Flow Control and Measurement
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Here’s a tip!
We can use smart fluids
to eliminate software,
computers, and
electronics!
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Creativity without a trip
Variations on a drip
Giving head loss the slip
Chemical doses that don’t dip
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Overview
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• Orifices and surface
tension
• Viscosity
12
Kinematic viscosity (mm^2/s)
• Floating bowl
• Hypochlorinators
in Honduras
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• Overflow tanks
• Marriot bottle
• Float valve
• AguaClara Flow
Controller
• Linear Flow Orifice
Meter
• AguaClara Linear
Dose Controller
• Extra
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• Fluids Review
• Applications of flow
control
• If you had electricity
• Constant head devices
Alum in distilled water
Alum Model
PACl in distilled water
PACl Model
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2
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0
200
400
Coagulant concentration (g/L)
600
Fluids Review
• What causes drag?
• Best orientation to reduce drag?
Streamlines
• Draw the streamlines that begin on the
upstream side of the object for these two
cases
• Which object has the larger wake?
• Which object has the lower pressure in the
wake? (if streamlines are bending hard at
the point of separation, then the streamlines
will be close together…)
Why is there drag?
• Fluid separates from solid body and forms a
2
2
p
v
p
v
1
recirculation zone
 z1  1  2  z2  2

2g 
2g
• Pressure in the recirculation zone must be
low because velocity in the adjacent flowing
fluid at the point of separation is high
• Pressure in recirculation zone (the wake) is
relatively constant because velocities in
recirculation zone are low
• Pressure behind object is low - DRAG
Fluids Review
•
•
•
•
Where should the luggage go?
Which equation for head loss?
Which process is inefficient?
Pipeline design
z1
Orifice Equation
p1
v12
p2
v22
 z1 

 z2 
g
2g  g
2g
Avc   vc Aor
Q  VA
Q   vc Aor 2 g h
𝛥ℎ
z2 = 0
mechanical
Bernoulli equation (no energy loss)
^
Area of the constricted flow
Continuity equation
Orifice Equation (memorize this!)
This equation applies to a horizontal orifice (so that the depth of
submergence is constant). For depth of submergence larger than
the diameter of the orifice this equation can be applied to vertical
orifices. There is a general equation for vertical orifices in the
AguaClara fluids functions.
Two kinds of drag
Two kinds of head loss
Drag (external flows)
• Skin (or shear) friction
• Shear on solid surface
• Classic example is flat plate
• Form (or pressure) drag
• Separation of streamlines
from solid surface and wake
results in a…
• Flow expansion (behind
object)
Head loss (internal flows)
• Major losses (hf – friction)
• Shear on solid surface
• Shear on pipe walls
• Minor losses (he – expansion)
• Separation of streamlines from
solid surface results in a…
• Flow expansion
hL  h f  he
Energy equation (NOT THE BERNOULLI EQUATION)
2
pin
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL
g
2g
g
2g
Head Loss in a Long STRAIGHT
Tube (due to wall shear)
• Laminar flow
hf 
Hagen–Poiseuille
32  LV 128 LQ

2
 gD
 g D 4
• Turbulent Flow
Re 
4Q
f 
D
Wall roughness
Flow proportional to hf
64
f 
Re
0.25
5.74

 
log


0.9

 3.7 D Re

Swamee-Jain



2
hf  f
8
g 2
LQ 2
D5
Darcy Weisbach
f for friction (wall shear)
Transition from turbulent to laminar occurs at about 2100
Frictional Losses in Straight Pipes
0.1
f
 D
f  Cp 
l 

0.25
  
5.74  
log

  3.7 D Re0.9  

 
2
0.05
0.04
0.03
friction factor
0.02
0.015
0.01
0.008
0.006
0.004

D
laminar
0.002
64
f
Re
0.001
0.0008
0.0004
0.0002
0.0001
0.00005
0.01
1E+03
smooth
1E+04
1E+05
1E+06
Re
1E+07
1E+08
Re 
VD

