# Chapter 10 Properties Of Steam¶

## Example 1 Page No:183¶

In [2]:
#Input data
mw=15      #Water steam
ms=185     #Dry steam

#Calculation
x=((ms)/(ms+mw))*100 #Dryness fuction of steam in %

#Output
print("Dryness fuction of steam=",x,"%")

Dryness fuction of steam= 92.5 %


## Example 2 Page No:183¶

In [3]:
#Input data
sps=150 #saturation pressure of the steam in degree celsius

#Output
P=4.76 #From steam table
print("saturation pressure=",P,"bar")

saturation pressure= 4.76 bar


## Example 3 Page No:184¶

In [4]:
#Input data
P1=28      #Absolute pressure in bar
P2=5.5     #Absolute pressure in MPa
P3=77      #Absolute pressure in mm of Hg

#Calcutation
ts1=230.05 #Saturation temperature in degree celsius
ts2=269.93 #Saturation temperature in degree celsius
ts3=45.83  #Saturation temperature in degree celsius

#Output
print("Saturation temperature= ",ts1,"degree celsius")
print("Saturation temperature= ",ts2,"degree celsius")
print("Saturation temperature= ",ts3,"degree celsius")

Saturation temperature=  230.05 degree celsius
Saturation temperature=  269.93 degree celsius
Saturation temperature=  45.83 degree celsius


## Example 4 Page No:185¶

In [12]:
#Input data
P=15              #Absolute pressure in bar
#From steam table (pressure basis at 15 bar)
ts=198.3          #In degree celsius
hf=844.7          #In KJ/Kg
hfg=1945.2        #In KJ/Kg
hg=2789.9         #In KJ/Kg
tsup=300          #In degree celsius
x=0.8
Cps=2.3
hg=2789.9

#Calculation
h1=hf+x*hfg        #Enthalpy of wet steam in KJ/KG
h=hg               #Enthalpy of dry and saturated steam in KJ/KG
h2=hg+Cps*(tsup-ts)#Enthalpy of superheated steam in KJ/KG

#Output
print("Enthalpy of wet steam= ",h1,"KJ/Kg")
print("Enthalpy of dry and saturated steam= ",h,"KJ/KG")
print("Enthalpy of superheated steam= ",h2,"KJ/Kg")

Enthalpy of wet steam=  2400.86 KJ/Kg
Enthalpy of dry and saturated steam=  2789.9 KJ/KG
Enthalpy of superheated steam=  3023.81 KJ/Kg


## Example 5 Page No:186¶

In [1]:
#Input data
ti=30         #Temperature in degree celsius
m=2           #Water in Kg
pf=8          #Steam at 8 bar
x=0.9         #Water to dry
tb=30
#From steam table at 30 degree celsius
hf=125.7
#h1=hf initial enthalpy of water
#From steam table at 8 bar
ts=170.4      #In degree celsius
hf1=720.9     #In KJ/KG
hfg=2046.6    #In KJ/KG
hg=2767.5     #In KJ/KG

#Calculation
h=hf1+(x*hfg) #Final Enthalpy of the steam in KJ/Kg
Qha=m*(h-hf)  #Quantity of the heat in KJ/Kg              #Calculation mistake m is not multiplied by (h-hf) in book

#Output
print("Final Enthalpy of the steam= ",h,"KJ/Kg")
print("Quantity of the heat= ",round(Qha,1),"KJ/Kg")

Final Enthalpy of the steam=  2562.84 KJ/Kg
Quantity of the heat=  4874.3 KJ/Kg


## Example 6 Page No:186¶

In [5]:
#Input data
IT=25               #Initial temperature
m=5                 #Heat required to generate steam in kg
pf=10               #Final pressure in bar
tsup=250            #Water temperature
#From steam table (temp basis)at 25degree celsius
#and at 10 bar(pressure basis)
hf=104.8            #In KJ/KG
h1=104.8            #In KJ/KG
ts=179.9            #In degree celsius
hf1=792.6           #In KJ/KG
hfg=2013.6          #In KJ/KG
hg=2776.2           #In KJ/KG
Cps=2.1

#Calculation
h=hg+Cps*(tsup-ts) #Enthalpy of superheated steam in KJ/Kg
H=m*(h-h1)         #Quantity of heat added in KJ/Kg

#Output
print("Enthalpy of superheated steam= ",h,"KJ/Kg")

Enthalpy of superheated steam=  2923.41 KJ/Kg
Quantity of heat added=  14093 KJ/Kg


## Example 7 Page No:188¶

In [6]:
#Input data
P=15               #Absolute pressure in bar
#From steam table (pressure basis at 15 bar)
ts=198.3+273       #In degree celsius
vg=0.1317          #In m**3/Kg
vf=0.001154        #In m**3/Kg
x=0.8
Tsup=300+273       #Degree celsius

#Calculation
v=(1-x)*vf+x*vg    #Volume of wet steam in m**3/Kg
vg=0.1317          #Dry and saturated steam in m**3/Kg
vsup=vg*(Tsup/ts)  #Volume of superheated steam m**3/Kg

