolds number.

ä To compare results with available theories and correlations.

ä To determine the influence of pipe fittings on pressure drop

ä To show the relation between flow area, pressure drop and loss as a function of flow rate for Venturi meter and Orifice meter.

3

Losses due to Friction Mechanical energy equation between locations 1 and 2 in the absence of

shaft work:

For flow in a horizontal pipe and no diameter change (V1=V2), then :

21 PP

F

Thus, the shear

stress at the wall

is responsible for

the losses due to

friction

Losshg

zzg

V

g

V

g

P

g

P

F)()

22()( 21

2

2

2

121

Hagen-Poiseuille Law-E5.10

4

oD

128x QF

Because of (5.4) :

D

xF w

4

(6.13) (6.14)

m

Qu

Fwhere

=Friction head/unit mass

Or F/g= hLoss= 128μQL/(πρgD4)

g

PP

21

LhF/g

OR F in J/kg

and

4

Total head loss , hL (=F/g), is regarded as the sum of

major losses and minor losses

hL major, due to frictional effects in fully developed flow in

constant area tubes,

hL minor, resulting from entrance, fitting, area changes,

and so on.

Losses due to Friction

5

Losses due to Friction/The Friction Factor In order to determine an expression for the losses due to friction we must

resort to experimentation.

D

V LF

2

By introducing the friction factor, f:

D.2

V f

2LF

where

)2/)(/( 2VDL

Ff

(6.15)

where L=length of the pipe,

D=diameter of the pipe,

V=velocity,

2/

)/(2V

LDPf

or

21 PP

F

f is called Darcy friction factor

6

Friction Factor

0)2

(2

Lwork PPV

gzP

The Darcy friction factor f is defined as

We know the wall shear stress

2/

)/(2V

LDPf

L

PDτw

4

2/2Vf w

(E6-21)

The friction factor is the ratio between wall shear stress

and flow inertial force.

orLmajorLL PPP min,,

now later

and

7

Major Loss and Friction Factor

0)2

(2

Lwork PPV

gzP

With the introduction of friction factor, we can calculate major loss by

(E6-22) 2

2

,

V

D

LfP MajorL

Friction factor Pipe geometry factor Dynamic pressure

Therefore, our job now is find the friction factor f for various flows.

orLmajorLL PPP min,,

8

Friction Factor - Laminar Pipe

Flow For a pipe with a length of L, the pressure gradient is constant, the pressure

drop based on Hagen-Poiseuille Law ,

Divided both sides by the dynamic pressure V2/2 and L/D

We have

2/32 DVLP

Re

64f

Re

6464

2/

/32

2/

)/(2

2

2

VDL

D

V

DVL

V

LDPf

9

The mechanical energy equation can be written in terms of heads :

(6.17)

Knowledge of the friction factor allows us to estimate the loss term in

the energy equation.

This head loss can also be calculated using H-B equation

Loss

shafth

gm

Wzz

g

V

g

V

g

P

g

P

)()

22()( 21

2

2

2

121

g

V

D

Lfh MajorL

2

2

,

Major Loss and Friction Factor

10

11

Example 1

losses in

12

13

Example 2

14

15

Case 2: Turbulent Flow When fluid flow at higher flow rates, the streamlines are not steady, not

straight and the flow is not laminar.

Generally, the flow field will vary in both space and time with fluctuations

that comprise "turbulence”

When the flow is turbulent the velocity and pressure fluctuate very

rapidly. The velocity components at a point in a turbulent flow field

fluctuate about a mean value.

n

R

r

V

u/1

max

1

Time-averaged velocity profile

can be expressed in terms of

the power law equation, n =7

is a good approximation.

16

Friction Factor-Turbulent Pipe Flow

•For a laminar flow, the friction factor can be analytically derived.

•It is impossible to do so for a turbulent flow so that we can only obtain the friction

factor from empirical results.

•For turbulent flow, it is impossible to analytically derive the friction factor f, which

can ONLY be obtained from experimental data.

• In addition, most pipes, except glass tubing, have rough surfaces.

