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

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