Soil Heat Flow and Temperature                  

By R.L. Snyder and K.T. Paw U

 

Copyright - Regents of the University of California

Created - June 22, 2000

Last Revision –June 13, 2001

 

 

Question: Why do we want to know the soil heat flow and temperature?

 

Answer: We need G at the surface to determining the energy balance

 

Neglecting small energy components, the energy balance on a surface can be expressed using the following equation, where Rn is net radiation, G is the soil heat flux density, H is the sensible heat flux density, and LE is the latent heat flux density. 

 

Net Radiation    =        Soil HFD   +  Sensible HFD   +      Latent HFD

 

Rn        =          G      +        H         +        LE

 

In this document, the methodology to estimate G at the surface is discussed.

 

 

DEFINITIONS

 

1)    Temperature (T) is a measure of the heat stored (oC or K)

2)    Upper temperature (T1) is the temperature at depth z1

3)    Lower temperature (T2) is the temperature at depth z2

4)    Volumetric Heat Capacity is the amount of heat required to raise the temperature of a unit volume by one Kelvin (J m-3 K-1)  

5)    Thermal Conductivity (C1) is the ratio of the heat flux density to the temperature gradient in (W m-1 K-1)

 

SOIL HEAT FLUX DENSITY

 

1)    Soil heat flux density (G) is the conduction of energy per unit area in response to a temperature gradient.  For small depth changes,

                                                                                  (1)

Here, the thermal conductivity C1 = KA, where K is the heat conductivity of the material in W m-3 K-1 and A is the surface area in m2, so for heat flow through a unit surface area C1 has the units W m-1K-1.  In Eq. 1, is positive when the temperature decreases with depth in the soil.  The negative sign is included to make G positive when heat is transferring downward.  Because of instrument limitations, it is not possible to accurately measure the  temperature gradient unless there is sufficient distance between the sensors.  Consequently, G is estimated as

                                                                            (2)

where z2 is sufficiently far below z1 to allow for a measurable difference between T2 and T1.  Equation 2 assumes that C1 is constant with depth in the soil.

 

2)    Thermal Diffusivity (k) is the ratio of the thermal conductivity to the volumetric heat capacity.

                                                                              (3)

 

C1=the thermal conductivity (W m-1 K-1)

rs = the apparent soil density (kg m-3)

Cp = the apparent mass specific heat capacity (J kg-1 K-1)

CV = volumetric heat capacity (J m-3 K-1)

k  = thermal diffusivity (m2 s-1)

 

 where

                                                                                      (4)

 

So the thermal conductivity in terms of diffusivity and volumetric heat capacity is

                                                                         (5)

 

Therefore, G can be expressed in terms of diffusivity, volumetric heat capacity, and the temperature gradient as:

 

                                                                                (6)

In terms of the temperature gradient, G is

 

                                                                       (7)

 

 

 

The variable k is useful as a measure of how fast the temperature of a soil layer changes.   The rate at which the heat content of a layer of soil changes depends on the volumetric heat capacity (CV) and the rate of temperature change of the soil volume per unit time.  For a unit surface area, the rate of change in heat storage within the soil layer is expressed as .  For a unit surface area, the rate of change in heat storage within the soil is also equal to the change in heat flux density through the soil layer  .  Assuming that the physical properties of the soil are constant with depth in the soil and equating these two expressions, we get   which simplifies to

 

                                                                (8)

 

Therefore, k is useful to determine the rate of temperature change of a soil layer.  Recall that k is directly proportional to C1 and inversely proportional to Cv.  Both C1 and Cv increase as the water content of the soil increases; however, C1 increases more rapidly with water content when the soil is dry and it slows as the soil becomes wet.  Cv continues to increase even when the soil is relatively wet.  Consequently, for a dry soil, k increases with water content, but it slows and sometimes decreases when the soil nears saturation.  As a result, the maximum change in temperature with time will occur at a water content below saturation.    

 

 

 

 


Question: How do you determine G= G1 at the surface?

                             

The change in heat storage of a layer of soil between the surface z1=0 and some depth z2 is given by

                                                    (9)

where G is positive downward into the soil.