Head Loss: Minor Losses
• Head (or energy) loss (hL) due to:
outlets, inlets, bends, elbows, valves, pipe
size changes
• Losses are due to expansions When V, KE  thermal
• Losses can be minimized by gradual
expansions
V2
• Minor Losses have the form he  K e 2 g
where Ke is the loss coefficient
and V is some characteristic velocity (could
be contracted flow or expanded flow)
Head Loss due to Sudden Expansion:
Conservation of Energy z
x
in
out
At centroid of control surface
Where is p measured?___________________________
2
pin
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL
g
2g
g
2g
2
pin  pout
Vout
 Vin2

 hex
g
2g
2
Relate Vin and Vout?
pin  pout Vin2  Vout
he 

g
2g
Relate pin and pout?
zin = zout
Mass
Momentum
Head Loss due to Sudden Expansion:
Conservation of Momentum
Aout
Ain
x
2
1
M1  M 2  W  Fp  Fp  Fss Apply in direction of flow
1
2
M 1 x  M 2 x  Fp  Fp
1x
M 1x   Vin2 Ain
Neglect surface shear
2x
2
M 2 x  Vout
Aout
2
 Vin2 Ain  Vout
Aout  pin Aout  pout Aout
pin  pout

g
2
Vout
 Vin2
g
Ain
Aout
Pressure is applied over all of
section 1.
Momentum is transferred over
area corresponding to upstream
pipe diameter.
Vin is velocity upstream.
Head Loss due to
Sudden Expansion
2
pin  pout Vin2  Vout

Energy he 
g
2g
2
2 Ain
Vout  Vin
Aout
Momentum pin  pout 
g
g
2
2 Vout
V

V
2 out 2 in
2
Vin
Vin2  Vout
he 

2g
2g
Ain
Vout
Mass A  V
out
in
2
Vout
 2VinVout  Vin2
he 
2g

Ain 


A
in
Vin  Vout 
K e  1 

1



he 
A
Aout 
out 


2g
Discharge into a reservoir?__________________
Loss coefficient = 1
2
2
in
V
he 
2g
2
2
Minor Loss Coefficient for an
Orifice in a Pipe (DOrifice < < Dpipe)
extra
e for expansion
he  K e
V
2g

V  Aout
he 
 1

2 g  Ain

2
out
K eorifice
2
Q
Minor loss coefficient h  K
e
e
2 gA2
2
Expansion losses
2
Vout
Ain
Aout
This is Vout, not Vin


APipe

 1
 A

vc
Orifice


2
hL
Vena contracta area
K eorifice
2


DPipe

 1
  D2

 vc Orifice

2
DOrifice
Vout
DPipe
Minor Loss Coefficient for an
Orifice in a pipe
Where is the expansion?
extra
extra
Equation for the diameter of an
orifice in a pipe given a head loss
2
V
he  K e
2g

D
he  
 D

2
Pipe
2
vc
Orifice
DOrifice 

 vc 

K eorifice

D

 D

2
Pipe
2
vc
Orifice

 1


2
2

8Q 2
 1
 g 2 D 4
Pipe

DPipe
2

D
he g
Pipe
8
Q

 1

Minor losses
dominate, thus
h e = hL
hl
Where do changes in pressure occur?
Where does head loss occur?
dorifice
Vout
dpipe
Applications of Constant Flow
• POU treatment devices (Point of Use)
• clay pot filters
• SSF (slow sand filters)
• Arsenic and fluoride removal devices
• Reagent addition for community treatment
processes
• Alum or Poly Aluminum Chloride (PACl) ____________
coagulation
• Calcium or sodium hypochlorite for ____________
disinfection
• Sodium carbonate for _____________
pH control
• A flow control device that maintains a constant
dose as the main flow varies
If you had electricity…
• Metering pumps (positive displacement)
•
•
•
•
Pistons
Gears
Peristaltic
Diaphragm
• Valves with feedback from flow sensors
• So an alternative would be to raise the per capita
income and provide RELIABLE electrical service to
everyone…
• But a simpler solution with fewer moving parts
would be better!
The Challenge of Chemical Metering
(Hypochlorinator)
What is the simplest
Access door to
hypochlorinator tank
representation that
captures the fluid
mechanics of this system?
Overflow
tube
Chlorine solution
Access door to
distribution tank
PVC valve
PVC pipe
Chlorine drip
Raw water entering
distribution tank
Hole in a Bucket
h
Orifice
Vena contracta
Avc  0.62 Aor
 vc  0.62
Q   vc Aor 2 g h
Q is flow rate [volume/time]
This is NOT a minor loss coefficient
It is the ratio of the vena contracta area to the orifice area
Use Conservation of Mass and
Minor Loss equation [Two unknowns (Q, h)]
Mass conservation on liquid in tank
volume Plan view area
ATank dh
dV
Q

