A friend was asking me about this but I couldn't remember what was safe/

not sure mate but i run around 1500 at the mo through one socket and a lot more in the winter when heaters are needed.
Hantshaze.

Nominally, around 3KW (3000W - 230Vx13A) - although if the load was all or mainly something with a powerful inductive kick like HID ballasts/lamps, you'd be best advised to set them up on a 'staggered' basis, so that they didn't all come on at the same time.

3000 watts, or 3kW, per plug top. but i wouldn't put 6kW oon a twin socket, since there may be other appliances elsewhere on the same circuit. bettter to spread it around a few sockets, if possible.

For resitive currents and DC yes, reactive AC circuits are far more complex and more mathematics is needed

Its 3kW maximum resistive current from a socket, when using inductive loads a good rule of thumb is to half the resistive rating to give the max inductive rating

With no calculus or geometry!!!

And not a sniff of sin(x) or cos(x)

Seriously I am spoiling you!

¿Would that be as simple of estimating the power factor as being 0.5 in that case?


e2a: ¿And does the 'half power' rule of thumb apply to the inductive appliances all the time, or just when they spike when they're first turned on?

true, i always try to stick to at most 1.5 kW, per outlet. more for peace of mind than anything else. i do electrical tests in about 5 - 8 houses a day, and measured nominal voltage ranges from about 232V to about 256V, so even with all the correct phase angle, power factor, reactance and calculus considered, there is still room for errors in calculations. always better to under utilise the installation, rather than push it to the linits.

Without the cos(x) we had P=VI so our current was simply P/V=I

P = VIcos(x)

Putting cos(x) = 1/2

P = VI/2

Rearrange for I

I = 2(P/V)

Your current just doubled!

The total power is the result of the resistive power and the inductive power right angled triangle pythagorous andll that shite

If the PF was 0.5 thats saying the total power (apparent power) divided by the real power (resistive) would give a figure of 0.5 (angle of 60 degrees for the observant!)

Its adjacent divide hypotenuse thats why its called cos!

By having the overall power (hypotenuse) and times this by power factor gives the adjacent side

The adjacent side is the real power the opposite is the inductive power the Hypotenuse is the total

Adj/Hyp = Cos(x) rearrange to give

Adj (real power) = (Hyp)x(cos(x))

basically yes scribs, that would be the case if the pf was 0.5, and it applies to ALL ac appliances, as every AC circuit contains capacitive/inductive components (as my example of cables in the other thread shows, even cables have inductance/capacitance) it's just in "purely" resistive circuits, the reactive component is so small it doesn't effect the current draw in the circuit

basically power factor is the ratio of resistive current to reactive current in the circuit, watts is a measure of resistive power only, so doesn't take the reactive component into consideration. so you divide the WATTS rating by whatever the power factor is (usually about 0.85 for HID with a cap or 0.5 without a cap or if the cap is damaged) and this will give you the VA (volts times amp) divide this by the voltage at the plug socket and you have AMPS


e2@: personally i would run no more than 2kw continuosly from one socket, and no more than 3.5kW from a double socket

yeah, nice one electric man, been studying hard i see , but i'm pretty sure both those posts were as clear as dogshit for people who haven't been to engineering college

Its something they teach throughout the land, I think it comes from the good guys at the IET

But its for ''when no information is available for discharge ligthting'' i.e when working with old installations where there is nothing printed on the existing lighting control gear, because all manufacturers are different the recommend that times the lamp rating by 1.8 to allow for the control gear which is just under doubling it

This IME gives a higher current than the real world value, which means you just overated everything

Not a bad thing

Nicely explained, Scraglor - cheers.

I should be able to work it out from what you've already told me, but if you staggered individual ballats/lamps so that they weren't all striking at once, ¿would it still be advisable to follow the 1/2 maximum rated power rule of thumb?

In other words, once the inductive appliances are chugging along, do you still have to worry about them being a greater load on the circuit than a resistive appliance.

Hope that makes sense.


e2a: Just seen your follow-up post, EM2 - cheers.

e3a: I did A-level Maths (P&A) & Physics so I'm not totally unfamiliar with calculus etc., but it doesn't half make my brain itch...

it's because 0.5 is the power factor of an uncorrected discharge lamp circuit

yeah, the power factor is taken into consideration during normal running, at start up, for the first few seconds they can draw MUCH higher current than normal running current, so this is why you stagger startup times.

