# Chapter 5:The Second Law of Thermodynamics¶

## Ex5.4:PG-112¶

In [3]:
# initialization of variables
Th=200+273.0 # higher temperture in kelvin
Tl=20+273.0 # lower temperture in kelvin
Wdot=15 # output of engine in kW

ef=1-(Tl/Th) # carnot efficiency

Qhdot=Wdot/ef # heat supplied by reservoir
print " The heat suppled by higher temperature reservoir is",round(Qhdot,2),"kW \n "
# using forst law
Qldot=Qhdot-Wdot # heat rejected to reservoir
print " The heat suppled by lower temperature reservoir is",round(Qldot,2),"kW"

 The heat suppled by higher temperature reservoir is 39.42 kW

The heat suppled by lower temperature reservoir is 24.42 kW


## Ex5.5:PG-113¶

In [7]:
# initialization of variables
TL1=-5+273.0 # lower temperature in kelvin for first situation
TH=20+273.0 # higher temperature in kelvin
TL2=-25+273.0 #lower temperature in kelvin for second situation

#solution

COP1=TL1/(TH-TL1) # carnot refrigerator COP for first situation
# Let Heat be 100 kJ
QL=100.0 # assumption
W1=QL/COP1 # work done for situation 1

# for situation 2
COP2=TL2/(TH-TL2) # COP carnot for second situation
W2=QL/COP2 # work done

Per=(W2-W1)*100/W1 # percentage increase in work done
#result
print" The perccentage increase in work is ",round(Per,1),"%"

 The perccentage increase in work is  94.5 %


## Ex5.6:PG-117¶

In [15]:
import math
# initialization of variables
T1=20+273 # initial temperature in kelvin
P=200 # pressure in kPa
V=2 # volume in m^3
R=0.287 # gas constant for air
W=720 # work done on air in kJ
Cv=0.717 # specific heat at constant volume for air

#solution
m=(P*V)/(R*T1) # mass of air

T2=T1+(W/(m*Cv))# final temperature in kelvin

delS=m*Cv*math.log(T2/T1) # ENROPY CHANGE FOR CONSTANT VOLUME PROCESS
print " The Entropy increase is",round(delS,3),"kJ/K "

 The Entropy increase is 1.851 kJ/K


## Ex5.7:PG-118¶

In [18]:
# initialization of variables
T1=350+273 # initial temperature in kelvin
P1=1200.0 # initial pressure in kPa
P2=140 # final pressure in kPa
k=1.4 # polytopic index for air
Cv=0.717 # specific heat at  constant volume for air
#solution
T2=T1*((P2/P1)**((k-1)/k))  # reversible adiabatic process relation

w=-Cv*(T2-T1) # work done by gases in reversible adiabatic process
print" The work done by gases is",round(w),"kJ/kg"

 The work done by gases is 205.0 kJ/kg


## Ex5.8:PG-120¶

In [22]:
import math
# initialization of variables
T1=20+273.0 # initial temperature in kelvin
P1=200.0 # pressure in kPa
V=2 #volume in m^3
R=0.287 # gas constant for air
W=-720 # negative as work is done on air in kJ

#solution

m=(P1*V)/(R*T1)# mass of air

u1=209.1 #specific internal energy of air at 293K and 200 kPa from table E.1
s1=1.678 # by interpolation from table E.1
# change  in internal energy= work done
u2=-(W/m)+u1 # final internal energy
T2=501.2# final temperature interpolated from table E.1 corresponding to value of u2
s2=2.222 # value of s from table E.3 by interpolating from corresponding to value of u2

P2=P1*(T2/T1) # final pressure in kPa

delS=m*(s2-s1-R*math.log(P2/P1))# entropy change
# result
print " The Entropy increase is",round(delS,3),"kJ/K "

 The Entropy increase is 1.855 kJ/K


## Ex5.9:PG-120¶

In [33]:
# initialization of variables
T1=350+273.0 # initial temperature in kelvin
P1=1200.0 # initial pressure in kPa
P2=140.0 # final pressure in kPa
k=1.4 # polytopic index for air

