power conversion

[clayton]

they are going to install a transformer here at work, fed from 100amp bussplug 3ph 480v to go to 120v. assume 3ph.

question being how much power do you have at 120volts??

lets say max power, i figured 83kva (i know not standard size but just for theory sake)

so do you get 400 amps 120v 3ph? and how much per leg? the 120v will be used as lighting and outlets so would you have to figure it like single phase?
as you can see im completely confused andy help and direction would be much appreciated

clayton.

 

[charlie b]

. . . so do you get 400 amps 120v 3ph?

You left out the factor of the square root of three (about 1.732). I agree with 83 KVA, it being 100 amps times 480 volts times 1.732. But if you take the same 83 KVA, divide it by 208 volts (the line-to-line voltage of the secondary side of the transformer), and divide that result by the 1.732 factor, you get 230 amps.

. . . and how much per leg?

The notion of "amps per leg" has no meaning. In this example, assuming the full 100 amp supply on the primary, the secondary Phase A current is 230 amps, and so is the Phase B current, and so is the Phase C current.

 

[LarryFine]

question being how much power do you have at 120volts??

lets say max power, i figured 83kva (i know not standard size but just for theory sake)

The power is the same at either voltage (ignoring losses and stuff.) As voltage lowers, ampacity raises proportionately. The available power will be the same (presuming properly-calculated OCP and wiring) at the new voltage.

You simply specify a transformer size for the load power requirements plus a little headroom (let's say 100kva), the primary voltage to suit your supply (480v Delta in this case), and the desired secondary voltage (208Y/120).

 

[mivey]

You left out the factor of the square root of three (about 1.732). I agree with 83 KVA, it being 100 amps times 480 volts times 1.732. But if you take the same 83 KVA, divide it by 208 volts (the line-to-line voltage of the secondary side of the transformer), and divide that result by the 1.732 factor, you get 230 amps.

The notion of "amps per leg" has no meaning. In this example, assuming the full 100 amp supply on the primary, the secondary Phase A current is 230 amps, and so is the Phase B current, and so is the Phase C current.

charlie, that just reminds me again to quit using "amps per phase". I had forgotten again. Old habits die hard.

 

[rattus]

charlie, that just reminds me again to quit using "amps per phase". I had forgotten again. Old habits die hard.

Isn't that the same thing as "phase current" in a wye?

 

[mivey]

Isn't that the same thing as "phase current" in a wye?

If we called it line current it might work better since line current and phase current are the same in a wye but different in a delta.

As you know, the classic way to analyze balanced 3-phase circuits is to simplify them into single phase circuits with line-line voltages and line currents. Having the impedances in ohms per phase, using VA per phase, and watts per phase, makes it easier to start saying "amps per phase" as well, even though one probably shouldn't.

I think another of charlie's issue is that "amps per phase" might make it sound like you could add up the "amps per phase" to get "total amps". It just is not the best terminology. I'm trying to reform to using line current and phase current but may be too old to change (quickly anyway).

 

[gar]

080708-2035 EST

mivey:

Line current seems quite clear that it is the current in one supply line.

By phase current I assume you are referring to the current in one primary or secondary winding. So when you have a delta connection the line and phase currents are different, but for a Y connection line and phase currents are exactly equal. Is this your definition?

.

 

[mivey]

080708-2035 EST

mivey:

Line current seems quite clear that it is the current in one supply line.

By phase current I assume you are referring to the current in one primary or secondary winding. So when you have a delta connection the line and phase currents are different, but for a Y connection line and phase currents are exactly equal. Is this your definition?

.

Yes.567890

 

[kingpb]

Is there a protective device in the tap? Is it 100% rated? This will make a difference on what the optimal size of transformer is to use.

Keep in mind you can always put in a transformer larger then you need, the protection will limit the use and therefore it's just not very cost effective.

