Ferrite Gen with a Classic 150

[Watt]

Thank you Flux.

Yes I do believe your comment has helped.  Couple of things I didn't REALLY take into consideration was the rpm and the fact he will not be coupling this to battery voltage, really blows every failled or successful combination I've ever tried out the window.  I should have read, reread and yet reread again before being so stubborn.  Thanks to all of you for helping yet again. 

There may be hope yet. 

[ChrisOlson]

And That would make a machine more resistant to burning up coils, hence the test bed results,
But possibly at the risk of run-away?

Glen, I've never seen one of these ferrite machines exhibit runaway tendencies.  But they do seem to lose some rpm/volt performance at higher amps, meaning it takes more rotor rpm than it should (or as compared to neos) to develop the amps as the amps increase.  This is actually good because it keeps the TSR of the rotor up and not near as much tendency to "stall" as a neo gen will cause.

This is all based on observation.  The dual stator two-phase ferrite gens on my 3.8 meter machines will easily push 100 amps in good wind with the bank around 29 volts.  They have the same rpm/volt as the single stator three phase neo gens I've used on them, and the same internal resistance (within a few 100th's - the neo is .39, the dual stator is .38).  But the dual stator ferrite is less "stiff" - the rotor runs much faster with it than it does with a neo gen.  And consequently it more often goes over 100 amps than the neo gens because the rotor efficiency is better in higher wind speeds due to the higher rpm.

I've never really determined the cause of that.  I can only surmise that it "bends" or "distorts" the weak flux at higher output.  I also surmise that this same phenomenon is why I have failed to burn one up on my test stand with the stator fully shorted.  I wish I had a better understanding of it but at this point all I can relate is what I have tested and observed with these things.  In the long run, the turbines with the dual stator ferrite generators have proven to be much more powerful than the neo gens, especially in higher wind speeds.  But at no time have I ever seen one even get close to runaway.

The two blade 4.0 meter machine I built got close once - but that's only because it hit 140 amps and tripped the breaker.  When the breaker tripped it went into runaway, and then when the breaker reset it loaded the turbine which caused it to furl and violently yaw out of the wind and it broke the tower mast.  I didn't lose it off the tower, but it came close and was leaning at a 30 degree angle on the tower by the time I got it shut down.  I since took that machine down and put up a 3.8 meter three-blader in it's place because it was too wild and I have to figure out a way to get rid of the terrible vibration it has when it yaws.

[Flux]

" I've never really determined the cause of that.  I can only surmise that it "bends" or "distorts" the weak flux at higher output.  I also surmise that this same phenomenon is why I have failed to burn one up on my test stand with the stator fully shorted.  I wish I had a better understanding of it but at this point all I can relate is what I have tested and observed with these things."

Yes that is right, you have a field produced in the stator by the current flowing in it and this field reacts with the field of the magnet. It is partly distorting and partly demagnetising.  It takes a lot of amp turns to produce much of a field in a gap with no iron so the effect on neo alternators is very small ( not true for iron cored machines with tiny air gaps).  When you get down to the weaker field of ferrite magnets the armature reaction starts to have a greater effect. On short circuit you will certainly get a significant effect and at high speed it may reach the point where you get a demagnetisation of the main magnets, but with air gaps several inches wide you would never reach that speed in normal working. The recoil line for ferrites is a straight line so as long as you don't go off the line the thing recovers perfectly as the current falls so demagnetising is not easy.

For direct loading this may act in your favour in avoiding stall. With the classic you will have much lower stator current and the reaction will be fairly small and you can ignore it but you still may have to consider the effect on clipper operation or braking circuits.

Flux

[ChrisOlson]

For direct loading this may act in your favour in avoiding stall. With the classic you will have much lower stator current and the reaction will be fairly small and you can ignore it but you still may have to consider the effect on clipper operation or braking circuits.

Indeed, and that's one area of research where I'm going to have to do a bunch of bench testing to develop a clipper load for this thing that works properly.  I've got it all worked out in theory.  But in practice I'm suspecting it might take more clipper load than what I have figured because it's very important for the clipper to pull the TSR of the rotor down when it engages.  If that "bending" or "distorting" of the flux becomes an issue at the high amp loads the clipper will require, it might end up being more of a shorting brake than an actual clipper.
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Chris

[rossw]

With three phase, each leg is basically handling the full load for 2/3's of the time.
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Chris

Stumbled on that comment, but it makes no sense to me.

