Kity 619 - standard plug?

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JulesN

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Hi all, I am interest in a kity 1619 with sliding table etc, but wondered if it ran on a standard 13amp circuit? Grateful for your advice! Many thanks
 
The thread is titled 619. The question aska about 1619. Could you please make your mind up.

1619 specifcations here:

https://web.archive.org/web/20080509154542/http://www.dbkeighley.co.uk/masters/kitty1619.htm

619 specifications here:

https://www.hoechsmann.com/en/article/45123/bench_saw/kity-bela_619.html

1619 says 1800W; 619 says 2200W. You can work on the basis of 4A per kilowatt.
Apologies, clearly too much time in the sun ! It’s a 619 stated on the plate
 
You can work on the basis of 4A per kilowatt.
Actually, you can't. Not with the induction motors used in these machines.
4A / kW would mean you could run a 3.25kW motor off a 13A plug. You most definately can't.

In my own experience, 1.5kW, maybe 1.8kW induction motors will run happily off a 13A plug.
2.2kW motors will sometimes run but they tend to age the fuse if they don't blow immediately. You tend to blow one every few months. This is not good practice.

The link for the Kity 619 (helpful, thanks) says it has a 2200W main motor. This would be fine on a French or German 16A Schuko plug but in the UK the sensible thing to do is wire in a 16A CE blue industrial socket and run you machine from that.

Cheers.
 
I have one of these and, NO, you definitely can't run off of a 13A plug as it'll blow the fuse and likely trip the breaker on startup.

You need a 16A Commando (BS EN 60309) blue (230VAC single phase range) plug & socket with a type C 16A breaker with appropriate cabling. The "type C" bit is important.

Standard/default breakers are type B and are used for loads that are largely resistive - they will trip with loads like motors.

Type C breakers are designed for inductive loads, e.g. motors, as they cater for the switch-on inrush current of these devices without tripping.

"Commando" is an MK product name for their implementation of this standard and as their use is common, the name has become generic in the industry (much like "Hoover" has for vacuum cleaners etc.).
 
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Many thanks for the advice - I will have to add 16amp circuit to the list! Thanks again, greatly appreciated
 
Actually, you can't.

Maybe I should have been more precise and said that 4A per kW is a good first approximation. If that shows you comfortably below 13A, then you are generally good to go. If the maths shows you approaching 13A, then more detailed analysis is needed.

Just to note, that below this reply, the forum shows "similar threads" and the 619 and 1619 feature there so they might have some useful information for the OP on other aspects of the machine.
 
Not sure if I’m missing something, but a 2,200W machine running on 230V would pull 9.56A (I=W/V), so you would be OK on a 13A plug top. The issue if you are blowing fuses may well be in rush current.
 
Not sure if I’m missing something, but a 2,200W machine running on 230V would pull 9.56A (I=W/V), so you would be OK on a 13A plug top. The issue if you are blowing fuses may well be in rush current.
Yes, you're missing something :)

An important part is the typically 5 to 8x inrush current. This is a brief but very big overload.

But induction motors aren't resistive loads so P=IV doesn't work as you think.
There's an additional current that flows in and out of these motors without being consumed. The power factor tells you how big this is and the full load current rating on the nameplate tells you the combined real and inductive current so you can select the proper overload protection (in the DOL starter or VFD).
In practice, woodwork machines with 1500-1600W induction motors are safe on a 13A plug, and brushed motors have none of these issues so your 2.2kW router and site saw are fine.
 
Many thanks for all the advice - is a 16amp supply costly? I have a fuse box in the shed, and the socket for the tablesaw is only 2 metres away.


Thanks again!
 
Not sure if I’m missing something, but a 2,200W machine running on 230V would pull 9.56A (I=W/V)
The part you are missing is called reactance, it consumes energy whilst delivering nothing in the way of useful energy because it is only resistance where energy is disipated. Your equation is ok for Dc and some Ac circuits providing you use RMS values. Capacitance and Inductance cause the voltage and current to be out of phase so this phase needs to be taken into considration when looking at total current used because these quantities are charged / discharged at twice the line frequency. Motors will show something called Cos Theta which is the ratio of active power(kW) to apparent power(kVA) which gives reactive power(kVAR). With a power factor of 1 your equation holds but with motors you get a PF of less than 1 because of the reactive component so more power is needed but this is not reflected in the output.

Many thanks for all the advice - is a 16amp supply costly?
All depends upon the current instalation and if the cables are sufficient to carry the current under both normal loads and fault conditions, hence why just swapping a type B device for a type C is a bad move unless you can measure the loop impedances.
 
LOL- second forum tonight that people have been told that AC and DC work differently- I did almost exactly the same explaination on another forum a couple of hours ago, that AC has to consider the power factor and voltage/current waveform phase shifts, that W=VxA doesn't work with inductive loads...
 
IF your power supply to the shed is up to it, as detailed in previous posts, then it is not expensive. You can buy wall sockets and plugs from Screwfix, then just wire them in with suitable cable. I have four dotted around my workshop for the kit that needs them. If you are not confident get an electrician.
 

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