 

Note: Soil heat flux density is positive downwards.  When DS = -(G2 – G1) is positive, then more heat is entering the top than leaving the bottom of the soil layer and the soil is warming.  If DS = -(G2 – G1) is negative, then more heat is leaving than entering the layer and the soil is cooling. 

 

Text Box: Examples:
Let G1 = 100 W m-2 and G2 = 50 W m-2, then 
DS= -(50-100)=50 W m-2 and the soil is warming

Let G1 = -100 W m-2 and G2 = -50 W m-2, then 
DS =-[-50-(-100)]= -50 W m-2 and the soil is cooling

Let G1 = -50 W m-2 and G2 = 10 W m-2, then 
DS = -[10-(-50)]=-60 W m-2 and the soil is cooling

Let G1 = 50 W m-2 and G2 = -10 W m-2, then 
DS = -[-10-50]=60 W m-2 and the soil is warming

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Accounting for the change in soil heat storage is important for estimating G at the surface.  This is clearly indicated in Fig. 1, which shows a plot of G2 + DS versus G2 for half hour measurements of G2 at 0.02 m depth under a cover of about 0.1 m tall, cool-season grass.  The regression equation indicates that the soil heat flux density at the surface is under-estimated by about 40% when not corrected for the change in soil heat storage.

 

 

Figure 1.  Surface soil heat flux density (G) calculated using a heat flux plate at 0.02 m depth (G2) and the change in heat storage (DS) estimated from temperature change at 0.01 m depth versus G2 alone using half hour data collected under 0.10 m tall, cool-season grass.  The soil was a Yolo clay loam soil that was irrigated once per week.
 
CHANGE IN HEAT STORAGE

 

The net change of heat storage per unit time (DQ/Dt) in J s-1 within a soil volume (V) is given by:

                                                         (10)

J s-1 = (J m-3 K-1) (K s-1) (m3)                                  

 

where Tf is the final temperature at time tf and Ti is the initial temperature at time ti.

 

Because the change in stored soil heat (DS) per unit surface area is equal to the net rate of heat storage divided by the soil surface area and V/A=Dz, we can calculate DS in W m-2 as

 

                               (11)

 

If G2 is measured and DS is calculated using Eq.11, then G (= G1 at the surface) is calculated using:

 

                                                                       (12)

 

Example:  Given a uniform soil with CV = rsCp = 1.677x106 J m-3 K-1, a soil heat flux plate placed at an 0.08 m depth, and soil-averaging temperature sensors located at 0.02 and 0.06 m depth to represent the change in heat storage between the surface and 0.08 m depth.  If the heat flux plate measurement averages G2 = 20 W m-2 and the average soil temperature from the two sensors increases by 1 K during a one-hour period, what is the soil heat flux density  (G = G1) at the surface?

 

Text Box:             
 

  G = G1 = G2 + DS =20 + 37.3 = 57.3 W m-2

 

 

 

 

 

 

 

 

 

Question:  Using the same criteria as above but now with the mean soil temperature dropping by 1 K, what is the soil heat flux density (G = G1) at the surface?

 

Text Box:  
 
G = G1 = G2 - DS =20 + (-37.3) = -17.3 W m-2

 

 

 

 

 

 

 

If the soil heat flux density at 0.08 m depth is G = 20 W m-2 and the temperatures 20oC and 20.8oC are measured on either side of the heat flux plate. What is the thermal conductivity (C1) if the temperature sensors are 0.02 m below and above the heat flux plate?  Which temperature is recorded above the heat flux plate?

T=20.8oC is the upper temperature because G>0

 

 

  W m-2K-1

 
 

 

 

 

 

 

 

 

 

 

 

 


Thermal Properties of Soils

 

Heat capacity depends on the mineral, organic matter, and water content of a soil.  The apparent heat capacity (Cv) on a volume basis has the units J m-3 K-1 and on a mass basis (Cp) it has the units J kg-1 K-1.  Equation 13 is used to express heat capacity.

