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

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ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 1 Chapter 6: Viscous Flow in Ducts 6.1 Laminar Flow Solutions Entrance, developing, and fully developed flow Le = f (D, V, ρ, μ) (Re) f D L theorem e i = Π f(Re) from AFD and EFD Laminar Flow: Re crit ~ 2000 Re < Re crit laminar Re > Re crit unstable Re > Re trans turbulent Re 06 . / D L e D D L crit e 138 ~ Re 06 . max = Max Le for laminar flow

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

Page 1: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 1

Chapter 6: Viscous Flow in Ducts 6.1 Laminar Flow Solutions Entrance, developing, and fully developed flow

Le = f (D, V, ρ, μ)

(Re)fDLtheorem e

i =→Π f(Re) from AFD and EFD

Laminar Flow: Recrit ~ 2000 Re < Recrit laminar Re > Recrit unstable Re > Retrans turbulent

Re06./ ≅DLe

DDL crite 138~Re06.max = Max Le for laminar flow

Page 2: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 2

Turbulent flow:

6/1Re4.4~/ DLe

(Relatively shorter than for

laminar flow)

Laminar vs. Turbulent Flow

Re Le/D 4000 18 104 20 105 30 106 44 107 65 108 95

Hagen 1839 noted difference in Δp=Δp(u) but could not explain two regimes

Laminar Turbulent Spark photo

Reynolds 1883 showed that the difference depends on Re = VD/ν

Page 3: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 3

Laminar pipe flow: 1. CV Analysis

Continuity:

.0 21 constQQdAVCS

==→⋅= ∫ ρρρ

1 2 1 2. . sin , ., avei e V V ce A A const and V Vρ= = = =

Momentum:

∑𝐹𝐹𝑥𝑥 = (𝑝𝑝1 − 𝑝𝑝2)�������Δ𝑝𝑝

𝜋𝜋𝑅𝑅2 − 𝜏𝜏𝑤𝑤2𝜋𝜋𝑅𝑅𝜋𝜋 + 𝛾𝛾𝜋𝜋𝑅𝑅2𝜋𝜋���𝑊𝑊

sin𝜙𝜙���Δ𝑧𝑧/𝐿𝐿

= �̇�𝑚(𝛽𝛽2𝑉𝑉2 − 𝛽𝛽1𝑉𝑉1)�����������=0

02 22 =∆+−∆ zRRLRp w γππτπ

RLzp wτγ 2

=∆+∆

Page 4: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 4

1 22( / ) w Lh h h p z

Rτγγ

∆ = − = ∆ + = or

)(

2

22

zpdxdRdxdhR

LhR

w

γ

γγτ

+−=

−=∆

=

For fluid particle control volume:

)(2

zpdxdr γτ +−=

i.e. shear stress varies linearly in r across pipe for either laminar or turbulent flow Energy:

LhzVg

pzVg

p+++=++ 22

2211

11

22α

γα

γ

RLhh w

L γτ2

==∆

∴ once τw is known, we can determine pressure drop In general,

),,,,( εµρττ DVww =

Πi Theorem

roughness

Page 5: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 5

)/,(Re82 Dffactorfrictionf

V Dw ε

ρτ

===

where υ

VDD =Re

g

VDLfhh L 2

2==∆ Darcy-Weisbach Equation

f (ReD, ε/D) still needs to be determined. For laminar flow, there is an exact solution for f since laminar pipe flow has an exact solution. For turbulent flow, approximate solution for f using log-law as per Moody diagram and discussed later. 2. Differential Analysis Continuity:

0=⋅∇ V

Use cylindrical coordinates (r, θ, z) where z replaces x in previous CV analysis

0)(1)(1=

∂∂

+∂∂

+∂∂

zvv

rvr

rrz

r θθ

Page 6: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 6

where 𝑉𝑉 = 𝑣𝑣𝑟𝑟𝑒𝑒𝑟𝑟� + 𝑣𝑣𝜃𝜃𝑒𝑒𝜃𝜃� + 𝑣𝑣𝑧𝑧𝑒𝑒𝑧𝑧�

Assume θv = 0 i.e. no swirl and fully developed

flow 0=∂∂

zvz , which shows rv = constant = 0 since

)(Rvr =0 ∴ 𝑉𝑉 = 𝑣𝑣𝑧𝑧𝑒𝑒𝑧𝑧� = 𝑢𝑢(𝑟𝑟)𝑒𝑒𝑧𝑧�

Momentum:

𝜌𝜌𝐷𝐷𝑉𝑉𝐷𝐷𝐷𝐷

= 𝜌𝜌𝜕𝜕𝑉𝑉𝜕𝜕𝐷𝐷

+ 𝜌𝜌𝑉𝑉 ⋅ ∇V = −∇(p + γz) + µ∇2𝑉𝑉

Where

𝑉𝑉 ⋅ ∇= 𝑣𝑣𝑟𝑟𝜕𝜕𝜕𝜕𝑟𝑟

+ 𝑣𝑣𝜃𝜃1𝑟𝑟𝜕𝜕𝜕𝜕𝜃𝜃

+ 𝑣𝑣𝑧𝑧𝜕𝜕𝜕𝜕𝑧𝑧

z equation:

uzpz

uVtu 2)( ∇++

∂∂

−=

∇⋅+∂∂ µγρ

)()(

1)(0

rfzfrur

rrzp

z

∂∂

∂∂

++∂∂

−= µγ

∴ both terms must be constant

Page 7: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 7

γµ

µ

µ

µ

+=++∂∂

=⇒

+∂∂

=∂∂

+∂∂

=∂∂

∂∂

=∂∂

∂∂

ppBrArzpu

Arzp

ru

Arzp

rur

zp

rur

rr

ˆlnˆ

41

ˆ21

ˆ21

ˆ)(

2

2

)0( =ru finite A = 0

u (r=R) = 0 dzpdRBˆ

4

2

µ−=

dzpdRrruˆ

4)(

22

µ−

= dzpdRuuˆ

4)0(

2

max µ−==

zpR

ru

yu

zpr

stressshearfluidru

ru

zv

Rryw

r

∂∂

−=∂∂

−=∂∂

=

∂∂

=

∂∂

=

∂∂

+∂∂

=

==

ˆ2

ˆ2

0

µµτ

µµτ

𝑦𝑦 = 𝑅𝑅 − 𝑟𝑟, 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑= 𝑑𝑑𝑟𝑟

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑟𝑟

= −𝑑𝑑𝑑𝑑𝑑𝑑𝑟𝑟

As per CV analysis

Page 8: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 8

42

max0

^1( )2

8 2

R R d pQ u r r dr u Rdz

ππ πµ

−= = =∫

2

max2

^12 8ave

Q R d pV uR dzπ µ

−= = =

Substituting V = Vave

28V

f w

ρτ

=

DV

RV

RVR aveave

wµµµτ 848

2 2 ==−

×−=

DDVf

Re6464

==ρ

µ

or D

wf

f

VC

Re16

421 2

===ρ

τ

VgD

LVg

VDL

DVgV

DLfhh L ∝=××===∆ 2

22 322

642 ρ

µρ

µ

for Vpz ∝∆→=∆ 0

Page 9: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 9 Both f and Cf based on V2 normalization, which is appropriate for turbulent but not laminar flow. The more appropriate case for laminar flow is:

==

==

64Re

16Re)(#

0

0

0 fP

CPPPoiseuille

f

fc f

for pipe flow

Compare with previous solution for flow between parallel plates with

^

xp

−=

2

max 1hyuu xphu

^

2

2

max µ−

=

−== xphhuq

^

32

34 3

max µ

𝑉𝑉𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑞𝑞2ℎ

= ℎ2

3𝜇𝜇(−𝑝𝑝𝑥𝑥�) = 2

3𝑢𝑢𝑚𝑚𝑎𝑎𝑥𝑥

hV

wµτ 3=

Page 10: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 10

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

Same as pipe other than constants!

32

9664

2416

0

0

0

0====

hf

f

hf

f

Donbasedchannel

pipe

Donbasedchannelc

pipec

P

P

P

P

Page 11: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 11

Exact laminar solutions are available for any “arbitrary” cross section for laminar steady fully developed duct flow

BVP

0)()(0

0^

=++−=

=

huuup

u

ZZYYx

x

µ

hyy /* = hzz /* = Uuu /* = )(^2

xphU −=µ

12 −=∇ u Poisson equation

u(1) = 0 Dirichlet boundary condition

Can be solved by many methods such as complex variables and conformed mapping, transformation into Laplace equation by redefinition of dependent variables, and numerical methods.

Related umax

Re only enters through stability and transition

Page 12: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 12

Page 13: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 13

Page 14: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 14

Page 15: Chapter 6: Viscous Flow in Ducts - University of Iowauser.engineering.uiowa.edu/.../Chapter6_part1-58160-2017.pdf · 2018. 10. 12. · ME:5160 Chapter 6-part1 Professor Fred Stern

ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 15