If 3 phase power is the RMS volts times the (per-phase current * sqrt(3)), how do you come up with the above statement?

Surely, if each of 3 phases were carrying "full load" for 2/3 the time and you have 3 of them, the vector sum would be 200% of the output!
(2/3 full load times 3 coils = 6/3 = 2)

*confused*

[ChrisOlson]

Maybe I didn't phrase that right.

If you have a three phase power supply rectified to DC and you have 30 amps DC from the rectifier, the RMS amp load on each leg of the three phase power supply is 20 amps.  For all practical purposes, since the peak power flows in any two of the three legs at any moment, each phase or leg is carrying the peak load 2/3's of the time.
--
Chris

[Flux]

If you take a conventional ac load then you are correct, power is root3. V line. I line cos phi  ( or 3 times power per phase)

When you feed the thing through a rectifier and clamp to a battery the rectifier conduction will at any instant be through two of the lines. when one phase volts drops the conduction will transfer to the next phase but still using 2 of the lines. The current at any time will be carried by two lines and that is why you save the cost of a wire for the same loss if you transmit the dc after the rectifier.

The power to the battery will be battery volts x dc mean current. The rms current in each ac line will be higher than 1/3 of the dc current and typically is about 0.76 of the dc value. The voltage on the input of the rectifier will be a clipped mess looking much more like a square wave so it will have a very strange rms value not far removed from battery volts.

Normally it is easiest to base power on the battery input as both volts and amps are meaningful and easily measured. If you want to get involved with ac readings you will need the rms current to measure stator loss. The actual power on the ac side is tricky, you need rms current and volts but you have an issue with power factor and as it is a non linear load the cos( phi) method doesn't work. You have to go back to the true definition of power factor ( real/apparent power). the whole thing gets very confusing.

Virtually all the conventional theory about machines is pretty useless for these strange machines clamped to a battery via a rectifier and that is why many strange ideas prove to be better in the end than the obvious ideas from conventional electrical engineering.

I hope Chris doesn't mind me intervening here, it must be a real challenge for a mechanical engineer to be confronted with this interesting question.

Flux

[rossw]

Maybe I didn't phrase that right.

If you have a three phase power supply rectified to DC and you have 30 amps DC from the rectifier, the RMS amp load on each leg of the three phase power supply is 20 amps.  For all practical purposes, since the peak power flows in any two of the three legs at any moment, each phase or leg is carrying the peak load 2/3's of the time.
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Chris

That doesn't match with any of the 3-phase theory I was taught.
Isn't root3 (1.732, near enough) the "magic" number?

So 30A output would require 17.3A/phase?.
And if peak power were "flowing in any two of the 3 legs at any moment" wouldn't that mean each was carrying 1/2 the current? (which doesn't work either)

I can't see any situation where any phase in a 3-phase installation would be carrying the equivalent of "2/3 full load all the time" even if averaged out.

[rossw]

When you feed the thing through a rectifier and clamp to a battery the rectifier conduction will at any instant be through two of the lines. when one phase volts drops the conduction will transfer to the next phase but still using 2 of the lines. The current at any time will be carried by two lines and that is why you save the cost of a wire for the same loss if you transmit the dc after the rectifier.

But surely in this situation, when we take time into account, the time when any given phases' voltage is high enough to have it's diodes conducting, will be somewhat less than 100%, so then if (by your statement) any *2* phases are supplying current, and the voltage is clamped to battery voltage, the current will be similar - so at any given time 2 out of 3 coils will be supplying 1/2 of the current each, and given that by inference each phase will only be "one of the 2" for 2/3 of the time, that makes each phases contribution (2/3) of (1/2) over time??

