Differences in Small Case Alternator Efficiencies

[newhaul]

Originally Posted by Supers0nic

Rod is helping me with my lithium battery installation. There is no way around it. An external rectifier is essential.

external rectifier or regulator?
What size bank ? What size alternator? What are the issues you are having?

[kenbo]

Originally Posted by tanglewood

Running the alternator at a higher voltage is indeed the best (only) solution, and I like the MPPT approach to dealing with highly varying RPM.


Anyone know the typical voltage rating for the diodes in an alternator? That is likely a limiting factor in stock alternators, but pretty easy to work around.

Typical automotive bridge rectifiers use 6-30V diodes, 3 pos and 3 neg. Balmar and others that supply the recreational marine market use 6-50V diodes. There are some 70V rectifiers available for the SI series of Delco Remy alternators. I use those bridge rectifiers to build remote set ups.

[kenbo]

Originally Posted by newhaul

external rectifier or regulator?
What size bank ? What size alternator? What are the issues you are having?

I've been building these for a few years for folks. I use 12SI cases for non yanmar engines and 10SI cases for Yanmars. There's a Chinese supplier for 80mm (3.14") Yanmar mounts but only in the 10SI format.

Internally regulated usually and external bridge rectifier using 210V rectifiers. I try to keep folks from over winding the alternator and typically use 105A stators. They run fairly cool as long as you don't go crazy on the stator winding. Built about a dozen so far and only one failure. Not sure but I think the guy had a short in one of his batteries.

[Supers0nic]

No real issues. I’m doing all the basic work myself: Fitting the large life PO battery, BMI controller, new pulleys and altenator, external rectifier. Those are all things I can do myself. Rod from Compass Marine will do all the wiring. It’s not a simple nor cheap installation but as a full time liveaboard I think it’s worth the investment. I will greatly benefit and will be helping to advance the technology.

[newhaul]

Originally Posted by Supers0nic

No real issues. I’m doing all the basic work myself: Fitting the large life PO battery, BMI controller, new pulleys and altenator, external rectifier. Those are all things I can do myself. Rod from Compass Marine will do all the wiring. It’s not a simple nor cheap installation but as a full time liveaboard I think it’s worth the investment. I will greatly benefit and will be helping to advance the technology.

when your done don't forget to post on the build/ cost thread .

http://www.cruisersforum.com/forums/...st-206421.html
And here
http://www.cruisersforum.com/forums/...ks-207565.html

[Jammer]

Originally Posted by sdj

The problem with alternators, is the low kick-in voltage required for them to usable at idle means a very high windings emf at high rpm. Mostly this problem is ignored, a larger cooling fan fitted, and the ((windings emf less the battery volts) squared divided by the windings resistance) is all dissipated as heat in the alternator. Heat = V^2 / R. As the terminal voltage rises linearly with RPM the heat dissipation rises with the square of it.

That isn't true, because the formula you show is the heat dissipation in a resistor (or resistive device or circuit), and the alternator, taken as a whole, isn't.

The ideal situation would be to run the alternator unregulated (or PWM the excitation) and put the output into your MPPT controller. Now the controller will track the alternators ideal operating point regardless of voltage, which should reduce dissipation dramatically.

It should be possible to double the output and halve the heat load doing this.

Alternators do not have the same characteristics as solar panels. There is improved performance with higher voltage up to a point, which is why 24 volt alternators perform somewhat better than 12 volt alternators.

Originally Posted by rgleason

So exactly what is dissipated as heat in the alternator?

There are I-squared-R losses in the stator windings, that is, proportional to the square of the current. The winding resistance also goes up with temperature resulting in greater losses as the alternator heats up.


By way of comparison, the loss in the diodes increases linearly with current, because the voltage drop is a relatively constant 0.7 volts.

Originally Posted by sdj

The heat (power in watts) dissipated in anything can be calculated with Watts = V^2 / R, or IxV. You should do your own spreadsheet with these formulas.

What many people do not understand, is the alternator output emf varies directly with the RPM, but is shunted down to the battery terminal voltage, BY the battery.

While it appears that the stator emf is the same as the battery voltage, and it is, because they are wired in parallel, substantial heat dissipation is occurring because of the current flowing BECAUSE of the difference between Vemf and Vterminal INSIDE the alternator. All this voltage multiplied by the current flowing is VxI = watts, and is dissipated as heat.

The amount of heat from the diodes is small in comparison.

edit: If you still don't follow, then study Kirchhoffs' laws, specifically his current law about equivalent resistances in a series circuit, and also Lenz's law for magnetic circuits. This is not really consumer territory however..

I don't think that explanation makes sense.

