How Much Power Can a Gulf Stream Turbine Produce?

            There is a great deal of information available on calculating the power output from wind-powered generators.  Although there is also some information available concerning the power output from water-driven turbines, the only information that I could find applies to the impulse turbines and the pressure or reaction turbines, which are very different from those that would be used for these underwater generators. 

            The impulse turbine is the type where a head of the supply water is converted into kinetic energy before it is directed through a nozzle as a high velocity jet onto the vanes or buckets that are mounted on the periphery of the turbine’s wheel.  In the pressure or reaction turbine, the wheel is provided with vanes into which water is directed by a series of guide vanes extending around the wheel’s periphery.  The water leaving these vanes is under pressure and it supplies energy partly in the kinetic form and partly in the pressure form.  Both of these water-driven turbines obtain their energy from a head of water, making them well suited for dam sites.  The underwater generators would not obtain any energy from a head of water and could be likened more to a child’s pinwheel that would be powered by water rather than air.

            Although the turbines on the generators would have more in common with the wind turbines than with the impulse and reaction water turbines, the water would have much more mass and would be moving more slowly.  A cubic unit of water weighs about 854 times the same cubic unit of air at sea level.  Also, unlike air, water cannot be compressed.  Because no information could be found on the type of water turbine that would be used on the Gulf Stream Turbines, the same method was used for calculating their power output as is used for wind turbines.  The amount of kinetic energy that passes through a turbine can be calculated using the formula: 

(KE = 1/2 x M x V2         M = mass       V = velocity)

The mass is the weight of the fluid that passes through the turbine’s rotor per second.  This can be obtained by calculating the blades’ sweep area and multiplying that quantity by the distance the fluid travels in one second.  This volume is then multiplied by the weight of the fluid per cubic unit to get the mass.  Because the mass passing through the blades in one second is a factor of the current’s velocity, the power produced by the current does not increase by the square of its velocity, but by its cube.  Therefore, the equation for the kinetic energy passing through a turbine can also be written:

(KE = 1/2 x A x D x V3   A = area swept    D = density/cu.m.    V = velocity)

            Though the density of air will vary due to differences in temperature, moisture content, and altitude, a cubic meter of air at sea level weighs approximately 1.2 kilograms (2.6 lb.).  A cubic meter of ocean water weighs approximately 1,025 kilograms (2,260 lb.). That means that a water turbine with a rotor diameter of 50 feet in a current of 5 mph would capture approximately the same kinetic energy as a wind generator having a rotor diameter of 262.5 feet in a 14-mph wind.        

            Because watts are a unit of power based on the MKS (metric, kilogram, second) system, all weights and measures were first converted to metric.  By converting everything to kilograms and meters, the kinetic energy figures produced are in joules per second, which is the same thing as watts.   

            When determining the amount of kinetic energy that can be captured by wind turbines, certain facts hold true.  The more energy that a turbine pulls from the air, the more that air is slowed.  If we tried to extract all the energy from the air, the air would move away from the turbine at zero speed.  In that case no energy would be extracted at all, since all the new air would be prevented from entering the turbine’s rotors.  At the other extreme, if all the air passed through the sweep area without being slowed, we would likewise not have extracted any energy from the air.  A German physicist, Albert Betz, in 1919 determined that, theoretically, no more than 59.3 percent of the kinetic energy in the wind can be converted to useful mechanical energy.

            The Betz’ law applies to wind turbines – not to water turbines.  The formulas that calculate the maximum kinetic energy obtainable from a wind turbine may understate the power obtainable from underwater turbines that are equipped with full-bladed rotors.  Although most of the energy being transferred to the water turbine could be in accordance to the Betz’ law, because water is not compressible, the total amount captured by a full-bladed turbine could be more than that coming from the kinetic energy of just that mass in the turbine’s sweep area.  This additional energy would come from the inertia of that uncompressible water that would be pushing from behind that water that is passing through the turbine.  In other words, the energy being applied to the turbine could be partly in the kinetic form and partly in the pressure form.  If there would be any additional energy coming in the pressure form, it would have the same affect as increasing the mass acting on the turbine.  It would be analogous to a mob of people pressing to get through a closed door.  The forces against the door can be much greater than that being applied by just those people having physical contact with that door.  If there were any additional energy input from the inertia of that water pushing from behind the water that is passing through the sweep area, the apparent efficiencies could be inflated above the limits of the Betz’ law.  This would not mean that the full-bladed turbines would be operating at those higher efficiencies, but rather that there is additional energy input in the pressure form that is not being considered.  If there would be any additional energy from the inertia of the non-compressible water, it should affect mostly those full-bladed rotors that cover almost all of the sweep area – not those with a few narrow blades.    

