subject:"\[Vo\]\:Report"


 Dear Gigi,

You wrote:

The pump absorbs from the grid a given amount of electrical power: for the 
sake of simplicity let's say 12 W. According to the data sheet 3 W are 
transformed into mechanical work and, eventually, transformed into heat inside 
the water. The other 9 W are directly dissipated into heat: part of this heat, 
as we measured is transferred to the water. It would be hard to say that if the 
pump wall near the water chamber is at 50°C the heat is not transferred to the 
water. I do not see any kind of thermal isolator in the disassembled pump. If 
1.5 W is transferred to the water everything is OK.

Your description above of how the input power is directed leads to a question.  
Have you contacted the vendor about the assumption that 3 watts of mechanical 
work is always occurring within the pump section of the device?  Does the 
amount of this heat generation depend upon the pressure that the pump must work 
into?  If so, that is what I assume may be occurring with your  substitution of 
a 5 mm tube when compared to the original 10 mm tubing.  The pressure is 16 
times higher in your case for the same water flow rate.

To be fair, you are suggesting that the system used by Mizuno is also forcing 
the pump to run at its maximum pressure due to the long tubing frictional loss. 
 This is in contrast to what Jed has been told by Mizuno.   I am not sure of 
the best way to prove your point short of connecting a new pipe that is 10 mm 
ID and 16 Meters long to the pump.

In a previous message I gave you a couple of links. In the second link, in the 
APPENDIX you will find the simulation

I sent you a second email where I mentioned that I have found the simulation 
and am reviewing it.  So far, I find it interesting.

I hope now you can remove the confusion in your mind.

That is a hopeless task Gigi. :-)

Regards,

Dave

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Excuse me Jed,

but I think that is very simple for you to say that I do not understand
calorimetry if you reply to a question that I did not ask.

The refrigerator example is quite evident, but is unfit to our situation,
by various causes. The main one is that there you have an abrupt *change *of
air temperature, while in the 18h test the air temperature is falling at a
modest rate of 0,36 °C/h that is very simple to follow for the calorimeter.

But the main question is that in your report you made a statement that is
not true. This is you:

*The temperature rose for 1.5 hours until it stabilized 0.6°C above room
temperature (Fig. 19.) It stabilized because heat losses equal the power
from the pump.*

If from now on the losses are equal to the pump power, since you have

Loss = K * deltaTand  PumpPower = loss = constant

and since K is valid over a broad range of deltaT you should have a
constant deltaT. Instead, deltaT keeps increasing for a few hours when
considering the missing file.

I fully  agree with your words, howewer. So in the case we both do not
understand anything of calorimetry.


So going back to the plots in the missing file you considered only the
first 1.5 hour only because just after the ambient temperature starts
decreasing. What have it happened if the ambient did not change for 5-6
hours? Can you answer this question? Where in the data do you see that 0.6
°C is the maximum? The true is that you simply stop there and are happy
with your data, but there is no theoretycal reason.

Please don't be contemptuous and dismissive; it is not the case. If someone
does't understand calorimetry it is not me.

By the way we performed a full simulation of this calibration: we got
exactly the same curve but at a much higher pump power. We shall show you
and Mizuno the results, hopefully at ICCF-19. Will you be there?

Regards

2015-01-13 16:52 GMT+01:00 Jed Rothwell jedrothw...@gmail.com:

 Gigi DiMarco gdmgdms...@gmail.com wrote:


 I could say that this is false but I will be fair and I will say that
 this is not true. From the missing file (Mizuno's data) we get the
 following situation for the difference between water and ambient temperature

 (4h 2.5°C) (5h 2.9°C) (6h 3.1°C) (7h 3.2°C) (8h 3.2°C) (9h 3.2°C) (10h
 3.1°C) (11h 3.1°C) (12h 3.1°C) (13h 3.0°C)
 (14h 2.9°C) (15h 2.8°C) (16h 2.7°C) (17h 2.6°C) (18h 2.5°C)

 were the temperatures are taken at the beginning of the hour.

 How do you explain this?


 The ambient temperature is falling. The reactor is well insulated so it
 takes longer to cool off than the room does. That is all there is to it.

 You can simulate this easily with the following steps:

 1. Fill a glass with warm water.
 2. Measure the temperature difference between the water and air.
 3. Move the glass to the refrigerator, and measure the difference between
 the water and the air in the refrigerator.
 4. The second temperature difference will be much larger, because the
 water does not instantly cool off. Does that mean there is a source of heat
 in the water? No.

 If you do not understand this, you are not qualified to do calorimetry.

 I pointed this out before. I will not point it out again, and I will not
 discuss this with you again.

 - Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

2015-01-13 Thread Jed Rothwell

Gigi DiMarco gdmgdms...@gmail.com wrote:


 The refrigerator example is quite evident, but is unfit to our situation,
 by various causes. The main one is that there you have an abrupt *change *of
 air temperature, while in the 18h test the air temperature is falling at a
 modest rate of 0,36 °C/h that is very simple to follow for the calorimeter.


No, it isn't. That is why a gap opens between the room temperature and the
calorimeter, and the gap persists until early morning.



 If from now on the losses are equal to the pump power, since you have

 Loss = K * deltaTand  PumpPower = loss = constant

 and since K is valid over a broad range of deltaT you should have a
 constant deltaT.


No, it isn't. See Newton's law of cooling.



 So going back to the plots in the missing file you considered only the
 first 1.5 hour only because just after the ambient temperature starts
 decreasing. What have it happened if the ambient did not change for 5-6
 hours? Can you answer this question?


Yes, I can. If ambient stays stable, the reactor and water temperature will
remain stable at 0.6 deg C above room



 Where in the data do you see that 0.6 °C is the maximum?


It goes no higher after 1.4 hours. You can see this in other data sets as
well, such as early in the morning with this data set. Whenever ambient
remains stable for a few hours or more, the reactor temperature always
settles 0.6 deg C warmer.




 Please don't be contemptuous and dismissive; it is not the case. If
 someone does't understand calorimetry it is not me.


You do not understand Newton's law of cooling and you cannot tell the
difference between ambient cooling and heat generation in a cell. In my
opinion, you are terribly confused and totally unqualified to do
calorimetry.

- Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Dear Gigi,

I have begun to analyze your report and find something that does not seem 
logical according to my understanding of heat flow.  On your figure A2 I see 
that you have overlaid your simulation results upon Jed's figure.  The 
correspondence between the curves is remarkable and you should be commended for 
your work.

The issue that I need to resolve is that the delta temperature between the 
Dewar and ambient is actually increasing during this time.  Also, the delta for 
the reactor is becoming less with time as I was expecting.  In order for the 
temperature delta to increase you would have to supply some form of heat power 
to that device.  The model that you are using is extremely simple and certainly 
does not suggest that anything more complex would be happening.

How do you explain that the delta is increasing?  Is there some process that is 
supplying extra power into the Dewar once the pump is turned off?

Regards,

Dave

 

 

 

-Original Message-
From: Jed Rothwell jedrothw...@gmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Jan 13, 2015 2:06 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised



Gigi DiMarco gdmgdms...@gmail.com wrote:

 



The refrigerator example is quite evident, but is unfit to our situation, by 
various causes. The main one is that there you have an abrupt change of air 
temperature, while in the 18h test the air temperature is falling at a modest 
rate of 0,36 °C/h that is very simple to follow for the calorimeter.






No, it isn't. That is why a gap opens between the room temperature and the 
calorimeter, and the gap persists until early morning.


 




If from now on the losses are equal to the pump power, since you have


Loss = K * deltaTand  PumpPower = loss = constant

and since K is valid over a broad range of deltaT you should have a constant 
deltaT.





No, it isn't. See Newton's law of cooling.


 



So going back to the plots in the missing file you considered only the first 
1.5 hour only because just after the ambient temperature starts decreasing. 
What have it happened if the ambient did not change for 5-6 hours? Can you 
answer this question?




Yes, I can. If ambient stays stable, the reactor and water temperature will 
remain stable at 0.6 deg C above room 


 


 Where in the data do you see that 0.6 °C is the maximum?




It goes no higher after 1.4 hours. You can see this in other data sets as well, 
such as early in the morning with this data set. Whenever ambient remains 
stable for a few hours or more, the reactor temperature always settles 0.6 deg 
C warmer.






 


Please don't be contemptuous and dismissive; it is not the case. If someone 
does't understand calorimetry it is not me.





You do not understand Newton's law of cooling and you cannot tell the 
difference between ambient cooling and heat generation in a cell. In my 
opinion, you are terribly confused and totally unqualified to do calorimetry.


- Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Dear Jed,

in your report you write:













*The temperature rose for 1.5 hours until it stabilized 0.6°Cabove room
temperature (Fig. 19.) It stabilized because heat losses equal the power
from thepump. In other words, with low input power after 1.5 hours, this
system acts as an isoperiboliccalorimeter. Based on the cooling curve after
the ambient temperature fell during the night, this0. 6°C difference
indicates that the pump delivers only ~0.4 W of heat to the circulating
water.More to the point, with this method of adiabatic calorimetry the
0.6°C temperature increaseover ambient is not included in the calculation
of excess heat, because the pump is left on all thetime, and it always does
the same amount of work, so the temperature is always 0.6°C aboveambient.
To be specific, with this method, the starting water temperature is
subtracted from theending water temperature, and the starting temperature
is already 0.6°C warmer than ambient.With other methods of calorimetry,
heat is measured by comparing the reactor temperature toambient. With these
methods, heat from the pump has to be subtracted from the total, or it
willbe mistaken for excess heat.*

I could say that this is false but I will be fair and I will say that this
is not true. From the missing file (Mizuno's data) we get the following
situation for the difference between water and ambient temperature

(4h 2.5°C) (5h 2.9°C) (6h 3.1°C) (7h 3.2°C) (8h 3.2°C) (9h 3.2°C) (10h
3.1°C) (11h 3.1°C) (12h 3.1°C) (13h 3.0°C)
(14h 2.9°C) (15h 2.8°C) (16h 2.7°C) (17h 2.6°C) (18h 2.5°C)

were the temperatures are taken at the beginning of the hour.

How do you explain this?

Thanks


2015-01-10 17:50 GMT+01:00 Jed Rothwell jedrothw...@gmail.com:

 I wrote:


 Where do you see that? At what hour? At hour 2.2 it reaches the peak. The
 water temperature is 23.3°C and ambient is 22.8°C.


 I meant to say: At hour 1.4 it reaches the peak. Taking the value at 2.2
 hours, the water temperature is 23.3°C and ambient is 22.8°C.

 At 2.2 hours the numbers are stable.

