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+// Display mode
+mode(0);
+// Display warning for floating point exception
+ieee(1);
+clear;
+clc;
+disp("Engineering Thermodynamics by Onkar Singh Chapter 4 Example 9")
+T1=1100;//temperature of high temperature reservoir in K
+T4=300;//temperature of low temperature reservoir in K
+disp("here W1:W2:W3=3:2:1")
+disp("efficiency of engine,HE1,")
+disp("W1/Q1=(1-(T2/1100))")
+disp("so Q1=(1100*W1)/(1100-T2)")
+disp("for HE2 engine,W2/Q2=(1-(T3/T2))")
+disp("for HE3 engine,W3/Q3=(1-(300/T3))")
+disp("from energy balance on engine,HE1")
+disp("Q1=W1+Q2=>Q2=Q1-W1")
+disp("above gives,Q1=(((1100*W1)/(1100-T2))-W1)=W1*(T2/(1100-T2))")
+disp("substituting Q2 in efficiency of HE2")
+disp("W2/(W1*(T2/(1100-T2)))=1-(T3/T2)")
+disp("W2/W1=(T2/(1100-T2))*(T2-T3)/T2=((T2-T3)/(1100-T2))")
+disp("2/3=(T2-T3)/(1100-T2)")
+disp("2200-2*T2=3*T2-3*T3")
+disp("5*T2-3*T3=2200")
+disp("now energy balance on engine HE2 gives,Q2=W2+Q3")
+disp("substituting in efficiency of HE2,")
+disp("W2/(W2+Q3)=(T2-T3)/T2")
+disp("W2*T2=(W2+Q3)*(T2-T3)")
+disp("Q3=(W2*T3)/(T2-T3)")
+disp("substituting Q3 in efficiency of HE3,")
+disp("W3/((W2*T3)/(T2-T3))=(T3-300)/T3")
+disp("W3/W2=(T3/(T2-T3))*(T3-300)/T3")
+disp("1/2=(T3-300)/(T2-T3)")
+disp("3*T3-T2=600")
+disp("solving equations of T2 and T3,")
+disp("we get,T3=(600+(2200/5))/(3-(3/5))in K")
+T3=(600+(2200/5))/(3-(3/5))
+disp("and by eq 5,T2 in K")
+T2=(2200+3*T3)/5
+disp("so intermediate temperature are 700 K and 433.33 K")