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2. 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
∴ + + =
3. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
4. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
5. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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 ?
6. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
7. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
8. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
9. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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=
10. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
11. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
12. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
( )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
13. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
14. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
=
15. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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∴ =
Pressure drop using Friction Factor
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 =
16. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
Pressure drop using Friction Factor
0 gauge
pressure
17. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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?
18. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Determination of Q or D
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
19. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Determination of Q or D
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
20. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
21. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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∴ =
22. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
∴ = +
23. IIT-Madras, Momentum Transfer: July 2005-Dec 2005
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
Editor's Notes
The formula is different for pipe with zero roughness (smooth pipe) vs rough pipe. Similarly another formula is available for flow in transition regime. Please refer to the text book for the actual formulas.
Notice that if the flow is laminar, then most of the pressure drop occurs in the valve and a very little in the pipe. Of course, the flow is not laminar, this is just to illustrate how to solve it if the flow is laminar
Again if the flow is laminar, one can solve it easily.