dt
dt
h0
Minor loss equation
he  K e
2
V
2g
Q  AValve
ATank
Q2
he  K e
2
2 gAValve
2he g
Ke
dh
 AValve
dt
Orifice in the PVC valve
2 gh
0
Ke
Integrate to get h as f(t)
Finding the chlorine depth as f(t)
 ATank
AValve
h
2g
Ke
 ATank
AValve
h 
2g
Ke

h0
dh

h
t
 dt
Separate variables
0
2  h1/ 2  h01/ 2   t
AValve
h0  t
2 Atank
2g
Ke
Integrate
Solve for height
Finding Q as f(t)
Q  AValve
2he g
Ke
Q  AValve
AValve
2g 
 h0  t
Ke 
2 Atank
he  h0  t
AValve
2 ATank
2g 

Ke 
2g
Ke
Q
Find AValve as function of initial target flow rate
AValve 
Q0
2h0 g
Ke
Set the valve to get desired dose initially
Surprise… Q and chlorine dose
decrease linearly with time!
Q  AValve
AValve
2g 
h

t
 0
Ke 
2 ATank
tQ0
Q
 1
Q0
2 ATank h0
AValve 
2g 

Ke 
Q0
2h0 g
Ke
Linear decrease in flow with time
Relationship between Q0 and ATank?
Assume flow at Q0 for time (tDesign) would empty reservoir
Q0t Design  ATank hTank
Q
1 t hTank
 1
Q0
2 t Design h0
Q0
hTank

ATank t Design
CCl2
CCl2
0
1 t hTank
 1
2 t Design h0
Reflections
Q
1 t hTank
 1
Q0
2 t Design h0
• Let the discharge to atmosphere be located
at the elevation of the bottom of the tank…
• When does the flow rate go to zero?
• What is the average flow rate during this
process if the tank is drained completely?
• How could you modify the design to keep Q
more constant?
Normalized flow
Normalized water depth
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
2
4
Normalized water depth
Case 1, h0=50 m,
htank = 1 m,
tdesign=4 days
Flow ratio (Actual/Target)
Effect of tank height above valve
Q2
h  2 h0
Q0
0
8
6
time (days)
Normalized flow
Normalized water depth
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
2
4
time (days)
6
0
8
Normalized water depth
Case 2, h0=1 m,
htank = 1 m,
tdesign=4 days
Flow ratio (Actual/Target)
h0
htank
A related tangent…
Design a drain system for a tank
h
AValve
h0  t
2 Atank
0
h 
h0  t Drain
AValve 
DValve 
AValve
2 ATank
2
 DValve
4
2g
Ke
Integrated results
giving h as f(t)
2g
Ke
Substitute valve diameter
8 LTankWTank  K e H Tank 


 t Drain
2
g


1
4
Here Ke is the total minor loss for the drain system
Brainstorm: Constant Flow
• Sketch a device that you could use to deliver
liquid bleach into the water leaving a water
treatment plant
• No electricity
• Must deliver a constant flow of bleach
• What are the desired properties of the
device that meters chemicals into a water
treatment plant?
• Why is this hard?
Constant Head: Floats
Head can be
varied by
changing
buoyancy of
float
h
orifice
Supercritical
open channel
flow!
VERY Flexible hose
Unaffected by downstream conditions!
Hypochlorinator with float design
What is the simplest
Access door to
hypochlorinator tank
representation that
captures the fluid
Float
mechanics of this system?
Orifice
Overflow
tube
Chlorine solution
Access door to
distribution tank
Transparent
flexible tube
PVC valve
PVC pipe
Chlorine drip
Raw water entering
distribution tank
Floating Bowl
• Adjust the flow by changing the rocks
Need to make
adjustments
(INSIDE) the
chemical tank
Rocks are
submerged in
the chemical
Safety issues
Constant Head: Overflow Tanks
h
Aor
Surface tension
effects here
What controls
the flow?
Q   vc Aor 2 g h
Constant Head:
Marriot bottle
hL
• A simple constant head
device
• Why is pressure at the top
of the filter independent of
water level in the Marriot
bottle?
• What is the head loss for
this filter?
• Disadvantage? batch
___________
system
2
pin
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL
g
2g
g
2g
Constant Head: Float Valve
htank
dorifice
hsubmerged
Float adjusts
opening to maintain
relatively constant
water level in lower
tank (independent
of upper tank level)
NOT Flow Control! ?
Force balance on float valve?
2
d 2float  hsubmerged  d orifice
 htank
dfloat
htank
hsubmerged

