Rapid Mix Chamber Design Program
Abstract
We need to invent clear names for each of the components in this design. Below are my proposals, but please feel free to revise and improve!
- 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 conduit that has a small water surface elevation drop (not a free fall waterfall) 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 AguaClara plant design needed to be extended to accommodate high flow rates such as those at Gracias (3000 L/min). 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 pipe connecting the entrance tank to the flocculator. The design used for smaller flow rates does not generate enough large 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 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.
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.
Check your outline structure. Not sure why it goes this deep! Maybe you should consider also adding a rendered CAD drawing, it may make it easier for a reader to see what the overall structure is.
Rapid Mix Chamber
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 orifice collector channel through the flow meter orifice. Once water passes through this orifice it travels to the end of the channel and then experiences a hydraulic drop. By the time the water reaches the bottom of the waterfall mixer, enough large scale mixing has occurred. Finally, as the water passes through the fine scale turbulence generating orifice of the rapid mix chamber, it experiences a contraction followed by an expansion. This expansion generates a high enough energy dissipation rate for turbulence to blend the solution to a small enough scale that molecular diffusion can finish the blending thereafter.
Design Assumptions and Specifications
Design Assumptions Design Program
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 is calculated to be 6.171 m 2. The length of the side of a square tank is 2.484 m.
Check with the Design team to find out how they are designing this tank. The tank will have the same length as the floc channels to keep the plant footprint compact.
Flow Meter Orifice
The flow meter orifice design is constrained by the maximum and minimum water height difference and the coefficient of discharge due to the vena contracta through the orifice. Taking these constraints into account and using the orifice equation the orifice area is calculated.
The derivation of the orifice area is overly complicated. The orifice equation is available in the Fluid functions and can calculate the area in a single step. The discussion of the minor loss coefficient is unnecessary here.
Given the flow meter orifice area requirement, the orifice height can be fixed and then subsequently the length can be calculated. The lowest measurable flow rate occurs when the orifice is barely submerged:
Unknown macro: {latex}
$$
Q_
Unknown macro: {Min}
= Pi_
Unknown macro: {VenaContractaOrifice}
A_
Unknown macro: {EtOrifice}
\sqrt {2gHW_
Unknown macro: {EtMin}
}
$$
According to the variable naming guide HW is the local water depth. The depth of water in the entrance tank is not the correct parameter for the equation above. HL.EtOrifice would be a reasonable variable name.
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:
Our goal is to give as large a measurement range as possible while making a reasonable design. The plant at Gracias has not yet been built.
In Figure 3 add a descriptive caption. Define the head loss ratio. How was the graph obtained and what does it mean? Show the equations.
This head loss ratio:
Unknown macro: {latex}
$$
Pi_
Unknown macro: {EtFlow}
= 0.2 \to heightratio = 0.08
$$
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:
Unknown macro: {latex}
$$
W_
Unknown macro: {EtOrifice}
= heightratio\left( {HW_
Unknown macro: {EtMin}
} \right)
$$
Knowing the width and area, the length is calculated:
Unknown macro: {latex}
\large
$$
L_
Unknown macro: {trueET}
= {{A_
Unknown macro: {EtOrifice}
} \over {W_
}}
$$
Where:
- A EtOrifice is the actual area of the orifice
- W EtOrifice is the width of the orifice
Upper Channel
The upper channel 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.
Describe the simplifying assumptions
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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\large
$$
h_
Unknown macro: {crit}
= \left( {{q \over g}2 } \right)
Unknown macro: {1/3}
$$
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.
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. 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!
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\large
$$
h_
Unknown macro: {crit}
= \left( {{
Unknown macro: {Q^2 }
\over
Unknown macro: {w^2 g}
}} \right)^1 \over 3
$$
Unknown macro: {latex}
\large
$$
\Pi _
Unknown macro: {OCC}
= {w \over {h_
Unknown macro: {crut}
}}
$$
Unknown macro: {latex}
\large
$$
h_
Unknown macro: {crit}
w^2 \over 3 = \left( {{
Unknown macro: {Q^2 }
\over g}} \right)^1 \over 3
$$
Unknown macro: {latex}
\large
$$
w = \Pi _
Unknown macro: {OCC}
^3 \over 5 \left( {{
Unknown macro: {Q^2 }
\over g}} \right)^1 \over 5
$$
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:
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\large
$$
\Delta x = \Delta y + {{V_1^2 } \over {2g - {{V_2^2 } \over
Unknown macro: {2g}
}} \over {S_f - S_o }}
$$
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
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 corresponding distance upstream in 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 upstream end of the channel entrance is was found.
Lower Hydraulic Drop Waterfall Mixer
The lower hydraulic jump waterfall mixer is the first mixing step after alum is added in the orifice collector channel 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 depth at the orifice collector 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
These values will change slightly with the square channel geometry.
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.
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\large
$$
HW_
Unknown macro: {LowerChannel}
= HL_
Unknown macro: {LowerDrop}
+ HL_
Unknown macro: {EntryOrifice}
+ HW_
Unknown macro: {Floc Start}
$$
Use variable naming guide nomenclature. This isn't clear if you are adding water heights or head losses or elevations.
Flocculator Entrance Orifice
The orifice that leads into the flocculator must provide enough energy dissipation so that turbulence will mix the alum with the water to a sufficiently small scale so that for molecular diffusion can finish the mixing process in a few seconds. 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 meet the diffusion requirements.
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\large
$$
{\varepsilon _
Unknown macro: {min }
= {{\pi {D
Unknown macro: {mix}
}4 \nu _{
Unknown macro: {water}
}^3 } \over {D_m^2 \tau _
Unknown macro: {diffusion}
^2 }}}
$$
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 0.673 cm 2/s; the diffusion coefficient for aluminum hydroxide
what is the value of the diffusion coefficient?
- tau diffusion is 10s; time scale allowed for diffusion to finish the mixing process
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.
It would be reasonable to put this energy dissipation rate into the expert users list in the VNG.
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\large
$$
V = \left( {{
Unknown macro: {2varepsilon sqrt Q }
\over {K_
Unknown macro: {exp }
}}} \right)^2 \over 7
$$
Unknown macro: {latex}
\large
$$
\vartheta = \sqrt {{Q \over
Unknown macro: {V^3 }
}}
$$
In order to design the dimensions of the flocculator entrance orifice, the velocity of the flow in the vena contracta is calculates. calculated. 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.
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\large
$$
V = \left( {{
Unknown macro: {2varepsilon sqrt Q }
\over {K_
Unknown macro: {exp }
}}} \right)^2 \over 7
$$
Where:
- Q is the flow rate
- K minor is 1.3; the minor head loss coefficient
This K was for the global mixing, not for the mixing down to the diffusion scale.
- V is the velocity in the vena contracta
- &Theta& is the resonance time
We have a port area equation (in the fluids functions) that can calculate the area of a port given the desired energy dissipation rate. I suggest you use that equation.
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.
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 - supply for 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:
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
Perhaps a note that higher concentrations (up to 500 g/L) could potentially be used.
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.