How to / DIY > Generators
Peak output and flywheels
Procrustes:
I'd like to understand better how flywheels help start induction motors. I understand that they store inertia and thus are well suited to the peaks that electric motors require. Why then does the SOM have a 2.5kW head? A standard 6/1 ought to be able to output 2.5kW continuous, so I don't see what advantage the oversized flywheels on the SOM confer, other than steadier output.
Also, I've heard that a 6/1 can output 3.8kW at sea level. Is this peak or continuous? If not, what is peak power? How would heavier flywheels affect peak output?
Rtqii:
Flywheels store energy. This energy can be unloaded very quickly as opposed to waiting for a power stroke from a piston. The faster the flywheel is spinning, the greater the rate of energy delivery to the load.
Peak power demands by a large load... Say a big induction motor on an air compressor can be extremely hard to start without a flywheel system. In order to start these kinds of loads, you either need to oversize the engine and the generator head, or you add flywheel mass. From an expense point of view, the cheapest place to add flywheel mass is on the generator head.
It is my view that the Lister orginal equipment, such as the SOM, were underated. These engines were really designed to serve industry. I do not have a 6/1, but it is my understanding that they can output 3.8kW continuous. Peak power on something like that can exceed 5kW for a very short period of time... Initially the load would be served by the flywheel, then the rack would open up.
Heavier flywheels permit higher peak power to be delivered. Faster spinning flywheels permit the power to be delivered at a faster rate.
Continous power is the rated HP output of the piston. Peak power is the rated HP output of the piston, plus the stored energy in the rotating mass: crankshaft, flywheels, generator rotor, and any additional masses such as pulleys or shafts.
I am currently designing a jackshaft for PTO from an engine system. The jackshaft will rotate at 1800 RPM and permit direct coupling of the generator head. The rest of the high speed shaft is available for mounting flywheel mass and for additional PTOs. Because of the higher speed, I am planning on boring and milling out junk steel flywheels from trucks and tractors and stacking them on the keyed jackshaft. This shaft will not increase the total horsepower, but the peak power would be increased dramatically, as well as the peak power delivery rate... Because of the stored energy in the shaft. Hard to start motors and other loads with high torque requirements can be started in many cases before the rack has a chance to respond.
One of the problems with having a generator without a flywheel in the system is with starting big motors and air conditioners, especially capacitor start electric motors. The generator may have a continous output rating sufficient to run this type equipment, and still be unable to start it. This really tears up the motor because they get very hot quickly if they are not brought up to speed rapidly. The commerical equipment that does not use flywheels get around this by running the engines and generators at higher speed, usually 3600 RPM. The increased shaft speed increases the delivery rate and makes starting big loads a little easier... But even at 3600 RPM, most generator set are not going to have a great deal of peak power available because they lack flywheel mass.
Another way to look at this is compare a Lister type engine with a gas or diesel modern engine running at higher speed, but with the same HP rating. The modern engine will have a faster delivery rate, but the Lister type will have greater peak power due to the flywheels. If you look at the SOM you will see the orginal equipment has added flywheel mass on the generator pulley. This provided an increase of peak power, and in delivery rate.
slowspeed1953:
--- Quote from: Procrustes on August 12, 2006, 07:49:04 AM ---I'd like to understand better how flywheels help start induction motors. I understand that they store inertia and thus are well suited to the peaks that electric motors require. Why then does the SOM have a 2.5kW head? A standard 6/1 ought to be able to output 2.5kW continuous, so I don't see what advantage the oversized flywheels on the SOM confer, other than steadier output.
Also, I've heard that a 6/1 can output 3.8kW at sea level. Is this peak or continuous? If not, what is peak power? How would heavier flywheels affect peak output?
--- End quote ---
Procrustes,
On average a 6/1 will carry 3300 watts with a st5 gen head, that represents a loss of 1174 watts in the transmission (belt) and gen head windings. Not taking into consideration the slight loss of the belt drive the st head is approximately 73% efficienct in converting mechanical energy into electricty.
If maximum efficiency is the main priority the most efficient way of producing electircty with the 4474 watts (6hp) available would be IMHO to use a dc permant magnet generator as they can be as efficient as 90% then convert to ac using an inverter as they average around 95% efficient. That would be a total loss of 15% rather than the 27% loss incured by the st head.
Or the totals for the two systems look like this.
ST5 = 3300 watts dc to inverter = 3803 (likely your 3800 #)
Peace&Love :D, Darren
GuyFawkes:
--- Quote from: slowspeed1953 on August 12, 2006, 07:17:50 PM ---
--- Quote from: Procrustes on August 12, 2006, 07:49:04 AM ---I'd like to understand better how flywheels help start induction motors. I understand that they store inertia and thus are well suited to the peaks that electric motors require. Why then does the SOM have a 2.5kW head? A standard 6/1 ought to be able to output 2.5kW continuous, so I don't see what advantage the oversized flywheels on the SOM confer, other than steadier output.
Also, I've heard that a 6/1 can output 3.8kW at sea level. Is this peak or continuous? If not, what is peak power? How would heavier flywheels affect peak output?
--- End quote ---
Procrustes,
On average a 6/1 will carry 3300 watts with a st5 gen head, that represents a loss of 1174 watts in the transmission (belt) and gen head windings. Not taking into consideration the slight loss of the belt drive the st head is approximately 73% efficienct in converting mechanical energy into electricty.
If maximum efficiency is the main priority the most efficient way of producing electircty with the 4474 watts (6hp) available would be IMHO to use a dc permant magnet generator as they can be as efficient as 90% then convert to ac using an inverter as they average around 95% efficient. That would be a total loss of 15% rather than the 27% loss incured by the st head.
Or the totals for the two systems look like this.
ST5 = 3300 watts dc to inverter = 3803 (likely your 3800 #)
Peace&Love :D, Darren
--- End quote ---
Cut the shit darren, you don't own a 6/1 so you haven't got a fucking clue how efficient they are.
How efficient is my start-o-matic when producing 2 Kw of AC?
bitsnpieces1:
It all goes back to E = M V2. For a flywheel, the amount of energy stored in it is applied to the gen head when you have a demand that is larger than the Listeroid can put out at that instant. Meaning when you have a momentary load (starting a large motor) the gen head draws its energy from the flywheel rather than the engine and then the engine picks up the load. Sorta like a capacitor in electrical loads. The M is the mass of the flywheel itself, the V is the velocity of the flywheel. Velocity is worked out at each distance from the center for what ever mass is at that distance.
Velocity (might as well call it rpm) supplies energy as its square, mass at a straight line. Whatever energy you dump from the flywheel has to be put back into it from the engine.
This is why I'm planning on integrating extra flywheel into my setup. On a jackshaft with its own bearings, setup wwith a clutch to take it out of the drivetrain when necessary.
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