heathicus
My son is participating in his school's 8th grade science fair. He wants to use electrolysis (aka "Electrolytic Cleaning") as his experiment and test how the change in electricity effects the speed of the process.
The plan is to set up 3 small tanks, with as identical as possible setups so we only have one variable. I've got some pieces of rebar that are pretty rusty and will be our test pieces. We'll have all three running at the same time, in the same size container, with anodes that are alike, an electrolyte solution that was made in one batch and distributed to all three, etc. I don't know if we can manage to set up a live demonstration of the process for the science fair, but if we can, we will.
They didn't give us much time. They first mentioned the science fair Friday, sent home an information packet Tuesday that didn't really give any guidelines, he had his project approved today (Thursday), and the fair us a week from tomorrow (Friday). Between his homework and me getting home late from work, we really only have this weekend to perform the experiment.
We've been planning on using" wall wart" DC transformers of various voltages with the same amperage and test the change in voltage. So I stopped by Goodwill today and picked up 3 transformers - 5v/550ma, 9v/500ma, 12v/500ma. But now I'm wondering if that is even enough amperage to get any results and if we should be testing change in amperage instead. What say you electrical experts?
And if that's the case, what is the best method, given our limited time, materials, and budget? Can we keep voltage the same and only change amperage? If so, how?
JPG
Current is what is doing the 'work'.
The following will affect the current.
Surface area of the anode and the cathode(sacrificial and derust electrodes).
The spacing of the electrodes.
The electrolyte.
The voltage.
The surface conductivity.
The type of voltage(pure dc vs pulsating dc[filtered vs unfiltered]).
Soooo the easiest way to vary the current(all else being equal) is to vary the separation.
That would allow using the same power supply for all three.
Make sure the combined load does not exceed the power supply rating.
As a control, make sure the three sets are indeed experiencing different current.
Take photos of before and at periodic time intervals.
Even with small currents, a week should provide enough time to demonstrate the effect of different currents.
Might be a good idea to document the current at each photo op. I predict as the surface changes, the current will also.(gut feel!)
P.S. When taking pix, capture the side of the cathode that faces the anode. The current flow is primarily 'line of sight'.
heathicus
Can I measure the current with a multimeter?
I understand that adjusting distance changes the current, but for the purposes of a scientific experiment, why wouldn't it be better to keep everything the same except for the current applied?
We can't really run the experiment for a week. We basically have just this weekend for it. We're just not going to have time next week. He will still have to assemble his display and write a report. That's another reason I wanted to run all 3 tanks simultaneously.
JPG
heathicus wroteCan I measure the current with a multimeter?
I understand that adjusting distance changes the current, but for the purposes of a scientific experiment, why wouldn't it be better to keep everything the same except for the current applied?
We can't really run the experiment for a week. We basically have just this weekend for it. We're just not going to have time next week. He will still have to assemble his display and write a report. That's another reason I wanted to run all 3 tanks simultaneously.
You do not apply current. You apply voltage and current results.
Since you are on a tight time schedule, you might consider using a car battery as a power source. Far greater current capability.
If the multimeter does not have a DC current range, inserting a small resistance and measuring the voltage drop across it would suffice. Small means less than 10 ohms. You could use one resistor across the meter leads for all three measurements(one at a time).
dusty
I would consider a battery charger for the power source. I use an old Sears charger when I do this. With the charger I use, there are at least two different charge rates (trickle and full).
heathicus
Well, we're not trying to achieve good cleaning, but to isolate the effect that increased voltage (or amperage) has on the speed of the process. I don't have a good enough battery charger. Just a trickle charger I use for my motorcycle and sometmes it works for electrolysis and sometimes doesn't, but I don't feel it gives me control over the variables. Plus, I don't have 3 of them to run simultaneously. With the wall warts, I can run them at the same time, and I know the exact voltage and amperage they provide. I can keep everything else the same.
My question is just whether 500mA is high enough to get results, and whether we should be testing change in amperage rather than voltage. If 500mA is not enough, how much is enough? If we should be testing amperage, is there an easy way to do that given our limitations and requirements? Wall warts of the same voltage but different amps? How many amps minimum? How much difference between steps up in amps?
dusty
I'm sorry, Heath, I can not answer your questions. I have never measured the current so I do not know. However, I believe that with only milliamps of current the process will work but very slowly and I do mean slowly.
I'd go set up and take some measurements but I can't do that either. I broke the bucket that I used and I don't have a suitable container to reassemble the setup.
Sorry. I have been of no help at all.
BuckeyeDennis
Heath, I don't know much about electrolysis, but JPG is correct that you can really only control the voltage (without a much more sophisticated setup). So you have "wall warts" at three different voltages. A good start.
Keep in mind that the wall warts are constant-voltage sources. If rated at 500mA, that means only that they can supply up to 500mA, without overheating and/or shutting down. It does not mean that they will supply 500 mA. The current you will get depends on the resistance of the load, as JPG said.
As to whether 500mA is enough juice to perform the desired electrolysis, I have not a clue. But there is one more factor that you can easily control: the size of the workpiece. I have a strong hunch that if you cut the workpiece in half, you will need only half the current to achieve the same rust-removal rate. So size the workpieces to match the current (and power) that you have available.
