# Chapter 2 : Gas Power Cycles¶

## Example 2.1 Page no : 55¶

In :
# Variables
P1 = 0.1;			#Pressure of air supplied in MPa
T1 = 308;			#Temperature of air supplied in K
rv = 8;			#Compression ratio
q1 = 2100;			#Heat supplied in kJ/kg
Cp = 1.005;			#Specific heat at constant pressure in kJ/kg-K
Cv = 0.718;			#Specific heat at constant volume in kJ/kg-K
R = 0.287;			#Universal gas constant in kJ/kg-K

# Calculations
y = Cp/Cv;			#Ratio of specific heats
n = (1-(1/(rv**(y-1))))*100;			#Cycle efficiency
v1 = (R*T1)/(P1*1000);			#Specific volume at point 1 in (m**3)/kg
v2 = v1/rv;			#Specific volume at point 2 in (m**3)/kg
T2 = T1*(rv**(y-1));			#Temperature at point 2 in K
T3 = (q1/Cv)+T2;			#Temperature at point 3 in K
P2 = P1*(rv**y);			#Pressure at point 2 in MPa
P3 = P2*(T3/T2);			#Pressure at point 3 in MPa
wnet = (q1*n)/100;			#Net workdone in J/kg
MEP = (wnet/(v1-v2))/1000;			#Mean effective pressure in MPa

# Results
print 'Maximum pressure of the cycle is %3.3f MPa  \
\nMaximum temperature of the cycle  is %3.0f K  \
\nCycle efficiency is %3.1f percent  \
\nMean effective pressure is %3.3f MPa'%(P3,T3,n,MEP)

Maximum pressure of the cycle is 9.434 MPa
Maximum temperature of the cycle  is 3632 K
Cycle efficiency is 56.4 percent
Mean effective pressure is 1.533 MPa


## Example 2.2 Page no : 57¶

In :
# Variables
d = 80;			#Bore in mm
L = 85;			#Stroke in mm
Vc = 0.06;			#Clearance volume in litre
n = 0.22;			#Actual thermal efficiency
y = 1.4;			#Ratio of specific heats

# Calculations
Vs = (3.147/4)*(d**2)*L;			#Stroke volume in mm**3
Vt = Vs+(Vc*(10**6));			#Total volume in mm**3
rv = Vt/(Vc*(10**6));			#Compression ratio
ni = (1-(1/(rv**(y-1))));			#Ideal thermal efficiency
nr = (n/ni)*100;			#Relative efficiency

# Results
print 'Relative efficiency of the engine is %3.1f percent'%(nr)

Relative efficiency of the engine is 38.8 percent


## Example 2.3 Page no : 57¶

In :
# Variables
d = 0.137;			#Bore in m
L = 0.13;			#Stroke in m
Vc = 280*(10**-6);			#Clearance volume in m**3
y = 1.4;			#Ratio of specific heats

# Calculations
Vs = (3.147/4)*(d**2)*L;			#Stroke volume in m**3
rv = (Vc/Vs)*100;			#Compression ratio
rvf = (Vs+Vc)/Vc;			#final compression ratio
n = (1-(1/rvf**(y-1)))*100;			#Cycle efficiency

# Results
print 'Clearance volume is %3.1f percent of swept volume  \
\nOtto cycle efficiency is %3.2f percent'%(rv,n)

Clearance volume is 14.6 percent of swept volume
Otto cycle efficiency is 56.15 percent


## Example 2.4 Page no : 58¶

In :
# Variables
rv = 9.5;			#Compression ratio
P1 = 100.;			#Air pressure in kPa
T1 = 290.;			#Air temperature in K
V1 = 600.*(10**-6);			#Volume of air in m**3
T4 = 800.;			#Final temperature in K
R = 287.;			#Universal gas constant in J/kg.K
Cv = 0.718;			#Specific heat at constant volume in kJ/kg.K
y = 1.4;			#Ratio of specific heats

# Calculations
T3 = T4*(rv**(y-1));			#Temperature at the end of constant volume heat addition in K
P2 = P1*(rv**y);			#Pressure at point 2 in kPa
T2 = T1*(rv**(y-1));			#Temperature at point 2 in K
P3 = P2*(T3/T2);			#Pressure at point 3 in kPa
m = (P1*1000*V1)/(R*T1);			#Specific mass in kg/s
Q = m*Cv*(T3-T2);			#Heat transferred in kJ
n = (1-(1/rv**(y-1)))*100;			#Thermal efficiency
Wnet = (n*Q)/100;			#Net workdone in kJ
MEP = Wnet/(V1*(1-(1/rv)));			#Mean effective pressure in kPa