d 2float
2
d orifice
Operator control of the constant
flow
• Head (or available energy to push fluid through the
flow resistance) (voltage drop)
• Flow resistance (resistor)
• Orifice
____________________________________
i.e. small hole or restriction
• Porous
____________________________________
media
• Long
____________________________________
straight small diameter tube
• Vary either the head or the flow resistance to vary
the flow rate (current)
• Vary head by adjusting the constant head tank elevation
or the outlet elevation of a flexible tube
• Vary orifice size by adjusting a valve
Variable Orifice
2 cm
Variable head or
variable
resistance?
How is flow
PVC
stem
varied?
Variable orifice
area
2.3 cm
1.5 cm
2 mm
0.5 cm
Rubber tip
4.4 cm
11.0 cm
8.3 cm
Float
mass:
6 grams
6.5 cm
5.2 cm
Barb tubing
adapter
5.6 cm
9.1 cm
IV roller
clamp
Housing Dimensions:
ID = 7.85 cm
OD = 8.8 cm
IV tubing
(~10 drops/mL)
2 mm
How would you figure out the required size of the orifice?
straight
Float valve and small^tube
(Gravity dosing system)
chemical stock tank
Flow control device
If laminar flow!
32 LV 128 LQ
Small diameter tubing
hf 

gD 2
g D 4
hf g D 4
Neglecting minor losses
Q
128 L
2
pin
Vin2
pout
Vout
 zin   in
 hP 
 zout   out
 hT  hL
g
2g
g
2g
hL  zin  zout
hL
Hypochorinator Fix
http://web.mit.edu/d-lab/honduras.htm
What is good?
How could you improve this system?
What might fail?
Safety hazards?
AguaClara Technologies
• “Almost linear” Flow Controller
• Linear Flow Orifice Meter
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• Linear Chemical Dose Controller
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• Orifice/valve based Chemical Dose Controller
AguaClara approach to flow
control
• Controlled variable head
• Float valve creates constant elevation of fluid at
inlet to flow control system
• Vary head loss by varying elevation of the end of
a flexible tube
• Head loss element
• Long straight small diameter tube
• We didn’t realize the necessity of keeping the
tube straight until 2012 and thus our early flow
controllers had curved dosing tubes
Early Flow
Controllers
Nonlinear relationship between
flow and height of end of tube
Air vent
1 L bottle
Tube from the stock tank
Float valve
The fluid level in the bottle
should be at the same level as the
0 cm mark
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hf g D 4
Q
128 L
Hagen–Poiseuille
Holes to
choose the
alum dose
hf
Flexible tube with length set to deliver target
maximum flow at maximum head loss
In our first flow controllers we
didn’t realize how important
“straight” was! We ignored minor
losses and minor losses weren’t
minor!
Raw water
Rapid Mix
Requirements for a Flow Controller
• Easy to Maintain
• Easy to change the flow in
using a method that does not
require trial and error
• Needs something to control
the level of the liquid (to get
a constant pressure)
• Needs something to convert
that constant liquid level into
a constant flow
Installing a Flow Controller for dosing
Chlorine
Hypochlorinator Fix
Access door to
hypochlorinator tank
Chlorine solution
Access door to
distribution tank
Raw water entering
distribution tank
Overflow
tube
Dose Controller:
The QC Control Problem
• How could we design a device that would maintain
the chemical dose (C) as the flow (Q) through the
plant varies?
• Somehow connect a flow measurement device to a
flow controller (lever!)
• Flow controller has a (mostly) linear response
• Need height in entrance tank to vary linearly with
the plant flow rate – solution is The Linear Flow
Orifice Meter (LFOM)
Linear Flow Orifice Meter
• Sutro Weir is difficult to machine
• Mimic the Sutro weir using a pattern of
holes that are easily machined on site
• Install on a section of PVC pipe in the
entrance tank
• Used at all AguaClara Plants
except the first plant (Ojojona)
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Invented by AguaClara team member David Railsback, 2007
Linear Orifice Meter
Photo by Lindsay France
Linear Flow Orifice Meter
• Holes must be drilled with a bit that leaves a
