Hot air gun, practical use of Faraday law

[Homo sapiens sapiens]

Hello, I have the following question, I am trying to build a home made hot air gun for electronics soldering, my approach was with a modified Microwave oven transformer, multifilar 12mmxmm copper cables, and a 15cm stainless steel tube, the air is pumped from a powerful aquarium air pump. The air gun works, but has some wire and transformer over heating problems. In my quest to understand why the wire and transformer heat so much I wanted to measure the power delivered to the load(this is the stainless steel tube, has high resistivity has far as I know), The only problem in measuring the power with a volt meter is that the current is far to high to be measured with a voltmeter, and measuring the voltage makes me had some doubts. Te secondary of the transformer is the tick copper cable, the voltage induced in the secondary circuit is about 3.3V if measured with no load. My question is, if I connect the multmeter to the extremes of the stainless steel tube to measure the voltage, am I measuring the voltage in the tube or am I measuring the voltage in the copper wire.
Putting it in a mathematical way: emf=R_load * I + R_wire * I, I know that emf is about 3.3V, I would like to know how to measure V_load and V_wire with a multimeter.

Thanks very much for reading and thinking about my problem!

 

[berkeman]

Hello, I have the following question, I am trying to build a home made hot air gun for electronics soldering, my approach was with a modified Microwave oven transformer, multifilar 12mmxmm copper cables, and a 15cm stainless steel tube, the air is pumped from a powerful aquarium air pump. The air gun works, but has some wire and transformer over heating problems. In my quest to understand why the wire and transformer heat so much I wanted to measure the power delivered to the load(this is the stainless steel tube, has high resistivity has far as I know), The only problem in measuring the power with a volt meter is that the current is far to high to be measured with a voltmeter, and measuring the voltage makes me had some doubts. Te secondary of the transformer is the tick copper cable, the voltage induced in the secondary circuit is about 3.3V if measured with no load. My question is, if I connect the multmeter to the extremes of the stainless steel tube to measure the voltage, am I measuring the voltage in the tube or am I measuring the voltage in the copper wire.
Putting it in a mathematical way: emf=R_load * I + R_wire * I, I know that emf is about 3.3V, I would like to know how to measure V_load and V_wire with a multimeter.

Thanks very much for reading and thinking about my problem!

Why did you use the transformer from a microwave oven?

 

[vk6kro]

If you are connecting the wires to the ends of the stainless steel tube, then you can measure the voltage across the length of the tube. This will be your load voltage.

Note that most cheap multimeters just connect a diode in series with the DC voltage to measure AC. So they read 0.6 volts low. If yours does this, you have to add 0.6 volts to the reading to be more accurate.

You should be able to get some reading of the resistance of the tube when no power is applied.
If it is too small to measure, then you probably should not be applying voltages to it.

It sounds very inefficient, though. A lot of the air passing down the middle of the tube will not even come in contact with the hot tube. Real hot air guns, and even hair dryers, put a heating element in the air path and force the air to pass through it.

 

[Homo sapiens sapiens]

I used a transformer from a microwave because the original primary winding can support large currents, and because it has a large case so I can rewind the secondary with tick multifilar copper wire.

I can read a voltage with the multimeter, just wasn´t sure if the voltage that I am reading is correspondent to the tube or the wire, in fact I think the last time I measured the voltage is about 2.2V, if the reading is correct this means there is a big voltage in the wire. As you can now imagine in order to have such a voltage in the cable the stainless steel tube must have a low resistance, and it does when measure it the multmeter is too low that I can't be sure about its value, the multimeter measures 0.2ohm, but the multimeter also measures 0.1ohm when the test probes are in short circuit. And tried to raise the resistance of the tube as high as I could, I tried to mold the outer diameter with a grinding stone, the best I got is 6.7mm outside and 5.8mm inside, I measured this with a digital calliper. Probably I should try to get a thinner tube, to get a higher resistance. I chose this design because was simple to manufacture, and it works, but I got to find a way to reduce the current, because the high current certainly induces high eddy currents in the transformer core, and in fact after 15 minutes turned on the transformer seem like it could grill a beef.
But my question was more, when I connect a multimeter in a circuit like this than only has two components (in my case resistors), how do I know what component is the voltage read correspondent to ??