# Results
print 'Maximum pressure of the cycle is %3.2f kPa  \
\nMaximum temperature of the cycle is %3.1f K  \
\nAmount of heat transferred is %3.2f kJ  \
\nThermal efficiency is %3.1f percent  \
\nMean effective pressure is %3.1f kPa'%(P3,T3,Q,n,MEP)

# rounding off error

Maximum pressure of the cycle is 6449.19 kPa
Maximum temperature of the cycle is 1968.7 K
Amount of heat transferred is 0.65 kJ
Thermal efficiency is 59.4 percent
Mean effective pressure is 718.3 kPa


## Example 2.5 Page no : 60¶

In :
# Variables
rv = 8.;			#Compression ratio
P1 = 95.;			#Pressure at point 1 in kPa
T1 = 300.;			#Temperature at point 1 in K
q23 = 750.;			#Heat transferred during consmath.tant volume heat addition process in kJ/kg
y = 1.4;			#Ratio of specific heats
Cv = 0.718;			#Specific heat at constant volume in kJ/kg-K
R = 287.;			#Universal gas constant in J/kg-K

# Calculations
T2 = T1*(rv**(y-1));			#Temperature at point 2 in K
P2 = P1*(rv**y);			#Pressure at point 2 in kPa
T3 = (q23/Cv)+T2;			#Temperature at point 3 in K
P3 = P2*(T3/T2);			#Pressure at point 3 in kPa
nth = (1-(1/(rv**(y-1))))*100;			#Thermal efficiency
Wnet = (nth*q23)/100;			#Net work output in kJ/kg
v1 = (R*T1)/(P1*1000);			#Speific volume at point 1 in (m**3)/kg
MEP = Wnet/(v1*(1-(1/rv)));			#Mean effective pressure in kPa

# Results
print 'Pressure at the end of heat addition process is %3.1f kPa'%P3
print 'Temperature at the end of heat addition process is %3.1f K'%T3
print 'Net work output is %3.2f kJ/kg'%Wnet
print 'Thermal efficiency is %3.2f percent'%nth
print 'Mean effective pressure is %3.0f kPa'%MEP

Pressure at the end of heat addition process is 4392.3 kPa
Temperature at the end of heat addition process is 1733.8 K
Net work output is 423.54 kJ/kg
Thermal efficiency is 56.47 percent
Mean effective pressure is 534 kPa


## Example 2.6 Page no : 61¶

In :
# Variables
rv = 14.;			#Compression ratio
c = 0.06;			#Cut-off percentage
y = 1.4;			#Ratio of specific heats

# Calculations
rc = 1.78;			#Cut-off ratio
nth = (1-(((rc**y)-1)/((rv**(y-1))*y*(rc-1))))*100;			#Thermal efficiency

# Results
print 'Air standard efficiency is %3.1f percent'%(nth)

Air standard efficiency is 60.4 percent


## Example 2.7 Page no : 62¶

In :
# Variables
rv = 16.;			#Compression ratio
P1 = 0.1;			#Pressure at point 1 in MPa
T1 = 288.;			#Temperature at point 1 in K
T3 = 1753.;			#Temperature at point 3 in K
y = 1.4;			#Ratio of specific heats
Cp = 1.005;			#Specific heat at constant pressure in kJ/kg-K
R = 0.287;			#Universal gas constant in kJ/kg-K

# Calculations
T2 = int(T1*(rv**(y-1)));			#Temperature at point 2 in K
rc = round(T3/T2,2);			#Cut-off ratio
q1 = Cp*(T3-T2);			#Heat supplied in kJ/kg
nth = (1-(((rc**y)-1)/((rv**(y-1))*y*(rc-1))))*100;			#Cycle efficiency
wnet = int((q1*nth)/100);			#Net work done in kJ/kg
v1 = round((R*T1)/(P1*1000),3);			#Speific volume at point 1 in (m**3)/kg
v2 = round(v1/rv,3);			#Speific volume at point 2 in (m**3)/kg
MEP = wnet/(v1-v2);			#Mean effective pressure in kPa

