# Chapter 4 : Second Law of Thermodynamics¶

### Example 4.1, Page no:90¶

In [21]:
# Variables
#Given:
T1 = 700.; 			#temperature of heat source(K)
T2 = 300.; 			#temperature of heat sink(K)

# Calculations
#To calculate the maximum efficiency
eff=((T1-T2)/T1); 			#efficiency of a heat engine

# Results
print 'Maximum efficiency of heat engine is %f'%eff

Maximum efficiency of heat engine is 0.571429


### Example 4.2, Page no:90¶

In [22]:
# Variables
#Given:
m = 1.; 			#mass of water(kg)
T1 = 300.; 			#temperature of surrounding(K)
T2 = 273.; 			#temperature of water(K)
Hf = 334.11; 			#latent heat of fusion of ice(kJ/kg)

# Calculations and Results
#To determine minimum amount of work and heat given upto surrounding
#(a)
Q2 = m*Hf; 			#heat absobed at temperature T2
W = ((Q2*(T1-T2))/T2); 			#minimumm amount of work required
print 'Minimum amount of work required is %f kJ'%W

#(b)
#Q1 is the heat given up the surrounding
Q1 = W+Q2;
print 'Heat given upto surrounding is %f kJ'%Q1

Minimum amount of work required is 33.043846 kJ
Heat given upto surrounding is 367.153846 kJ


### Example 4.3, Page no:90¶

In [23]:
# Variables
#Given:
P_out = 4.5; 			#output power(hp)
P_in = 6.25; 			#input power(kW)
T1 = 1000.; 			#source temperature(K)
T2 = 500.; 			#sink temperature(K)

# Calculations and Results
#To determine efficiency of proposed engine
ep = ((P_out*745.7)/(P_in*1000)); 			#proposed efficiency
print 'Efficiency of proposed engine is %f'%ep

em = ((T1-T2)/T1); 			#maximum efficiency
print 'The maximum efficieny is %f'%em
print 'Hence the claim of the proposed engine is impossible'

Efficiency of proposed engine is 0.536904
The maximum efficieny is 0.500000
Hence the claim of the proposed engine is impossible


### Example 4.4, Page no:93¶

In [24]:
# Variables
#Given:
P = 500.; 			#pressure of dry saturated steam(kPa)

#From steam tables
Hv = 2106.; 			#latent heat of vaporisation(kJ/kg)
T = 425.; 			#saturation temperature(K)

# Calculations
#To calculate the entropy of evaporation
#By equation 4.25 (Page no. 93)
Sv = (Hv/T); 			#entropy change accompanying vaporisation

# Results
print 'Entropy of evaporation is %f kJ/kg K'%Sv

Entropy of evaporation is 4.955294 kJ/kg K


### Example 4.5, Page no:94¶

In [25]:
# Variables
#Given:
m = 2.; 			#mass of gas(kg)
T1 = 277.; 			#initial temperature(K)
T2 = 368.; 			#final temperature(K)
Cv = 1.42; 			#specific geat at constant volume(kJ/kg K)

# Calculations
import math
#Using equation 4.31 (Page no. 94)
S = (m*Cv*math.log(T2/T1)); 			#change in entropy(kJ/K)

# Results
print 'Change in entropy is %f kJ/K'%S

Change in entropy is 0.806746 kJ/K


### Example 4.6, Page no:94¶

In [26]:
#Given:
T = 300.; 			#temperature in K
P1 = 10.; 			#initial pressure(bar)
P2 = 1.; 			#final pressure(bar)
R = 8.314; 			#ieal gas constant

# Calculations
import math
#To calculate the entropy change
#Using equation 4.33(Page no. 94)
S = (R*math.log(P1/P2)); 			#(kJ/kmol K)

# Results
print 'Entopy change is %f kJ/kmol K'%S

Entopy change is 19.143692 kJ/kmol K


### Example 4.7, Page no:94¶

In [27]:
# Variables
#Given:
T1 = 335.; 			#initial temperature in K
T2 = 300.; 			#final temperature in K
P1 = 10.; 			#initial pressure(bar)
P2 = 1.; 			#final pressure(bar)
Cp = 29.3; 			#specific heat constant at constant pressure(kJ/kmol K)
R = 8.314; 			#ideal gas constant

# Calculations
import math
#To determine change in entropy
#Using equation 4.30 (Page no. 94)
S = ((Cp*math.log(T2/T1))-(R*math.log(P2/P1))); 			#entropy change(kJ/kmol K)

