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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+clc
+clear
+//INPUT DATA
+t1=300;//temperature in K
+t3=1300;//temperature in K
+cp=5.22;//specific pressure
+cv=3.13;//specific volume
+g=1.688;//for helium as working medium
+
+//CALCULATIONS
+Rc=((t3/t1)^(1/(2*(1.4-1))));//compression ratio
+no1=(1-((1/Rc)^(1.4-1)))*100;//efficiency of air
+Rcn=((t3/t1)^(1/(2*(g-1))));//compression ratio
+no2=(1-((1/Rcn)^(g-1)))*100;//efficiency of helium
+
+//OUTPUT
+printf('(a)air as working medium efficiency is %3.2f percentage \n (b)Helium as working medium efficiency is %3.2f percentage \n Hence the change in efficiency is zero',no1,no2)
+