Volume and temperature rise in tissue

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  • #1
roam
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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:

7WGGsfm.png


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? :confused:

Any suggestions would be greatly appreciated.
 

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

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

Yes, my samples are much thicker than the laser penetration depth. In fact, the wavelength of my laser only causes superficial burns.
 
  • #4
A.T.
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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.
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
roam
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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
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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
roam
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Where in the sample is the temperature being measured? The temperature is going to be varying with spatial position and time within the sample.

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
mjc123
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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).
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
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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
roam
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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
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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?
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.

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

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

Semi-infinite solid (linear flow): with prescribed initial and zero surface temperature, 58–62, 273, 357; with prescribed surface temperature, 62–64, 305, 357; with harmonic surface temperature, 64–67, 317–19; with periodic surface temperature, 68–70, 401; with radiation at the surface, 70–74, 305, 358–9 ; application to determination of thermal conductivity, 77 ; with prescribed flux at the surface, 75–77; with heat generated within it, 78-80, 307-8; with the surface in contact with well-stirred fluid or perfect conductor, 306–7 ; semi-infinite composite solid, 319–23, 326.

Semi-infinite solid (general case): with arbitrary initial temperature and the surface at zero, 276, 370; 'heated over portion of its surface, 214–17, 263-5, 461; with radiation at the surface, 371; with surface diffusion, 374–5; with pre-scribed temperature over its surface, 166, 370.

Which one of these cases should I use? :confused:

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.

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
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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? :confused:
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?


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

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.

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.

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
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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?

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

Can you please draw a schematic diagram of the sample and the laser hitting it, showing approximate dimensions of the sample?
I'm not sure how you derived this but it doesn't show the inverse relationship between ##V## and ##\Delta T##.
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
LURCH
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This may be an overly obvious question, but do these plant stems change color as they mature?
 
  • #17
roam
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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:

3c8H6Ez.png


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.

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?

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

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

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  • #18
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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:

View attachment 230659
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)?
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?
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
roam
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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)?

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.

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.

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.

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.

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
roam
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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:

Temperature at ##x## at time ##t## due to a source at ##x'## at time ##\tau##:

$$\frac{1}{2\sqrt{\pi\kappa\left(t-\tau\right)}}\left[\exp\left(-\frac{\left(x-x'\right)^{2}}{4\kappa\left(t-\tau\right)}\right)-\exp\left(-\frac{\left(x+x'\right)^{2}}{4\kappa\left(t-\tau\right)}\right)\right]$$

Integrating to give the temperature at ##x## at time ##t##:

$$\frac{1}{2\sqrt{\pi\kappa t}}\intop_{0}^{\infty}f(x')\left[\exp\left(-\frac{\left(x-x'\right)^{2}}{4\kappa\left(t-\tau\right)}\right)-\exp\left(-\frac{\left(x+x'\right)^{2}}{4\kappa\left(t-\tau\right)}\right)\right]dx'+\frac{x}{2\sqrt{\pi\kappa}}\intop_{0}^{t}\phi(\tau)\frac{\exp\left(-\frac{x^{2}}{4\kappa(t-\tau)}\right)}{\left(t-\tau\right)^{3/2}}$$

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? :confused:
 
  • #21
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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.
It's just an approximation for the distance beyond which the temperature is undisturbed from its initial value.
Another question: is not more appropriate to use cylindrical geometries for (roughly) modeling the stem?
In my judgment, yes. That's what I would use.
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 last term is the rate of heat generation per unit volume. If essentially all the heat is released at the surface, this translates into a heat flux at the surface, with no heat released below the surface.
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? :confused:
They must be solving for the steady state temperature variation, obtained after a long time. I don't have a copy of c&j, so I don't know what's on p. 223. But, in any case, for your situations, you should be first looking at the 1D case, where the heated area is the entire cross sectional area of the stem. This will give you a good first-order picture of what is happening. After that, you can start to consider cases where the heated area is less than the area of the stem (and the sides of the stem are insulated).

There are other things that can be done and approximations that can be used, but it is hard to tell you how to proceed because of your very limited experience with heat transfer. Like I said, start by doing calculations for the 1D transient case with a constant heat flux at the surface and see what that predicts. For the part after the heating time, just add a negative flux to the solution at that time, so that the total flux is zero beyond that time; basically time superposition of the solutions.
 
  • #22
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Please look up in C&J the solution to the 1D transient heat conduction equation for a semi-infinite medium with constant heat flux at x = 0. What do they give for the temperature at the surface as a function of time?

NEVER MIND. The solution to this problem is given in Transport Phenomena by Bird, Stewart, and Lightfoot, Chapter 12, Problem 12B.3. For the surface temperature rise, they give:

$$\Delta T=\frac{q}{k}\sqrt{\frac{4\alpha t}{\pi}}=2q\sqrt{\frac{t}{\pi k \rho C}}$$where k is the thermal conductivity, ##\rho## is the density, C is the heat capacity, and q is the (constant) surface heat flux.

