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
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
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
=∆+∆
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
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
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
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
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
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=
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
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
ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 12
ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 13
ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 14
ME:5160 Chapter 6-part1 Professor Fred Stern Fall 2017 15