Cross Section sizing calculations

[ianganderton]

I'm trying to understand how to calculate a couple of things. I'm trying to create my own wire sizing spreadsheet so that I understand properly the maths behind it


I've got the target resistance calculated at =A4-(A4*A2)/A6
The CSA equation is =A14/A12

For 166A I'm expecting a CSA of around 35mm2 or AWG2 @ 33.6mm2

What have i done wrong?

I have a table with metric and AWG cable specs I'll use to draw on the answers

 

[svetz]

...What have i done wrong?

Don't really understand your spreadsheet as the formulas aren't shown and don't know how to do it. But, let's see what Mr. Google says...

Resistance = pL / A, where L is the length, A is the cross sectional area, and p is the resistivity of the metal (Copper is 1.7 × 10-8 Ω m).​

Cross sectional area depends on shape. Square wire would be length x width. Round wire would be πr2.​

Mr. google also knows copper 12 gauge wire has a diameter of 0.0808" and a resistance of 1.59Ω /1000'

Convert feet to meters: 1000 ft x 0.3048 m/f = 304.8 m​

Convert inches to meter: 0.0808 in x 1 ft/12 in x 0.3048 m/f = 0.00205232 m​

Calculate A = π x ( 0.00205232 / 2)^2 = 3.3064336452e-6 m^2​

Calculate R = 1.7× 10-8 Ω m x 304.8 m / 3.3064336452e-6 m^2 = 1.57 Ω ​

​

Hope that helps!

 

[Zil]

It is already here; http://circuitwizard.bluesea.com/

 

[RCinFLA]

R_dc = rho * length / Area

length in meters, cross section area in meters^2

Metal	rho
Silver	1.5860E-08	ohm-m
Copper	1.6780E-08	ohm-m
Gold	2.2400E-08	ohm-m
Aluminum	2.6548E-08	ohm-m
Nickel	6.8400E-08	ohm-m
Brass	8.1000E-08	ohm-m
Tin	1.1000E-07	ohm-m

Stranded wire gets complicated. I have a spreadsheet for wire based on NEMA wire resistance numbers, and another sheet for bars/straps based on above calculation. Some gauge wires have many variants of stranded subwires gauge and number of strands. Each have slightly different resistance for same overall wire gauge number.

NEC current carrying limits on wire is based on wattage loss generated per foot. Their concern is heating for insulation degradation and fire hazard. Total voltage drop is up to you. A 12v battery cable cannot tolerate much % voltage drop compared to a 240vac line. I made a wire spreadsheet that takes current and length and generates total voltage drop, wattage per foot of wire, and total wattage loss for length for all gauges. Makes a convenient chart to scan down list of gauges and type of wire to pick what you need.

A watt or two of wire heating per foot is a noticably warm wire. Heavy insulation reduces the ability to dissipate heat. This is why SOOW cable commonly used for generator power cords have lower maximum amperage rating then normal hookup house wire of same gauge.

I don't like using % voltage drop as it has little meaning to relate actual need or ramifications. A short piece of wire may have small % voltage drop but still have a lot of self heating, It is called a fuse.

 

[ianganderton]

OK, its cross sectional area equation I want to focus on

I know that from the following information its possible to calculate the cross sectional area (CSA) in mm2

1. Planned acceptable voltage drop (typically between 1% and 5%
2. System voltage (12V, 24V, 48V)
3. Planned max amps for the cable
4. Length of cable in meters
5. Resistivity, for copper this is 0.017 (I think this is ohm.mm2/m but I'm getting confused in my units - e.g. millimeters, meters, kilometers)

The formula for voltage drop is


from Redarc here

I can't remember my school maths which is really frustrating, I should know how to manipulate this formula to switch it around to give me the area.

If the planned drop is 2%, the voltage is 12V, the ammps are 166 and the length is 1m I know that the cable size should be 35 mm² or 2 AWG

DC Cable Sizing Tool - Use The Correct Sized Cables - Free Calculator

Use our free cable-sizing tool to establish the correct size of cable for your DC power system cables. This is especially important for low voltage, high current systems.

www.solar-wind.co.uk

Keeping the units aligned is also proving challenging

Here is the link to a spreadsheet on DROPBOX

I really appreciate folks taking the time to help

The reason I want to go down this path is so that I understand the calculation. The other reason is lots of the online calculators dont do what I need them to do and that's work between metric and imperial or they don't have the right parameters (e.g. the one suggested above is imperial and doesn't have 48V as an option)

 

[RCinFLA]

Write down the units your numbers are in will help prevent mixing up.

copper (annealed) resistivity as 0.0171 units is in ohm * mm²/m. (as rho would be 1.71 x 10^-8 ohm*meter)

Need area in square mm's and length in meters..

