Posted by Cheezy on September 13, 2005, 9:18 pm
I have a 1/2 hp Craftsman Convertible Jet Pump configured to 110V and
am wanting to run it off of an inverter. I did some research and
determined that 1/2 hp corresponds to about 375W operating power. So I
figured that 2x that would be a good starting place for an inverter,
but the startup on a 750W inverter blows it out. Assuming that this
inverter has a max 1500W, I'm amazed that a little 375W motor would
need more than 1500W startup. I briefly plugged it into a 1000/2000W
inverter (not a reliable unit unfortunately) with the same results.
Is my math/reasoning off here, or does someone with a similar 1/2 hp
pump setup see the same behavior with required wattage to get the thing
going? I'm thinking of buying next a 1200 W inverter but it seems like
math should determine what I should buy and not trial and error.
Thanks in advance.
Posted by John Beardmore on September 13, 2005, 9:48 pm
Some pumps and compressors can have a start up surge current of 20 times
the normal load current.
Peak value and duration probably depend on the motor size (power and
weight and mass distribution of rotating parts), mechanical load, and
any 'soft start' electronics that may be incorporated.
Ask the manufacturer or use a current clamp and a scope to find out ?
Posted by BobG on September 13, 2005, 9:58 pm
Starting surge on an induction motor is x5 to x8 current for "a few
mains cycles" (epanorama). The inverter is putting out a modified
square wave of 140 volts peak and 120 volts average. (We all remember
from EE class that the RMS is the same as the average for a square
wave). You have 0 gauge wires to the battery? Put a voltmeter on the
12volts. Its probably sagging big time. Try it with a charger on it.
Posted by nicksanspam on September 13, 2005, 10:23 pm
is the same as the average for a square wave...
The average would be zero.
Posted by Solar Flare on September 14, 2005, 1:23 am
LOL. Good math there.
Average of the absolute voltage = RMS = peak