Originally Posted by sdj

Favorite thing to try at the moment, Delco Remi 55si 24V 275A brushless (6.6KVA) running at 48V (now 13VKA!!) This is a $2000 alternator. The manufacturer, whilst being extremely helpful and forthcoming, does not condone such warranty-voiding activities, and has no data.

You'll eventually reach either magnetic saturation or the breakdown voltage of the winding insulation. The magnetic losses do go up when you increase the field voltage (or RPM) in order to do that, which is another source of heat.

[sdj]

Originally Posted by Jammer

That isn't true, because the formula you show is the heat dissipation in a resistor (or resistive device or circuit), and the alternator, taken as a whole, isn't.

The alternator windings don't have resistance? That's wrong. The diodes don't have forward biased resistance? Thats wrong.

Originally Posted by Jammer

Alternators do not have the same characteristics as solar panels. There is improved performance with higher voltage up to a point, which is why 24 volt alternators perform somewhat better than 12 volt alternators.

No, that's wrong. The same frame 24V alternator with the same output current WILL have double the output watts as a 12V alternator, not just 'somewhat' better performance.

Originally Posted by Jammer

There are I-squared-R losses in the stator windings, that is, proportional to the square of the current. The winding resistance also goes up with temperature resulting in greater losses as the alternator heats up.

Sure, but that's not what I referred to. I'm talking about the Vemf induced in the windings (see lentz laws) less the battery terminal voltage (kitchhoffs laws) giving rise to heating. Look it up yourself before you post erroneous consumer junk that isn't science.

Originally Posted by Jammer

I don't think that explanation makes sense.

No, you dont, because you dont understand it so you publicly deny it. Sadly, that's religion, not science.

Originally Posted by Jammer

You'll eventually reach either magnetic saturation or the breakdown voltage of the winding insulation. The magnetic losses do go up when you increase the field voltage (or RPM) in order to do that, which is another source of heat.

How does this work? Quote sources, laws, research, equations. Typical insulation breakdown on that alternator should be at least 500V, probably 2KV, which is well outside the topic so your explantion is bunkum.

Post facts backed by science, not religious opinion.

[newhaul]

Originally Posted by sdj

The alternator windings don't have resistance? That's wrong. The diodes don't have forward biased resistance? Thats wrong.

No, that's wrong. The same frame 24V alternator with the same output current WILL have double the output watts as a 12V alternator, not just 'somewhat' better performance.

Sure, but that's not what I referred to. I'm talking about the Vemf induced in the windings (see lentz laws) less the battery terminal voltage (kitchhoffs laws) giving rise to heating. Look it up yourself before you post erroneous consumer junk that isn't science.

No, you dont, because you dont understand it so you publicly deny it. Sadly, that's religion, not science.

How does this work? Quote sources, laws, research, equations. Typical insulation breakdown on that alternator should be at least 500V, probably 2KV, which is well outside the topic so your explantion is bunkum.

Post facts backed by science, not religious opinion.

all of this is well and good but the real question is how long can I expect to get 100 amps out of my 135 amp small frame GM 10DN, 12 SI alternator feeding 200ah Lfp bank?

[kenbo]

Originally Posted by newhaul

all of this is well and good but the real question is how long can I expect to get 100 amps out of my 135 amp small frame GM 10DN, 12 SI alternator feeding 200ah Lfp bank?

Do you have a 10DN or 12SI, their cooling capacities are different.

[newhaul]

Originally Posted by kenbo

Do you have a 10DN or 12SI, their cooling capacities are different.

sorry typo there was supposed to be an or in there .
I can get either case . The other one is in putting it on a westerbeke m18

[kenbo]

My opinion, no test data but have lots of experience with both. The 10DN is rugged as hell and requires an external regulator. But it was designed by Delco Remy to handle about 62A. They get hot, the diodes are pressed into the SRE case and because of that you need REALLY good diodes. I think Motorola still makes 50V press fit diodes and if you can find them use them. 100A continuous would be a stretch, 75A to 80A and you should be able to go all day.

The 12SI cases/fan were designed to handle 94A. If you use an external bridge rectifier you can go all day. Use a solid pulley and a good notched belt, well aligned to keep heat off the DE case and bearing. Use a good quality NTN bearing on the DE and an INA bearing on the SRE end and your set.

Since you mentioned a 135A derated to 100A I'm assuming you will be using a 135 stator. I have had better results with the 105 stator, YMMV.

[tanglewood]

Originally Posted by sdj

The heat (power in watts) dissipated in anything can be calculated with Watts = V^2 / R, or IxV. You should do your own spreadsheet with these formulas.

What many people do not understand, is the alternator output emf varies directly with the RPM, but is shunted down to the battery terminal voltage, BY the battery.

