h1.
Automated Materials List
The goal of the Automated Materials List is to calculate the volume of concrete needed, areas of components that will be made out of bricks (these areas can be divided by the area of one brick to see how many bricks are needed), lengths and sizes of pipes, amount of plastic sheets for the lamella, etc. The final Materials List should act as a rough outline for on-site construction and facilitate the job of the engineers and planners. This program requires inputs from the [user|https://confluence.cornell.edu/display/AGUACLARA/User+Inputs+Design+Program] and from the [design assumptions|https://confluence.cornell.edu/display/AGUACLARA/Design+Assumptions+Design+Program] to compute the necessary calculations. Currently, the List calculates the total materials needed for construction and various dimensions for different components of the plant.
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The output of the [Design Tool|AGUACLARA:AguaClara Design Tool] now includes a list of several variables comprising the Materials List. Engineers designing plants will now be able see how much material is needed to construct the components of the plant, including the walls of the various tanks and channels, lengths of PVC pipe, corrugated plastic sheets for the lamella, and ferrous cement for the floc baffles. The Materials List is conveniently located at the bottom of the Design Values excel spreadsheet that is returned to the user with all the outputs for the plant. This will allow them to have a better of idea of what is needed to construct an AguaClara plant. The Materials List will continue to be updated based on feedback from the engineers in Honduras so that it provides information that is most useful to them in the construction process.
h2. Materials List Program Algorithm
[Automated Materials List Program Inputs|Automated Materials Program Inputs]
[Automated Materials List Program Outputs]
h3. Algorithm
h5.
The first step in calculating the materials was to determine the geometry of the different components of the plant and which parts of the plant were constructed out of which types of materials (e.g. concrete, ferrous cement, corrugated plastic, PVC pipe). Then the volumes and quantities were calculated from the dimensions used in each components design algorithm.
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h5. Entrance Tank
The Entrance tank volume was found using the design specifications from the automated entrance tank design program and the user inputs. The tank's volume is calculated by considering the four walls lengths, widths and thicknesses. The algorithm also accounts for the entrance to\\ !Eqn207.gif!\\ !Eqn208.gif!\\ !Eqn209.gif!\\ !Eqn210.gif!\\ !Eqn211.gif!\\
h5. Flocculation Tank
The total volume for the flocculation tank bydepends subtractingon the volume of the channel from onewidth and total number of floc channels and the length of the entrancesedimentation tank walls.
!Eqn101.gif|align=left!
Below is a diagram of This information, along with the Entranceuser Tankinputs asand seendesign fromassumptions, above.provides Noteall that the height,necessary notinformation shownto incalculation the diagram, is H.Ettank's volume.
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\\ !floc tank.PNG!\\
The total volume areaof concrete of the walls of the entranceflocculation tank can be found by adding up the areasdepends on the length of the individualflocculation wallstank, andthe consequentlythickness subtractingof the areaplant wall, the height of the inletflocculation channeltank, andthe thenumber floorof areaflocculation canchannels, beand foundthe bywidth multiplyingof the effectiveflocculation widthchannels. andSince lengthall of the entrance tank:
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!Eqn10.gif!
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Finally, volume of the entrance tank floor is simply the product of the area andwalls in the flocculation tank are the same thickness, the volume can be easily calculated by multiplying the area of the floc walls by the thickness of the entranceplant tank floor:
\\ !Eqn3.gifwall (T.PlantWall).
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!Eqn27.gif!\\
!Eqn32.gif|align=left!\\
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h5.The Flocculation Tank
The total volume forarea of the floor of the flocculation tank depends on the width and total numberlength of flocthe channelsflocculation andtank, the lengththickness of the sedimentationplant tank.wall, Thisthe information,number alongof with the user inputsflocculation channels, and design assumptions, provides all the necessary information to calculationwidth of the tank'sflocculation volume.
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\\ !floc tank.PNG!\\
The total volume of concretechannels. The volume of the wallsfloor of the flocculation tank dependsis onsimply the lengthproduct of the flocculation tank, thearea and thickness of the plant wall, the heightfloor:
\\ !Eqn25.gif! !Eqn45.gif|align=left!\\
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The volume of the flocculation tank,floc baffles is dependent on the numberwidth of flocculationthe channels, and the widththickness of the flocculationbaffle, channels.the Sincelength all of the wallsbaffle, inand the number of flocculationbaffles tankper arechannel. theThe samenumber thickness,of thebaffles volumeper canchannel beis easilya calculated byvariable based multiplyingon the arealength of the flocflocculation wallstank byand the perpendicular thicknessspacing ofbetween thebaffles, plantwhich wall (T.PlantWall).
