- 09 Nov '15 21:25 / 1 editI'm trying to bring a Steel wire in a plating line that has a diameter "D", and line velocity "v", through a temp differential "δT" over a distance "L"

Assuming Lumped capacitance ( neglecting temperature gradients in the wire), and heat transferred through the differential surface is predominantly accomplished via radiation with some heat transferred by way of convection, I apply a power balance to a differential element of wire and the relationship follows:

µ*Ac*v*c*dT = σ*ε*dAs*(Ts^4-T^4) + h*dAs*(Ts - T)

dT = 1/(µ*π*D^2/4*v*c)*(σ*ε*(Ts^4-T^4) + h*dAs*(Ts - T))π*D*dl

dT = 4/(µ*π*D*v*c)*(σ*ε*(Ts^4-T^4) + h*dAs*(Ts - T))*dl

The above can be numerically integrated to find T(l) (Temp as a function of position)

The question I have is how much power does it take to accomplish this for a wire at a distance "y" from the radiation source. I know the solution involves view factor, which is inversely proportional to y², but I'm pretty confused on how to apply it. Any help in clearing these ideas up would be appreciated.

Intuitively, the closer the wire is to the radiation source the less power I need to bring it through the temp change. Is the above relation a minimum if the wire were virtually touching the radiation source? - 10 Nov '15 00:23

Is your source of radiation point-like or something like a plate? It's only in the former case that the inverse square law applies. If you've got some sort of plate then if you have y <~ R where R is the shortest diameter of the plate then you can ignore edge effects and the heat flux will be more or less constant as you vary y within that region. For a point source of radiation and assuming the wire is side on to the source. you want a formula on the lines of:*Originally posted by joe shmo***I'm trying to bring a Steel wire in a plating line that has a diameter "D", and line velocity "v", through a temp differential "δT" over a distance "L"**

Assuming Lumped capacitance ( neglecting temperature gradients in the wire), and heat transferred through the differential surface is predominantly accomplished via radiation with some heat transferred by ...[text shortened]... ange. Is the above relation a minimum if the wire were virtually touching the radiation source?

Power impinging on wire = Pw = L* D * P/(4*pi*y^2)

where P is the total power emitted by the source.

For a plate source of area A with the wire a short distance away from it (y < sqrt(A)):

Pw = L*D*P/A

But you also have to take into account what fraction of the incident radiation the wire actually absorbs and how much is reflected. You also need to take into account losses. - 10 Nov '15 02:50

The source of radiation is not yet determined. I am still somewhat confused on this issue:*Originally posted by DeepThought***Is your source of radiation point-like or something like a plate? It's only in the former case that the inverse square law applies. If you've got some sort of plate then if you have y <~ R where R is the shortest diameter of the plate then you can ignore edge effects and the heat flux will be more or less constant as you vary y within that region. For ...[text shortened]... the wire actually absorbs and how much is reflected. You also need to take into account losses.**

Lets say the there is one wire passing through the an enclosed tube with all four interior sides maintained at temperature T_s, and the wire enters the at T_i and exits at T_L.

From Stefan Boltzmann's equation the enclosure should emit power to the wire by;

q"_rad = σ*ε*(T_s^4-T^4) ...correct?

It bothers me that this equation is independent of distance from the source, that is you can set the power knowing the body's emissivity,wire temp, and radiating surface temp.

What if I increase the size of the surroundings, and maintain the same surface temp... using the relationship above would indicate no effect?

This is where the view factor "F" comes into play. F α 1/R², and yet I never see the equation in the form:

q"_rad α 1/R²*σ*ε*(T_s^4-T^4)

If the Surface and Body are perfect blackbodys does the veiw Factor always = 1? Is this what the familiar equation represents? - 10 Nov '15 03:34

The power emitted by the radiator is a property of the radiator alone. You get the total emitted power using that. The radiation field strength (*) at some point distant from the source is the radiation going through a unit area. The area of a sphere is 4*pi*r^2, so you get the power per unit area of P/4*pi*r^2.*Originally posted by joe shmo***The source of radiation is not yet determined. I am still somewhat confused on this issue:**

Lets say the there is one wire passing through the an enclosed tube with all four interior sides maintained at temperature T_s, and the wire enters the at T_i and exits at T_L.

From Stefan Boltzmann's equation the enclosure should emit power to the wire by;

...[text shortened]... fect blackbodys does the veiw Factor always = 1? Is this what the familiar equation represents?

One thing you may have missed is that the Stefan Boltzmann law gives the power emitted per unit surface area of the source, not the total power emitted. Also, why do you have a T_s^4 - T^4? The temperature of the environment is irrelevant to the radiation emitted. To be in equilibrium with it's environment (ignoring conduction effects) we need other radiators for it to absorb radiation from.

