I brought a 12 vol,t 2500 to 5000 Watt IPower Inverter pure sine wave $269. This will not run my Caravan wren 1800watt AC unit
My 2000watt Honda Gen runs the AC unit all night
This 2500 Watt IPower Inverter pure sine wave will not start the AC unit motor
FYI
C00P said
01:13 PM Feb 9, 2017
The problem is probably that the starting power for the AC unit is too large. Our 2kW Honda generator will run our AC, but you will notice a brief "overload" indication as it starts. Running an 1800W AC unit from batteries via an inverter would flatten the average caravan battery very quickly. 1800W/12V = 150Amps. So a 120Ah caravan battery would be flat within an hour. At that load level it wouldn't last a fraction of that time, as the battery voltage drops faster the higher the load. Battery capacity is usually calculated based on a load that would flatten it in 20 hours. So a 120Ah battery would carry a 6A load for 20 hours. However, trying to draw 150A from it would result in a much reduced capacity. That load is similar to the load imposed when cranking an engine. How long would your battery last if you kept cranking your engine without it starting? It simply isn't practical to run heavy loads like AC's off a battery, unless you have a very, very big battery! Cheers
C00P
-- Edited by C00P on Thursday 9th of February 2017 01:19:49 PM
DeBe said
01:13 PM Feb 9, 2017
With that size inverter the current draw on 12V would be increadable, & the batteries probably wouldn't handle the surge current. That size inverter should be 48V to keep the current draw to a respectable limit.
dragonfly1 said
01:13 PM Feb 9, 2017
Make sure your hot water is not switched on and how big are your battries
Dragonfly1
C00P said
01:50 PM Feb 9, 2017
Found a neat calculator for determining battery discharge time here: www.batterystuff.com/kb/tools/calculator-for-load-specific-run-time.html You need good information about battery capacity at two discharge rates in order to use it. It calculates how long it would take to drag your battery down to 50% capacity (discharging it more than this will seriously shorten the battery life). Using figures I obtained for my batteries (two 120Ah batteries in parallel = 240Ah capacity at the 20 hour rate) they would be 50% discharged in 54 minutes under a 150A load!!
Cheers
C00P
-- Edited by C00P on Thursday 9th of February 2017 01:51:05 PM
Peter_n_Margaret said
02:09 PM Feb 9, 2017
josu wrote:
I brought a 12 vol,t 2500 to 5000 Watt IPower Inverter pure sine wave $269. This will not run my Caravan wren 1800watt AC unit
My 2000watt Honda Gen runs the AC unit all night
This 2500 Watt IPower Inverter pure sine wave will not start the AC unit motor
FYI
It should start and run the A/C without any problem.
Either the voltage drop from the batteries is too high or the inverter specifications are "slightly exaggerated".
Cheers,
Peter
Mike Harding said
04:46 PM Feb 9, 2017
An 1800W motor will have a starting current somewhere between three and six times it's running load ie. 5.4kW+ so you're already above your inverter rating.
It gets worse:
As Current = Watts / Voltage (I = W/V) - 5400 /12.5 = 432 amps. This is what you battery is being required to supply during the start phase. Actually it's worse because an inverter is not 100% efficient but let's not worry about that.
It get even worse:
As Voltage = Current * Resistance (V = IR) - a 0.01 ohm resistance in your battery terminals, fuses, wires or inverter terminals will drop that 12.5V at the battery to:
12.5 - 4.32 = 8.18V at the inverter.
And you will have much more than a 0.01R resistance in the system I described above.
Don't use inverters for anything above about 500W maximum even then you'll be drawing close to 50A from the battery so you'll need a big battery in excellent condition and excellent cables/connectors/fuses.
MH
C00P said
04:33 PM Feb 10, 2017
Thanks for that post, Mike. I knew the starting current was greater, but didn't know by how much.
Peter, you need to go back to basics. The formula for power in electrical circuits is Power (in Watts) = Voltage x Current. (you can use this for Alternating or Direct current, it isn't exactly right for Alternating current, but it's near enough).
So, if you want a lot of power, you either have lots of current or lots of voltage (or both).
With 240 volts, the current to run the 1800W Airconditioner only needs to be 1800/240= 7.5A. That's why you can use relatively thin wires in a house (or van) to run it.
But with only 12 volts, you need much more current. In theory this will be about 1800/12=150A. Hence, you will need massive cables and good (low resistance) connectors to carry that current, as Mike said.
