Jed, is it possible to calculate the amount of power that is being added to the
water by looking at the system? I assume that the water is not moving just
prior to being accelerated to finally reach the speed that it is moving inside
the pipe. That may allow you to calculate the kinetic energy that must be
imparted to it which would be deposited into the far tank. The frictional
losses within the pipe would have to be supplied by the pump as well. That
heat would likely end up within the water instead of conducting through the
hose surface.
A good first start would be to calculate the kinetic portion of the pump power.
It does make me curious as to where they assume the 3 watts is being
dissipated. I find it difficult to believe that this much power is always
being delivered regardless of the load configuration. In a very short pipe the
losses must be kinetic.
Do you have information concerning the mass of water pumped per second and its
velocity at exit? Energy is equal to 1/2 * mass * velocity squared. The power
can then be determined by dividing by the time during which that energy is
imparted. Give it a shot!
Dave
-Original Message-
From: Jed Rothwell jedrothw...@gmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Nov 18, 2014 5:45 pm
Subject: [Vo]:Heat from the pump would not be a problem even if we could detect
it
Some people are still confused about the input power from the pump in Mizuno’s
calorimetry. Let me point out two things about this:
1. While there has to be some heat from the pump, with this configuration,
that heat is too small and too close to the noise to be detected with this
equipment. This is obvious from the data I uploaded.
2. If the heat could be detected, it would still not be a problem. It
would simply be included in the baseline. In calorimetry you sometimes see
input power from the instruments themselves, or from something like a
circulation fan in a Seebeck calorimeter. The pump runs under the same
conditions at all times so power is stable and it would be easy to subtract.
Let me discuss these two points in more detail.
Some people seem to have difficulty grasping the notion that heat can be too
small to measure with a given instrument. I suppose the heat from this pump is
on the order of ~0.2 W. Based on other data I think ~0.5 W is barely detectable
with this system. The pump heat cannot be measured because it is close to the
noise from ambient temperature changes. With any calorimeter it is always more
difficult to measure at the bottom of the scale down in the noise. You can
measure the difference between 3.0 W and 3.2 W more easily than the difference
between 0.0 W and 0.2 W.
Mizuno left the pump running for a day to see whether he could detect heat from
it. Looking at the water temperature for the day he did not see an elevation
above ambient. No doubt there was one, but he could not see it. Ambient
temperature changes swamped it. One minute the room is warmer than the water by
0.2°C. A few minutes later the room heater turns off and the reactor is soon
warmer by 0.1°C. This is what I observed on October 23 when we did not conduct
testing until afternoon and I left the Omega thermometer in the T1-T2
comparative mode. That means heat from the room is sloshing into and out of the
water, albeit at a very low rate thanks to the insulation. Still, it is
apparently doing that enough to hide the effects of the pump. Once the water is
heated above ambient, the heat sloshes out only.
After the heating and air-conditioning in Mizuno’s lab is upgraded, it may be
possible to detect a slight average temperature rise above ambient caused by
the pump. If that happens we can then subtract that difference from the
temperature readings. That is what I mean by included in the baseline.
A low level of input power will cause a persistent average higher temperature
compared to ambient. It will not cause the temperature to climb higher and
higher indefinitely, until you can see it. The temperature instead reaches a
peak where losses equal input. In other words, after a while the system
functions as an isoperibolic calorimeter, not an adiabatic one. Because
insulation is not perfect.
That seems to confuse people. Let me go it over it again with an example. On
October 21 the average power measured with the reactor metal and water is
roughly 4.7 W. That is 1.4 W from the resistance heater pulses plus 3.3 W of
anomalous power, ignoring losses. (If you want to estimate losses, which I
figure are ~1 W, they should all be added to the anomalous power by this
method.)
The temperature rises throughout the day as you see in Fig. 7. In Fig. 9 we
zoom in, and you can measure the water and wall temperature increase from hour
1.0 to hour 2.0. This increase is 0.3°C, which means the power during this time
is ~3.5 W (ignoring losses). That was all anomalous power; by hour 1.0, the
effect of the
