# Chapter 9 : Gas Power Cycles¶

## Example 9.1 Page No : 462¶

In :
Rc = 7.; 			#Compression Ratio Rc = v2/v3
k = 1.4;  			#It is apparent incerease in compression ratio yields an increased cycle efficiency
notto = (1-(1/Rc)**(k-1))*100; 			#Efficiency of an otto engine
print "The efficiency of the otto cycle is %.2f percentage"%(notto);
#For the carnot cycle,
#Nc = 1-(T2/T4) 			#efficiency for the carnot cycle 			#T2 = lowest temperature 			#T4 = Highest temperature

T2 = 70.+460; 			#for converting to R 			#Conversion of unit
#At 700 F
T4 = 700.+460; 			#temperatures converted to absolute temperatures;
nc = (1-(T2/T4))*100; 			#efficiency of the carnot cycle
print "When peak temperature is 700 fahrenheit, efficiency of the carnot cycle is %.2f percentage"%(nc);

#At 1000 F
T4 = 1000.+460; 			#temperatures converted to absolute temperatures;
nc = (1-(T2/T4))*100; 			#efficiency of the carnot cycle
print "When peak temperature is 1000 fahrenheit, efficiency of the carnot cycle is %.2f percentage"%(nc);

#At 3000 F
T4 = 3000.+460; 			#temperatures converted to absolute temperatures;
nc = (1-(T2/T4))*100; 			#efficiency of the carnot cycle
print "When peak temperature is 3000 fahrenheit, efficiency of the carnot cycle is %.2f percentage"%(nc);

The efficiency of the otto cycle is 54.08 percentage
When peak temperature is 700 fahrenheit, efficiency of the carnot cycle is 54.31 percentage
When peak temperature is 1000 fahrenheit, efficiency of the carnot cycle is 63.70 percentage
When peak temperature is 3000 fahrenheit, efficiency of the carnot cycle is 84.68 percentage


## Example 9.2 Page No : 463¶

In :
from numpy.linalg import inv

cv = 0.172; 			#Unit:Btu/(lbm*R) 			#Specific heat constant
Rc = 7; 			#Compression Ratio Rc = v2/v3
k = 1.4; 			#It is apparent incerease in compression ratio yields an increased cycle efficiency
T2 = 70+460; 			#for converting to R 			#Conversion of unit
#For 1000 F
T4 = 1000+460; 			#temperatures converted to absolute temperatures;
T3byT2 = Rc**(k-1); 			#Unit less
T3 = T3byT2*T2;
qin = cv*(T4-T3); 			#Unit:Btu/lbm 			#Heat added
#Qr = cv*(T5-T2)*(T5/T4) = (v2/v3)**(k-1)
Qr = (inv([[Rc]]))**(k-1); 			#Unit:Btu/lbm 			#Heat rejected
T5 = T4*Qr;
Qr = cv*(T5-T2); 			#Unit:Btu/lbm 			#Heat rejected
print "The net work out is %.2f Btu/lbm"%(qin-Qr);
notto = ((qin-Qr)/qin)*100; 			#The efficiency of otto cycle
print "The efficiency of otto cycle is %.2f percentage"%(notto);
#The value agrees with the results of problem 9.1

The net work out is 28.44 Btu/lbm
The efficiency of otto cycle is 54.08 percentage


## Example 9.3 Page No : 464¶

In :
cv = 0.7186; 			#Unit:kJ/(kg*K) 			#Specific heat constant for constant volume process
Rc = 8.; 			#Compression Ratio Rc = v2/v3
k = 1.4;  			#It is apparent incerease in compression ratio yields an increased cycle efficiency
T2 = 20.+273; 			#20 C converted to its kelvin value
qin = 50.; 			#Heat added 			#Unit:kJ
T3byT2 = Rc**(k-1);
T3 = T3byT2*T2; 			#Unit:K
#qin = cv*(T4-T3) 			#heat added 			#Unit:kJ
T4 = (qin/cv)+T3; 			#The peak temperature of the cycle 			#Unit:K
print "The peak temperature of the cycle is %.2f Kelvin i.e. %.2f Celcius"%(T4,T4-273);

