AIC Calculations & It's Significance

[mbrooke]

Yes. You may find situations where something like the 800 amp service in question here is heavily loaded for say 6-8 hours a day , but is lightly loaded the rest of the day - this gives transformer time to cool down again - they factor that in when selecting a transformer. Now if that same service is heavily loaded 24 hours five days at a time they also consider that and will likely size it much closer to actual KVA being used.

Yup- and in both those scenarios I am willing to bet you will max out at 200 to 400amps peak.

 

[mbrooke]

For the History- How pole pigs are loaded

For the History- How pole pigs are loaded

Since we are on the subject I might as well post this for future generations- at least we'll have something none political to read when our conscience is down loaded to the internet. If mods feel like this is veering off to much for this thread they have my permission ahead of time to move it to new thread.

Anyway a few months ago a member here shared concern about a 50kva pole pig serving 11 homes. Here were my PMs:

50kva pig for 11 homes


Because the average home draws only a few amps when people are at work, the pole pig has time for the oil to cool. With cool oil a pole pig can take a 300% overload for about 1 hour or 200% overload for about 3 hours. Assuming the peak load will take place for about 1 1/2 hours when people are coming home and cooking dinner/doing laundry/turning the AC down, ect we will do a 250% overload.

250% of 50= 125 Kva

125Kva / 11 homes= 11.3kva per home

11.3kva/240volts= 47amps


That would make for a #6 AL service drop. #6 AL is rated 50amps at 75*C in the NEC, but because its in open air you can do 80 amps. 110amps if the peak load is expected in the dead of winter. Some POCO guys might push it even higher.

http://www.southwire.com/ProductCata...prodcatsheet34

Yes the catalog say 70/85, but those values tend to be conservative when compared to most POCOs.

Most homes do not draw steady loads, even during the hottest days. Those AC units are cycling on and off, and at any given point in time (random snap shot) some ACs will be on and others will be off as they cycle relative to each other to maintain a set temperature. The same holds true for refrigerators, electric heat, even cooking during thanksgiving. So on a cool day that pole pig might only see 1/4 the AC units drawing current at any given point in time even if all of them are "on" (cycling) trying to maintain comfortable temperatures, and 1/2 to 2/3 current on a blistering hot day at any given point in time. Thus when everyone is away at work, the worse case assumption looks something like this:


1. 150-350 VA refrigerator (when the compartments are cold the compressor draw is actually a lot lower then the label. In fact sometimes the label is actually accounting the defrost heater(s) as a higher load x 125% or more %. Today's refrigerators are also more efficient for many regards, ie the case and door gasket heating is accomplished by running a portion of the condenser around the edge instead of a separate electric heater)

2. 150 VA Standby and miscellaneous loads (forgotten lights, charging electronics, appliance clocks, wifi, standby of TVs, game cubes, routers, printers, garage door opener, ect ect)

3. 2,500 VA exterior AC condensing unit (this will be higher or lower for many reasons even for the same ton units, but often less than 1/2 the MCA. Remember the MCA is already worse case x 125%)

4. 450 va air handler blower

So:

150VA x 11= 1,650 VA

450+2,500+150= 3,100 x 5.5 (since only half the units will be running at any point in time) = 17,050

= 18,700kVA on an August day with everyone at work

For a summer weekend with people home in between meals; vaccuming going on, some laundry, lights, TVs, game counciles, ect we might see 24kva. An extra electric clothes dryers or two could bump it up to 28kva for 15-20 minutes, then the heater cycles as the clothes loose mositure. A winter load (minus 2,500 VA ACs) would be 5,000 VA with 8,000VA in between meals when people are home.

