CHE315 Pressure Drop and Friction Loss 2.10 Design Equations for Laminar and Turbulent Flow in...

CHE315 Pressure Drop and Friction Loss

2.10 Design Equations for Laminar and Turbulent Flow in Pipes

friction loss in Pipes SF

SkinFriction

Contraction

Friction

Expansion

Friction

FittingsFriction

http://www.google.com/imgres?imgurl=http://blog.nialbarker.com/wp-content/uploads/2010/03/laminar_turbulent_flow.gif&imgrefurl=http://blog.nialbarker.com/page/2&usg=__Hd-ulrbQCQnylJfN-Rf4fv59P_4=&h=243&w=323&sz=12&hl=en&start=19&zoom=1&um=1&itbs=1&tbnid=Xj5fDfHR81UCtM:&tbnh=89&tbnw=118&prev=/images?q=laminar+and+turbulent&start=18&um=1&hl=en&safe=active&sa=N&ndsp=18&tbs=isch:1


2.10 Design Equations for Laminar and Turbulent Flow in Pipes

0

2

1 1212

21

22

FW

ppzzgvv savav

ff

ppF 21



2.10B PRESSURE DROP AND FRICTION LOSS IN LAMINAR FLOW

20

32D

L

ppv L

2

1221

32

D

LLvppp ff

Example.

2.10-1



Fanning friction factor f for friction loss in laminar flow

f is defined as drag force per wetted surface unit area (ts at the surface area) divided by the product of density times the velocity head (r v2/2):

22vf s

The drag force The wetted surface unit area 2Rp f LR 2

L

Dp

L

Rp

LR

Rp fffs

422

2



2

1

4 2vL

Dpf f

2

42v

D

Lfp f

24

2v

D

Lf

pF ff

For laminar flow only, combining equations:

24

2v

D

Lfp f

2

1232

D

LLvp f

and vDNf

1616

Re

Eq (2.10-

2)

Eq (2.10-

5)




SkinFriction

Contraction

Friction

Expansion

Friction

FittingsFriction

24

2v

D

Lfp f

2

1232

D

LLvp f

Re

16

Nf

f fromFigure 2.10-3

Lam

inar

Turb

ule

nt


CHE315

CHE315

Example 2.10-

3

NOTE: If velocity and diameter are both unknown, Solution should be Trial –and-error soln. (or computerized).

Pressure Drop and Friction Loss

CHE315

Step by step procedure for trial-and error solution

1.Assume an initial value for the velocity (or the

diameter)

2.Calculate the Reynolds number (and e/D)

3.From the figure 3.10-3, read the corresponding

friction factor f

4.Substitute f into:

5.Compare the obtained value with the initial one

6.If values are different repeat from step2 with the

obtained value

24

2v

D

LfFf


CHE315

Example 2.10-

4Example

2.10.5



For gases, the equation 24

2v

D

Lfp f

can be rewritten as

follows:

avf D

LGfpp

2

4 2

21

vG

Pressure drop and friction factor in flow of gas

RT

MpRT

MpRT

M

mnRTpV

Ideal Gas Law:


DM

RTLGfpp

22

22

1

4

How did we arrive to this equation?

av

f D

LGfpp

2

4 2

21

RT

MpRT

MpRT

M

mnRTpV

Using Ideal Gas Law;

The following equation can be obtained

CHE315

Contraction

Friction

Expansion

Friction

kg

JvKh s

2

2

AL AS

vL vS

2

1

L

sex A

AK

L

sc A

AK 155.0

exhhchh

FittingsFriction 2

21vKh ff


Kf :Turbulent: Table

2.10-1Laminar: Table

2.10-2

CHE315

22224

21

222 vK

vK

vK

v

D

LfF f

sc

sex


SkinFriction

Contraction

Friction

Expansion

Friction

FittingsFriction

24

2v

D

Lf

2

2s

ex

vK 2

21vK f

2

2s

c

vK


CHE315

Example

2.10.6Example

2.10.7


CHE315

FRICTION LOSSES IN NONCIRCULAR CONDUITS

The equivalent diameter D is defined as four times the hydraulic radius

rH, defined as the ratio of the cross-sectional area of the channel to the

wetted perimeter of the channel.:

channel ofperimeter wetted

channel of area sectionalcross4

HrD


CHE315

Calculate the equivalent diameter?


http://ce-review.com/2010/02/15/hydraulics-ce-board-exam-nov-1998-29/figure-1-403/

http://ce-review.com/2010/02/15/hydraulics-ce-board-exam-nov-1998-29/solution-488/

CHE315 Pressure Drop and Friction Loss 2.10 Design Equations for Laminar and Turbulent Flow in...

Documents

Transcript of CHE315 Pressure Drop and Friction Loss 2.10 Design Equations for Laminar and Turbulent Flow in...