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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 5 Example 18")
+T1_a=800;//temperature of reservoir a in K
+T1_b=700;//temperature of reservoir b in K
+T1_c=600;//temperature of reservoir c in K
+T2=320;//temperature of sink in K
+W=20;//work done in KW
+Q2=10;//heat rejected to sink in KW
+disp("let heat supplied by reservoir at 800 K,700 K,600 K be Q1_a , Q1_b , Q1_c")
+disp("here Q1-Q2=W")
+disp("so heat supplied by source(Q1)=W+Q2 in KW")
+Q1=W+Q2
+disp("also given that,Q1_a=0.7*Q1_b.......eq 1")
+disp("Q1_c=Q1-(0.7*Q1_b+Q1_b)")
+disp("Q1_c=Q1-1.7*Q1_b........eq 2")
+disp("for reversible engine")
+disp("Q1_a/T1_a+Q1_b/T1_b+Q1_c/T1_c-Q2/T2=0......eq 3")
+disp("substitute eq 1 and eq 2 in eq 3 we get, ")
+disp("heat supplied by reservoir of 700 K(Q1_b)in KJ/s")
+disp("Q1_b=((Q2/T2)-(Q1/T1_c))/((0.7/T1_a)+(1/T1_b)-(1.7/T1_c))")
+Q1_b=((Q2/T2)-(Q1/T1_c))/((0.7/T1_a)+(1/T1_b)-(1.7/T1_c))
+disp("so heat supplied by reservoir of 800 K(Q1_a)in KJ/s")
+disp("Q1_a=0.7*Q1_b")
+Q1_a=0.7*Q1_b
+disp("and heat supplied by reservoir of 600 K(Q1_c)in KJ/s")
+disp("Q1_c=Q1-1.7*Q1_b")
+Q1_c=Q1-1.7*Q1_b
+disp("so heat supplied by reservoir at 800 K(Q1_a)")
+Q1_a
+disp("so heat supplied by reservoir at 700 K(Q1_b)")
+Q1_b
+disp("so heat supplied by reservoir at 600 K(Q1_c)")
+Q1_c=-Q1_c
+disp("NOTE=>answer given in book for heat supplied by reservoir at 800 K,700 K,600 K i.e Q1_a=61.94 KJ/s,Q1_b=88.48 KJ/s,Q1_c=120.42 KJ/s is wrong hence correct answer is calculated above.")
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