h2. Abstract
The AguaClara plant design needed to be -altered- +extended+ to accommodate high flow rates such as those at [Gracias] (3000 L/min). -This meant- +One of the new design issues was+ that we could no longer rely on having the alum adequately mixed with the raw water in the -first sector- +pipe connecting the entrance tank to the+ -of the- flocculator -like in previous plants- . -This previous- +The+ design +used for smaller flow rates+ does not generate enough large -scaling- +scale+ mixing to evenly distribute alum throughout the flow or enough energy dissipation to ensure that alum is mixed at the molecular level. Consequently, a new entrance tank was designed to accommodate -for- a rapid mix chamber that could appropriately mix alum with raw water before flocculation. The entrance tank and rapid mix chamber were designed to be used in conjunction with the [Nonlinear Chemical Dose Controller] between the maximum and minimum expected plant flow rates. The rapid mix chamber uses a combination of orifices and hydraulic drops to achieve desired even mixing.
{panel} I would prefer a generic design procedure and an assessment of the range of flows for which this design is applicable. The specific design for a flow rate of 50 L/s could be included in its own section after the general procedure is documented.{panel}
h2. -Gracias- Entrance Tank Design
h3. Design Schematic
h4. Rapid Mix Chamber
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!RapidMixer.JPG|width=600px,align=top!
h5. Figure 1: Rapid Mix Chamber design (Dimensions are in meters)
{float}
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The rapid mix chamber is assumed to be made of concrete and attached to one side of the entrance tank. The water within the entrance tank enters the rapid mix chamber's upper channel through a rectangular orifice. The dimensions of the rectangular orifice were based on the minimum flow rate required by the plant. Once water passes through this orifice into the upper channel it travels to the end of the channel and then experiences a hydraulic drop. By the time the water reaches the bottom of the hydraulic drop, enough large scale mixing has occurred. Finally, as the water passes through the flocculator entrance orifice of the rapid mix chamber, it experiences a contraction due to the orifice's small area. This contraction allows for the water to experience a jet-like flow that generates enough energy dissipation for molecular diffusion.
h2. Design Assumptions and Specifications
The majority of the assumptions made for this design were the same as in past designs.
h3. Entrance Tank Dimensions
The upflow velocity of the entrance tank is assumed to be 700 m/day. This design constraint is based on the [dissolved air flotation of flocs |Dissolved Air Flotation of Flocs] research and is based on the velocity used to backwash a filter. Using this assumption, along with the flow rate through the plant, the cross sectional area of the tank is calculated to be 6.171 m ^2^. The length of the side of a square tank is 2.484 m.
h3. Entrance Tank Orifice
The entrance tank orifice design assumes that the maximum height of the water from the center of the entrance tank orifice's -width- +height+ is 20 cm and that the coefficient of discharge due to the vena contracta through the orifice is 0.62.
It is assumed that major head loss is negligible at the upper level of the entrance tank. Therefore, the minor head loss equation along with the orifice flow equation are used to calculate the orifice's area. In the minor head loss equation, the velocity -variable- is -substituted- +replaced+ with -V = QA- +V = Q/A+ relationship in order to solve for the vena contracta area of the orifice:
{latex}
$$
A_{EtOrificeVenaContracta} = {{Q_{Plant} } \over {\sqrt {2gHL_{EtMax} } }}
$${latex}
Where:
* Q ~Plant~ is the plant's total maximum flow rate
* HL ~EtMax~ is the height of the water from the center of the entrance tank orifice's width to the surface
* g is the gravitational force
This area, though, is the cross sectional area of the contracted water as it passes through the orifice. To calculate the actual orifice area, the vena contracta area is divided by the coefficient of discharge:
{latex}
\large
$$
A_{EtOrifice} = {{A_{EtOrificeVenaContracta} } \over {Pi_{VenaContractaOrifice} }}
$${latex}
Where:
* A ~EtOrificeVenaContracta~ is the vena contracta area of the orifice
* Pi ~VenaContractaOrifice~ is the coefficient of discharge due to the vena contracta
Given the area requirement, the width or the length can be fixed and then subsequently the length or width, respectively, can be solved for. It is important to note that the lowest attainable minimum flow rate is proportional to half the width of the orifice (lowest water height allowed) also known as the lowest attainable head loss:
{latex}
$$
Q_{Min} = Pi_{VenaContractaOrifice} A_{EtOrifice} \sqrt {2gHW_{EtMin} }
$${latex}
Where:
* A ~EtOrifice~ is the actual orifice area
* Pi ~VenaContractaOrifice~ is the coefficient of discharge due to the vena contracta
* g is the gravitational force
* HW ~EtMin~ is the lowest water height allowed also known as the lowest attainable head loss
In Gracias, the plant had to operate at 20% of the maximum flow rate which corresponds to a specific head loss ratio as shown in the graph below:
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!Graph for RapidMixer.png|width=600px,align=top!
