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
ave d w
f 8u 64/Re
2 =
= ρ
τ
Re = ≤ 2000
υ d u
aved
A turbulent flow “approximate” solution can be
obtained simply by computing u
avebased on log law.
v B yu u
u* = 1 ln * +
κ
Where
r R y u
B y
u
u = ( );κ = 0.41; = 5; * = τw / ρ; = −
+ −
=
+
=
=
=
∫
κ κ
κ π π
2 3 2 ln
2 1
2 1 ln
1
* * 0
* * 2
v B u Ru
dr r v B
u yu R
A u Q
V ave R
Or:
34 . 1 ln
44 .
2 *
* = +
v Ru u
V
8 . 0 ] log[Re
2
02 . 1 ] log[Re
99 . 1
2 / 1
2 / 1 2
/ 1
−
=
−
− =
f f f
d d
Since f equation is implicit, it is not easy to see dependency on ρ, μ, V, and D
4 /
Re
1316 . 0 )
( pipe =
−Df
g v D f L h
fp
2
=
2= ∆ γ
Turbulent Flow: ∆ p = 0 . 158 L ρ
3/4µ
1/4D
−5/4V
7/475 . 1 75 . 4 4 / 1 4 /
241
3.
0 L D
−Q
= ρ µ
Laminar flow: ∆ p = 8 µ LQ / π R
4EFD 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)
V
2p
∆
(turbulent) decreases more sharply with D than
p
∆
(laminar) for same Q; therefore, increase D for smaller
p
∆
. 2D decreases
∆pby 27 for same Q.
Ru B u
r u u
u = = +
=
κ υ
*
*
max* ( 0) 1 ln
Combine with
κ υ
κ 2
ln 3
1 *
* = Ru + B−
u V
V u V
u u u
u V u u
V
κ κ
κ 2
1 3 2
3 2
3 * max *
* max max
* = − ⇒ = − ⇒ = +
⇒
Also
8 8 /
/ 1 8
/ 1
* 2
*2 2
*2 f
V u V
f u f V
and
u w
w = = ⇒ = ⇒ =
ρ ρ ρ
ρ τ τ
f V f
u V
u /8 1 1.3
2 1 3 2
1 3 *
max = + = + = +
⇒ κ κ
Or:
For Turbulent Flow: ( )1
max
3 . 1
1+ −
= f
u V
Recall laminar flow:
5 . 0 /umax = V
2. Turbulent Flow in Rough circular pipe )
, (
+ ++
= f y k
U
f = f (Red,k /d)) (
1 ln + +
+ = y + B− ∆B k
U κ
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 / 2 1
/ 1
7 . 3
/ Re
9 . log 6 8 . 1
~
Re
51 . 2 7
. 3 log / 2
d k
f d
f k
d
d
Log law shifts downward
Moody diagram Approximate explicit formula
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 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
f L V2
h f h
= D g = ∆
( )
−
+
−
=
∆ 1 2 γ1 γ2
p z p
z
h
=
+
∆ p z
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 / 2 1
/
f 1 f
D / L
gh
V 2 ⋅ −
=
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 / 2 1
/ 2 3
/
1 2
=
L D gh
f f
ν
scale on Moody Diagram
1) compute Ref1/2 and ks/D 2) read f
3) solve V from
g 2 V D f L
hf = 2
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 2 f
2 f
gh LQ
D 8 ⋅
= 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 L Vg2 Dhf f
= 2 .
Minor losses in pipe flow are a major part in calculating the flow, pressure, or energy reduction in piping systems. Liquid moving through pipes carries momentum and energy due to the forces acting upon it such as pressure and gravity. Just as certain aspects of the system can increase the fluids energy, there are components of the system that act against the fluid and reduce its energy, velocity, or momentum. Friction and minor losses in pipes are major contributing factors
Abrupt Expansion
Consider the flow from a small pipe to a larger pipe. Would like to know hL = hL(V1,V2). Analytic solution to exact problem is extremely difficult due to the occurrence of flow separations and turbulence. However, if the assumption is made that the pressure in the
separation region remains approximately
constant and at the value at the point of separation, i.e., p1, an approximate solution for hL is possible:
Apply Energy Eq from 1-2 (α1 = α2 = 1)
L 22
2 2 12
1 1 h
g 2 z V p
g 2 z V
p + + +
= γ +
γ +
Momentum eq. For CV shown (shear stress neglected)
∑Fs = p1A2 −p2A2 −Wsinα = ∑ρuV⋅A
=ρV1(−V1A1)+ρV2(V2A2)
=ρV22A2 −ρV12A1 Wsinα
next divide momentum equation by γA2
L L z A2 ∆ γ
÷ γA2
( )
−
=
−
=
− γ −
γ − 1
A A A
A g V A
A g V g
z V p z
p
2 1 2
2 1 1 2
2 1 2 1
2 2 2 1
1
from energy equation ⇓
2 2 1 2 1
L 2 12
22
A A g V g
h V g 2 V g
2
V − + = −
− +
=
2 2 1
2 1
L 2 A
A 1 2
g 2 V g
2 h V
+ −
=
2 2 1 2 1
2 1 2
L A
V A 2 V
g V 2 h 1
−2V1V2
L
[
V2 V1]
2g 2
h = 1 −
If V2 << V1,
L 1 12
h = V
2g
continuity eq.
V1A1 = V2A2
1 2 2
1 V
V A
A =
