Diesel Emergency Generator Amp Load

[lexus31rus]

Hi there!
I have here a CAT Standby 650 ekW 813 kVA 60 Hz 1800 rpm 480 Volts generator with CAT 3412C TA Diesel Engine.

Im trying to come up with maximum apm load which can be connected to this generator.
What formula should I use for that?
simple one W=V*A, where A=W/V good enought? Or there is some "underwater" things i should keep in mind?

 

[psparky]

I'm just guessing here...but here's my guess.

Lets just talk about one load...a motor for example.

The generator should be able to power a motor that puts out 650KW at the shaft of the motor. It will take the full 813KVA to power the motor...the "extra power" or reactive power is in the magnetic field of the motor.

That being said...you should be able to power 813KW worth of lights.

So to answer your question...813KVA=480*1.73*I
I= 979 amps per phase.

Three 3/C 400 MCM cables should do the trick. Or just go 3-3/C 500 MCM's to keep it simple and to keep your cables from ever being stressed.

Just a guess...but it sounds reasonable. You prob know this...waiting for the "underwater" things.

 

[lexus31rus]

I'm just guessing here...but here's my guess.

Lets just talk about one load...a motor for example.

The generator should be able to power a motor that puts out 650KW at the shaft of the motor. It will take the full 813KVA to power the motor...the "extra power" or reactive power is in the magnetic field of the motor.

That being said...you should be able to power 813KW worth of lights.

So to answer your question...813KVA=480*1.73*I
I= 979 amps per phase.

Three 3/C 400 MCM cables should do the trick. Or just go 3-3/C 500 MCM's to keep it simple and to keep your cables from ever being stressed.

Just a guess...but it sounds reasonable. You prob know this...waiting for the "underwater" things.

But then, if I do my reverse engeenering,
979 Amp * 480 V ≈ 470 kW.
Not 650 kW.. Did I miss some adjustments for 3 Phase?
If i do 470 kW * 1.73 =813 kVA, but where is 650 kW?

 

[psparky]

480*1.73*979=813KVA

This generator apparently has a built in power factor of .8

813KVA * .8 = 650KW

979 Amp * 480 V ≈ 470 kW...this is a bogus statement. You just stated the power that a SINGLE phase generator could produce. The gigantic generator you spec'd is certainly THREE phase.

Keep in mind this is almost no difference between a generator and a motor. Same...same. Think about it. One is driven by a force at the shaft and produces electric power...the other one takes the electric power and converts it to force at the shaft.

 

[iceworm]

Im trying to come up with maximum apm load which can be connected to this generator...

I'd look at the nameplate first. There will be a number on it. However, as psparky noted, the generator is 3 phase and his number is probably what it will show.

If that number is all you needed, then the rest of this post is just noise. I won't feel a bit bad if you put it in the ignore column.

... Or there is some "underwater" things i should keep in mind?

Yes - lots

1. The gen is rated at 80% power factor. So the load can not pull rated current at unity pf. The driver will run out of horsepower (or kw if you are not in North America). So for a unity pf load the best it can do is 780A (that coresponds to 650kw)

2. The Gen is "Standby" rated. That means if the load is carefully teased up to a max with a stable load bank it will put out the rated kw with a unity pf load (straight resistive) or rated kva with a .8pf lagging (inductive) load.

However, very few loads are stable load banks:
Motors loads come with inrush, locked rotor startup currents, and lagging power factor.
Transformer loads come with inrush currents.
HID lighting can draw high 3rd Harmonic currents.
SCR controlled heaters jerk the current and kw all over the map.
Some inverters have a leading reactive input.
Some VFDs have a leading reactive input.

And then one can get really weird (read "uncommon but possible):
Low leading pf load which can sent the voltage regulator unstable.
Low lagging pf load which can overheat the alternator winding even though the current is below the max.

And since you didn't mention it, I'll leave out the issues dealing with parallel operation.

