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# A = L x w = 18 X 4 = 72 m2

 tarix 25.06.2016 ölçüsü 35 Kb.
 OIT PRACTICE MATH QUIZ 6 DETAILED CALCULATIONS ANSWERS 1. Calculate the surface area of a rectangular tank 18 m long and 4 m wide. A = L x W = 18 x 4 = 72 m2 2. Calculate the surface area of a circular sand filter that has a diameter of 15 m. A = π R2 or A = π D2 = 3.14 x 7.52 4 = 176.71 m 2 = 3.14 x 152 = 177 m2 4 = 176.71 m2 = 177 m2 3. Calculate the volume of a raw water intake crib that is 8 m long, 3 m wide and 6 m deep. V = LxWxD = 8 x 3 x 6 = 144 m3 4. What is the volume of a cylindrical storage tank that is 7 m in diameter and 15 m high? V = π R2 x H V = π D2 x H = 3.14 x 3.52 x 15 4 = 577.26m3 = 3.14 x 72 x 15 = 577 m3 4 = 2309.07 4 = 577.26 m3 = 577 m3 5. What is the volume of water contained in 84 m of a 10 cm diameter pipe? a) in m3 - .660 m3 V = π R2 x H V = π O2 x H = 3.14 x .052 x 84 OR 4 b) in L – 660 L = .660 m3 = 3.14 x .102 x 84 .660 m3 x 1000 = 660 L 4 = 2.639 4 = .660 m3 x 1000 = 660 L if a pump delivers 1.44 m3in 20 min, what is the pumping rate expressed in: a) L/s? - 1.2 L/s a) 1.44 x 1000 = 1440 L divide 20 = 72 L/min 72 L = 1.2 L/s b) m3/d? – 103.68 m3/d 60 sec. b) 1.44 = .072 m3/min. 20 .072m3/min x 1440 = 103.68 m3 **NOTE PROBLEM 6 HAS SEVERAL SOLUTIONS** How many m3 of water will a 5L/s pump deliver in 5 min? 1000 L = 1 m3 5 L = .005 m3/s .005 m3 x 60 sec. x 5 min. 1.5 m3/5 min A 12 m3 storage tank supplies alum for coagulation at a rate of 330 mL/min. How often will the tank need refilling? 12 m3 x 1000 L = 12,000 L m3 12,000 L = 3636.36 m .330 mL/min The prechlorination chamber at a water treatment plant has a volume of 225 m3. if the flow rate out of the tank is 11 L/s, what is the average detention time in hours. 225 m3 x 1000 = 225000 L 225000 L = 20454.545 s 11 L/s 340.91 min 5.68 hrs or 5.7 hrs 5 hr. 60 min. x .7 = 42 min = 5 hr 42 min How many kg of chlorine are required each day to treat 18,000 m3 with chlorine at 5.0 mg/L? 5.0 mg/L = 5 kg/1000 m3 18000 m3 x 5 kg 1000 m3 1 90 kg 5mg 18000 m3 1000L 1kg L day m3 10000000 mg 11. A gas chlorinator treats 2,700 m3 with 2 kg of chlorine each day. Calculate the dosage rate. the residual is measured at 0.27 mg/L. What is the chlorine demand? a) in mg/L - .74 mg/L 2700 x 1000 = 2700000 L 2 kg x 10000000 = 20,000000 mg b) in kg/d 20 mg = .74 mg/L 2 kg = 1 kg 27 L 2700 m3 X .74 mg/L – 0.27 mg/L = .74 mg/L 2 kg x = 2700 kg/m3 X = 2700 m3 kg 1m1.L = 1 kg/m3 2 kg .74 mg/L = .74 kg/m3 X = 1350 m3 12. A liquid solution with a total mass of 97 kg contains 84 kg of water and the remainder is alum a) Calculate the percent that is water. – divide water by total multiply by 100 b) Calculate the percent that is alum. – 100 % - % of water = % of alum. a) 84 x 100 = 8400 = 86.6 % water 97 1 97 100 % - 86.6 % = 13.4 % alum A mixture of water and powdered carbon is to be 85% water. if the total volume of solution required is 3.6 m3, what is the mass of the powdered carbon? 3.6 m3 = 100% 3.6 m3 x 85% = 3.06 m3 water 3.6 m3 – 3.06 m3 = .54 m3 or 100% - 85% = 15% 3.6 m3 x .15 = .54 m3 14. A hypochlorite solution contains 12% available chlorine. if 3 kg of availible chlorine are needed to disinfect a main: a) how many kg of solution are required? b) How many litres of solution are required 3 kg = 12% a) 12% = 3 300 divide 12 = 25 kg required X b) 1 kg = 1 L – 25 kg = 25 L

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