Design details for a butter knife warmer

[Shay10825]

Hi everyone!
I need help on how to build a butter knife that gets warm. I know I need to put a nichrome wire in the blade but would anything go at the end? Would I connect the wire to a battery? How do I find out what size battery to use and make it so the blade gets to a specific heat?

Any help would be greatly appreciated.

~Thanks

 

[philocrazy]

batteries won't last long
nichrome wire might not work at low voltage,there
are other resistance/heating wires available for low voltage applications

 

[philocrazy]

i suggest you consult an electrician about this cause it could get
dangerous if you don't know what youre doing!

 

[Ivan Seeking]

You need to read this.

http://www.sci-journal.org/index.ph.../v3n1k44.html&link=reports/home.php&c_check=1

and this
http://www.the12volt.com/ohm/ohmslaw.asp

By controlling the watts [power produced] you will control the heat produced. More watts means more heat. You don't want to mess with AC power [110 volts] especially without help from an adult who is knowledgeable about electricity.

I was thinking that you may not even need the nichrome wire. Maybe you could just use a resistor. Why don't you try this as a place to start. Get a 10 ohm, 1/4 watt resistor. You can get these at Radio Shack. Use one D cell to power the resistor. This will overload the battery a bit so don't run it very long; say no longer than one, five, ten, then thirty seconds at a time at first. Check the battery frequently and make sure that it doesn't get hot. You will probably see the battery voltage drop as a result of the overload. The resistor will get hot and should produce about 0.2 watts of power. See if this will do the job. Then you can think about your next step.

You should work with an adult who is knowledgeable about electricity. For one, the battery can overheat and even explode. Always wear safety goggles.

EDIT: Does anyone know the internal resistance [ESR] of a typical D cell?

 

[Shay10825]

Would I have the end of the resistor just touching the battery and the other end not touching anything?

The D cell has an internal resistance of about 3 Ω

 

[Cliff_J]

Shay the resistor is the heating element. It needs to have current passing through it to heat up. So the resistor is on the butter knife and then wires connect each end of it to the battery.

To find the power, consult ohm's law. Use algerbra to solve the following equations:

Voltage = Current * Resistance
Power = Voltage * Current

Then you could weigh the butter knife and find the grams of steel in it. Now what is the heat capacity of the steel in the butterknife? How about the butter that will coat the blade (could probably use water to get close) as its cutting? That should be in your report as to how you determined the resistance and power level.

Where did you get the ESR figure from for a D cell? Sounds wayyyyyyy too high as I found 3 miliohms in a quick google search.
Cliff

 

[Shay10825]

Since I'm using a D cell battery how would I connect the wires to the battery? For example if the D cell battery was in the battery compartment of a flash light the ends of the battery would be touching the metal spring and the metal plate. Would I connect the resistor wires to the spring and the plate?

Would it work with a D cell battery? Is there any way I could use a AA battery?

The resistor will be in the middle of two blades put together to make one blade. Will the resistor have to touch anything while in the blade? Since both ends of the resistor have to touch the battery the resistor will touch the battery, go up the blade, turn around, and touch the other end of the battery?

I found the ESR from a google search too. Mine could be wrong.

 

[Cliff_J]

Battery sizes roughly give a clue of the power available.

You'll notice an AAA, AA, C, D battery are all 1.5V so voltage doesn't change. Instead its the current they can deliver that changes (determined by the ESR) and how long they can deliver that current (determined by the capacity).

Look at this chart:
http://www.techlib.com/reference/batteries.html

According to this if you use a D cell and used 12 amps (12,000 miliamps) it should last 1 hour, or if you used 6 amps it should last 2 hours. Mathimatically that's correct but remember our ESR from above? It will drastically reduce that and the more current you pull the less accurate the formula. A lot of batteries are rated for capacity at 20 hours so capacity at 1 hour its probably overrated by 5-10x. So with 12A from a D cell I'd guess maybe 6-12 mins before its dead and I might be optimistic.

Ok, back up from the construction for a second. Answer these questions:
How much heat power do you need?
How much voltage do you have with 1 cell? How about 2 cells?
How much resistance do you need to flow the correct amount of current from that voltage?
How long will each battery last flowing that current?

Yes, you will need the resistor to touch the blade. You'll likely want to cover the resistor leads with heatshrink or electrical tape. Then maybe some heatsink compound to assist in the heat transfer from resistor to blade.

Oh, and make sure the resistor has a rating higher than the number of watts you plan to use.

