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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+// Example 9_8
+clc;funcprot(0);
+// Given data
+r=10;// The compression ratio
+P_1=200;// kPa
+T_1=100+273;// K
+T_3=600+273;// K
+R=0.287;// kJ/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+v_1=(R*T_1)/P_1;// m^3/kg
+T_4=T_3;// K
+v_4=(T_4/T_1)*v_1;// m^3/kg
+v_2=v_4/r;// m^3/kg
+T_2=T_1;// K
+P_2=(R*T_2)/v_2;// kPa
+P_3=P_2;// kPa
+v_3=(R*T_3)/P_3;// m^3/kg
+w_out=(R*T_1*log(v_2/v_1))+(P_2*(v_3-v_2))+(R*T_3*log(v_4/v_3))+(P_1*(v_1-v_4));// The work output in kJ/kg
+T_L=T_1;// K
+T_H=T_3;// K
+n=1-(T_L/T_H);// The thermal efficiency
+q_in=w_out/n;// The heat input in kJ/kg
+printf("\nThe work output,w_out=%3.0f kJ/kg \nThe heat input,q_in=%3.0f kJ/kg",w_out,q_in);