Chapter 12 : Vapour Power Cycles

Example 12.1 Page no : 548

In [11]:
import math

# Variables
p1 = 60.; 			#bar; Inlet to turbine
p2 = 0.1; 			#bar; Exit from turbine
p3 = 0.09; 			#bar; Exit from condenser 
p4 = 70.; 			#bar ; Exit from pump
p5 = 65.; 			#bar; Exit from boiler

t1 = 380.; 			#0C
t5 = 400.; 			#0C
x2 = 0.9; 			#Quality at exit from turbine
C = 200.; 			#m/s; Velocity at the exit from turbine 

# Calculations and Results

#At 60 bar 380 0C, From steam tables
h1 = 3123.5; 			#kJ/kg; By interpolation
h_f2 = 191.8; 			#kJ/kg
h_fg2 = 2392.8; 			#kJ/kg
x2 = 0.9; 

h2 = h_f2+x2*h_fg2;
m_s = 10000./3600; 			#Rate of stem flow in kg/s

P = m_s*(h1-h2);
print ("(i)Power output of the turbine  = %.0f")% (P), ("kW")


h_f4 = 1267.4; 			#kJ/kg
h_a = 3167.6; 			#kJ/kg
Q1 = 10000*(h_a - h_f4); 
print ("(ii)Heat transfer per hour in the boiler  = %.2e")% (Q1), ("kJ/h")

h_f3 = 183.3; 			#kJ/kg
Q2 = float("%.2e"%(10000*(h2-h_f3)));
print ("Heat transfer per hour in the condenser  = %.2e")% (Q2), ("kJ/h")

print ("(iii) Mass of cooling water circulated per hour in the condenser")
c_pw = 4.18;
t2 = 30.;
t1 = 20.;

m_w = Q2/(c_pw*(t2-t1));
print ("m_w = %.3e")% (m_w), ("kg/h")
print ("This is the exact answer.")

print ("(iv) Diameter of the pipe connecting turbine with condenser")

v_g2 = 14.67; 			#m**3/kg

d = math.sqrt(m_s*x2*v_g2*4/math.pi/C)*1000;
print ("Diameter  = %d")% (d), ("mm")

# Note : Answer for part 3 in book is incorrect.

(i)Power output of the turbine  = 2162 kW
(ii)Heat transfer per hour in the boiler  = 1.90e+07 kJ/h
Heat transfer per hour in the condenser  = 2.16e+07 kJ/h
(iii) Mass of cooling water circulated per hour in the condenser
m_w = 5.167e+05 kg/h
This is the exact answer.
(iv) Diameter of the pipe connecting turbine with condenser
Diameter  = 483 mm

Example 12.2 Page no : 550

In [2]:
# Variables
p1 = 15.; 			#bar
x1 = 1.;
p2 = 0.4; 			#bar

#At 15 bar
t_s1 = 198.3; 			#0C
h_g1 = 2789.9; 			#kJ/kg
s_g1 = 6.4406; 			#kJ/kg K

#At 0..4 bar
t_s2 = 198.3; 			#0C
h_f2 = 317.7; 			#kJ/kg
h_fg2 = 2319.2; 			#kJ/kg
s_f2 = 1.0261; 			#kJ/kg K
s_fg2 = 6.6448; 			#kJ/kg K
T1 = 471.3; 			#K
T2 = 348.9; 			#K

# Calculations and Results
n_carnot = (T1-T2)/T1;
print ("Carnot efficiency = %.3f")%(n_carnot)

x2 = (s_g1 - s_f2)/s_fg2;
h2 = h_f2+x2*h_fg2;

n_rankine = (h_g1-h2)/(h_g1-h_f2);
print ("Rankine efficiency = %.3f")%(n_rankine)

Carnot efficiency = 0.260
Rankine efficiency = 0.236

Example 12.3 Page no : 551

In [3]:
# Variables
p1 = 20.; 			#bar
p2 = 0.08; 			#bar

#At 20 bar, 360 0C
h1 = 3159.3; 			#kJ/kg
s1 = 6.9917; 			#kJ/kg K

#At 0.08 bar
h_f2 = 173.88; 			#kJ/kg
s_f2 = 0.5926; 			#kJ/kg K

h_fg2 = 2403.1; 		#kJ/kg
s_g = 8.2287; 			#kJ/kg K
v_f = 0.001008; 		#m**3/kg
s_fg = 7.6361; 			#kJ/kg K

# Calculations and Results
x2 = (s1-s_f2)/s_fg;

h2 = h_f2+x2*h_fg2;

W_pump = v_f*(p1-p2)*100; 			#kJ/kg
W_turbine = h1-h2;

W_net = h1-h2;
print ("Net work done = %.3f")% (W_net), ("kJ/kg")

h_f4 = W_pump+h_f2;
Q1 = h1-h_f4;

n_cycle = W_net/Q1;
print ("Cycle efficiency = %.3f")% (n_cycle)

Net work done = 971.607 kJ/kg
Cycle efficiency = 0.326

Example 12.4 Page no : 552

In [4]:
# Variables
n_turbine = 0.9;
n_pump = 0.8;
p1 = 80.; 			#bar
p2 = 0.1; 			#bar
v_f = 0.0010103; 			#m**3

#At 80 bar, 600 0C
h1 = 3642.; 			#kJ/kg
s1 = 7.0206; 			#kJ/kg K
s_f2 = 0.6488; 			#kJ/kg K
s_fg2 = 7.5006; 			#kJ/kg K
h_f2 = 191.9; 			#kJ/kg
h_fg2 = 2392.3; 			#kJ/kg

