Post on 03-Feb-2016
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HEADLOSS
ESTIMATION Mekanika Fluida 1
HST
1
Friction Factor : Major losses
• Laminar flow
• Hagen-Poiseuille
• Turbulent (Smooth, Transition, Rough)
• Colebrook Formula
• Moody diagram
• Swamee-Jain
2
Hagen-Poiseuille
Darcy-Weisbach
Laminar Flow Friction Factor
3
2
32
lhDV
L
f 2
32 LVh
gD
2
f f2
L Vh
D g
2
2
32f
2
LV L V
gD D g
64 64f
ReVD
Slope of ___ on log-log plot
f 4
128 LQh
gD
-1
Turbulent Pipe Flow Head Loss
• ___________ to the length of the pipe
• Proportional to the _______ of the velocity (almost)
• ________ with surface roughness
• Is a function of density and viscosity
• Is __________ of pressure
4
Proportional
Increases
independent 2
f f2
L Vh
D g
square
(used to draw the Moody diagram)
Smooth, Transition, Rough
Turbulent Flow
• Hydraulically smooth
pipe law (von Karman,
1930)
• Rough pipe law (von
Karman, 1930)
• Transition function for
both smooth and
rough pipe laws
(Colebrook)
5
1 Re f2log
2.51f
1 2.512log
3.7f Re f
D
1 3.72log
f
D
2
f2
f
L Vh
D g
Moody Diagram
6
0.01
0.1
1E+03 1E+04 1E+05 1E+06 1E+07 1E+08 Re
fric
tio
n fa
cto
r
laminar
0.05
0.04
0.03
0.02
0.015
0.01 0.008 0.006
0.004
0.002
0.001 0.0008
0.0004
0.0002
0.0001
0.00005
smooth
f p
DC
l
D
Swamee-Jain
• 1976
• limitations
• /D < 2 x 10-2
• Re >3 x 103
• less than 3% deviation
from results obtained
with Moody diagram
• easy to program for
computer or calculator
use
7
5/ 2 f
3/ 2 f
1.782.22 log
3.7
ghQ D
L D ghD
L
0.044.75 5.2
21.25 9.4
f f
0.66LQ L
D Qgh gh
2
0.9
0.25f
5.74log
3.7 ReD
no f
Swamee-Jain gets an f
• The challenge that S-J solved was deriving explicit
equations that are independent of the unknown
parameter.
• 3 potential unknowns (flow, head loss, or diameter): 3
equations for f
• that can then be combined with the Darcy Weisbach
equation
8
2
f 2 5
8f
LQh
g D
2
f f2
L Vh
D g
2
0.9
0.25f
5.74log
3.7 ReD
Colebrook Solution for Q
9
1 2.512log
3.7f Re f
D
2
f 2 5
8f
LQh
g D
2
2 5
f
1 1 8
f
LQ
h g D
Re 4Q
D 2 5
f 2
4Re f
8
Q g Dh
D LQ
3
f21Re f
gh D
L
2
1 2.514 log
f 3.7 Re f
D
f
2 5 2
8f
h g
D LQ
Colebrook Solution for Q
10
2
2
2 5 3f f
1 8 2.514 log
3.7 21
LQ
h g D D gh D
L
5/ 2 3f f
2 2.51log
3.7 21
L Q
gh D D gh D
L
5/ 2 f
3
f
log 2.513.7 22
gh LQ D
L D gh D
Swamee D?
11
0.045 1/ 4 5 1/5
2 2 2 21.250.66
Q Q Q QD
g g Q g g
1/ 251/5 1/ 4 1/5
2 2 25/ 40.66
Q Q QD
g g Q g
2
f 2 5
8f
LQh
g D
25
2
8f
QD
g
25
2
64f
8
QD
g
1/51/ 4 1/5
2 25/ 4
2
64f
Q Q
g Q g
1/52
2
64f
8
QD
g
1/51/5
1/ 4 1/52 2 2
5/ 4
8
Q Q QD
g g Q g
1/51/ 4 1/52 2 2
5/ 41f
4 4
Q Q
g Q g
Pipe Roughness
12
pipe material pipe roughness (mm)
glass, drawn brass, copper 0.0015
commercial steel or wrought iron 0.045
asphalted cast iron 0.12
galvanized iron 0.15
cast iron 0.26
concrete 0.18-0.6
rivet steel 0.9-9.0
corrugated metal 45
PVC 0.12
Solution Techniques
13
find head loss given (D, type of pipe, Q)
find flow rate given (head, D, L, type of pipe)
find pipe size given (head, type of pipe,L, Q) 0.04
4.75 5.22
1.25 9.4
f f
0.66LQ L
D Qgh gh
2
2 5
8ff
LQh
g D2
0.9
0.25f
5.74log
3.7 ReD
Re 4Q
D
5/ 2 f
3
f
log 2.513.7 22
gh LQ D
L D gh D
Exponential Friction Formulas
• Commonly used in commercial and industrial settings
• Only applicable over _____ __ ____ collected
• Hazen-Williams exponential friction formula
14
f
n
m
RLQh
D=
C = Hazen-Williams coefficient
range of data
Head loss:
Hazen-Williams Coefficient
C Condition
150 PVC
140 Extremely smooth, straight pipes; asbestos cement
130 Very smooth pipes; concrete; new cast iron
120 Wood stave; new welded steel
110 Vitrified clay; new riveted steel
100 Cast iron after years of use
95 Riveted steel after years of use
60-80 Old pipes in bad condition
15
Hazen-Williams vs
Darcy-Weisbach • Both equations are empirical
• Darcy-Weisbach is dimensionally correct, and ________.
