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Pipe flow April 8 and 15, 2008

ME 390 – Fluid Mechanics 1

Pipe FlowPipe Flow

Larry CarettoMechanical Engineering 390

Fluid MechanicsFluid Mechanics

April 8 and 15, 2008

2

Outline• Laminar and turbulent flows• Developing and fully-developed flows• Laminar and turbulent velocity profiles:

effects on momentum and energy• Calculating head losses in pipes

– Major losses from pipe only– Minor losses from fittings, valves, etc.

• Noncircular ducts

3

Piping System

• System consists of– Straight pipes– Joints and valves– Inlets and outlets– Work input/output

Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore OkiishiCopyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

4

What We Want to Do• Determine losses from friction forces in

straight pipes and joints/valves– Will be expressed as head loss or

“pressure drop” hL = ΔP/γ• Will show that this is head loss in energy equa-

tion if variables other than pressure change– Losses in straight pipes are called “major”

losses– Losses in fittings, joints, valves, etc. are

called “minor” losses– Minor losses may be greater than major

losses in some cases

5

Pipe Cross Section• Most pipes have circular cross section

to provide stress resistance• Main exception is air conditioning ducts• Consider round pipes first then extend

analysis to non-circular cross sections– Extension based on using same equations

as for circular pipe by defining hydraulic diameter = 4 (area) / (perimeter), which is D for circular cross sections

6

The Pipes are Full• Consider only flows where the fluid

completely fills the pipe• Partially filled pipes are considered

under open-channel flow

Driving force is gravity

Driving force is pressure

Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.



7

Laminar vs. Turbulent Flow

• Laminar flows have smooth layers of fluid • Turbulent flows

have fluctuationsFundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

8

Laminar vs. Turbulent Flow II• Most flows of engineering interest are

turbulent– Analysis relies mainly on experimentation

guided by dimensional analysis– Even advanced computer models, called

computational fluid dynamics (CFD) rely on “turbulence models” that have large degree of empiricism

• Can get some (very limited) analytical results for laminar flows

9

Laminar vs. Turbulent Flow III• Condition of flow as laminar or turbulent

depends on Reynolds number• For pipe flows

– Re = ρVD/μ < 2100 is laminar– Re = ρVD/μ > 4000 is turbulent– 2100 < Re < 4000 is transition flow

• Other flow geometries have different characteristics in Re = ρVLc/μ and different values of Re for laminar and turbulent flow limits

10

Flow Development


11

Developing Flows• Entrance regions and bends create

changing flow patters with different head losses

• Once flow is “fully developed” the head loss is proportional to the distance

• Entrance pressure drop is complex– Complete entrance region treated under

minor losses– Will not treat partial entrance region here

12

Developing Flows II• Entrance regions and bends create

changing flow patters with different head losses

• Once flow is “fully developed” the head loss is proportional to the distance




13

Developing Flows III• After development region, pressure

drop (head loss) is proportional to pipe length

• Equations for entrance region length, ℓe

– Laminar flow:

– Turbulent flow:

– Turbulent flow rule of thumb ℓe ≈ 10D

Re06.0=Del

61Re4.4=Del

14

Fluid Element in Pipe Flow

• Look at arbitrary element, with length ℓ, and radius r, in fully developed flow

• What are forces on this element?Fundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

15

Fully Developed Flow

• Pressure drop is due to viscous stresses

( ) 0212

12 =−Δ−−=∑ lrpprprFx πτππ

Flow Direction

rp lτ=Δ

2


No change in momentum

16

Extend Relation to Wall

• Have Δp = 2τℓ/r for any r: 0 < r < R = D/2• For wall r = R = D/2 and τ = τw = wall

shear stress: Δp = 2τwℓ/R = 4τwℓ/DFundamentals of Fluid Mechanics, 5/E by Bruce Munson, Donald Young, and Theodore Okiishi. Copyright © 2005 by John Wiley & Sons, Inc. All rights reserved.