# Results
print 'Cut-off ratio is %3.2f  \
\nHeat supplied is %3.1f kJ/kg\
\nCycle efficiency is %3.1f percent  \
\nMean effective pressure is %3.2f kPa'%(rc,q1,nth,MEP)

# rounding off error

Cut-off ratio is 2.01
Heat supplied is 884.4 kJ/kg
Cycle efficiency is 61.3 percent
Mean effective pressure is 699.35 kPa


## Example 2.8 Page no : 64¶

In :
# Variables
d = 0.15;			#Bore in m
L = 0.25;			#Stroke in m
Vc = 400*(10**-6);			#Clearance volume in m**3
V2 = Vc;			#Clearance volume in m**3
c1 = 0.05;			#Cut-off percentage 1
c2 = 0.08;			#Cut-off percentage 2
y = 1.4;			#Ratio of specific heats

# Calculations
Vs = (3.147/4)*(d**2)*L;			#Stroke volume in m**3
V31 = V2+(c1*Vs);			#Volume at the point of cut-off in m**3
rc1 = V31/V2;			#Cut-off ratio 1
rv = (Vc+Vs)/Vc;			#Compression ratio
nth1 = (1-(((rc1**y)-1)/((rv**(y-1))*y*(rc1-1))))*100;			#Air standard efficiency 1
V32 = V2+(c2*Vs);			#Volume at the point of cut-off in m**3
rc2 = V32/V2;			#Cut-off ratio 2
nth2 = (1-(((rc2**y)-1)/((rv**(y-1))*y*(rc2-1))))*100;			#Air standard efficiency 2
pl = nth1-nth2;			#Percentage loss in efficiency

# Results
print 'Air standard efficiency at 5 percent cut-off is %3.2f percent\
\nAir standard efficiency at 8 percent cut-off is %3.2f percent\
\nPercentage loss in efficiency is %3.2f percent'%(nth1,nth2,pl)

Air standard efficiency at 5 percent cut-off is 59.36 percent
Air standard efficiency at 8 percent cut-off is 57.40 percent
Percentage loss in efficiency is 1.95 percent


## Example 2.9 Page no : 65¶

In :
# Variables
e = 7.5;			#Expansion ratio
c = 15.;			#Compression ratio
P1 = 98.;			#Pressure at point 1 in kN/(m**2)
P4 = 258.;			#Pressure at point 4 in kN/(m**2)
T1 = 317.;			#Temperature at point 1 in K
y = 1.4;			#Ratio of specific heats

# Calculations
T4 = T1*(P4/P1);			#Temperature at point 4 in K
T3 = T4*(e**(y-1));			#Temperature at point 3 in K
t3 = T3-273;			#Temperature at point 3 in oC
T2 = T1*(c**(y-1));			#Temperature at point 2 in K
n = (1-((T4-T1)/(y*(T3-T2))))*100;			#Thermal efficiency

# Results
print 'Maximum temperature attained during the cycle is %3.1f oC  \
\nThermal efficiency of the cycle is %3.1f percent'%(t3,n)

# rounding off error

Maximum temperature attained during the cycle is 1595.4 oC
Thermal efficiency of the cycle is 60.3 percent


## Example 2.10 Page no : 66¶

In :
# Variables
rv = 20.;			#Compression ratio
P1 = 95.;			#Pressure at point 1 in kPa
T1 = 293.;			#Temperature at point 1 in K
T3 = 2200.;			#Temperature at point 3 in K
y = 1.4;			#Ratio of specific heats
R = 287.;			#Universal gas constant in J/kg-K
Cp = 1.005;			#Specific heat at constant pressure in kJ/kg-K

# Calculations
P2 = P1*(rv**y);			#Pressure at point 2 in kPa
T2 = T1*(rv**(y-1));			#Temperature at point 2 in K
v2 = (R*T2)/(P2*1000);			#Specific volume at point 2 in (m**3)/kg
v3 = v2*(T3/T2);			#Specific volume at point 3 in (m**3)/kg
rc = v3/v2;			#Cut-off ratio
nth = (1-(((rc**y)-1)/((rv**(y-1))*y*(rc-1))))*100;			#Thermal efficiency
q23 = Cp*(T3-T2);			#Heat flow between points 2 and 3 in kJ/kg
wnet = (nth*q23)/100;			#Net workdone in kJ/kg
MEP = wnet/(v2*(rv-1));			#Mean effective pressure in kPa

