I am no expert, but since nobody else has answered I thought I could help.
I understood that the C (coloumb) rating of a battery indicated the rate of discharge that was advised. For instance, a 5C battery can be discharged at up to 5 times the rated AH of the battery. So my understanding is that a 10C 12v battery with 150 AH could have an instantaneous draw of 1500 amps on the battery.
I also understand that the actual capacity of a battery does depend on the draw. For instance, I understand that lead acid batteries tend to last longer AND exhibit greater total capacity if the discharge is small (e.g., 500 milliamps). Lead acid batteries that are discharged quickly (e.g., 500 amps) tend to discharge a lot faster AND exhibit much lower total capacity. The level to which this is true depends on the battery chemistry, from what I've read.
For example, if I was drawing 1 amp from your 150AH battery, I may get 150 hours of usage. On the other hand, if I was drawing 150 amps from your battery, I may only get 30 minutes of usage. The numbers were randomly picked, but you get the idea.
Hopefully that cleared up your confusion. Yes, your C5, C10, and C20 batteries all offer 150AH but only under specific load conditions as you said in your question.
Not only do higher C ratings on batteries result in a higher battery cost, but, in general, high amperage charge/discharge of lead acid batteries tends to crush a battery's life significantly. This could mean that with a high enough amperage on your charge/discharge, you may only see 10 cycles to the battery before its useful life is depleted.
[1]. No battery is really better, I think. This really depends on your use case. For example, you could do one of the two extremes:
[A]. Buy just a small number of C20 batteries for your system. They're designed with high charge and discharge rates in mind. A C20 rating would guarantee that you could get the 150AH out of the battery despite a high discharge. You'd get the same total AH rating but in a smaller package. You'll probably get fewer charge/discharge cycles before the batteries need to be replaced. If you were only planning on using this solar system as a backup when the power went out (you did say UPS), then this is the better option because you would buy fewer batteries (fewer dollars). But you may need to replace the whole thing sooner.
[B]. Buy a lot more C5 batteries. They're each designed to only handle a small charge/discharge rate to give you the same total AH rating. You'd need to have a lot more of them since you're maximum demand of amps probably couldn't be met with just a few C5 rated batteries. Also, the total number of amps flowing into the batteries from the charger would need to be lower for C5 rated batteries when compared with C20. You'll probably get a lot more charge/discharge cycles out of this setup before the batteries need to be replaced. You'd probably need to consider this option if you wanted to use the battery pack for more than just a UPS. Assuming you used these batteries daily, you'd definitely want them to have a much longer lifetime/cycles than, say, option A.
In either case, the maximum discharge and the charging rate would need to be considered when you decide how many batteries and which C rating you'll need.
[2]. Lower C ratings are probably better for solar power use because most people who install a solar system plan to use it on a daily basis (unlike a UPS). Even though the lower C rating translates to a need to have more total batteries, the cost is probably a lot lower and the total lifetime in cycles is probably a lot higher.
Here's a couple examples:
Assuming you need a maximum demand of 1000 amps at 12v (12000 watts), then here are some options:
With C20 batteries rated at 150AH, the maximum amp discharge is 3000 amps. So you could definitely run the load with a single battery since 1000 amps < 3000 amps. However, your useful running time with a single battery would only be, roughly, 150/1000 * 60 = 9 minutes.
With a C5 battery rated at 150AH, the maximum amp discharge is 750 amps. So even though you probably could run the load with a single battery, it wouldn't actually give you 150 AH (and wouldn't be advised). Instead, you'd want at least 2 batteries. By spreading the load across 2 batteries, the demand from a single battery (assuming constant voltage) is only 500 amps. In this scenario, you'd be pulling (150/500 * 60) * 2 batteries = 36 minutes. In addition, the C5 battery is probably less money.
When planning this system, remember to take into consideration heat losses in your wiring which could be high depending on the gauge and length. Also remember that your inverter is only 80-95% efficient depending on how much you spend on that component. Remember that the total efficiency of what you're powering may also be affected by the quality of the sine wave (pure sine wave or the cheaper jagged/stair step kind) being output by the inverter. In the case of a computer where the power supply will simply convert the AC back to DC again, my hunch is that the efficiency of the total load will not be improved by using a pure sine wave inverter.
If I'm horribly wrong on any of these points, please somebody more knowledgable speak up! I'd be happy to know I was wrong.
Hopefully this makes some sense and is on target.
In any case, have fun and good luck with your project!
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
The formula assumes that the temperature is constant. If the battery will be subject to temperature changes that you cannot control, then the safest approach is to assume the worst case temperature (and the corresponding battery capacity) for the whole period. If you have some control and can predict a temperature profile, then you can use that profile to calculate an average battery capacity based on the temperature and the time at that temperature (this can be done piecewise if the profile is not too complicated but you could use an integral if you need more accuracy). By the way, do you know that the current draw of your device remains constant with temperature? If not, you should also take this into account.