[ChrisOlson]

So 30A output would require 17.3A/phase?.
And if peak power were "flowing in any two of the 3 legs at any moment" wouldn't that mean each was carrying 1/2 the current? (which doesn't work either)

Ross, this is a little outside my area of engineering expertise.  And Flux can explain it much better.  But, no, each leg does not carry half the current.  Any pair of the three legs (and the generator phases too) are carrying the full current 2/3's of the time.  And when I've measured current on a rectified three-phase using a DC shunt type ammeter and my Fluke meter on the AC side, that's what I see too - 2/3's of the current that is flowing on the DC side of the rectifier is flowing in each generator leg.

Again, Flux is able to explain this much better than I can.  I understand why it does what it does in basic terms.  But I don't understand the theory behind it.
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Chris

[scoraigwind]

that makes each phases contribution (2/3) of (1/2) over time??

OK you need 2 wires to complete a circuit.  At any given time you could be using 2 out of the 3 wires that you have.  In such (theoretical) cases you would use the full current in any given stator wire for 2/3 of the time.  So the average current is 2/3 of the DC,  and the rms would be root(2/3)= 82%.  In reality they share to a degree so the rms is reduced accordingly, but the exact amount depends on impedance in the stator side of the circuit.

In reality the rms can be assumed to be under 24 amps for a 30 amp DC current out.

[kenl]

Chris,

 Any chance you have a shot of the inside of the gear case? I would really like to see the insides. Also what are you using to hold the bearings, it almost looks like a floor flange from the pictures. Cool project, I can't wait to see some of the test results hooked up to the classic. After messing around with a F&P alt with different voltage battery banks I've come to the conclusion that most alternators don't seem to be run to their full potential. And the main cause of that in my case at least is to low of battery bank voltage and using an alt>rectifiers>battery bank setup. I really believe to course your on is the way to go.

kenny

[ChrisOlson]

Kenny,

the gearcase is just hollow right now - just shafts thru it while I test assemble and get the stator mount built and stuff.  I'll take photos of it during the final assembly.

The gearcase is the same, albeit different shaft layout, than my previous geared turbines.

I did discover one problem last night - the blade hub is still too close to the gearbox and the root airfoil section on the blades will hit the PTO shaft bearing as they go around.  The root airfoil section on the blades I'm using are a little over 10" wide with 18 degree pitch.  I either have to build a slightly longer input shaft or move the front bearing on the PTO shaft to the inside of the case.

So I still have a lot of work to do on it.
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Chris

[Flux]

Timed out and i had to log in again in which time hugh has basically answered the question. His 24A agrees very closely with my measurements in the past .

Here was my reply anyway.

Probably easiest to think of a star connected winding.

At any instant two phases in series are feeding the rectifier and the  current in those two conducting windings will be 30A.  As the conduction is for 2/3 of the time the mean current in each line ( and phase since we chose star) will be 20A.

You can now help out with your maths, I can't do it, but you have a mean current of 20A flowing for 2/3 of the time and it is something approaching a rectangular block of current. I know from measurement that this comes out to something close to 23A rms but I can't calculate the precise rms value. You can assume there is no overlap in the rectifier ( there isn't).

There is no way that the rms current can be 30/root3.  Manufactures ( I won't name them ) have been known to get the line losses badly wrong and supply under size fuses for machines when they have failed to understand  this.

Probably the easiest way to get the ac power in the windings is to assume it is battery power plus stator loss. You are dealing with undefined waveforms for current and voltage and although they are in phase the power factor due to waveform error is not unity.

If you consider a delta winding then in theory the phase current will be line current/root3 but we are not questioning that. We are considering the mean and rms currents into the rectifier.

Flux

[Watt]

Is there any chance this proposed current could also be limited by actual waveform?  I mean, variations from sinusoidal.  Not saying I see this in this case only that these are homebrew stators and rotors. 

[Watt]

Kenny,

the gearcase is just hollow right now - just shafts thru it while I test assemble and get the stator mount built and stuff.  I'll take photos of it during the final assembly.

The gearcase is the same, albeit different shaft layout, than my previous geared turbines.

I did discover one problem last night - the blade hub is still too close to the gearbox and the root airfoil section on the blades will hit the PTO shaft bearing as they go around.  The root airfoil section on the blades I'm using are a little over 10" wide with 18 degree pitch.  I either have to build a slightly longer input shaft or move the front bearing on the PTO shaft to the inside of the case.

