How is it analog? The portion of water flowing down through the boilers is proportional to the available power of A minus B. Since A always has power, if there is only 50% power available to B (main grid) half the water flows through the boilers and the result is half max heated water. The 'bootstrap' pumps provide the amplification. At 50% power available from the steam engines, the pumps operate at 50% and the huge array of steam generators output 50% of their max. All you need to do is add some dummy load to the bootstrap line so that the max output of the op-amp is exactly the power drawn. I found it easier to limit the number of boilers on the op-amp
This works because pumps provide 30 water/s (proportional to the available power) and steam engines consume up to 6 water/s [see below edit], regardless of the temperature of the water.
More advanced circuits could be made, but my mission is accomplished and quest completed.
Edit: If you want the power system to turn on before the accumulators run out, place an accumulator filter between power pole B and the main power grid.
Important explanation of water consumption mechanics of steam engines: Steam engines consume water proportional to the amount of the available power they are producing. To make the steam engines consume exactly 6 water/s all the time, they must either have dummy loads attached (increasing power draw) or have colder water (fewer boilers to reduce available power). For instance, if 5 steam engines has 1 boiler (390 kW available power) and has 4 small pumps (90 kW) attached as a power draw, each steam engine will consume 2.3 water/s. (total of about 11.5 water/s)