##### Transcript

BRAD NEWBOLD 0:09

Hello, everyone, and welcome to Thermal Properties of Stony Soils: How to Get the Right Answer in a Soil/Rock Mix. Today’s presentation will be about 20 minutes, followed by about 10 minutes of Q&A with our presenter Dr. Gaylon Campbell, whom I’ll introduce in just a moment. But before we start a couple of housekeeping items. First, we want this to be interactive, so we encourage you to submit any and all questions in the Questions pane. And we’ll be keeping track of these for the Q&A session toward the end. Second, if you want us to go back or repeat something you missed, no worries. We’re recording the webinar and we’ll send around a link to that recording and the slides via email within the next three to five business days. Alright, let’s get started. Today we’ll hear from Dr. Gaylon Campbell, who will discuss how to combine the conductivity of rock and soil to get the right thermal conductivity or thermal resistivity value for the soil profile. Dr. Gaylon S. Campbell has been a research scientist and engineer at METER for over 20 years, following nearly 30 years on faculty at Washington State University. His first experience with environmental measurement came in the lab of Sterling Taylor at Utah State University, making water potential measurements to understand plant water status. Dr. Campbell is one of the world’s foremost authorities on physical measurements in the soil plant atmosphere continuum. His book written with Dr. John Norman on environmental biophysics provides a critical foundation for anyone interested in understanding the physics of the natural world. He has written three books, over 100 refereed journal articles and book chapters and has several patents. So without further ado, I’ll hand it over to Dr. Campbell to get us started.

GAYLON CAMPBELL 1:46

Okay, thank you. Today’s seminar will have a little tighter focus than we usually have. Rather than talking about the general problem of measuring or modeling thermal properties of soil, today we’ll show how to determine the thermal conductivity of the soil profile that contains rocks. Now, if you had the responsibility of determining heat transfer, say from a buried electrical cable through a soil profile like the one in this picture, I don’t think you’d find that very challenging. The water content might differ from layer to layer as you go deeper in the soil, but you could measure or model those values and do the calculation for the whole soil profile without too much difficulty. But what if the profile looked like this one? Where and how do you make the measurements of conductivity? The material is extremely heterogeneous with rocks having a high density and a conductivity, and those are mixed with the lower density and lower conductivity soil in between the rocks. The heat will flow through the whole soil profile. But how do you assign a conductivity value for that profile? Now to get an idea of what we’re dealing with, here are some densities and conductivities of a few types of rock. You can see that the densities are reasonably consistent among the different types but the thermal conductivities vary over an enormous range with quartz at 8.8 watts per meter Kelvin and limestone at 1.3. Some of the other typical rocks you might find are more around two or three watts per meter Kelvin. Now the thermal conductivity of rocks can vary pretty widely and if the rocks are porous, the water content can vary and that will change the thermal conductivity of the rock. These are complications that you need to be aware of, you need to look out for. But for the purposes our purposes here, we’ll assume that the conductivity of the rock can be measured and is constant with water content. So all material between the rock typically has a dry density around half that of the rock, and the thermal conductivity varies almost an order of magnitude from dry to wet. Here’s a typical relation jump between thermal conductivity and water content with dry soil having a conductivity less than .2 watts per meter Kelvin, and a saturated thermal conductivity of 1.2 or greater. And so all of these lower than the value for the rocks that we saw on the previous slide, and sometimes a lot lower, since these conductivities are so much smaller than the rock, temperature gradients will mainly be in the soil between the rocks. So we have a soil profile made up of rocks and soil. How do we get the thermal conductivity that applies for that whole profile? Do we take an average of those things? And if we take an average, what sort of average that we take?

