1. Smooth circular pipe - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part4-58160-2020.pdf058:0160 Chapter 6-part4 Professor Fred Stern Fall 2020 5 2. Turbulent Flow

058:0160 Chapter 6-part4

Professor Fred Stern Fall 2020 1

Chapter 6: Viscous Flow in Ducts

6.4 Turbulent Flow in Pipes and Channels using mean-

velocity correlations

1. Smooth circular pipe

Recall laminar flow exact solution

d

ave

w

uf Re/64

82

2000Re

duaved

A turbulent-flow “approximate” solution can be

obtained simply by computing uave based on log law.

Bv

yu

u

u

*

*ln

1

Where

rRyuByuu w ;/;5;41.0);( *

32ln

2

2

1

2ln11

**

0

**

2

Bv

Ruu

drrBv

yuu

RA

QuV

R

ave

Or:



34.1ln44.2*

*

v

Ru

u

V

8.0]log[Re2

02.1]log[Re99.1

2/1

2/12/1

f

ff

d

d

Since f equation is implicit, it is not easy to see

dependency on ρ, μ, V, and D

4/1Re316.0)( Dpipef

g

v

D

Lf

ph f

2

2

Turbulent Flow: 4/74/54/14/3158.0 VDLp

75.175.44/14/3241.0 QDL

Laminar flow: 4/8 RLQp

EFD Adjusted constants

f only drops by a factor of 5 over 104 < Re < 108

4000 < ReD < 105

Blasius (1911) power law

curve fit to data

Nearly linear Only slightly

with μ

Drops weakly with

pipe size

Near quadratic

(as expected)

V2



p (turbulent) decreases more sharply with D than

p (laminar) for same Q; therefore, increase D for smaller

p . 2D decreases p by 27 for same Q.

BRu

u

ru

u

u

*

**

max ln1)0(

Combine with

2

3ln

1 *

* B

Ru

u

V

V

u

V

uuuV

u

u

u

V

2

31

2

3

2

3 *

max

*

max*

max

*

Also

8/8/18/1

*

2

2*

2

2* fV

u

V

uf

Vfandu w

w

ffV

u

V

u3.118/

2

31

2

31

*

max

Or:

For Turbulent Flow: 1

max

3.11

fu

V

Recall laminar flow: 5.0/ max uV





2. Turbulent Flow in Rough circular pipe

),( kyfU )/,(Re dkff d

)(ln1 kBByU

which leads to three roughness regimes:

1. k+ < 4 hydraulically smooth

2. 4 < k+ < 60 transitional roughness (Re dependence)

3. k+ > 60 full rough (no Re dependence)

11.1

2/1

2/1

7.3

/

Re

9.6log8.1~

Re

51.2

7.3

/log2

dk

f

dkf

d

d

There are basically four types of problems involved with

uniform flow in a single pipe:

1. Determine the head loss, given the kind and size of pipe

along with the flow rate, Q = A*V

2. Determine the flow rate, given the head, kind, and size of

pipe

3. Determine the pipe diameter, given the type of pipe, head,

and flow rate

4. Determine the pipe length, given Q, d, hf, ks, , g

Log law shifts downward

Moody diagram

Approximate explicit

formula



1. Determine the head loss

The first problem of head loss is solved readily by obtaining f

from the Moody diagram, using values of Re and ks/D

computed from the given data. The head loss hf is then

computed from the Darcy-Weisbach equation.

f = f(ReD, ks/D)

2

2f

L Vh f h

D g

2121

ppzzh

=

z

p

ReD = ReD(V, D)

2. Determine the flow rate

The second problem of flow rate is solved by trial, using a

successive approximation procedure. This is because both

Re and f(Re) depend on the unknown velocity, V. The

solution is as follows:

1) solve for V using an assumed value for f and the Darcy-

Weisbach equation

2/12/1

f fD/L

gh2V

known from note sign

given data



2) using V compute Re

3) obtain a new value for f = f(Re, ks/D) and reapeat as

above until convergence

Or can use Re

2/12/3

2/12

L

ghDf

f

scale on Moody Diagram

1) compute 2/1Re f and ks/D

2) read f

3) solve V from g2

V

D

Lfh

2

f

4) Q = VA

3. Determine the size of the pipe

The third problem of pipe size is solved by trial, using a

successive approximation procedure. This is because hf, f,

and Q all depend on the unknown diameter D. The solution

procedure is as follows:

1) solve for D using an assumed value for f and the Darcy-

Weisbach equation along with the definition of Q

5/1

5/1

f2

2

fgh

LQ8D

known from

given data



2) using D compute Re and ks/D

3) obtain a new value of f = f(Re, ks/D) and repeat as above

until convergence

4. Determine the pipe length

The four problem of pipe length is solved by obtaining f from

the Moody diagram, using values of Re and ks/D computed

from the given data. Then using given hf ,V, D, and

calculated f to solve L from f

Dh

V

gL

f

2

2 .