How does this compare with your observed temperature rise?
 
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  • #23
roam
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Hi Chestermiller,

Please look up in C&J the solution to the 1D transient heat conduction equation for a semi-infinite medium with constant heat flux at x = 0. What do they give for the temperature at the surface as a function of time?

NEVER MIND. The solution to this problem is given in Transport Phenomena by Bird, Stewart, and Lightfoot, Chapter 12, Problem 12B.3. For the surface temperature rise, they give:

$$\Delta T=\frac{q}{k}\sqrt{\frac{4\alpha t}{\pi}}=2q\sqrt{\frac{t}{\pi k \rho C}}$$where k is the thermal conductivity, ##\rho## is the density, C is the heat capacity, and q is the (constant) surface heat flux.

How does this compare with your observed temperature rise?

Is there any way that we can express this equation so that it takes into account the overall size of the object?

In my last post, I quoted C&J's solution to the 1D semi-infinite medium heated over the whole cross-sectional area. I am not 100% sure if it's the right solution I should be looking at (it's different from the solution you quoted from "Transport Phenomena") but it does depend on the overall volume of the object.

Similarly, for the 3D case, they give a solution which involves a triple integration over the volume (here is a photo of the relevant page in my C&J textbook). Do you believe this would be enough to demonstrate that there is a relationship between the temperature and the overall volume? And do you think this solution is applicable to my experiment?

Another question: when they say "at z=0 there is radiation into medium" is that similar to the medium being irradiated by a laser beam?

P. S. Here is a photo of the page giving a solution for a semi-infinite cylinder. Yes, they say these are all for the steady-state.
 
  • #24
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Hi Chestermiller,



Is there any way that we can express this equation so that it takes into account the overall size of the object?

In my last post, I quoted C&J's solution to the 1D semi-infinite medium heated over the whole cross-sectional area. I am not 100% sure if it's the right solution I should be looking at (it's different from the solution you quoted from "Transport Phenomena") but it does depend on the overall volume of the object.

Similarly, for the 3D case, they give a solution which involves a triple integration over the volume (here is a photo of the relevant page in my C&J textbook). Do you believe this would be enough to demonstrate that there is a relationship between the temperature and the overall volume? And do you think this solution is applicable to my experiment?

Another question: when they say "at z=0 there is radiation into medium" is that similar to the medium being irradiated by a laser beam?

P. S. Here is a photo of the page giving a solution for a semi-infinite cylinder. Yes, they say these are all for the steady-state.
Hi Roam,

I looked over your photos of the C&J pages, and they don't seem relevant to your situation. So let's start over.

In doing modeling, the most important and basic principle is to START SIMPLE. There are numerous reasons for doing this. First, if you can't solve the simple problem, you will never be able to solve the more complicated version. Second, getting a solution to a simpler version of the problem gives you a quick start, and allows you to have an answer under your belt in short order. Third, the answer to the simpler version provides you with a basis of comparison for the more complicated solution to the problem.

So we are going to start out by first looking at solutions to the 1D problem, which assumes that the area of the laser beam is equal to the area of the stem. Then we can start looking at more complicated 2D version in which the area of the laser beam is less than the area of the stem.

For the 1D problem, we have already looked at the case where the stem is infinitely long, and obtained the answer for the maximum temperature rise given in post #22. This will also be the solution for the finite length stem at short times (where the heat has not had enough time to penetrate to the far end of the stem). Therefore, it provides us with a basis of comparison for the results that we will obtain for the finite length stem.

Next, let's solve for the 1D problem with a finite length stem. Let ##\theta (t,z)## be the temperature rise at time t as a function of z, and let q be the heat flux at z = 0 (for all times). The differential equation and boundary conditions for this situation are:
$$\frac{\partial \theta}{\partial t}=\alpha\frac{\partial^2 \theta}{\partial z^2}\tag{1}$$
##-k\frac{\partial \theta}{\partial z}=q\tag{2}## at z = 0

##-k\frac{\partial \theta}{\partial z}=0\tag{3}## at z = L

where k is the thermal conductivity, ##\alpha=\frac{k}{\rho C}## is the thermal diffusivity, and L is the length of the stem. The initial condition is ##\theta = 0## for all z.