#Output
print("Volume of wet steam= ",round(v,4),"m**3/Kg")
print("Dry and Saturated Steam= ",vg,"m**3/Kg")
print("volume of superheated steam= ",round(vsup,4),"m**3/Kg")


Volume of wet steam=  0.1056 m**3/Kg
Dry and Saturated Steam=  0.1317 m**3/Kg
volume of superheated steam=  0.1601 m**3/Kg


## Example 8 Page No:188¶

In [21]:
#Input data
P=25           #Absolute pressure
ts=223.9       #Volume
#Frome steam table (pressure basis at 25 bar)
vf=0.001197    #In m**3/Kg
vg=0.0799      #In m**3/Kg
v=8            #In m**3/Kg

#Calculation
m=v/vg         #Mass of steam in Kg

#Output
print("Mass of steam= ",round(m,3),"Kg")

Mass of steam=  100.125 Kg


## Example 9 Page No:190¶

In [12]:
#Input data
P=12*10**5             #Absolute pressure
#From steam table (pressure basis at 12 bar)
ts=188+273             #In degree celsius
vf=0.001139            #In m**3/Kg
vg=0.1632              #In m**3/Kg
hf=798.4               #In KJ/Kg
hfg=1984.3             #In KJ/Kg
hg=2782.7              #In KJ/Kg
x=0.94
Cps=2.3
tsup=350+273           #In degree celsius

#Calcuation
h=hf+x*hfg             #Enthalpy of wet steam in KJ/Kg
v=(1-x)*vf+x*vg        #Volume of wet steam m**3/Kg
u=h-((P*v)/10**3)      #Internal Energy in KJ/Kg
hg=2782.7              #Enthalpy of dry & saturated steam in KJ/Kg
v1=vg                  #Volume of dry & saturated steam  m**3/Kg
u1=hg-((P*vg)/10**3)   #Internal Energy in KJ/Kg
h1=hg+Cps*(tsup-ts)    #Enthalpy of superheated steam in KJ/Kg
vsup=vg*(tsup/ts)      #Volume of superheated steam in m**3/Kg
u2=h1-((P*v)/10**3)    #Internal Energy in KJ/Kg

#Output
print("Enthalpy of wet steam= ",h,"KJ/Kg")
print("Volume of wet steam= ",round(v,5),"m**3/Kg")
print("Internal Energy= ",round(u,2),"KJ/Kg")
print("Enthalpy of dry & saturated steam= ",hg,"KJ/Kg")
print("Volume of dry & saturated steam= ",v1,"m**3/Kg")
print("Internal Energy= ",u1,"KJ/Kg")
print("Enthalpy of superheated steam= ",round(h1,1),"KJ/Kg")
print("Volume of superheated steam= ",round(vsup,3),"m**3/Kg")
print("Internal Energy= ",round(u2,1),"KJ/Kg")

Enthalpy of wet steam=  2663.642 KJ/Kg
Volume of wet steam=  0.15348 m**3/Kg
Internal Energy=  2479.47 KJ/Kg
Enthalpy of dry & saturated steam=  2782.7 KJ/Kg
Volume of dry & saturated steam=  0.1632 m**3/Kg
Internal Energy=  2586.8599999999997 KJ/Kg
Enthalpy of superheated steam=  3155.3 KJ/Kg
Volume of superheated steam=  0.221 m**3/Kg
Internal Energy=  2971.1 KJ/Kg


## Example 10 Page No:191¶

In [23]:
#Input data
P1=10*10**5                  #Pressure of steam in bar
tsup1=300+273                #Temperature of steam n degree celsius
P2=1.4*10**5                 #Internal energy of steam
x2=0.8                       #Dryness fraction
Cps=2.3
#from steam table properties of saturated steam (temp basis)
#at 25 degree celsius and at 10 bar(pressure basis)
ts1=179.9+273
vf=0.001127                  #In m**3/Kg
vg=0.1943                    #In m**3/Kg
hf=762.6                     #In KJ/Kg
hfg=2013.6                   #In KJ/Kg
hg1=2776.2                   #In KJ/Kg
#at 1.4 bar
ts=109.3                     #In degree celsius
vf1=0.001051                 #In m**3/Kg
vg1=1.2363                   #In m**3/Kg
hf1=458.4                    #In KJ/Kg
hfg1=2231.9                  #In KJ/Kg
hg=2690.3                    #In KJ/Kg

#calculation
h1=hg1+Cps*(tsup1-ts1)       #Enthalpy of superheated steam in KJ/Kg
v1=vg*(tsup1/ts1)            #Volume of superheated steam in m**3/Kg
u1=h1-((P1*v1)/10**3)        #Internal energy in KJ/Kg
h2=hf1+x2*hfg1               #Enthalpy of wet steam in KJ/Kg
Vwet=(1-x2)*vf1+x2*vg1       #Volume of wet steam in m**3/Kg
u2=h2-((P2*Vwet)/10**3)      #Internal energy in KJ/Kg
DeltaU=u1-u2                 #Change of Internal energy in KJ/Kg