clean hole with a sharp entrance (hole saws
are not a good choice)
• The sharp entrance into the hole is critical
because that defines the point of flow
separation for the vena contracta
• The zero point for the LFOM is the bottom
of the bottom row of orifices
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Flow Controller
What must the operator do if
the plant flow rate decreases?
Wall height of entrance tank
Max water level in entrance tank
Linear Flow
Orifice Meter
Min water level in entrance tank
Max water level in floc tank
Bottom of entrance tank
5 cm
20 cm
5 cm
? cm
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Raw water
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Linear Dose Controller
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• Combine the linear flow controller and the linear
flow orifice meter to create a Linear Dose
Controller
• Flow of chemical proportional to flow of plant
(chemical turns off when plant turns off)
• Directly adjustable chemical dose
• Can be applied to all chemical feeds (coagulant and
chlorine)
• Note: This is NOT an automated dose device
because the operator still has to set the dose
Constant
head tank
Stock Tank
of coagulant
0
H
5
Dosing tubes
Purge valves
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
0
10
Float valve
Open channel
supercritical
flow
Drop tube
LFOM
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Float
Decrease dose
0
H
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
0
10
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Increase dose
0
H
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
0
10
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
H
Half flow
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Plant off
0
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Linear Flow
Orifice Meter
Lever
Constant
Head
Tanks
Atima doser: Three 1/8” diameter tubes, 1.93 m long, 187.5
g/L stock solution of PACl, 5 to 48 mg/L dose range.
Coagulant
Stock Tank
Flow calibration
column
Lever
Constant
Head
Tank
Chemical Dose Controller
• The operator sets the
dose directly
• No need for
calculations
• Visual confirmation
• A key technology for
high performing
plants
• We need an elegant
design!
Dose Controller and LFOM
(lacking the straight dose tubes)
Chemical
Feed
Tank
Design specifications
for the float valve
The float valve has an orifice that restricts the
flow of the chemical and that affects the head
loss. Different float valves have different sizes
of orifices.
h
Q   vc Aor 2 g h
Aor 
 vc
Q
2 g h
Why not
increase h?
Dor2
Aor  
4
Why are we concerned with the orifice head loss?
4Q
Dor 
Where else is there head loss?
 vc 2 g h
Is this a minimum or a maximum orifice diameter?
Constraints on flow controller
Dosing tube design
• Flow must be laminar (Re<2100)
• Minor losses must be small (small V!)
• It took us a while to discover how critical this
constraint is!
• Dosing tube must “be reasonable” length
which might mean shorter than an available
wall in the plant. How can I get a shorter tube?
• Designing dose controllers over a wide
range of chemical flow rates requires good
engineering
Nonlinearity Error Analysis:
The problem caused by minor losses
128 QL
hf 
g D 4
8Q 2
he 
g 2 D 4
K
e
Actual head loss
 128 L
8Q
hL (Q )  

4
g 2 D 4
 g D

K
 e Q

No minor losses
With minor losses
Linearized model
 128 L
8Q
hl (Q)  

4
g 2 D 4
 g D
 128 L 8QMax
hLinear (Q )  

4
g

D
g 2 D 4

y=
m

K
 e Q

x
Head loss (cm)
15
Linearized model of head loss
Calibrate at max flow
K
e

Q

10
5
0
 128 L 8QMax
hLinear (Q)  

4
g

D
g 2 D 4

0
0.5
1
Flow rate (mL/s)
1.5
K
e

Q

2
Relationship between Q and L
given error constraint
 Error
1   Error
hLinear  hL
hL

 1
hLinear
hLinear
 128 L
8Q

 g D 4 g  2 D 4

 128 L 8QMax
 g D 4  g  2 D 4

1   Error 
 Error

 128 L 
K
 e
 g D 4 



  128 L 8QMax
K
 e   g D 4  g  2 D 4
 
8QMax
128 L

1




Error
g D 4
g 2 D 4
8QMax
128 L

1



Error 
g D 4
g 2 D 4
 1   Error
L
  Error
Set a limit on the error
caused by nonlinearity (we
use 0.1)
 QMax

