How to change gear speed on a multi-planetary gear system based on electricty request?

Thread Starter

humptydumptyaac

Joined Nov 3, 2021
16
Hi, newbie here.

I want to build an electricity generator.

It would be based on a water wheel, to which I'd connect a multi-planetary gear system, to which I'd connect a rod magnet.

I'd found out that when you begin to extract electricity it creates a sort of resistance, and it lowers the RPM of the magnet/water wheel.

I'm trying to find out if there's a way to correlate the amount of wattage extracted to the resistance created, and hopefully estimate how much that would lower the RPM.

For example, if the magnet is spinning at 2000 RPM, with no appliance connected to the generator, then I'd connect an appliance rated at 2000W and begin using it - what resistance would that create, and what would be the RPM of the magnet?

I understand that as a newbie I may ask stupid or incomplete questions, but help me out here please;

For slightly more context: I'm ultimately trying to find out what force would need to be applied on the water wheel, so that when extracting electricty it would keep spinning despite the resistance created.

For example, let's say the generator is capable of supplying 10 outlets at 220V AC, with about 2000W each, or 20,000W total.

Well, I'd want to find out, in case I use all 10 outlets at once, powering 20,000W of appliances - how much resistance would be created, and what force/water pressure would need to be applied to the water wheel to prevent it from stopping?

Now this may be 2 questions in 1, I only provided the second one as bit of context.

Mainly I'd want to know how to calculate the resistance created.

Thank you.
 

Yaakov

Joined Jan 27, 2019
4,074
Welcome to AAC.

Watts are interchangeable with horsepower, both are measurements of work. 20kW is about 26.8hp but there are also losses in the system so the actual work done by the waterwheel will have to be somewhat higher to generate a usable 20kW of electrical power.

I can't help you directly with your question, I don't even know what units you would expect the answer in, but the formula for calculating horsepower is:

1636011549289.png

Where T is torque in foot-pounds, and N is the rotational speed in RPM.
 

Thread Starter

humptydumptyaac

Joined Nov 3, 2021
16
Hi, newbie here.

I want to build an electrity generator.

It would be based on a water wheel, to which I'd connect a multi-planetary gear system, to which I'd connect a cylindrical rod maget spinning inside copper.

I want the gears to be able to change speeds, much like an automatic transmission in a car, but based on how much electricity is needed.

Let's assume, for example, that the multi-planetary gear system has 10 planetary gears, each set of planetary gears would increase the RPM by 100.

So if no gears are involved - the magnet spins at the same speed as the water wheel, let's assume 1 RPM.

and with the fastest gear involved, the magnet would spin at 1000 RPM.

Now, let's assume this generator has 10 outlets of 200V AC to it.

If no appliance is plugged into an outlet, and no electricity is requested - then no gear is involved and the magnet spins at the same speed as the water wheel.

If I plug in one appliance - a signal would be transmitted from the outlet to the multiplanetary gear system to engage gear 1/speed 1 - and increase the RPM by 100.

If 5 outlets are in use - a signal would be transmitted to engage speed 5 - and the RPM is then 500.

I'm looking for a mechanical/electrical way to send these signals, so that it would not involve any raspberry Pi and coding or anything similar to it. (But if a mechanical way is not possible I'd like to know a computer-based way)

Also, I gave the example above with 1 oulet to 1 gear speed to make it clearer, but it would be ideal if this increase is based on the wattage of the appliances, not the outlets themselves. For example: Let's assume that speed 1 can generate 2000W of electricity, and I plug in 2 appliances at 1000W each, then 2 outlets would be used, but ideally the speed would stay at 1/gear 1. As there's no need to enagage speed 2 if speed 1 produces enough electricity for both devices.

As you can imagine - if this can be achieved as I described it - it would solve the problem of storing electricity, since the generator would produce only the needed electricity, and not more (assuming the max. produced electricity is also the max. used durnig peak times)

Thank you.
 