• Hazen-Williams can be considered valid only over the
range of gathered data.
• Hazen-Williams can’t be extended to other fluids without
further experimentation.
16 1.852
f 4.8704
10.675 SI units
L Qh
D C
2
f 2 5
8f
LQh
g D
preferred
Minor Losses
• Most minor losses can not be obtained analytically, so
they must be measured
• Minor losses are often expressed as a loss coefficient, K,
times the velocity head.
17
2
2l
Vh K
g=
( )geometry,RepC f=
2
2C
V
ghlp
g
Vh pl
2C
2
High Re
Head Loss due to Sudden Expansion:
Conservation of Energy
18
1 2
ltp hHg
Vz
pH
g
Vz
p
22
2
222
2
2
2
111
1
1
lhg
VVpp
2
2
1
2
221
g
VVpphl
2
2
2
2
121
z1 = z2
What is p1 - p2?
Apply in direction of flow
Neglect surface shear
Divide by (A2 )
Head Loss due to Sudden Expansion:
Conservation of Momentum
19
Pressure is applied over all of section 1. Momentum is transferred over area corresponding to upstream pipe diameter. V1 is velocity upstream.
sspp FFFWMM 2121
1 2
xx ppxx FFMM2121
1
2
11 AVM x 2
2
22 AVM x
22212
2
21
2
1 ApApAVAV
g
A
AVV
pp 2
12
1
2
2
21
A1 A2
x
Energy
Head Loss due to
Sudden Expansion
20
g
VVpphl
2
2
2
2
121
g
A
AVV
pp 2
12
1
2
2
21
1
2
2
1
V
V
A
A
g
VV
g
V
VVV
hl2
2
2
2
11
22
1
2
2
g
VVVVhl
2
2 2
121
2
2
g
VVhl
2
2
21
2
2
1
2
1 12
A
A
g
Vhl
2
2
11
A
AK
Momentum
Mass
Contraction
• losses are reduced with a gradual contraction
21
V1 V2
EGL
HGL
vena contracta
g
VKh
cc2
2
2
Expansion!!!
Entrance Losses
• Losses can be reduced
by accelerating the flow
gradually and eliminating
the vena contracta
22
Ke 0.5
Ke 1.0
Ke 0.04
he Ke
V 2
2g
reentrant
Head Loss in Valves
• Function of valve type and valve
position
• The complex flow path through
valves often results in high head
loss
• What is the maximum value that Kv
can have? _____
23
hv Kv
V 2
2g
How can K be greater than 1?
Example
24
D=40 cm L=1000 m
D=20 cm L=500 m
valve 100 m
Find the discharge, Q.
Use S-J on small pipe
ltp hHg
Vz
pH
g
Vz
p
22
2
222
2
2
2
111
1
1
cs1
cs2
2
21002
l
Vm h
g= +
Non-Circular Conduits:
Hydraulic Radius Concept
• A is cross sectional area
• P is wetted perimeter
• Rh is the “Hydraulic Radius” (Area/Perimeter)
• Don’t confuse with radius!
25
2
f2
f
L Vh
D g=
2
f f4 2h
L Vh
R g=
2
44
h
DA D
RP D
p
p= = = 4 hD R=
For a pipe
We can use Moody diagram or Swamee-Jain with D = 4Rh!
GENERAL
CONSIDERATION
HGL-EGL DRAWING
26
EGL & HGL for a Pipe System
• Energy equation
• All terms are in dimension of length (head, or energy per unit weight)
• HGL – Hydraulic Grade Line
• EGL – Energy Grade Line
• EGL=HGL when V=0 (reservoir surface, etc.)
• EGL slopes in the direction of flow
22
22
211
21
122
zp
g
Vhz
p
g
VL
27
zp
HGL
g
VHGLz
p
g
VEGL
22
22
EGL & HGL for a Pipe System
• A pump causes an abrupt rise in EGL (and
HGL) since energy is introduced here
28
EGL & HGL for a Pipe System
• A turbine causes an
abrupt drop in EGL
(and HGL) as energy
is taken out
• Gradual expansion
increases turbine
efficiency
29
EGL & HGL for a Pipe System
• When the flow passage changes diameter,
the velocity changes so that the distance
between the EGL and HGL changes
• When the pressure becomes 0, the HGL
coincides with the system
30
EGL & HGL for a Pipe System
• Abrupt expansion into reservoir causes a
complete loss of kinetic energy there
31
EGL & HGL for a Pipe System
• When HGL falls below the pipe the
pressure is below atmospheric pressure
32
Example (1)
33
Example (2)
34
Example (3)
35