# Results
print 'Thermal efficiency is %3.1f percent  \
\nMean effective pressure is %.f kPa'%(nth,MEP)

# rounding off error

Thermal efficiency is 63.5 percent
Mean effective pressure is 933 kPa


## Example 2.11 Page no : 68¶

In :
# Variables
rv = 21.;			#Compression ratio
re = 10.5;			#Expansion ratio
y = 1.4;			#Ratio of specific heats

# Calculations
rc = rv/re;			#Cut-off ratio
nth = (1-(((rc**y)-1)/((rv**(y-1))*y*(rc-1))))*100;			#Air standard efficiency

# Results
print 'Cut-off ratio is %3.0f  \
\nAir standard efficiency is %3.2f percent'%(rc,nth)

Cut-off ratio is   2
Air standard efficiency is 65.36 percent


## Example 2.12 Page no : 69¶

In :
# Variables
rv = 16.;			#Compression ratio
rp = 1.5;			#Pressure ratio
y = 1.4;			#Ratio of specific heats
cp = 8;			#Cut-off percentage

# Calculations
rc = 2.2;			#Cut-off ratio
ntd = (1-((rp*(rc**y)-1)/((rv**(y-1)*((rp-1)+(y*rp*(rc-1)))))))*100;			#Dual cycle efficiency

# Results
print 'Ideal efficiency of engine is %3.1f percent'%(ntd)

Ideal efficiency of engine is 61.5 percent


## Example 2.13 Page no : 69¶

In :
# Variables
d = 0.2;			#Bore in m
L = 0.5;			#Stroke in m
c = 0.06;			#Cut-off percentage
y = 1.4;			#Ratio of specific heats
rv = 15.;			#Compression ratio
rp = 1.4;			#Pressure ratio

# Calculations
Vs = (3.147/4)*(d**2)*L;			#Stroke volume in m**3
DV = c*Vs;			#Difference in volumes at points 4 and 3
V3 = Vs/(rv-1);			#Specific volume at point 3 in m**3
V4 = V3+DV;			#Specific volume at point 4 in m**3
rc = V4/V3;			#Cut-off ratio
ntd = (1-((rp*(rc**y)-1)/((rv**(y-1)*((rp-1)+(y*rp*(rc-1)))))))*100;			#Ideal efficiency

# Results
print 'Ideal efficiency of the engine is %3.1f percent'%(ntd)

Ideal efficiency of the engine is 62.2 percent


## Example 2.14 Page no : 70¶

In :
# Variables
d = 0.2;			#Bore in m
L = 0.3;			#Stroke in m
c = 0.04;			#Cut-off percentage
y = 1.4;			#Ratio of specific heats
rv = 8.;			#Compression ratio
P1 = 1.;			#Pressure at point 1 in bar
P3 = 60.;			#Pressure at point 3 in bar
T1 = 298.;			#Temperature at point 1 in K
R = 287.;			#Universal gas constant in J/kg
Cv = 0.718;			#Speific heat at constant volume in kJ/kg-K
Cp = 1.005;			#Speific heat at constant pressure in kJ/kg-K

# Calculations
Vs = (3.147/4)*(d**2)*L;			#Stroke volume in m**3
V2 = Vs/(rv-1);			#Specific volume at point 2 in m**3
V3 = V2;			#Specific volume at point 3 in m**3
V1 = V2+Vs;			#Specific volume at pont 1 in m**3
V5 = V1;			#Specific volume at pont 5 in m**3
P2 = P1*(rv**y);			#Pressure at point 2 in bar
T2 = T1*(rv**(y-1));			#Temperature at point 2 in K
T3 = T2*(P3/P2);			#Temperature at point 3 in K
V4 = V3+(c*(V1-V2));			#Specific volume at point 4 in m**3
T4 = T3*(V4/V3);			#Temperature at point 4 in K
T5 = T4*((V4/V5)**(y-1));			#Temperature at point 5 in K
q1 = (Cv*(T3-T2))+(Cp*(T4-T3));			#Heat added in kJ/kg
q2 = Cv*(T5-T1);			#Heat rejected in kJ/kg
nth = (1-(q2/q1))*100;			#Thermal efficiency
m = (P1*V1*(10**5))/(R*T1);			#Mass of air supplied in kg
W = m*(q1-q2);			#Workdone in kJ/cycle

