Research About Hydrology

According to Google Maps, the lake has an estimated size of 5325.81m2, Terry's pond has an area of 961.53m2, and the stream has an area of 5.95m2. The entire drainage area is 6293.29m2 (about 1.56 acres).

2-hour rainfall over a 25-year period = 18mm = 0.71 inches

Assuming that the runoff coefficient remains constant during the storm and that rainfall is evenly distributed across the drainage region (Elsebaie, 2012), the modified rational technique is used to compute the peak runoff, Q:

Q= CIA

C= 0.25 Average I = 0.71 inches/ 2 hours = 0.355 inches/hour Area of drainage = 1.56 acres Tc = 15 minutes Storm duration= 2 hours Q = runoff rate (cubic feet per second) = CIA

C = Rational Method runoff coefficient; I = rainfall intensity (inches per hour); A = drainage area (acres); D = storm duration (hours)
Q= 0.25 × 0.355 inches/ hour × 1.56 acres = 0.13845 CFS
The runoff volume, V, is given by:
V = peak runoff × storm duration = Q × D
V = 0.13845 cubic feet per second × 2 hours × 3600 sec/ hour = 996.84 cubic feet = 0.0229 acre
feet
Annual water balance is original volume ± change in volume (MA, 2014)
∆〈S〉 = 〈P〉 - 〈ET〉 - 〈Q〉 + 〈R〉
∆〈S〉 = 〈P〉 - 〈ET〉 - 〈Q〉 + 996.84 cubic feet

Part B - Environmental Engineering

Water treatment is done to make water palatable and potable irrespective of its use. The best technique to apply in this case is a combination of sediment filtration and chlorine disinfection. Its steps are as follows: screening, coagulation, flocculation, sedimentation, filtration, disinfection, storage and finally distribution (Mohammad & Hankins, 2014).

Surface water supply Coagulation flocculation settling tank filtration & disinfection treated water distribution To find the dimensions of the sediment tank:
Assuming the university needs 500000litres = 500m3, detention time = 4 hours, water depth = 4m, then
Quantity of water to be treated in a day = 500m3/day = 20.83m3/ hour
Volume of the tank = Q × detention time = 20.83 × 4 = 83.33m3
Horizontal flow velocity = 0.15m/min
Required tank length = 0.15 × 4 × 60 = 36m
Tank's cross section area = 83.33/36 = 2.31m2
Depth of tank = 4m
Sludge depth = 0.8m
Water depth = 4 - 0.8 = 3.2m
Thus, the width of the tank is 2.31/ 3.2 = 0.8m
Overflow rate = 500000 / (36 x 0.8 x 24) = 723.38 litres/hour/m3
The dimensions of the sedimentation tank are: 36m x 3.2m x 0.8m
To find the area of slow sand filter:
Maximum daily demand = 500000litres
Assuming the slow sand filters with an infiltration rate of 100litres/hour/m2 are being used, then the required filter area will be:
500000/ (100 x 24) = 210m2. This can be subdivided into 3 units each being 70m2 (7m by 10m).
Slow sand filters are advantageous over rapid sand filters because they don't require skilled supervision, they have low depreciation and they are very efficient in turbid and color removal. Its drawback is that it has high initial deployment cost.
To design the under-drainage system:
Under drainage system is made up of laterals and manifolds.
Assume total area of perforations in the under drainage to be 0.3% of the area of filters
0.3/100 x 7 x 10 = 0.21m2 = 2100cm2
Area of laterals = 2 x area of perforations = 2 x 0.21m2 = 0.42m2
Area of manifold = 2 x 0.42 = 0.84m2
If lateral spacing is 10cm, then
Number of laterals = 10 x 100 / 10 = 100 laterals
Total laterals = 100 x 2 = 200
Length of each lateral = (7 - 0.55)/2 = 3.225m
If each perforation is 12mm, total number of perforations = 1857/20 = 10
Area of perforation per lateral = 10 x π x (1.2)2 / 4 =11 cm2
11 x 2 = 22cm2
Dia of the lateral = (22.x 4÷ π )1/2 = 5.4cm
Hence 200 laterals each measuring 10cm by 5.4cm and having 10 12mm dia are needed.
Length of lateral/ dia of lateral = 3.225 x 100 ÷ 5.4 = 59. 59 < 60 hence the design is ok.
The volume of water used by the university is high. This filtration and chlorine disinfection method is advantageous because it is economical in treatment of large volumes of water. The use of chlorine to kill bacteria is also cost-effective as compared to other methods such as boiling. At the same time, once the treatment system is set up, very minimal human involvement is required to keep the process going. Additionally, slow sand filters used in this technique are very efficient in removal any coloring in water.
The drawback of this technique however is that it is expensive to build. Moreover, the chlorine that is used in disinfection leaves a residue. Intake of this chlorine residue over a long period of time is detrimental to human health. Although effective in killing bacteria, the chlorine doesn't kill protozoa that is in water and thus the university could have a protozoan outbreak if the water is ever contaminated with protozoa. Also, the slow sand filters that are used in this design require a lot of space and land for the biological reliant process to be seamless.

Disinfection

Chlorination is the last stage just before distribution, the residual chlorine should be less than 2.0mg/L almost immediately after it has been added to water. Since water is also being used for drinking, the smaller the residual chlorine the more natural the water will taste. Sample 1 with residual chlorine of 0.14mg/L is therefore best for this design.

References

Elsebaie, I. (2012). Developing rainfall intensity-duration-frequency relationship for two regions in Saudi Arabia. Journal Of King Saud University - Engineering Sciences, 24(2), 131-140. http://dx.doi.org/10.1016/j.jksues.2011.06.001

MA, Z. (2014). Analysis of Short Duration Rainfall Intensity Data of Makoran Region-Iran. Irrigation & Drainage Systems Engineering, 3(2). http://dx.doi.org/10.4172/2168-9768.1000123

Mohammad, A., & Hankins, N. (2014). The Journal of Water Process Engineering. Journal Of Water Process Engineering, 1, 1. http://dx.doi.org/10.1016/j.jwpe.2014.04.005