In your example, you have a cylinder which emits radiation into its central cavity. Radiation emitted internally will be focused inwards, so*all*the heat emitted by the radiator will impinge on the wire. This problem does not depend on the radius of the container. In fact it doesn't depend strongly on the the geometry of the container as you have an instant conduction assumption in the wire. It only depends on the total internal surface area of the container and the temperature. You'll need to assume that the radiation is either absorbed by the wire or reabsorbed by the container walls after one transit of the container (I'm not sure about this point) otherwise the problem becomes incredibly complex. Also the problem is a lot easier if you ignore air in the container. The two corrections are in opposite directions so the simplifying assumptions probably don't do too much damage.

(*) Which I'm using to mean power per unit area and not the electromagnetic field strength (E and B). - 10 Nov '15 14:03
*Originally posted by DeepThought***The power emitted by the radiator is a property of the radiator alone. You get the total emitted power using that. The radiation field strength (*) at some point distant from the source is the radiation going through a unit area. The area of a sphere is 4*pi*r^2, so you get the power per unit area of P/4*pi*r^2.**

One thing you may have missed is tha ...[text shortened]... hich I'm using to mean power per unit area and not the electromagnetic field strength (E and B).One thing you may have missed is that the Stefan Boltzmann law gives the power emitted per unit surface area of the source, not the total power emitted. Also, why do you have a T_s^4 - T^4? The temperature of the environment is irrelevant to the radiation emitted.

I thought the (Ts^4 - T^4) gives the net radiation absorbed by the control volume?

αG(Ts)*dA - εσT^4*dA = dM/dt*c*dT

I was assuming that α = ε, and G(Ts) = σTs^4

To be in equilibrium with it's environment (ignoring conduction effects) we need other radiators for it to absorb radiation from.

The body (wire) isn't in thermal equilibrium with the surroundings. - 10 Nov '15 20:03

Look, I'm not following your algebra. I don't know what you are expecting the overall terms in your equations to do. What I need to know is what:*Originally posted by joe shmo*One thing you may have missed is that the Stefan Boltzmann law gives the power emitted per unit surface area of the source, not the total power emitted. Also, why do you have a T_s^4 - T^4? The temperature of the environment is irrelevant to the radiation emitted.

I thought the (Ts^4 - T^4) gives the net radiation absorbed by the cont ...[text shortened]... b radiation from.[/quote]

The body (wire) isn't in thermal equilibrium with the surroundings.

αG(Ts)*dA

represents - is this the amount of radiation absorbed by the wire? So I may have the wrong end of the stick in the following.

What you seem to be trying to do is run the Stefan Boltzmann equation in reverse to work out absorption. This isn't a terrible idea since as a microscopic process the one is the reverse of the other. The problem is that won't work due to entropy considerations.

We can think of the absorber as being in a radiation bath. The radiation bath's temperature is determined by its frequency and only that. The amount of radiation available to be absorbed by the wire is given by the energy density of the radiation. So if there's only one photon impinging on the wire every second there simply cannot be enough power available to satisfy Stefan's law in reverse.

What you can do is assume the wire absorbs all the radiation impinging on it. Then:

dU = Q_in - Q_out.

Change in internal energy over some time dt is heat in during that time less heat out during that time.

Q_in = ε_s * σ * T_s^4 * A_s * dt

Where dt is the duration of the time interval. The heat going into the wire is equal to the radiation emitted by the container. T_s is the temperature of the container and is a constant.

Q_out = ε_w * σ * T_w(t)^4*A_w

The heat radiated by the wire in time interval dt is Q_out, notice that T_w(ire) is a function of time.

Dropping the _w subscripts, so the only subscripted variables are the container's, and using M for the mass of the wire and C for it's heat capacity we get:

M*C*dT / dt = σ * ( ε_s * A_s * T_s^4 - ε * A * T^4)

There is no reason to believe that the epsilons are the same for the wire as the container (unless it's the same material), the surface areas are also different. - 11 Nov '15 01:25 / 1 edit

I think we are close to reaching an understanding on this...let me try to be more thorough in explaining the set up, and adjust the equations (again thank you for your patience, this is made more difficult by the lack of pictures and the fact that I am grappling with the main concepts of radiation).*Originally posted by DeepThought***Look, I'm not following your algebra. I don't know what you are expecting the overall terms in your equations to do. What I need to know is what:**

αG(Ts)*dA

represents - is this the amount of radiation absorbed by the wire? So I may have the wrong end of the stick in the following.

What you seem to be trying to do is run the Stefan Boltzmann eq ...[text shortened]... the wire as the container (unless it's the same material), the surface areas are also different.

