Science Project help... dam energy?

You are a hydrologic engineer asked to do a feasibility study of building a hydroelectric dam on the Hiwassee River in the Narrows. The surface elevation of the reservoir is limited to 1180 feet above sea level. The Hiwassee has a 1210 mi2 drainage basin upstream from the Narrows. The average rainfall in the basin is 45 in/yr. The Hiwassee River is a 5th order stream. The average area runoff is 30%. 70% of precipitation will infiltrate the ground or evaporate before it flows through the penstock. The Narrows is deep enough to create a large upstream reservoir.The concrete dam will have one penstock that can have a discharge (Q) equal to the average natural discharge of the river into the reservoir. Electrical generation in this plant will maintain 75% efficiency. The big question you are attempting to answer is if this power plant will be able to supply total electrical energy needs for a minimum of 30,000 average households and how much direct revenue this plant will make per year

Answer:
Here's an example of the power output of a hydro-electric power plant for you to plug in your own figures.
What the average consumption per household will be, I've no idea.
But, the following should supply at least the 30,000 households.

A hydroelectric power station is 150 m below the level of water in a reservoir.
Calculate the power output when water flows to the turbines at 32 m³/s and the overall efficiency of the plant is 75%. (Water density = 1000 kg/m3)
In this example, the PE of the water is converted to KE to drive the turbines.Therefore:

(Mass = Density x Volume) The mass of water flowing per second = 32 x 1000 kg
PE of water = mgh = 32 x 1000 x 9.81 x 150 (Joules)
Power is the rate of transfer of energy, measured in
Watts (J/s)

Assuming 100% efficiency : P = 32 x 1000 x 9.81 x 150 J/s (W) = 47.09 MW

However, efficiency = 75%. Therefore:
Power output = 47.09 x 0.75 = 35.32 MW.

.
Questions for you.
What is your definition of average energy consumption by a household?.
What is the selling price per KWH for the energy?
What is the elevation drop from the water intake to the power plant turbine?

Need these answers to solve problem
One penstock? You'd be hard pressed to turn on a nightlight.
Feasibility study :

Reservoir level = 1180 ft = 359.756 m. from m.s.l..
Reservoir water INTAKE per year from catchment area=
1210 x 45 x 2.54 x 0.3 / 100 = 415 cubic meter/year = 1.3673 cubic meter/hr.(avg)

Head available = 1180 ft = 359.75 m from M.S.L

Electrical power generation capability =
1.3673x 359.75x 0.75x1000 / (3600 x 75 x 1.35 .)
= 1.0121114 kw. ( MAXIMUM)

Since generation will be very small, the plant can not be run continuously over the year due to water shortage.
But it can be run seasonally a few months ( in rainy season )only and have to be run as peaking power station in conjunction with a thermal power station to be built .

For a community containing 30,000 households, the average energy consumption will be 30,000 x 1.5 = 45000 kwhr / hr.
i.e. a 45 MW TPS is required to meet the demand.
The hydal station capacity will be too small for that and not a feasible project considering the cost involved and revinue earning capability with the hydal station.

The answers post by the user, for information only, FunQA.com does not guarantee the right.



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