The peak temperature of the cycle is 742.72 Kelvin i.e. 469.72 Celcius


## Example 9.4 Page No : 465¶

In :
#For an Otto cycle,
rc = 7.; 			#Compression Ratio Rc = v2/v3
q = 50.; 			#Unit:Btu/lbm 			#Heat added
p2 = 14.7; 			#Unit:psia 			#pressure at point 2
T2 = 60.+460; 			#temperatures converted to absolute temperatures; 			#Unit:R
cp = 0.24; 			#Unit:Btu/(lbm*R) 			#Specific heat constant for constant pressure process
cv = 0.171; 			#Unit:Btu/(lbm*R) 			#Specific heat constant for constant volume process
R = 53.3; 			#Unit:ft*lbf/lbm*R 			#constant of proportionality
k = 1.4;  			#It is apparent incerease in compression ratio yields an increased cycle efficiency
#Refering to figure 9.9,
#At (2),we need v2.
#p2*v2 = R*T2
v2 = (R*T2)/(p2*144); 			#Unit:ft**3/lbm 			#1ft**2 = 144 in**2 			#specific volume at point 2
print "At point 2, specific volume v2 = %.2f ft**3/lbm"%(v2);
#For The isentropic path (2)&(3),p3*v3**k = p2*v2**k,so
#So,p3 = p2*(v2/v3)**k;
p3 = p2*rc**k; 			#Unit:psia 			#pressure at point 3
print "At path2&3";
print "pressure p3 = %.2f psia"%(p3);
v3 = v2/rc; 			#Unit:ft**3/lbm 			#specific volume at point 3
print "specific volume v3 = %.2f ft**3/lbm"%(v3);
T3 = (p3*v3*144)/R; 			#Unit:R 			#1ft**2 = 144 in**2 			#temperature at point 3
print "temperature T3 = %.2f R"%(T3);
print "At point4"
#To obtain the values at (4),we note
v4 = v3; 			#Unit:ft**3/lbm 			#specific volume at point 4
print "specific volume v4 = %.2f ft**3/lbm"%(v4);
#qin = cv*(T4-T3)
T4 = T3+(q/cv); 			#Unit:R 			#temperature at point 4
print "temperature T4 = %.2f R"%(T4);
#For p4,
p4 = (R*T4)/(144*v4); 			#Unit:psia 			#1ft**2 = 144 in**2 			#pressure at point 4
print "pressure p4 = %.2f psia"%(p4);
#The last point has the same specific volume as (2),giving
print "At last point"
v5 = v2; 			#Unit:ft**3/lbm 			#specific volume at point 5
print "specific volume v5 = %.2f ft**3/lbm"%(v5);
#The isentropic path equation,p5*v5**k = p4*v4**k,so
p5 = p4*(v4/v5)**k; 			#Unit:psia 			#pressure at point 5
print "pressure p5 = %.2f psia"%(p5);
T5 = (p5*v5*144)/(R); 			#Unit:R 			#1ft**2 = 144 in**2 temperature at point 5
print "temperature T5 = %.2f R"%(T5);
n = (((T4-T3)-(T5-T2))/(T4-T3))*100; 			#The efficiency of the cycle
print "The efficiency of the cycle is %.2f percentage"%(n);

At point 2, specific volume v2 = 13.09 ft**3/lbm
At path2&3
pressure p3 = 224.11 psia
specific volume v3 = 1.87 ft**3/lbm
temperature T3 = 1132.51 R
At point4
specific volume v4 = 1.87 ft**3/lbm
temperature T4 = 1424.91 R
pressure p4 = 281.97 psia
At last point
specific volume v5 = 13.09 ft**3/lbm
pressure p5 = 18.50 psia
temperature T5 = 654.26 R
The efficiency of the cycle is 54.08 percentage


## Example 9.7 Page No : 468¶

In :
#For four cycle engine,
#Using the results of problem 9.6,
pm = 1000.; 			#Unit:kPa 			#mean effective pressure 			#Unit:psia
N = 4000./2; 			#Power strokes per minute 			#2L engine 			#Unit:rpm
LA = 2.     			#Mean 			#Unit:liters
hp = (pm*LA*N)/44760; 			#The horsepower 			#Unit:hp
print "The horsepower is %.2f hp"%(hp);

The horsepower is 89.37 hp


## Example 9.9 Page No : 469¶

In :
# variables.
c=0.2;      #clearance equal to 20% of its displacement
#An otto engine
#Using results of problem 9.8,
rc = (1+c)/c; 			#The compression ratio
print "The compression ratio is %.2f"%(rc);

The compression ratio is 6.00


## Example 9.10 Page No : 470¶

In :
import math

#For four cycle,six cylinder engine,
#Using the results of problem 9.5,
hp = 100; 			#Horsepower 			#Unit:hp
L = 4./12; 			#Unit:ft 			#stroke is 4 in.
A = (math.pi/4)*(3)**2*6; 			#Cylinder bore is 3 in.
N = 4000/2; 			#Power strokes per minute 			#2L engine 			#Unit:rpm
#hp = (pm*LA*N)/33000;
pm = (hp*33000)/(L*A*N); 			#The mean effective pressure 			#psia

# Results
print "The mean effective pressure is %.2f psia"%(pm);

The mean effective pressure is 116.71 psia


## Example 9.11 Page No : 470¶

In :
#six cylinder engine,with print lacement 3.3L
#Using the results of problem 9.5,
hp = 230; 			#Horsepower 			#Unit:hp
#3.3L*1000 cm**3/L*(in/2.54 cm)**3
LA = 3.3*1000*(1/2.54)**3; 			#mean 			#in**3
N = 5500/2; 			#Power strokes per minute 			#2L engine 			#Unit:rpm
#hp = (pm*LA*N)/33000;
pm = (hp*33000*12)/(LA*N); 			#1ft = 12inch 			#The mean effective pressure 			#psia
print "The mean effective pressure is %.2f psia"%(pm);