GAYLON CAMPBELL 5:58

Now, to illustrate what I’m talking about with that question, here’s an extreme example of that. Assume we have a mixture of air and marble that each takes up half of the total volume. Now, we could arrange these in different ways. On the left we have the air and marble paths in parallel. Now, to get the conductivity of that arrangement, we’d just take weighted arithmetic average of the two. We assume the marble conductivity is 2.5. The air conducts very little of the heat, and so the average conductivity would be about half of the marble or 1.25. But we could also arrange the components in series, like those on the right. When we have a series arrangement then we want to take the harmonic mean. The reciprocal of the conductivity of the mixture is the sum of the reciprocals of the conductivities of the components. And here the air layer determines the amount of heat that can flow, and the conductivity turns out to be about twice the conductivity of air or about .05 watts per meter degree. The temperature gradient is almost all in the air layer. Now with soil that has rocks in it, would be somewhere between those two cases. It isn’t all parallel; it isn’t all series. So how do we model something that’s in between those? In fact, this problem is very similar to the problem of modeling heat flow in soil or any other porous medium. We can treat soil as a mixture of highly conducting spheroids, the mineral part of the soil suspended in poorer conductive fluids like air or water. In fact, this is exactly what what D.A. De Vries did to develop the most successful of the models that we use for describing soil thermal conductivity. His model is based on an earlier model for dielectric constant mixtures. For dry sand, he postulated continuous phase of air in which sterile idle inclusions of the sand particles were suspended. And we’ll use the same model. But instead of sand grains in air, our mixture will be rocks suspended in soil. The soil will form the continuous phase, and the rocks will be the inclusions. In order to do the calculation, we’ll need to know the thermal properties of the soil, the rocks and their volume fractions. Now here is the debris model. You can see in the top equation that the conductivity of the mixture is just a weighted sum of the conductivities of the two components in the mixture, the soil and the rock. The subscripts s and r indicate soil and rock. The k is the thermal conductivity and x is the volume fraction of each of those constituents. The ξ, the Greek letter xi, represents the weighting factor that we use and that’s calculation is shown as the equation on the bottom. Now, g in that equation is called a shape factor. And we assume that the rocks are spheroids with axes a, b, and c, but we’ll assume that A and B are equal to each other. So just deal with a and c, then if the inclusions are spheres, then the shape factors for all three will be .33. If we have elongated stones, then GA becomes smaller, GC bigger, but the three of those have to add up to one.

GAYLON CAMPBELL 10:59

Now we also need to know the volume fractions. But what we’re likely to actually know are the mass fractions, and the densities of the soil so and the rock. And so this equation allows us to convert from mass fraction to volume fraction, ρ soil or ρ rock and ρ soil are the densities of the rock and of the— this is the wet density now of the soil and then the ratio M rock or M soil to M rock is the mass of soil divided by the mass of the rock in the sample. And though that’s the mass ratio, the volume fractions of the soil and rock have to add to one, so if we know the fraction of— the volume fraction of rock we can calculate the volume fraction of soil. Now let’s talk a little bit about how we get the thermal conductivity measurements for the calculation we want to do. As I said earlier, some porous rocks have water content dependent thermal conductivity. And if you’re dealing with rocks like those, you will need to determine that dependence. The rock we have here has very little porosity, so a single value for its thermal conductivity will suffice. The panel on the left shows us drilling a hole in the rock with a rotor hammer. METER’s RK3 probe is ideal for measuring thermal conductivity of rocks and concrete, but it needs a hole to make the measurement. We want the probe to fit in the hole as tightly as possible, since the contact resistance between the probe and that rock is the main source of error in these measurements. The second panel shows the hole that we’ve drilled in the rock. It’s ready now to insert the probe. The central panel shows the RK3 probe along with a tube of thermal grease that we’ll use to minimize contact resistance. The grease is used both to coat the needle and to fill the hole. In the fourth panel, the probe is shown in the rock, ready for the measurement. It’s connected to the TEMPOS handheld reader in the fifth panel, and a reading is shown on the screen. The reading is made by monitoring the temperature of the probe while heat is applied to a heater inside the needle. The point of this series of shots is to show you that it’s pretty easy to find out the thermal conductivity of the rock that’s in that soil profile. Now there’s a minimum size of raw rock that you need to work with for this since rock has a pretty high thermal conductivity. The radius of the material around the rock has to be at least a few centimeters so that the heat pulse will be contained within the sample. The soil thermal conductivity can be measured in the field as I’m showing in the picture on the right. In this picture, our older KD2 Pro was used. The TR3 needle is— or TR1 needle is the one that you would use for making this measurement. It’s 10 centimeters long, you just poke that in to the soil and take a reading. It takes about a minute to get a thermal conductivity measurement with this. Now there are other options. One is to take samples from the field, take those to the laboratory and make the measurements in the laboratory. More typically, the samples are taken to the laboratory, or repacked to the densities specified in the engineering design. And then a dry out function is determined. The relationship between water content and thermal conductivity that I showed in an earlier slide is determined. Then from that water content from that curve and a water content measurement in the field, we can know what the thermal conductivity of the soil is.

