# I Volume and temperature rise in tissue

1. Sep 11, 2018

### roam

I am looking for an explicit equation that shows that a sample with lesser volume will reach higher temperatures when irradiated than a sample with a larger volume.

The samples in my project are biological tissues which are being irradiated by a laser. In my experiment, using a thermal camera I found that smaller samples (less volume) reached higher temperatures. Since water is the most abundant constituent, as a first-order approximation we could model the tissue as a volume of water. Here is a diagram I made:

If the sample has an initial temperature $T_{i}$ and final temperature $T_{f}$ and the temperature rise is $\Delta T$, we can write:

$$T_{f}=T_{i}+\Delta T. \tag{1}$$

Therefore we need to show that $\Delta T$ decreases if the volume increases.

1. The only thing that my thermodynamics textbook states is that the heat capacity will be larger the more of a substance you have. But I really need the relationship between (solid) volume and temperature rise in the form of an explicit formula.

2. I have found an equation (source) that gives the value for $\Delta T$ for an irradiated sample:

$$\underbrace{E}_{\text{deposited energy}}=\underbrace{m}_{\text{mass}}\overbrace{c}^{\text{heat capacity}}\Delta T=\left(\underbrace{\rho}_{\text{density}}\overbrace{L}^{\text{height}}\underbrace{A}_{\text{beam area}}\right)c\Delta T. \tag{2}$$

However, they say that this is when the beam is absorbed in a distance of $L$ beneath the surface in a piece of material of mass $m$. Therefore the volume that appears in the equation above ($LA$) is the volume in which the laser power is absorbed, not the overall volume of the material. I don't know how the overall volume comes into it...

3. The one-dimensional general heat conduction equation which gives the change in temperature over time at point a specific point in the tissue:

$$\frac{\partial T\left(z,\ t\right)}{\partial t}=K\ \frac{\partial^{2}T\left(z,\ t\right)}{\partial z^{2}}+\frac{Q\left(z,\ t\right)}{\rho c}. \tag{3}$$

This equation takes into account conduction (typically away from the irradiated spot) since the first term on the RHS describes heat conduction. The diffusivity is given by $K=k/\rho c$. And $Q$ is the heat source term describing the rate at which heat is generated (this depends on the intensity of the laser and the absorptance of the material).

But how does the overall volume of the sample factor into this?

Any suggestions would be greatly appreciated.

2. Sep 11, 2018

### A.T.

Note that this will give the average change in temperature in the absorbing volume. The surface temperature that the camera mainly pics up might be higher.

Are your samples thicker than the laser penetration depth?

3. Sep 11, 2018

### roam

Yes, my samples are much thicker than the laser penetration depth. In fact, the wavelength of my laser only causes superficial burns.

4. Sep 11, 2018

### A.T.

When do you measure the temperature? At equilibrium, when the temperature doesn't rise anymore? The bigger sample has a greater surface area to give the heat away, so it will reach equilibrium at a lower temperature.

5. Sep 11, 2018

### roam

I irradiated the samples for 10 seconds and recorded the entire process. Then I extracted some temperature curves from the thermal camera that look like this picture.

Smaller samples reached a higher maximum temperature. So I believe the reason is that more extended samples (having more volume) would heat up slower. Is it possible to show that this hypothesis is true mathematically using some of the thermodynamic laws?

As I said in my first post the textbook says the heat capacity will be larger as the material gets larger. But they don't offer any proofs, and I am not sure how to relate it to equation (1).

6. Sep 11, 2018

### Staff: Mentor

Where in the sample is the temperature being measured? The temperature is going to be varying with spatial position and time within the sample.

7. Sep 11, 2018

### roam

The curves show the change in temperature of the hottest point in the frame of the thermal camera. This is really the point where the laser hits the sample.

