Chapter 3 : Properties of Pure Substances

Example 3.1 page no : 76

In [1]:
# Variables
m_s = 50. 			#kg
m_w = 1.5; 			#kg

# Calculations
x = m_s/(m_s+m_w);

# Results
print ("dryness fraction = %.3f")%(x)

dryness fraction = 0.971

Example 3.2 page no : 76

In [2]:
# Variables
V = 0.6; 			#m**3
m = 3.0; 			#kg
p = 5.; 			#bar
v = V/m;

# At 5 bar: From steam tables
v_g = 0.375; 			#m**3/kg
v_f = 0.00109; 			#m**3/kg

# Calculations
v_fg = v_g - v_f;
x = 1-((v_g - v)/v_fg);

# Results
print ("(i) Mass and volume of liquid")
m_liq = m*(1-x);
print ("mass of liquid = %.3f")%(m_liq),("kg")
V_liq = m_liq*v_f;
print ("volume of liquid = %.3f")%(V_liq),("m**3")

print ("(ii) Mass and volume of vapour")
m_vap = m*x;
print ("mass of vapour = %.3f")%(m_vap),("kg")
V_vap = m_vap*v_g;
print ("volume of vapour = %.3f")%(V_vap),("m**3")

(i) Mass and volume of liquid
mass of liquid = 1.404 kg
volume of liquid = 0.002 m**3
(ii) Mass and volume of vapour
mass of vapour = 1.596 kg
volume of vapour = 0.598 m**3

Example 3.3 page no : 76

In [4]:
# Variables
V = 0.05; 			#m**3
m_f = 10.; 			#kg
# From steam tables corresponding to 245 0C
p_sat = 36.5; 			#bar
v_f = 0.001239; 			#m**3/kg
v_g = 0.0546; 			#m**3/kg
h_f = 1061.4; 			#kJ/kg
h_fg = 1740.2; 			#kJ/kg
s_f = 2.7474; 			#kJ/kg.K
s_fg = 3.3585; 			#kJ/kg.K

# Calculations and Results
print ("(i) The pressure = "),(p_sat),("bar")

print ("(ii) The mass")
V_f = m_f*v_f;
V_g = V - V_f;
m_g = V_g/v_g;
m = m_f+m_g;
print ("The total mass of mixture = %.3f")%(m),("kg")

print ("(iii) The specific volume")
v_fg = v_g-v_f;
x =  m_g/(m_g+ m_f);
v = v_f+x*v_fg;
print ("specific volume = %.3f")%(v),("m**3/kg")

print ("(iv)The specific enthalpy")
h = h_f+x*h_fg;
print ("specific enthalpy = %.3f")%(h),("kJ/kg")

print ("(v)The specific entropy")
s = s_f+x*s_fg;
print ("specific entropy  = %.3f")%(s),("kJ/kg.K")

print ("(vi)The specific internal enegy")
u = h-(p_sat*v*10**2); 			#kJ/kg
print ("specific internal energy = %.3f")%(u),("kJ/kg")

(i) The pressure =  36.5 bar
(ii) The mass
The total mass of mixture = 10.689 kg
(iii) The specific volume
specific volume = 0.005 m**3/kg
(iv)The specific enthalpy
specific enthalpy = 1173.545 kJ/kg
(v)The specific entropy
specific entropy  = 2.964 kJ/kg.K
(vi)The specific internal enegy
specific internal energy = 1156.471 kJ/kg

Example 3.4 page no : 77

In [5]:
# Variables
m_w = 2.; 			#kg
t_w = 25.; 			#0C
p = 5.; 			#bar
x = 0.9;
c_pw = 4.18;
# at 5 bar; from steam tables
h_f = 640.1; 			#kJ/kg
h_fg = 2107.4; 			#kJ/kg

# Calculations and Results
h = h_f+x*h_fg;

Qw = c_pw*(t_w-0);
print ("Sensible heat associated with 1kg of water, Qw = %.3f")%(Qw),("kJ")

Q = h-Qw;
print ("Net quantity of heat to be supplies per kg of water, Q = %.3f")%(Q),("kJ")

Q_total = m_w*Q;
print ("Total amount of heat supplied, Q_total = "),(Q_total),("kJ")

Sensible heat associated with 1kg of water, Qw = 104.500 kJ
Net quantity of heat to be supplies per kg of water, Q = 2432.260 kJ
Total amount of heat supplied, Q_total =  4864.52 kJ