 

[vk6kro]

When you have resistors in series, and put a voltage across all of them, the voltages across each one depends on its resistance, and all the voltages add up to the supply voltage.

So, you have 3 things in series, One side of the wire, then the stainless steel pipe, then the other side of the wire.

The meter will read the voltage difference between the points where its probes touch.
(Don't forget to add 0.6 volts to all readings.)

You should read 3 voltages. One across each side of the wire (from where it leaves the transformer to where it joins onto the stainless steel pipe.) and one across the pipe.

 

[Homo sapiens sapiens]

Come on, you are running away from the question I made, I don't know the answer but I don't make up messy explanations. And I also don't know if is correct to measure the voltage in a circuit where dB/dt is different from zero with a multimeter. Because of the transformer winding dB/dt is not zero, this means that the line integral of E.dl , is not zero (l means distance, recall Faraday law, that is very different from Kirchoff law), I measured the value of E.dl with the multimeter, with the multimeter as the load of the transformer. I called this the emf, and I know that E.dl is equivalent to the voltage, if I calculate E.dl for the closed loop that the circuit of the transformer and the wire and the tube is, then this is the sum of all voltages in the circuit, now this is emf=V_load+V_wire <=> emf=R_load*I + R_wire*I, in my case the wire inside the transformer is exactly the same wire that connects to the tube, e re-winded the secondary of the transformer, so all the wire is the same and continuous, the only real terminals are the ones connected to the tube.
If I measure R_load and R_wire with the multimeter, that is a simple and possible thing to do, and if I insert the multimeter in series in the circuit to measure the current, then I can calculate V_load and V_wire, and I am sure the values are correct. The only problem for me is that I can't measure the current because I know the current is too high.
Now my problem with measuring directly the voltage is that the two terminals of the tube are also the two terminal of the wire, so I don't know what voltage I am reading, but I know that if R_load is different from R_wire, am I'm sure that the values of V_load and V_wire will be different, just don't know how to measure them with a multimeter.

 

[vk6kro]

It is a lot simpler than you seem to be saying.

Imagine you have 3 light bulbs in series across a battery.

If you measure across the battery with a multimeter (on a DC volts range) then you measure the voltage of the battery but also the total voltage across the 3 lamps.

If you put the meter across any of the lamps, you measure the voltage across that lamp.

The 3 lamp voltages add up to the battery voltage.

Your heat gun has an AC supply, but it is no different. The length of the wire doesn't matter at this frequency (60 Hz) because a wavelength is 5000 km, and your wire from the transformer to the pipe is just two resistors in series with the pipe.

There is an AC voltage coming from a transformer and you can measure the voltage at that point. It doesn't matter where this voltage comes from, just like it didn't matter if the light bulbs were supplied with a battery or a power supply. If the voltage is there, the rest of the circuit will behave the same.

[PLAIN]http://dl.dropbox.com/u/4222062/HEAT%20GUN.PNG

If you measure the voltage between A and D you get the supplied voltage. If you measure the voltage between B and C, you get the voltage across the load, which is the pipe.

The difference between these two voltages is the voltage drop across the wires.

 

[Carl Pugh]

3.3 volts seems awfully high. Are you taking more than one turn through the core?

Take one turn (?maybe 2 turn), keep the conductor as short as possible.

Keep the conductor as close together as possible.
You can have considerable inductance if the conductors are not kept close together.

Stainless steel pipe should have the thinnest wall practical.

 

[Homo sapiens sapiens]

Ok, let me explain my problem.
[PLAIN]http://dl.dropbox.com/u/4352530/Circuit_battery.png
If the circuit is a battery and the resistors(2 cables and the tube), because dB/dt=0, then you can apply Kirshoff law(that is what you get when you put in Farady law dB/dt=0), that says that the sum of the voltages in a close circuit is zero, let's apply Kirshoff law tho the circuit, I will sum all the marked voltages traversing the circuit in the clockwise direction, I will start at point A. V_wire1+V_tube+V_wire2-V=0, if I measure with a multimeter the voltage drop V_BC=V_B-V_C=V_tube, this is the same as V_tube=V-V_wire1-V_wire2.