 I refer to Fig. 19 on p. 25 here:

 http://lenr-canr.org/acrobat/RothwellJreportonmi.pdf

 - Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

*I missed the simulation for some reason.  Where can I find that?  Sorry if
I overlooked it.*

In a previous message I gave you a couple of links. In the second link, in
the *APPENDIX *you will find the simulation



*The fact that you measure 4.5 watts versus a specification of 3 watts
maximum suggests that something is wrong with your procedure.  How do you
explain that difference?*
Instead, I think it suggests that you do not read carefully what we write
since you seem able to understand what we write

The pump absorbs from the grid a given amount of electrical power: for the
sake of simplicity let's say 12 W. According to the data sheet 3 W are
transformed into mechanical work and, eventually, transformed into heat
inside the water. The other 9 W are directly dissipated into heat: part of
this heat, as we measured is transferred to the water. It would be hard to
say that if the pump wall near the water chamber is at 50°C the heat is not
transferred to the water. I do not see any kind of thermal isolator in the
disassembled pump. If 1.5 W is transferred to the water everything is OK.



*I hope now you can remove the confusion in your mind.*

2015-01-13 0:07 GMT+01:00 David Roberson dlrober...@aol.com:

 I missed the simulation for some reason.  Where can I find that?  Sorry if
 I overlooked it.

 Do you have data that shows the mass flow rate when a 10 mm tube is
 attached to the pump output?  I assume that a large pipe is on the suction
 port.

 You need to attach a full length 10 mm tube to the pump and measure the
 flow rate and heating as a main step.  There are far too many variables
 associated with operation of the pump with the 5 mm pipe.  I have pointed
 out several problems that need to be addressed.  If you do this and also
 measure the AC power into the pump and then clean up the pump bearings so
 that the frictional losses are low then that will go a long way toward
 proving your position.

 Do you have any method of verifying that the frictional losses are as low
 as those of the pump used by Mizuno?  The fact that you measure 4.5 watts
 versus a specification of 3 watts maximum suggests that something is wrong
 with your procedure.  How do you explain that difference?

 Also, the difference between what you measure and what Mizuno and Jed
 measures may be nothing more than those associated with operation in a
 different pump pressure range and a damaged pump.  These types of questions
 remain unanswered.

 Dave



  -Original Message-
 From: Gigi DiMarco gdmgdms...@gmail.com
 To: vortex-l vortex-l@eskimo.com
 Sent: Mon, Jan 12, 2015 5:34 pm
 Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

   Dave,

  you said nothing about simulations that should be a confirmation of our
 experiments. But I think that we can do something more: what will convince
 you that we are right and Mizuno is wrong?

  Regards

 2015-01-12 23:17 GMT+01:00 David Roberson dlrober...@aol.com:

 Dear Giancarlo,

 Thanks for publishing your report in English so that many of us that do
 not speak Italian can understand it.  There is no disagreement between the
 method that I used to calculate the kinetic transport power and what you
 would have calculated with the same numbers since we used the same basic
 principles.  I relied upon the information from Jed about the mass flow
 rate of the pump where he stated that Mizuno had told him that it was 8
 liters per second.  If you match that rate with your 5 mm pipe as you have
 stated as a plan for replication of Mizuno's experiment then you will
 obtain my results.

 I do not have a pump and 16 meters of 10 mm inside diameter tubing before
 me to determine exactly what flow rate is obtained.   It is going to be
 necessary for you to either obtain a matching pipe or for us to verify
 exactly what flow rate is being measured by Mizuno before a final answer
 can be established.   Jed apparently believes that the friction within the
 16 meter tubing is not sufficient to reduce the unloaded pump fluid flow
 rate to a value that is anywhere close to the 2.31 liters per minute that
 you are proposing.   In your report, you state that you are matching the
 performance seen by Mizuno as far as fluid flow rate is concerned but I
 strongly doubt that this is occurring.

 If you make additional calculations you will see that the pressure
 required at the pump output is (10 mm/5 mm)^4 or 16 times as large when
 achieving the same flow rate for a 5 mm tube as compared to a 10 mm  tube.
 This is a dramatic difference and you find that you quickly run out of head
 room when using the 5 mm tube for your test.   Just this reason alone
 should be sufficient for you to realize that your replication attempt is
 failed.  And, as further supporting evidence, the pumping power needed to
 reach the 8 liters per minute flow rate when using a 10 mm tube is only
 .192 watts which is well within the operational range of the MD-6.

 We can approach the power required

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

2015-01-13 Thread Jed Rothwell

Gigi DiMarco gdmgdms...@gmail.com wrote:


 I could say that this is false but I will be fair and I will say that this
 is not true. From the missing file (Mizuno's data) we get the following
 situation for the difference between water and ambient temperature

 (4h 2.5°C) (5h 2.9°C) (6h 3.1°C) (7h 3.2°C) (8h 3.2°C) (9h 3.2°C) (10h
 3.1°C) (11h 3.1°C) (12h 3.1°C) (13h 3.0°C)
 (14h 2.9°C) (15h 2.8°C) (16h 2.7°C) (17h 2.6°C) (18h 2.5°C)

 were the temperatures are taken at the beginning of the hour.

 How do you explain this?


The ambient temperature is falling. The reactor is well insulated so it
takes longer to cool off than the room does. That is all there is to it.

You can simulate this easily with the following steps:

1. Fill a glass with warm water.
2. Measure the temperature difference between the water and air.
3. Move the glass to the refrigerator, and measure the difference between
the water and the air in the refrigerator.
4. The second temperature difference will be much larger, because the water
does not instantly cool off. Does that mean there is a source of heat in
the water? No.

If you do not understand this, you are not qualified to do calorimetry.

I pointed this out before. I will not point it out again, and I will not
discuss this with you again.

- Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Gigi,

I just recalculated the combined thermal time constants and now I believe you 
have them right.  I must have performed that calculation 5 times and kept 
getting a different answer!  The thermal K's that you used are inverted from 
the normal R's(resistors) that I always use when calculating time constants.  
Since one value is so much larger than the other the inverses are vastly 
different.

Please check yourself to ensure that my latest figure is correct.  I finally 
get a thermal time constant of 5.84 hours.  There is little doubt that the 
power pulses and any resulting LENR power will influence the output with that 
large of a time constant.  This is particularly true since you begin your 
analysis only about 2 hours after the last pulse.

In the case of the dead pump, the problem is multiplied by the extreme time 
constant of the Dewar which is 36.11 hours according to my latest calculation 
using your values.  When the pump stops, most of the energy contained within 
the Dewar is locked in place.  The same is not true for the body of the power 
vessel which only has a time constant of 4.2 hours.

Dave

 

 

 

-Original Message-
From: David Roberson dlrober...@aol.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Jan 13, 2015 4:46 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised


Take your time Gigi, we want to get to the facts.  I am very impressed by the 
simulations that you have shown and how well they match the curves made by Jed. 
 Now, we need to verify that things add up as they should by combining thermal 
resistance and capacities of two parts to get to the whole.

They must combine according to normal physical laws.  My first attempt using 
your models did not seem to match properly.  That is what I want you to show.

And, if the thermal time constant is large, then average power due to the test 
itself will show up.  As you know, you are assuming that there is nothing 
except for the input drive signal of 20 watts and the pump leakage power.  I 
want to see how those parameters impact your results.  Then, we need to 
determine whether or not we can match the curves of Jed with a actual input 
signal due to LENR using your model.

I also still believe that the pump power is less than you are considering.  
Nevertheless, I will keep an open mind and give your model an opportunity to 
show its strength.

Thanks,

Dave

 

 

 

-Original Message-
From: Gigi DiMarco gdmgdms...@gmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Jan 13, 2015 4:35 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised


Ok, we need some time to perform the full set of simulations. But please do not 
take for sure that the pump only test had exactly the same configuration than 
the test run had. We will present them as soon as we are confident that all the 
problems have been settled.



2015-01-13 22:18 GMT+01:00 David Roberson dlrober...@aol.com:

Gigi,

I have another issue that you might be able to discuss.  You made two 
independent simulations of the behavior of the Dewar temperature and reactor 
body over time.  In one case the pump was working and in the other if had 
failed.  I took the values that you calculated for the thermal components of 
your model and get very different results for the combined time constant than 
what you have shown for the individual ones.

The thermal masses should be added in parallel directly which you seem to have 
done.  I combined the thermal impedances in what I consider the proper manner.  
When a final calculation of the time constant is computed by using the two 
others, the number does not come very close to matching what you are using for 
the first case.

Please take time to perform that combination on this forum for us to view and 
analyze.  Also, please turn that answer into an actual hour figure for us.  
This will be very important as we attempt to understand the impact of residual 
drive signal and any additional due to LENR activity.

Begin with the time constant you calculate in hours.

Regards,

Dave

 

 

 

-Original Message-
From: David Roberson dlrober...@aol.com
To: vortex-l vortex-l@eskimo.com

Sent: Tue, Jan 13, 2015 3:27 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised


I may have answered my own question below.   The drop in ambient acts much like 
a negative signal as I have proposed before.  Eventually the delta will become 
zero for both signals. 

Forget the first question and concentrate upon the next one.

I notice that in both curves that you use to determine the pump power leakage 
actual true signal power due to the 20 watt drive and any device LENR power was 
contributing to the total.   The shape of the temperature curves with time 
clearly show an initial rise during the first few hours that affect your final 
answer.

What have you done to subtract this effect from your determination of the 
constant average power presumed to be leaking from

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

According to Jed the pump is never turned off. So this is the real fact:
there is no excess heat, only the pump.
Plus the calorimeter external and internal time constant
(capacity+resistance)

We can overlaid the experimental figures only by using the calorimeter
parameter and an estimated pump power [in line with our measurements].
That's all

Sorry about that.

2015-01-13 20:49 GMT+01:00 David Roberson dlrober...@aol.com:

 Dear Gigi,

 I have begun to analyze your report and find something that does not seem
 logical according to my understanding of heat flow.  On your figure A2 I
 see that you have overlaid your simulation results upon Jed's figure.  The
 correspondence between the curves is remarkable and you should be commended
 for your work.

 The issue that I need to resolve is that the delta temperature between the
 Dewar and ambient is actually increasing during this time.  Also, the delta
 for the reactor is becoming less with time as I was expecting.  In order
 for the temperature delta to increase you would have to supply some form of
 heat power to that device.  The model that you are using is extremely
 simple and certainly does not suggest that anything more complex would be
 happening.

 How do you explain that the delta is increasing?  Is there some process
 that is supplying extra power into the Dewar once the pump is turned off?

 Regards,

 Dave



  -Original Message-
 From: Jed Rothwell jedrothw...@gmail.com
 To: vortex-l vortex-l@eskimo.com
 Sent: Tue, Jan 13, 2015 2:06 pm
 Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

   Gigi DiMarco gdmgdms...@gmail.com wrote:


The refrigerator example is quite evident, but is unfit to our
 situation, by various causes. The main one is that there you have an abrupt 
 *change
 *of air temperature, while in the 18h test the air temperature is
 falling at a modest rate of 0,36 °C/h that is very simple to follow for the
 calorimeter.


  No, it isn't. That is why a gap opens between the room temperature and
 the calorimeter, and the gap persists until early morning.



   If from now on the losses are equal to the pump power, since you have

  Loss = K * deltaTand  PumpPower = loss = constant

 and since K is valid over a broad range of deltaT you should have a
 constant deltaT.


