Dear Dave,
Your suggestion is a "down to earth" solution to the problem. Its like finding the distance from one city to another by actually travelling the distance and watching the odometer register the actual distance travelled. But my problem calls for a calculation where the "travelling" has yet to happen. Just like when you have the info that the distance by land from one place to another is about 400km and you want to know how long it will take you to reach that place (without actually travelling). In the same way, you want to know (by calculation) how much power a certain welding machine will require without actually measuring the current draw-off during welding and while idle. In my example, everything is given already (assumed to be from the machine's name plate or from the manufacturer's manual); the current, the voltage, phase, etc. But again, I'd like to re-iterate that it is not as simple as multiplying the voltage and the current and some factor to get the power draw-off. You would note that in my first posting I included the formula:
Power, kw/hr=(460*37)*(120/400)*0.8*1.73/1000
You will notice that everything is there. I was expecting someone to ask what the (120/400)*0.8 stand for and why the units is kw/hr. Again, if I may repeat, the simple formula Power, kw=(460*37)*1.73/1000 is giving us absurd result. If you use this, you will arrive at 29+kva; and if you have 10 of this machine, you will be needing a 290kva generator which is too much!
Our company is engaged in electro-mechanical projects and for that we have an electrical/instrumentation department keeping all sorts of measuring tools. As a matter of policy, we kept our fleet of welding machines calibrated every now and then and we cannot do that without amprobe. A different amprobe is used for the primary and secondary side respectively.
But again, we don't want our engineers energizing and measuring the output of every machine that comes. Our stocks of w.machines comes in rated capacity of various sizes (250A, 300, 350, 450, 500, 600A), not to mention the multi-operator units; and these are of different make (miller, lincoln, etc) which give different input current for the same capacity. We want our engineers to be able to plan and decide "on the table" the power requirements for any combination of these machines... that's what engineers do... that's what I want them to be able to do. That's why I posted the question in this forum because I myself is not sure how to go about it. I developed the above formula which I presume is correct.. but I needed a vote of confidence to use it. In fact i was expecting somebody to contest it and probably with more heads, we can come up with a good solution. It was my fault from the start that I did not specifically ask for a good and reliable formula than what I have posted above. But that is what I was really asking for: a formula. I am sorry if I have created confusion.
Yes, You do need to explane what the units stand for as You have left us to guess. I guess that You are trying to interpolate input power at other than rated output. I guess 460 is the line voltage. I guess 37 is the amperage at rated output. I guess that 400 amps is the rated output of the machine, from line 3 in Your first post. I guess 120 is the output amperage while welding, but that differs from the 110 You mention. NOW I AM REALLY GUESSING that the .8 might be the power factor since You are working in KW. From the 1.73 I guess You are using 3 phase, and the dividing by 1000 to get us to Kilo. I think that for sizing the equipment You should be using KVA as opposed to KW. If You use KVA this formula works out pretty close to the 9 KVA which was My guess which I mentioned in a previous post. I must say that You are "shooting at a moving target" so to speak. What I see as the biggest problem is that there is no way to predict how many welders may be welding at a given time, this is independant of duty cycle or an operator factor. I think You need to size for the peak load, not the average. This is why I suggested you use the actual KVA input of each machine at the output it will be used at and just add them up. This will undoubtedly give a much higher number than the average load. So I agree pretty much [except for the KW/KVH part] with Your theory. Theory only goes so far, it needs to be tested, so that is why I still suggest using the amprobe on a few machines representative of what You will have in the field. If You run some of the constant [lighting & ventilation etc.] loads on the same gensets as the welders a smaller percentage of the total is variable and You can run at a greater percentage of capacity.
Ok Dave, I stand corrected and I'm sorry. I pasted the formula hastily from Excel without editing. Allow me to correct as follows:
----------------------
W.Machine Specs:
Rated output: 400A
Duty Cycle: 60%
Input current: 37A
Input voltage:460V
Phase: 3-Phase
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Electrode Current: 120A
----------------------
Power, kva==(460*37)*(120/400)*0.8*1.73/1000
This will result to about 7kva. The 0.8 is not for power factor but for genset efficiency. The 1.73 factor means it's a 3-phase connection. The 1000 is a conversion to kilo.
I am not an electrical engineer but I can safely say that kw and kva units are "approximately" the same.
Why did I say I arrive at 42 KW/HR daily consumption? I assumed that the w.machine will be operating at load (120A during welding) only for 6hrs, though the working time is 8hrs, thus 7kva x 6hs/day=42KW-HR/day.
The actual arc-on time of welder maybe less than 6hrs esp on SMAW and GTAW but this will be a conservative value considering other loads.
The 120/400 is a factor applied to power due to rated input volt/amp on the "principle" that the actual welding amp is only 120A and "perhaps" the rated input volt/amp of 460V/37A will be true only if you run the w.machine full capacity at 400A.
The above formula worked fine with me and "all" of the licensed electrical engineers I have shown it (bewildered as they are) accepted the approach, otherwise they have to come up with a practical/alternative solution... and I haven't got one.
If I cannot get a different approach from this forum, then perhaps I could say that I have somewhat contributed something useful for other visitors who might have the same problem as stated. The subject is still open for suggestion though.
I am not an electrical engineer either, but I offer these criticisms. Your .8 factor makes no sense to Me, as it lowers the load figure from the welding machine. You don't explane where it came from, and I can't guess. KW and KVA are the same in a resistive load, but they differ in an inductive load. Transformers are an inductive load. Amps x Volts will give You KVA, You then multiply that by the power factor to get KW. A random example, a Miller Goldstar 452 draws 35.5 KVA and 23.3 KW at 450 amps output as per the company literature. The KW and KVA differ less on a CV machine. The ratings the manufacturers provide are in fact at the rated load, in the case of Your example the 37 A @ 460 V is at 400 A output, using the actual output as a ratio of the rated output will give an aproxamate and acurate enough for Your purposes value for primary load. It is My feeling that KW-Hr/day is more usefull for estimating fuel useage as per the formulas I previously mentioned than for sizing the genset. Having said all this, If You are sucessfully using Your formula, who am I to argue with sucess?
Good morning Dave!
Again im sorry for being vague. This may not be correct but when I used the 0.8 factor, I had the following in mind:
1. For the kva units, it does not function as power factor rather, as I said, for engine efficiency. I did not elaborate that most of our generator sets are over their useful life already so I had to underrate them. The 0.8 factor also takes care of some other loads such as grinding tools, ovens, and some lighting fixtures in the area. This factor and the formula that goes with it is for sizing of the genset (no power factor required).
2. For the kw-hr/day units, the 0.8 factor was carried over in the computation to take care of the power factor. This is for the computation of monthly energy consumption that a unit of welding machine will cost if supplied from the electric company (not by geneators) as I previously mentioned for budgetary purposes.
As I've said, this is not a valid formula/approach that's why I offered it for criticism.
Thanks Dave! I'm learning from you ;-).
I think You are making a mistake multiplying the load by .8 to compensate for the wear in the gensets. This is going in the wrong direction, understating the load, causing You to actually over load the gensets. You would multiply the genset rating by .8 to get a reduced rating, or multiply the actual load by a number greater than 1, perhaps 1.25 or divide the actual load by .8, overstating the actual load to compensate for the wear on the gensets. I would be interested to hear how this works out in the field when You have fine tuned the formula.
You got me again Dave! I see your point. I got it really mixed up. I'm going back to my files again. Thanks a lot!
I think You will be on th right track now. Good luck with it.
Thanks! Happy new year to you and your family!