4.4 Soil Thermal Properties .1 Heat Capacity
4.4.2 Thermal Conductivity
The conduction of heat within a material is due to the interactions of energy carriers.
The energy carriers can be different depending on the material. For instance, the energy carriers in metals are electrons; in gases, they are gas molecules; in a liquid fluid like water they are the molecules of the liquid. Thermal conductivity is therefore proportional to the number and volume density of energy carriers. The thermal conductivity of pure metals is very high because the electron density is high. On the other hand, gas thermal conductivity is small because the number density of gas molecules in a gas is small (Nellis and Klein, 2009).
Soil is a three-phase system composed of solids, liquids and gases. Its thermal con- ductivity is therefore determined by the contribution of the different phases, which have different material properties and therefore energy carriers. Since the proportion of these components varies with space and time, the thermal conductivity of soil depends on properties that vary with space and time, such as its bulk density, water content, quartz content and organic matter content. At low water content, the air space controls the thermal conductivity. At high water content, the thermal conductivity of the solid phase becomes more important. The large differences in thermal conductivity among different materials are the result of differences in bulk density and composition. The transition from low to high conductivity occurs at low water content in sands and at high water content in soils high in clay.
Thermal conductivity model
Here we present a model to compute thermal conductivity from bulk density, water content, clay content and temperature. The model was originally published by Campbell et al.(1994) and it is a modification of the original de Vries (1963) model. The model is based on the assumption that the thermal conductivity of soil is given by a weighted sum of the thermal conductivities of its components:
λ= kwθλw+kgφgλg+ksφsλs
kwθ+kgφg+ksφs
(4.30) where kg, ks and kw are weighting factors for gas, solid and water. λ is the thermal conductivity andλg,λs andλware thermal conductivities of gas, solid and water. The volumetric fractions of gas, solid and water areφg,φsandθ. The thermal conductivity of the gas phase accounts for both the conductivity of airλaand the latent heat component given by evaporation and condensation. In most soil models, thermal conductivity is
Soil Thermal Properties 67 considered to be independent of temperature. However, Campbellet al.(1994) showed thatλis temperature-dependent, with the dependence being mostly due to the variation of the latent heat transport with temperature. With respect to the de Vries (1963) model, the model of Campbellet al.(1994) provides a more rational treatment of the transition from high to low vapour contributions as the soil dries. Indeed, the de Vries (1963) model accounted for the thermal conductivity of continuous phases: when the soil is saturated, the continuous phase is the liquid water, whereas when the soil is dry, the continuous phase is gas. However, the de Vries (1963) model did not account for the intermediate conditions of partial saturation occurring in soils. This was introduced by Campbellet al.(1994) by defining a continuous function through a ‘fluid’ thermal con- ductivity variable that can be used over the whole range of water contents. The authors defined a variable called ‘fluid’ thermal conductivity:
λf =λg+fw(λw–λg) (4.31) wherefwis an empirical parameter computed by
fw= 1 1 +
θ θ0
–q (4.32)
ranging from 0 in dry soil to 1 in saturated soil. The parametersθ0 andqare specific soil properties that determine when water content begins to affect thermal conductivity and the rate of the transition from air- to water-dominated conductivity. The parameters qandθ0 were found to be highly correlated to clay content and the following regression equations were presented:
q= 7.25my+ 2.52 (4.33)
θ0 = 0.33my+ 0.078 (4.34)
wheremyis the fractional clay content [0–1]. Using eqn (4.31), the weighting factors are then computed:
kg = 1 3
⎡
⎢⎢
⎣ 2 1 +
λg
λf
– 1
ga
+ 1
1 + λg
λf
– 1
gc
⎤
⎥⎥
⎦ (4.35)
ks = 1 3
⎡
⎢⎢
⎣ 2 1 +
λs
λf
– 1
ga
+ 1
1 + λs
λf
– 1
gc
⎤
⎥⎥
⎦ (4.36)
68 Soil Temperature and Heat Flow
kw= 1 3
⎡
⎢⎢
⎣ 2 1 +
λw
λf
– 1
ga
+ 1
1 + λw
λf
– 1
gc
⎤
⎥⎥
⎦ (4.37)
where ga and gc are shape factors. The value of ga for mineral soils is 0.088 and gc= 1 – 2ga. Therefore it is sufficient to obtain only one shape factor.
The thermal conductivity of the gas phase,λg, is given by λg=λa+LvfwρˆaDv
P–ea (4.38)
whereλais the thermal conductivity of air (reported in Table 4.1),Lvis the latent heat of vaporization,is the slope of the saturation vapour pressure function,Dvis the vapour diffusivity for soil,ρˆais the molar density of air,Pis the atmospheric pressure andeais the actual vapour pressure. The second term in eqn (4.38) is the latent heat term and is responsible for almost all the temperature dependence of soil thermal conductivity.
Below we present the program to estimate thermal conductivity for the theory de- scribed above. The project is calledPSP_thermalConductivityand it is made of three files:
1. main.py
2. PSP_thermalCond.py 3. PSP_readDataFile.py
The file PSP_readDataFile.py was already described in Appendix A and is used to read a row of soil temperatures. The file main.py contains instructions to read temperature data from the file soilTemperature.txt; the user is then asked to input a bulk density and a clay content value. Then incremental values of water content are created in the first for loop, while in the second for loop, the function thermalConductivityis called. The last instructions are written to plot the results.
