OK, old friend I understand what you sya, the energy of the pump is consumed, is money spent for making the generator to work. No connection with heat balance of the system- but goes to expenses. Right? Peter
On Wed, Apr 20, 2011 at 6:42 PM, Jones Beene <jone...@pacbell.net> wrote: > Dear Peter, > > > > I do not understand the problem. There are two systems involved: heat and > electricity > > At the system level P-out is thermal and refers to net heat. The > calorimetry determines P-out for heat. > > > > P-in for the system, not for the calorimetry, is determined by the sum of > all the electrical inputs. The pump must be included as it is necessary. > > > > There is nothing in the calorimetry loop which is used to determine **system > P-in**. > > > > Yes the heat loop itself may have it own designation for P-in and P-out, > but that is not for the system; that is why I believe you could be > conflating the two issues. > > > > > > Jones > > > > > > *From:* Peter Gluck > > > > Want I wanted to say- the pump is part of the cooling circuit to which the > heat produced is transferred. Has nothing to do with the heat produced. > > peter > > On Wed, Apr 20, 2011 at 6:15 PM, Jed Rothwell <jedrothw...@gmail.com> > wrote: > > Peter Gluck wrote: > > Fortunately the inlet temperature of water is measured and this includes > or, if you wish excludes the effect of the pump/motor. > But he effect is negligible- and not on the side of Pin- it is at Pout. > > > > No, not Pout. The heat from the pump shows up past Pout, at the place where > the water stops moving. That would be either in your drain pipe, or -- if > you recycle the water -- in the reservoir tank. As long as the water is > moving at the same speed as it did when it left the pump, the energy has not > yet converted to heat. > > - Jed > > > > > -- > Dr. Peter Gluck > > Cluj, Romania > > http://egooutpeters.blogspot.com > > > -- Dr. Peter Gluck Cluj, Romania http://egooutpeters.blogspot.com
