Abstract
The AguaClara plant design needed to be altered to accommodate high flow rates such as those at Gracias (3000 L/min). This meant that we could no longer rely on having the alum adequately mixed with the raw water in the first sector of the flocculator like in previous plants. This previous design does not have enough large scaling mixing to evenly distribute alum throughout the flow or have 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.
Gracias Entrance Tank Design
Design Schematic
Rapid Mix Chamber
Rapid Mix Chamber design (Dimensions are in meters)
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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 horizontal channel through a rectangular orifice. The dimensions of the rectangular orifice were set based on the minimum flow rate needed to be acquired by the plant. Once water passes through this orifice into a horizontal channel it travels to the end of the channel and then experiences the second water fall into a vertical channel. By the time the water reaches the bottom of this free fall, large scale mixing has occurred. Finally, as the water passes through the last orifice of the rapid mix chamber, it experiences 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 to mix the alum and water completely.
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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$$
A_
Unknown macro: {EtOrificeVenaContracta}
= {{Q_
Unknown macro: {Plant}
} \over {\sqrt {2gHL_
Unknown macro: {EtMax}
} }}
$$
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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\large
$$
A_
Unknown macro: {EtOrifice}
= {{A_
Unknown macro: {EtOrificeVenaContracta}
} \over {Pi_
Unknown macro: {VenaContractaOrifice}
}}
$$
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 can be set and then subsequently solve for the length or width, respectively. 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) or the lowest attainable head loss:
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$$
Q_
Unknown macro: {Min}
= Pi_
Unknown macro: {VenaContractaOrifice}
A_
Unknown macro: {EtOrifice}
\sqrt {2gHW_
Unknown macro: {EtMin}
}
$$
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 in turn gave us a specific width:
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$$
Pi_
Unknown macro: {EtFlow}
= 0.2 \to heightratio = 0.08
$$
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$$
W_
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= heightratio\left( {HW_
Unknown macro: {EtMin}
} \right)
$$
Given the width and area, the length was calculated:
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$$
L_
Unknown macro: {trueET}
= {{A_
Unknown macro: {EtOrifice}
} \over {W_
}}
$$
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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\large
$$
\Delta x = \Delta y + {{V_1^2 } \over {2g - {{V_2^2 } \over
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}} \over {S_f - S_o }}
$$
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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\large
$$
S_f = f{{V^2 } \over {8gR_h }}
$$
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
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Figure 2: Cross Sectional View of Upper Channel
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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\large
$$
h = K{{V^2 } \over {2g}}
$$
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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\large
$$
{\varepsilon _
Unknown macro: {min }
= {{\pi {D
Unknown macro: {mix}
}4 \nu _{
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}^3 } \over {D_m^2 \tau _
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^2 }}}
$$
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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\large
$$
\vartheta = \sqrt {{Q \over
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}}
$$
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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\large
$$
V = \left( {{
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\over {K_
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}}} \right)^2 \over 7
$$
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
As with previous designs, the Gracias design called 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, we needed to design the tanks so that they are big enough for the operator to have enough time to mix the chemical before the other tank has drained. It was decided that a drainage time of 30 hours would be adequate. The following formula was then used to determine what the maximum flow of alum out of the stock tank would need to be:
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- 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 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.