Electric Race Car: What Gauge Wire & Fuse/Circuit Breaker?

AI Thread Summary
For building a human-sized electric race car powered by five HR22-12 batteries in series, the optimal wire gauge is crucial to prevent overheating and excessive losses. The batteries will provide a total of 60V at 20AH, with expected currents between 12 and 20 amps. Calculating wire resistance and losses is essential, with a focus on using copper wire to minimize heat generation. Additionally, a suitable fuse or circuit breaker must be selected to ensure safety during the endurance race. Local sourcing in Tampa, FL, for cabling and protective devices is also recommended.
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I am in the process of building an human sized electric race car, and am wondering what gauge wire would best benefit me.

The car will be powered by 5 HR22-12, data sheet attached, batteries in series running at 12 V 20AH each for a total of 60V 20AH. This is an endurance race lasting one hour. Based on this time frame and the battery profile, these batteries should be able to supply somewhere between 12 and 20 amps.

I am wondering what gauge wire would I be able to use, so as to not burn out the wire or have excessive losses from to large a wire.

Also these races require the use of a fuse or circuit breaker. I am wondering which would be the best option for me.

The HR22-12 have a type B1 terminal. I am wondering what is the best way to connect the terminals with the cable.

I am located in Tampa FL, and am looking to get the cabling and fuse/circuit breaker locally.

I greatly appreciate any help given.
 

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james138698 said:
I am wondering what gauge wire would I be able to use, so as to not burn out the wire or have excessive losses from to large a wire.
While you wait for a more concrete answer to your query, let me help you along with the basics:

Any loss in the wire will be due to the ohmic resistance, R, in ohms

R = ρL/A

where ρ is the resistivity of the material (for copper, ρ = 1.724 x 10-8 Ωm at room temperature), L is the length in meters, and A is the wire cross-sectional area in m2.

If you know how long your wire needs to be, you can calculate the loss P = I2R in that wire segment. For example, a 2 meter long AWG10 (diameter = 0.1019 in = 0.259 cm = 0.00259 m) copper wire would have a resistance of

R = 1.724 x 10-8 Ωm x 2 m / (0.001295 m2) = 6.5 mΩ

The loss (in the form of heat) due to a current of 20A through that wire would then be

P = (20A)2 x 6.5mΩ = 2.6 W

Or, in terms of voltage that would equivalent with a voltage drop over that wire segment:

V = IR = 20A x 6.5mΩ = 0.13V
 
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