# Chapter3:PROPERTIES OF A PURE SUBSTANCE¶

## Ex3.1:pg-57¶

In [2]:
#example 1
#determinig the phase of water

print"from the table,we find that at 120C,saturation pressure of water is 198.5 kPa.But here we have pressure of 500 kPa.hence,water exists as a compressed liquid here."
print"also at 120C,vf=0.00106 kg/m^3 and vg=0.89186 kg/m^3.given v=0.5 m^3/kg i.e. vf<v<vg,so we have two phase mixture of liquid and vapor."

from the table,we find that at 120C,saturation pressure of water is 198.5 kPa.But here we have pressure of 500 kPa.hence,water exists as a compressed liquid here.
also at 120C,vf=0.00106 kg/m^3 and vg=0.89186 kg/m^3.given v=0.5 m^3/kg i.e. vf<v<vg,so we have two phase mixture of liquid and vapor.


## Ex3.2:pg-58¶

In [3]:
#example 2
#determinig the phase

print"from the table,we find that at 30C,saturation pressure of ammonia is 1167 kPa.But here we have pressure of 1000 kPa.hence,ammonia exists in superheated vapor state."
print"for R-22 at 200 kPa,vg=0.1119 kg/m^3.given v=0.15 m^3/kg i.e. v>vg,so the state is superheated vapor"

from the table,we find that at 30C,saturation pressure of ammonia is 1167 kPa.But here we have pressure of 1000 kPa.hence,ammonia exists in superheated vapor state.
for R-22 at 200 kPa,vg=0.1119 kg/m^3.given v=0.15 m^3/kg i.e. v>vg,so the state is superheated vapor


## Ex3.3:pg-59¶

In [13]:
#example 3
#determining the quality and specific volume

v1=0.5 #given specific volume in m^3/kg
vf=0.001073 #specific volume when only liquid phase is present in m^3/kg
vfg=0.60475 #in m^3/kg
x=(v1-vf)/vfg #quality
print"For water at a pressure of 300 kPa,the state at which v1 is 0.5 m^3/kg is seen to be in the liquid-vapor two-phase region,at which T=133.6 C and the quality is",round(x,3)

v2=1 #given specific volume in m^3/kg

# using the method of interplotation
T=((400-300)*(1.0-0.8753))/(1.0315-0.8753)+300 #temperature of the water
print"For water at a pressure of 300 kPa,the state at which v2 is 1 m^3/kg is seen to be in the liquid-vapor two-phase region,the temperature is",round(T,1)

For water at a pressure of 300 kPa,the state at which v1 is 0.5 m^3/kg is seen to be in the liquid-vapor two-phase region,at which T=133.6 C and the quality is 0.825
For water at a pressure of 300 kPa,the state at which v2 is 1 m^3/kg is seen to be in the liquid-vapor two-phase region,the temperature is 379.8


## Ex3.4:pg-60¶

In [7]:
 #example 4
#percentage of vapor

vliq=0.1 #volume of saturated liquid in m^3
vf=0.000843 #in m^3/kg
vvap=0.9 #volume of saturated vapor R-134a in equilbrium
vg=0.02671 #in m^3/kg
mliq=vliq/vf #mass of liquid in kg
mvap=vvap/vg #mass of vapor in kg
m=mliq+mvap #total mass in kg
x=mvap*100/m #percentage of vapor on mass basis
print"hence,% vapor on mass basis is",round(x,1),"%"

hence,% vapor on mass basis is 22.1 %


## Ex3.4E:pg-60¶

In [1]:
 #example 5
#percentage of vapor

vliq=0.1 #volume of saturated liquid in ft^3
vf=0.0136 #in ft^3/lbm
vvap=0.9 #volume of saturated vapor R-134a in equilbrium
vg=0.4009 #in ft^3/lbm
mliq=vliq/vf #mass of liquid in lbm
mvap=vvap/vg #mass of vapor in lbm
m=mliq+mvap #total mass in lbm
x=mvap*100/m #percentage of vapor on mass basis
print"hence,% vapor on mass basis is",round(x,1),"%"

hence,% vapor on mass basis is 23.4 %


## Ex3.5:pg-61¶

In [7]:
#example 5

v1=0.14922 #specific volume of sautrated ammonia in m^3/kg
print"Since the volume does not change during the process,the specific volume remains constant.therefore"
v2=v1 #in m^3/kg
print"Since vg at 40C is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that"
P2=945 #final pressure in kPa
print"hence,the final pressure is 945 kPa"

Since the volume does not change during the process,the specific volume remains constant.therefore
Since vg at 40C is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that
hence,the final pressure is 945 kPa


## Ex3.5E:pg-61¶

In [8]:
#example 5

v1=2.311 #specific volume of sautrated ammonia in ft^3/lbm
print"Since the volume does not change during the process,the specific volume remains constant.therefore"
v2=v1 #in ft^3/lbm
print"Since vg at 120f is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that"
P2=145 #final pressure in lbf/in^2
print"hence,the final pressure is 145 lbf/in^2"

Since the volume does not change during the process,the specific volume remains constant.therefore
Since vg at 120f is less than v2,it is evident that in the final state the Ammonia is superheated vapor.By interplotation,we find that
hence,the final pressure is 145 lbf/in^2


