65A on 8AWG 1.5m length (hot?)

[rg12]

65A CONSTANT current for about 20 minutes, could see 125A for 10-15 second bursts.
Will 8AWG be enough?
* Length is 1.5 meters

The pack will have 2 positive and 2 negative wires that are 8AWG giving 130A constant for 20 minutes with 250A bursts for 10-15 seconds.
Nickels are 6.75mm wide, 0.2mm thick (32P) so they will see about 4A constant and 8A bursts for 10-15 seconds, all connected to a 10mm wide, 1.5mm thick copper bus bar (the length of the whole 32P pack having each cell connected to it one next to another on a 65cm wide pack (32 cells width with cell holders).
The two 8AWG wires will be soldered in two different spots on each busbar for even current distribution.
The busbar mm2 is equal to about 5.5AWG wire but doesnt have much length.

 

[john61ct]

Blue Sea's Circuit Wizard.

 

[flippy]

what voltage are you running that requires such amps? who not increase the voltage so you can lower the amps?

 

[rg12]

what voltage are you running that requires such amps? who not increase the voltage so you can lower the amps?

48V, must be 48V due to regulations and stuff beyond my control for this project

 

[john61ct]

Bit of trivia - been looking at rectifiers in exactly that ballpark, about what's required to get full power output from an iCharger 4010-Duo.

 

[flangefrog]

You've got two things to consider here. One is the voltage drop at the load. This will be based on wire resistance (size) * length. Even a 16AWG wire will probably have allowable voltage drop if it's only a few mm long. But the longer the wire, the bigger it will need to be.

The other is the heat output. Generally you will just look at the temperature of the wire, but it is sometimes needed to look at the total heat output (dissipated power) over the whole length.

Once you know your allowable voltage drop, you can calculate the wire resistance needed via ohms law. If you want to cheat you can use this calculator: http://www.ohmslawcalculator.com/ohms-law-calculator. Enter the max current and allowable voltage drop at that current and you can see the max wire resistance and total power/heat output.

Here are a couple of links to find the resistance of a certain size of wire. If the unit is something like ohms/m you will need to multiply it by the length of your wire.
http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/wirega.html
https://chemandy.com/calculators/round-wire-resistance-calculator.htm

Calculating the temperature of the wire is a lot harder. The bus bars may help conduct heat away from the wires and you also need to consider the thermal resistance of different sized wires. I would recommend you look at a wire ampacity chart. This is a chart that shows the recommended max amps for various wire sizes for a specific insulation type, bundle configuration, ambient temp and environment.

Note that these are often geared towards wires installed in houses and thus are very conservative. Silicone wire will increase the maximum conductor temperature allowed and if this is an ebike application, there may be air flow to help draw heat away.

Here is one such ampacity chart you can use: https://www.allaboutcircuits.com/textbook/direct-current/chpt-12/conductor-ampacity/

You will need to calculate both the minimum wire size for allowable voltage drop, and the minimum wire size for the allowable temperature, and then use the biggest of the two in your application.

If this is not a safety critical application I would say that 8AWG is more than enough, especially if it has silicone insulation. It might be the solder joints that require more attention. Generally a well crimped ring terminal like used in house wiring would be much better and safer.

 

[flippy]

what voltage are you running that requires such amps? who not increase the voltage so you can lower the amps?

48V, must be 48V due to regulations and stuff beyond my control for this project

i would double the 8awg cables. that should make it easyer to manage then a single 4awg.
also use crimp connectors, do not solder this stuff. with these kinds of amps it will get hot.

 

[fechter]

8ga will be fine. I doubt it will even get warm if it has any kind of air movement around it.

 

[spinningmagnets]

48V and 65A equals 3100W, and that pack should be able to power a 120V AC inverter during a power outage...just a thought.

 

[rg12]

You've got two things to consider here. One is the voltage drop at the load. This will be based on wire resistance (size) * length. Even a 16AWG wire will probably have allowable voltage drop if it's only a few mm long. But the longer the wire, the bigger it will need to be.

The other is the heat output. Generally you will just look at the temperature of the wire, but it is sometimes needed to look at the total heat output (dissipated power) over the whole length.

Once you know your allowable voltage drop, you can calculate the wire resistance needed via ohms law. If you want to cheat you can use this calculator: http://www.ohmslawcalculator.com/ohms-law-calculator. Enter the max current and allowable voltage drop at that current and you can see the max wire resistance and total power/heat output.