Example 3.5 page no : 78

In [6]:
# Variables
m = 4.4; 			#kg
p = 6.; 			#bar
t_sup = 250.; 			#0C
t_w =  30.; 			#0C
c_ps = 2.2; 			#kJ/kg
c_pw = 4.18;
# At 6 bar, 250 0C; From steam tables
t_s = 158.8; 			#0C
h_f = 670.4; 			#kJ/kg
h_fg = 2085; 			#kJ/kg

# Calculations and Results
h_sup = h_f+h_fg+ c_ps*(t_sup-t_s);

Qw = c_pw*(t_w-0);
print ("Amount of heat added per kg of water, Qw = "),(Qw)

Q = h_sup-Qw;
print ("Net amount of heat required to be supplied per kg, Q = "),(Q)

Q_total = m*Q;
print ("Total amount of heat required, Q_total = "),(Q_total),("kJ")

Amount of heat added per kg of water, Qw =  125.4
Net amount of heat required to be supplied per kg, Q =  2830.64
Total amount of heat required, Q_total =  12454.816 kJ

Example 3.6 page no : 78

In [7]:
# Variables
v = 0.15; 			#m**3
p = 4.; 			#bar
x = 0.8;
# At 4 bar: From steam tables
v_g = 0.462; 			#m**3/kg
h_f =  604.7; 			#kJ/kg
h_fg = 2133.; 			#kJ/kg

# Calculations and Results
density = 1/x/v_g;

m = v*density;
print ("mass of 0.15 m**3 steam, m = %.3f")%(m),("kg")

Q = density*(h_f+x*h_fg);
print ("Total heat of 1 m3 of steam which has a mass of 2.7056 kg, Q = %.3f")%(Q),("kJ")

mass of 0.15 m**3 steam, m = 0.406 kg
Total heat of 1 m3 of steam which has a mass of 2.7056 kg, Q = 6252.976 kJ

Example 3.7 page no : 78

In [8]:
# Variables
m = 1000.; 			#kJ/kg.K
p = 16.; 			#bar
x = 0.9;
T_sup = 653.; 			#K
T_w = 30.; 			#0C
c_ps = 2.2; 			#kJ/kg
c_pw = 4.18;
# At 16 bar:From steam tables
T_s = 474.4; 			#K
h_f = 858.6; 			#kJ/kg
h_fg = 1933.2; 			#kJ/kg

# Calculations and Results
H = m*((h_f+x*h_fg)-c_pw*(T_w-0));
print ("(i) Heat supplied to feed water per hour to produce wet steam is given by"),(H),("kJ")

Q = m*((1-x)*h_fg+c_ps*(T_sup-T_s));
print ("(ii) Heat absorbed by superheater per hour, Q = "),(Q),("kJ")

(i) Heat supplied to feed water per hour to produce wet steam is given by 2473080.0 kJ
(ii) Heat absorbed by superheater per hour, Q =  586240.0 kJ

Example 3.8 page no : 79

In [9]:
print ("(i) at 0.75 bar, between 100°C and 150°C")

# Variables
# At 100 °C
T1 = 100.; 			#°C
h_sup1 = 2679.4; 			#kJ/kg
# At 150 °C
T2 = 150.; 			#°C
h_sup2 = 2778.2; 			#kJ/kg

# Calculations and Results
c_ps = (h_sup2-h_sup1)/(T2-T1);
print ("mean specific heat = "),(c_ps)

print ("(ii) at 0.5 bar, between 300°C and 400°C")
T1 = 300; 			#°C
h_sup1 = 3075.5; 	#kJ/kg
T2 = 400; 			#°C
h_sup2 = 3278.9; 	#kJ/kg

c_ps = (h_sup2-h_sup1)/(T2-T1);
print ("mean specific heat c_ps = "),(c_ps)

(i) at 0.75 bar, between 100°C and 150°C
mean specific heat =  1.976
(ii) at 0.5 bar, between 300°C and 400°C
mean specific heat c_ps =  2.034

Example 3.9 page no : 79

In [10]:
# Variables
m = 1.5; 			#kg
p = 5.; 			#bar
x1 = 1.;
x2 = 0.6;
p1 = 5.*10**5; 			#N/m
# At 5 bar: From steam tables
t_s = 151.8; 			#0C
h_f = 640.1; 			#kJ/kg
h_fg = 2107.4; 			#kJ/kg
v_g = 0.375; 			#m**3/kg
v_g1 = 0.375*10**(-3);