# Results
print 'Amount of heat added is %3.1f kJ/kg  \
\nAmount of heat rejected is %3.2f kJ/kg  \
\nWorkdone per cycle is %3.2f kJ/cycle  \
\nThermal efficiency is %3.2f percent'%(q1,q2,W,nth)

# rounding off error

Amount of heat added is 1742.1 kJ/kg
Amount of heat rejected is 772.91 kJ/kg
Workdone per cycle is 12.23 kJ/cycle
Thermal efficiency is 55.63 percent


## Example 2.15 Page no : 72¶

In :
# Variables
P1 = 1.;			#Pressure at point 1 in bar
P3 = 70.;			#Pressure at point 3 in bar
T1 = 310.;			#Temperature at point 1 in K
rv = 10.;			#Compression ratio
y = 1.4;			#Ratio of specific heats
qin = 2805.;			#Heat added in kJ/kg
m = 1.;			#Mass of air in kg
R = 287.;			#Universal gas constant in J/kg
Cv = 0.718;			#Speific heat at constant volume in kJ/kg-K
Cp = 1.005;			#Speific heat at constant pressure in kJ/kg-K

# Calculations
V1 = (m*R*T1)/(P1*(10**5));			#Volume at point 1 in m**3
T2 = T1*(rv**(y-1));			#Temperature at point 2 in K
P2 = P1*(rv**y);			#Pressure at point 2 in K
T3 = T2*(P3/P2);			#Temperature at point 3 in K
q23 = Cv*(T3-T2);			#Heat supplied at constant volume in kJ/kg
q34 = qin-q23;			#Heat supplied at constant pressure in kJ/kg
T4 = (q34/Cp)+T3;			#Temperature at point 4 in K
V2 = V1/rv;			#Volume at point 2 in m**3
V4 = V2*(T4/T3);			#Volume at point 4 in m**3
V5 = V1;			#Volume at point 5 in m**3
T5 = T4*((V4/V5)**(y-1));			#Temperature at point 5 in K
qout = Cv*(T5-T1);			#Heat rejected in kJ/kg
nth = (1-(qout/qin))*100;			#Thermal efficiency
W = qin-qout;			#Workdone in kJ/kg
Vs = V1*(1-(1/rv));			#Swept volume in (m**3)/kg
MEP = (W/Vs)/100;			#Mean effective pressure in bar

# Results
print 'Mean effective pressure is %3.2f bar\
\nThermal efficiency is %3.2f percent'%(MEP,nth)

Mean effective pressure is 19.78 bar
Thermal efficiency is 56.48 percent


## Example 2.16 Page no : 74¶

In :
# Variables
P1 = 1.;			#Pressure at point 1 in bar
T1 = 298.;			#Temperature at point 1 in K
P2 = 3.;			#Pressure at point 2 in bar
T3 = 923.;			#Temperature at point 3 in K
y = 1.4;			#Ratio of specific heats
Cp = 1.005;			#Speific heat at constant pressure in kJ/kg-K

# Calculations
x = (y-1)/y;			#Ratio
rp = P2/P1;			#Pressure ratio
nth = (1-(1/(rp**x)))*100;			#Cycle efficiency
T2 = T1*(rp**x);			#Temperature at point 2 in K
q1 = Cp*(T3-T2);			#Heat supplied in kJ/kg
Wout = (nth*q1)/100;			#Work output in kJ/kg
q2 = q1-Wout;			#Heat rejected in kJ/kg
T4 = T3*((1/rp)**x);			#Temperature at point 4 in K

# Results
print 'Cycle efficiency is %3.2f percent  \
\nHeat supplied to air is %3.1f kJ/kg  \
\nWork available at the shaft is %3.2f kJ/kg\
\nHeat rejected in the cooler is %3.2f kJ/kg  \
\nTurbine exit temperature is %3.2f K'%(nth,q1,Wout,q2,T4)

Cycle efficiency is 26.94 percent
Heat supplied to air is 517.7 kJ/kg
Work available at the shaft is 139.47 kJ/kg
Heat rejected in the cooler is 378.23 kJ/kg
Turbine exit temperature is 674.34 K


## Example 2.17 Page no : 75¶

In :
import math

# Variables
T1 = 283.;			#Temperature at point 1 in K
T3 = 1353.;			#Temperature at point 3 in K
y = 1.41;			#Ratio of specific heats
Cp = 1.007;			#Specific heat constant pressure in kJ/kg-K