17

Fully Developed Laminar Flow• Can get

exact equation for pressure drop

4128

DQp

πμ

=Δl

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

2

1Rruu c

• Laminar velocity profile


R

uc

18

Fully Developed Laminar Flow II

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=

2

1Rruu c

• Laminar shear stress profile found from


R

rD

uR

rudrdu c

c 2282 μ

=μ=μ=τdrdu

μ=τ



19

Fully Developed Laminar Flow III

∫∫∫ π⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−=π==π==

R

c

R

A

rdrRrurdruudARVVAQ

0

2

0

2 212

• What is centerline velocity, uc?

42

4222

2

02

42

0 02

3 RuRrrudr

RrrdruQ c

R

c

R R

c π=⎥⎥⎦

⎤

⎢⎢⎣

⎡−π=

⎥⎥⎦

⎤

⎢⎢⎣

⎡−π= ∫ ∫

VRRV

RVA

RQuRuQ cc 2222

2 2

2

22

2=

ππ

=π

=π

=⇒π=

Centerline uc is twice the mean velocity, V 20

Effect of Velocity Profile• Momentum and kinetic energy flow for

mean velocity, V– FlowMomentum = V = ρVAV = ρV2(πR2)– FlowKE = V2/2 = ρVAV2/2 = ρV3(πR2)/2

• Accurate representation uses profile

m&

m&

AVrdrRruudAuFlow

R

cA

Momentum2

0

2

2

2

3421 ρ=π

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−ρ=ρ= ∫∫

2221

21

2

3

0

3

2

22 VArdrRruuudAFlow

R

cA

KE ρ=π⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−ρ=ρ= ∫∫

21

Turbulent Flow• For laminar and turbulent flows, the

velocity at the wall is zero– This is called the no-slip condition– Momentum is maximum in the center of the

flow and zero at the wall• Laminar flows: momentum transport from wall

to center is by viscosity, τ = μdu/dr• Turbulent flows: random fluctuations exchange

eddies of high momentum from the center with low momentum flow from near-wall regions

22

Turbulent Flow Quantities


Velocities at one point as a function of time

u(t) = instantaneous velocityu’ = velocity fluctuation = u –

∫+

=Tt

t

dttuT

u0

0

)(1

u

23

Momentum Exchange

Laminar flow –random molecular motion

Turbulent flow –eddies have structure

Fundamentals of Fluid

Mechanics, 5/E by Bruce

Munson, Donald Young, and

Theodore Okiishi.

Copyright ©2005 by John Wiley & Sons, Inc. All rights

reserved.


24

Turbulence Regions/Profiles

• Very thin viscous sublayer next to wall– 0.13% of R = 3 in for H20 at = 5 ft/s

• Flat velocity profile in center of flow

( )drud

η+μ=τ

turbulent eddy

viscosity, η

u




25

Profilen

c Rr

Vu 1

1 ⎟⎠⎞

⎜⎝⎛ −=

Turbulent velocity profiles with n a

function of Reynolds number

n = 6: Re = 1.5x104; Vc/V = 1.264n = 8: Re = 4x105; Vc/V = 1.195

n = 10: Re = 3x106; Vc/V = 1.155Laminar: Vc/V = 2 V = Q/A


26

Effect of Velocity Profile• Analysis similar to one used for laminar

flow profile– Determine momentum and kinetic energy

flow for mean velocity– Correction factor multiplies average V

results to give integrated u2 and u3 values

1.0311.0113x106101.0461.0164x10581.0771.0271.5x1046

KEMomentumRen

27

Pipe Roughness• Effect of rough walls on pressure drop

may depend on surface roughness of pipe

• Typical roughness values for different materials expressed as roughness length, ε, with units of feet or meters

• Only turbulent flows depend on roughness length, laminar flows do not

28

Pipe roughness effects in viscous sublayer affects pressure drop in turbulent flow

No effect on laminar flow


29

Use this table (p 433 of text) to find ε


30

Energy Equation• Energy equation between inlet (1) and

outlet (2)

Ls hhg

Vpzg

Vpz −++γ

+=+γ

+22

211

1

222

2

• Previous applications allowed us to compute the head loss from all other data in this equation– Call this the measured head loss

