Manifolds

```Inlet and Outlet Manifolds and
Plant Hydraulics
In which Kinetic
Energy BECOMES
SIGNIFICANT
(Thanks to A.A.
Milne)
Nomenclature: a start
Symbol
Description
Sub
Q
Flow
P
Port
A
Area
M
Manifold
H
D
Diffuser
hL
HGL
EGL
Cp
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?
1
2
n-1
n
• 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?
1
2
n-1
n
Will the flow be the same? NO!
An example to illustrate the concepts
Dh
Long
K=1
p1
V12
p2
V22
 z1 

 z2 
 hL
g
2g  g
2g
K=0.5
K=0.2
Short
K=1
Head loss for long route = head loss for short route if KE is ignored
Q for long route< Q for short route
Flow Division Analysis
V2
hl  C p
2g
hl  C p
C pShort Q
L V2
hf  f
D 2g
L
Cp   K  f
D
2
8Q
g 2 D 4
hl  C p Q
2
Short
V2
he  K
2g
Short path
2
hl  C pShort QShort
2
 C pLong Q
2
Long
PQ 
Long path
2
hl  C pLong QLong
QLong
QShort

C pShort
C pLong
How did the flow divide?
Long
PQ 
K=1
Improve this?
Short
K=0.2
PQ 
QLong
QShort

C pShort
C pLong
K Control 
C pShort  K Control
C pLong  K Control
2
PQ
C pLong  C pShort
Set PQ to 0.95
2
1  PQ
2

0.2
 0.26
3
K.Control
P.Q  C.P Long C.P Short
2
1  P.Q
 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 into
sludge 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
•
•
•
p
z
g
controls the port flow
2

a



p


g



V


Port flow
• based on orifice
_______ equation
Piezometric head change (DH) across port
• flow expansion
In manifold
Piezometric head change (DH) between ports
• Darcy-Weisbach and Swamee-Jain
Sudden Expansion
2
pin  pout Vin2  Vout

Energy hex 
g
2g
2
2 Ain
Vout  Vin
Aout
Momentum pin  pout 
g
g
hex
2
2 Vout
V

V
2 out 2 in
2
Vin
Vin2  Vout


2g
2g
hex
Ain
Vout
Mass A  V
out
in
2
Vout
 2VinVout  Vin2

2g

Ain 


A
in
Vin  Vout 
K ex   1 

hex
1



hex 
A
Aout 
out 


2g
Discharge into a reservoir?_________
Kex=1
2
2
in
V

2g
2
2
Inlet Manifold
Vin  Vout  Vout

pout  pin

g
g
pin  pout

g
DH expansion
EGL
2
out
V
V
2
in
Ain
Aout
g
Vin  Vout  Vout


g
Pressure recovery
HGL
1
2
n-1
n
What is total SDHexpansion as a function
of n?
DH expansion
n 1
 DH
i 1
Vin  Vout  Vout


g
QM  n  i  QM 1

nAM
g
i 1 nAM
n1
expansion
n 1
 DH expansion 
i 1
n 1
 DH
i 1
expansion
2 n 1
M
V
g

n  i
i 1
VM2 n  1

g 2n
n2
Vin  Vout
QP

AM
QP 
Vouti
QM
n
n  i  QM


nAM
VM2
Approaches
for large n
2g
_______________
All kinetic energy is recovered for very gradual expansion.
Outlet Manifold (Launder)
Flow contractions, thus no significant minor loss!
EGL
HGL
VM2
All of the changes at the ports sum to
2g
1
2
n-1
n
Head Loss in a Manifold (same for
inlet or outlet) between first and last
ports
L V2
hf  f
D 2g
Define manifold length as
2
LP VM i
hfi  f i
DM 2 g
VM i 
QM
nAM
n  i
LM  LP  n  1
EGL
HGL
1
LM 1  QM 
h

f

fi
i


DM 2 g  AM 
i 1
n 1
n 1
n  i
i 1
n 1
 hfi  fi
i 1
2

2
1
 n  1 n 2
n  n  1 2n  1
n  i
2
 2n  1
n-1
n
2
i 1
n1
1
2


( n  1)  n i  1
6
LM 1  QM 


DM 2 g  AM 
n 1
2
2
( n  i) simplify 
1
3

1
6 n
EGL
HGL
1
2
n-1
6n
Head loss in a manifold is 1/3
__ of the head loss with constant Q.
n
EGL
in an Outlet Manifold
LM 1  QM 
hfi  f i



DM 2 g  AM 
i 1
n
2
DH total
2
6n
 2n  1  1
6n
1
2
n-1
VM2
2g
 2n  1
1  QM   LM


 fi
2 g  AM   DM
HGL


Total change in
C plong
Note: We have factored out the friction factor knowing
0.25
that f 
and thus f is not constant
2
  
5.74  
log  3.7 D  Re0.9  

 
n
EGL
HGL
Change in Piezometric
LM 1  QM 
hfi  f i



DM 2 g  AM 
i 1
n
2
DH total
2
1
n 1
 DH
 2n  1
i 1
6n
LM
1  QM   n  1


f
i

 
2 g  AM   n
DM
2
expansion
n-1
VM2 n  1

2g n
 2n  1 
6n


C plong
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.
