Chapter 12 : Vapour Power Cycles

Example 12.1 Page No : 475

In [1]:
# Variables
# Part (a)
P1 = 1e05;
P2 = 10e05;
vf = 0.001043;

# Calculation and Results
Wrev = vf*(P1-P2);
print "The work required in saturated liquid form is",Wrev/1000,"kJ/kg"

# Part (b)
h1 = 2675.5; 
s1 = 7.3594;
s2 = s1;
h2 = 3195.5;
Wrev1 = h1-h2;
print "The work required in saturated vapour form is",Wrev1,"kJ/kg"
The work required in saturated liquid form is -0.9387 kJ/kg
The work required in saturated vapour form is -520.0 kJ/kg

Example 12.2 Page No : 476

In [4]:
# Variables
h1 = 3159.3;
s1 = 6.9917;
h3 = 173.88; 
s3 = 0.5926; 
sfp2 = s3; 
hfp2 = h3;
hfgp2 = 2403.1; 
sgp2 = 8.2287;
vfp2 = 0.001008; 
sfgp2 = 7.6361;

# Calculation and Results
x2s = (s1-sfp2)/(sfgp2);
h2s = hfp2+(x2s*hfgp2);

# Part (a)
P1 = 20e02; 
P2 = 0.08e02;
h4s = vfp2*(P1-P2)+h3 ; 
Wp = h4s-h3;
Wt = h1-h2s;
Wnet = Wt-Wp;
Q1 = h1-h4s;
n_cycle = Wnet/Q1;
print "Net work per kg of steam is %.2f kJ/Kg"%Wnet
print "Cycle efficiency is %.3f"%n_cycle

# Part (b)
n_p = 0.8; 
n_t = 0.8;
Wp_ = Wp/n_p;
Wt_ = Wt*n_t;
Wnet_ = Wt_-Wp_;
P = 100*((Wnet-Wnet_)/Wnet) ;
n_cycle_ = Wnet_/Q1;
P_ = 100*((n_cycle-n_cycle_)/n_cycle);
print "Percentage reduction in net work per kg of steam is %.1f %%"%P
print "Percentage reduction in cycle efficiency is %.1f %%"%P_
Net work per kg of steam is 969.60 kJ/Kg
Cycle efficiency is 0.325
Percentage reduction in net work per kg of steam is 20.1 %
Percentage reduction in cycle efficiency is 20.1 %

Example 12.3 Page No : 477

In [3]:
# Variables
P1 = 0.08; 			# in bar
sf = 0.5926; 
x2s = 0.85; 
sg = 8.2287; 

# Calculation
s2s = sf+(x2s*(sg-sf));
s1 = s2s;
P2 = 16.832; 			# by steam table opposite to s1 in bar
h1 = 3165.54;
h2s = 173.88 + (0.85*2403.1);
h3 = 173.88;
vfp2 = 0.001;
h4s = h3 + (vfp2*(P2-P1)*100);
Q1 = h1-h4s;
Wt = h1-h2s;
Wp = h4s-h3;
n_cycle = 100*((Wt-Wp)/Q1);
Tm = (h1-h4s)/(s2s-sf);

# Results
print "The greatest allowable steam pressure at the turbine inlet is",P2,"bar"
print "Rankine cycle efficiency is %.2f %%"%n_cycle
print "Mean temperature of heat addition is %.2f K"%Tm
The greatest allowable steam pressure at the turbine inlet is 16.832 bar
Rankine cycle efficiency is 31.68 %
Mean temperature of heat addition is 460.66 K

Example 12.4 Page No : 478

In [8]:
# Variables
h1 = 3465.;
h2s = 3065.; 
h3 = 3565.;
h4s = 2300.; 
x4s = 0.88; 
h5 = 191.83;
v = 0.001;
P = 150.; 			# in bar

# Calculation
Wp = v*P*100;
h6s = 206.83;
Q1 = (h1-h6s)+(h3-h2s);
Wt = (h1-h2s)+(h3-h4s);
Wnet = Wt-Wp;
n_cycle = 100*Wnet/Q1;
sr = 3600/Wnet;

# Results
print "Quality at turbine exhaust is",0.88
print "Cycle efficiency is %.1f %%"%n_cycle
print "steam rate is %.2f Kg/kW h"%sr
Quality at turbine exhaust is 0.88
Cycle efficiency is 43.9 %
steam rate is 2.18 Kg/kW h