If starting from wire diameter don't forget area is PI *( Diameter/2)^2

 

[ianganderton]

Write down the units your numbers are in will help prevent mixing up.

copper (annealed) as 0.0171 units is in ohm * mm²/m.

Need area in square mm's and length in meters..

Yes, I'm trying to do that but the numbers don't seem to be working out

 

[ianganderton]

So firstly is this the correct equation and units for voltage drop?

Volts(V)= Length(m) x Current(A) x 0.017(ohm*mm²/m)/Area(mm²)

And if so there should be a "therefore" Area(mm²)=?

Really sorry for all the questions, I feel really dumb, like I'm staring at these equations but my brain is mud. I used to be top of the class for this stuff at school ?

 

[RCinFLA]

So firstly is this the correct equation and units for voltage drop?

Volts(V)= Length(m) x Current(A) x 0.017(ohm*mm²/m)/Area(mm²)

And if so there should be a "therefore" Area(mm²)=?

Really sorry for all the questions, I feel really dumb, like I'm staring at these equations but my brain is mud. I used to be top of the class for this stuff at school ?

Equation is correct if using correct units. I verfied from published wire resistance charts.

If starting from wire diameter don't forget area is PI *( Diameter/2)^2

 

[ianganderton]

So firstly is this the correct equation and units for voltage drop?

Volts(V)= Length(m) x Current(A) x 0.017(ohm*mm²/m)/Area(mm²)

And if so there should be a "therefore" Area(mm²)=?

Really sorry for all the questions, I feel really dumb, like I'm staring at these equations but my brain is mud. I used to be top of the class for this stuff at school ?

Damn

So Ive just rearanged the equation to

area = length x current x 0.017 / volts

But that doesnt take into account allowed voltage drop percentage ?

 

[svetz]

...But that doesnt take into account allowed voltage drop percentage ?

Ohms Law... V = I R

So you first calculate the resistance for a given gauge and length.

Now let's do a little thought experiment to help clarify things. If resistance is 0 what is voltage? V = IR, so V = 0. When measuring a short circuit current flow (e.g., ISC), if you measured the voltage it should be zero. What about infinite resistance? Resistance resists current flow, so infinite resistance means I=0, this is what you measure for example with VOC.

So, one more hint....

Resistance = pL / A. For a given diameter and wire type p and A are constants. So, resistance (R) is a function of length and can be written R = ZL where Z is the constant p/A.

V=IR, R= ZL. So, V= I ZL. So for a constant 10 amps at length one (L1): V1 = 10 x ZL1. For L2 at 10 amps, V2 = 10 x ZL2. What's the voltage drop between distances L1 and L2 assuming a constant current of 10 amps?

 

[ianganderton]

Ohms Law... V = I R

So you first calculate the resistance for a given gauge and length.

Now let's do a little thought experiment to help clarify things. If resistance is 0 what is voltage? V = IR, so V = 0. When measuring a short circuit current flow (e.g., ISC), if you measured the voltage it should be zero. What about infinite resistance? Resistance resists current flow, so infinite resistance means I=0, this is what you measure for example with VOC.

So, one more hint....

Resistance = pL / A. For a given diameter and wire type p and A are constants. So, resistance (R) is a function of length and can be written R = ZL where Z is the constant p/A.

V=IR, R= ZL. So, V= I ZL. So for a constant 10 amps at length one (L1): V1 = 10 x ZL1. For L2 at 10 amps, V2 = 10 x ZL2. What's the voltage drop between distances L1 and L2 assuming a constant current of 10 amps?

I’m sorry but that’s really confusing because while you might know what each of the letters mean for the variables I’m not quite sure.

OK I'm going to try and work through this. 0236hrs here, this is actually keeping me up at night its so frustrating

 

[ianganderton]

Ohms Law... V = I R

Voltage (V) in volts = Current (I) in amps x Resistance I) in Ohms

These are the variables I'm going to use because this is the situation I have
Volts = 12V, Amps = 166A, distance = 2 meters (this isnt true but I dont want to use 1, I want to see change), cable cross section is 35mm2

I also know the resistivity of copper ( HERE )
Resistivity has the symbol r - (rho to rhyme with snow) and its units are ohm metres.
The resistivity of pure copper is 1.7 × 10-8 W, this is a very small number – 0.000 000 017 W m.

resistance = resistivity × length / area


Now I'm not quite sure on units here, its easy to get things mixed up. The resistivity for instance is measured in ohm meters but the cross section I have is millimeters squared. There are 1,000,000 mm2 in a m2

So my resistance = 0.000 000 017 x 2 / 3.5e-5 (which I think is 0.000 035 m2) = 9.714285714285714e-4 = 0.000 9714285 ohms
To get milliohms I'm assuming you x1000 same as a millimeter therefore we have 0.1 milliohms is the resistance of 2m of 35mm2 cable

So you first calculate the resistance for a given gauge and length. - done I think

Now let's do a little thought experiment to help clarify things. If resistance is 0 what is voltage? V = IR, so V = 0. When measuring a short circuit current flow (e.g., ISC), if you measured the voltage it should be zero. What about infinite resistance? Resistance resists current flow, so infinite resistance means I=0, this is what you measure for example with VOC.

