diy solar

diy solar

Cross Section sizing calculations

Step 3: Calculate Resistance for 166 amps
ΔV = I x R, so R = ΔV / I =

Does it matter which way this is calculated because

R = I / ΔV = 166 / 0.18V = 0.001084337 Ω which seems more viable. Thinking about it to rearrange the equation it would have needed to be divided by ΔV not by I

Then

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 ÷ 0.001084337 = 1.56778E-05m2 = 15.67mm2 I think which makes more sense but is still very small compared to the 35mm2 / 2/0AWG recommended by other calculators
 
... this seems very wrong...
Don't worry, I do that all the time.

So, let's start by getting answer from an online calculator to validate the math...
chart to the right shows an AWG of 5... so... let's see what the math says!

Step 4: ΔV from Voltage Loss where V is 12 volts and acceptable loss is 3% RT
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?) ❌
At least you knew it was wrong! Looks like you inverted the result somehow.
0.18V / 166 = 1.0845986985e-3Ω
1598714801323.png

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 ❌ GIGO!
A = 1.7 × 10-8 Ω m x 1 m ÷ 1.0845986985e-3Ω = 1.5677777778e-5 m^2, convert m^2 to mm^2-> 15.68 mm^s

Step 1: Calculate minimum Diameter of Wire (Just adding this in to check the result against the calculator above)
Area is πr^2
Solve for d where, r = Sqrt(A/π), and d=2r
d = 2 x Sqrt(A/π) = 2 x Sqrt(15.68 mm^2/π) = 4.47 mm

From this chart awg 5 is the next rounded up value at 4.6 mm^2, so matches the calculator.
✅
 
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