h5. Figure 3: Head loss Ratio
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This head loss ratio:
{latex}
$$
Pi_{EtFlow} = 0.2 \to heightratio = 0.08
$${latex}
Where:
* Pi ~EtFlow~ is the flow rate ratio (Q ~min~ / Q ~max~)
* heightratio is the head loss or water height ratio (W ~EtOrifice~ / HL ~EtMax~)
in turn gives us a specific width:
{latex}
$$
W_{EtOrifice} = heightratio\left( {HW_{EtMin} } \right)
$${latex}
Knowing the width and area, the length is calculated:
{latex}
\large
$$
L_{trueET} = {{A_{EtOrifice} } \over {W_{EtOrifice} }}
$$ {latex}
Where:
* A ~EtOrifice~ is the actual area of the orifice
* W ~EtOrifice~ is the width of the orifice
h3. Upper Channel
The upper channel is modeled as an open channel flow conduit. The water enters through the entrance tank orifice on the side of the channel and flows towards the hydraulic drop into the flocculator. The length of the channel is dependent on the length of the entrance tank orifice and the channel must be longer than this orifice. This approximation is conservative because it assumes that more water enters the channel at first than it actually does. The height of water along the channel is iteratively solved for by using the direct step method described below.
The flow at the top of the hydraulic drop is critical\- potential energy and kinetic energy are in balance and the height of water in the tank can be solved for using the equation:
{latex}
\large
$$
h_{crit} = \left( {{q \over g}^2 } \right)^{1/3}
$${latex}
Where:
* h ~crit~ is the critical depth
* q is the flow per unit width
* g is the acceleration -dues- +due+ to gravity
The width of the channel is based on the baffle spacing in the first section of the flocculator.
{panel} We need to invent clear names for each of the components in this design.
* Flow meter orifice - the rectangular orifice used to measure the plant flow rate
* Orifice collector channel - Channel that collects the water from the flow meter orifice
* Waterfall mixer - Vertical channel that has a small water fall at the top and the orifice to the flocculator at the bottom
* Fine scale turbulence generating orifice - the orifice that delivers water into the flocculator from the waterfall mixer.
The new layout includes a right angle turn in the orifice collector channel so that the waterfall mixer is inside the entrance tank. The flocculator first section baffle spacing is no longer a constraint on the orifice collector channel width. Thus we need another logical constraint for the orifice collector channel width. The orifice collector channel depth contributes to total head loss through the plant since the water surface in the channel must be below the bottom of the orifice and the bottom of the channel must be above the water level in the waterfall mixer. Thus the water depth in the channel should be kept small. It doesn't work to specify a depth for the channel since that will not work over a wide range of flow rates. It would be better to specify an aspect ratio (channel width/water depth). I propose that we use an aspect ratio of 1 for this design. That will make the flow geometry square. I estimate that the orifice collector channel for Gracias would be 19 cm deep by 19 cm wide. Check my calculations! {panel}
+{latex}$$
\eqalign{
& h_{crit} = \left( {\frac{{Q^2 }}
{{w^2 g}}} \right)^{\frac{1}
{3}} \cr
& \Pi _{OCC} = \frac{w}
{{h_{crut} }} \cr
& h_{crut} = \frac{w}
{{\Pi _{OCC} }} \cr
& h_{crit} w^{\frac{2}
{3}} = \left( {\frac{{Q^2 }}
{g}} \right)^{\frac{1}
{3}} \cr
& \frac{{w^{\frac{5}
{3}} }}
{{\Pi _{OCC} }} = \left( {\frac{{Q^2 }}
{g}} \right)^{\frac{1}
{3}} \cr
& w = \Pi _{OCC} ^{\frac{3}
{5}} \left( {\frac{{Q^2 }}
{g}} \right)^{\frac{1}
{5}} \cr}
$${latex}+
Given this initial height in the channel, the height of water in the rest of the channel can be -theoretically- determined. The energy equation for an open channel flow can be rearranged to give the relationship:
{latex}
\large
$$
\Delta x = {{\Delta y + {{V_1^2 } \over {2g}} - {{V_2^2 } \over {2g}}} \over {S_f - S_o }}
$${latex}
Where:
* S ~o~ is the slope of the channel which is zero in this case
* S ~f~ is the friction slope
* y is the depth along the channel
* x is the distance from the hydraulic drop to the end of the channel
The friction slope (Figure 2) was found using the equation:
{latex}
\large
$$
S_f = f{{V^2 } \over {8gR_h }}
$${latex}
Where:
* f is the friction factor
* V is velocity
* g is acceleration due to gravity
* R ~h~ is the hydraulic radius defined as the cross sectional area over the wetted perimeter
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!directstep.jpg|width=490px|align=bottom!
