Volume and temperature rise in tissue

[roam]

Thank you very much for your input.

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

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

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?

No. It is the spatial part of the asymptotic temperature variation at long times.

So $\theta_{\infty}(t,z)$ is valid for long times t, and also for long z (far removed from heat source $q$)?

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)

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:

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

 

[Chestermiller]

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.

Yes.

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?

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 $\xi>1$.

So $\theta_{\infty}(t,z)$ is valid for long times t, and also for long z (far removed from heat source $q$)?

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, 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:

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

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.

 

[roam]

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?

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.

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

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 $\xi>1$.

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?

No. It is the actual temperature.

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?

Thank you very much for the help.

 

[Chestermiller]

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?

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.

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

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.

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?

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.

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?

Yes.

 

[roam]

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:

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

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?

 

[Chestermiller]

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:

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

This equation should read $$Q=-\partial I/\partial z=\alpha I$$