  No, it isn't. See Newton's law of cooling.



   So going back to the plots in the missing file you considered only the
 first 1.5 hour only because just after the ambient temperature starts
 decreasing. What have it happened if the ambient did not change for 5-6
 hours? Can you answer this question?


  Yes, I can. If ambient stays stable, the reactor and water temperature
 will remain stable at 0.6 deg C above room



   Where in the data do you see that 0.6 °C is the maximum?


  It goes no higher after 1.4 hours. You can see this in other data sets
 as well, such as early in the morning with this data set. Whenever ambient
 remains stable for a few hours or more, the reactor temperature always
 settles 0.6 deg C warmer.




  Please don't be contemptuous and dismissive; it is not the case. If
 someone does't understand calorimetry it is not me.


  You do not understand Newton's law of cooling and you cannot tell the
 difference between ambient cooling and heat generation in a cell. In my
 opinion, you are terribly confused and totally unqualified to do
 calorimetry.

  - Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

I may have answered my own question below.   The drop in ambient acts much like 
a negative signal as I have proposed before.  Eventually the delta will become 
zero for both signals. 

Forget the first question and concentrate upon the next one.

I notice that in both curves that you use to determine the pump power leakage 
actual true signal power due to the 20 watt drive and any device LENR power was 
contributing to the total.   The shape of the temperature curves with time 
clearly show an initial rise during the first few hours that affect your final 
answer.

What have you done to subtract this effect from your determination of the 
constant average power presumed to be leaking from the pump?   It would be 
useful if you include this heat input into your model and see how that would 
modify the average power that is coming form the pump.  Since you know the 
shape of this heat signal and you assume that no additional power is added by 
the LENR effect, it should be easy to model it as an additional heat source.  
Spice would handle this nicely.

Dave


 

-Original Message-
From: David Roberson dlrober...@aol.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Jan 13, 2015 2:49 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised


Dear Gigi,

I have begun to analyze your report and find something that does not seem 
logical according to my understanding of heat flow.  On your figure A2 I see 
that you have overlaid your simulation results upon Jed's figure.  The 
correspondence between the curves is remarkable and you should be commended for 
your work.

The issue that I need to resolve is that the delta temperature between the 
Dewar and ambient is actually increasing during this time.  Also, the delta for 
the reactor is becoming less with time as I was expecting.  In order for the 
temperature delta to increase you would have to supply some form of heat power 
to that device.  The model that you are using is extremely simple and certainly 
does not suggest that anything more complex would be happening.

How do you explain that the delta is increasing?  Is there some process that is 
supplying extra power into the Dewar once the pump is turned off?

Regards,

Dave

 

 

 

-Original Message-
From: Jed Rothwell jedrothw...@gmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Jan 13, 2015 2:06 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised



Gigi DiMarco gdmgdms...@gmail.com wrote:

 



The refrigerator example is quite evident, but is unfit to our situation, by 
various causes. The main one is that there you have an abrupt change of air 
temperature, while in the 18h test the air temperature is falling at a modest 
rate of 0,36 °C/h that is very simple to follow for the calorimeter.






No, it isn't. That is why a gap opens between the room temperature and the 
calorimeter, and the gap persists until early morning.


 




If from now on the losses are equal to the pump power, since you have


Loss = K * deltaTand  PumpPower = loss = constant

and since K is valid over a broad range of deltaT you should have a constant 
deltaT.





No, it isn't. See Newton's law of cooling.


 



So going back to the plots in the missing file you considered only the first 
1.5 hour only because just after the ambient temperature starts decreasing. 
What have it happened if the ambient did not change for 5-6 hours? Can you 
answer this question?




Yes, I can. If ambient stays stable, the reactor and water temperature will 
remain stable at 0.6 deg C above room 


 


 Where in the data do you see that 0.6 °C is the maximum?




It goes no higher after 1.4 hours. You can see this in other data sets as well, 
such as early in the morning with this data set. Whenever ambient remains 
stable for a few hours or more, the reactor temperature always settles 0.6 deg 
C warmer.






 


Please don't be contemptuous and dismissive; it is not the case. If someone 
does't understand calorimetry it is not me.





You do not understand Newton's law of cooling and you cannot tell the 
difference between ambient cooling and heat generation in a cell. In my 
opinion, you are terribly confused and totally unqualified to do calorimetry.


- Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Jed,

I think you should study heat transfer. I suggest you the book by Incropera
et al.

In one comment you say that the loss is equal to the pump power and the
system stay constant; in the following comment you do not remember this and
start speaking about the Newton's law of cooling. You present your
arguments as a priest would present his Bible readings. Jumping from one
chapter to the other.

Could you tell us why the equation loss = supplied power doesn't hold
during the steady state? Have you a good argument or only fatwas?

The calorimeter loss equation doesn't imply that the room temperature stay
constant. Why do you insist on that? Ask Dave...
Maybe you can trust him better than me.

Regards

2015-01-13 20:05 GMT+01:00 Jed Rothwell jedrothw...@gmail.com:

 Gigi DiMarco gdmgdms...@gmail.com wrote:


 The refrigerator example is quite evident, but is unfit to our situation,
 by various causes. The main one is that there you have an abrupt *change
 *of air temperature, while in the 18h test the air temperature is
 falling at a modest rate of 0,36 °C/h that is very simple to follow for the
 calorimeter.


 No, it isn't. That is why a gap opens between the room temperature and the
 calorimeter, and the gap persists until early morning.



 If from now on the losses are equal to the pump power, since you have

 Loss = K * deltaTand  PumpPower = loss = constant

 and since K is valid over a broad range of deltaT you should have a
 constant deltaT.


 No, it isn't. See Newton's law of cooling.



 So going back to the plots in the missing file you considered only the
 first 1.5 hour only because just after the ambient temperature starts
 decreasing. What have it happened if the ambient did not change for 5-6
 hours? Can you answer this question?


 Yes, I can. If ambient stays stable, the reactor and water temperature
 will remain stable at 0.6 deg C above room



 Where in the data do you see that 0.6 °C is the maximum?


 It goes no higher after 1.4 hours. You can see this in other data sets as
 well, such as early in the morning with this data set. Whenever ambient
 remains stable for a few hours or more, the reactor temperature always
 settles 0.6 deg C warmer.




 Please don't be contemptuous and dismissive; it is not the case. If
 someone does't understand calorimetry it is not me.


 You do not understand Newton's law of cooling and you cannot tell the
 difference between ambient cooling and heat generation in a cell. In my
 opinion, you are terribly confused and totally unqualified to do
 calorimetry.

 - Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Gigi,

I have another issue that you might be able to discuss.  You made two 
independent simulations of the behavior of the Dewar temperature and reactor 
body over time.  In one case the pump was working and in the other if had 
failed.  I took the values that you calculated for the thermal components of 
your model and get very different results for the combined time constant than 
what you have shown for the individual ones.

The thermal masses should be added in parallel directly which you seem to have 
done.  I combined the thermal impedances in what I consider the proper manner.  
When a final calculation of the time constant is computed by using the two 
others, the number does not come very close to matching what you are using for 
the first case.

Please take time to perform that combination on this forum for us to view and 
analyze.  Also, please turn that answer into an actual hour figure for us.  
This will be very important as we attempt to understand the impact of residual 
drive signal and any additional due to LENR activity.

Begin with the time constant you calculate in hours.

Regards,

Dave

 

 

 

-Original Message-
From: David Roberson dlrober...@aol.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Jan 13, 2015 3:27 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised


I may have answered my own question below.   The drop in ambient acts much like 
a negative signal as I have proposed before.  Eventually the delta will become 
zero for both signals. 

Forget the first question and concentrate upon the next one.

I notice that in both curves that you use to determine the pump power leakage 
actual true signal power due to the 20 watt drive and any device LENR power was 
contributing to the total.   The shape of the temperature curves with time 
clearly show an initial rise during the first few hours that affect your final 
answer.

What have you done to subtract this effect from your determination of the 
constant average power presumed to be leaking from the pump?   It would be 
useful if you include this heat input into your model and see how that would 
modify the average power that is coming form the pump.  Since you know the 
shape of this heat signal and you assume that no additional power is added by 
the LENR effect, it should be easy to model it as an additional heat source.  
Spice would handle this nicely.

Dave


 

-Original Message-
From: David Roberson dlrober...@aol.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Jan 13, 2015 2:49 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised


Dear Gigi,

I have begun to analyze your report and find something that does not seem 
logical according to my understanding of heat flow.  On your figure A2 I see 
that you have overlaid your simulation results upon Jed's figure.  The 
correspondence between the curves is remarkable and you should be commended for 
your work.

The issue that I need to resolve is that the delta temperature between the 
Dewar and ambient is actually increasing during this time.  Also, the delta for 
the reactor is becoming less with time as I was expecting.  In order for the 
temperature delta to increase you would have to supply some form of heat power 
to that device.  The model that you are using is extremely simple and certainly 
does not suggest that anything more complex would be happening.

How do you explain that the delta is increasing?  Is there some process that is 
supplying extra power into the Dewar once the pump is turned off?

Regards,

Dave

 

 

 

-Original Message-
From: Jed Rothwell jedrothw...@gmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Jan 13, 2015 2:06 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised



Gigi DiMarco gdmgdms...@gmail.com wrote:

 



The refrigerator example is quite evident, but is unfit to our situation, by 
various causes. The main one is that there you have an abrupt change of air 
temperature, while in the 18h test the air temperature is falling at a modest 
rate of 0,36 °C/h that is very simple to follow for the calorimeter.






No, it isn't. That is why a gap opens between the room temperature and the 
calorimeter, and the gap persists until early morning.


 




If from now on the losses are equal to the pump power, since you have


Loss = K * deltaTand  PumpPower = loss = constant

and since K is valid over a broad range of deltaT you should have a constant 
deltaT.





No, it isn't. See Newton's law of cooling.


 



So going back to the plots in the missing file you considered only the first 
1.5 hour only because just after the ambient temperature starts decreasing. 
What have it happened if the ambient did not change for 5-6 hours? Can you 
answer this question?




Yes, I can. If ambient stays stable, the reactor and water temperature will 
remain stable at 0.6 deg C above room 


 


 Where in the data do you see that 0.6 °C is the maximum?




It goes

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Ok, we need some time to perform the full set of simulations. But please do
not take for sure that the pump only test had exactly the same
configuration than the test run had. We will present them as soon as we are
confident that all the problems have been settled.

2015-01-13 22:18 GMT+01:00 David Roberson dlrober...@aol.com:

 Gigi,

 I have another issue that you might be able to discuss.  You made two
 independent simulations of the behavior of the Dewar temperature and
 reactor body over time.  In one case the pump was working and in the other
 if had failed.  I took the values that you calculated for the thermal
 components of your model and get very different results for the combined
 time constant than what you have shown for the individual ones.

