Drinking-Water Plant Design in El Paso, TX
Posted: August 27th, 2021
Drinking-Water Plant Design in El Paso, TX
Executive Summary
Designing a drinking water plant in El Paso, Texas based on the informationon the place’s population density. The population density of the Elpaso is relevant in estimating the design steps coupled with systematic calculationsinvolving each unit of the plant. Similarly, the applied design illustrations are useful in determining the flow rates based on the demand for water among the inhabitants of Elpaso. The paper’s objective entails evaluating the water demand by presenting the design steps coupled with calculations for each unit of the treatment plant. Categorically, the design unit of treatment plant processes comprises preliminary treatment and sedimentation, rapid mix, flocculation, granular media filtration, disinfection, and pumping. The water flow rates and the population of the place are6,200 cubicmetersper second and 951,000 people, respectively. However, the water flow rate is sufficient at 6,520 cubic meters per second in regions around Rio Grande Basin along the Gulf of Mexico. With assigned calculations of the parameters, there is a tabulation of data from the Elpaso field.
Background and Scope of the Work
With the official development and launching of this WTP in 1996, the Texas Water Development Board has initiated some program plans meant to determine the water plant’s economical use. Notably, the plans have featured collection and water treatment measures to meet the present demand for water from the increased population (Hutchison 11). Therefore, the whole design units and processes have been re-developed to ensure that the system is more effective in meeting the increased population size requirements based on flow rates in 2020. Water treatment units and processes involve the design of sources of surface water. Specifically, water treatment plant (WTP) involves all the necessary steps from pre-sedimentation, coagulation, adsorption, filtration, disinfection, storage, and pumping the already treated water for consumption(Aziz and Mustafa 803). Indeed, the WTP proposal involves an average population size of 951,000 inhabitants who would possibly consume a per capita water use rate of 130 gallons per day (Aziz and Mustafa 805). Therefore, both the minimum and maximum discharge flow rates are estimated at 40% and 180%, respectively, as follows.
Average discharge (Q. Ave) = 200,000 x 300 LPCD = 60,000,000 L/day = 60,000m3/day = 0.694m3/s.
Minimum discharge (Q min) = 60,000 x 0.4 = 24,000 m3/day = 0.278 m3/s.
Maximum discharge (Q max) = 60,000 x 1.8 = 108,000 m3/day = 1.25 m3/s.
Notably, this WTP system’s regular daily flow rate regards an approximate per capita production rate of wastewater at 91.2 GMCD and 4 people per household(Aziz and Mustafa 807). The system is planned to pass peak flows that seem 4 times the approximate flow rates per day. Therefore, the treatment plant has a peaking factor of 4, which corresponds to the 2-hour adjusted peak flow illustrated below (Table 2).
| Flow Parameters MGD = a million gallons per day | 2020 |
| Average Daily Flow (MGD) | 0.59 |
| Treatment Plant Design Flow (MGD) | 0.73 |
| Peak flow (MGD) | 2.37 |
The suggested WTP collection system is meant to serve both the original and extended project areas encompassing a gravity sewer of about 78,087 LF. The design’s size varies from a diameter of 6 to 18 inches, together with a 500 LF diameter force main of about 12 inches. Similarly, the proposed wastewater treatment plant has a capacity of approximately 734,000 gallons per day, coupled with extended aeration and oxidation ditch type regarding the water treatment system(Aziz and Mustafa 807). Subsequently, the water facility involves an influent pump station, screenings and grit removal subsections, a broad aeration facility, a secondary clarifier, chlorination facilities, sludge drying beds, and a 200 feet pumping facility to Tornillo Drain.
