Abstract
The AguaClara plant design needed to be altered to accommodate high flow rates such as those at Gracias (3000 L/min). More specifically, alum was adequately mixed with raw water in the first section of the flocculator in the previous plants. This design does not have enough large scaling mixing to evenly distribute alum throughout the flow or energy dissipation ensure that alum is mixed at the molecular level. A new entrance tank has been designed to appropriately mix alum with raw water before flocculation occurs. The entrance tank is designed to be used in conjunction with the Nonlinear Chemical Dose Controller between the maximum and minimum expected plant flow rates. The entrance tank uses a combination of orifices and hydraulic drops achieve desired even mixing.
Gracias Entrance Tank Design
Design Schematic
Rapid Mix Chamber
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Rapid Mix Chamber design (Dimensions are in cm.)
The AutoCAD file for this preliminary design is attached to this page.
The rapid mix chamber is made of concrete and is attached to one side of the entrance tank. The water within the entrance tank enters the horizontal channel through a rectangular orifice. This rectangular orifice dimensions were set based on the maximum and minimum flow rates through the plant. Water passes through this orifice into a horizontal channel. At the end of the horizontal channel water falls into a vertical channel. At the bottom of this free fall is where large scale mixing occurs. Finally, before entering the flocculator water will be contracted through a rectangular orifice. The energy dissipated as the flow travels through this orifice mixes alum on the scales of molecular diffusion.
Design Assumptions and Specifications
The majority of the assumptions made for this design were the same as in past designs.
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 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 should be 6.171 m 2. The length of the side of a square tank is 2.484 m.
Entrance Tank Orifice
The entrance tank orifice design assumed that the maximum height of the water from the center of the entrance tank orifice's width is 20 cm and that the orifice coefficient due to the vena contracta through the orifice is 0.62.
It was assumed that major headloss is negligible at the upper level of the entrance tank. Therefore, the minor headloss equation along with the orifice flow equation was used to calculate the orifice area. We substituted the velocity variable in the minor headloss equation with the V=QA relationship to be able to solve for the vena contracta area of the orifice:
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$$
A_
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= {{Q_
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} \over {\sqrt {2gHL_
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} }}
$$
.
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
- g is the gravitational force
This area, though, is the cross sectional area of the contracted water as it is passes through the orifice. To calculate the actual orifice area we divide the vena contracta area by the vena contracta coefficient:
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$$
A_
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= {{A_
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} \over {Pi_
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}}
$$
Where:
- A~EtOrificeVenaContracta~ is the vena contracta area of the orifice
- Pi~VenaContractaOrifice~ is the orifice loss-coefficient due to the vena contracta
Given the area requirement, the width or the length could be set and then subsequently solve for the length or width, respectively. It was important to note that the minimum flow rate that could be acquired by the plant depended on the width of the orifice since this is was also the lowest attainable head loss:
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In Gracia's particular case, the plant had to operate at 20% of the maximum flow rate which corresponded to a specific head loss ratio as shown in the graph below:
This head loss ratio I turn gave us a specific width:
Given the width and area, the length was calculated:
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$$
L_
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= {{A_
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} \over {W_
}}
$$
Where:
- A~EtOrifice~ is the actual area of the orifice
- W~EtOrifice~ is the width of the orifice
Upper Channel
The upper channel was modeled as an open channel flow conduit. The channel was modeled as though water entered from its end and flowed toward the hydraulic drop into the flocculator. This is a conservative assumption, water enters the horizontal channel along one side. The length of this channel is determined by the length of the entrance tank orifice; the channel must be longer than this orifice. This approximation is conservative because it assumes that more water enters the channel at first than actually does. The height of water along the channel was iteratively solved for used 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:
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$$
h_
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= \left( {{q \over g}2 } \right)
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$$
Where:
- h crit is the critical depth
- q is the flow per unit width
- g is the acceleration dues to gravity
The width of the channel was based on the baffle spacing in the first section of the flocculator.
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 open channel flow can be rearranged to give the relationship:
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$$
\Delta x = \Delta y + {{V_1^2 } \over {2g - {{V_2^2 } \over
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}} \over {S_f - S_o }}
$$
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:
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$$
S_f = f{{V^2 } \over {8gR_h }}
$$
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
Finally to solve for the heights, the critical depth was used as the initial V 1 and a change in depth of 0.0001 cm was used to find V 2. The modified energy equation was 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 was repeated over the entire length of the channel until the height of water at the channel entrance was found.
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 it was 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:
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$$
h = K{{V^2 } \over {2g}}
$$
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
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 headloss through the flocculator entrance orifice(described below). The height of the channel includes the water level and the lower hydraulic drop height.
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$$
h_
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= h_
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+ h_
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+ h_
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$$
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.
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$$
{\varepsilon _
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= {{\pi {D
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}4 \nu _{
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}^3 } \over {D_m^2 \tau _
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^2 }}}
$$
The minimum energy dissipation is calculated to be approximately to be 0.733 W/kg. 0.8 W/kg was used as a conservative estimate of energy dissipation in the flocculator entrance orifice.
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$$
V = \left( {{
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\over {K_
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}}} \right)^2 \over 7
$$
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$$
\vartheta = \sqrt {{Q \over
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}}
$$
In order to design the dimensions of the flocculator entrance orifice, the velocity of the flow in the vena contracta will need to be determined. After the flow through the vena contracta has been determined, the area of the flow through the vena contracta will be calculated with the flowing equation.
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$$
V = \left( {{
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\over {K_
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}}} \right)^2 \over 7
$$
By using a constant related to the vena contracta , 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.
Alum Stock Tank
A parameter of interest is how big the alum stock tank will need to be in order to have a drainage time of 30 hours. The design calls for two tanks so that the operator is able to mix a batch of alum without interrupting plant operation. A 30 hour drainage time ensures that the operator will have enough time to do this. The concentration of alum in the stock tank is assumed to be 120 gm/L. The maximum concentration of alum needed in the flocculator has been determined to be 90 mg/L. The maximum flow rate of alum can be calculated with the following equation:
From this equation it was determined that the maximum flow of alum to be 2.25 L/min. The volume required to sustain this flow rate for 30 hours is 4050 L.