iit-madras, momentum transfer: july 2005-dec 2005
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IIT-Madras, Momentum Transfer: July 2005-Dec 2005
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Background1. Energy conservation equation
2
.2
P Vgh Const
If there is no friction
21Kinetic energy
2mV
2
What is ?2
V
21 Kinetic energy
2 Unit massV
2 Total energy
2 Unit mass
P Vgh
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2. If there is frictional loss , then
Frictional loss
Unit mass
P
2 2 Frictional loss
2 2 Unit massinlet outlet
P V P Vgh gh
In many cases
outlet inleth h
outlet inletV V
Background
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 propertiesproperties
Background
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
250 valveP Pa
50 L m
0 gauge pressure
Example
What should be the pressure at inlet ?
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Solution : taking pressure balance
0inlet valve pipeP P P
2 * . 2piper P rL
Example (continued)
For pipe, Force balance
Hence we can find total pressure drop
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
2valveP K V
2
32Laminar flow . pipe
VP L
D
2Turbulent flow , , , , ,pipe nP f L V e D
Flow properties
Empirical
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
( )12
Define f DimensionlessV
In general we want to find
f is a measure of frictional loss
higher f implies higher friction
This is Fanning-Friction factor ff
Friction Factor: Definition
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
So we write
,......pipe nP f
,......pipe nP f f
22
1 .2
2
f rLV
r
2
.2 rL
r
2 .f LV
r
Friction factor
This is for pipe with circular cross section
2 .2
f LV
D
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 VLP
D
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
1 2 3log( )avV 0where K, , are depends on the , , ,....
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Equation relating shear stress and average velocity,
and implicit nis i
Because original equation
*where
VV
V
*.yVy
* 0V
5.5 2.5ln( )V Y
Equation for Friction Factor
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
10
14log Re 0.4f
f
2
In the implicit equation itself,
1substitute for with , and we get
2f V
r R
V
y
2
21m
rV V
R
This is equivalent of laminar flow equation relating f and Re (for turbulent flow in a smooth pipe)
Equation for Friction Factor
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2
2 mV rV
r R
2 m
r R
VV
r R
21. 22 av mf V V R
Friction Factor: Laminar Flow
2 2 4 81.2
m av avav
V V Vf V
R R D
2
16 16 16
Reav
av av
Vf
V D V D
1
2av mV VFor laminar flow
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
21.2valve avP K V
?pipeP
ReDV
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?
1 av
mV
s
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
For turbulent flow
10
14log Re 0.4f
f
We can solve for f, once you know f, we can get shear
21.2
f V
Pressure drop using Friction Factor
Once you know shear , we can get pressure drop
2 * . 2piper P rL
If flow is laminar , ( i.e. Re < 2300 ), we use 16
Ref
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
2 22
1 1 2. .2 2
rLP K V f V
r
21.2 pipeP K V P
22
1 2.2
rLP K V
r
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
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?
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 (Re2f)
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
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
Determination of Q or D
Method (ii) 2
2
1 2.2
rLP f V
r
22
P Df
L V
ReDV
2 22
2 22Re
2
D P Df
L
V
V
3 2
2 2
D P
L
From the plot of f vs Re,
plot Re vs (Re2f)
From the known parameters, calculate Re2f
From the plot of Re vs (Re2f), determine ReCalculate Vav
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
KP V P
2 5 22250 250 5*10V V f
22
2
2
K rLP V
r
2 21 2.
2 2
K LP V f V
r
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
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
If flow is laminar, you can actually solve the equation
22250 250 40V V
22
322250 250
4
VLV
r
2
32pipe
VLP
D
240 40 4*2250*250
2*250V
2.92 /V m s
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
If you are given pressure drop and Q , we need to find D
221 2
. .2 2 / 2
V LP K f V
D
2
.2 pipe
VP K P
2
2
2.
2
V rLP K
r
2 2
2 2
2
2 2 / 24 4
K Q f Q LP
DD D
2 2
2 4 2 5
8 32K Q fL QP
D D
4 5
0.4 159.842250
f
D D
IIT-Madras, Momentum Transfer: July 2005-Dec 2005
4 5
0.4 1.59842250
D D
52250 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