# Calculations and Results
h1 = h_f+h_fg;
V = m*v_g;
u1 = h1-p1*v_g1;
v_g2 = V/m/x2; 			#m**3/kg

# From steam table corresponding to 0.625 m**3/kg
p2 = 2.9; 			#bar
print ("Pressure at new state  = "),(p2),("bar")

t_s = 132.4; 			#0C
print ("Temperature at new state  = "),(t_s),("°C")
h_f2 = 556.5; 			#kJ/kg
h_fg2 = 2166.6; 			#kJ/kg
u2 = (h_f2+x2*h_fg2)-p2*x2*v_g2*10**2;

Q = u2-u1; 			#heat transferred at consmath.tant volume per kg

Q_total = m*Q;
print ("Total heat transfered,Q_total = "),(Q_total),("kJ")

Pressure at new state  =  2.9 bar
Temperature at new state  =  132.4 °C
Total heat transfered,Q_total =  -1218.435 kJ

Example 3.10 page no : 80

In [6]:
# Variables
V = 0.9; 			#m**3
p1 = 8.; 			#bar
x1 = 0.9;
p2 = 4.; 			#bar
p3 = 3.; 			#bar
v_g1 = 0.24; 		#m**3/kg

print ("(i) The mass of steam blown off :")
m1 = V/x1/v_g1;
h_f1 = 720.9; 			#kJ/kg
h_fg1 = 2046.5; 		#kJ/kg
h_f2 = 604.7; 			#kJ/kg
h_fg2 = 2133; 			#kJ/kg
v_g2 = 0.462; 			#m**3/kg

# Calculations and Results
h1 = h_f1+x1*h_fg1; 			#The enthalpy of steam before blowing off
h2 = h1;
x2 = (h1-h_f2)/h_fg2;
m2 = x1/(x2*v_g2);

m = m1-m2;
print ("Mass of steam blown off  = %.3f")%(m),("kg")

print ("(ii) Dryness fraction of steam in the vessel after cooling")
v_g3 = 0.606; 			#m**3/kg
x3 = x2*v_g2/v_g3;
print ("dryness fraction  = %.4f")%(x3)
x3 = 0.699

print ("(iii) Heat lost during cooling")
h_f3 = 561.4; 			    #kJ/kg
h_fg3 = 2163.2; 			#kJ/kg
h3 = h_f3+x3*h_fg3;
u2 = h2-p2*x2*v_g2*10**2; 			#kJ/kg
u3 = h3-p3*x3*v_g3*10**2; 			#kJ/kg
Q = m*(u3-u2);
print ("Heat lost during cooling = %.3f")%(-Q),("kJ")

(i) The mass of steam blown off :
Mass of steam blown off  = 2.045 kg
(ii) Dryness fraction of steam in the vessel after cooling
dryness fraction  = 0.6998
(iii) Heat lost during cooling
Heat lost during cooling = 913.322 kJ

Example 3.11 page no : 82

In [12]:
# Variables
p = 8*10**5; 			#Pa
x = 0.8; 
v_g = 0.240; 			#m**3/kg
h_fg = 2046.5; 			#kJ/kg

# Calculations and Results
print ("(i) External work done during evaporation")
W = p*x*v_g/10**3; 			#kJ
print ("W = "),(W),("kJ")

print ("(ii) Internal latent heat")
Q = x*h_fg-W;
print ("Q = "),(Q),("kJ")

(i) External work done during evaporation
W =  153.6 kJ
(ii) Internal latent heat
Q =  1483.6 kJ

Example 3.12 page no : 82

In [13]:
p1 = 10; 			#bar
import math 
p2 = 10; 			#bar
x1 = 0.85;
V1 = 0.15; 			#m**3
t_sup2 = 300; 			#0C
t_sup1 = 179.9; 			#0C
c_ps = 2.2; 			#kJ/kg.K
v_g1 = 0.194; 			#m**3/kg
m = V1/(x1*v_g1);
h_fg1 = 2013.6; 			#kJ/kg
Q = (1-x1)*h_fg1+c_ps*(t_sup2-t_sup1);
Q_total = m*Q;

print ("Total heat supplied = %.3f")%(Q_total),("kJ")

v_sup2 = v_g1*(t_sup2+273)/(t_sup1+273)
W = p1*(v_sup2 - (x1*v_g1))*10**2;
Percentage = W/Q*100;

print ("Percentage of total heat supplied = %.3f")%(Percentage),("%")

