Chapter 5 : Properties of Steam

Example 5.1 Page No : 6

In [1]:
# Variables : 
deltaQ = 1000.;			#KJ
T = 1073.;			#Kelvin
T0 = 20.+273;			#Kelvin

# Calculations and Results
deltaS = deltaQ/T;			#KJ/K
A = deltaQ-T0*deltaS;			#KJ
print "Available energy in KJ : %.2f"%A

UA = T0*deltaS;			#KJ
print "Unavailable energy in KJ : %.2f"%UA

Available energy in KJ : 726.93
Unavailable energy in KJ : 273.07

Example 5.2 Page No : 6

In [2]:
import math 
			
# Variables :
m = 2.;			#Kg
T1 = 300.+273;			#Kelvin
T2 = 150.+273;			#Kelvin
T0 = 20.+273;			#Kelvin
Cp = 0.45;			#KJ/KgK

# Calculations and Results
deltaQ = m*Cp*(T1-T2);			#KJ
deltaS = m*Cp*math.log(T1/T2);			#KJ/K
A = deltaQ-T0*deltaS;			#KJ
print "Reversible work or Available energy in KJ : %.2f"%A

UA = T0*deltaS;			#KJ
print "Irreversibility in KJ : %.2f"%UA
#Irreversibilty is not calculated in the book and asked in the question.

Reversible work or Available energy in KJ : 54.96
Irreversibility in KJ : 80.04

Example 5.3 Page No : 6

In [3]:
import math 
			
# Variables :
m = 5.;			#Kg
p = 1.;			#bar
T0 = 20.+273;			#Kelvin
T1 = 23.+273;			#Kelvin
T2 = 227.+273;			#Kelvin
Cp = 1.005;			#J/KgK

# Calculations
deltaS = Cp*math.log(T1/T2);			#KJ/KgK
deltaQ = Cp*(T2-T1);			#KJ
A = m*(deltaQ+T0*deltaS);			#KJ

# Results
print "Increase in availability due to heating in KJ : %.2f"%A

Increase in availability due to heating in KJ : 253.24

Example 5.4 Page No : 7

In [4]:
			
# Variables :
Q1 = 400.;			#KJ
T1 = 1227.+273;			#Kelvin
T2 = 27.+273;			#Kelvin

# Calculations and Results
A = Q1-T2*Q1/T1;			#KJ
print "Availability of the system in KJ : ",A

UA = Q1-A;			#KJ
print "Unavailable energy in KJ : ",UA

Availability of the system in KJ :  320.0
Unavailable energy in KJ :  80.0

Example 5.5 Page No : 7

In [4]:
			
# Variables :
P = 1.;			#KW or KJ/s
Q = 6.;			#MJ/hr
Q = Q*1000./3600;			#KJ/s
T1 = 26.+273;			#Kelvin
T2 = 3.+273;			#Kelvin

# Calculations
COP = T1/(T1-T2);
W = Q/COP;			#KJ/s or KW

# Results
print "Work required to pump heat in KJ/s or KW : %.3f"%W
print ("As P>W, required condition can be maintained.")

Work required to pump heat in KJ/s or KW : 0.128
As P>W, required condition can be maintained.

Example 5.6 Page No : 8

In [6]:
			
# Variables :
T = 727.+273;			#Kelvin
T0 = 17.+273;			#Kelvin
deltaQ = 4000.;			#KJ

# Calculations and Results
deltaS = deltaQ/T;			#KJ/K
A = deltaQ-T0*deltaS;			#KJ
print "Availability of heat energy in KJ : ",A
UA = T0*deltaS;			#KJ
print "Unavailable heat energy in KJ : ",UA

Availability of heat energy in KJ :  2840.0
Unavailable heat energy in KJ :  1160.0

Example 5.7 Page No : 8

In [5]:
			
# Variables :
deltaQ = 850.;			#KJ
T = 180+273.;			#Kelvin
T0 = 22+273.;			#Kelvin

# Calculations
deltaS = deltaQ/T;			#KJ/K
A = deltaQ-T0*deltaS;			#KJ

# Results
print "Available energy in KJ : %.2f"%A

Available energy in KJ : 296.47

Example 5.8 Page No : 8

In [6]:
			