 

                                                 (13)

 

J m-3K-1 = (kg m-3)(J kg-1K-1)   

 

where rs is the apparent density in kg of moist soil per m3, rb is the bulk density of the soil in kg of dry soil per m3, cpav is the average heat capacity on a mass basis for the solid constituents of the soil in J per kg of dry soil per Kelvin, qm is the water content on a mass basis in kg of water per kg of dry soil, and cpw is the heat capacity of water on a mass basis in J per kg of H20 per Kelvin. 

 

The heat capacity of the solid constituents of the soil (Cpav) depends on the amount of sand, clay, silt, and organic matter in the soil.  For most mineral soils,

 

                                                                   J kg-1K-1  

 

The heat capacity of the water component of the soil (qmcpw) depends on the water content and the heat capacity for water.

 

               J kg-1K-1

 

 

The volumetric heat capacity is rwcpw, which is approximated as

 

                                                            J m-3K-1

 

and the volumetric water content qv in m3 H2O per m3 of dry soil is

 

 

Therefore, the heat capacity of the soil component is approximated as

 

 

and the heat capacity of the water component is approximated as

 

 

Substitution into Eq. 13 gives the following approximation for Cv.

 

                          J m-3 K-1             (15)

 

Because we are interested in heating and cooling the soil, it is useful to have an expression for the amount of energy (Q) needed to raise or lower the temperature of a known volume (V) of soil from Ti to Tf .  The equation is

 

                         J = J m-3 K-1 (K) m3                       (16)


Problems:

 

 

1)    Given a wet soil at Ti = 18oC with qv = 0.23 and rb = 1300 kg m-3.  Find the amount of heat required to raise a 1 m3 volume of soil to a temperature Tf = 20oC for a surface area of 1 m2 to a depth of 1 m.

 

Text Box: Q = (837 rb + 4.19´106 qv) (Ti – Tf) V 

Q = (837´1300 + 4.19´106 ´ 0.23) (20 –18) (1.0 m3)

Q = (1.088100´106 + 0.963700´106) (2) (1) = 4.10´106 J = 4.10 MJ

 

 

 

 

 

 

 

 

2)    Given a wet soil with heat flux G = 25 W m-2 into the soil surface (qv = 0.18, rb = 1300 kg m-3 and Q = G A t). Find the average temperature increase that occurs in one day in the top 1 m of soil assuming that the heat is evenly distributed in the 1 m soil layer.

 

Text Box: Q = Cv V DT 

Q = G A t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3)    Given a wet soil with qv = 0.18, rb = 1300 kg m-3, Q = G A t, and the average daily net radiation is Rn = 150 W m-2, find the ratio G/Rn if the increase in average soil temperature to 1 m depth over a month’s time is 10oC.

 

 

From problem 2, we know that Cv.V = 1.84´106 J m-3 K-1

 

 

 

 

4)    Given a soil with the following temperatures and properties, find the quantity of heat stored in the soil per unit area (Q/A), for the period from 0500 to 1400 hrs.

 

 

Soil Depth

Ti

Tf

rb

qv

m

0500 h

1400 h

kg m-3

 

0.00-0.05

15

35

1.0

0.05

0.05-0.20

20

30

1.1

0.10

0.20-0.60

18

25

1.2

0.20

0.60-1.20

17

18

1.3

0.25

 

 

Text Box: Q/A = [(837      ´      rb   +   4.19´106 ´qv) ´ V  ´ DT] / A
         (J kg-1K-1)    (kg m-3)    (J m-3K-1)         (m3)  (K)  (m2)

0.00-0.05 m 	
Q/A=[(837´1000+4.19´106´0.05) ´ 0.05 ´ 20]/1=1.048´106=1.05 MJ m-2
0.05-0.20 m 	
Q/A=[(837´1100+4.19´106´0.10) ´ 0.15 ´ 10]/1=2.013´106=2.01 MJ m-2
0.20-0.60 m 
Q/A=[(837´1200+4.19´106´0.20) ´ 0.40 ´ 7]/1=5.165´106=5.17 MJ m-2
0.60-1.20 m 
Q/A=[(837´1300+4.19´106´0.25) ´ 0.60 ´ 1]/1=1.283´106=1.28 MJ m-2

Total heat stored = 1.05 + 2.01 + 5.17 + 1.28 = 9.51 MJ m-2