# Calculations
x = (y-1)/y;	            		#Ratio
rpmax = ((T3/T1)**(1/x));			#Maximum pressure ratio
rpopt = math.sqrt(rpmax);			#Optimum pressure ratio
T2 = T1*(rpopt**x);	        		#Temperature at point 2 in K
T4 = T2;			                #Maximum temperature at point 4 in K
Wmax = Cp*((T3-T4)-(T2-T1));			#Maximum net specific work output in kJ/kg
nth = (Wmax/(Cp*(T3-T2)))*100;			#Thermal efficiency
WR = nth/100;           			#Work ratio
nc = ((T3-T1)/T3)*100;	    		#Carnot efficiency

# Results
print 'Optimum pressure ratio is %3.2f  \
\nMaximum net specific work output %3.0f kJ/kg  \
\nThermal efficiency %3.0f percent  \
\nWork ratio is %3.2f  \
\nCarnot efficiency is %3.0f percent'%(rpopt,Wmax,nth,WR,nc)

# rounding off error. please check.

Optimum pressure ratio is 14.74
Maximum net specific work output 401 kJ/kg
Thermal efficiency  54 percent
Work ratio is 0.54
Carnot efficiency is  79 percent


## Example 2.18 Page no : 76¶

In :
# Variables
Tmin = 300.;			#Minimum temperature in K
Tmax = 1073.;			#Maximum temperature in K
Cp = 1.005;			#Specific heat at constant pressure in kJ/kg-K

# Calculations
Wmax = Cp*((math.sqrt(Tmax)-math.sqrt(Tmin))**2);			#Maximum work output in kJ/kg
nB = (1-math.sqrt(Tmin/Tmax))*100;			#Brayton cycle efficiency
nC = (1-(Tmin/Tmax))*100;		        	#Carnot efficiency
r = nB/nC;	                    		    #Ratio of brayton cycle efficiency to carnot efficieny

# Results
print 'Maximum work per kg of air is %3.2f kJ/kg  \
\nCycle efficiency is %3.0f percent\
\nRatio of brayton cycle efficiency to carnot efficieny is %3.3f'%(Wmax,nB,r)

Maximum work per kg of air is 239.47 kJ/kg
Cycle efficiency is  47 percent
Ratio of brayton cycle efficiency to carnot efficieny is 0.654


## Example 2.19 Page no : 77¶

In :
# Variables
T1 = 291.;			#Temperature at point 1 in K
P1 = 100.;			#Pressure at point 1 in kN/(m**2)
nC = 0.85;			#Isentropic efficiency of compressor
nT = 0.88;			#Isentropic effficiency of turbine
rp = 8.;			#Pressure ratio
T3 = 1273.;			#Temperature at point 3 in K
m = 4.5;			#Mass flow rate of air in kg/s
y = 1.4;			#Ratio of speciifc heats
Cp = 1.006;			#Specific heat at constant pressure in kJ/kg-K

# Calculations
x = (y-1)/y;			#Ratio
T2s = T1*(rp**x);			#Temperature at point 2s in K
T2 = T1+((T2s-T1)/nC);			#Temperature at point 2 in K
t2 = T2-273;			#Temperature at point 2 in oC
T4s = T3*((1/rp)**x);			#Temperature at point 4s in K
T4 = T3-((T3-T4s)*nT);			#Temperature at point 4 in K
t4 = T4-273;			#Temperature at point 4 in oC
W = m*Cp*((T3-T4)-(T2-T1));			#Net power output in kW
nth = (((T3-T4)-(T2-T1))/(T3-T2))*100;			#Thermal efficiency
WR = W/(m*Cp*(T3-T4));			#Work ratio

# Results
print 'Net power output of the turbine is %3.0f kW  \
\nThermal efficiency of the plant is %3.0f percent\
\nWork ratio is %3.3f'%(W,nth,WR)

Net power output of the turbine is 1014 kW
Thermal efficiency of the plant is  32 percent
Work ratio is 0.446


## Example 2.20 Page no : 79¶

In :
# Variables
P1 = 0.1;			#Pressure at point 1 in MPa
T1 = 303.;			#Temperature at point 1 in K
T3 = 1173.;			#Temperature at point 3 in K
rp = 6.; 			#Pressure ratio
nC = 0.8;			#Compressor efficiency
nT = nC;			#Turbine efficiency
e = 0.75;			#Regenerator effectiveness
y = 1.4;			#Ratio of specific heats
Cp = 1.005;			#Specific heat at constant pressure in kJ/kg-K

