Originally Posted by
joeavaerage
Hi,
about 2kW is the effective maximum for a 230V single phase household supply. Even then you will need a circuit with at least 2.5mm2
wire and a 20-25A fuse.
VFDs are notorious for drawing large currents relative to their output power. You might imagine that 230V X 10A would supply 2.3kW to your spindle....
in fact it will take nearly DOUBLE the current, ie 20A, to deliver 2.3 kW to a spindle. This phenomenon is called 'power factor'. A plain VFD
will have a power factor of about 0.5 to 0.6. In an ideal world the power factor would be 1.0......that is almost never achieved. If you fit line reactors you could
improve the power factor to about 0.8 to 0.9, or you could possibly use active power factor correction and get close to ideal, say 0.95. Active power
factor correction will cost as much or more than the VFD.
Output power= Input Amps X input Voltage X Power factor
As an example a plain VFD with a 0.55 power factor:
output power=230 X 10 X 0.55
=1.265 kW
Highspeed spindles such as you have linked to have low torque and will struggle to cut steel. You might get away with very light cuts with small diameter
(3mm) tools. The trouble is with steel that you need to spin the tool slowly or risk having it overheat and destroy itself. These spindles do not like
going at slow rpm....they tend to overheat. Water cooling is almost mandatory if you want to slow to a few thousand rpm.
I have a 750W 24000 rpm spindle that I thought would be adequate for steel....its not....not by a WIDE margin. I made another spindle for steel/stainless
steel based on a 6Nm (cont) 3500rpm Allen Bradley servo, it works rally well. The highspeed spindle is good for aluminum and engraving etc but hopeless
when you need torque to spin a tool slowly to cut steel.
Craig