Will a Victron Quattro-II 48/5000/70-50 be happy with the inrush current of a 3HP table saw motor or a Hammer A331 planer with a 3kW motor?
According to this https://www.youtube.com/watch?v=4SXtGIx0x5w, a single-phase motor can draw up to 6x its rated power to start up. There is no load on these machines when they start up, so 6x the rated power seems like a lot to me. I have no idea how much current they actually draw obviously.
Is it safe to try once, to see what the inverter does? I assume it’ll shut off if it goes over its threshold?
I know the obvious solution is not to feed them through the inverter, but I am trying to figure out if it is worth doing the essential/non-essential load split thing.
It’d be nice to be able to carry on working in the workshop during load-shedding or from solar.
I had 150l Compressor with 1.5kW motor…inrush was 38A
Sold and got a high flow 100l with three motors(750W each) which was rewired to start sequentially a second apart.
My radial arm is 1.5kW and starts quite happily on my 3kVA Victron.
3HP is 2.2kW and should be fine on the Quattro
The Victron inverters (except for the new Multi-RS, which is an HF design) can do 200% of the rated power for a second, which is enough to start an induction motor, if 5 times the nominal rating of the motor does not exceed 200% of the inverter rating. Depends, is the motor loaded at startup, or free running (like a table saw).
I think the 5kVA (peak 10kW for one second) will struggle to start a 3kW motor, but you may have a chance with the 3HP (2.2kW) motor.
Also remember, the battery needs to be able to handle the surge without the voltage dropping out completely.
Edit: For completeness sake, the Multi RS can do 150% for three seconds. Which is also nice.
The issue is in the “up to 6 times” estimate people make for an induction motor. That is a worst case scenario. For my pool pump, it certainly is up there (around 4-5 times), because the pump starts under load (it is already filled with water), but for something like a table saw or a planer, where the motor is essentially spinning freely and the load only comes when you start pushing material through the machine, that estimate is going to be way over.
So I am not too surprised, and very happy that it worked out for you.
It’s more dependent on the motor design, and not necessarily intuitive like you describe… Like with your pool pump, it does not really start under load, the load is proportional to the square of the rpm, so it should be very comparable to an unloaded motor for most of the start-up time.
Of course. Also to the pressure. Also, I have a small air leak in one of the heating mats on the roof, so the pipes always drain over the course of 24 hours, and the air has to be ejected from the system when it starts up, which means it isn’t even immediately under full pressure either.
But certainly I expect a pool pump to have a higher consumption in the first second because it does pick up pressure and start doing work almost immediately, even if there is still a ramp involved.
What I absolutely agree with, is that the rule of thumb is just that… and not necessarily intuitive.
I assume you measured your pump with a scope? Then you will have seen that the current is pretty much constant for the startup cycles, it does not change as the speed changes (for >90% of the sequence). The rotor is essentially a shorted secondary of a transformer before it reaches it’s synchronous speed (with slip).
Here is an example of an 1800W lawnmower, the startup current is 42A RMS.
So you’re telling me the peak current isn’t worse if the motor starts up under load? I sort of knew that, I probably just expressed it poorly.
Yeah… I kinda did say it.
What I mean is that a motor that isn’t driving anything gets up to speed a lot faster, so this part is narrower.
Since the energy (well, the charge in this case, amps on the vertical scale) is the area covered, and this is the bit that eventually overloads the inverter (200% for 1 second), I expect an unloaded motor to be started easier than a loaded one.
On an electric lawnmower running during LS, using a non-scientific analysis without a scope, I can atest under oath that if you lift the front wheels up by pressing down on the handles when you start it, that said lawnmower starts up much easier and faster than if you were to have left it standing flat on the grass i.e. the motor starting under “load”.
I just could not resist when I saw the word “expect”.
. I had the sparky tie into my DB board today and while he was here, I tested the machines. The inverter didn’t even flinch! 