GAYLON CAMPBELL 16:05

So to make sure that you understand the things that I’ve covered here, let’s work through a problem. Assume that we have a soil profile like the one I showed you in the second picture. The start of the seminar will separate the soil material into that less than two millimeters diameter and call that the soil, and greater than two millimeters, we’ll call that the rock or the stones. And let’s assume that we did this measurement, and we determined that 60% of the mass of the sample was stones. Now let’s say that the engineering specification calls for a final density of 1.8 Mega grams per cubic meter for our fill. We’ll assume that the stones are granite, we measure their conductivity and we find that it’s three watts per meter Kelvin. The soil is moist, has a conductivity of .5 watts per meter Kelvin. We’ll assume that the stones are elongated. And so our shape factor will use as a ga of 0.1. So here’s the calculation that we do. The volume fraction of rock is calculated a little differently here than we showed in the previous formula, because in this case, we know the final density. And so we do that calculation and determine that the volume fraction of rock is .41, so it was 60% by weight. And it’s 41% by volume now, because the rocks are so dense.

GAYLON CAMPBELL 18:14

We compute the weighting factor, and we come out with the value of .51 for that. And then we compute the thermal conductivity of the mixture of soil and rock. And we come out with with a value of 1.15 watts per meter Kelvin for that mixture. Now, at this point, we’d like to know whether that is a reasonable number. And so let’s do a couple of other calculations, just to see how that comes out. There isn’t any way to make a direct measurement to determine whether we’ve gotten the right number, at least not without an enormous budget and with an awful lot of work. But we can do a couple of calculations to see the number’s reasonable first. The rock has a conductivity of 3 and soil has a conductivity of .5, then so if we assume that those two are in parallel, we would with the volume fractions that we have there, we get from that top calculation of conductivity of the mix of 1.53. Now if we had the soil and the rock in series, we could do that calculation too that’s shown on the bottom and we’d get a value of .76. We’d expect the parallel one to be higher than the actual soil, the series one to be lower. And that’s exactly how they come out. And so we can conclude that that value is reasonable one, that 1.15 that we got. Now, is there any advantage to having a soil profile that’s dominated by rock, the way our calculation shows here? Well, of course there is. When we bury a cable, and it heats up, the heat that it gives off will drive water away from the cable. And so the tendency is for the soil to dry out around the cable. If we had just soil around the cable, and it dried out, we saw in that early graph, the thermal conductivity would go down to something like .2 watts per meter Kelvin. What would it go to in this mixture that we have here. We can do that calculation by setting the thermal conductivity of the soil material to .2. And if we do that, we come out with a conductivity of the mixture of .74. So we can see that having the rock in the mixture maintains a much higher thermal conductivity than we would get just with the soil itself. So let me conclude with these thoughts. It isn’t possible to directly measure the thermal conductivity of a soil unless you have a large budget, willing to do a lot of work. Probes can only measure the thermal conductivity of homogeneous media. So the only practical way forward is to model it. Now fortunately, a good model is available that just requires measurements of thermal properties and volume fractions of the constituents. By making that calculation, we arrive at a composite thermal conductivity that is likely at least as reliable as any direct measurement we might have made with a lot less work and expense. Thank you.

BRAD NEWBOLD 22:32

All right. Thank you, Dr. Campbell. So we’d like to take the next 10 minutes or so to take some questions from the audience. Thanks again for all those who have submitted questions, there’s plenty of time to submit your questions. I did want to mention that if we do not get your questions here during this live session, we will have them recorded and Dr. Campbell or somebody else from our METER Environment team will be able to get back to you via email to answer your question directly. And so we’ll try to get to as many as we can before we finish. There was one question that came up during the presentation asking about converting units to non metric. And we do have that at least in METER’s TEMPOS system, there is the the ability to measure in both metric and non metric units. So there’s BTU hours— was it per feet per, sorry— anyway, but we do have that ability for those of you, especially working in the United States. Okay, so let’s see. Here’s another question. Dr. Campbell, do I need to measure the thermal conductivity of the rocks as they are in the soil, or can I just find out what they are, find out their values from the internet and just use those values?

GAYLON CAMPBELL 24:05

Well, certainly taking into account the fact that there is both soil and rock will give you a lot better number than if you don’t take that into account. And so just if you have a good geologist identify the minerals that you’re working with that will go a ways, but it’s always the most reliable thing to do is to make a measurement. The measurement is an easy one to do. And then you can know what what you’re dealing with thermal conductivity.

BRAD NEWBOLD 24:45

Do we have any issues or have you noticed any issues in working with stones that have a varied lithology?

GAYLON CAMPBELL 24:58

Well that, yeah, if the question is, if I’ve had experience with that, I have.

BRAD NEWBOLD 25:07

What if, in general.

GAYLON CAMPBELL 25:09

Yeah, but if, I think, again, that would come down to having some measurements. So you know how much variation exists if you had a lithology that where you have had big differences in thermal conductivity spatially, then you would have to do the calculations to take that into account. But I think an approach like the one we’re talking about here would still serve you well to do that, that this, the mixture can contain more than just two constituents the way I showed it here. The approach will work with numerous, with as many constituents as you want.

