6. What is the frequency of the voltage created by one generator? The shaft makes 107 revolutions per minute, but there are 12 magnets attached; so each time the shaft turns, there are 6 cycles of North/South/North. The speed at which the magnetic field oscillates is 107 rev 6 cycles 1 min --------- * --------- * ------- = 10.7 cycles/sec min rev 60 sec 7. How many turns, N, of wire must there be inside a single generator? Each generator creates a voltage V(RMS) = 13,000 volts; the peak voltage V(peak) = V(RMS) * sqrt(2) = 18,400 volts. V(peak) = N * A * B * omega where N = number of turns A = area of coils = 9 m^2 B = strength of mag field = 0.1 T omega = angular frequency of generator = 2*pi*(10.7 cycles/sec) = 67 radians/sec So, we can re-arrange to find N: 18,400 volts N = ------------------------------ = approx 300 turns (9 m^2) * (0.1 T) * (67 rad/s) 8. How large is the current running through the coils of wire? The power generated is P = 52,000,000 Watts, and the RMS voltage is V(RMS) = 13,000 volts. 52,000,000 W P = V(RMS) * I ---> I = ------------ = 4,000 Amps 13,000 V 9. How long is the wire? The coils are roughly squares of length 3 m. Their area is thus (3 m)*(3 m) = 9 m^2, and the distance around one coil is (3 m) + (3 m) + (3 m) + (3 m) = 12 m. The total length of wire is roughly L = (300 turns) * 12 m = 3600 m 10. All that current running through the wire could heat up the wire quite a bit -- wasting lots of the energy generated. If the wire is made of pure copper, and circular in cross section, what is the minimum possible diameter it can have? The resistance of the wire is (resistivity) * (length) R = ------------------------- (cross-section area) Now, if ALL the generated power (52 MW) is dissipated as heat within this wire, we're in trouble -- our generator will produce no real output (just a lot of heat). So we must make the wire big enough that the resistive heating is much less than the generated power; say, only 10%. Then we'd still get lots of power coming out of the generator. So, what resistance would give rise to a loss of about 5 MW as heat? (dissipated power) = I^2 * R 5 MW ---> R < ----------- = 0.31 ohm (4000 A)^2 Okay, the wire's total resistance must be less than 0.31 ohm. So, solve for the cross-section area: (resistivity) * (length) area > ------------------------- (resistance) (1.72 x 10^(-8) ohm-m) * (3600 m) > --------------------------------- 0.31 ohm > 0.00020 m^2 Assuming a circular cross-section, this corresponds to a wire which has at least radius r > 0.008 m = 8 mm And so diameter d > 0.016 m = 16 mm 11. What is the mass of all that copper? mass = (volume) * (density) = (area) * (length) * (density) = (0.00020 m^2) * (3600 m) * (8900 kg/m^3) = 6400 kg 12. Is it really possible to pack the required number of turns of wire, of the required diameter, into the space inside one of these generators? Remember that we need N = 300 turns. How many coils could we pack into a single row, side-by-side, which was one meter wide? 1 m (coils per meter) = ------------- = 62 0.016 m/coil So, if we arrange the coils in a package which is 62 rows wide wide and 5 columns high, we can make the required number of turns. This means that the wire would have to be arranged in a space around 1 meter wide and 8 cm high (or any other equivalent combination of width and height). Looking at the pictures of the giant generators, I conclude that there is easily this much room available.