n
Calculating the Control (Orifice)
Pressure Coefficients
0
K Control 
2
PQ
C plong  C pshort
2
1  PQ
2
PQ
C plong
K Control 
2
1  PQ
EGL
EGL
HGL
HGL
1
C plong
For a manifold the short
path head loss is zero (not
including the flow control
2
n-1
n
 LM  2n  1 
 1
fi
 DM 6n

1
2
n-1
n
 n 1
LM  2n  1 
f


n
D
6
n
M


Minor Loss Coefficient for an
Orifice Port (in or out)
heP  K eP
Ke has a value of 1 for an exit and is
close to 1 for an entrance
nQP  QM
Vvc2
2g
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
nP vc DP2
VM
QM AP P vc


2
Vvc
AM QP
DM
heP  K eP
2
 D
 VM2

2 
n
P
D
vc
P  2g

2
M
K Control
Vvc  VM
KControl
2
DM
nP vc DP2
 D

 K eP 
2 
n
P
D
vc P 

2
M
2
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
EGL
Guidelines
• Recommended port velocity is 0.46 to 0.76
HGL
1
2
n-1
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? _____
NO!
Q  P vc AOrifice 2 g Dh
1  VPort 
Dh 


2 g  P vc 
2
n
Design Constraints
• For sed tank Inlet Manifold the port velocities
•
and the manifold diameter are set by the
_____________________________________
energy dissipation rate in the flocculator
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)
EGL
Design for Outlet
Launder
HGL
1
2
n-1
n
• Given target head loss between sed tank and
clear water channel (5 cm for AguaClara)
8QM 2 CPTotal
hl 
g 2 DM 4
DM
 8QM 2 CPTotal 


2
g

h
l


CPTotal  CPLong  K Control
Minor loss equation
1
4
Solve the minor loss
0
equation for the manifold
diameter
2
PQ
C pLong  C pShort
K Control 
2
1  PQ
2
PQ
C pLong
K Control 
2
1  PQ
Outlet Launder
Diameter: Iterative
EGL
HGL
1
solution for DM
CPTotal  CPLong  K Control
CPTotal  C pLong
f
C pLong
DM

P Q2
1 
2
1

P
Q

  C pLong
  
2
1

P
Q
 



K Control 
2
n-1
2
PQ
C pLong
2
1  PQ
0.25
  
5.74  
log

  3.7 D Re0.9  

 
2
 LM  2n  1 
 f
 1
 DM 6n

 8QM 2 C pLong

 g 2 h 1  P 2
l
Q





1
4
The iterative solution
will converge quickly
because f varies slowly
with Re.
n
Example Code for Iteration
y0 
First guess at solution
Error ← 1
MaxError ← _____
While Error > MaxError
Set error to be large to ensure
that loop executes once
a  f  y0 
y1  f  a 
Error 
y0  y1
Return y1
y0  y1
y0  y1
Improved guess
Dimensionless error
Launder Diameter (Approximate
Solution)
 8QM 2 C pLong 
DM   2
 g h 1  P 2 
l
Q 

 8QM 2
1
DM   2
 g h 1  P 2
l
Q




1
4
1
4
In this equation the head loss is
the total head loss for both the
orifices and the pipe flow
C pLong
 LM  2n  1 
 f
 1
 DM 6n

Here we are omitting the
loss contribution
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
 8QM 2
1
DM  
 g 2 h 1  P 2
l
Q




1
4
50
Q 
8
h l   5cm
L
s
 6 .25
L
s
1
4
 8 Q2

1
  1 1.6cm
DM   

 g   2 h 1  P 2 
l
Q 

Example: Launder
C pLong
 LM  2n  1 
 f
 1
 DM 6n

• 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?
V↑ = 1 mm/s
3
0.05
m
s
LSed  
8
m
1m 0 .00 1
s
 6 .25m
Example: Launder
C pLong
 LM  2n  1 
 f
 1
 DM 6n

• n is the number of orifices (ports). If the port
spacing is 10 cm how many are there?
