 Notice. New forum software under development. It's going to miss a few functions and look a bit ugly for a while, but I'm working on it full time now as the old forum was too unstable. Couple days, all good. If you notice any issues, please contact me. |
Forum Index : Other Stuff : Gearing and power output.
| Author | Message | ||||
| Davo99 Guru  Joined: 03/06/2019 Location: AustraliaPosts: 1578 |
One for the mechanical engineers..... I have tried googling this question but no matter how I phrase it , I always get irrelevant answer's to what I think should be straightforward for those in the know. I found a Diesel engine today up the back I forgot I even had. It's 10 HP continuous @ 2400 RPM. I have an 8 KW gen head which is 1500 RPM. My belief is it takes double the engine HP to drive an alternator / Generator for every KW the alternator is. Going by this rule of thumb, the engine does not have the power to drive the gen head to full load. My question is if I reduce the driven speed of the engine output with the appropriate size pulleys and have the engine doing 2400 rpm and the alternator geared down to 1500 rpm, will the effective power the alternator sees be increased say to 12 or whatever HP it works out to be? As I understand, power is function of torque Vs. Speed but I do not know how this applies in this case. I would think that at the given speed, the alternator needs increased torque as the speed needs to be fixed so the answer should be Yes, the engine will have more effective power or if that is not the right term, Gearing the faster engine to the slower alternator will in fact increase it's ability to drive a higher load on the alternator at a fixed RPM. Can anyone confirm that gearing the engine down will effectivel increase its HP at the lower RPM the alt would see? Thanks in advance. |
||||
phil99 Guru Joined: 11/02/2018 Location: AustraliaPosts: 2134 |
Power = Rotation Speed x Torque Units:- Power - Watts Rotation Speed - Radians / Second (Rad/S = RPM x 2 x π / 60) Torque - Metre-Newtons (9.8 Newtons = 1kg force) When you gear the speed down the torque is geared up by the same proportion so the power remains unchanged - ignoring losses. Getting more power requires more fuel. Fuel and air are the only inputs to the system, everything else consumes power. |
||||
| Davo99 Guru Joined: 03/06/2019 Location: AustraliaPosts: 1578 |
OK, so in this instance with the faster engine and the slower generator, would I be able to put more effective load on the generator than the engines rating or not? If the engine is 10 HP and it's 2 HP per kw, that should at engine rating allow me to generate 5KW out of the alt. By gearing the speed of the engine down, am I going to be able to get more than 5 KW out of that same engine due to the multiplication of the touque? |
||||
| phil99 Guru Joined: 11/02/2018 Location: AustraliaPosts: 2134 |
The power is produced by burning the fuel. You can't get more out than you put in. Again Power = Rotation Speed x Torque Reducing the speed reduces the power by the same amount that increasing the torque increases it. Only with magic can you get more out than you put in. |
||||
Revlac Guru Joined: 31/12/2016 Location: AustraliaPosts: 1022 |
Found a handy tool. https://www.blocklayer.com/pulley-belt Cheers Aaron Off The Grid |
||||
| jeffj Regular Member  Joined: 04/02/2012 Location: AustraliaPosts: 84 |
The diesel will supply 10 hp at 2400 rpm 1 hp =746 watts so at 2400 rpm you should get 7.46 kw at the shaft The generator is rated at 8 kw at 1500 rpm so you will need to reduce the speed of the diesel to 1500 rpm with belts. The belts will absorb a bit of the power The other consideration is the frequency of the generator which is proportional to the speed. |
||||
 Print this page |
