2 files changed, 48 insertions, 49 deletions
diff --git a/3434/CH14/EX14.7/Ex14_7.sce b/3434/CH14/EX14.7/Ex14_7.sce
index c3edd8ba7..21f42aa8c 100644
--- a/3434/CH14/EX14.7/Ex14_7.sce
+++ b/3434/CH14/EX14.7/Ex14_7.sce
@@ -1,26 +1,23 @@
-clc
-//given data
-i=12/100.0 // interest rate 
-n=10 // time in years
-
-time=100.0 // days geyser is used in year
-effi=0.9 // efficiency of geyser
-w=100.0 // weight of water in kg
-C=4.2 // heat capacity in kJ/kg-degree C
-theta=60-15 // temperature difference in C
-cost=4 // cost of electricity per kWh
-
-Elec=(1/effi)*w*C*theta/3600.0 // electricity used in kWh/day
-
-
-
-A=Elec*time*cost // annual saving in Rs
-
-P=A*(((1+i)**n)-1)/(i*((1+i)**n)) // final amount in rs
-
-printf("The final amount after 10 years is Rs %i",P)
-
-// the answer is slightly different in textbook due to approximation while in scilab answers are precise
-end
-
-
+clc
+//given data
+i=12/100.0 // interest rate 
+n=10 // time in years
+
+time=100.0 // days geyser is used in year
+effi=0.9 // efficiency of geyser
+w=100.0 // weight of water in kg
+C=4.2 // heat capacity in kJ/kg-degree C
+theta=60-15 // temperature difference in C
+cost=4 // cost of electricity per kWh
+
+Elec=(1/effi)*w*C*theta/3600.0 // electricity used in kWh/day
+
+
+
+A=Elec*time*cost // annual saving in Rs
+
+P=A*(((1+i)**n)-1)/(i*((1+i)**n)) // final amount in rs
+
+printf("The final amount after 10 years is Rs %i",P)
+
+// the answer is slightly different in textbook due to approximation while in scilab answers are precise
+\ No newline at end of file
diff --git a/3434/CH9/EX9.1.iii/Ex9_1_iii.sce b/3434/CH9/EX9.1.iii/Ex9_1_iii.sce
index 6387a3e1d..5b1af1f5e 100644
--- a/3434/CH9/EX9.1.iii/Ex9_1_iii.sce
+++ b/3434/CH9/EX9.1.iii/Ex9_1_iii.sce
@@ -1,23 +1,25 @@
-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)
+clc
+// given data
+G=39.0 // temperature gradient in K/km.
+h2=10.0 // depth in km
+rhow = 1000;
+rhor=2700.0 // kg/m^3
+cr=820.0 // in J/kg-K
+cw=4200.0; // specific heat of water in J/kg-K 
+QbyA=0.5;
+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
+t=25 // time in years
+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)
+\ No newline at end of file