This doesn't make sense, it can't be nothing because if there is electricity there has to be some watts and watts are watts and watts = volts x amps. If volts was 0 then watts would = 0 and that cant be because watts are watts

So, one more hint....

Resistance = pL / A. For a given diameter and wire type p and A are constants. So, resistance (R) is a function of length and can be written R = ZL where Z is the constant p/A.

I don't understand this at all, this is where I get completely lost

V=IR, R= ZL. So, V= I ZL. So for a constant 10 amps at length one (L1): V1 = 10 x ZL1. For L2 at 10 amps, V2 = 10 x ZL2. What's the voltage drop between distances L1 and L2 assuming a constant current of 10 amps?

I'm lost here

 

[ianganderton]

Resistance = pL / A. For a given diameter and wire type p and A are constants. So, resistance (R) is a function of length and can be written R = ZL where Z is the constant p/A.

With my constants R = pL (2 meters)/Area ( 0.000 035 m2 ) therefore Z = 2/ 0.000 035 = 57,142.857

Damn, I'm lost at this point, I dont see where this is going how I get to Area =

 

[ianganderton]

I'm just going to go back to this

resistance = resistivity × length / area

Therefore to get area

resistance (x area) = resistivity × length / area (x area)

area x resistance (/resistance) = resistivity × length (/resistance)

area = resistivity × length /resistance

hmmmmm but I dont know what the resistance is before I know what the area is

 

[ianganderton]

The constants I know when starting my cable sizing calculation are

Acceptable loss in %
Current in amps
Nominal battery voltage
length in meters

Edit - I know the resistivity for copper too (at 20 degC)

Stuck back at this point

 

[svetz]

...V=IR, R= ZL. So, V= I ZL. So for a constant 10 amps at length one (L1): V1 = 10 x ZL1. For L2 at 10 amps, V2 = 10 x ZL2. What's the voltage drop between distances L1 and L2 assuming a constant current of 10 amps?

I'm lost here

Rather than letters let through make up some numbers. Let's not worry about you calculate R for now, I think you have that.

The goal is to calculate the voltage drop over a given length of wire of a known resistance for a known current.

That is, we want to solve for V given I and R. Let's set the current (I) at 10 amps.
Next we need to solve for R. For simplicity we'll say R is 1 ohm per foot (if you were doing it for real R=pL/A)"

In the equations above "L" is length.... lets use two lengths, L1=1 foot and and L2=2 feet. So, R1 = 1 and R2 = 2 ohms.​

​

Now we can solve for V1 and V2:

V1 = 10 x R1 = 10 x 1 = 10 V​

V2 = 10 x R2 = 10 x 2 = 20 V​

So, the voltage drop between 1 foot of our mythical wire is V2 - V1 = 10V.

You don't really need two lengths, if the first had been a length of 0 the voltage would have been 0.

Hope that helps!

 

[ianganderton]

Rather than letters let through make up some numbers. Let's not worry about you calculate R for now, I think you have that.

The goal is to calculate the voltage drop over a given length of wire of a known resistance for a known current.

That is, we want to solve for V given I and R. Let's set the current (I) at 10 amps.
Next we need to solve for R. For simplicity we'll say R is 1 ohm per foot (if you were doing it for real R=pL/A)"

In the equations above "L" is length.... lets use two lengths, L1=1 foot and and L2=2 feet. So, R1 = 1 and R2 = 2 ohms.​

​

Now we can solve for V1 and V2:

V1 = 10 x R1 = 10 x 1 = 10 V​

V2 = 10 x R2 = 10 x 2 = 20 V​

So, the voltage drop between 1 foot of our mythical wire is V2 - V1 = 10V.

You don't really need two lengths, if the first had been a length of 0 the voltage would have been 0.

Hope that helps!

I don’t understand where you got the 1 ohm from Without knowing the area

 

[ianganderton]

The goal I’m aiming for is to

Calculate the cross sectional area given acceptable loss, nominal voltage, known current, length and resistive

I know voltage drop is part of that because of acceptable loss but I just can’t see how

 

[svetz]

I don’t understand where you got the 1 ohm from Without knowing the area

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.

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?

Cross sectional area

...The goal I’m aiming for is to...Calculate the cross sectional area ...

In R = pL/A, A is the cross sectional area. You get that from the wire gauge and a chart to look up the diameter. The area is πr2. That is divide the diameter by two, square it, multiply by pi. In order to back-calculate it, you must first understand the equations to manipulate it by working it forward. So go through all of those equations first to make sure you understand how everything works, then work the math backwards.

Voltage Drop
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.