h5. Figure 2: Cross Sectional View of Upper Channel
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Finally, to solve for the heights, the critical depth is used as the initial V ~1~ and a change in depth of 0.0001 cm is used to find V ~2~. The modified energy equation is solved to find the length of the channel, given the calculated friction slope that corresponds to a change in water depth of 0.0001 cm. This process -is- +was+ repeated -over the entire length of the channel- until the +upstream end of the channel was reached and the+ height of water at the channel entrance -is- +was+ found.
h3. -Lower Hydraulic Drop-
The lower hydraulic jump is the first mixing step after alum is added to the rapid mix. The hydraulic drop provides a region of energy dissipation for large scale mixing. The minor loss coefficient can be directly related to the mixing length and in this case it is assumed that a minor loss coefficient of 1.3 is adequate for large scale eddies to evenly mix the alum. The velocity can be found with the critical height at horizontal channel exit and channel width. The equation for minor loss is then used to determine the height of the free fall necessary to obtain the desired minor loss:
{latex}
\large
$$
h = K{{V^2 } \over {2g}}
$${latex}
Where:
* V is 1.07 m/s; the critical velocity of the drop
* K is 1.3; the minor loss coefficient
* g is acceleration due to gravity
h3. First Baffle Section
The hydraulic drop deposits water into the first baffle section. The water level in this section is determined by the water level in the flocculator and the head loss through the flocculator entrance orifice(described below). The height of the channel includes the water level and the lower hydraulic drop height.
{latex}
\large
$$
h_{LowerChannel} = h_{LowerDrop} + h_{EntryOrifice} + h_{flocculator}
$${latex}
h3. Flocculator Entrance Orifice
The orifice that leads into the flocculator must provide enough energy dissipation for molecular diffusion to evenly spread alum in the water. Therefore, energy dissipation rate from minor head loss has to be greater than diffusion requirements: ε = 0.5-1.0W/kg. The following equation is used to determine the minimum energy dissipation required to overcome the diffusion requirements.
{latex}
\large
$$
{\varepsilon _{\min } = {{\pi _{D_{mix} }^4 \nu _{^{water} }^3 } \over {D_m^2 \tau _{diffusion} ^2 }}}
$${latex}
Where:
* pi ~Dmix~ is 2.4; the ratio between the diffusion length scale and the Kolmogrov length scale
* nu ~water~ is the kinematic viscosity of water
* D ~m~ is the diffusion
* tau ~diffusion~ is 10s; time scale for diffusion
The minimum energy dissipation is calculated to be approximately 0.733 W/kg. This number is then rounded to 0.8 W/kg (conservative estimate of energy dissipation) in the flocculator entrance orifice equation.
{latex}
\large
$$
V = \left( {{{2\varepsilon \sqrt Q } \over {K_{\exp } }}} \right)^{{2 \over 7}}
$${latex}
{latex}
\large
$$
\vartheta = \sqrt {{Q \over {V^3 }}}
$${latex}
In order to design the dimensions of the flocculator entrance orifice, the velocity of the flow in the vena contracta is calculates. Once the flow through the vena contracta has been determined, the area of the flow through the vena contracta is calculated with the following equation.
{latex}
\large
$$
V = \left( {{{2\varepsilon \sqrt Q } \over {K_{\exp } }}} \right)^{{2 \over 7}}
$${latex}
Where:
* Q is the flow rate
* K ~minor~ is 1.3; the minor head loss coefficient
* V is the velocity in the vena contracta
* Theta is the resonance time
By using the coefficient of discharge, the area of the orifice was calculated. A rectangular orifice is determined to be practical for this design. An orifice height of 13 cm is assumed, but will eventually need to be less than the width of the baffle spacing. Based on the calculated orifice area, the length of the rectangular flocculator entrance orifice will be determined.
After final calculations, the rectangular orifice was determined to have a height of 13 cm and a width of 51.59 cm.
h3. Alum Stock Tank
As with previous designs, the Gracias design calls for two alum stock tanks. This way, while one of the tanks is operating, the other tank may be filled with the appropriate amount of chemicals. Therefore, the tanks are designed to be big enough for the operator to have enough time to mix the chemical before the other tank has drained. It is assumed that a drainage time of 30 hours is adequate. The following formula is then used to determine what the maximum flow of alum out of the stock tank needs to be:
{include:Q.FcmMax}
Where:
* C ~FcmDoseMax~ is 90 mg/L; the maximum concentration needed for flocculation
* C ~FcmAlum~ is 120 gm/L; the concentration of alum in the stock tank
From this equation, the maximum flow of alum is determined to be 2.25 L/min; and the volume required to sustain this flow rate for 30 hours is 4050 L.
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