Differences in Small Case Alternator Efficiencies

[tanglewood]

Originally Posted by 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

See attached paper from Delco Remy quantifying the sources of power loss in a typical alternator.


Re the math, someone else described it earlier, but I'll recap. The voltage drop across the rectified diodes is a constant, nominally 0.7V. But it's a full wave rectified (diodes to both the positive and negative DC lines), so it's really a total of 1.4V. With a constant voltage, the power loss in the diode is P=IV where I is the output current and V is 1.4V. Of note is that this varies linearly with output current, and remains a constant % of output power. Double the current, double the output power, and double the loss.


In contrast, with stator power losses if you double the current you double the output power, but quadruple the power lost to heating of the stator.

[sdj]

Originally Posted by tanglewood

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.

No. This indicates you didn't look up anything I said at all. Go and read Kitchhoff and Lentz.

Originally Posted by tanglewood

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.

Isn't it? Where did it go then? If that output winding is running at 100V unloaded and you load it to 24V, WHERE did the remaining voltage GO? Maybe it evaporated? Did it sublimate? Maybe it went into the wire and hid there somewhere, only to come out later as some new electrical concept you just mentioned and now existed since you said so?

Originally Posted by tanglewood

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.

Is it? Demonstrate and prove that with a kirchhoffs equivalent circuit.

Originally Posted by tanglewood

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.

You're allowed to be wrong, the internet is littered with it.

[sdj]

A few snippets from that Delco PDF ;

"The largest source of loss within an alternator is the ohmic loss occurring within the stator windings.This is the familiar i2R loss that occurs when current flows through a resistance."

On page 20 there is a graph - where losses increase linearly with RPM. Why is that? Did the diode loss increase with rpm? Did the resistance of the wire change with rpm? Did the magnetic properties change and corresponding losses increase with rpm?

On page 24 there is an efficiency map, where the efficiency peak is directly above the cut-in rpm. Why is the efficiency +70% just above cut-in, and 30% at high rpm? What CHANGED on the alternator OTHER than rpm? What caused that change?

What is your explanation that losses increased linearly with rpm? Reminder to quote sources.

[kenbo]

Originally Posted by 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

Bill,

For me it the absolutely easy user serviceability of the SI series. I believe the CS130D to be a better alternator as am OEM but the SI series can be beat for customization by the user.

That said, I am still a GC2 battery user and I find the value proposition of the SI series can't be beat for FLA and Gel batteries.

[tanglewood]

Originally Posted by sdj

No. This indicates you didn't look up anything I said at all. Go and read Kitchhoff and Lentz.


Isn't it? Where did it go then? If that output winding is running at 100V unloaded and you load it to 24V, WHERE did the remaining voltage GO? Maybe it evaporated? Did it sublimate? Maybe it went into the wire and hid there somewhere, only to come out later as some new electrical concept you just mentioned and now existed since you said so?

Is it? Demonstrate and prove that with a kirchhoffs equivalent circuit.

You're allowed to be wrong, the internet is littered with it.

My, so grumpy.


Page 20 in the paper summarized the sources of power loss. Which of those includes the loss you assert?


I'm talking about the first one - Stator winding loss. It's calculated a I^R where R is the resistance of the winding wire, and I is the output current. Just like I described.



You are describing the voltage of a single node in two different circuit configurations, namely open circuit, and connected to a battery.


But you are treating them as the voltages of two nodes in the same circuit connected by a resistor (P=V^2R), and that doesn't exist in an alternator. If you look at an alternator connected to a battery, the alternator's open circuit voltage doesn't exist anywhere in that circuit. Now if you change the circuit such that it's open circuit, then the voltage could rise to 100V (using your example), but now you have zero current, so no power loss in the stator (or the rectifier).

[rgleason]

Originally Posted by tanglewood

See attached paper from Delco Remy quantifying the sources of power loss in a typical alternator.


Re the math, someone else described it earlier, but I'll recap. The voltage drop across the rectified diodes is a constant, nominally 0.7V. But it's a full wave rectified (diodes to both the positive and negative DC lines), so it's really a total of 1.4V. With a constant voltage, the power loss in the diode is P=IV where I is the output current and V is 1.4V. Of note is that this varies linearly with output current, and remains a constant % of output power. Double the current, double the output power, and double the loss.


In contrast, with stator power losses if you double the current you double the output power, but quadruple the power lost to heating of the stator.

Great summary. Thank you.

On page 23 of the document are good quantitative charges and a list of the losses.

[kenbo]

But the only heat source that is removeable is the bridge rectifier. An adder benefit but not easily quantifiable, is the added cooling effect of the increased airflow now that the rectifier is not blocking the vent holes.

Maybe I missed it but the original question was which small case alternator wound run cooler a 135A run at 80A or a 100A run at 80A and why. Would one set up vs the other reduce stator heat losses?

[tanglewood]

Originally Posted by kenbo

Maybe I missed it but the original question was which small case alternator wound run cooler a 135A run at 80A or a 100A run at 80A and why. Would one set up vs the other reduce stator heat losses?