So I still have a lot of work to do on it.
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Chris

Or 3, grind on the blade.  What??? What is wrong with a little humor? 

[ChrisOlson]

Or 3, grind on the blade.  What??? What is wrong with a little humor? 

Actually, I might do that.  It would be the first set of blades I've ever had with a bearing clearance notch cut in the root section of the airfoil with a hacksaw.   

--
Chris

[Watt]

Or 3, grind on the blade.  What??? What is wrong with a little humor?  

Actually, I might do that.  It would be the first set of blades I've ever had with a bearing clearance notch cut in the root section of the airfoil with a hacksaw.    

--
Chris

   Are you also part REDNECK...     Heck, justify the cut with power tools.   

[ChrisOlson]

I spin tested this new gen about an hour ago and got 119.1 volts @ 1,000 rpm.  That works out to 8.39 rpm/volt.  So I'm getting more volts out of this scrunched down outfit than I got from the other 16 pole I built.

But one difference is the air gap.  I got the gap at 18.5 mm on this one where I ran 20 mm on the other one.  The stator on this one is a lot thinner and I got lots of room between the mags and the stator on this one.  I could could take two 1 mm shims out and drop the air gap to 16.5 mm on this one and still have good room between the mags and the stator.

And any rate, I'm pretty happy with the spin test numbers.
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Chris

[Watt]

I spin tested this new gen about an hour ago and got 119.1 volts @ 1,000 rpm.  That works out to 8.39 rpm/volt.  So I'm getting more volts out of this scrunched down outfit than I got from the other 16 pole I built.

But one difference is the air gap.  I got the gap at 18.5 mm on this one where I ran 20 mm on the other one.  The stator on this one is a lot thinner and I got lots of room between the mags and the stator on this one.  I could could take two 1 mm shims out and drop the air gap to 16.5 mm on this one and still have good room between the mags and the stator.

And any rate, I'm pretty happy with the spin test numbers.
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Chris

So hook it up to the classic already.   

[fabricator]

Or 3, grind on the blade.  What??? What is wrong with a little humor?  

Actually, I might do that.  It would be the first set of blades I've ever had with a bearing clearance notch cut in the root section of the airfoil with a hacksaw.    

--
Chris

   Are you also part REDNECK...     Heck, justify the cut with power tools.   

Part redneck? That neck aint never been any other color.

[ChrisOlson]

Well, this is a photo of the rotor I'm using on this machine, take a couple years ago when I took it off a different turbine



The blades mount on the rear of the tapered bore hub and there's a sandwich plate that goes on the rear that pilots on the center of the hub.  The root section of the airfoil makes a radical sweep to the rear right at the hub and the trailing edge of that airfoil just nicks the front PTO shaft bearing.

Somewhat of an oversight on my part when I calculated how long the input shaft had to be.  I figured for 30 mm clearance between the airfoil and the front of the gearcase.  But forgot all about that bearing protruding out the front of the gearcase.
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Chris

[scoraigwind]

The amount of max safe current for magnet wire in coils is given by the cross section squared times 4869.48.  13 AWG wire is .072" diameter.  So it's max safe current is 25.2 amps. 

This is going back a few days.  I have been studying Chris's figures and summarising the project here http://scoraigwind.co.uk/?p=1148 and I noticed the above.

I am not sure where Chris got this "cross section squared times 4869.48" but its a rule of thumb at best.  The temperature rise is going to depend on the geometry as well as the heat generated in each wire.  13 gauge wire will work OK at this loading in a thin stator but if the stator thickness were doubled (for example) then the situation would be a lot different.

I prefer to compare stators based on the watts of heat produced per unit of exposed coil area.  I work out how much area of copper is at the surface of the stator and the power it has to dissipate.  The cooling (ambient temperature and air flow) are also considerations.  I like not to exceed 0.5 watts or so per square cm of coil surface like that.  Using 13 gauge wire in this situation gives about that same maximum dissipation as it happens, so it looks OK.  In this case each coil will be dumping about 40 watts and the exposed surface is about 44 sq cm on each side.