[PLAIN]http://dl.dropbox.com/u/4352530/Circuit_transformer.png
If the circuit is my heat gun, dB/dt in not zero, by Faraday law (closed line integral of) E.dl=-(open surface integral) (dB/dt) . dA ,I measured the equivalent value of (closed line integral of) E.dl, with a multimeter, that is with the transformer working but with the multimeter connected as the load (a multimeter reading voltage has a huge resistance so the current in the circuit is zero, so there is no voltage drop in the wires, so the voltage read in the multimeter is the induced emf). Now another way to measure the induced emf is applying ohm law to the elements of the circuit that are only resistors, I know from Faraday law that the sum of the voltages in the circuit is equal to the induced emf. So if I sum the voltages traversing the circuit in the clockwise direction, starting at point A, I get V_{half1_wire}+V_tube+V_{half2_wire}, from farady law V_{half1_wire}+V_tube+V_{half2_wire}=emf, this is V_tube=emf- V_{half1_wire}-V_{half2_wire}, from this equation clearly V_tube is different from V_{half_wire1}+V_{half_wire2} (the only way they would be the same value, is if R_tube=R_wire, but that is not the case).
But if I want to measure V_tube I connect the multimeter in point B,C, if I want to measure V_{half_wire1}+V_{half_wire2} I connect the multimeter in point B,C. So I have exactly the same procedure to measure two things that I know that are different, how can that be??

 

[vk6kro]

[PLAIN]http://dl.dropbox.com/u/4222062/heat%20gun%202.PNG

In the above diagram, the wire which is wound on the transformer's iron core has a voltage induced in it (shown as an AC EMF symbol) and the wire inside the transformer has some resistance.
So it can be represented as in the box at left.

Faraday's Law does not apply outside the transformer as there is no varying flux outside the transformer.

At the moment when the EMF polarity is as shown, the voltages around the resistor network are as shown. These add up to the same voltage as the EMF if you include the drop across the internal resistance of the transformer winding.

Have we met before?

 

[Homo sapiens sapiens]

I don't think we have met, I just placed this question in the forum for anyone that would like to answer. Is just that you are trying so much to put the source of the voltage in a box so that you can apply Kirchhoff law, so that the voltage drop in two components with the same two terminals is equal, but I think that is not really the case I described, because I never talked about a voltage source or battery, and if I really want to be pratical and measure the voltage in the wire with a voltmeter(so that I can determine the energy losses in the wire) I would have to make a hole in the isolation of the wire in the position correspondent to the half of the wire, and even if I did so I would have exactly the same doubt I am having right now, but just in a different place of the circuit. Thanks very much for thinking about my doubt!

 

[vk6kro]

If you want to analyse the transformer, you need a lot more information. So, it is usual to just put the transformer in a box that gives an output voltage. In this case I gave a Thevenin equivalent of an EMF in series with an internal resistance.

From the point of view of the hot air gun it doesn't matter what happens in the box as long as you get a known voltage out.

Kirchoff just stated the obvious. Voltages add up to the supply voltage. Currents don't disappear. Yay!

If you consider what happens in the transformer, where do you stop? What effect does your hot air gun have on the mains supply, or the street transformer or the power station?

 

[Homo sapiens sapiens]

I studied that model for the transformer in one of the classes of my course, and also briefly studied a procedure to measure experimentally the values of the components in the model, the model is a way of modeling the transformer with basic components, and if you analyze the circuit with that model, then yes the sum of the voltages in the secondary circuit add up to zero. But anyway it's a model that can be very useful in calculating power losses in the core of the transformer and on the windings of the transformer. But I don't thing it is elucidating for my doubt, anyway the core of my doubt was can you measure voltage drops with a voltmeter in a circuit were dB/dt is not zero and the source of the magnetic field is created by a component that is not in the circuit?. And in case it's possible how do you measure it?

A way to better understand the source of my doubt would be to watch the video from “MIT opencourseware” on Induced EMFs, that is very well explained and elucidating.
http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-16-electromagnetic-induction/
The explanation begins in minute 35 from the start.

 

[vk6kro]

Why not explain the reason you feel that a multimeter can't measure an AC voltage?

I assure you it can, and I won't be downloading any 35 minute (or more) videos.