Post on 21-Feb-2016
description
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
Closed Conduit FlowCEE 332
Closed Conduit Flow
Energy equationEGL and HGLHead loss
major lossesminor losses
Non circular conduits
Conservation of Energy
Kinetic, potential, and thermal energy
hL =
Ltp hhzg
Vphzg
Vp 2
22
22
1
21
11
22
hp =ht =
head supplied by a pumphead given to a turbinemechanical energy converted to thermal
Cross section 2 is ____________ from cross section 1!downstream
Point to point or control volume?Why ? _____________________________________
irreversibleV is average velocity, kinetic energy 2V
Energy Equation Assumptions
hp
p1
1
V12
2g z1 hp
p2
2
V22
2g z2 ht hL
hydrostatic
densitySteady
kinetic
Pressure is _________ in both cross sectionspressure changes are due to elevation only
section is drawn perpendicular to the streamlines (otherwise the _______ energy term is incorrect)
Constant ________at the cross section_______ flow
EGL (or TEL) and HGL
velocityhead
elevationhead (w.r.t.
datum)
pressurehead (w.r.t. reference pressure)
zg
VpEGL
2
2
zγpHGL
downward
lower than reference pressure
The energy grade line must always slope ___________ (in direction of flow) unless energy is added (pump)
The decrease in total energy represents the head loss or energy dissipation per unit weight
EGL and HGL are coincident and lie at the free surface for water at rest (reservoir)
If the HGL falls below the point in the system for which it is plotted, the local pressures are _____ ____ __________ ______
Energy equation
z = 0pump
Energy Grade LineHydraulic G
Lvelocity head
pressure head
elevation
datum
z
2gV2
p
Ltp hhzg
Vphzg
Vp 2
22
22
1
21
11
22
static headWhy is static
head important?
Bernoulli Equation Assumption
constp
g
Vz
2
2
densitySteady
streamline
Frictionless_________ (viscosity can’t be a significant parameter!)
Along a ________________ flowConstant ________No pumps, turbines, or head loss
Why no Does direction matter? ____Useful when head loss is small
point velocityno
Pipe Flow: Review
2 21 1 2 2
1 1 2 22 2p t Lp V p Vz h z h h
g g
dimensional analysis
We have the control volume energy equation for pipe flow.
We need to be able to predict the relationship between head loss and flow.
How do we get this relationship? __________ _______.
Flow Profile for Delaware Aqueduct
Rondout Reservoir(EL. 256 m)
West Branch Reservoir(EL. 153.4 m)
70.5 km
Sea Level
(Designed for 39 m3/s)
2 21 1 2 2
1 1 2 22 2p t lp V p Vz H z H h
g g
Need a relationship between flow rate and head loss
1 2lh z z
Ratio of Forces
Create ratios of the various forcesThe magnitude of the ratio will tell us
which forces are most important and which forces could be ignored
Which force shall we use to create the ratios?
Inertia as our Reference Force
F=maFluids problems (except for statics) include
a velocity (V), a dimension of flow (l), and a density ()
Substitute V, l, for the dimensions MLT
Substitute for the dimensions of specific force
F a F a
f f M
L T2 2
L l T M
fi
lV
l3
Vl
2
Dimensionless Parameters
Reynolds Number
Froude Number
Weber Number
Mach Number
Pressure/Drag Coefficients (dependent parameters that we measure experimentally)
Re Vlrm
=
Fr Vgl
=
2
2C p
pV
lVW
2
cVM
AVd
2
Drag2C
2fuVl
fg g
2fl
2
fvE
cl
r=
2
fiVl
( )p g zrD + D
Problem solving approach
1. Identify relevant forces and any other relevant parameters2. If inertia is a relevant force, than the non dimensional Re,
Fr, W, M, Cp numbers can be used3. If inertia isn’t relevant than create new non dimensional
force numbers using the relevant forces4. Create additional non dimensional terms based on
geometry, velocity, or density if there are repeating parameters
5. If the problem uses different repeating variables then substitute (for example d instead of V)
6. Write the functional relationship
Pipe Flow: Dimensional Analysis
What are the important forces?______, ______,________. Therefore ________number and _______________ .
