Inverter calculations

[Shaneyj]

Hello All.

A scenario came up at work today and my EE could not give me a concrete answer. After an attempt at research, neither could google.

I'm installing an inverter on a unit for a customer. This is a refurb, and the customer is upgrading from a 1500W to a 5000W inverter.
The existing wiring is 4awg supplied by (4) 12VDC batteries wired in parallel.

The inverter specs call out either parallel 1/0 or single 4/0 to supply the inverter.

I wanted to figure DC current for my own education and curiosity, but after some inquiry to my EE and google, I found conflicting information.

Several online ac to dc current calculators seemed to be using the simple 1:1 inverse relationship of V to I found in transformer calcs.** using this the 5kW @ (120VAC)= 41A. 12VDC primary to 120VAC secondary is a 1:10 ratio, so the DC current would calculate to be 10*41A= 410A DC.

I did find more than one source that quoted dc current to be 63% of ac current, but this seems to be going the wrong direction.

I hope this is not too disjointed... On to my questions.

I have been unsuccessful in tracking down a formula to calculate the DC current with respect to the current of the AC load. Does one exist where I can plug in my variables, (i. e., source voltage, load voltage, etc.) to determine whatever unknown exists?

Also, are there literature that specifies the differing ampacity ratings of conductors when dealing with DC as opposed to AC?

Thanks in advance to all who weigh in.

P. S. I've been reading this forum for over a year and this is my first post... It has added to my education. Thanks to all who support it.

 

[electrofelon]

Hello All.

A scenario came up at work today and my EE could not give me a concrete answer. After an attempt at research, neither could google.

I'm installing an inverter on a unit for a customer. This is a refurb, and the customer is upgrading from a 1500W to a 5000W inverter.
The existing wiring is 4awg supplied by (4) 12VDC batteries wired in parallel.

The inverter specs call out either parallel 1/0 or single 4/0 to supply the inverter.

I wanted to figure DC current for my own education and curiosity, but after some inquiry to my EE and google, I found conflicting information.

Several online ac to dc current calculators seemed to be using the simple 1:1 inverse relationship of V to I found in transformer calcs.** using this the 5kW @ (120VAC)= 41A. 12VDC primary to 120VAC secondary is a 1:10 ratio, so the DC current would calculate to be 10*41A= 410A DC.

I did find more than one source that quoted dc current to be 63% of ac current, but this seems to be going the wrong direction.

I hope this is not too disjointed... On to my questions.

I have been unsuccessful in tracking down a formula to calculate the DC current with respect to the current of the AC load. Does one exist where I can plug in my variables, (i. e., source voltage, load voltage, etc.) to determine whatever unknown exists?

Also, are there literature that specifies the differing ampacity ratings of conductors when dealing with DC as opposed to AC?

Thanks in advance to all who weigh in.

P. S. I've been reading this forum for over a year and this is my first post... It has added to my education. Thanks to all who support it.

The reason you will likely get varying answers on the DC current is for three reasons:
1. Inefficiency in the inverter
2. variations in what voltage is used to calculate the current from the wattage
3. Variations in what wattage is used to calculate the current.

To elaborate on a few of these, lets look at #2: The battery voltage will "sag" under load, especially high loads at the top of the inverter range. Also, it is common to periodically bring a 12V system down below 12V even under no load. Thus the extreme case would be to calculate the current at the lowest possible battery voltage. Regarding #3, most inverters have several "surge" ratings above their continuous rating. One should consider these higher values.

Note, for our purposes, DC and AC ampacity is the same.

If I may comment on the system, a 5000 watt inverter off 4 12V batteries is not a good design.

 

[Shaneyj]

The reason you will likely get varying answers on the DC current is for three reasons:
1. Inefficiency in the inverter
2. variations in what voltage is used to calculate the current from the wattage
3. Variations in what wattage is used to calculate the current.

To elaborate on a few of these, lets look at #2: The battery voltage will "sag" under load, especially high loads at the top of the inverter range. Also, it is common to periodically bring a 12V system down below 12V even under no load. Thus the extreme case would be to calculate the current at the lowest possible battery voltage. Regarding #3, most inverters have several "surge" ratings above their continuous rating. One should consider these higher values.

