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+// Example 16_6
+clc;funcprot(0);
+// Given data
+p_os=1.00;// MPa
+T_os=20.0+273.15;// K
+k=1.40;// The specific heat ratio
+p=0.1013;// MPa
+g_c=1;// The gravitational constant
+R=286;// J/kg.K
+
+// Solution
+// (a)
+p_r=p/p_os;// The pressure ratio
+M=((2/(k-1))*(((p_os/p)^((k-1)/k))-1))^(1/2);// The exit Mach number
+// (b)
+T=(T_os/(1+(((k-1)*M^2)/2)))-273.15;// The exit temperature in °C
+// (c)
+V=M*sqrt(k*g_c*R*(T+273.15));// The exit velocity in m/s
+// (d)
+p_throat=p_os*[2/(k+1)]^(k/(k-1));// The pressure at the throat of the nozzle in MPa
+// (e)
+T_throat=T_os*[2/(k+1)];// The temperature at the throat of the nozzle in K
+T_throat=T_throat-273.15;// The temperature at the throat of the nozzle in °C
+printf("\n(a)The exit Mach number,M=%1.2f \n(b)The exit temperature,T=%3.0f°C \n(c)The exit velocity,V=%3.0f m/s \n(d)The pressure at the throat of the nozzle,p_throat=%0.3f MPa \n(e)The temperature at the throat of the nozzle,T_throat=%2.1f°C",M,T,V,p_throat,T_throat);