# Chapter 9 : Properties of Steam and Thermodynamic Cycles¶

## Example 9.1 Page No : 217¶

In :
import math

# Variables
print ("Part (i) : For dry saturated steam at 17.8 bar")
p= 17.8;			# in bar
p1= 17.5;			# in bar
p2= 18.0;			# in bar
Vs1= 0.1135;			# in litre/kg
Vs2= 0.1104;			# in litre/kg
Hs1= 2796.1;			# in kJ/kg
Hs2= 2796.4;			# in kJ/kg
L1= 1918;			# in kJ/kg
L2= 1912;			# in kJ/kg
phi_s1= 6.389;			# in kJ/kg K
phi_s2= 6.379;			# in kJ/kg K

# Calculations and Results
Vs= Vs1-(Vs2-Vs1)/(p2-p1)*(p-p1);			# in litre/kg
Hs= Hs1+(Hs2-Hs1)/(p2-p1)*(p-p1);			# in kJ/kg
L= L1- (L1-L2)/(p2-p1)*(p-p1);			# in kJ/kg
phi_s= phi_s1- (phi_s1-phi_s2)/(p2-p1)*(p-p1);			# in kJ/kg K
print ("Part (i) : For dry saturated steam at 17.8 bar")
print "The specific volume in litre/kg is : ",Vs
print "The enthalpy in kJ/kg is : ",Hs
print "The latent heat in kJ/kg is : ",L
print "The entropy in kJ/kg K",phi_s
print ("Part (ii) : For superheated steam at 16 bar and 340°C")

T= 340.;			# in K
T1= 300.;			# in K
T2= 350.;			# in K
Vsup1= 0.1585;			# in m**3/kg
Vsup2= 0.1743;			# in m**3/kg
Hsup1= 3030;			# in kJ/kg
Hsup2= 3142;			# in kJ/kg
phi_sup1= 6.877;			# in kJ/kg K
phi_sup2= 7.063;			# in kJ/kg K
Vsup= Vsup1+(Vsup2-Vsup1)/(T2-T1)*(T-T1);			# in m**3/kg
Hsup= Hsup1+(Hsup2-Hsup1)/(T2-T1)*(T-T1);			# in kJ/kg
phi_sup= phi_sup1+(phi_sup2-phi_sup1)/(T2-T1)*(T-T1);			# in kJ/kg
print "The specific volume in m**3/kg is : ",Vsup
print "The enthalpy in kJ/kg is : ",Hsup
print "The entropy in kJ/kg K is : ",phi_sup

Part (i) : For dry saturated steam at 17.8 bar
Part (i) : For dry saturated steam at 17.8 bar
The specific volume in litre/kg is :  0.11536
The enthalpy in kJ/kg is :  2796.28
The latent heat in kJ/kg is :  1914.4
The entropy in kJ/kg K 6.383
Part (ii) : For superheated steam at 16 bar and 340°C
The specific volume in m**3/kg is :  0.17114
The enthalpy in kJ/kg is :  3119.6
The entropy in kJ/kg K is :  7.0258


## Example 9.2 Page No : 219¶

In :
# Variables
h_sen = 798.43;			# in kJ/kg
L = 1984.3;			# in kJ/kg
H_total_wet = 2665.7;

# Calculations and Results
# H_total_wet= h_sen+x*L
x = (H_total_wet - h_sen)/L;
print "The value of x is : %.3f"%x

# Part (b)
h_total_sup= 2961;			# in kJ/kg
Cps= 2.112;			# in kJ/kg
H_total_dry= 2782.7;			# in kJ/kg
# Let deltaT= T_sup-T_sat
# h_total_sup = h_sen+L+h_sup = H_total_dry +Cps*deltaT
deltaT= (h_total_sup-H_total_dry)/Cps;			# in °C
print "Degree of superheat in °C is : %.2f"%deltaT

The value of x is : 0.941
Degree of superheat in °C is : 84.42


## Example 9.3 Page No : 219¶

In :
# Variables
H2 = 3055;			# in kj per kg
H3 = 2550;			# in kj per kg
fie_1 = 7.15;			# kj per kg k
fie_2 = 7.57;			# kj per kg k

# Calculations and Results
d_fie= fie_2 - fie_1;			# in kj per kg k
print "Change in entropy during throttling process in kJ/kg-K is :",d_fie

dH = H2 - H3;			# in kj per kg
print "in enthalpy during isentropic process in kJ/kg is : ",dH

Change in entropy during throttling process in kJ/kg-K is : 0.42
in enthalpy during isentropic process in kJ/kg is :  505


## Example 9.4 Page No : 227¶

In :
# Exa 9.4
import math

# Variables
H_w = 670.4;			# in kJ/kg
L = 2085;			# kJ per kg
T_sat = 158.8;			# in degree c
m = 4;			# in kg
x = 0.5;
h_sen = 670.4;			# in kJ/kg