62
 2n  1 
6n
1
3
LM  2n  1
f

DM 6n
C pLong
For large n
6 .25m 1
0 .02
  0 .35 9
0 .11 6m 3
 LM  2n  1 
 f
 1  1.36
 DM 6n

f
0.25
  
5.74  
log  3.7 D  Re0.9  

 
2
More exact
solution…
 8QM 2 C pLong 
DM   2
 g h 1  P 2 
l
Q 

0 .02
 8QM C pLong 
DM   2
 g h 1  P 2 
l
Q 

2
6 .94m 1
  0 .42 1
0 .11m 3
1
4
1
What diameter launder do you recommend?
6 inches
4
 8 Q2
1 .36 

DM 

 1 1.8cm
 g  2 h 1  P 2 
l
Q 

1
4
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?______________
Equal orifice flow
Why do we even need a launder pipe?
For uniform flow distribution between (and within)
___________________________________________
plate settlers
___________
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? __________
2 mm/s
What is the horizontal velocity above the
plate settlers without a launder?
HSup er   2 5cm
mm
VU p   1
s
LSed   6m
LSed
HSup er
LSed VU p
 24
mm
VSup er  
 24
HSup er
s
2
VSuper
2g
 29m
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
Approach to Find Port
Diameter
EGL
HGL
1
• Calculate the head loss in
L
1  QM 
hf  f M


DM 2 g  AM 
the manifold
• Subtract 50% of that head
loss (5 cm) to estimate
• Calculate the port
diameter directly using
the orifice equation
2
2
n-1
 2n  1
6n
QP
AP 
P vc 2 g Dh
Dorificio
4Q

P vc 2 g Dh
n
Manifold Design?
EGL
HGL
1
2
n-1
• 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
n
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!
 Max

VPort 
 Jet P 
vc 

DPort P vc
3
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
C pLong
EGL
HGL
1
 n 1
LM  2n  1 

f

n
D
6
n
M


K Control
0
PQ 
n-1
 D

 K eP 
2 
n
P
D
vc D 

2
M
n
2
Control resistance
by orifice
C pShort  K Control
C pLong  K Control
2
PQ 
2
2
 DM

K eP 
2 
 nP vc DD 
2
2
 n 1
 DM

LM  2n  1 
f

  K eP 
2 
n
D
6
n
n
P
D
M
vc
D 



DM2
VD

2
nP vc DD VM
n 1
n
?
f
LM
DM
What can we play with to get a better flow distribution?
 2n  1
6n
Area ratio if the DM
and DD cause the same
Max
6
7
AM
n
1
7
1
Area ratio to achieve equal ε.Max


4 Jet 
 QM
1
7

 
3
6
 Max P vc
AM



6
AP

7
4 Jet 
QM

n
1
7



3
6
 n Max P vc

0.9
0.8
0.7
0.6
0.5
0
20
40
60
80
100
Number of ports per manifold
But apparently energy dissipation rate doesn’t matter!
EGL
Importance of Area A
A
Ratio
HGL
1
M
2
n-1
n
Ratio of actual port flow to average port flow
D
 0 .55
 0 .6 
 0 .65
 0 .7 


1.2
Area
Area
Area
Area
1.1
Ratio
Ratio
Ratio
Ratio
of
of
of
of
0.55
0.6
0.65
0.7
AM
7
 6 6
36
 
 2 0
 1 2
 
ports
Effect of
pressure
recovery
1
0.9
 Max 
0.8
0.7
0
0.2
0.4
0.6
0.8
Normalized distance along man ifold
1
3
 JetVJet 
DJet
1 .2  1 .72 8
3
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
Pérdida de Carga Acu
20
Head loss in an AguaClara Plant
10
50
(cm)(cm)
loss
Cumulative
0
 0.01
40
40
30
30
20
20
0.495
1
Orificio
de laOrifice
Mezcla Rápida
Rapid Mix
Tubo
la Mezcla
RapiddeMix
Pipe Rápida
Flocculator
Launder
Tubo
de Recolección
Settled water
Vertedero
de Agua
• Why isn’t there much head loss between
10
10
•
0
0
 0.01 0.01
the flocculator and the launder pipe?
How do we ensure that the flow divides
equally
between
sedimentation tanks?
0.495
1
0.495
1
10 50
Orificio de la Mezcla Rápida
Orificio de la Mezcla Rápida
L/sTubo de la Mezcla Rápida
Settled Water Weir: Controls the Plant Level
H is water level measured from the top of the weir
2
P vcW 2 g H 3/2 With a maximum H of 5
3
cm the sedimentation tank
water level can change a
3
Q
W
total of 10 cm! Launders
3/2
2 P vc 2 g H
have 5 cm of head loss
also.
Q
3
3
W
2
 0.62 
m
0.05
s
m
3/2

2  9.8 2   0.05m 
s 

3
Q  0 .05
m
s
H   5cm
W 
3
Q
2
3
Kv c 2g  H
2
 2 .44 3m
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.
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