Copper pipe, 1/2-inch. What is max current capacity?

Ample

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Inspired by one of Will's videos, I hammered flat some 1/2-inch copper pipe (the standard hard stuff at a home building supply store (not the curly soft stuff)) to use as bus bars.

How much current will this crude bus bar be able to handle?

So far, I have been careful to keep my loads to be at 100A or lower because that was the max current that my single 100Ah battery could provide.

Recently, I've upgraded to 3 batteries and thus theoretically 300A max current. I don't think I will ever need to use more than 2400W (200A). If this homemade bar can't handle 200A, could I stack 2 (or more) together to increase the amount of copper and thus increase capacity that way?

I am using a small sledgehammer and an anvil of sorts and thus the bars are very flat. There is no issue getting good contact with standard cable lugs.
 
Short Answer: ~120 Amps if the resistivity of type M hard pipe were the same as copper. I suggest slowly ramping up the current on your system 10 amps at a time and watch the busbar temperature to confirm. Let us know what you find.

Long answer:
It would be easier to just measure it with an Ohm meter.... but...I just worked the first part of the math for wire this morning for @ianganderton here.
It depends on the thickness of the copper pipe and the hardness (which affects resistivity). There's 4 types of "hard" copper I know of (e.g., K, L/ACR, M, and DWV).

Example math:
1/2 Pipe, type M: 120mm long, 15.8 mm outer diameter, .7 mm thick wall, resistivity 1.7e-8 Ω m (?)
If you pound the pipe flat, it would be 1.4 mm high and width = π x d/2, so 24.8 mm.

Ω = 1.7e-8 Ω m L/A​
A = 1.4 x 24.8 = 34.27 mm^2 = 3.427e-5 m^2​
Ω = 1.7e-8 Ω m 0.12m / 3.427e-5 m^2 = .059 mΩ​

Joules Law: H = I^2 x R (ignoring T for DC)

The temperature of the busbar will rise until the heat leaving equals the heat generated. How much heat leaving the busbar depends on the shape, ambient conditions, airflow, etc.

I was about to attempt to figure this out when I realized probably only @ianganderton wanted to know that much detail and I'd spoil his fun by just spelling it out (that and I don't know ;-).

Fortunately there's a shortcut. Knowing the area is 34.27 mm^2, we can calculate the diameter as if it were wire since A = π r^2, so d = 2r = 6.6mm or ~ gauge 3 (rounded down). From this ref, the max amps for AWG 3 in air is 153 amps. I know we already rounded down once, but since it's not pure copper let's take off 20% for safety... call it 120 amps?

It'll be interesting to see what those who have a lot of experience with busbars think of this.... @fhorst, @Steve_S and @Haugen (I know there are more, these are just the ones that popped into my head) -- linking you all so you can put some practice to this theory.
 
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One way to check connections. I use a multi-meter and measure voltage across the connection while on a medium to heavy load. It is a quick way to find a bad connection. Measuring voltage drop.
 
I imagine that people smash or hammer the copper into inconsistent thicknesses. That also effects the contacts and how well the bolts hold.
I've got a bunch of bus bars to make. I'm thinking copper bar is probably worth the extra cost. Especially considering that it will be easier to stack them in a vice to drill the holes. At least that's what I'm thinking now. I have some time before I have to make them.
 
I have 4 separate copper pipe busbars on a midsized 24v system with a 3000W multiplus. I periodically check for heat and my busbars have never even gotten warm. That said, I've never loaded my system with more than about 90A and peak average is more like 50A.
 
Monitoring for heat is an excellent way to discern if your conductors are adequate. (just double them up if you see un-acceptable temp rise) That said, I have seen copper tubing conduct as much as 25Kw of energy in AM antenna systems. We always looked for hot spots - Impedance = energy wasted in heat.
 

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By the way, I'm planning to eventually invest in a Flir heat monitor. I think that's going to be the best way to quickly check for something bad happening in the system.
You can't measure temp of shinny objects like chrome plated buss bars....put take or paint them. You can search emissivity or read link below. Those welding advertisements where they are using a ir temp gun are BS....

