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clear;
clc;
disp('Example 1.9');
// Given values
m_dot = 20.4; // mass flowrate of petrol, [kg/h]
c = 43; // calorific value of petrol, [MJ/kg]
n = .2; // Thermal efficiency of engine
// solution
m_dot = 20.4/3600; // [kg/s]
c = 43*10^6; // [J/kg]
// power output
P_out = n*m_dot*c; // [W]
mprintf('\n The power output of the engine is = %f kJ\n',P_out*10^-3);
// power rejected
P_rej = m_dot*c*(1-n); // [W]
P_rej = P_rej*60*10^-6; // [MJ/min]
mprintf('\n The energy rejected by the engine is = %f MJ/min \n',P_rej);
//End
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