# Results
print 'Entropy change in the process is %f kJ/kmol K'%S

Entropy change in the process is 15.910494 kJ/kmol K


### Example 4.8, Page no:95¶

In [28]:
# Variables
#Given:
m1 = 10.; 			#mass of water at 375 K (kg)
m2 = 30.; 			#mass of water at 275 K (kg)
c = 4.2; 			#specific heat of water (kJ.kg K)

# Calculations
import math
#To determine the change in entropy
#Let T be the final temperature(K)
T = ((m1*375)+(m2*275))/(m1+m2);
#S1 be change in entropy for hot water
S1 = (m1*c*math.log(T/375)); 			#[kJ/K]
#S2 be the change in entropy for cold water
S2 = (m2*c*math.log(T/275)); 			#[kJ/K]
#S be the total entropy change
S = S1+S2;

# Results
print 'The total entropy change is %f kJ/K'%S

The total entropy change is 1.591404 kJ/K


### Example 4.9, Page no:95¶

In [29]:
# Variables
#Given:
m1 = 35.; 			#mass of steel in kg
m2 = 150.; 			#mass of oil in kg
T1 = 725.; 			#temperature of steel(K)
T2 = 275.; 			#temperature of oil(K)
c1 = 0.88; 			#specific heat of steel (kJ/kg K)
c2 = 2.5; 			#specific heat of oil(kJ/kg K)

# Calculations
import math
#To calculate the total entropy change
#Let T be the final temperature
T = (((m1*c1*T1)+(m2*c2*T2))/((m1*c1)+(m2*c2)));
#S1 be the in entropy for steel
S1 = (m1*c1*math.log(T/T1)); 			#[kJ/K]
#S2 be the change in entropy for oil
S2 = (m2*c2*math.log(T/T2)); 			#[kJ/K]
#S be the total entropy change
S = S1+S2;

# Results
print 'The total entropy change is %f kJ/K'%S

The total entropy change is 17.649827 kJ/K


### Example 4.10¶

In [30]:
# Variables
#Given:
n1 = 0.21; 			#volume % of oxygen in air
n2 = 0.79; 			#volume % of nitrogen in air
R = 8.314; 			#ideal gas constant

# Calculations
import math
#To calculate entropy of 1 kmol of air
#Using equation 4.35 (Page no. 96)
S = (-R*(n1*math.log(n1)+n2*math.log(n2))); 			#[kJ/kmol K]

# Results
print 'The total entropy change is %f kJ/kmol K'%S

The total entropy change is 4.273036 kJ/kmol K


### Example 4.11¶

In [31]:
# Variables
H = -2.8318*10**5; 			#heat of reaction (J/mol)
T = 298.; 			#temperature of reaction in K
#Absolute entropies for CO, O2, CO2 are (in J/mol K)
S_CO = 198.;
S_O2 = 205.2;
S_CO2 = 213.8;

# Calculations and Results
# To determine the change in entropy for the reaction
# Referring equation 4.36 (Page no. 96)
S_reactant = S_CO + 0.5*S_O2; 			#entropy change for reactants
S_product = S_CO2; 			#entropy change for products
S = S_product-S_reactant; 			#total entropy change
print 'The total entropy change for the reaction is %f J/mol'%S
print 'Since the reaction is highly irreversible, entropy change cannot be calculated as the ratio of heat of reaction to the temperature'

#The energy available for useful work is the difference between heat of reaction and entropy energy due to ireversible nature of the process
W_useful = -H+(T*S); 			#energy available for useful work (J)
print 'Energy available for useful work is %3.2e J'%W_useful

The total entropy change for the reaction is -86.800000 J/mol
Since the reaction is highly irreversible, entropy change cannot be calculated as the ratio of heat of reaction to the temperature
Energy available for useful work is 2.57e+05 J


### Example 4.13¶

In [32]:
# Variables
#Given:
H_steam = 2923.5; 			#enthalpy of superheated steam (kJ/kg)
S_steam = 6.71; 			#entropy of superheated steam (kJ/kg K)
H_liquid = 845.; 			#enthalpy of saturated liquid (kJ/kg)
S_liquid = 2.32; 			#entropy of saturated liquid (kJ/kg K)
T = 300.; 			#temperature of system (K)

# Calculations
#To calculate change in entropy and check whether the process is reversible
S_system = S_liquid-S_steam; 			#change in entropy of steam

#Let Q be the heat given out during condensation
Q = -(H_liquid-H_steam);
S_surrounding = Q/T; 			#change in entropy of the surrounding
S_total = S_system+S_surrounding; 			#total entropy change

