Difference between revisions of "Solar Systems - Wiring and Fitting"
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Like other electric components, accessories (particularly cables) cause losses. Since energy is expensive, these should be minimized. The power loss within a copper wired cable depends on the current I, the cable length L (back and forth) and the cross-section A of the wire as follows: | Like other electric components, accessories (particularly cables) cause losses. Since energy is expensive, these should be minimized. The power loss within a copper wired cable depends on the current I, the cable length L (back and forth) and the cross-section A of the wire as follows: | ||
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ΔP [W] = 0.018 ∗ I2<font size="3">[A2] </font>∗ L [m] / A [mm2] | ΔP [W] = 0.018 ∗ I2<font size="3">[A2] </font>∗ L [m] / A [mm2] | ||
Revision as of 15:19, 24 June 2009
Many system failures are caused by unsuitable or improperly installed cables, connectors, switches or sockets. Their current, voltage and power ratings should be in line with the system characteristics.
Unlike AC, in DC systems switching the poles (+) and (-) will cause problems. The application of special DC fittings is recommended. Otherwise, high-quality AC items should be applied. The current ratings of AC-switches can be applied only for 12 V/ 24 V DC systems. In high-voltage DC systems, they have to be higher.
The current drawn by an appliance or circuit can be calculated by dividing the power rating by the system voltage.
Example:A 1 kW inverter in a 24 V system will draw a current of 42 A (1,000 W / 24 V). Add safety margins and take overload capacity into account.
Like other electric components, accessories (particularly cables) cause losses. Since energy is expensive, these should be minimized. The power loss within a copper wired cable depends on the current I, the cable length L (back and forth) and the cross-section A of the wire as follows:
ΔP [W] = 0.018 ∗ I2[A2] ∗ L [m] / A [mm2]
Example:For a generator current of 5 A and a cable of 1.5 mm2 'with a length of 10 m (oneway) the power loss will be 0.018 ∗ (5 A)'2∗ (2∗10 m) / 1.5 mm'2 '= 6 W In a 12 V system, this would be 10% of the whole generator power.
Losses can be reduced by increasing the cross-section of wires or by higher voltage levels. Keep the losses in the generator circuit as well as in the load circuits below 5% (better: 3%) and in the controller- battery circuit below 0.5%. The necessary cross-section of the cable is
A [mm] = 0.018 ∗ I ∗ L / U / 0.05 (=5%)
Use appropriate outdoor cables to connect the modules, protect them against sunlight and damage (e.g. by rodents) and lay cablesin conduits where appropriate.
