# Chapter8:ENTROPY¶

## Ex8.1:pg-290¶

In :
#example 1
#coefficient of performance of refrigerator

Th=60 #temperature at which heat is rejected from R-134a
Tl=0 #temperature at which heat is absorbed into the R-134a
s1=1.7262 #specific entropy at 0 Celsius
s2=s1 #process of state change from 1-2 is isentropic
s3=1.2857 #specific entropy at 60 celsius
s4=s3 #process of state change from 3-4 is isentropic
print"if Pressure is 1400 kPa,then s=1.7360 kJ/kg-K and if P=1600 kPa,then s=1.7135 kJ/kg-K.Therefore"
P2=1400+(1600-1400)*(1.7262-1.736)/(1.7135-1.736) #pressure after compression in kPa
B=(Th+273.15)/(Th-Tl) #coefficient of performance of refrigerator
print" \n hence,pressure after compression is ",round(P2,1),"kPa"
print"\n and coefficient of performance of refrigerator is",round(B,2)

if Pressure is 1400 kPa,then s=1.7360 kJ/kg-K and if P=1600 kPa,then s=1.7135 kJ/kg-K.Therefore

hence,pressure after compression is  1487.1 kPa

and coefficient of performance of refrigerator is 5.55


## Ex8.2:pg-290¶

In :
#example 2
#heat transfer in a given process

u1=87.94 #specific internal energy of R-12 at state 1 in kJ/kg
u2=276.44 #specific internal energy of R-12 at state 2 in kJ/kg
s1=0.3357 #specific entropy at state 1 in kJ/kg-K
s2=1.2108 #specific entropy at state 2 in kJ/kg-K
V=0.001 #volume of saturated liquid in m^3
v1=0.000923 #specific volume in m^3/kg
m=V/v1 #mass of saturated liquid in kg
T=20 #temperature of liquid in celsius
Q12=m*(T+273.15)*(s2-s1) #heat transfer in kJ to accomplish the process
W12=m*(u1-u2)+Q12 #work required to accomplish the process
print" \n hence,work required to accomplish the process is",round(W12,1),"KJ"
print" \n and heat transfer is",round(Q12,1),"KJ"


hence,work required to accomplish the process is 73.7 KJ

and heat transfer is 277.9 KJ


## Ex8.3:pg-293¶

In :
#example 3
#entropy change
import math
C=4.184 # specific heat of water in kJ/kg-K
T1=20 #initial temperature of water in celsius
T2=90 #final temperature of water in celsius
dS1=C*math.log((T2+273.2)/(T1+273.2)) #change in entropy in kJ/kg-K
dS2=1.1925-0.2966 #in kJ/kg-K using steam tables
print"\n hence,change in entropy assuming constant specific heat is",round(dS1,4),"KJ/kg-k"
print"\n using steam table is",round(dS2,4),"KJ/kg-k"

 hence,change in entropy assuming constant specific heat is 0.8958 KJ/kg-k

using steam table is 0.8959 KJ/kg-k


## Ex8.4:pg-295¶

In :
#example 4
#entropy change with different assumptions
import math
T1=300 #initial temperature in kelvins
T2=1500 #final temperature in kelvins
P1=200.0 #initial pressure in kPa
P2=150.0 #final pressure in kPa
R=0.2598 # in kJ/kg-K
Cp=0.922 #specific heat in kJ/kg-K at constant pressure
dsT2=8.0649 #in kJ/kg-K
dsT1=6.4168 #in kJ/kg-K
dS1=dsT2-dsT1-R*math.log(P2/P1) #entropy change calculated using ideal gas tables
dS3=Cp*math.log(T2/T1)-R*math.log(P2/P1) #entropy change assuming constant specific heat in kJ/kg-K
dS4=1.0767*math.log(T2/T1)+0.0747 #entropy change assuming specific heat is constant at its value at 990K
print"\n hence,change in entropy using ideal gas tables is",round(dS1,4),"KJ/kg-k"
print"\n hence,change in entropy using the value of specific heat at 300K is",round(dS3,4),"KJ/kg-k"
print"\n hence,change in entropy assuming specific heat is constant at its value at 900K is ",round(dS4,4),"KJ/kg-k"

 hence,change in entropy using ideal gas tables is 1.7228 KJ/kg-k

hence,change in entropy using the value of specific heat at 300K is 1.5586 KJ/kg-k

hence,change in entropy assuming specific heat is constant at its value at 900K is  1.8076 KJ/kg-k


