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

Tags: 100c, end1000, free, length, require, wire
 P: 21 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/ i really need to know, please help
 Mentor P: 11,589 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.
 P: 21 the temperature is 20 c, room temperature. and no air flow. so natural convection
 Mentor P: 11,589 Require length of a wire in free air with one hot end(1000 c) to 100c 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.
 P: 21 why black body radiation? i thought the main heat transfer will be conduction, and heat loss by convection?
 Mentor P: 11,589 Blackbody radiation is easier to model ;). And it does not require to know details about the environment.
 P: 21 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 ?
Mentor
P: 11,589
Blackbody radiation is a part of the total cooling process - adding more cooling mechanisms will decrease the required length.
 but give you average total heat transfer of a hot body to surrounding
?
 whats the wavelength you use?
I don't use any wavelength.

 if possible can i see you working?
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).
 P: 21 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?
Mentor
P: 11,589
 Quote by Askara 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?
Right.

 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?
750 W.

 back to the question,λ is thermal conductivity?
Right.

 can you copy and paste(or upload) the formulae in excel to here?
See attachment.
Attached Files
 temp.xlsx (10.3 KB, 2 views)
Engineering
HW Helper
Thanks
P: 6,948
 Quote by mfb I did a quick simulation with blackbody radiation alone, and got a length of ~20cm. That looks short.
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.
 Mentor P: 11,589 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.
 Mentor P: 12,071 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.
 Engineering Sci Advisor HW Helper Thanks P: 6,948 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.
 P: 21 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?
Mentor
P: 11,589
 Quote by Redbelly98 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.
I used 1, as I neglected all other losses.

 1. Try finer divisions than 1 cm, to make sure the solution has converged
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.
 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.
Sorry, I forgot to mention this here. All simulations included that.

 Quote by Askara T increase as x increase? should T = 100 at x 0.18 or atleast close?
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).
 if is possible to write it as a function such as T=something*x eg. i sub in x to get T.
WolframAlpha did not get a reasonable result.
 why is it [second derivative]
Heat flow depends on the first derivative, but we need the derivative of this heat flow (to set it equal to blackbody radiation).
 also you used 2 \pi r should it not be pi*R^2?
Blackbody radiation (leaving the material) occurs at the surface only.
 P: 21 but 2*pi*r is the circumference and pi*R^2 is the cross section area of the wire?
Mentor
P: 12,071
 Quote by Askara but 2*pi*r is the circumference and pi*R^2 is the cross section area of the wire?
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).

 Related Discussions Introductory Physics Homework 3 Introductory Physics Homework 2 Introductory Physics Homework 3