Power Circuit for Raspberry pi portable

...hold about the same number of AH

But there's the catch! A big one:

The mAh rating for the SLA battery is at 12 V so that equates to:

12 V * 22 Ah = 264 Whr (Watt*hour).

But for the powerbank it usually is at 3.7 V (the nominal voltage of a Li-Ion based cell), almost all powerbanks use a 3.7 V battery and a DCDC boostconverter to make the 5 V output. The manufacturers want to advertise with a large numger of mAh so they use the actual battery capacity but "forget" to tell you that that number is at 3.7 V and not 5 V.

So at 5 V you will not get 20.1 Ah but (ideal maximum) 3.7 V/5 V * 20.1 Ah = 14.9 Ah. That exclude the losses of the 3.7 V to 5 V conversion. But it gets worse, the actual energy stored is:

3.7 V * 20.1 Ah = 74.4 Whr

So the SLA battery has about 3.5 times more energy than the powerbank.

So your assumption that the same mAh rating means they also contain the same amount of energy is false!

Combined with an efficient DCDC converter to convert 12 V into 5 V my guess is that the SLA battery can easily charge a phone at least 4 times more than the power bank. In practice that 4 times might even be conservative as down conversion (12 V to 5 V) is generally more efficient than upconverting (3.7 V to 5 V).

There are two fundamental values you need to determine before much of anything else is possible:

1. How much energy the batteries have to be able to provide between charges.

2. How much power the pump needs when running.

Either way, you start with the electrical specs to the water pump. You seem to have already decided that you will use a 12 V pump. But, how much current does the pump draw when running?

This tells you two things. The current the battery has to be able to deliver regardless of how much energy it stores, and the rate of energy consumption when the pump is on.

Let's say for example that the pump draws 10 A at 12 V when running. That means your battery system must be able to supply 10 A. A car battery can certainly do that, for example. The power is (10 A)(12 V) = 120 W. Now you multiply that by the time you want the pump to be able to run with no sunlight until the battery is depleted. That yields the energy the battery must be able to store.

Let's say you use a car battery rated for 12 V and 50 A-h. The pump draws 10 A, so in theory, this battery can run the pump for 5 hours from a full charge. However, car batteries are damaged if run too low. I'd plan not to use more than half the capacity. You therefore would get 2.5 hours run time from a fully charged battery.

The solar panel needs to be sized according to how fast you want to be able to fully recharge the battery under whatever you consider your nominal or worst case conditions. This really up to you, but you also have to take local climate conditions into account.

Let's say, for example, that you want 6 hours of full sun to be able to fully charge the battery. We've already said that a drained battery is down by 25 A-h. There is some inefficiency in charging, so let's say the solar panel has to put out 35 A-h during the 6 hours of full sun. That means 5.8 A with full sun. The charging voltage of a typical "12 V" lead-acid battery is 13.6 V. The power into the battery is therefore 80 W. Figuring the power conversion circuitry between the panel and the battery is 80% efficient, so the solar panel output must be 100 W.