While it appears that the stator emf is the same as the battery voltage, and it is, because they are wired in parallel, substantial heat dissipation is occurring because of the current flowing BECAUSE of the difference between Vemf and Vterminal INSIDE the alternator. All this voltage multiplied by the current flowing is VxI = watts, and is dissipated as heat.

The amount of heat from the diodes is small in comparison.

edit: If you still don't follow, then study Kirchhoffs' laws, specifically his current law about equivalent resistances in a series circuit, and also Lenz's law for magnetic circuits. This is not really consumer territory however..

Maybe I'm not following you correctly, but I think you misunderstand this.


The alternator outputs current. How that manifests itself as voltage and current depends on the load, which is nominally a battery being charged. Initially the voltage rises, but then the battery's resistance drops and it starts accepting current. In a heavily discharged battery, this ultimately balances out at the alternator's max charge rate. Over time the battery voltage climbs and ultimately reaches a point where the regulator starts to back off on the field current which reduces alternator current output. Because the regulator is modulating field current based on alternator output voltage, it's the output current that reduces.


Power loss in the stator is more easily expressed as I^2R. It can also be expressed as V^2/R. But V is not the difference between the max possible voltage of the alternator and the actual operating voltage, and I think that's what you are saying. For power loss, V is the voltage drop through the windings, exclusive of induced voltage. In other words, it's IR. Plug that into your formula, and you get I^2R.


So if at some RPM an alternator could output 50V at 10A, but because it is driving a battery load it is instead outputting 25V at 20A, the power loss difference is not (50V-25V)^2R, or 625R. The power loss difference is 20A^2R - 10A^2R, or 300R.


So increasing voltage and lowering current does indeed reduce stator winding losses at the same power output level, but not as much as you are saying, and not for the reasons you are saying.

[wsmurdoch]

Among small case alternators, why is there the attachment to the 10SI and 12SI alternators in preference to the newer CS130D alternators with their dual internal fans, open frame design, and diode pack external to the alternator case?

Bill

[rgleason]

Tanglewood thank you for untangling that so clearly.


Can you comment on this statement and perhaps quantify it in some way?


"The amount of heat from the diodes is small in comparison"


So what is the approximate relationship/ratio between diode heat which you quantified earlier, and the heat from the stator and rotor (with its range through the charge cycle)?


Thanks

Originally Posted by tanglewood

Maybe I'm not following you correctly, but I think you misunderstand this.


The alternator outputs current. How that manifests itself as voltage and current depends on the load, which is nominally a battery being charged. Initially the voltage rises, but then the battery's resistance drops and it starts accepting current. In a heavily discharged battery, this ultimately balances out at the alternator's max charge rate. Over time the battery voltage climbs and ultimately reaches a point where the regulator starts to back off on the field current which reduces alternator current output. Because the regulator is modulating field current based on alternator output voltage, it's the output current that reduces.


Power loss in the stator is more easily expressed as I^2R. It can also be expressed as V^2/R. But V is not the difference between the max possible voltage of the alternator and the actual operating voltage, and I think that's what you are saying. For power loss, V is the voltage drop through the windings, exclusive of induced voltage. In other words, it's IR. Plug that into your formula, and you get I^2R.


So if at some RPM an alternator could output 50V at 10A, but because it is driving a battery load it is instead outputting 25V at 20A, the power loss difference is not (50V-25V)^2R, or 625R. The power loss difference is 20A^2R - 10A^2R, or 300R.


So increasing voltage and lowering current does indeed reduce stator winding losses at the same power output level, but not as much as you are saying, and not for the reasons you are saying.

[rgleason]

I understand these points are focused on small case alternators, but I was wondering what your experience has been with the larger Delco SI28 (160a and 200a) which has dual fans and I think has good cooling features, and what I think is the more recent AD244 which has just one fan, but is rated at 244amps and more. It is my impression that the SI28 inherently has better cooling and would be best for use in a derated continuous high output situation as is found in marine use. What are your thoughts? Are there ways one could improve airflow in these alternators? - New blades? - Bigger openings somehow??

Originally Posted by kenbo

My opinion, no test data but have lots of experience with both. The 10DN is rugged as hell and requires an external regulator. But it was designed by Delco Remy to handle about 62A. They get hot, the diodes are pressed into the SRE case and because of that you need REALLY good diodes. I think Motorola still makes 50V press fit diodes and if you can find them use them. 100A continuous would be a stretch, 75A to 80A and you should be able to go all day.

The 12SI cases/fan were designed to handle 94A. If you use an external bridge rectifier you can go all day. Use a solid pulley and a good notched belt, well aligned to keep heat off the DE case and bearing. Use a good quality NTN bearing on the DE and an INA bearing on the SRE end and your set.

Since you mentioned a 135A derated to 100A I'm assuming you will be using a 135 stator. I have had better results with the 105 stator, YMMV.