# Chapter 6: Thermodynamic Properties of Pure Substance¶

## Example 2, page no. 182¶

In [3]:
#Variable Declaration:
p1 = 10 #Pressure at which steam is entering(in MPa):
p2 = 0.05 #Pressure at which steam is coming out(in MPa):
T = 100 #Temperature of the steam(inC):
h2 = 2682.5 #Enthalpy of superheated steam at 0.05 MPa and 100 C(in kJ/kg): #From steam tables:
hf10 = 1407.56
hfg10 = 1317.1

#Calculation:
h1 = h2 #Due to throttling:
x1 = (h1-hf10)/hfg10 #Dryness fraction:

#Results:
print "Dryness fraction: ",round(x1,3)

Dryness fraction:  0.968


## Example 3, page no. 183¶

In [4]:
#Variable Declaration:
p = 12 #Pressure(in MPa):
v = 0.017 #Specific volume(in m**3/kg):
h = 2848 #Enthaply(in kJ/kg):

#Calculation:
u = h-p*10**3*v #Internal energy(in kJ/kg):

#Results:
print "Internal energy: ",round(u),"KJ/Kg"

Internal energy:  2644.0 KJ/Kg


## Example 4, page no. 183¶

In [5]:

from math import log

#Variable Declaration:
m = 5      #Mass of steam(in kg):
p = 2    #Pressure(in MPa):
Tss = 300+273.15 #Temperature of superheated steam(in K):
Cps = 2.1  #Specific heat of super heated steam(in kJ/kg.K):
Cpw = 4.18  #Specific heat of water(in kJ/kg.K):
hfg = 1890.7  #From steam tables:

#Calculations:
Tsat = 212.42+273.15  #Saturation temperature(in K):
s = Cpw*log(Tsat/273.15)+hfg/Tsat+Cps*log(Tss/Tsat) #Entropy of unit mass of superheated steam with reference to absolute zero(in kJ/kg.K):
S = m*s    #Entropy of 5 kg of steam(in kJ/K):

#Result:
print "Entropy of steam: ",round(S,2),"KJ/K"

Entropy of steam:  33.23 KJ/K


## Example 5, page no. 183¶

In [6]:
#Variable Declaration:
Tb = 110 #Boiling point(inÂ°C):
p =  143.27 #Pressure at which it boils(in kPa): #From steam table:
Tsat =  108.866 #From steam table this temperature(in Â°C): #Boiling point at this depth  =  Tsat at 138.365

#Calculation:
p1 = p-9.81*0.50 #Pressure at 50 cm depth(in kPa):

#Result:
print "Boiling point :",round(Tsat,2),"°C"

Boiling point : 108.87 °C


## Example 6, page no. 184¶

In [7]:
#Variable Declaration:
T = 100 #Temperature of the water vapor mixture(in Â°C):
V = 0.5 #Volume of the rigid vessel(in m**3):
v2 = 0.003155 #Specific volume at state 2(in m**3/kg): #From steam tables:
vf = 0.001044
vg = 1.6729

#Calculations:
v1 = v2 #Specific volume at state 1(in m**3/kg):
x1 = (v1-vf)/vg #Dryness fraction:
m = V/v2 #Total mass of fluid(in kg):
v = m*vf #Volume of water(in m**3):

#Results:
print "Mass of water :",round(m,2),"Kg"
print "Volume of water :",round(v,4),"m**3"

Mass of water : 158.48 Kg
Volume of water : 0.1655 m**3


## Example 7, page no. 184¶

In [8]:
#Variable Declaration:
p = 2 #Pressure(in MPa):
T = 500+273.15 #Temperature(in K):

#Calculation:
s = T #Slope of isobar:(dh/ds)at constant pressure = T:

#Result:
print "Slope :",s

Slope : 773.15


## Example 8, page no. 185¶

In [9]:
#Variable Declaration:
x = 0.10 #Dryness fraction:
p = 0.15 #Pressure(in MPa):
hf  =  467.11 #From steam tables:(at 0.15 MPa):
hg  =  2693.6
vf  =  0.001053
vg  =  1.1593
sf  =  1.4336
sg  =  7.2233

