6

This is an excellent question, and one that has not yet been fully answered by people working in this area of research. Let's look at some of the bigger projects and see what their proposals are: ITER, the tokamak (giant doughnut) being built in France is supposed to be the first reactor to ever beat the break even point, unless someone either beats them to ...


6

The current generated by a conductor moving through a magnetic field is related to the length of the conductor and the magnetic field strength. A coil is simply a convenient way to get a longer length of conductor within a given space. The cross sectional area doesn't actually matter in terms of generating power except insofar as it determines the current ...


6

But I'm thinking it may be better (more efficient maybe? less vibration maybe?) to run it in three-phase mode, with the load balanced among the three hot terminals. Correct. Figure 1. The load is constant through the generator cycle. Source: T. Davies - website not found. My thinking, based on my coarse understanding of generators, is that in three ...


5

As you add bulbs in parallel, the load impedance drops and to light all the bulbs you have to pedal harder. When the load impedance becomes less than the output impedance of the generator, then more work is expended in heating the generator coils than in lighting the bulbs. In the limit of near-zero load impedance (short circuit), all the pedal work is lost ...


4

Generators don't use fuel at all. You are apparently asking about a generator coupled to some sort of chemical engine, like a "genset". Usually these engine/generator combinations have a regulator that tries to keep the speed reasonably constant. This supplies as much fuel to the engine as necessary to maintain that speed. With lower electrical load on ...


4

At a lower load they run at the same speed but the generator becomes easier to turn which means the throttle to the engine driving it can be reduced to save fuel. The throttle is controlled by a system designed to maintain the correct speed no matter the load. The efficiency Vs load curve is going to be generator specific.


4

If you have horizontal water flow (no head) you are extracting its kinetic energy (from its velocity) not its potential energy (from its head). So you need to know the velocity of that flow as well as the flow rate to calculate the energy available. (You can calculate it from the flow rate and channel dimensions) You cannot extract all the available ...


3

the iron structure(s) inside a generator serve the purpose of minimizing flux leakage, which maximizes the utilization of the magnetic field in generating electricity. One of the ways in which they accomplish this is by furnishing what is called a return path for the lines of force, so as to "close" the magnetic circuit inside the device. For things ...


3

Electronics can be quite sensitive to the quality of the power supply, particularly in terms of providing a stable voltage and the right sort of AC waveform. There are two obvious solutions for backup power. The first is to use a generator powered by an IC engine. The complication here is that the engine output needs to be matched to the load by regulating ...


3

You can have brushless DC as well, but its all a balancing game of cost over its lifetime. Efficiency only matters when the entirety of the process is considered. One high efficiency process can be useless if it forces you into a low efficiency one later. AC can be generated directly off something that rotates. If you smooth it out to DC, you lose some ...


3

TL;DR:IMHO, the driver for keeping the electrical generation closer to the hub is for primarily for eliminating -as much as possible- moving parts for safety concerns. The secondary benefit is the reduced losses. IMHO the problem is the distance and the complexity. For the following examples I will use as an example a Wind turbine with the following ...


2

The most practical and simplest is to review the breaker box. Below is an example breaker box for a typical US single family house. Note: US base houses are power using 120 VAC. Add up total current for circuits that are of interest, which essentially will help the estimate the total current. Then calculate the total wattage for the generator. Also ...


2

Obtain a cheap plastic hand crank generator or several and play. That will show you much of what you need to know. If you turn the handle 1/8th turn (45 degrees) in 1/4 second then its RPM is 60/time_to_turn x fraction of turn = 60s/0.25s x 1/8 turn ~~= 30 RPM. In engineering 30 ~~~= 100 in many cases :-). But if you want closer to 100 RPM then a 3 or 4 ...


2

what you are describing here is basically a reversed heat pump that uses a temperature difference to run a heat engine in which the working fluid is so chosen as to undergo a phase change during the engine's power extraction cycle, and the high temperature source is a solar collector of some sort and the low temperature sink is night-time ambient. Systems ...


2

While the other answers here are compelling, they're only half the truth. If you consult your manual you'll find that the generator load windings are reconfigured when using the mode selection switch to one of three output modes. In either mode, the full output of the generator is available, either as 104A@120V(1ph, one circuit), 52A@120/240(1ph/split, two ...