## Ex3.6:pg-61¶

In [19]:
#example 6
#Determinig the missing property

T1=273-53.2 #given temperature in K
P1=600 #given pressure in kPa
print"This temperature is higher than the critical temperature (critical temp. at P=600 kPa) is 96.37 K.Hence,v=0.10788 m^3/kg"
T2=100 #given temp. in K
v2=0.008 #given specific volume in m^3/kg
vf=0.001452 #in m^3/kg
vg=0.0312 #in m^3/kg
Psat=779.2 #saturation pressure in kPa
vfg=vg-vf #in m^3/kg
x=(v2-vf)/vfg #quality
print"\n hence, the pressure is",round(Psat,1),"kPa"
print"\n and quality is",round(x,4),"%"

This temperature is higher than the critical temperature (critical temp. at P=600 kPa) is 96.37 K.Hence,v=0.10788 m^3/kg

hence, the pressure is 779.2 kPa

and quality is 0.2201 %


## Ex3.7:pg-62¶

In [21]:
#example 7
#determining the pressure of water

vg=0.12736 #specific volume in m^3/kg for water at 200C
v=0.4 #specific volume in m^3/kg
P1=500 #in kPa
v1=0.42492 #specific volume at P1 in m^3/kg
P2=600 #in kPa
v2=0.35202 #specific volume at P2 in m^3/kg
P=P1+(P2-P1)*(v-v1)/(v2-v1) #calculating pressure by interplotation
print "hence,the pressure of water is",round(P,1)," kPa"

 hence,the pressure of water is 534.2  kPa


## Ex3.8:pg-66¶

In [27]:
#example 8
#calculating mass of air

P=100 #pressure in kPa
V=6*10*4 #volume of room in m^3
R=0.287 #in kN-m/kg-K
T=25 #temperature in Celsius
m=P*V/(R*(T+273.1)) #mass of air contained in room
print"\n hence, mass of air contained in room is",round(m,1),"kg"

 hence, mass of air contained in room is 280.5 kg


## Ex3.9:pg-67¶

In [30]:
#example 9
#calculating pressure inside tank

V=0.5 #volumr of tank in m^3
m=10 #mass of ideal gas in kg
T=25 #temperature of tank in Celsius
M=24 #molecular mass of gas in kg/kmol
Ru=8.3145 #universal gas constant in kN-m/kmol-K
R=Ru/M #gas constant for given ideal gas in kN-m/kg-K
P=m*R*(T+273.2)/V #pressure inside tank
print"\n hence,pressure inside tank is",round(P),"kpa"

 hence,pressure inside tank is 2066.0 kpa


## Ex3.10:pg-67¶

In [40]:
#example 10
#mass  flow rate

dt=185 #time period in seconds over which there is incrrease in volume
dV=0.75 #increase in volume in 0.75 in m^3
V=dV/dt #volume flow rate in m^3/s
P=105 #pressure inside gas bell kPa
T=21 #temperature in celsius
R=0.1889 #ideal gas constant in kJ/kg-K
m=P*V/(R*(T+273.15)) #mass flow rate of the flow in kg/s
print"\n hence,mass flow rate is",round(m,5),"kg/s"
print"\n and volume flow rate is",round(V,5),"m^3/s"
#The answer of volume flow rate in the book is wrong.

 hence,mass flow rate is 0.00766 kg/s

and volume flow rate is 0.00405 m^3/s


## Ex3.12:pg-71¶

In [51]:
#example 12
#determining specific using diffenet laws

T=100.0 #given temp.in 100 celsius
P=3.0 #given pressure in MPa
v1=0.0065 #specific volume in m^3/kg using table
print"\n hence,the specific volume for R-134a using R-134a tables is",round(v1,3),"m^3/kg"
M=102.3 #molecular mass in kg
R=8.3145 #in kJ/K
Ru=R/M #in kJ/K-kg
v2=Ru*(T+273)/(P*1000) #specific volume assuming R-134a to be ideal gas in m^3/kg
print"\n hence,the specific volume for R-134a using R-134a the ideal gas laws is",round(v2,5),"m^3/kg"
Tr=373.2/374.2 #reduced temperature using generalized chart
Pr=3.0/4.06 #reduced pressure using generalized chart
Z=0.67 #compressibility factor
v3=Z*v2 # specific volume using generalized chart in m^3/kg
print"\n hence,the specific volume for R-134a using the generalized chart is",round(v3,5),"m^3/kg"

 hence,the specific volume for R-134a using R-134a tables is 0.006 m^3/kg

hence,the specific volume for R-134a using R-134a the ideal gas laws is 0.01011 m^3/kg

hence,the specific volume for R-134a using the generalized chart is 0.00677 m^3/kg


## Ex3.13:pg-71¶

In [44]:
#example 13
#calculating mass of gas

Pc=4250 #critical pressure of propane in kPa
Tc=369.8 #critical temperature in K
T=15 #temperature of propane in celsius
Tr=T/Tc #reduced temperature
Prsat=0.2 # reduced pressure
P=Prsat*Pc #pressure in kPa
x=0.1 #given quality
Zf=0.035 #from graph
Zg=0.83 #from graph
Z=(1-x)*Zf+x*Zg #overall compressibility factor
V=0.1 #volume of steel bottle in m^3
R=0.1887 #in kPa-m^3/kg-K
m=P*V/(Z*R*(T+273)) #total propane mass in kg
print"\n hence,the total propane mass is",round(m,2),"kg"
print"\n and pressure is",round(P,2),"kPa"

 hence,the total propane mass is 13.66 kg

and pressure is 850.0 kPa