To give an answer to this problem it is essential to know what volume of water is to be pumped per unit time (second, hour, day etc.).
The pump would need to be lowered Inside the well under water level, so preferably a submersible pump.
The approximately calculation formula is: Pump power P = (Height (meters) X Volume per second (meter cube per second) X 1000 (density of water in Kg per meter cube) X 9.81 (gravity constant) X efficiency of pump and motor (I suggest as a ballpark figure take 0.75)) divided by 1000 to get kilowatts of pump rating.
Remember that a solar panel produces direct current. And we need an inverter to get alternating current needed for AC motors.
I don't have any idea about capacity/volume of water, I just need to fill a 2000 ltrs tank. My supply pipe is one and half inches diameter. Does this information help?
The simple answer is that you need to lift 2000 kgs of water 70 metres Energy required is expressed in the formula E = Mass x Height x gravitational constant E = 2000 x 70 x 9.8 = 140,000 J. Power (Watts) = E/s. If we assume a flow rate of 0.5 l/s pumping would be complete in 1 hour. I'm not sure if this is accurate for a 1.5" pipe but will do for a rough estimate. Watts = 140,000/3600 = 39. taking into account DC/AC conversion losses, motor efficiency losses etc. I would add a 50% factor. Therefore you should consider at least a 60W pump.
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