#main.py
from _ _future_ _ import print_function, division import matplotlib.pyplot as plt
import PSP_thermalCond as soil from PSP_readDataFile import * def main():
A, isFileOk = readDataFile("soilTemperature.txt", 0, ’,’, False) if (isFileOk == False):
print ("Incorrect format") return()
Soil Thermal Properties 69 soilTemperature = A[0]
print ("Temperatures = ", soilTemperature) nrTemperatures = len(soilTemperature) myStr = "bulk density [kg/m^3]: "
bulkDensity = float(input(myStr)) myStr = "clay [0 - 1]: "
clay = float(input(myStr)) particleDensity = 2650
porosity = 1 - (bulkDensity / particleDensity) step = 0.02
nrValues = int(porosity / step) + 1 waterContent = np.zeros(nrValues)
thermalConductivity = np.zeros(nrValues) for i in range(nrValues):
waterContent[i] = step*i fig = plt.figure(figsize=(10,8))
plt.xlabel(’Water Content [m$^{3}$ m$^{-3}$]’, fontsize=20,labelpad=8)
plt.ylabel(’Thermal Conductivity [W m$^{-1}$C$^{-1}$]’, fontsize=20,labelpad=8)
plt.tick_params(axis=’both’, which=’major’, labelsize=20,pad=8) plt.tick_params(axis=’both’, which=’minor’, labelsize=20,pad=8)
#plt.xlim(0, 0.7)
for t in range(nrTemperatures):
for i in range(nrValues):
thermalConductivity[i] = soil.thermalConductivity (bulkDensity, waterContent[i], clay,
soilTemperature[t])
if (t == 0): plt.plot(waterContent, thermalConductivity,’k’) if (t == 1): plt.plot(waterContent, thermalConductivity,’--k’) if (t == 2): plt.plot(waterContent, thermalConductivity,’-.k’) if (t == 3): plt.plot(waterContent, thermalConductivity,’:k’) plt.show()
main()
The filePSP_thermalCond.pycontains the program with thethermalConduct- ivity function. It takes as arguments bulk density, water content, clay content and temperature. Therefore, from knowledge of four basic soil properties, it is possible to obtain an estimate of thermal conductivity. The functionheatCapacity( ) computes the soil heat capacity.
70 Soil Temperature and Heat Flow
#PSP_thermalCond.py
from _ _future_ _ import division from math import exp
import numpy as np GEOMETRIC = 0 LOGARITHMIC = 1 bulkDensity = 1300
def kMean(meanType, k1, k2):
if (meanType == GEOMETRIC):
k = np.sqrt(k1 * k2)
elif (meanType == LOGARITHMIC):
if (k1 == k2):
k = k1 else:
k = (k1-k2) / np.log(k1/k2) return k
def thermalConductivity(bulkDensity, waterContent, clay, temperature):
ga = 0.088
thermalConductivitysolid = 2.5 atmPressure = 100
q = 7.25 * clay + 2.52 xwo = 0.33 * clay + 0.078
solidContent = bulkDensity / 2650 porosity = 1 - solidContent
gasPorosity = max(porosity - waterContent, 0.0) temperatureK = temperature + 273.16
Lv = 45144 - 48 * temperature
svp = 0.611 * exp(17.502 * temperature / (temperature + 240.97)) slope = 17.502 * 240.97 * svp / (240.97 + temperature)**2.0 Dv = 0.0000212 * (101.3 / atmPressure) * (temperatureK / 273.16)
**1.75
rhoair = 44.65 * (atmPressure / 101.3) * (273.16 / temperatureK) stcor = max(1 - svp / atmPressure, 0.3)
thermalConductivitywater = 0.56 + 0.0018 * temperature if waterContent < 0.01 * xwo :
wf = 0 else:
#empirical weighting function D[0,1]
wf = 1 / (1 + (waterContent / xwo)**(-q))
thermalConductivitygas = (0.0242 + 0.00007 * temperature + wf * Lv * rhoair * Dv * slope / (atmPressure * stcor))
Numerical Implementation 71 gc = 1 - 2 * ga
thermalConductivityfluid = (thermalConductivitygas + (thermalConductivitywater - thermalConductivitygas) * (waterContent / porosity)**2.0)
ka = (2 / (1 + (thermalConductivitygas / thermalConductivityfluid - 1) * ga) + 1 /
(1 + (thermalConductivitygas / thermalConductivityfluid - 1) * gc)) / 3
kw = (2 / (1 + (thermalConductivitywater /
thermalConductivityfluid - 1) * ga) + 1 / (1 + (thermalConductivitywater /
thermalConductivityfluid - 1) * gc)) / 3 ks = (2 / (1 + (thermalConductivitysolid /
thermalConductivityfluid - 1) * ga) + 1 / (1 + (thermalConductivitysolid /
thermalConductivityfluid - 1) * gc)) / 3 thermalConductivity = ((kw * thermalConductivitywater *
waterContent + ka * thermalConductivitygas * gasPorosity + ks * thermalConductivitysolid * solidContent) / (kw * waterContent + ka * gasPorosity + ks * solidContent))
return(thermalConductivity)
def heatCapacity(bulkDensity, waterContent):
return (2.4e6 * bulkDensity / 2650 + 4.18e6 * waterContent) Figure 4.5 shows the dependence of thermal conductivity on water content for four different temperatures, generated using a bulk density of 1300 kg m–3and a clay content of 0.3. Note that the temperature dependence is mainly from the latent heat transport, which decreases as the soil dries. At about 65◦C, the apparent thermal conductivity from latent heat transport is equal to the thermal conductivity of water, so increasing water content does not affect the soil conductivity, at water content above 0.3 m3m–3.