#Calculations:
h = hf+x*(hg-hf) #Enthalpy(in kJ/kg):
v = vf+x*(vg-vf) #Specific volume(in m**3/kg):
s = sf+x*(sg-sf) #Entropy(in kJ/kg.K):

#Results:
print "Enthalpy :",h,"KJ/Kg"
print "Specific volume :",v,"m**3/kg"
print "Entropy :",s,"KJ/Kg.K"

Enthalpy : 689.759 KJ/Kg
Specific volume : 0.1168777 m**3/kg
Entropy : 2.01257 KJ/Kg.K


## Example 9, page no. 185¶

In [10]:
#Variable Declaration:
#Initial State:
p1 = 1 #Pressure(in MPa):
V1 = 0.05 #Volume(in m**3):
x1 = 0.80 #Dryness fraction:
#Final state:
p2 = 1 #Pressure(in MPa):
V2 = 0.2 #Volume(in m**3):

#From steam table:(at state 1):
vf  =  0.001127                 #(m3/kg)
vg  =  0.19444                  #(m3/kg)
uf  =  761.68                   #(kJ/kg)
ufg  =  1822                    #(kJ/kg)

#Calculations:
W = p1*10**3*(V2-V1) #Work done(in kJ):
v1 = vf+x1*(vg-vf) #Specific volume at state 1(in m**3/kg):
m = V1/v1 #Mass of system(in kg):
v2 = V2/m #Specific volume at state 2(in m**3/kg):
Tf = 1077.61 #Temperature at final state(in C):
u2 = 4209.6 #Internal energy at final state(at 1077.61 C):
u1 = uf+x1*ufg #Internal energy at initial state(in kJ/kg):
Q = m*(u2-u1)+W #Heat added(in kJ):

#Results:

Heat added : 788.83 kJ


## Example 10, page no. 186¶

In [11]:
#Variable Declaration:
p1 = 800 #Presure of the steam(in kPa):
T = 200 #Temperature(in °C)
Tsat = 170.43 #Saturation temp(in °C):
#From steam tables:
v1 = 0.2404 #Specific volume(in m**3/kg):
vgI = 0.2168
vgII = 0.2428
TI = 175
TII = 170
PI = 892
PII = 791.7

#Calculations:
T2 = TI - (TI-TII)*(v1-vgI)/(vgII-vgI) #Final temperature after interpolation (in °C):
p2 = PI - (PI-PII)*(v1-vgI)/(vgII-vgI) #Final pressure after interpolation (in kPa):

#Results:
print "Pressure :",round(p2,2),"kPa"
print "Temperature :",round(T2,2),"°C"

Pressure : 800.96 kPa
Temperature : 170.46 °C


## Example 11, page no. 187¶

In [12]:
#Variable Declaration:
T = 30 #Temperature of water(in C):
p = 200 #Pressure(in kPa):
p1 = 4.25 #Corresponding pressure at 30 C(in kPa): #From steam tables:
v1 = 0.001004 #Specific volume(in m**3):

#Calculations:
dh = v1*(p-p1) #Enthalpy change(in kJ/kg):

#Result:
print "Enthalpy change :",dh,"KJ/Kg"

Enthalpy change : 0.196533 KJ/Kg


## Example 12, page no. 187¶

In [13]:

#Variable Declaration:
V1 = 3./5*2 #Volume occupied by water(in m**3):
V2 = 2./5*2 #Volume occupied by steam(in m**3):
#From steam table
vf  =  0.001091 #(m**3/kg)
vg  =  0.3928 #(m**3/kg)

#Calculations:
mf = V1/vf #Mass of water(in kg):
mg = V2/vg #Mass of steam(in kg):
mt = mf+mg #Total mass(in kg):
x = mg/mt #Dryness fraction:

#Results:
print "Mass :",round(mt,2),"kg"
print "Quality :",round(x,6)

