A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s) with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 °C is 0.27 d^-1.
The answer therefore would be the number 0.27 divided by two and then square while getting the square you would make it a binomial.
I wont give the answer but the steps
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Answer:
Electroosmotic velocity will be equal to 
Explanation:
We have given applied voltage v = 100 volt
Length of capillary L = 5 mm = 0.005 m
Zeta potential of the capillary surface 
Dielectric constant of glass is between 5 to 10 here we are taking dielectric constant as 
Viscosity of glass is 
Electroosmotic velocity is given as 
So Electroosmotic velocity will be equal to
Answer:
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Explanation:
