Cross Section sizing calculations

[ianganderton]

I made it up since you seem to understand the math for calculating the resistance (r=pL/A), so I avoided that altogether to try and simplify things by saying it was 1 ohm per foot of wire.

But, if you want to do a real example calculate R1 & R2 for two given distances along the same piece of wire. Then calculate V1 and V2 for a given current at those distances. Then the voltage drop for the distance is just V2 - V1. Once you've got that, hopefully you'll see that if L1 is 0, V1 is zero... so you can simplify the math to calculate the voltage drop by leaving out L1 and R1.

First off I really appreciate you taking the time to go though this and sincerely apologise for being so stupid

I understand the maths for resistance but I don’t understand how to get it if I don’t know the area (cable size)

 

[svetz]

First off I really appreciate you taking the time to go though this and sincerely apologise for being so stupid

I understand the maths for resistance but I don’t understand how to get it if I don’t know the area (cable size)

You're not stupid. It took me years of math to get this incredibly boring.

So, sorry... but you hopefully know how I am by now and did about 20 edits to post #20. Take a look at again, then post the math going forward one step at a time as if you had the wire gauge and wanted to calculate voltage drop. Then work the math backwards to see if you get the same gauge. If something doesn't seem right stop and ask. Next time I'm online I'll go through it.

 

[ianganderton]

So I’m obviously no clever enough to work this out. It’s so frustrating

Can you please put me out of my misery and give me the calculation for cross sectional area using acceptable voltage drop, nominal battery voltage, current, cable length and resistive for copper.

If I have the formula I’ll see the relationship between the different components

I’m not learning much at the moment. I’m just smashing my head against a wall

 

[svetz]

... I’m just smashing my head against a wall...

Hmmm, that always worked for me. So, you're not sure how to work the math going forward from wire diameter to voltage drop?

 

[ianganderton]

Hmmm, that always worked for me. So, you're not sure how to work the math going forward from wire diameter to voltage drop?

Wire cross sectional area to voltage drop is easy. I can do that

It’s acceptable voltage drop to cross sectional area I can’t do

 

[ianganderton]

How do I calculate cross sectional area of cable required when I have the following information

Acceptable loss in %
Nominal battery voltage
Current
Resistive for the conducting material
Length

 

[svetz]

...How do I calculate cross sectional area of cable required when I have the following information...

Let's start by working the math forward.... first... a caveat...

...The goal to solving any of this is to break things down one bit a time and track units like a hawk. The dimensions on the left side have to balance out with the dimensions of the right side. For example, in r=pL/A, you can tell the units of p are in (or need to be converted to) ohm-meters because R is in ohms, L is in meters, and area is meter squared. Getting rid of the numbers, the equation looks like: Ω = Ω m x m / (m x m). See how the meters (m) cancel out?

Facts 12 gauge wire has a diameter of 2.0523mm [ref] and the resistance is 1.59Ω /1000' Let's use an online wire calculator to confirm the calculations below, the image is to the right. Step 1: Calculate Cross sectional area Area is πr^2; radius (r) = diameter / 2 Area = π x (2.0523mm/2)^2 = 3.309 mm^2 = 3.3093777279e-6 m^2 (units conversion from mm^2 to m^2 is x1e-6)​ Step 2: Calculate Resistance for 1000' of 12 gauge wire R = pL ÷ A; p for copper is 1.7 × 10-8 Ω m R = 1.7 × 10-8 Ω m x (1000 feet x 0.3048 meters/foot) ÷ 3.3093777279e-6 m^2 = 1.57Ω matches our facts within round-off errors!​

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Step 3: Calculate Voltage Drop for 1000' of 12 gauge wire at 5 amps
V= IR; ΔV = I x (R2 - R1) where R1 is 0 for zero length and R2 is calculated in step 2:
ΔV = 5 x 1.57Ω = 7.83 Volts. Or round trip is 2000' and a voltage drop of 15.66 V

Step 4: Voltage Loss if V0 is 400 volts
The starting voltage is 400 volts, the voltage drop is 7.83 volts, the ending voltage is 400-15.66=384.3 .
The %loss is 15.66V / 400V = 3.9%

Let's pause here and make sure that all makes sense... If it does make sense... perhaps you'd like to take a stab at working it backwards to calculate wire gauge?

 

[ianganderton]

Let's start by working the math forward.... first... a caveat...

Facts 12 gauge wire has a diameter of 2.0523mm [ref] and the resistance is 1.59Ω /1000' Let's use an online wire calculator to confirm the calculations below, the image is to the right. Step 1: Calculate Cross sectional area Area is πr^2; radius (r) = diameter / 2 Area = π x (2.0523mm/2)^2 = 3.309 mm^2 = 3.3093777279e-6 m^2 (units conversion from mm^2 to m^2 is x1e-6)​ Step 2: Calculate Resistance for 1000' of 12 gauge wire R = pL ÷ A; p for copper is 1.7 × 10-8 Ω m R = 1.7 × 10-8 Ω m x (1000 feet x 0.3048 meters/foot) ÷ 3.3093777279e-6 m^2 = 1.57Ω matches our facts within round-off errors!​

View attachment 21189​

Step 3: Calculate Voltage Drop for 1000' of 12 gauge wire at 5 amps
V= IR; ΔV = I x (R2 - R1) where R1 is 0 for zero length and R2 is calculated in step 2:
ΔV = 5 x 1.57Ω = 7.83 Volts. Or round trip is 2000' and a voltage drop of 15.66 V

Step 4: Voltage Loss if V0 is 400 volts
The starting voltage is 400 volts, the voltage drop is 7.83 volts, the ending voltage is 400-15.66=384.3 .
The %loss is 15.66V / 400V = 3.9%

Let's pause here and make sure that all makes sense... If it does make sense... perhaps you'd like to take a stab at working it backwards to calculate wire gauge?

Yeap with you there

 

[svetz]

...with you there...

I almost hate giving it to you as I think you were very close. Most just use a calculator and don't even think about what's behind it, just so you know, that's pretty cool

Okay, and now without further ado... sdrawkcab!

What we want to do is to calculate the wire diameter. Even better, we know if we use all the same numbers the answer should be 12 gauge or 2.053 mm diameter. Steps are numbered so they coincide with the steps numbered in post #27 since we're working backwards from there.

Step 4: Voltage drop from Voltage Loss where V is 400 volts and acceptable loss is 3.9%
Voltage drop is 400 x 3.9% / 2 = 7.8V

Step 3: Calculate Resistance for 5 amps
ΔV = I x R, so R = ΔV / I = 7.8 / 5 = 1.56 Ω

Step 2: Calculate Cross Sectional Area for 1000' distance
R = pL ÷ A; p for copper is 1.7 × 10-8 Ω m
Solve for area: A = pL ÷ R;
A = 1.7 × 10-8 Ω m x (1000 feet x 0.3048 meters/foot) ÷ 1.56Ω = 3.32e-6 m^2; or 3.32 mm^s

Step 1: Calculate minimum Diameter of Wire
Area is πr^2
Solve for d where, r = Sqrt(A/π), and d=2r
d = 2 x Sqrt(A/π) = 2 x Sqrt(3.32 mm^2/π) = 2.055 mm

 

[ianganderton]

I almost hate giving it to you as I think you were very close. Most just use a calculator and don't even think about what's behind it, just so you know, that's pretty cool

Okay, and now without further ado... sdrawkcab!

What we want to do is to calculate the wire diameter. Even better, we know if we use all the same numbers the answer should be 12 gauge or 2.053 mm diameter.

Step 4: Voltage drop from Volage Loss where V0 is 400 volts and loss is 3.9%
Voltage drop is 400 x 3.9% / 2 = 7.8V

Step 3: Calculate Resistance for 5 amps
ΔV = I x R, so R = ΔV / I = 7.8 / 5 = 1.56 Ω

Step 2: Calculate Area for 1000'
R = pL ÷ A; p for copper is 1.7 × 10-8 Ω m
Solve for area: A = pL ÷ R;
A = 1.7 × 10-8 Ω m x (1000 feet x 0.3048 meters/foot) ÷ 1.56Ω = 3.32e-6 m^2; or 3.32 mm^s

The bit I dont understand is you are using a resistance you have calculated from knowing the cross sectional area

In a wire size calculator you dont have the cross sectional area so you cant have that resistance

When calculating a conductor size (it could be a busbar so only need mm2) calculators use only the following numbers

acceptable loss as a %
Nominal battery voltage
Planned current
Resistive for material
length of conductor

So here's what I'm trying to work out
I have a 2000W inverter 2 meters away from a 12V battery, therefore, my current is 166A. Best practice says I should have a maximum loss of 3%

From that how do I calculate the mm2 cross sectional area of the conductor needed

I know you can see it in the calculations you are showing me but I'm getting lost in all the radar clutter of variables I haven't got

 

[svetz]

...The bit I dont understand is you are using a resistance you have calculated from knowing the cross sectional area...

Which step and variable are you referring to?