•The pipe surface roughness is quantified by a dimensionless number, relative pipe

roughness (ε / D ), where ε is pipe roughness and D is pipe diameter.

•For laminar pipe flow, the flow is dominated by viscous effects hence surface

roughness is not a consideration.

•However, for turbulent flow, the surface roughness may protrude beyond the

laminar sublayer and affect the flow to a certain degree. Therefore, the friction factor

f can be generally written as a function of Reynolds number and pipe relative

roughness

There are several theoretical models available for the prediction of shear

stresses in turbulent flow.

17

Surface Roughness Additional dimensionless group /D

need to be characterize

Thus more than one curve on friction factor-

Reynolds number plot

Fanning diagram or Moody diagram

Depending on the laminar region.

If, at the lowest Reynolds numbers, the laminar portion

corresponds to f =16/Re Fanning Chart

(or f = 64/Re Moody chart)

18

Friction Factor and Pipe Roughness

• Most pipes, except glass tubing, have rough

surfaces

- Pipe surface roughness,

- Relative pipe roughness,

• The surface roughness may affect the friction

factor. Generally, we have

D/

)(Re,D

f

19

Pipe Surface Roughness

20

Friction Factor of Turbulent Flow

If the surface protrusions are

within the viscous layer, the pipe

is hydraulically smooth;

If the surface protrusions extend

into the buffer layer, f is a

function of both Re and /D;

For large protrusions into the

turbulent core, f is only a

function of /D.

)(Re, Df

)( Df

1/ 4

0.316

Ref

21

Friction Factor for Smooth, Transition,

and Rough Turbulent flow

1

f 4.0 * log Re* f 0.4

Smooth pipe, Re>3000

1

f 4.0 * log

D

2.28

Rough pipe, [ (D/ε)/(Re√ƒ) <0.01]

1

f 4.0 * log

D

2.28 4.0 * log 4.67

D /

Re f1

Transition function

for both smooth and

rough pipe

f P

L

D

2U 2

f 0.079Re0.25Or

22

Fanning Diagram

f =16/Re

1

f 4.0 * log

D

2.28

1

f 4.0 * log

D

2.28 4.0 * log 4.67

D /

Re f1

23

Friction Factor – The Moody Chart

24

Example: Comparison of Laminar or

Turbulent pressure Drop

• Air under standard conditions flows through a 4.0-mm-

diameter drawn tubing with an average velocity of V = 50 m/s.

For such conditions the flow would normally be turbulent.

However, if precautions are taken to eliminate disturbances to

the flow (the entrance to the tube is very smooth, the air is dust

free, the tube does not vibrate, etc.), it may be possible to

maintain laminar flow.

• (a) Determine the pressure drop in a 0.1-m section of the tube

if the flow is laminar.

• (b) Repeat the calculations if the flow is turbulent.

Straight and horizontal pipe and same diameters give same velocity:

Z1=Z2=0 V1=V2 and thus

Losshgp /

25

Solution1/2

flowTurbulent700,13.../VDRe

Under standard temperature and pressure conditions

=1.23kg/m3, μ=1.7910-5Ns/m

The Reynolds number

kPaVD

fp 179.0...2

1 2

If the flow were laminar and using Darcy friction

f=64/Re=`…=0.00467

kPaVD

fp 179.0...2

14 2

F=16/Re=`…=0.001167

If the flow were laminar and using Fanning friction

26

Solution2/2

kPaVD

fp 076.1...2

1 2

If the flow were turbulent

From Moody chart f=Φ(Re, smooth pipe) =0.028

From Fanning chart f=Φ(Re, smooth pipe) =0.007

kPaVD

fp 076.1...2

14 2

27

Example

Straight and horizontal pipe and same diameters give

same velocity:

Z1=Z2=0 V1=V2 and thus

pressurepumpghp Loss

28

29

30

Example

0.04 m

31

32

33

f 0.079Re0.25

for fanning

34

Example: Determine Head Loss

• Crude oil at 140°F with γ=53.7 lb/ft3 and μ= 810-5

lb·s/ft2 (about four times the viscosity of water) is

pumped across Alaska through the Alaska pipeline, a

799-mile-along, 4-ft-diameter steel pipe, at a

maximum rate of Q = 2.4 million barrel/day = 117ft3/s,

or V=Q/A=9.31 ft/s. Determine the horsepower

needed for the pumps that drive this large system.

