Well, the LV has a full 13kW output. The Solark only has, what, 8kW inverting?
"AC Output Power 9kW On-Grid & Off-Grid"
"Surge AC Power 10sec 16,000VA L-L (240V)
Surge AC Power 100ms 25,000VA L-L(240V)"
We expect 67A for 10 seconds.
(would be 20 HP for 10 seconds if 100% efficient)
It seems the inverter is not putting out rated current.
I said my 12K reported an F18 AC over-current fault. That was the only fault it reported in multiple shut-down events when starting any of these motors.
The manual contains this wording that I find bizarre:
F18 | Tz_Ac_OverCurr_Fault | Overloaded the Load Output, reduce loads. Wiring Short on the AC Side can also cause this error. Overloads
can result in F15, F18, F20, or F26. |
So the reported fault is AC over-current.
Motor windings when stopped are something like a short-circuit, with wire resistance and inductance due to coil and laminations providing impedance.
If LRA < 67A we expect inverter to try for a while before quitting. Maybe it is unhappy driving inductance, which shoves current back at it out of phase.
Is this with or without PV production? The architecture has inverter fed by HV capacitor bank. Battery voltage is boosted to capacitors, and PV has MPPT buck converter to charge capacitors. Maybe, boost converter current is limited (or slow to ramp up) and inverter quits when capacitor voltage drops too low. Maybe MPPT keeps capacitor bank at higher voltage. Boost converter adds to capacitor bank when voltage gets lower, and operates in reverse direction to charge battery when voltage gets higher. (Charging from AC would require additional smarts.)
Ordinarily we keep wires to a motor short. I wonder if an extension cord, adding resistance and reducing peak current, would help? Having another load already powered, so inverter is producing some current when this load is applied?
I've tried paralleling an induction motor while feeding a VFD, to see if it would reduce current distortion seen by my inverter, but it did not. Was hoping spinning mass would act as a generator to supply the peaks. In the case of a suddenly applied load, if a resistive load was already present, when voltage dropped less current would be taken by resistive load leaving more for motor.