h2. 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. TheA new entrance tank has been designed to ensure that alum is appropriately mixed with raw water before flocculation occurs. The entrance tank uses a combination of orifices and hydraulic drops to mix the alum. Additionally 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.
h2. Gracias Entrance Tank Design
h3. Design Schematic
The entrance tank uses a combination of orifices and hydraulic drops achieve desired even mixing.
h2. Gracias Entrance Tank Design
h3. Design Schematic
h4. Rapid Mix Chamber
{float:left|border=2px solid black|width=200px}
!Rapid Mix.JPG|thumbnail,align=left!
h5. Rapid Mix Chamber design (Dimensions are in cm.)
{float}
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. This At the bottom of this free fall is the impetus forwhere large scale mixing ofoccurs. alum in the rapid mix. Finally, beforeFinally, 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.
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
When designingThe upflow velocity of the entrance tank, weis willassumed assumeto that the upflow velocity is be 700 m/day. This design constraint is based on the [dissolved air workflotation of otherflocs subteams,|Dissolved asAir itFlotation isof approximatelyFlocs] the velocityresearch and is based on the velocity used to backwash a filter. Using this assumption, along with the flow rate through the plant, the itcross wassectional determinedarea thatof the tank should have a cross sectional area of be 6.171 m ^2^. IfThe welength thenof assumethe thatside theof tanka has square dimensions,tank weis find that the length of one of the sides of the tank is 2.484 m.
2.484 m.
h3. Entrance Tank Orifice
For the design of the The entrance tank orifice, wedesign will assumeassumes that the maximum height of the water from the center of the entrance tank orifice's width, {latex}
\large
$$
HL_{EtMax}
$${latex} is 20cm. SinceThe weorifice arecoefficient dealingdue withto ana orifice,vena we will also assume that the cross sectional area of the water, as it passes contracta through the orifice, is reduced by a factor, {latex}
\large
$$
Pi_{VenaContractaOrifice}
$${latex} ofis 0.62. Finally, we will assume that the smallest width, {latex}
\large
$$
W_{EtOrifice}
$${latex}, that is practical to build given that the tank will be built out of cement, is 4cm.
AssumingIt was assumed that major headloss is negligible at the upper level of the entrance tank. Therefore, we used 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 {latex}
\large
$$
A_{EtOrificeVenaContracta} = {{Q_{Plant} } \over {\sqrt {2gHL_{EtMax} } }}
$${latex}. 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 {latex}
\large
$$
A_{EtOrifice} = {{A_{EtOrificeVenaContracta} } \over {Pi_{VenaContractaOrifice} }}
$${latex}. Given the width constraint we can calculate the length of the orifice by diving the area of the orifice by the width {latex}
\large
$$
L_{trueET} = {{A_{EtOrifice} } \over {W_{EtOrifice} }}
$$ {latex}.
h3. Upper Channel
The upper channel was modeled as an open channel flow conduit. The channel was modeled as though water entered from theits end of the channel and flowed intoward the hydraulic directiondrop ofinto the hydraulic dropflocculator. This is incorrecta waterconservative willassumption, bewater entering throughenters the entrancehorizontal tankchannel orifice onalong one side. ThisThe approximationlength shouldof bethis conservationchannel thoughis becausedetermined itby assumesthe thatlength thereof isthe moreentrance watertank inorifice; the endchannel ofmust thebe channellonger than therethis actually isorifice. TheThis flowapproximation atis theconservative topbecause ofit theassumes hydraulicthat dropmore iswater criticalenters meaningthe thatchannel Froudeat numberfirst isthan oneactually anddoes. thatThe potential energy and kinetic energy are equal. The length of the channel must be greater than the entrance tank orifice length.
The height of height of water along the channel was iteratively solved for used the direct step method described below.
The heightflow ofat waterthe righttop beforeof the hydraulic drop wasis solved critical- potential energy and kinetic energy are in balance and the height of water in the tank can be solved for using the equation for critical depth.:
{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 to gravity
The width of the channel was chosen to minimize the critical depth and lower hydraulic drop height (described below). The width used is 40 cm.
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:
{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 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
TheFinally 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 changeof inthe lengthchannel, given the calculated friction slope where the depth of, that corresponds to a change in water increaseddepth byof 0.0001 cm. This process was repeated forover the entire length of the channel until the height of water in the channel was found. Theuntil the height of water inat the channel includingentrance the critical depth is 13.166 cm.
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 isbe directly related to the mixing length. and Itit iswas assumed that a minor loss coefficient of 1.3 is adequate for large scale eddies to evenly mix the alum. The critical height at the top of the hydraulic drop and channel width are used to solve for critical velocity at the top of the dropvelocity 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 waterfall 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 height of hydraulic drop deposits water into the hydraulicfirst drop is 7.592 cm.
h3. Lower Channel
baffle section. The water level in thethis lower channelsection 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.
{latex}
\large
$$
h_{LowerChannel} = h_{LowerDrop} + h_{EntryOrifice} + h_{flocculator}
$${latex}
The height of the lower channel is 2.36 m.
h3. Flocculator Entrance Orifice
The design of the orifice that leads into the flocculator must provide the necessaryenough energy dissipation to achieve the requiredfor molecular diffusion to ofevenly thespread alum in the water. Therefore, Energyenergy 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}
The minimum energy dissipation is calculated to be approximately to be 0.733 W/kg. We assume a minimum energy dissipation of 0.8 W/kg iswas requiredused atas thea flocculationconservative entranceestimate orificeof toenergy bedissipation safein duringthe ourflocculator designentrance processorifice.
{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 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.
{latex}
\large
$$
V = \left( {{{2\varepsilon \sqrt Q } \over {K_{\exp } }}} \right)^{{2 \over 7}}
$${latex}
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.
h3. 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. This time should be adequate to mix the required amount of alum in the other tank so that there is always at least one tank with alum flowing through it. 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:
{include:Q.FcmMax}
. 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. | 