cve 372 hydromechanics - 2 flow in closed conduits 2

CVE 372 Hydromechanics 1/54

Dr. Bertu AkntuDepartment of Civil EngineeringMiddle East Technical University

Northern Cyprus Campus

CVE 372CVE 372HYDROMECHANICSHYDROMECHANICS

FLOW IN CLOSED CONDUITSFLOW IN CLOSED CONDUITSIIII


2. FLOW IN CLOSED CONDUITS2. FLOW IN CLOSED CONDUITS

Overview

2.2 Fully Developed Flow in Closed Conduits

2.2.1 Derivation of Darcy-Weisbach Equation

2.2.2 Laminar Flow in Pipes

2.2.3 Turbulent Flow in Pipes

2.2.4 Moody Chart



Fully Developed Flow in Closed Conduits

2.1.1 Derivation of Darcy-Weisbach Equation Consider a steady fully developed flow in a prismatic pipe

Assumptions:- Fully developed flow (uniform) - Incompressible fluid- Circular tube (pipe) - Constant diameter- Steady Flow

V1=V2=VA1=A2=A1=2=11=2=1




2.1.1 Derivation of Darcy-Weisbach Equation(a) Relationship between wall shear stress

and head loss Continuity Equation

Momentum Equation

where and

constant2211 ==== VAAVAVQ

( )11222211 sin VVQFWApAp f =+

)(sinsin

21 zzAALW

==

PLF wf =P: wetted perimeter

0, since 1= 2




2.1.1 Derivation of Darcy-Weisbach Equation Momentum Equation gives

where RH is hydraulic radius

( )H

ww

w

RL

ALPpzpz

PLzzAApAp

==+

=+2

21

1

212211 0

)(sinsin

21 zzAALW

==

24

42 RDD

DPARH ===

P: wetted perimeter




2.1.1 Derivation of Darcy-Weisbach Equation Momentum Equation gives

Energy Equation gives

Note that the above equation is applicable for both laminar and turbulent flows and for open channel flows as well.

Lhpzpz =+

22

11

H

w

RLpzpz

=+

22

11

RLh

RLh

wf

H

wf

2=

=

DLh wf

4=




2.1.1 Derivation of Darcy-Weisbach Equation(b) Relationship between wall shear stress

and velocity Dimensional Analysis

),,,,( DVFw =

2Vfw =

( )

=

=

DVD

Vw

,

,

2

321

Roughness Relative

Number Reynolds

pressure dynamicstressshear

3

2

21

=

==

==

D

VDVD

fV

w




2.1.1 Derivation of Darcy-Weisbach Equation(c) Relationship between head loss and velocity

==D

Re,function2

2 fg

VDLfhf

DLVf

DLh

VfDV

f

wf

w

w

2

2

2

44

Re,function

===

==

let 8f'=f and g=/



Fully Developed Flow in Closed Conduits2.2.2 Laminar Flow in Pipes

Local acceleration is zero. Convective acceleration is zero as well as a result of fully developed flow

Motion of a cylindrical fluid element within a pipe.

Assumptions:- Fully developed flow (uniform) - Incompressible fluid- Circular tube (pipe) - Constant diameter- Steady Flow




Free-body diagram of a cylinder of fluid

Local acceleration is zero. Convective acceleration is zero as well as a result of fully developed flow

rlp

rlrpprp

2

02)( 212

1

==

Momentum Equation




Shear stress distribution within the fluid in a pipe (laminar or turbulent flow) and typical velocity profiles.

Dlp

Dr ww 42 == Small shear stress can produce large pressure difference if l/D is large

(relatively long pipe)

Applicable to both laminar and turbulent flow in pipes




2.2.2 Laminar Flow in PipesFor laminar flow of a Newtonian fluid, the shear stress is

simply proportional to the velocity gradient (Section 1.6) = (du/dr). For our pipe flow,

The negative sign is included to give > with du/dr < 0 ( the velocity decreases from the pipe centerline to the pipe wall)

drdur

lp ==

2




integration yields

We can evaluate the integration constant using the no slip condition at the pipe wall so that u = 0 at r=D/2 and C1=(pD2/(16l)). Hence

Velocity:

rl

pdrdu

= 2 1

2)4

()( Crl

pru +=

=

=

222 212116

)(DrV

Dr

lpDru c

Centerline (max) velocityVc=(pD2/(16l)).




