#example 1
#Calculating reversible work
#Form the Steam Tables,the inlet and the exit state properties are
hi=171.95 #initial specific heat of enthalpy in kJ/kg
si=0.5705 #initial specific entropy in kJ/kg-K
se=2.1341 #final specific entropy in kJ/kg-K
he=765.34 #final specific heat of enthalpy in kJ/kg-K
m=5 #mass flow rate of feedwater in kg/s
q1=900/m #heat added by one of the sources in kJ/kg
q2=he-hi-q1 #second heat transfer in kJ/kg
To=25+273.3 #Temp. of the surroundings in K
T1=100+273.2 #temp. of reservoir of one of the source in K
T2=200+273.2 #temp. of reservoir of second source in K
wrev=To*(se-si)-(he-hi)+q1*(1-To/T1)+q2*(1-To/T2) #reversible work in kJ/kg
print"\n Hence, the irreversibility is",round(wrev,1),"kJ/kg"
#example 2
#Calculating reversible work
import math
#Form the Steam Tables,the inlet and the exit state properties are
hi=298.6 #initial specific heat of enthalpy in kJ/kg
si=6.8631 #initial specific entropy in kJ/kg-K
se=7.4664 #final specific entropy in kJ/kg-K
he=544.7 #final specific heat of enthalpy in kJ/kg-K
q=-50 #heat lost to surroundings in kJ/kg
w=hi-he+q #work in kJ/kg
To=25+273.2 #Temp. of the surroundings in K
P1=100 #Pressure of ambient air in kPa
P2=1000 #Final pressure of air after compression in kPa
R=0.287 #Universal gas constant in kJ/kg-K
wrev=To*(se-si-R*math.log(P2/P1))-(he-hi)+q*(1-To/To)#reversible work for the given change of state in kJ/kg
i=wrev-w #irreversibility in kJ/kg
print"\n Hence, the irreversibility is",round(i,1),"kJ/kg"
#example 3
#Calculating reversible work and irreversibility
#Form the Steam Tables at state 1
u1=1243.5 #initial specific internal energy in kJ/kg
s1=4.4819 #initial specific entropy in kJ/kg-K
v1=28.895 #initial specific volume in m^3/kg
v2=2*v1 #final specific volume in kg/m^3
u2=u1 #initial specific internal energy in kJ/kg
#These two independent properties, v2 and u2 , fix state 2.The final temp. is calculated by interplotation using the data for T2=5C and v2,x=0/3928 and u=948.5 kJ/kg. For T2=10C and v2, x=0.5433 and u=1317 kJ/kg
T2=9.1+273.2 #final temp. in K
x2=0.513 #quality in final state
s2=4.644 #final specific entropy in kJ/kg
V1=1 #volume of part of A in m^3
m=V1/v1 #mass flow rate in kg/s
To=20+273.2 #Room temperature in K
Wrev=To*m*(s2-s1) #reversible work in kJ
I=Wrev #irreversibility of the process
print"\n The irreversibility is ",round(I,3),"kJ"