# Calculations
x = (y-1)/y;    			#Ratio
T2s = T1*(rp**x);			#Temperature at point 2s in K
T4s = T3/(rp**x);			#Temperature at point 4s in K
DTa = (T2s-T1)/nC;			#Difference in temperatures at point 2 and 1 in K
DTb = (T3-T4s)*nT;			#Difference in temperatures at point 3 and 4 in K
wT = Cp*DTb;	    		#Turbine work in kJ/kg
wC = Cp*DTa;		    	#Compressor work in kJ/kg
T2 = DTa+T1;			    #Temperature at point 2 in K
q1 = Cp*(T3-T2);			#Heat supplied in kJ/kg
nth1 = ((wT-wC)/q1)*100;			#Cycle efficiency without regenerator
T4 = T3-DTb;		    	#Temperature at point 4 in K
T5 = T2+(e*(T4-T2));			#Temperature at point 5 in K
q2 = Cp*(T3-T5);			#Heat supplied with regenerator in kJ/kg
nth2 = ((wT-wC)/q2)*100;			#Cycle efficiency with regenerator
p = ((nth2-nth1)/nth1)*100;			#Percentage increase due to regeneration

# Results
print 'Percentage increase in the cycle efficiency due to regeneration is %3.2f percent'%(p)

# rounding off error. please check.

Percentage increase in the cycle efficiency due to regeneration is 41.41 percent


## Example 2.21 Page no : 80¶

In :
# Variables
P1 = 1.;			#Pressure at point 1 in atm
P3 = 5.;			#Pressure at point 3 in atm
T1 = 288.;			#Temperature at point 1 in K
T4 = 1143.;			#Temperature at point 4 in K
y = 1.4;			#Ratio of specific heats
Cp = 1.005;			#Specific heat at constant pressure in kJ/kg-K

# Calculations
rp = P3/P1;			#Pressure ratio
x = round((y-1)/y,3);			#Ratio
rpx = round(rp**x,2)
T3 = round(T1*(rpx));			#Temperature at point 3 in K
T5 = T4-(T3-T1);			#Temperature at point 5 in K
T6 = T4/(rpx);			#Temperature at point 6 in K
C6 = math.sqrt(2000*Cp*(T5-T6));			#Velocity of air leaving the nozzle in m/s

# Results
print 'Velocity of air leaving the nozzle is %3.1f m/s'%(C6)

# rounding error. Please check. there is rounding off error in book

Velocity of air leaving the nozzle is 712.5 m/s


## Example 2.22 Page no : 81¶

In :
# Variables
C1 = 280.;			#Velocity of aircraft in m/s
P1 = 48.;			#Pressure at point 1 kPa
T1 = 260.;			#Temperature at point 1 in K
rp = 13.;			#Pressure ratio
T4 = 1300.;			#Temperature at point 4 in K
Cp = 1005.;			#Specific heat at constant pressure in J/kg
y = 1.4;			#Ratio of specific heats

# Calculations
x = (y-1)/y;			#Ratio
T2 = T1+((C1**2)/(2*Cp));			#Temperature at point 2 in K
P2 = P1*((T2/T1)**(1/x));			#Pressure at point 2 in kPa
P3 = rp*P2;			#Pressure at point 3 in kPa
P4 = P3;			#Pressure at point 4 in kPa
T3 = T2*(rp**x);			#Temperature at point 3 in K
T5 = T4-T3+T2;			#Temperature at point 5 in K
P5 = P4*((T5/T4)**(1/x));			#Pressure at point 5 in kPa
P6 = P1;			#Pressure at point 6 in kPa
T6 = T5*((P6/P5)**x);			#Temperature at point 6 in K
C6 = math.sqrt(2*Cp*(T5-T6));			#Velocity of air at nozzle exit in m/s
W = (C6-C1)*C1;			#Propulsive power in J/kg
Q = Cp*(T4-T3);			#Total heat transfer rate in J/kg
nP = (W/Q)*100;			#Propulsive efficiency

# Results
print 'Pressure at the turbine exit is %3.1f kPa  \
\nVelocity of exhaust gases are %3.1f m/s  \
\nPropulsive efficiency is %3.1f percent'%(P5,C6,nP)

Pressure at the turbine exit is 374.2 kPa
Velocity of exhaust gases are 933.5 m/s
Propulsive efficiency is 26.9 percent