#Output
print("Enthalpy of superheated steam= ",h1,"KJ/Kg")
print("Volume of superheated steam= ",round(v1,4),"m**3/Kg")
print("Internal energy= ",round(u1,1),"KJ/Kg")
print("Enthalpy of wet steam= ",h2,"KJ/Kg")
print("Volume of wet steam= ",round(Vwet,5),"m**3/Kg")
print("Internal energy= ",round(u2,1),"KJ/Kg")
print("Change of Internal energy= ",round(DeltaU,1),"KJ/Kg")

Enthalpy of superheated steam=  3052.43 KJ/Kg
Volume of superheated steam=  0.2458 m**3/Kg
Internal energy=  2806.6 KJ/Kg
Enthalpy of wet steam=  2243.92 KJ/Kg
Volume of wet steam=  0.98925 m**3/Kg
Internal energy=  2105.4 KJ/Kg
Change of Internal energy=  701.2 KJ/Kg


## Example 11 Page No:193¶

In [26]:
#Input data
import math
P=15                         #Absolute pressure
#From steam table (pressure basis at 15 bar)
ts=198.3+273                 #In degree celsius
Sf=2.3145                    #In KJ/KgK
Sfg=4.1261                   #In KJ/KgK
Sg=6.4406                    #In KJ/KgK
tsup=300+273
Cps=2.3
x=0.8

#calculation
S=Sf+x*Sfg                    #Entropy of wet steam in KJ/Kg
S1=Sg                         #Entropy of superheated steam in KJ/Kg
S2=Sg+Cps*(math.log(tsup/ts)) #Entropy of superheated steam in  KJ/Kg

#Output
print("Entropy of wet steam",round(S,3)," KJ/Kg")
print("Entropy of dry and saturated steam",S1," KJ/Kg")
print("Entropy of superheated steam",round(S2,2)," KJ/Kg")

Entropy of wet steam 5.615  KJ/Kg
Entropy of dry and saturated steam 6.4406  KJ/Kg
Entropy of superheated steam 6.89  KJ/Kg


## Example 12 Page No:194¶

In [31]:
#Input data
#Input data
import math
m=1.5                  #Entropy of the steam
P=10*10**5             #Absolute pressure in bar
#From steam table properties of saturated steam
#(pressure basis)at 10 bar
ts=179.9+273           #Indegree celsius
vf=0.001127            #In m**3/Kg
vg=0.1943              #In m**3/Kg
hf=762.6               #In KJ/Kg
hfg=2013.6             #In KJ/Kg
hg=2776.2              #In KJ/Kg
Sf=2.1382              #In KJ/KgK
Sfg=4.4446             #In KJ/KgK
Sg=6.5828              #In KJ/Kg
Cps=2.3
tsup=250+273

#Calculation
#(1)Enthalpy of dry and saturated steam

h=hg                   #Enthalpy of dry and saturated steam
EODS=hg*m              #Enthalpy of 1.5Kg of dry and saturated steam
v=vg                   #volume of dry and saturated steam
u=h-((P*v)/10**3)      #Internal Energy
IES=u*m                #Internal energy of the steam
s=6.5858               #Entropy of dry and saturated steam
EODSS=s*m              #Entropy of 1.5Kg dry and saturated steam
x=0.75
#(2)Enthalpy of wet steam
h1=hf+x*hfg            #Enthalpy of wet steam
EWS=h1*m               #Enthalpy of1.5Kg of wet steam
Vwet=x*vg              #Volume of steam
u1=h1-((P*Vwet)/10**3) #Internal energy
IES1=u1*m              #Internal energy of1.5Kg of the steam
s1=Sf+x*Sfg            #Entropy of wet steam
EWS1=s1*m              #Entropy of1.5Kg of wet steam

#(3)Enthalpy of superheated steam
h2=hg+Cps*(tsup-ts)    #Enthalpy of superheated steam
EOSHS=h2*m             #Enthalpy of 1.5Kg of superheated steam
Vsup=vg*(tsup/ts)      #Volume of superheated steam
u2=h2-((P*Vsup)/10**3) #Internal energy
IES2=u2*m              #Internal energy of 1.5Kg of the steam
s2=Sg+Cps*(math.log(tsup/ts))#Entropy of superheated steam
EOSHS1=s2*m            #Entropy of 1.5Kg of superheated steam

#Output
print("Enthalpy of dry and saturated steam=  ",h,"KJ/Kg")
print("Enthalpy of 1.5Kg of dry and saturated steam=  ",round(EODS,2),"KJ")
print("volume of dry and saturated steam=  ",v,"m**3/kg")
print("Internal Energy=  ",round(u,2),"KJ/Kg")
print("Internal energy of the steam=  ",round(IES,2),"kJ")
print("Entropy of dry and saturated steam = ",s,"KJ/KgK")
print("Entropy of 1.5kg of dry and saturated steam=  ",EODSS,"KJ/K")

print("Enthalpy of wet steam=  ",round(h1,2),"KJ/Kg")
print("Enthalpy of1.5Kg of wet steam=  ",EWS,"KJ")
print("Volume of steam= ",Vwet,"m**3/Kg")
print("Internal energy=   ",u1,"KJ/Kg")
print("Internal energy of1.5Kg of the steam=  ",round(IES1,2),"KJ")
print("Entropy of wet steam=  ",round(s1,2),"KJ/KgK")
print("Entropy of 1.5Kg of wet steam=  ",EWS1,"KJ/K")