# Calculations and Results
H_totalwet = m * ( h_sen + (x *L));			# in kJ
x1 = 0.95
H_totalwet1 = m *( h_sen + (x1 *L));			# in kJ
Q1 =  H_totalwet1 - H_totalwet;			# in kJ
print "Part (i) The quantity of heat in case first in kJ is : ",Q1

# Part (b)
x2 = 1;
H_totaldry = m *( h_sen + (x2 *L));			# in kJ
Q2 = H_totaldry - H_totalwet;			# in kJ
print "Part (ii) The quantity of heat in case second in kJ is : ",Q2

# Part (c)
H_totalsup = 3062.3;			# in kJ per kg
H_totalsup = m * H_totalsup;			# in kJ
Q3 = H_totalsup - H_totalwet;			# in kJ
print "Part (iii) The quantity of heat in case third in kJ is : ",Q3

# Part (d)
H_totalsup = 2950.4;			# in kj per kg
H_totalsup = m * H_totalsup;			# in kj
Q4 = H_totalsup - H_totalwet;			# in kj
print "(iv) The quantity of heat in case forth in kJ is : ",Q4

Part (i) The quantity of heat in case first in kJ is :  3753.0
Part (ii) The quantity of heat in case second in kJ is :  4170.0
Part (iii) The quantity of heat in case third in kJ is :  5397.6
(iv) The quantity of heat in case forth in kJ is :  4950.0


## Example 9.5 Page No : 228¶

In :
# Variables
p1 = 2.5;       			# Mpa
p1 = p1 * 10**6;			# in pa
p1 = p1 * 10**-5;			# in bar
p2 = 10;		        	# in kpa
p2 = p2 * 10**3;			# in pa
p2 = p2 * 10**-5;			# in bar
H1 = 2878;		        	# in kJ/kg at 25 bar and 250°C
H2 = 2583.9;		    	# in kJ/kg at 0.1 bar for dry saturated steam
AHD= H1-H2;			        # actual heat drop in kJ/kg
H2_desh = 2110;			    # in kj per kg

# Calculations and Results
IHD = H1 - H2_desh;			# Isentropic heat drop in kJ/ kg
Eta_Isentropic = (AHD/IHD) * 100;			# in %
print "Isentropic efficiency in %% is : %.2f"%Eta_Isentropic

# H1 + v1**2/2 + g*z1 + Q = H2 + v2**2/2 + g*z2 + W
W = H1 - H2;			# in kJ/kg (as v1=v2, z1= z2 and Q=0)
print "Turbine work is in kJ/kg is :",W

Isentropic efficiency in % is : 38.29
Turbine work is in kJ/kg is : 294.1


## Example 9.6 Page No : 231¶

In :
# Variables
p1 = 11;			# in bar
p2 = 1.2;			# in bar
H_w1 = 781.1;			#in kJ/kg
L1 = 2000;			# in kJ/kg
t1 = 120;			# in degree c
t1 = t1 + 273;			# in K
t2 = 104.81;			# in degree c
t2 = t2 + 273;			# in K
H_dry2 = 2683.4;			# in kJ/kg
C_p = 2.607;			# in kJ/kgK

# Calculations and Results
# From Hw1+x*L1 = H_dry2+Cp*(t1-t2)
x = (H_dry2 + (C_p * (t1 - t2)) - H_w1) / L1;
print "Dryness fraction of steam is : %.2f"%x

x1 = (H_dry2 - H_w1)/L1;
print "Maximum dryness fraction of steam is : %.3f"%x1

Dryness fraction of steam is : 0.97
Maximum dryness fraction of steam is : 0.951


## Example 9.7 Page No : 234¶

In :
# Variables
W = 21.; 			# in kg
w_wp = 2.;			# in kg
h1 = 781.15;			# in kJ/kg
L1 = 1998.5;			# in kJ/kg
m = 2.;	    		# in kg
h2 = 420.5;			# in kJ/kg
L = 2255.9;			# in kJ/kg
t_sat = 100.4;			# in degree c
t1 = 110.;			# in degree c
C_ps = 2.;			# in kJ/kgK

# Calculations
x1 = W / (W + w_wp);
x2 = (h2 + L + m * (t1-t_sat) - h1)/ L1;
x = x1 * x2;

# Results
print "The dryness fraction for sample steam is : %.3f"%x

The dryness fraction for sample steam is : 0.875


## Example 9.8 Page No : 235¶

In :
# Variables
h_sen = 417.4;			# in kJ/kg
h_totaldry = 2675.4;			# in kJ/kg
L = 2258;			# in kJ/kg
v = 5;			# in m**3
v_v = 4.95;			# in m**3

