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authorpriyanka2015-06-24 15:03:17 +0530
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+clc
+// given data
+G=39.0 // temperature gradient in K/km.
+h2=10.0 // depth in km
+rhor=2700.0 // kg/m^3
+cr=820.0 // in J/kg-K
+
+h1=120/G // T1-T0=120 K is given
+h21=h2-h1 // in km
+E0byA=(rhor*(1000**3)*G*cr*h21**2)/2 // in J/km^2 Heat content per square km
+
+thetao=G*h21/2.0 // in degree K
+tau=rhor*cr*h21*(1000**3)/(QbyA*rhow*cw) // in seconds
+tau=tau/(2*60*60*24*365) // in years
+theta=thetao*exp(-t/tau) // in degree Kelvin
+
+Heatinitial=E0byA/(60*60*365*24*tau)/1000000 // intial heat extraction rate in MW /km^2
+
+Heat25=Heatinitial*exp(-t/tau) // heat extraction rate after 25 years in MW /km^2
+
+printf( "Initial Heat extraction rate is %.2f MW/km^2",Heatinitial)
+
+printf(" \n Final Heat extraction rate is %.2f MW/km^2",Heat25)