# Chapter 2:Properties of Pure Substances¶

## Ex2.1, PG-28¶

In [4]:
# initialization of variables

m=10;  #mass of saturated water in kg

# All the necessary values are taken from table C.2

# part (a)

P=0.001; # Pressure in MPa
vf=0.001; #specific volume of saturated liquid at 0.001 Mpa in Kg/m^3
vg=129.2; # specific volume of saturated vapour at 0.001 Mpa in Kg/m^3
deltaV=m*(vg-vf) # by properties of pure substance
# result
print "The Volume change at pressure ",(P)," MPa is",round(deltaV)," m^3 \n"

# part (b)

P=0.26;  # Pressure in MPa
vf=0.0011; #   specific volume of saturated liquid at 0.26 MPa( it is same from at 0.2 and 0.3 MPa upto 4 decimals)
vg=(P-0.2)*(0.6058-0.8857)/(0.3-0.2)+0.8857;  # specific volume of saturated vapour by interpolation of Values at 0.2 MPa and 0.3 MPa
deltaV=m*(vg-vf) # by properties of pure substance
# result
print "The Volume change at pressure ",(P)," MPa is",round(deltaV,2)," m^3 \n"

# part (c)
P=10;  # Pressure in MPa
vf=0.00145; # specific volume of saturated liquid at 10 MPa
vg=0.01803; # specific volume of saturated vapour at 10 MPa
deltaV=m*(vg-vf) # by properties of pure substance
# result
print "The Volume change at pressure ",(P)," MPa is",round(deltaV,4)," m^3 \n"

The Volume change at pressure  0.001  MPa is 1292.0  m^3

The Volume change at pressure  0.26  MPa is 7.17  m^3

The Volume change at pressure  10  MPa is 0.1658  m^3



## Ex2.2, PG-29¶

In [8]:
#    initialization of variables
m=4.0 # mass of water in kg
V=1.0 # volume in m^3
T=150 # temperature of water in degree centigrade

# TABLE C.1 is used for values in wet region
# Part (a)
P=475.8 # pressure in KPa in wet region at temperature of 150 *C
print " The pressure is",round(P,2)," kPa \n"

# Part (b)
#first we determine the dryness fraction
v=V/m # specific volume of water
vg=0.3928 #  specific volume of saturated vapour @150 degree celsius
vf=0.00109 # specific volume of saturated liquid @150 degree celsius
x=(v-vf)/(vg-vf); # dryness fraction
mg=m*x; # mass of vapour
print " The mass of vapour present is",round(mg,3)," kg \n"

# Part(c)
Vg=mg*vg; # volume of vapour
print " The volume of vapour is",round(Vg,3)," m^3"

 The pressure is 475.8  kPa

The mass of vapour present is 2.542  kg

The volume of vapour is 0.998  m^3


## Ex2.3, PG-29¶

In [10]:
#    initialization of variables
m=2 # mass of water in kg
P=220 # pressure in KPa
x=0.8 # quality of steam

# Table C.2 is used for values

vg=(P-200)*(0.6058-0.8857)/(300-200)+0.8857 # specific volume of saturated vapour @ given pressure by interpolating
vf=0.0011 # specific volume of saturated liquid @ 220 KPa
v=vf+x*(vg-vf)# property of pure substance
V=m*v # total volume
#result
print "The Total volume of the mixture is ",round(V,3)," m^3"

The Total volume of the mixture is  1.328  m^3


## Ex2.4, PG-30¶

In [12]:
#   initialization of variables
m=2 # mass of water in kg
P=2.2 # pressure in Mpa
T=800 # temperature in degree centigrade
# Table C.3 is used for values
v=0.2467+(P-2)*(0.1972-0.2467)/(2.5-2) # specific volue by interpolatin between 2 and 2.5 MPa
V=m*v # final volume
print "The Final Volume is",round(V,3)," m^3"

The Final Volume is 0.454  m^3


## Ex2.5, PG-32¶

In [14]:
#    initialization of variables
V=0.6 # volume of tyre in m^3
Pgauge=200 # gauge pressure in KPa
T=20+273 # temperature converted to kelvin
Patm=100 # atmospheric pressure in KPa
R=287 # gas constant in Nm/kg.K
Pabs=(Pgauge+Patm)*1000 # calculating absolute pressue in Pa

m=Pabs*V/(R*T)# mass from ideal gas equation
print "The Mass of air is",round(m,2)," Kg"

The Mass of air is 2.14  Kg


## Ex2.6, PG-33¶

In [23]:
import math
#    initialization of variables
T=500+273 # temperature of steam in kelvin
rho=24.0 # density in Kg/m^3
R=0.462 # gas constant from Table B.2
v=1/rho # specific volume and density relation
# PART (a)
P=rho*R*T # from Ideal gas equation
print " PART (a) The Pressure is ",int(P)," KPa \n"
# answer is approximated in textbook

# PART (b)
a=1.703 #  van der Waal's constant a value from Table B.7
b=0.00169 # van der Waal's constant b value from Table B.7
P=(R*T/(v-b))-(a/v**2) # Pressure from van der Waal's equation
print " PART (b) The Pressure is ",int(P)," KPa \n"
# answer is approximated in textbook

# PART (c)
a=43.9 # van der Waal's constant a value from Table B.7
b=0.00117 # van der Waal's constant b value from Table B.7

P=(R*T/(v-b))-(a/(v*(v+b)*math.sqrt(T))) # Redlich-Kwong equation
print " PART (c) The Pressure is ",int(P)," KPa \n"
# answer is approximated in textbook

# PART (d)
Tcr=947.4 # compressibilty temperature from table B.3
Pcr=22100 # compressibility pressure from table B.3

TR=T/Tcr # reduced temperature
PR=P/Pcr # reduced pressure
Z=0.93 # from compressiblility chart
P=Z*R*T/v # Pressure in KPa
print " PART (d) The Pressure is ",int(P)," KPa \n"
# answer is approximated in textbook

# PART (e)
P=8000 # pressure from steam table @ 500*c and v= 0.0417 m^3
print " PART (e) The Pressure is ",int(P)," KPa \n"
# answer is approximated in textbook

 PART (a) The Pressure is  8571  KPa

PART (b) The Pressure is  7952  KPa

PART (c) The Pressure is  7934  KPa

PART (d) The Pressure is  7971  KPa

PART (e) The Pressure is  8000  KPa