 The thermal masses should be added in parallel directly which you seem to
 have done.  I combined the thermal impedances in what I consider the proper
 manner.  When a final calculation of the time constant is computed by using
 the two others, the number does not come very close to matching what you
 are using for the first case.

 Please take time to perform that combination on this forum for us to view
 and analyze.  Also, please turn that answer into an actual hour figure for
 us.  This will be very important as we attempt to understand the impact of
 residual drive signal and any additional due to LENR activity.

 Begin with the time constant you calculate in hours.

 Regards,

 Dave



  -Original Message-
 From: David Roberson dlrober...@aol.com
 To: vortex-l vortex-l@eskimo.com
 Sent: Tue, Jan 13, 2015 3:27 pm
 Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

  I may have answered my own question below.   The drop in ambient acts
 much like a negative signal as I have proposed before.  Eventually the
 delta will become zero for both signals.

 Forget the first question and concentrate upon the next one.

 I notice that in both curves that you use to determine the pump power
 leakage actual true signal power due to the 20 watt drive and any device
 LENR power was contributing to the total.   The shape of the temperature
 curves with time clearly show an initial rise during the first few hours
 that affect your final answer.

 What have you done to subtract this effect from your determination of the
 constant average power presumed to be leaking from the pump?   It would be
 useful if you include this heat input into your model and see how that
 would modify the average power that is coming form the pump.  Since you
 know the shape of this heat signal and you assume that no additional power
 is added by the LENR effect, it should be easy to model it as an additional
 heat source.  Spice would handle this nicely.

 Dave


  -Original Message-
 From: David Roberson dlrober...@aol.com
 To: vortex-l vortex-l@eskimo.com
 Sent: Tue, Jan 13, 2015 2:49 pm
 Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

  Dear Gigi,

 I have begun to analyze your report and find something that does not seem
 logical according to my understanding of heat flow.  On your figure A2 I
 see that you have overlaid your simulation results upon Jed's figure.  The
 correspondence between the curves is remarkable and you should be commended
 for your work.

 The issue that I need to resolve is that the delta temperature between the
 Dewar and ambient is actually increasing during this time.  Also, the delta
 for the reactor is becoming less with time as I was expecting.  In order
 for the temperature delta to increase you would have to supply some form of
 heat power to that device.  The model that you are using is extremely
 simple and certainly does not suggest that anything more complex would be
 happening.

 How do you explain that the delta is increasing?  Is there some process
 that is supplying extra power into the Dewar once the pump is turned off?

 Regards,

 Dave



  -Original Message-
 From: Jed Rothwell jedrothw...@gmail.com
 To: vortex-l vortex-l@eskimo.com
 Sent: Tue, Jan 13, 2015 2:06 pm
 Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

   Gigi DiMarco gdmgdms...@gmail.com wrote:


The refrigerator example is quite evident, but is unfit to our
 situation, by various causes. The main one is that there you have an abrupt 
 *change
 *of air temperature, while in the 18h test the air temperature is
 falling at a modest rate of 0,36 °C/h that is very simple to follow for the
 calorimeter.


  No, it isn't. That is why a gap opens between the room temperature and
 the calorimeter, and the gap persists until early morning.



   If from now on the losses are equal to the pump power, since you have

  Loss = K * deltaTand  PumpPower = loss = constant

 and since K is valid over a broad range of deltaT you should have a
 constant deltaT.


  No, it isn't. See Newton's law of cooling.



   So going back to the plots in the missing file you considered only the
 first 1.5 hour only

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Take your time Gigi, we want to get to the facts.  I am very impressed by the 
simulations that you have shown and how well they match the curves made by Jed. 
 Now, we need to verify that things add up as they should by combining thermal 
resistance and capacities of two parts to get to the whole.

They must combine according to normal physical laws.  My first attempt using 
your models did not seem to match properly.  That is what I want you to show.

And, if the thermal time constant is large, then average power due to the test 
itself will show up.  As you know, you are assuming that there is nothing 
except for the input drive signal of 20 watts and the pump leakage power.  I 
want to see how those parameters impact your results.  Then, we need to 
determine whether or not we can match the curves of Jed with a actual input 
signal due to LENR using your model.

I also still believe that the pump power is less than you are considering.  
Nevertheless, I will keep an open mind and give your model an opportunity to 
show its strength.

Thanks,

Dave

 

 

 

-Original Message-
From: Gigi DiMarco gdmgdms...@gmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Jan 13, 2015 4:35 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised


Ok, we need some time to perform the full set of simulations. But please do not 
take for sure that the pump only test had exactly the same configuration than 
the test run had. We will present them as soon as we are confident that all the 
problems have been settled.



2015-01-13 22:18 GMT+01:00 David Roberson dlrober...@aol.com:

Gigi,

I have another issue that you might be able to discuss.  You made two 
independent simulations of the behavior of the Dewar temperature and reactor 
body over time.  In one case the pump was working and in the other if had 
failed.  I took the values that you calculated for the thermal components of 
your model and get very different results for the combined time constant than 
what you have shown for the individual ones.

The thermal masses should be added in parallel directly which you seem to have 
done.  I combined the thermal impedances in what I consider the proper manner.  
When a final calculation of the time constant is computed by using the two 
others, the number does not come very close to matching what you are using for 
the first case.

Please take time to perform that combination on this forum for us to view and 
analyze.  Also, please turn that answer into an actual hour figure for us.  
This will be very important as we attempt to understand the impact of residual 
drive signal and any additional due to LENR activity.

Begin with the time constant you calculate in hours.

Regards,

Dave

 

 

 

-Original Message-
From: David Roberson dlrober...@aol.com
To: vortex-l vortex-l@eskimo.com

Sent: Tue, Jan 13, 2015 3:27 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised


I may have answered my own question below.   The drop in ambient acts much like 
a negative signal as I have proposed before.  Eventually the delta will become 
zero for both signals. 

Forget the first question and concentrate upon the next one.

I notice that in both curves that you use to determine the pump power leakage 
actual true signal power due to the 20 watt drive and any device LENR power was 
contributing to the total.   The shape of the temperature curves with time 
clearly show an initial rise during the first few hours that affect your final 
answer.

What have you done to subtract this effect from your determination of the 
constant average power presumed to be leaking from the pump?   It would be 
useful if you include this heat input into your model and see how that would 
modify the average power that is coming form the pump.  Since you know the 
shape of this heat signal and you assume that no additional power is added by 
the LENR effect, it should be easy to model it as an additional heat source.  
Spice would handle this nicely.

Dave


 

-Original Message-
From: David Roberson dlrober...@aol.com
To: vortex-l vortex-l@eskimo.com
Sent: Tue, Jan 13, 2015 2:49 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised


Dear Gigi,

I have begun to analyze your report and find something that does not seem 
logical according to my understanding of heat flow.  On your figure A2 I see 
that you have overlaid your simulation results upon Jed's figure.  The 
correspondence between the curves is remarkable and you should be commended for 
your work.

The issue that I need to resolve is that the delta temperature between the 
Dewar and ambient is actually increasing during this time.  Also, the delta for 
the reactor is becoming less with time as I was expecting.  In order for the 
temperature delta to increase you would have to supply some form of heat power 
to that device.  The model that you are using is extremely simple and certainly 
does not suggest that anything more

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

2015-01-12 Thread Jed Rothwell

Bob Cook frobertc...@hotmail.com wrote:


 I am not sure how Mizuno measured the 10.8 Watts of power used by the pump.


It says in the report: Mizuno used the WattChecker watt meter to measure
the electric power consumed by the pump, which is 10.8 W.



   I think the pump specifications indicate the pump uses about 22 watts.


No, as I told you, the specifications are written on the side of the pump,
and they are:

Iwaki Co., Magnet Pump MD-6K-N
Maximum capacity: 8/9 L/min
Maximum head: 1.0/1.4
100V 12W/60Hz, 12W/50Hz

This is also in the report.



 I plan to talk with the pump vendor technical staff to better understand
 the performance of this type of pump and the wattage vs voltage/amperage
 specs and the efficiency.


Please do not waste their time. The heat from the pump cannot possibly
affect the calorimetry, for the reasons I stated here and in the report.



 I do not have the same report that you have  identifying the pump
 specifications on page 24.  My version of your report, dated November 14,
 2014, does not include the specification you state exists . . .


For goodness sake download the latest version!

http://lenr-canr.org/acrobat/RothwellJreportonmi.pdf

If you do not see the November 14 version, click on Reload this page. Web
browsers sometimes fail to see they are not accessing the latest version of
a page. As a general rule for anything on the web, when in doubt, press
Reload.



 I think by baseline you mean a condition at which the energy introduced
 into the circulating system by the pump creates a temperature of the
 reactor and water bath and all the reactor internals that is the same and
 in equilibrium with a non-changing differential temperature between the
 ambient atmosphere and the water bath.


Exactly. This is shown in Fig. 19. Heat is measured based on the difference
shown with the purple arrow in Fig. 7. The bottom of that arrow points to
the temperature of the water which already includes heat from the pump.



 In any case a good description of baseline conditions is warranted.


What is the matter with the description on p. 24? It does not seem
complicated to me.

- Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

2015-01-12 Thread Jeff Driscoll

I have not followed this debate closely, but I assume Jed is correct.
So Dave, how do you address this statement:

The steady state baseline includes the heat from the pump, any diversion
from the baseline indicates excess heat.

On Mon, Jan 12, 2015 at 3:44 PM, Gigi DiMarco gdmgdms...@gmail.com wrote:

Dave,

as promised and while you still insist saying that we were deeply wrong,
we have put on-line two different updates

1)
https://gsvit.wordpress.com/2015/01/12/further-measurements-on-the-md-6k-n-pump-used-by-tadahiko-mizuno/

2)
https://gsvit.wordpress.com/2014/12/10/analysis-of-jed-rothwells-report-about-his-calorimetry-performed-on-mizunos-cell/

The first one shows how you are terribly wrong with your calculations
based on the kinetic energy only. We show that your assumption are
completely wrong just referring to usual pump working diagram. In the pump
under test you can not have simultaneously maximum head and maximum flow
rate; the working point we chose was such that we had almost the same
working conditions Mizuno had. Please take your time to read our post
before commenting. The major result is that we measured 43°C in the pump
body very close to the water so it is really easy to understand that,
despite what Jed says, the pump motor delivers a lot of heat to the water;
it is this the power we measure and it is by far much more that the
mechanical power (3 W maximum from the data sheet).

But, let me say that the second link is even more interesting [you have to
go to the end of the article, the Appendix]: we set up a software
simulation tools and were able to replicate by simulation the Mizuno's
measurement. It was enough to evaluate the overall thermal transmittance of
the system that is constant at least for the considered temperature range.
If we simulate the Mizuno's curve starting from a time instant when the
reactor is no more generating excess heat, it is possible to evaluate the
only source of heat: the pump. We have to use only the room temperature as
provided by Mizuno's data and the system starting temperature. The pump
power turns out to be about 4 W.

So we get comparable results by using very different methods

1) Pump theory and data sheet

2) Experiment

3) Simulations

All the rest are only free words.