[ChrisOlson]

I am not sure where Chris got this "cross section squared times 4869.48" but its a rule of thumb at best.

It is kind of interesting.  I got that equation from engineering at Essex Wire, who makes the magnet wire I use.
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Chris

I should quantify that by mentioning that the engineer that gave me that equation to figure amps for a particular wire size said the equation will only give the max safe amps that the wire should carry until the rating temperature of the insulation is reached.  Once the wire gets to its rated temperature, then the max safe amps has to be de-rated for temperature.  And they also sent me a chart that shows the de-rate factor that has to be used, based on temperature of the wire.

So, basically, you can design for a duty cycle in the generator.  With proper or adequate cooling you might get 100% duty cycle using the equation.  With high ambient temperature or less than adequate cooling the full load duty cycle might be less, depending on the temperature of the wire.

[boB]

I am not sure where Chris got this "cross section squared times 4869.48" but its a rule of thumb at best.

It is kind of interesting.  I got that equation from engineering at Essex Wire, who makes the magnet wire I use.
--
Chris

I should quantify that by mentioning that the engineer that gave me that equation to figure amps for a particular wire size said the equation will only give the max safe amps that the wire should carry until the rating temperature of the insulation is reached.  Once the wire gets to its rated temperature, then the max safe amps has to be de-rated for temperature.  And they also sent me a chart that shows the de-rate factor that has to be used, based on temperature of the wire.

So, basically, you can design for a duty cycle in the generator.  With proper or adequate cooling you might get 100% duty cycle using the equation.  With high ambient temperature or less than adequate cooling the full load duty cycle might be less, depending on the temperature of the wire.

I assume this equation is also based on an ambient temperature of  25 degrees C /  77 F  (?)

boB

[ChrisOlson]

I assume this equation is also based on an ambient temperature of  25 degrees C /  77 F  (?)

boB, from what I understood when I talked to the guy was that could be used at any temperature up to the limit for the insulation on it.  Then once the max temp for the wire insulation is reached it has to be de-rated to keep the insulation below its safe temperature.  Like I told Ross the other day, this is a little outside my area of expertise, but that equation has worked well for me in deciding what wire to use in generators.
--
Chris

[scoraigwind]

Intending no disrespect to Essex wire, the temperature that a wire will reach in a winding also depends on the relation between the surface area of the winding and the mass of wire inside it.  Also as previously stated it will depend on the ambient air and the rate at which it circulates.  There will be some cooling due to radiation of heat too that depends on the colour of the finish but this is secondary.  Often the steel laminations provide a pathway for sinking heat too.

I must assume that his equation makes assumptions, which may or may not be applicable to our situation.  It's probably safer to compare the watts per square inch of the stator surface with successful examples that you have built and tested.  It's not hard to compute.  And it seems to lead to a similare conclusion.  But bear in mind that a thicker stator will have less ability to cool itself.  (Also I have seen that a very thick stator with a thick layer of fibreglass/resin on it in a tropical situation is a very strong candidate for burn-out.)

Like you I don't always accept what I am told. I need it to make sense to me first.

[ChrisOlson]

I must assume that his equation makes assumptions, which may or may not be applicable to our situation.

I think it does too, and I agree with what you say.  I assume that calculation is slanted towards windings in conventional air-over cooled electric motors and generators, and may not apply 100% to the axial style generators we (mostly) all build.  And it may not apply to some stators where they are very thick, like you say.  I have always tried to make the stator as thin as possible and these high speed gear driven generators I have used for quite awhile have less cooling problems than direct drives do.  The rotors on these things move a lot of air because they spin so fast, and I think that helps some.  And I usually include holes in the rotors so the magnets, which act as centrifugal fan blades, can draw more air in from the center to blow it out.

I have run various versions of these at very high outputs on my test stand without overheat problems so I am pretty confident in them installed on the turbine where more air is flowing over them than on a test stand.