Jog the fuk on cock head!

Been studying well hard actually, I just love it I am specialising in the mathematics side fourier harmonic analysis partial differential equations etc etc, its great to see just how basic those equations they showed you at college really are.

Thing is Troll boy the whole time I have been at 420 I ALWAYS have avoided using technical language to people who don't understand it, its not my game to try and impress with technical language like you used in the contactor thread!

But for your information I was talking specifically to scribble who without doubt is more than intelligent enough to understand

Now trot on and find another bridge to troll

hahaha, "its not my game to try and impress with technical language" yeah, of course

Now now, chaps - you've both explained highly technical matters related to electricity and power etc. really well to lots of people including myself, which I am most grateful for, both today and in the past - I am only able to Google so much of the relevant data and understand it properly - ¿how about we keep it sweet that way, and let's not allow things to devolve into a degenerate cock-waving competition?

Tut tut tut

I thought power factor was real power divided by apparent power!

i.e ratio between real power and apparent power!

Reactive power over resistive power is opposite/ adjacent = tangent of the phase angle!!!

Surely you already knew that though?

Yeah agreed

I have better things to do with my saturday afternoon!, but you gtta feel sorry for some folk

UK cultivator's gone and school holidays aren't quite finished

what are you on about? it is the RATIO, between the currents and of course the power, but it's irrelevant, we are interested in current ratings, i never mentioned the specific relationship, and as shown by the simple calculation at the bottom, you divide the watts by pf to give VA, divide VA by V to give amps. nice and simple

From what I can gather, Scraglor was just simplifying the relationship between the two, so as not to have to then go into a complete sub-explanation of what exactly 'real power' and 'apparent power' actually are.

I think that most people on this site who have had dealings with HID ballasts/lights are likely to already be familiar with the concepts of resistive and inductive (reactive) power, and this makes his simplified explanation far easier to grasp for us and is still relevant and accurate enough to be able to help understand the issue.

I'm *positive* that Scraglor wouldn't answer test/qual questions in that manner, if the terms aren't strictly speaking *technically* correct.

tbh, if i took an exam now, i'd probably fail all that's needed for the real world nowadays are basic equations and an understanding mostly

yep, it is better i find to simplify things, unless someone asks for a more detailed explanation. and all of this discussion, altough it may be technically correct, doesn't consider the things which may in reality affect how many watts you can safely load a socket with.

the calculations are correct for perfect conditions, or for textbooks. but look at the average domestic installation, and there are a lot of other things to consider in the real world. like, standard of the installation. is it a fairly new, or decent quality socket, is it wired with good quality correctly sized conductors, not old perished VIR cables, is it protected by an rcd, mcb,rcbo, cartridge fuse, or maybe an old rewirable fuse with a bit of 1.0 mm copper in place of fuse wire. also rating of the protective device. is the socket even earthed, has some DIY'er connected the socket reverse polarity, are you using extension leads or adaptors, plugged into yet more extension leads and adaptors, when was it last tested? the main cutout fuse and cablehead, is it a modern series 7 type, or an ancient PILC cable with rewirable cutout? still very common in the uk, along with undersized mains tails. what else is plugged into other socket outlets on the same circuit?

all of these things are as important as the calculations, which should be based on measured values, nominal voltage (in the uk) is a bit varied these days, since the eu harmonisation which was supposed to improve things, you cant assume 230 volts when making calculations, although equipment manufacturers do just that.

Cheers guys.

Hey guys, I have a bit of a problem. I am running 1kW load through ballast > relay (mains/timer) in testing my newly wired ringmains sockets.

I created 4 gang ringmains that terminates at the fusebox to a 16A fuse. (4 double sockets) The fuse keeps tripping when i switch my relay on. I have 2kW on two relays (1kW each). Any ideas? 16A surely cant be too less?

It stays on when i power my intake and exhaust (2x 8" ruck fan/cf, 1x4" rvk intake)

Thanks in advance !! Cheers!

It sounds like the startup pulse from the 2x1KW ballasts/lamps is what's tripping your fuse - try starting them staggered, with 1 or 2 minutes in-between each one.

Spot on! It all started fine, however im down by one 600whps since it toasted itself :/

Cheers for that Scribble, your a star!

How you doing?