Yaakov

Joined Jan 27, 2019
4,074
It seems the shifting would be based on demand for torque not on the demand for current. It would be the same as if you were trying to go uphill in a vehicle, you would need to downshift to produce more power at a loss of forward speed and you wouldn't do it by measuring your angle of inclination but by the demand on the engine.

Your case is complicated by the fact that the generator must be optimized for the output and you will need to maintain a particular speed for that, you won't be speeding up and slowing down based on load, you will just be turning harder. The voltage will be proportional to the speed of rotation so you won't be running it slowly for a light load and faster for more load.
 

Thread Starter

humptydumptyaac

Joined Nov 3, 2021
16
Welcome to AAC.

Watts are interchangeable with horsepower, both are measurements of work. 20kW is about 26.8hp but there are also losses in the system so the actual work done by the waterwheel will have to be somewhat higher to generate a usable 20kW of electrical power.

I can't help you directly with your question, I don't even know what units you would expect the answer in, but the formula for calculating horsepower is:

View attachment 251836

Where T is torque in foot-pounds, and N is the rotational speed in RPM.
Thank you for taking the time to help.

If I understand correctly (or at least close to correctly) - the force of the water acting on the water wheel would need to be 26.8 horse power (or closer to 30hp to account for losses) in order to be able to extract 20kW of electricity?

But the water pressure may be different from HP. I assume the force of the water would be measured in pressure units like PSI.

Is there a way to convert or at least estimate what water pressure would create how much horse power?
 

Yaakov

Joined Jan 27, 2019
4,074
I would suggest you build a small model of your generator and experiment to get an idea of actual behavior. I have the sense oyu are laboring under some misconceptions that would be best corrected by practical experience and a scale model would be one of the best ways to do that. Scale down the size and voltage, and try to make something work.
 

Yaakov

Joined Jan 27, 2019
4,074
Thank you for taking the time to help.

If I understand correctly (or at least close to correctly) - the force of the water acting on the water wheel would need to be 26.8 horse power (or closer to 30hp to account for losses) in order to be able to extract 20kW of electricity?

But the water pressure may be different from HP. I assume the force of the water would be measured in pressure units like PSI.

Is there a way to convert or at least estimate what water pressure would create how much horse power?
Is it an undershot or overshot wheel? What is the average rate of flow of the water? What is the size if the wheel and its paddles or buckets? You have to know how much torque will be generated by the weight of the water acting on the wheel's perimeter over time to decide that.
 

BobTPH

Joined Jun 5, 2013
4,322
You talk about pressure. Are you talking about a gravity powered water wheel or a turbine, which is powered by the kinetic energy of the water coming in?

Bob
 

Thread Starter

humptydumptyaac

Joined Nov 3, 2021
16
Is it an undershot or overshot wheel? What is the average rate of flow of the water? What is the size if the wheel and its paddles or buckets? You have to know how much torque will be generated by the weight of the water acting on the wheel's perimeter over time to decide that.
it would be an overshot water wheel. I assumed the sizes and water flow would be determined by the HP needed to produce the electricity, since it's not already built, I'm just designing it now. But let's assume some sizes I had in mind. For example, a water wheel with 30 CM on the outer diameter, and 20 CM on the inner diameter. It would have a circumeference of 62.8 CM, divided by 30 paddles, and give us each paddle with a volume capacity of 400 grams of water. The water wheel would be covered in such a way that the water would not spill to the sides or out until it reaches one third of the way down, so that 10 paddles would be filled with water at any given time. 400 grams times 10 paddles would give us 4 liters of water, or a little over 1 gallon. So that's how much water would be on the wheel at any given time (but again, nothing is built yet, so I can easily decide on a bigger sized wheel as needed)
 

BobTPH

Joined Jun 5, 2013
4,322
The power can then be calculated from the mass of the the water and the drop in height.

If 1 Kg of mass drops 1m every second, it would produce 9.8W of power, theoretically. That would be if there is no change in velocity if the water, which is the ideal case. The velocity will certainly differ, so to get a better estimate you need to subtract the difference in kinetic energy in and out.

