Answered step by step
Verified Expert Solution
Link Copied!

Question

1 Approved Answer

To evaluate the effect of sludge recirculation, the sludge pump was turned off, but aeration was continued. Next, the Page 377 sludge pump was turned

image text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribed
image text in transcribedimage text in transcribedimage text in transcribedimage text in transcribedimage text in transcribed
To evaluate the effect of sludge recirculation, the sludge pump was turned off, but aeration was continued. Next, the Page 377 sludge pump was turned back on, and aeration was discontinued. Finally, both processes were discontinued. Results obtained during the test periods were averaged and compared to the values obtained when both processes were operational. The results obtained from the four operating modes showed that the hardness decreased by 4.7%% when both processes were in operation (i.e., sludge recirculation and aeration). When only sludge was recirculated, the reduction was 3.8%. There was no reduction due to aeration only, or when there was neither aeration nor recirculation. For turbidity, the reduction was 28% when both recirculation and aeration were used. The reduction was 18% when neither aeration nor recirculation was used. The reduction was also 18%% when aeration alone was used, which means that aeration alone was of no benefit for turbidity reduction. With sludge recirculation alone, the turbidity reduction was only 676, meaning that sludge recirculation alone actually resulted in an increase in turbidity-the difference between 18% and 6%. Since aeration and sludge recirculation did cause readily identifiable effects on treated water quality (some good and others bad), the cost-effectiveness of each process for turbidity and hardness reduction was investigated. The calculations are based on the following data: Aerator motor = 40 hp Aerator motor efficiency = 90% Sludge recirculation motor = 5 hp Recirculation pump efficiency = 90% Electricity cost = 9 (/KWh (previous analysis) Lime cost = 7.9 c/kg Lime required = 0.62 mg/L per my/L hardness Coagulant cost = 16.5 0/kg Days/month = 30.5 As a first step, the costs associated with aeration and sludge recirculation were calculated. In each case, costs are independent of flow rate.Costs associated with turbidity and hardness removal are a function of the chemical dosage required and the water flow rate. The calculations below are based on a design flow of 53 m-/min. As stated earlier, there was less turbidity reduction through the primary clarifier without aeration than there was with it (28% vs. 6"). The extra turbidity reaching the flocculators could require further additions of the coagulating chemical. If it is assumed that, as a worst case, these chemical additions would be proportional to the extra turbidity, then 22% more coagulant would be required. Since the average dosage before discontinuation of aeration was 10 mg/L, the incremental chemical cost incurred because of the increased turbidity in the clarifier effluent would be (10 x 0.22) mg/L x 10-6 kg/mg x 53 m /min x 1000 L/m: x 0.165 $/kg x 60 min/h x 24 h/day = $27.70/day or $845/month Similar calculations for the other operating conditions (i.e., aeration only, and neither aeration nor sludge recirculation) reveal that the additional cost for turbidity removal would be $469 per month in each case, as shown in column 5 of Table 13-1(). Changes in hardness affect chemical costs by virtue of the direct effect on the amount of lime required for water softening. With aeration and sludge recirculation, the average hardness reduction was 12.1 mg/L (i.e., 258 my/L X 4.7%). However, with sludge recirculation only, the reduction was 9.8 mg/L, resulting in a difference of 2.3 mg/L attributed to aeration. The extra Page 376 cost of lime incurred because of the discontinuation of aeration, therefore, was 2.3 mg/L x 0.62 my/L lime * 10-6 kg/mg x 53m /min x 1000 L/m: * 0.079 $/kg x 60 min/h x 24 h/day = $8.60/day or $262/monthAeration cost: 40 hp x 0.75 kW/hp x 0.09 5/1:Wh * 24 h/day = 0.90 = $72 per day or $2196 per month Sludge recirculation cost: 5 hp x 0.75 kW/hp x 0.09 $/KWh x 24 h/day = 0.90 =$9 per day or $275 per month The estimates appear in columns 1 and 2 of the cost summary in Table 13-1 { TABLE 13-1 Cost Summary in Dollars per Month Savings from Extra Cost for Discontinuation of Removal of Total Not Alternative Aoration Rocirculation Total Savings Hardness Turbidity Extra Cost Savings Alt. I.D. Description (1) (2) (3) = (1) + (2) (4) (5) (6) = (4) + (5) (7) = (3) - (6) Sludge recirculation Normal operating condition and aeration 2 Aeration only 275 275 1380 469 1849 -1574 3 Sludge recirculation 2196 2196 262 845 1107 +1089 only 4 Neither aeration nor 2196 275 2471 1380 469 1849 +622 sludge recirculation Costs associated with turbidity and hardness removal are a function of the chemical dosage required and the water flow rate. The calculations below are based on a design flow of 53 m-/min.Background 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. Information This study was conducted at a 106 m /min watertreatment plant where, under normal operating circumstances, sludge from the secondary clarifiers is returned to the aerator and subsequently removed in the primary clarifiers. Figure 13-12 () is a schematic of the process. Chemical Flash additions Flocculation mix Primary Secondary clarifier clarifier Acrator Filter To clear well CanalWhen sludge recirculation was discontinued, there was no hardness reduction through the clarifier, so that the extra lime cost would be $1380 per month. The total savings and total costs associated with changes in plant operating conditions are tabulated in columns 3 and 6 of Table 13-1 42, respectively, with the net savings shown in column 7. Obviously, the optimum condition is represented by "sludge recirculation only." This condition would result in a net savings of $1089 per month, compared to a net savings of $622 per month when both processes are discontinued and a net cost of $1574 per month for aeration only. Since the calculations made here represent worst-case conditions, the actual savings that resulted from modifying the plant operating procedures were greater than those indicated. In summary, the commonly applied water treatment practices of sludge recirculation and aeration can significantly affect the removal of some compounds in the primary clarifier. However, increasing energy and chemical costs warrant continued investigations on a case-by-case basis of the cost-effectiveness of such practices. Case Study Exercises 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 C/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

Step by Step Solution

There are 3 Steps involved in it

Step: 1

blur-text-image

Get Instant Access to Expert-Tailored Solutions

See step-by-step solutions with expert insights and AI powered tools for academic success

Step: 2

blur-text-image

Step: 3

blur-text-image

Ace Your Homework with AI

Get the answers you need in no time with our AI-driven, step-by-step assistance

Get Started

Recommended Textbook for

Principles of Economics

Authors: Gregory Mankiw

7th edition

128516587X, 978-1285165875

More Books

Students also viewed these Economics questions

Question

Evaluate the following limits. 2t cos 2t i 4 sin tj + lim T/2 k TT

Answered: 1 week ago

Question

1. What will happen in the future

Answered: 1 week ago

Question

3. Avoid making mistakes when reaching our goals

Answered: 1 week ago