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Assuming the fan is 12 V, and you know the amperage, you can figure 80% usable capacity of a sealed lead-acid battery:

LATITUDERESERVE TIME0° to 30° (N or S) 144 to 168 hr 30° to 50° (N or S) 288 to 336 hr 50° to 60° (N or S) 432 hr

You can pretty much figure the capacity of battery required based on this rule of thumb.

Say for example your fan is 12 V, 0.2 A. Since Portugal's latitude is around 40° N, you would aim to have a usable capacity of between 288 and 336 hours.

Multiply the reserve time by the current drain divided by 0.8 (80%) and you get the required battery Ah for your setup.

(0.2 A * 288 hr) / 0.8 = 72 Ah

And for any other configuration not just a fan you can plug in the numbers and arrive at that value. Hope this helps.

Cheers

Your calculation is not entirely correct:

A 72ah 12v battery with 0.2a (2.4w) of current required gives 288h (with coefficient 0. for a complete discharge.

For a 50% discharge: 144h, i.e. 6 dayswithout any solar charge.

A 20w solar panel will produce 1.12a of charge current, from which we subtract 0.2a to run the motor and 0.006a absorbed by the charge controller. So we have +/- 0.8a of available charging current. With a 40w panel we have a balance of 2a of charge current. In this case a 72ah battery is greatly oversized. A 72ah 12v battery weighs 20kg and costs +/- 190€.

But I could be wrong.

Rule of thumb is 2 weeks of reserve so you eliminate any possibility of lost power especially during the winter when you’ve got sometimes weekly cloud decks and the charge efficiency of only10-15%. But that’s more so rule of thumb for legit automated weather station sites.

Cheers]]>

Assuming the fan is 12 V, and you know the amperage, you can figure 80% usable capacity of a sealed lead-acid battery:

LATITUDERESERVE TIME0° to 30° (N or S) 144 to 168 hr 30° to 50° (N or S) 288 to 336 hr 50° to 60° (N or S) 432 hr

You can pretty much figure the capacity of battery required based on this rule of thumb.

Say for example your fan is 12 V, 0.2 A. Since Portugal's latitude is around 40° N, you would aim to have a usable capacity of between 288 and 336 hours.

Multiply the reserve time by the current drain divided by 0.8 (80%) and you get the required battery Ah for your setup.

(0.2 A * 288 hr) / 0.8 = 72 Ah

And for any other configuration not just a fan you can plug in the numbers and arrive at that value. Hope this helps.

Cheers

Your calculation is not entirely correct:

A 72ah 12v battery with 0.2a (2.4w) of current required gives 288h (with coefficient 0. for a complete discharge.

For a 50% discharge: 144h, i.e. 6 days

A 20w solar panel will produce 1.12a of charge current, from which we subtract 0.2a to run the motor and 0.006a absorbed by the charge controller. So we have +/- 0.8a of available charging current. With a 40w panel we have a balance of 2a of charge current. In this case a 72ah battery is greatly oversized. A 72ah 12v battery weighs 20kg and costs +/- 190€.

But I could be wrong.]]>

LATITUDE | RESERVE TIME |

0° to 30° (N or S) | 144 to 168 hr |

30° to 50° (N or S) | 288 to 336 hr |

50° to 60° (N or S) | 432 hr |

You can pretty much figure the capacity of battery required based on this rule of thumb.

Say for example your fan is 12 V, 0.2 A. Since Portugal's latitude is around 40° N, you would aim to have a usable capacity of between 288 and 336 hours.

Multiply the reserve time by the current drain divided by 0.8 (80%) and you get the required battery Ah for your setup.

(0.2 A * 288 hr) / 0.8 = 72 Ah

And for any other configuration not just a fan you can plug in the numbers and arrive at that value. Hope this helps.

Cheers]]>