So what is the max current draw? I don't know. What is the application?

So where am I going?
A 650kw, 480V, 3ph generator, protective relay settings and load capability are non-trivial design applications. If you are picking protective relay settings, OCPDs, conductor sizes, then you are the "Engineer of Record". If you are not comfortable with that, then I'd really recommend getting some help.

The worm

 

[lexus31rus]

the other one takes the electric power and converts it to force at the shaft.

Thanks god its diesel!
Ok, waiting for underwater things...

 

[jim hardy]

Im trying to come up with maximum apm load which can be connected to this generator.
What formula should I use for that?
simple one W=V*A, where A=W/V good enought? Or there is some "underwater" things i should keep in mind?

somebody answered but it mighta not been apparent :

here's some tuning to your formula

W = V X A X √3 X power factor ---(the square root of 3 acconts for its being three phase.)
if you don't know power factor use instead the generator's VA rating (813KVA) instead of its watts

VA = V X A X √3

for max AMP load you should use VA instead of watts - because not all the amps go into making watts some of them go into something called "VARS" and that's what power factor accounts for.

Power factor is 1 for resistive load like heaters and incandescent lamps
but for motors and transformers and ballast lamps it's always somewhat less than 1( because of the inductance - they have wire wrapped around an iron core), and 0.8 is a typical number.

so VA = V X A X √3 and Psparky's answer of 979 amps looks mighty near right for 813KVA at 480V..even though i got 978...

when you're buying electric motors for your rig you should be aware of something called "KVA CODE" for motors. Whan starting up, motors pull a lot of current and the power factor can be really low. KVA code tells you how severe this starting current is.
it's not much of an issue when starting from the grid where there's really substantial current available but ought to be considered if you're starting big motors from a local generator.

http://www.joliettech.com/nema_codes-ac_motors.htm
Lower letter means lower starting amps.
look around your rig and see what code motors the designers bought.

 

[iceworm]

Whoops -
I missed the "emergency" part of the title. Some jurisdictions have a limit for the voltage drop to an electric motor driven fire pump. If one of those is part of your application, that could really limit the available current.

the worm

 

[psparky]

Great points by Iceworm and Jim...

In general, you could say it takes approximately 20 HP of electricity to run a 15 HP motor at steady state. Transient or start up is always much higher for a few cycles or more. Yes, that is loosely translated from KVA...

KVA is figured off of full load amps. Yes, the actual amperage measured from an ampmeter will give you your KVA when multiplied by 480V and 1.73 in this case.

And yes, you can correct power factor on a motor or in factories which they obviously do. How is this achieved? A current lagging the voltage is inductive...like in motors. Picture the current vector of your motor sitting at let's say roughly -20 degrees...were as your voltage is sitting at 0 degrees. (-20 degrees is approximate...whatever .8 is.)

Ok...so we have a vector at -20 degrees. How can we correct this. Well, let's take a capacitor and put it in parallel. What does the current vector look like when you put a voltage across a capacitor? Hmm...reactance of a capacitor is 1/(JWC)...do the math and you get a current vector pointing straight up at 90 degrees. So what does this do for us? Simple...the motor sees the EXACT same voltage and current (if the motor's voltage and current change...it won't work!)...However, the power company now sees the ADDITION of the capacitor current...and the motor current. Add a 90 degree vector with the correct magnitude to the -20 degree vector...and now summation of the vector gets much closer to zero...and the total magnitude of the current vector gets gets smaller. So the result is a smaller total current delivered at an improved power factor...which means less coal burned at the power plant!

And yes, the motor sees the exact same voltage and amperage as before. If you are a student and trying to get a grasp on power factor and KVA and all that stuff...take the time to draw a circuit with a motor at .8 power factor at 50 KVA or whatever...with a 480 voltage source hooked to it. Find the amps.

