Pipe Flow Friction factor in fluid mechanics

IIT-Madras, Momentum Transfer: July 2005-Dec 2005

Background
1. Energy conservation equation
2
.
2
P V
gh Const
ρ
+ + = If there is no friction
21
Kinetic energy
2
mV −
2
What is ?
2
V
21 Kinetic energy
2 Unit mass
V −
2
Total energy
2 Unit mass
P V
gh
ρ
∴ + + =

2. If there is frictional loss , then
Frictional loss
Unit mass
P
ρ
∆
∴ =
2 2
Frictional loss
2 2 Unit massinlet outlet
P V P V
gh gh
ρ ρ
   
+ + = + + + ÷  ÷
   
In many cases
outlet inleth h=
outlet inletV V=
Background

Q. Where are all frictional loss can occur ?
• in pipe, in valves, joints etc
• First focus on pipe friction
In pipe, Can we relate the friction to other properties ?
Flow properties
Fuid properties
properties Z
]
Background

Example for general case:
At the normal operating condition given following data
Shear stress = 2 Pa
250
50
0.1
1 /
valveP Pa
L m
r m
V m s
τ
∆ =
=
=
=
250valveP Pa∆ =
50L m=
0 gauge
pressure
Example
What should be the pressure at inlet ?

Solution : taking pressure balance
0inlet valve pipeP P P∆ = +∆ +∆
( ) ( )2
* . 2piper P rLπ τ π∆ =
Example (continued)
For pipe, Force balance
Hence we can find total pressure drop

We have said nothing about fluid flow properties
valve pipeP and P∆ ∆However , Normally we do not know the
Usually they depend on flow properties and fluid
properties
?pipeP∆ =
21
2
valveP K V ρ∆ =
2
32
Laminar flow .pipe
V
P L
D
µ
∆ =
( )2
Turbulent flow , , , , ,pipe nP f L V e Dµ ρ∆ =
Flow properties
Empirical

2
( )
1
2
Define f Dimensionless
V
τ
ρ
=
In general we want to find τ
f is a measure of frictional loss
higher f implies higher friction
This is Fanning-Friction factor ff
Friction Factor: Definition

So we write
( ),......pipe nP f τ∆ =
( ),......pipe nP f f∆ =
2
2
1 .2
2
f rL
V
r
π
ρ
π
=
2
.2 rL
r
τ π
π
=
2 .f L
V
r
ρ=
Friction factor
This is for pipe with circular cross section
2 .
2
f L
V
D
ρ=

Here f is function of other parameters
For laminar flow , don’t worry about f , just use
2
32 VL
P
D
µ
∆ =
For turbulent flow , Is it possible to get expression for shear ?
Friction factor: Turbulent Flow
Using log profile
1 2 log( )V K K Y+ +
= +
1 2 2log( )V α α α= +
1 2 3log( )avV β β β= +
0where K, , are depends on the , , ,....α β µ ρ τ

Equation relating shear stress and average velocity,
and implicit nis iρ µ τ
Because original equation
*
where
V
V
V
+
=
*
.y V
y
ρ
µ
+
=
* 0
V
τ
ρ
=
5.5 2.5ln( )V Y+ +
= +
Equation for Friction Factor

( )10
1
4 log Re 0.4f
f
= −
2
In the implicit equation itself,
1
substitute for with , and we get
2
f Vτ ρ
r R
V
y
τ µ
=
∂
=−
∂
2
2
1m
r
V V
R
 
= − ÷
 
This is equivalent of laminar flow equation relating f and Re
(for turbulent flow in a smooth pipe)
Equation for Friction Factor

2
2 mV rV
r R
−∂
=
∂
2 m
r R
VV
r R=
−∂
=
∂
21
. 2
2
av mf V V Rτ ρ∴ = =−
Friction Factor: Laminar Flow
2 2 4 81
.
2
m av av
av
V V V
f V
R R D
µ µ µ
ρ∴ = = =
2
16 16 16
Re
av
av av
V
f
V D V D
µ µ
ρ ρ
∴ = = =
1
2
av mV V=For laminar flow

21
.
2
valve avP K Vρ∆ =
?pipeP∆ =
Re
DV ρ
µ
=
Use of f is for finding effective shear stress and
corresponding “head loss” or “ pressure drop”
What is ?valveP∆
K 0.5valve =
In the original problem, instead of saying “normal operating
condition” we say
Pressure drop using Friction Factor
Laminar or turbulent?
1av
m
V
s
=

For turbulent flow
( )10
1
4 log Re 0.4f
f
= −
We can solve for f, once you know f, we can get shear
21
.
2
f Vτ ρ∴ =
Once you know shear , we can get pressure
drop
( ) ( )2
* . 2piper P rLπ τ π∆ =
If flow is laminar , ( i.e. Re < 2300 ), we use 16
Re
f =