If you mean Step 3 in the backwards calculations; R is calculated from the voltage drop, not the cross sectional area.

 

[svetz]

...I have a 2000W inverter 2 meters away from a 12V battery, therefore, my current is 166A. Best practice says I should have a maximum loss of 3%... Oh! I thought you wanted to understand the calculations. Just use an online calculator if that's all you want to know.

​

 

[ianganderton]

Oh! I thought you wanted to understand the calculations. Just use an online calculator if that's all you want to know.

View attachment 21195​

I absolutely want to understand the calculations.

The reason I’m giving those variables is so I can see it calculated from there. At the moment it’s going round the houses and I’m getting lost. You know when you are blindfolded and someone spins you round to disorientate you before you try and whack the piñata, well that’s what’s happening to me.

I want to see variables I recognise and understand the relationships between. I’m looking at low voltage high amp systems at the moment. As soon as the voltage goes up to 400V it’s like everything goes green, I can’t recognise anything

 

[svetz]

...you try and whack the piñata, well that’s what’s happening to me...

At least you get a stick in that game and can hit things!

...I absolutely want to understand the calculations....

You had no problems with the math in #27 and understood those steps ... hmmm... well... you need to help me out with the first step in #29 you didn't understand as I don't understand where you got lost.... the steps are numbered in #29 so they're easy to refer to. Just start with step #4 and work your way to step #1, when something doesn't make sense copy paste it and tell me why it doesn't make sense....

 

[ianganderton]

At least you get a stick in that game and can hit things!


You had no problems with the math in #27 and understood those steps ... hmmm... well... you need to help me out with the first step in #29 you didn't understand as I don't understand where you got lost.... the steps are numbered in #29 so they're easy to refer to. Just start with step #4 and work your way to step #1, when something doesn't make sense copy paste it and tell me why it doesn't make sense....

The bit I don’t understand (I put it in bold if you expand it) is the resistance 1.57ohm I think is then used in other calculations

I understand the calculations but I don’t understand how they relate to the formula I’m trying to solve with the data I’ve got.

I’ve still got this question how do I get from the data I have to knowing the cross sectional area of the conductor

 

[ianganderton]

Please be patient with me. I really appreciate your time.

Can you just show me in the simplest steps how a cable sizing calculator works using just the data available. Just up to the point of cross sectional area. I understand it all from there

 

[svetz]

...Can you just show me in the simplest steps how a cable sizing calculator works using just the data available. Just up to the point of cross sectional area. I understand it all from there

It's all in post #29. If you can't tell me what step lost you I don't see how I can help you. #29 isn't really that hard, it just takes the steps from #27 and works them backwards.

 

[ianganderton]

It's all in post #29. If you can't tell me what step lost you I don't see how I can help you. #29 isn't really that hard, it just takes the steps from #27 and works them backwards.

The calculations didn’t lose me. I’m using the wrong terminology, sorry

I guess I’m trying to say the relevance to my aim/task lost me. I don’t understand how these equations help me calculate my specific aim

 

[svetz]

...I guess I’m trying to say the relevance to my aim/task lost me. I don’t understand how these equations help me calculate my specific aim...

No worries!

...When calculating a conductor size (it could be a busbar so only need mm2) calculators use only the following numbers: acceptable loss as a % Nominal battery voltage Planned current Resistive for material length of conductor

Guess I'm not understanding what you want... from what you said to the left I thought you wanted to calculate wire diameter or cross sectional area from voltage, voltage loss, current, distance (conductor length), resistivity of material.... which is what #29 shows. If not, what are you trying to do?

 

[ianganderton]

OK, so first off I want to apologise most sincerly, I wasn't looking properly at post #29. That will teach me to look at this stuff in the early hours of the morning.

I now see what was there and I'm just trying to work through it but I seem to be getting an error

This the calculation for the wires for my inverter.
3% acceptable loss
12 Vdc
166 Amps
1m length
0.000000017 resistive for copper

Step 4: Voltage drop from Voltage Loss where V is 12 volts and acceptable loss is 3%
Voltage drop is 12 x 3% / 2 = 0.18V

Step 3: Calculate Resistance for 166 amps
ΔV = I x R, so R = ΔV / I = 0.18V / 166 = 922.22 Ω (this looks wrong?)

Step 2: Calculate Cross Sectional Area for 1m distance
R = pL ÷ A; p for copper is 1.7 × 10-8 Ω m
Solve for area: A = pL ÷ R;
A = 1.7 × 10-8 Ω m x 1 ÷ 922.22 Ω = 1.84337E-11 m^2 - this seems very wrong