Mass : 1101.95 kg
Quality : 0.001848


## Example 13, page no. 188¶

In [14]:
#Variable Declaration:
p = 4 #Pressure of the steam(in MPa):
T1 = 300  #Temperature of steam entering(in °C):
T2 = 50  #Temperature of steam at turbine exit(in °C):
#From steam tables:
h1 = 2886.2                 #kJ/kg
s1 = 6.2285                 #kJ/kg.K
hf  =  209.33               #kJ/kg
sf  =  0.7038               #kJ/kg.K
hfg  =  2382.7              #kJ/kg
sfg  =  7.3725              #kJ/kg.K

#Calculation:
s2 = s1  #Assumed
x2 = round((s2-sf)/sfg,4)   #Dryness fraction:
h2 = hf+x2*hfg #Enthalpy at state 2(in kJ/kg):
W = h1-h2  #Turbine work(in kJ/kg):

#Results:
print "Turbine output: ",round(W,2),"kJ/kg"

Turbine output:  891.27 kJ/kg


## Example 14, page no. 188¶

In [15]:
#Variable Declaration:
m1 = 100 #Mass of steam(in kg):
p1 = 100 #Initial pressure(in kPa):
p2 = 1000 #Final pressure(in kPa):
x1 = 0.5 #Dryness fraction:
p3 = 2000 #Pressure of dry saturated steam(in kPa):

#From steam tables:
hf100kPa  =  417.46                     #kJ/kg
uf100kPa  =  417.36                     #kJ/kg
vf100kPa  =  0.001043                   #m**3/kg
hfg100kPa  =  2258                      #kJ/kg
ufg100kPa  =  2088.7                    #kJ/kg
vg100kPa  =  1.6940                     #m**3/kg
vg2000kPa  =  0.09963                   #m**3/kg
ug2000kPa  =  2600.3                    #kJ/kg
hg2000kPa  =  2799.5                    #kJ/kg
hf1000kPa  =  762.81                    #kJ/kg,
hfg1000kPa  =  2015.3                   #kJ/kg
vf1000kPa  =  0.001127                  #m3/kg
vg1000kPa  =  0.19444                   #m3/kg

#Calculations:
v1 = vf100kPa+x1*(vg100kPa-vf100kPa) #Initial specific volume(in m**3/kg):
h1 = hf100kPa+x1*hfg100kPa #Enthalpy at 1(in kJ/kg):
V = m1*x1*v1 #Volume of vessel(in m**3):
v2 = vg2000kPa*v1/(vg2000kPa+v1) #Final specific volume(in m**3/kg):
x2 = (v2-vf1000kPa)/(vg1000kPa-vf1000kPa)#Final dryness fraction:
h2 = hf1000kPa+x2*hfg1000kPa #Final enthalpy(in kJ/kg):
m = m1*(h1-h2)/(h2-hg2000kPa) #Mass of dry steam at 2000kPa(in kg):
U1 = m*(uf100kPa+x1*ufg100kPa) #Internal energy in the beginning(in kJ):

#Results:
print "Mass of dry steam at 2000 kPa to be added: ",round(m,2),"kg"
print "Quality of final mixture: ",round(x2,3)

Mass of dry steam at 2000 kPa to be added:  11.97 kg
Quality of final mixture:  0.455


## Example 15, page no. 190¶

In [17]:
#Variable Declaration:
r =  71.5  #Recorded condenser vacuum(cm of Mercury)
br =  76.8 #Barometer reading(cm of Mercury)
Tc =  35  #Temperature of condensation(°C)
Thw =  27.6  #Temperature of hot well(°C)
mc =  1930  #Mass of condensate per hour()Kg
mcw =  62000    #Mass of cooling water per hour(Kg)
Ti =  8.51 #Inlet temperature (°C)
To =  26.24    #Outlet temperature(°C)
#From steam tables:
hf  =  146.68 #kJ/kg
hfg  =  2418.6 #kJ/kg

