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Is using a calorimeter a good way to verify the difference in total power output of 2 test circuits and results of a meter?

Posted on 2006-05-12
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Is using a calorimeter a good way to verify the difference in total power output of 2 test circuits.
The proticol that I followed---Provide a known amount of power to a variable resistive load that converts electrical energy into heat energy inside of insulated container. Using a probe from a therometer, track heat over time from starting temp back to "starting temp". Graph results on paper and count squares. If one  graph provides more squares than other, means more "total" power was provided to the load from that circuit...

The actual power supply used on all test is a 6f capacitor charged to 12v and discharged to 8v..
if this helps

Is my logic correct, if so help me prove it......

Thanks Brad

Note: This test is used to verify results of a meter
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Question by:whiskers1
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LVL 27

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by:aburr
ID: 16672625
"Is using a calorimeter a good way to verify the difference in total power output of 2 test circuits."
no

too many uncontrolled variables.

Voltage and current measurements are better. Meters can be callibrated if necessary.
Two data loggers and a precision resistor should do the trick Total cost less than $700 using Pasco data loggers.
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Author Comment

by:whiskers1
ID: 16672720
aburr As suggested from my last posting, I called PASCO today told them what I needed- The tech. said that they had nothing that could accuratly do the job- their stuff is mainly for educational purposes. This is the reason for my other open question concerning power analysers.

To aburr only--By the way the caloriimeter tests I have already performed matched almost exactly to the results, proportionaly, from my hydrolysis experiments,  in the form of additional heat produced when using the "special circuit". Also please note that all of this is to verify the results of a Fluke Power Quality Analyzer  (college equipment) test performed by an electrical engineering instructor-who wouldn't accept his own test results,  blaiming equipment/ human error. My goal was to eliminate that possibility though other test based on sound scientific principles......

Thanks Brad
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by:grg99
ID: 16673562
Homemade calorimiters have too many sources of error, you're unlikely to be able to get better than 10% accuracy.

 I'd use two meters and a clipboard and pencil.  Use a high value resistor so the discharge takes place over many minutes, so you can write down many readings.

Easy to get 1% accuracy, even .1% with good meters.

For even better accuracy, ditch the capacitors, use a regulated power supply so you only really need to read the amp meter.



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by:whiskers1
ID: 16674275
I agree grg99- the homemade calorimeter is very inefficent (based on the coffee cup calorimeter from a university chemistry lab website), but would not these same iinefficiencies effect both test A & B equaly? ie. Heat loss rate of caloiremeter, thermometer calibration,  etc etc ........


Why am I stuck on the capacitors--We went down that road with the Electrical engineer ( he also had full knowledge of how the circuit worked). Again mentioned on my other open question- the caps were used  to eliminate his theory that the additonal power came from the input power supply only and that the "special circuit" had nothing to do with the additional work performed. After 3 different test with the same results under his supervision and using his equipment, he still could not give me a straight answer nor would he accept his own results .  AND THIS GUY HAS TRAIN OTHER PEOPLE TO BE ELECTRICAL ENGINEERS  FOR OVER 20 YEARS!!!!! I  gave up on him and his theories......

Almost forgot, we went down the high value resistor, mid value, low value.... road too, per the engineers instructions and supervison--SAME RESULTS each time.....

In closing, Im not trying to be a jerk, Im asking questions because I want the truth and to learn.....

Thanks  Brad
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by:d-glitch
ID: 16674457
This question has ancient roots:
      http://www.experts-exchange.com/Miscellaneous/Math_Science/Q_21331896.html

I think the experts all agree:

       A calorimeter measures energy not power.
       A home-made calorimeter is much good for anything.
       A capacitor bank is not a power source.
       Charging and discharging a capacitor is not a precise way to deliver energy.
==========================================================

The energy stored in an ideal  capacitor is ½CV²

      So you start out with 6F @ 12V ==> 432 Joules
      You end up with 6F @ 8V         ==> 192 Joules

So you have to account for the difference ==> 240 joules.

How big is difference in energy between your two circuits?
What is the peak current you draw from the capacitor?
How long does it take to discharge the capacitor to 8V with each of the test circuits?
How do you turn the circuit off at exactly 8V?
What is the model number of the capacitors in the bank?

Have you done any control experiments on the cap bank?
       For example, if your test typically runs for 60 seconds (implying a load of 10 ohms),
       a wonderful test would be to measure the energy output from bank using the
       Fluke Power Quality Analyzer with load resistances of  2, 5, 10, 20, and 50 ohms.

Do you think you would or should get the same amount of energy in each case?
Do the test and explain the results, and you will be well on your way to understanding.
============================================================  

The energy stored in an ideal  capacitor is ½CV².  But your electrolytic capacitors are far from ideal.