 16
K
e
K
e
 Ke 
0

K
 e

Plug in head loss
equations. Take
the limit as Q→0.
128 L
g D 4
Solve for L
Relationship between minimum tube length and maximum
flow from the error constraint (both unknowns – we need
another equation between Q and L!)
Relationship between Q and L
given head loss constraint
2
 128 LQMax 8QMax
hL  

4
2 4
g

D
g

D

 1   Error
L
  Error
 QMax

 16
K
 2nd equation relating Q and L
 Ke  is total head loss

1st equation relating Q and L
– error constraint
e
 8 K e  1   2
hL   2 4 
Combine two equations
  QMax
 g D   Error  
2hL g  Error
 D 2 2hL g  Error
VMax 
Solve for Max Q
QMax 
4
 Ke
 Ke
 8 KQ 2 
D
2

h
g

 Error l

1
4
This is the minimum D that we can use
assuming we use the shortest tube possible.
Dosing Tube Lengths
 1   Error
L
  Error
 QMax

 16
1   Error  D 2

L
64
major
K
QMax 
e
 D2
4
2hL g  Error
 Ke
This is the shortest tube that can
be used assuming that the velocity
is the maximum allowed.
2hL g  K e
 Error
minor
2
 128 LQMax 8QMax
hL  

4
2 4
g

D
g

D

 ghLMax  D 4 QMax
L

 128 QMax 16


 Ke 


 Ke 

This is for laminar flow!
If the tube isn’t operating at its maximum
flow, then use this equation.
Does tube length increase or decrease if you decrease
the flow while holding head loss constant?
Reynolds Continuity

4Q
V
 D2
Is this a min or max D?
4Qmax
D
 Re max
2100
Minor Loss Errors
 8 KQ

D
2

h
g

 Error l

2
hl  20cm
 gh  D
Q
L L
 Max
 128 QMax 16
1
4

 Ke 

laminar
minimum error
design diameter
6
4
2
0
4
0
2
6
4
6




3

 in
Davailable   4 
 5.44  32


 8 
2
 Error  0.1
Viscosity = 1.0 mm2/s, ΣKe = 2
8
Flow Rate (mL/s)
L
Tube Length (m)
Re 
VD
Tube Diameter (mm)
Tube Diameter for Flow/Dose
Controller (English tube sizes)
1   Error  D 2
64
10
12
2hL g  K e
 Error
4
2
Minimum Length
Design Length
0
0
2
4
6
8
Flow Rate (mL/s)
10
12
Dose Controller Accuracy
0
hStockTank
• Float valves only attenuate the
hSubmergedFloat  2
fluctuations in level of the fluid
d Float
 Lever
2
surface
dOrifice
• Surface tension effects at the location
where the fluid switches from closed
conduit to open channel flow (the
end of the dosing tube) could cause
errors of a few millimeters
• The weight of the tubing and slide
supported by the lever will apply
different amounts of torque to the
lever depending on the dose chosen.
This determines the required
diameter of the float
5
H
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
0
10
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
Maximum Plant Flow Rate
Using Linear Flow Controller
• Assume the PACl concentration is 160 g/L,
the maximum PACl dose is 30 mg/L, and the
maximum flow in a 5 mm diameter dosing
tube is about 8.7 mL/s.
QC CC  QP CP
CC
QP  QC
CP
Mass conservation
P = Plant
C = Chemical Feed
g

