# Chapter 3:PROPERTIES OF A PURE SUBSTANCE¶

## Example 3.1, Page No:77¶

In [1]:
#Variable declaration

V=0.01;                               # Volume of water in a rigid vessel in m^3
m=4.5;                                # Mass of water+ steam in a rigid vessel in kg
T=35;                                 # Temperature of water in a rigid vessel in degree celcius
#Calculation

# (a)
v=V/m;                                # specific volume of water
# From steam table
vf=0.001006; vg=25.22;                # specific volume in m^3/kg
x=(v-vf)/(vg-vf);                     # Quality of steam
x1=1-x;                               # Quality of water
mg=x*m;                               # Mass of steam
mf=x1*m;                              # Mass of water
#Result for (a)

print "(a).\n","Quality of steam in a rigid vessel = ",x,"\nQuality of water in a rigid vessel = ",x1
print "Mass of steam in a rigid vessel = ",round(mg,8),"kg\n","Mass of water in a rigid vessel = ",round(mf,9),"kg\n"
# (b)
#Result for (b)

vc=0.003155;                          #  Crictical volume for water in m^3/kg
print "(b).\n","The level of liquid water will rise in the vessel. Since v < vc and refer figure 3.21\n"
# (c)
#Result for (c)

print "(c)\n","The final temperature after heating is 370.04 oC.  Because it is constant volume process and refer figure 3.21\n"
#  (d)
#Calculation for (d)

m1=0.45;                             # Mass of water in kg
v1=V/m;                              # specific volume of water
#Result for (d)

print"(d).\n", "Level of liquid drops to bottom (v1 > vc).  Temperature on reaching saturation state is 298.5 oC and refer figure 3.21"

(a).
Quality of steam in a rigid vessel =  4.82264368762e-05
Quality of water in a rigid vessel =  0.999951773563
Mass of steam in a rigid vessel =  0.00021702 kg
Mass of water in a rigid vessel =  4.499782981 kg

(b).
The level of liquid water will rise in the vessel. Since v < vc and refer figure 3.21

(c)
The final temperature after heating is 370.04 oC.  Because it is constant volume process and refer figure 3.21

(d).
Level of liquid drops to bottom (v1 > vc).  Temperature on reaching saturation state is 298.5 oC and refer figure 3.21


## Example 3.2, Page No:79¶

In [2]:
#Variable declaration for (a)
# (a) Ammonia 26 oC and 0.074 m^3/kg
# From saturation table of ammonia at 26 oC
v=0.074;                      # specific volume of ammonia in m^3/kg
vf=0.001663; vg=0.1245;       # specific volume  of ammonia in m^3/kg

#Calculation for (a)
x=(v-vf)/(vg-vf);             # Quality of vapour  since v<vg

#Result for (a)
print "(a) Ammonia 26 oC and 0.074 m^3/kg\n","The Quality of ammonia = ",round(x,3)

#Variable declaration for (b)
# (b).Ammonia 550kPa and 0.31m^3/kg
# From saturation table of ammonia at 550 kPa
v=0.31;                      # specific volume of ammonia in m^3/kg
vg=0.23;                     # specific volume  of ammonia in m^3/kg

#Calculation for (b)
# v > vg . Since from superheated table by interpolation for 550kPa and v
T=82.1;                      # Temperature of ammonia in degree celcius

#Result for (b)
print "\n(b).Ammonia 550kPa and 0.31m^3/kg\n","Temperature of ammonia = ",T,"oC\n"

#Variable declaration for (c)
# (c).Freon 12, 0.35MPa and 0.036 m^3/kg
# From saturation table of Freon 12 at 0.35MPa
v=0.036;                     # specific volume of Freon 12 in m^3/kg
vf=0.000722; vg=0.049329;    # specific volume  of Freon 12 in m^3/kg

#Calculation for (c)
x=(v-vf)/(vg-vf);            # Quality of vapour since v<vg

#Result for (c)
print "(c).Freon 12, 0.35MPa and 0.036 m^3/kg\n","The Quality of Freon 12  = ",round(x,2)

