Chapter 6: Viscous Flow in Ducts - University of...

058:0160 Chapter 6-part4 Professor Fred Stern Fall 2018 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

dave

wu

f Re/6482 ==

ρτ

2000Re ≤=υ

duaved

A turbulent-flow “approximate” solution can be obtained simply by computing uave based on log law.

Bv

yuuu

+=*

* ln1κ

Where

rRyuByuu w −===== ;/;5;41.0);( * ρτκ

−+=

+=== ∫

κκ

πκπ

32ln221

2ln11

**

0

**

2

Bv

Ruu

drrBv

yuuRA

QuVR

ave

Or:


34.1ln44.2*

* +=v

RuuV

8.0]log[Re2

02.1]log[Re99.12/1

2/12/1

−=

−=−

fff

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

vDLfphf 2

2=

∆=

γ

Turbulent Flow:

4/74/54/14/3158.0 VDLp −=∆ µρ 75.175.44/14/3241.0 QDL −= µρ Laminar flow: 4/8 RLQp πµ=∆

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.

BRuuru

uu

+==

=υκ

*

**max ln1)0(

Combine with

κυκ 23ln1 *

* −+= BRuuV

Vu

VuuuV

uu

uV

κκκ 231

23

23 *

max*

max*max

* +=⇒−=⇒−=⇒ Also

8/8/18/1

*

2

2*

2

2* fVu

Vuf

Vfandu w

w =⇒=⇒==ρ

ρρτρτ

ffV

uV

u 3.118/231

231

*max +=+=+=⇒

κκ Or:

For Turbulent Flow: ( ) 1

max

3.11−

+= fuV

Recall laminar flow:

5.0/ max =uV


2. Turbulent Flow in Rough circular pipe

),( +++ = kyfU )/,(Re dkff d=

)(ln1 +++ ∆−+= kBByUκ

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

2/12/1

7.3/

Re9.6log8.1~

Re51.2

7.3/log2

dk

fdkf

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, µ, g

Log law shifts downward

Moody diagram

Approximate explicit formula

058:0160 Chapter 6-part4 Professor Fred Stern Fall 2018 6 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

2fL Vh f hD g

= = ∆ ( )

−+−=∆γγ

2121

ppzzh

=

+∆ zp

γ

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

2/12/32/1 2

=

LghDf f

ν

scale on Moody Diagram

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

3) solve V from g2

VDLfh

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

5/15/1

f2

2f

ghLQ8D ⋅

π=

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 fDh

VgL f2

2= .


3. Concept of hydraulic diameter for noncircular ducts

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

definitions of 28V

f wρτ

= and 22V

C wf ρ

τ= are required.

Define average wall shear stress

dsP

Pww ∫=

0

1 ττ ds = arc length, P = perimeter

Momentum:

0=

∆+−∆

Lz

WALPLpA w γτ

( )PALzph w

//

γτγ =+∆=∆

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

2/RLh w

γτ

=∆ )


Energy:

PALhh w

L /γτ

==∆

( )

−=

+−=

−=

∆=

dxpd

PA

dxzpd

PA

dxdh

PA

Lh

PA

w

^γγγτ non-circular duct

Recall for circular pipe:

dxpdD

dxpdR

wˆ

4ˆ

2−=−=τ

In analogy to circular pipe:

PADD

PA

dxpdD

dxpd

PA

hhh

w4

4

^

4

^=⇒=⇒

−=

−=τ

For multiple surfaces such as concentric annulus P and A based on wetted perimeter and area

υε

ρτ h

DhDw VD

DfV

fhh=== Re)/,(Re8

2

gV

DLf

RLfV

RLhh

hhh

wL 28

22

====∆γ

ργτ

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

0 fP

CPPPoiseuille

f

fc f

Recall for channel flow:

hD

hhVhf

Re42 Re

96Re4824

===ρµ

hD

hhf

f

VhC

fC

Re42 Re

24Re126

4/

===

⇒=

ρµ

==

==

96Re

24Re)(#

0

0

0

h

hf

Df

Dfc

fP

CPPPoiseuille

Therefore:

32

9664

2416

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

dYBuyhuh

Vh

+

−= ∫ υκ

*

0

* )(ln11

−+=κυκ1ln1 *

* Bhuu

hhB

hBPAD

Bh 442

)2(4lim4=

+==

∞→ h= half width

Define υυhVVDh

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 −=− fchannelf

hD

Define Deffective hf

fh D

channelPcircleP

D24)(

16)(~64.0

0

0

==

=

(therefore, improvement on Dh is)

υ

effeffD

VD=Re

hC

Ch

f

feff D

circularnonP

circlePD

circularnonPcircleP

Df

f

)(

)(

)()(

0

0

0

0

−=

−=

Or

hC

hf

eff DcircularnonP

DcircularnonP

Df

)(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

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