2.2.2 Laminar Flow in PipesIntegration of u(r) over the pipe area gives volume flowrate

by definition, the average velocity is the V=Q/A = Q/R2

Above equation is referred to as Poiseuilles Law. Recall that all of these results are restricted to laminar flow ina horizontal pipe.

22)(

2

0

cR VRrdrruudAQ ===

lpDQ

128

4====l

pDVRVRV cc

3222

2

2

2




2.2.2 Laminar Flow in Pipes

Flowrate:

Proportional to the pressure drop Inversely proportional to viscosity Inversely proportional to the pipe length Proportional to the pipe diameter to the fourth power

lpDQ

128

4=




For Non-horizontal pipe, the adjustment is required by replacing p by (p-l sin) where is the angle between the pipe and the horizontal.

rllp 2sin =

lDlpV

32

)sin( 2=l

DlpQ

128)sin( 4=

Free-body diagram of a fluid cylinder for flow in a nonhorizontal pipe.




Example 3: An oil with a viscosity of = 0.40 N.s/m2 and density = 900 kg/m3 flow in a pipe of diameter D = 0.020 m.a) What pressure drop, p1-p2, is needed to produce a flowrate of Q = 2.0 x 10-5 m3/s if the pipe is horizontal with x1=0 and x2=10 m?

b) How steep a hill, , must the pipe be on if the oil is to flow through the pipe at the same rate as in part (a), but with p1 = p2?

c) For the conditions of part (b), if p1 = 200 kPa, what is the pressure at section x3 = 5 m, where x is measures along the pipe?

Solved in the class room




2.2.2 Laminar Flow in PipesFrom dimensional analysis: Assume that p is a function of

# of variables = 5 and # of dimensions = 3 (M,L,T)According to the result of dimensional analysis(CVE 371), this flow can be described in terms of 5-3=2 dimensionlessGroup. One such represents

A further assumption: the pressure drop is directly proportional to the pipe length

),,,( DlVfp =

)(Dl

VpD =

=DCl

VpD

=

2DVC

lp

lpDCAVQ

4)4/( ==



Fully Developed Flow in Closed Conduits2.2.2 Laminar Flow in PipesThe same functional form as theory implies

Recall average velocity was found to be

We can divide both sides by the dynamic pressure (recall from Chapter 3)

lpDCAVQ

4)4/( ==

=

==Dl

Dl

VDV

DlV

V

pRe6464

21

/32

21 2

2

2

lpDV 32

2=2

21 V




2.2.2 Laminar Flow in PipesThis is often written as

where the dimensionless quantity

Darcy friction factor

For a fully developed laminar flow the friction factor is

Alternate expression as a dimensional wall shear stress 28V

f w=

2

2VDlfp =

= 22VlDpf

Re64=f



Fully Developed Flow in Closed Conduits2.2.2 Laminar Flow in PipesEnergy Consideration

Recall from Chapter 5 (in the absence of energy sources)

where alpha values (always >=1) compensate for the fact that velocity profile across the pipe is not uniform (Chapter 5).

For fully developed flow velocity profile is constant

Hence

LhzgVpz

gVp +++=++ 22

2

22

11

2

11

22

21 =

Lhzpzp =

+

+ 2211

gVgV 2/2/ 2222

11 =




2.2.2 Laminar Flow in PipesRemember

Using the shear stress at the pipe wall, we have

rlp

rlrpprp

2

02)( 212

1

==

Lhzpzp =++ )()( 2211 r

lhh fL 2==

Dlh wf 4= Friction loss is proportional to the shear stress, which is proportional to viscosity.

No minor loss




2.2.2 Laminar Flow in PipesIn summaryVelocity:

Average Velocity:

maxuVc =

gVgV 2/2/ 2222

11 =

=

=

222 21)(or2116

)(DrVru

Dr

lpDru c

velocitymax.

lpDVV c 322

2==




2.2.2 Laminar Flow in PipesIn summaryWall Shear Stress:

or

Shear Stress:

ordrdu =

Dr

w2 =

lpD

w 4=

DV

w 8=




2.2.2 Laminar Flow in PipesIn summaryFlow Rate:

Friction Loss:

Friction Factor:

Dlh wf 4=

lpDVAQ

128

4==

2

32D

Vlhf =

gV

Dlfhf 2

2

=

Re64=f 28Vf

w

=




Example 4: The flow rate Q of corn syrup through the horizontal pipe shown in the figure is to be monitored by measuring the pressure difference between sections (1) and (2). The variations of the syrups viscosity and density with temperature are given in the following table

a) Determine the wall shear stress and the pressure drop p=p1-p2 for Q=14 lt/sfor T=40C.b) For the condition of part (a), determine the net pressure force and the net shear force on the fluid within the pipe between the section (1) and (2).

