OK i dont claim to be an electrician but i have his job, and would like to see if you guys can help me size some of the projects i have cuz i found the formulas i need for sizing transformers ie: VxA=VA, VA/V=A, W/V=A. So with this new found info i begin my journey as a technician. I have 480VAC input to a fused disconnect switch with (3)30amp fuses. L1,L2,L3 goto terminal blocks, then split and goto a motor starter with blower(5.5amps), and to a 240/480VPri. 120Vsec. 1000VA transformer. I have 18 total pilot lights(.02a each), a daul solenoid valve(45VA / 120v = .375a?), 12 solenoid valves(8.1Watts each / 120V = 0.0675a X 12=.81a), 6 flame monitors at 1a a piece. This is all running on the 120v line from the transformer except the motor starter and blower. Is my transformer sized correctly?
.36 amps in pilot lights? 18 total @.02a each? .375 amps for dual valve? 45VA .81 amps total for single solenoids @ 8.1watts each 6. amps in flame monitors ------------------------------- 7.545amps total? X 120v ------------------ 905.4VA?
Maybe. You left out the coil for the motor starter. Things with coils have 2 ratings, a "Sealed" rating which is the power consumption of the coil when it is holding something, and an "Inrush" rating, which is the power it draws for an instant when you first turn it on. So all of your solenoids will have that, as well as the motor starter coil. One quick and dirty rule of thumb is to add up all of the sealed and resistive VA + the largest inrush, then round up to the nearest standard control transformer size. So in your case you already have the sealed at 905.4VA, but you need to add the inrush VA of the motor starter coil. If you have 30A fuses, it is likely a Size 1 starter and the coil inrush could be over 200VA. But check it out before going to the next size up.
BTW, in your opening statement you said W/V=A, which is not necessarily true. Watts and VA are not the same thing unless it is a purely resistive load, i.e. a heater element. Power supplies and coils are inductive and have a "power factor", the measurement of the lag between voltage and current. So Watts = V x A x pf, which means the VA is often higher than the watts.
Thanx for the help, how can i figure out what the power factor is? And when im wiring the transformer i use L1 and L3, is this ok or should i be using differnt legs? Also is there a rule of thumb on how much the inrush will be based off of the constant amps the unit pulls? And i can never tell what the amp rating for the coil is on the motor starter, so for example im using an AB 120v coiled 3phaze starter with 1 N/O contact part# 100C12D10.
[This message has been edited by JLC (edited 12-01-2006).]
Thanx for the help, how can i figure out what the power factor is?
Not easy, probably not important anyway. if it is a motor, it may say so on the nameplate, if it doesn't, assume .8 for making calculations. if it is a coil, there is not way to know, but it doesn't matter much anyway, just find out the inrush VA from the manufacturer.
And when im wiring the transformer i use L1 and L3, is this ok or should i be using differnt legs?
The convention is to use L1 and L2, but technically it doesn't matter. The only reason to follow that convention is for the convenience of someone in the future who would expect to find it there. In practice though, sometimes you need to balance loads, so it is acceptable to connect anywhere.
Also is there a rule of thumb on how much the inrush will be based off of the constant amps the unit pulls?
No. None whatsoever. It has to do with the iron in the coil, the winding design, the magnetic force it needs to exert etc. Every coil is going to be different. For that reason, manufacturers always make that information available to you. The web is great for that now.
And i can never tell what the amp rating for the coil is on the motor starter, so for example im using an AB 120v coiled 3phaze starter with 1 N/O contact part# 100C12D10.
Forget the coil's part number, that is just for ordering a replacement. Look in the AB catalog in the back of the contactor section for the "Technical Data", it will be there cross referenced to the part number of the starter or contactor. Here's an example if it is a "Buletin 509" starter; scroll down to the section that says "Coil Data" A-B Bul 509 Technical data link
I might be stating the obvious here but most control ladders don't have all of the relays and pilot devices active at the same time. For instance, I won't see, "OFF", "ON SUPPLY", and "ON EXHAUST" active at the same time. Same with high and low speed or time delay and run modes. You just have to consider your highest current draw condition and make sure you have enough ctl transformer to handle it. While it usually doesn't matter which 2 phases you pick, it is a good practice to monitor the health of the phase you aren't using. Before phase loss monitors (and voltage and rotation) from Time Mark, Diversified Products, Square D , Et,al. became common, relays were used to monitor the other primary combinations. Joe
Thanx again for all your help guys, as stated by JoeTestingEngr, not all my lights will be on at the same time, probably about half at full operation. I was only giving the part number of the motor starter so i can show you what i was using for a reference and i saved that bulletin on my favs for future ref. If this same setup was supplied by a smaller trans. say 500VA, what would be the out come? Wont work at all, will work but with short future, will work but....
Also is there a rule of thumb on how much the inrush will be based off of the constant amps the unit pulls
FLA (full load amps) * 1.25 = Inrush, sort of. Actually, 125% should cover a little more than the actual inrush. As the motor gets older and a little stickier, it may require more inrush, hopefully never exceeding the 125 %, until the day it actuall locks up and quits working. Also, you could do 125 % on the motor, to allow for inrush current, and total it with the total of the other circuits, doodads, etc. Take that sum total and make it 80% of whatever feeds it. The reciprocal of that 80% is 125%. So, (((motor fla)*1.25) + (total other loads))*1.25 = equals ample current supply, with of course, fuses where you need them.
If this same setup was supplied by a smaller trans. say 500VA, what would be the out come?
Ideal Transformer Law Power in = Power out. Example 75 kVA xfmr 480v P / 208 V S. Primary current is 90 A. Secondary current is 208 A. (figures are rounded) You can use that example to see what minimum size xfmr you can get away with. Also, by looking at this, you can see why the load side wire in this type of xfmr is bigger than the line side wire. And vice versa, a voltage step-up xfmr would have smaller wire on the load side.
[This message has been edited by rws (edited 12-07-2006).]