Bob
 

BobTPH

Joined Jun 5, 2013
4,322
To be clear, the kinetic energy would add to the power if the water exits with less velocity than it enters. Not sure how big this effect is either way.

Bob
 

Thread Starter

humptydumptyaac

Joined Nov 3, 2021
16
For 20kW the wheel would have to be huge.
Here's a calculator.
Thanks for the calculator link.

Tell me please this, is it possible to leverage pressure from a tank instead of from the water wheel, and therefore have a smaller wheel?

For example: Let's assume there's a stream/river from which we get the water, under the river stream we place a water tank with the capacity of 1,500 litres of water, and under the water tank we place a relatively small water wheel, up to 1 meter in diameter or smaller.

We'd first block the exit from the water tank to the water wheel, we let the water tank fill up, and when it's filled we let the water fall from the water tank, onto the water wheel.

Now, assuming the exit flow from the water tank and the entering flow from the stream into the water tank is the same - the water tank would be constantly filled, and it would constantly apply the pressure of 1,500 litres of water onto the water wheel.

Can you please tell me if and how I'm mistaken?
 

BobTPH

Joined Jun 5, 2013
4,322
The energy comes from the change in height of the water. Putting a tank in between that fiills and empties at the same rate makes no difference at all.

Where do you think the additional energy is coming from?

Edit: I was hoping you would do the calculation yourself and discover what is needed. If you prefer to work in horsepower, 1hp is 550 ft pounds per second.

That means that you get 1 hp by capturing all the energy from 550 pounds of water dropping 1 foot every second or 55 pounds of water dropping 10 feet every second.

Bob
 
Last edited:

Yaakov

Joined Jan 27, 2019
4,074
The energy comes from the change in height of the water. Putting a tank in between that fiills and empties at the same rate makes no difference at all.

Where do you think the additional energy is coming from?

Bob
You would have to use power to pump it up higher to add potential energy, inevitably at a loss, so...
 

Irving

Joined Jan 30, 2016
2,424
If this is a real-world water-wheel driven generator the "a magnet spinning inside copper" isn't going to be a practical solution. You need to understand how much power the wheel generates and match that to a real generator. Power output is not related directly to speed, but voltage is - as is AC frequency, Typically a generator will want to run at a fixed speed, eg 1800 or 3000rpm at which point it will generate an appropriate voltage and frequency output. At that point, ignoring losses, the generator will exert a torque on the driving shaft.

Power (W) = Torque (Nm) x rotational speed (rads/sec)​

so, rearranging,

torque (Nm) = 10 * Power (W)/speed (rpm) (the x 10 is a rough conversion from rpm to radians/sec)​

Your gearbox is there to maintain the motor revs at the required speed whilst matching the needed torque based on power demand. Ideally this is a continuously variable gearbox. If its stepped then you will need some additional input speed control, either a brake on the water wheel (or input shaft) or more commonly a controlled water flow (sluice gate).

You can estimate the output power of a top-fed water wheel by:

Pout = g x F x D x eff​
where​
g = 9.81 gravitational constant​
F = water flow in litres/sec​
D = diameter of wheel in metres​
eff = efficiency of wheel, approx 0.85 for a steel one, 0.7 for a wooden one​
So a 5m diameter wooden wheel driven by 100litres/sec water flow will generate 9.81 * 100 * 5 * .7 = 3.4kW, of which you'll actually get around 70-80% of at the generator output, so 2.6kW generated approx.​
The wheel is working most efficiently when every bucket is filled and no water is wasted. An N bucket top-fed wheel (which has 1/2 the buckets in active use), with each bucket holding b litres will therefore revolve at:​
R = 2F/(N x b) revs/sec​
where F is the flow rate. eg a 20 bucket wheel holding 10l per bucket will rotate at 2 * 100/(20 * 10) = 1rps or 60 rpm, which will need gearing up by 30:1 for a 1800rpm generator or 50:1 for a 3000rpm generator.​
 
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