Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Need help with sizing a line reactor for my application

Status
Not open for further replies.

I'mClueless

New Member
Hello,

I need help with selecting the correct line reactor for my application. I'm dealing with up to 270A/100V DC that is derived from AC current produced with an automotive alternator and fed through a 300A/1000V bridge rectifier. If I have correctly comprehened what I have read then I need a swinging type of reactor to cope with the variable A/V.

I have been reading as much as I can find on the subject and the following is as close as I have found to offering a formula. The problem is that my lack of formal education in electrons prevents me from being able to interpret the characters in the equation that are not found on a standard keyboard. I tried to research 2¶FL which is used in the Rockwell Automation equation below and all I came up with was ads featuring products that are packaged in 2 fluid ounce volumes.

It is widely publicized that 3% is a constant used for smoothing and 5% for dampening harmonics. Using that value I have calculated the amount impedance to be entered into the Rockwell Automation equation below.

100V / 270A = 0.370 Ohms
3% * .370 = .011 impedance

This is the equation that I used to calculate the frequency of the voltage ripple:

Alternator revolving at 6000 RPM / 60 = 100 Revolutions Per Second
100 RPS * (6 North and 6 South poles) = 1200
1200 alternating currents * 3 phases = AC @3600Hz

I would appreciate assistance with interpreting the equation in the following:

Copied from:

Line Reactors and AC Drives
Rockwell Automation
Mequon Wisconsin

Sizing a reactor:

The first rule is make sure you have a high enough amp rating. In terms of the impedance value, you will usually find that 3% to 5% is the norm with most falling closer to 3%. A 3% reactor is enough to provide line buffering and a 5% reactor would be a better choice for harmonic mitigation if no link choke is present. Output reactors, when used, are generally around 3%. This % rating is relative to the load or drive where the reactor impedance is a % of the drive impedance at full load. Thus a 3% reactor will drop 3% of the applied voltage at full rated current. To calculate the actual inductance value we would use the following formula. L =XL/(2¶FL) Where L is inductance in Henrys, XL is inductive reactance or impedance in Ohms (0.370 Ohms) and F is the frequency (3600Hz). In general Frequency will be the line frequency for both input and output reactors.

Your drive distributor should be able to help you size a reactor for use with a drive. If you wish to calculate the value yourself, the following example may be helpful. If a 3% reactor was required for a 100 amp 480 volt drive, a 100 amp or larger current rating would be required. The drive impedance would be: Z=V/I or 480/100 = 4.8 ohms. 3% X 4.8 ohms = 0.114 ohms inserting this 0.114 impedance in the equation for inductance we get a value of about 300 Microhenrys.

Yep,
I'mClueless
 
Last edited:
If you are up for a bit of experimenting try taking a piece of solid round stock about 2.5 to 4 inches in dia and about 12 - 15 inches long and put 20 - 30 turns of welding cable around it then make a weld.
If the weld has poor penetration characteristics add another 10 turns and try again. If it has too much penetration and the weld puddle is too wet take 5 turns off and try another weld.

Congratulations, You just made yourself a farmer Fred engineered welder reactor. (assuming you can use a tape measure and can count of course!) ;)

It may sound hokey but an old school pipeline welder showed me this many years ago. Apparently back in the old days when almost all engine drive welders used brushed commutator based generators they tended to have a bit of an individual welding personality to them from one to the next and some needed a bit of added inductance at times to smooth out a weld on some odd metals while using questionable rods.

Apparently the right number of turns of welding cable on any old chunk of mild steel can make a nice adjustable arc stabilizer if tuned properly. I hear old steel wheel rims also work well too! :)
 
Beware if you're wearing steel toe capped shows, when you weld your feet might be drawn towards the huge electromagnet you've just made.:D
 
Beware if you're wearing steel toe capped shows, when you weld your feet might be drawn towards the huge electromagnet you've just made.:D

ZERO999,

Please allow me to apologize for the PM, I simply did not see the bottom of the page where your admonishment was. I figured out your equation for calculating capacitance and I thank you for publishing it.

After the ripping that you gave me I hardly find your comment to be helpful or humorous. If you don't have anything intellient offer then keep your smart remarks to yourself. DO YOU UNDERSTAND?

Yep,
I'mClueless
 
Be nice now. People have a sense of humor here and tend to use it on a whim.
Besides its an open forum. It sort of like a bar where anyone could walk in and join the conversation at any time.

Some days you get the town scientist or engineer or repair man. Other times you get the town drunk or village idiot or the local snot nosed know it all kid. :p

Most of the time it hard to tell which one is which though.:D
 
ZERO999,

Please allow me to apologize for the PM, I simply did not see the bottom of the page where your admonishment was. I figured out your equation for calculating capacitance and I thank you for publishing it.
Apology accepted.

I'm also sorry if my PM offended you, perhaps I shouldn't have been so abrupt.

