BS 1361 main fuse short circuit capacity/overload at ka ratings explained.

[Joe9876543210]

Hi.
I am an apprentice electrician, and was wondering if anyone could help me as I'm getting confused with something that's probably quite simple.

If I am correct to work out the amps running though an appliance you do w/voltage. So 10kw/240v = approx 42amps. Of I'm correct in this?

My boss recently at work was trying to explain that a bs1361 100amp fuse has a short circuit capacity of about 50ka. What does This mean of it's already 100amps anyway?

In a flat there was a bs1361 60amp, but I presumed you would need higher, more like a 100amp as collectively the property had two 10.5kw showers, an electric cooker, a kitchen ring with dryer, fridge, freezer and also a 9.5kw water heater so collectively that equates to roughly 35kw if every appliance is used at once? But then surely 35000/240 = 145amps? On a 60amp fuse. Something isn't right here.

I'm very confused... Please can someone explain to me in a simple and understandable format how it works? It's really playing on my mind.

thank you.

 

[Simonslimline]

Not all the loads will be on at the same time meaning you will never reach anywhere near your figure. Read up about diversity.

Again with overload, the main fuse will happily pull more than its rated 60amps. To figure out how much and for how long you need to look at the Time current Graphs in BS7671.

Again the 50KA is the max fault current that the fuse can interrupt safely, eg ... without exploding

 

[NickD]

If I am correct to work out the amps running though an appliance you do w/voltage. So 10kw/240v = approx 42amps. Of I'm correct in this?

Pretty much. However these days because of harmonisation with Europe we work off 230V (single phase) so you'd calculate 10kW/230V. Obviously it doesn't make a huge difference to the calculation but some lecturer or examiner would pick you up on it if you work off 240V

Yes power = volts x current (and therefore current = power / volts) is correct, but be aware this is is only totally accurate for a purely resistive load like a resistor or simple heater element.

[At risk of overcomplicating (and if this blows your mind then don't worry about it yet), when the load contains significant inductance (for example a motor) or capacitance, the sine wave of the current waveform gets advanced/delayed a bit in time compared to the voltage sine wave - if you draw or imagine it, the peaks/troughs of the voltage sine wave no longer exactly line up with the peaks/troughs of the current sine wave. When that happens, power consumption can be worked out as volts x current x power factor, where power factor is a value between 0 and 1. As I say, if this blows your mind then don't worry about it. Just trying to give you a complete answer.]

 

[mechelec]

Pretty much. However these days because of harmonisation with Europe we work off 230V (single phase) so you'd calculate 10kW/230V. Obviously it doesn't make a huge difference to the calculation but some lecturer or examiner would pick you up on it if you work off 240V

Yes power = volts x current (and therefore current = power / volts) is correct, but be aware this is is only totally accurate for a purely resistive load like a resistor or simple heater element.

[At risk of overcomplicating (and if this blows your mind then don't worry about it yet), when the load contains significant inductance (for example a motor) or capacitance, the sine wave of the current waveform gets advanced/delayed a bit in time compared to the voltage sine wave - if you draw or imagine it, the peaks/troughs of the voltage sine wave no longer exactly line up with the peaks/troughs of the current sine wave. When that happens, power consumption can be worked out as volts x current x power factor, where power factor is a value between 0 and 1. As I say, if this blows your mind then don't worry about it. Just trying to give you a complete answer.]

You are right the declared voltage is 230V +10% -6% and that should be used in calculations. In reality it is nearly always around 240V. This is unlikely to change in the near future.

 

[Joe9876543210]

Simon, thank you for the response.. makes things quite clear.
Nick D. That is quite complicated, however I'm half way in my 2nd year now so I should probably start to be understanding it by now.
I have a copy of amendment 2 bs7671 17th so I will be looking this up to understand more.
thanks for your help guys really appreciate it x

 

[PEG]

I have given you a like for three reasons. Your a fellow Mancunian,your spelling and grammar are good,but mostly for that little kiss at the end....you made me feel special :saddam:

 

[telectrix]

I have given you a like for three reasons. Your a fellow Mancunian,your spelling and grammar are good,but mostly for that little kiss at the end....you made me feel special :saddam:

a manc. special? special needs, more like LOL. :21:

sent from the Kop End.

 

[Joe9876543210]

I have given you a like for three reasons. Your a fellow Mancunian,your spelling and grammar are good,but mostly for that little kiss at the end....you made me feel special :saddam:

Ha ha!
No problem.. As long as your wearing red socks & definetely not blue !

 

[11kstanton]

Pretty much. However these days because of harmonisation with Europe we work off 230V (single phase) so you'd calculate 10kW/230V. Obviously it doesn't make a huge difference to the calculation but some lecturer or examiner would pick you up on it if you work off 240V

Yes power = volts x current (and therefore current = power / volts) is correct, but be aware this is is only totally accurate for a purely resistive load like a resistor or simple heater element.

[At risk of overcomplicating (and if this blows your mind then don't worry about it yet), when the load contains significant inductance (for example a motor) or capacitance, the sine wave of the current waveform gets advanced/delayed a bit in time compared to the voltage sine wave - if you draw or imagine it, the peaks/troughs of the voltage sine wave no longer exactly line up with the peaks/troughs of the current sine wave. When that happens, power consumption can be worked out as volts x current x power factor, where power factor is a value between 0 and 1. As I say, if this blows your mind then don't worry about it. Just trying to give you a complete answer.]

Mate love that answer as its not an easy one... what about the situation we are facing now though in this day and age.. 3 level house, every child wants to use the 9kw shower at the same time potentially... on seperate circuits set up correctly at average temperatures, plus mums washings on.. of course 30 degrees eco nothing to worry about but shes kaneing the dryer for the bedsheets in the utility... youve then got the average joe having to plug in to this charger and that charger.. my toddlers new word is charge when one of our devices isnt working.. this is the world we are living in

 

[11kstanton]

Mate love that answer as its not an easy one... what about the situation we are facing now though in this day and age.. 3 level house, every child wants to use the 9kw shower at the same time potentially... on seperate circuits set up correctly at average temperatures, plus mums washings on.. of course 30 degrees eco nothing to worry about but shes kaneing the dryer for the bedsheets in the utility... youve then got the average joe having to plug in to this charger and that charger.. my toddlers new word is charge when one of our devices isnt working.. this is the world we are living in

luckily we have transformers and LeDs.. We need to get to smart electric showers and cookers

 

[11kstanton]

Pretty much. However these days because of harmonisation with Europe we work off 230V (single phase) so you'd calculate 10kW/230V. Obviously it doesn't make a huge difference to the calculation but some lecturer or examiner would pick you up on it if you work off 240V

Yes power = volts x current (and therefore current = power / volts) is correct, but be aware this is is only totally accurate for a purely resistive load like a resistor or simple heater element.

[At risk of overcomplicating (and if this blows your mind then don't worry about it yet), when the load contains significant inductance (for example a motor) or capacitance, the sine wave of the current waveform gets advanced/delayed a bit in time compared to the voltage sine wave - if you draw or imagine it, the peaks/troughs of the voltage sine wave no longer exactly line up with the peaks/troughs of the current sine wave. When that happens, power consumption can be worked out as volts x current x power factor, where power factor is a value between 0 and 1. As I say, if this blows your mind then don't worry about it. Just trying to give you a complete answer.]

imagine if you could get to an efficient power source that produces energy and sustains longevity, its not going to happen though anytime soon in our lifetime as its not going to make any shareholder money