Now play with a capacitor in parallel...and draw all the vectors and see how they add. It will take some time...but when the bell rings...you will be johnny on the spot when it comes to power factor! Draw the power factor triangle before and after as well. Does the voltage or amperage change across the motor? Of course not...the motor has 480V across it whether there is a capacitor in parallel or not!

Once you grasp that...keep in mind that the vectors are rotating. But that 90 degree capacitor current vector will always be 110 degrees out of phase with that -20 motor current pulling it up to the voltage vector and keeping it close to unity power factor seen by the power plant.

 

[lexus31rus]

Whoops -
I missed the "emergency" part of the title. Some jurisdictions have a limit for the voltage drop to an electric motor driven fire pump. If one of those is part of your application, that could really limit the available current.

the worm

You got me! the water pump-is what bring that question in my mind. I have a fire pump as a part of this installation. I have about 235 amp been used, my question is can I add a load of 200 amp more to this generator? how much of "reserve" this water pump will need?
another limiting factor is the system 800 Amp circuit breaker.
here is a pic of my fire pump name plate http://i42.tinypic.com/zkoeoy.jpg

 

[psparky]

What? Heck you you can add it. Why didn't you state your load in the first place!

The generator may shake a little on start up...but heck ya...you are fine.

Add two of them if you like.

Always, always state your load when asking a question. Lesson learned.

And on the bright side, we got a little power factor correction refresher course:)

 

[iceworm]

... the water pump-is what bring that question in my mind. I have a fire pump as a part of this installation. I have about 235 amp been used, my question is can I add a load of 200 amp more to this generator? how much of "reserve" this water pump will need?...

Good we finally got the issue. The short answer is, "I still don't know." I can't help with a design of a Legally Required Emergency Standby from my side of your keyboard. But maybe I can get you started looking in the right direction. Most of the issues are not going to be if the generator can handle the additional load, but rather if the generator and connected system meet the applicable regulatory codes.

A 650kw emergency standby generator equates to a large enough facility that it will be under regulatory scrutiny - fire marshal, electrical inspector, government building inspector - somebody like that. And if by some fluke the install is not subject to any regulatory agency, I would still insist that "good engineering practice" be followed - and that means meeting the same criteria as if it were under regulatory authority.

Marginal design and installation of Emergency Standby Systems is just is not a good idea. So get ready to do the load studies and have the backup paper trail to show the design criteria meets applicable codes. And if the installation is in area open to public access, the AHJ usually wants to see the drawings stamped.

First, is the installation subject to the NEC? Generally that would be in the US or a US military controlled installation. If not, I can't help at all.

However, if the install is subject to the NEC, look at NEC (2011) Article 700, Emergency Systems, and art 695, Fire Pumps. Art 695.7 requires the VD be no more than 15% during motor startup. I've always figured the VD at locked rotor. And that will be with everything else on-line.

Good Luck with your project. I'll be interested in how it turns out.

The Worm

 

[psparky]

My gut feeling is that your pump will run just fine.

However, if you need proof like the Worm says...you will simply need to hire a local engineering consulting firm to make the drawing with calculations and "stamp it" like he says.

So all the information in the world here will just give you some insight...nothing more.

But again...that pump is going to run just fine.

 

[jim hardy]

""look around your rig and see what code motors the designers bought.""

iceworm speaks truth.
Look carefully at starting loads.
Starting 235 amps worth of code G motors all at once could overload you, and i'd think 435 surely would.
our big motors got started in sequence not all at once.

we're pullin for you -

 

[psparky]

A 3 phase, 240 full load amp motor at 480 V calls for a 400 amp breaker looking at General Electric or Square D motor charts. You state you only need 200 amps. Assuming your former load is worst case all motors at 235 amps...this would also require 400 amp breaker. You have a 800 breaker available.

Would a staggered start be necessary...absolutely...and that is easily accomplished.

So again, I state that the motor will fire and run just fine.

