Generating electricity using a DC motor

[AyobA]

Hi all.

Am new to the field and just started my degree.

I have a project I am interested in trying.
I am looking into generating about 300 watts of electricity using a wind turbine.

In theory, would I be able to use a 12v DC motor rated at 300 watts. Then use the wind to turn this motor?
Would it generate 300 watts of 12v electricy that I can use to charge some deep cycle batteries?

 

[BrianE22]

The output voltage will vary depending on shaft speed. If it is not a permanent magnet motor you will have to energize the field. To get out 300 watts you'd have to put in greater that 300 watts of mechanical power so the shaft load would be operating a bit higher than designed - probably o.k.

 

[IRstuff]

In theory, would I be able to use a 12v DC motor rated at 300 watts. Then use the wind to turn this motor?
Would it generate 300 watts of 12v electricy that I can use to charge some deep cycle batteries?

The answer is basically no. There are electrical and mechanical losses that will require excess mechanical power to overcome the losses. However, on top of that, there is no DC motor that outputs pure DC when spun up. In order for a DC motor to work correctly, the winding power connections require commutation. In a brushed motor, therefore, you'll get a chopped voltage that must be rectified to get DC, and that will incur additional losses. In a brushless motor, you'll get nothing without removing the electrical commutation circuitry, since that circuitry operates in one direction only. Removal of the commutation circuitry will then require you to convert each motor phase into DC using rectification, etc, again incurring losses.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert!

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[LionelHutz]

Try looking at otherpower.com

 

[AyobA]

Hi guys. Thanks for answers.

Ok I understand that there are many losses of energy.

But lets take a closed system where we remove the effects of losses due to friction etc.

Lets say we use a 300w 12v dc motor to winch up a mass to a height of 100 meters using a cord. And it takes 4 hours for the mass to reach the top.

Could we then say that if we released that mass so that the rope turns the same motor over 4 hours until it reaches the bottom, would that motor then generate 12v of electricity at 300w?

Sorry if it sounds a bit far fetched but it is something that seems would be the case due to conservation of energy in a closed system

 

[IRstuff]

Lets say we use a 300w 12v dc motor to winch up a mass to a height of 100 meters using a cord. And it takes 4 hours for the mass to reach the top.

Could we then say that if we released that mass so that the rope turns the same motor over 4 hours until it reaches the bottom, would that motor then generate 12v of electricity at 300w?

Asked and answered, which is NO. A "closed system" is NOT friction-free; that's not real physics. If this is strictly a homework problem, then sure, go ahead and ignore real physics, friction, resistance, etc.

In a physically realizable 12V 300W motor does not output 300 W of mechanical power, so you will not get back anywhere close to 300 W.

TTFN (ta ta for now)
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[danw2]

Is there some reason not to use an automotive 12V alternator or generator?

 

[waross]

However, on top of that, there is no DC motor that outputs pure DC when spun up. In order for a DC motor to work correctly, the winding power connections require commutation. In a brushed motor, therefore, you'll get a chopped voltage that must be rectified to get DC, and that will incur additional losses.

A DC motor versus a DC generator.
The difference between a typical DC motor and a typical DC generator is either a couple of Volts or a couple of RPM.

For a permanent magnet DC motor:
Imagine a DC machine on a test stand.
The test stand is able to drive the machine or to brake the machine.
The motor is rated 1000 RPM at 12 Volts.
The motor is driven at 1000 RPM and 12 Volts is applied.
The speed is varied slightly until the current drops to zero.
At this point the applied voltage matches the back EMF and machine is neither motoring nor generating.
Now slow the speed. As the driven speed slows, the machine becomes a motor and the current increases as the speed is reduced.

Now hold the sped constant at the null point and vary the voltage.
As the voltage is increased, the machine becomes a motor and tries to accelerate, drawing current.
Now drop the voltage below the null point voltage and the machine becomes a generator and drive current back into the source.

The voltage is almost directly proportional to the speed.

Over-speed will generate over-voltage on an open circuit.
The voltage limit with increasing speed is the lower of
1. insulation failure
and
2. Flash-over between the commutator bars.
Over current may also cause flash-over between the commutator bars.
Flash-over will generally occur first.
Anecdote alert.
Test and commissioning on a large walking drag-line excavator.
There were four 1300 HP motors in series with four 1300 HP generators.
Due to a wrong setting in the controls during testing, All four generators flashed-over the commutator bars.
This is called ringing the Comm'.
Four simultaneous rings of fire around the commutators of four 1300 HP generators was spectacular.



--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[waross]

In a brushed motor, therefore, you'll get a chopped voltage that must be rectified to get DC,

Not quite.
The voltage is not chopped, but there is a ripple, generally not a problem.
Take the number of commutator segments between a positive brush and the closest negative brush.
For example, 8 segments.
Each segment is (180 degrees / 8 = 22.5 electrical degrees).
The ripple will be comprised of the top 22.5 degrees of a sign wave, in the first instance.
Any inductive or capacitance will tend to smooth that out.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[waross]

In a physically realizable 12V 300W motor does not output 300 W of mechanical power, so you will not get back anywhere close to 300 W.

Motors are rated in mechanical Watts and kiloWatts,
The expected output of a 300W motor is 300 mechanical Watts.

The output will be directly related to speed.
At slightly above rated speed, the motor will put out 300 Watts at rated voltage.
At 1/2 rated speed, the motor will put out 150 Watts.
The limit is current times terminal voltage.
An old automotive generator would often be run at 300%, 400% or even 500% of the speed needed to produce the rated voltage.
The field would be weakened to control the voltage.
Many years ago, there was an after market circuit for early alternators.
The control was bypassed and the engine run at a fixed, high speed.
With higher voltage rated diodes, the output voltage was high enough to run 120 Volt rated power tools.

The challenge with using a DC motor for a wind generator is the variable speed.
I recently worked on a small windmill installation.
This charged 24 Volt batteries to run a well pump to supply drinking water for cattle.
There was motion detector that started the pump to put water in the trough when the cattle came near to drink.
As the wind speed increased, the current increased.
The speed was limited by the load imposed by charging the batteries.
If the wind became to strong, the current and the voltage would rise to an unsafe level.
A protection circuit detected this and stopped the windmill by shorting the output.



--------------------
Ohm's law
Not just a good idea;
It's the LAW!