If all you have is voltmeter you can still do this test:
       Wrap a wire around each of the meter probes to make a short circuit.
       Charge up the cap bank to 12V and disconnect the charging circuit.
       Put the short across the bank briefly.
       You will get a spark, and the meter will read 0V.
       Take the wire off the meter leads, and measure the voltage on the cap bank again.
       Would you be surprised or amazed if it isn't zero?
       How much energy is it and where does it come from?

Good science has to be skeptical, quantitative, nearly obsessive.
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by:whiskers1
ID: 16675895
d-glitch thanks for jumping in on this, I will try to get back to you on this within 24 hrs and give you some answers. Thanks for your time Brad
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Author Comment

by:whiskers1
ID: 16676500

d-glitch
I tried to answer all questions to the best of my ability..I hope this will give you some insight,  
You guys are making me think ----GOOD JOB!!!!!!!


"A calorimeter measures energy not power. "

I agree, so if circuit A produces more squares on the graph paper than circuit B, would this mean that more energy was available to use with circuit A when compared to circuit B?

       "A home-made calorimeter is much good for anything."

As mentioned earlier, although the homemade calorimeter is very inefficient, would not these same inefficiencies effect the results of both test A & B equally?

      " A capacitor bank is not a power source."
not always true--- examples: memory backup, c mos circuitry


       "Charging and discharging a capacitor is not a precise way to deliver energy."
From our electrolysis experiments, discharging the cap from any voltage to a lower voltage produces proportionally the (approximate) same amount of gas - test were performed dozens of times, always the same results—seems to verify Faraday’s Law of electrolysis-
 
The first law states that the amount of chemical change produced by a current is proportional to the quantity of electricity passed, ie equal currents produce equal amounts of decomposition.

In our case, if discharging our capacitor is not a precise way to deliver energy would the volume of gas production not be as consistent as it is?  

My thoughts are- if we are wrong, then Faraday was wrong




The energy stored in an ideal  capacitor is ½CV²

      So you start out with 6F @ 12V ==> 432 Joules
      You end up with 6F @ 8V         ==> 192 Joules

So you have to account for the difference ==> 240 joules.

"How big is difference in energy between your two circuits?"
Should be the same- will verify with meter when available

"What is the peak current you draw from the capacitor?"
.15 amps at 12 volts

"How long does it take to discharge the capacitor to 8V with each of the test circuits?"
 Test 1:  1m 38 s     Test 2:    6 m  (Not what you think- no trickle charge) Cant go into detail  proprietary info

"How do you turn the circuit off at exactly 8V?"  
Manually disconnect while monitoring meter. Cap discharges slowly for both test- easy to do

What is the model number of the capacitors in the bank?
Bose audio system 3f stiffeng capacitor…..

Have you done any control experiments on the cap bank?
       For example, if your test typically runs for 60 seconds (implying a load of 10 ohms),      
 No

       "A wonderful test would be to measure the energy output from bank using the
       Fluke Power Quality Analyzer with load resistances of  2, 5, 10, 20, and 50 ohms."
                   Been there, done that with the electrical engineer

"Do you think you would or should get the same amount of energy in each case?"
Theoretically yes, but if to low-- “screwdriver across car battery effect”
 KA-BOOM!!!!!!

Do the test and explain the results, and you will be well on your way to understanding.      
  Huh?
============================================================  

The energy stored in an ideal  capacitor is ½CV².  But your electrolytic capacitors are far from ideal.

If all you have is voltmeter you can still do this test:
       Wrap a wire around each of the meter probes to make a short circuit.
       Charge up the cap bank to 12V and disconnect the charging circuit.
       Put the short across the bank briefly.
       You will get a spark, and the meter will read 0V.
       Take the wire off the meter leads, and measure the voltage on the cap bank again.
       
"Would you be surprised or amazed if it isn't zero?"
No
       "How much energy is it ?"
 would have to measure to give you an accurate answer

        “where does it come from”?
Nature of the beast-- referred to as the “battery effect”

”When you eliminate the impossible, no matter how improbable, what remains is the truth”
Sherlock Holmes

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Expert Comment

by:grg99
ID: 16677546
Again, a real scientist would use the best experimetal setup and tools, NOT intentionally go with the less accurate setup.

The  errors are going to be really high:

Electrolysis cell:  unlikely to be better than 20% accurate

capacitor:  initial build tolerance:  (read it off the side): never better than 20%, can be as high as -20 to +200% !!

capacitor:  variation with voltage, temperature, previous history:  30%

calorimiter:  unlikely to be better than 30%

-----------------

By the time you add up all those tolerances, your experiment is unlikely to be any better than 20-50% accurate.

----------------
A better experiment would use cheap digital meters, good to better than 0.5%, a power supply for a source, a resistor for a load.
With that arrangement you shoul dbe a ble to measure power within 1% easily.