So even during the hottest days the pig has sufficently reduced load and enough time to cool in between peak periods. If the stove and dryer are gas (I assumed they were both electric like some homes with a gas service) and there is good insulation in those walls, you could size that pig at 37.5kva. And yes, 37.5kva pigs do exist :thumbsup::

https://upload.wikimedia.org/wikiped...y_stepdown.jpg

If those hosues were all electric then I would say 4 homes would do it. Maybe 5 at 1200 sqft or good insulation. Worse case winter:

General: 150x4= 600
Refrigeration: 150x2= 300

8 VA pr square foot heating: 8 x 1,500 x 2 (assume half the heating is running at any given time)=24,000kva

So 24,900 kVA. 25kva is more then enough load relief to allow sufficient cooling ESPECIALLY in the dead of winter. Then we can do a 250% overload as people start to shower around the same time/ wash dishes/do laundry {4 water heaters @ 18.75 amps each = 18kVA}, crank up the electric heat, make toast, preheat the oven, skillet, laundry, lights, hair dryer, ect, ect.


Of note, I am aware that its often 10va per foot when calculated by the electrician, however, as you oversize electric heat the diversity increases. So instead of multiplying by two, we could do 1.8. Hence why I decided on 8va from the start- kept the two.

Also, in regards to load diversity, this is actually the reason why POCO can not just "flip the switch" after a blackout or major long duration outages. The technical term is cold load pickup:

https://www.google.com/search?source....0.7rAr0efjHyQ


Everything that was cycling like refrigersators and HVAC all start calling at once, and on top of that when power is restored, everyone starts useing everything at once. It is for this reason that POCOs will restore feeder by feeder, or even in 500-1000 customer chunks over the course of hours (or even days after major blackouts) when power is finally restored to a particular substation.

The absolute nightmare in all this is electric heat. Most of electric heat is grossly over-sized in the real world, which is fine when you are mainintng temps. But after 15va per square foot start calling, for 20,000 1500 sqft homes...

 

[kwired]

Yup- and in both those scenarios I am willing to bet you will max out at 200 to 400amps peak.

Single load services like a pumping station can be the exception. Irrigation wells around here they look at HP of the well motor and practically ignore any other load that may be associated - but most other load is usually very minimal. Grain storage facilities - the aeration fans are usually the only thing they want to know, that is what can run 24/7 for days at a time. Most other loads are pretty light in comparison and especially on the farm sites don't coincide with aeration anyway - commercial storage sites - may have more grain moving activity year round, on the farm - they fill them at harvest, when they empty them can vary depending on a lot of circumstances, but they often sit empty until the next harvest.

 

[Ingenieur]

For a real fault study with accurate numbers, you do.

Its the other way around- I am asking questions because you have not shown a solid understanding of utility practices hence why I am trying to gauge your knowledge level while simultaneously trying to evoke critical insight. Much like me asking a person "then why does code only require #6 to (a ground rod) but a much larger gauge for else-where in article 250" who claims a ground rod's purpose is to open an OCPD.

Again- you took 800 amps and automatically derived 300kva. The math is 100% correct in terms amps x volts= VA, but its far away from how utilities actually size transformers and that is the point I am trying to get across.

Service size ≠ to transformer size. Period.

no you do not
you need software

it is ok to not understand the basic concepts
your inability to perform a 5 minute excercise has shown that

do it in steps
10 mva fault source
13.2 kva
x/r 12
what is source Z (hint: r + j x, x = 12r)
??
litterally a 20 sec problem

you ask a lot of question but never have answered on
deflect, ignore, chane the subject or answer with a question
show me you have a better inderstanding than it appears

 

[topgone]

no you do not
you need software

it is ok to not understand the basic concepts
your inability to perform a 5 minute excercise has shown that

do it in steps
10 mva fault source
13.2 kva
x/r 12
what is source Z (hint: r + j x, x = 12r)
??
litterally a 20 sec problem

you ask a lot of question but never have answered on
deflect, ignore, chane the subject or answer with a question
show me you have a better inderstanding than it appears

10 MVA fault source? Where is that?

 

[Phil Corso]

Ingenieur...

You've omitted the calc to determine pre-fault source voltage!!!

Phil Corso

 

[Ingenieur]

10 MVA fault source? Where is that?

at the source
it provides fault power

from the info you can determine:
system Z
xfmr Z (you can est an x/r or consider it high enough to assume inductive)
series loop: 13.2 kv v source, sys Z, xfmr Z reflected to primary
i fault prim = v / (sum Z)

 

[Ingenieur]

Ingenieur...

You've omitted the calc to determine pre-fault source voltage!!!

