# How would you solve this problem?

1. Sep 10, 2006

### John Galt

I am sizing an electric heating element to produce a steady supply of heat. The heating element consists of a ceramic tube, coiled on the outside with resistance wire. This tube is encapsulated by another larger ceramic tube and the space between them is evacuated to help direct the heat to the air column in the tube wound with resistance wire. A fan then blows this heated air to where it is needed.

What I know is:

1) The diameter of the resistance wire.
2) The wall thickness and inside diameter of the ceramic tube coiled with resistance wire.
3) The resistivity of the wire in Ohms/ft.
4) The voltage and current running through the resistance wire.

What I don't know is:

1) How many feet of wire to wind around the ceramic tube to reach a desired temperature that remains constant.
2) Which CFM fan rating should be used to blow past the element to deliver the air at constant temperature.

I need somebody to help me figure out what I don't know from the information I do know. Am I missing anything? Any and all help is appreciated.

Last edited: Sep 10, 2006
2. Sep 10, 2006

### Danger

Hi, John. Unfortunately, I can't help you. This is out of my area. I would still recommend, however, that you post the values that you do know, rather than just state that you know them. They're essential to working out your solution.

3. Sep 11, 2006

### Integral

Staff Emeritus
Danger,
Not really. Generally actual numbers are not very important at this stage. They tend to distract from the essence of the problem.

Since you know the voltage and current, you know the resistance, therefore given the resistivity of the wire you can compute the length. I suspect that you really do not know the current, but rather a max current for your power supply. You would be well off to design your heating element to use say 80% of that. You may want to look at other constraints to see if there is some upper limit on the power you can handle.

Once you have your heater working you can then determine at what rate the air needs to be moving to get your final temperature.

To the best of my knowledge getting the final air temperature analytically is very difficult.

4. Sep 11, 2006

### Staff: Mentor

Trying to figure out how the convective heat transfer works is very difficult, but you can probably skip all that by making the assumption that all of the heat produced goes into heating the air. While that's true, the catch is that the temperature of the heating element will depend on the effectiveness of the heat transfer, and that can affect its operation - resistance is temperature dependent. It is also tough to know how "forgiving" the heating element is (we don't want to burn it out). But when your approach temperature is so large, big changes in the temperature of your output air don't have a very big impact on the heating element's temperature.

All that said, the answers to the two questions are dependent on each other. Your given #4 is your wattage, so if you pick a temp or airflow, you can calculate the other.

Last edited: Sep 11, 2006
5. Sep 11, 2006

### Danger

Sorry 'bout that.
I was thinking that the desired temperature and the air flow rate at which it can be sustained would be the starting point. As I said, I don't know much about this stuff.

6. Sep 11, 2006

### Artman

This looks like a homework problem. Look at the two questions: They are both looking for points where the temperature would remain constant. This is true of any amount of wire and any amount of air. The constant value changes only if one of the other values is changed, wraps of wire or CFM.

7. Sep 11, 2006

### Qleap1000

Very good point Artman. Determine the unit in equilibrium, then the heat flow and distribution is another set of equations based on heat transfered to the air volume in time (flow or rate.)

8. Sep 11, 2006

### Artman

Yup. I believe it would be harder to determine a specified point (number of wraps required to produce so much temperature rise, inlet to outlet, at how much CFM).

What he asked is more like a trick question on a test.

A1) Any number of wraps (feet) of wire can produce a temperature that remains constant.

A2) Any CFM can also produce a constant heat rise as long as the CFM remains constant.

Last edited: Sep 11, 2006
9. Sep 11, 2006

### Qleap1000

The stumpers questions

-and again those are very valid points that we hope John Galt is looking at.

I have stumped many a teacher with their own "BADLY" stated questions.

So, keeping all perspectives in mind is a good way to go, especially if it is an essay question which when there is enough reportoire with the teacher/proffesor you might be able to discuss the answers with the instructor afterwards in any case.

10. Sep 11, 2006

### Integral

Staff Emeritus
In my DIRECT experience, the resistance of Nichrome is not extremely temperature dependent. That is how you are able to reliably design a heating element. If it were otherwise it would be very difficult to maintain a temperature.

A fellow I worked for a few years back was certain that there would be a strong temperature dependence in Nichrome. To test it we spooled of ~30ft (10m) of Nichrome and measured its resistance cold, then but current to it and measured a voltage drop when it was hot. There was barely a measurable change in the resistance. We then designed our furnace with a constant resistivity in mind, and it worked exactly as designed. Note that I have specified the Heater wire we used, this is pretty commonly used and I will bet that other heater element wire is of a similar nature.

11. Sep 12, 2006

### Artman

I'll let someone else give the means for calculating the amount of wire to give a specific wattage, it sounds like Integral and some of the others have a better handle on that than I would, and just help out with the heat rise of the air.

If your air is entering at or near standard conditions, this calculation will give you the temperature rise of the air.

TD = MBH/CFM/.00108

TD = Temperature difference
MBH = 1000 btu = 3412 Watts

Say for a moment you have enough resistance to give 3 KW,
this would be 3 x 3412watts = 10236, or 10.236 MBH
and 500 CFM

500/10.236/.00108 = 18.96 deg F TD.

Lower the CFM to 400 and the calculation becomes:

400/10.236/.00108 = 23.7 deg F TD.

Or raise the length of element to deliver 5.0 KW and the calculation using 500 CFM becomes:

500/17.06/.00108 = 31.6 deg F TD.

You see how changing either the CFM or the KW results in a change in the TD?

12. Sep 12, 2006

### Staff: Mentor

Good to know.