2 m

10 cm


0.0011104170

0.0044104660

0.0211105150

0.1819105740

0.9097106225

1.9152106715

(N.s/m2) (kg/m3)T (C)




2.2.3 Turbulent Flow in Pipes Transition from Laminar to Turbulent Flow

We slowly increase the flowrate in a long section of a pipe.




2.2.3 Turbulent Flow in Pipes Transition from Laminar to Turbulent Flow

The time-averaged, , and fluctuating, , description of a parameter for tubular flow.

u(t): instantaneous velocity in the x-directionu(t): fluctuating part of u(t)




2.2.3 Turbulent Flow in Pipes Turbulent Shear Stress

'uuu +=We can write the velocity vector as

Average velocity:

Fluctuation (time average of fluctuations is zero)

+

=Tt

t

o

o

dttzyxuT

u ),,,(1Fluctuations are equally distributed on either side of the average. However, the square of fluctuation is always greater than zero.

0)(1)'( 22 >= +Tt

t

o

o

dtuuT

u




2.2.3 Turbulent Flow in PipesTurbulent Shear Stress

u

dtuuT

uu

Tt

t

o

o

2

2

2)(1

)'(

=+

The structure and characteristics of turbulence may vary from one flow situation to another. A measure of turbulence is called turbulence intensity

As this parameter increases, the fluctuations of the velocity increases. Well designed wind tunnels have typical values of 0.01 or smaller.




2.2.3 Turbulent Flow in Pipes Turbulent Shear StressThe random velocity components result in momentum transfer in turbulent flow resulting in an additional term in the shear stress expression:

turblam'' +== vudyud

If the flow is laminar, then fluctuations vanish and we recover the viscosity expression for Newtonian fluids. The second term is called the turbulent shear stress and it is always positive. Hence the shear stress in turbulent flow is always greater than shear stress in laminar flow.





Structure of turbulent flow in a pipe. (a) Shear stress. (b) Average velocity.

lam is dominant

turb is dominant





Typically turbulent shear stress is 100 to 1000 times greater than the shear stress in the laminar region, while the converse is true in the viscous sublayer.

Note that an accurate model of turbulent flow requires the knowledge of Reynolds stresses which require the knowledge of velocity fluctuations which can not be solved for most turbulent flow problems.




2.2.3 Turbulent Flow in Pipes Turbulent Velocity Profile

Fully developed turbulent flow in a pipe can be broken into three regions which are characterized by their distances from the wall: Viscous sublayer: The viscous shear stress is dominant

compared with the turbulent (or Reynolds) stress and the random (eddying) nature of flow is absent. In this layer fluid viscosity is important parameter.

Overlap region: Transition region Outer turbulent layer: The Reynolds stress is dominant, and

there is considerable mixing and randomness of the flow. In this layer density is important parameter.



Fully Developed Flow in Closed Conduits2.2.3 Turbulent Flow in Pipes Turbulent Velocity Profile

In the viscous sublayer:

where

= yuuu

*

y

R

r

u wu =

5/0 yu

rRy == the time av. x component of the velocity

friction velocity

Law of the wall, validvery near the smooth wallfor

y = s: thickness of the viscous sublayer

*5us =





In the overlap region:

where

0.5ln5.2* +

=

yu

uu

y

R

r

u

wu =

5/0 yu

rRy == the time av. x component of the velocity

friction velocity

Law of the wall, validvery near the smooth wallfor





In the turbulent region:

or

=

yR

uuVc ln5.2*

n

c Rr

Vu /11

= Figure 8.17Exponent, n, for power-law velocity profiles.

Power-Law velocity profile





Typical laminar flow and turbulent flow velocity profiles.




2.2.3 Turbulent Flow in PipesExample 5: Water at 20C (=998 kg/m3 and =1.004 x 10-6 m3/s) flows

through a horizontal pipe of 0.1 m diameter with a flowrate of Q=4x10-2 m3/s and a pressure gradient of 2.59 kPa/m.

a) Determine the approximate thickness of the viscous sublayer.

b) Determine the approximate centerline velocity, Vc.

c) Determine the ratio of the turbulent to laminar shear stress, turb/ lam at a point midway between the centerline and the pipe wall (r=0.025 m).