After the ripping that you gave me I hardly find your comment to be helpful or humorous. If you don't have anything intellient offer then keep your smart remarks to yourself. DO YOU UNDERSTAND?
I was only joking, hence the :D smilie.

The same was also true of RCinFLA's comment.

I know it isn't easy telling if people are being series on an online forum but I think you need to take a deep breath and think before taking offence.
 
tcmtech,

Once again you're the only one to respond with a solution. So if I fabricate a guided reel for my welding cable then I could just draw the cable from the reel until my weld is just right. Novel idea but it does not appeal to me.

My high frequency inductor coil has 20 wraps of welding cable with 20 wraps of 18AWG/30kV around an air core. With the HF installed I might not need to be concerned with a reactor. Is that possible?

Yep,
I'mClueless
 
Last edited:
You need the high inductance of the iron to make it work. Air has very little inductance so it works well at very high frequency's.

What you are going for requires a considerable amount of magnetic energy in motion for lack of a better simplified term. Basically a big magnetic flywheel effect of sorts.

Once you have your basic turns ratios figured out then you may want to build an actual reactor out of a big transformer core and heavy wire unless you have one around already.

Given your higher operating frequency and it being from a three phase source the ripple is already low. All you need is the right inductance to make the welds better.

As I had mentioned in your other thread many years ago I just used a old battery chargers high current low voltage secondary as a reactor and it worked rather well.
You may likely just need something larger due to the higher amperage capacity that you can produce. It could become as simple as some heavy gauge wire on a large microwave oven transformer core or similar larger transformer or large lighting ballast core.
likely it will need to be something in the 1 - 2 KVA range though. Microwaves are usually 1 - 1.5 KVA on the larger ones so you may be close enough.

Really all of this comes down to what type of weld quality is right for your aplication and how much time and money you're going to spend. At some point you will better off just buying a used welder.
There is no point other than the learning experience if you are going to spend $1000 building something that can be bested by a used factory built machine that could be found for around half of that and would give you other usefull functions as well.
 
There is no point other than the learning experience if you are going to spend $1000 building something that can be bested by a used factory built machine that could be found for around half of that and would give you other usefull functions as well.

Yes once again you are correct. I have already reach the point where I have to turn the articles upside down in order to read them. The only way to get upside up is to make this work.

What I lack in knowledge I make up for with determination. There are very few people that could write a full feature and function CNC machine tool control program, let alone doing it as a first project for learning how to write software.

Yep,
I'mClueless
 
"100 RPS * (6 North and 6 South poles) = 1200"

Please correct me if I'm wrong, but my understanding of motors and generators are that frequency is dictated by pole pairs, not by individual north and south poles.
Thus your frequency should be half that value.
 
"100 RPS * (6 North and 6 South poles) = 1200"

Please correct me if I'm wrong, but my understanding of motors and generators are that frequency is dictated by pole pairs, not by individual north and south poles.
Thus your frequency should be half that value.

Hi, I'mClueless and I'm open to correction if I have misunderstood. I thought that alternating current was generated with opposing poles. If I'm wrong then I definitely want to be corrected.

Does anyone else have input on this?

Yep,
I'mClueless
 
Electric motors, generators, and alternators are classified by each individual pole not the pairs. A 2 pole motor essentially has one north and one south pole. A 12 pole alternator has 6 north and 6 south poles.
 
Electric motors, generators, and alternators are classified by each individual pole not the pairs. A 2 pole motor essentially has one north and one south pole. A 12 pole alternator has 6 north and 6 south poles.
Of course...No argument here.
But the question here is: are the frequency and rotation related to individual poles or to pole pairs?
 
Of course...No argument here.
But the question here is: are the frequency and rotation related to individual poles or to pole pairs?

Yes, the frequency is equal to the number of pole pairs multiplied by the rotational frequency, measured in Hz.
 
schmitt trigger,

Does the current reverse when the polarity changes from north to south or south to north?

Does the reversal of the current constitute a cycle?

Yep,
I'mClueless
 
Last edited:
"Does not the current reverse when the polarity changes from north to south or south to north?"
That is called a half-cycle
Period consists of two half-cycles.
And the frequency is the reciprocal of the period.

Now, re-reading your original post.
You are mentioning that you'll use a full wave 3 phase rectifier.
In that instance, the output ripple frequency is 6 times your input frequency, because both half-cycles are rectified.....

You arrive at the same result.......but you just need to get the basic electrical facts straight.
 
You arrive at the same result.......but you just need to get the basic electrical facts straight.

Please clarify which facts you are referring to. I'mClueless and I'm hear to learn, not profess.

Yep,
I'mClueless
 
Last edited:
A cycle consists of a positive half-cycle and a negative half-cycle.
The voltage reverses, and the current follows it, plus or minus the power factor angle.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top