What size or capacity a genset should be

Date 12-22-2007 04:23

What do You really need to know, 1) the size of a generator that would power all of the machines working at rated capacity at the same time. 2) the size of a generator that would power all of the machines @110A at the same time. 3) the size of a generator that would power the machines @ 110A if everybody took turns so that the minimum number would have an arc lit at the same time. 4) the KW hours consumed by the machines in a work day. In a real world situation the overload capacity of the generator will play a big part of the equasion. In any case a conservative aproach is a good idea, as there may be times such as right after a break when everybody starts working at the same time, that all of the machines are under load. At another time the machines might be working at a higher output. If You don't have enough capacity You have to do load sharing / load shedding, and there is danger to all equipment from running at reduced voltage and in the case of the generator from running at overload. Alternatly running a genset at less than 50% load for extended periods is poor economy and at extremely light loading over long periods results in cylinder glazing. So You are damned if You do and damned if You dont. Grid power is a wonderfull thing.

By Tommyjoking (****)

Date 12-22-2007 09:12

Dave you have no idea how glad I am that you are here man.

By toddler (**)

Date 12-23-2007 10:40

The scenario is this:

At a pre-planning stage, i.e, prior to mobilization of welders and equipment at project site, I am asked by the project director to identify the number of welding machines (rectifier & engine type) based on the work area and projected number of welders running from150 to 300 peak. This is typical of a petrochemical and power generating projects. In most cases, there will be no power supplied from the grid that's why we have to supply our own, in the form of electric generators ranging from 100kva to 200kva (for lighting, tools, w.mach, etc). Say we have identified a total of 85 transformer rectifier welding machines (varying rated cap:250A~350A), the next question is: How many generators will be required and what are the capacities?

Another scenario is when you are putting up a training school/laboratory and you need to know the estimated monthly energy consumption for a 20-booth:20 welding machines. This is required for budgetary analysis/proposal.

In each case, the computation should take into account that the welders may or may not be welding at the same time and maybe drawing off (primary) currents equivalent to (secondary) currents required for electrodes/fillerwire of various sizes.

The generator selected should mot be underrated nor overrated (economy wise) and in the case of power consumption, the computed monthly consumption should neither be too high nor too small (for the budget to be viable).

This computation also come into play when you need to know in advance the ratings of circuit breakers you will need for the mains as well as for the distribution.

By DaveBoyer (*****)

Date 12-24-2007 06:43

You will need a mind boggeling ammount of power just to light a project the size You are talking about. You will probably end up using much larger than 100-200 KVA units, probably You will need some serious gas turbines. I would suggest You size the transformer/rectifires to the work being done. You might look into invertor power suplies as they are much more efficient. I think You are on the right track calculating the load at actual output rather than the rated output on the machines, but I think You need to be able to run them all at the same time as this is something that could happen if only once in a blue moon.

By toddler (**)

Date 12-24-2007 08:24

Dear Mr. Dave,

Thanks for the suggestion. Just to make myself a bit clearer, I am not looking for a one-unit power source to energize 85 welding machines as mentioned in my example. A typical power plant or refinery project will normally cover a wide area say for the pipework, storage tanks, and steel structures. Stationary welding machines (x'former) are distibuted in these areas in groups of 8, 10, 12, etc depending on the capacity of genset available. I mentioned 100~200kva because experience-wise, these are the optimum capacities that we have used so far. If for example you have computed that a 150kva genset can supply 12 w.machines, then you will be needing at least 7 units of that genset. The calculation is I believe different from the simple P=EI as there are two currents involve i.e., primary and secondary. I am not an electrical engineer but using either the primary or secondary current straight in the formula yields absurd results. I need a firm or reliable method or approach of calculation so I can include them in my lessons to our young welding engineers under training.

Regards!