#Variable declaration for (d)
# (d).Methane 0.5MPa and 1.0 m^3/kmol
v=1;                         # specific volume of Methane in m^3/kmol
# From table at 0.5 MPa molar values are
vf=0.04153; vg=2.007;        # specific volume  of Methane in m^3/kmol

#Calculation for (d)
x=(v-vf)/(vg-vf);            # Quality of vapour since v<vg

#Result for (d)
print "\n(d).Methane 0.5MPa and 1.0 m^3/kmol","\nThe Quality of Methane  = ",round(x,4)

(a) Ammonia 26 oC and 0.074 m^3/kg
The Quality of ammonia =  0.589

(b).Ammonia 550kPa and 0.31m^3/kg
Temperature of ammonia =  82.1 oC

(c).Freon 12, 0.35MPa and 0.036 m^3/kg
The Quality of Freon 12  =  0.73

(d).Methane 0.5MPa and 1.0 m^3/kmol
The Quality of Methane  =  0.4877


## Example 3.3, Page No:83¶

In [3]:
#Variable declaration
V=300;        # Volume of air in the room in m^3
p=1;          # Atmospheric pressure in bar
T=25;         # Temperature of air in Degree Celcius
R=287;        # Characteristic constant of Air in J/kg k

#Calculation
m=(p*10**5*V)/(R*(T+273)); # Ideal gas equation

#Result
print "Mass of air in room = ",round(m,2),"kg"

Mass of air in room =  350.0 kg


## Example 3.4, Page No:84¶

In [4]:
#Variable declaration
D=20;                      # Diameter of the sphere in cm
m=2.54;                    # Mass of gas filled in sphere in gram
p=10;                      # Pressure of gas in bar
T=25;                      # Temperature of gas in Degree Celcius
R=8.3144*10**3;            # Universal gas constant in J/kmol K

#Calculation
V=(3.14*(D*10**-2)**3)/16; # Volume of das in sphere in m^3
M=(m*10**-3*R*(T+273))/(p*10**5*V); # Molecular weight of the gas

#Result
print "Molecular weight of the gas = ",round(M,3),"\nTherefore gas in sphere is Helium (unless mixture of two or more gases)"

Molecular weight of the gas =  4.008
Therefore gas in sphere is Helium (unless mixture of two or more gases)


## Example 3.5, Page No:85¶

In [5]:
#Variable declaration
p2=2.5;                        # Pressure of air in the cylinder in bar
T1=430;                        # Temperature of air in cylinder in Degree Celcius
V1=1.2;                        # Volume of cylinder in m^3
V2=0.6;                        # Volume of cylinder upto end stops in m^3

#Calculation for (a)
# (a) Temperature of air when the piston reaches the stops
T2=(T1+273)*(V2/V1);           # constant pressure process

#Result for (a)
print "(a).Temperature of air when the piston reaches the stops = ",round(T2,1),"K"

#Calculation for (b)
# (b) The pressure of air when its temperature equals to 25 oC
T3=25;                        #Room temperature in Degree Celcius
p3=p2*((T3+273)/T2);          # constant volume process

#Result for (b)
print "\n(b).The pressure of air when its temperature equals to 25 oC = ",round(p3,2),"bar"

(a).Temperature of air when the piston reaches the stops =  351.5 K

(b).The pressure of air when its temperature equals to 25 oC =  2.12 bar


## Example 3.7, Page No:92¶

In [6]:
import math
from __future__ import division

#Variable declaration
p=6000;                   # Pressure of nitrogen gas in kPa
T=150;                    # Temperature of nitrogen gas in kelvin
V=250;                    # Volume of tank in litres
R_1=8.3143;               # Universal gas constant in kJ/kmol K
M=28.1013;                # Molecular mass