We are going to apply the simulation to all the Mizuno's experiments to
see if we can get those curves without any excess heat.

Regards and take it easy.

Please, consider to read all the articles in our site concerning the
Mizuno's experiment.

Gigi aka Giancarlo

2015-01-12 19:09 GMT+01:00 David Roberson dlrober...@aol.com:

Bob,

You have uncovered a pump specification that proves that the replication
work by Gigi and allies is not accurate. They report to have determined
that approximately 4.5 watts of thermal power is being absorbed by the
circulating water under their test condition. This amount of reported
power is clearly more than the pump should add and they need to explain why
we should accept their data as accurate.

Also, I have performs extensive calculations within a spreadsheet that is
based upon the lift head versus fluid flow rate of this model pump. It is
capable of delivering less than 1 watt of fluid power into the water
coolant under the best of conditions. My actual calculation is .75 watts
at 6 liters per minute which I rounded off for convenience to 1 watt. I
included both potential as well as kinetic energy related powers.

Any additional power imparted to the water must come from pump friction
and thermal leakage through the construction materials. Without further
careful measurements we or Gigi can not assume that the pump used by
Mizuno is operating at its specification limit of 3 watts. Of course the
measurement of 4.5 watts by Gigi is certainly not representative of a pump
that is in good condition.

The pump manual has several warnings about how easy it is to damage it
and that strongly suggests that Gigi and his team has done just that in
order to obtain their non representative performance. No one but Mizuno
knows the status of his pump during those tests so the only conclusion that
can conservatively be drawn is that the skeptical report by Gigi and team
should not be considered valid.

The pump manual states that the water reservoir must be at least 1 foot
above the pump input port in order to prevent possible air intake along
with the coolant water. Operation under conditions that do not meet this
requirement can damage the pump according to the manual. Unfortunately, in
both of the cases being discussed this was not done. The setup used by
Gigi very clearly shows the pump mounted above the Dewar by several
inches. The same appears true for Mizuno's experiment.

Dave

-Original Message-
From: Bob Cook frobertc...@hotmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Mon, Jan 12, 2015 12:15 pm
Subject: Re: [Vo]:Report on Mizuno's

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

2015-01-12 Thread Daniel Rocha

 To: vortex-l vortex-l@eskimo.com
 Sent: Mon, Jan 12, 2015 12:15 pm
 Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

  Jed--

 I have researched the pump characteristics further and find that this
 pump has a low efficiency and would use  at most about 3 watts of power in
 heating the circulating water.  This is consistent with what you have
 stated.

 I am not sure how Mizuno measured the 10.8 Watts of power used by the
 pump.  I think the pump specifications indicate the pump uses about 22
 watts.  However, The specifications for the amperage and voltage during
 operation would indicate the 29 watts I suggested some time ago.  I plan to
 talk with the pump vendor technical staff to better understand the
 performance of this type of pump and the wattage vs voltage/amperage specs
 and the efficiency.  I will report on what I find.  However, it would
 appear the pump is only about 15% efficient at best in converting
 electrical energy into the mechanical energy causing the circulation. At
 low circuit frictional pressure drop (low heads) it appears even less
 efficient.  I was wrong in assuming an efficient pump.

 I do not have the same report that you have  identifying the pump
 specifications on page 24.  My version of your report, dated November 14,
 2014, does not include the specification you state exists on the side of
 the pump body. In addition I do not think I have the same description of a
 baseline that your make reference to.

 I think by baseline you mean a condition at which the energy
 introduced into the circulating system by the pump creates a temperature of
 the reactor and water bath and all the reactor internals that is the same
 and in equilibrium with a non-changing differential temperature between the
 ambient atmosphere and the water bath. This would allow a reasonable
 determination of the average thermal resistance of the insulation and
 hence a measure of the approach to a desired adiabatic condition of the
 test setup.  In any case a good description of baseline conditions is
 warranted.

 In addition, if you have information as to when it was determined that
 excess reaction heat was produced in the reactor, this would be helpful in
 comparing temperature profiles with rates of change, compared to times when
 there was no excess energy input to the system.  For example, when is the
 excess energy produced with respect to the time the spikes of electrical
 heat are applied to the electrodes?  In this regard it seems that the
 excess energy production, if any, does not continue indefinitely, since the
 temperature increase levels  off and then decrease without the spikes of
 electrical input to the electrodes.  However, does it continue in the time
 frame between spikes of input energy to the electrodes.

 The temperature of the system and water bath should return to the
 baseline with time, if the only input is the energy  was from the
 pump. If excess energy form a reaction continues the temperature should
 level out at somewhat above the baseline.  This would be nice confirmation
 of excess energy.

 I summary I have the following additional questions:

 What is the date of your latest report of the Mizuno test?  Does it
 exist on-line: If so, what is the link?  Is there any information from the
 Mizuno testing as to when excess energy from an unknown reaction starts and
 stops? Is there a good definition of baseline?

 Bob

 - Original Message -

 *From:* Jed Rothwell jedrothw...@gmail.com
 *To:* vortex-l@eskimo.com
 *Sent:* Saturday, January 10, 2015 8:18 PM
 *Subject:* Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

  Bob Cook made two large mistakes here. I wish he -- and others --
 would

  The Iwaik pump, if running, would have added heat at about 29 watts
 per the pump specification.

  In my report, p. 24, I list the pump specifications. Mizuno measured
 the pump input power with the watt meter. It is 10.8 W, not 29 W. However,
 only a tiny fraction of this power is delivered to the water. Mizuno
 measured how much is delivered. It was only ~0.4 W. If you do not think so,
 explain why Fig. 19 is wrong.

  You can confirm that nearly all the electric power converts to heat at
 the pump motor. Touch a pump and you will feel the heat radiating. Many
 pumps have fans that blow the hot air out of the motor. With a good pump,
 the water is at the other end away from the motor, and very little heat
 transfers to it.

This was more than enough to raise the temperature without any
 reactor heat source given the recorded decrease of 1.7 watts when nothing
 was running or reacting.

  Suppose this is true. Suppose it was 1.7 W and suppose that raises the
 temperature by 4 deg C. Pick any temperature rise you like: suppose it
 raises the temperature by 10 deg C, or 20 deg C. Here is the point, which I
 have made again and again:

  THE TEMPERATURE WAS ALREADY that much higher when the test began. The
 pump runs all the time. Using

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

2015-01-12 Thread Jed Rothwell

Gigi DiMarco gdmgdms...@gmail.com wrote:


The major result is that we measured 43°C in the pump body very close to
 the water so it is really easy to understand that, despite what Jed says,
 the pump motor delivers a lot of heat to the water . . .


You are wrong. This is not what I say. This is what Fig. 19 proves. If your
graphs show something else, your experiment is different. Perhaps you are
using a different kind of pump, or more pressure in the tubes, or perhaps
you have confused the effects of falling ambient temperature with rising
water temperature, as you did before.

In the second paper you wrote:

GSVIT-1) We do not agree at all. The pump was not stopped during the test
and, as Rothwell says, we are speaking about a differential temperature
increase equal to +2.5°C. . . .

No one said the pump is stopped during the test. It runs all the time. If
it were stopped, the test would fail because the heat from the reactor
would no longer be collected.


The pump power turns out to be about 4 W.


Suppose, for the sake of argument, that is true. And suppose that raises
the temperature by about 6°C. (Obviously that cannot be true because
nowhere do we see a 6°C elevation above ambient, but let us pretend it is
true.) In that case, all of the excess heat calculations must begin at a
baseline 6°C above ambient, because the pump is always left on. Therefore
this has absolutely no impact on the excess heat measurement.

- Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Dave,

as promised and while you still insist saying that we were deeply wrong, we
have put on-line two different updates

1)
https://gsvit.wordpress.com/2015/01/12/further-measurements-on-the-md-6k-n-pump-used-by-tadahiko-mizuno/

2)
https://gsvit.wordpress.com/2014/12/10/analysis-of-jed-rothwells-report-about-his-calorimetry-performed-on-mizunos-cell/

The first one shows how you are terribly wrong with your calculations based
on the kinetic energy only. We show that your assumption are completely
wrong just referring to usual pump working diagram. In the pump under test
you can not have simultaneously maximum head and maximum flow rate; the
working point we chose was such that we had almost the same working
conditions Mizuno had. Please take your time to read our post before
commenting. The major result is that we measured 43°C in the pump body very
close to the water so it is really easy to understand that, despite what
Jed says, the pump motor delivers a lot of heat to the water; it is this
the power we measure and it is by far much more that the mechanical power
(3 W maximum from the data sheet).

So we get comparable results by using very different methods

1) Pump theory and data sheet

2) Experiment

3) Simulations

All the rest are only free words.

We are going to apply the simulation to all the Mizuno's experiments to see
if we can get those curves without any excess heat.

Regards and take it easy.

Please, consider to read all the articles in our site concerning the
Mizuno's experiment.

Gigi aka Giancarlo

2015-01-12 19:09 GMT+01:00 David Roberson dlrober...@aol.com:

Bob,

The pump manual has several warnings about how easy it is to damage it and
that strongly suggests that Gigi and his team has done just that in order
to obtain their non representative performance. No one but Mizuno knows
the status of his pump during those tests so the only conclusion that can
conservatively be drawn is that the skeptical report by Gigi and team
should not be considered valid.

Dave

-Original Message-
From: Bob Cook frobertc...@hotmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Mon, Jan 12, 2015 12:15 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Jed--

I have researched the pump characteristics further and find that this pump
has a low efficiency and would use at most about 3 watts of power in
heating the circulating water. This is consistent with what you have
stated.

I am not sure how Mizuno measured the 10.8 Watts of power used by the
pump. I think

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

2015-01-12 Thread Jeff Driscoll

Jed is correct, when the pump is turned on and everything reaches steady
state, (using his example) the pump is putting in 4 watts of power to the
tubing, the reservoir and the LENR chamber and all these tubes and the LENR
chamber emit 4 watts of thermal power to the ambient at steady state. Then
when the LENR experiment is turned on, any delta T can be attributed to the
LENR device, not the pump (assuming the pump doesn't change speed).

On Mon, Jan 12, 2015 at 4:10 PM, Jed Rothwell jedrothw...@gmail.com wrote:

 Gigi DiMarco gdmgdms...@gmail.com wrote:


 The major result is that we measured 43°C in the pump body very close to
 the water so it is really easy to understand that, despite what Jed says,
 the pump motor delivers a lot of heat to the water . . .


 You are wrong. This is not what I say. This is what Fig. 19 proves. If
 your graphs show something else, your experiment is different. Perhaps you
 are using a different kind of pump, or more pressure in the tubes, or
 perhaps you have confused the effects of falling ambient temperature with
 rising water temperature, as you did before.

 In the second paper you wrote:

 GSVIT-1) We do not agree at all. The pump was not stopped during the test
 and, as Rothwell says, we are speaking about a differential temperature
 increase equal to +2.5°C. . . .

