To induce more current (increase output) it requires more energy to spin the alternator. That energy comes from the engine which ultimately comes from the gasoline. This page has a short description of how they work (in history/principles section): http://www.batee.com/corvette/dcrg/readerstips/larsstuff/this_tech_paper_will_discuss_sim.htm If you want more information, just let me know. I'll be happy to provide it but it might take a while to type!
That's nice but..... The claim is to produce the increased current, the alternator increases the load to the engine. I'm asking how?:Smarty: I claim once the alt is rotating, the load the engine sees (defined as the force needed to drive the alt) is relatively the same regardless of the electrical output. I see two different frictions, one from the 2 bearings on the rotator shaft and the other are the brushes used for electrical contact. Tell me the specific force (forces, or friction) that is responsible, or capable of slowing down the belt that I did NOT mention?o No: So explain (or prove) how it takes more HP to drive the alt if the output is 55 amps instead of 45 amps?uzzled: I grant you that if you removed the alt from the engine, yes you would see a real HP increase. But if two cars are exact in every detail (except one has a 55 amp and the other 45 amp alt), how much HP is lost because of the slightly different alt? o No: Almost forgot, I'll take that bet Jim, a 12 pack?:beer
First ignore the frictional losses which, as you say, are due to the bearings and the brushes. These are negligible. You can see this by turning the car off and taking the belt off. The alt spins easily by hand. Now there are the rotating coil and the stationary coils. The stationary coils are hooked up to the cars electrical system and are what charge the battery, etc. The spinning coil is energized, through the brushes, to create an electromagnet. This electromagnet is spun by the engine. The force the engine has to overcome in spinning the alt is this: the spinning magnet induces a current in the stationary coils. The current in the stationary coils (remember V=IR) is related to how much the car's electrical system is drawing at that moment. So if the current is higher in the stationary coil that means it is harder for the engine to turn the electromagnet. This means, as the electrical load is increased, it takes more HP to turn the alt to keep up with the load. To make this point again: the electromagnet turning (by the engine) induces current in the stationary coils. The induced current is directly tied to how much the car's electrical system is drawing. It takes energy to induce a current in coil and the amount is proportional to the electrical load. Again, this energy is coming from the engine. Now for your example of a 45 vs. 55 amp alternator: if both are under the same load they will be drawing the same amount of HP from the engine. So the 45 amp alt with a 30 amp load takes as much to turn as the 55 amp with a 30 amp load. The only difference is that the 55 amp alt can supply a maximum of 10 amps more than the 45. A 45 amp alt trying to supply 55 amps wouldn't keep up and would either overheat (coils) or drop in voltage output. I hope this helps. To get more technical starts to get into the physics of electricity and magnetism: the how's and why's of induced currents and other fun stuff. But this forum isn't really the appropriate setting to discuss those topics in detail. Just keep in mind that it takes energy to induce a current.
If you want to see an electrical load in action start an emergency generator driven by an 8 hp engine with no load. Then add a heavy load like a hot water heater to the unit. You will definitely hear the engine load down. If an 45 amp alternator and 55 amp alternator are supplying the same electrical loads there should be no difference in power requirements ignoring efficiencies in the alternators. The rating of the alternator is what it is capable of performing when you start adding all the loads the car may have on board. If the two alternators are suppling their rated outputs there would definately be a difference in the power necessary to drive the alternator.
Loyd You made some good points, and I was aware of them. The higher output alt will have a stronger (and I assume more mass) magnet to spin. There will also be slightly more mass on the stationary windings..... The original claim was that if a mechanically driven fan (which will obviously vary the load to the engine with varying RPM's) was removed, and electrical driven ones installed, it would be a wash, meaning no HP gain or loss. I just don't (can't) see it that way. Who has dyno'ed and measured the difference?