8. Sep 11, 2018

### mjc123

Be careful that you understand the definition of heat capacity that is being used. "Heat capacity" of a sample means the amount of heat required to raise its temperature by 1K. It has units of energy/temperature, e.g. J/K. "Specific heat capacity" of a substance is the heat capacity per unit mass, and has units like J/g/K. The heat capacity of a sample (of uniform composition) is the mass times the specific heat capacity. Heat capacity is often denoted by a capital C, and specific heat capacity by a small c, but not always. And specific heat capacity is sometimes referred to as "heat capacity" for short (in your equation 2, for example). Heat capacity (properly speaking) increases with the size of the sample - it is an extensive property - whereas specific heat capacity does not (it is an intensive property). (There is also molar heat capacity, but you probably don't need that here.)
Now what do you use for this problem? If your sample were absorbing heat uniformly, it would be simple. But the sample is thicker than the laser penetration depth, and the laser spot has a limited area, so that is not the case. Heat is absorbed at the front face, and raises its temperature. Heat then diffuses into the bulk sample by conduction (and maybe convection, if it is water).You measure the temperature of the front face with a thermal camera - but the temperature is not uniform through the sample. Basically, you have to solve equation 3. The "overall volume of the sample" doesn't factor simply into this, because T varies over the sample volume. You have to integrate over the sample volume.

9. Sep 11, 2018

### Staff: Mentor

And the energy released by the laser is distributed over the depth over which the laser flux is attenuated? So the release of the energy, as you have shown in your transient equation is a function of depth. You would have to solve the transient heat conduction equation to determine the temperature distribution. Can you approximate the behavior by assuming that all the heat is released at the surface? If so, then there is probably a solution in Carslaw and Jaeger, Conduction of Heat in Solids that addresses this.

10. Sep 11, 2018

### roam

Hi Chestermiller,

The laser is infrared so it doesn't burn deep into the tissue (i.e. not a volumetric absorption). The absorption coefficient is much larger than the scattering coefficient. So we can assume that all the heat is released at the surface because that is where all the light absorption occurs.

I will look into Carslaw & Jaeger (I don't have it at hand right now). Do they give a solution to Eq. (3) such that it takes into account the overall volume of the object?

Also, I wrote the equation in 1-D for simplicity but it is a function of coordinate not (just depth):

$$\frac{\partial T\left(x,\ y,\ z,\ t\right)}{\partial t}=K\nabla^{2}T\left(x,\ y,\ z,\ t\right)+\frac{Q\left(x,\ y,\ z,\ t\right)}{\rho c}.$$

Hi mjc123,

Thanks for the explanation. So do I need to solve Eq. 3 for $T$ and then integrate the expression over the volume of the object?

Would I end up with an expression that shows the dependency of temperature rise (or final temperature) on the overall volume of the object?

11. Sep 11, 2018

### Staff: Mentor

The equation will change, such that the generation term is removed and replaced by a constant heat flux boundary condition. C & J will have the solution for a semi-infinite medium, which might be all that you need. During the time that the laser is on, the penetration of the solution might be small, so that this might be a good approximation.

This could, of course, be solved numerically. If the laser is on a short time, you can integrate to get the total heat added, and then determine the final temperature (after re-equilibration) by recognizing that, in the end, this is distributed uniformly over the volume.
If the laser is on only a short time, the maximum temperatures calculated and measured during the time the laser is on will not depend on the overall sample volume if the sample is thick.

12. Sep 13, 2018

### roam

Hi Chestermiller,

I found a copy of the textbook by Carslaw & Jaeger. But I'm a bit swamped by all the info in the book. Can I ask: why do you use a semi-infinite medium rather than a finite medium (which is what the samples are)?

Also, here is the book's subject index on semi-infinite medium:

Which one of these cases should I use?

The laser is on for 10 seconds. I think this is long enough for the heat transfer processes and the overall volume to be non-negligible.

I need the simplest possible analytical expression that shows a relationship between overall volume and temperature rise.

13. Sep 13, 2018

### Staff: Mentor

You would use the appropriate reference for the general case. Can you give us more information on the geometry of your sample and the area impinged upon by the laser?

This thinking is based on actual calculations and estimations? As I said, it would help to know the geometry of the sample and the laser area.
The simplest is that the temperature rise averaged over the volume of the sample is equal to the rate of energy flux in the laser divided by the product of mass times heat capacity of the sample.

14. Sep 13, 2018

### roam

My samples are stems of plants in different growth stages. The larger plants (stems with more volume) definitely do not get as hot.

Foliar tissue and dermal area of the plant are on average ~67% water. The pith in the middle is also predominantly water. That's why I thought it might be simpler to model the sample as an equivalent volume of water.

I think the difference in overall volume is a possible explanation for the discrepancy. I just need an equation that explicitly shows the relationship.

So is it the following?

$$\frac{\Delta T}{V}=\frac{E}{m.C}.$$

I'm not sure how you derived this but it doesn't show the inverse relationship between $V$ and $\Delta T$.