Note, for our purposes, DC and AC ampacity is the same.

If I may comment on the system, a 5000 watt inverter off 4 12V batteries is not a good design.

So is the ratio of primary voltage:secondary voltage the multiplier for determining primary current?
For instance, ignoring inefficiencies or voltage variations as you mentioned, using the scenario I provided, was the calculation for the DC current accurate?
And if so, because the only spec mentioned in the literature (term used loosely, it was more of a leaflet) that came with the inverter was for 4/0 conductors and 4/0 is rated at 230-260A, could it be assumed that the inverter is only 50-60% efficient?

Can you elaborate on your comment regarding bad design?
I did not mention that the inverter supplys a receptacle in a control cabin. Theoretically, it will only be drawing a load when the unit is running (2 600HP cummins engines).

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[ggunn]

One thing to consider is that battery inverters, unlike grid tied solar inverters, are capable of delivering more than their nameplate output current for short periods of time, and therefore drawing more current from the batteries than you would calculate based on the nameplate output power.

 

[electrofelon]

So is the ratio of primary voltage:secondary voltage the multiplier for determining primary current?

Ignoring inefficiencies, yes. However again the question is what voltage to use for the input. Pulling 5000 watts off a small 12 volt battery bank (I admit I am assuming - you didnt say you big your batteries were) will cause the voltage to sag well into the low 11's

For instance, ignoring inefficiencies or voltage variations as you mentioned, using the scenario I provided, was the calculation for the DC current accurate?
And if so, because the only spec mentioned in the literature (term used loosely, it was more of a leaflet) that came with the inverter was for 4/0 conductors and 4/0 is rated at 230-260A, could it be assumed that the inverter is only 50-60% efficient?

Your inverter should have a "low voltage cut out" setting, usually this is somewhere around 11 V. I would use that as you voltage from which to calculate current.

Can you elaborate on your comment regarding bad design?
I did not mention that the inverter supplys a receptacle in a control cabin. Theoretically, it will only be drawing a load when the unit is running (2 600HP cummins engines).

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Just that sized inverter on a small (again assuming your battery size) 12V system. Maybe more details about the load profile would change my opinion.

 

[GoldDigger]

The biggest problems with trying to run a 5000W inverter off a nominal 12V supply are:
1. At that voltage the problem of voltage drop requires inconveniently large wires and terminals.
2. Running DC at 400-500A is not safe for your typical consumer environment.
When you double the voltage you reduce the percent voltage drop by a factor of four for the same wire size.
A 5000W inverter really should be fed from 48V. Now would be a good time to make that change.
I cannot be sure, but for common 12V lead acid batteries a drain of 100A is going to cut severely into battery life. If your customer is really going to use that 5000W capability he needs to have a much larger battery bank and a correspondingly larger charging system.
What sort of load in a control cabin would be drawing 5000W? A window A/C?


mobile

 

[ggunn]

Those 12V batteries are not automotive batteries, are they? If they are they will die very soon.

 

[Shaneyj]

The biggest problems with trying to run a 5000W inverter off a nominal 12V supply are:
1. At that voltage the problem of voltage drop requires inconveniently large wires and terminals.
2. Running DC at 400-500A is not safe for your typical consumer environment.
When you double the voltage you reduce the percent voltage drop by a factor of four for the same wire size.
A 5000W inverter really should be fed from 48V. Now would be a good time to make that change.
I cannot be sure, but for common 12V lead acid batteries a drain of 100A is going to cut severely into battery life. If your customer is really going to use that 5000W capability he needs to have a much larger battery bank and a correspondingly larger charging system.
What sort of load in a control cabin would be drawing 5000W? A window A/C?


mobile

The customer asked for that size but claimed it would only be servicing cell phone chargers and laptops.
Because they didn't want the extra expense of adding the 4/0 cable, we installed a 80 amp breaker to feed the inverter.