However, if there are some underlining codes that state a certain amount of this or that...then that will have to be looked into.

But yes gosh darnit...that generator will fire that motor!

 

[lexus31rus]

Thank you a lot everybody!
Of course I will hire a professional to prepare all paper work for annual fire inspection. The main purpose of my question-is self-understanding of process. I would like to know and understand, what I’m dealing with. So, is there any formula, I can use for estimate generator loads in the future? I learn that I can't exceed more than 80% of generator capacity, is that sound like a true? But having an 800 Amp breaker-then it would be 80% of 800?
In my position I just want to be able to answer my boss question "Hey, can we put this equipment on the emergency power?" and be able to support it with calculations.

 

[psparky]

Thank you a lot everybody!
Of course I will hire a professional to prepare all paper work for annual fire inspection. The main purpose of my question-is self-understanding of process. I would like to know and understand, what I’m dealing with. So, is there any formula, I can use for estimate generator loads in the future? I learn that I can't exceed more than 80% of generator capacity, is that sound like a true? But having an 800 Amp breaker-then it would be 80% of 800?
In my position I just want to be able to answer my boss question "Hey, can we put this equipment on the emergency power?" and be able to support it with calculations.

80% is a good guesstimate that works nice. Also works nice for this application since 800 amps is roughly 80% of 973 amps. An 800 amp breaker can handle 800 amps if need be. Breakers that big are also typically adjustable. The breaker is there to protect the wire to the load it is feeding. If talking about a single load for example...you could use the entire 800 amp breaker...but the wire would have to be at 80%. In other words...you need a cable rated at 1,000 amps...take 80% of that and you have your 800 amps. Tricky...I know.

As you can see...that wire would never be in danger of being overloaded. But you are right, typically on breaker panels...80% of the breaker load is typically used. But if you want to use the full breaker you can...just jump up a wire size or two. You can also adjust "start up current" on todays fancy breakers, the bigger breakers anyways...Keeps the breaker from popping during the first few cylcles of start up.

What I explained above is for basic industrial wiring. Any fire codes or what not in your application... Worm will have to comment on.

 

[iceworm]

... So, is there any formula, I can use for estimate generator loads in the future? ...

first a couple of basics:
1. The generator is rated at 650kw. That is the rating of the driver. If the gen has a 615kw load on it, the throttle will be wide open. If more load is added, the driver (engine) will slow down. This has nothing to do with the ratings of the alternator end.

2. the generator is rated for "Standby" duty. Thsi means the mfg has put together as thin a package as they can as still get it to put out 650kw. Usually this means one must use a very stable load bank and slowly tease the load up to 650kw. They won't put out anything close to 650kw continuous. For example, if the gen were loaded to 550kw and you started a 125hp motor (that's 93kw) the driver would stumble and probably even trip off - under freq.

If the gen were loaded to 550kw and you turned on 100kw of tungsten lighting - the gen would probably trip.

3. Part of the standby rating also includes time under load. This one you will have to look up. It could easily be rated for no more than 10 hours per month. Yep - that little.

So this is a good place for
Rule 1: For standby rated machines: If the gen is to be run more than actual emergencies load to only 75% continuous. If the gen is run only during actual emergencies, load to only 80%. These are real power numbers - KW.

Rule 2: If you are starting large motors, say .3x to .5x of the gen rating, then that may be all the gen can do - unless one employs some crafty load sheding.

Now let's look at the alternator end. The KVA rating is essentially the heat rejection capability. The alternator is rated at 978A at an 80% power factor - lagging. That is because for the last 150 years the standard industrial load was mostly motors, which are inductive, which is called lagging power factor. Don't confues this "80%" with any kind of a derate for the driver - it is not. It is a measure of how much of an inductive load the alternator can take without overheating. So the .8pf rating tells you that with any industry standard motor load, the alternator will not overheat. The driver will run out of power first.

So now you should see that the gen is driver limited - unless there are really weird loads.

