**What are the rules for designing the cross-section of power supply wires?**

The proper sizing of the wires and cables for the DC power supply on our boats is not an occult science. By knowing some basic rules of electrical engineering, every skipper himself can calculate the right cross-sections for connecting wire and cables.

First of all, it is important to know that the maximum current density, i.e.; the current I [A] divided by the cable's cross-section Q [mm^{2}] should not become too large. Otherwise there is a risk of overheating the wire. This overheating has to be preventend in any circumstance. If disregarded, overheating can result in a fire on board. Therefore every electric line has to be protected with a suitable fuse to avoid the excess of the maximum allowed current density. Anyone who connects an electrical line directly with the boat's battery without applying a properly sized fuse is risking an **on-board fire.**

We know this maximum current density I/Q already from our home installation. In general, not more than 16 A are allowed for a copper wire with a cross section of 1.5 mm

^{2}. With a standard wire's cross-section of 1.5 mm

^{2}in our houses, the circuit breakers are designed for a maximum current of 16 A.

The DC supply on our boats, with its 12 or 24 volt only, requires a second rule for calculating the maximum cable length for a given wire cross-section; or for calculation the approriate cross-section for a given cable length. This is because the effective supply voltage of a consumer is reduced by the resistance of the applied connecting wire. A loss of power of the connected device is the result. As a rule of thumb this loss of power should be less than 5 percent.

**A bilge pump with a maximum power P of 120 watts shall be operated with 12 volts. The pump shall be installed into the engine's room. The pump shall be connected with a two-wire cable with a length of 5 meters. The cable consists of two wires with a cross section of 1.5 mm**

An example:

An example:

^{2}. Therefore the total length of the supply wire results as 2 x 5 = 10 meters.

- Question 1: How many volts are available at the bilge pump?
- Question 2: What is the effective remaining power of the bilge pump ?

The calculation is rather simple:

120 Watts of total power require an electrical current of 120 W / 12 V = 10.00 Amps. The wire's resistance of a 10 meter long copper line with a cross section of 1.5 mm^{2} amounts to 0.12 Ohms. With a current of 10 Amps this results in a voltage drop of: 0.12 Ohms by 10.00 A = 1.2 Volts. Thus the pump is actually supplied with 12.0 - 1.2 = 10.8 Volts. Thus the pump can produce 10.8 V x 10 A = 108 Watts of power. 12 Watts of electric energy are waisted via the line from the battery to the pump. 12 Watts of power dissipation equals to 10 % of the total power. In other words, 10 % loss of power is equivalent to an efficiency of 90 %. This efficiency is the bases for cable design. In general, the goal value should stay above 95%.

The following input form allows to determine the required wire cross section and the resulting efficiency of the cable. U [V] denotes the supply voltage. Enter a 12 or 24. I [A] equals the electrical current through the line and the connected consumer. P [W] denotes the total power output in Watts. It is calculated by the product of U by I. The term used for the cross section of the wire is Q [mm^{2}]. The length of the wire is denoted as L [m]. From these two input values the wire's resistance RL [Ohm] is resulting. The efficiency is denoted by the abbreviation E [%]. It indicates the percentage of the available power at the connected device.

All values are calculated automatically after your input. Once you enter a value of more than 0.00 m for the cable length L, the resistance and the resulting voltage drop dV [V] and the dissipation dP [W] is computed. Select a larger wire's cross-section, if the current density becomes too large, or when efficiency drops below 95%.

As first exercise you may start with the example of the bilge pump from above.

If you want to determine the maximum length of an extension cord (cable drum) for 230 Volts, you can also use this form. Enter 230 for the voltage and you can see that a cable drum with a length of 50 meters and a wire's cross-section of 1.5 mm^{2} delivers an efficiency of 95.09 %, while driven at 10 Amps. Remember to enter the value 100 for a 50 m cable (forward and backward wire). In this example the power loss dP result as 107.31 Watts. Therefore alway operate a cable drum in rolled-up state! **A wish to you in the end**

Next time you should apply your newly acquired knowledge to inform your boat neighbor: Whenever he operates his 1000 Watts heater without unwinding his cable drum. Probably you contribute to an improved safety on the water ...

Incorrect sizing of cables, wires and fuses is one of the most important fire causes on our sport boats!