Total heat supplied = 515.094 kJ
Percentage of total heat supplied = 14.224 %

Example 3.13 page no : 83

In [14]:
# Variables
p = 18.; 			#bar
x = 0.85;
h_f = 884.6; 			#kJ/kg
h_fg = 1910.3; 			#kJ/kg
v_g = 0.110; 			#m**3/kg
u_f = 883.; 			#kJ/kg
u_g = 2598.; 			#kJ/kg

# Calculations and Results
v = x*v_g;
print ("Specific volume of wet steam = "),(v),("m**3/kg")

h = h_f+x*h_fg;
print ("Specific enthalpy of wet steam = "),(h),("kJ/kg")
u = (1-x)*u_f+ x*u_g;
print ("Specific internal energy of wet steam  = "),(u),("kJ/kg")

Specific volume of wet steam =  0.0935 m**3/kg
Specific enthalpy of wet steam =  2508.355 kJ/kg
Specific internal energy of wet steam  =  2340.75 kJ/kg

Example 3.14 page no : 83

In [15]:
# Variables
p = 7.; 			#bar
h = 2550.; 			#kJ/kg
h_f = 697.1; 			#kJ/kg
h_fg = 2064.9; 			#kJ/kg
v_g = 0.273; 			#m**3/kg
u_f = 696.; 			#kJ/kg
u_g = 2573.; 			#kJ/kg

# Calculations and Results
x = (h-h_f)/h_fg;
print ("(i) Dryness fraction = %.3f")%(x)

v = x*v_g;
print ("(ii) Specific volume of wet steam  = %.3f")%(v),("m**3/kg")

u = (1-x)*u_f+ x*u_g;
print ("(iii) Specific internal energy of wet steam = %.3f")%(u),("kJ/kg")

(i) Dryness fraction = 0.897
(ii) Specific volume of wet steam  = 0.245 m**3/kg
(iii) Specific internal energy of wet steam = 2380.291 kJ/kg

Example 3.15 page no : 84

In [16]:
# Variables
p = 120.; 			#bar
v = 0.01721; 			#m**3/kg

T = 350.; 			#°C
print ("Temperature = "), (T),("°C")

h = 2847.7; 			#kJ/kg
print ("specific enthalpy = "), (h),("kJ/kg")

u = h-p*v*10**2; 			#kJ/kg
print ("Internal energy = "), (u),("kJ/kg")

Temperature =  350.0 °C
specific enthalpy =  2847.7 kJ/kg
Internal energy =  2641.18 kJ/kg

Example 3.16 page no : 84

In [18]:
# Variables
p = 140.; 			#bar
h = 3001.9; 			#kJ/kg
T = 400; 			#0C

# Calculations and Results
print ("Temperature = "),(T), ("°C")

v = 0.01722; 			#m**3/kg
print ("The specific volume %.3f")%(v), ("m**3/kg")

u = h-p*v*10**2;
print ("specific internal energy = "),(u),("kJ/kg")

Temperature =  400 °C
The specific volume 0.017 m**3/kg
specific internal energy =  2760.82 kJ/kg

Example 3.17 page no : 85

In [19]:
# At 10 bar: From steam table for superheated steam

# Variables
h_sup = 3051.2; 			#kJ/kg
T_sup = 573; 			#K
T_s = 452.9; 			#K
v_g = 0.194; 			#m**3/kg
v_sup = v_g*T_sup/T_s;
p = 10.; 			#bar

# Calculations and Results
u1 = h_sup-p*v_sup*10**2; 			#kJ/kg
print ("Internal energy of superheated steam at 10 bar =  %.3f")%(u1), ("kJ/kg")

# At 1.4 bar: From steam tables
p = 1.4; 			#bar
h_f = 458.4; 			#kJ/kg
h_fg = 2231.9; 			#kJ/kg
v_g = 1.236; 			#m**3/kg
x = 0.8;
h = h_f+x*h_fg;
u2 = h-p*x*v_g*10**2; 			#kJ
du = u2-u1;
print ("Change in internal energy = %.3f")%(du),("kJ")

Internal energy of superheated steam at 10 bar =  2805.755 kJ/kg
Change in internal energy = -700.267 kJ