## Ex8.5:pg-296¶

In :
#example 5
#entropy change
import math
Cp=1.004 #specific heat at constant pressure in kJ/kg-K
R=0.287 #gas constant in kJ/kg-K
P1=400.0 #initial pressure in kPa
P2=300.0 #final pressure in kPa
T1=300 #initial temperature in K
T2=600 #final temperature in K
dS1=Cp*math.log(T2/T1)-R*math.log(P2/P1) #entropy change assuming constant specific heat
s1=6.8693 #specific entropy at T1
s2=7.5764 #specific entropy at T2
dS2=s2-s1-R*math.log(P2/P1) #entropy change assuming variable specific heat
print"\n hence,entropy change assuming constant specific heat is",round(dS1,4),"KJ/kg-k"
print"\n and assuming variable specific heat is",round(dS2,4),"KJ/kg-k"

 hence,entropy change assuming constant specific heat is 0.7785 KJ/kg-k

and assuming variable specific heat is 0.7897 KJ/kg-k


## Ex8.6:pg-297¶

In :
#example 6
#work done by air
import math
T1=600 #initial temperature of air in K
P1=400.0 #intial pressure of air in kPa
P2=150.0 #final pressure in kPa
u1=435.10 #specific internal energy at temperature T1 in kJ/kg
sT1=7.5764 #specific entropy at temperature T1 in kJ/kg-K
R=0.287 #gas constant in kJ/kg-K
ds=0
sT2=ds+sT1+R*math.log(P2/P1) #specific entropy at temperature T2 in kJ/kg-K
print"we know the values of s and P for state 2.So,in order to fully determine the state,we will use steam table"
T2=457 #final temperature in K
u2=328.14 #specific internal energy at temperature T2 in kJ/kg
w=u1-u2 #work done by air in kJ/kg
print"\n hence,work done by air is",round(w,4),"KJ/kg"

we know the values of s and P for state 2.So,in order to fully determine the state,we will use steam table

hence,work done by air is 106.96 KJ/kg


## Ex8.7:pg-300¶

In :
#example 7
#work and heat transfer

P2=500  #final pressure in cylinder in kPa
P1=100 #initial pressure in cylinder in kPa
T1=20+273.2 #initial temperature inside cylinder in Kelvins
n=1.3
T2=(T1)*(P2/P1)**((n-1)/n) #final temperature inside cylinder in K
R=0.2968 #gas constant in kJ/kg-K
w12=R*(T2-T1)/(1-n) #work in kJ/kg
Cvo=0.745 #specific heat at constant volume in kJ/kg-K
q12=Cvo*(T2-T1)+w12 #heat transfer in kJ/kg
print" \n hence,work done is",round(w12,2),"KJ/kg"
print"\n and heat transfer are",round(q12,1),"KJ/kg"


hence,work done is -130.47 KJ/kg

and heat transfer are -32.2 KJ/kg


## Ex8.8:pg-308¶

In :
#example 3
#calculating increase in entropy

m=1 #mass of saturated water vapour
sfg=6.0480 #in kJ/K
T=25 #temperature of surrounding air in celsius
dScm=-m*sfg #change in entropy of control mass in kJ/K
hfg=2257.0 #in kJ/kg
Qtosurroundings=m*hfg #heat transferred to surroundings in kJ
dSsurroundings=Qtosurroundings/(T+273.15) #in kJ/K
dSnet=dScm+dSsurroundings #net increase in entropy in kJ/K
print" hence,net increase in entropy of water plus surroundings is ",round(dSnet,3),"KJ/k"

 hence,net increase in entropy of water plus surroundings is  1.522 KJ/k


## Ex8.9:pg-309¶

In :
#example 9
#entropy generation

Qout=1.0 #value of heat flux generated by 1kW of electric power
T=600.0 #temperature of hot wire surface in K
Sgen=Qout/T #entropy generation in kW/K
print"\n hence,entropy generation is",round(Sgen,5),"KW/k"

 hence,entropy generation is 0.00167 KW/k


## Ex8.10:pg-310¶

In :
#example 10
#Determiining the entropy generated

B=4.0#COP of air conditioner
W=10.0 #power input of air conditioner in kW
Qh=B*W #in kW
Ql=Qh-W #in kW
Thigh=323 #in Kelvin
Tlow=263 #in Kelvin
SgenHP=(Qh*1000/Thigh)-(Ql*1000/Tlow) #in W/K
Tl=281 # in K
Th=294 #in K
SgenCV1=Ql*1000/Tlow-Ql*1000/Tl #in W/K
SgenCV2=Qh*1000/Th-Qh*1000/Thigh #in W/K
SgenTOT=SgenCV1+SgenCV2+SgenHP #in W/K
print" \n Hence,Total entropy generated is",round(SgenTOT,1),"W/k"


Hence,Total entropy generated is 29.3 W/k