# Calculations
x2 = (s1-s_f2)/s_fg2;
h2 = h_f2+x2*h_fg2;

W_turbine = n_turbine*(h1-h2);
W_pump = v_f*(p1-p2)*10**2;

W_actual = W_pump/n_pump; 			#Actual pump work

W_net = W_turbine - W_actual;

# Results
print ("Specific work  = %.3f")% (W_net), ("kJ/kg")

h_f4 = h_f2+W_actual;
Q1 = h1-h_f4;

n_thermal = W_net/Q1; 			#Thermal efficiency
print ("Thermal efficiency  = %.3f")% (n_thermal)

Specific work  = 1265.955 kJ/kg
Thermal efficiency  = 0.368

Example 12.5 Page no : 553

In [5]:
# Variables
p1 = 28.; 			#bar
p2 = 0.06; 			#bar

#At 28 bar
h1 = 2802.; 			#kJ/kg
s1 = 6.2104; 			#kJ/kg K

#At 0.06 bar
h_f2 = 151.5; 			#kJ/kg
h_f3 = h_f2;
h_fg2 = 2415.9; 		#kJ/kg
s_f2 = 0.521; 			#kJ/kg K
s_fg2 = 7.809; 			#kJ/kg K
v_f = 0.001; 			#m**3/kg

# Calculations
x2 = (s1-s_f2)/s_fg2;
h2 = h_f2 + x2*h_fg2;
W_turbine = h1-h2;
W_pump = v_f*(p1-p2)*100; 			#kJ/kg
h_f4 = h_f2+W_pump;
Q1 = h1-h_f4;
W_net = W_turbine - W_pump;
n_cycle = W_net/Q1;

# Results
print ("cyclic efficiency  = %.3f")% (n_cycle)

ratio = W_net/W_turbine; 			#Work ratio
print ("Work ratio  = %.3f")% (ratio)

S = 3600/W_net; 			#Specific steam combustion
print ("Specific steam combustion = %.3f")% (S), ("kg/kWh")

cyclic efficiency  = 0.335
Work ratio  = 0.997
Specific steam combustion = 4.056 kg/kWh

Example 12.6 Page no : 554

In [6]:
# Variables
p1 = 35.; 			#bar
x = 1.;
p2 = 0.2; 			#bar
m = 9.5; 			#kg/s

#At 35 bar
h1 = 2802.; 			#kJ/kg
h_g1 = h1;
s_g1 = 6.1228; 			#kJ/kg K

#At0.26 bar
h_f = 251.5; 			#kJ/kg
h_fg = 2358.4; 			#kJ/kg
v_f = 0.001017; 		#m**3/kg
s_f = 0.8321; 			#kJ/kg
s_fg = 7.0773; 			#kJ/kg K

# Calculations and Results
print ("(i) The pump work")
W_pump = v_f*(p1-p2)*100; 			#kJ/kg
P = m*W_pump; 			#power required
print ("Power required to drive the pump %.3f")% (P), ("kW")

print ("(ii) The turbine work")
x2 = (s_g1-s_f)/s_fg;
h2 = h_f+x2*h_fg;
W_turbine = m*(h1-h2);
print ("Turbine work = %.3f")% (W_turbine), ("kW")

print ("(iii) The Rankine efficiency")
n_rankine = (h1-h2)/(h1-h_f);
print ("rankine efficiency = %.3f")% (n_rankine)

print ("(iv) The condenser heat flow :")
Q = m*(h2-h_f);
print ("The condenser heat flow = %.3f")% (Q), ("kW")

print ("(v) The dryness at the end of expansion = %.3f")% (x2)

(i) The pump work
Power required to drive the pump 33.622 kW
(ii) The turbine work
Turbine work = 7480.838 kW
(iii) The Rankine efficiency
rankine efficiency = 0.309
(iv) The condenser heat flow :
The condenser heat flow = 16748.912 kW
(v) The dryness at the end of expansion = 0.748

Example 12.7 Page no : 555

In [7]:
# Variables
dh = 840.;   			#kJ/kg; Adiabatic enthalpy drop
h1 = 2940.; 			#/kJ/kg;
p2 = 0.1; 	    		#bar
h_f2 = 191.8; 			#kJ/kg

# Calculations and Results
n_rankine = (dh)/(h1-h_f2)*100;
print ("rankine efficiency = %.3f")% (n_rankine)

S = 3600/dh; 			#Specific steam combustion
print ("Specific steam combustion = %.3f")% (S), ("kg/kWh")

rankine efficiency = 30.565
Specific steam combustion = 4.286 kg/kWh

Example 12.8 Page no : 555

In [2]:
# Variables
IP = 35.; 			# Power developed by the engine in kW
S = 284.; 			#Steam combustion in kg/h
p2 = 0.14; 			#Condenser pressure in bar
p1 = 15.; 			#bar

h1 = 2923.3; 			#kJ/kg
s1 = 6.709; 			#kJ/kg K

h_f = 220.; 			#kJ/kg
h_fg = 2376.6; 			#kJ/kg
s_f = 0.737; 			#kJ/kg K
s_fg = 7.296; 			#kJ/kg K

# Calculations and Results
x2 = (s1-s_f)/s_fg;
print ("(i) Final condition of steam  = %.3f")% (x2)

h2 = h_f+x2*h_fg;
n_rankine = (h1-h2)/(h1-h_f);
print ("(ii) Rankine efficiency = %.3f")% (n_rankine)


n_thermal = IP/(S/3600)/(h1-h_f);

n_relative = n_thermal/n_rankine;
print ("(iii)relative efficiency = %.3f")% (n_relative)