Plant Summary and Plan
Layout
The suggestedWTP has the following summary plan and schematic diagram
| No. | Unit description | No. of | Shape of unit | Dimensions of units | Notes | ||
| unit | |||||||
| The diameter of suction pipe=0.4m | |||||||
| 1 | Intake | 4 | circular | Diameter=5.15m | The diameter of the raw water gravity pipe | ||
| Depth=10m | = 0.5m | ||||||
| Diameter=3m | G=300s-1 | ||||||
| 2 | Coagulation | 2 | circular | P=21KW | |||
| Depth=3m | |||||||
| t= 1min | |||||||
| Width=4.17m | G=20, 40, and 60 s-1 | ||||||
| 3 | Flocculation | 6 | Rectangle | Length=3*4.17=12.51m | P=84, 336..02, and 756.04 watt | ||
| Depth=4m | t= 30min | ||||||
| Width=4.17m Length=3*4.17=12.51m | Vs=0.3594mm/s | ||||||
| 4 | Clarification | 4 | rectangle | Vc=64.7mm/s | |||
| Depth=4m | |||||||
| dp= 0.02mm | |||||||
| Width=4.5m | Filtration=6.94m/hr | ||||||
| 5 | Filtration | 10 | Rectangle | Length=8m | Rate= | ||
| Depth= 3m | Backwashing=124.92m/hr | ||||||
| Width=12.5m | Time=30min | ||||||
| 6 | Disinfection | 1 | Rectangle | Length=25m | |||
| Consumed chlorine=21.6kg/d | |||||||
| Depth= 4m | |||||||
| Storage | Q=2500m3/hr | ||||||
| 7 | 3pumps | V=1.57m/s | Dosage | ||||
| &pumping | |||||||
| Dpipe=0.75m | |||||||
Figure 1. Plain view or site layout of WTP in El Paso
Figure 2. Unit process flow of El Paso WTP
Process Design
Preliminary Treatment
The intake structure’s primary function is to help safely withdraw water from the Rio Grande snowmelt runoff in Southern Colorado as well as Northern New Mexico and discharge this water into the withdrawal conduit (typically called intake conduit) through which it flows up to a WTP [13,16-17]. The water is diverted through a raw water gravity pipe into the wet well (intake).
The average discharge (Q avg.) used in the design of the intake is described as follows:
Q avg. = 0.694 m3/s
When four pipes were used to convey raw water;
Q per one gravity pipe = 0.694 / 4 = 0.17 m3/s;
Velocity inside the gravity pipe = 1 m/s;
Area (A) = Discharge (Q) / velocity (v);
Diameter (D) of each raw water gravity pipe = 0.47–0.5 m;
No. of circular wells = 4;
Detention time (t) = 20 min.
Therefore, Q. average = 0.694 m3/s, whereas the Q. min = 10.41 m3/min
V = Q × t = 10.41 × 20 = 208.2 m3
Figure 2. Plan of the intake design
From figure 2, the design of the circular suction pipe is estimated at Q = 0.17m3/s and V = 1.5m/s. Therefore, the cross-section area of the suction pipe is obtained as follows.
A = Q/V = 0.17/1.5 = 0.06 m2
Figure 3. Plan of the wet wells
Based on the provided data, the high-water level (HWL) is 6.3 m, the LWL is 3.4 m, and the groundwater level is 290 m above the sea level. The following criteria are considered, including the total discharge (Q) = 0.694 m3/s and velocity through strainer (v) = 0.15 m/s
Preliminary Sedimentation
Rapid Mix or Coagulation
It is the process of adding a coagulant to water to destabilize colloidal suspensions and the steps of the design criteria of the coagulation tank as follows:
Q = 0.694 m3/s = 41.64 m3/min
Therefore, the discharge determination is estimated by employing two flash mixers at the rate of 41.64 m3/min for two flash mixers. Thus, the rapid mix rate is 20.82 m3/min for a singles mixer at the set time of t = 1 min.
Volume of flash mixer (V) = 20.82 m3/min × 1 min= 20.82 m3
Furthermore, with a tank depth of 3 m, obtaining the cross-sectional area is expressed as follows.’
A = V / d = 20.82 m3/ 3 m = 6.94 m2
Besides, establishing the value of each paddle requires the application of a powder formula as follows.
P = G2µV
Most importantly, it is easy to find the expression using the following mathematical expressions, especially for rapid mixing, P at G = 60/s.
P = (60)2 × (1.0087 × 10−3) × (208.2) = 756. 04 Watt For (G=40/s)
Medium mixing P = (40)2 × (1.0087 × 10−3) × (208.2) = 336.02 Watt For (G=20/s)
Slow mixing P = (20)2× (1.0087 × 10−3) × (208.2) = 84 Watt
Flocculation
Flocculation is the process of slow mixing that can be achieved in a basin known as a flocculator. It is an essential operation designed to agitate force in fluid and coagulation. The design criteria of the flocculation tank in that Q = 41.64 m3/min at t = 30 min.
V = 41.64 m3/min × 30 min = 1,249.2 m3
Moreover, there is employment of six flocculation parallel tanks as illustrated by the expression
= 1,249.2 / 6 = 208.2 m3
With diameter, d = 4 m
A = V / d = 208.2 / 4 = 52.05m2
Therefore, finding the dimension of a square tank is obtained by the following expression.
Area = width × length
Using L = 3 W
A=W×3W
52.05 = 3 W2
W = 4.17 m
Figure 4. Plan and side view of the flocculation tank
Sedimentation
Sedimentation is a process that involves the removal of solid particles by use of gravity. The design of a sedimentation tank entails obtaining Q as 0.694 m3/s in the WTP while considering the design of a single tank would involve the following expressions.