To determine just what size workpiece this might be, I'd start with the 12V wall wart. Set up a test unit, monitoring the current with your multimeter. You will need to set it up for DC current measurement, which usually means selecting the right multimeter mode and moving the red test lead to a different socket. Then connect the meter in series between the wall wart and the electrolysis tank.
Play with the size of the workpiece until the current draw is in the 300 to 400 mA range. That will give you a bit of safety margin, so that you don't overload the wall wart. Then use that size workpiece for all three experiments. The other tests should draw less current, because the supply voltage is less.
If your multimeter cannot measure DC current, do as JPG suggested and wire a resistor in series with each wall wart. These serve as current-sense resistors, and you can measure the DC voltage across them to determine the current. I'd suggest using 1 Ohm resistors, for two reasons. First, they will give you an easily measurable voltage level (with a decent meter), but not drop the voltage so much as to affect your experiment. Second, the voltage measured by the multimeter will be exactly the same as the amperage (current) through the resistor -- no calculations or conversions required.
Of course, it remains to be seen what the rust-removal rate will be. But hey, this is science, right!
BuckeyeDennis
A couple more thoughts:
1. Make sure that any 1 Ohm current-sense resistors are rated for at least 1/4 Watt. A 1/2 or 1 Watt resistor would be preferrable (so that they won't get hot enough to burn you).
2. Make sure that your wall warts all have DC outputs. A lot of them have AC outputs.
dusty
BuckeyeDennis wroteA couple more thoughts:
1. Make sure that any 1 Ohm current-sense resistors are rated for at least 1/4 Watt. A 1/2 or 1 Watt resistor would be preferrable (so that they won't get hot enough to burn you).
2. Make sure that your wall warts all have DC outputs. A lot of them have AC outputs.
A 1ohm resistor in series with a 12volt supply will draw 12amps of current or said another way will dissipate 12watts of power.
E=IR while P=IE Where E = volts dc, P= power in watts, I=current in amps and R=resistance in ohms.
Example: I=E/R = 12v/1ohm = 12 amps
What am I missing?
BuckeyeDennis
dusty wroteA 1ohm resistor in series with a 12volt supply will draw 12amps of current or said another way will dissipate 12watts of power.
E=IR while P=IE Where E = volts dc, P= power in watts, I=current in amps and R=resistance in ohms.
Example: I=E/R = 12v/1ohm = 12 amps
What am I missing?
Yes, if the 1 Ohm resistor were the ONLY load, it would draw 12A. And 12A x 12V = 144W =
SMOKE!. :eek:
But the electrolysis tank is also in series with the 1 Ohm resistor, and the plan is to tune it to draw less than 500mA, as that is the most that the wall warts can supply. So the worst case power dissipation in the resistor is 0.5A * 0.5A * 1.0 Ohm = 0.25W.
JPG
dusty wroteA 1ohm resistor in series with a 12volt supply will draw 12amps of current or said another way will dissipate 12watts of power.
E=IR while P=IE Where E = volts dc, P= power in watts, I=current in amps and R=resistance in ohms.
Example: I=E/R = 12v/1ohm = 12 amps
What am I missing?
The resistor is connected in series between the source and the load.
Supply+........Resistor.......Anode......Electrolyte.......Cathode.......Supply-
The voltage drop across the resistor will be determined by the source voltage and the ratio of the resistor and load.
VR = Vs x (RΩ / RΩ+LΩ)
V=Voltage
R=Resistor
L=Load
Ω=resistance
s=source
JPG
BuckeyeDennis wroteYes, if the 1 Ohm resistor were the ONLY load, it would draw 12A. And 12A x 12V = 144W = SMOKE!. :eek:
But the electrolysis tank is also in series with the 1 Ohm resistor, and the plan is to tune it to draw less than 500mA, as that is the most that the wall warts can supply. So the worst case power dissipation in the resistor is 0.5A * 0.5A * 1.0 Ohm = 0.25W.
Put another way,
P = IE
E = IR
Soooo P = I²R
Also I = E/R
Soooo P = E²/R
Just make sure all those parameters are all relevant to the same component(s).
heathicus
I REALLY wish I understood electronics better!
I should have added to the original post, "Explain it to me like I'm 5."
But, I think I've gathered some important points. Multimeter in series with the electrolysis set, then adjust the distance between anode and cathode until mA reading is ~400 or so. That makes sense to me. "P = IE" just makes me hungry for pie.
I'm going to read all the above posts a dozen more times and hope more of it starts to click.
frank81
Have you considered making the power source one of the controls rather than choosing the most difficult variable? You could change your variable to the amount of water in the tank (dispersal), amount of sodium carbonate (concentration), or changing the catalyst?
Electrolysis can occur with any salt, doesn't have to be washing soda that just gives the quickest and cheapest result for removing rust from ferrous metals. And you can choose any two conductive materials for the electrodes. Just be careful you don't choose something that creates a toxic byproduct, like stainless steel. Also don't forget, the form of electrolysis you are planning gives off hydrogen gas bubbles so careful of ventilation or leaving the experiment running in a classroom overnight.