(i) Final condition of steam  = 0.819
(ii) Rankine efficiency = 0.280
(iii)relative efficiency = 0.585

Example 12.9 Page no : 556

In [9]:
# Variables
P = 5000.; 			#kW
C = 40000.; 			#kJ/kg
n_rankine = 0.5;
n_turbine = 0.9;
n_heat_transfer = 0.85;
n_combustion = 0.98;

# Calculations
m_f = P/n_turbine/(C*n_heat_transfer*n_combustion*n_rankine);

# Results
print ("Fuel oil combustion = %.3f")% (m_f), ("kg/s")

Fuel oil combustion = 0.333 kg/s

Example 12.10 Page no : 558

In [10]:
# Variables
p2 = 2.; 			#bar
p3 = 1.1; 			#bar
IP = 1.;
m = 12.8/3600; 			#kg/kWs
n_mech = 0.8; 			#Mechanical efficiency
h1 = 3037.6; 			#kJ/kg
v1 = 0.169; 			#m**3/kg
s1 = 6.918; 			#kJ/kg K
t_s2 = 120.2; 			#0C
h_f2 = 504.7; 			#kJ/kg
h_fg2 = 2201.6; 		#kJ/kg
s_f2 = 1.5301; 			#kJ/kg K
s_fg2 = 5.5967; 		#kJ/kg K
v_f2 = 0.00106; 		#m**3/kg
v_g2 = 0.885; 			#m**3/kg
t_s3 = 102.3; 			#0C
h_f3 = 428.8; 			#kJ/kg
h_fg3 = 2250.8; 		#kJ/kg
s_f3 = 1.333; 			#kJ/kg K
s_fg3 = 5.9947; 		#kJ/kg K
v_f3 = 0.001; 			#m**3/kg
v_g3 = 1.549; 			#m**3/kg

# Calculations and Results
x2 = (s1-s_f2)/s_fg2;
h2 = h_f2+x2*h_fg2;
v2 = x2*v_g2+(1-x2)*v_f2;

W = (h1-h2) + (p2-p3)*v2*100; 			#kJ/kg
print ("(i)Ideal work = %.3f")% (W), ("kJ/kg")


n_rankine = W/(h1-h_f3);
print ("(ii) Rankine engine efficiency = %.3f")% (n_rankine)


print ("(iii) Indicated and brake work per kg")
W_indicated = IP/m;
print ("Indicated worK  = %.3f")% (W_indicated), ("kJ/kg")

W_brake = n_mech*IP/m;
print ("Brake work  = %.3f")% (W_brake), ("kJ/kg")

n_brake = W_brake/(h1-h_f3);
print ("(iv) Brake thermal efficiency = %.3f")% (n_brake)


print ("(v) Relative efficiency :")

n1 = W_indicated/W; 			#on the basis of indicated work
print ("Relative efficiency on the basis of indicated work = %.3f")%(n1)

n2 = W_brake/W; 			#on the basis of brake work
print ("Relative efficiency on the basis of brake work = %.3f")%(n2)

(i)Ideal work = 490.119 kJ/kg
(ii) Rankine engine efficiency = 0.188
(iii) Indicated and brake work per kg
Indicated worK  = 281.250 kJ/kg
Brake work  = 225.000 kJ/kg
(iv) Brake thermal efficiency = 0.086
(v) Relative efficiency :
Relative efficiency on the basis of indicated work = 0.574
Relative efficiency on the basis of brake work = 0.459

Example 12.11 Page no : 560

In [11]:
# Variables
p2 = 0.75; 			#bar
p3 = 0.3; 			#bar
h1 = 3263.9; 			#kJ/kg
v1 = 0.307; 			#m**3/kg
s1 = 7.465; 			#kJ/kg K
T_s2 = 369.7; 			#K
h_g2 = 2670.9; 			#kJ/kg
s_g2 = 7.3954; 			#kJ/kg K
v_g2 = 1.869; 			#m**3/kg
h_f3 = 289.3; 			#kJ/kg
v_g3 = 5.229; 			#m**3/kg
cp = 2.1;

# Calculations and Results
print ("(i) Quality of steam at the end of expansion")
T_sup2 = T_s2*(math.e**((s1-s_g2)/cp));
t_sup2 = T_sup2-273;
print ("t_sup2 = %.3f")% (t_sup2), ("°C")

h2 = h_g2+cp*(T_sup2-366.5);

print ("(ii) Quality of steam at the end of consmath.tant volume operation, x3 :")
v2 = v_g2/T_s2*T_sup2;
v3 = v2;
x3 = v3/v_g3;
print ("x3 = %.3f")% (x3)


print ("(iii) Power developed")
P = (h1-h2) + (p2-p3)*v2*100;
print ("P = %.3f")%(P), ("kW")


ssc = 3600./P;
print ("(iv) Specific steam consumption  = %.3f")% (ssc), ("kg/kWh")


n_mR = ((h1-h2)+(p2-p3)*v2*100)/(h1-h_f3);
print ("(v) Modified Rankine cycle efficiency  = %.3f")% (n_mR)

(i) Quality of steam at the end of expansion
t_sup2 = 109.158 °C
(ii) Quality of steam at the end of consmath.tant volume operation, x3 :
x3 = 0.369
(iii) Power developed
P = 647.057 kW
(iv) Specific steam consumption  = 5.564 kg/kWh
(v) Modified Rankine cycle efficiency  = 0.218