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by:d-glitch
ID: 16678768
The nominal energy you use in your test is 240 Joules.
What is the performance difference you observe between your two test circuits?

A 10% difference in performance could be the result of a 10% difference in energy extraction, or 24 Joules.

==================
An electrolytic capacitor is complicated electrical and mechanical system.  
It has internal series and parallel (leakage) resistances, and exhibits peculiar dielectirc effects.

If you discharge your cap bank too quickly, you will waste energy in the series resistance and
leave energy in the dielectric.

If you dischage it too slowly, you will wate energy in the leakage resistance.

Your cap bank will have an optimum dischage time.  What is it?
You have to do the control experiments to figure it out.

I have absolutely no trouble believing that a 6 minute discharge is more efficient than a 1.5 minute one.
======================= ============================================

Just using round numbers, one of your tests take 90 s, the other takes 360 s (4x longer).

That suggests that your loads are 15 and 60 ohms respectively, and that your special box
has an impedance of 45 ohms.

So for the electrolysis cell:
        In one case, the applied voltage varies from 12 to 8 volts.
        In the other, the applied voltage varies from 3 to 2 volts.

I have absolutely no trouble believing that the lower voltage is more efficient for electrolysis.
===============================================================

But that doesn't mean I wouldn't to the necessary control experiment.

            Run the electrolyis for 98 s at with a 12.0 V power supply to model the first case.

            Run it again for 360 s with a 3.37 V power supply to model the second.

           3.37 V  =  (360/98) * 12 V
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Expert Comment

by:aburr
ID: 16679902
The last idea presented by  d-glitch
is excellent
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Author Comment

by:whiskers1
ID: 16679953
Guys you are for real, Its almost 1 AM, I will reread and talk to you tomorrow-
Before I go, thanks for not treating me like a hack, and are serious with helping--A million thanks

Brad
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Expert Comment

by:d-glitch
ID: 16682109
>> An electrolytic capacitor is complicated electrical and mechanical system.

     I meant to say " a complicated electrical and chemical system."
     Fortunately, it's not that complicated mechanically.
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Author Comment

by:whiskers1
ID: 16686995
OK guys following d-glitch's instructions "almost" (refer to note below) to the letter---here is the results from d-g;itch's test:


note: we could not run the test more than 3 min. at 12v because volume of gas would have exceeded the volume of the test tube. For ease of  calculation, we decided at 1 min.
Test was performed 3 times-same results each time


At time of testing, electrolysis device/ with dielectric measured 1003 ohms
The resistance of special circuit can not be given due to proprietary info.


12v at 60s          approx  1.8 cc volume of total gas

3.37v at 360s     approx  .5 cc volume of  total gas

ALMOST A 4 TO 1 RATIO!!!!!!!

What does this mean??

Thanks Brad



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Author Comment

by:whiskers1
ID: 16687052
Almost forgot, anything above 3.5 cc gas volume, water would have been below uninsulated portion of the electrodes--ceasing electrolysis

Sorry Brad
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Author Comment

by:whiskers1
ID: 16687130
Correction correction

Should  have read 3.37 at 240 sec !!!!!  Sorry - I was too excited with the test results
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Author Comment

by:whiskers1
ID: 16687814
"water would have been below uninsulated portion of the electrodes"--ceasing electrolysis ---- AND CAUSE AN EXPLOSION....
DAHHHHH--  : )  Brad
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LVL 22

Assisted Solution

by:grg99
grg99 earned 200 total points
ID: 16689259
You have to measure the voltage and current during actual electrolysis.

In theory we could deduce the current from the resistance, but in this case
 the resistance isnt much help as that changes dramatically as electrolysis progresses.

The results are roughly as expected-- you applied about 1/4 the voltage, ergo about 1/4 the current, therefore 1/16th the power, for 4 times as long.  So you should expect about 1/4 of the volume, which is quite close to what you got.  
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Author Comment

by:whiskers1
ID: 16690447
grg99
If I understand correctly, what you are saying is that this again verified Faradays law.
Also, the results seems to disprove the theory that a slower discharge rate would increase the efficiency of the electrolysis experiment.....

Is my logic sound????


Thanks Brad
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Expert Comment

by:grg99
ID: 16691697
Well, since you didnt measure the current, we don't have a very good basis for any conclusions.

If we *Assume* the cell stayed at 1004 ohms, which is mighty shaky, a simple spreadsheet shows:

      volts      ohms      current       power                 time      watt-secs      volume      cc/watt-sec      
                                                            
      12      1000      0.012         0.144               60            8.64             1.8      0.208333333      1.041666667
                                                            
      3.37      1000      0.00337      0.0113569            240      2.725656      0.5      0.183442078      0.917210389


So there's about 10% less gas at the low amp rate.  Which means nothing as your volume of "0.5" looks mighty fishy, like you rounded it off to the nearest tenth, so a 10% error is to be expected.

