batteriescapacitychargedischargewatts
Extremely simple question for you. Probably beneath all of you to answer apologies for that. My camera's flash is rated in Watt seconds. My battery is rated in mAh .
If my battery can output 1000 milliamp hours and outputs at 110 Volts. And my camera flash draws 100 Watt seconds per fire. Assuming there are no other variables and it's a perfect system, how many camera flashes could I fire.
You're not really looking for the C rating (maximum discharge current in multiples of nominal capacity), you're looking for the adjusted capacity at your nominated discharge current.
A 1.8Ah lithium battery can theoretically give 1.8A for 1hour, or 3A for 1.8/3h = 36 minutes. HOWEVER the capacity for a battery is traditionally quoted for a 20 hour discharge. That is, a capacity rating of 1.8Ah means the battery delivered 90mA for 20 hours in testing.
The relationship between continuous current and time-to-full-discharge is NOT linear. You have discovered that when discharged at 3A, your 1.8Ah battery is delivering much less capacity (only about a quarter!) than a linear interpolation of the amp-hour rating would suggest. This is not unusual.
The better batteries will give rated capacity at several discharge time samples, or even a graph of current vs capacity. A battery intended for remote control uses will probably give better performance, as quick discharge is the intended application of these batteries.
(I use a 1500mAh 3-cell (11.1v) 25C lithium pack to power a 3A LED bike light, and I get around an hour, which is close to rated capacity given a reasonably efficient buck regulator).
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.
simulate this circuit – Schematic created using CircuitLab
Best Answer
$$1000\mathrm{\ mA \cdot h}\times110\mathrm{\ V} = 396\mathrm{\ kJ}$$
$$100\frac{\mathrm{\ W \cdot s}}{\text{flash}} = 100\frac{\mathrm{\ J}}{\text{flash}}$$
$$\frac{396\mathrm{\ kJ}}{100\frac{\mathrm{\ J}}{\text{flash}}} = 3960\text{ flashes}$$