Alternator issues

[Sir Nibble]

May I ask what size the engine pulley is? The alternator is goosed but the story so far suggests it may have been running too slow.

[Mikexx]

10 minutes ago, Sir Nibble said:

May I ask what size the engine pulley is? The alternator is goosed but the story so far suggests it may have been running too slow.

 

The direct connection to the field diodes would indicate a history of alternator 'starting up' issues.

[Sir Nibble]

15 minutes ago, Mikexx said:

 

The direct connection to the field diodes would indicate a history of alternator 'starting up' issues.

Yes that's pretty well what I meant. Think I will build a slow speed machine ready in case.

[Nick1969]

The engine alternater drive pulley is 6 and a half inch diameter, is this small or large

would the alternater recomended be ok or is a special one required 

many thanx

Edited April 30, 2017 by Nick1969

[Sir Nibble]

I believe that is the small pulley and your alternator will run slow. Bizzard will know. If I build a slow unit it will be identical to what you have but with higher output at slow speed at the cost of reduced output at the higher revs you wouldn't achieve anyway. First consideration should be a bigger pulley.

[Nick1969]

Thanx sir nibble, glad I held off from buying one this weekend. Isnt getting hold of and fitting a bigger pulley a big affair compared to another alternater.

4 hours ago, Sir Nibble said:

I believe that is the small pulley and your alternator will run slow. Bizzard will know. If I build a slow unit it will be identical to what you have but with higher output at slow speed at the cost of reduced output at the higher revs you wouldn't achieve anyway. First consideration should be a bigger pulley.

thanx

nick

[bizzard]

22 hours ago, Nick1969 said:

Thanx sir nibble, glad I held off from buying one this weekend. Isnt getting hold of and fitting a bigger pulley a big affair compared to another alternater.

thanx

nick

Your alternator drive pulley on the SR engine is on the camshaft which rotates at only half the speed of the crank shaft unlike most other engines which DO drive from the crank shaft. Therefore, as you have the standard 6'' pulley the alternator will really rotate too slowly for a decent charge output, unless the engine is revving virtually flat out. A large diameter pulley of 9, 10, or preferably a 12'' will greatly speed up the alternator to give a decent charge even at tick over. But you would need to check and measure around and below the existing pulley to make sure that there is enough room to swing it with nothing solid and un-moveable in the way.  Various size Taperlock pulleys are available to fit your keyway'd shaft which should be 1.1/2'' diameter. The pulleys are a light interference fit on the shaft held by a grub screw. You would of course need a larger drive belt too.

Edited May 3, 2017 by bizzard

[Sir Nibble]

I don't know what that horrible thing is in his avatar but there's the word from it's mouth. As Bizz says the pulley is on the camshaft at half engine speed. Your alternator pulley will be around 2" so one and a half times engine speed. About half what's wanted on a higher revving 3 or 4 cylinder jap engine.  Measure up and see how big a pulley will fit, Google for belt drive suppliers and tell them you have a 1 1/2" shaft and want a 12", 9" or whatever size pulley will fit in. Specify an "A" type belt. Then a standard replacement alternator will work happily. In the event it is not possible to fit a bigger pulley in we can explore  a slow speed alternator. It won't be as good but a damned sight better than what you have.

Marine engines really belong on yachts as an auxiliary or in launches and going tump....tump....tump at 3mph just doesn't spin the alternator fast enough. A usual, and perfectly acceptable dodge to get the alternator to cut in slower is to put a bigger bulb in the warning light, and bigger, and the end of that process is the direct bulbless connection you had. It will cut the alternator in at a lower speed but it still won't improve the charge once it is going but saves you blipping the throttle to cut it in every time you get under way. Best, proper engineering solution is the bigger pulley, but I would quite like to test one of my slow alternators against the problem. 

Edited May 3, 2017 by Sir Nibble

[bizzard]

The bore of the pulley will need a 3/8'' wide key slot through it.

[Tony Brooks]

SirN, I am just wondering how the low speed and slower cut in speed squares with the reduced cooling air flow with a slow alternator speed or is the output so reduced there is no chance of the alternator overheating. Have you any feed back about longevity for the one you did for a forum member way back?

Nick, Getting a higher alternator speed benefits the alternator cooling as well as giving a higher output.

[Mikexx]

3 hours ago, Tony Brooks said:

SirN, I am just wondering how the low speed and slower cut in speed squares with the reduced cooling air flow with a slow alternator speed or is the output so reduced there is no chance of the alternator overheating. Have you any feed back about longevity for the one you did for a forum member way back?

Nick, Getting a higher alternator speed benefits the alternator cooling as well as giving a higher output.

If all things being equal, ie half the diameter of wire, twice as many turns, 1/2 the current the copper losses will be the same, but the diode losses will be 1/2.

If SirN uses high temp enamelled wire and the diode pack will have half the dissipation I would have thought flow would still be sufficient. Engine bay temps are likely to be lower than under bonnet temperatures on a hot day whilst stationary. BICBW

[Sir Nibble]

4 hours ago, Tony Brooks said:

SirN, I am just wondering how the low speed and slower cut in speed squares with the reduced cooling air flow with a slow alternator speed or is the output so reduced there is no chance of the alternator overheating. Have you any feed back about longevity for the one you did for a forum member way back?

Nick, Getting a higher alternator speed benefits the alternator cooling as well as giving a higher output.

The diodes have an easier time of it, the bit that gets hammered is the stator. Let's say the varnish is well baked but the machine appears to be reliable.

[Tony Brooks]

Thanks, good to know.

[Mikexx]

On 04/05/2017 at 12:53, Sir Nibble said:

The diodes have an easier time of it, the bit that gets hammered is the stator. Let's say the varnish is well baked but the machine appears to be reliable.