160

L 
L
QP  0.0087
g
s
0.03


L


L
QP  46
s
Multiple tubes
What is the solution for larger plants? ___________
Dose Control Summary
• Laminar tube flow and linear flow orifice meters
• Coagulant dosing for plants with flow rates less than about
200 L/s
• Chlorine dose controllers even for larger plants
• These devices are robust AND a good design requires
excellent attention to details
•
•
•
•
•
•
•
No small parts to lose
No leaks
Compatible with harsh chemicals
Locally sourced materials if possible
Dosing tubes must be straight
Must account for viscosity of the chemical
Minor losses must be minor!
Extras!
extra
Extending the range of the flow
controllers
• A single laminar flow controller will not be able
to deliver sufficient alum for a large plant
• Could you design a turbulent flow flow
controller?
• Could we go non linear with both flow
measurement and flow control to get a simple
design for larger water treatment plants?
• Clogging will be less of an issue with larger
flows
• What are our options for relationships between
flow rate and head loss?
extra
Closed Conduit Flow options for
Flow Controllers
Governing equation
Q
Range Limitations
• Laminar flow in a tube
128 LQ
hf 
 g D 4
Low flow rates
• Turbulent flow in a tube
hf  f
8
g
2
LQ 2
D5
• Orifice flow
Q   vc Aor 2 g h
High flow rates to
achieve constant f
Valid for both
laminar and
turbulent flow!
extra
Open Channel Flow Relationships
Sharp-Crested Weir
Q
Explain the exponents of H!
2
CdW 2 g H 3/2
3
V-Notch Weir
8
   5/ 2
Q  Cd 2 g tan   H
15
2
Broad-Crested Weir
2 
Q  Cd W g  H 
3 
3/2
Sluice Gate (orifice)
Q  CdWy g 2 gy1
V  2 gH
extra
Dose Controller with nonlinear scale
QC  K C hCnC
QP  K P hPnP
CC QC
CP 
QP
CC K C hCnC
CP 
K P hPnP
hC  K L hP
General Flow – Head Loss relationships
for chemical feed and plant flow
Concentration of the chemical in the plant
Connect the two heads with a lever,
therefore the two heads must be proportional
CC K C K LnC hPnC
CP 
Is the plant concentration constant as the
K P hPnP
plant flow rate changes?
Constant dose with changing
plant flow requirement
extra
CC K C K LnC hPnC
CP 
K P hPnP
K LnC hPnC
CP 
hPnP
What must be true for Cp to be constant as
head loss, hp changes? QP  K P hPn
P
What does this mean?
What are our options?
extra
0
Dose scale on the lever arm?
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
0
10
CC K C K LnC hPnC
CP 
K P hPnP
CP  K L
CC K C
KP
KL is the ratio of the height
change of the float to the height
change of the flow controller
This relationship makes it difficult to
accurate control a wide range of alum
dosages on a reasonable length lever
K L  CP
Pivot Point
8
10
12
14
Alum dose (mg/L)
16
18
20
extra
AguaClara Dose Control History
control
measurement
• Laminar tube flow and simple orifice
• Laminar tube flow and linear flow orifice
meter
• Dose controller that combines laminar tube
flow and linear flow orifice meter
• Dose controller with orifice flow for both
flow control and measurement
• Dose controller with variable valve and
simple orifice
extra
Variable dosing valve:
For coagulant dosing for plant flows above 100 L/s
Photo courtesy of Georg Fischer
Piping Systems
This method has not yet been attempted. The dose will be set by turning the valve.
The valve will replace the head loss tube. The transition to open channel flow will be
on a lever that tracks plant flow rate but no slider will be required. The exit from the
entrance tank will be a simple orifice system. The system will still respond to plant
flow changes correctly.
extra
Surface Tension
Will the droplet drop?
Is the force of gravity stronger than surface tension?
4 r 3
Fg 
g
3 2
Fs= 2rs
4 r 3
 g   g h  r 2   2 rs
3 2
2

g

h

r


F p=
h
extra
4 r 3
 g   g h  r 2   2 rs
3 2
Solve for height of water
required to form droplet
4 r 3
2 rs 
g
3 2
h 
 g  r 2 
2s 2r
h 

 gr 3
Surface tension (N/m)
Surface Tension can prevent flow!
0.080
0.075
0.070
0.065
0.060
0.055
0.050
0 20 40 60 80 100
Temperature (C)
h f  20cm
LTube  1m
extra
A constraint for flow control
devices: Surface Tension
Diameter (mm)
4
3
2
1
0
Delineates the
boundary between
stable and unstable
Flow control devices
need to be designed
to operate to the
right of the red line!
0
2
4
6
8
Flow rate (mL/s)
100
head required to produce droplet (mm)
2s 2r
h 

 gr 3
Tube flow
Orifice flow
h
2s/gr
10
1
0.1
1
droplet radius (mm)
10
extra
2
Kinematic Viscosity (m /s)
Kinematic Viscosity of Water
2.0E-06
1.5E-06
1.0E-06
5.0E-07
0.0E+00
0
20
40
60
80
Temperature (C)
This is another good reason to have a building around AguaClara facilities!
100
 Alum
extra
Kinematic Viscosity of Coagulants
Kinematic viscosity (mm^2/s)
12
Alum in distilled water
Alum Model
PACl in distilled water
PACl Model
10
8
6
4
2
0
0
200
400
600
2.289


  Coagulant concentration (g/L) 