2.2.3 Turbulent Flow in Pipes Most turbulent pipe flow analyses are based on

experimental data and semi-empirical formulas which are expressed in dimensionless forms.

We need to determine the head loss. For convenience, we will consider two types of energy losses; minor (local) and major (friction) losses

minormajor LLL hhh +=Note that major and minor losses do not necessarily reflect the magnitude of the energy losses




2.2.3 Turbulent Flow in PipesConsider the pipe flow again. Pressure drop is a function of a number of physical and geometrical parameters:

),,,,,( lDVFp =

This time we included a parameter which is a measure of the roughness of the pipe wall (unit is length)




2.2.3 Turbulent Flow in PipesThere are seven parameters and three

reference dimensions.

),,(~2/1 2 DD

lVDV

p

=

Dynamic pressureReynolds number




2.2.3 Turbulent Flow in Pipes

)(Re,2/1 2 DD

lV

p =

Lets assume that pressure drop is proportional to the length of the pipe:

Recall that 2/2Vl

pDf = is the friction factor. Then we have )(Re,

Df =

The energy equation for steady, incompressible flow is given by

LhzgVpz

gVp +++=++ 22

222

11

211

22




2.2.3 Turbulent Flow in PipesFor constant diameter, horizontal pipe with fully developed flow (alphas are equal)

gV

DlfhL 2

2

major =

For nonhorizontal pipes

gV

Dlfzzhzzpp L 2

)()(2

21major2121 +=+=

This is called Darcy-Weisbach equation



For laminar flow:f=64/Re


2.2.3 Turbulent Flow in PipesThe Moody Chart.

for large values, f is independent of Re

+=

fD

f Re51.2

7.3/log21

For turbulent flow:




2.2.3 Turbulent Flow in PipesSmooth Pipe and Hydraulically Smooth Flow

Colebrook White Transition Flow

Rough Pipe Hydraulically Rough Flow

=

ff Re51.2log21

+=

fD

f Re51.2

7.3/log21

=7.3

/log21 Df




2.2.4 Moody Diagram



2.2.4 Moody DiagramLoss coefficient for a sudden expansion.

3311 VAVA =

)( 13333331 VVVAApAp = Conservation of momentum

Conservation of mass

Conservation of energy

LhgVp

gVp ++=+

223

231

21

Note that: )2//( 21 gVhK LL =

2

2

1 )1(AAKL =





Example 6:A soft drink with the properties of 10C water is sucked through a 4 mm diameter, 0.25 m long straw at a rate of 4 cm3/s. Is the flow at the outlet of the straw laminar? Is it fully developed?





Example 7:

Given Q=3.6 lt/min

a) Determine the state of the flow and velocity in the pipe.

b) Draw E.G.L. and H.G.L. and determine the kinematics viscosity.


0.5 m

D=10 mm

Entrance K=0.5L=2 m




Exercise 1:Water at 20C flows in a 15 cm diameter pipe with a flowrate of 60 lt/s.

a) Determine the centerline velocity. (Ans: 4.06 m/s)

b) What is the approximate velocity at a distance 5 cm away from the wall? (Ans: 3.86 m/s)




Exercise 2:For a smooth pipe of diameter 75 mm, the head loss for a distance of 150 m is 21 m. When the flowrate is 8.5 lt/s. Is the flow laminar or turbulent? (Ans: laminar)




Exercise 3:The pressure heads measured in a 2 cm diameter circular pipe arep1/ = 22 m and p2/ = 21.5 m. The distance between two measuring points is 1500 m. Taking =9810 N/m3, = 1000 kg/m3, =1x10-6 m2/s determine:a) State of the flowb) The equation of velocity and shear stress profilesc) Maximum velocityd) Velocity and shear stress at r=5 mm and y=4 mm.e) Discharge

ANS: (a) laminar, (b) u=0.0816-816r2, =1.635r, (c)=0.0816 m/s, (d) u(r=5mm)=0.0612 m/s, (r=5mm)=0.008175 N/m2

u(y=4mm)=0.0522 m/s, (y=4mm)=0.00981 N/m2,(e) 1.282 x 10-5 m3/s

cve 372 hydromechanics - 2 flow in closed conduits 2

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