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Inlet and Outlet Manifolds and Plant Hydraulics. In which Kinetic Energy BECOMES SIGNIFICANT (Thanks to A.A. Milne). Nomenclature: a start. 1. The Problem. How can we deliver water uniformly into the bottom of the sedimentation tank and - PowerPoint PPT Presentation
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In which Kinetic Energy BECOMES SIGNIFICANT(Thanks to A.A. Milne)
Inlet and Outlet Manifolds and Plant Hydraulics
Nomenclature: a startSymbol Description SubQ Flow P PortA Area M ManifoldH Piezometric head D DiffuserhL Total Head Loss
HGL Hydraulic Grade LineEGL Energy Grade LineCp Pressure Coefficient (includes shear and expansion
effects)Pvc Area of the vena contracta divided by the orifice area =
0.62
D Diameter
P Dimensionless ratio
n Number of ports
The Problem• How can we deliver water uniformly • between sed tanks • into the bottom of the sedimentation tank and• between StaRS layers• Within layers of the StaRS?
• Extract water uniformly • from above the plate settlers and
• How can we make it so that the water doesn’t preferentially take the easy path?
n-12 n1
• Draw manifold in lake picture• Define Pi.Q= Qp1/Qpn• Define H average (=vjet^2/2g) and
deltaH (vpipe^2/2g) showing manometers• Note that shape of inlet (pitot tube)
matters• H proportional to 1/A^2 (from orifice
eq)• How do you get Pi.Q = 1?• Hbar> dH• Ajet<Apipe
How can we make water choose equally between
several paths?• Draw a manifold with ports that you
think would give unequal flow
• Draw a manifold with ports that you think would give equal flow
• What do you think is important?n-1n-122 nn11
Will the flow be the same?
Dh
Long
Short
Head loss for long route = head loss for short route if KE is ignoredQ for long route< Q for short route
K=1K=1
K=1
K=0.2K=0.5 K=1
2 21 1 2 2
1 22 2 Lp V p Vz z hg g g g
NO!
An example to illustrate the concepts
Flow Division Analysis
2
2 4
8l p
Qh Cg D
2
2l pVh C
g
2
2eVh K
g
2
2fL Vh fD g
pLC K fD
2l ph C Q
2Shortl p Shorth C Q 2
Longl p Longh C Q
2 2Short Longp Short p LongC Q C Q Short
Long
pLongQ
Short p
CQQ C
P
Short path Long path
How did the flow divide?
Short
Long
pLongQ
Short p
CQQ C
P
0.2 0.263
Long
Short
K=1K=1
K=1
K=0.2
Improve this? Short
Long
p ControlQ
p Control
C KC K
P
2
21Long ShortQ p p
ControlQ
C CK
P
PSet PQ to 0.95
K.ControlP.Q
2 C.PLong C.PShort
1 P.Q2
25.718
Plant Flow Distribution
• Equal flow between sed tank bays?• Equal flow through diffusers into
sed tank?• Equal flow between
plate settlers?• Equal flow
through ports intosludge drain
Where can we use flow restrictions? ___________________After flocs are removed
Terminology
• Flow into tank (out of manifold) – Inlet Manifold• Flow out of tank (into manifold) –
Outlet launder• Overflow Weir• Submerged pipe with orifices (head
loss through orifices is set to be large relative to construction error in level of weir
Ease of construction, avoid floating flocs
Manifold: Flow Calculations
• We will derive equations in terms of Hydraulic Grade Line (HGL) because piezometric head controls the port flow
• Port flow• based on _______ equation
• Piezometric head change (DH) across port• flow expansion
• Piezometric head change (DH) between ports• Darcy-Weisbach and Swamee-Jain
orifice
In manifold
p zg
2p a g V
Head Loss due to Sudden
Expansion
2 22 2
2
outout in
in in outex
VV VV V V
hg g
2 22
2out in out in
exV V V V
hg
2
2in out
ex
V Vh
g
22
12
in inex
out
V Ahg A
2
1 inex
out
AK
A
in out
out in
A VA V
Discharge into a reservoir?_________
Energy
Momentum
Mass
Kex=1
2 2
2in out in out
exp p V Vh
g g
2 2 inout in
in out out
AV Vp p A
g g
2 2
2
Inlet Manifold
1 n-12 n
EGL
2 2 inout in
in out out
AV Vp p A
g g
in out outout in V V Vp pg g
in out outexpansion
V V VH
g
D
HGL
Major head loss
Pressure recovery
What is total SDHexpansion as a function of n?
in out outexpansion
V V VH
g
D in ouM
tP
AV V Q
i
Mout
M
nV
i QnA
MPQ Q
n 1 1
1 1
1n nMM
expansioni i M M
n i QQHnA nA g
D
21 1
21 1
n nM
expansioni i
n iVHg n
D
21
1
12
nM
expansioni
V nHg n
D Approaches for large n
2
2MVg
_______________ is recovered for very gradual expansion.All kinetic energy
Outlet Manifold (Launder)
1 n-12 n
All of the changes at the ports sum to
Flow contractions, thus no significant minor loss!
