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Diffstat (limited to '572/CH9/EX9.6/c9_6.sce')
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diff --git a/572/CH9/EX9.6/c9_6.sce b/572/CH9/EX9.6/c9_6.sce new file mode 100755 index 000000000..dfbfa1cc9 --- /dev/null +++ b/572/CH9/EX9.6/c9_6.sce @@ -0,0 +1,42 @@ +//(9.6) Reconsider Example 9.4, but include in the analysis that the turbine and compressor each have an isentropic efficiency of 80%. Determine for the modified cycle (a) the thermal efficiency of the cycle, (b) the back work ratio, (c) the net power developed, in kW.
+
+//solution
+
+
+etat = .8 //turbine efficiency
+etac = .8 //compressor efficiency
+//part(a)
+wtdots = 706.9 //The value of wtdots is determined in the solution to Example 9.4 as 706.9 kJ/kg
+//The turbine work per unit of mass is
+wtdot = etat*wtdots //in kj/kg
+
+wcdots = 279.7 //The value of wcdots is determined in the solution to Example 9.4 as 279.7 kJ/kg
+//For the compressor, the work per unit of mass is
+wcdot = wcdots/etac //in kj/kg
+
+h1 = 300.19 //h1 is from the solution to Example 9.4, in kj/kg
+h2 = h1 + wcdot //in kj/kg
+
+h3 = 1515.4 //h3 is from the solution to Example 9.4, in kj/kg
+qindot = h3-h2 //The heat transfer to the working fluid per unit of mass flow in kj/kg
+eta = (wtdot-wcdot)/qindot //thermal efficiency
+printf('the thermal efficiency is: %f',eta)
+
+//part(b)
+bwr = wcdot/wtdot //back work ratio
+printf('\nthe back work ratio is: %f',bwr)
+
+//part(c)
+mdot = 5.807 //in kg/s, from example 9.4
+Wcycledot = mdot*(wtdot-wcdot) //The net power developed by the cycle in kw
+printf('\nthe net power developed, in kW. is: %f',Wcycledot)
+
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