A MOVING FLUID’S KINETIC ENERGY

__
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 V^{2} 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 V^{3} 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:

** **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 = ----------------------

5252

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

RPM

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

RPM

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.

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
(