Selection of branch breaker

[tahir mehmood]

I am confused about the oven mentioned in the attachment. how i have to select a breaker as per NEC.

Oven Specifications:


Voltage: 380V 2N – 60Hz220V – 60 Hz
kW =6,4kW
kPa 100-200 (1.0-2.0 bar)

 

[kwired]

Looks like it is rated for 6kW @ 380 volts or 4 kW @ 220 volts.

Which voltage are you connecting it to?

 

[tahir mehmood]

Oven Name Plate

Oven Name Plate

check the oven name plate

 

[kwired]

check the oven name plate

I am still sticking with what I said earlier - it is intended to be connected to 220 volts or 380 volts. At 220 volts it will draw 4 kW, at 380 volts it will draw 6 kW. My question was which voltage will you be utilizing? You need to know that before you will know which power rating to use to find circuit amps.

 

[retirede]

I am still sticking with what I said earlier - it is intended to be connected to 220 volts or 380 volts. At 220 volts it will draw 4 kW, at 380 volts it will draw 6 kW. My question was which voltage will you be utilizing? You need to know that before you will know which power rating to use to find circuit amps.

Be careful, in Europe a comma is a decimal point. It could be 6.4 KW.
And the ratios 380/220 vs 6/4 don't really jive, either. If it draws 4@220, I'd expect closer to 7@380.
But I struggle to visualize how a combination of elements would be connectable for the same KW at both of those voltages. :?
A connection diagram would be helpful - if the connection is identical for both voltages, then you are right! :thumbsup:

 

[retirede]

Found this in a manual online:



https://www.angelcommercial.co.uk/wp-content/uploads/2015/05/Instruction_manual_EKF_411_UD.pdf

 

[MAC702]

Be careful, in Europe a comma is a decimal point. It could be 6.4 KW. ...

This.

According to that diagram, it is also making the other-than-America assumption that we have 230V-to-neutral. I don't think it should matter for this application. Land one hot on the "N" and split the other between the "L1" and "L2."

 

[Ingenieur]

380/sqrt3 = 220
same v either way
1 ph 220 l-n, 1 ph is paralleled to L1 and L2
3 ph 380/sqrt3 x 2 phase-neut, one to each L1 and L2
15 Ohm elements x 2 (I looked at the manual earlier)

220/15= 15 a per element
if 1 ph 35 a 1 pole
if 3 ph 20 a 2 pole

 

[Ingenieur]

conductor ampacity
1 ph line at least 35 a, same for neut
3 ph lines at least 20 a, neut vector sum of 2 x 15 a phases x 1.25???

edit 3 ph neut same as a line

 

[kwired]

Be careful, in Europe a comma is a decimal point. It could be 6.4 KW.
And the ratios 380/220 vs 6/4 don't really jive, either. If it draws 4@220, I'd expect closer to 7@380.
But I struggle to visualize how a combination of elements would be connectable for the same KW at both of those voltages. :?
A connection diagram would be helpful - if the connection is identical for both voltages, then you are right! :thumbsup:

Connection diagram submitted did help a lot. I was questioning myself what that "2N" actually meant - but now I see it.

 

[tahir mehmood]

calculation for oven

calculation for oven

380/sqrt3 = 220
same v either way
1 ph 220 l-n, 1 ph is paralleled to L1 and L2
3 ph 380/sqrt3 x 2 phase-neut, one to each L1 and L2
15 Ohm elements x 2 (I looked at the manual earlier)

220/15= 15 a per element
if 1 ph 35 a 1 pole
if 3 ph 20 a 2 pole

Note: since the power is 6.4 KW, so for single phase the current will be I= 6400/220 = 29 Amps L-N, 1 Ph is paralleled to L1 and L2

For three phase : 6400/1.73 x 380 = 10 amperes

but You multiplied by a factor 2, is this because of two neutral lines?

 

[Ingenieur]

Note: since the power is 6.4 KW, so for single phase the current will be I= 6400/220 = 29 Amps L-N, 1 Ph is paralleled to L1 and L2

For three phase : 6400/1.73 x 380 = 10 amperes

but You multiplied by a factor 2, is this because of two neutral lines?

you are using only 2 legs out of a 3 phase source, each ref to neut thru the htg elements
so each element sees 380/1.73 / 15 ohm per leg ~ 15 a
hence a 2 poie 20 a
the source 220/380/3 wye

 

[retirede]

Note: since the power is 6.4 KW, so for single phase the current will be I= 6400/220 = 29 Amps L-N, 1 Ph is paralleled to L1 and L2

For three phase : 6400/1.73 x 380 = 10 amperes

but You multiplied by a factor 2, is this because of two neutral lines?

Another way to look at it is that you have 2 elements. In either configuration, they each have 220 volts applied to them. 220 applied to 15 ohms gives 14.7 A. Withe the 380 connection, that's your load. The 220 connection parallels them so the current is 2 X 14.7.

 

[kwired]

Note: since the power is 6.4 KW, so for single phase the current will be I= 6400/220 = 29 Amps L-N, 1 Ph is paralleled to L1 and L2

For three phase : 6400/1.73 x 380 = 10 amperes

but You multiplied by a factor 2, is this because of two neutral lines?

Not that other reply was wrong but another way to look at it -

you are not connected to all three phases, if you were and 6400 was your total kVA, then that would be correct.

What you have is two 3200 watt elements that operate at 220 volts each. Connect them to 380 volt wye with two phases and a neutral and they still see 220 volts across each element because of the phase angle. Current will be 3200/220 = 14.55 amps on each phase conductor, because of the 120 degree phase angle the neutral will also carry approximately same current level.

 

[topgone]

Not that other reply was wrong but another way to look at it -

you are not connected to all three phases, if you were and 6400 was your total kVA, then that would be correct.

What you have is two 3200 watt elements that operate at 220 volts each. Connect them to 380 volt wye with two phases and a neutral and they still see 220 volts across each element because of the phase angle. Current will be 3200/220 = 14.55 amps on each phase conductor, because of the 120 degree phase angle the neutral will also carry approximately same current level.

Well said!