2 2
2
1 1 2
. .
2 2
rL
P K V f V
r
π
ρ ρ
π
 
= + ÷
 
21
.
2
pipeP K V Pρ= +∆
2
2
1 2
.
2
rL
P K V
r
π
ρ τ
π
= +
And original equation becomes,
In above equation the value of f can be substitute from laminar and
turbulent equation
Laminar flow – straight forward
Turbulent flow – iterative or we can use graph
0 gauge
pressure

Determination of Q or D
Given a pipe (system) with known D and a specified flow rate
(Q ~ V), we can calculate the pressure needed
i.e. is the pumping requirement
We have a pump: Given that we have a pipe (of dia D), what
is flow rate that we can get?
OR
We have a pump: Given that we need certain flow rate, of
what size pipe should we use?

We have a pump: Given that we have a pipe (of dia D), what
is flow rate that we can get?
To find Q
i.e. To find average velocity (since we know D)
Two methods: (i) Assume a friction factor value and
iterate (ii) plot Re vs (Re2
f)
Method (i)
Assume a value for friction factor
Calculate Vav from the formula relating ∆P and f
Calculate Re
Using the graph of f vs Re (or solving equation), re-estimate f; repeat

Method (ii) 2
2
1 2
.
2
rL
P f V
r
π
ρ
π
 
∆ = ÷
 
2
2
P D
f
L Vρ
∆
= Re
DV ρ
µ
=
2 22
2 2
2
Re
2
D P D
f
L
V
V
ρ
µ ρ
∆
=
3 2
2
2
D P
L
ρ
µ ρ
∆
=
From the plot of f vs Re,
plot Re vs (Re2
f)
From the known parameters, calculate Re2
f
From the plot of Re vs (Re2
f), determine Re
Calculate Vav

We take original example , assume we know p, and need to find V and
Q
Let us say 2250
0.5
0.1
What is ?
P Pa
K
r
V
=
=
=
2
2
pipe
K
P V Pρ= +∆
2 5 2
2250 250 5*10V V f= +
2
2
2
2
K rL
P V
r
π
ρ τ
π
= +
2 21 2
.
2 2
K L
P V f V
r
ρ ρ 
= + ÷
 
Iteration 1: assume f = 0.001 gives V = 1.73m/s , Re = 3.5x105
, f = 0.0034
Iteration 2: take f = 0.0034 gives V = 1.15m/s , Re = 2.1x105
, f = 0.0037
Iteration 3: take f = 0.0037 gives V = 1.04 m/s , Re = 2.07x105
, f = 0.0038

If flow is laminar, you can actually solve the equation
2
2250 250 40V V= +
2
2
32
2250 250
4
VL
V
r
µ
= +
2
32
pipe
VL
P
D
µ
∆ =
2
40 40 4*2250*250
2*250
V
− ± +
=
2.92 /V m s∴ =

If you are given pressure drop and Q , we need to find D
2
21 2
. .
2 2 / 2
V L
P K f V
D
ρ
ρ 
= + ÷
 
2
.
2
pipe
V
P K Pρ= +∆
2
2
2
.
2
V rL
P K
r
ρ π
τ
π
= +
2 2
2 2
2
2 2 / 2
4 4
K Q f Q L
P
DD D
ρ ρ
π π
    
 ÷ ÷  ÷
 ÷= + ÷  ÷
 ÷ ÷  ÷
 ÷    
2 2
2 4 2 5
8 32K Q fL Q
P
D D
ρ ρ
π π
∴ = +
4 5
0.4 159.84
2250
f
D D
∴ = +

4 5
0.4 1.5984
2250
D D
∴ = +
5
2250 0.4 1.5984 0D D∴ − − =
0.24
0.69 /
Re 160000
0.0045
D
V m s
f
=
=
=
;
Iteration 1: Assume f = 0.01
Iteration 2: take f = 0.0045 and follow the same procedure
Solving this approximately (how?), we get

Pipe Flow Friction factor in fluid mechanics

Empfohlen

Empfohlen

Weitere ähnliche Inhalte

Was ist angesagt?

Was ist angesagt? (20)

Ähnlich wie Pipe Flow Friction factor in fluid mechanics

Ähnlich wie Pipe Flow Friction factor in fluid mechanics (20)

Mehr von Usman Shah

Mehr von Usman Shah (20)

Kürzlich hochgeladen

Kürzlich hochgeladen (20)

Pipe Flow Friction factor in fluid mechanics

Hinweis der Redaktion