print("Enthalpy of superheated steam=  ",h2,"KJ/Kg")
print("Enthalpy of 1.5Kg of superheated steam=  ",round(EOSHS,1),"KJ")
print("Volume of superheated steam= ",round(Vsup,4),"m**3/Kg")
print("Internal energy=  ",round(u2,4),"")
print("Internal energy of1.5Kg of the steam=  ",round(IES2,1),"KJ")
print("Entropy of superheated steam=  ",round(s2,4),"KJ/KgK")
print("Entropy of 1.5Kg of superheated steam=  ",round(EOSHS1,2),"KJ/K")

Enthalpy of dry and saturated steam=   2776.2 KJ/Kg
Enthalpy of 1.5Kg of dry and saturated steam=   4164.3 KJ
volume of dry and saturated steam=   0.1943 m**3/kg
Internal Energy=   2581.9 KJ/Kg
Internal energy of the steam=   3872.85 kJ
Entropy of dry and saturated steam =  6.5858 KJ/KgK
Entropy of 1.5kg of dry and saturated steam=   9.8787 KJ/K
Enthalpy of wet steam=   2272.8 KJ/Kg
Enthalpy of1.5Kg of wet steam=   3409.2 KJ
Volume of steam=  0.145725 m**3/Kg
Internal energy=    2127.075 KJ/Kg
Internal energy of1.5Kg of the steam=   3190.61 KJ
Entropy of wet steam=   5.47 KJ/KgK
Entropy of 1.5Kg of wet steam=   8.207475 KJ/K
Enthalpy of superheated steam=   2937.43 KJ/Kg
Enthalpy of 1.5Kg of superheated steam=   4406.1 KJ
Volume of superheated steam=  0.2244 m**3/Kg
internal energy=   2713.0562
Internal energy of1.5Kg of the steam=   4069.6 KJ
Entropy of superheated steam=   6.9138 KJ/KgK
Entropy of 1.5Kg of superheated steam=   10.37 KJ/K


## Example 13 Page No:196¶

In [40]:
#Input data
V=0.04                  #Volume of vessel in m**3
x=1
t=250+273               #Saturated steam temp in degree celsius
mw=9                    #Mass of liquid in Kg
#From steam table(temp basis,at t=250)
P=39.78*10**5               #in bar
Vf=0.001251             #In m**3/kg
Vg=0.05004              #In m**3/Kg
hf=1085.7               #KJ/Kg
hfg=2800.4              #KJ/Kg
hg=1714.7               #KJ/Kg

#Calculation
Vw=mw*Vf                #Volume occupied by water in m**3
Vs=V-Vw                 #Volume of waterin m**3
ms=Vs/Vg                #Volume of dry and saturated steam in Kg
m=mw+ms                 #Total mass of steam in Kg
x=ms/(ms+mw)            #Dryness fraction of steam
Vwet=(1-x)*Vf+x*Vg      #Specific volume of steam in m**3/Kg
h=hf+x*hfg              #Enthalpy of wet steam in KJ/Kg
EOWS=h*m                #Enthalpy of 9.574 Kg of wet steam KJ
u=h-((P*Vwet)/10**3)    #Internal Energy in KJ/Kg
IEOS=u*m                #Internal energy of 9.574 Kg of steam in KJ

#Output
print("Volume occupied by water= ",round(Vw,5),"m**3")
print("Volume of water= ",round(Vs,5),"m**3")
print("Volume of dry and saturated steam=  ",round(ms,3),"Kg ")
print("Total mass of steam= ",round(m,3),"Kg")
print("Dryness fraction of steam= ",round(x,2),)
print("Specific volume of steam= ",round(Vwet,6)," m**3/Kg")
print("Enthalpy of wet steam= ",round(h,1),"KJ/Kg")
print("Enthalpy of 9.574 Kg of wet steam= ",round(EOWS,),"KJ")
print("Internal Energy= ",round(u,1),"KJ/Kg")
print("Internal energy of 9.574 Kg of steam= ",round(IEOS),"KJ")

Volume occupied by water=  0.01126 m**3
Volume of water=  0.02874 m**3
Volume of dry and saturated steam=   0.574 Kg
Total mass of steam=  9.574 Kg
Dryness fraction of steam=  0.06
Specific volume of steam=  0.004178  m**3/Kg
Enthalpy of wet steam=  1253.7 KJ/Kg
Enthalpy of 9.574 Kg of wet steam=  12003 KJ
Internal Energy=  1237.1 KJ/Kg
Internal energy of 9.574 Kg of steam=  11844 KJ


## Example 14 Page No:197¶

In [7]:
#Input Data
P=7                 #Absolute pressure in bar
t=200               #Absolute temperature
ts=165              #In degree celsius from steam table

#Calculation
dos=t-ts            #Degree of superheat in degree celcius

#Output
print("Degree of superheat= ",dos,"degree celcius")