 No one said the pump is stopped during the test. It runs all the time. If
 it were stopped, the test would fail because the heat from the reactor
 would no longer be collected.


 The pump power turns out to be about 4 W.


 Suppose, for the sake of argument, that is true. And suppose that raises
 the temperature by about 6°C. (Obviously that cannot be true because
 nowhere do we see a 6°C elevation above ambient, but let us pretend it is
 true.) In that case, all of the excess heat calculations must begin at a
 baseline 6°C above ambient, because the pump is always left on. Therefore
 this has absolutely no impact on the excess heat measurement.

 - Jed




-- 
Jeff Driscoll
617-290-1998

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Dear Giancarlo,

Thanks for publishing your report in English so that many of us that do not 
speak Italian can understand it.  There is no disagreement between the method 
that I used to calculate the kinetic transport power and what you would have 
calculated with the same numbers since we used the same basic principles.  I 
relied upon the information from Jed about the mass flow rate of the pump where 
he stated that Mizuno had told him that it was 8 liters per second.  If you 
match that rate with your 5 mm pipe as you have stated as a plan for 
replication of Mizuno's experiment then you will obtain my results.

I do not have a pump and 16 meters of 10 mm inside diameter tubing before me to 
determine exactly what flow rate is obtained.   It is going to be necessary for 
you to either obtain a matching pipe or for us to verify exactly what flow rate 
is being measured by Mizuno before a final answer can be established.   Jed 
apparently believes that the friction within the 16 meter tubing is not 
sufficient to reduce the unloaded pump fluid flow rate to a value that is 
anywhere close to the 2.31 liters per minute that you are proposing.   In your 
report, you state that you are matching the performance seen by Mizuno as far 
as fluid flow rate is concerned but I strongly doubt that this is occurring.

If you make additional calculations you will see that the pressure required at 
the pump output is (10 mm/5 mm)^4 or 16 times as large when achieving the same 
flow rate for a 5 mm tube as compared to a 10 mm  tube.  This is a dramatic 
difference and you find that you quickly run out of head room when using the 5 
mm tube for your test.   Just this reason alone should be sufficient for you to 
realize that your replication attempt is failed.  And, as further supporting 
evidence, the pumping power needed to reach the 8 liters per minute flow rate 
when using a 10 mm tube is only .192 watts which is well within the operational 
range of the MD-6.

We can approach the power required to match Mizuno's flow rate from another 
direction if you wish.  The mathematics implies that the power required to 
drive a certain ratio of flow rates varies as that ratio to the third power.  
In your case that means (8/2.31)^3 or 41.53 times less than to reach 8 liters 
per minute.  To take your example: 41.53 * .074 watts = 3.07 watts.  (your 
numbers).  So again, you would need to have 3.07 watts of pumping power 
delivered to the water stream in order to reach 8 liters per minute of mass 
flow rate just as I have shown.

Giancarlo, you are the one that must defend your procedure to show that it 
truly replicates the experiment conducted by Mizuno.  I am merely demonstrating 
why you have failed to do so.  Unless you can prove that you are not damaging 
the operation of the pump in some manner by your technique then you can not 
expect me or anyone else to take seriously your claim that you have proven that 
there is no additional power being generated by Mizuno's device.

Why are we expected to accept the notion that a pump that is being driven into 
overload by high pressure operation per your demonstration is not adding 
significant additional power into the water stream?  The forces acting upon the 
pump are very much increased by your choice of pipe diameter and it does not 
take much imagination to expect the internal bearings to overload in a manner 
that generates significant heating as a consequence.

I can not say with certainly that your technique is completely without merit, 
but you are also left with many issue to resolve before you can claim a good 
reproduction of the cooling system used by Mizuno.  And, since you see powers 
that fail to match those derived from the experiment, it suggests that you are 
making some major error.

If we continue to discuss this subject in additional dept, I believe that we 
will eventually come to a mutual understanding with respect to your effort.  I 
remain neutral in my acceptance of whether or not excess power is being 
generated by the Mizuno experiment and I hope that you remain flexible.

I await your response to this posting and perhaps we should begin considering 
additional tests that you can perform to help verify the facts.  I like the 
horizontal flow demonstration that you used to measure the mass flow rate for 
the 5 mm tubing.  Can you do the same with 10 mm as a beginning step?

Best Regards,

Dave

 

 

 

-Original Message-
From: Gigi DiMarco gdmgdms...@gmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Mon, Jan 12, 2015 3:44 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised









Dave,


as promised and while you still insist saying that we were deeply wrong, we 
have put on-line two different updates

1) 
https://gsvit.wordpress.com/2015/01/12/further-measurements-on-the-md-6k-n-pump-used-by-tadahiko-mizuno/

2) 
https://gsvit.wordpress.com/2014/12/10/analysis-of-jed-rothwells-report-about-his-calorimetry-performed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Jeff,

I could agree entirely with you. I've have some problems with the internal
and external calorimeter time constants that are too short. But let's go on
and assume that what you say is completely right.

Now can you tell me where in the Mizuno's results (excel files and figures)
you see this behaviour? I do not see it, so if you tell me which is the
right curve we can discuss about it.

2015-01-12 22:58 GMT+01:00 Jeff Driscoll jef...@gmail.com:

 Jed is correct, when the pump is turned on and everything reaches steady
 state, (using his example) the pump is putting in 4 watts of power to the
 tubing, the reservoir and the LENR chamber and all these tubes and the LENR
 chamber emit 4 watts of thermal power to the ambient at steady state. Then
 when the LENR experiment is turned on, any delta T can be attributed to the
 LENR device, not the pump (assuming the pump doesn't change speed).

 On Mon, Jan 12, 2015 at 4:10 PM, Jed Rothwell jedrothw...@gmail.com
 wrote:

 Gigi DiMarco gdmgdms...@gmail.com wrote:


 The major result is that we measured 43°C in the pump body very close to
 the water so it is really easy to understand that, despite what Jed says,
 the pump motor delivers a lot of heat to the water . . .


 You are wrong. This is not what I say. This is what Fig. 19 proves. If
 your graphs show something else, your experiment is different. Perhaps you
 are using a different kind of pump, or more pressure in the tubes, or
 perhaps you have confused the effects of falling ambient temperature with
 rising water temperature, as you did before.

 In the second paper you wrote:

 GSVIT-1) We do not agree at all. The pump was not stopped during the
 test and, as Rothwell says, we are speaking about a differential
 temperature increase equal to +2.5°C. . . .

 No one said the pump is stopped during the test. It runs all the time. If
 it were stopped, the test would fail because the heat from the reactor
 would no longer be collected.


 The pump power turns out to be about 4 W.


 Suppose, for the sake of argument, that is true. And suppose that raises
 the temperature by about 6°C. (Obviously that cannot be true because
 nowhere do we see a 6°C elevation above ambient, but let us pretend it is
 true.) In that case, all of the excess heat calculations must begin at a
 baseline 6°C above ambient, because the pump is always left on. Therefore
 this has absolutely no impact on the excess heat measurement.

 - Jed




 --
 Jeff Driscoll
 617-290-1998

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Dave,

you said nothing about simulations that should be a confirmation of our
experiments. But I think that we can do something more: what will convince
you that we are right and Mizuno is wrong?

Regards

2015-01-12 23:17 GMT+01:00 David Roberson dlrober...@aol.com:

 Dear Giancarlo,

 Thanks for publishing your report in English so that many of us that do
 not speak Italian can understand it.  There is no disagreement between the
 method that I used to calculate the kinetic transport power and what you
 would have calculated with the same numbers since we used the same basic
 principles.  I relied upon the information from Jed about the mass flow
 rate of the pump where he stated that Mizuno had told him that it was 8
 liters per second.  If you match that rate with your 5 mm pipe as you have
 stated as a plan for replication of Mizuno's experiment then you will
 obtain my results.

 I do not have a pump and 16 meters of 10 mm inside diameter tubing before
 me to determine exactly what flow rate is obtained.   It is going to be
 necessary for you to either obtain a matching pipe or for us to verify
 exactly what flow rate is being measured by Mizuno before a final answer
 can be established.   Jed apparently believes that the friction within the
 16 meter tubing is not sufficient to reduce the unloaded pump fluid flow
 rate to a value that is anywhere close to the 2.31 liters per minute that
 you are proposing.   In your report, you state that you are matching the
 performance seen by Mizuno as far as fluid flow rate is concerned but I
 strongly doubt that this is occurring.

 If you make additional calculations you will see that the pressure
 required at the pump output is (10 mm/5 mm)^4 or 16 times as large when
 achieving the same flow rate for a 5 mm tube as compared to a 10 mm  tube.
 This is a dramatic difference and you find that you quickly run out of head
 room when using the 5 mm tube for your test.   Just this reason alone
 should be sufficient for you to realize that your replication attempt is
 failed.  And, as further supporting evidence, the pumping power needed to
 reach the 8 liters per minute flow rate when using a 10 mm tube is only
 .192 watts which is well within the operational range of the MD-6.

 We can approach the power required to match Mizuno's flow rate from
 another direction if you wish.  The mathematics implies that the power
 required to drive a certain ratio of flow rates varies as that ratio to the
 third power.  In your case that means (8/2.31)^3 or 41.53 times less than
 to reach 8 liters per minute.  To take your example: 41.53 * .074 watts =
 3.07 watts.  (your numbers).  So again, you would need to have 3.07 watts
 of pumping power delivered to the water stream in order to reach 8 liters
 per minute of mass flow rate just as I have shown.

 Giancarlo, you are the one that must defend your procedure to show that it
 truly replicates the experiment conducted by Mizuno.  I am merely
 demonstrating why you have failed to do so.  Unless you can prove that you
 are not damaging the operation of the pump in some manner by your technique
 then you can not expect me or anyone else to take seriously your claim that
 you have proven that there is no additional power being generated by
 Mizuno's device.

 Why are we expected to accept the notion that a pump that is being driven
 into overload by high pressure operation per your demonstration is not
 adding significant additional power into the water stream?  The forces
 acting upon the pump are very much increased by your choice of pipe
 diameter and it does not take much imagination to expect the internal
 bearings to overload in a manner that generates significant heating as a
 consequence.

 I can not say with certainly that your technique is completely without
 merit, but you are also left with many issue to resolve before you can
 claim a good reproduction of the cooling system used by Mizuno.  And, since
 you see powers that fail to match those derived from the experiment, it
 suggests that you are making some major error.

 If we continue to discuss this subject in additional dept, I believe that
 we will eventually come to a mutual understanding with respect to your
 effort.  I remain neutral in my acceptance of whether or not excess power
 is being generated by the Mizuno experiment and I hope that you remain
 flexible.

 I await your response to this posting and perhaps we should begin
 considering additional tests that you can perform to help verify the
 facts.  I like the horizontal flow demonstration that you used to measure
 the mass flow rate for the 5 mm tubing.  Can you do the same with 10 mm as
 a beginning step?