Phil Corso

source >>>> load or fault
xx Gva >>>> 10 mva, fault will have very little impact, 1000 order of mag
assume constant, not unreasonable
you do have the info to calc v at load
v load = v source - i fault x sys Z

very simple excercise
EE second level power course stuff

hint
sys Z = v^2 / fault S
sys Z = ((12 r)^2 + r^2) you know Z , calc r, then X = 12 r

 

[Phil Corso]

Injunear...

Oh sure... now you mention it !

Phil Corso

 

[mbrooke]

source >>>> load or fault
xx Gva >>>> 10 mva, fault will have very little impact, 1000 order of mag
assume constant, not unreasonable
you do have the info to calc v at load
v load = v source - i fault x sys Z

very simple excercise
EE second level power course stuff

hint
sys Z = v^2 / fault S
sys Z = ((12 r)^2 + r^2) you know Z , calc r, then X = 12 r

I can't see this working on an L-G fault on the secondary, only 3 phase.

 

[mbrooke]

you ask a lot of question but never have answered on
deflect, ignore, chane the subject or answer with a question
show me you have a better inderstanding than it appears

As you refuse to acknowledged what I know. Its a two way street. The fact you do not size a transformer by the service has been deflected for several pages now.

 

[topgone]

at the source
it provides fault power

from the info you can determine:
system Z
xfmr Z (you can est an x/r or consider it high enough to assume inductive)
series loop: 13.2 kv v source, sys Z, xfmr Z reflected to primary
i fault prim = v / (sum Z)

Doesn't ring a bell to me. Utilities have SCAs in the hundreds.
Maybe you just made that figure up. A fault available of 437A at 13.2 kV is not gonna happen.

 

[mivey]

The fault current value of 56,000 at the utility transformer is likely not realistic.

If part of a larger service serving more than the 800A load it could be. A bank of 250 kVA transformers or larger could easily get there and the primary source would not likely prevent that.

An 800A service would not necessarily require a dedicated transformer bank. For that, a bank of 100kVA transformers would be likely and you would be near the 30 kA range.

 

[Ingenieur]

Doesn't ring a bell to me. Utilities have SCAs in the hundreds.
Maybe you just made that figure up. A fault available of 437A at 13.2 kV is not gonna happen.

happens all the time, very common
in the op's case 20 mva at the sec
likely served by 12.47 or 13.2
isc prim ~800 A range

 

[Ingenieur]

good comparison of various methods
https://appanet.cms-plus.com/files/eando/appa-module_6-fault_current_analysis.pdf

mva method ~pg 40

 

[mivey]

happens all the time, very common
in the op's case 20 mva at the sec
likely served by 12.47 or 13.2
isc prim ~800 A range

I would call 437A at 13.2kV uncommon.

 

[Ingenieur]

I would call 437A at 13.2kV uncommon.

not according to utilities
on many systems the source is not the limit, the xfmr is

in the op's case it is 20 mva or 870 A at 13.2
even if we assume it is off by a factor of 2 it is >430 A

 

[mivey]

not according to utilities
on many systems the source is not the limit, the xfmr is

in the op's case it is 20 mva or 870 A at 13.2
even if we assume it is off by a factor of 2 it is >430 A

If you meant at least 437A or greater you are correct. 437A is uncommonly low.

 

[Ingenieur]

If you meant at least 437A or greater you are correct. 437A is uncommonly low.

yes
>400 (at 15 kv range) is common
it was stated that an avail fault mva of 10 is not common
as in it's usually lower

Doesn't ring a bell to me. Utilities have SCAs in the hundreds.

Maybe you just made that figure up. A fault available of 437A at 13.2 kV is not gonna happen.

 

[electrofelon]

If part of a larger service serving more than the 800A load it could be. A bank of 250 kVA transformers or larger could easily get there and the primary source would not likely prevent that.

An 800A service would not necessarily require a dedicated transformer bank. For that, a bank of 100kVA transformers would be likely and you would be near the 30 kA range.

Certainly possible. However I cant remember the last time I saw a commercial service fed from a non-dedicated bank. Dwellings sure, they do that all the time.