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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
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