For any time t, we can get the equation for the average temperature rise of the sample (over the length of the sample) by integrating the differential equation from z = 0 to z = L to obtain: $$\rho CL \frac{d\bar{\theta} }{dt}=q\tag{4}$$where $$\bar{\theta}=\frac{1}{L}\int_0^L{\theta(t,z) dz}\tag{5}$$The solution to this differential equation subject to the initial condition is: $$\bar{\theta}(t)=\frac{q}{\rho C L}t\tag{6}$$
At very long times, the solution to the differential equation and boundary conditions will approach ##\theta_{\infty}(t,z)##, given by $$\theta_{\infty}(t,z)=\bar{\theta}(t)+f(z)\tag{7}$$The next step in our solution is to determine the function f(z). If we substitute Eqns. 6 and 7 into our differential equation, we obtain:
$$\frac{q}{kL}=\frac{d^2f}{dz^2}$$Integrating this from z = 0 to arbitrary z (subjection of Eqn. 2) yields:
$$-\frac{q}{k}\left(1-\frac{z}{L}\right)=\frac{df}{dz}\tag{8}$$Note that this relationship also satisfies the boundary condition represented by Eqn. 3. If we integrate this again between z = 0 and arbitrary z, we obtain:
$$f(z)=f(0)-\frac{q}{k}z+\frac{q}{2kL}z^2\tag{9}$$
The constant of integration f(0) must be such that the average of the function f(z) over the interval from z = 0 to z = L must be equal to zero (since the average temperature is already ##\bar{\theta}##. This requires that ##\int_0^L{f(z)}dz=0\tag{10}##
From this, it follows that $$f(0)=\frac{qL}{3k}\tag{11}$$ and $$f(z)=\frac{qL}{3k}\left[1-3\frac{z}{L}+\frac{3}{2}\left(\frac{z}{L}\right)^2\right]\tag{12}$$So the asymptotic solution to this problem at long times is given by:
$$\theta_{\infty}(t,z)=\frac{q}{\rho C L}t+\frac{qL}{3k}\left[1-3\frac{z}{L}+\frac{3}{2}\left(\frac{z}{L}\right)^2\right]\tag{13}$$
And the maximum temperature rise at the boundary is given by ##\frac{q}{\rho C L}t+\frac{qL}{3k}##. Note that, as expected, these results depend on the length L (i.e., the volume of the sample V = AL).

So now, for the 1 D problem, we have obtained the solution for the temperature rise at very short times and at very long times.

I think I'll stop here an give you a chance to digest this. How do your experimental results compare to these results at short times and long times?
 
  • #25
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Before leaving the asymptotic 1D solutions at short- and long times, I'd like to add a few comments.

Our solutions for these cases can be reduced to dimensionless form by introducing the following dimensionless variables:

Dimensionless temperature rise = ##\theta^*=\frac{\theta k}{qL}##

Dimensionless time = Fourier Number = ##\xi=\frac{\alpha t}{L^2}##

In terms of these parameters, the dimensionless maximum temperature rise at short times and long times is given by:

$$\theta^*=\frac{2}{\sqrt{\pi}}\sqrt{\xi}\tag{short times}$$
$$\theta^*=\frac{1}{3}+\xi\tag{long times}$$

Next, let's see how large the dimensionless time parameter ##\xi## can get in your experiments. The thermal diffusivity of the stem is roughly 0.15 cm^2/sec, the heating time is about 10 sec., and the length of the sample is roughly 5 cm. So, for these values, the highest that the ##\xi## parameter reaches in the experiments is $$\xi =\frac{(0.15)(10)}{5^2}=0.06$$
DIMENSIONLESS TEMPERATURE.png


The short time asymptote looks very much like the experimental data plot you presented earlier.
 

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  • #26
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The difference in basic character between the calculated temperature histories from the model and the experimental data in post #5 (for the region above 80 C) strongly suggests that an important physical mechanism has been omitted from your model. Can you guess what that physical mechanism might be?
 
  • #27
roam
1,271
12
This is very helpful. Thank you so much for your time, I really appreciate that.

The difference in basic character between the calculated temperature histories from the model and the experimental data in post #5 (for the region above 80 C) strongly suggests that an important physical mechanism has been omitted from your model. Can you guess what that physical mechanism might be?

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
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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.
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.
I am still working through the equations. Is your equation 13 also given in the "Transport Phenomena" textbook?
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.
 
  • #29
roam
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12
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?

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
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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?
The equation in post #22 is just what this equation predicts at z = 0.
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 actual temperature.
2. ##f## is the initial distribution of temperature at ##z##, i.e., ##\theta\left(0, z\right)=f(z)##—is this correct?
No. It is the spatial part of the asymptotic temperature variation at long times.
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.
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)
 
  • #31
roam
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12
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:

$$\gamma=\underbrace{\alpha}_{\text{absorption coefficient}}+\underbrace{\beta}_{\text{scattering coefficient}}$$
 
  • #32
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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.
 
  • #33
roam
1,271
12
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?

Chestermiller said:
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.
 
  • #34
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5,205
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.
 
  • #35
roam
1,271
12
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?
 

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