WATER TREATMENT PLANT PROCESS COSTS

Question

Answer 1

Sometimes, there is not a definitive answer to a case study exercise. Here are example responses.
Sometimes, there is not a definitive answer to a case study exercise. Here are example responses.1. Savings = 40 hp * 0.75 kw/hp * 0.12 $/kwh * 24 hr/day * 30.5 days/month ÷ 0.90
= $2928 per month

2. A decrease in the efficiency of the aerator motor renders the selected alternative of “sludge recirculation only” more attractive, because the cost of aeration would be higher, and, therefore the net savings from its discontinuation would be greater.

3. If the cost of lime increased by 50%, the lime costs for “sludge recirculation only” and “neither aeration nor sludge recirculation” would increase by 50% to $393 and $2070, respectively. Therefore, the cost difference would increase.

4. If the efficiency of the sludge recirculation pump decreased from 90% to 70%, the net savings between alternatives 3 and 4 would decrease. This is because the $262 saved by not recirculating with a 90% efficient pump would increase to a monthly savings of $336 by not recirculating with a 70% efficient pump.

5. If hardness removal were discontinued, the extra cost for its removal (column 4 in Table 13-1) would be zero for all alternatives. The favored alternative under this scenario would be alternative 4 (neither aeration nor sludge recirculation) with a total savings of $2,471 – 469 = $2002 per month.

6. If the cost of electricity decreased to 8¢/kwh, the aeration only and sludge recirculation only monthly costs would be $244 and $1952, respectively. The net savings for alternative 2 would then be $-1605, alternative 3 would save $845, and alternative four would save $347. Therefore, the best alternative continues to be number 3.
Sometimes, there is not a definitive answer to a case study exercise. Here are example responses.1. Savings = 40 hp * 0.75 kw/hp * 0.12 $/kwh * 24 hr/day * 30.5 days/month ÷ 0.90
= $2928 per month

2. A decrease in the efficiency of the aerator motor renders the selected alternative of “sludge recirculation only” more attractive, because the cost of aeration would be higher, and, therefore the net savings from its discontinuation would be greater.

3. If the cost of lime increased by 50%, the lime costs for “sludge recirculation only” and “neither aeration nor sludge recirculation” would increase by 50% to $393 and $2070, respectively. Therefore, the cost difference would increase.

4. If the efficiency of the sludge recirculation pump decreased from 90% to 70%, the net savings between alternatives 3 and 4 would decrease. This is because the $262 saved by not recirculating with a 90% efficient pump would increase to a monthly savings of $336 by not recirculating with a 70% efficient pump.

5. If hardness removal were discontinued, the extra cost for its removal (column 4 in Table 13-1) would be zero for all alternatives. The favored alternative under this scenario would be alternative 4 (neither aeration nor sludge recirculation) with a total savings of $2,471 – 469 = $2002 per month.

6. If the cost of electricity decreased to 8¢/kwh, the aeration only and sludge recirculation only monthly costs would be $244 and $1952, respectively. The net savings for alternative 2 would then be $-1605, alternative 3 would save $845, and alternative four would save $347. Therefore, the best alternative continues to be number 3.7. (a) For alternatives 1 and 2 to breakeven, the total savings would have to be equal to
the total extra cost of $1,849. Thus,

1,849/ 30.5 = (5)(0.75)(x)(24) / 0.90
x = 60.6 cents per kwh

(b) 1107/ 30.5 = (40)(0.75)(x)(24) / 0.90
x = 4.5 cents per kwh

(c) 1,849/ 30.5 = (5)(0.75)(x)(24) / 0.90 + (40)(0.75)(x)(24) / 0.90
x = 6.7 cents per kwh