2
2MVg
EGL
HGL
Head Loss in a Manifold (same for inlet or outlet)
between first and last ports2
f f2
L VhD g
2
f f2
i
i
MPi
M
VLhD g
21 1
2f 2
1 1
1 1f2 1i
n nM M
ii iM M
L Qh n iD g A n n
i
MM
M
QV n inA
12
1
1 2 16
n
i
n n nn i
21
f1
2 11f2 6i
nM M
ii M M
nL QhD g A n
1M PL L n
11 n-1n-122 nn
EGL
HGL
11 n-1n-122 nn
EGLHGL
Define manifold length as
1
n 1( ) n2 1
n 1
i
n i( )2
simplify13
16 n
Head loss in a manifold is __ of the head loss with constant Q.1/3
Change in Piezometric Head in an Outlet Manifold
22 11 f 1
2 6M M
total iM M
nQ LHg A D n
D
11 n-1n-122 nn
EGL
HGL
2
f1
2 11f2 6i
nM M
ii M M
nL QhD g A n
longpC
Note: We have factored out the friction factor knowing that and thus f is not constant
2
0.9
0.25f5.74log
3.7 ReD
2
2MVg
Total change in piezometric head
Change in Piezometric Head in an Inlet Manifold
11 n-1n-122 nn
EGLHGL
21
1
12
nM
expansioni
V nHg n
D
22 11 1 f
2 6M M
total iM M
nQ LnHg A n D n
D
This equation gives the difference in piezometric head between the first port and the last port. Since the two terms have opposite signs the maximum difference could be at an intermediate port. We need to determine if one of these terms dominates to see if the maximum difference really is between the first and last ports.
2
f1
2 11f2 6i
nM M
ii M M
nL QhD g A n
longpC
Calculating the Control (Orifice) Pressure
Coefficients2
21long shortQ p p
ControlQ
C CK
P
PFor a manifold the short path head loss is zero (not including the flow control head loss)
2
21longQ p
ControlQ
CK
P
P
longpC 2 1f 1
6M
iM
nLD n
2 11 f6
M
M
nLnn D n
11 n-1n-122 nn
EGL
HGL
11 n-1n-122 nn
EGLHGL
0
Minor Loss Coefficient for an Orifice Port (in or out)
2
2P P
vce e
Vh K
g
But this V is the vena contracta velocity. The control coefficient analysis normalizes everything to the maximum velocity in the manifold. So let’s get the velocity ratio
2
2P vc vc PM M
vc M P M
A n DV QV A Q D
P P
22 2
2 2P P
M Me e
vc P
D Vh Kn D g
P
Ke has a value of 1 for an exit and is close to 1 for an entrance P MnQ Q
ControlK
22
2P
MControl e
vc P
DK Kn D
P
2
2M
vc Mvc P
DV Vn D
P
Solution Path• The length of the manifold will be
determined by the plant geometry• The spacing of the ports will be set
by other constraints• We need to determine the diameter
of the manifold and the diameter of the ports
Launder: Traditional Design Guidelines• Recommended port velocity is 0.46
to 0.76 m/s (Water Treatment Plant Design 4th edition page 7.28) • The corresponding head loss is 3 to 8
cm through the orifices• How do you design the diameter of the
launder? (coming up…)• Would this work if head loss through the
manifold were an additional 10 cm? _____
11 n-1n-122 nn
EGL
HGL
2vc OrificeQ A g hP D
21
2Port
vc
Vhg
D P
NO!
Design Constraints
• For sed tank Inlet Manifold the port velocities and the manifold diameter are set by the _____________________________________
• For the launder that takes clear water from the top of the sed tank bays the goal will be to keep head loss low and greater than construction errors in level of weir (we aim for about 5 cm)
• For Outlet Manifold that takes sludge from the bottom of the sed tank bays the goal is to be able to drain the tanks in a reasonable length of time (perhaps 30 minutes) (this means that the initial flow rate would be able to drain the tank in 15 minutes: remember the hole in a bucket analysis)
energy dissipation rate in the flocculator
Design for Outlet Launder
• Given target head loss between sed tank and clear water channel (5 cm for AguaClara)2
M2 4
8 TotalPl
M
CQhg D
12 4
2
8 TotalPMM
l
CQDg h
Solve the minor loss equation for the manifold diameter
Minor loss equation
Total LongP P ControlC C K
2
21Long ShortQ p p
ControlQ
C CK
P
P
0
2
21LongQ p
ControlQ
CK
P
P
11 n-1n-122 nn
EGL
HGL
Outlet Launder Diameter: Iterative
solution for DMTotal LongP P ControlC C K 2
21LongQ p
ControlQ
CK
P
P2
2 211 1
Long
Total Long
pQP p
Q Q
CC C
P P P
12 4
2 2
81
LongpMM
l Q
CQDg h P
2 1f 1
6Long
Mp
M
nLCD n
11 n-1n-122 nn
EGL
HGL
2
0.9
0.25f5.74log
3.7 ReD
The iterative solution
will converge quickly because f varies slowly with Re.