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!Eqn27.gif!\\
!Eqn32.gif|align=left!\\
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The area of the floor of the flocculation tank depends on the length of the flocculation tank, the thickness of the plant wall, the number of flocculation channels, and the width of the flocculation channels. The volume of the floor of the flocculation tank is simply the product of the area and thickness of the floor:
\\ !Eqn25.gif! !Eqn45.gif|align=left!\\
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The volume of the floc baffles is dependent on the width of the channels, the thickness of the baffle, the length of the baffle, and the differs per channel. The baffles are also split between upper and lower baffles. Since the floc channel begins with a lower baffle, the equation is set-up so an odd number of baffles results in an extra lower baffle. If there are an X number of floc channels in a floc tank, the N.FlocChannelBaffles, L.FlocBaffleUpper, and L.FlocBaffleLower are X value arrays, which allow us to use a simple dot product to find the total length of the tank's baffles, rather than calculating each channel individually.
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To see how the number of baffles per channel. Theis numberdetermined ofand baffleswhy perthere channelare isupper aand calculatedlower variablebaffles, basedplease onconsult the lengthflocculator ofalgorithm. theThe flocculationvolume tankneeded and(of theplastic perpendicularor spacingconcrete, betweendepending baffles,on whichthe differssize perof channel.the Theplant) bafflesfor arethe alsothe splitbaffles betweenis upperunder andthe lowerdiagram baffles. Since the floc channel begins with a lower baffle, the equation is set-up so an odd number of baffles results in an extra lower baffle. If there are an X number of floc channels in a floc tank, the N.FlocChannelBaffles, L.FlocBaffleUpper, and L.FlocBaffleLower are X value arrays, which allow us to use a simple dot product to find the total length of the tank's baffles, rather than calculating each channel individually.
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To see how the number of baffles per channel is determined and why there are upper and lower baffles, please consult the flocculator algorithm. The volume needed (of plastic or concrete, depending on the size of the plant) for the the baffles is under the diagram below.
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{latex}
\large
$$
Vol_{FlocBafflesTotal} = W_{FlocChannel} \cdot T_{FlocBaffle} \left[ {L_{FlocBaffleLower} \left( {ceil\left( {{{N_{FlocChannelBaffles} } \over 2}} \right)} \right) \cdot L_{FlocBaffleUpper} \left( {floor\left( {{{N_{FlocChannelBaffles} } \over 2}} \right)} \right)} \right]
$$
{latex}
The totalbelow.
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{latex}
\large
$$
Vol_{FlocBafflesTotal} = W_{FlocChannel} \cdot T_{FlocBaffle} \left[ {L_{FlocBaffleLower} \left( {ceil\left( {{{N_{FlocChannelBaffles} } \over 2}} \right)} \right) \cdot L_{FlocBaffleUpper} \left( {floor\left( {{{N_{FlocChannelBaffles} } \over 2}} \right)} \right)} \right]
$$
{latex}
The total surface area of the flocculation baffles can be found be dividing the total volume of all the flocculation baffles by the thickness of one flocculation baffle.
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h5. Inlet and Exit Channels
The volumes of the exit and inlet channels are derived from the user-input dimensions for each component. Using the widths, lengths, and thicknesses of each wall, the volume is computed by direct multiplication. The inlet channel also considers the dimensions of the sedimentation manifold entrance and the number of sedimentation inlet pipes. The quantity that is removed from the inlet channel due to the sedimentation inlet pipes is the volume of a single inlet pipe multiplied by the number of pipes.
\\ !34534312123115.PNG|align=left!\\
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The wall and floor volumes of the inlet and exit channels similarly depends on their respective length, width, height, and thickness:
!Eqn30.gif!
!Eqn51.gif!\\
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!Eqn50.gif!\\
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The surface area of the flocculationwalls bafflesand canthe be found be dividingfloor of the totalinlet volumeand ofexit allchannels thealso flocculationdepend baffles byon the thickness of one flocculation baffle.respective width, length, height, and thickness:
\\ !Eqn12.gif!