Example 12.12 Page no : 564

In [12]:
# Variables
h1 = 3100.; 			#kJ/kg
h2 = 2100.; 			#kJ/kg
h3 = 2500.; 			#kJ/kg
h_f2 = 570.9; 			#kJ/kg
h_f5 = 125.; 			#kJ/kg
h_f2 = 570.9; 			#kJ/kg
a = 11200.; 			#Quantity of bled steam in kg/h

# Calculations
m = (h_f2-h_f5)/(h2-h_f5);
S = a/m; 			#Steam supplied to the turbine per hour
W_net = (h1-h3) + (1-m)*(h3-h2);
P = W_net*S/3600.; 			#Power developed by the turbine

# Results
print ("Power developed by the turbine = %.3f")% (P), ("kW")

Power developed by the turbine = 12535.426 kW

Example 12.13 Page no : 565

In [13]:
# Variables
#At 30bar, 400 0C
h1 = 3230.9; 			#kJ/kg
s1 = 6.921; 			#kJ/kg
s2 = s1;
s3 = s1;
#At 5 bar
s_f1 = 1.8604;
s_g1 = 6.8192; 			#kJ/kg K
h_f1 = 640.1; 			#kJ/kg
t2 = 172. 			    #0C
h2 = 2796.; 			#kJ/kg

#At 0.1 bar
s_f3 = 0.649; 			#kJ/kg K
s_fg3 = 7.501; 			#kJ/kg K
h_f3 = 191.8; 			#kJ/kg
h_fg3 = 2392.8; 		#kJ/kg

# Calculations and Results
x3 = (s2-s_f3)/s_fg3;
h3 = h_f3+x3*h_fg3;

h_f4 = 191.8; 			#kJ/kg
h_f5 = h_f4;

h_f6 = 640.1; 			#kJ/kg
h_f7 = h_f6;
s7 = 1.8604; 			#kJ/kg K
s4 = 0.649; 			#kJ/kg K
m = (h_f6-h_f5)/(h2-h_f5);
W_T = (h1-h2) + (1-m)*(h2-h3);
Q1 = h1-h_f6;

n_cycle = W_T/Q1;
print ("(i) Efficiency of cycle  = %.3f")% (n_cycle)

SR = 3600/W_T; 			#Steam rate
print ("Steam rate  = %.3f")% (SR), ("kg/kWh")


T_m1 = (h1-h_f7)/(s1-s7);

T_m1r = (h1-h_f4)/(s1-s4); 			#Without regeneration

dT_m1 = T_m1-T_m1r;
print ("Increase in T_m1 due to regeneration = %.3f")% (dT_m1), ("0C")

W_Tr = h1-h3; 			#Without regeneration
SR1 = 3600/W_Tr; 			#Steam rate without regeneration
dSR = SR-SR1; 
print ("Increase in steam rate due to regeneration = %.3f")% (dSR), ("kg/kWh")

n_cycle1 = (h1-h3)/(h1-h_f4); 			#without regeneration
dn_cycle = n_cycle-n_cycle1;
print ("Increase in cycle efficiency due to regeneration %.3f")% (dn_cycle*100), ("%")

(i) Efficiency of cycle  = 0.361
Steam rate  = 3.852 kg/kWh
Increase in T_m1 due to regeneration = 27.405 0C
Increase in steam rate due to regeneration = 0.385 kg/kWh
Increase in cycle efficiency due to regeneration 1.902 %

Example 12.14 Page no : 567

In [14]:
# Variables
#At 3 bar
t_s1 = 133.5; 			#0C
h_f1 = 561.4; 			#kJ/kg

#At 0.04 bar
t_s2 = 29.; 			#0C
h_f2 = 121.5; 			#0C

h0 = 3231.; 			#kJ/kg
h1 = 2700.; 			#kJ/kg
h2 = 2085.; 			#kJ/kg

t1 = 130.; 			#0C
t2 = 27.; 			#0C
c = 4.186;

# Calculations and Results
print ("(i) Mass of steam used")
m1 = c*(t1-t2)/(h1-h_f2);
print ("m1 = %.3f")% (m1), ("kg")

print ("(ii) Thermal efficiency of the cycle")
W = (h0-h1)+(1-m1)*(h1-h2);
Q = h0-c*t1;

n_thermal = W/Q;
print ("n_thermal = %.3f")% (n_thermal)

(i) Mass of steam used
m1 = 0.167 kg
(ii) Thermal efficiency of the cycle
n_thermal = 0.388

Example 12.15 Page no : 569

In [15]:
# Variables
h0 = 3115.3; 			#kJ/kg
h1 = 2720.; 			#kJ/kg
h2 = 2450.; 			#kJ/kg
h3 = 2120.; 			#kJ/kg

h_f1 = 640.1; 			#kJ/kg
h_f2 = 417.5; 			#kJ/kg
h_f3 = 173.9; 			#kJ/kg

# Calculations and Results
m1 = (h_f1-h_f2)/(h1-h_f1);
print ("m1 = %.3f")% (m1), ("kJ/kg")

m2 = ((h_f2-h_f3)-m1*(h_f1-h_f3))/(h2-h_f3);
print ("m2 = %.3f")% (m2), ("kJ/kg")

