Effective Current calculation

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JB78

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I am reviewing plans for a floating marina project and need some clarification. The drawings provide a chart for the power pedestal branch circuit calculations that seems to use "effective current" to size overcurrent protection. To be brief, an example single phase 120/240v circuit has a load of 270 amps (9 x 30a receptacles). After the adjustments from NEC 555.12 (80% load factor & 90% meter factor) I arrive at 194.4 amps. The chart lists an effective current of 108 amps and a 125 amp circuit breaker. I am questioning the effective current. Is the value correct? And should that be used here?

Thank you,
JB78
 
Effective current concerns RMS, DC, and conversion formulas.

It really does not apply here.

I played with the numbers a bit and can get 108 from 270, none of which make sense NEC values correctly.

I looked at table in 555 and see the .8 adjustment and I think that is correct.

Where did you come up with this .9 for meters? Never heard of it before.

The 108 value seems to me that some fat fingered the calculator or doubled the adjustment factor. 108 is 40% of 270.
 
Ingenieur said:
not saying it is correct but I've seen it down
Click to expand...

Sorry, just getting back to this now. Thanks for all the input. I was trying to attach part of the chart that I was referring to but was having difficulty. Let me elaborate a bit. I stopped just short of calculating the "effective current" (at 194.4a) in my original question. My understanding is that the calculation for "effective current" is I/sqrt(2). If that's correct, eff. current would be 137.5 amps. The person creating this chart would be way off with the 108a. Also, proving the chart further, I find that the "meter factor" of 90% was only used sometimes. Actually, if you drop the meter factor in the example I provided and just use 270a x 0.8 = 216a, it looks like he divided the current by (2), arriving at 108a, and not the square root of (2). I don't know for sure but there are (14) circuits and the pattern is the same. I am just trying to understand this chart before pointing-out any discrepancies. Does this help at all?

Thanks,
JB78
 
180115-0920 EST

JB78:

Nothing makes particular sense from what you have said.

The wording "effective" current is usually used in a classroom setting to try to have a student understand that for a particular AC waveform that a simple multiplier can be used to relate that AC waveform to a DC current based on having the same heating effect in a resistor. As the AC waveform changes that multiplier changes. For a sine wave a ratio is Ipeak * 0.707 = Irms. Theoretically one can calculate this ratio with calculus. The 0.707 value is an approximation of the exact value of sq-root of 2 divided by 2.

In general I would use the words "effective" and RMS as being identical.

If you have a split phase system based on a center tapped 240 V secondary, your loads are 120 V, and you try to approximately balance the loads on you source, then one phase is loaded to 4*30 A and the other to 5*30 A. So you would use an RMS load current of 150 A. The 150 is the greater of the two split loads and will determine breaker and wire sizes.

.
 
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If the marina is going to be busy, you really need to keep the Information Note in mind.
555.12 ... Informational Note: These demand factors may be inadequate in areas of extreme hot or cold temperatures with loaded circuits for heating, air-conditioning, or refrigerating equipment.
Click to expand...
 
gar said:
180115-0920 EST

JB78:

Nothing makes particular sense from what you have said.

The wording "effective" current is usually used in a classroom setting to try to have a student understand that for a particular AC waveform that a simple multiplier can be used to relate that AC waveform to a DC current based on having the same heating effect in a resistor. As the AC waveform changes that multiplier changes. For a sine wave a ratio is Ipeak * 0.707 = Irms. Theoretically one can calculate this ratio with calculus. The 0.707 value is an approximation of the exact value of sq-root of 2 divided by 2.

In general I would use the words "effective" and RMS as being identical.

If you have a split phase system based on a center tapped 240 V secondary, your loads are 120 V, and you try to approximately balance the loads on you source, then one phase is loaded to 4*30 A and the other to 5*30 A. So you would use an RMS load current of 150 A. The 150 is the greater of the two split loads and will determine breaker and wire sizes.

.
Click to expand...

I agree. It doesn't make sense. I have never seen this done before. My main reason for questioning it, along with the calculated load being reduced so far. You mentioned splitting the loads for 120v. Maybe that has something to do with it?? One other thing about this pedestal feeder: as the chart indicates, it is (9) 120/240v 30a receptacles for the load calculation. But each of the (5) pedestals has a 20a/120v GFCI receptacle that is dropped from the calculation, as allowed by NEC table 555.12, note-1. Any thoughts on this additional info?
Again, I want to understand why he chose this method, as I may have to split-up or add some circuits and want to stay consistent.

Thank you,
JB78
 
180115-1229 EST

Are these 240 V receptacles? If so then 9*30 would be the number to use. But, still if the source is 270 A, and the receptacles are 30 A, then protection is needed at each receptacle.

.
 
gar said:
180115-1229 EST

Are these 240 V receptacles? If so then 9*30 would be the number to use. But, still if the source is 270 A, and the receptacles are 30 A, then protection is needed at each receptacle.

.
Click to expand...

Right. Five pedestals on this circuit. Four of them have (2) 30a, 120/240v recept. and (1) 20a/120v. The fifth pedestal has (1) 30a, 120/240v and (1) 20a/120v. That's where we get (9) 30a, 120/240v for calculations - the 20a receptacles are not counted (table 555.12, note-2). All receptacles have individual OC protection in the pedestal.

JB78
 
he is only using 1 30 from each pedestal
not 2 if it has 2
5 x 30 x 0.8 x 0.9 = 108 A

the note is not clear
it says use the largest
it does not address if you have 2 'largest'
 
Ingenieur said:
he is only using 1 30 from each pedestal
not 2 if it has 2
5 x 30 x 0.8 x 0.9 = 108 A

the note is not clear
it says use the largest
it does not address if you have 2 'largest'
Click to expand...
Although it does not say it explicitly, my impression is that the reason for only counting the largest circuit is the assumption that if there are two circuits of different size per pedestal that only one of them will be used at a time and the two circuits are to accommodate different connection needs for different boats, not that two boats will be connected to each pedestal.

If the OP's design is for 35 pedestals for a 70 boat marina, then he would have to count both before applying any diversity factor.
 
GoldDigger said:
Although it does not say it explicitly, my impression is that the reason for only counting the largest circuit is the assumption that if there are two circuits of different size per pedestal that only one of them will be used at a time and the two circuits are to accommodate different connection needs for different boats, not that two boats will be connected to each pedestal.

If the OP's design is for 35 pedestals for a 70 boat marina, then he would have to count both before applying any diversity factor.
Click to expand...

that is logical
 
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