The fact that the inverter is rated at 5000 watts is irrelevant if the battery and cables cannot supply the required current. Such an inverter might be used in a household power setup where you have a large battery bank with many batteries linked in parallel. The current is then divided across all the batteries(not just one or two) making it possible for each battery to supply the current with minimal loss of voltage. Even then, you would probably want to have several battery banks wired to provide 48 or 60 volts rather than 12, in order to limit the current, and keep the size of your cables practical.
I brought a 12 vol,t 2500 to 5000 Watt IPower Inverter pure sine wave $269. This will not run my Caravan wren 1800watt AC unit
My 2000watt Honda Gen runs the AC unit all night
This 2500 Watt IPower Inverter pure sine wave will not start the AC unit motor
FYI
The problem is probably that the starting power for the AC unit is too large. Our 2kW Honda generator will run our AC, but you will notice a brief "overload" indication as it starts.
Running an 1800W AC unit from batteries via an inverter would flatten the average caravan battery very quickly.
1800W/12V = 150Amps. So a 120Ah caravan battery would be flat within an hour. At that load level it wouldn't last a fraction of that time, as the battery voltage drops faster the higher the load. Battery capacity is usually calculated based on a load that would flatten it in 20 hours. So a 120Ah battery would carry a 6A load for 20 hours. However, trying to draw 150A from it would result in a much reduced capacity. That load is similar to the load imposed when cranking an engine. How long would your battery last if you kept cranking your engine without it starting?
It simply isn't practical to run heavy loads like AC's off a battery, unless you have a very, very big battery!
Cheers
C00P
-- Edited by C00P on Thursday 9th of February 2017 01:19:49 PM
Make sure your hot water is not switched on and how big are your battries
Dragonfly1
Found a neat calculator for determining battery discharge time here:
www.batterystuff.com/kb/tools/calculator-for-load-specific-run-time.html
You need good information about battery capacity at two discharge rates in order to use it.
It calculates how long it would take to drag your battery down to 50% capacity (discharging it more than this will seriously shorten the battery life).
Using figures I obtained for my batteries (two 120Ah batteries in parallel = 240Ah capacity at the 20 hour rate) they would be 50% discharged in 54 minutes under a 150A load!!
Cheers
C00P
-- Edited by C00P on Thursday 9th of February 2017 01:51:05 PM
It should start and run the A/C without any problem.
Either the voltage drop from the batteries is too high or the inverter specifications are "slightly exaggerated".
Cheers,
Peter
An 1800W motor will have a starting current somewhere between three and six times it's running load ie. 5.4kW+ so you're already above your inverter rating.
It gets worse:
As Current = Watts / Voltage (I = W/V) - 5400 /12.5 = 432 amps. This is what you battery is being required to supply during the start phase. Actually it's worse because an inverter is not 100% efficient but let's not worry about that.
It get even worse:
As Voltage = Current * Resistance (V = IR) - a 0.01 ohm resistance in your battery terminals, fuses, wires or inverter terminals will drop that 12.5V at the battery to:
12.5 - 4.32 = 8.18V at the inverter.
And you will have much more than a 0.01R resistance in the system I described above.
Don't use inverters for anything above about 500W maximum even then you'll be drawing close to 50A from the battery so you'll need a big battery in excellent condition and excellent cables/connectors/fuses.
MH
Peter, you need to go back to basics. The formula for power in electrical circuits is Power (in Watts) = Voltage x Current. (you can use this for Alternating or Direct current, it isn't exactly right for Alternating current, but it's near enough).
So, if you want a lot of power, you either have lots of current or lots of voltage (or both).
With 240 volts, the current to run the 1800W Airconditioner only needs to be 1800/240= 7.5A. That's why you can use relatively thin wires in a house (or van) to run it.
But with only 12 volts, you need much more current. In theory this will be about 1800/12=150A. Hence, you will need massive cables and good (low resistance) connectors to carry that current, as Mike said.
The fact that the inverter is rated at 5000 watts is irrelevant if the battery and cables cannot supply the required current. Such an inverter might be used in a household power setup where you have a large battery bank with many batteries linked in parallel. The current is then divided across all the batteries(not just one or two) making it possible for each battery to supply the current with minimal loss of voltage. Even then, you would probably want to have several battery banks wired to provide 48 or 60 volts rather than 12, in order to limit the current, and keep the size of your cables practical.
Cheers
C00P