# Calculations
x = v_v/v;
Q = h_totaldry -(h_sen +x*L);			# in kJ/kg

# Results
print "Heat transfered per kg in kJ/kg is : ",Q

Heat transfered per kg in kJ/kg is :  22.58


## Example 9.9 Page No : 235¶

In :
# Variables
m = 1;      			# in kg
p = 10;		        	# in bar
p = p * 10**2;			# in kpa
x = 0.94;
h_sen = 762.61;			# in kJ/kg
L = 2013.6;			    # in kJ/kg
v_s = 0.1942;			# in m**3 per kg

# Calculations and Results
w_ext = p * x * v_s * m;			# in kJ/kg
print "The work done during evaporation in kJ/kg is : %.2f"%w_ext

# Part (b)
L_internal = (x * L) - w_ext;			# in kJ/kg
print "Internal latent heat in kJ/kg is : %.2f"%L_internal

# Part (c)
U_wet = h_sen+x*L-p*x*v_s;			# in kJ/kg
print "Internal energy in kJ/kg is : %.2f"%U_wet

The work done during evaporation in kJ/kg is : 182.55
Internal latent heat in kJ/kg is : 1710.24
Internal energy in kJ/kg is : 2472.85


## Example 9.10 Page No : 236¶

In :
import math

# Variables
T_sat = 179.88;			# in degree c
T_sat = T_sat + 273;			# in k
T_sup = 200;			# in degree c
T_sup = T_sup + 273;			# in k
L = 2013.6; 			# in kJ/kg
C_ps = 2.326;
C_pw = 1;
x = 0.8;

# Calculations and Results
phi_wet = C_pw *math.log(T_sat/273) + ( (x * L)/T_sat);			# in kJ/kg-K
print "Entropy of wet steam in kJ/kg-K is : %.3f"%phi_wet

# Part (b)
phi_dry =C_pw * math.log(T_sat/273)+L/T_sat;		        	# in kJ/kg
print "Entropy of dry and saturated steam in kJ/kg-K is : %.3f"%phi_dry

# Part (c)
phi_sup = phi_dry+C_ps *math.log(T_sup/T_sat);			        # in kJ/kg
print "Entropy at 200°C in kJ/kg-K is : %.3f"%phi_sup

Entropy of wet steam in kJ/kg-K is : 4.063
Entropy of dry and saturated steam in kJ/kg-K is : 4.952
Entropy at 200°C in kJ/kg-K is : 5.053


## Example 9.11 Page No : 236¶

In :
# Variables
m = 1;			        # in kg
x = 0.9;
p = 1;	        		# N/mm**2
p = p * 10**1;			# in bar
p = p * 10**2;			# in kPa
h_sen  = 762.61;		# in kJ/kg
L = 2013.6;			    # in kJ/kg
v_s = 0.1944;			# in m**3 per kg

# Calculations and Results
H_totalwet = h_sen + x*L;			# in kJ/kg
U_wet = H_totalwet - (p * x * v_s);			# in kJ/kg
I = U_wet / H_totalwet;			# internal energy as a fraction of total heat
I = I * 10**2;			# in %
print "The internal energy in %% is : %.2f"%I

# Part (b)
v_s = 0.1542;	    		# in m**3/kg
h_sen = 815;	    		# in kJ/kg
L = 1972;   	    		# in kJ/kg
H_totaldry = 2787;			# in kJ/kg
C_ps = 2.199;
t_sup = 250;		    	# in °C
t_sup = t_sup + 273;		# in K
t_sat = 190.74;			    # in °C
t_sat = t_sat + 273;		# in K
p1 = 13 * 10**2;			# in kPa
v_ssup = v_s*t_sup/t_sat;			# in m**3/kg
print "The volume of superheated steam in m**3/kg is : %.4f"%v_ssup

# Part (c)
t_sat = t_sat - 273;			# in °C
t_sup = t_sup - 273;			# in °C
U_sup =h_sen+L+C_ps*(t_sup-t_sat)-p1*v_ssup;			# in kJ
del_U =  U_sup -  U_wet;			# in kJ
print "Change in internal energy in kJ is : %.1f"%del_U

# rounding off error

The internal energy in % is : 93.21
The volume of superheated steam in m**3/kg is : 0.1739
Change in internal energy in kJ is : 291.3


## Example 9.12 Page No : 238¶

In :
# Variables
m = 0.5;			# in kg
M = 6.6;			# in kg
x1 = M / (M+m);
h_dry = 2683;			#in kJ/kg
C_p = 2.1;
h_sen = 814.5;			#in kJ/kg
L = 1973;			# in kJ/kg
t_sup = 120;			# in °C
t_sat = 104.8;			# in °C

# Calculations
x2 =(h_dry+C_p*(t_sup - t_sat)-h_sen)/ L;
x = x2 * x1;

# Results
print "the dryness fraction of steam is : %.4f"%x

the dryness fraction of steam is : 0.8954