# Require length of a wire in free air with one hot end(1000 c) to 100c

1. Apr 26, 2013

let say i have a horizontal wire of alumel(or steel or what ever), the thermal conductivity is 30 w/m^2/k. let say D=0.4mm

one end of the wire at 1000 degree c ( so like submerge in a cement which is at 1000 c). how long do i need wire to horizontaly in free air so that the cold end is at 100 degree c?

image of problem here
http://postimg.org/image/b8pt33u75/

2. Apr 26, 2013

### Staff: Mentor

It depends on the cooling situation of the room - do you have air flow? Do you get convection? Do you have some other cooling mechanisms?

The wire is quite hot, an approximation via blackbody radiation only might give some reasonable result. You need the temperature of the environment, however.

3. Apr 27, 2013

the temperature is 20 c, room temperature. and no air flow. so natural convection

4. Apr 27, 2013

### Staff: Mentor

I did a quick simulation with blackbody radiation alone, and got a length of ~20cm. That looks short.

The idea: neglect radial temperature differences, T is a function of length x only. For every part of the wire, set blackbody radiation equal to the inflowing heat due to temperature differences. This leads me to
$$\lambda A \frac{\partial^2 T}{\partial x^2} = 2 \pi r \sigma T^4$$
Where $\sigma$ is the Stefan-Boltzmann-constant.

It assumes that the whole environment is at 20°C. The hot concrete will change this, of course, but then we have to simulate the whole setup I think.

5. Apr 28, 2013

why black body radiation? i thought the main heat transfer will be conduction, and heat loss by convection?

Last edited: Apr 28, 2013
6. Apr 28, 2013

### Staff: Mentor

Blackbody radiation is easier to model ;). And it does not require to know details about the environment.

7. Apr 28, 2013

the heat transfer by conduction, convection and radiation.

so blackbody radiation is not a method that just take into account of radiation. but give you average total heat transfer of a hot body to surrounding? whats the wavelength you use?

if possible can i see you working? i mean how do i integrate it ?

8. Apr 28, 2013

### Staff: Mentor

Blackbody radiation is a part of the total cooling process - adding more cooling mechanisms will decrease the required length.
?
I don't use any wavelength.

I plugged in all values and simulated the wire in steps of 1cm in excel. Nothing mysterious.

After fixing the units for the given thermal conductivity, $$c:=\frac{2 \pi r \sigma}{\lambda A} = 1.9 \cdot 10^{-6} \frac{1}{m^2K^3}$$
and
$$\frac{\partial^2 T}{\partial x^2} = c T^4$$
I started with the point of 100°C and assumed that the first derivative of the temperature is 0 there (as no heat can be conducted away any more).

9. Apr 28, 2013

so say i have a cylinder heated 1000 degree c. i use the black body formulae q = σ*(T^4)*A and got a value of say 500 watt. this the the total value of heat transfer from cylinder to surrounding?

or use convective heat transfer of say 5watt/m^2/k, so power=h*(delta T)* A. if i got a value of say 250 watt using this formulae. would i say the total heat dissipated by the cylinder be 500 watt or 500+250watt?

back to the question,λ is thermal conductivity? can you copy and paste(or upload) the formulae in excel to here?

or is the formala = T=0.5*x*c*T^4?

10. Apr 28, 2013

### Staff: Mentor

Right.

750 W.

Right.

See attachment.

#### Attached Files:

• ###### temp.xlsx
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11. Apr 28, 2013

### AlephZero

The length would increase for thicker wire, siice the radiation is proportional to the diameter but the thermal resisttance of the wire depends on its area (diameter squared).

20cm doesn't sound crazy for the OP's very thin wire. For a metal bar of say 1cm diameter the result would be very different.

12. Apr 28, 2013

### Staff: Mentor

The constant c is proportional to the inverse diameter (r/A), and the length scale is roughly proportional to $\sqrt{c}$. Increasing the diameter by a factor of 2.5 increases the length by a factor of ~1.6 to ~26cm.

13. Apr 28, 2013

### Redbelly98

Staff Emeritus
mfb, I think you need to use a lower emissivity -- it looks like you have assumed it is 1. For a shiny wire, 0.1 might be more realistic, though in the real world I would worry about oxidation & darkening at the hot end, depending on the wire material.

Also, would it might make sense to:
1. Try finer divisions than 1 cm, to make sure the solution has converged
2. Account for radiation absorption from the 20 C surroundings? Not a big deal, but it does make the power loss about 38% less at the 100 C end.

14. Apr 28, 2013

### AlephZero

THis isn't exactly the same as the OP's situation, but FWIW we have a test rig at work that heats up a small object to about 1000C and then maintains it at constant temperature. The object is mounted on a thin metal rod, with a force transducer at the other end which can only stand 200C. There is no special cooling in place (just convection and radiation), the rod is significantly shorter than 20cm, and we don't have any problems with exceeding the 200C llimit (monitored with a thermocouple to prodect the force transducer) at the "cold" end.

So without getting into the details of mfb's calculation, I think the order of magnitude is right.

15. Apr 29, 2013

hmmm still kinda confused about the formula in excel.

i mean i understand what does those number mean? T increase as x increase? should T = 100 at x 0.18 or atleast close?

x T dT/dx d^2T/dx^2
0.18 1391.989962 86890.92844 14517931.07

if is possible to write it as a function such as T=something*x eg. i sub in x to get T.

why is it

$$\lambda A \frac{\partial^2 T}{\partial x^2} = 2 \pi r \sigma T^4$$
Where $\sigma$ is the Stefan-Boltzmann-constant.

not

$$\lambda A \frac{\partial T}{\partial x} = 2 \pi r \sigma T^4$$
Where $\sigma$ is the Stefan-Boltzmann-constant.

also you used 2 \pi r should it not be pi*R^2?

Last edited: Apr 29, 2013
16. Apr 29, 2013

### Staff: Mentor

I used 1, as I neglected all other losses.

Does not change the result by more than 1-2cm (even with that bad integration scheme I used), and other errors are far more significant. A better resolution reduces the length a bit.
Sorry, I forgot to mention this here. All simulations included that.

x=0 is the cold end of the wire. It is easier to calculate in that direction, as I can directly include the boundary condition (dT/dx=0).
WolframAlpha did not get a reasonable result.
Heat flow depends on the first derivative, but we need the derivative of this heat flow (to set it equal to blackbody radiation).
Blackbody radiation (leaving the material) occurs at the surface only.

17. Apr 30, 2013

but 2*pi*r is the circumference and pi*R^2 is the cross section area of the wire?

18. Apr 30, 2013

### Redbelly98

Staff Emeritus
The radiation occurs at the surface of the wire, so the amount of power radiated is proportional to the surface area.

The surface area of a wire is (circumference)x(length).

19. Apr 30, 2013

### Staff: Mentor

Right. Why "but"?
Blackbody radiation happens at the circumference of the cross-section (the surface of the wire).