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Chestermiller

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- #26

Chestermiller

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- #27

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I think the start of my plots looks similar to the short time asymptote (it gradually increases like the graph of a radical function ##\sqrt{\xi}##). The middle sections look like the long time asymptote (linear increase). The final part looks like the short time asymptote again (the spikes at the end).

Do you know what could be happening? What is the mechanism that has been neglected? I am not sure.

Here is the picture of the temperature curves I extracted for 3 different species. For the first plot, the maximum temperature of my camera is ~250°C, so the 280°C readings are definitely saturated. All the readings are from the hottest spot where the laser hits the tissue like this.

I am still working through the equations. Is your equation 13 also given in the "Transport Phenomena" textbook?

- #28

Chestermiller

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Evaporation (phase change) must be occurring at these temperatures. There will almost certainly be an interface between a dry region and a wet region within the pores of the stem. The leveling off of the curves in the middle region of the temperature-time curves is a strong indication of this, particularly since it is close to the boiling point of water.This is very helpful. Thank you so much for your time, I really appreciate that.

I think the start of my plots looks similar to the short time asymptote (it gradually increases like the graph of a radical function ##\sqrt{\xi}##). The middle sections look like the long time asymptote (linear increase). The final part looks like the short time asymptote again (the spikes at the end).

Do you know what could be happening? What is the mechanism that has been neglected? I am not sure.

No. BSL doesn't solve this particular problem. This is my own solution to the equations (which I have total confidence in). However, BSL does describe a similar approach to analyzing the asymptotic heat transfer behavior for laminar flow in a tube being heated by a constant heat flux along its length.I am still working through the equations. Is your equation 13 also given in the "Transport Phenomena" textbook?

- #29

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I had a few questions that I forgot to ask:

I have found a copy of the BSL textbook. In problem 12b.3, they give the solution as follows:

$$\Delta T=\frac{q}{k}\left(\sqrt{\frac{4\alpha t}{\pi}}\exp\left(\frac{z^{2}}{4\alpha t}\right)-\frac{2z}{\sqrt{\pi}}\int_{z/\sqrt{4\alpha t}}^{\infty}\exp(-u^{2})du\right).$$

This is different from the form you used in your post #22. Why are we justified in ignoring those other terms in the equation?

I also had some questions regarding the terms you used in your derivation (post #24):

1. In Eqn. 1 for heat diffusion in 1D, does ##\theta## represent temperature rise in dimensionless form (i.e., ##\theta(t,z) =\frac{T\left(t,z\right)-T_{0}}{T_{max}-T_{0}}##)?

2. ##f## is the initial distribution of temperature at ##z##, i.e., ##\theta\left(0, z\right)=f(z)##—is this correct?

3. Is it possible to express the constant flux at the surface as ##q=Q \times \text{Volume}##? Here ##Q## is the rate of energy generation per unit volume (the term in the transient equation that we ignored), and according to this paper, it is equal to ##I \alpha##, where ##\alpha## is the absorption coefficient and ##I## is the intensity of the laser.

- #30

Chestermiller

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The equation in post #22 is just what this equation predicts at z = 0.Hi Chestermiller,

I had a few questions that I forgot to ask:

I have found a copy of the BSL textbook. In problem 12b.3, they give the solution as follows:

$$\Delta T=\frac{q}{k}\left(\sqrt{\frac{4\alpha t}{\pi}}\exp\left(\frac{z^{2}}{4\alpha t}\right)-\frac{2z}{\sqrt{\pi}}\int_{z/\sqrt{4\alpha t}}^{\infty}\exp(-u^{2})du\right).$$

This is different from the form you used in your post #22. Why are we justified in ignoring those other terms in the equation?

No. It is the actual temperature.I also had some questions regarding the terms you used in your derivation (post #24):

1. In Eqn. 1 for heat diffusion in 1D, does ##\theta## represent temperature rise in dimensionless form (i.e., ##\theta(t,z) =\frac{T\left(t,z\right)-T_{0}}{T_{max}-T_{0}}##)?

No. It is the spatial part of the asymptotic temperature variation at long times.2. ##f## is the initial distribution of temperature at ##z##, i.e., ##\theta\left(0, z\right)=f(z)##—is this correct?

q is the integral of Q over the length of the stem. That is ##q=\int_0^L{Qdz}## (Don't forget to include the attenuation of I)3. Is it possible to express the constant flux at the surface as ##q=Q \times \text{Volume}##? Here ##Q## is the rate of energy generation per unit volume (the term in the transient equation that we ignored), and according to this paper, it is equal to ##I \alpha##, where ##\alpha## is the absorption coefficient and ##I## is the intensity of the laser.

- #31

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I see. I guess we used z=0 because at short times the heat hasn't really traveled much into the stem.The equation in post #22 is just what this equation predicts at z = 0.

If water is substituted for tissue (a zero-order approximation), do you know for what sort of timescales each of the equations in post #25 are valid?

So ##\theta_{\infty}(t,z)## is valid for long times t, and also for long z (far removed from heat source ##q##)?No. It is the spatial part of the asymptotic temperature variation at long times.

So, when we write ##Q=I \alpha## doesn't that already account for the attenuation? Isn't this what you mean by the attenuation of I?q is the integral of Q over the length of the stem. That is ##q=\int_0^L{Qdz}## (Don't forget to include the attenuation of I)

The attenuation coefficient of a sample is related to the absorption coefficient through:

$$\gamma=\underbrace{\alpha}_{\text{absorption coefficient}}+\underbrace{\beta}_{\text{scattering coefficient}}$$

- #32

Chestermiller

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Yes.Thank you very much for your input.