It is not the mass but a greater resistance cause by the stator turning through a stronger electro magnetic field. I agree with Danhei's explanation except if we are dealing with alternators as opposed to generators E= IR (cosine of theta)with theta being the phase shift between the voltage and the current. Also alternators technically should be rated in KVAs as opposed to watts. Generators have brushes whereas alternators have slip rings. Now imagine a running Buick GS 455 sitting on a chassis dyno producing 100 hp at the rear wheels. (Assume for simplicity of the numbers there is no drive train loss and the output at the fly wheel is also 100 hp.) Now further assume a belt driven mega alternator that is driven off the crank shaft pulley with a rating of 74,600 VA (74.6 KVA) For simplicity we will ignore phase shift and equate that to watts as you would have in a pure DC system. One horsepower= 746 watts so this is an system capable of a rated 100 horsepower. (Keep the votage at 12 which in the real world you would never do and assume no energy loss in the following set up. ) Imagine this alternator tied to our Buick via the belt drive is wired to an electrical motor also rated at 100 hp at 12 volts. We slowly start apply a brake load to the motor requiring it to work to a output of 50 hp. ( To correlate to the original example imagine the brake load is in the form of a giant fan) The current from the alternator is going to rise to satisfy the increased demand of the motor coming under load and the 455 is going to share its hp out between the rear wheels and what is required to spin the stator through the stronger electromagnetic field. Now the chassis dyno will show 50 hp at the rear wheels because in a perfect system it took 50 hp of what was available to rear wheels to drive the alternator. If you want to keep 100 hp on the chassis dyno meter you would have to add more accelerator pedal so the the total engine output is now 150 hp with only 100 making it to the rear wheels. Simple thermodynamics which I dare say have not changed since I last took a course a mere 33 years ago.
Ok, now I understand the argument: I know of no dyno tests for these results, but the following argument may be made: The electric motor or motors on the electric fan have a current requirement to produce a rated CFM of air movement. That is all the current they are going to draw. Of course that needs to be made up by the alternator. This load is limited while the load to drive a mechanical fan will continue to increase with higher engine RPMs as you pointed out. With forward motion of the car making air easier to move the current requirement of the electric motors may decrease, at a higher rate than a mechanical fan. The main benefit is the electric fan does not need to continuously operate since there are switches on the engine (either AC or high temperature) or computer to operate the fan when needed. The remainder of the time the forward motion of the car should be sufficient to control the temperature. I know on the turbo Regals the AC clutch is released for wide open throttle, but am not aware if the low speed fan operation controlled by the computer is jettisoned as well. Of course a clutch fan should do basically the same thing.
Lloyd you are correct that the original point I was making is that there was no free energy lunch from going to the electric fan motor set up to cool versus a mechanical fan clutch.
Oh well.... Some good points. Let us 1st assume we are looking at a GM 1971, 55 amp and a 37 amp alts to compare. These were the largest and smallest supplied by them. And don't forget, our cars are DC, not A/C (even though the alt produces A/C, it also converts it to DC before it leaves the alt. P (input)= P(output) + PL (internal losses, maybe we can find a chart, solving this is way beyond what we can discuss here). The difference (55-37) is 18 amps output current. One can measure the output voltage, I'll assume 18 v, it is typically less. P=VI=Volt x Amps= Watts, 1Watt=746HP Output Power difference=Po=VoIo=18x18=324w, or .43 HP I thought I had my power book here at work, but I guess I took it back home, so I can't look anything up. But I think I showed the output power difference supplied is quite low, and my guess is that alts are very efficeint. Lets assume 95% eff, that increases .43 to about .46 HP. KVA stands for (1000)x(V)x(A), and is used in reference to transformer capacities mostly, instead of Watts. Maybe I'll get sometime this weekend to dust off a few books, and look up a few things.:bglasses:
Simply stated, the alternator is a magnetic device. Current from the field creates a magnetic field around it. This field is inductively coupled to the stator, which is connected directly to the battery via the rectifiers. More magnetic field strength is necessary to create a larger charging current. This strength is created by using larger windings, or more of them, on the field, which will draw more from the engine due to drag. The draw, however, is not linear. Twice the charging current is created by an additional 10-15% additional drag. Ray