3. Concept of hydraulic diameter for noncircular

ducts

For noncircular ducts, τw= f(perimeter); thus, new

definitions of 2

8

Vf w

and

2

2

VC w

f

are required.

Define average wall shear stress

dsP

P

ww 0

1 ds = arc length, P = perimeter

Momentum:

0

L

z

W

ALPLpA w

PA

Lzph

w

//

A/P =Rh= Hydraulic radius (=R/2 for circular pipe and 2/R

Lh w

)



Energy:

PA

Lhh

w

L/

dx

pd

P

A

dx

zpd

P

A

dx

dh

P

A

L

h

P

Aw

^

non-circular duct

Recall for circular pipe:

dx

pdD

dx

pdRw

ˆ

4

ˆ

2

In analogy to circular pipe:

P

AD

D

P

A

dx

pdD

dx

pd

P

Ah

hhw

4

4

^

4

^

For multiple surfaces such as concentric annulus P and A

based on wetted perimeter and area

h

DhD

w VDDf

Vf

hh Re)/,(Re

82

g

V

D

Lf

R

LfV

R

Lhh

hhh

w

L28

22

However, accuracy not good for laminar flow (40%) and

marginal turbulent flow (15%).

Hydraulic

diameter



a. Accuracy for laminar flow (smooth non-circular

pipe)

Recall for pipe flow:

64Re

16Re)(#

0

0

0fP

CPPPoiseuille

f

fc f

Recall for channel flow:

hD

hhVhf

Re

42 Re

96

Re

4824

hD

hh

f

f

VhC

fC

Re

42 Re

24

Re

126

4/

96Re

24Re)(#

0

0

0

h

hf

Df

Dfc

fP

CPPPoiseuille

Therefore:

3

2

96

64

24

16

0

0

0

0

hf

f

hf

f

Donbasedchannel

pipe

Donbasedchannelc

pipec

P

P

P

P



Thus, if we could not work out the laminar theory and

chose to use the approximation 64Re hDf or 16Re

hDfC ,

we would be 33 percent low for channel flow.

For laminar flow, 0P

varies greatly,

therefore it is better to

use the exact solution



b. Accuracy for turbulent flow(smooth non-circular

pipe)

For turbulent flow, Dh works much better especially if

combined with “effective diameter” concept based on

ratio of exact laminar circular and noncircular duct P0

numbers, i.e. fCP 0/16 or fP0/64 .

First recall turbulent circular pipe solution and compare

with turbulent channel flow solution using log-law in both

cases

Channel Flow

dYBuyh

uh

Vh

*

0

* )(ln

11

1ln

1 ** B

huu

hhB

hB

P

AD

Bh 4

42

)2(4lim

4

h= half width

Define

hVVDh

Dh

4Re

Y=h-y wall coordinate



19.1Relog2 2/12/1 ffhD

Therefore error in Dh concept relatively smaller for

turbulent flow.

Note 8.0Re64.0log2)( 2/12/1 fchannelfhD

Define Deffective h

f

f

h DchannelP

circlePD

24)(

16)(~64.0

0

0

(therefore, improvement on Dh is)

eff

effD

VDRe

h

C

C

h

f

f

eff DcircularnonP

circlePD

circularnonP

circlePD

f

f

)(

)(

)(

)(

0

0

0

0

Or

h

C

h

f

eff DcircularnonP

DcircularnonP

D

f)(

16

)(

64

00

Very nearly the same as circular pipe

7% to large at Re = 105

4% to large at Re = 108

From exact laminar solution

Laminar solution



































Documents

1. Smooth circular pipe - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part4-58160-2020.pdf058:0160 Chapter 6-part4 Professor Fred Stern Fall 2020 5 2. Turbulent Flow