The enclosure is a four sided rectangular box with a length "L" with two open ends. It will eventually have gas/electric powered heaters/burners inside to create a high temperature atmosphere/surroundings for the wire to pass through. The wire is moving through the box/furnace with a constant velocity. My goals are as follows:

1) Find the temperature distribution T(x) in the wire as it passes through the enclosure that is maintained at a constant surface temp Ts.

2) Find the power needed to maintain the enclosure at Ts

The first objective:

Take a differential element of the wire of length "dx" and perform an energy balance on it as it is passing through the enclosure with velocity "v" in the x direction

Again, assuming radiation is the dominant mode of heat transfer

dq_rad = dM/dt*C*[(T+dT)-T]

dq_rad = p*Ac*v*C*dT

p = density of steel

Ac = cross-sectional area of wire

v = velocity of wire as it passes through the enclosure

C = specific heat of the steel

dq_rad = dq_incident - dq_emitted

dq_incident = αG(Ts)*dA ---> this is the radiation impinging on the differential wire element (irradiation), it is to be approximated as emission from a black body at "Ts"

dq_incident = σ * T_s^4 * dA

dq_emitted = σ * ε * T^4 * dA ---> the radiation from the differential element to the surroundings

dA = pi*D*dx ---> differential elements surface area

The energy balance should be as follows:

(p*Ac*v*C)*dT = (σ * T_s^4) * dA - (σ * ε * T^4) * dA

(p*Ac*v*C)*dT = (σ * T_s^4) * pi*D*dx - (σ * ε * T^4) * pi*D*dx

dT/dx = pi * D / (p * Ac * v * C) * σ * (T_s^4 - ε * T^4) ...(1)

from (1) I can find T(x) for a given set of parameters.

The power needed to maintain the enclosure surface at that temperature Ts is dependent of the size of the enclosure and the environment, etc... so I will have to perform another power balance to obtain that.

What I cant figure out is if I make the enclosure very large with the wire running through its center two things should happen: The power needed to maintain the enclosure at Ts will increase, and the rate of change of the wires internal energy per unit length will decrease. This is problematic because I have to change the wires Temp by a large designated amount, in a finite distance, during a very short time period as the wire passes through the furnace/enclosure.

I need high intensity radiation to do this, which is certainly a function of the distance of the wire to the radiation source. So what size enclosure will get me there?

Hope this is a better explanation, or maybe I'm still goofing these concepts up? - 11 Nov '15 03:29 / 1 editHi Joe, I'm about to read your post, but you should have a look at the Wikipedia page as it presents a calculation for the sun/earth system. Note that they make a number of assumptions that are relevant to the earth/sun system rather than your enclosure and wire set up.

https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

Edit: Ok, I've read it, but I've also been drinking Chianti so a properly considered answer will have to wait until tomorrow. But one impression I have is that your dM/dt term is odd. I don't think that's right, but need to think it through with an unbefuddled mind. - 11 Nov '15 12:31

Haha, no problem enjoy your wine!*Originally posted by DeepThought***Hi Joe, I'm about to read your post, but you should have a look at the Wikipedia page as it presents a calculation for the sun/earth system. Note that they make a number of assumptions that are relevant to the earth/sun system rather than your enclosure and wire set up.**

https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

Edit: Ok, I've read ...[text shortened]... term is odd. I don't think that's right, but need to think it through with an unbefuddled mind.

I wouldn't worry to much about that dM/dt term, I pulled that whole side out of my Heat Transfer text. The application is the same except it is a fluid element flowing down a long pipe picking up heat along the way via convection. Seems to be pretty much identical with the exception of the mode of heat transfer, but I'm sure you'll come to that conclusion when you think it over.

Also, I looked over the wiki page. It seems to me that taking the distance from the source into account is going to complicate this analysis. Unless I can figure our some reasonable simplifying assumptions of my own. - 11 Nov '15 15:41 / 3 edits

Seems like it would be simpler to just build a model and plug it in.*Originally posted by joe shmo***Haha, no problem enjoy your wine!**

I wouldn't worry to much about that dM/dt term, I pulled that whole side out of my Heat Transfer text. The application is the same except it is a fluid element flowing down a long pipe picking up heat along the way via convection. Seems to be pretty much identical with the exception of the mode of heat transfer, but I' ...[text shortened]... icate this analysis. Unless I can figure our some reasonable simplifying assumptions of my own.

Is this an oven? What is the app here?

Ah, fluid in a pipe being heated by the elements which presumably do not come in contact with the fluid. Is the fluid a liquid to be heated or is is drawing heat out of the device for temperature control?

Your mention of 'lumped capacitance', does that mean the power applied is RF, like one of the medical frequencies, like 13.56 Mhz? - 11 Nov '15 15:47 / 1 edit

Its an oven. Its not fluid, I'm trying to heat a steel wire passing through a relatively small furnace from T1 - T2, where T2 >>T1 in the short time that it passes through said furnace.*Originally posted by sonhouse***Seems like it would be simpler to just build a model and plug it in.**

Is this an oven? What is the app here?

