Question

A city wants to build and operate a wastewater treatment facility over a very long service life as “perpetual” infrastructure for the city. A consulting engineer has estimated that the plant will cost $40 million to construct. She further estimates it will require $12 million every 20 years to replace major unit operations equipment (bar screens, aeration equipment, dewatering centrifuges, etc.) and $21 million every 40 years for major renovation of tanks and other structural elements. Operations and maintenance costs are estimated to be $3 million/year. In order to pay for the construction, the city plans to take out a bond to repay both the interest and principal in equal annual payments. Due to a recent recession, the discount rate and the bond interest rate are both 3%.

1. The city plans to use sewage fees to pay for the construction and the capitalized costs of the plant (O&M, equipment and renovation costs). The average wastewater flow rate in the city sewers is 8,000 kgal/day. What is the rate the city utility should charge users in $/kgal?
2. As an alternative, the city is considering just collecting sewer fees to pay for initial construction, O&M, equipment replacement at year 20 and major renovation at year 40. Given the same demand, what should the break-even sewer rates be ($/kgal)?
3. Which rate structure would you select? Explain.

Answer 1

In first case where we have to calculate break even sewer rate of O&M, equipment and renovation costs , we have to skip Initial construction cost in order to achieve break evere price.

Overall cost for Equipment (40 years )

= 12 million $ × 2

= 24 million $

Overall cost for renovation (40 years )

= 21 million $

Overall cost for operation and management (40 years )

= 3 million $ × 40 = 120 million dollar

Thus Total O&M+ Renovation + Equipment cost

= 120 + 21+24

= 165 million $

Now 3% discount applied on face value of bond i.e 165 million dollor

= 165 - 4.95 million dollar

= 160.15 million dollar

Again intrest calculated for 40 years @3%/annum on face value of bond

= 160.15+ 160.15×3×40/100

= 192.18 million dollar

Now per day rate of 8000 kgal =

Overall cost /8000×365×40

= 192.18 million dollar / 8000×365×40

= 1.645 (approx) Kgal $ per litre.

In second case where construction cost also included , we have to add initial construct cost BUT BEFORE APPLYING BOND DISCOUNT AND INTREST

= O&M + Renovation+ Equipment + Construction cost

= 165 million $ + 40 million $

= 205 million $

Now bond discount @3 %

= 205 - 6.15 million $

= 198. 85 million $

Intrest rate on face value of bond @3 % per annum =

= 198.85 +198.85 × 3× 40 /100

= 238.62 million dollar

Break-even sewer rates be ($/kgal)

= 238.62 million dollar / 8000 × 365 × 40

= 2.04 Kagl $ per liter

Rate structure used here is Longmedow Current rate system ,Longmeadow’s current water and sewer rate system relies on a single meter in each home. That meter records only the incoming utilization of water. There is no metering system for wastewater leaving the home on an individual basis. So, over the years, a rate structure has been developed that makes certain assumptions about how much of the metered water that arrives at each home is thereafter removed for treatment by the Sewers Ratepayers are billed on the basis of water metered and a sewer charge that is related to water usage.