Cliff

 

[Ivan Seeking]

Thanks for jumping in Cliff. I am fighting time right now...darned work! I have PF to attend to.

Oh, and make sure the resistor has a rating higher than the number of watts you plan to use.

But not so much higher that the resistor never gets hot [hence the 0.2 watt target]. Also, the knife will heat sink the resisitor so I was thinking that with good thermal coupling he may be able to overdrive things a bit. I was thinking that staying just below the rated power is a good place to start in order to insure a high max temperature.

Note: I just wanted to be sure that we weren't close to the ESR with a 10 ohm resistor. I thought it was more like 3 ohms for a D cell...hmmm...learn something new every day. Now, wait, that can't be right. At 1.5 volts and an ESR of milliohms, the total current available is way too high.

 

[Shay10825]

Ok, back up from the construction for a second. Answer these questions:
How much heat power do you need?
How much voltage do you have with 1 cell? How about 2 cells?
How much resistance do you need to flow the correct amount of current from that voltage?
How long will each battery last flowing that current?

Cliff

Is this correct?:
~~~~~~~

Since the specific heat of silver, cal/g * degrees C, is .056 you change that into watts which is .234 W. So is .234 W the amount of heat power I need?
~~~~~~~

With one cell I have 1.5 volts, and with two clees I have 3 volts.
~~~~~~~

P=VC
.234= 1.5C
C=.156

V=CR
1.5=.156R
R=9.615

For 1.5 V I need 9.615 ohms of resistance?
~~~~~~~

P=VC
.234=3C
C=.078

V=CR
3=.078R
R=38.462

For 3 V I need 38.462 ohms of resistance?
~~~~~~~

With 1.5 V the battery will last 77 hours?
With 3 V the battery will last 154 hours?
~~~~~~~
Is any of this correct?

 

[Cliff_J]

Good, you looked up specific heat. That was the point of the question, you can describe that different materials heat differently. You'll likely want to use steel instead of silver. I did a quick search and this site came up, pretty handy comparison:
http://www.ex.ac.uk/trol/dictunit/notes5.htm

Note that the level for water is much higher. You need to factor in how long you can wait and how much heat will be transmitted to the air from the knife to be exact, unnecessary in all likelyhood. But if instead you could heat up the butter knife first and then cut the butter you can take advantage of the thermal mass of the butter knife. Let's assume .2W is a nice starting point, its a safe level to begin experimenting.

Your math looks fine for all the rest. (As a side note, on the internet most people will use the standard letters E=voltage and I=current and R=resistance. Its a long story why...)

You can find resistors of sizes close to those values, like the 10 ohm mentioned by Ivan above or a 39 ohm resistor if you choose to use two cells. As you can also tell now that you've done the math (and should have a spiffy report with) you can see a 1/4 watt resistor will work fine.

As another side note, you can see that for the same power level a doubling of voltage means 4x the resistance is need to stay at the same power level. Or if you double the voltage and keep the resistance the same, you have 4x the power.

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

Ivan - 3 ohms still sounds too high, even at .2A the drop is .6V? Sure .003 sounds way too low (how good is a unknown source on the internet) but maybe somewhere in the middle like 1 ohm? Maybe find time to crack open a flashlight this weekend and test a couple batteries.

Cliff

 

[Ivan Seeking]