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!Eqn52.gif!\\
!Eqn14.gif!\\
h5. Inlet and Exit Channels!Eqn53.gif!
The volumesinlet of theand exit and inlet channels are derivedattached fromto thetanks user-inputcontaining dimensionsthe forweirs eachthat component.connect Usingto the widths, lengths, and thicknesses of each wall, the volume is computed by direct multiplication. The inlet channel also considers the dimensions of the sedimentation manifold entrance and the number of sedimentation inlet pipes. The quantity that is removed from the inlet channel due to the sedimentation inlet pipes is the volume of a single inlet pipe multiplied by the number of pipes.
\\ !34534312123115.PNG|align=left!\\
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The wall and floor volumes of the inlet and exit channels similarly depends on their respective length, width, height, and thickness:
!Eqn30.gif!
!Eqn51.gif!\\
\\ !Eqn28.gif!
!Eqn50.gif!\\
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The surface area of the walls and the floor of the inlet and exit channels also depend on the respective width, length, height, and thickness:
\\ !Eqn12.gif!
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!Eqn52.gif!\\
!Eqn14.gif!\\
!Eqn53.gif!
The inlet and exit channels are attached to tanks containing the weirs that connect to the sedimentation tank. Since the plant weirs are being recoded, these equations will change in the future, but for now, these dimensions were calculated to be built around the nominal diameter of the plant weir and the spacing required between the elbows.
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The total length of weir pipes in the inlet and exit channel tanks is derived from the user-input values that are found upon calculating the necessary height of the pipe for the determined water velocity. This height multiplied by the number of weir pipes yields the total length of pipe necessary.
{latex}
\large
$$
L_{Weir} = 2 \cdot H_{PltWeir}
$$
{latex}
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The inlet channel is also attached to drop chimneys that run to the sedimentation tank. These chimneys are rectangular blocks of concrete molded around a pipe, to leave a hole for the water to flow through. We first calculated the volume of this pipe based on it's diameter and the height of the chimney. The height of the chimney is equal to the height of the sedimentation tank minus the inlet channel and wall and also accounting for the space at the bottom of the sedimentation tank for the sludge drain.\\
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h5. Sedimentation Tank
The volume of the sedimentation tank depends on user-input values. The dimensions of the sludge drain must then be subtracted from the overall volume to account for the design of the drainage system.
\\ !tank2131234124.PNG!
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The area of the floor of the sedimentation tank depends on the length and width of the sedimentation tanks as well as the number of sedimentation tanks. The volume of the floor of the sedimentation tank is simply the product of the area and thickness of the floor. There has been some question as to whether the thickness of the inner walls of the sedimentation tank (T.PltWall) and that of the outer walls (T.PlantWall) is the same. Assuming they are not the same, the equations for the area and volume of the floor of the sedimentation tank are as follows:
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!Eqn44.gif|align=left!\\
!Eqn54.gif!
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!Eqn55.gif|align=left!\\
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The volume of the sedimentation inlet slopes is derived from the user input dimensions for the sedimentation slope plate and the slope manifold. There is one inlet slope per sedimentation port.
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h1.
The total number of launders is determined by the number of launders per sed bay and the total number of sed bays.
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!Eqn24.gif!
The total length of launder pipe needed is dependent on the individual length of one launder and the total number of launders.
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The size of the PVC pipe will depend on the user-input ND.SedLaunder.
!89.gif!
The launder also requires a coupling to go through the concrete wall of the sedimentation tank. The number of couplings needed is equal to the number of launders in the plant. !90.gif!
h1. !launders.PNG|width=603,height=294!
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The number of valves and valve couplings needed to drain the tank are found from the sedimentation design specifications with one valve per sedimentation bay.
\\ !91.gif! !92.gif!
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The necessary output to construct the plate settlers is the number of corrugated sheets used. This is determined by calculating the total number of plate settlers, how many plate settlers would fit on a given sheet of plastic, and and dividing to find the total number of required sheets.
{latex}
\large
$$
N_{SedPlatesTotal} = N_{SedPlates} \cdot N_{SedBays} \cdot N_{SedTanks}
$$
{latex}
{latex}
\large
$$
N_{SedPlateSheets} = ceil\left( {{{N_{SedPlatesTotal} } \over {N_{SedPlatesPerSheet} }}} \right)
$$
{latex}
The plate settlers are supported by a sedimentation plate frame constructed of PVC pipes running across the width and length of the sedimentation tank. The length of PVC pipe required to construct this module was determining using the dimensions of the sedimentation tank and the center-to-center distance between each parallel pipe.