#solution
# The values are taken from table E.1
Pr660=23.13# relative pressure @ 660K
Pr620=18.36# relative pressure @ 620K
Pr1=((Pr660-Pr620)*3/40)+Pr620 # relative pressure by interpolation

Pr2=Pr1*(P2/P1) # relative pressure at state 2

Pr340=2.149 # relative pressure @ 340K
Pr380=3.176 # relative pressure @ 380K
T2=((Pr2-Pr340)/(Pr380-Pr340))*40+340 # interpolating final temperature from table E.1

# now interpolating u1 AND u2 from table E.1
u620=451.0# specific internal energy @ 620k
u660=481.0# specific internal energy @ 660k
u1=(u660-u620)*(3/40.0)+u620 # initial internal energy

u380=271.7 #specific internal energy @ 380k
u340=242.8 #specific internal energy @ 340k
u2=((Pr2-Pr340)/(Pr380-Pr340))*(u380-u340)+u340 # final internal energy

w=u2-u1 # work= change in internal energy
print " The work done by gas is",int(w),"kJ/kg"
# The answer is slightly different as values are approximated in textbook

 The work done by gas is -209 kJ/kg


## Ex5.10:PG-123¶

In [37]:
# initialization of variables
T1=300+273.0 # initial temperature in kelvin
P1=600 # initial pressure in kPa
P2=40 # final pressure in kPa

#solution
#please refer to steam table for values
v1=0.4344 # specific volume from steam table @ 573k and 600 kPa
v2=v1 # rigid container
u1=2801.0 # specific internal energy from steam table @ 573k and 600 kPa
s1=7.372 # specific entropy @ 600 kPa and 573 K

vg2=0.4625 # specific volume of saturated vapour @ 40 kPa and 573 K
vf2=0.0011 # specific volume of saturated liquid @ 40 kPa and 573 K
sf2=1.777 # specific entropy of saturated liquid @ 40 kPa and 573 K
sg2=5.1197 # specific entropy of saturated vapour @ 40 kPa and 573 K
x=(v2-vf2)/(vg2-vf2)# quality of steam using pure substance relation

s2=sf2+x*sg2 # overall specific enthalpy at quality 'x'
delS=s2-s1 # entropy change
print" The entropy change  is",round(delS,3),"kJ/kg.K \n "

#heat transfer
uf2=604.3 #specific internal energy of saturated liquid @ 40 kPa and 573 K
ug2=1949.3 #specific internal energy of saturated vapour @ 40 kPa and 573 K
u2=uf2+x*ug2 #specific internal energy @ quality x
q=u2-u1 # heat transfer in kJ/kg from first law as W=0
print " The heat transfer is",int(q),"kJ/kg"
# result
# the answers are approximated in textbook but here they are precise thus minute difference is there

 The entropy change  is -0.787 kJ/kg.K

The heat transfer is -366 kJ/kg


## Ex5.11:PG-126¶

In [49]:
import math
# initialization of variables
v1=0.5 # assumed as air is filled in half of the tank
v2=1.0 # final volume when partition is removed
R=0.287 # gas contant for air
#solution
q=0 # heat transfer is zero
w=0 # work done is zero
# temperatue is constant as no change in internal energy by first law
dels=R*math.log(v2/v1)# change in entropy when temperature is constant
print "The change in specific entropy is",round(dels,3),"kJ/kg.K"

The change in specific entropy is 0.199 kJ/kg.K


## Ex5.12:PG-127¶

In [53]:
# initialization of variables
T1=400+273.0 # initial temperature in kelvin
P=600 # pressure in kPa
Tsurr=25+273.0 # surrounding temperature in K
m=2 # mass of steam in kg

#solution
#please refer to steam table for values
s1=7.708 # specific entropy of steam @ 400 degree celsius and 0.6 MPa
s2=1.9316# specific enropy of condensed water @ 25 degree celsius and 0.6 MPa
delSsys=m*(s2-s1) # entropy change in system i.e of steam

h1=3270 # specific enthalpy of steam @ 400 degree celsius and 0.6 MPa
h2=670.6#specific enropy of condensed water @ 25 degree celsius and 0.6 MPa

