# Ex9.1:pg-336¶

In [5]:
#example 1
#work done by steam

hi=3051.2 #initial specific heat of enthalpy of steam in kJ/kg
si=7.1228 #initial specific entropy of steam in kJ/kg-K
Pe=0.15 #final pressure in MPa
se=si #specific entropy in final state in kJ/kg-K
sf=1.4335 #in kJ/kg-K
sfg=5.7897 #in kJ/kg-K
vi=50.0 #velocity with which steam enters turbine in m/s
ve=200.0 #velocity with which steam leaves the turbine in m/s
xe=(se-sf)/sfg #quality of steam in final state
hf=467.1 #in kJ/kg
hfg=2226.5 #in kJ/kg
he=hf+xe*hfg #final specific heat of enthalpy of steam in kJ/kg
w=hi+vi**2/(2*1000)-he-ve**2/(2*1000) #work of steam for isentropic process in kJ/kg
print"\n hence, work per kilogram of steam for this isentropic process is",round(w,1),"KJ/kg"

 hence, work per kilogram of steam for this isentropic process is 377.5 KJ/kg


## Ex9.2:pg-337¶

In [9]:
#example 2
#exit velocity of steam from nozzle

hi=3051.2 #initial specific heat of enthalpy in kJ/kg
si=7.1228 #initial specific entropy in kJ/kg-K
se=si #final specific entropy
Pe=0.3 #final pressure in MPa
print"from steam table,various properties at final state are"
he=2780.2 #final specific heat of enthalpy in kJ/kg-K
Te=159.1 #final temperature in celsius
vi=30.0 #velocity with which steam enters the nozzle in m/s
ve=((2*(hi-he)+(vi**2/1000))*1000)**0.5 #final velocity of steam with which it exits in m/s
print"\n hence,exit velocity of the steam from the nozzle is",round(ve),"m/s"

from steam table,various properties at final state are

hence,exit velocity of the steam from the nozzle is 737.0 m/s


## Ex9.2E:pg-339¶

In [7]:
#example 2E
#exit velocity of steam from nozzle

hi=1279.1 #initial specific heat of enthalpy in Btu/lbm
si=1.7085 #initial specific entropy in Btu/lbm R
se=si #final specific entropy
Pe=40 #final pressure in lbf/in^2
he=1193.9 #final specific heat of enthalpy in Btu/lbm
Te=314.2 #final temperature in F
vi=100.0 #velocity with which steam enters the nozzle in ft/s
ve=((2*((hi-he)+(vi**2/(32.17*778)))*(32.17*778)))**0.5 #final velocity of steam with which it exits in ft/s
print"\n hence,exit velocity of the steam from the nozzle is",round(ve),"ft/s"

 hence,exit velocity of the steam from the nozzle is 2070.0 ft/s


## Ex9.3:pg-340¶

In [10]:
#example 3
#violation of second law

print"from R-134a tables"
se=1.7148 #specific entropy in final state in kJ/kg-K
si=1.7395 #initial specific entropy in kJ/kg-K
print"therefore,se<si,whereas for this process the second law requires that se>=si.The process described involves a violation of the second law and thus would be impossible."

from R-134a tables
therefore,se<si,whereas for this process the second law requires that se>=si.The process described involves a violation of the second law and thus would be impossible.


## Ex9.4:pg-340¶

In [12]:
#example 4
#calculating required specific work

Cp=1.004 #specific heat of air at constant pressure in kJ/kg-K
Ti=290 #initial temperature in kelvins
Pi=100 #initial pressure in kPa
Pe=1000 #final pressure in kPa
k=1.4
Te=Ti*(Pe/Pi)**((k-1)/k) #final temperature in kelvins
we=Cp*(Ti-Te) #required specific work in kJ/kg
print"\n hence,specific work required is",round(we),"kJ/kg"

 hence,specific work required is -271.0 kJ/kg


## Ex9.4E:pg-341¶

In [34]:
#example 4E
#calculating required specific work

Cp=0.24 #specific heat of air at constant pressure in Btu/lbm R
Ti=520 #initial temperature in R
Pi=14.7 #initial pressure in lbf/in^2
Pe=147 #final pressure in lbf/in^2
k=1.4
Te=Ti*(Pe/Pi)**((k-1)/k) #final temperature in R
we=Cp*(Ti-Te) #required specific work in Btu/lbm
print"\n hence,specific work required is",round(we,2),"Btu/lbm"

 hence,specific work required is -116.15 Btu/lbm


## Ex9.5:pg-342¶

In [15]:
#example 5
#entropy generation

h1=2865.54 #specific heat of enthalpy at state 1 in kJ/kg
h2=83.94 #specific heat of enthalpy  at state 2 in kJ/kg
h3=2725.3 #specific heat of enthalpy at state 3 in kJ?kg
s1=7.3115 #specific entropy at state 1 in kJ/kg-K
s2=0.2966 #specific entropy at state 2 in kJ/kg-K
s3=6.9918 #specific entropy at state 3in kJ/kg-K
m1=2 #mass flow rate at state 1 in kg/s
m2=m1*(h1-h3)/(h3-h2) #mass flow rate at state 2 in kg/s
m3=m1+m2 #mass flow rate at state 3 in kg/s
Sgen=m3*s3-m1*s1-m2*s2 #entropy generation in the process
print"\n hence,entropy generated in this process is ",round(Sgen,3),"kW/K"