15. Sep 13, 2018

### Staff: Mentor

Can you please draw a schematic diagram of the sample and the laser hitting it, showing approximate dimensions of the sample?
If you do the math right, $$V\Delta T=\frac{E}{\rho C}$$where $\rho$ is the density and C is the specific heat capacity.

16. Sep 14, 2018

### LURCH

This may be an overly obvious question, but do these plant stems change color as they mature?

17. Sep 14, 2018

### roam

Hi Chestermiller,

I haven't measured them but I would say the plant stems are about 0.5 mm to 1 cm in diameter (more mature plants are larger). The laser beam is ~3 mm. I made a diagram of the geometry of the setup:

Could you please explain where the equation comes from? And do you think this is as relevant to the problem as the equations in Carslaw & Jaeger?

Not in this case. I've looked at their spectrophotometric curves and the absorption characteristics are the same at the wavelength of the laser.

18. Sep 14, 2018

### Staff: Mentor

1. Are you saying that, in some cases, the diameter of the laser beam is greater than the stem diameter?
2. What is the length of the sample (since you are assuming the sample has finite volume)?
3. Over what depth would you say the laser is attenuated (or what is the absorption coefficient)?
This equation is the same as your Eqn. 2 in post #1. It would not be relevant to the semi-infinite case in C&J.

Here is a rough calculation. The thermal diffusivity of water is about 0.15 cm^2/sec near room temperature. Assuming that the diameter of the laser beam exactly matches the diameter of the stem, the heat transfer will be one-dimensional (also assuming no heat escapes from the sides of the stem during the heating period). Also assume that all the energy from the laser is released at the surface. Assuming a semi-infinite medium, the depth of penetration of the temperature profile into the sample (predicted by the transient heat conduction equation) will be on the order of about $$\delta=4\sqrt{\pi \alpha t}$$For a 10 second exposure time, the would work out to $$\delta=4\sqrt{(3.14)(0.15)(10)}=9\ cm$$If the sample length is more than this, the sample can be treated as semi-infinite (during the heating time). If the sample length is less than this, the finite length of the sample must be taken into consideration.

19. Sep 15, 2018 at 10:21 AM

### roam

1. Sorry, there was a typo in my post. I meant to say ~0.5–1 cm, so the diameter of the beam is always less than the stem diameter. Older plants are thicker so more of the sample resides outside of the beam.

2. The average length would be ~5 cm. Plants in early growth stages are slightly shorter.

3. Nearly all of the infrared light gets absorbed in the surface (epidermis of the plant). There is hardly any penetration into the tissue. It doesn't burn a deep hole like visible lasers do (i.e. not a volumetric absorption). The heat from the surface gets transferred to other areas only through conduction. I haven't calculated the absorption coefficient, but the absorptance is in excess of 95% according to the spectrophotometric measurements.

But the volume in Eq. 2 is the volume in which the light gets absorbed, not the overall volume of the specimen. This equation only shows that a wavelength with more absorption gets absorbed in a smaller volume thus creating more heat. It doesn't take into account the heat that leaves the area through conduction. I am trying to show that $\Delta T$ would be different when we are using the same wavelength but different sample sizes.

Thanks a lot for that.

My samples are less than 9 cm (not taking into account the root system which is the hidden half of the plant). Would it be simpler to treat this as a finite or a semi-infinite medium?

20. Sep 19, 2018 at 7:05 AM

### roam

Hi Chestermiller,

Could you please explain where you found the quantity $\delta$? The term $1/\sqrt{\pi\kappa t}$ does appear in some of the solutions, but they never explain its physical significance (C&J use $\kappa$ rather than $\alpha$ for diffusivity).

It seems that there is not always a factor of 4 in front of the expression. For instance, for the one-dimensional case that you've suggested, Carslaw and Jaeger give:

Another question: is not more appropriate to use cylindrical geometries for (roughly) modeling the stem?

C&J solve the heat transfer equation in cylindrical coordinates:

$$\frac{\partial T}{\partial t}=K\left[\frac{\partial^{2}T}{\partial r^{2}}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{\partial^{2}T}{\partial z^{2}}\right]+\frac{Q}{\rho c}$$

Although for some reason, the last term (heat source term) is absent from their analysis. The solutions that they give (e.g. for a semi-infinite cylinder, p. 223) have no time-dependence. Is it not possible to find time-dependent solutions like the one I quoted above?