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//(10.5) Reconsider Example 10.4, but include in the analysis that the compressor and turbine each have an isentropic efficiency of 80%. Determine for the modified cycle (a) the net power input, in kW, (b) the refrigeration capacity, in kW, (c) the coefficient of performance, and interpret its value.
//solution
funcprot(0)
//part(a)
wcdots = 99.99 //work per unit mass for the isentropic compression determined with data from the solution in Example 10.4 in kj/kg
mdot = 1.807 //mass flow rate in kg/s from 10.4
etac = .8 //isentropic efficiency of compressor
Wcdot = mdot*wcdots/etac //The power input to the compressor in kw
//Using data form the solution to Example 10.4 gives
wtdots =81.19 //in kj/kg
etat = .8 //isentropic efficiency of turbine
Wtdot = mdot*etat*wtdots //actual turbine work in kw
Wdotcycle = Wcdot-Wtdot //The net power input to the cycle in kw
printf('the net power input in kw is: %f',Wdotcycle)
//part(b)
h3 = 300.19 //in kj/kg
h4 = h3 -Wtdot/mdot
//from table A-22
h1 = 270.11 //in kj/kg
Qindot = mdot*(h1-h4) //refrigeration capacity in kw
printf('\nthe refrigeration capacity in kw is: %f',Qindot)
//part(c)
beta = Qindot/Wdotcycle //coefficient of performance
printf('\nthe coefficient of performance is: %f',beta)
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