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// Given:-
p1 = 30.0 // pressure of entering steam in bar
t1 = 400.0 // temperature of entering steam in degree celcius
v1 = 160.0 // velocity of entering steam in m/s
t2 = 100.0 // temperature of exiting saturated vapor in degree celcius
v2 = 100.0 // velocity of exiting saturated vapor in m/s
W = 540.0 // rate of work developed in kj per kg of steam
Tb = 350.0 // the temperature on the boundary where heat transfer occurs in kelvin
T0 = 25.0 // in degree celcius
p0 = 1.0 // in atm
// From table A-4
h1 = 3230.9 // in kj/kg
s1 = 6.9212 // in kj/kg.k
// From table A-2
h2 = 2676.1 // in kj/kg
s2 = 7.3549 // in kj/kg.k
// From example 6.6
Q = -22.6 // in kj/kg
// Calculations
DELTAef = (h1-h2)-(T0+273)*(s1-s2)+(v1**2-v2**2)/(2*1000)
// The net exergy carried in per unit mass of steam flowing in kj/kg
Eq = (1-(T0+273)/Tb)*(Q) // exergy transfer accompanying heat in kj/kg
Ed = ((1-(T0+273)/Tb)*(Q))-W+(DELTAef) // The exergy destruction determined by rearranging the steady-state form of the exergy
// rate balance
// Results
printf( ' Balance sheet')
printf( ' Net rate of exergy %f kJ/kg,',DELTAef)
printf( ' Disposition of the exergy:')
printf( '* Rate of exergy out')
printf( ' Work %f kJ/kg.',W)
printf( ' Heat transfer %f',-Eq)
printf( '• Rate of exergy destruction %f kJ/kg.',Ed)
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