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+clc
+// Given that
+F = 680 // Net brake load in N
+N = 360 //
+d = 10// Bore in cm
+L = 15 // Stroke in cm
+T = 58 // Torque in Nm
+v = 300 // Speed in m/min
+n_m = 0.8 // Mechanical efficiency
+n_th = 0.4 // Indicated thermal efficiency
+c_v = 44 // Calorific value of gasoline in MJ/kg
+printf("\n Example 20.3\n")
+N = v/(2*L*(10^(-2)))
+BP = (2*%pi*T*N)/60000
+IP = BP/n_m
+p_m = (IP*60)/(L*(%pi/4)*(d^2)*N*10^(-6))
+m_f = (IP*3600)/(n_th*c_v*1000)
+bsfc = m_f/BP
+printf("\n Indicated power = %f kW\n Indicate mean effective pressure = %f kN/m^2\n Fuel consumption per kWh on brake power output = %f Kg/kWh",IP,p_m,bsfc)
+
+