 Best Regards,

 Dave



  -Original Message-
 From: Gigi DiMarco gdmgdms...@gmail.com
 To: vortex-l vortex-l@eskimo.com
 Sent: Mon, Jan 12, 2015 3:44 pm
 Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Dave,

  as promised and while you still insist saying

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

2015-01-12 Thread Jeff Driscoll

Ill have to leave that to you and others,

I assumed Jed was making a point that Dave didn't understand.

I don't know the details of Mizuno's experiment.



On Mon, Jan 12, 2015 at 5:32 PM, Gigi DiMarco gdmgdms...@gmail.com wrote:

 Jeff,

 I could agree entirely with you. I've have some problems with the internal
 and external calorimeter time constants that are too short. But let's go on
 and assume that what you say is completely right.

 Now can you tell me where in the Mizuno's results (excel files and
 figures) you see this behaviour? I do not see it, so if you tell me which
 is the right curve we can discuss about it.

 2015-01-12 22:58 GMT+01:00 Jeff Driscoll jef...@gmail.com:

 Jed is correct, when the pump is turned on and everything reaches steady
 state, (using his example) the pump is putting in 4 watts of power to the
 tubing, the reservoir and the LENR chamber and all these tubes and the LENR
 chamber emit 4 watts of thermal power to the ambient at steady state. Then
 when the LENR experiment is turned on, any delta T can be attributed to the
 LENR device, not the pump (assuming the pump doesn't change speed).

 On Mon, Jan 12, 2015 at 4:10 PM, Jed Rothwell jedrothw...@gmail.com
 wrote:

 Gigi DiMarco gdmgdms...@gmail.com wrote:


 The major result is that we measured 43°C in the pump body very close to
 the water so it is really easy to understand that, despite what Jed says,
 the pump motor delivers a lot of heat to the water . . .


 You are wrong. This is not what I say. This is what Fig. 19 proves. If
 your graphs show something else, your experiment is different. Perhaps you
 are using a different kind of pump, or more pressure in the tubes, or
 perhaps you have confused the effects of falling ambient temperature with
 rising water temperature, as you did before.

 In the second paper you wrote:

 GSVIT-1) We do not agree at all. The pump was not stopped during the
 test and, as Rothwell says, we are speaking about a differential
 temperature increase equal to +2.5°C. . . .

 No one said the pump is stopped during the test. It runs all the time.
 If it were stopped, the test would fail because the heat from the reactor
 would no longer be collected.


 The pump power turns out to be about 4 W.


 Suppose, for the sake of argument, that is true. And suppose that raises
 the temperature by about 6°C. (Obviously that cannot be true because
 nowhere do we see a 6°C elevation above ambient, but let us pretend it is
 true.) In that case, all of the excess heat calculations must begin at a
 baseline 6°C above ambient, because the pump is always left on. Therefore
 this has absolutely no impact on the excess heat measurement.

 - Jed




 --
 Jeff Driscoll
 617-290-1998





-- 
Jeff Driscoll
617-290-1998

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

 this was not done.  The setup used by
 Gigi very clearly shows the pump mounted above the Dewar by several
 inches.  The same appears true for Mizuno's experiment.

 Dave



  -Original Message-
 From: Bob Cook frobertc...@hotmail.com
 To: vortex-l vortex-l@eskimo.com
 Sent: Mon, Jan 12, 2015 12:15 pm
 Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

  Jed--

 I have researched the pump characteristics further and find that this
 pump has a low efficiency and would use  at most about 3 watts of power in
 heating the circulating water.  This is consistent with what you have
 stated.

 I am not sure how Mizuno measured the 10.8 Watts of power used by the
 pump.  I think the pump specifications indicate the pump uses about 22
 watts.  However, The specifications for the amperage and voltage during
 operation would indicate the 29 watts I suggested some time ago.  I plan to
 talk with the pump vendor technical staff to better understand the
 performance of this type of pump and the wattage vs voltage/amperage specs
 and the efficiency.  I will report on what I find.  However, it would
 appear the pump is only about 15% efficient at best in converting
 electrical energy into the mechanical energy causing the circulation. At
 low circuit frictional pressure drop (low heads) it appears even less
 efficient.  I was wrong in assuming an efficient pump.

 I do not have the same report that you have  identifying the pump
 specifications on page 24.  My version of your report, dated November 14,
 2014, does not include the specification you state exists on the side of
 the pump body. In addition I do not think I have the same description of a
 baseline that your make reference to.

 I think by baseline you mean a condition at which the energy
 introduced into the circulating system by the pump creates a temperature of
 the reactor and water bath and all the reactor internals that is the same
 and in equilibrium with a non-changing differential temperature between the
 ambient atmosphere and the water bath. This would allow a reasonable
 determination of the average thermal resistance of the insulation and
 hence a measure of the approach to a desired adiabatic condition of the
 test setup.  In any case a good description of baseline conditions is
 warranted.

 In addition, if you have information as to when it was determined that
 excess reaction heat was produced in the reactor, this would be helpful in
 comparing temperature profiles with rates of change, compared to times when
 there was no excess energy input to the system.  For example, when is the
 excess energy produced with respect to the time the spikes of electrical
 heat are applied to the electrodes?  In this regard it seems that the
 excess energy production, if any, does not continue indefinitely, since the
 temperature increase levels  off and then decrease without the spikes of
 electrical input to the electrodes.  However, does it continue in the time
 frame between spikes of input energy to the electrodes.

 The temperature of the system and water bath should return to the
 baseline with time, if the only input is the energy  was from the
 pump. If excess energy form a reaction continues the temperature should
 level out at somewhat above the baseline.  This would be nice confirmation
 of excess energy.

 I summary I have the following additional questions:

 What is the date of your latest report of the Mizuno test?  Does it
 exist on-line: If so, what is the link?  Is there any information from the
 Mizuno testing as to when excess energy from an unknown reaction starts and
 stops? Is there a good definition of baseline?

 Bob


 - Original Message -

 *From:* Jed Rothwell jedrothw...@gmail.com
 *To:* vortex-l@eskimo.com
 *Sent:* Saturday, January 10, 2015 8:18 PM
 *Subject:* Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

  Bob Cook made two large mistakes here. I wish he -- and others --
 would


  The Iwaik pump, if running, would have added heat at about 29 watts
 per the pump specification.


  In my report, p. 24, I list the pump specifications. Mizuno measured
 the pump input power with the watt meter. It is 10.8 W, not 29 W. However,
 only a tiny fraction of this power is delivered to the water. Mizuno
 measured how much is delivered. It was only ~0.4 W. If you do not think so,
 explain why Fig. 19 is wrong.

  You can confirm that nearly all the electric power converts to heat
 at the pump motor. Touch a pump and you will feel the heat radiating. Many
 pumps have fans that blow the hot air out of the motor. With a good pump,
 the water is at the other end away from the motor, and very little heat
 transfers to it.



This was more than enough to raise the temperature without any
 reactor heat source given the recorded decrease of 1.7 watts when nothing
 was running or reacting.


  Suppose this is true. Suppose it was 1.7 W and suppose that raises
 the temperature by 4 deg C. Pick any

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

Jed,

I think you did not catch the importance of time constants in your
calorimeter.
I do not know how to explain it in more details. You will continue to say
no forever.


Do you think that simulation are a valid tools as far as they reproduce
exactly the experiments?

2015-01-12 22:10 GMT+01:00 Jed Rothwell jedrothw...@gmail.com:

 Gigi DiMarco gdmgdms...@gmail.com wrote:


 The major result is that we measured 43°C in the pump body very close to
 the water so it is really easy to understand that, despite what Jed says,
 the pump motor delivers a lot of heat to the water . . .


 You are wrong. This is not what I say. This is what Fig. 19 proves. If
 your graphs show something else, your experiment is different. Perhaps you
 are using a different kind of pump, or more pressure in the tubes, or
 perhaps you have confused the effects of falling ambient temperature with
 rising water temperature, as you did before.

 In the second paper you wrote:

 GSVIT-1) We do not agree at all. The pump was not stopped during the test
 and, as Rothwell says, we are speaking about a differential temperature
 increase equal to +2.5°C. . . .

 No one said the pump is stopped during the test. It runs all the time. If
 it were stopped, the test would fail because the heat from the reactor
 would no longer be collected.


 The pump power turns out to be about 4 W.


 Suppose, for the sake of argument, that is true. And suppose that raises
 the temperature by about 6°C. (Obviously that cannot be true because
 nowhere do we see a 6°C elevation above ambient, but let us pretend it is
 true.) In that case, all of the excess heat calculations must begin at a
 baseline 6°C above ambient, because the pump is always left on. Therefore
 this has absolutely no impact on the excess heat measurement.

 - Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

I agree completely with Jed as long as the ambient is kept at a constant 
temperature.  Any constant source of power introduced into the system will 
eventually result in a fixed delta between the device coolant temperature and 
the ambient.  The time constant associated with the transient delta is quite 
long according to Jed's data but if enough time constants pass, the temperature 
will settle down at a fixed value and remain constant.

Any new power applied as a step will result in a ramp to the temperature curve 
much as is seen during his testing.  I do have a concern about what occurs when 
the ambient changes.

I consider the change in ambient as being the equivalent of an input power 
application who's value is proportional to the ratio of the rapid change in 
ambient degrees to the total change above the normal stable ambient.  For a 
simple example assume that 1 watt of power is leaking into the test system as a 
result of pump power.  When settled out we can assume that the coolant resides 
at a temperature that is 4 degrees greater than ambient due to the long term 
application of  the 1 watt leakage.

Now if the ambient rapidly changes by 1 degree I believe that this is exactly 
the same as a signal appearing that is 1 watt * (1 C/ 4 C) = .25 watts.  The 
pulse nature of the input drive power and the resulting LENR heating is spread 
over a relatively large duty cycle enhances the effect of the input.  In this 
case for a 10% drive duty cycle our leakage would behave like a 2.5 watt valid 
signal.

I may be mistaken in this model and perhaps Jed can clear up any errors.

Dave

 

 

 

-Original Message-
From: Jeff Driscoll jef...@gmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Mon, Jan 12, 2015 3:53 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised



I have not followed this debate closely, but I assume Jed is correct.
So Dave, how do you address this statement:

The steady state baseline includes the heat from the pump, any diversion from 
the baseline indicates excess heat.





On Mon, Jan 12, 2015 at 3:44 PM, Gigi DiMarco gdmgdms...@gmail.com wrote:








Dave,


as promised and while you still insist saying that we were deeply wrong, we 
have put on-line two different updates

1) 
https://gsvit.wordpress.com/2015/01/12/further-measurements-on-the-md-6k-n-pump-used-by-tadahiko-mizuno/

2) 
https://gsvit.wordpress.com/2014/12/10/analysis-of-jed-rothwells-report-about-his-calorimetry-performed-on-mizunos-cell/


The first one shows how you are terribly wrong with your calculations based on 
the kinetic energy only. We show that your assumption are completely wrong just 
referring to usual pump working diagram. In the pump under test you can not 
have simultaneously maximum head and maximum flow rate; the working point we 
chose was such that we had almost the same working conditions Mizuno had. 
Please take your time to read our post before commenting. The major result is 
that we measured 43°C in the pump body very close to the water so it is really 
easy to understand that, despite what Jed says, the pump motor delivers a lot 
of heat to the water; it is this the power we measure and it is by far much 
more that the mechanical power (3 W maximum from the data sheet).