The mean effective pressure is 164.47 psia


## Example 9.12 Page No : 478¶

In :
from numpy.linalg import inv

rc = 16; 			#Compression Ratio Rc = v2/v3
v4byv3 = 2; 			#Cutoff ratio = v4/v3
k = 1.4; 			#with the cycle starting at 14 psia and 100 F  			#It is apparent incerease in compression ratio yields an increased cycle efficiency
T2 = 100+460; 			#temperatures converted to absolute temperatures;
ndiesel = 1-((inv([[rc]]))**(k-1)*(((v4byv3)**k-1)/(k*(v4byv3-1)))); 			#The efficiency of the diesel engine
print "The efficiency of the diesel engine is %.2f percentage"%(ndiesel*100);
# T3/T2 = rc**k-1 and T5/T4 = (1/re**k-1) 			#re = expansion ratio = v5/v4
#But T4/T3 = v4/v3 = rc/re
#So,
T5 = T2*(v4byv3)**k; 			#The temperature of the exhaust of the cycle 			#Unit:R
print "The temperature of the exhaust of the cycle is %.2f R i.e. %.2f F"%(T5,T5-460);

The efficiency of the diesel engine is 61.38 percentage
The temperature of the exhaust of the cycle is 1477.85 R i.e. 1017.85 F


## Example 9.13 Page No : 479¶

In :
#Now,in problem 9.12,
rc = 16; 			#Compression Ratio Rc = v2/v3
v4byv3 = 2; 			#Cutoff ratio = v4/v3
k = 1.4; 			#with the cycle starting at 14 psia and 100 F  			#It is apparent incerease in compression ratio yields an increased cycle efficiency
T2 = 100; 			#Unit:F 			#temperature
T5 = 1018; 			#Unit:F 			#Found in 9.12 			#The temperature of the exhaust of the cycle 			#Unit:R
ndiesel = 0.614 			#Efficiency of the diesel engine 			#Found in 9.12
#Now,in problem 9.13,
cp = 0.24; 			#Unit:Btu/(lbm*R) 			#Specific heat constant for constant pressure process
cv = 0.172; 			#Unit:Btu/(lbm*R) 			#Specific heat constant for constant volume process

Qr = cv*(T5-T2); 			#Heat rejected 			#Unit:Btu/lbm
#ndeisel = 1-(Qr/qin); 			#Efficiency = ndeisel 			#qin = heat added
qin = Qr/(1-ndiesel); 			#Unit:Btu/lbm
J = 778; 			#J = Conversion factor
networkout = J*(qin-Qr); 			#(ft*lbf)/lbm 			#Net work out per pound of gas
print "Net work out per pound of gas is %.2f ft*lbf)/lbm"%(networkout);
#The mean effective pressure is net work divided by (v2-v3):
mep = networkout/((16-1)*144); 			#1ft**2 = 144 in**2 			#Unit:psia  			#The mean effective pressure
print "The mean effective pressure is %.2f psia"%(mep);

Net work out per pound of gas is 195403.25 ft*lbf)/lbm
The mean effective pressure is 90.46 psia


## Example 9.14 Page No : 489¶

In :
from numpy.linalg import inv

#A Brayton cycle
rc = 7; 			#Compression Ratio Rc = v2/v3
k = 1.4;  			#It is apparent incerease in compression ratio yields an increased cycle efficiency
cp = 0.24; 			#Unit:Btu/(lbm*R) 			#Specific heat constant for constant pressure process
T3 = 1500; 			#(unit:fahrenheit) 			#peak tempeature
p1 = 14.7; 			#Unit:psia 			#Initial condition
T1 = 70+460; 			#temperatures converted to absolute temperatures; 			#Initial condition
R = 53.3; 			#Unit:ft*lbf/lbm*R 			#constant of proportionality
nBrayton = 1-(0.1428571**(k-1)); 			#A Brayton cycle efficiency
print "A Brayton cycle efficiency is %.2f percentage"%(nBrayton*100);
#If we base our calculation on 1 lbm of gas and use subscripts that corresponds to points (1),(2),(3) and (4) of fig.9.22,we have
v1 = (R*T1)/p1; 			#Unit:ft**3/lbm 			#specific volume at point 1
#Because rc = 7 then,
v2 = v1/rc; 			#Unit:ft**3/lbm 			#specific volume at point 2
#After the isentropic compression, T2*v2**k-1  =  T1*v1**k-1
T2 = T1*(v1/v2)**(k-1); 			#Unit:R 			#temperature at point 2
T2 = T2-460; 			#Unit:fahrenheit 			#temperature at point 2
qin = cp*(T3-T2); 			#Heat in 			#Unit:Btu/lbm
print "The heat in is %.2f Btu/lbm"%(qin);
#Because efficiency can be stated to be work out divided by heat in,
wbyJ = nBrayton*qin; 			#The work out 			#Unit:Btu/lbm
print "The work out is %.2f Btu/lbm"%(wbyJ); 			#Answer is wrong in the book.cause they have taken efficiency value wrong
print "The heat rejected is %.2f Btu/lbm"%(qin-wbyJ); 			#Anser is affected because of value of wbyJ

A Brayton cycle efficiency is 54.08 percentage
The heat in is 193.37 Btu/lbm
The work out is 104.58 Btu/lbm
The heat rejected is 88.79 Btu/lbm