• We can compute it, but we have no way of knowing its cause



31

Pressure Drop/Head Loss• We now seek a design calculation for hL

• Use level pipe (z1 = z2) with constant area (V1 = V2) and no shaft head (hs = 0)

γΔ

=γ

−γ

=ppphL

21

• Calculated Δp for z1 = z2, V1 = V2, and hs= 0 gives hL for more general flows

Ls hhg

Vpzg

Vpz −++γ

+=+γ

+22

211

1

222

2

32

Pressure Drop/Head Loss• We now seek a design calculation for hL

• Use level pipe (z1 = z2) with constant area (V1 = V2) and no shaft head (hs = 0)

γΔ

=γ

−γ

=ppphL

21

• Will use friction factor f for Δp in such flows, but we are really getting hL– Extend to more general flows later

33

Head Loss in Pipes• Dimensional analysis shows that

dimensionless pressure drop, Δp/ρV2, is a function of Reynolds number, ρVD/μ, the ℓ/D ratio and relative roughness, ε/D

• Expressed in terms of friction factor, f

⎟⎟⎠

⎞⎜⎜⎝

⎛ εμ

ρ=

ρ

Δ=

DVDf

VD

pf ,

21 2l

gV

Dfh

VD

hf LL

221

2

2

ρ=⇒

ρ

γ=

l

l34

How do we get f?• Have said that f = f(Re, ε/D)• What is form of this relationship?• For laminar flow we will later show that

f = 64/Re• Relationship determined experimentally

with empirical fit to equations for turbulent flows

• Results expressed as Moody diagram

35

Moody Diagram



36

Moody Diagram Equations• Colebrook equation

(turbulent) ⎟⎟⎠

⎞⎜⎜⎝

⎛+

ε−=

fD

f Re51.2

7.3log0.21

10

• Laminar

Re64644

256

21

128

21 2

23

2

4

2=

μρ

=ρ

ππ

μ

=ρ

πμ

=ρ

Δ= VDV

DVD

VD

DQ

VD

pfl

l

l

• Haaland equation (turbulent) ⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎠⎞

⎜⎝⎛ ε+−≈

11.1

10 7.3Re9.6log8.11 D

f



37

Wholly Turbulent Flows• Large Reynolds numbers: f independent

of Re depends only on ε/D

4

2

22

2

2 16

42 D

QVD

QAQVV

Dfp

π=⇒

π==

ρ=Δ

l

5

2

25

2

24

2

22 8816

22 Dmf

DQf

DQ

DfV

Dfp

ρπ=

ρπ

=π

ρ=

ρ=Δ

&llll

• Pressure drop varies as D-5

– Similar to D-4 dependence in laminar flow

38

Pressure Drop Problems• Find the pressure drop given fluid data,

pipe dimensions, ε, and flow (volume flow, mass flow, or velocity)– Get A = πD2/4– Get V = Q/A or V = /ρA if not given V– Find ρ and μ for fluid at given T and P– Compute Re = ρVD/μ and ε/D– Find f from diagram or equation

• Laminar f = 64/Re; Colebrook for turbulent– Compute Δp = f (ℓ/D) ρV2/2

m&

39

Sample Problem• You have been asked to size a pump

for an airport fuel delivery system. JP-4 fuel (ρ = 1.50 slug/ft3, μ = 1.2x10-5

slug/ft·s) has to travel 0.5 mi through commercial steel, schedule 40 pipe with a nominal 6 in diameter. The flow rate is 5 slug/s. What is the head loss?