35

Solution1/2

The energy equation between points (1) and (2)

Assume that z1=z2, p1=p2=V1=V2=0 (large, open tank)

ftg

V

Dfhh PL 17700...

2

2

L2

2

22P1

2

11 hzg2

Vphz

g2

Vp

hP is the head provided to the

oil by the pump.

Minor losses are negligible because of the large length-

to-diameter ratio of the relatively straight, uninterrupted

pipe.

f=0.0124 from chart ε/D=(0.00015ft)/(4ft), Re=…..

36

Solution2/2

The actual power supplied to the fluid

power=Q∆P =Qρgh

unit of power (SI): Watt =N.m/s=J/s

hpslbft

hpgQhP 202000

/550

1...Power

37

Minor Losses 1/5

Most pipe systems consist of

considerably more than

straight pipes. These

additional components

(valves, bends, tees, and the

like) add to the overall head

loss of the system.

Such losses are termed

MINOR LOSS.

The flow pattern through a valve

38

900 Bend

900 Elbow 450 Elbow

Tees-line flow Tees-branch flow

39

Globe valve Angle valve Gate valve

Ball valve

Valves

40

Minor Losses 2/5

The theoretical analysis to predict the details of flow

pattern (through these additional components) is not,

as yet, possible.

The head loss information for essentially all

components is given in dimensionless form and based

on experimental data.

The most common method used to determine these

head losses or pressure drops is to specify the loss

coefficient, KL

41

Minor Losses 3/5

2

L2

2

L

L V2

1Kp

V2

1

p

g2/V

hK ormin

Re),geometry(KL

f

DK

g

V

Df

g

VKh

Leq

eq

LL or

22

22

minMinor losses are sometimes

given in terms of an

equivalent length eq

The actual value of KL is strongly dependent on the

geometry of the component considered. It may also

dependent on the fluid properties. That is

42

Minor Losses 4/5

For many practical applications the Reynolds number is large enough so that the flow through the component is dominated by inertial effects, with viscous effects being of secondary importance.

In a flow that is dominated by inertia effects rather than viscous effects, it is usually found that pressure drops and head losses correlate directly with the dynamic pressure.

This is the reason why the friction factor for very large Reynolds number, fully developed pipe flow is independent of the Reynolds number.

43

Minor Losses 5/5

This is true for flow through pipe components.

Thus, in most cases of practical interest the loss

coefficients for components are a function of

geometry only,

)geometry(KL

44

Minor Losses Coefficient

For Example Entrance flow

Entrance flow condition

and loss coefficient

(a) re-entrant, KL = 0.8

(b) sharp-edged, KL =

0.5

(c) slightly rounded,

KL = 0.2

(d) well-rounded,

KL = 0.04

45

Summary of Minor Losses • Major losses: Associated with the friction in the straight portions of the

pipes

• Minor losses: Due to additional components (pipe fittings, valves,

bends, tees etc.) and to changes in flow area (contractions or

expansions)

Method 1: We try to express the head loss due to minor losses in

terms of a loss coefficient, KL:

g2

VK

g

F 2

L

lossesminor

Values of KL can be found in the literature (for example, see Table next

ppt, for losses due to pipe components.

