Weight Driven Generator Calculations

[OhioRay]

Hello! I am in the beginning stages of designing a system that would drive a generator using gravity - similar to an old grandfather clock. I have a couple of questions about my assumptions/design that I am stuck on:

- I found a motor which is rated 6v-12v DC, 188 RPM @ 12V with no load, and a max no-load current of 0.53mA. Does this mean that if I were to drive the motor at 188 RPM, I would see 12V of output, with a max current draw of 0.53 mA?

- For my gear system right now I am calculating that in order to drive @ 188 RPM I will need to have a four gear system, with the 2nd and 3rd gears on the same shaft in order to step the gear ratio twice. Basically I am looking at this: Weight drives Gear 1 (100 Teeth), Gear 1 drives Gear 2 (16 teeth), Gear 2 is on the same shaft as Gear 3 (100 teeth) so they have the same rotational speed. Gear 3 drives Gear 4 (16 teeth) which is connected to the motor shaft to be driven at 188 RPM. In order to achieve the proper output and the drop rate that I want, I calculated the first gear needs to rotate at 0.5 RPM, which gives me a drop rate of 1.2 inches/min - or 6 feet over the course of an hour. What I am unsure of is how to calculate the weight that I need to get this rate! How would I go about calculating the friction/other forces in the system which will create an equilibrium with a certain weight at this drop rate?

Any help on either of these 2 items is much appreciated!

Thanks.

 

[Baluncore]

Welcome to PF.
You can expect an efficiency of about 50% in the motor and 50% in the generator. So the generator voltage will be the same as the motor but the current generated will be about one quarter of the motor current.

Consider that you are converting potential energy E = mass * gravity * height, into electrical energy, E = volts * amps * time.
Therefore; volts * amps * time = mass * gravity * height
Hence; required mass = (volts * amps * time) / (gravity * height)

Use the MKSA units. One hour is 3600 seconds and 6 feet is 1.8288 metres.
Therefore; mass = (12V * 0.53A * 3600 sec) / ( 9.8 ms^-2 * 1.8288 m)
mass = 1277.5 kg, or about one and a quarter tonnes.
Then apply the 25% efficiency by increasing the mass by four times to 5110 kg or 11,265 pounds.

 

[Danger]

a motor which is rated 6v-12v DC, 188 RPM @ 12V with no load

I'm very confused by the rest of your post, and the reply, after this statement. Doesn't the no-load rpm figure of a motor assume that there is no gear train attached?

 

[OhioRay]

Welcome to PF.
You can expect an efficiency of about 50% in the motor and 50% in the generator. So the generator voltage will be the same as the motor but the current generated will be about one quarter of the motor current.

Consider that you are converting potential energy E = mass * gravity * height, into electrical energy, E = volts * amps * time.
Therefore; volts * amps * time = mass * gravity * height
Hence; required mass = (volts * amps * time) / (gravity * height)

Use the MKSA units. One hour is 3600 seconds and 6 feet is 1.8288 metres.
Therefore; mass = (12V * 0.53A * 3600 sec) / ( 9.8 ms^-2 * 1.8288 m)
mass = 1277.5 kg, or about one and a quarter tonnes.
Then apply the 25% efficiency by increasing the mass by four times to 5110 kg or 11,265 pounds.

That is...substantial. So to power the lamp for ~1 minute would require a 187 lb weight... Or to power the lamp for an hour using a 25 kg weight I would need a 374 m tall ceiling. Sounds like this would work perfectly for the Empire State building! Yea... I don't think that'll work! Thanks for the help!