Making an assumption that the 135A alternator has beefier winding wire to carry the higher current, i.e. lower winding resistance, then the 135A alternator would run cooler at least as far as the stator goes. But you would need to check the specs for each alternator to see if the winding resistance is really any different. This also ignores whether there are any air flow and related cooling differences between the 135A and 100A units that might sway things one way or another. But if I had the choice of the two, and no other info, I'd pick the 135A.

[kenbo]

Originally Posted by tanglewood

Making an assumption that the 135A alternator has beefier winding wire to carry the higher current, i.e. lower winding resistance, then the 135A alternator would run cooler at least as far as the stator goes. But you would need to check the specs for each alternator to see if the winding resistance is really any different. This also ignores whether there are any air flow and related cooling differences between the 135A and 100A units that might sway things one way or another. But if I had the choice of the two, and no other info, I'd pick the 135A.

Thanks and that would be my thoughts as well. But I have built identical alternators, one fitted with a 145A stator and one with 105A and the 105A has run cooler.

Now there are variables that I don't know about like the setting on the Balmar VR and the SOC of the different batteries as well as the internal resistance of the batteries. But the 105 does consistently perform at a cooler temp.

Would be an interesting test to control all the variables and then measure temp.

[newhaul]

Originally Posted by kenbo

Thanks and that would be my thoughts as well. But I have built identical alternators, one fitted with a 145A stator and one with 105A and the 105A has run cooler.

Now there are variables that I don't know about like the setting on the Balmar VR and the SOC of the different batteries as well as the internal resistance of the batteries. But the 105 does consistently perform at a cooler temp.

Would be an interesting test to control all the variables and then measure temp.

now the 105 amp stator runs cooler but at what amp output to the battery?
Same question the 145 amp stator ran hotter at what amp output ?
And last what was the actual temperature difference?
Amp output difference?

[tanglewood]

Originally Posted by kenbo

Thanks and that would be my thoughts as well. But I have built identical alternators, one fitted with a 145A stator and one with 105A and the 105A has run cooler.

Now there are variables that I don't know about like the setting on the Balmar VR and the SOC of the different batteries as well as the internal resistance of the batteries. But the 105 does consistently perform at a cooler temp.

Would be an interesting test to control all the variables and then measure temp.

Interesting. Any sense what's different between the two stators? And when comparing temps, is that with each running at it's rated output, or both limited to the same output? If both are running at their respective rated outputs, then I'd expect the 145A to run hotter.


On my last boat I had a 200A, 24V large frame alternator and would frequently run it for hours at full output. Man did it get hot - over 300F at times. But Leece Neville assured me it was OK. That was a direct temp reading off the stator. The rectifier stayed pretty cool and never got hot enough to trigger the Balmar regulator temp control. This is why external rectifiers don't excite me, but lower currents do. But different alternators may have different weak links. Mine just weren't the rectifiers.

[tanglewood]

Originally Posted by newhaul

now the 105 amp stator runs cooler but at what amp output to the battery?
Same question the 145 amp stator ran hotter at what amp output ?
And last what was the actual temperature difference?
Amp output difference?

Exactly. That could account for lots of temp difference.


The OP was asking about a very specific set of conditions where the output is de-rated on both alternators to 80A. That's where my money would be on the higher rated alternator running cooler.

[kenbo]

Originally Posted by newhaul

now the 105 amp stator runs cooler but at what amp output to the battery?
Same question the 145 amp stator ran hotter at what amp output ?
And last what was the actual temperature difference?
Amp output difference?

Sorry, I don't have those answers I just sold the alternators and didn't set up the regulators. When I was on the boats a bit to check the owners installation I shot both alternators with an IR meter and I remember the 105A being much cooler but so what I didn't have the other details that you need to make an informed judgement.

The 105A stator is wound with more wire but smaller gauge so it does perform better at lower rpm.

Like I said in an earlier post, it would be a valuable test if all the variables were controlled.

[kenbo]

[QUOTE=tanglewood;2969245]Interesting. Any sense what's different between the two stators? And when comparing temps, is that with each running at it's rated output, or both limited to the same output? If both are running at their respective rated outputs, then I'd expect the 145A to run hotter.


Two different boats two different systems, they both had FLA GC2's but different brands. There were no controlled variables but I know both owners did derate the power output, to what exact level I don't know.

[wsmurdoch]

Originally Posted by tanglewood

Making an assumption that the 135A alternator has beefier winding wire to carry the higher current, i.e. lower winding resistance, then the 135A alternator would run cooler at least as far as the stator goes. But you would need to check the specs for each alternator to see if the winding resistance is really any different. This also ignores whether there are any air flow and related cooling differences between the 135A and 100A units that might sway things one way or another. But if I had the choice of the two, and no other info, I'd pick the 135A.

To rewind a 100A stator to produce 135A, would you not increase the number of turns in the stator coils? If the slots of the 100A alternator were full, the only way to add more turns would be to use smaller diameter wire which would have higher resistance, produce more heat at the same output current, and be less beefy.

Bill