Q=m*(h1-h2)# heat transfer at constant pressure
delSsurr=Q/Tsurr # entropy change in surroundings

sigma=delSsys+delSsurr # net entropy change

print "The net entropy production is",round(sigma,1),"kJ/K"

The net entropy production is 5.9 kJ/K


## Ex5.13:PG-130¶

In [56]:
# initialization of variables
T1=600+273 # initial temperature in kelvin
P1=2 # initial pressure in MPa
P2=10 # final pressure in kPa
mdot=2 # mass flow rate in kg/s

#solution
#please refer to steam table for values
h1=3690 # specific enthalpy in kJ/kg @ 2MPa and 600 degree celsius
s1=7.702 #specific entropy in kJ/kg.K @ 2MPa and 600 degree celsius
s2=s1 # Reversible adiabatic process thus entropy is constant
sf2=0.6491 #specific entropy of saturated liquid from steam table @ 10 kPa
sg2=8.151 #specific entropy of saturated vapour from steam table @ 10 kPa

x2=(s2-sf2)/(sg2-sf2) # quality of steam at turbine exit

h2f=191.8 #specific enthalpy of saturated liquid from steam table @ 10 kPa
h2g=2584.8 #specific enthalpy of saturated vapour from steam table @ 10 kPa
h2=h2f+x2*(h2g-h2f) # specific enthalpy @ quality 'x'

WdotT=mdot*(h1-h2)# from work done in adiabatic process
# result
print " The maximum power output is",int(WdotT),"kJ/s"

 The maximum power output is 2496 kJ/s


## Ex5.14:PG-130¶

In [59]:
# initialization of variables

T1=600+273 # initial temperature in kelvin
P1=2 # initial pressure in MPa
P2=10 # final pressure in kPa
mdot=2 # mass flow rate in kg/s
EffT=0.8 # efficiency of turbine
WdotT=2496 # theoritical power of turbine in kW

#solution
Wdota=EffT*WdotT # actual power output of turbine
h1=3690 # specific enthalpy @ 2MPa and 600 degree celsius
h2=h1-(Wdota/mdot) # final enthalpy from first law of thermodynamics

T2=((h2-2688)/(2783-2688))*(150-100)+100 # by interpolating from steam table @ P2= 10 kPa, h2=2770
s2=8.46 # final specific entropy by interpolation from steam table

print "The temperature by interpolation is",round(T2)," degree celsius \n"
print "The final entropy by interpolation is",round(s2,2),"kJ/kg.K"
# The temperature and entropy are found by interpolation from steam table and cannot be shown here.

The temperature by interpolation is 102.0  degree celsius

The final entropy by interpolation is 8.46 kJ/kg.K


## Ex5.15:PG-131¶

In [60]:
# initialization of variables

T2=250.0 # temperature of steam in degree celsius
mdot2=0.5 # mass flow rate of steam in kg/s
T1=45 # temperature of water in degree celsius
mdot1=4 # mass flow rate of water in kg/s
P=600.0 # pressure in kPa

mdot3=mdot1+mdot2 # by mass balance

h2=2957 # specific enthalpy in kJ/kg of steam @ 600 Kpa from steam table
h1=188.4 # specific enthalpy in kJ/kg  of water @ 600 Kpa from steam table

h3=(mdot1*h1+mdot2*h2)/mdot3 # specific enthalpy in kJ/kg at exit

# by interpolation from saturated steam table
T3=(h3-461.3)*10/(503.7-461.3)+110 # temperature of mixture

sf3=1.508 # entropy of saturated liquid in kJ/kg.K at 600Kpa and T3 temperature from steam table
s3=sf3  # by interpolating sf
s2=7.182 # entropy of superheated steam in kJ/kg.K @ 600Kpa from steam table
s1=0.639 # entropy of entering water in kJ/kg.K at T= 45 degree celsius

sigmaprod=mdot3*s3-mdot2*s2-mdot1*s1
# result
print "The rate of entropy production is",round(sigmaprod,3),"kW/K "

The rate of entropy production is 0.639 kW/K