Is this correct?:
~~~~~~~

Since the specific heat of silver, cal/g * degrees C, is .056 you change that into watts which is .234 W. So is .234 W the amount of heat power I need?

A couple of more details here. The specific heat is in calories per gram per degree centigrade. So, after converting calories to joules [not watts], you need 0.234 joules, per gram of silver, per degree C of temperature change for the silver mass.

Now, power - watts - is the number of joules of energy per second. The 0.234 watts calculated would raise one gram of silver, one degree C per second. Two grams of silver would heat at a rate of 0.5 degrees C per second, and so on. Also, as Cliff pointed out, this does not account for the heat lost.

Ideally we would want to include the mass and specific heat of the butter as well. This, plus the heat lost to the atmosphere would give us the total heat load, but you can probably ignore this for now. It should be mentioned in your report though.

 

[Ivan Seeking]

If we see you selling this on QVC, Cliff and I will be waiting for our royalty checks.

Oh yes, there is another issue of how fast the resistor tranfers heat to the metal. As mentioned earlier, use heat sink paste [available at Radio Shack], but also make sure that the resistor [not the leads] is in direct contact with the metal blade. This will be a little tedious to set up properly but it is very important.


You may also find that the metal blade is able to dissipate the heat [hence power] faster than you can supply it. If this happens you will need to produce more power.

 

[BoulderHead]

I get around this 'hard butter' problem by avoiding refrigeration.

Good luck

 

[Ivan Seeking]

I get around this 'hard butter' problem by avoiding refrigeration.

Good luck

Actually, if you use real butter you can get minor food poisoning this way. Most such cases are misinterpreted as minor influenza or just a bad stomach.

 

[BoulderHead]

You mean it wasn't really the flu making my belly hurt all these years?

I hadn't known that, but generally use the imitation stuff, if any at all.

 

[megashawn]

hey, instead of making a heated knife, why not make a knife heater?

You know, some lil thing just barely wide enough to slide a standard butter knife in it. Then, use some heating elements to warm the knife. This way, you can warm any knife, instead of one special knife.

 

[Ivan Seeking]

I hadn't known that, but generally use the imitation stuff, if any at all.

I hate to guess what I've spent on Kaopectate due to my long, room temp butter habbit. It took me years to make the connection. I only realized the potential connection when I saw a woman from the National Centers for Disease Control who stated that something like 80% of all minor, alleged cases of influenza are really minor food poisoning.

 

[Shay10825]

If I use two D cell batteries an tthe 39 ohm resistor will the blade heat up faster?

The 10 ohm .25 watt resistor is short in length. How would I have both ends of it touching the battery if it is soshort. Would I connect some kind of wire t theend of the resistor to make it longer?

 

[Ivan Seeking]

If I use two D cell batteries an tthe 39 ohm resistor will the blade heat up faster?

No. In this case the power is about the same. Just try multiplying the voltage and current for each. In order to increase the power you would need less resistance or more voltage. Note that you can also get 1/2 watt resistors and higher.

The 10 ohm .25 watt resistor is short in length. How would I have both ends of it touching the battery if it is soshort. Would I connect some kind of wire t theend of the resistor to make it longer?

Yes. Make sure that you don't cause a short circuit. You might want to consider using an old flashlight for a battery holder.

 

[Ivan Seeking]

hey, instead of making a heated knife, why not make a knife heater?

You know, some lil thing just barely wide enough to slide a standard butter knife in it. Then, use some heating elements to warm the knife. This way, you can warm any knife, instead of one special knife.

You may not have seen the original thread in GD where some similar suggestions were made. Shay10825 is hooked on a heated blade for now.

 

[Gokul43201]

A couple of more details here. The specific heat is in calories per gram per degree centigrade. So, after converting calories to joules [not watts], you need 0.234 joules, per gram of silver, per degree C of temperature change for the silver mass.

Now, power - watts - is the number of joules of energy per second. The 0.234 watts calculated would raise one gram of silver, one degree C per second. Two grams of silver would heat at a rate of 0.5 degrees C per second, and so on. Also, as Cliff pointed out, this does not account for the heat lost.

Ideally we would want to include the mass and specific heat of the butter as well. This, plus the heat lost to the atmosphere would give us the total heat load, but you can probably ignore this for now. It should be mentioned in your report though.

Hmm...completely missed this thread, or I would have jumped in earlier.

Okay I don't want to be a spoiler, but here's a problem I raised in the original thread that I'll rephrase in the current context.

Say you have 0.23 W of power (1.5V battery with 10 ohm resistor).

For stainless, C ~ 0.5 J/K-gm.
For a blade that is pretty small (2"long, 1/2" wide and 1mm thick), the mass is about 4 gms (density ~ 8 gm/cc).
Assuming we need no more than a 10K rise in temperature, the required heat is H ~ 4*0.5*10 ~ 20 J

With losses (and keeping in mind that stainless is a pretty lousy conductor ~ 16 W/K-m) it looks like you'll need over 3 minutes to get the knife hot enough.

Is that okay ?

 

[Shay10825]

