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The orifice between the rapid mix and flocculation tanks is designed to produce a difference in water level high that can then be sensed by a float which would then change the flow rate of aluminum sulfate:

{latex}$$
\Delta H = K_{_{orifice} } {{V_{jet} ^2 } \over {2*g}}
$$
{latex}

where

• Latex
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  {latex}$$ \Delta H $${latex}

  is the difference in head loss between the rapid mix and flocculation tank
• K _orifice is the required minor loss coefficient through the orifice
• V _jet is the velocity in the dosing tube

This head loss was then used to determine the velocity of the water through the orifice and the residence time. Using the following equations:

Velocity of Jets:

{latex}
$$V_{jet} = {Q \over {C_d *A_{orifice} }}
$${latex}

where

• V _jet is the velocity of the jet
• Q is the flow rate through the system
• C _d is the vena contracta coefficient for exit condition in orifice
• A _orifice is the area of the orifice

Residence time :

{latex}$$
\theta = {{d_{orifice} } \over {V_{jet} }}
$${latex}

where

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  {latex} $$ \theta $$ {latex}

  is the residence time

• d _orifice is the diameter of the orifice
• V _jet is the velocity of the jet

Once these values were determined, we were able to calculate the energy dissipation rate using the following equation:

{latex}$$
\varepsilon = {{g*\Delta H} \over \theta }
$${latex}

where

• Latex
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  {latex}$$ \varepsilon $${latex}

  is the energy dissipation rate
• g is gravity

• Latex
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  {latex}$$ \Delta H $${latex}

  is the head loss

• Latex
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  {latex}$$ \theta $${latex}

  is the residence time

We sought to keep the energy dissipation rate between .5 and 1 W/kg so that molecular scale diffusion works and in order for small scale turbulent mixing to be effective.

...

We first must determine the size of the counterweight on the doser arm in order to ensure that the dosage will only be a function of the difference in water height in the flocculation and rapid mix tanks. The mass of the weight is calculated by determining the mass of the doser when full.

{latex}$$
m_{doserful} = [({{.375in} \over 2})^2 *\pi *25cm + ({{D_{actual} } \over 2})^2 *\pi *1m]\rho _{water} + m_{doser}
$${latex}

where

• D _actual is the difference between the given diameter of the dosing tube and the measured diameter of the dosing tube

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  {latex}$$ \rho _{water} $${latex}

  is the density of water

• Latex
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  {latex}$$ m_{doser} $${latex}

  is the mass of the doser empty

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