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[Shaneyj]

Those 12V batteries are not automotive batteries, are they? If they are they will die very soon.

They're HD, 1000CCA.

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[Ingenieur]

Power in Pi = Power out Po + losses L

L = Pi x (1 - e)
Where e = eff x pf (eff = unit efficiency and pf = power factor)

Plug/rearrange
Pi = Po + Pi (1 - e)
Po = Pi - Pi + e Pi = e Pi

Define
Po = Vo Io e
Pi = Vi Ii

Plug/rearrange
Po = e Vi Ii
Ii = (Po) / (e Vi)

assume
Vi = 12 Vdc
Po = 5000 W
eff 0.9
pf 0.9
so e = 0.81 (81%)

Ii = 5000/(0.81 12) = 515 A
Pi = 12 515 = 6180 W
L = 6180 - 5000 = 1180
L % = 1180/6180 = 19% or e = (1 - 0.19) = 0.81 or 81%

getting values for eff and pf is the hard part

 

[electrofelon]

The customer asked for that size but claimed it would only be servicing cell phone chargers and laptops.
Because they didn't want the extra expense of adding the 4/0 cable, we installed a 80 amp breaker to feed the inverter.



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If it is truly just some phone chargers and laptops, I dont know why they want such a big inverter. Maybe a case of someone not knowing what they are doing and wanting "a couple circuits" and figuring a 20 amp circuit is typical? I would be looking at one or two of these: Great little inverter, super low standby losses.

http://www.morningstarcorp.com/products/suresine/

No really sure what is going on here, maybe there are bigger loads at play......

 

[GoldDigger]

Make sure that the 80A breaker you get is rated for DC. Most molded case breakers have no DC interrupting rating.

 

[Shaneyj]

If it is truly just some phone chargers and laptops, I dont know why they want such a big inverter. Maybe a case of someone not knowing what they are doing and wanting "a couple circuits" and figuring a 20 amp circuit is typical? I would be looking at one or two of these: Great little inverter, super low standby losses.

http://www.morningstarcorp.com/products/suresine/

No really sure what is going on here, maybe there are bigger loads at play......

Thanks for your input!

As far as why, the EE and I both thought it to be suspicious. But then again, this is a chemical pumper picked up from auction for $100,000 which is pennies on the dollar but they don't want to spend the extra $500 for cable to allow them an air conditioner when they're in a 10'x8' box in west Texas in August for 12 hours at a time.
Add to that this company has ordered upward of $10M worth of equipment to be delivered in the first 3 quarters of 2017...
Several things on this particular refurb have not made sense.

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[Shaneyj]

Power in Pi = Power out Po + losses L

L = Pi x (1 - e)
Where e = eff x pf (eff = unit efficiency and pf = power factor)

Plug/rearrange
Pi = Po + Pi (1 - e)
Po = Pi - Pi + e Pi = e Pi

Define
Po = Vo Io e
Pi = Vi Ii

Plug/rearrange
Po = e Vi Ii
Ii = (Po) / (e Vi)

assume
Vi = 12 Vdc
Po = 5000 W
eff 0.9
pf 0.9
so e = 0.81 (81%)

Ii = 5000/(0.81 12) = 515 A
Pi = 12 515 = 6180 W
L = 6180 - 5000 = 1180
L % = 1180/6180 = 19% or e = (1 - 0.19) = 0.81 or 81%

getting values for eff and pf is the hard part

Exactly what I was seeking. Thank you, sir.

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[Shaneyj]

Make sure that the 80A breaker you get is rated for DC. Most molded case breakers have no DC interrupting rating.

These are the high current DC breakers we use.

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[drcampbell]

... we installed a 80 amp breaker to feed the inverter.

That will limit the power to 960 watts, (12 volts x 80 amps) so there's nothing to be gained by replacing the 1500-watt inverter with a larger one.

 

[GoldDigger]

That will limit the power to 960 watts, (12 volts x 80 amps) so there's nothing to be gained by replacing the 1500-watt inverter with a larger one.

And instead you will lose, because of the greater idle current of the larger inverter (typically).