What are the important geometric parameters? _________________________Create dimensionless geometric groups
______, ______Write the functional relationship
C p f
Re, ,l
D D
Inertial
diameter, length, roughness height
Reynolds
l/D
viscous
/D
22CV
pp
Other repeating parameters?
pressurePressure coefficient
Dimensional Analysis
How will the results of dimensional analysis guide our experiments to determine the relationships that govern pipe flow?
If we hold the other two dimensionless parameters constant and increase the length to diameter ratio, how will Cp change?
,RepDC fl D
f ,RepDC fl D
22CV
pp
Cp proportional to l
f is friction factor
, ,ReplC fD D
Dimensional Analysis
Darcy-Weisbach equation
Pressure Coefficient and Head Loss
2
2C pp
V
2
2C lp
ghV
f2
2f gh DV L
2
f f2
L VhD g
lh p
Always true (laminar or turbulent)
Assume horizontal flow
More general
lh p h
f pDCl
Definition of f!
Darcy-Weisbach
Friction Factor : Major losses
Laminar flowHagen-Poiseuille
Turbulent (Smooth, Transition, Rough) Colebrook FormulaMoody diagramSwamee-Jain
Hagen-Poiseuille
Darcy-Weisbach
Laminar Flow Friction Factor
2
32lhDV
L
f 2
32 LVhgD
2
f f2
L VhD g
2
2
32 f2
LV L VgD D g
64 64fReVD
Slope of ___ on log-log plot
f 4
128 LQhgD
-1
Turbulent Pipe Flow Head Loss
___________ to the length of the pipeProportional to the _______ of the velocity
(almost)________ with surface roughnessIs a function of density and viscosityIs __________ of pressure
Proportional
Increases
independent
2
f f2
L VhD 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)
1 Re f2log2.51f
1 2.512log3.7f Re f
D
1 3.72logf
D
2
f2f
L VhD g
Moody Diagram
0.01
0.1
1E+03 1E+04 1E+05 1E+06 1E+07 1E+08Re
fric
tion
fact
or
laminar
0.050.040.03
0.020.015
0.010.0080.0060.004
0.002
0.0010.0008
0.0004
0.00020.00010.00005smooth
f pDCl
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
5/ 2 f
3/ 2 f
1.782.22 log3.7
ghQ DL D ghD
L
0.044.75 5.221.25 9.4
f f
0.66 LQ LD Qgh gh
2
0.9
0.25f5.74log
3.7 ReD
no f
Each equation has two terms. Why?
fghL
L hf
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
2
f 2 5
8f LQhg D
2
f f2
L VhD g
2
0.9
0.25f5.74log
3.7 ReD
Colebrook Solution for Q
1 2.512log3.7f Re f
D
2
f 2 5
8f LQhg D
2
2 5f
1 1 8f
LQh g D
Re 4QD
2 5
f 2
4Re f8
Q g DhD LQ
3f21Re f gh D
L
21 2.514 logf 3.7 Re f
D
f2 5 2
8f h gD LQ
Colebrook Solution for Q2
2
2 5 3f f
1 8 2.514 log3.7 21
LQh g D D gh D
L
5/ 2 3f f
2 2.51log3.7 21
L Qgh D D gh D
L
5/ 2 f3
f
log 2.513.7 22
gh LQ DL D gh D
Swamee D?0.045 1/ 4 5 1/52 2 2 2
1.250.66 Q Q Q QDg g Q g g
1/ 251/5 1/ 4 1/52 2 25/ 40.66 Q Q QD
g g Q g
2
f 2 5
8f LQhg D
25
2
8f QDg
25
2
64f8QDg
1/51/ 4 1/52 25/ 4
2
64f Q Qg Q g
1/52
2
64f8QDg
1/51/51/ 4 1/52 2 25/ 4
8Q Q QDg g Q g
1/51/ 4 1/52 2 25/ 41f
4 4Q Qg Q g
Pipe Roughness
pipe material pipe roughness (mm)glass, drawn brass, copper 0.0015commercial steel or wrought iron 0.045asphalted cast iron 0.12galvanized iron 0.15cast iron 0.26concrete 0.18-0.6rivet steel 0.9-9.0corrugated metal 45PVC 0.12
Solution Techniques
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)0.044.75 5.22
1.25 9.4
f f
0.66 LQ LD Qgh gh
2
2 5
8ffLQh
g D2
0.9
0.25f5.74log
3.7 ReD
Re 4QD
5/ 2 f3
f
log 2.513.7 22
gh LQ DL D gh D
Exponential Friction Formulas
f
n
m
RLQhD
=
units SI 675.10
units USC727.4
n
n
C
CR
1.852
f 4.8704
10.675 SI unitsL QhD C
æ ö= è ø
C = Hazen-Williams coefficient
range of data
Commonly used in commercial and industrial settings
Only applicable over _____ __ ____ collected
Hazen-Williams exponential friction formula
Head loss:Hazen-Williams Coefficient
C Condition150 PVC140 Extremely smooth, straight pipes; asbestos cement130 Very smooth pipes; concrete; new cast iron120 Wood stave; new welded steel110 Vitrified clay; new riveted steel100 Cast iron after years of use95 Riveted steel after years of use60-80 Old pipes in bad condition
Hazen-Williams vs
Darcy-Weisbach
1.852
f 4.8704
10.675 SI unitsL QhD C
2
f 2 5
8f LQhg D
preferred
Both equations are empiricalDarcy-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.
