# Chapter 13 : Gas Power Cycles¶

## Example 13.1 Page No : 543¶

In :
# Variables
T1 = 273.+35;
P1 = 100e03; 			# in kN/m2
Q1 = 2100.;
R = 0.287;
v1 = 0.884
v2 = 0.11
v3 = v2;
rk = 8.
g = 1.4; 			# gamma

# Calculation
n_cycle = 1-(1./rk**(1.4-1));
v12 = 8; 			# v1./v2
v1 = (R*T1)/P1;
v2 = v1/8;
T2 = T1*(v1/v2)**(g-1);
cv = 0.718;
T3 = Q1/cv + T2
P21 = (v1/v2)**g;
P2 = P21*P1;
P3 = P2*(T3/T2);
Wnet = Q1*n_cycle;
Pm = Wnet/(v1-v2);

# Results
print "Maximum pressure is %.2f mPa"%(P3/1e06)
print "Temperature of the cycle is %.2f K"%T3,"K"
print "Cycle efficiency is %.2f %%"%(n_cycle*100)
print "Mean effective pressure is %.3f MPa"%(Pm/1e06)

Maximum pressure is 9.43 mPa
Temperature of the cycle is 3632.39 K K
Cycle efficiency is 56.47 %
Mean effective pressure is 1.533 MPa


## Example 13.2 Page No : 544¶

In :
# Variables
rk = 14.;
k = 0.06
rc = k*(14-1)+1;
g = 1.4;

# Calculation
n_diesel = 1-((1./g))*(1./rk**(g-1))*((rc**(g-1))/(rc-1));

# Results
print "Air standard efficiency is %.f %%"%(n_diesel*100)

# note : rounding off error. please check.

Air standard efficiency is 60 %


## Example 13.3 Page No : 544¶

In :
# Variables
rk = 16.;
T1 = 273.+15;
P1 = 100.; 			# in kN/m2
T3 = 1480.+273;
g = 1.4; 			# gamma
R = 0.287;
T2 = 288*(rk**(g-1));

# Calculation
rc =  T3/T2 ;
cp = 1.005; cv = 0.718;
Q1 = cp*(T3-T2);
T4 = T3*((rc/rk)**(g-1));
Q2 = cv*(T4-T1);
n = 1-(Q2/Q1); 			# cycle efficiency
n_ = 1-((1./g))*(1./rk**(g-1))*((rc**(g-1))/(rc-1)); 			# cycle efficiency from another formula
Wnet = Q1*n;
v1 = (R*T1)/P1 ;
v2 = v1/rk;
Pm = Wnet/(v1-v2);

# Results
print "cut-off ratio is %.2f"%rc
print "Heat supplied per kg of air is %.1f kJ/Kg"%Q1
print "Cycle efficiency is %.1f %%"%(n*100)
print "Mean effective pressure is %.2f Kpa"%Pm

# rounding off error is there.

cut-off ratio is 2.01
Heat supplied per kg of air is 884.3 kJ/Kg
Cycle efficiency is 61.3 %
Mean effective pressure is 699.97 Kpa


## Example 13.4 Page No : 546¶

In :
# Variables
T1 = 273.+50;
rk = 16.;
g = 1.4; 			# gamma
P3 = 70.;
cv = 0.718;
cp = 1.005;
R = 0.287;

# Calculation
T2 = T1*((rk**(g-1)));
P1 = 1; 			# in bar
P2 = P1*(rk)**g;
T3 = T2*(P3/P2);
Q23 = cv*(T3-T2);
T4 = (Q23/cp)+T3;
v43 = T4/T3; 			# v4/v3
v54 = rk/v43; 			# v5/v4 = (v1./v2)*(v3/v4)
T5 = T4*(1./v54)**(g-1);
P5 = P1*(T5/T1);
Q1 = cv*(T3-T2)+cp*(T4-T3);
Q2 = cv*(T5-T1);
n_cycle = 1-(Q2/Q1);
v1 = (R*T1)/P1;
v12 = (15./16)*v1; 			# v1-v2
Wnet = Q1*n_cycle;
Pm = Wnet/(v12);

# Results
print "Efficiency of the cycle is %.2f %%"%(n_cycle*100)
print "Mean effective pressure is %.2f"%Pm,"bar"

Efficiency of the cycle is 66.31 %
Mean effective pressure is 4.76 bar


## Example 13.5 Page No : 547¶

In :
# Variables
P1 = 0.1e06;
T1 = 303.;
T3 = 1173.;
PR = 6.; 			# Pressure ratio
rp = 6.;
nt = 0.8;
nc = 0.8;
g = 1.4;
cv = 0.718;
cp = 1.005;
R = 0.287;

# Calculation
j = (PR)**((g-1)/g);
T2s = j*T1;
T4s = T3/j;
T21 = (T2s-T1)/nc ; 			# T2-T1
T34 = nt*(T3-T4s); 			# T3-T4
Wt = cp*T34;
Wc = cp*T21;
T2 = T21+T1;
Q1 = cp*(T3-T2);
n = (Wt-Wc)/Q1;
T4 = T3-375;
T6 = 0.75*(T4-T2) + T2 ;
Q1_ = cp*(T3-T6);
n_ = (Wt-Wc)/Q1_;
I = (n_-n)/n ;

# Results
print "The percentage efficiency in cycle efficiency due to regeneration is %.f %%"%(I*100)

# note : rounding off error is there.

The percentage efficiency in cycle efficiency due to regeneration is 42 %


## Example 13.6 Page No : 549¶

In :
import math

# Variables
cp = 1.005;
Tmax = 1073.
Tmin = 300.;

# Calculation
Wnet_max = cp*(math.sqrt(Tmax)-math.sqrt(Tmin))**2;
n_cycle = 1-math.sqrt(Tmin/Tmax);
n_carnot = 1-(Tmin/Tmax);
r = n_cycle/n_carnot;