 

[weressl]

they are going to install a transformer here at work, fed from 100amp bussplug 3ph 480v to go to 120v. assume 3ph.

question being how much power do you have at 120volts??

lets say max power, i figured 83kva (i know not standard size but just for theory sake)

so do you get 400 amps 120v 3ph? and how much per leg? the 120v will be used as lighting and outlets so would you have to figure it like single phase?
as you can see im completely confused andy help and direction would be much appreciated

clayton.

The simple way to think about transformers: voltage up, current down and vice versa. Power in=power out.

Primary voltage 480(277), secondary voltage 208(120). First set of voltages is between phase, the second is between pahse and neutral(grounded conductor). Whichever set you take the ratio will be the same. so either 480/208V or 277/120V your multiplier will be ~2.31. So based on that your secondary current will be 2.31 times higher than your primary current.

But....

A 100A busplug is not likely to serve a transformer that requires the full 100A and the transformer primary protection is required to be larger than its nameplate FLA(full load amperes). In other words it can be anywhere from 125% to 250%, so your actual available power will be determined by that. So the transformer can be anywhere between 60 to 30kVA.

 

[kingpb]

I would have to disagree that a transformer protection is required to be greater. The Code states the "maximum" protection. You can always go smaller.

Which is why I asked if the tap is 100% rated, because if it is, then the transformer and conductors does not need protection that is 125% of the current carrying capacity.

 

[weressl]

I would have to disagree that a transformer protection is required to be greater. The Code states the "maximum" protection. You can always go smaller.

Which is why I asked if the tap is 100% rated, because if it is, then the transformer and conductors does not need protection that is 125% of the current carrying capacity.

Yeah you can go smaller and you can also go and pop your fuses after a couple of years and nuissance trip your circuit breaker without overloads. There is a reason why they ALLOW you to size they protection that way. The allowance is there to accomodate sound enegineering practice. Sure you can go smaller. You can have it protected at 10Amps, but then that is your power limitation. The 100% rating of the tap is a different issue and it is normally used wheather the protective device can be used to it's full rating, ex. where a 100A device is only allowed to provide CONTINOUS rating of 80%.

 

[clayton]

thanks for the replys.
i understand about continous and full rated current ratings.
i was just using the hypothetical instance of the 100amps maximum mathmaticaly speaking to keep things simple,

so far there is no transformer yet, to even check nameplate data,

where im a little foggy is how to go form 480 3ph to the 120 3ph and be sure of how much power is available i guess im looking for the equation(s) to do that.

this is what i think i know( which words are probably debatable in themselves)

480v 100amps 3ph works out to 83.04KVA

then 83040va/208v * 1.73 = 230 amps then

83040va/120v * 1.73 = 400 amps

if this is correct then there would be (theoretically) 400amps available 120v power which would be used in a single phase set up??

thanks

 

[weressl]

thanks for the replys.
i understand about continous and full rated current ratings.
i was just using the hypothetical instance of the 100amps maximum mathmaticaly speaking to keep things simple,

so far there is no transformer yet, to even check nameplate data,

where im a little foggy is how to go form 480 3ph to the 120 3ph and be sure of how much power is available i guess im looking for the equation(s) to do that.

this is what i think i know( which words are probably debatable in themselves)

480v 100amps 3ph works out to 83.04KVA

then 83040va/208v * 1.73 = 230 amps then

83040va/120v * 1.73 = 400 amps

if this is correct then there would be (theoretically) 400amps available 120v power which would be used in a single phase set up??

thanks

No it is not.

But 79674va/{[120v * SQRT(3)]*SQRT(3)}= 221.32 amps is. Therefore you would have - theoretically - 221.32 A available in each phases whether some or all of each flows between pahse-to-phase or phase-to-neutral is immaterial.

The first SQRT(3) is to convert phase-to-neutral voltage to phase-to-phase and the second one is to convert single phase to the effective momentary value of the three phases as the voltage factor in the formula.

