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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+// Example 5_6
+clc;funcprot(0);
+// Given data
+T_H=500;// K
+T_L=300;// K
+P_1=80*10^3;// Pa
+v_4=10;// m^3/kg
+R=287;// J/kg.K
+k=1.4;// The specific heat ratio
+
+// Calculation
+n=(1-(T_L/T_H))*100;// %
+T_1=T_L;// K
+T_2=T_H;// K
+v_1=(R*T_1)/P_1;// m^3/kg
+v_2=v_1*(T_1/T_2)^(1/(k-1));// m^3/kg
+T_4=T_L;// K
+T_3=T_H;// K
+v_3=v_4*(T_4/T_3)^(1/(k-1));// m^3/kg
+q_H=(R/10^3)*T_H*log(v_3/v_2);// kJ/kg
+w=(n/100)*q_H;// kJ/kg
+printf("\nThe thermal efficiency,n=%2.0f percentage \nThe work output,w=%3.0f kJ/kg",n,w);