 
I get a 12c temperature rise at 250amps 48v through 1/2 refrigeration soft copper pipe, at 120amps the copper is just a bit warm . Bluesea buss bars are rated up to a 40c rise in temperature.
 
I think 1/2" copper pipe is similar to 2AWG copper wire. But it does eliminate the connective lugs - that helps.
 
You can't measure temp of shinny objects like chrome plated buss bars....put take or paint them. You can search emissivity or read link below. Those welding advertisements where they are using a ir temp gun are BS....

I’d be looking at terminal connections, not at the bus bar itself.
 
Which.. has anyone put cable into the pipe then flattened one end as a lug?

Interesting idea if one has a lot of undersized pipe. But the pipe then needs to be covered - unlike most wire. It would be almost twice as much work.
 
Which.. has anyone put cable into the pipe then flattened one end as a lug?

Something like this?

Also, to get the ends flat, you don't need to get the entire thing flat - just the part that makes contact with the terminal. For that, one trick I sometimes use is to slightly clamp a washer with a slightly larger diameter than the terminal with a vise on the copper. Also, use annealed copper (the curly soft stuff as @Ample calls it). It has better conductivity and is much easier to work with.
 
You can't measure temp of shinny objects like chrome plated buss bars....put take or paint them. You can search emissivity or read link below. Those welding advertisements where they are using a ir temp gun are BS....


Absolutely correct! Here is in fact an example: I have 8 resistors, 6 of them have current flowing, the two in the center do not. They are bolted to a shiny 4 mm thick aluminum plate. The resistors body is aluminum that has been anodized to a gold mate color (but clearly radiates well in the infrared). Here is my FLIR photos of the side of the resistors. They appear to be at ~ 93 C.The ones with no current have basically the same temperature as the ones with thanks to the good conductivity of the Al plate. The Al plate APPEARS to be colder (38 C, but obviously isn't). On the other side of the Al plate I added black insulating tape to some spots. The spots with the tape appear to be at 98 C, the area without tape at 39 C (which again, it is not correct). The visible photos show the setup.
 

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Short Answer: ~120 Amps if the resistivity of type M hard pipe were the same as copper. I suggest slowly ramping up the current on your system 10 amps at a time and watch the busbar temperature to confirm. Let us know what you find.

Long answer:
It would be easier to just measure it with an Ohm meter.... but...I just worked the first part of the math for wire this morning for @ianganderton here.
It depends on the thickness of the copper pipe and the hardness (which affects resistivity). There's 4 types of "hard" copper I know of (e.g., K, L/ACR, M, and DWV).

Example math:
1/2 Pipe, type M: 120mm long, 15.8 mm outer diameter, .7 mm thick wall, resistivity 1.7e-8 Ω m (?)
If you pound the pipe flat, it would be 1.4 mm high and width = π x d/2, so 24.8 mm.

Ω = 1.7e-8 Ω m L/A​
A = 1.4 x 24.8 = 34.27 mm^2 = 3.427e-5 m^2​
Ω = 1.7e-8 Ω m 0.12m / 3.427e-5 m^2 = .059 mΩ​

Joules Law: H = I^2 x R (ignoring T for DC)

The temperature of the busbar will rise until the heat leaving equals the heat generated. How much heat leaving the busbar depends on the shape, ambient conditions, airflow, etc.

I was about to attempt to figure this out when I realized probably only @ianganderton wanted to know that much detail and I'd spoil his fun by just spelling it out (that and I don't know ;-).

Fortunately there's a shortcut. Knowing the area is 34.27 mm^2, we can calculate the diameter as if it were wire since A = π r^2, so d = 2r = 6.6mm or ~ gauge 3 (rounded down). From this ref, the max amps for AWG 3 in air is 153 amps. I know we already rounded down once, but since it's not pure copper let's take off 20% for safety... call it 120 amps?

It'll be interesting to see what those who have a lot of experience with busbars think of this.... @fhorst, @Steve_S and @Haugen (I know there are more, these are just the ones that popped into my head) -- linking you all so you can put some practice to this theory.