**********************
Now let's look at the distribution - that's everything from the alternator terminals out toward the loads. Starts with the first circuit breaker, and the conductors out to the panels.

First I have no clue why the first CB is an 800A. That is nutzz - completely bonkers.

Unless you have special stuff, the circuit breakers are only rated to carry 80% continuous. There's that pesky "80%" again. Nope it is not related to power factor or driver derate. So with an 800A CB for the first overcurent protection, the most continuous load would be .8 x 800 = 640A, which is 530kw - less if the loads are standard motors - maybe 90% of that or 480kw. If one needed to run the gen WFO for an emergency - couldn't do it, the CB would trip. The norm would be a 1200A CB.

But you said the existing load was 235A. Assuming a .9pf that's 175kw. So the existing install is a 650kw genset, limited to 480kw bythe distribution, and only has a 175kw load on it. Stranger and stranger.

That is as much as I can infer from what you have told up - and I could be alll wet.

... In my position I just want to be able to answer my boss question "Hey, can we put this equipment on the emergency power?" and be able to support it with calculations.

First, is this a creeping thing where every now and then the boss wants to add load to the Emergency? If so, "Does the equipment meet the criteria for needing to be on the "Legally Required Emergency Standby" generator? If "No", then, "No".

Or is this a redesign of the emergency standby system? If this is so, then you need to learn how to do load studies, protective relays, NEC compliant design - because you are the engineer of record.

I can give you some hints:
The concept that an Emergency generator can stumble or quake on startup is not a good idea. Emergency generators need to be pretty rock solid. Some emergency loads require that the gen be running and loaded within 10 seconds from power out.

Figure out why this has a 650kw gen. The loads are such that a Standby rated 250kw would have done fine - unless there are a couple of 100hp motors. I can't open your fire pump sheet so I don't know what you have.

Wonder why I'm not giving you any better (more definitive) answers? Designing robust, stable, cost effective, emergency systems is difficult enough if one is standing there. It's impossible from my side of your keyboard.

So sayeth the Worm

 

[psparky]

I disagree with your 80% breaker rating. Are you saying that a 800 amp breaker will trip at 640 amps? I think not. Today's breakers that size are adjustable. If you set it to 800 amps...by golly it will not trip till 800 amps. Now the wire sizes are only meant to run at 80%. There's a difference.

If I have a typical house breaker that is 20 amps...are you saying I can't run that baby all day long at 19 amps with a #10 wire? I say you can. It would be silly to not replace it with a 30 amp breaker...but still...that 20 amp breaker aint poppin till 20 amps.

 

[jim hardy]

now we're going throough the looking glass to what's behind design of a power system.

i'd suggest tabulating all the loads on your generator,
with both their running current (nameplate) and running KW
and their starting current (from KVA code) and estimate starting KW at 2X running,

tote them all up to get a feel for what is being asked of this generator.
I'm sure that's above and beyong the call for duty
but look what you'll learn...

Be aware that incandescent lights pull 10X rated on energization and that can wreak havoc with computers when voltage sags.
I don't know a thing about ballasted lights startup characteristic, maybe Ice or Sparky does.

The grid is a stout source for a distribution system
a local generator is far more weak-kneed
and one can innocently build a system that's fail-sure.
my industry learned this with electronic inverters.

lastly - is this generator ever paralleled with the grid or a larger generator?
if so double check the cables used for your current metering. We had armored cables with metal sheath grounded at both ends, and that messed up current indication. We wondered why exhaust manifolds glowed red at indicated full load...
if your cables are armored be sure sheath is grounded at only one end, or that sheath is routed OUTSIDE the CT's else it can cause KW indication to read low. we noticed phase was off on CT's, which eventually led to findiing trouble.

old jim

 

[russ_watters]

I disagree with your 80% breaker rating. Are you saying that a 800 amp breaker will trip at 640 amps?

