Check my math, please.

delcrossv

If I have ,say, 3 radiant loops off a single circulator (no zoning), piped off a common header, I would add head heights like a parallel circuit?

Head total = 1/ loop1 +1/loop2 +1/loop3 ?

I'd need circuit setters to balance them so the long loop would run and add those restrictions ? Or are those ignorable?

Jamie Hall

No. The formula you quote up there is for impedance n parallel for electronics. Water doesn't work that way.

If you are seeking the maximum head in a system of parallel loops, you simply look for the loop with the highest head loss. You do NOT add the heads in the various loops in any way. You DO add the flows to get the total flow.

What is somewhat analogous electronics is that the flows in the various loops will divide so that each loop will have the same loss, just as in electronics each parallel path will have the same voltage drop.

Jamie Hall

Let me add to the above. The formula you posted is, of course, correct for determining the impedance in electrical circuits. The reason it does NOT work for hydraulic circuits is not, perhaps, obvious: in electrical circuits, the impedance is independent of the current flow, although, of course, there is a nice relationship (dear old Ohm) between impedance, current, and voltage — and the relationship is linear (let's not talk about alternating currents here…). In hydraulics this is not the case. The resistance to flow in an hydraulic circuit is not constant, but is related to flow velocity. In turbulent flow, the head loss is more or less proportional to the square of the velocity (and hence the square of the flow) (the best fit empirical exponent is 1.83, not 2). Contrast this with electrical circuits, where the voltage drop (analogous to head loss) is linearly proportional to the current (analogous to flow).

EBEBRATT-Ed

What @Jamie Hall said