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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);