# Variables :
deltaQ = 850.;			#KJ
T1 = 1400.+273.;			#Kelvin
T2 = 250.+273.;			#Kelvin
T0 = 20.+273.;			#Kelvin
Q = -1000.; 			#KJ

# Calculations and Results
deltaS1 = Q/T1;			#KJ/K(-ve as heat leaving)
deltaS2 = abs(Q)/T2;			#KJ/K(+ve Q as steam receives heat)
deltaS = deltaS1+deltaS2;			#KJ/K
print ("Part (i) As energy leaves the hot gases : ");
A = (T1-T0)*deltaS1;			#KJ
UA = T0*deltaS1;			#KJ
print "Available energy in KJ : %.2f"%A
print "Unavailable energy in KJ : %.2f"%UA
print ("Part (ii) As energy enters the system : ");

A = (T2-T0)*deltaS2;			#KJ
UA = T0*deltaS2;			#KJ
print "Available energy in KJ : %.2f"%A
print "Part (iii) Unavailable energy in KJ : %.2f"%UA

Part (i) As energy leaves the hot gases : 
Available energy in KJ : -824.87
Unavailable energy in KJ : -175.13
Part (ii) As energy enters the system : 
Available energy in KJ : 439.77
Part (iii) Unavailable energy in KJ : 560.23

Example 5.9 Page No : 9

In [7]:
import scipy
from scipy.integrate import quad 
			
# Variables :
deltaQ = 850.;			#KJ
T1 = 523.;			#Kelvin
T2 = 873.;			#Kelvin
T0 = 288.;			#Kelvin
dQ_by_dT = 100.;			#KJ/K


# Calculations and Results
def f3(T): 
	 return 100/T

deltaS =  quad(f3,T1,T2)[0]


def f4(T): 
	 return 100

deltaQ =  quad(f4,T1,T2)[0]

print "Total heat abstracted in KJ : ",deltaQ

A = deltaQ-T0*deltaS;			#KJ
print "Availability in KJ : %.1f"%A

Loss = deltaQ-A;			#KJ
print "Loss of availability in KJ : %.1f"%Loss

Total heat abstracted in KJ :  35000.0
Availability in KJ : 20244.2
Loss of availability in KJ : 14755.8

Example 5.10 Page No : 10

In [8]:
import math 

# Variables :
p0 = 1.;    			#bar
T0 = 17.+273;			#Kelvin
T1 = 1817.+273;			#Kelvin
Cp = 1.;	    		#KJ/KgK

# Calculations
deltaQ = Cp*(T1-T0);			#KJ/Kg
deltaS = Cp*math.log(T0/T1);			#KJ/KgK
deltaS_fluid = -deltaS;			#KJ/KgK(As deltaS_surrounding = 0)
A = deltaQ-T0*deltaS_fluid;			#KJ

# Results
print "Availability of hot products in KJ : %.2f"%A

Availability of hot products in KJ : 1227.24

Example 5.11 Page No : 10

In [11]:
# Variables :
T1 = 1200.;			#Kelvin
T2 = 400.;			#Kelvin
T0 = 300.;			#Kelvin
Qsource = -150.;			#KJ/s
Qsystem = 150.;			#KJ/s

# Calculations and Results
deltaS_source = Qsource/T1;			#KJ/sK
deltaS_system = Qsystem/T2;			#KJ/sK
deltaS_net = deltaS_source+deltaS_system;			#KJ/sK
print "Net change in entropy in KJ/sK : ",deltaS_net

A1 = (T1-T0)*-deltaS_source;			#KJ/s
print "Available energy of heat source in KJ/s : ",A1

A2 = (T2-T0)*deltaS_system;			#KJ/s
print "Available energy of system in KJ/s : ",A2

E_decrease = A1-A2;			#KJ/s
print "Decrease in available energy in KJ/s : ",E_decrease

Net change in entropy in KJ/sK :  0.25
Available energy of heat source in KJ/s :  112.5
Available energy of system in KJ/s :  37.5
Decrease in available energy in KJ/s :  75.0