Example Code for Iteration
0a f y
1y f a
0y
Error ← 1
While Error > MaxErrorMaxError ← _____
0 1
0 1
y yError
y y
0 1y y
First guess at solution
Improved guess
Return y1
Dimensionless error
Set error to be large to ensure that loop executes once
Launder Diameter (Approximate Solution)
12 4
2 2
81
LongpMM
l Q
CQDg h
P
2 1f 1
6Long
Mp
M
nLCD n
12 4
2 2
8 11
MM
l Q
QDg h
P
Here we are omitting the major (wall shear) head loss contribution
In this equation the head loss is the total head loss for both the orifices and the pipe flow
Example: Launder
What is the minimum launder diameter for a plant flow rate of 50 L/s divided between 8 bays if we use 5 cm of head loss? For an approximate solution you can omit the effect of the major losses. Use a value of 0.8 for the minimum flow ratio between the last and first orifice
PQ 0.8 hl 5cm
Q50
Ls
86.25
Ls
DM8 Q2
g 2 hl
1
1 PQ2
1
4
11.6 cm
12 4
2 2
8 11
MM
l Q
QDg h
P
Example: Launder
• What is the effect of the shear force?• How can we estimate the length of
the launder? We will assume that the sed bay has a width of 1 m.• What is the length of the
sedimentation tank?
2 1f 1
6Long
Mp
M
nLCD n
V↑ = 1 mm/s
LSed
0.05m3
s
8
1m 0.001ms
6.25m
Example: Launder
• n is the number of orifices (ports). If the port spacing is 10 cm how many are there?
2 1f 1
6Long
Mp
M
nLCD n
62 2 1
6n
n
For large n
2 1f
6M
M
nLD n
2 1f 1
6Long
Mp
M
nLCD n
1.36
13
2
0.9
0.25f5.74log
3.7 ReD
0.026.25m0.116m
13 0.359
More exact solution…
12 4
2 2
81
LongpMM
l Q
CQDg h
P
12 4
2 2
81
LongpMM
l Q
CQDg h
P
What diameter launder do you recommend?
0.026.94m0.11m
13 0.421
DM8 Q2
g 2 hl
1.36
1 PQ2
1
4
11.8cm
6 inches
Why is the launder
diameter so large?
• (50L/s /9) launder of 6 inches• The head loss in the launder is small and it
would be tempting to use a smaller pipe• Why is such a large pipe necessary?
______________• Why do we even need a launder pipe?
___________________________________________ ___________
• What is the max velocity above the plate settlers given a 1 m wide tank, 25 cm of water above the plates, a single launder? __________
Equal orifice flow
For uniform flow distribution between (and within) plate settlers
2 mm/s
What is the horizontal velocity above the plate settlers
without a launder?HSuper 25cm VUp 1
mms
LSed 6m LSedHSuper
24
VSuperLSed VUp
HSuper24
mms
This velocity is very large compared with the head loss through the plate settlers (about 1 m) and thus elimination of the launder would result in preferential flow through the plate settlers closest to the exit
VSuper2
2g29m
Approach to Find Port Diameter
• Calculate the head loss in the manifold• Subtract 50% of that
head loss from the target head loss (5 cm) to estimate the port head loss• Calculate the port
diameter directly using the orifice equation
2
f
2 11f2 6
M M
M M
nL QhD g A n
2P
Pvc
QAg h
P D
11 n-1n-122 nn
EGL
HGL
42orificio
vc
QDg h
P D
What about Inlet Manifold Design?
• Total head loss is not a constraint (it will be VERY small)• Energy dissipation rate at the inlet of
the manifold determines the manifold diameter• Energy dissipation rate at the inlet to
the diffuser pipes will set the diffuser diameter• Available pipe sizes for inlet manifold
and for the diffusers is a constraint
11 n-1n-122 nn
EGLHGL
Schulz and Okun guidelines:
Note these cause floc breakup!• VPort = 0.2 to 0.3 m/s (assumes no
diffusers)• “The velocity through the ports
should be 4x higher than any approaching velocities.” (but to prevent sedimentation approach velocities need to be 0.15 m/s which would give velocities of 0.6 m/s!)These guidelines result in
extremely high energy dissipation rates!
