Water Disinfection: Methods, Design, and Management
Water Disinfection
Disinfection is the selective destruction or inactivation of pathogenic organisms. Sterilization, on the other hand, is the complete destruction of all organisms, which is not necessary in water treatment and is also quite expensive.
Methods of Water Disinfection
- Basic strategies: Keeping microbiota out of the water source.
- Physical methods:
- Ultraviolet (UV) rays
- Heat
- Chemical methods:
- Iodine
- Bromine
- Base
Factors Influencing Disinfection Ability
- Contact time
Disinfection Mechanisms
Disinfection mechanisms vary depending on the method used. For example, UV rays damage the DNA of microorganisms, while chlorine disrupts their metabolic processes.
Chlorine Demand and Residual
The chlorine demand curve on the graph was obtained for drinking water for a 1-hour contact time. Determine the daily amount of NaOCl to be applied to this water to produce a combined residual of 0.4 mg/L and a free residual of 0.5 mg/L after a contact time of 1 hour in a flow of 24,000 m3/d.
The dose of chlorine required to obtain a free residual of 0.5 mg/L is:
1.1 mg/L + 0.5 mg/L = 1.6 mg/L
The total residual is 0.58 mg/L at this dosage.
Due to the following reactions:
NaOCl → Na+ + OCl–
Cl2 + H2O → HOCl + HCl
HOCl ↔ H+ + OCl–
Disinfection Design Using the CT Concept
Example of disinfection design according to the CT concept
Calculate the quantity of chlorine consumed on a daily basis and the detention time required for 99.9% reduction of G. lamblia according to the SWTR rule. The minimum temperature of the water is 5˚C, the free chlorine residual in the effluent from the basin is to be 2 mg/L, the decay rate for chlorine is assumed to be 0.2 h-1, and the flow rate is 1.50 × 104 m3/d. The pH of the water varies between 7.2 and 8.0. Make the calculations for both a CSTR and a PFR.
The highest pH requires the largest CT value. From table 6.1, CT is 243.
The effective contact time, t10 = 243/2 = 122 min.
If the basin is a PFR, then tR = t10 =122 min.
From 
We get 
The quantity of chlorine consumed is
If the basin is a CSTR, then tR = t10/0.105 = 122min/0.105 = 19.4h.
From 
We get 
The quantity of chlorine consumed in this case is
The benefits of a PF regime are apparent from both the size of the basins and the quantity of chlorine required.
Professional and Ethical Responsibilities in Water Supply Systems
Find out the professional and ethical responsibilities as well as standards in dealing with the water supply system? Regulatory bodies or Regulation for Water Qualities e.g.
- Safe Drinking Water Act by US EPA
- World Health Organization (WHO)
- Different state guidelines for US EPA
- DW Guideline for Ontario Ministry of Environment, Ontario in Canada
Safe Drinking Water Act in the USA
- Sets Drinking Water (DW) Standards to protect public health
- Watershed management to protect drinking water sources
World Health Organization (WHO) Drinking Water Guidelines
- To improve the quality of drinking-water and provide significant benefits to health.
Population Forecasting Methods
- Arithmetic increase method
- Geometrical increase method
- Incremental increase method
- Graphical method
Water Resources in KSA
Conventional:
- Surface runoff
- Springs
Non-conventional:
- Desalinated water
- Treated wastewater
Water Distribution Systems
Water distribution systems consist of complex interconnected pipes. Water demand is highly variable, whereas supply is normally constant.
Wastewater Management
Wastewater from domestic and industrial use can be the main non-conventional water resource, especially in countries where water demand for these two sectors is high.
Activated Sludge Process Design
Design an activated sludge process to yield an effluent BOD of 20 mg/l and suspended solids of 25 mg/l. The influent BOD following primary clarification is 160 mg/l. Assume Y= 0.65, Kd = 0.05 and sludge retention time 10 days. The waste flow is 10 m3/min.
Trickling Filter Design
Design a one-stage high rate trickling filter to produce a BOD effluent of 50 mg/L, given the following data: Q = 10,000 m3/d, Influent BOD = 400 mg/L, Temperature (T) = 20 oC. Primary sedimentation tanks will be used before the trickling filter.
BOD entering T.F = 0.67 x 400 = 268 mg/L
BOD load to the trickling filter = 10,000 x 268 x 10-6 x103 = 2680 kg BOD/d
The required efficiency of the T.F is:
E = (268 -50) / 268 = 0.813 = 81.3%
Go to figure (), the 81.3% efficiency can be achieved by following combinations:
R = 1.0 BOD load = 500 g BOD/m3.d
R = 2.0 BOD load = 620 g BOD/m3.d
R = 3.0 BOD load = 700 g BOD/m3.d
Note: any recycle (R) less than 1.0 is not accepted to achieve 81.3% efficiency because it will give a BOD load less than 500 gBOD/m3.d while the accepted range for one stage high rate trickling filter is 500 – 1500 g BOD/m3.d
So use R = 2.0 BOD load = 620 g BOD/m3.d
Volume of the filter media = Daily BOD load / volumetric BOD load
= 2680 x103/ 620 = 4323 m3
The surface loading rate (Typical range 10-30 m3/m2.d). Take it as 20 m3/m2.d
The surface area = 3 x 10,000 / 20 = 1500 m2
Assume that we want to use one T.F only, so the diameter of T.F is π D2 / 4 = 1500, D= 43.7 m
The max diameter should not exceed 35 m. So we use 2 Trickling filters each has D = {(1500/2)x4/T}0.5 = 31 m
The volume of each = 4323/2 = 2161.5 m3
The area of each T.F = π (31)2 / 4 = 755 m2
The depth of each = 2161.5/ 755 = 2.86 m Standard depth range 1.5 to 3.5 m.