# Chapter 7 : Thermodynamic property relations¶

### Example 7.10 Page No : 259¶

In [1]:
import math

# Variables
T2 = 150.   			 #temperature at which water it is desired to boil water in degree celsius
P1 = 0.10133			 #ambient pressure in MPa
T1 = 100.	    		 #temperature at which water boils corresponding at pressure P1 in degree celsius
del_hv = 2256.94		 #enthalpy of vaporization in kJ/kg
R = 8.314;		    	 #universal gas constant in J/molK
M = 18*10**-3;			 #molar mass of water in kg/mol

# Calculations
T1 = T1+273.15
T2 = T2+273.15
P2 = P1*(math.exp(((del_hv*10**3*M)*((1./T1)-(1./T2)))/(R)))

# Results
print " The approximate pressure at which the boiler is to be operated = %0.3f MPa"%(P2);

 The approximate pressure at which the boiler is to be operated = 0.476 MPa


### Example 7.11 Page No : 259¶

In [3]:
# Variables
m = 60.         			 #mass of the person who wants to skate in kg
T = -2.			             #temperature of the ice in degree celsius
A = 15.			             #area of contact between the skate edges and ice in mm**2
vs = 1.091*10**-3;			 #specific volume of ice in m**3/kg (at Tref)
vf = 1.0*10**-3;			 #specific volume of water in m**3/kg (at Tref)
del_hf = 6.002;		    	 #enthalpy of melting of ice in kJ/mol
g = 9.81;			         #accleration due to gravity in m/s**2
Tref = 0.       			 #reference temperature at which the specific enthalpy of ice and water are taken in degree celsius

# Calculations
Tref = Tref+273.15
del_P = ((m*g)/(A*10**-6))*10**-6
del_v = (vf-vs)*(18*10**-3);
del_T = (del_P*10**6)/((del_hf*10**3)/(Tref*del_v));

# Results
print " The temperature of ice originally  =  %d degree celsius "%(T);
print " The reduction in melting point of ice due to the additional pressure,computed using the Clayperon equation  =  %0.2f degree celsius "%(del_T);
if del_T<T :
print  " The ice can melt due to the additional pressure and therefore it will be possible to skate "
else:
print  " The ice will not melt and therefore it will be difficult to skate "

 The temperature of ice originally  =  -2 degree celsius
The reduction in melting point of ice due to the additional pressure,computed using the Clayperon equation  =  -2.93 degree celsius
The ice can melt due to the additional pressure and therefore it will be possible to skate


### Example 7.12 Page No : 261¶

In [4]:
# Variables
T1 = 100.	    		 #temperature of water in degree celsius
del_hv1 = 2256.94;			 #enthalpy of vaporization at T1 in kJ/kg
T2 = 150.   			 #temperature at which the enthalpy of vaporization is to be determined in degree celsius
Cp_f = 4.26	    		 #isobaric heat capacity of liquid in kJ/kgK
Cp_g = 1.388			 #isobaric heat capacity of vapour in kJ/kgK

# Calculations
del_hv2 = ((Cp_g-Cp_f)*(T2-T1))+del_hv1

# Results
print " The enthalpy of vaporization at 150 degree celsius = %0.2f kJ/kg"%(del_hv2);

 The enthalpy of vaporization at 150 degree celsius = 2113.34 kJ/kg


### Example 7.13 Page No : 261¶

In [6]:
# Variables
T1 = 100.           			 #temperature of water in degree celsius
del_hv1 = 2256.94   			 #enthalpy of vaporization at T1 in kJ/kg
T2 = 150.   		        	 #temperature at which the enthalpy of vaporization is to be determined in degree celsius
del_hv_kirchoff = 2113.34			 #enthalpy of vaporization predicted by the Kirchhoff relation taken from Example 7.12 for comparison, in kJ/kg
del_hv_steam_tables = 2113.25			 #enthalpy of vaporization taken from the steam tables corresponding to T2,for comparison, in kJ/kg
Tc = 647.3          			 #critical temperature of water in K

# Calculations
T1 = T1+273.15
T2 = T2+273.15
Tr1 = T1/Tc
Tr2 = T2/Tc
del_hv2 = del_hv1*(((1-Tr2)/(1-Tr1))**0.38)

# Results
print " The enthalpy of vaporization at 150 degree celsius using ";
print " Watson correlation \t  =  %f kJ/kg"%(del_hv2);
print " Kirchhoffs relation \t  =  %f kJ/kg"%(del_hv_kirchoff);
print " From steam tables \t  =  %f kJ/kg"%(del_hv_steam_tables);

 The enthalpy of vaporization at 150 degree celsius using
Watson correlation 	  =  2090.687877 kJ/kg
Kirchhoffs relation 	  =  2113.340000 kJ/kg
From steam tables 	  =  2113.250000 kJ/kg


### Example 7.14 Page No : 262¶

In [7]:
import math
# Variables
T = 373.15  			 #normal boiling point of water in K (temperature at which the enthalpy of vaporization is to be determined)
Pc = 221.2	    		 #critical pressure of water in bar
Tc = 647.3		    	 #critical temperature of water in K
R = 8.314			     #universal gas constant in J/molK
del_hvn_steam_tables = 2256.94			 #enthalpy of vaporization at the normal boiling point taken from the steam tables, for comparison, in kJ/kg

# Calculations
Tbr = T/Tc
del_hvn = ((1.093*R*Tc*(Tbr*((math.log(Pc)-1.013)/(0.930-Tbr))))*10**-3)/(18*10**-3);
err = abs((del_hvn-del_hvn_steam_tables)/del_hvn_steam_tables)*100

# Results
print " The enthalpy of vaporization at the normal boiling point ";
print " Using Riedels correlation \t  =  %f kJ/kg"%(del_hvn);
print " From the steam tables \t \t  =  %f kJ/kg"%(del_hvn_steam_tables);
print " Error \t \t \t \t  =  %f %% "%(err);

 The enthalpy of vaporization at the normal boiling point
Using Riedels correlation 	  =  2337.179008 kJ/kg
From the steam tables 	 	  =  2256.940000 kJ/kg
Error 	 	 	 	  =  3.555212 %