C
5  C PACl
 1  4.225 106  Alum   H 2O



1

2.383

10
 kg
PACl
kg



 3 
 3


 m  

 m
1.893







 H 2 O


extra
Holding container
(bucket or glass
column)
Sealing pipe
Pong pipe
Driving head for sand column
Sand column
HJR
Upflow prevents trapped air
(keyword: “prevent”)!
Porous media as resistance element
Porous Media Head Loss: Kozeny
equation
V pore 
hf 
Velocity of fluid above the porous media
Va

Dynamic viscosity
32  LV pore
 gd
hf
 36k
L
extra
Analogy to laminar flow in a pipe
2
pore
1   
3
Kinematic viscosity
2
 Va
2
gd sand



k = Kozeny constant
Approximately 5 for
most filtration conditions
Orifice Image
extra
 2 


3

 in
Davailable   4 
 5.44  32


 8 
hl  20cm
 Error  0.1
Viscosity = 1.8 mm2/s, ΣKe = 2
laminar
minimum error
design diameter
6
4
2
0
0
2
6
Tube Length (m)
• Increasing the
viscosity decreases
the length of the
tubes
• Higher flow rates can
be attained
Tube Diameter (mm)
Tube Diameter for Flow/Dose
Controller (English tube sizes)
4
6
8
10
12
Flow Rate (mL/s)
4
2
Minimum Length
Design Length
0
0
2
4
6
8
Flow Rate (mL/s)
10
12
extra
Tube Diameter (mm)
Tube Diameter for Flow/Dose
Controller (Metric tube sizes)
6
4
laminar
minimum error
design diameter
2
0
0
2
Tube Length (m)
6
hl  20cm
 Error  0.1
Viscosity = 1.0 mm2/s, ΣKe = 2
4
6
8
10
12
Flow Rate (mL/s)
4
2
Minimum Length
Design Length
0
0
2
4
6
8
Flow Rate (mL/s)
10
12
extra
• It should be possible
to achieve flows of
8.7 mL/s using the 5
mm diameter tubes
Tube Diameter (mm)
Tube Diameter for Flow/Dose
Controller (Metric tube sizes)
6
4
laminar
minimum error
design diameter
2
0
0
2
Tube Length (m)
6
hl  20cm
 Error  0.1
Viscosity = 1.8 mm2/s, ΣKe = 2
4
6
8
10
12
Flow Rate (mL/s)
4
2
Minimum Length
Design Length
0
0
2
4
6
8
Flow Rate (mL/s)
10
12
extra
Design Algorithm
1.
Calculate the maximum flow rate through each available dosing tube
diameter that keeps error due to minor losses below 10%. Q   D 2h g
4
K
Calculate the total chemical flow rate that would be required by the
treatment system for the maximum chemical dose and the maximum
allowable stock concentration.
Calculate the number of dosing tubes required if the tubes flow at
maximum capacity (round up)
 ghL D 4 QMax

L

Ke 
Calculate the length of dosing tube(s) that

 128 QMax 16

correspond to each available tube diameter.
Select the longest dosing tube that is shorter than the maximum tube
length allowable based on geometric constraints.
Select the dosing tube diameter, flow rate, and stock concentration
corresponding to the selected tube length.
2
L
Error
Max
2.
3.
4.
5.
6.
e
extra
Making Minor Losses Minor
• Eliminate curvature in the dosing tubes;
keep them straight and taut (in tension)
• Use fittings that have a larger ID than the
tube; it will be necessary to stretch the
tubing to get it on the larger diameter
barbed fittings
Doser Calibration Steps
(whenever the dosing tubes are replaced or elevations are varied)
• Make sure the lever is level at zero flow
(adjust the length of the cable to the float)
• The doser is a linear device and thus when
we calibrate it we have ____
2 adjustments
• __________
the zero flow setting (adjust the
Intercept
height of the constant head tank)
• __________
maximum flow rate (we will adjust
Slope
this by adjusting the length of the dosing tubes
if needed)
extra
extra
Float Valve Head Loss
• What orifice diameter is required for a flow
of 7.5 mL/s if we don’t want more than 30
cm of head loss?
4Q
Dor 
 vc 2 g h

m3 
4  0.0000075 
s 

Dor 
 2.5mm
m

  0.62  2  9.8 2   0.3m 
s 

http://www.cdivalve.com/products/float-valves/pvc-float-valves