Question on Overunity Device

[mansoon]

Related Question:
Assumusing that it does not use fuel or other external power after disconneting from battery, system outputs 100w power for long hours,  however out of 100w output, 50w will be used to self-maintain system and remaining 50w made available for external use. Is this still covered under same OU concept or does it become free-energy. Few post stated that OU means output power should be morethan input or what system consumes.

I'm trying to clear-up my understanding of OU, accordingly we can place our system in respective category.
thx

[Creativity]

it is ou.Half of input power is supplied in known energy form(the same as output type) and another 50W comes from undefined (yet) energy source. Now if it for example uses ambient heat then it is free energy device also. Howewer use of 50Hz radio waves emitted by power lines would not make it free energy device anymore.

[mansoon]

thanks for the clarification. we are in the process of system measurements (i/o) under all conditions. but i know sure this is not ambient heat. shall post details later once we are completely ready.
ManSoon

[mansoon]

Hi,
Here is the test data:

We wanted to test the system under different conditions like==
1. Connect battery to system and test system till battery dies (test1)
2. Connect battery to start the system and after brief period of time switch off battery and perform the same
   test procedure (test2)

By doing so we will have some understanding on contineous battery connected power runtime (includes wattage) and
system itself how long it can work that exposes heat dissipation, noise helps us to measure when battery disconnected test


Test1
===========
In this test we do not have much expectation until something major output achieved.

Input: 48V, 40Ah Lead Acid Battery (connected till battery discharged full)

Observation: Minimum 50 times system has been switched on and off for short period of 5 to 10 mins which seem to
have taken much of battery current during start of the system as surge current. (all the time 300W bulbs were on)

Observed Output: AC 300W (bulbs) @ 220V for contineous 1 hour (total of 4 hour duration). After this voltage indicator
started dropping contineously and battery died.

We could not keep the system on for more than 1 hour as it was getting heated after 45min also. We know the reason
for the heat that's being produced by system which will be modified in our next system.

In this test, we have not put any meters to monitor input DC voltage/amp and output. But output we could notice by lighting 300W bulb and we have kept AC Voltage indicator on the output port which connected to bulbs. In this complete testing we never charged battery.

reason for us to put these results is to better understand measurement if we are doing right  with your suggestions.

Test2
===========
to follow includes meter monitored.
Thx,,,

[Creativity]

good that u started with testing.It is a systematic approach that u r willing to follow, as such u will have clear proofs to present and no speculation.Great!

Lead acid  batteries are not well surviving full discharge :| The closer to discharge the more voltage drop also.It can make ur results inaccurate in longer runs,as battery life will be shorter and shorter every time(permanently lost capacity). Wouldn't it be better to use a wattage meter or voltage meter + amperage meter and a stopwatch to get more accurate results?U could get more repeatable results from each run of the test(battery will not suffer lost capacity).In ur set up it would mean to change test1 into a similar one,but to keep ur eye on the voltage and not let it to drop under let say 44V limit.

is battery voltage a real 48V or higher when fully charged?

Results are quite interesting.
let us calculate a bit:

1)battery if discharged continuously at 1/10th of the 40A will be capable of a work:
48V*40Ah=1,92 kWh (under best conditions for lead accid batteries)

2)50 times * 6 min = 5 hours of work done in first stage of test1
To power 300W for 5h it would drain 1,5kWh out of battery and other sources.

3)in the next step (without recharging of the battery if i understand well)
4h * 300W would drain additional 1,2kWh from battery and other sources

4)as a rough calculation result, ur device could use input of 1,92kWh to output around 2,7 kWh worth electric energy.Not bad at all

As u did no precise measurements,all calculations above are just a first approximation of what could be happening.We will wait for further results.

Is the output a standard AC?i ask because the multimeters are designed to measure a 50/60 Hz sinus wave and r behaving badly with other wave forms.For other than standard output only oscilloscope can tell the real voltage.
Next weak point in the chain is the bulb rating.It is an aproximation given by a manufacturer.Again appriopirate amperage metter...(testing of the bulbs resistance will not work fine,because it changes with the heat and so does the amperage).
Readings from the input side should be quite problemless if ur device takes a DC current.

PS:check what class of meters r u using.With this knowledge error range can be calculated on ur readings.