Degree of superheat=  35 degree celcius


## Example 15 Page No:197¶

In [8]:
#Input data
P=15          #Absolute pressure in bar
#From steam table (pressure basis at 15 bar)
h=1950       #In KJ/Kg
ts=198.3     #In degreee celsius
hf=844.7     #In KJ/Kg
hfg=1945.2   #In KJ/Kg
hg=2789.9    #In KJ/Kg

#calculation
x=((h-hf)/hfg)  #Enthalpy of wet steam

#Output
print("Enthalpy of wet steam= ",round(x,3),"")

Enthalpy of wet steam=  0.568


## Example 16 Page No:197¶

In [9]:
#Input data
P=15                      #Absolute pressure in bar
#From steam table (pressure basis at 15 bar)
h=3250                    #In KJ/Kg
ts=198.3                  #In degree celsius
hf=844.7                  #In KJ/Kg
hfg=1945.2                #In KJ/Kg
hg=2789.9                 #In KJ/Kg
Cps=2.3

#Calculation
tsup=(h-hg+(Cps*ts))/2.3 #Enthalpy of superheated steam in degree celsius
dos=tsup-ts              #Degree of superheated in degree celsius
##The value of ts in not used according to data in book instead of ts=198.3 author used ts=165

#Output
print("Enthalpy of superheated steam=  ",round(tsup,2),"degree celcius")
print("Degree of superheated=  ",round(dos,2),"degree celcius")

Enthalpy of superheated steam=   398.34 degree celcius
Degree of superheated=   200.04 degree celcius


## Example 17 Page No:198¶

In [10]:
#Input data
P=7                  #Absolute pressure in bar
v=0.2                #Specific volume in m**3/Kg
#from steam table (pressure basis at 7 bar)
ts=165               #In degree celsius
vf=0.001108          #In m**3/Kg
vg=0.2727            #In m**3/Kg

#calculation
x=v/vg               #Volume of steam dryness fraction

#Output
print("Volume of steam dryness fraction= ",round(x,3),)

Volume of steam dryness fraction=  0.733


## Example 18 Page No:198¶

In [11]:
#Input data
P=7                             #Absolute pressure in bar
v=0.3                           #Specific volume in m**3/Kg
#From steam table (pressure basis at 7 bar)
ts=165+273                      #In degree celsius
vf=0.001108                     #In m**3/Kg
vg=0.2727                       #In m**3/Kg

#Calculation
#v=vg*tsup/ts
tsup=((v/vg)*ts)-273            #Temp of superheated steam  in degree celsius
DOS=tsup+273-ts                 #Degree of superheated  in degree celsius

#Output
print("Temp of superheated steam= ",round(tsup,2),"degree celsius")
print("Degree of superheated=  ",round(DOS,2),"degree celsius")

Temp of superheated steam=  208.85 degree celsius
Degree of superheated=   43.85 degree celsius


## Example 19 Page No:198¶

In [12]:
#Input data
m=2        #steam of vessel in Kg
V=0.1598   #volume of vessel in M**3
P=25       #Absolute pressure of vessel in bar

#Calculation
v=V/m      #Quality of steam in m**3/Kg

#Output
print("Quality of steam",v," m**3/Kg")

Quality of steam 0.0799  m**3/Kg


## Example 20 Page No:200¶

In [13]:
#Input data
P=10*10**2                #Absolute pressure in bar
x1=0.9                    #Dryness enters
tsup2=300+273             #Temperature in degree celsius
#From steam table at 10 bar
ts=179.9+273              #In degree celsius
Vg=0.1943                 #In m**3/Kg
hf=762.6                  #In KJ/Kg
hfg=2013.6                #InK/Kg
hg=2776.2                 #In KJ/Kg

#Calculation
h1=hf+x1*hfg              #Initial enthalpy of steam in KJ/Kg
V1=x1*Vg                  #Initial specific volume of steam
u1=h1-P*V1                #Initial internal energy of steam in KJ/Kg
h2=hg+Cps*(tsup2-ts)      #Final enthalpy of steam in KJ/Kg
V2=Vg*(tsup2/ts)          #Final specific volume of steam in m**3/Kg
u2=h2-P*V2                #Final internal energy of steam in KJ/K
deltah=h2-h1              #Heat gained by steam in KJ/Kg
deltaU=(u2-u1)            #Change in internal energy in KJ/Kg

#Output
print("Initial enthalpy of steam= ",h1,"KJ/Kg")
print("Initial specific volume of steam= ",V1,)
print("Initial internal energy of steam= ",round(u1,2),"KJ/Kg")
print("Final enthalpy of steam= ",h2,"KJ/Kg")
print("Final specific volume of steam= ",round(V2,4),"m**3/Kg")
print("Final internal energy of steam= ",round(u2,3),"KJ/Kg")
print("Heat gained by steam= ",round(deltah,2),"KJ/Kg")
print("Change in internal energy=  ",round(deltaU,2),"KJ/Kg")