46

Minor Losses Component KL

Elbows

Regular 90°, flanged 0.3

Regular 90°, threaded 1.5

Long radius 90°, flanged 0.2

Long radius 90°, threaded 0.7

Long radius 45°, flanged 0.2

Regular 45°, threaded 0.4

180° return bends

180° return bend, threaded 0.2

180° return bend, flanged 1.5

Tees

Line flow, flanged 0.2

Line flow, threaded 0.9

Branch flow, flanged 1.0

Branch flow, threaded 2.0

Component KL

Union, threaded 0.8

Valves

Globe, fully open 10

Angle, fully open 2

Gate, fully open 0.15

Gate, ¼ closed 0.26

Gate, ½ closed 2.1

Gate, ¾ closed 17

Ball valve, fully open 0.05

Ball valve, 1/3 closed 5.5

Ball valve, 2/3 closed 210

47

Minor Losses The mechanical energy equation can be written:

g

VK

g

V

D

Lf

gm

Wzz

g

V

g

V

g

P

g

P

L

shaft

22

)()22

()(

22

21

2

2

2

121

(6.22)

48

Minor Losses Method 2:

Using the concept of equivalent length, which is the

equivalent length of pipe which would have the same

friction effect as the fitting.

Values of equivalent length (L/D) can be found in the literature

g

V

D

Lf

gm

Wzz

g

V

g

V

g

P

g

P

equiv

shaft

2 )()

22()(

2

.

21

2

2

2

121

(6.23)

49

Example: Determine Pressure Drop

• Water at 60°F flows from the basement to the second

floor through the 0.75-in. diameter copper pipe (a drawn

tubing) at a rate of Q = 12.0 gal/min (= 0.0267 ft3/s) and

exits through a faucet of diameter 0.50 in. as shown in

Figure

Determine the pressure at

point (1) if: (a) all losses

are neglected, (b) the

only losses included are

major losses, or (c) all

losses are included.

50

Solution1/4

The energy equation

45000/VDReft/slb1034.2

ft/slug94.1s/ft70.8...A

QV

25

3

1

1

s/ft6.19...A/QV

)jetfree(0p,ft20z,0z

22

221

Lhzg

V

g

pz

g

V

g

p 2

2

221

2

11

22

The flow is turbulent

L

2

1

2

221

21 h)VV(zp

Head loss is different

for each of the three

cases.

51

Solution2/4

(a) If all losses are neglected (hL=0)

45000Re108D/000005.0 5

psi7.10ft/lb1547...)VV(zp 22

1

2

221

21

f = 0.0215

(b) If the only losses included are the major losses, the head

loss is

g

V

DfhL

2

2

1 chart

psiftlbV

D

ftfVVzp 3.21/3062...

2

)60(4)( 2

2

12

1

2

221

21

52

Solution3/4

(c) If major and minor losses are included

]2)5.1(410[2

)/70.8()/94.1(3.21

23.21

22)(

23

2

1

22

12

1

2

221

21

sftftslugspsi

VKpsip

VK

g

V

DfVVzp

L

L

psi5.30psi17.9psi3.21p1

53

54

55

Noncircular Ducts

The empirical correlations for pipe flow may be used for computations involving noncircular ducts, provided their cross sections are not too exaggerated.

The correlation for turbulent pipe flow are extended for use with noncircular geometries by introducing the hydraulic diameter, Dh, defined as

P

A4Dh

Where A is cross-

sectional area, and

P is wetted perimeter.

56

Noncircular Ducts 3/4

The friction factor can be written as f=C/Reh, where the

constant C depends on the particular shape of the duct, and

Reh is the Reynolds number based on the hydraulic diameter.

The hydraulic diameter is also used in the definition of the

friction factor, , and the relative

roughness /Dh. )g2/V)(D/(fh 2

hL

hD V

Re

57

Non circular conduits

Define hydraulic diameter, Dh:

perimeter wetted

area) sectional (crossx 4Dh

Then use Reynolds number based on hydraulic diameter, Dh:

hD V

Re

58

Non circular conduits

• Pipe of circular cross-section

DD4

D4

D

2

h

• Annulus (inside diameter D1, outside D2)

12

12

2

1

2

2

h DDDD

4

D

4

D4

D

• Rectangular conduit (area ab)

ba

ab2

b2a2

)ab(4Dh

59

Example

Solved before

60

61

Multiple Pipe Systems

• Pipes in series

321BA LLLL

321

FFFF

QQQ

62

Multiple Pipe Systems

• Parallel pipes

321 LLL

321

FFF

QQQQ