#Calculations:
pc = (br-r)/73.55*101.325 #Condensor pressure(in kPa):
ps =  5.628 #Partial pressure of steam corresponding to 35Â°C from steam table(in kPa):
x = (mcw*(To-Ti)*4.18-mc*hf+mc*4.18*To)/(mc*hfg) #Dryness fraction:

#Results:
print "Dryness fraction of the steam entering:",round(x,2)

Dryness fraction of the steam entering: 0.97


## Example 16, page no. 191¶

In [19]:
from math import pi
#Variable Declaration:
D = 0.2 #Diameter of the vessel(in m):
d = 0.02 #Depth(in m):
T = 150 #Temperature(in °C):
F = 10 #Force applied(in kN):
Q = 600 #Heat supplied(in kJ):
#From steam tables:
hf = 612.1
hfg = 2128.7
vg = 0.4435
h2 = 1582.8

#Calculations:
p = F/(pi*D**2)*4+101.3 #Pressure at which process is taking place(in kPa):
V1 = pi*D**2*d/4 #Volume of water contained(in m**3):
m = V1*1000 #Mass of water(in kg):
x = (Q-hf*m+4.18*T*m)/(hfg*m) #Dryness fraction:
U1 = m*4.18*T-p*V1 #Internal energy of water initially(in kJ):
V2 = m*x*vg #Final volume(in m**3):
U2 = m*h2-p*V2 #Internal energy at state 2(in kJ):
dU = U2-U1 #Change in internal energy(in kJ):
W = p*(V2-V1) #Work done(in kJ):

#Results:
print "Dryness fraction of the steam produced: "  ,round(x,3)
print "Change in internal energy: ",round(dU,2), "kJ"
print "Work done: ",round(W,2),"kJ"

Dryness fraction of the steam produced:  0.456
Change in internal energy:  547.54 kJ
Work done:  53.01 kJ


## Example 17, page no. 192¶

In [20]:
#Variable Declaration:
ms = 40   #Mass of steam passed(in kg):
mw = 2.2  #Mass of water passed(in kg):
p1 = 1.47 #Initial pressure of steam(in MPa):
T = 120  #Temperature after throttling(in C):
p2 = 107.88 #Pressure after throttling(in kPa):
s = 2.09 #Specific heat of superheated steam(in kJ/kg.K):
hf = 840.513 #From steam tables:
hfg = 1951.02
h1 = 2673.95

#Calculations:
ds = T-101.8 #Degree of superheat(in C):
h2 = h1+ds*s #Enthalpy of superheated steam(in kJ/kg):
x2 = (h2-hf)/hfg #Dryness fraction after throttling:
x1 = (ms-mw)/ms #Dryness fraction before throttling:
x = x1*x2 #Overall dryness fraction:

#Results:
print "Dryness fraction ",round(x,4)

Dryness fraction  0.9065


## Example 18, page no. 192¶

In [21]:
#Variable Declaration:
Va = 0.4 #Initial volume in part A(in m**3):
pa = 10 #Pressure(in bar):
V = 0.4 #Initial volume in part B(in m**3):
p1 = 10 #Pressure in part B(in bar):
p2 = 15 #Final pressure in part B(in bar):
#From steam tables:
hf = 762.83
hfg = 2015.3
h2 = 2792.2

#Calculations:
Q = V*(p2-p1)*10**2 #Heat added(in kJ):
x1 = (h2-Q-hf)/hfg #Dryness fraction:

#Results:
print "Initial quality: ",round(x1,3)

Heat added:  200.0 kJ
Initial quality:  0.908


## Example 19, page no. 193¶

In [23]:
#Variable Declaration:
m = 3 #Mass of wet steam(in kg):
p1 = 1.4 #Initial pressure(in bar):
V1 = 2.25 #Initial volume(in m**3):
T = 400 #Final temperature of steam(in °C):
V2 = 4.65 #At 400 °C,volume of steam(in m**3):
#From steam tables:
vg = 1.2455
hf = 457.99
hfg = 2232.3
h2 = 3276.6
uf = 457.84
ufg = 2059.34
u2 = 2966.7