2

Here is the cheap way to proceed. First, find you a battery-powered weedeater that is second-hand, or a discard. disassemble and test the motor for operation. If it operates, now you have a generator that has almost exactly the right power output for your application, because in its previous lifetime it was a motor in the same application as the little gas ...


1

@Transistor's answer is quite complete and correct, but I feel a bit more complex than what you require. The only difference in how you pull power off of your generator is in how the output windings are connected to a plug. Each of the three windings will have the same rating whether you pull 1, 2, or all 3 phases. So if your total load (the bunch of ...


1

My question was based on an appearance that the thin winding wire seemed to violate the laws of physics, so there might be some other factors at play. Neil_UK answers this in generator coils: max power given wire gauge on the Electrical Engineering site. The bottom line is no. It's mainly a matter of using multiple coils in parallel to manage the current ...


1

Note: Small Gauge mean the wire can carry a significant amount of current. Example per the attach AWG 4-5 can carry 100A. There are lot other factors need to be considered before selecting a wire. Below is guide line help you appreciate wire gauge vs current. This is purely a guideline. I suggest searching American Wire Gauge Charts vs Electrical Current ...


1

It will vary between 0 and maximum depending on the demands of the drive system - ie the engine output is shared whatever the wheels need and the remainder to the generator to charge the batteries.


1

At the very basic level, this is how steam turbines or an organic rankin cycle in solar thermal power plants work. https://www.barber-nichols.com/sites/default/files/wysiwyg/images/rankine_cycle_thermodynamic_diagram_large.jpg However you would be limited/constrained by the volume of the lower container. By increasing its volume it takes longer to be ...


1

It’s a pointless design since all you do is decrease the amount of magnetic coupling between the rotor and stator. You may as well just reduce the current - it’ll have the same effect. You’d have to explain why you think that a wildly varying load would somehow benefit from having less of a motor available (when the rotor is not all the way in). What sort of ...


1

I think you may be confusing vertical axis wind turbine based on lift mechanism as in an airplane wing, with the low efficiency models that are used to show the wind speed on rooftops as part of weather vane. For example in a sail ship as opposed to intuition, the best wind is not the back wind which pushes the ship forward. it at best will push the ship ...


1

your circuit requires (3 volts x 10 microamps) = 30 microwatts to operate. in a 24 hour day it will consume (30 uW x 24hr) = 720 uW-hours of energy. your pump produces (1 oz-in x 100RPM) units of power (we'll worry about the unit conversion later). in a day it will produce (1 oz-in x 100RPM x 60 minutes) of mechanical energy. Now go to "Unit Conversions" ...


1

You're supplying a power of only 30 microwatt, so in a day 2.6 joules will be generated. That's very little. 1 oz-in is about 0.0071 newton meter, 100rpm is about 10.5 rad/s. 0.0071Nm at 10.5rad/s multiplied is 0.074 Watt. With 2.6 Joule you'll be able to do that for about 35 seconds. And that's assumed there are no losses, anywhere. But there will be. ...


1

AC generators are simpler. Their inherent construction property is that the sine wave of AC is directly derived from the rotary motion of the rotor. There is no commutator, which is a part that suffers wear over the course of use of a dynamo (a DC generator). Even with a dynamo, the current isn't exactly DC, just an approximation. Neither is "better" or "...


1

There is a back EMF in a motor, due to current flow in a magnetic field. No. A motor develops back EMF due to conductors cutting across magnetic field lines. This has nothing to do with current. So, in a generator, is there a reverse rotation (opposed to original rotation) due to a current flow in magnetic field? Reverse rotation doesn't make sense. A ...


1

Just like gear ratios, if you change the diameter you change the torque. The power will be the same, but if you decrease the primary gear the RPM will go up and the torque will increase. I always remember it from riding a bicycle.When you change from the larger sprocket on the pedal gear to the smaller it is easier for you to pedal but you go slower. In ...


1

No expert, but the problem is interesting. Assumptions: Generate $6.5V_{RMS}$ at 100 rpm with 9 stator coils. $V_{MAX} = 9.2V$ 100rpm = 1.67rps if radius to center of magnets/coils = 0.1m $C = 2 \pi r = 2 \pi \times 0.1m = 0.628m$ at 1.67 rps, v = 1.05m/s N48 Neodymium Bar Magnets 1 in x 1/2 in x 1/4 in L = 0.0254m, W = 0.0127m, D = 0.00635m. ...


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