Although a mass of moving water might produce more power from a full-bladed rotor than an equal mass of moving air, because of the water not being compressible all the calculations that follow make no allowance for this possibility because I could find no proof that this additional force actually exists.  Using the same formulas that are used for wind turbines and assuming operating efficiencies of 42 percent, the amount of kilowatts and horsepower equivalents that could be produced with single water turbines having various rotor diameters would be as follows: 

Submersible Generators in Various Currents 

            Torque is a measure of force on a rotating object and it is equal to the force, times the distance that force is from the fulcrum of the object.  If a force of one pound were applied to the rim of a wheel having a two-foot diameter, the torque would be equal to one foot-pound because the distance between where the force is applied and the wheel’s center would be one foot (one-half the wheel’s two-foot diameter), times the force of one pound.  If the wheel were to turn at 60 RPM, a point on the wheel’s rim would travel one circumference in one second.  Because the circumference of the wheel is Pi times the 2-foot diameter, that distance would be 6.2832 feet.  The power turning that wheel would be one foot-pound, times the 6.2832 feet that the rim traveled in one second, or 6.2832 foot-pounds.  Because one horsepower is equal to 550 foot-pounds per second, which is same as 33,000 foot-pounds per minute, if we divide the 6.2832 foot-pounds of work done per revolution into the 33,000 foot-pounds that equal one horsepower, we find that one horsepower would be required to spin the wheel at 5252 RPM.  This is the formula: 

                                                                              Torque    x    RPM

                                                Horsepower          =           ----------------------             


Because one kW is equal to 1.014 horsepower, the formula can be changed to:      

                                                                              Torque   x    RPM                       Torque   x    RPM

                                Kilowatt                =           -----------------------          =         ---------------------

                                                                                 5252   x   1.341                                  7042.9

            And this equation can be easily converted algebraically to calculate the torque by dividing both side of the equation by the RPM and multiplying both sides by the 7042.9

                                            7042.9 kW

                                                      ----------------------        =        Torque   


            Because rotor blades’ tip speeds will be roughly equal to the speeds of the current, the calculations to determine the rotor’s RPM are based on the assumption that those speeds are equal.  The power captured by the turbine’s rotor increases with the cube of the current’s velocity and with the square of the sweep area’s diameter.  If we lengthen the rotor’s blades to increase the sweep area to increase the kinetic energy that the rotor can capture, the tip speeds will remain about the same as the current’s velocity.  However, because the both the tip speeds be slower and the circumference of the rotor’s sweep area will be larger, the RPM will decrease in reverse proportion to the diameters, times the percent reduction in the current’s velocity.  The following graphs show the torque, rotor diameters, and RPM required for generating the listed kilowatts of electricity from those current velocities that are stated along the bottom of each graph. 




            Because power is equal to the torque times the RPM, the torque will vary inversely with the RPM for any amount of power.   

                                                                              Torque   x    RPM                       Torque   x    RPM

                                Kilowatt                =           -----------------------          =         ---------------------

                                                                                 5252   x   1.341                                 7042.9

            This equation can be easily converted algebraically to calculate the torque by dividing both side of the equation by the RPM and multiplying both sides by the 7042.9.   

                                              7042.9 kW

                                                      ----------------------        =        Torque   



Large Wind Turbines Also Produce High Torque

            Though the water turbines must be rather large to be efficient, just because the wind turbines have been growing ever-larger does not mean that water turbines must also be huge in order to produce low-cost electricity.  Very large wind turbines are more efficient than the smaller turbines because there are economies of scale.  The costs of the foundations, road building, electrical grid connections, plus a number of components in the turbine (the electronic control system, etc.) are somewhat independent of the size of the machine.  The larger machines are particularly well suited for offshore wind power because the costs of the foundations do not rise in proportion to the size of the machines.  Maintenance costs are also largely independent of the size of the machines.  The larger wind turbines also have an advantage over the smaller machines in that they have taller towers and their longer blades place their sweep areas farther from the ground where the winds are both stronger and less turbulent.   

            Many of the larger wind turbines have diameters of 80 meters (262.4 feet) and are rated to produce 2,500 kW in winds of between 15 mph and 17 mph.  Based on the assumption that the rotors’ tip speeds will equal the wind velocities, the tip speeds would be between 22 and 24.9 feet per second.   Because the circumference of the rotor would be 824.35 feet, the rotors would require between 33.1 and 37.47 seconds per revolution, rotating at between 0.5516 and 0.6245 RPM.  The torque forces produced by these large turbines would be between 23,156,525.2 and 26,216,914.4 foot-pounds.

Torque = (7042.9 x 2500) / .6245 RPM = 28,190,552.4 foot-pounds

Torque = (7042.9 x 2500) / .5516 RPM = 31,920,322.7 foot-pounds

Present wind turbine blade technology is based on advanced processing methods and high-performance materials systems that enable the mass production of large blades. The largest blades currently in serial production are about 40-50 meters (131.23-164.04 feet) and their sizes are expected to increase considerably in the future.

 Blade in transport

The above illustration shows a rotor blade being installed on a new turbine that will have a generating capacity 5,000 kilowatts, owned by REpower Systems AG (Hamburg, Germany).  The rotor’s diameter will be 410 feet (125 meters), giving it a sweep area of 132,302,025 square feet – an area equal to two-and-a-half soccer fields.    The circumference of the rotor’s sweep area is 1,288.052 feet, which is only 31.5 feet less than a quarter mile.  Assuming that the rotor’s tip speed will be the same as a 15-mph wind, the tips would move at a speed of 22 feet per second and require 36.66 seconds to complete just one revolution.  The speed of rotation would be .6111 RPM.  The torque produced would be (7042.9 x 5,000 kW) / .6111 RPM) or 57,624,775 foot-pounds.  The strengths of those shafts and gears required to transmit the equivalent of 3,728.5 HP at these very low rotation speeds would be truly mind-boggling.