Lets make sure we agree on the terms. Then we can proceed to discuss the details. See below.
-----Original Message----- From: Joshua Cude <[email protected]> To: vortex-l <[email protected]> Sent: Sat, Nov 19, 2011 12:45 am Subject: Re: [Vo]:High school physics says > 1 GJ excess energy for the Oct. 28 demo On Fri, Nov 18, 2011 at 10:54 PM, David Roberson <[email protected]> wrote: Ok, I just did some calculating about the 1% power regulation you insist upon and it is bogus. Do you wish to prove your point? >If the output is dry steam, and the flow rate is constant, which would be the >case if the heating element is exposed, then the output power is >(dm/dt)(c1*deltaT1 + L + c2*deltaT2) Here I would like to ensure that we have an agreement. The input flow rate is constant, but I do not think that the output flow rate would be. My contention is that the output flow rate depends upon the instantaneous power output of the devices. This most likely drops throughout the test. >where c1 is the specific heat of liquid water (1 cal/gK), deltaT1 is the >change in temperature of the water (about 80C), L is the latent heat of >vaporization (540 cal/g), c2 is the specific heat of steam (0.5 cal/gK), and >deltaT2 is the temperature change >of the steam. >So, that means the power is proportional to >(620 + .5 deltaT(in C)) Please modify your formula to take into consideration that the output flow rate is not constant and varies. >Now, if you look at the temperature graph, after boiling is reached, it is >pretty well between 100C and 110C, for a fluctuation of +/- 5C. Actually the >std dev is quite a bit smaller. >And a fluctuation of +/- 5 C results in a fluctuation in the power of +/- >2.5/620 or about +/- 0.5 % for temperature stable within a range of 1%. Ok, if that is the way you want to define the range, I might be able to agree to it. The accuracies seem a bit too tight for a first look and I may have to see if it seems correct. But first, we have to account for the fact that the output flow rate is not constant. The power output is also not going to be constant. One additional factor which needs to be considered is the fact that there is a direct connection between the dry vapor and the boiling water within the ECATs. A valve is all that separates them and it is open on all of the units I presume. How did you take this into consideration? My belief is that this valve will control the vapor temperature as if the vapor were in some contact with the liquid within the ECAT. What do you think of this assumption? >Now, if the heating elements are submerged, and the output flow rate varies >with power, then the level is bound within a tight range, meaning, as I argued >before that the average flow rate would have to be matched to the power to an >accuracy of 1% >to avoid either exposing the heating element or sending liquid >out of the ecat. Here the 1% comes from the fact that the ecat is filled 11 >times during the test, and assuming that you have to fill it to about 90% to >cover the heating elements. Here I do not agree that the ECAT is filled 11 times during the test. I obtain 30000 grams/ECAT / 1.7539 grams/seconds = 17105 seconds/ECAT. This is 4.75 hours to empty one cat. That is only a bit more than one refill in the 5.5 hour period. Do you agree? We need to make sure we are talking the same language. Explain where you get the figure of 11 times? It is my opinion that the ECAT core or heat sink is submerged in water during the entire test period.
