Testing the TK Tar Baby

[mrsean2k]

@tk


The whole power vs energy discussion is a distraction anyway now though, isn't it?


Her rhetorical tics aside, I think that when she concentrates shed be able to follow how the energy is integrated through the course of the run she claims shows COP > whatever. It doesn't really matter that she doesn't really get it or employ terms in their commonly acknowledged way.


That in itself - with some generous approximations concerning the temperature of the water - is enough to show the claims are baseless?

[TinselKoala]

@s2k

Really? Do you think so?

Here is how she calculated from the data in the experiment described in Blog Posts 117 and 118, which I hope you've read.
The blog:

http://newlightondarkenergy.blogspot.com/2011/05/117-this-test-took-water-to-boil-with.html

And her calculation of the results from that experiment:

NOW.  Let's look at your 'self-runner' demands.  We have never recharged those batteries - with one exception.  Two caught fire and BOTH were fully recharged.  We've had those batteries since January 2010.  We've been running them since August 2010.  I've now FINALLY checked their rated capacities.  They're 40 ampere hours each.  We've used 6 of them continually since that time.  According to this rating they are each able, theoretically to dissipate 12 volts x 40 amps x 60 seconds x 60 minutes x 1 hour x 6 batteries.  That gives a work potential - a total potential output of 10 368 000 JOULES.

According to what has been carefully established it takes 4.18 Joules to raise 1 gram of water by 1 degree centigrade.  We've taken a little under 900 grams of water to 82 degrees centigrade.  We ran that test for 90 minutes.  Then we upped the frequency and took that water up a further 20 degrees to 104.  We ran that part of the test for 10 minutes.  Ambient was at 16.  Joules = 1 watt per second.  So.  Do the math.  4.18 x 900 grams x (82 - 16) 66 degrees C = 248 292 joules per second x 90 minutes of the test period = 22 342 280 joules.  Then ADD the last 10 minutes where the water was taken to boil and now you have 4.18 x 900 grams x (104 - 16) 88 degrees C = 331 156 joules per second x 10 minutes = 3 310 560 Joules.  Then add those two values 22 342 280 + 3 310 560 = 25.6 Million Joules.  All 5 batteries maximum potential output - available for work - is 10.3 Million Joules. In that test alone the battery outperformed its watt hour rating.  And that was just one test.  Now.  Over the 10 month period that those batteries have been running at various outputs - which, when added to the output on just this one test - then I think its safe to say that the evidence is conclusive.  Those batteries have outperformed. They are still at OVER 12 volts EACH.  They are all of them still FULLY CHARGED.

Perhaps you'd like to "do the math" and check her result and the claim made as a consequence of the result.     Here's a little quiz: How many batteries were used?

Here's another one, perhaps a "tad" less egregious:

In any event it has now been running for 67 hours.  Therefore it's dissipated 10 x 60 x 60 x 67 = 2 412 000 watts. Sorry I've overstated this.  It's been running since Friday 10.30am therefore only 54 hours.  Therefore 1 944 000 watts dissipated. It's rated capacity is 60 ah's = 60 x 60 x 6 batteries @ 12 volts each = 1 296 000 watts. Technically it's already exceeded its watt hour rating at absolutely NO EVIDENT LOSS OF POTENTIAL DIFFERENCE.

And yet another one showing the continual confusion of units and the effect it has on conclusions, not to mention conversations, and also showing that there isn't the slightest bit of sanity-checking of results. Her calculator says 22 x 220 is 2200 and she believes it because it's digital and digital equipment never errs and of COURSE Ainslie herself never errs, and simply seeing, as any eight-grader would,  that 220 times TEN is 2200 is beyond her ability.

Let's say that our utility supply is feeding current into an element on an electric stove to a temperature of say 260 degrees centigrade.
. Let's say that the element is has a resistance of 10 Ohms.  The source voltage is 220 volts.  The applied current is therefore 220/10 = 22 amps.
. Therefore the wattage delivered is 22 amps * 220 volts - which, according to my calculator is 2 200 watts.
. Now I assure you.  While that temperature over that resistor stays at that constant output of 260 degrees - there is no reduction in the rate of current flow.
. In other words our utility supplier both measures and charges for us for a wattage that they compute at 2 200 watts
. every second
. for every minute
. of each of those six hours
. giving a staggering product of 2200 x 60 x 60 x 6 hours being 47 520 000 watts.

Yes, these are all UNEDITED by me, except cropping the ends back to get to the bones.

