Calculate sulfur dioxide emission rate for a coal sample – without scrubbing and

Question

Calculate sulfur dioxide emission rate for a coal sample – without scrubbing and with scrubbing that is 80% effective.
Background
“Clean Air” legislation restricts the sulfur dioxide emissions of coal-burning facilities to about one pound per million Btu’s of energy generated. Factors affecting emissions rates include the energy content and sulfur content of the coal and the use of scrubbing, a reprocessing of the combustion gases to remove sulfur.

Your program should calculate the number of pounds of sulfur dioxide emissions per million Btu’s of energy for a given coal sample. The program should reflect the fact that complete oxidation of n pounds of sulfur emissions levels both without and with scrubbing, assuming that scrubbing would remove 80% of the sulfur dioxide emissions.
Use an oxidation factor of 2.0 and scrub efficiency of 0.8; oxidizing n pounds of sulfur produces about 2n pounds of sulfur dioxide and the scrubbing combustion gases reduces sulfur dioxide by 80%.

Implementation
Prompt user for coal sample number, percent of sulfur in sample and Btu’s of energy from one pound of the sample.
Calculate emissions rate for burning this sample
Calculate emissions rate with scrubbing
Display results

Output Example

Enter sample identification number => 541
What percent of the sample is sulfur => 3
How many Btu’s of energy from one pound of the sample =>12000
Sulfur dioxide emissions of coal sample 541 with sulfur content of 3 percent and energy content of 12000 Btu per pound:

Before scrubbing: 5 pounds per million Btu
After scrubbing: 1 pounds per million Btu

Explanation / Answer

#include 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 > sampleID; cout > sulfurContent; cout > energyContent; //Calculate emissions rate for burning this sample double poundsDioxide = sulfurContent / 100.0 * OXIDATION_FACTOR; double emissions = poundsDioxide * 1.0e6 / energyContent; //calculate emissions rate with scrubbing double withScrubbing = emissions * (1 - SCRUB_EFFICIENCY); //TO PROGRAM - Display results cout

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