Difference between revisions of "Solar Systems - Wiring and Fitting"

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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.
 
  
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Unlike [[Alternating current|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.
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= Overview<br/> =
  
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The current drawn by an appliance or circuit can be calculated by dividing the power rating by the system voltage.
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Many system failures are caused by unsuitable or improperly installed cables, connectors, switches or sockets.
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<pre>Example: A 1 kW inverter in a 24 V system will draw a current of 42 A (1,000 W / 24 V).
 
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Add safety margins and take overload capacity into account.</pre>
 
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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:
 
  
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ΔP [W] = 0.018 x I² [A²] x L [m] / A [mm²]
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Their current, voltage and power ratings should be in line with the system characteristics.
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<pre>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:
 
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0.018 x (5 A)² x (2x10 m) / 1.5 mm² = 6 W. In a 12 V system, this would be 10% of the whole generator power.</pre>
 
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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
 
  
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A [mm] = 0.018&nbsp;x I&nbsp;x L / U / 0.05 (= 5%)
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<br/>
  
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Use appropriate outdoor cables to connect the modules, protect them against sunlight and damage (e.g. by rodents) and lay cablesin conduits where appropriate.
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= How to Improve<br/> =
  
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<br>
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Unlike alternating current (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.
  
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The current drawn by an [[Lamps_and_Electric_Appliances|electric appliance]] or circuit can be calculated by dividing the power rating by the system voltage.
  
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<br/>
  
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Source: [http://www.gtz.de/de/dokumente/en-electricity-from-sunlight.pdf Klaus Haars, GATE-ID technical brief: '''Electricity from Sunlight, '''Solar Energy Supply for Homes and Buildings, Eschborn, 2002.]
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'''Example:'''
  
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A 1 kW inverter in a 24 V system will draw a current of 42 A (1,000 W / 24 V).<br/>Add safety margins and take overload capacity into account.
  
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|}
  
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'''⇒ '''[[Solar Main Page|'''Back to Solar Section''']]
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<br/>
  
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Like other electric components, accessories (particularly cables) cause losses. Since [[Access_to_Modern_Energy|energy]] is expensive, these should be minimized.
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<u>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:</u>
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ΔP [W] = 0.018 x I² [A²] x L [m] / A [mm²]
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<br/>
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{| border="1" cellspacing="1" cellpadding="1" style="width: 100%"
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|-
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| style="background-color: rgb(204, 204, 204)" |
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'''Example:'''
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For a generator current of 5 A and a cable of 1.5 mm² with a length of 10 m (one way) the power loss will be:<br/>0.018 x (5 A)² x (2x10 m) / 1.5 mm² = 6 W.
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In a 12 V system, this would be 10% of the whole generator power.
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|}
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<br/>
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*Losses can be reduced by increasing the cross-section of wires or by higher voltage levels.
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*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 x I x L / U / 0.05 (= 5%).
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*Use appropriate outdoor cables to connect the modules, protect them against sunlight and damage (e.g. by rodents) and lay cables in conduits where appropriate.
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<br/>
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= Typical Problems with Wiring and Fittings<br/> =
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Metals can oxidate if exposed to oxygen and moisture. This corrosion typically produces oxide(s) and/or salt(s) of the original metal. Corrosion can also refer to other materials than metals, such as ceramics or polymers, although in this context, the term degradation is more common. In other words, corrosion is the wearing away of metals due to a chemical reaction.
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Because corrosion is a diffusion controlled process, it occurs on exposed surfaces. As a result, methods to reduce the activity of the exposed surface, such as passivation and chromate-conversion, can increase a material's corrosion resistance. However, some corrosion mechanisms are less visible and less predictable.<ref name="Gate, GTZ: Haars, Klaus: Electricity from Sunlight, Solar Energy Supply for Homes and Buildings, Eschborn, 2002">Gate, GTZ: Haars, Klaus: Electricity from Sunlight, Solar Energy Supply for Homes and Buildings, Eschborn, 2002</ref>
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<br/>
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= Further Information<br/> =
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*For further information on wiring and fitting for [[Solar_Home_Systems_(SHS)|solar home systems]] see the EnDev wiki page on [[Standards_for_System_Installation_and_Wiring|standards for system installation and wiring]].
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*[[Portal:Solar|Solar Portal on energypedia]]<br/>
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= References<br/> =
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<references />
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[[Category:Solar_Home_Systems_(SHS)]]
 
[[Category:Solar]]
 
[[Category:Solar]]

Latest revision as of 12:11, 15 August 2014

Overview

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.


How to Improve

Unlike alternating current (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 electric 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 x I² [A²] x 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 (one way) the power loss will be:
0.018 x (5 A)² x (2x10 m) / 1.5 mm² = 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 x I x 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 cables in conduits where appropriate.


Typical Problems with Wiring and Fittings

Metals can oxidate if exposed to oxygen and moisture. This corrosion typically produces oxide(s) and/or salt(s) of the original metal. Corrosion can also refer to other materials than metals, such as ceramics or polymers, although in this context, the term degradation is more common. In other words, corrosion is the wearing away of metals due to a chemical reaction.

Because corrosion is a diffusion controlled process, it occurs on exposed surfaces. As a result, methods to reduce the activity of the exposed surface, such as passivation and chromate-conversion, can increase a material's corrosion resistance. However, some corrosion mechanisms are less visible and less predictable.[1]


Further Information


References

  1. ↑ Gate, GTZ: Haars, Klaus: Electricity from Sunlight, Solar Energy Supply for Homes and Buildings, Eschborn, 2002
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