Q per tank = Q / 4 = 0.17 m3/s = 10.2 m3/min
However, where the detention time (t) equals 2 hours, the volume of the discharge at the sedimentation phase is expressed as follows.
Volume = discharge × time
V = 10.2 m3/min × 120 min
V = 1,224 m3
With the depth of sedimentation regarded as 4 m, the finding cross-section area of the sedimentation tank is obtained as expressed below.
Area = Volume / depth
A = 1,224 m3 / 4 m
A = 306 m2
Therefore, finding the shape of the circular sedimentation entails determining the velocity based on the particles’ dynamic viscosity, where Vs. is the terminal velocity of the solid particle in m/s. Whereas g is the gravitational acceleration in m/s2, Gs is the specific gravity of particles, and Gw is the specific gravity of water. Therefore, ds is the diameter of a particle in meter, and ʋ regarded as the dynamic viscosity of water in square meter per second.
Figure 5. Plan of the sedimentation
Granular Media Filtration
The filtration process intends to remove the deferred solids inside the sedimentation unit. Sometimes the removal of suspended particles might take a long time, especially outside the basin. Consequently, it is essential to design a rapid sand filtration tank.
Q avg. = 0.694 m3/s = 2,498.4 m3/h
Therefore, the filter bed flow: Q filter = 7 m3/h/m2, Area of the filter bed = 356.9 m2.Specifically, a total number of 10 filters were applied to obtain A as 357/10 = 35.7 m2. The width of 4.5 m of one filter unit was used. Therefore, to find length = 35.7 / 4.5 = 7.93 m. Checking filtration rate:
Total Area = 4.5 m × 8 m × 10 Nos.= 360 m2
Figure 6. Detail of one filter unit
The required size and the uniformity coefficient of the filter tanks are estimated at0.5 and 1.6correspondingly.However, the filter media’s suggested depth is 50 cm, implying that various supporting layers of about 20 and 30 cm as recommended values for the filter tank measurements (WHO 16). Furthermore, the head of the water above the filter media is considered 2 meters high, with the backwash process suggested for an average of 15 min. Therefore, the total number of minutes for backwashing is 30, while the filter run time is presumed 24 h.
Disinfection System
Since there might be bacteria or disease-causing micro-organisms in the filtered water, it is essential to disinfect the filter units. Carrying out disinfection would ensure the possibility of removing all the pathogens that might cause waterborne disease (Aziz and Mustafa 9). Therefore, disinfection encompasses a series of procedural methods.First, there is a granular activated carbon tank for re-carbonization after chemical clarification. Indeed, chlorine disinfection happens to stabilize the nutrients content of water, as illustrated by figure 7 below. Disinfection design also entails the use of UltravioletLight (WHO 16). In this case, the use of chlorine is reliable in disinfecting the water. The reason is that the applied method is inexpensive as it is easy for a person to handle it based on the required safety measures.
Figure 7. CT disinfection design details of WTP in El Paso
With the demand for water estimated at 60,000 m3 per day, the needed chlorine and residual chlorine are estimated at 0.36mg/L and 0.2 mg/L levels.
Chlorine demand = 0.36 mg/L – 0.2 mg/L = 0.16 mg/L
Consumed chlorine = 0.36 mg/L × (1/ 10 6) × 60,000 × 1000 = 21.6 kg/day
The time required to complete the disinfection performed in a storage tank is 0.5 h.
Q = Volume/time
Volume = Q × time
Volume = 60,000 m3 per day × (1/24) × 0.5 h = 1,250 m3
The use of an effective depth is 4 m with the length, L expressed as follows;
= 2× width (W)
A = 1,250 m3 / 4 m = 312.5 m2
L × W = 312.5 m2
2 W × W = 312.5 m2
W2 = 156.25 m2
W = 12.5 m, L = 2 × 12.5 = 25 m
Velocity = distance / time
Velocity = 25 m/0.5
Therefore, h = 50 m/h = 0.0139 m/s
Clearwell
With the completion of the final stages of the WTP process, water is distributed to the consumers or storage tanks using high lift pumps. Consequently, the treated water is used for drinking among households in Texas.
Figure 8. Plan
of disinfection and storage tank
Works Cited
Aziz, Shoukr, and Jwan Mustafa. “Step-By-Step Design and Calculations for Water Treatment Plant Units.” Advances in Environmental Biology, vol. 1, no (2019): pp. 802-811.
Hutchison, William. “Conceptual Evaluation of Surface Water Storage in El Paso County.” EPWU Hydrogeology, Report 08, 2008.
World Health Organization (WHO). “Portable Reuse – Guidance for Producing Safe Drinking Water.” 2017, pp. 1-131.
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