Example 3.18 page no : 85

In [20]:
# Variables
m = 1.; 			#kg
p = 20.; 			#bar
T_sup = 400.; 			#0C
x = 0.9;
c_ps = 2.3; 			#kJ/kg.K

print ("(i) Internal energy of 1 kg of superheated steam")
# At 20 bar: From steam tables
T_s = 212.4; 			#0C
h_f = 908.6; 			#kJ/kg
h_fg = 1888.6; 			#kJ/kg
v_g = 0.0995; 			#m**3/kg

# Calculations and Results
h_sup  =  h_f+h_fg+c_ps*(T_sup-T_s);
v_sup = v_g*(T_sup+273)/(T_s+273);
u = h_sup-p*v_sup*10**2;
print ("Internal energy = %.3f")%(u),("kJ/kg")

print ("(ii) Internal energy of 1 kg of wet steam")
h = h_f+x*h_fg;
u = h-p*x*v_g*10**2;
print ("Internal energy = %.3f")%(u),("kJ/kg")

(i) Internal energy of 1 kg of superheated steam
Internal energy = 2952.769 kJ/kg
(ii) Internal energy of 1 kg of wet steam
Internal energy = 2429.240 kJ/kg

Example 3.19 page no : 86

In [21]:
# Variables
h_g1 = 2797.2; 			#kJ/kg
c_ps =  2.25;
T_sup = 350.; 			#0C
T_s = 212.4; 			#0C

# Calculations
h1 = h_g1+c_ps*(T_sup-T_s);
h_f2 = 908.6; 			#kJ/kg
h_fg2 = 1888.6; 			#kJ/kg

# Main:20 bar, 250 0C
T_sup = 250.; 			#0C
Q = 2*(h_g1+c_ps*(T_sup-T_s));
x2 = (Q-h1-h_f2)/h_fg2;

# Results
print ("Quality of steam %.3f")%(x2)

Quality of steam 0.926

Example 3.20 page no : 87

In [22]:
import math

# Variables
m = 1.; 			#kg
p = 6.; 			#bar
x = 0.8;
T_s = 473.; 			#K
h_fg = 2085.; 			#kJ/kg
c_pw = 4.18;

# Calculations
s_wet = c_pw*math.log(T_s/273)+x*h_fg/T_s;

# Results
print ("Entropy of wet steam = %.3f")%(s_wet),("kJ/kg.K")

Entropy of wet steam = 5.824 kJ/kg.K

Example 3.21 page no : 87

In [23]:
# Variables
p1 = 10.; 			#bar
t_sup = 400.; 		#0C
p2 = 0.2; 			#bar
x2 = 0.9;
h_sup = 3263.9; 			#kJ/kg
s_sup = 7.465;  			#kJ/kg
h1 = 3263.9; 			#kJ/kg
s1 = s_sup;
h_f2 = 251.5; 			#kJ/kg
h_fg2 = 2358.4; 		#kJ/kg
s_f2 = 0.8321; 			#kJ/kg.K
s_g2 = 7.9094; 			#kJ/kg.K

# Calculations and Results
s_fg2 = s_g2-s_f2;
h2 = h_f2+x2*h_fg2;
s2 = s_f2+x2*s_fg2;

print ("(i) Drop in enthalpy")
dh = h1-h2;
print ("Drop in enthalpy  =  %.3f")%(dh),("kJ/kg")

print ("(ii) Change in entropy")
ds = s1-s2;
print ("Change in entropy  =  %.3f")%(ds),("kJ/kg.K")

(i) Drop in enthalpy
Drop in enthalpy  =  889.840 kJ/kg
(ii) Change in entropy
Change in entropy  =  0.263 kJ/kg.K

Example 3.22 page no : 88

In [8]:
import math 

# Variables
m = 1.; 		    	    #kg
p = 12.; 		    	#bar
T_sup = 523.; 			#K
c_ps = 2.1; 			#kJ/kg.K
T_s = 461.; 			#K
h_fg = 1984.3; 			#kJ/kg
c_pw = 4.18;

# Calculations
s_sup = c_pw*math.log(T_s/273)+h_fg/T_s+c_ps*math.log(T_sup/T_s);

# Results
print ("Entropy  = %.3f")%(s_sup),("kJ/kg.K")

Entropy  = 6.759 kJ/kg.K

Example 3.23 page no : 88

In [25]:
# Variables
m = 3.; 			#kg
v1 = 0.75; 			#m**3/kg
v2 = 1.2363; 			#m**3/kg
x = v1/v2;
h_f = 458.4; 			#kJ/kg
h_fg = 2231.9; 			#kJ/kg
h_s = m*(h_f+x*h_fg); 			#kJ
v_sup = 1.55; 			#m**3/kg
p = 2; 			#bar
t_s = 120.2; 			#0C
t_sup = 400; 			#0C
h = 3276.6; 			#kJ/kg
U = 1708.; 			#kJ/kg