Initial enthalpy of steam=  2574.84 KJ/Kg
Initial specific volume of steam=  0.17487
Initial internal energy of steam=  2399.97 KJ/Kg
Final enthalpy of steam=  3052.43 KJ/Kg
Final specific volume of steam=  0.2458 m**3/Kg
Final internal energy of steam=  2806.606 KJ/Kg
Heat gained by steam=  477.59 KJ/Kg
Change in internal energy=   406.64 KJ/Kg


## Example 21 Page No:201¶

In [14]:
#Input data
m=4                        #Steam in Kg
P=13                       #Absolute pressure in bar
tsup1=450                  #Absolute temp in degree celsius
deltaH=2.8*10**3           #loses in MJ
#from steam table at 13 bar
ts=191.6                   #In degree celsius
Vg=0.1511                  #In m**3/Kg
hf=814.7                   #In m**3/Kg
hfg=1970.7                 #In  KJ/Kg
hg=2785.4                  #In KJ/Kg

#Calculation
h1=hg+Cps*(tsup1-ts)       #Initial enthalpy of steam in KJ/Kg
Deltah=deltaH/m            #Change in enthalpy/unit mass in KJ/Kg
h2=h1-Deltah               #Final enthalpy of steam in KJ/Kg
x2=(h2-hf)/hfg             #wet & dryness fraction

#Output
print("Initial enthalpy of steam= ",round(h1,2)," KJ/Kg")
print("Change in enthalpy/unit mass= ",Deltah,"KJ/Kg")
print("Final enthalpy of steam= ",round(h2,2),"KJ/Kg")
print("wet & dryness fraction= ",round(x2,4),)

Initial enthalpy of steam=  3379.72  KJ/Kg
Change in enthalpy/unit mass=  700.0 KJ/Kg
Final enthalpy of steam=  2679.72 KJ/Kg
wet & dryness fraction=  0.9464


## Example 22 Page No:202¶

In [15]:
#Input data
m=2                           #Steam in Kg
x=0.7                         #Initial dryness
P=15                          #Constant pressure in bar
#V2=2V1
#from steam table properties of
#saturated steam(pressure basis) at 15 bar
Ts=198.3+273                  #In degree celsius
Vg=0.1317                     #In m**3/Kg
hf=844.7                      #In KJ/Kg
hfg=1945.2                    #In KJ/Kg
hg=2789.9                     #In KJ/Kg
Cps=2.3

#Calculation
V1=x*Vg                       #Initial specific volume of steam in m**3/Kg
V2=2*V1                       #Final specific volume of steam in m**3/Kg
Tsup=(V2/Vg)*Ts               #Steam is superheated in degree celsius
FSS=Tsup-Ts                   #Degree of superheated in degree celsius
h1=hf+x*hfg                   #Initial enthalpy of steam in KJ/Kg
h2=hg+Cps*(Tsup-Ts)           #Final enthalpy of steam in KJ/Kg
Q=(h2-h1)*m                   #Heat transferred in the process in KJ
W1=P*(m*V2-m*V1)              #Work transferred in the process in KJ

#Output
print("Initial specific volume of steam= ",round(V1,4),"m**3/Kg")
print("Final specific volume of steam= ",round(V2,4),"m**3/Kg")
print("Steam is superheated=  ",round(Tsup,2),"K")
print("Degree of superheated= ",round(FSS,2),"degree celsius")
print("Initial enthalpy of steam= ",h1,"KJ/Kg")
print("Final enthalpy of steam= ",round(h2,2),"KJ/Kg")
print("Heat transferred in the process= ",round(Q,2),"KJ")
print("Work transferred in the process= ",round(W1,3),"KJ")

Initial specific volume of steam=  0.0922 m**3/Kg
Final specific volume of steam=  0.1844 m**3/Kg
Steam is superheated=   659.82 K
Degree of superheated=  188.52 degree celsius
Initial enthalpy of steam=  2206.34 KJ/Kg
Final enthalpy of steam=  3223.5 KJ/Kg
Heat transferred in the process=  2034.31 KJ
Work transferred in the process=  2.766 KJ


## Example 23 Page No:203¶

In [16]:
#Input data
ms=1000                      #Steam in Kg/h
P=16                         #Absolute pressure in bar
x2=0.9                       #Steam is dry
t1=30+273                    #temperature in degree celsius
tsup=380                     #tmperature rised in degree celsius

#from steam table(pressure basis at 16 bar)
h1=125.7                     #in KJ/Kg
ts=201.4                     #In degree celsius
hf=858.5                     #in kJ/Kg
hfg=1933.2                   #in kJ/Kg
hg=2791.7                    #in kJ/Kg
Cps=2.3

#Calculation
h2=hf+x2*hfg                 #Final enthalpy of wet steam in KJ/Kg
Q1=(ms*(h2-h1))*10**-3       #Constant pressure process in KJ/h
h3=hg+Cps*(tsup-ts)          #Final enthalpy of superheated steam in KJ/g
Q2=(ms*(h3-h2))*10**-3       #Suprheated steam in KJ/h

#Output
print("Final enthalpy of wet steam= ",round(h2,1),"KJ/Kg ")
print("Constant pressure process= ",round(Q1,1)," KJ/h ")
print("Final enthalpy of superheated steam= ",round(h3,1)," KJ/g")
print("Suprheated steam= ",round(Q2,1),"KJ/h")