I see. I guess we used z=0 because at short times the heat hasn't really traveled much into the stem.

It depends on the specific physical properties. As a rule of thumb, the short time solution is good approximation for ##\xi<1## and the long time solution is good for ##\xi>1##.If water is substituted for tissue (a zero-order approximation), do you know for what sort of timescales each of the equations in post #25 are valid?

As I said, it's good for values of ##\xi>1##. For a finite sample like this, it applies to all values of z at long times.So ##\theta_{\infty}(t,z)## is valid for long times t, and also for long z (far removed from heat source ##q##)?

I was neglecting scattering. In that case, $$\frac{dI}{dz}=-\alpha I$$and $$q=I(0)e^{-\alpha z}$$So, $$q=I_0\int_0^{\infty}{\alpha e^{-\alpha z}dz}=I_0$$This assumes that all the attenuation takes place a short distance from the interface, and neglects the finite length of the stem.So, when we write ##Q=I \alpha## doesn't that already account for the attenuation? Isn't this what you mean by the attenuation of I?

The attenuation coefficient of a sample is related to the absorption coefficient through:

$$\gamma=\underbrace{\alpha}_{\text{absorption coefficient}}+\underbrace{\beta}_{\text{scattering coefficient}}$$

- #33

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I had just a few more questions.

I was looking at another textbook, and they define the Fourier Number differently from your post #25. It is instead given as:

$$\xi=\frac{\alpha t}{l_c^{2}},$$

where ##l_c## is the "characteristic length" given by:

$$l_{c}=\frac{\text{Volume}}{\text{Surface area}}.$$

For a plate of thickness ##L##, we have ##l_c \simeq L/2##.

Do you think this is applicable to my situation?

Yes. And according to the reference, the scattering coefficient term will not contribute to the heating of the tissue. But is the minus sign there a typo? In the paper, they give this as: ##Q=\partial I/\partial z = \alpha I##.I was neglecting scattering. In that case, $$\frac{dI}{dz}=-\alpha I$$and $$q=I(0)e^{-\alpha z}$$So, $$q=I_0\int_0^{\infty}{\alpha e^{-\alpha z}dz}=I_0$$This assumes that all the attenuation takes place a short distance from the interface, and neglects the finite length of the stem.

I see. For some reason, my temperature curves always look like the short-time asymptote — even when the plant is a seedling (##L \approx 1 \ cm##, so ##\xi>1##). Could this be because we neglected the width of the plant?It depends on the specific physical properties. As a rule of thumb, the short time solution is good approximation for ##\xi<1## and the long time solution is good for ##\xi>1##.

To make sure I've understood the notations correctly:Chestermiller said:No. It is the actual temperature.

##\theta\left(t,z\right)=T(t,z)##

##\theta_{\infty}(t,z)=\Delta T(t,z)\ \text{(none-dimensionless)}##

##\theta^{*}(t)=\Delta T(t)\ \text{(dimensionless)}##

Is that right?

Thank you very much for the help.

- #34

Chestermiller

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No. The selection of the appropriate length to use depends on the specific physical situation. This is at the discretion of the analyst. In your problem, it is most appropriate to use the length of the sample.Hi Chestermiller,

I had just a few more questions.

I was looking at another textbook, and they define the Fourier Number differently from your post #25. It is instead given as:

$$\xi=\frac{\alpha t}{l_c^{2}},$$

where ##l_c## is the "characteristic length" given by:

$$l_{c}=\frac{\text{Volume}}{\text{Surface area}}.$$

For a plate of thickness ##L##, we have ##l_c \simeq L/2##.

Do you think this is applicable to my situation?

This depends on which direction you choose for z. If z is measured in the opposite direction of the photon flow, then the positive sign is appropriate. If z is measured in the same direction as the photon flow, then the negative sign is appropriate.Yes. And according to the reference, the scattering coefficient term will not contribute to the heating of the tissue. But is the minus sign there a typo? In the paper, they give this as: ##Q=\partial I/\partial z = \alpha I##.

I think it's more that you have neglected the effect of water vaporization. I would try some experiments where I reduced the intensity of the laser so that the temperature did not rise so much and see how the results compare.I see. For some reason, my temperature curves always look like the short-time asymptote — even when the plant is a seedling (##L \approx 1 \ cm##, so ##\xi>1##). Could this be because we neglected the width of the plant?

Yes.To make sure I've understood the notations correctly:

##\theta\left(t,z\right)=T(t,z)##

##\theta_{\infty}(t,z)=\Delta T(t,z)\ \text{(none-dimensionless)}##

##\theta^{*}(t)=\Delta T(t)\ \text{(dimensionless)}##

Is that right?

- #35

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The energy absorption in a differential distance ##dz## is:

$$dI=-\alpha I dz,$$

so using your convention for measuring z:

$$Q=\partial I/\partial z=-\alpha I.$$

To find the heat flux ##q## we will integrate this between the limits x=0 and x=L:

$$q=\int_{0}^{L}Q\ dz=-\alpha \int_{0}^{L} I\ dz=-\alpha I_{0} \int_{0}^{L}e^{-\alpha z}\ dz,$$

$$q=I_{0}\left[e^{-\alpha L}-1\right],$$

which for ##L\to\infty## reduces to ##-I_{0}## and not ##I_{0}## that you obtained in your post. Why does this negative sign mean?

- #36

Chestermiller

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This equation should read $$Q=-\partial I/\partial z=\alpha I$$I also need a little help to walk me through the calculation of ##q## in your post #32.

The energy absorption in a differential distance ##dz## is:

$$dI=-\alpha I dz,$$

so using your convention for measuring z:

$$Q=\partial I/\partial z=-\alpha I.$$

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