Gracias Entrance Tank Design

Abstract

As the flow rate through an AguaClara plant is increased, the plants design will need to be altered. More specifically, to achieve the required amount of energy dissipation for adequate alum mixing, we would need to redesign the entire entrance tank. Since the flow is so much greater than that of previous plants, the orifice exiting the entrance tank will need to be much larger. This presents a problem as the alum orifice will not be as large, and thus we would require a greater amount of mixing. This problem will be achieved by using two hydraulic drops.

Gracias Entrance Tank Design

Design Schematic

Rapid Mix Chamber

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 initially enters the horizontal channel through a rectangular orifice. This rectangular orifice will help provide the necessary energy dissipation to create the large eddies that are necessary for global mixing. The water that enters this rectangular orifice will experience free fall into the horizontal channel. Then, another waterfall occurs when the water flows from the horizontal channel into the vertical channel. The water that flows into the vertical channel will be in free fall. The water will experience will experience a contraction at the rectangular orifice that leads into the flocculation tank. This will allow for the necessary energy dissipation to achieve the molecular diffusion of the alum.

Design Assumptions and Specifications

The majority of the assumptions made for this design were the same as in past designs.

Entrance Tank Dimensions

When designing the entrance tank, we will assume that the upflow velocity is 700 m/day. This design constraint is based on the work of other subteams, as it is approximately the velocity used to backwash a filter. Using this assumption, along with the flow rate through the plant, it was determined that the tank should have a cross sectional area of 6.171 m ². If we then assume that the tank has square dimensions, we find that the length of one of the sides of the tank is 2.484 m.

Entrance Tank Orifice

For the design of the entrance tank orifice, we will assume that the maximum height of the water from the center of the entrance tank orifice's width,

is 20cm. Since we are dealing with an orifice, we will also assume that the cross sectional area of the water, as it passes through the orifice, is reduced by a factor,

of 0.62. Finally, we will assume that the smallest width,

, that is practical to build given that the tank will be built out of cement, is 4cm.

Assuming major headloss is negligible at the upper level of the entrance tank, we used the minor headloss equation along with the orifice flow equation 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

. 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

. Given the width constraint we can calculate the length of the orifice by diving the area of the orifice by the width

.

Upper Channel

The upper channel was modeled as an open channel flow conduit. The channel was modeled as though water entered from the end of the channel and flowed in the direction of the hydraulic drop. This is incorrect water will be entering through the entrance tank orifice on one side. This approximation should be conservation though because it assumes that there is more water in the end of the channel than there actually is. The flow at the top of the hydraulic drop is critical meaning that Froude number is one and that potential energy and kinetic energy are equal. The length of the channel must be greater than the entrance tank orifice length.
The height of water along the channel was iteratively solved for used the direct step method described below.
The height of water right before the hydraulic drop was solved for using the equation for critical depth.

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.
The energy equation for open channel flow can be rearranged to give the relationship:

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:

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

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 change in length given the friction slope where the depth of water increased by 0.0001 cm. This process was repeated for the entire length of the channel until the height of water in the channel was found. The height of water in the channel including the critical depth is 13.166 cm.

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 mixing. The minor loss coefficient is directly related to the mixing length. It is 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 drop. The equation for minor loss is then used to determine the height of the waterfall:

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

The height of the hydraulic drop is 7.592 cm.

Lower Channel

The water level in the lower channel is determined by the water level in the flocculator and the headloss through the flocculator entrance orifice. The height of the channel includes the water level and the lower hydraulic drop height.

The height of the lower channel is 2.36 m.

Flocculator Entrance Orifice

The design of the orifice that leads into the flocculator must provide the necessary energy dissipation to achieve the required molecular diffusion of the alum in the water. 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.

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 is required at the flocculation entrance orifice to be safe during our design process.

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.

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

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:

. 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.