No, it's a code requirement.

 

[psparky]

No, it's a code requirement.

A month ago, I would totally agree with the 80% rating and the whole nine yards. And if I ever designed something, I would clearly stick to the 80% rule.

What actually started this conversation is this...a little while back I had a 208 volt electric grill for a kitchen. It was rated at 29 amps. Simple to set up...double 40 amp breaker with a #8 wire. My boss was looking over the work...and he suggested I use a 30 amp breaker with a #8 wire to protect the grill. I assume his thoughts are that the unit was going to be used intermintantly so the long term trip would never be used. Strange I know...

Also, when talking with our E-tap guy, (he makes and studies the curves of typically larger breakers during start up and steady state) he said most breakers are rated for 80% of their load. When heated past the 80%...they pop. He then said...however, they do make 100% rated breakers which can be used at their full capacity...as long as the wire size connected to that 100% breaker is 80% of the wire's full rated load.

Also, I have seen some specs that show a full load amp rating of 18 amps...and they spec a 20 amp breaker. Obvioulsy, you would use a #10 wire. I think there are some exceptions to the rule.

So again I totally hear what you are saying...just wanted to bring these other comments up to see what people think.

 

[psparky]

Great points by Iceworm and Jim...

In general, you could say it takes approximately 20 HP of electricity to run a 15 HP motor at steady state. Transient or start up is always much higher for a few cycles or more. Yes, that is loosely translated from KVA...

KVA is figured off of full load amps. Yes, the actual amperage measured from an ampmeter will give you your KVA when multiplied by 480V and 1.73 in this case.

And yes, you can correct power factor on a motor or in factories which they obviously do. How is this achieved? A current lagging the voltage is inductive...like in motors. Picture the current vector of your motor sitting at let's say roughly -20 degrees...were as your voltage is sitting at 0 degrees. (-20 degrees is approximate...whatever .8 is.)

Ok...so we have a vector at -20 degrees. How can we correct this. Well, let's take a capacitor and put it in parallel. What does the current vector look like when you put a voltage across a capacitor? Hmm...reactance of a capacitor is 1/(JWC)...do the math and you get a current vector pointing straight up at 90 degrees. So what does this do for us? Simple...the motor sees the EXACT same voltage and current (if the motor's voltage and current change...it won't work!)...However, the power company now sees the ADDITION of the capacitor current...and the motor current. Add a 90 degree vector with the correct magnitude to the -20 degree vector...and now summation of the vector gets much closer to zero...and the total magnitude of the current vector gets gets smaller. So the result is a smaller total current delivered at an improved power factor...which means less coal burned at the power plant!

And yes, the motor sees the exact same voltage and amperage as before. If you are a student and trying to get a grasp on power factor and KVA and all that stuff...take the time to draw a circuit with a motor at .8 power factor at 50 KVA or whatever...with a 480 voltage source hooked to it. Find the amps.

Now play with a capacitor in parallel...and draw all the vectors and see how they add. It will take some time...but when the bell rings...you will be johnny on the spot when it comes to power factor! Draw the power factor triangle before and after as well. Does the voltage or amperage change across the motor? Of course not...the motor has 480V across it whether there is a capacitor in parallel or not!

Once you grasp that...keep in mind that the vectors are rotating. But that 90 degree capacitor current vector will always be 110 degrees out of phase with that -20 motor current pulling it up to the voltage vector and keeping it close to unity power factor seen by the power plant.

No arguments on my power factor theory? Are you guys suggesting I finally got something right?

 

[lexus31rus]

Oh, my favorite stuff!
The dipper we go-the more questions I have :)
But thanks a lot every one, I can only imagine how much in $/hr I have to pay to our electric engeener to explain me all that! Viva la internet ;)
p.s. pic of my fire pump name plate http://i42.tinypic.com/zkoeoy.jpg doesn't opens?
It said Volts 230/460 and AMPs 154/77, I thought it may clear how much of my amps i have to reserve for fire pump. As long as I know-its only one motor connected to my generator. the rest are critical HVAC systems and office stuff, such as servers...