By DaveBoyer (*****)

Date 12-25-2007 05:13

OK, I think I see where You are coming from. When You mentioned petrochemical and power plant I was thinking allong the lines of constructing a refinery or power plant complete from scratch. Bechtel built a nuke plant near Me, I got tours through the project 2 times while it was under construction. It was a huge job that went on round the clock for many years. I think You have in mind smaller projects. What You need to provide is the primary power for the welding transformer/rectifiers. The volts, amps, KVA and KW listed on the machine tag is the power needed at the rated output. The actual load will be less or greater depending on weather the machine is putting out less [the usual case] or more than rated load. From Your earlier post I think You understand this. The machines usually list KVA on the plate that is the aparent power [at rated load]. To determine the required power at other outputs You could take an educated guess if You were familliar with the machines used. The only sure way I know to determine what a machine actually draws doing a particular job is to measure it with an amprobe while in use.

By DaveBoyer (*****)

Date 12-25-2007 09:14

Toddler: Here is some more food for thought. If You have a generator with an average output of 100 KW it is going to burn 300 gallons of diesel every 24 hours. If You have a fleet of these gensets scattered around a site You have at least 1 shift a day to staff for the guy with the fuel truck. The logic behind this is as follows: You need 2 HP from the engine per KW output on the generator, and a diesel makes 16 HP per gallon per hour. Using Your 110 amp stick welder example from Your earlier post My guess is that this (1) operator would need about 9 KVA from the genset to power His transformer/rectifier machine at this output or probably less than 7 KVA with an inverter. I have no idea how to estimate the lighting requirements for such a job, and I think the power requirement for power tools for the welders will be insignificant in the overall scheme, as they will be using them while not welding.

By toddler (**)

Date 12-25-2007 10:23

Thanks Dave!

The lighting and tool power requirement is insignificant in this situation since the area covered by one genset is small (w/in 100ft from the genset, ave length of w.mach lead cable). I really appreciate your inputs but as a matter of practice.. well how would i say it.. I can't tell or teach my students to "guess" or look up to the ceiling for the answer. If one says the answer is 10, another will ask: how did you arrive to that; show me your solution.

How would I show the correct solution if I were to give my students this test?:

Given:
12 units 300A x'former welding machines; 37A/460V/3-Ø
Application: SMAW using 3.2mmØ E7018 electrodes, say @110A, 10hrs a day work
Find:
The size of genset in kva required (ignore power for tools and other accessories)

Further, we do have fuel trucks going around the site in the morning and in the afternoon to supply diesel to the gensets round the clock (including engine driven welding machines).

By DaveBoyer (*****)

Date 12-26-2007 04:30 Edited 12-26-2007 04:35

A transformer will draw aproxametly porportional to it's output. There are some cases where there are significant exceptions to this rule, like the 600 amp power factor corected machine I spoke of in another post that draws 9 KVA while doing nothing. You can check the spec plate on various machines, or look up the power requirements in the literature. I said "You could take an educated guess if You were familliar with the machines", I understand that Your students havn't got the education yet. My suggestion to check a machine with an amprobe while someone is welding is a good method and a valuable thing to teach. It is better to know than to believe.

By DaveBoyer (*****)

Date 12-26-2007 06:19

To elaborate on My suggestion to check how much a machine draws with an amprobe, on a single phase machine all You have to do is multiply the measured amps X the measured volts, and You have the KVA. On a 3 phase system You need to multiply the volts X amps X 1.732 to get KVA. 1.732 is the square route of 3, 3 being the number of phases. If You don't have an amprobe ask to borrow one from the electrical maimtainance staff, or buy one Yourself. Harbour Freight has cheap ones if You don't want to put a lot of Your own money on the line.

By toddler (**)

Date 12-26-2007 11:16

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.

By DaveBoyer (*****)

Date 12-27-2007 05:11

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.

By toddler (**)

Date 12-27-2007 06:38

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
----------------------
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.

By DaveBoyer (*****)

Date 12-28-2007 04:49

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?

By toddler (**)

Date 12-29-2007 03:38

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 ;-).

By DaveBoyer (*****)

Date 12-29-2007 07:05

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.

By toddler (**)

Date 12-29-2007 10:11

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!

By DaveBoyer (*****)

Date 12-30-2007 05:01

I think You will be on th right track now. Good luck with it.

By toddler (**)

Date 12-30-2007 08:05

Thanks! Happy new year to you and your family!