Calculation of true thickness of the bed

 From the experimental studies, values for Crustal rocks are density $ρ=2650 kg/m^3$, thermal conductivity $K_c= 2.51 W /m K$ and Specific heat $Cp =500 J/Kg. K$. Approximate the penetration depth (The daily temperature variation = $7.27 ×10^{-5} /s$).

Solution:

Density $\rho = 2650 kg/m^3$

Thermal Conductivity $K_c=2.5 W/m K$

Specific Heat $C_p=700 J/Kg. K$

To Calculate the Thermal diffusivity $K_d$ is

$K_d= \frac{K_c}{\rho c_p}$

$K_d= \frac{2.5}{2650 \times 700}$

$K_d= 1.35 \times 10^{-6} \frac{m^2}{s}$

Decay depth (or) Penetration depth (d) is

$d =\sqrt{\frac{2 K_d}{\omega}}$

$d =\sqrt{\frac{2\times 1.35\times 10^{-6}}{7.27\times 10^{-5}}}$

$d=0.192 m$

$d= 19.2 cm$

 

 

 

Extra information:

 

Heat flux (q) is defined as the amount of heat passing through a unit area in a unit time.

$q= \frac{J}{s m^2}$

$q=\frac{W}{m^2}$

Derive the Work units,

$Work =Force \times Displacement$

$Work = mass \times acceleration \times Displacement$

$Work = mass \times \frac{Velocity}{time} \times Displacement$

$Work = kg \times \frac{m}{s^2} \times m$

$Work = kg \times \frac{m^2}{s^2}$

$kg \times \frac{m^2}{s^2} = Joules$

Watts is defined as 

$Watts= \frac{Joules}{second}$

 

Finally derive the units of Thermal conductivity $K_c$ is

$K_c=\frac{q}{\frac{dT}{dZ}}$

q= Heat flux

$\frac{dT}{dZ}$=geothermal gradient 

$\frac{dT}{dZ} units= \frac{K}{m}$

$K_c units=\frac{Joules}{m. K}$

Reference : Fundamental of Geophysics, William Lowrie.