The Power Equation      P = QehK

The "power" (P) of a hydropower unit is its capability. To use a car as an analogy, its engine is rated at a certain horsepower (hp). The car has this much capability whether it is running or not. The hydropower plant has its rated power whether it is running or not. The basic unit of measure is watt, but normally it is expressed as kilowatts (1000 watts) or, more often, megawatts (1 million watts). That is to say that 1 megawatt (MW) is 1,000,000 watts. Most normally hydropower output is expressed in MW. In an automobile, one of the big factors which determines horsepower is engine displacement. In a hydropower unit, the two BIG factors which determine this are flow rate (Q) and head (h). (careful this link is from England and speaks metric)

The "flow rate" (Q) is a measure of the volume of water passing through the turbine per unit of time. For hydropower in the United States the units used are cubic feet per second (cfs). At home, flow rates from your home system are measured in gallons per minute (gpm). But both of these are, in the truest sense, volume of water per time. A fact that should be rather obvious is that as the flow rate increases then the power output increaes. By the way, for a particular location on a stream the possible flow rate through the unit is controlled to a great degree by "water availability".

The "head" (h) is a measure of the vertical difference in elevation between the reservoir (lake) and the discharge point of the hydropower turbine. Head is clearly a surrogate for pressure. Above it was stated that as flow rate increases the power output increases. It is also true that as pressure of the water at the turbine increases, power output increases. And head (h) is merely a surrogate for pressure. In fact, for each foot of head available the pressure increaes by 0.433 pounds per square inch (psi). So at 100 feet of additional head, the pressure increases by 43.3 psi. By the way, in mountainous or semi-mountainous regions (as compared to flat lands) more head is possible, because of the "topography of the land".

The "coefficient" (k) is the the least important of these factors to understand. In fact, in an engineering equation like the power equation, the units must "work out". And without k, the units would be Watts = cfs * feet (because e if unitless). Clearly cfs * feet is feet to the fouth power per second (which is rather meaningless) , NOT Watts. So k is conversion factor. If you have 15 inches of rope then you have 1.25 feet of rope because the conversion factor is 1 foot / 12 inches. By the way, in England they use the metric system, so their k is different than our (US) k because their units are different.