 hence,entropy generated in this process is  0.072 kW/K


## Ex9.6:pg-344¶

In [18]:
#example 6
#work required to fill the tank
import math
T1=17+273 #initial temperature of tank in Kelvins
sT1=6.83521 #specific entropy in kJ/kg-K
R=0.287 #gas constant in kJ/kg-K
P1=100 #initial pressure in kPa
P2=1000 #final pressure in kPa
sT2=sT1+R*math.log(P2/P1) #specific entropy at temperature T2 in kJ/kg-K
T2=555.7 #from interplotation
V1=0.04 #volume of tank in m^3
V2=V1 #final volume is equal to initial volume
m1=P1*V1/(R*T1) #initial mass of air in tank in kg
m2=P2*V2/(R*T2) #final mass of air in tank in kg
Min=m2-m1 #in kg
u1=207.19 #initial specific heat of enthalpy in kJ/kg
u2=401.49 #final specific heat of enthalpy in kJ/kg
hin=290.43 #in kJ/kg
W12=Min*hin+m1*u1-m2*u2 #work required to fill the tank in kJ
print"\n hence,the total amount of work required to fill the tank is",round(W12,1),"kJ"

 hence,the total amount of work required to fill the tank is -31.9 kJ


## Ex9.7:pg-347¶

In [21]:
#example 7
#work required to pump water isentropically

P1=100 #initial pressure in kPa
P2=5000 #final pressure in kPa
v=0.001004 #specific volume in m^3/kg
w=v*(P2-P1) #work required to pump water isentropically
print"\n hence,work required to pump water isentropically  is ",round(w,2),"kJ/kg"

 hence,work required to pump water isentropically  is  4.92 kJ/kg


## Ex9.8:pg-348¶

In [24]:
#example 8
#Velocity in exit flow

print"From Steam Tables, for liquid water at 20 C"
vf=0.001002 #in m^3/kg
v=vf
Pi=300 #Line pressure in kPa
Po=100 #in kPa
Ve=(2*v*(Pi-Po)*1000)**0.5 #velocity in the exit flow
print" \n Hence, an ideal nozzle can generate upto ",round(Ve),"m/s"

From Steam Tables, for liquid water at 20 C

Hence, an ideal nozzle can generate upto  20.0 m/s


## Ex9.9:pg-351¶

In [29]:
#example 9
#Rate of Entropy Generation

print"From R-410a tables,we get"
hi=280.6 #in kJ/kg
he=307.8 #in kJ/kg
si=1.0272 #in kJ/kg
se=1.0140 #in kJ/kg
m=0.08 #flow rate of refrigerant in kg/s
P=3 #electrical power input in kW
Qcv=m*(he-hi)-P #in kW
To=30 #in Celsius
Sgen=m*(se-si)-Qcv/(To+273.2) #rate of entropy generation
print"\n Hence,the rate of entropy generation for this process is",round(Sgen,5),"kW/K"

From R-410a tables,we get

Hence,the rate of entropy generation for this process is 0.00166 kW/K


## Ex9.10:pg-353¶

In [32]:
#example 10
#turbine efficiency

hi=3051.2 #initial specific heat of enthalpy in kJ/kg
si=7.1228 #initial specific entropy in kJ/kg-K
sf=0.7548 #in kJ/kg-K
sfg=7.2536 #in kJ/kg-K
ses=si #final specific entropy is same as the initial
xes=(si-sf)/sfg #quality of steam when it leaves the turbine
hf=225.9 #in kJ/kg
hfg=2373.1 #in kJ/kg
hes=hf+xes*hfg #final specific heat of enthalpy in kJ/kg
ws=hi-hes #work output of turbine calculated ideally  in kJ/kg
wa=600 #actual work output of turbine in kJ/kg
nturbine=wa/ws #efiiciency of turbine
print"\n hence,efficiency of the turbine is",round(nturbine*100,1),"%"

 hence,efficiency of the turbine is 80.9 %


## Ex9.11:pg-355¶

In [36]:
#example 11
#turbine inlet pressure
import math
hi=1757.3 #initial specific heat of enthalpy of air in  kJ/kg
si=8.6905 #initial specifc entropy of airin kJ/kg-K
he=855.3 #final specific heat of enthalpy of air in kJ/kg
w=hi-he #actual work done by turbine in kJ/kg
n=0.85 #efficiency of turbine
ws=w/n #ideal work done by turbine in kJ/kg
hes=hi-ws #from first law of isentropic process
Tes=683.7 #final temperature in kelvins from air tables
ses=7.7148 #in kJ/kg-K
R=0.287 #gas constant in kJ/kg-K
Pi=100/math.e**((si-ses)/-R) #turbine inlet pressure in kPa
print"\n hence,turbine inlet pressure is",round(Pi),"kPa"

 hence,turbine inlet pressure is 2995.0 kPa


## Ex9.12:pg-357¶

In [44]:
#example 12
#required work input

Pe=150.0 #final pressure of air in kPa
Pi=100.0 #initial presure of air in kPa
k=1.4
Ti=300.0 #initial temperature of air in kelvis
Tes=Ti*(Pe/Pi)**((k-1)/k) #from second law
ws=1.004*(Ti-Tes) #from first law of isentropic process
n=0.7 #efficiency of automotive supercharger
w=ws/n #real work input in kJ/kg
Te=Ti-w/1.004 #temperature at supercharger exit in K
print"\n hence,required work input is ",round(w),"kJ/kg"
print"\n and exit temperature is ",round(Te,1),"K"

 hence,required work input is  -53.0 kJ/kg

and exit temperature is  352.6 K