Example 12.5 Page No : 479

In [1]:
# Variables
h1 = 3230.9; 
s1 = 6.9212; 
s2 = s1; 
s3 = s1;
h2 = 2796; 
sf = 0.6493; 
sfg = 7.5009;

# Calculation
x3 = (s3-sf)/sfg;
h3 = 191.83 + x3*2392.8;
h4 = 191.83; h5 = h4;
h6 = 640.23; h7 = h6;
m = (h6-h5)/(h2-h5);
Wt = (h1-h2)+(1-m)*(h2-h3);
Q1 = h1-h6;
n_cycle = 100*Wt/Q1;
sr = 3600/Wt;
s7 = 1.8607; s4 = 0.6493;
Tm = (h1-h7)/(s1-s7);
Tm1 = (h1-h4)/(s1-s4); 			# With out regeneration
dT = Tm-Tm1;
Wt_ = h1-h3;
sr_ = 3600/Wt_;
dsr = sr-sr_;
n_cycle_ = 100*(h1-h3)/(h1-h4);
dn = n_cycle-n_cycle_;

# Results
print "Efficiency of the cycle is %.2f %%"%n_cycle
print "Steam rate of the cycle is %.2f kg/kW h"%sr
print "Increase in temperature due to regeneration is %.1f degree centigrade"%dT
print "Increase in steam rate due to regeneration is %.1f kg/kW h"%dsr
print "Increase in Efficiency of the cycle due to regeneration is %.2f %%"%dn

# note: rounding error is there because of decimal points.
Efficiency of the cycle is 36.07 %
Steam rate of the cycle is 3.85 kg/kW h
Increase in temperature due to regeneration is 27.4 degree centigrade
Increase in steam rate due to regeneration is 0.4 kg/kW h
Increase in Efficiency of the cycle due to regeneration is 1.90 %

Example 12.6 Page No : 481

In [4]:
# Variables
h1 = 3023.5; 
s1 = 6.7664; 
s2 = s1; 
s3 = s1; 
s4 = s1;
t_sat_20 = 212.;
t_sat_1 = 46.;

# Calculation
dt = t_sat_20-t_sat_1;
n =3; 			# number of heaters
t = dt/n;
t1 = t_sat_20-t;
t2 = t1-t;

# 0.1 bar
hf = 191.83; 
hfg = 2392.8; 
sf = 0.6493;
sg = 8.1502;
# At 100 degree
hf100 = 419.04; 
hfg100 = 2257.0; 
sf100 = 1.3069; 
sg100 = 7.3549;
# At 150 degree
hf150 = 632.20; 
hfg150 = 2114.3; 
sf150 = 1.8418; 
sg150 = 6.8379;
x2 = (s1-sf150)/4.9961;
h2 = hf150+(x2*hfg150);
x3 = (s1-sf100)/6.0480;
h3 = hf100+(x3*hfg100);
x4 = (s1-sf)/7.5010;
h4 = hf+(x4*hfg);
h5 = hf; h6 = h5;
h7 = hf100; h8 = h7;
h9 = 632.2; h10 = h9;
m1 = (h9-h7)/(h2-h7);
m2 = ((1-m1)*(h7-h6))/(h3-h6);
Wt = 1*(h1-h2)+(1-m1)*(h2-h3)+(1-m1-m2)*(h3-h4);
Q1 = h1-h9;
Wp = 0 ; 			# Pump work is neglected
n_cycle = 100*(Wt-Wp)/Q1;
sr = 3600/(Wt-Wp);

# Results 
print "Net work per kg of stem is %.2f kJ/Kg"%Wt
print "Cycle efficiency is %.2f %%"%n_cycle
print "Stream rate is %.2f Kg/kW h"%sr

# rounding off error would be there.
Net work per kg of stem is 798.64 kJ/Kg
Cycle efficiency is 33.40 %
Stream rate is 4.51 Kg/kW h