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Accepted Solution

by:
d-glitch earned 1800 total points
ID: 16691909
Let me go through my calculations more carefully:

The votlage on an RC circuit will decay by 1/e in T=RC seconds.    ==> 1/e = 36.8%
In your tests, the capactior discharges from 12 to 8 volts               ==>  66.7%           66.7/36.8 =  0.55 RC

===========================================================================
Experiment 1 with no Special Circuit runs for 98 seconds and genrerates    EXP1? cc of gas.

                    0.55 RC = 98 s       ==>  R = 27.7 ohms            

But this is inconsistant with your measurements of the cell resistance (1003 ohms) and the peak current (015 A).

===========================================================================
Experiment 2 with the Special Circuit runs for 360 seconds and genrerates    EXP2? cc of gas.

                    0.55 RC = 360 s       ==>  R = 109.1 ohms            

This suggests that your Special Circuit adds some series resistance in series with the cell.

What is special about the relationship between the quanitities of gas in the two experiments ( EXP1 and EXP2).
What do you think your special circuit does?

It certainly lowers the current through the cell.  The longer discharge time demonstrates that.

Is the resistance of the cell linear with current?  
I assumed that it was when I told you to set the power supply to 3.37 volts.  But that may not be correct.

===========================================================================
The average voltage on the cell in Experiment 1 is 10 V.

So run Control 1 without the Special Circuit.  
Set the power supply set to 10 V and run it for 98 seconds and measure CON1 cc of gas.
Also monitor and record the voltage across the electrolyis cell V_C1.

               Compare the quantities of gas produced in  EXP1 and CON1.  They should be very close.

               How does VC_1 behave during the test?

===========================================================================
Next run Control 2 with the Special Circuit.  
Set the power supply set to 10 V and run it for 360 seconds and measure CON2 cc of gas.

Once again, monitor and record the voltage across the electrolyis cell V_C2.  <== This is very important.

              Compare the quantities of gas produced in EXP2 and CON2.  They should also be very close.

              How does VC_2 behave during the test?    Does it change smoothly or eratically?
              Calculate the average value of V_C2?    <== This is very important.
              How do VC_1 and VC_2 compare?

===========================================================================
Now you aready to run the final test.

Run Control 3 without the Special Circuit.  
Set the power supply set to the average of V_C2 you calculated in the last test,
run it for 360 seconds, and measure CON3 cc of gas.

              Compare the quantities of gas produced in CON2 and CON3.  I expect them to be very close.
===========================================================================
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Author Comment

by:whiskers1
ID: 16692178
I got my work ahead of me ,, I will let you you know the results when I have it... Thanks for the input,  this could prove to very helpful in finding what is realy happening


Again thanks for your time

Brad
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Author Comment

by:whiskers1
ID: 16692230
grg99--"So there's about 10% less gas at the low amp rate.  Which means nothing as your volume of "0.5" looks mighty fishy, like you rounded it off to the nearest tenth, so a 10% error is to be expected"--

nay nay

I was using a test tube that has hash marks printed on it in increments of .1 cc --I just reported what I saw, nothing more....
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by:grg99
ID: 16693779
>I was using a test tube that has hash marks printed on it in increments of .1 cc --I just reported what I saw, nothing more....

Are you saying you're reporting the volume to the nearest 0.1cc?  If so you're adding an additional 10% to the measurement error.  ( half of 0.1cc is 0.05cc,  that's 10% of your measurement)

You may need to run this experiment for a longer time in order to collect enough gas so your readout error isnt so big a part of the total.

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Author Comment

by:whiskers1
ID: 16695430
That the reason I used the word approximate. I chose the mark that I thought was closest to that water level. To be technical the  3.37 v results was somewhere between .475 and .5 (around .and the 12v results was closer to 1.775 or just a hair below 1.8 Again estimates- even still this still shows an almost 4 to 1 ratio... If i had access to a chemistry lab Im sure I could have been more accurate...
Thanks Brad
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Author Comment

by:whiskers1
ID: 16785992
After performing d-glitchs last experiment and studying the results, it is is time to seek independent verification of my claims.... I would list the results, but I feel that it would not be believed ---so whats the use.

I commend both d-glitch and grg99 for their input and verification of the testing procedures of the Electical engineer that oversaw our earlier testing..

Although maintaining heathy skepticism, d-glitch provided the best most thought-out answers/testing protocol and took this question seriously- again, I commend his professionalism.. for that he will receive the lion's share of points

Happy trails and thanks for your help and time,, Now Im off to find a good chemistry professor...

Brad
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