Table 7

First we determine the capacity of chlorinator with flow 25,000 gal/day → 100.3 m^{3}/dayAssume peak daily factor for treatment plant was 3 and maximum required chlorine dosage set by state regulation is 20 mg/L Cl _{2} kg/day = 20 g/m^{3} × 100.3 m^{3}/day × 1 kg/10^{3}g= 2 kg/day Although the peak capacity will not be required during most of the day, it must be available to meet the chlorine requirements at peak flow. So, estimation of daily consumption of chlorine at average dosage of 10 mg/L Cl _{2} kg/day = 10 g/m^{3} × 100.2 m^{3}/day × 1 kg/10^{3}g= 1 kg/day Chlorine contact tank design is as follows: Assume some trial cross sectional dimensions for chlorine contact tank based on design flow and chlorine dosage: Width = 4.2 ft (1.28 m), depth = 2.8 ft (0.85 m) Desired dispersion number = 0.15 Detention time at peak flow of 100.2 m ^{3}/day is 60 minSo length of CCT = Q × contact time/width × depth = 100.2 m ^{3}/d × 60 min/1440 min/d × 1.28 × 0.85= 3.8 m (12.5 ft) Check velocity at peak flow V = Q/width × depth = 100.2 m ^{3} /d/1440 min/d × 60 sec/min (1.28 × 0.85)= 0.0011 m/sec Computing the dispersion number a: Reynolds number = N _{R} = 4VR/v_{o}V = 0.0011, R = A/ P = 1.28 × 0.85/2(1.28 + 0.85) = 0.25v _{o} = velocity in open channel = 1.003 × 10^{−6}By putting values, we get N _{R} = 1120D = coefficient of dispersion = 1.01v _{o} (N_{R})^{0.87} = 4.55 × 10^{−4}Dispersion number = D·t/L ^{2} = 4.55 × 10^{−4} (60 min × 60 min/sec)/(3.8)^{2} = 0.11 < 0.15Because the computed dispersion number was below the desired dispersion number, the given design was valid for chlorination |

First we determine the capacity of chlorinator with flow 25,000 gal/day → 100.3 m^{3}/dayAssume peak daily factor for treatment plant was 3 and maximum required chlorine dosage set by state regulation is 20 mg/L Cl _{2} kg/day = 20 g/m^{3} × 100.3 m^{3}/day × 1 kg/10^{3}g= 2 kg/day Although the peak capacity will not be required during most of the day, it must be available to meet the chlorine requirements at peak flow. So, estimation of daily consumption of chlorine at average dosage of 10 mg/L Cl _{2} kg/day = 10 g/m^{3} × 100.2 m^{3}/day × 1 kg/10^{3}g= 1 kg/day Chlorine contact tank design is as follows: Assume some trial cross sectional dimensions for chlorine contact tank based on design flow and chlorine dosage: Width = 4.2 ft (1.28 m), depth = 2.8 ft (0.85 m) Desired dispersion number = 0.15 Detention time at peak flow of 100.2 m ^{3}/day is 60 minSo length of CCT = Q × contact time/width × depth = 100.2 m ^{3}/d × 60 min/1440 min/d × 1.28 × 0.85= 3.8 m (12.5 ft) Check velocity at peak flow V = Q/width × depth = 100.2 m ^{3} /d/1440 min/d × 60 sec/min (1.28 × 0.85)= 0.0011 m/sec Computing the dispersion number a: Reynolds number = N _{R} = 4VR/v_{o}V = 0.0011, R = A/ P = 1.28 × 0.85/2(1.28 + 0.85) = 0.25v _{o} = velocity in open channel = 1.003 × 10^{−6}By putting values, we get N _{R} = 1120D = coefficient of dispersion = 1.01v _{o} (N_{R})^{0.87} = 4.55 × 10^{−4}Dispersion number = D·t/L ^{2} = 4.55 × 10^{−4} (60 min × 60 min/sec)/(3.8)^{2} = 0.11 < 0.15Because the computed dispersion number was below the desired dispersion number, the given design was valid for chlorination |

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