On Fri, 16 Jan 2009 18:11:49 +0000, Martin Phillips <[email protected]> wrote:
>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. Ok 1 out of 3 is not too bad <G> Probably a University entrance pass in our current educational system <G> > >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. I assume from your numbers that you are working absolute. Therefore the initial pressure in the pot is 1 bar and that rises to 3 bar at pump cutout, otherwise I cannot see why you use 1/(2+1) L. > >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. Actually I thought that that was what I said? apart from the "three times" I look forward to being eddykated. -- Richard
