Can't power 240v devices in campervan.

UK_Bantam

Hi all

Newbie here - treat me kind. 

The set up is this (allegedly!) - three leisure batteries, a 2000w pure sine inverter and a solar panel (not sure of capacity tbh).

I'm finding that if I plug a 240v device in (household kettle, or a small Nespresso machine for example) , I get a near instant drop of power to the device stops heating.

The MT50 gives me the following -

Solar - 20.1v /4.4 amps
Battery - 13.2v /6.7amp
Load - 0 v /0 amp

The inverter remote shows single green, 2 x amber before device turn-on

The basic multi-switch controller shows 12.7 volts.

I have no idea why I am getting a power failure. 

Please help! 

BB.

Welcome to the forum UKB.

With solar--Details matter. Starting with you loads...

For example, in the UK, the "typical" electric kettle draws around 2,300 to 3,000 Watts (2x more Watts than we see in the USA on 120 VAC).

You can purchase an inexpensive 23s0 VAC Power Meter to measure how many Watts your electric kettle uses:

https://www.amazon.co.uk/s?k=energy+meter+uk+plug

Now, for the other side of the setup--Your Solar hardware. As you can see, the 2,300-3,000 Watts is more power than your 2,000 Watt (2 kWatt) AC inverter is rated for (again, guessing about your specific electric kettle). That is the first issue (kettle vs AC inverter power ratings).

Next--Understanding the math to generate 230 VAC at XXX Watts from 12 VDC. Let's say that you have a 3,000 Watt kettle and a 3kW+ inverter. The basic math would be:

• Assume 10.5 VDC for 12 volt battery cutoff voltage (for the AC inverter)
• Power = Voltage * Current
• Current = Power/Voltage
• Current = 3,000 Watts * 1/0.85 AC inverter eff * 1/10.5 VDC minimum battery voltage = 336 Amps @ 12 VDC battery bus

That is actually a huge amount of current... Something like 3-5x more current than your typical car takes to start. And while the starter cranks for 5 seconds, the kettle takes several minutes to boil the water--Much harder for the battery bank.

For a lead acid battery bank, a typical rule of thumb is for every 1,000 Watts of load, the battery should be at least 400 AH @ 12 volts of capacity per 1,000 Watts of load... The math again:

• 400 AH * 3,000 Watt load * 1/1,000 Watts = 1,200 AH @ 12 volt battery bank (for a cabin would suggest 2x larger AH battery battery bank for reliability)

The typical RV battery is around 100 AH @ 12 volt per battery--Or around 300 AH @ 12 volt battery bank.

Another stopping point... Depending on your Battery Bank type and AH capacity--Would not expect a 300 AH @ 12 VDC battery bank to reliably supply 2.3 to 3.0 kWatts reliably.

The last issue is that a typical RV solar panel is around 140 Watts (can be between 100 and 300 Watts typical for solar panel)... I won't bother with the rest of the "solar math" at this point as I have made so many guesses about your setup at this point that this is just going to be more confusing than clarifying at this point--Just sufficient to say that, typically, you would want the same size solar array as your AC inverter... I.e., if you have a 2,000 Watt inverter, you would want (very roughly) a 2,000 Watt solar array--Or 14x 140 Watt panels--Most RVs do not have that sort of roof space (assuming that you want to power your system mostly from solar--And not run an AC genset/use shore power very much).

The battery is the "heart" of your system--Need to design it to power your required loads (voltage and AH rating). Design the solar array to keep up with your loads (and hours of sun/location of RV/seasonal usage). And finally the solar charging and AC inverter hardware.

In summary, my first guesses are that your AC load is too large for both your AC inverter and your DC battery bank. And depending on how much energy per day (Watts*Hours of use = Watt*Hours)--Your solar array will not supply enough Watts/Watt*Hours per day to keep the battery bank charged to your expectations.

Solar system do not supply very much power (Watts) or energy (Watt*Hours). You need to keep your loads as low/most energy efficient as you can. And design the solar system to reliably power those loads. 

For example, as a guess, your ~300 AH @ 12 volt battery bank would supply at most, a 750 Watt load. Time wise:

• 750 Watts * 1/0.85 AC inverter eff * 1/12 volts = 73.5 Amp load @ 12 volts
• 300 AH battery bank * 0.50 "usable capacity" (for long battery life) = 150 AH of available capacity
• 150 AH battery bank capacity / 73.5 Amp (aka 750 Watt load) = 2.0 hours to power loads (with your bank and 750 max continuous load)

I will stop here... I know that this is a lot of information. Your details and questions?

-Bill

UK_Bantam

Wow! Thanks Bill.  Let me read that a few more times to understand it and I will come back to you! 

BB.

You are very welcome Bantam.

Please feel free to ask for clarification on any of the above "math"... Or other assumptions I made.

The rules of thumbs we use to size systems is a very rough first estimate when sizing a system. The numbers are "approximate" to get you, quickly, somewhat close on your hardware needs. And can be refined when we know more about your exact needs and how much room you have in the RV, how much money you wish to spend, etc..

-Bill