Example 12.7 Page No : 483

In [17]:
import math 

# Variables
Ti = 2000.;
Te = 450.;
T0 = 300.;
Q1_dot = 100e03; 			# in kW
cpg = 1.1;

# Calculation
wg = Q1_dot/(cpg*(Ti-Te));
af1 = wg*cpg*T0*((Ti/T0)-1-math.log(Ti/T0));
af2 = wg*cpg*T0*((Te/T0)-1-math.log(Te/T0));
afi = af1-af2;
h1 = 2801; h3 = 169; h4 = 172.8; h2 = 1890.2;
s1 = 6.068; s2 = s1; s3 = 0.576; s4 = s3;
Wt = h1-h2;
Wp = h4-h3;
Q1 = h1-h4;
Q2 = h2-h3;
Wnet = Wt-Wp;
ws = Q1_dot/2628;
afu = 38*(h1-h4-T0*(s1-s3));
I_dot = afi-afu;
Wnet_dot = ws*Wnet;
afc = ws*(h2-h3-T0*(s2-s3));
n2 = 100*Wnet_dot/af1;

# Results
print "The second law efficiency is %.1f %%"%n2
The second law efficiency is 47.3 %

Example 12.8 Page No : 485

In [25]:
import math 

# Variables
h1 = 2758.;      # kJ/kg
h2 = 1817.; 
h3 = 192.
h4 = 200.;
cpg = 1.1

# Calculation and Results
Wt = h1-h2; 
Wp = h4-h3;
Q1 = h1-h4; 
Wnet = Wt-Wp;
n1 = Wnet/Q1;
WR = Wnet/Wt;
Q1_ = 100.;
PO = n1*Q1_;
cp = 1000.;
wg = (Q1_/(833.-450));
EIR = 59.3
n2 = PO/EIR ;
print ("Part (a)")
print "n1 is %.1f %%"%(n1*100)
print "n2 is %.1f %%"%(n2*100)
print "Work ratio is %.3f"%WR

# Part (b)
h1b = 3398.; 
h2b = 2130.
h3b = 192.; 
h4b = 200.;
Wtb = 1268.; 
Wpb = 8.; 
Q1b = 3198.;
Wt = 1268.
Wp = 8.
n1b = (Wt-Wp)/Q1b;
WRb = (Wt-Wp)/Wtb;
EIRb = 59.3;
Wnetb = Q1b*n1b;
n2b = 36.5/EIRb;
print ("\nPart (b)")
print "n1 is %.1f %%"%(n1b*100)
print "n2 is %.1f %%"%(n2b*100)
print "Work ratio is %.3f"%WRb

# Part (c)
h1c = 3398.; 
h2c = 2761.; 
h3c = 3482.; 
h4c = 2522.; 
h5c = 192.; 
h6c = 200.;
Wt1 = 637.; 
Wt2 = 960.; 
Wtc = Wt1+Wt2; Wpc = 8.;
Wnetc = 1589.; 
Q1c = 3198+721.;
n1c = Wnetc/Q1c;
WRc = Wnetc/Wtc;
POc = Q1_*n1c;
EIRc = 59.3;
n2c = POc/EIRc;
print ("\nPart (c)")
print "n1 is %.1f %%"%(n1c*100)
print "n2 is %.1f %%"%(n2c*100)
print "Work ratio is %.3f"%WRc

# Part (d)
T3 = 318.8; T1 = 568.;
n1d = 1-(T3/T1);
Q1d = 2758-1316.;
Wnet = Q1d*n1d;
Wpd = 8.; 
Wtd = 641.;
WRd = (Wtd-Wpd)/Wtd;
POd = Q1_*0.439;
EIRd = (Q1_/(833-593.))*cpg*((833.-300)-300*(math.log(833./300)));
n2d = POd/94.6;
print ("\nPart (d)")
print "n1 is %.1f %%"%(n1d*100)
print "n2 is %.1f %%"%(n2d*100)
print "Work ratio is %.3f"%WRd
Part (a)
n1 is 36.5 %
n2 is 61.5 %
Work ratio is 0.991

Part (b)
n1 is 39.4 %
n2 is 61.6 %
Work ratio is 0.994

Part (c)
n1 is 40.5 %
n2 is 68.4 %
Work ratio is 0.995

Part (d)
n1 is 43.9 %
n2 is 46.4 %
Work ratio is 0.988

Example 12.9 Page No : 488

In [9]:
# Variables
hfg = 2202.6;
Qh = 5.83;
ws = Qh/hfg;
eg = 0.9; 			# efficiency of generator
P = 1000;
Wnet = 1000/0.9;
nbrake = 0.8;

