Question

a power plant treats 2 ML/d of feedwater to remove total organic carbon (TOC). The source water contains 30 mg/L of TOC. The treatment system consists of a square coagulation and flocculation tank that with a removal efficiency of 30% TOC (because organic matter sticks to floc). The coagulation tank is followed by a long (20 m), narrow(2 m) ozonation channel of 100,000 L with TOC degradation rate of 1.73/hour by reaction with ozone. What is the TOC concentration at the end of the complete treatment system?

Answer 1

Solution:

Given:

Capacity of power plant: 2 ML/day = 2 * 10^6 L/day

Amount of TOC = 30 mg/L

Removal efficiency = 30% TOC.

Degradation rate 1.73%/hour or 1.73mg/hour (Not mentioned in question)

Removal efficiency of tank is 30% TOC.

30mg/L * 30% = 9mg/L. Therefore, 21mg/L will be remaining after feedwater leaves tank.

Step 2

Volume of channel is = length * width = 20 * 2 = 40 m^3= 40000 L

Time required for 100,000 L water is = 100,000/40000 = 2.5 hours.

So, it will take 5 hours to pass all the feedwater to end of the complete treatment system.

Step 3

Degradation rate is 1.73% per hour. Therefore, for 5 hours 1.73*5 = 8.65%= 1.8165 mg/L

Remaining concentration at the end of complete treatment system is =21 – 1.8165

                                                                                                                                          = 19.18 mg/L

                                                                                or

Degradation rate is 1.73mg per hour. Therefore, for 5 hours 1.73*5 = 8.65mg/L

Remaining concentration at the end of complete treatment system is =21 – 8.65

                                                                                                                                          = 12.35 mg/L