Here are a couple of links to find the resistance of a certain size of wire. If the unit is something like ohms/m you will need to multiply it by the length of your wire.
http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/wirega.html
https://chemandy.com/calculators/round-wire-resistance-calculator.htm

Calculating the temperature of the wire is a lot harder. The bus bars may help conduct heat away from the wires and you also need to consider the thermal resistance of different sized wires. I would recommend you look at a wire ampacity chart. This is a chart that shows the recommended max amps for various wire sizes for a specific insulation type, bundle configuration, ambient temp and environment.

Note that these are often geared towards wires installed in houses and thus are very conservative. Silicone wire will increase the maximum conductor temperature allowed and if this is an ebike application, there may be air flow to help draw heat away.

Here is one such ampacity chart you can use: https://www.allaboutcircuits.com/textbook/direct-current/chpt-12/conductor-ampacity/

You will need to calculate both the minimum wire size for allowable voltage drop, and the minimum wire size for the allowable temperature, and then use the biggest of the two in your application.

If this is not a safety critical application I would say that 8AWG is more than enough, especially if it has silicone insulation. It might be the solder joints that require more attention. Generally a well crimped ring terminal like used in house wiring would be much better and safer.

Thank you so much for taking the time to write all this!
Now for the questions
In the wire resistance calculator, when I input 190cm length and a diameter of 3.2mm (8AWG) it shows 0.004ohm, but when I input 6.4mm (as I will be using two 8AWG wires paralleled) it doesn't show half the ohms but 0.0001ohm
Why is that?

Now about the voltage drop and power dissipation, I input 0.004ohm and 130A which is the constant amp draw for this pack and it shows a 0.52V drop and 67W power.
The question is, how much heat does 67W cause in a 190cm long wire.
Sounds like quite alot to me but consider I will be using two 8AWG wires paralleled, so it will be cut in half, but still I will need to figure how much heat will about 34W cause in a 190cm 8AWG wire.

 

[flangefrog]

In the wire resistance calculator, when I input 190cm length and a diameter of 3.2mm (8AWG) it shows 0.004ohm, but when I input 6.4mm (as I will be using two 8AWG wires paralleled) it doesn't show half the ohms but 0.0001ohm
Why is that?

You need to think in terms of cross sectional area rather than diameter. Paralleling two wires will double the area but only slightly increase the diameter of the equivalent single wire.

Now about the voltage drop and power dissipation, I input 0.004ohm and 130A which is the constant amp draw for this pack and it shows a 0.52V drop and 67W power.
The question is, how much heat does 67W cause in a 190cm long wire.
Sounds like quite alot to me but consider I will be using two 8AWG wires paralleled, so it will be cut in half, but still I will need to figure how much heat will about 34W cause in a 190cm 8AWG wire.

67W is a lot when it's continuous, but if it's only for 10-15s the wire should have enough thermal capacity to absorb this without increasing the temperature much. As power is current squared * resistance, a little less current will make a big difference in the power loss. At 65A there will only be 17W of heat generated. Calculating the actual temperature for a steady state current (assuming the pulses are short you can use an estimated RMS average) requires looking at the thermal resistance of everything involved in the thermal path including wire to ambient resistance. The following links can help explain the basics and calculate this. When dealing with wires in a practical application there are many factors that will influence the temperature so that is why I recommend to simply look at an ampacity chart instead.

https://www.petervis.com/electronics/Voltage_Regulator_Heatsink/Heatsink_for_TO-220_Voltage_Regulator.html
https://www.allaboutcircuits.com/tools/heat-sink-calculator/

 

[flippy]

problem is that the average current is already pretty high, having even higher pulses of 60W+ means you will get very toasty wires when the cables get heat soaked.

when there is no reason to put in a bigger cable or doubling up a thinnner one to divide the heating between 2 cables it is just good practice to do so, but with these kinds of currents and usage case it would be poor practice to have one very hot cable when you dont have to.
i am sure that with these currents the connectors will be impossible to touch due tot heat.

 

[rg12]

So that raises the question of what is an acceptable gauge for my continuous current of 130A considering I use two paralleled silicone wires?

It always looked weird to me...
I will be using 32pcs 6.75mm wide 0.2mm thick pure nickel, meaning it's 4A per nickel which is great.
Then I take lets say two 5AWG wires paralleled which have a cross sectional area 33.6mm2 while the 32pcs nickels amount to 43.2mm2 BUT nickel is 22% as conductive as copper so it's actually worth 9.5mm2 of copper.
I understand the length part but still, 9.5mm2 vs. 43.2mm2.
The pack is 13S and the nickels that will hover between the cells (each + and -) will be about 2cm max which multiplies by 14 pieces meaning it's 28cm length of nickel.
Actually, the part of the nickel that is in the air and doesn't touch the cell is much less, so it's maybe even half that length.

