Kennebunk Light and Power


KLPD Robbing Peter to Pay Paul

KLPD is proposing to increase its electric delivery rates. Not needed!

If KLP fixes the generators over the next 6 years, the 3 dams will provide a revenue of $695,654 (March 15, 2016 Wright-Pierce (W-P) Alternative #4 Cease Operation; “Value of generated electricity”)

Diverting hydropower revenue from maintenance since 2011 results in depreciation with a loss of income.  The result is the July letter from KL&PD for a rate increase in the “delivery rates”.

In 2011 our hydropower dams generated      1,495,760 kWh providing $122,054.

In 2012 (from KLPD HYDROPOD 1.PDF)         1,201,640 kWh providing $ 98,053.

In 2013 (revenue equals kWh X $.0816)        1,521,420 kWh providing $124,147.

In 2014 (kWh used by Wright-Pierce costs)   1,391,880 kWh providing $113,577.

In 2015 (2 months 3 dams not generating)      957,720 kWh providing $  78,150.

Total Revenue generated since 2011                        TOTAL        $535,980.

In 2011 the W-P Draft Hydropower Facility Alternative Assessment Study recommended deferring maintenance because the Net Revenue vs. Cost to “Seek New License to Continue Operation” projected a loss over 46 years of $(6,400,000).  W-P did not present a loss of $(11,000,000) over 46 years to Cease Operations and remove the dams.

A New Alternative without fish passage (for 9 Fish) projects a profit of $ 4,400,000.

The average kWh over the past 13 years was 1,795,000 kWh providing   $ 146,472.

In 2006 the KLPD records recorded dams generated 2,627,000 kWh providing $214,363.

Kennebunk Light & Power District does not need to increase its electric delivery rates!  KLPD is shifting (robbing) revenue from dams to revenue from ratepayers!

Albert Kolff, Kennebunk, ME.            7/25/2016

(207) 985-0335


How much coal, natural gas, or petroleum is used to generate a kilowatt hour of electricity?


The amount of fuel used to generate electricity depends on the efficiency or heat rate of the generator (or power plant) and the heat content of the fuel. Power plant efficiencies (heat rates) vary by types of generators, power plant emission controls, and other factors. Fuel heat contents also vary.


Two formulas can be used to calculate the amount of fuel used to generate a kilowatthour (kWh) of electricity:

  • Amount of fuel used per kWh = Heat rate (in Btuper kWh) / Fuel heat content (in Btu per physical unit)
  • Kilowatthour generated per unit of fuel used = Fuel heat content (in Btu per physical unit) / Heat rate (in Btu per kWh)

Calculation examples using these two formulas and the assumptions below:

  • Amount of fuel used to generate 1 kWh:
  • Coal = 0.00052 short tons or 1.04 pounds
  • Natural gas = 0.01011 Mcf (an Mcf equals 1,000 cubic feet)
  • Petroleum = 0.00173 barrels (or 0.07 gallons)


Using the 1,400,000 kwh/year used in the Wright-Pierce cost analysis Alternatives;

If we use the 1,400,000 kWh generated by our 3 KLPD dams in 2014, that comes to 3,800 kWh per day

3,800 kWh/day X Coal @ 1.04lb/kWh  = 3,952 lb/day of Coal burned

3,800 kWh/day X N Gas@ 1,000 cf/kwh = 3,800,000 cubic feet/day of Natural Gas  burned

3,800 kWh/day X Petroleum @ .07gallons/kwh = 266 gallons/day of Petroleum burned

Albert Kolff 6/24/2016