3
PortJet
vcMax
Port vc
V
D
P
P
Schulz and Okun famous quote…
“In practice, one can rarely meet all four basic requirements because they conflict with one another; thus a reasonable compromise must be attained.”
Conclusion of inlet design for sedimentation tanks.Page 135 in Surface Water Treatment for Communities in Developing Countries
Flow Distribution Equation for Inlet Manifold
2 11 f6Long
Mp
M
nLnCn D n
11 n-1n-122 nn
EGLHGL
Short
Long
p ControlQ
p Control
C KC K
P
22
2
22
2
2 11 f6
P
P
Me
vc DQ
M Me
M vc D
DKn D
nL Dn Kn D n n D
P P
P
22
2P
MControl e
vc D
DK Kn D
P
Control resistance by orifice
0
What can we play with to get a better flow distribution?
2
2M D
vc D M
D Vn D V
P
2 11 ? f6
M
M
nLnn D n
Area ratio if the DM and DD cause
the same Max67
1 73 6
67
1 73 6
4
4
JetM
Max vcM
P
JetM
Max vc
Q
AA
Qnn
P P
17M
D
A nA
0 20 40 60 80 1000.5
0.6
0.7
0.8
0.9
1
Number of ports per manifold
Are
a ra
tio to
ach
ieve
equ
al ε.
Max
But apparently energy dissipation rate doesn’t matter!
Importance of Area Ratio
M
D
AA
11 n-1n-122 nn
EGLHGL
M
D
AA
Effect of pressure recovery
0.55
0.6
0.65
0.7
7 66
36
20
12
0 0.2 0.4 0.6 0.8 10.7
0.8
0.9
1
1.1
1.2Area Ratio of 0.55Area Ratio of 0.6Area Ratio of 0.65Area Ratio of 0.7
Normalized distance along manifold
Rat
io o
f act
ual p
ort f
low
to a
vera
ge p
ort f
low ports
3Jet JetMax
Jet
VD
1.23 1.728
One more Issue: Vena Contracta with High
Velocity Manifold• The vena contracta at each port
must be much more pronounced (small Pvc) when the velocity inside the manifold is high.• If the vena contracta, Pvc, is smaller,
then the velocities are higher and the energy dissipation rate is higher. • This requires further investigation
Manifold Conclusions• Outlet manifolds (launder) require an
iterative design to get the manifold diameter
• Inlet manifold design has complex constraints…• Avoid breaking flocs• Don’t let flocs settle (ignore if ports are on
bottom)• Distribute flow uniformly• Eliminate horizontal velocity in the sed tank• Produce jets to resuspend flocs to form floc
blanket
Head loss in an AguaClara Plant
• Why isn’t there much head loss between the flocculator and the launder pipe?
• How do we ensure that the flow divides equally between sedimentation tanks?
0.01 0.495 10
10
20
30
40
Orificio de la Mezcla RápidaTubo de la Mezcla RápidaFloculadorTubo de RecolecciónVertedero de Agua Decantada
Pérd
ida
de C
arga
Acu
mul
ada
(cm
)
0.01 0.495 10
10
20
30
40
Orificio de la Mezcla RápidaTubo de la Mezcla RápidaFloculadorTubo de RecolecciónVertedero de Agua Decantada
Pérd
ida
de C
arga
Acu
mul
ada
(cm
)
0.01 0.495 10
10
20
30
40
50
Orificio de la Mezcla RápidaTubo de la Mezcla RápidaFloculadorTubo de RecolecciónVertedero de Agua Decantada
Pérd
ida
de C
arga
Acu
mul
ada
(cm
)
10 50L/s
Rapid Mix OrificeRapid Mix PipeFlocculatorLaunderSettled water weir
Cum
ulat
ive
head
loss
(cm
)
Settled Water Weir: Controls the Plant Level
3/22 23 vcQ W g H P
3/2
32 2vc
QWgH
P
3
3/22
0.0532
0.62 2 9.8 0.05
msW
m ms
Q 0.05m3
s
H 5cm
W32
Q
Kvc 2g H
3
2
2.443m
With a maximum H of 5 cm the sedimentation tank water level can change a total of 10 cm! Launders have 5 cm of head loss also.
H is water level measured from the top of the weir
Hydraulic Conclusions
• The water level in the plant is set by the settled water weir
• The most significant head loss in the sedimentation tank is the orifices in the launder
• The water level increases through the flocculator.
• The entrance tank water level is significantly higher than the flocculator due to head loss in the rapid mix orifice
• The stock tanks have to be even higher to be able to flow by gravity thru the chemical doser and into the entrance tank.