• Schedule 40 pipe: OD = 6.625 in; thickness = 0.280 in; ID = 6.065 in

40

Sample Problem Solution( ) ( ) 222 2006.05054.0

445054.0

12065.6 ftftDAft

inftinD =

π=

π===

( ) sft

ftftslug

sslug

AmV 61.16

2006.050.1

5

33

==ρ

=&

( )41001005.1

102.1

5054.061.1650.1

Re 65

3>=

⋅

=μ

ρ= − x

sftslugx

fts

ftftslug

VD

Since Re > 4,100, flow is turbulent

41

Sample Problem Solution II000297.0

5054.000015.0

==ε

ftft

Dε = 0.00015 for commercial steel (Table 8.1, page 433)

0155.0)000297.0,1005.1(Re 6 ==ε= Dxf

Check f value with Colebrook equation

Find f from Moody diagram (page 434)

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=⎟

⎟⎠

⎞⎜⎜⎝

⎛+

ε−=

0155.01005.151.2

7.3000297.0log0.2

Re51.2

7.3log0.21

61010 xfD

f

0156.0005.81005.81

2 ==⇒= ff

Use f = 0.0156 42

Sample Problem Solution III22

3

2 61.16150.152805054.0

5.02

0156.02

⎟⎠⎞

⎜⎝⎛

⋅

⋅=

ρ=Δ

sft

ftslugslb

ftslug

mift

ftmiV

DfP fl

psiin

lb

ft

lbP ff 2.117

2.1171687622 ===Δ

• For shaft head to overcome this lead loss

ρΔ

≥⇒ρΔ

=γΔ

=≥=PmW

gPPhh

gm

W

innetshaftLs

innetshaft

&&&

&

hplbftshp

ftslug

ft

lbs

slugPmW

f

f

innetshaft 102

55050.1

168765

3

2=

⋅⋅

=ρΔ

≥&&



43

Pressure Drop Problems II• Find the diameter for a given pressure

drop given fluid data, ε, and flow (volume flow, mass flow, or velocity)– Find ρ and μ for fluid at given T and P– Guess D; get A = πD2/4– Get V = Q/A or V = /ρA if not given V– Compute Re = ρVD/μ and ε/D– Find f from diagram or equation

• Laminar f = 64/Re; Colebrook for turbulent– Compute Δpcalculated = f (ℓ/D) ρV2/2– Iterate on D until Δpcalculated = Δprequired

m&

44

A Harder Problem• You have a 200 hp pump to deliver 5

slug/s of JP-4 fuel (ρ = 1.50 slug/ft3, μ = 1.2x10-5 slug/ft·s) over 0.5 mi. What diameter of commercial steel, schedule 40 pipe should be used?

• Compute required Δp

( )2

3 330005

55020050.1

ft

lb

sslug

shplbft

hpftslug

m

Wp f

f

innetshaft

required =⋅

⋅

=

ρ

=Δ&

&

45

Iterative Solution• The calculation we just did for D = 6.065

in gave Δp = 16876 psf an error of 16876 psf – 33000 psf = –16124 psf

1Count

–16124168766.065Error (psf)Δpcomputed (psf)Dguess (in)

• Take second guess of D = 5 in and repeat calculations done previously to find Δpcomputed

46

Iterative Problem Solution( ) ( ) 222 1364.04167.0

444167.0

125 ftftDAft

inftinD =

π=

π===

( ) sft

ftftslug

sslug

AmV 45.24

1364.050.1

5

33

==ρ

=&

( )41001027.1

102.1

1364.045.2450.1

Re 65

3>=

⋅

=μ

ρ= − x

sftslugx

fts

ftftslug

VD

Since Re > 4,100, flow is turbulent

47

Iterative Problem Solution II00036.0

4167.000015.0

==ε

ftft


0159.0)000297.0,1027.1(Re 6 ==ε= Dxf



⎟⎟⎠

⎞⎜⎜⎝

⎛+−=⎟

⎟⎠

⎞⎜⎜⎝

⎛+

ε−=

0159.01027.151.2

7.300036.0log0.2

Re51.2

7.3log0.21

61010 xfD

f

0160.0894.71894.71

2 ==⇒= ff

Use f = 0.0160 48

Iterative Problem Solution III22

3

2 45.24150.152804167.0

5.02

0160.02

⎟⎠⎞

⎜⎝⎛

⋅

⋅=

ρ=Δ

sft

ftslugslb

ftslug

mift

ftmiV

DfP fl

psiin

lb

ft

lbP ff 4.316

4.3164556422 ===Δ

• We now have two iterations

1256445564521

Count–16124168766.065

Error (psf)Δpcomputed (psf)Dguess (in)