# Results
print 'The total entropy change is %f kJ/kg'%S_total
print 'Since total entropy change is positive,the process is irreversible'

The total entropy change is 2.538333 kJ/kg
Since total entropy change is positive,the process is irreversible


### Example 4.14¶

In [33]:
# Variables
#Given:
V = 1.; 			#volume of each compartment in cubic meters
P_sat = 683.6; 		#pressure of saturated steam (kPa)
P_steam = 101.3; 	#pressure of supereated steam (kPa)
T_sat = 437.2; 		#temperature of system (K)

#Referring steam tables
#For saturated steam at pressure 683.6 kPa and temp 437.2 K
H_sat = 2761.; 			    #enthalpy of saturated steam (kJ/kg)
S_sat = 6.7133; 			#entropy of saturated steam (kJ/kg K)
spvol_sat = 278.9*10**-3; 	#specific volume of saturated steam (cubic m/kg)
U_sat = 2570.4; 			#specific internal energy of saturated steam (kJ/kg)

#For superheated steam at 101.3 kPa and 437.2 K
H_steam = 2804.; 			    #enthalpy of superheated steam (kJ/kg)
S_steam = 7.6712; 			    #entropy of superheated steam (kJ/kg K)
spvol_steam = 1976.2*10**-3; 	#specific volume of superheated steam (cubic m /kg)
U_steam = 2603.3; 			    #specific internal energy of superheated steam (kJ/kg)

# Calculations
#To determine the change in entropy of system
m_sat = V/spvol_sat; 			#mass of satureated steam(kg)
m_steam = V/spvol_steam; 		#mass of superheated steam (kg)
m_sys = m_sat+m_steam; 			#mass of system (kg)
spvol_sys = (2.*V)/m_sys; 		#specific volume of system (cubic m/kg)
#Since no heat exchange and work interaction occurs so internal energy after mixing remains the same
U1_sat = m_sat*U_sat; 			    #internal energy of saturated steam (kJ)
U1_steam = m_steam*U_steam; 	    #internal enegy of superheated steam (kJ)
U_sys = (U1_sat+U1_steam)/m_sys; 	#specific internal energy of system (kJ/kg)

#Referring steam tables
#At calculated U_sys and spvol_sys
S_sys = 6.9992; 			               #specific entropy of system (kJ/kg K)
Si = ((m_sat*S_sat)+(m_steam*S_steam));    #initial entropy of system (kJ/K)
Sf = (m_sys*S_sys); 			           #final entropy of system (kJ/K)
S = Sf-Si; 			                       #change in entropy

# Results
print 'The change in entropy of the system is %f kJ/K'%S
print 'Since entropy change is positive, the process is irrevresible'

The change in entropy of the system is 0.685052 kJ/K
Since entropy change is positive, the process is irrevresible


### Example 4.15¶

In [34]:
# Variables
#Given:
V = 1.; 			#volume of each compartment in cubic m
T = 300.; 			#temperature of ideal gas in 1st compartment (K)
P = 200.; 			#pressure of ideal gas in 1st compartment (kPa)
R = 8.314; 			#ideal gas constant

# Calculations
#To calculate entropy change
#Let n be the number of moles of gas
n = ((P*V)/(R*T));
#Since gas in vessel exchanges no heat and work with surrounding so internal energy remains same
#This implies temperature after mixing is same as that before mixing

#Final conditions:
Tf = 300.; 			#final temperature (K)
Vf = 2.; 			#final volume (cubic m)
Pf = 100.; 			#final pressure (kPa)

#Initial conditions:
Ti = 300.; 			#initial temperature (K)
Vi = 1.; 			#initial volume (cubic m)
Pi = 200.; 			#initial pressure (kPa)
import math
#Using equation 4.33 (Page num 94)
S = n*R*math.log(Vf/Vi); 			#entropy change of system (kJ/K)
#Since entropy of surrounding does not change
S_total = S; 			#total entropy change

# Results
print 'The change in total entropy is %f kJ/K'%S_total

The change in total entropy is 0.462098 kJ/K


### Example 4.16¶

In [35]:
# Variables
#Given:
m_oil = 5000.; 			#mass flow rate of oil (kg/h)
Tin_oil = 500.; 			#inlet temperature of oil (K)
Tin_water = 295.; 			#inlet temperature of water (K)
c_oil = 3.2; 			#specific heat of oil (kJ/kg K)
c_water = 4.2; 			#specific heat of water (kJ/kg K)
import math
#To calculate entropy change in the process
#Assuming oil is cooled to minimum permissible temperature
Tout_oil = 305.; 			#exit temperature of oil (K)
Tout_water = 490.; 			#exit temperature of water (K)