According to MY Texan calculator, 220 volts x 22 amps is 4840 Watts, not 2200, just to start off with.

And a Watt is a Joule per second, so ... 4840 Joules per second TIMES 60 seconds PER minute TIMES 60 minutes PER hour TIMES 6 hours is... an even more staggering 104,544,000 JOULES. And yes, you are charged by the ENERGY USED, so that is what you will have to pay for. But we pay in units of kWh,  which is just a lot of Watt-seconds... that is, Joules... all bunched together.  So you don't need to go all the way to Joules to find out your energy cost. Just to the kiloWatt-hours. So you are using 4840 Watts continuously for 6 hours. That is 4840 x 6 = 29040 Watt-hours or 29 kWh.

Here in Texas we are billed for electricity by the kilowatt-hour. And a kWh costs something like 6 cents.  So, Ainslie claims
47 520 000 watts of ?power? or ?what?  times 6 hours is 285120000 ?watt-hour? or 285120 kiloWhaat-hours, at 6 cents per kWh... that makes about seventeen THOUSAND DOLLARS worth of oxtail soup. GIGO , iow.

But knowing what a Watt and a Joule are, 29 kWh costing .06 dollars per kWh is only a dollar and seventy four cents worth of ENERGY, which was dissipated at a RATE of 4840 Watts... or 4840 Joules PER second, every second for the whole six hours.

[mrsean2k]

@tk


Ha, well, maybe I'm being very optimistic. I'd only really registered the calculation that was out by a very obvious factor of 60 from your previous dissection, and hadn't realised there were so many other howlers.


But looking at the content of her posts, as time goes on, it looks as if she's making some sort of attempt to gain a slightly deeper understanding of the objections. I  don't know that the algebra or mathematics would improve, but some of the apparent errors of comprehension have faded a bit.


Maybe not enough to calculate the energy balance accurately herself, but perhaps enough to follow someone else's correct calculation and accept it?

[mrsean2k]

Rosemary,


Last on this particular point I think.


Wikipedia isn't an infallible, automatic, electronic source of knowledge; it's a meat based computer, the product of the efforts of many, many fallible human beings. The entry you lean on to describe power and energy as synonymous is no exception.


Looking at the Wikipedia article where they are described as "synonyms" I've traced back through the history and find that that particular nugget of information was added to the article on 7th April 2011, by user PaulTanenbaum:


http://en.wikipedia.org/w/index.php?title=Energy&diff=422793403&oldid=421132955




I've asked him what he intended to illustrate or convey when he added that paragraph to the article. If I get an answer, I'll reproduce it here, if it doesn't disrupt any other actual test-related posts too much.


I can't imagine it's possible for you to object to the author as the canonical source to resolve any ambiguity on how his own post should be interpreted and applied.

[TinselKoala]

You see, Ainslie, the UNITS are important. Several of us have stressed this but you don't understand it. Not just numbers are treated algebraically for cancellation and multiplication and division, the UNITS have to follow the same rules.

So when you want to convert, say, from miles per hour to feet per minute, you include the units and make sure they cancel properly and that you are left with the correct units in your answer. So if our input is in MILES per HOUR and our output is in FEET per MINUTE, those units must obey the same algebraic rules during the computation as the numbers themselves do

So, (MILES/HOUR) x (FEET/MILE) == FEET/HOUR... the units of "miles" cancel. And (FEET/HOUR) x (HOUR/MINUTES) ==  FEET/MINUTE and again, the units of "hours" cancel and we are left with only the correct units in our answer. 
(This is how GL was able to track down his error, an admirable exercise and a really significant one. You, Ainslie, would benefit greatly from studying what Groundloop did and how he did it.)

So, 10 Miles per hour x 5280 feet per mile == 52800 feet per hour.
And 52800 feet per hour x 1 hour per 60 minutes = 880 feet per minute. (see, I make math errors too, but I correct them, I had a different figure here at first).

And to CHECK OUR WORK we work backwards. 880 f/min x 60 min/hr = 52800 feet per hour , which is how I caught my original error which is now corrected.

And 52800 feet/hr  divided by 5280 feet/mile == 10 miles/hour. Numbers check out, units check out, so I'm _probably_ right, until someone proves me wrong with a better calculation.

Now... please do the same thing with your electrical calculations. When you put a number into an equation, also put in the units and make sure that the units are consistent throughout the calculations. This will prevent a lot of errors, please take my word for it.