49

Iterative Problem Solution IV• Use linear interpolation to get new guess,

Di+1 that sets error ei+1 to zero

1256445564521

Count–16124168766.065

Error (psf)Δpcomputed (psf)Dguess (in)

( )iiii

iiii ee

eeDDDD −

−−

+= +−

−+ 1

1

11

( ) 466.51612412564

065.651256451

11 =

−−−

−=−−

−=−

−+

ii

iiiii ee

DDeDD0

1

1

−

−

−−

−=ii

iiii ee

DDeD

50

Iterative Problem Solution V• Continue iterations until error is “small”

–1329995.323672330725.3215

–823321765.3494–4219287805.4663125644556452

1Count

–16124168766.065Error (psf)Δpcomputed (psf)Dguess (in)

51

Iterations and Reality• Commercial pipe and tubing only comes

in fixed sizes– Extra iterations not needed once the

minimally acceptable commercial size is found

– In this case available nominal diameters are 5 in and 6 in with actual inside diameters of 5.047 in and 6.065 in (for Schedule 40)

– Only choice is 6 in (nominal)52

Pressure Drop Problems III• Find the flow rate for a given pressure

drop given fluid data, ε, and diameter– Get A = πD2/4– Find ρ and μ for fluid at given T and P– Guess V– Compute Re = ρVD/μ and ε/D– Find f from diagram or equation

• Laminar f = 64/Re; Colebrook for turbulent– Compute Δpcalculated = f ℓ/D ρV2/2– Iterate on V until Δpcalculated = Δprequired– Compute Q or as desiredm&

53

Different Friction Factors• The friction factor definition we are

using here is the common one– Called the Darcy friction factor if

clarification is needed• Another definition, called the Fanning

friction factor is sometimes seen– Fanning factor = τw / (ρV2/2)

• From the relationship that τw = DΔp/4ℓ we get the result that the Fanning factor is one fourth of the Darcy factor

54

Minor Losses• Determine pressure drop (head loss) in

a variety of flow passages– Entrance into a piping system– Exit from a piping system– Expansion in a piping system– Contraction in a piping system– Valves of various types (with different

opening fractions)– Fittings (elbows, tees, bends, unions)



55

Minor Losses• Fittings in pipe systems modeled as

loss coefficients, KL

222

222 VKpg

VKgp

gVKh LLL

LLL

ρ=Δ⇒=

ρΔ

⇒=

• KL depends on geometry and Re– For flows dominated by inertia effects KL is

a function of geometry only• Alternative process, not given here,

uses equivalent length for minor losses56

Entrance Losses

Reentrant: KL = 0.8

Sharp edged: KL = 0.5

Slightly rounded: KL = 0.2 Well rounded:

KL = f(r/D)

r

D

2

2VKp LLρ

=Δ V = Pipe velocity


57

Rounded Inlet KL


Slightly rounded KL= 0.2 for r/D = 0.055

r/D = 0 is square inlet

58

Full KE loss cannot be recovered in sharp-edged entrance


59

KL = 1 for all exit flows

Reentrant

Slightly rounded

Sharp edged

Well rounded


60

New AreaSudden contraction (left)

For sudden expansion (right) KL = ( 1 –A1/A2)2



61


62


63

Problem with Minor Losses• 4 kg/s of oil with SG = 0.82 and μ = 0.05

kg·m/s2 is pumped from one tank to another. The line of 2-in Schedule-40 pipe has a total length of 40 m, with two gate valves and six elbows (regular flanged 90o). The entrance is rounded with an r/D ratio of 0.1.