Head Loss: Minor Losses
potential thermalVehicle drag Hydraulic jump
Vena contracta Minor losses!
Head loss due to outlet, inlet, bends, elbows, valves, pipe size changes
Flow expansions have high losses Kinetic energy decreases across expansion Kinetic energy ________ and _________ energy Examples – ________________________________
__________________________________________ Losses can be minimized by gradual transitions
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.
2
2lVh K
g=
( )geometry,RepC f=
2
2CV
ghlp
g
Vh pl2
C2
High Re
Head Loss due to Sudden Expansion:Conservation of Energy
1 2
ltp hHg
VzpHg
Vzp 22
22
222
22
111
1
1
lhgVVpp
2
21
2221
gVVpphl 2
22
2121
z1 = z2
What is p1 - p2?
Apply in direction of flowNeglect surface shear
Divide by (A2 )
Head Loss due to Sudden Expansion:Conservation of Momentum
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
12
11 AVM x 22
22 AVM x
222122
212
1 ApApAVAV
gAAVV
pp 2
121
22
21
AA11
AA22
xx
Energy
Head Loss due to Sudden Expansion
g
VVpphl2
22
2121
gAAVV
pp 2
121
22
21
1
2
2
1
VV
AA
gVV
gVVVV
hl 2
22
211
221
22
g
VVVVhl 22 2
1212
2
gVVhl 2
221
2
2
12
1 12
AA
gVhl
2
2
11
AAK
Momentum
Mass
Contraction
V1 V2
EGL
HGL
vena contracta
gVKh cc 2
22
losses are reduced with a gradual contraction
Expansion!!!
Entrance Losses
Losses can be reduced by accelerating the flow gradually and eliminating the vena contracta
Ke 0.5
Ke 1.0
Ke 0.04
he Ke
V2
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? _____
hv Kv
V2
2g
How can K be greater than 1?
Questions
What is the head loss when a pipe enters a reservoir?
Draw the EGL and HGL
V
g
V
2
2
EGLHGL
2
2
11
AAK
Questions
Can the Darcy-Weisbach equation and Moody Diagram be used for fluids other than water? _____Yes
No
Yes
Yes
What about the Hazen-Williams equation? ___ Does a perfectly smooth pipe have head loss?
_____ Is it possible to decrease the head loss in a
pipe by installing a smooth liner? ______
Example
D=40 cmL=1000 m
D=20 cmL=500 m
valve100 m
Find the discharge, Q.What additional information do you need?Apply energy equationHow could you get a quick estimate? _________________Or spreadsheet solution: find head loss as function of Q.
Use S-J on small pipe
ltp hHg
VzpHg
Vzp 22
22
222
22
111
1
1
cs1
cs2
22100
2 lVm h
g= +
Non-Circular Conduits:Hydraulic Radius Concept
A is cross sectional areaP is wetted perimeterRh is the “Hydraulic Radius” (Area/Perimeter)Don’t confuse with radius!
2
f2f
L VhD g
=
2
f f4 2h
L VhR g
=2
44h
DA DRP D
p
p= = = 4 hD R=
For a pipe
We can use Moody diagram or Swamee-Jain with D = 4Rh!
Quiz
In the rough pipe law region if the flow rate is doubled (be as specific as possible) What happens to the major head loss? What happens to the minor head loss?
Why do contractions have energy loss? If you wanted to compare the importance of minor
vs. major losses for a specific pipeline, what dimensionless terms could you compare?