W = h0-h1 + (1-m1)*(h1-h2) + (1-m1-m2)*(h2-h3);
Q = h0-h_f1;

n = W/Q;
print ("Thermal Efficiency of the cycle = %.3f")% (n)

m1 = 0.107 kJ/kg
m2 = 0.085 kJ/kg
Thermal Efficiency of the cycle = 0.365

Example 12.16 Page no : 570

In [16]:
# Variables
h0 = 2905.; 			#kJ/kg
h1 = 2600.; 			#kJ/kg
h2 = 2430.; 			#kJ/kg
h3 = 2210.; 			#kJ/kg
h4 = 2000.; 			#kJ/kg

h_f1 = 640.1; 			#kJ/kg
h_f2 = 467.1; 			#kJ/kg
h_f3 = 289.3; 			#kJ/kg
h_f4 = 137.8; 			#kJ/kg

# Calculations and Results
print ("(i) Mass of bled steam")
m1 = (h_f1-h_f2)/(h1-h_f1);
print ("m1 = %3f")% (m1), ("kJ/kg")

m2 = ((h_f2-h_f3) - (m1*(h_f1-h_f2)))/(h2-h_f2);
print ("m2 = %.3f")% (m2), ("kJ/kg")

m3 = ((h_f3-h_f4)-(m1+m2)*(h_f2-h_f4))/(h3-h_f4);
print ("m3 = %.3f")% (m3), ("kJ/kg")

W = (h0-h1) + (1-m1)*(h1-h2)+(1-m1-m2)*(h2-h3) + (1-m1-m2-m3)*(h3-h4);
Q = h0-h_f1;

n_thermal = W/Q;
print ("(ii) Thermal efficiency of the cycle = %.3f")%(n_thermal)

n_rankine = (h0-h4)/(h0-h_f4);
print ("(iii) Thermal efficiency of Rankine cycle  = %.3f")% (n_rankine)

gain = (n_thermal-n_rankine)/(n_thermal);
print ("(iv) Theoretical gain due to regenerative feed heating  = %.3f")%(gain)

S1 = 3600./W;
print ("(v) Steam consumption with regenerative feed heating  = %.3f")% (S1), ("kg/kWh")

S2 = 3600./(h0-h4);
print ("Steam consumption without regenerative feed heating  = %.3f")% (S2), ("kg/kWh")

quantity1 = S1*(1-m1-m2-m3)*50000;
print ("(vi) Quantity of steam passing through the last stage of a 50000 kW turbine with \
regenerative feed-heating  = %.3f")%(quantity1), ("kg/h")

quantity2 = S2*50000;
print ("quantity of steam without regeneration  = %.3f")% (quantity2), ("kg/h")

(i) Mass of bled steam
m1 = 0.088270 kJ/kg
m2 = 0.083 kJ/kg
m3 = 0.046 kJ/kg
(ii) Thermal efficiency of the cycle = 0.356
(iii) Thermal efficiency of Rankine cycle  = 0.327
(iv) Theoretical gain due to regenerative feed heating  = 0.082
(v) Steam consumption with regenerative feed heating  = 4.462 kg/kWh
Steam consumption without regenerative feed heating  = 3.978 kg/kWh
(vi) Quantity of steam passing through the last stage of a 50000 kW turbine with regenerative feed-heating  = 174693.345 kg/h
quantity of steam without regeneration  = 198895.028 kg/h

Example 12.17 Page no : 573

In [3]:
# Variables
h1 = 3460.; 			#kJ/kg
h2 = 3460.; 			#kJ/kg
h3 = 3111.5; 			#kJ/kg
h4 = 3585.; 			#kJ/kg
h5 = 3207.; 			#kJ/kg
h6 = 2466.; 			#kJ/kg
h7 = 137.8; 			#kJ/kg
h8 = 962.; 			    #kJ/kg
h9 = 670.4; 			#kJ/kg
h10 = 962.; 			#kJ/kg

p1 = 100.; 			#bar
p2 = 95.; 			#bar
p3 = 25.; 			#bar
p4 = 22.; 			#bar
p5 = 6.; 			#bar
p6 = 0.05; 			#bar
n_mech = 0.9;
n_gen = 0.96;
n_boiler = 0.9;

# Calculations
P = 120.*10**3; 			#kW
m1 = (h10-h9)/(h3-h8);
m2 = (h9-m1*h8-(1-m1)*h7)/(h5-h7);
W_IP = (1-m1-m2)*(p5-p6)*0.001*10**2;
W_HP = (p1-p5)*0.001*10**2;
W_total = (W_IP+W_HP)/n_mech;
W_indicated = (h2-h3) + (1-m1)*(h4-h5) + (1-m1-m2)*(h5-h6);
Output = (W_indicated - W_total)*n_mech*n_gen; 			#net electrical output
rate = P*3600/Output;

amt1 = m1*rate; 			#Amounts of bled off, surface(high pressure) heater

# Results
print ("Amounts of bled off, surface(high pressure) heater  = %.3f")%(amt1),("kg/h")

amt2 = m2*rate; 			#Amounts of bled off, surface(low pressure) heater
print ("Amounts of bled off, surface(low pressure) heater %.3f")% (amt2), ("kg/h")


Q_boiler = (h1-h10)/n_boiler;
Q_reheater = (h4-h3)/n_boiler;

n_overall = Output/(Q_boiler+Q_reheater)*100;
print ("(iii)Overall thermal efficiency  = %.3f")% (n_overall), ("%")


ssc = rate/P; 			#Specific steam consumption
print ("(iv) Specific steam consumption  = %.3f")% (ssc), ("kg/kWh")

Amounts of bled off, surface(high pressure) heater  = 56375.368 kg/h
Amounts of bled off, surface(low pressure) heater 56974.301 kg/h
(iii)Overall thermal efficiency  = 31.486 %
(iv) Specific steam consumption  = 3.463 kg/kWh

