1. How much energy does a single chunk of water, 
      of mass 1 kg, gain as it drops over the falls?

              it loses GPE = m * g * H  =  (1 kg) * (9.8 m/s^2) * (54 m)

                           =  529  Joules        for the American falls
        

   2. If only half of that energy is captured by the generator, 
      how many kilograms of water must flow flow 
      through each generator's turbine each second?

              If each kilogram of water produces only half of 529 J,
              then it contributes about 265 Joules.

              Each generator produces 70,000 hp = 52,000,000 Watts, so

                                            52,000,000 J/s
                  kilograms per second  =  ---------------  =  196,000 kg/s
                                              265 J/kg

   3. How does this compare to the mass of all 
      the water actually spilling over the falls each second?

              The American Falls have about 600,000 kg/sec of water,
              so one generator requires about 1/3 of the current flow.
             

   4. There are about twenty generators combined in the US and 
      Canadian hydropower plants. What is the total mass of 
      water flowing through the turbines?

              If each generator requires about 200,000 kg/sec, then
              together they need about 4,000,000 kg/sec of water.
              That is more water than currently flows over the falls ...
              which means that less than half the Niagara River actually
              reaches the falls!
           

   5. How fast is the water going as it runs through the turbine?

              Easy:  the GPE lost by water turns into KE.  

                       KE = 0.5 * m * v^2 =  m * g * H

                       -->  v = sqrt (2 * g * H) =  33 m/s    (American)

              Hard:  intake tubes are 10.5 ft = 3.2 m in diameter.
                     If water moves through the tube with speed v m/s, 
                     then in 1 sec, a sausage of length v meters will
                     pass through the tube.  The KE of this object
                     will be 

                       KE = 0.5 * (mass) *                         (speed)^2

                          = 0.5 * (area) * (length) * (density) *  (speed)^2

                          = 0.5 * (8.0 m^2) * v     * (1000 kg/m^3) * v^2

                          =  4020 * v^3  Joules      in 1 sec

                     If half this energy is transferred to the generator,
                     which yields 52,000,000 Joules each second, then we find
 
                                 [  52,000,000 J  ] 1/3
                            v =  [ -------------- ]       =  30 m/s  
                                 [ 0.5 * 4,020 J  ]

              So the two different methods yield roughly the same result...