Is 4x50 busbar inside electrical panel is over-enough for 100a ?, @220v, with artificial ventilation, no heat-shrink, ambient temperature is 35 degree-celcius

 

Is 4x50 busbar inside electrical panel is over-enough for 100a ?, @220v, with artificial ventilation, no heat-shrink, ambient temperature is 35 degree-celcius

Artificial Ventilation is 12v DC Fan inside panel

 

Current carrying capacity? You do not mention Copper, Aluminum or Brass material? The link will point to the applicable standards.

Ron

 

Current carrying capacity? You do not mention Copper, Aluminum or Brass material? The link will point to the applicable standards.

Ron

Pure Copper

 

Also, is the "4x50" dimension in millimeters or centimeters or micro-furlongs?

 

Also, is the "4x50" dimension in millimeters or centimeters or micro-furlongs?

4mm x 50mm, please don't answer question if you can't answer, I see you has been reply many question with question, stop it

 

If you provided ALL the relevant information in your original question members would not need to ask so many questions.
So the logical conclusion is that ask a question in future without all the required information you want no one to reply.

Les.

 

If you provided ALL the relevant information in your original question members would not need to ask so many questions.
So the logical conclusion is that ask a question in future without all the required information you want no one to reply.

Les.

Les is correct! In addition, without enough information, any answer is only a guess, based on assumptions. Since most of the responders want to provide useful answers, we often seek additional information.
4x50 mm =200 square mm, and I am not sure what wire size that equates to.

 

If you provided ALL the relevant information in your original question members would not need to ask so many questions.
So the logical conclusion is that ask a question in future without all the required information you want no one to reply.

Les.

@MisterBill2 Problem is you have answer many question with answer; you can think 4 x 50 micro meter is unusual, also you can think is 4 x 50 centimeter is also not usual, you can imagine if 4 x 50 micro meter is for 100a, also you can answer for each possibility nor very usual possibility

 

Normally I do not consider metric dimensions relative to power circuits. So please excuse my asking for units when the presumption is that you were asking for a correct answer instead of a guess based on assumptions.
There are some individuals in my acquaintance who often speak of dimensions in "mills", which actually is thousandths of an inch. When talking with them I know what they mean, but not everybody knows. Specifying units is not that difficult and it reduces the probability of incorrect understandings.

 

Everyone has the right idea here, except telling the TS what they are doing.
Need to now:
Material
Cross-sectional area.

I'd agree, use wire tables and find the ampacity of the same cross-sectional area. We know your in Japan, but there are still other issues to worry about.

In the US, you would go for 1.20x of what you need. In other words, it can carry 80% of the load continuously.

Not too sure if the size and number of bolts.

If I was really worried, I'd assume that there are no bolts and look at the cross sectional area of that, not of the bus bar.

You can go back to fundamentals where R=ρL/A where ρ is resistivity, L = length and A= cross-sectional area as if there were holes. You might want to look at a 3% drop, but then it depends on where the feed is located. Ends or center? Now add temperature dependence. Someof the formulas have a 2L term. That might matter too. Resistivity is a materiil property with a temperature dependence.

 

Here is a good site for busbar calculations,
According to your dimensions it will take 240A.

https://www.electrical4u.net/calculator/busbar-current-calculator-online/

 

Everyone has the right idea here, except telling the TS what they are doing.
Need to now:
Material
Cross-sectional area.

I'd agree, use wire tables and find the ampacity of the same cross-sectional area. We know your in Japan, but there are still other issues to worry about.

In the US, you would go for 1.20x of what you need. In other words, it can carry 80% of the load continuously.

Not too sure if the size and number of bolts.

If I was really worried, I'd assume that there are no bolts and look at the cross sectional area of that, not of the bus bar.

You can go back to fundamentals where R=ρL/A where ρ is resistivity, L = length and A= cross-sectional area as if there were holes. You might want to look at a 3% drop, but then it depends on where the feed is located. Ends or center? Now add temperature dependence. Someof the formulas have a 2L term. That might matter too. Resistivity is a materiil property with a temperature dependence.

The cros sectional area is easy: 50x4=200 mm**2. At 240 amps that would be 2.4 amps per square mm, so that does seem like a reasonable number.

 

There are some individuals in my acquaintance who often speak of dimensions in "mills", which actually is thousandths of an inch.

IMO That is a most confusing expression, why can't it simply be referred to a thou's?
Especially when you have millimeters in the mix.
It took me a while to get used to it in N.A.
Fortunately the UK dropped it in 1960

 

This is the link I left out of post #3. Darn, I hate it when I do that.

Ron

 

This is the link I left out of post #3. Darn, I hate it when I do that.

Ron

If it's 500a so... no problem, I need's 200-500% than continuous load, for further expansion

 

Everyone has the right idea here, except telling the TS what they are doing.
Need to now:
Material
Cross-sectional area.

I'd agree, use wire tables and find the ampacity of the same cross-sectional area. We know your in Japan, but there are still other issues to worry about.

In the US, you would go for 1.20x of what you need. In other words, it can carry 80% of the load continuously.

Not too sure if the size and number of bolts.

If I was really worried, I'd assume that there are no bolts and look at the cross sectional area of that, not of the bus bar.

You can go back to fundamentals where R=ρL/A where ρ is resistivity, L = length and A= cross-sectional area as if there were holes. You might want to look at a 3% drop, but then it depends on where the feed is located. Ends or center? Now add temperature dependence. Someof the formulas have a 2L term. That might matter too. Resistivity is a materiil property with a temperature dependence.

Why not 200% ?, I am against 125% rules, we use 200%-500%, but project's were we works is need's highly available to expasion, so... 125% rules is forbidden

 

Most cases the 10 A/mm2 is current density causing the heat output to markable but acceptable degree. 5A/mm2 is better value for badly cooled obstacles or if ultra-bad like in the middle of winding for transformers only 3...3.5, but in excessively well ventilated areas like hanging wires without insulation on the poles up to the 20 A/mm2. Thus, the busbar 4x50=200 mm2 and Your load makes 0.5 A/mm2 what is shameless small load.

 

Why not 200% ?, I am against 125% rules, we use 200%-500%, but project's were we works is need's highly available to expasion, so... 125% rules is forbidden

You did ask what would work. You would never melt copper, but voltage drop is what your looking for. Standard is <3% in the US. There's temperature effects.

The buss bar really is a piece of swiss cheese. The cross-sectional area really should not include the holes.

Resistivity is a material property.
Resistivity is temperature dependent.
Hole size is unknown.
Where along the bus bar you bring in power matters too. Using the center would be different than an edge, Your bar is 1/2 the circuit. because a bar, hopefully is used in the return path, so you need to consider both of them. One drops 3%/2 and the other 3%/2 maximum. Now 3% for the bar's contribution, isn't right either because it;s 3% total drop from the distribution. So, pick a number. Is it 1%, 0.5%. Maybe a good number is < 1% of the expected feed current considering a reasonable length from the supply.

The fundamental equations are R=ρL/A and V=IR; ρ is in units of ohm-cm or ohm-length

BUT, There are tables that use the wire gauge which relates directly to the cross-sectional area. The thermodynamics equations I'm not familiar with. Your not using the bar as a heating element.

So, a 1cm width bar, 2cm wide with a 5mm hole is effectively 1.5cm wide, not 2cm.

Confirm everything by testing.

 

At one time heating of the conductor wa the main concern, because rubber insulation and paper insulation can burn. Open bars in a steel enclosure with ceramic insulator material can live with the heat.