main circuit

[stev11]

hi

I Have 10 pcs of 20 Watt DC solenoid for each power supply, and each power supply has the following 230v, 6 A ,and the inrush current is 50A .
So
Breaker size =1.25*6 = 7.5A then the slandered breaker size will be 10A
Slandered breaker size = 10 A .
But I have 12 pcs of power supply with similar loads
So
main circuit breaker = 10*12 = 120A.

Slandered main circuit breaker = 125 A .
is this correct???? Also does the inrch current will make any for the main breaker???

 

[augie47]

Sounds like a ton of overkill...
If you have (10) 20 watt loads per power supply, you have a load of 200 watts.
Your power supply input is 6 amps @ 230v or 1380 watts... Taking into account
losses inherit with the power supply, it seems way over-sized.
I can not see the power supply drawing the full 6 amp load with only 200 watts of DC load.
Even if you wanted to look at it as being full load, 12 power supplies at 6 amps would be 72 amps and IF that load was continuous you would be looking at 90 amps.

In real life I would think your load would be more in the 30 amp range, but what do I know.

 

[iceworm]

hi

I Have 10 pcs of 20 Watt DC solenoid for each power supply, and each power supply has the following 230v, 6 A ,and the inrush current is 50A .
So
Breaker size =1.25*6 = 7.5A then the slandered breaker size will be 10A
Slandered breaker size = 10 A .
But I have 12 pcs of power supply with similar loads
So
main circuit breaker = 10*12 = 120A.

Slandered main circuit breaker = 125 A .
is this correct???? Also does the inrch current will make any for the main breaker???

I think you have it. 50A inrush on each 10A cb and 500A inrush on the 125A CB. Is that correct?

If you are curious, you may wish to get the Time Current Curves for 10A CB and the 125A CB. Lay out the inrush currents and expected duration. Hand sketch is fine. This way you will know the inrush will not trip the CBs.

ice

 

[stev11]

Sounds like a ton of overkill...
If you have (10) 20 watt loads per power supply, you have a load of 200 watts.
Your power supply input is 6 amps @ 230v or 1380 watts... Taking into account
losses inherit with the power supply, it seems way over-sized.
I can not see the power supply drawing the full 6 amp load with only 200 watts of DC load.
Even if you wanted to look at it as being full load, 12 power supplies at 6 amps would be 72 amps and IF that load was continuous you would be looking at 90 amps.

In real life I would think your load would be more in the 30 amp range, but what do I know.

Thank you, yes, its big but I have 0.65 power factor ,0.7 efficiency and 0.75 temp derating(@60C) ,so the final power will be =6*230*0.65*0.7*0.75 = 470 watt.

Also load is continuous but i need the calculation to acceptable to nec code???

 

[stev11]

I think you have it. 50A inrush on each 10A cb and 500A inrush on the 125A CB. Is that correct?

If you are curious, you may wish to get the Time Current Curves for 10A CB and the 125A CB. Lay out the inrush currents and expected duration. Hand sketch is fine. This way you will know the inrush will not trip the CBs.

ice

thank you I will look to curves but do you think breakers sizes are acceptable for nec code and real life???

 

[iceworm]

thank you I will look to curves but do you think breakers sizes are acceptable for nec code and real life???

Yes. However, for the first time you specify something like this, it is a really good to lay it out and verify for yourself. Excellent learning tool. You will know for ever exactly what you are dealing with.

And I certainly would not trust free internet engineering - expecially from me:roll:

ice

 

[stev11]

Yes. However, for the first time you specify something like this, it is a really good to lay it out and verify for yourself. Excellent learning tool. You will know for ever exactly what you are dealing with.

And I certainly would not trust free internet engineering - expecially from me:roll:

ice

ok, I will check again.