//
// Calculate sulfur dioxide emission rate for a coal sample - without
// scrubbing and with scrubbing that is 80% effective
//
#include <iostream>
using namespace std;

const double OXIDATION_FACTOR = 2.0; // oxidizing n pounds of sulfur
// produces about 2n pounds of sulfur dioxide
const double SCRUB_EFFICIENCY = 0.8; // scrubbing combustion gases
// reduces sulfur dioxide emissions by 80%

int main()
{
   int sampleId; // input - sample identification number
   double sulfurContent; // input - % of sample that is sulfur
   double energyContent; // input - number of Btu per pound

   // Get data on coal sample
   cout << "Enter sample identification number => ";
   cin >> sampleId;
   cout << "What percent of the sample is sulfur? => ";
    
   cin >> sulfurContent;
   cout << "How many Btu's of energy from one pound of the sample? => ";
   cin >> energyContent;

   // Calculate emissions rate for burning this sample
   double poundsDioxide = sulfurContent / 100.0 * OXIDATION_FACTOR;
   double emissions = poundsDioxide * 1000000.0 / energyContent;

   // Calculate emissions rate with scrubbing
   double withScrubbing = emissions - emissions * SCRUB_EFFICIENCY;

   // Display results
   cout << "Sulfur dioxide emissions of coal sample " << sampleId <<
   " with sulfur content of " << endl;
   cout << " " << sulfurContent << " percent and energy content of "
   << energyContent << " Btu per pound:" << endl;
   cout << " Before scrubbing: " << emissions <<
   " pounds per million Btu" << endl;
   cout << " After scrubbing: " << withScrubbing <<
   " pounds per million Btu" << endl;

   return 0;
}