• Find pressure loss with both valves open• 2-in schedule 40 pipe has OD = 2.375 in

and thickness = 0.154 in, so ID = 2.067 in = 0.05250 m

64

Problem Solution( )

2

22

002165.0

05250.044

m

mDA

=

π=

π=

( ) sm

mm

kgskg

AmV 26.2

002165.02.819

4

33

==ρ

=&

( )21001940050.0

05250.026.22.819

Re3

<=

⋅

=μ

ρ=

smkg

ms

mm

kgVD

Since Re < 2,100, flow is laminar

( )

skg

skg

SG ref2.819999)82.0( =

=ρ=ρ

65

Problem Solution IIFind Δpmajor directly from laminar flow equation

( ) ( )

( )kPa

mN

ms

mmm

sN

DQpmajor 369.5252369

05250.0

004883.04005.012812824

3

2

4 ==π

⋅

=πμ

=Δl

sm

kgm

skgmQ

33 04883.02.819

4==

ρ=&

Minor losses coefficients: rounded entrance (r/D = 0.1), KL = 0.12; exit, KL = 1; fully open gate valve, KL = 0.15; 6 elbows, KL = 6(0.3) = 1.8. Total KL = 0.12 + 1 + 0.15 + 1.8 = 3.07

Could also use f = 64/Re

66

Problem Solution IIIΔpminor = (Loss coefficient sum) times ρV2/2

( ) 2

22

3

2 397,6126.2819207.3

2 mN

mkgsN

sm

mkgVKp L =

⋅⋅

⎟⎠⎞

⎜⎝⎛=

ρ=Δ ∑minor

22397,6369,52m

Nm

Nppp majortotal +=Δ+Δ=Δ minor

kPam

Nptotal 8.58766,582 ==Δ



67

Noncircular Ducts• Define hydraulic diameter, Dh = 4A/P

– A is cross-sectional area for flow– P is wetted perimeter– For a circular pipe where A = πD2/4 and P

= πD, Dh = 4(πD2/4) / (πD) = D• For turbulent flows use Moody diagram

with D replaced by Dh in Re, f, and ε/D• For laminar flows, f = C/Re (both based

on Dh) – see next slide for C values68

2

2VD

fPh

ρ=Δ

l

PADh

4=

μρ

=VD

hRe


42

hDQCP lμ

π=Δ

69

Problem• An 10-in-square, commercial steel air

conditioning duct contains air at 80oF and atmospheric pressure and has a flow rate of 125 ft3/min. Find the pressure drop per unit duct length

• Property data at 80oF (Table B.3) ρ = 0.002286 slug/ft3; ν = 1.69x10-4 ft2/s

• Solution: find Reh to see if flow is laminar or turbulent then find f and Δp

70

Solution

ftin

LLL

PADh

8333.010444 2

==

===

( )5

24 1078.11069.1

8333.03

Re x

sftx

ftsft

VDhh ==

ν= −

( ) sft

inftin

sft

AQV 3

14410

60min

min125

2

22

3

===

00018.08333.000015.0

==ε

ftft


Turbulent flow for Reh > 4100

71

Solution II

0172.0)00018.0,1027.1(Re 6 ==ε= Dxf



⎟⎟⎠

⎞⎜⎜⎝

⎛+−=⎟

⎟⎠

⎞⎜⎜⎝

⎛+

ε−=

0172.01078.151.2

7.300018.0log0.2

Re51.2

7.3log0.21

51010 xfD

f

0173.0611.71611.71

2 ==⇒= ff Use f = 0.0173

3

522

3

2 1078.13100229.08333.0

12

0173.02

1ft

lbxsft

ftslugslb

ftslug

ftV

DfP ff

−

=⎟⎠⎞

⎜⎝⎛

⋅

⋅=

ρ=

Δl

72

Recommended Air Velocity

6.5 - 152.0 - 4.5Ventilation ducts (offices)5.9 - 131.8 - 4Ventilation ducts (hospitals)66 - 9820 - 30Compressed air pipe26 - 498 - 15Vacuum cleaning pipe

2.6 - 3.30.8 - 1.0Warm air for house heating3.3 - 9.81 - 3Air inlet to boiler room40 - 6612 - 20Combustion air ducts

ft/sm/sAir Velocity

Air Ducts

http://www.engineeringtoolbox.com/flow-velocity-air-ducts-d_388.html

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