1. In the following problem, you may assume that the river parameters are: The d
ID: 1714795 • Letter: 1
Question
1. In the following problem, you may assume that the river parameters are: The deoxygenation rate constant k,-0.208/day The reaeration rate constant &,-0.416day. The river speed is 9 km/day The saturation value of the dissolved oxygen 10.5 mg/L a) The river upstream of a beer brewery has a BOD concentration of 5.0 mg/L and a flow of 9.5x10 m day. The effluent flow from the facility is 0.5x10 m'/day containing a BOD of 505 mg/L. Assuming complete mixing what is the BOD concentration Le in the river at the location of the beer brewery (x-0)? b) The differential equation that describes the BOD concentration L as a function of dt downstream distance x is: ud--ka L. Find the BOD 30 km. downstream. c) Assuming that the initial oxygen deficit is zero, find the oxygen deficit at that location. d) The differential equation for the oxygen deficit D= DO-DO is dD kL-k, D. dx Describe the physical significance of each term in the equation e) At 30 kilometers downstream, which is greater the oxygen being consumed by bacteria or the oxygen entering the river from the atmosphere? e) Find the minimum value of dissolved oxygen D What must Lo be to guarantce that the dissolved oxygen is always above 5.0 mg/L? 8) value?Explanation / Answer
Answer:
E) At 30km downstream, Oxygen being consumed by bacteria is greater than the oxygen entering the river from atmosphere since it is evident from the fact that there is a oxygen deficit of 34.5% at 30km downstream.
E) The minimum value of dissolved oxygen is 10.5mg/L. It is evident from the fact that saturation level of dissolved oxygen is 10.5mg/L as given in the question. Since saturation level is the minimum value required, then both are same.
F) The guarantee L0, at which the BOD will be 5mg/L is 37.5km downstream.
Where, Kd = 0.208, u = 9,
i.e., -.208*37.5/9 = -0.867,
BOD = L0-dL/dx = 37.5 *(1-0.867) = 5 mg/L
G) The removal efficiency of this waste water treatment facility that provides this value is 100%, given be the formula below,
Efficiency = (Infulent - Effulent)/ Infulent x 100%
= (505-500/5)*100%
= 1*100%
= 100%
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