FSH 2709.15 - HYDROELECTRIC HANDBOOK
                             2/87 WO



06 - CALCULATION OF OUTPUTS, AND VALUES.  Exhibits 1, 2, 3, and 4 
show the user how to analyze the tradeoffs associated with 
various alternatives and understand the developer's interest in a 
project.  Perform the calculations for projects being evaluated 
when necessary or desirable.
                         
                         
                            Exhibit 1

                  Value of the Energy Resource


        kW = (Q)(H)(e)
              11.8

  Where kW =  Power in kilowatts.
         Q =  Volume of flow in cubic feet per second (cfs).
         H =  Usable head in feet. (static head less friction
              losses in penstock).
     11.8  =  Factor to convert units of potential energy
              into kilowatts.
        e  =  Average efficiency factor:  70 percent efficiency 
              for a small generating system; 80 percent 
              efficiency for large systems.
                           
                           
                            Exhibit 2

                Maximum Capacity of a Power Plant


Calculate the installed or maximum capacity for a hypothetical 
project with the following parameters:

Given:

    Maximum diversion flow = 100 ft3/s
    Average annual diversion flow = 20 ft3/s (during operating  
    season).
    Plant operates 10 months per year.
    Penstock = 10,000 linear feet and 48-inch diameter.
    Gross Head = 400 feet.
    Head loss @ 100 ft3/s = 60 feet.
    Net head @ max. flow = 340 feet.
    Average plant efficiency = 70 percent.

Calculation:

Installed capacity in kilowatts =

     (Q)(H)(e)  = kW
       11.8

    (100)(340)(.70) = 2,016 kW or 2.02 MW
         11.8
                          
                          
                            Exhibit 3

                         Changes in Head


In order to evaluate the monetary effects of locating the 
proposed structure or powerhouse up or down the stream, calculate 
the value of 1 foot of head for the hypothetical project, using 
the givens in exhibit 2 and assuming the selling price to a 
utility is $.05 per kWh, as follows (note that relocation 
distance is measured by the amount of vertical change in feet):

Annual Value:

(.70 eff)(20 ft3/s)(1 ft)(24 h/d)(30.5 d/mo)(10 mo/yr)($.05/kWh)
                                11.8

    = $434.24/year for 1-foot change in head.

Consider this change in project benefits along with other changes 
in costs due to changes in penstock length or size of generating 
equipment, ease of construction, and changes in volume of 
diversion flow.

The present worth (PW) of a 25-foot change in head for this 
project over 20 years @ 10 percent interest is:

PW = (SPWf of 10% @ 20 yrs.)($434/ft. x 25 ft.) = $92,372.
                           
                           
                            Exhibit 4

                    Hydroelectric Comparisons


Where:

SPWf is the uniform series present worth factor from the 
formula SPWf = (1 + i) n-1
                i(1+i) n

Where:

    i = interest rate
    n = number of payments
    note:  if n is in months, then i must be a monthly interest 
           rate.

    60-100 watts = typical light bulb.
    1,000 watts = 1 kilowatt (kW).
    1,000,000 watts = 1,000 kW = 1 megawatt (MW).
    1,000 MW = 1 Gigawatt (GW) = 1 billion watts.
    1 Kilowatt-hour (kWh) = 1,000/watt-hours (Wh) = energy 
    consumed by a 100-watt bulb lit for 10 hours.
    1 Gigawatt-hour (Gwh) = 1,000 MWh.
    1.5 MW = about 2,000 horsepower.

The following information may help to put the units of generated 
capacity into perspective:

    1.  The average home in California uses 15 Kwh/day, or 5,475 
Kwh per year.

    2.  The town of Yreka (population 5,916) consumes 12,500,000 
Kwh per year (12.5 Gwh).  Discounting peak energy demands, this 
is equivalent to the power produced annually by a 1.5 MW (1,500 
KW) plant (typical small hydro project size).

    3.  12.5 Gwh is equivalent to the power produced in a thermal 
plant burning 21,500 barrels (903,000 gallons) of crude oil per 
year (42 gallons equals 1 barrel).

    4.  At $30 per barrel, the value of 21,500 barrels of oil 
would be $645,000 per year.


07 - REFERENCES

    1.  U.S. Congress, Federal Power Act, 16 U.S.C. 797 and 
Federal Power Commission, U.S. Government Printing Office, April 
1, 1975, stock number 015-000-00324-3.

    2.  Code of Federal Regulations, Title 18, Chapter 1, U.S. 
Government Printing Office.  (Contains Federal Energy Regulatory 
Commission rules and regulations.)

    3.  U.S. Department of Agriculture, Agriculture Handbook 478, 
National Forest Landscape Management, Volume 2, Chapter 2, 
"Utilities," 1975.

    4.  U.S. Department of Agriculture and the Interior 
Environmental Criteria for Electric Transmission System, 1970.  
U.S. Government Printing Office, Washington, D.C.  20402, stock 
number 404-9320-70-2.

    5.  U.S. Congress, Federal Land Policy and Management Act of 
1976, Public Law 94-579 (90 Stat. 2743 - 2794).