Example 12.18 Page no : 579

In [18]:
# Variables
p1 = 15.; 			#bar
p2 = 4.; 			#bar
p4 = 0.1; 			#bar
h1 = 2920.; 			#kJ/kg
h2 = 2660.; 			#kJ/kg
h3 = 2960.; 			#kJ/kg
h4 = 2335.; 			#kJ/kg

# Calculations and Results
W = h1-h2+h3-h4;
print ("work done per kg of steam"), (W), ("kJ/kg")

h_reheat = h3-h2;
print ("Amount of heat supplied during reheat  = "), (h_reheat), ("kJ/kg")

h_4a = 2125.;  			#kJ/kg

W1 = h1-h_4a;
print ("Work output without reheat  = "), (W1), ("kJ/kg")

work done per kg of steam 885.0 kJ/kg
Amount of heat supplied during reheat  =  300.0 kJ/kg
Work output without reheat  =  795.0 kJ/kg

Example 12.19 Page no : 580

In [19]:
# Variables
h1 = 3450.; 			#kJ/kg
h2 = 3050.; 			#kJ/kg
h3 = 3560.; 			#kJ/kg
h4 = 2300.; 			#kJ/kg

h_f4 = 191.8; 			#kJ/kg

# Calculations and Results
#From mollier diagram
x4 = 0.88;
print ("(i)Quality of steam at turbine exhaust  = "), (x4)

n_cycle = ((h1-h2) + (h3-h4))/((h1-h_f4) + (h3-h2));
print ("(ii) Cycle efficiency  = %.3f")%(n_cycle)

SR = 3600/((h1-h2) + (h3-h4));
print ("(iii) Steam rate in kg/kWh  = %.3f")% (SR), ("kg/kWh")

(i)Quality of steam at turbine exhaust  =  0.88
(ii) Cycle efficiency  = 0.441
(iii) Steam rate in kg/kWh  = 2.169 kg/kWh

Example 12.20 Page no : 581

In [20]:
# Variables
h1 = 3250.; 			#kJ/kg
h2 = 2170.; 			#kJ/kg
h_f2 = 173.9; 			#kJ/kg

# Calculations and Results
W = h1-h2;
Q = h1-h_f2;

n_thermal = W/Q;
print ("Thermal effifciency = %.3f")% (n_thermal);

x2 = 0.83; 			#From mollier chart
print ("x2 = %.3f")% (x2)


print ("Second case")

h1 = 3250.; 			#kJ/kg
h2 = 2807.; 			#kJ/kg
h3 = 3263.; 			#kJ/kg
h4 = 2426.; 			#kJ/kg
h_f4 = 173.9; 			#kJ/kg
W = h1-h2+h3-h4;
Q = h1-h_f4+h3-h2;

n_thermal = W/Q;
print ("Thermal effifciency = %.3f")% (n_thermal);

x4 = 0.935; 			#From mollier chart
print ("x4 = %.3f")% (x4)

Thermal effifciency = 0.351
x2 = 0.830
Second case
Thermal effifciency = 0.362
x4 = 0.935

Example 12.21 Page no : 582

In [7]:
print ("Part (c)")

# Variables
h1 = 3580.; 			#kJ/kg
h2 = 3140.; 			#kJ/kg
h3 = 3675.; 			#kJ/kg
h4 = 2335.; 			#kJ/kg
h5 = 191.8; 			#kJ/kg

# Calculations and Results
P = 15.*10**3; 			#kW
a = 0.104; 			#moisture content in exit from LP turbine

p = 40.; 			#bar; From the mollier diagram
print ("(i)Reheat pressure = "),  (p), ("bar")


W = h1-h2+h3-h4;
Q = h1-h5+h3-h2;
n_th = W/Q*100; 
print ("(ii) Thermal efficiency"),("n_th = %.3f")% (n_th), ("%")

sc = P/W;			#steam consumption
ssc = sc*3600./P; 			#specific steam consumption
print ("(iii)Specific steam consumption = %.3f")% (ssc), ("kg/kWh")

rate = sc*0.15;
print ("(iv) Rate of pump work  = %.3f")%(rate),("kW")

Part (c)
(i)Reheat pressure =  40.0 bar
(ii) Thermal efficiency n_th = 45.371 %
(iii)Specific steam consumption = 2.022 kg/kWh
(iv) Rate of pump work  = 1.264 kW

Example 12.22 Page no : 588

In [10]:
# Variables
h_l = 355.988; 			#kJ/kg
s_l = 0.5397; 			#kJ/kg K 
s_f = 0.0808; 			#kJ/kg K
s_g = 0.6925; 			#kJ/kg K
h_f = 29.98; 			#kJ/kg
h_g = 329.85; 			#kJ/kg
p1 = 4.; 			    #bar
p2 = 0.04; 			    #bar
v_f2 = 76.5*10**(-6); 	#m**3/kg
h1 = 2789.9; 			#kJ/kg
s1 = 6.4406; 			#kJ/kg
h_f = 121.5; 			#kJ/kg
h_fg = 2432.9; 			#kJ/kg
s_f = 0.432; 			#kJ/kg K
s_fg2 = 8.052; 			#kJ/kg K
p4 = 15; 			    #bar
p3 = 0.04; 			    #bar
v_f = 0.0001; 			#kJ/kg K
h_f4 = 123; 			#kJ/kg
h_m = 254.88; 			#kJ/kg
h_fn = 29.98; 			#kJ/kg
h_fk = 29.988; 			#kJ/kg