 

[jim hardy]

""p.s. pic of my fire pump name plate""


well i can't quite make out the hp
but it's clearly code F
so it'll draw 5 to 5.6 KVA per horsepower starting.
http://www.engineeringtoolbox.com/locked-rotor-code-d_917.html

 

[lexus31rus]

""p.s. pic of my fire pump name plate""


well i can't quite make out the hp
but it's clearly code F
so it'll draw 5 to 5.6 KVA per horsepower starting.
http://www.engineeringtoolbox.com/locked-rotor-code-d_917.html

It's 60. so i go 60*5.6=336 kVA

But, that fire is already connected to that generator, and should be include in this 235 existing Amp! or...they never start fire pump, while doing investigative report and come up with 235 existing amps...

 

[psparky]

If you do this job...perhaps a new nameplate might be a good idea?

 

[jim hardy]

336 kva at 480 volts would be 336,000/(480 * √3) amps = 404 amps while starting

and the 813 KVA generator should be good for 978 amps
404 + 235 = 639, well below the 800 amp breaker
so it looks like there's room to start your fire pump even if there's 235 amps of other load.

a prudent fellow would find starting current for those other loads
and find whether they can ever all be asked to start simultaneously,
and if so will it embarass the generator.

 

[psparky]

You ever notice you can ask advice from three very experienced engineers and get three completely different answers?

At that point...there's really only one question left. What would you do?

 

[jim hardy]

Indeed !
that's why one has to start with the basics and build from there.
Else he's just taking somebody else's word.

We now know what's starting characteristic of fire pump
but not of those HVAC units and computers.

Then there's question what're overload characteristics of generator - often the voltage regulator/exciter has boost windings from current transformers to help it hold up voltage during a brief overload...
and how fast is that 800 amp breaker...

one has to be thorough or Mother Nature will drop Thor's silver hammer on him.
( old Beatles Tune - Maxwell Edison...)

Lexus - are you sorry now you asked?
but it's fun learning. Thanks to all who contributed -- I've learned things.

I'm sure you'll find the system was well designed.
What a prudent fellow has to do is be sure he doesn't degrade it with his modifications.
To that end he needs to understand the design.

 

[lexus31rus]

You ever notice you can ask advice from three very experienced engineers and get three completely different answers?

At that point...there's really only one question left. What would you do?

I will defiantly not going to invest my money in this project! :)
But if seriously-I will go thru all what we learn, squeeze all juice out of it and send one big and smart e-mail to my boss. thanks god I'm not making desigions here, I'm only a "junior".

Lexus - are you sorry now you asked?
but it's fun learning. Thanks to all who contributed -- I've learned things.

I'm sure you'll find the system was well designed.
What a prudent fellow has to do is be sure he doesn't degrade it with his modifications.
To that end he needs to understand the design.

I like to learn new things too. And I'm really glad that here is a people who would help. I’m really appreciate that. I'm try to stay a "prudent fellow" and not going to mess it up by myself.
And... I didn't get your question..

 

[psparky]

I will defiantly not going to invest my money in this project! :)
But if seriously-I will go thru all what we learn, squeeze all juice out of it and send one big and smart e-mail to my boss. thanks god I'm not making desigions here, I'm only a "junior".

And above else...don't forget to tell him you got all your info from the Physics Forums! lol

 

[jim hardy]

""And... I didn't get your question..""

""Sorry i asked" is, over here, a humorous response to a long-winded answer.

I'm hoping you aren't mad at us for the long winded answers.

Have Fun !
and make the responsible design organization be prudent fellows too.

What part of the world are you in?

old jim

 

[lexus31rus]

What part of the world are you in?

old jim

Hawaii. So, Mahalo and Aloha!