BRAD NEWBOLD 26:06

Okay, this next question is asking in early on, you described the weighting factor. And they just wanted to know just a little bit more about what is the physical significance of this weighting factor?

GAYLON CAMPBELL 26:23

Well, if you think about the example that I gave between parallel and a series arrangement of the conducting elements, you can think of those as end members. And this weighting factor is something that tells you where you are between those two end members. And I mean, we’d have to go deeper than I can go just in our session here today to show mathematically how that works. But if you think of it as those the series in parallel being the end members and you’re just looking for something between those two extremes. That’s where the weighting factor comes from.

BRAD NEWBOLD 27:25

Okay, so maybe Dr. Campbell will be able to get back to you via email and, and lay out the math behind that. Our next question here, What if you’re dealing with a mixture of rocks of different types within your soul column?

GAYLON CAMPBELL 27:43

And then you need to expand this equation to where it includes those additional factors. So instead of just soil and rock, you’d have soil and some different kinds of rock, but the approach would still be the same that you would need to get the volume fractions of each of the constituents and the thermal conductivities of each of the constituents and then put them together in a weighted average the way we’ve shown here.

BRAD NEWBOLD 28:21

Okay. Let’s see. What about what if the water content of the soil changes?

GAYLON CAMPBELL 28:32

Now that’s assumed through all of this, that you would need to redo the calculation as those changes occur. So you’d need to monitor the water content if you know the drought curve for that soil or you would need to monitor the thermal conductivity of the soil and then as the water content changes, you would need to redo the calculation to show the change or the new thermal conductivity.

BRAD NEWBOLD 29:05

Okay, and then along those same lines, what if, with water content, what if your rock is porous? How do you deal with that kind of change with water content?

GAYLON CAMPBELL 29:15

Then you would need to characterize the rock too and and know how its thermal conductivity changes with water content. And then the calculation itself becomes kind of complicated because the equilibrium between the water in the rock and the water in the soil depends on the water potential in the two and so the water content of each fraction will depend on the moisture release curve of each of the components and the water potential of the components. So at that point the calculation gets quite a bit more complicated. But it can be done.

BRAD NEWBOLD 30:11

All right. We’ll take one or two more questions here. This next one, so with when drilling the hole for the probe and the application of the grease, does the grease affect that measured conductivity of the rock to any degree?

GAYLON CAMPBELL 30:31

Well, that’s always a challenge to make these measurements, the contact resistance. What we choose is grease that has a thermal conductivity, say, a thermal conductivity that’s high compared to the rock. And then we try to make the hole fit as tightly as we can. And one way, if we get those things right, then that grease assures us of a good thermal conductivity measurement. But it’s, what should I say, it’s an art. You have to pay close attention to have that, I mean, the measurement without the grease would be way, way worse than it is with, but I wouldn’t stick my neck out so far as to say that it doesn’t have any effect. It certainly minimizes the errors.

BRAD NEWBOLD 31:42

So you do want to drill a nice clean hole that has some good contact with the probe. And so at least that you’re using the minimal amount of grease that you would likely run to. Okay. Finally, this is our last question here. We discussed dealing with soil that has different types of rocks in it, what if they’re dealing with rocks whose diameter is too small to work with the probe?

GAYLON CAMPBELL 32:13

I’ve puzzled over that a little bit. And I think the approach I would take is to try to identify the mineralogy of the rocks that we’re dealing with and find some large enough rocks that we can make a reliable measurement in them, that have the same mineralogy as the smaller rocks that are in the soil. So that would require you to be a good enough geologist to to identify the mineralogy of the rocks that you’re dealing with. And then make your assignments according to the rocks, the smaller rocks that you find that are similar to the large ones that you measure. I can’t think of any other way to do that measurement.

BRAD NEWBOLD 33:06

Okay. All right. Thank you. That’s gonna wrap it up for us today. Thanks again for joining us. And we hope you enjoyed this discussion as much as we did. Again, thank you for the questions that have been submitted. There are a few that we did not get to, again, Dr. Campbell, or somebody else from our METER Environment team will be able to get back to you and answer your question directly via email. Thanks again for all your great questions. And please answer the survey. There’s a short survey that will appear after the end of this webinar, just to tell us what kinds of webinars you’d like to see in the future. And for more information on what you’ve seen today, visit us at metergroup.com. Finally, look for a link to the recording of this webinar as well as the slides in your email. And stay tuned for future METER webinars. Stay safe and have a great day.