# Calculations and Results
m = (h1-h_f4)/(h_m-h_fn); 	#The amount of mercury circulating for 1kg of steam in the bottom cycle
Q1 = m*(h_l-h_fk); 			#total
x2 = (s1-s_f)/(s_fg2);
h2 = h_f+x2*h_fg;
W_T = m*(h_l-h_m)+(h1-h2); 		#total
n_overall = W_T/Q1; 			#W_P may be neglected
print ("(i) Overall thermal efficiency "),(" = %.3f")% (n_overall)

A = 48000.; 			#kg/h
m_Hg = m*A;
print ("(ii) Flow through mercury turbine = %.3f")% (m_Hg), ("kg/h")


W_total = A*W_T/3600; 
print ("(iii) Useful work in binary vapour cycle = %.3f")% (W_total), ("kW")



n_Hg = 0.84;
n_steam = 0.88;

W_Hg = n_Hg*101.1;
h_m1 = h_l-W_Hg;
m1 = (h1-h_f4)/(h_m1-h_fn);

h_g = 3037.6; 			#kJ/kg
s_g = 6.918; 			#kJ/kg
s_f2 = 0.423; 			#kJ/kg K
s_fg2 = 8.052; 			#kJ/kg K

Q1 = m1*(h_l - h_fk) + (h_g-h1);

x2 = (s_g-s_f2)/s_fg2;
h2 = h_f+x2*h_fg;

W_steam = n_steam*(h_g-h2);

W_total = m1*W_Hg + W_steam;

n_overall = W_total/Q1;
print ("(iv) Overall efficiency under new conditions "),("= %.3f")% (n_overall)

(i) Overall thermal efficiency   = 0.531
(ii) Flow through mercury turbine = 569191.641 kg/h
(iii) Useful work in binary vapour cycle = 27358.201 kW
(iv) Overall efficiency under new conditions  = 0.462

Example 12.23 Page no : 591

In [23]:
# Variables
p1 = 60.; 			#bar
t1 = 450.; 			#0C
p2 = 3.; 			#bar
p3 = 0.07; 			#bar; p3 = (760-707.5)/760*1.013
n_turbine = 0.87;
n_boiler = 0.86;
n_alt = 0.94;
n_mech = 0.97;
P = 22500.; 			#kW
h1 = 3300.; 			#kJ/kg
h2 = 2607.; 			#kJ/kg

# Calculations and Results
h2a = h1-n_turbine*(h1-h2);
h3 = 2165.; 			#kJ/kg
h3a = h2a-n_turbine*(h2a-h3);
h_f4 = 163.4; 			#kJ/kg
h_f5 = 561.4; 			#kJ/kg

print ("(i) The steam bled per kg of steam supplied to the turbine")
m = (h_f5-h_f4)/(h2a-h_f4);
print ("m = %.3f")% (m), ("kJ/kg")

print ("(ii) Steam generated per hour")
W = (h1-h2a) + (1-m)*(h2a-h3a); 			#Work developed per kg of steam in the turbine
W_act = P/n_alt/n_mech; 			#actual work 

steam = W_act/W*3600./1000; 			#tonnes/h
print ("Steam generated = %.3f")%(steam), ("tonnes/h")

print ("(iii) The overall efficiency of the plant")
P_avail = P*(1-0.09); 			#Net power available deducting pump power
Q = steam*1000*(h1-h_f5)/n_boiler/3600.; 			#kW

n_overall = P_avail/Q
print ("n_overall = %.3f")% (n_overall)

(i) The steam bled per kg of steam supplied to the turbine
m = 0.157 kJ/kg
(ii) Steam generated per hour
Steam generated = 89.451 tonnes/h
(iii) The overall efficiency of the plant
n_overall = 0.259

Example 12.24 Page no : 593

In [24]:
# Variables
t1 = 350.; 			#0C
t_s = 350.; 		#0C
p2 = 7.; 			#bar
p3 = 7.; 			#bar
p4 = 0.4; 			#bar
t3 = 350.; 			#0C
h1 = 2985.; 			#kJ/kg
h2 = 2520.; 			#kJ/kg
h3 = 3170.; 			#kJ/kg
h4 = 2555.; 			#kJ/kg
h_f2 = 697.1; 			#kJ/kg
h_f4 = 317.7; 			#kJ/kg


# Calculations and Results
P = 110.*10**3; 			#kW
print ("(i) The ratio of steam bled to steam generated")
m = (h_f2-h_f4)/(h2-h_f4);
ratio = 1/m;
print ("ratio = %.3f")% (ratio)

m_s = P/(h1-h2+(1-m)*(h3-h4))*3600/1000.; 			#tonnes/hour
print ("(ii) The boiler generating capacity  = %.3f")% (m_s), ("tonnes/hour")

n_thermal = ((h1-h2) + (1-m)*(h3-h4))/((h1-h_f2)+(1-m)*(h3-h2));
print ("(iii) Thermal efficiency of the cycle  = %.3f")%(n_thermal)

(i) The ratio of steam bled to steam generated
ratio = 5.805
(ii) The boiler generating capacity  = 406.549 tonnes/hour
(iii) Thermal efficiency of the cycle  = 0.345

Example 12.25 Page no : 595

In [25]:
# Variables
h1 = 3315.; 			#kJ/kg
h2 = 2716.; 			#kJ/kg
h3 = 3165.; 			#kJ/kg
h4 = 2236.; 			#kJ/kg
h_f2 = 697.1;			#kJ/kg
h_f6 = h_f2;
h_f4 = 111.9; 			#kJ/kg
h_f5 = h_f4;