Example 5.12 Page No : 11

In [9]:
import math 

# Variables :
Tg1 = 1127.+273;			#Kelvin
Tg2 = 527.+273;			#Kelvin
T2 = 250.+273;			#Kelvin
T0 = 27.+273;			#Kelvin
Cpg = 1.;			#KJ/KgK
mw = 5.;			#Kg/s
hfg = 1716.2;			#KJ/Kg

# Calculations and Results
#mg*Cpg*(Tg1-Tg2) = mw*hfg
mg = mw*hfg/Cpg/(Tg1-Tg2);			#Kg/s
print "Mass flow rate of gases in Kg/s : %.1f"%mg

deltaSg = mg*Cpg*math.log(Tg2/Tg1);			#KJ/sK
print "Entropy change of gases in KJ/sK : %.4f"%deltaSg

deltaSw = mw*hfg/T2;			#KJ/sK
print "Entropy change of water in KJ/sK : %.4f"%deltaSw

deltaSnet = deltaSg+deltaSw;			#KJ/sK
print "Net Entropy change in KJ/sK : %.4f"%deltaSnet

Q1 = mw*hfg;			#KJ/s
Sa_sub_Sb = -deltaSg;			#KJ/sK
A1 = Q1-T0*(Sa_sub_Sb);			#KJ/s
print "Availability of hot gases in KJ/s : %.2f"%A1

A2 = Q1-T0*deltaSw;			#KJ/s
print "Availability of water in KJ/s : %.2f"%A2

UA1 = T0*(Sa_sub_Sb);			#KJ/s
print "Unavailable energy of hot gases in KJ/s : %.2f"%UA1

UA2 = T0*deltaSw;			#KJ/s
print "Unavailable energy of water in KJ/s : %.2f"%UA2

E_increase = T0*deltaSnet;			#KJ/s
print "Increase in unavailable energy in KJ/s : %.2f"%E_increase

Mass flow rate of gases in Kg/s : 14.3
Entropy change of gases in KJ/sK : -8.0034
Entropy change of water in KJ/sK : 16.4073
Net Entropy change in KJ/sK : 8.4038
Availability of hot gases in KJ/s : 6179.97
Availability of water in KJ/s : 3658.82
Unavailable energy of hot gases in KJ/s : 2401.03
Unavailable energy of water in KJ/s : 4922.18
Increase in unavailable energy in KJ/s : 2521.15

Example 5.13 Page No : 12

In [10]:
import math 

			
# Variables :
mg = 5.;			#Kg
p1 = 3.;			#bar
T1 = 500.;			#Kelvin
Q = 500.;			#KJ
Cv = 0.8;			#KJ/Kg
T0 = 300.;			#Kelvin
T = 1300.;			#Kelvin

# Calculations
#Q = mg*Cv*(T2-T1)
T2 = Q/mg/Cv+T1;			#Kelvin
A1 = Q-T0*Q/T;			#KJ
deltaSg = mg*Cv*math.log(T2/T1);			#KJ/K
Ag = Q-T0*deltaSg;			#KJ
Loss = A1-Ag;			#KJ

# Results
print "Loss of Availability due to heat transfer in KJ : %.1f"%Loss

Loss of Availability due to heat transfer in KJ : 152.4

Example 5.14 Page No : 13

In [11]:
import math 
			
# Variables :
m = 3.;			#Kg
p1 = 3.;			#bar
T1 = 450.;			#Kelvin
Q = 600.;			#KJ
Cv = 0.81;			#KJ/Kg
T0 = 300.;			#Kelvin
T = 1500.;			#Kelvin

# Calculations
deltaSsource = Q/T;			#KJ/K
#Q = m*Cv*(T2-T1)
T2 = Q/m/Cv+T1;			#Kelvin
A1 = Q-T0*deltaSsource;			#KJ
deltaSg = m*Cv*math.log(T2/T1);			#KJ/K
A2 = Q-T0*deltaSg;			#KJ
Loss = A1-A2;			#KJ

# Results
print "Loss in available energy due to heat transfer in KJ : %.1f"%Loss

Loss in available energy due to heat transfer in KJ : 198.9