# Results
print "Maximum work done per kg of air is %.2f kJ/Kg"%Wnet_max
print "cycle efficiency is %.0f %%"%(n_cycle*100)
print "ratio of brayton and carnot efficiency is %.3f"%r

# note : rounding off error is there.

Maximum work done per kg of air is 239.47 kJ/Kg
cycle efficiency is 47 %
ratio of brayton and carnot efficiency is 0.654


## Example 13.7 Page No : 549¶

In :
# Variables
rp = 6.;
g = 1.4;
cv = 0.718;
cp = 1.005;
R = 0.287;
T1 = 300.;
T3 = 1100.;
T0 = 300.;

# Calculation
n_cycle = 1-(1./rp**((g-1)/g));
j = rp**((g-1)/g);
T2 = T1*j;
T4 = T3/j;
Wc = cp*(T2-T1);
Wt = cp*(T3-T4);
WR = (Wt-Wc)/Wt;
Q1 = 100; 			# in MW
PO = n_cycle*Q1;
m_dot = (Q1*1e06)/(cp*(T3-T2));
R = m_dot*cp*T0*((T4/T0)-1-math.log(T4/T0));

# Results
print "The thermal efficiency of the cycle is %.1f %%"%(n_cycle*100)
print "Work ratio is %.3f"%WR
print "Power output is %.1f MW"%PO
print "Energy flow rate of the exhaust gas stream is %.2f MW"%(R/1e06)

# rounding error is there.

The thermal efficiency of the cycle is 40.1 %
Work ratio is 0.545
Power output is 40.1 MW
Energy flow rate of the exhaust gas stream is 20.53 MW


## Example 13.8 Page No : 550¶

In :
# Variables
nc = 0.87;
nt = 0.9;
T1 = 311.;
rp = 8.; 			# P2/P1
P1 = 1.;
P2 = 8.;
P3 = 0.95*P2;
P4 = 1;
T3 = 1100.;
g = 1.4;
cv = 0.718;
cp = 1.005;
R = 0.287;

# Calculation
# With no cooling
T2s = T1*((P2/P1)**((g-1)/g));
T2 = T1 + (T2s-T1)/0.87;
T4s = T3*(P4/P3)**((g-1)/g);
n = (((T3-T4s)*nt)-((T2s-T1)/nc))/(T3-T2);
# With cooling
n_cycle = n-0.05;
x = 0.13;
r = 0.13/1.13;

# Results
print "Percentage of air that may be taken from the compressor is %.1f%%"%(r*100)

Percentage of air that may be taken from the compressor is 11.5%


## Example 13.10 Page No : 555¶

In :
# Variables
T1 = 233.;
V1 = 300.;
cp = 1.005;
g = 1.4;

# Calculation and Results
T2 = T1+((V1**2)/(2*cp))*1e-03 ;
P1 = 35;
P2 = P1*(T2/T1)**(g/(g-1));
rp = 10; 			# Pressure ratio
P3 = rp*P2;
T3 = T2*(P3/P2)**((g-1)/g);
T4 = 1373;
T5 = T4-T3+T2;
P4 = P3;
P5 = P4*(T5/T4)**(g/(g-1));
print "Temperature at the turbine exit is %.2f K"%T5
print "Pressure at the turbine exit is %.2f kPa"%P5
P6 = P1;
T6 = T5*(P6/P5)**((g-1)/g);
V6 = (2*cp*1000*(T5-T6))**0.5 ;

print "Velocity of the gas at the nozzle exit is %.1f m/s"%V6
w = 50.;
Ve = V6; Vi = 300.;
Wp_dot = w*Vi*(Ve-Vi);
h4 = 1373.;
h3 = 536.66;
Q1 = w*cp*(h4-h3); 			# in kJ/kg
np = Wp_dot/(Q1*1000);
print "The propulsive efficiency of the cycle is %.1f %%"%(np*100)

Temperature at the turbine exit is 1114.47 K
Pressure at the turbine exit is 312.00 kPa
Velocity of the gas at the nozzle exit is 1020.3 m/s
The propulsive efficiency of the cycle is 25.7 %


## Example 13.11 Page No : 556¶

In :
# Variables
Ta = 288.;
rp = 8.; 			# Pb/Pa
g = 1.33;
g1 = 1.44;
cv = 0.718;
cpa = 1.005;
cpg = 1.11;
R = 0.287;

# Calculation
Tb = Ta*(rp)**((g1-1)/g1);
Tc = 1073.;
Tm = 800.+273;
Tmin = 100.+273;
Td = Tc/(rp**((g-1)/g));
Wgt = cpg*(Tc-Td)-cpa*(Tb-Ta);
Q1 = cpg*(Tc-Tb);
Q1_ = cpg*(Tc-Td);
h1 = 3775.; h2 = 2183.; h3 = 138.; h4 = h3;
Q1_st = h1-h3; 			# Q1'
Q_fe = cpg*(Tm-Tmin);
was = Q1_st/Q_fe; 			# wa/ws
Wst = h1-h2;
PO = 190e03; 			# in kW
ws = PO/(was*Wgt+Wst);
wa = 322.5;
CV = 43300.; 			# in kJ/kg
waf = round(CV/(1095.),1)
FEI = (wa/waf)*CV;
noA = PO/FEI;

# Results
print "Air fuel ratio is %.2f"%waf
print "Overall efficiency of combined plant is %.1f %%"%(noA*100)

Air fuel ratio is 39.50
Overall efficiency of combined plant is 53.7 %