One of the bigger problems for my climate is how to keep snow out of it.  If the turbine is running in a snow storm with the generator warm, and snow blows in it it melts and freezes on cold metal things like magnets, builds up in there, and locks it up when the ice gets too thick between the mags and the stator.  I have to figure something out for this new machine because it's got big holes in the rotors where snow will blow in and not very much space around the magnets to let it back out.  I can see where it could build up ice on the inside of the magnets, then a piece of ice come loose and lock it up.  Or if it stops when the stator is warm, it will melt the ice that accumulates on the inside of the magnets, the water will run down between the mags and stator, freeze, and then it won't turn.  I might just duct tape the front holes shut for winter time operation.
--
Chris

[ChrisOlson]

One thing I decided to do on this new machine is put steel bushings in the stator mounts.  The bushings are 1/2" ID so I can use 1/2" bolts and torque that stator down to 65-70 lb-ft of torque without crushing, splitting or cracking the stator, and it won't come loose.  The stator on this machine is not adjustable with threaded rod.  It bolts right to the stator mount.

Since the stator is solidly mounted I machined a spacer that slides over the PTO shaft and butts up against the lock on the rear PTO bearing.  Then the inner (front) rotor goes on up against that spacer.  Then the stator is bolted on and there's a 3 mm gap between the face of the magnets and the stator on the front rotor.

I machined a nominal spacer that slides on the shaft between the rotors and sets the magnet face to stator clearance on the rear rotor to zero.  I added five 1 mm shims on the shaft with the nominal spacer and that sets the air gap to 18.5 mm.  I can pull the rear rotor add or remove 1 mm shims and change the gap accordingly.


--
Chris

[bj]

    The spacers are a nice idea Chris.  No worries about fasteners letting go. 

[Flux]

Normally the maximum temperature of a winding is limited by the class of insulation. Given suitable materials elsewhere that ultimately means a limit on the wire coating.

Normally I wouldn't see any limit on current as long as that wire coating temperature is not exceeded but your information seems to imply that there is a practical limit even when the wire coating temperature is not reached.

It probably implies that gross overload stresses the wire coating in some way that is detrimental even if you terminate it before the final temperature reaches the coating limit. During severe overload there will be rapid expansion that will cause stress and probably other factors.

For many years, here (UK) wire tables gave current ratings at 1000A/ square inch. This was very conservative and even class A machines normally ran above this. Typical class F transformers and small machines tend to run at about 3000A/sqi.  Class H welders and similar run about 4000 for fairly continuous rating but can run to about 6000 for intermittent duty.

Your .072 diameter wire is running at near 6000A/sqi for 25A so the makers probably think that is about the practical limit. Certainly with Kapton tape insulation you can reach a state where the copper oxidises if the current is too high so there is an absolute limit somewhere.

Relating this to what a wire can carry in a winding is rather different except for short term rated machines. With these type stators the limiting temperature is reached on the wire at the centre of the coil where heat can't get away easily. Stator thickness is a big factor here even if there is air cooling over the surface, the centre turns will be much hotter than those near the surface.

Flux

[ChrisOlson]

Relating this to what a wire can carry in a winding is rather different except for short term rated machines. With these type stators the limiting temperature is reached on the wire at the centre of the coil where heat can't get away easily. Stator thickness is a big factor here even if there is air cooling over the surface, the centre turns will be much hotter than those near the surface.

One ones I've burned out in the past usually seem to want to fail at a solder joint on an interconnect.  I used to twist the wires together, then solder them.  I decided that method (probably) stretches the wire a bit below the twist and makes it thinner, and that's why it tends to fail there.

So what I did instead, is that now I get some crimp-type butt connectors and cut the plastic sleeve off them.  I slide a short length of heat shrink tubing over one wire, butt the ends together in the butt connector, crimp both sides, solder it, then slide the heat shrink tube over the joint while it's still hot, which causes the tubing to shrink and seal it.  Since going to that method I have never had another problem with burning interconnects off.
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Chris

[Flux]

If you are running at the limit of the wire coating for 200C wire you are actually above the melting point of eutectic solder. Having the crimp in addition to the solder is a good idea.

For class H operation you would be better off using silphos if you have it over there , it is a self fluxing braze material that let's you join the wires without having to strip the enamel, it is often used in motor manufacture as it saves a lot of time when there are lots of connections, it does need oxy propane or oxy acetylene to do it effectively. Probably not worth it for one or two connections unless you have it around.

Flux