Question: Aeration and sludge recirculation have been practiced for many years at municipal and industrial water treatment plants. Aeration is used primarily for the physical removal

Aeration and sludge recirculation have been practiced for many years at municipal and industrial water treatment plants. Aeration is used primarily for the physical removal of gases or volatile compounds, while sludge recirculation can be beneficial for turbidity removal and hardness reduction. When the advantages of aeration and sludge recirculation in water treatment were first recognized, energy costs were so low that such considerations were seldom of concern in treatment plant design and operation. With the huge increases in electricity cost that have occurred in some localities, however, it became necessary to review the cost-effectiveness of all water treatment processes that consume significant amounts of energy. This study was conducted at a municipal water treatment plant for evaluating the cost-effectiveness of the pre-aeration and sludge recirculation practices.


1. What will be the monthly savings in electricity from discontinuation of aeration if the cost of electricity is now 12 ¢/kWh?

2. Does a decrease in the efficiency of the aerator motor make the selected alternative of sludge recirculation only more attractive, less attractive, or the same as before?

3. If the cost of lime were to increase by 50%, would the cost difference between the best alternative and second-best alternative increase, decrease, or remain the same?

4. If the efficiency of the sludge recirculation pump were reduced from 90% to 70%, would the net savings difference between alternatives 3 and 4 increase, decrease, or stay the same?

5. If hardness removal were to be discontinued at the treatment plant, which alternative would be the most cost-effective?

6. If the cost of electricity decreased to 8 ¢/kWh, which alternative would be the most cost-effective?

7. At what electricity cost would the following alternatives just break even?

(a) Alternatives 1 and 2,

(b) Alternatives 1 and 3,

(c) Alternatives 1 and 4.


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1 Savings 40 hp 075 kwhp 012 kwh 24 hrday 305 daysmo 090 2928 per month 2 A decrease in the efficiency of the aerator motor renders the selected alter... View full answer

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