#Calculation for (a)
# (a).Beattie - Bridgeman equation of state
# Constants for nitrogen gas
c=4.2*10**4; Ao=136.2315; a=0.02617; Bo=0.05046; b=-0.00691;
# By substituting these values in the following equation
# p=(R_1*T/v^2)*(1-(c/(vT^3)))*(v+Bo*(1-(b/v)))-(Ao/v^2*(1-(a/v)))
# By trial and error we get
v=0.1222;                # specific volume in m^3/kmol
m=(M*V/1000)/v;          # Mass of nitrogen gas

#Result for (a)
print "(a).Beattie - Bridgeman equation of state","\nMass of nitrogen gas = ",round(m,2),"kg"
print "specific volume of nitrogen gas = ",round(v,4),"m^3/kmol"

#Calculation for (b)
# (b).Nitrogen tables
# From property table of nitrogen fas
v=0.004413;            # specific volume in m^3/kg
m=(V/1000)/v;          #  Mass of nitrogen gas

#Result for (b)
print "\n\n(b).Nitrogen tables","\nMass of nitrogen gas = ",round(m,2),"kg","\nspecific volume of nitrogen gas = ",round(v,6),"m^3/kg"

#Calculation for (c)
# (c).Ideal gas equation of state
m=(p*V/1000)/(R_1*T/M);  #Mass of nitrogen gas

#Result for (c)
print "\n\n(c).Ideal gas equation of state","\nMass of nitrogen gas = ",round(m,2),"kg"

#Calculation for (d)
# (d).Generalized compressibility chart
# The crictical properties for nitrogen gas
Tc=126.2;              # Temperature in kelvin
Pc=3.349;              # Pressure in MPa
# Reduced properties are
Pr=p/Pc; Tr=T/Tc;
z=0.6;                 # From chart
m=(p*V/1000)/(z*R_1*T/M); #Mass of nitrogen gas

#Result for (d)
print "\n\n(d).Generalized compressibility chart","\nMass of nitrogen gas = ",round(m,2),"kg"

#Result for (e)
#(e).Arrangement the methods in order of percentage error
print " \n\n(e).Arrangement the methods in order of percentage error : "
print "Nitrogen tables","\nBeattie - Bridgeman equation of state","\nGeneralized compressibility chart","\nIdeal gas equation of state"

(a).Beattie - Bridgeman equation of state
Mass of nitrogen gas =  57.49 kg
specific volume of nitrogen gas =  0.1222 m^3/kmol

(b).Nitrogen tables
Mass of nitrogen gas =  56.65 kg
specific volume of nitrogen gas =  0.004413 m^3/kg

(c).Ideal gas equation of state
Mass of nitrogen gas =  33.8 kg

(d).Generalized compressibility chart
Mass of nitrogen gas =  56.33 kg

(e).Arrangement the methods in order of percentage error :
Nitrogen tables
Beattie - Bridgeman equation of state
Generalized compressibility chart
Ideal gas equation of state


## Example 3.8, Page No:97¶

In [7]:
#Variable declaration
T=-58.7;              #Normal boling point of CF3Br in Degree Celcius
Tc=340.9;             # Crictical temperature of CF3Br in K
pc=4.05;              # Crictical pressure of CF3Br in MPa
M=148.9;              # Moleclar mass of CF3Br
p=1.01325*10**5;      # Atmospheric pressure in N/m^2
R1=8314.4;            # Universal gas constant in J/kmol K
R=R1/M;               # Gas constant of CF3Br

#Calculation
a=(0.42748*R**2*Tc**2.5)/(pc*10**6); # Constant of Redlich-Kwong equation of state
b=(0.08664*R*Tc)/(pc*10**6); # Constant of Redlich-Kwong equation of state
vi=(R*(T+273))/p; # Ideal gas volume for assigning initial value
# By substituting these values in the Redlich-Kwong equation of state
vi_1=(R*(T+273)/p)+b-((a/(p*(273+T)**0.5*vi))*((vi-b)/(vi+b))); #and solving it by trial and error method we get

#Result
print "Saturated vapour volume = ",round(vi_1,5),"m^3/kg   (roundoff error)"

Saturated vapour volume =  0.1145 m^3/kg   (roundoff error)