The calculation I made earlier was for Sterling Silver. The Silver was what had the .23 W not stainless steal. I don't think I am going to be able to use silver so would it still take 3 min if I used Stainless Steal? Would I still be able to use the .25 W 10 ohm resistor for the steal or would I have to get a different resistor for it to heat up faster?

 

[Ivan Seeking]

Okay I don't want to be a spoiler, but here's a problem I raised in the original thread that I'll rephrase in the current context.

My degree is in physics. That's an engineering problem.

I thought 1/4 watt was a nice place to start - that we can "see where to go from there".

 

[Ivan Seeking]

The calculation I made earlier was for Sterling Silver. The Silver was what had the .23 W not stainless steal. I don't think I am going to be able to use silver so would it still take 3 min if I used Stainless Steal? Would I still be able to use the .25 W 10 ohm resistor for the steal or would I have to get a different resistor for it to heat up faster?

Have you tried connecting the resistor to the battery yet? No knife, just the resistor and battery?

Oh yes, do you have access to a voltmeter?

 

[Shay10825]

I have the .25 W 10 ohm resistor, copper wire, electrical tape,
and the D cell battery. No I have not conected the resistor to the battery becase I don't really know how I would do this. Can I touch the resistor when it is connected to the battery?

No I don't have access to a voltmeter.

 

[Ivan Seeking]

I have the .25 W 10 ohm resistor, copper wire, electrical tape,
and the D cell battery. No I have not conected the resistor to the battery becase I don't really know how I would do this. Can I touch the resistor when it is connected to the battery?

No I don't have access to a voltmeter.

How much money can you spend? We need to talk about your budget: $10, $20, $50, $5000?

You really must find some safety goggles as well. As for the battery, you can feel free to touch things with the understanding that the resistor will get warm, or even hot. If you make a mistake [like if you use the wrong resistor] the resistor could even go fizzzzz..poooof. If they smoke they can really burn so always be careful - really! I had a glass diode vaporize between my fingers once and I still have a scar over twenty years later.

Check the battery frequently. This should not get hot; slightly warm is okay but no more.

You should look up the RC Color code - that is the Resistor Capacitor color code. Learn how to use it. This is a required first step for any resistor work.

Is there any chance of using a voltmeter at school?

Make sure that you understand ohms law. Do you?

Gokul43201 makes a great point. Be thinking about ways to minimize the mass of the blade. We can increase the power but we are limited. The batteries are the real limit here. How many do you think is reasonable? For example, twenty D cells may be a bit cumbersome for a knife; this would allow for a very powerful butterknife though! :tongue2:

 

[Gokul43201]

I hate to say this again, but with a knife-holder-heater, there are several advantages :

1. More room for more batteries, so more power. Also more room, in general, for everything else.

2. More room for resistance wire avoids complications of embedding in the blade and the drawbacks of having an external circuit running outside the blade

3. Heater is largely not exposed to air, so lower losses

4. Heater wire can be made to impart heat to only the front (cutting) edge of the blade. Because stainless is such a poor conductor, this will work.

5. You can simply use AC from a wall outlet, and have no lack of power at all. There will be enough power to heat the knife in less than half a minute.

 

[Ivan Seeking]

Ohm's law tells you that the voltage measured across your resistor is equal to the current flowing through the resistor multiplied its resistance: Volts = Amps X Ohms. If you know any two values you can calculate the third.

The watts of power produced - in this case the rate that heat energy is produced in the resistor - is calculated by multiplying the voltage measured across the resistor by the value of the current flowing through it: Watts = Volts X Amps. If you know any two the third can be calculated.

It is important that these points are clear. This allows you to start calculating alternative values to try.

If you are producing up to but no more than 1/2 watt of power in the resistor, you need a 1/2 watt resistor. You can also get 1 watt, 5 watt, 10 watt, and up to resistors capable of handling tens of thousands of watts. So, the batteries and the blade become the real issues.

 

[Ivan Seeking]

I hate to say this again, but with a knife-holder-heater, there are several advantages :

I think Shay10825 needs to make some engineering decisions here.

 

[Gokul43201]

Also, I can't say I really think this is going about things the right way.

Call me a whiner, but I think it's a bad idea to go about buying stuff and trying to build things without a working plan at hand.

Maybe, Ivan, you've got a plan...but does Shay have one or know what your plan is ?

For instance, I'm curious where the heater wire actually runs, physically, with respect to the blade, and how a thermal short is provided.

And I don't think it's a great idea to use a resistor (instead of a resistance wire). That's a point source of heat, while a line source will be much better. A point sorce introduces a time constant and increases inefficiency.

If I'm whining too much, just ask me to shut up. I really don't want to be such a wet blanket, you know.

 

[Ivan Seeking]

Come on Shay the pressure's on. The world is waiting!

 

[Ivan Seeking]

I appreciate your point.

Is it fair for us to completely engineer the project for him? A dollar's worth of reistors and batteries are the extent of spending so far.

 

[Gokul43201]

Ivan, what say you ?

I don't want to look like someone that just walked in and threw a wrench in the works.

 

[Ivan Seeking]

Not at all. I'm glad you're jumping in.