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+clear;
+clc;
+disp('Example 1.14');
+
+
+
+//  Given values
+P_out = 500; //  output of power station, [MW]
+c = 29.5; //  calorific value of coal, [MJ/kg]
+r=.28; 
+
+//  solution
+
+//  since P represents only 28 percent of energy available from coal
+P_coal = P_out/r; //  [MW]
+ 
+m_coal = P_coal/c; //  Mass of coal used, [kg/s]
+m_coal = m_coal*3600; //  [kg/h]
+
+//After one hour
+m_coal = m_coal*1*10^-3; //  [tonne]
+mprintf('\n Mass of coal burnt by the power station in 1 hour is  =  %f  tonne \n',m_coal);
+
+//  End