A quick example...ocean water + aluminum hull boat + steel bolts to hang the motor - any insulation from eachother = warranty claim..errr I mean electrolysis.
In the end you just need a science experiement he can write a report on to show he understands it...the teacher doesn't care about rust removal like we do!
spiderclimber
I may be off base here, but I thought you were looking for a way to adjust the voltage using one source. The best way I know to do this is by using a computer power supply. You have 3 v 5 v and 12 v coming out of this. You also have a few other odd voltage wires in there as well. It has a built in short circuit breaker that shuts it down if you have a short or an issue. Just a thought as you can get one pretty cheap if not free. It will take you about 10 minutes to hack it into a usable supply. You can't regulate the amps this way but the voltage you can.
your wires you need are yellow, orange and red for your voltage. the black will be the neutral for each. You will also need to trick the CPU into thinking it has a motherboard attached. Take the one green wire and pick any black wire in the harness and connect them. Should give you fairly consistent and even voltage to your project. As well, most supplies are rated to 10 to 15amps so you get better output than a wall wart.
Hope this helps. Probably won't be back on here in time to see a question before your weekend is over, so good luck. Youtube hacking a computer power supply if you need help doing it.
JPG
OK I hit you with a lot of ifo.
Here is the simple approach as I see it.
Keep everything except electrode separation the same.
Common power supply.
Common container(3 of them)
Common electrolyte mixture and volume of each
Identical test pieces and sacrificial parts.(3 each)
For current setting and monitoring, 'momentarily' insert a low value resistor in series with each test sample(either + or - as long as multimeter polarity is observed when connecting the meter across the resistor).
The current should be variable by changing the electrode spacing.
Remember (-) to test sample, (+) to sacrificial part.
I leave the power supply choice to you, but wall warts are likely too small.
Battery charger is likely safer than a car battery, but having both connected will produce the fastest results.
BTW Oxygen is also being released as well as hydrogen.
heathicus
I had considered a computer power supply. I actually have a couple stashed back just for the purpose of converting to a "lab" power supply. But, different amps are supplied with the different voltages and I felt that would invalidate the test. If differences were notable, then was it the voltage or the amperage that made the difference? So we thought it best to make sure one of the values stayed the same across the experiment. We had decided on voltage as the variable and amperage as a constant, but I'm still unsure about that decision. Time to change it is just about out, though.
I did present some of the other variables to him as possibilities for the experiment, such as those you mentioned, frank81, but this is what he wanted to test.
I'm going to ruminate on everything that's been said here.
Quick question, though. How do I tell which wire from the wall warts is positive and which is negative?
JPG
heathicus wroteI had considered a computer power supply. I actually have a couple stashed back just for the purpose of converting to a "lab" power supply. But, different amps are supplied with the different voltages and I felt that would invalidate the test. If differences were notable, then was it the voltage or the amperage that made the difference? So we thought it best to make sure one of the values stayed the same across the experiment. We had decided on voltage as the variable and amperage as a constant, but I'm still unsure about that decision. Time to change it is just about out, though.
I did present some of the other variables to him as possibilities for the experiment, such as those you mentioned, frank81, but this is what he wanted to test.
I'm going to ruminate on everything that's been said here.
Quick question, though. How do I tell which wire from the wall warts is positive and which is negative?
Doing that will prove that the voltage will not affect the result as long as the current is the same!!! i.e. There will be only incidental differences in the results of all three samples.
ALL the work is performed by the current. More current = more results etc.
The hardest thing to do when determining what to 'do' is to make sure only one parameter is varied. You already got the list of things(parameters) to control/vary between samples.
Put yer multimeter on a high dc voltage range. If analog touch the leads to the output terminals/wires BRIEFLY. Observe needle deflection. You can figure it out from there. If digital, it will indicate polarity or an error.
Only do that after determining the wall wart does not tell you on a label.
The +12v portion of a desktop computer supply should provide more than enough current.
If not. the +5 volt portion will.
All else being equal, either the voltage or the spacing will affect the current. By using a common power supply (voltage) only the spacing will affect the current. Even if the voltage varies during the test, all three samples will be equally affected and any current changes will share a common ratio to each other.
I do think as the rust is depleted, the current on each sample will vary, but that will also be a constant for each sample, just not at the same time interval.
dusty
heathicus wroteI had considered a computer power supply. I actually have a couple stashed back just for the purpose of converting to a "lab" power supply. But, different amps are supplied with the different voltages and I felt that would invalidate the test. If differences were notable, then was it the voltage or the amperage that made the difference? So we thought it best to make sure one of the values stayed the same across the experiment. We had decided on voltage as the variable and amperage as a constant, but I'm still unsure about that decision. Time to change it is just about out, though.
I did present some of the other variables to him as possibilities for the experiment, such as those you mentioned, frank81, but this is what he wanted to test.
I'm going to ruminate on everything that's been said here.
Quick question, though. How do I tell which wire from the wall warts is positive and which is negative?
Just be aware that if you change the voltage the current will change, if you change the solution the current changes, if you change the electrodes the current changes.