# Calculations and Results
m = (h_f2-h_f4)/(h2-h_f4);
print ("(i) Amount of steam bled off for feed heating  = %.3f")% (m), ("steam bled off is 22.5% of steam generated by the boiler.")


amt = 100-m*100;
print ("(ii) Amount of steam supplied to L.P. turbine  = %.3f")%(amt), ("77.5% of the steam generated by the boiler.")

print ("(iii) Heat supplied in the boiler and reheater")
Q_boiler = h1-h_f6;
print ("Q_boiler = %.3f")% (Q_boiler), ("kJ/kg")

Q_reheater = (1-m)*(h3-h2);
print ("Q_reheater = %.3f")% (Q_reheater), ("kJ/kg")

Qs = Q_boiler+Q_reheater;

print ("(iv) Cycle efficiency")
W = h1-h2 + (1-m)*(h3-h4);

n_cycle = W/Qs;
print ("n_cycle = %.3f")% (n_cycle)

print ("(v) Power developed by the system")
ms = 50.; 			#kg/s
Power = ms*W/1000; 			#MW
print ("Power = %.3f")% (Power), ("MW")

(i) Amount of steam bled off for feed heating  = 0.225 steam bled off is 22.5% of steam generated by the boiler.
(ii) Amount of steam supplied to L.P. turbine  = 77.528 77.5% of the steam generated by the boiler.
(iii) Heat supplied in the boiler and reheater
Q_boiler = 2617.900 kJ/kg
Q_reheater = 348.100 kJ/kg
(iv) Cycle efficiency
n_cycle = 0.445
(v) Power developed by the system
Power = 65.962 MW

Example 12.26 Page no : 597

In [26]:
# Variables
h1  =  3578.; 			#kJ/kg
h2  =  3140.; 			#kJ/kg
h3  =  3678.; 			#kJ/kg
h4  =  3000.; 			#kJ/kg
h5  =  2330.; 			#kJ/kg
h_f1 = 1611.; 			#kJ/kg
h_f2 = 1087.4; 			#kJ/kg
h_f4 = 640.1; 			#kJ/kg
h_f5 = 191.8; 			#kJ/kg
h_f6 = h_f5;

# Calculations and Results
print ("(i) Fraction of steam extracted from the turbines at each bled heater  = ")

m1 = (h_f2-h_f4)/(h2-h_f4);
print ("closed feed heater %.3f")% (m1), ("kg/kg of steam supplied by the boiler")


m2 = (1-m1)*(h_f4-h_f5)/(h4-h_f6);
print ("open feed heater %.3f")% (m2), ("kg/kg of steam supplied by the boiler")

print ("(ii) Thermal efficiency of the system")

W_total = (h1-h2) + (1-m1)*(h3-h4) + (1-m1-m2)*(h4-h5);
p1 = 150.; 			#bar
p2 = 40.; 			#bar
p4 = 5.; 			#bar
p5 = 0.1; 			#bar
v_w1 = 1./1000; 			#m**3/kg
v_w2 = v_w1;
v_w3 = v_w1;
W_P1 = v_w1*(1-m1-m2)*(p4-p5)*100; 		#kJ/kg
W_P2 = v_w2*(1-m1)*(p1-p4)*100; 		#kJ/kg
W_P3 = v_w3*m1*(p1-p2)*100; 			#kJ/kg
W_P = W_P1+W_P2+W_P3; 			        #Total pump work
W_net = W_total-W_P;
Q = (1-m1)*h_f1 +m1*(h_f1); 			#Heat of feed water extering the boiler
Qs1 = h1-Q;
Qs2 = (1-m1)*(h3-h2);
Qst = Qs1+Qs2;

n_thermal = W_net/Qst*100;
print ("n_thermal = %.3f")% (n_thermal), ("%")

(i) Fraction of steam extracted from the turbines at each bled heater  = 
closed feed heater 0.179 kg/kg of steam supplied by the boiler
open feed heater 0.131 kg/kg of steam supplied by the boiler
(ii) Thermal efficiency of the system
n_thermal = 59.898 %

Example 12.27 Page no : 599

In [27]:
# Variables

p_min = 10.; 			#bar
print ("(i) The minimum pressure at which bleeding is necessary = "), (p_min), ("bar")
h1 = 3285.; 			#kJ/kg
h2 = 2980.; 			#kJ/kg
h3 = 3280.; 			#kJ/kg
h4a = 3072.5; 			#kJ/kg
h5 = 2210.; 			#kJ/kg
h5a = 2356.6; 			#kJ/kg
h_f6 = 163.4; 			#kJ/kg
h_f8 = 762.6; 			#kJ/kg
h2a = 3045.6; 			#kJ/kg

# Calculations and Results
m = (h_f8-h_f6)/(h4a-h_f6);
print ("(ii) The quantity of steam bled per kg of flow at the turbine inlet  = %.3f")% (m), ("kg of steam flow at turbine inlet.")

n_cycle = ((h1-h2a)+(h3-h4a)+(1-m)*(h4a-h5a))/((h1-h_f8) + (h3 - h2a))*100;
print ("(iii) Cycle efficiency  = %.3f")% (n_cycle), ("%")

(i) The minimum pressure at which bleeding is necessary =  10.0 bar
(ii) The quantity of steam bled per kg of flow at the turbine inlet  = 0.206 kg of steam flow at turbine inlet.
(iii) Cycle efficiency  = 36.830 %