 

[DogDipstick]

Live on the edge. 250 circular mil / amp. ( 700 for buildings lol) Short runs, and remember 2 (10ga) > 1 (8 ga).

 

[rg12]

So that raises the question of what is an acceptable gauge for my continuous current of 130A considering I use two paralleled silicone wires?

It always looked weird to me...
I will be using 32pcs 6.75mm wide 0.2mm thick pure nickel, meaning it's 4A per nickel which is great.
Then I take lets say two 5AWG wires paralleled which have a cross sectional area 33.6mm2 while the 32pcs nickels amount to 43.2mm2 BUT nickel is 22% as conductive as copper so it's actually worth 9.5mm2 of copper.
I understand the length part but still, 9.5mm2 vs. 43.2mm2.
The pack is 13S and the nickels that will hover between the cells (each + and -) will be about 2cm max which multiplies by 14 pieces meaning it's 28cm length of nickel.
Actually, the part of the nickel that is in the air and doesn't touch the cell is much less, so it's maybe even half that length.

Anyone?

 

[john61ct]

Again, Blue Sea's Circuit Wizard, exposes all the relevant variables.

 

[rg12]

Again, Blue Sea's Circuit Wizard, exposes all the relevant variables.

Just tried it but it doesn't let me choose a voltage above 32V (it has only 3 options and not an open text field).
I heard that voltage doesn't affect heat, am I right?
The main factor that affects the gauge in this calculator is the max insulation temperature so when I select 200c it says 4AWG and when I choose 60c it shows 1AWG.
Does it mean that if I choose 60c that the wire will heat to 60c max or to 60c? because I rather go with 4AWG with silicone insulation (180c) but I don't want the wire to heat even close to these levels.

Can you please address my message about what I wrote about the nickel total mm2 equivalent of copper? how come it's accepted when 130A spread around a 9.5mm2 equivalent of a copper wire?

Thanks

 

[john61ct]

I would if I knew, only used boat cable myself so far.

Go to other online calculators then, yes Amps is what causes heat but voltage drop is a big concern.

You do realize the cladding heat rating is a "never approach" maximum right? Fuse for CP if there's any chance current will get the wire that hot.

The insulation rating is a given, that helps define your gauge, best to always round up (thicker) with the gauge for safety rather than relying on more heat-tolerant cladding just to reduce wire thickness.

 

[fechter]

Most wire size calculators assume continuous operation. By that, they mean 24/7. In practice with a EV, the battery will be discharged in less than an hour at full current, which is hard to do. Much time is spent at less than the maximum also, so heating will be a lot less than a truly continuous situation. You also have to consider wire cooling. If the wires are wrapped in thermal insulation, there's no place for the heat to go. On a bike, you have lots of air blowing around that will cool the wires.

My Sur-ron has 8ga battery wires and runs around 75A max. At the end of a long ride, the wires are barely warm.

 

[rg12]

Most wire size calculators assume continuous operation. By that, they mean 24/7. In practice with a EV, the battery will be discharged in less than an hour at full current, which is hard to do. Much time is spent at less than the maximum also, so heating will be a lot less than a truly continuous situation. You also have to consider wire cooling. If the wires are wrapped in thermal insulation, there's no place for the heat to go. On a bike, you have lots of air blowing around that will cool the wires.

My Sur-ron has 8ga battery wires and runs around 75A max. At the end of a long ride, the wires are barely warm.

Yeah, for ebikes I use 8AWG for even 150A since nothing is constant there but this pack is for an electric car (its 6pcs 13S 32p modules so it can have even a 20 minute of constant power of 130A).
Any chance you have an answer to my question about the nickel equivalence of 9.5mm2 that I asked about two messages ago?

 

[flangefrog]

I haven't actually looked at your calculations but I would guess that it is because the nickel is relatively short (less total power dissipation) and has good heat sinking in the form of the battery cells.

 

[rg12]

I haven't actually looked at your calculations but I would guess that it is because the nickel is relatively short (less total power dissipation) and has good heat sinking in the form of the battery cells.

I haven't thought about the heatsinking thing, that's probably it then...
About the length, why is more length equal more heat? I know it's something with the voltage drop across the wire but why does voltage drop create heat?

 

[john61ct]

The energy is being converted into heat by the extra resistance, causing the additional V drop in the process.

 

[rg12]

The energy is being converted into heat by the extra resistance, causing the additional V drop in the process.

Thanks

 

[slaphappygamer]

The energy is being converted into heat by the extra resistance, causing the additional V drop in the process.

Is the additional V drop, what I see when taking off from a stop? I'll see voltage drop at that instance. Then if I change my phase wires from 14 AWG to 10AWG, I'd see less voltage loss? Or would this "additional V drop" only be present/measurable in the phase wires.