Final enthalpy of wet steam=  2598.4 KJ/Kg
Constant pressure process=  2472.7  KJ/h
Final enthalpy of superheated steam=  3202.5  KJ/g
Suprheated steam=  604.1 KJ/h


## Example 24 Page No:204¶

In [1]:
#Input data
FB=15                    #First boiler in bar
SB=15                    #Second boiler in bar
tsup1=300                #Temperature of the steam in degree celsius
tsup2=200                #Temperature of the steam in degree celsius
#From steam table (pressure basis at 15 bar )
ts=198.3                 #In degree celsius
hf=844.7                 #In KJ/Kg
hfg=1945.2               #In KJ/Kg
hg=2789.9                #In KJ/I
Cps=2.3

#Calculation
h1=hg+Cps*(tsup1-ts)     #Enthalpy of steam of first boiler in KJ/Kg
h3=hg+Cps*(tsup2-ts)     #Enthalpy of steam in steam main in KJ/Kg
h2=2*h3-h1               #Energy balance in KJ/Kg
x2=(h2-hf)/hfg           #Enthalpy of wet steam

#OUTPUT
print("Enthalpy of steam of first boiler= ",round(h1,1),"KJ/Kg")
print("Enthalpy of steam in steam main= ",round(h3,1),"KJ/Kg")
print("Energy balance= ",round(h2,1),"KJ/Kg")
print("Enthalpy of wet steam= ",round(x2,3),)

Enthalpy of steam of first boiler=  3023.8 KJ/Kg
Enthalpy of steam in steam main=  2793.8 KJ/Kg
Energy balance=  2563.8 KJ/Kg
Enthalpy of wet steam=  0.884


## Example 25 Page No:205¶

In [2]:
#Input data
V=0.35                     #Capacity of vessel in m**3
P1=10*10**2                #Absolute pressure in bar
tsup1=250+273              #Absolute temperature in degree celsius
P2=2.5*102                 #Absolute pressure in the vessel fall in bar

#From steam table (pressure basis at 10 bar)
ts1=179.9+273              #In degree celsius
Vg1=0.1943                 #In m**3/Kg
hf1=762.6                  #In KJ/Kg
hfg1=2013.6                #In KJ/Kg
hg1=2776.2                 #In KJ/Kg

#From steam table(pressure basis at 2.5 bar)
V2=0.2247                  #In m**3/Kg
ts2=127.4                  #In degree celsius
Vg2=0.7184                 #In m**3/Kg
hf2=535.3                  #In KJ/Kg
hfg2=2181.0                #In KJ/Kg
hg2=2716.4                 #In KJ/Kg

#Calculation
V1=Vg1*(tsup1/ts1)         #Initial specific volume of steam in m**3/Kg
m=V/V1                     #Initial mass of steam in Kg
x2=V2/Vg2                  #Final condition of wet steam
h1=hg1+Cps*(tsup1-ts1)     #Initial enthalpy of steam in KJ/Kg
u1=h1-P1*V1                #Initial internal energy of steam in KJ/Kg
h2=hf2+x2*hfg2             #Final enthalpy of steam in KJ/Kg
u2=h2-P2*V2                #Final internal energy of steam in KJ/Kg
deltaU=(u2-u1)*m           #Change in internal energy in KJ

#Output
print("Initial specific volume of steam= ",round(V1,4),"m**3/Kg")
print("Initial mass of steam= ",round(m,4),"Kg")
print("Final condition of wet steam= ",round(x2,4),)
print("Initial enthalpy of steam= ",h1,"KJ/Kg")
print("Initial internal energy of steam= ",round(u1,2),"KJ/Kg")
print("Final enthalpy of steam= ",round(h2,1)," KJ/Kg")
print("Final internal energy of steam= ",round(u2,3),"KJ/Kg")
print("Change in internal energy= ",round(deltaU,1),"KJ")

Initial specific volume of steam=  0.2244 m**3/Kg
Initial mass of steam=  1.5599 Kg
Final condition of wet steam=  0.3128
Initial enthalpy of steam=  2937.43 KJ/Kg
Initial internal energy of steam=  2713.06 KJ/Kg
Final enthalpy of steam=  1217.5  KJ/Kg
Final internal energy of steam=  1160.171 KJ/Kg
Change in internal energy=  -2422.3 KJ


## Example 26 Page No:207¶

In [3]:
#Input data
m=1.5                          #Saturated steam in Kg
x1=1
x2=0.6
P1=5*10**5                     #Absolute  pressure in bar
#From steam table at pressure basis 5 bar
hg1=2747.5                     #In KJ/Kg
Vg1=0.3747                     #In m**3/Kg
V1=0.3747                      #In m**3/Kg
V2=0.3747                      #In m**3/Kg
#From steam table at Vg2 is 2.9 bar
P2=2.9*10**5                   #Absolute pressure in bar
t2=132.4                       #In degree celsius
hf2=556.5                      #In KJ/Kg
hfg2=2166.6                    #In KJ/Kg