Ah, fluid in a pipe being heated by the elements which presumably do not come in contact with the fluid. Is the fluid a liquid to be heated or is is drawing heat out of the device for temperature control?

Lumped capacitance assumes that the body you are trying to heat has no internal temperature gradients...it basically means that any heat added to the body causes the entire temperature of the body to uniformly change. This criteria being valid is dependent on the physical properties of the body and environment the body is in. things like: thermal conductivity,surface area,convection,radiation coefficients etc... - 11 Nov '15 19:59

I think you are forced to consider conduction in the wire, which makes the whole thing a little painful. You have to track elements of wire as they move through the tube which is reminiscent of the way fluid dynamics problems are addressed.*Originally posted by joe shmo***Haha, no problem enjoy your wine!**

I wouldn't worry to much about that dM/dt term, I pulled that whole side out of my Heat Transfer text. The application is the same except it is a fluid element flowing down a long pipe picking up heat along the way via convection. Seems to be pretty much identical with the exception of the mode of heat transfer, but I' ...[text shortened]... icate this analysis. Unless I can figure our some reasonable simplifying assumptions of my own.

The cross-section of the tube doesn't affect the radiation calculation, but it is important for losses from the tube (the smaller cross-section of the tube relative to its length the smaller the proportionate losses will be). You also need the x sectional area of the tube as big as possible to increase the power emitted. So this is sounding like an optimization problem

The really painful part is taking account of conduction within the wire. Suppose you have some sort of steady state, so that the temperature of the bit of wire a distance x from the entrance of the tube is T and unchanging (each element of wire gets to a temperature T as it reaches position x). The function T(x) probably won't be linear so we should avoid making that assumption. I'll try doing the calculation and get back to you to see if I can get (basically) the same answer that you have. - 12 Nov '15 02:19
*Originally posted by DeepThought***I think you are forced to consider conduction in the wire, which makes the whole thing a little painful. You have to track elements of wire as they move through the tube which is reminiscent of the way fluid dynamics problems are addressed.**

The cross-section of the tube doesn't affect the radiation calculation, but it is important for losses from the ...[text shortened]... e calculation and get back to you to see if I can get (basically) the same answer that you have.I think you are forced to consider conduction in the wire, which makes the whole thing a little painful. You have to track elements of wire as they move through the tube which is reminiscent of the way fluid dynamics problems are addressed.

Yeah, that is going to muck it up (they neglected axial conduction in the fluid example)

The cross-section of the tube doesn't affect the radiation calculation

I think the distance and geometries most definitely affect the radiation calculation via the View Factor...Incidentally I think this is what I was looking for. I don't see a readily computed factor for what my specific geometry will be, but follow the link to see what I am referring to.

http://webserver.dmt.upm.es/~isidoro/tc3/Radiation%20View%20factors.pdf

The radiation impinging on the wire will be scaled according to the view factor associated with the specific geometry. I'll have to revise the power balance to include this.

(p*Ac*v*C)*dT = F(1,2)*(σ * T_s^4) * dA - F(2,1)*(σ * ε * T^4) * dA

The view factors F(1,2) + F(2,1) = 1, F(1,2) will most likely have a large impact on the power needed to achieve this goal, unfortunately in the less inviting direction.

. - 12 Nov '15 11:49

If the wire is metal, I assume it is, why can't you just use electricity to directly heat the wire like an incandescent light bulb? Is there some background gas in the pipe that you want to do some chemical thing like low O2 or some such?*Originally posted by joe shmo***Its an oven. Its not fluid, I'm trying to heat a steel wire passing through a relatively small furnace from T1 - T2, where T2 >>T1 in the short time that it passes through said furnace.**

Lumped capacitance assumes that the body you are trying to heat has no internal temperature gradients...it basically means that any heat added to the body causes the enti ...[text shortened]... y is in. things like: thermal conductivity,surface area,convection,radiation coefficients etc...

Another way to heat the wire would be just with a coil around the wire fed with RF. That would induce currents in the wire without touching it and probably saving a lot of energy that would otherwise be needed to heat up the entire inside of the pipe. - 13 Nov '15 00:23

There is no special atmosphere other than a hot one. Iv'e been told that electromagnetic induction heating is expensive, but its not out of the realm of possibilities yet.*Originally posted by sonhouse***If the wire is metal, I assume it is, why can't you just use electricity to directly heat the wire like an incandescent light bulb? Is there some background gas in the pipe that you want to do some chemical thing like low O2 or some such?**

Another way to heat the wire would be just with a coil around the wire fed with RF. That would induce currents in th ...[text shortened]... saving a lot of energy that would otherwise be needed to heat up the entire inside of the pipe.