The rotor would still be dissipating 50W or so.

Having said that, with my limited experience, I have only known a single wiring fail and that was a crimped stator star point that had come unstuck. Most have been brushes and occasionally regulators/diodes.

Edited May 5, 2017 by Mikexx

[Sir Nibble]

17 hours ago, Mikexx said:

The rotor would still be dissipating 50W or so.

Having said that, with my limited experience, I have only known a wiring fail and that was a crimped stator star point that had come unstuck. Most have been brushes and occasionally regulators/diodes.

I think the heating effect is partly down to out of phase forces. I've not really thought about it.

It does put increased load on the belt though.

[Mikexx]

5 hours ago, Sir Nibble said:

I think the heating effect is partly down to out of phase forces. I've not really thought about it.

It does put increased load on the belt though.

The two main losses should be copper losses and iron losses. Copper losses should be a function of current and resistance, and iron losses a function of field and rpm.

I would have thought the torque, all else being same, would be the same but start drawing a load at a lower rpm?

[Sir Nibble]

It's an interesting comparison between electrical and mechanical power. More electrical power is generated at lower rpm. Therefore more mechanical input is demanded which with rpm fixed can only come from meeting an increased torque demand.

Copper and iron losses. Copper slightly up due to longer winding and reduced XL from lower frequency. This effect is significant enough to show 5 or 6 amps more output than dividing the output of the standard delta winding by root 3.would give.

[Mikexx]

18 minutes ago, Sir Nibble said:

It's an interesting comparison between electrical and mechanical power. More electrical power is generated at lower rpm. Therefore more mechanical input is demanded which with rpm fixed can only come from meeting an increased torque demand.

Copper and iron losses. Copper slightly up due to longer winding and reduced XL from lower frequency. This effect is significant enough to show 5 or 6 amps more output than dividing the output of the standard delta winding by root 3.would give.

 

If we establish that we are trying to 1/2 the cut in speed through twice the number of stator turns with 1/2 the wire diameter, then: Instead of a 100Amp alternator we have a 50Amp.

1/2 the amount of power is being 'created' at 12V at 1/2 speed. So the torque should be the same at the new cut in rpm as per old cut-in rpm (give or take).

If the copper diameter is 1/2, then I^2 R losses will be the same as current will now be 1/2 the old current. I am neglecting thickness of insulation which will be nearly the same for either winding. The inductance of the stator will be 4 x the old inductance (from twice the number of turns), but twice the emf at any rpm, so above cut-in speed, such we should have half the phase current in each winding.

I am aware that delta windings have significant third harmonic at the star point, such there are sometimes two further diodes are added to 'catch' this extra energy at high rpms.

I would also have thought you would keep the star configuration if it was already there, or make your new alternator star and reap the advantage?

[Sir Nibble]

I don't wind a stator for this. I change the existing one delta to star. Your analysis above is flawed as the entire point is to get more power at slow speed. For instance a standard unit may give 10A at a speed where the modified unit makes 30. At that speed torque will be X3.

[Mikexx]

8 hours ago, Sir Nibble said:

I don't wind a stator for this. I change the existing one delta to star. Your analysis above is flawed as the entire point is to get more power at slow speed. For instance a standard unit may give 10A at a speed where the modified unit makes 30. At that speed torque will be X3.

 

My apologies, I had believed you might have rewound them. I would have thought that star would be the default winding as you need less turns for the same voltage-current charateristics.

By either using more turns, or changing to star, you are inherently creating a higher emf, which will provide a lower cut-in rpm and so start charging at a lower rpm. I'm not sure of the flaw in my analysis.

After conversion, at higher revs a star stator should be producing 0.57 [1/sqrt(3)] of the current that would be produced in a delta configuration.

[jddevel]

Compliments to the participants in this "discussion" sort of informative even if a bit over my head. Further confirmation of the need to often take problems to the experts rather than fiddle yourself. If you know the right one!!! Saves stress.

[The Ents]

I am no electronic engineer.

On Balthazar we have a Lister sr3, originally with a standard camshaft pulley. We suffered from a lack of good charge.

On advice we changed the cam pulley to a 12" one and problems solved. We can even run the washing machine when under way, no problems.

Mike.

 

[WotEver]

On 06/05/2017 at 01:57, Mikexx said:

I'm not sure of the flaw in my analysis.

Doesn't this explain the flaw?

On 05/05/2017 at 16:12, Sir Nibble said:

a standard unit may give 10A at a speed where the modified unit makes 30. At that speed torque will be X3

 

[Mikexx]

1 hour ago, WotEver said:

Doesn't this explain the flaw?

 

I don't believe so.

The original quote " Your analysis above is flawed as the entire point is to get more power at slow speed".

I have already said the 'cut-in' would be lower which means the alternator will be producing current at lower revs where it wouldn't have before. Given power is proportional to current at a fixed voltage I still don't see the 'flaw'.

 

[Sir Nibble]

12 hours ago, Mikexx said:

I don't believe so.

The original quote " Your analysis above is flawed as the entire point is to get more power at slow speed".

I have already said the 'cut-in' would be lower which means the alternator will be producing current at lower revs where it wouldn't have before. Given power is proportional to current at a fixed voltage I still don't see the 'flaw'.

 

You compared two maximum outputs at two different speeds. Think of it like this. The alternator spins at a speed S. Output is around 10A at 13V, 130W. Reconfigure the same alternator and at speed S it gives 30A at 14V, 420W. The extra power has to come from torque because S is the same in both cases. The potential for the original alternator to produce more at higher speed is irrelevant since not only is increased speed a contributor to overall power but is not achievable anyway, which is why the alternator has been modified.