WATER POLLUTION CIVL270-Maraqa Precipitation on sea The Hydrologic Cycle 458x10 km 3 Precipitation on land 119x103 km3 Evapotranspiration Evaporation 72x103 km3 505x103 km3 Runoff 47x103 km3 CIVL270-Maraqa Water Resources Resources Water Conventional Conventional Surface water water Surface CIVL270-Maraqa Groundwater Groundwater Unconventional Unconventional Desalination Desalination Treated Treated wastewater wastewater Water Resources About 2.2 billion people in developing countries lack access to safe drinking water About 2.7 billion people in developing countries lack access to safe sanitation services Water Use Use Water Domestic Domestic Industrial Industrial Agricultural Agricultural Percent of withdrawal 100 Domestic Industrial Agricultural 80 60 40 20 Low Mid High Income level Water withdrawals by sector in low, mid and high income countries (Environmental Science: A Global Concern, W.P Cunningham and B.W Saigo, 3rd Ed Wm Brown Pub © 1995) CIVL270-Maraqa Water Usage Household Water Usage Liters Standard toilet, per flush 10-30 Shower head, per minute 20-30 Dishwasher, per load 50-120 Washing car with running water, 20 minute 400-800 Uncovered 60m2 pool, per day 100-400 Agricultural Items One egg 150 Glass of milk 380 One kg of rice 4240 CIVL270-Maraqa Water pollution pollution Water Physical Physical Radiological Radiological •Thermal •Solids Chemical Chemical Biological Biological •Hardness •Pathogens •Heavy metals •Nutrients •Pesticides •Oxygen demanding waste •Volatile organic compounds CIVL270-Maraqa •Radio-isotopes •Thermal Pollution Causes a drop in the dissolved oxygen due to higher metabolic rate and lower DO solubility at higher temperature •Solids Could be suspended (causing turbidity) or dissolved causing salinity) Total dissolved solids (TDS) in water is the sum of all cations and anions present expressed in mg/L Water use Drinking TDS, mg/L mg/L (ii) DOi-DO5>2 mg/L CIVL270-Maraqa P Example 5.2 A test bottle containing dilution water has its DO level drop by 1.0 mg/L in a five-day test A 300-ml BOD bottle filled with 15 ml of wastewater and the rest dilution water experiences a drop of 7.2 mg/L in the same time period What is the BOD5 of the waste? Solution P= Vsample Vsample + Vdilution BOD5 = BOD5 = = 15 = 0.05 15 + 285 (DO i − DO5 ) diluted sample − (DOi − DO5 ) dilution (1 − P) P (7.2) − (1.0)(1 − 0.05) = 125 mg / L 0.05 CIVL270-Maraqa Problem 5.4 with modification A BOD test is to be run on a sample of wastewater that has a BOD5 of approximately 250 mg/L If the initial DO of a mix of dilution water and wastewater is mg/L What is the best dilution factor to be used in order to determine the exact BOD5 of the sample? Solution Assume dilution water is free of organics BOD5 =  Two conditions have to be met in determining P: (i) DO5 > mg/L (ii) DOi-DO5>2 mg/L BOD5 = DOi − DO5 8−2 ⇒ 250 = ⇒ P = / 41.7 P P BOD5 = DOi − DO5 ⇒ 250 = ⇒ P = / 125 P P The best P value is the average of the two values CIVL270-Maraqa DO i − DO5 P P=1/83 Example 6.4 A groundwater sample has 100 mg/L Ca and 10 mg/L Mg Classify the water from a hardness point-of-view Solution mg Concentration of X (mg / L) × 50 (mg CaCO / meq) of X as CaCO = L Equivalent weight of X (mg / meq) mg 100 (mg / L) × 50 (mg CaCO / meq) of Ca as CaCO = = 250 L 20 (mg / meq) mg 10 (mg / L) × 50 (mg CaCO / meq) of Mg as CaCO = = 41 L 12.2 (mg / meq) So, the total hardness of the water is 291 mg/L as CaCO3 and the water is classified as hard CIVL270-Maraqa [...]