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The issue I'm facing is calculating current draw from different household items. Like, if a fan is rated at 75 Watts, its current draw would be A = W/V so A = 75/220 and A = .34 amperes.

That doesn't sound right to me, because I have two 150 A-h batteries connected in series to an inverter/charger. Assuming that I use up only 60% of the batteries' capacity to prolong their life, I have 180 A-h of battery capacity. And also if the inverter is only 80% efficient, I have 144 A-h of usable battery capacity. If the fan does indeed draw .34 amperes of current, that would mean that this battery setup should run the fan for around 424 hours.

I know I'm doing something wrong, I just don't know what it is. Could you guys help me out here? What's the backup time for a 75 watt load drawing power from an inverter connected to two 150 A-h batteries in series?

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    \$\begingroup\$ What's the voltage of the battery? Not taking the voltage difference sounds like where you're going wrong. \$\endgroup\$ – PeterJ May 19 '13 at 7:25
  • \$\begingroup\$ What is the C-rating of the batteries? \$\endgroup\$ – Ignacio Vazquez-Abrams May 19 '13 at 7:30
  • \$\begingroup\$ @PeterJ Each one of the batteries is rated for 12 Volts. Is that what you were asking? \$\endgroup\$ – Person09 May 19 '13 at 8:23
  • \$\begingroup\$ @IgnacioVazquez-Abrams These batteries didn't have a C-rating since they were made by a small local company. \$\endgroup\$ – Person09 May 19 '13 at 8:24
  • \$\begingroup\$ @Person09, yes, see the answer below. Because the voltage is lower the current is higher. So if you use the A = W/V formula again using 24V if they are in series you'll see the current on the battery side is higher so the expected time will go down. \$\endgroup\$ – PeterJ May 19 '13 at 8:27
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Your 75W fan required 220 Volts, it will not operate at typical battery voltages such as 12V.

A 75W fan operating (indirectly) from a 12V supply would draw 6.25 A not 0.34 A.

Even with a theoretical 100% efficient invertor, the current on the 12V side is greater than the current on the 220V side to deliver the same power across the system from battery to Fan.

Note that the reason that power is delivered through the grid at high voltage is to keep currents low. losses and cable sizes are proportional to (a function of) current. High V are used to keep currents low.

You convert 12V DC into 220 V AC using your inverter. Depending on capacity and utilisation the efficiency of an inverter might be the 80% you estimate or it may be worse - 50% Ref So your 6.25A may become 13A drawn from the 12V battery.

Therefore your 300 Ah battery-set might power a 75W fan for less than 24 hours.


Update: for 2 x nominal 12V SLA battery, the following diagram may clarify what is happening.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ "connected ... to an inverter/charger" \$\endgroup\$ – Ignacio Vazquez-Abrams May 19 '13 at 7:40
  • \$\begingroup\$ @Ignacio: Good point. I'll revise the Answer. \$\endgroup\$ – RedGrittyBrick May 19 '13 at 7:44
  • \$\begingroup\$ @RedGrittyBrick Oh! I understand now. That is what I was missing. I should have been using the battery voltage when calculating the amperage for the fan. But, like I said in my question, the batteries are both 12 Volts, so the input to the inverter is 24 Volts. Does that make the fan's amperage 3.12 A? \$\endgroup\$ – Person09 May 19 '13 at 8:28
  • \$\begingroup\$ @Person09: See diagram added to answer. \$\endgroup\$ – RedGrittyBrick May 19 '13 at 10:04
  • \$\begingroup\$ @RedGrittyBrick Ah! Thank you for taking the time to add the diagram. It makes the situation easier to understand. \$\endgroup\$ – Person09 May 19 '13 at 10:38

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