# Calculations and Results
Degree = t_sup-t_s;
h_sup = m*h;

Q_added = h_sup - h_s;
print ("Heat added  = %.3f")%(Q_added),("kJ")

U_s = m*U;
U_sup = m*(h-p*v_sup*10**2);
dU =  U_sup - U_s;
W = Q_added - dU;
print ("work done  =  %.3f")%(W),("kJ")

Heat added  = 4392.661 kJ
work done  =  616.861 kJ

Example 3.24 page no : 91

In [26]:
# Variables
p = 5.; 			#bar
m = 50.; 			#kg
T1 = 20.; 			#0C
m_s = 3.; 			#kg
T2 = 40.; 			#0C
m_eq = 1.5; 			#kg
h_f = 640.1; 			#kJ/kg
h_fg = 2107.4; 			#kJ/kg
c_pw = 4.18;

# Calculations
m_w = m+m_eq;
x = ((m_w*c_pw*(T2-T1))/m_s + c_pw*T2 - h_f)/h_fg;

# Results
print ("Dryness fraction of steam %.3f")%(x)

Dryness fraction of steam 0.457

Example 3.25 page no : 91

In [27]:
# Variables
p = 1.1; 			#bar
x = 0.95;
c_pw = 4.18;
m1 = 90.; 			#kg
m2 = 5.25; 			#kg
T1 = 25.; 			#0C
T2 = 40.; 			#0C

# Calculations
m = m1+m2;
h_f = 428.8; 			#kJ/kg
h_fg =  2250.8; 			#kJ/kg
m_s =  (m*c_pw*(T2-T1))/((h_f + x*h_fg) - c_pw*T2)

# Results
print ("Mass of steam condensed = %.3f")%(m_s),("kg")

Mass of steam condensed = 2.489 kg

Example 3.26 page no : 93

In [28]:
# Variables
p1 = 8.; 			#bar
p2 = 1.; 			#bar
T_sup2 = 115.; 		#0C
T_s2 = 99.6; 		#0C
h_f1 = 720.9; 		#kJ/kg
h_fg1 = 2046.5; 	#kJ/kg
h_f2 = 417.5; 		#kJ/kg
h_fg2 = 2257.9; 	#kJ/kg
c_ps = 2.1;

# Calculations
x1 = (h_f2+h_fg2+c_ps*(T_sup2-T_s2)-h_f1)/h_fg1;

# Results
print ("Dryness fraction of the steam in the main  =  %.3f")%(x1)

Dryness fraction of the steam in the main  =  0.971

Example 3.27 page no : 94

In [29]:
# Variables
m_w = 2.; 			#kg
m_s = 20.5; 		#kg
t_sup = 110.; 		#0C
p1 = 12.; 			#bar
p3 = 1.; 			#bar
p2 = p1;
h_f2 = 798.4; 		#kJ/kg
h_fg2 = 1984.3; 	#kJ/kg
T_s = 99.6; 		#0C
h_f3 = 417.5; 		#kJ/kg
h_fg3 = 2257.9; 	#kJ/kg
T_sup = 110.; 		#0C
c_ps = 2.; 			#kJ/kg.K

# Calculations
x2 = (h_f3+h_fg3 + c_ps*(T_sup-T_s) - h_f2)/h_fg2;
x1 = x2*m_s/(m_w+m_s);

# Results
print ("Quality of steam supplied  =  %.3f")%(x1)

Quality of steam supplied  =  0.871

Example 3.28 page no : 95

In [30]:
# Variables
p1 = 15.; 			#bar
p2 = p1;
p3 = 1.; 			#bar
t_sup3 = 150.; 		#0C
m_w = 0.5; 			#kg/min
m_s = 10.; 			#kg/min
h_f2 = 844.7; 		#kJ/kg
h_fg2 = 1945.2; 	#kJ/kg
h_sup3 = 2776.4; 	#kJ/kg

# Calculations
x2 = (h_sup3 - h_f2)/h_fg2;
x1 = x2*m_s/(m_s + m_w);


# Results
print ("Quality of steam supplied  =  %.3f")%(x1)

Quality of steam supplied  =  0.946