# Calculation
h1_2s = Wnet/(ws*nbrake); 			# h1-h2s
n_internal = 0.85;
h12 = n_internal*h1_2s;
hg = 2706.3; 
h2 = hg;
h1 = h12+h2;
h2s = h1-h1_2s;
hf = 503.71;
x2s = (h2s-hf)/hfg;
sf = 1.5276; 
sfg = 5.6020;
s2s = sf+(x2s*sfg);
s1 = s2s;
P1 = 22.5; 			# in bar from Moiller chart
t1 = 360;

# Results
print "Temperature of the steam is",t1,"degree"
print "Pressure of the steam is ",P1,"bar"
Temperature of the steam is 360 degree
Pressure of the steam is  22.5 bar

Example 12.10 Page No : 489

In [3]:
# Variables
h1 = 3037.3; h2 = 561+(0.96*2163.8);
s2 = 1.6718+(0.96*5.3201);
s3s = s2;

# Calculation
x3s = (s3s-0.6493)/7.5009;
h3s = 191.83+(x3s*2392.8);
h23 = 0.8*(h2-h3s); 			# h2-h3
h3 = h2-h23;
h5 = 561.47; h4 = 191.83;
Qh = 3500.; 			# in kJ/s
w = Qh/(h2-h5);
Wt = 1500.;
ws = (Wt+w*(h2-h3))/(h1-h3); 
ws_ = 3600.*ws ; 			# in kg/h
h6 = ((ws-w)*h4+w*h5)/ws;
h7 = h6;
n_boiler = 0.85;
CV = 44000.; 			# in kJ/kg
wf = (1.1*ws_*(h1-h7))/(n_boiler*CV);

# Results
print "Fuel buring rate is %.2f Kg/h"%wf
print "Wf = %.2f tonnes/day"%(wf*24/1000)
Fuel buring rate is 756.64 Kg/h
Wf = 18.16 tonnes/day

Example 12.11 Page No : 491

In [21]:
# Variables
h1 = 3285.; 
h2s = 3010.
h3 = 3280.; 
h4s = 3030.;

# Calculation
h4 = h3-0.83*(h3-h4s);
h5s = 2225;
h5 = h4-0.83*(h4-h5s);
h6 = 162.7; h7 = h6;
h8 = 762.81;
h2 = h1-0.785*(h1-h2s);
m = (h8-h7)/(h4-h7);
n_cycle = ((h1-h2)+(h3-h4)+(1-m)*(h4-h5))/((h1-h8)+(h3-h2))

# Results
print "Steam flow at turbine inlet is %.3f Kg/s"%m
print "cycle efficiency is %.2f %%"%(n_cycle*100)
Steam flow at turbine inlet is 0.206 Kg/s
cycle efficiency is 35.92 %

Example 12.12 Page No : 493

In [4]:
# From table and graph 

# Variables
h1 = 2792.2;
h4 = 122.96;
hb = 254.88;
hc = 29.98;
ha = 355.98;
hd = hc;
h2 = 1949.27;

# Calculation and Results
m = (h1-h4)/(hb-hc); 			# Amount of mercury circulating
Q1t = m*(ha-hd);
W1t = m*(ha-hb) + (h1-h2);
Nov = W1t/Q1t ;
print "Overall efficiency of the cycle %.2f %%"%(Nov*100)

S = 50000.; 			# Stem flow rate through turbine in kg/h
wm = S*m;
print "Flow through the mercury turbine is %.2e Kg/h"%wm

Wt = W1t*S/3600;
print "Useful work done in binary vapour cycle is %.2e kW"%Wt

nm = 0.85; 			# Internal efficiency of mercury turbine
ns = 0.87; 			# Internal efficiency of steam turbine
WTm = nm*(ha-hb);
hb_ = ha-WTm; 			# hb'
m_ = (h1-h4)/(hb_-hc); 			# m'
h1_ = 3037.3; 			# h'
Q1t = m_*(ha-hd)+(h1_-h1);
x2_ = (6.9160-0.4226)/(8.47-0.4226);
h2_ = 121+(0.806*2432.9);
WTst = ns*(h1_-h2_);
WTt = m_*(ha-hb_)+WTst;
Nov = WTt/Q1t;

print "Overall efficiency is %.1f %%"%(Nov*100)

# note : rounding off error.
Overall efficiency of the cycle 52.80 %
Flow through the mercury turbine is 5.93e+05 Kg/h
Useful work done in binary vapour cycle is 2.84e+04 kW
Overall efficiency is 46.2 %