But, let me say that the second link is even more interesting [you have to go 
to the end of the article, the Appendix]: we set up a software simulation tools 
and were able to replicate by simulation the Mizuno's measurement. It was 
enough to evaluate the overall thermal transmittance of the system that is 
constant at least for the considered temperature range.

If we simulate the Mizuno's curve starting from a time instant when the reactor 
is no more generating excess heat, it is possible to evaluate the only source 
of heat: the pump. We have to use only the room temperature as provided by 
Mizuno's data and the system starting temperature. The pump power turns out to 
be about 4 W.


So we get comparable results by using very different methods


1) Pump theory and data sheet


2) Experiment


3) Simulations


All the rest are only free words.


We are going to apply the simulation to all the Mizuno's experiments to see if 
we can get those curves without any excess heat.


Regards and take it easy.


Please, consider to read all the articles in our site concerning the Mizuno's 
experiment.


Gigi aka Giancarlo




2015-01-12 19:09 GMT+01:00 David Roberson dlrober...@aol.com:

Bob,

You have uncovered a pump specification that proves that the replication work 
by Gigi and allies is not accurate.  They report to have determined that 
approximately 4.5 watts of thermal power is being absorbed by the circulating 
water under their test condition.  This amount of reported power is clearly 
more than the pump should add and they need to explain why we should accept 
their data as accurate.

Also, I have performs extensive calculations within a spreadsheet that is based 
upon the lift head

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

It appears as if Jed and Mizuno have the situation under control.  I will be 
relieved when the variations in ambient are taken out of the picture since that 
will make analysis of the system much simpler.

You might be correct that I got the direction of the pseudo input wrong.  It is 
so easy to get mixed up when you think about these types of systems.  Allow me 
to present the logic that I used to derive the behavior observed when an 
ambient temperature step takes place.

First, it is assumed that the ambient is steady and the temperature of the 
coolant has settled down and no longer is changing value with time.  If power 
is leaking into the system from the outside such as by means of the pump 
network then the coolant must obtain a temperature above the static ambient.  
This is required in order for the heat to flow outwards from the test system 
into the outside world.

In this case heat is flowing across the delta in temperature at a rate that is 
proportional to that delta.  Since the coolant is hotter than the ambient, heat 
is flowing from the coolant to the ambient.  If the ambient now undergoes a 
step upwards the difference between the static coolant and the new ambient is 
less than before.  Since a lower delta is now measured, less heat flows 
outwards from the coolant to the ambient.

Before the step, all of the heat associated with the pump power was flowing 
through the insulation and to the ambient environment.  Now, once that step 
takes place, the delta become smaller and less heat flows outwards.  The 
difference in heat flowing is directed into the thermal capacity of the coolant 
and test device.  This then should cause the temperature of those components to 
rise.  I believe this behavior would be the same as would be observed by a real 
signal adding its heat currents into the calorimeter.

A check to this thought is established by considering where the ultimate 
coolant and device system temperature stabilizes.  Eventually, that combined 
temperature rises until the same temperature delta as before the step is 
reached.  If that happened to be 4 degrees before a +1 degree step, the coolant 
temperature should move in the same direction as the step beginning at the 
hypothetical 3 degrees of delta slowly upwards to 4 degrees where it stabilizes.

Perhaps I am overlooking something in my mental model that someone will point 
out.

Dave

 

 

 

-Original Message-
From: Jed Rothwell jedrothw...@gmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Mon, Jan 12, 2015 6:55 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised



David Roberson dlrober...@aol.com wrote:


I agree completely with Jed as long as the ambient is kept at a constant 
temperature.


When ambient changes a great deal over a short time, calorimetry becomes too 
complicated. You need to throw away those results. Or use them for a limited 
purpose. Mizuno has reduced fluctuations and I hope he soon eliminates the 
overnight temperature change. 




I consider the change in ambient as being the equivalent of an input power 
application who's value is proportional to the ratio of the rapid change in 
ambient degrees to the total change above the normal stable ambient.


Do you mean when ambient temperature falls? This would look like input power . 
. . but only if you do not record the ambient temperature! As long as you see 
it is falling, you know this is not actually input power. The hard part is when 
you have actual power plus a falling ambient. It is difficult to separate them 
out. It is not worth the effort. Just toss out the data.


If ambient temperature rises (and you don't notice) it looks like heat 
vanishing in a magic endothermic reaction.


 

Now if the ambient rapidly changes by 1 degree I believe that this is exactly 
the same as a signal appearing that is 1 watt * (1 C/ 4 C) = .25 watts.


Well, it is not the same because we have a record of the ambient temperature. 
As I said, suppose you take a glass of warm water and put it outside in 
January. The difference between ambient and the water suddenly increases by 20 
deg C. That does not mean a heat source appeared out of nowhere. It means the 
water does not instantly cool off.


With this system, if I have derived Newton's cooling coefficient right, 1 W 
going into the water should produce a 1.5 deg C temperature difference.


- Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

I missed the simulation for some reason.  Where can I find that?  Sorry if I 
overlooked it.

Do you have data that shows the mass flow rate when a 10 mm tube is attached to 
the pump output?  I assume that a large pipe is on the suction port.

You need to attach a full length 10 mm tube to the pump and measure the flow 
rate and heating as a main step.  There are far too many variables associated 
with operation of the pump with the 5 mm pipe.  I have pointed out several 
problems that need to be addressed.  If you do this and also measure the AC 
power into the pump and then clean up the pump bearings so that the frictional 
losses are low then that will go a long way toward proving your position.

Do you have any method of verifying that the frictional losses are as low as 
those of the pump used by Mizuno?  The fact that you measure 4.5 watts versus a 
specification of 3 watts maximum suggests that something is wrong with your 
procedure.  How do you explain that difference?

Also, the difference between what you measure and what Mizuno and Jed measures 
may be nothing more than those associated with operation in a different pump 
pressure range and a damaged pump.  These types of questions remain unanswered.

Dave

 

 

 

-Original Message-
From: Gigi DiMarco gdmgdms...@gmail.com
To: vortex-l vortex-l@eskimo.com
Sent: Mon, Jan 12, 2015 5:34 pm
Subject: Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised



Dave,


you said nothing about simulations that should be a confirmation of our 
experiments. But I think that we can do something more: what will convince you 
that we are right and Mizuno is wrong?


Regards



2015-01-12 23:17 GMT+01:00 David Roberson dlrober...@aol.com:

Dear Giancarlo,

Thanks for publishing your report in English so that many of us that do not 
speak Italian can understand it.  There is no disagreement between the method 
that I used to calculate the kinetic transport power and what you would have 
calculated with the same numbers since we used the same basic principles.  I 
relied upon the information from Jed about the mass flow rate of the pump where 
he stated that Mizuno had told him that it was 8 liters per second.  If you 
match that rate with your 5 mm pipe as you have stated as a plan for 
replication of Mizuno's experiment then you will obtain my results.

I do not have a pump and 16 meters of 10 mm inside diameter tubing before me to 
determine exactly what flow rate is obtained.   It is going to be necessary for 
you to either obtain a matching pipe or for us to verify exactly what flow rate 
is being measured by Mizuno before a final answer can be established.   Jed 
apparently believes that the friction within the 16 meter tubing is not 
sufficient to reduce the unloaded pump fluid flow rate to a value that is 
anywhere close to the 2.31 liters per minute that you are proposing.   In your 
report, you state that you are matching the performance seen by Mizuno as far 
as fluid flow rate is concerned but I strongly doubt that this is occurring.

If you make additional calculations you will see that the pressure required at 
the pump output is (10 mm/5 mm)^4 or 16 times as large when achieving the same 
flow rate for a 5 mm tube as compared to a 10 mm  tube.  This is a dramatic 
difference and you find that you quickly run out of head room when using the 5 
mm tube for your test.   Just this reason alone should be sufficient for you to 
realize that your replication attempt is failed.  And, as further supporting 
evidence, the pumping power needed to reach the 8 liters per minute flow rate 
when using a 10 mm tube is only .192 watts which is well within the operational 
range of the MD-6.

We can approach the power required to match Mizuno's flow rate from another 
direction if you wish.  The mathematics implies that the power required to 
drive a certain ratio of flow rates varies as that ratio to the third power.  
In your case that means (8/2.31)^3 or 41.53 times less than to reach 8 liters 
per minute.  To take your example: 41.53 * .074 watts = 3.07 watts.  (your 
numbers).  So again, you would need to have 3.07 watts of pumping power 
delivered to the water stream in order to reach 8 liters per minute of mass 
flow rate just as I have shown.

Giancarlo, you are the one that must defend your procedure to show that it 
truly replicates the experiment conducted by Mizuno.  I am merely demonstrating 
why you have failed to do so.  Unless you can prove that you are not damaging 
the operation of the pump in some manner by your technique then you can not 
expect me or anyone else to take seriously your claim that you have proven that 
there is no additional power being generated by Mizuno's device.

Why are we expected to accept the notion that a pump that is being driven into 
overload by high pressure operation per your demonstration is not adding 
significant additional power into the water stream?  The forces acting upon the 
pump are very much

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised

2015-01-12 Thread Jed Rothwell

David Roberson dlrober...@aol.com wrote:

I agree completely with Jed as long as the ambient is kept at a constant
 temperature.


When ambient changes a great deal over a short time, calorimetry becomes
too complicated. You need to throw away those results. Or use them for a
limited purpose. Mizuno has reduced fluctuations and I hope he soon
eliminates the overnight temperature change.


I consider the change in ambient as being the equivalent of an input power
 application who's value is proportional to the ratio of the rapid change in
 ambient degrees to the total change above the normal stable ambient.


Do you mean when ambient temperature falls? This would look like input
power . . . but only if you do not record the ambient temperature! As long
as you see it is falling, you know this is not actually input power. The
hard part is when you have actual power plus a falling ambient. It is
difficult to separate them out. It is not worth the effort. Just toss out
the data.

If ambient temperature rises (and you don't notice) it looks like heat
vanishing in a magic endothermic reaction.



 Now if the ambient rapidly changes by 1 degree I believe that this is
 exactly the same as a signal appearing that is 1 watt * (1 C/ 4 C) = .25
 watts.


Well, it is not the same because we have a record of the ambient
temperature. As I said, suppose you take a glass of warm water and put it
outside in January. The difference between ambient and the water suddenly
increases by 20 deg C. That does not mean a heat source appeared out of
nowhere. It means the water does not instantly cool off.

With this system, if I have derived Newton's cooling coefficient right, 1 W
going into the water should produce a 1.5 deg C temperature difference.

- Jed

Re: [Vo]:Report on Mizuno's Adiabatic Calorimetry revised