Thanks, @svetz I really appreciate you doing the math! (I'm getting different results though; see below)

Firstly, I didn't know that there were different types of 1/2-inch copper pipe. I just walked into the big orange store and bought one of the precut lengths they had in the bin.

Without going back to the store, I believe (from a quick web search) that it's: 1/2 inch x 3 ft. Copper Type L Hard Temper Straight Pipe. In other words, Type "L".

I believe "L" is thicker than "M", from: https://en.wikipedia.org/wiki/Copper_tubing#Sizes

As far as purity of copper, I believe nearly all copper pipe sold in N. America, from ASTM B 42 - Standard Specification for Seamless Copper Pipe, Standard Sizes, would be at least 99.9% pure copper.

So hopefully we can remove the 20% safety margin.

Here's my calculation. TL;DR is that 1/2-inch Type L cable is a slightly better conductor than AWG 1.

===========

From the Wikipedia article, Type L has an Outside diameter of 15.875mm and an Inside Diameter of 13.843.

Following in your footsteps for doing the math, the area of this OD would be:
A = π r^2 = 3.14159 * (15.875 / 2)^2 = 3.14159 * 63.0 = 197.9 sq. mm

Similarly, the area of the Inside Diameter would be 3.14159 * (13.843 / 2)^2 = 3.14159*47.9 = 150.5 sq. mm

Subtracting the area of the inside from that of the outside should give us the cross-sectional area of the copper itself.
197.9 - 150.5 = 47.4 sq mm.

Let'd double-check using a different method since I'm old and it's been decades since I've done any math:

Using good old "Length x Width". Let's take length as circumference and width as the thickness of the copper.

I'm going to calculate an Avg circumference to try to get a "better" circumference of the copper pipe for calculating the cross-sectional area. Hopefully more accurate than using just the OD or ID.

Circumference OD = 2 pi R = 2 * 3.14159 * (15.875 / 2) = 49.87mm
Circumference ID = 2 pi R = 2 * 3.14159 * (13.843 / 2) = 43.49mm

Avg circumference = (49.87 + 43.49) / 2 = 93.359/2 = 46.68mm

What is the thickness of the copper in 1/2-inch Type L copper pipe?

OD = 15.874 mm
ID = 13.843 mm

The difference between OD and ID should be 2 x thickness.
Therefore Thickness = (15.874 - 13.843) / 2 = 2.03 / 2 = 1.016mm.

Area = Thickness x circumference = 1.016 x 46.68 = 47.40 sq. mm

Yay, the 2 methods agree!

If this area (47.40 sq mm) were a circle, what would its diameter be?

D = 2 x r = 2 x sqrt(A / pi) = 2 x sqrt (47.40 / 3.14159) = 2 x sqrt(15.09) = 2 x 3.884 = 7.77mm
D = 7.77mm

That is, I propose that 1/2-inch Type L copper pipe is similar to a copper rod of diameter 7.77mm.

Using the AWG wire chart referenced by Svetz, the closest-without-going-over wire gauge is AWG 1 at 7.348mm, with a Max current in air of 211A.

I therefore submit, with a sigh of relief--because I can use the bus bars I had already made--that my crude banged-out-of-copper-pipe bus bar is better than an AWG 1 cable and can thus safely handle my need to conduct 200A.

Since this has been a long comment already, I'll go even further and show the "real-world" situation.

I have a spare piece of the raw hammered-out pipe. And I using a micrometer, the width of the bar is just under 24mm and the thickness is just over 2mm. The approximate cross-sectional area is thus about 24 x 2 = 48mm, pretty close to the theoretical 47.4mm. I say "approximate because it's not perfectly flat in any direction. This piece was from an early attempt. The ones I've ended up using were done with a small 3lb sledgehammer onto a kind of an anvil and those are quite flat on one side (and I use that flat side to contact the lugs of cables).

Bus bar from 1-inch copper pipe, width.jpgBus bar from 1-inch copper pipe, thickness.jpg

If anyone sees any problems, let me know. I don't want to start a fire. However, I doubt I'll ever get near 200A.

The biggest consumer of power in my rig is my induction burner at 1800W on max (I intentionally never run it at max). I'd have to do something silly like run a toaster and the induction burner at the same time.
 
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