208/1 and 480/1 load calcs?

[CTAEEEI]

I've been in consulting engineering for almost 4 years now and still am having troubles with something that should be simple. When entering wattages/amperage into panel schedules, I seem to always get confused with 208/1 and 480/1 loads when I have three phase service. Could someone please help me with this? It's been described to me before, but just has never clicked.

For example:
Load of 2228W on at 208V/1 circuit. So it's a 20/2 pole breaker on two of three phases. Do I:
1) 2228W/208V=10.7A and put 10.7A on the two phase.
2) 2228W/120V=18.56A and then divided by two to have 9.28 on the two phases.

I'm pretty sure, based on the company panel schedule calculations, I do the second method; but I'm getting confused as to why.

Thanks.

 

[LarryFine]

#1 if the load is line-to-line, and #2 if the load is line-to-neutral-to-line.

As in the Voltage / Ampere thread, a given load (resistance) will not produce the same wattage on different voltages.

 

[infinity]

The single phase 2228 watt, 208 volt load should have no neutral load. The current would be 2228/208=10.7 amps on each phase of the 2 pole CB.

 

[al hildenbrand]

Yet another way to point at the answer, is to ask yourself, as you consider each single phase load on a panel schedule: Is a neutral required to be connected to the load? If so, there may be a combination of line to neutral and line to line loads on the the branch circuit.

 

[CTAEEEI]

Thank You

Thank You

Thank you guys for your reply. My next question would be to confirm an error in the excel file I'm using. This panel schedule has me insert the amperage for phase instead of watts per phase into each pole. For the total connected VA, it's adding the total Amps per phase and mulitplying it by the Line to Neutral voltage. This method will not produce the wattage of 2228W when using 10.7Amps per phase of the two pole breaker. I'm trying to think of a formula to use that will produce the correct VA, but can't think of a simple formula to use. Is there a simple formula that I'm over looking?

It would be so much easier if the panel used watts instead of amps for load values.

Thanks

 

[al hildenbrand]

For a combination load comprised of line to neutral and line to line loads, your spreadsheet is inadequate. Start with a formula that asks for two currents, the line to ground and the line to line, and have their default value an zero. The final VA for the schedule will be the sum of the product of the two E*I, line to neutral and line to line.

 

[kingpb]

The correct answer is 9.3A. You simply divide the total power by the number of lines ( or for some of us, phases), and then divide by the line to ground voltage, i.e. 120V.



PΦ = VΦ x IΦ x cos(θ)

and,

Pll = 2 x (VΦ x IΦ x cos(θ))

Assume cos(θ) = 1, therefore;

Pll = 2 x (VΦ x IΦ)

VΦ = Vll / √3

Substituting,

Pll = 2 x (Vll / √3) x IΦ

Solving for IΦ yields;

IΦ = Pll
(2 x Vll /√3)

The formula for three phase is:

P3 = 3 x (VΦ x IΦ x cos(θ))

and in it's final form:

P3 = √3 x Vll x IΦ x cos(θ)


If you look at the formula's above, they explain why you are not able to get the VA to match with the current simply dividing by 208 V, because that yields an incorrect amperage; off by a factor of (2/√3), or 1.155.

 

[infinity]

The correct answer is 9.3A. You simply divide the total power by the number of lines ( or for some of us, phases), and then divide by the line to ground voltage, i.e. 120V.



PΦ = VΦ x IΦ x cos(θ)

and,

Pll = 2 x (VΦ x IΦ x cos(θ))

Assume cos(θ) = 1, therefore;

Pll = 2 x (VΦ x IΦ)

VΦ = Vll / √3

Substituting,

Pll = 2 x (Vll / √3) x IΦ

Solving for IΦ yields;

IΦ = Pll
(2 x Vll /√3)

The formula for three phase is:

P3 = 3 x (VΦ x IΦ x cos(θ))

and in it's final form:

P3 = √3 x Vll x IΦ x cos(θ)


If you look at the formula's it explains why you are not able to get the VA to match with the current. Simply dividing by 208 V yields an incorrect amperage, that is off by a factor of (2/√3), or 1.155.

Why would the square root of 3 be involved if the load is single phase and not three phase?

 

[rattus]

King,

Is this not a single line to line load? Then the line current and phase currents are the same and equal to,

Iline = Iphase = P/Vll = 2228W/208V = 10.7A

Now if we consider apparent power we have phase angles to consider, and we have argued that point ad nauseum a few weeks back.

Am I missing something?

 

[kingpb]

Why would the square root of 3 be involved if the load is single phase and not three phase?

Single phase 208V, is 120V x sqrt 3 = 208V.

The load will theoretically divide equally between the phases (lines), and the current you are calculating is the current flowing in each 120V leg.


To illustrate, if you simply take the 208V, single phase load of 2228W to derive the phase current, you get:

2228W/208V = 10.7A

but the load has to divide equally on each phase so,

10.7A x 120V = 1285W per phase, or 1285W x 2 = 2570W

That would mean your load is actually 2570W? How is it possible to have line current that calculates out to a higher wattage then the load? The answer is it can't. Load is based on each phase, not line to line.

 

[kingpb]

King,

Is this not a single line to line load? Then the line current and phase currents are the same and equal to,

Iline = Iphase = P/Vll = 2228W/208V = 10.7A

The line current and phase current would be the same, and would still be the same if a three phase load was connected in wye.

Using that same thought process, and because the load (watts) splits evenly between phases, the load has to be 1114W per phase, and the line voltage is the phase-to-phase voltage divided by the sqrt 3.

 

[infinity]

King,

Is this not a single line to line load? Then the line current and phase currents are the same and equal to,

Iline = Iphase = P/Vll = 2228W/208V = 10.7A

Now if we consider apparent power we have phase angles to consider, and we have argued that point ad nauseum a few weeks back.

