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//(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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