{latex}
\large
$$
L_{PipeSedPLateFrameTotal} = N_{SedTanks} \left[ {\left( {{{W_{Sed} } \over {B_{SedPlateFramePipes} }} - 1} \right) \cdot L_{Sed} + \left( {{{L_{Sed} } \over {B_{SedPLateFramePipes} }} - 1} \right) \cdot W_{Sed} } \right]
$$
{latex}sedimentation tank. Since the plant weirs are being recoded, these equations will change in the future, but for now, these dimensions were calculated to be built around the nominal diameter of the plant weir and the spacing required between the elbows.
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The total length of weir pipes in the inlet and exit channel tanks is derived from the user-input values that are found upon calculating the necessary height of the pipe for the determined water velocity. This height multiplied by the number of weir pipes yields the total length of pipe necessary.
{latex}
\large
$$
L_{Weir} = 2 \cdot H_{PltWeir}
$$
{latex}
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h5. Sedimentation Tank
The volume of the sedimentation tank depends on user-input values. The dimensions of the sludge drain must then be subtracted from the overall volume to account for the design of the drainage system.
\\ !tank2131234124.PNG!
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The area of the floor of the sedimentation tank depends on the length and width of the sedimentation tanks as well as the number of sedimentation tanks. The volume of the floor of the sedimentation tank is simply the product of the area and thickness of the floor. There has been some question as to whether the thickness of the inner walls of the sedimentation tank (T.PltWall) and that of the outer walls (T.PlantWall) is the same. Assuming they are not the same, the equations for the area and volume of the floor of the sedimentation tank are as follows:
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!Eqn44.gif|align=left!\\
!Eqn54.gif!
\\ !eqn46.gif|align=left!\\
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!Eqn55.gif|align=left!\\
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The area and volume of the sedimentation slopes is determined as follows:\\ !Eqn201.gif!
!Eqn202.gif!
!Eqn203.gif!
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\\ !126456456456.PNG|align=left,width=713,height=252!\\
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h1.
The total number of launders is determined by the number of launders per sed bay and the total number of sed bays.
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!Eqn24.gif!
The total length of launder pipe needed is dependent on the individual length of one launder and the total number of launders.
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The size of the PVC pipe will depend on the user-input ND.SedLaunder.
!89.gif!
The launder also requires a coupling to go through the concrete wall of the sedimentation tank. The number of couplings needed is equal to the number of launders in the plant. !90.gif!
h1. !launders.PNG|width=603,height=294!
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The number of valves and valve couplings needed to drain the tank are found from the sedimentation design specifications with one valve per sedimentation bay.
\\ !91.gif! !92.gif!
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The necessary output to construct the plate settlers is the number of corrugated sheets used. This is determined by calculating the total number of plate settlers, how many plate settlers would fit on a given sheet of plastic, and and dividing to find the total number of required sheets.\\ !Eqn204.gif!\\ !Eqn205.gif!\\ !Eqn206.gif!\\
The plate settlers are supported by a sedimentation plate frame constructed of PVC pipes running across the width and length of the sedimentation tank. The length of PVC pipe required to construct this module was determining using the dimensions of the sedimentation tank and the center-to-center distance between each parallel pipe.
!sdfhdfg.PNG!
!Lsed.PNG!
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h5. Stock Tank
The function of the stock tank to add the alum coagulant to the raw water during rapid mix. The volume of the stock tank is constrained by flow rate and retention time. More specifically, the volume of the stock tank is the maximum flow rate through the aluminum stock tank multiplied by the duration of alum stock tank at maximum alum dose and plant flow rate.
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\\ !Eqn19.gif!
The only other variable that needs be specified by the user to determine the other dimensions of the stock tank is diameter of the stock tank. Once the diameter has been set, the radius is simply half the diameter, and the corresponding required height can subsequently be calculated.
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!Eqn20.gif!
!Eqn21.gif!
Finally, the total area and the floor area of the stock tank are determined by the previously calculated height and radius:
\\ !Eqn22.gif!
!Eqn23.gif! |