#Calculation
Vg2=V2/x2                       #Constant volume process in m**3/Kg
u1=hg1-((P1*Vg1)/1000)          #Initial internal energy in KJ/Kg
u2=(hf2+x2*hfg2)-((P2*V2)/1000) #Final internal energy in KJ
deltaU=(u1-u2)*m                #Heat supplied in KJ

#Output
print("Constant volume process= ",round(Vg2,4),"m**3/Kg")
print("Initial internal energy= ",u1,"KJ/Kg")
print("Final internal energy= ",round(u2,1),"KJ")
print("Heat supplied= ",round(deltaU,2),"KJ")

Constant volume process=  0.6245 m**3/Kg
Initial internal energy=  2560.15 KJ/Kg
Final internal energy=  1747.8 KJ
Heat supplied=  1218.53 KJ


## Example 27 Page No:208¶

In [4]:
#Input data
P1=20            #Initial steam in bar
x1=0.95          #dryness throttled
P2=1.2           #Absolute pressure in bar

#From steam table (pressure basis at 20 bar)
ts=212.4         #In degree celsius
hf=908.6         #In KJ/Kg
hfg=1888.6       #In KJ/Kg
hg=2797.2        #In KJ/Kg
#From steam table (pressure basis at 1.2 bar)
h2=h1            #In KJ/Kg
ts2=104.8        #In degree celsius
hf2=439.3        #In KJ/Kg
hfg2=2244.1      #In KJ/Kg
hg2=2683.4       #In KJ/Kg
Cps=2.3

#Calculation
h1=hf+x1*hfg     #Enthalpy of steam in KJ/Kg
tsup2=((h1-hg2)/Cps)+ts2 #Enthalpy of wet steam in degree celsius
DOS=tsup2-ts2    #Degree of superheat in degree celsius

#Output
print("Enthalpy of steam= ",h1,"KJ/Kg")
print("Enthalpy of wet steam= ",round(tsup2,2),"degree celsius")
print("Degree of superheat= ",round(DOS,2),"degree celsius")

Enthalpy of steam=  2702.77 KJ/Kg
Enthalpy of wet steam=  113.22 degree celsius
Degree of superheat=  8.42 degree celsius


## Example 28 Page No:209¶

In [5]:
#Input data
P1=12           #Throttled steam
x1=0.96         #Dryness is brottled
x2=1            #Constant enthalpy process
#From steam table at12 bar
ts=188          #In degree celsius
hf=798.4        #In KJ/Kg
hfg=1984.3      #In KJ/Kg
hg=2782.7       #In KJ/Kg

#Calculation
h1=hf+x1*hfg    #Enthalpy of the steam  in KJ/Kg
h2=h1           #Enthalpy after throttling in KJ/Kg

#Output
print("Enthalpy of the steam=  ",round(h1,2),"KJ/Kg ")
print("Enthalpy after throttlin= ",round(h2,2),"KJ/Kg ")

Enthalpy of the steam=   2703.33 KJ/Kg
Enthalpy after throttlin=  2703.33 KJ/Kg


## Example 29 Page No:210¶

In [6]:
#Input data
import math
P1=15                         #Initial steam in bar
tsup1=250+273                 #Temperature of steam in degree celsius
P2=0.5                        #Steam turbine in bar

#From steam table at 15 bar
ts1=198.3+273                 #In degree celsius
hg1=2789.9                    #In KJ/Kg
sf1=2.3145                    #In KJ/KgK
sfg1=4.1261                   #In KJ/KgK
sg1=6.4406                    #In KJ/KgK
#From steam table at 0.5 bar
ts2=81.53                     #In degree celsius
sf2=1.0912                    #In KJ/Kg
sfg2=6.5035                   #In KJ/Kg
sg2=7.5947                    #In KJ/Kg
hf2=340.6
Cps=2.3
hfg2=2646

#Calculation
S1=sg1+Cps*(math.log(tsup1/ts1)) #Entropy  of superheated steam in KJ/KgK
S2=S1                            #Entropy after isentropic processes in KJ/KgK
x2=(S2-sf2)/sfg2                 #Enthalpy of wet steam
h1=hg1+Cps*(tsup1-ts1)           #Enthalpy of steam at 15 bar
h2=hf2+x2*hfg2                   #Enthalpy of wet steam at 0.5 bar
WOT=h1-h2                        #Work output of the turbine

#OUTPUT
print("Entropy  of superheated steam=  ",round(S1,2),"KJ/KgK")
print("Entropy after isentropic processes= ",round(S2,2),"KJ/KgK")
print("Enthalpy of wet steam= ",round(x2,2),"")
print("Enthalpy of steam= ",h1,"KJ/Kg")
print("Enthalpy of wet steam= ",round(h2,2),"KJ/Kg")
print("Work output of the turbine= ",round(WOT,2),"KJ/Kg")

Entropy  of superheated steam=   6.68 KJ/KgK
Entropy after isentropic processes=  6.68 KJ/KgK
Enthalpy of wet steam=  0.86
Enthalpy of steam=  2908.81 KJ/Kg
Enthalpy of wet steam=  2614.45 KJ/Kg
Work output of the turbine=  294.36 KJ/Kg