... initial at 20 oC DOi DO5 Sample Sample Dissolved oxygen after 5 days Case 2: Diluted sample with blank free from organics 5 days at 20 oC DOi DO5 Diluted sample Diluted sample CIVL270-Maraqa DOi − DO 5 BOD5 = P Vsample P= Vsample + Vdilution Dilution factor Case 3: Diluted sample with blank that has organics 5 days 5 days at 20 oC at 20 oC DOi DO5 DOi DO5 Diluted sample Diluted sample Dilution water. .. Classification based on intended use: insecticides, herbicides, fungicides, etc Classification based on chemical structure: organochloroines (ex DDT), organophosphate, and carbamates •Nutrients Include N, P, S, C, Ca, K, Fe, etc Allow algal growth •Volatile Organic Compounds (VOCs) Chlorinated compound, PCE, TCE, DCE, vinyl chloride, TCA Aromatic hydrocarbons, BTEX CIVL270-Maraqa Biochemical... CIVL270-Maraqa DO i − DO5 P P=1/83 Example 6.4 A groundwater sample has 100 mg/L Ca and 10 mg/L Mg Classify the water from a hardness point-of-view Solution mg Concentration of X (mg / L) × 50 (mg CaCO 3 / meq) of X as CaCO 3 = L Equivalent weight of X (mg / meq) mg 100 (mg / L) × 50 (mg CaCO 3 / meq) of Ca as CaCO 3 = = 250 L 20 (mg / meq) mg 10 (mg / L) × 50 (mg CaCO 3 / meq) of Mg as CaCO 3 = =... What is the BOD5 of the waste? Solution P= Vsample Vsample + Vdilution BOD5 = BOD5 = = 15 = 0.05 15 + 285 (DO i − DO5 ) diluted sample − (DOi − DO5 ) dilution (1 − P) P (7.2) − (1.0)(1 − 0.05) = 125 mg / L 0.05 CIVL270-Maraqa Problem 5.4 with modification A BOD test is to be run on a sample of wastewater that has a BOD5 of approximately 250 mg/L If the initial DO of a mix of dilution water and wastewater... Dilution water BOD5 = (DOi − DO5 ) diluted sample − (DOi − DO5 ) dilution (1 − P) How to select P? Two conditions have to be met: (i) DO5 > 2 mg/L (ii) DOi-DO5>2 mg/L CIVL270-Maraqa P Example 5.2 A test bottle containing dilution water has its DO level drop by 1.0 mg/L in a five-day test A 300-ml BOD bottle filled with 15 ml of wastewater and the rest dilution water experiences a drop of 7.2 mg/L in the same... Oxygen Demand (BOD) BOD: The amount of O2 needed by the microorganisms to oxidize organic wastes Organic matter + O2 Microorganisms CO2 + H2O +……… BOD5: Amount of oxygen consumed in 5 days per liter of solution BOD BOD CBOD CBOD Carbonaceousoxygen oxygendemand demand Carbonaceous CIVL270-Maraqa NBOD NBOD Nitrogeneousoxygen oxygendemand demand Nitrogeneous Case 1: No Dilution BOD5 = DOi - DO5 5 days Dissolved... × 50 (mg CaCO 3 / meq) of Ca as CaCO 3 = = 250 L 20 (mg / meq) mg 10 (mg / L) × 50 (mg CaCO 3 / meq) of Mg as CaCO 3 = = 41 L 12.2 (mg / meq) So, the total hardness of the water is 291 mg/L as CaCO3 and the water is classified as hard CIVL270-Maraqa ... wastewater is 8 mg/L What is the best dilution factor to be used in order to determine the exact BOD5 of the sample? Solution Assume dilution water is free of organics BOD5 =  Two conditions have to be met in determining P: (i) DO5 > 2 mg/L (ii) DOi-DO5>2 mg/L BOD5 = DOi − DO5 8−2 ⇒ 250 = ⇒ P = 1 / 41.7 P P BOD5 = DOi − DO5 2 ⇒ 250 = ⇒ P = 1 / 125 P P The best P value is the average of the two values