#Calculations:
v1 = V1/m #Specific volume of wer steam in cylinder(in m**3/kg):
x1 = v1/vg #Dryness fraction of initial steam:
h1 = hf+x1*hfg #Initial enthalpy of wet steam(in kJ/kg):
v2 = V2/m #At 400°C specific volume of steam(in m**3/kg):
p2 = 0.20 #Actual pressure(from steam table)(in MPa):
ds = T-120.23 #Finally the degree of superheat(in °C): #Saturation temp at this pressure  =  120.23Â°C
Q = m*(h2-h1) #Heat added during the process(in kJ):
u1 = uf+x1*ufg #Internal energy of initial wet steam(in kJ/kg):
dU = m*(u2-u1) #Change in internal energy(in kJ):
W = Q-dU #Work done(in kJ):

#Results:
print "Heat transfer: ",round(Q,2),"kJ"
print "Work transfer : ",round(W,2),"kJ"
print "____Note: Please check the value of x1 calculated and used: (calculated is 0.602 and used is 0.607 hence there is a difference in answer)____"

Heat transfer:  4423.17 kJ
Work transfer :  616.8 kJ
____Note: Please check the value of x1 calculated and used: (calculated is 0.602 and used is 0.607 hence there is a difference in answer)____


## Example 20, page no. 194¶

In [24]:

#Variable Declaration:
p1 = 10 #Pressure of the steam(in bar):
T = 500 #Temperature(in °C):
p2 = 1 #Final pressure(in bar):
#From steam tables:
h10bar500  =  3478.5            #kJ/kg
s10bar500  =  7.7622            #kJ/kg.K
v10bar500  =  0.3541            #m**3/kg
h1bar400  =  3278.2             #kJ/kg
h1bar500  =  3488.1             #kJ/kg
v1bar500  =  3.565              #m**3/kg
v1bar400  =  3.103              #m**3/kg
s1bar500  =  8.8342             #kJ/kg.K
s1bar400  =  8.5435             #kJ/kg.K
h2 = h10bar500

#Calculations:
T2 = (h2-h1bar400)*(T-400)/(h1bar500-h1bar400)+400 #Final temperature(in °C):
s2 = s1bar400+(s1bar500-s1bar400)/(T-400)*(T2-400) #Final entropy(in kJ/kg.K):
ds = s2-s10bar500  #Change in entropy(in kJ/kg.K):
v2 = v1bar400+(v1bar500-v1bar400)/(T-400)*(T2-400) #Final specific volume(in m**3/kg):
p = v10bar500/v2*100 #Percentage volume occupied by steam:

#Results:
print "Final temperature: ",round(T2,2),"°C"
print "Change in entropy: ",round(ds,4),"kJ/Kg K"
print "Percentage of vessel volume initially occupied by steam: ",round(p,2),"%"

Final temperature:  495.43 °C
Change in entropy:  1.0587 kJ/Kg K
Percentage of vessel volume initially occupied by steam:  9.99 %


## Example 21, page no. 195¶

In [25]:
#Variable Declaration:
#Steam entering:
p1 = 2.5     #Pressure(in MPa):
T1 = 350   #Temperature(in °C):
#Steam rejected:
p2 = 20   #Pressure(in kPa):
x2 = 0.92  #Dryness fraction:
p3 = 30  #Pressure of one quater of intial steam(in kPa):
T0 = 30+273 #Temperature(in K):
m1 = 1
m2 = 0.25
m3 = 0.75
Q = -10 #Heat lost during expansion(in kJ):
#From steam tables:
h1 = 3126.3                 #kJ/kg
s1 = 6.8403                 #kJ/kg.K
h2 = 2878.6                 #kJ/kg
s2 = 8.5309                 #kJ/kg.K
h3  =  2421.04              #kJ/kg
s3  =  7.3425               #kJ/kg.K
hf  =  251.40               #kJ/kg
hg  =  2609.7               #kJ/kg
sf  =  0.8320               #kJ/kg.K
sfg  =  7.0766              #kJ/kg.K
h0 = 125.79
s0 = 0.4369