# Calculations
#Let m_water be the mass flow rate of water
#By enthalpy balance
m_water = ((m_oil*c_oil*(Tin_oil-Tout_oil))/(c_water*(Tout_water-Tin_water)));  			#(kg/h)
S_oil = m_oil*c_oil*math.log(Tout_oil/Tin_oil); 			#entropy change of oil (kJ/K)
S_water = m_water*c_water*math.log(Tout_water/Tin_water); 			#entropy change of water (kJ/K)
S_tot = S_oil+S_water; 			#total entropy change

# Results
print 'The total entropy change in the process is %f kJ/K'%S_tot

The total entropy change in the process is 210.139407 kJ/K


### Example 4.17¶

In [36]:
# Variables
#Given:
To = 275.; 			#temperature of quenching oil (K)

# Calculations
S_steel = -26.25; 			#change in entropy os casting (kJ/K)
S_oil = 43.90; 			#change in entropy of oil (kJ/K)
S_tot = S_steel+S_oil; 			#total entropy change
#Let W be loss in capacity for doing work
W = To*S_tot; 			#(kJ)

# Results
print 'The loss in capacity for doing work is %f kJ'%W

The loss in capacity for doing work is 4853.750000 kJ


### Example 4.18¶

In [37]:
import math
#Given:
m_oil = 5000.; 			    #mass flow rate of hydrocarbon oil (kg/h)
Tin_oil = 425.; 			#inlet temperature of oil (K)
Tout_oil = 340.; 			#exit temperature of oil (K)
m_water = 10000.; 			#mass flow rate of water (kg/h)
Tin_water = 295.; 			#inlet temperature of water (K)
c_oil = 2.5; 			    #mean specific heat of oil (kJ/kg K)
c_water = 4.2; 			    #mean specific heat of water (kJ/kg K)

#To determine total change in entropy and available work

# Calculations and Results
#(a)
#By energy balance
Tout_water = ((m_oil*c_oil*(Tin_oil-Tout_oil))/(m_water*c_water))+295; 			#exit temperature of water (K)
S_oil = m_oil*c_oil*math.log(Tout_oil/Tin_oil); 			#change in entropy of oil (kJ/K)
S_water = m_water*c_water*math.log(Tout_water/Tin_water); 			#change in entropy of water (kJ/K)
S_tot = S_oil+S_water; 			#total entropy change
print 'The total entropy change is %f kJ/K'%S_tot

#(b)
To = 295.; 			#temperature at which heat is rejected to surrounding (K)
#Let Q be heat given out by the oil on cooling
Q = m_oil*c_oil*(Tin_oil-Tout_oil);
#Heat rejected to the surrounding at To by the Carnot Engine is given by
#Q2 = To(Q/T) = -To*S_oil
Q2 = -To*S_oil; 			#(kJ)
#Let W be the work output of engine
W = Q-Q2;
print 'The work output of the engine would be %4.3e kJ'%W

The total entropy change is 666.266812 kJ/K
The work output of the engine would be 2.397e+05 kJ


### Example 4.19¶

In [38]:
# Variables
#Given:
T = 10.; 			#temperature of metal (K)
Cp = 0.45; 			#molar heat capacity at 10 K (J/mol K)

# Calculations
#To determine the molar entropy of metal
#Entropy of solid at 10 K is calculated using first integral in equation 4.55 (Page no. 108)
S = Cp/3;

# Results
print 'Molar entropy of meatl at 10 K is %f J/mol K'%S

Molar entropy of meatl at 10 K is 0.150000 J/mol K


### Example 4.20¶

In [39]:
# Variables
#Given:
T = 473.; 			#temperature at entropy is to be determined (K)
Tf = 273.; 			#base temperature (K)
Tb = 373.; 			#boiling temperature (K)
Cpl = 4.2; 			#avearge heat capacity of water (kJ/kg K)
Cpg = 1.9; 			#avearge heat capacity of water vapour between 373 K and 473 K
Hv = 2257.; 			#latent heat of vaporisation at 373 K (kJ/kg)

# Calculations
import math
S = (Cpl*math.log(Tb/Tf))+(Hv/Tb)+(Cpg*math.log(T/Tb));

# Results
print 'Absolute entropy of water vapour at 473 K and 101.3 kPa is %f kJ/kg K'%S
print 'It compares favourably with the value reported in steam tables'

Absolute entropy of water vapour at 473 K and 101.3 kPa is 7.813068 kJ/kg K
It compares favourably with the value reported in steam tables