Peter,
The Rossi effect is controlled in a narrow range by balancing heat removal
and heat addition.
It will not work reliably without constant heat removal.
Therefore, power input related to the proper operation must be included as
P-in.
Jones
From: Peter Gluck
Dear
Dear Jones,
If the power has to be included, it has to be measured. But only a part of
the energy consumed by the motor of the pump is used to make the water to
moveand this produces a small heating of water due to friction,
So the reverse is true- the power of the motor has to be subtracted from
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
Want I wanted to say- the pump is part of the cooling circuit to which the
heat produced is transfered. 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
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
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,
From: Peter Gluck
OK, old friend I understand what you say, 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?
Dear Peter,
Yes, but we can take it further. As a
Dear Jones,
make some calculations please. for this case. OK?
The word constrictors is terrific (Boa)
Peter
On Wed, Apr 20, 2011 at 7:18 PM, Jones Beene jone...@pacbell.net wrote:
*From:* Peter Gluck
OK, old friend I understand what you say, the energy of the pump is
consumed, is money
a) The pump does add a tiny amount of heat to the water passing
through it : the input temperature should be measured AFTER the pump.
b) There is FRICTIONAL loss in a pipe
http://www.efunda.com/formulae/fluids/calc_pipe_friction.cfm
(Though in that calculation it's expressed as pressure drop).
yes, good point Jones, the system input power includes the power to operate the
pump and the resistive heaters.
Harry
Alan J Fletcher wrote:
a) The pump does add a tiny amount of heat to the water passing
through it : the input temperature should be measured AFTER the pump.
It is always measured after the pump. It would be rather difficult to
measure it before the pump. Offhand, I can't imagine how you would
hmm if water flow is required then could it be powered by the e-cat through
convective heating.
Harry
From: Jones Beene jone...@pacbell.net
To: vortex-l@eskimo.com
Sent: Wed, April 20, 2011 10:34:59 AM
Subject: [Vo]:Pump power must be included
Peter,
The “Rossi effect” is controlled
Harry Veeder wrote:
yes, good point Jones, the system input power includes the power to
operate the pump and the resistive heaters.
That is incorrect. Please review the messages I have posted. The input
power does not include the pump any more than it includes the overhead
lights or an
Whether or not the water flow is powered by the pump
or a waterfall, the kinetic energy of the flow may be a factor.
A lack of water movement may explain why some PF type cells failed to perform
in
the past.
They depended on the fickle nature of convection to spring to life.
Harry
yes,
Harry Veeder wrote:
Whether or not the water flow is powered by the pump
or a waterfall, the kinetic energy of the flow may be a factor.
No, it may not. That's out of the question. I have operated many flow
calorimeters of all sizes and types, and there is absolutely no way you
can detect
the exiting water after measurement into same
diameter pipe as the source and measure with an identical flow rate meter?
Fran
From: Harry Veeder [mailto:hlvee...@yahoo.com]
Sent: Wednesday, April 20, 2011 3:11 PM
To: vortex-l@eskimo.com
Subject: EXTERNAL: Re: [Vo]:Pump power must be included
Whether
Roarty, Francis X wrote:
I agree the energy utilized should be subtracted from
the output but how much of the pressure or flow rate is actually
removed from the system?
No measurable amount is removed. I guarantee that.
-- the differential measurements are only for temp
In reply to Alan J Fletcher's message of Wed, 20 Apr 2011 09:37:23 -0700:
Hi,
[snip]
If we assume 100 psi for the mains pressure, then a flow rate of 1 L /s equates
to a total power of 724 W, assuming all the power in the water gets used. This
would raise the temperature of that water by 0.173
Jed Rothwell wrote:
Harry Veeder wrote:
Whether or not the water flow is powered by the pump
or a waterfall, the kinetic energy of the flow may be a factor.
No, it may not. That's out of the question. I have operated many flow
calorimeters of all sizes and types, and there is absolutely no
From: Roarty, Francis X francis.x.roa...@lmco.com
To: vortex-l@eskimo.com vortex-l@eskimo.com
Sent: Wed, April 20, 2011 3:34:37 PM
Subject: RE: EXTERNAL: Re: [Vo]:Pump power must be included
Harry,
I agree the energy utilized should be subtracted from the
output
but how much
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