 

[mivey]

83040va/208v * 1.73 = 230 should have been
83040VA/(208V * 1.73) = 230 OR
83040VA/208V/1.73 = 230

83040va/120v * 1.73 = 400 amps should have been:
83040VA/(120v * 1.73 * 1.73) = 230 OR
83040VA/(120v * 1.73)/1.73 = 230

Use:
kVA(3-phase) / kV(line-line) / sqrt[3] = amps

VA(3-phase) / V(line-line) / sqrt[3] = amps

VA(3-phase) / [V(line-ground)*sqrt[3]] / sqrt[3] = amps

VA(1-phase) / V = amps

 

[kingpb]

Yeah you can go smaller and you can also go and pop your fuses after a couple of years and nuissance trip your circuit breaker without overloads. There is a reason why they ALLOW you to size they protection that way. The allowance is there to accomodate sound enegineering practice. Sure you can go smaller. You can have it protected at 10Amps, but then that is your power limitation. The 100% rating of the tap is a different issue and it is normally used wheather the protective device can be used to it's full rating, ex. where a 100A device is only allowed to provide CONTINOUS rating of 80%.

So the sound engineering which you speak of is why people, including engineers blindly use the NEC to design by, when the NEC's intent is clearly for safety, not as an engineering cookbook. I can easily show on a TCC curve that a breaker sized at 250% on the HV side of a transformer in many instances will lie above the transformer damage curve.v Hence, why these are guideline maximums, not mandatory minimums.

Could you please share/expand on the reason why it is allowed, and how sound engineering is accomodated?

 

[mivey]

So the sound engineering which you speak of is why people, including engineers blindly use the NEC to design by, when the NEC's intent is clearly for safety, not as an engineering cookbook. I can easily show on a TCC curve that a breaker sized at 250% on the HV side of a transformer in many instances will lie above the transformer damage curve.v Hence, why these are guideline maximums, not mandatory minimums.

Could you please share/expand on the reason why it is allowed, and how sound engineering is accomodated?

I think what Laszlo may have meant was that a properly sized transformer would require the protection curve above the load level.

Did he claim that the NEC was used to set the high-side protection? The idea is to get the curve above the load and below the damage curve and engineers like to hug the damage curve to allow for overloads, cold load pickup etc. If the transformer were grossly over-sized, I would back down from the damage curve.

 

[clayton]

why would it only be 230 amps 120v

sorry im so slow. would that be 230amps per 120v leg phase A to N
phase B to N and phase C to N?

 

[mivey]

why would it only be 230 amps 120v

sorry im so slow. would that be 230amps per 120v leg phase A to N
phase B to N and phase C to N?

It is 230 amps in each leg but be careful that you do not say the total is 3x230 or 690 amps because that would not make sense. If you want to talk about the split between legs, talk about splitting the power. We are splitting the load between three phases.

Start with the fundamental calcs:
83040VA/208V/1.73 = 230 OR
83040VA/(120v * 1.73)/1.73 = 230

This is 230 amps but we have split 83040VA between three phases.

You could also look at it from a per phase perspective. So we have 83040/3 or 27680 VA per phase. That yields: 27680VA/120V=230 amps

 

[cripple]

power conversion

To find the current of a transformer you need to use the right formula, which is as follows:
Single phase 120 Volts off a 120Y/208 system (A to neutral B to neutral or C to neutral)
I = (Kva x 1000/3)/voltage
I = (83 Kva x 1000/3)/120
I = (83000/3)/120
I = 27666/120
I = 230 Amps
Single phase 208 Volts
I = (Kva x 1000/3) x 2/voltage
I = (83 Kva x 1000/3) x 2 /208 x 1.732
I = (83000/3) x 2/ 208
I = 27666 x 2/208
I = 55332/208
I = 266 Amps
Three phase 208
I = Kva x 1000/voltage x 1.732
I = 83000/208 x 1.732
I = 83000/360
I = 230