On Fri, 2009-01-16 at 13:15 +0000, Electricky Dicky wrote: > If the initial pressure is next to damn all for your system (water > tank head only) lets assume it is 0.1bar assume the accumulator has a > volume of 1000cc. When YOUR pump kicks out at 2 bar the air volume in > the accumulator will be 1000/(2/0.1)= 50cc now assume water is drawn > from the system until the cut in pressure of the pump is reached at > say 1 bar. The air volume at this point will be 1000/(1/0.1) = 100 cc. > The difference between 50cc and 100cc is the water displaced. > > Now do the same calculations based upon a diapragm type accumulator > pressurised to 1 bar. > > Initial volume at 0.1 bar (water pressure) is 1000cc. When the pump > kicks out at 2 bar the volume will change from 1000 x 1bar to 500 at 2 > bar we now have 500cc of water retained in the accumulator. 10 times > the amount in the unpressurised pot or bottle!
Doh! Right conclusion, wrong sums, wrong reason. Consider firstly the air-filled accumulator, of volume 1 litre. Initial pressure in it when installed is atmospheric, 0 bar(g). Suppose the pump cuts in at 1 bar and goes up to 2 bar. On first use it goes to 2 bar(g) at which point its internal air volume has gone down to 1/(2+1) l, and the water content is 2/3 l. The pump cuts back in at 1 bar(g), at which point the internal volume of water and air are both 1/2 l. Hence, the buffering effect of the accumulator is 2/3 - 1/2 = 1/6 litre. Secondly, same assumptions but a diaphragm accumulator preset to 1 bar(g) (to make the sums easier). Further, assume that the diaphragm stretches to fill the accumulator when no water pressure is present [1]. On first connection, the pump takes it to 2 bar(g) so the volume in the air side of the accumulator is now 1/2 l. Release water, and the pump won't kick in until 1 bar at which point the diaphragm again fills the accumulator. So, the buffering effect is now 1/2 l, or three times that of the plain air accumulator. [1] Without knowledge of the internal construction of an accumulator, I'd be wary of assuming that the diaphragm moves that much when no external pressure is applied, so the sums presented are probably an upper limit to the difference in the two scenarios. Now, consider the effect of solubility of air. Assume that the plain accumulator is in a position where its connection is at the bottom. On first connection to water, it's air-filled, and water goes up and down in it. Every time some fresh water goes in, some of the air will dissolve. Every time a tap is run, that water with the dissolved air will leave. Over time, the air volume in the accumulator will decrease. Unless a cunning way is present of adding air (e.g. by draining the system in the winter or running the tank dry), the buffering effect of the accumulator will gradually diminish. Solubility of air in water at 25 C and 2 bar(g) is about 0.068 g/l. The initial mass of air in the accumulator is about 1.2 g. So, it would require 1.2 / 0.068 = 18 l of water to flow in and out of the accumulator before there was no air left. That's not many pump cycles. QED. Wassail!