Am I missing something?

I must be missing something too. A 2 wire circuit with a voltage of 208 volts and a load of 2228 watts.

I=P/E

2228w/208v= 10.7 amps

 

[kingpb]

Look at it the other way-

You are at a panel, and you have a 2-pole load. You whip out your clamp on ammeter and measure the current on each 120V phase, and it is approx. 9.3 A.

What is the phase load, and what is the total load?

Phase load - 120V x 9.3A = 1114 VA (assuming pf=1)

Total load - 2 x 1114 VA = 2228 VA

If you were to multiply the 9.3A by 208V, the load would only be 1934 VA

(BTW: off by a factor of 2/sqrt 3)

 

[infinity]

Look at it the other way-

You are at a panel, and you have a 2-pole load. You whip out your clamp on ammeter and measure the current on each 120V phase, and it is approx. 9.3 A.

Why would the load on each phase be 9.3 amps? In a simple 2 wire 208 volt circuit the current would be 2228/208=10.7 amps on each phase. Where are you getting the 120 volt component of this equation?

 

[steve66]

Look at it the other way-

You are at a panel, and you have a 2-pole load. You whip out your clamp on ammeter and measure the current on each 120V phase, and it is approx. 9.3 A.

What is the phase load, and what is the total load?

Phase load - 120V x 9.3A = 1114 VA (assuming pf=1)

Total load - 2 x 1114 VA = 2228 VA

If you were to multiply the 9.3A by 208V, the load would only be 1934 VA

(BTW: off by a factor of 2/sqrt 3)

But you wouldn't measure 9.3 amps, you would measure 10.7 amps, like Infinity's post above. The current doesn't "divide" between the two phases, it is the same current (we are talking about a single phase load with no neutral).

Power/Voltage = Current. 2228/208 = 10.7 amps.

Steve

 

[kingpb]

Where are you getting the 120 volt component of this equation?

I'm not trying to be a smart@; but to try and and prove the point, may I ask how many conductors did you put your hypothetical amprobe around, when you measured the current? Was it one or two?

My guess would be one, right?

 

[steve66]

I've been in consulting engineering for almost 4 years now and still am having troubles with something that should be simple. When entering wattages/amperage into panel schedules, I seem to always get confused with 208/1 and 480/1 loads when I have three phase service. Could someone please help me with this? It's been described to me before, but just has never clicked.

For example:
Load of 2228W on at 208V/1 circuit. So it's a 20/2 pole breaker on two of three phases. Do I:
1) 2228W/208V=10.7A and put 10.7A on the two phase.
2) 2228W/120V=18.56A and then divided by two to have 9.28 on the two phases.

I'm pretty sure, based on the company panel schedule calculations, I do the second method; but I'm getting confused as to why.

Thanks.

The actual current flowing on each phase would be 10.7 amps.

However, say you want to calculate the total load on the panel, or the total load on one phase. The currents you are labeling are going to be misleading - because they won't add together right. Thats why most engineers do calculations with Watts or KVA.

Lets take your example of 2228 Watts on phases A and B. Now lets add another 2228 Watt load on phases B and C. You labeled 10.7 amps on phase B from the AB load, and another 10.7 amps on B from teh BC load. So what is the total current flowing to the panel on the B phase? you might think it is 10.7 + 10.7 or 21.1 amps. But thats not the case. The two currents you labeled on the B phase are actually out of phase.

So if you are trying to calculate total loads, the thing to do is to take the 2228 Watts on phase AB, and put half of it on phase A, and half on phase B. Then if you add all the Watts on Phase A, and divide by 120 volts, you will get the Input current to the panel on phase A.

Steve

 

[kingpb]

So if you are trying to calculate total loads, the thing to do is to take the 2228 Watts on phase AB, and put half of it on phase A, and half on phase B. Then if you add all the Watts on Phase A, and divide by 120 volts, you will get the Input current to the panel on phase A.

Steve

Steve, when you split the 2228W, and put half on phase A and half on phase B, (1114W ea) and divide by 120V, as you stated above. Now assume all other load on phase A is zero. What is your current? According to my calculator it's 9.3A. So if that is the case, which you just described, how is it possible to have different amount of current on the panel phase versus the load phase?

For power calculation purposes, you always, always, always, go back to the line current and use the line voltage. You cannot use the line current and line-line voltage.

 

[rattus]

The line current and phase current would be the same, and would still be the same if a three phase load was connected in wye.

Using that same thought process, and because the load (watts) splits evenly between phases, the load has to be 1114W per phase, and the line voltage is the phase-to-phase voltage divided by the sqrt 3.

Agreed. Then in the original problem,

Iphase = 2228W/[2x120Vxcos(30)] = 10.7A

Now are we talking KW or KVA? In this unbalanced case, they don't compare.

 

[kingpb]

Agreed. Then in the original problem,

Iphase = 2228W/[2x120Vxcos(30)] = 10.7A

Now are we talking KW or KVA? In this unbalanced case, they don't compare.

I was not wanting to get into symmetrical components becasue it would complicate the issue.

I think I came to the realization on where people may be having difficulty. It is because of the differences in a 208Y/120V system versus a 120/240V system. With a standard single phase 120/240V system, you can take the total load, and divide by the line-to-line voltage. The answer will be the same as if you take 1/2 the total load and divide by the line-to-neutral voltage.

This is NOT the case with three phase systems such as described herein, because the line-to-line voltage is not double the line-neutral voltage, and thusly you have to use the phase load, divided by the line-to-neutral voltage.

The answer is 9.3A. (assuming pf=1.0 and symmetrical components are not invoked due to an unbalanced system, thanks Rattus )