#Calculations:
A1 = (h1-h0)-T0*(s1-s0)     #Availability of steam entering turbine(in kJ/kg):
A2 = (h2-h0)-T0*(s2-s0)     #Availability of steam leaving turbine at state 2(in kJ/kg):
A3 = (h3-h0)-T0*(s3-s0)     #Availability of steam leaving turbine at state 3(in kJ/kg):
Wmax = m1*A1-m2*A2-m3*A3    #Maximum work per kg of steam entering turbine(in kJ/kg):
I = T0*(m2*s2+m3*s3-m1*s1)-Q#Irreversibilty(in kJ/s):

#Results:
print "Maximum work",round(Wmax,2),"kJ/kg"
print "Irreversibility",round(I,2),"kJ/s"

Maximum work 833.06 kJ/kg
Irreversibility 252.19 kJ/s


## Example 22, page no. 196¶

In [26]:
#Variable Declaration:
p1 = 6 #Initial pressure(in MPa):
p2 = 5 #Final pressure(in MPa):
T1 = 400 #Initial temperature(in °C):
patm = 100 #Atmospheric pressure(in kPa):
Ta = 20+273 #Atmospheric temperature(in °K):
#From steam tables:
h1 = 3177.2             #kJ/kg
s1 = 6.5408             #kJ/kg.K
h2 = h1
T2 = 392.7              #°C(by interpolation)
s2 = 6.6172             #kJ/kg.K(#By interpolation Entropy)
h0 = 83.96              #kJ/kg
s0  =  0.2966           #kJ/kg

#Calculations:
A1 = (h1-h0)-Ta*(s1-s0) #Availability at state 1(in kJ/kg):
A2 = (h2-h0)-Ta*(s2-s0) #Availability at state 2(in kJ/kg):
dA = A2-A1 #Change in availibilty(in kJ/kg):

print "Change in availability: ",round(-dA,1),"kJ/kg decrease",

Change in availability:  22.4 kJ/kg decrease


## Example 23, page no. 198¶

In [27]:
#Variable Declaration:
TH1 =  95 #Temperature of the hot water entering(in Â°C):
TH2 =  50 #Temperature of the hot water at exit(in Â°C):
mH  =  0.8 #Mass flow rate(in kg/s):
TC1 =  15+273 #Temperature of cooling water entering(in Â°K):
TC2 =  45+273 #Temperature of cooling water at exit(in Â°K):
T0 = 25+273 #Temperature of dead state(in K):
#From steam tables:
h0 = 104.89                     #kJ/kg
s0 = 0.3674                     #kJ/kg.K
hH1 = 397.96                    #kJ/kg
sH1 = 1.2500                    #kJ/kg.K
hH2 = 209.33                    #kJ/kg.K
sH2 = 0.7038                    #kJ/kg.K
hC2 = 188.45                    #kJ/kg.K
sC2 = 0.6387                    #kJ/kg.K
hC1 = 62.99                     #kJ/kg.K
sC1 = 0.2245                    #kJ/kg.K

#Calculations:
mC = mH*(TH1-TH2)/(TC2-TC1)	#Mass flow rate of cooling water(in kg/s):
AH1 = mH*((hH1-h0)-T0*(sH1-s0))	#Exergy entering through hot water stream(in kJ/s):
dAc = mC*((hC2-hC1)-T0*(sC2-sC1))#Rate of exergy increase in cold stream(in kJ/s):
n = dAc/AH1*100 #Second law efficiency:
dAh = mH*((hH1-hH2)-T0*(sH1-sH2))#Rate of exergy loss in hot stream(in kJ/s):
dA = dAh-dAc #Exergy destruction(in kJ/s):

#Results:
print "Second law efficincy: ",round(n,2),"%"
print "Exergy destruction: ",round(dA,2),"kJ/s"

Second law efficincy:  10.12 %
Exergy destruction:  18.26 kJ/s