Chapter 13 : Gas Power Cycles

Example 13.1 Page no : 606

In [1]:
# Variables
T1 = 671.; 			#K
T2 = T1;
T3 = 313.; 			#K
T4 = T3;
W = 130.; 			#kJ

 
    n_th = (T2-T3)/T2;
print ("(i) Engine thermal efficiency  = %.3f")% (n_th)

Q = W/n_th;
print ("(ii) Heat added  = %.3f")% (Q), ("kJ")


Q_rejected = Q-W;
dS = Q_rejected/T3;
print ("(iii) The entropy changes during heat rejection process"),(" = %.3f")% (dS), ("kJ/K")

(i) Engine thermal efficiency  = 0.534
(ii) Heat added  = 243.659 kJ
(iii) The entropy changes during heat rejection process  = 0.363 kJ/K

Example 13.2 Page no : 607

In [2]:
import math

# Variables
cv = 0.721; 			#kJ/kg K
cp = 1.008; 			#kJ/kg K
m = 0.5; 			    #kg
n_th = 0.5;
Q_isothermal = 40.; 	#kJ
p1 = 7.*10**5; 			#Pa
V1 = 0.12; 			    #m**3
R = 287.; 			    #J/kg K


# Calculations and Results
print ("(i) The maximum and minimum temperatures")
T1 = p1*V1/m/R;
print ("Maximun temperature  = %.3f")% (T1), ("K")

T2 = (1-n_th)*T1;
print ("Minimum temperature  = %.3f")% (T2), ("K")


V2 = V1*math.e**(Q_isothermal*10**3/m/R/T1);
print ("(ii) The volume at the end of isothermal expansion  = %.3f")% (V2), ("m**3")

print ("(iii) The heat transfer for each of the four processes")

Q1 = Q_isothermal;
print ("Isothermal expansion %.3f")% (Q1), ("kJ")

Q2 = 0;
print ("Adiabatic reversible expansion"), (Q2)

Q3 = -Q_isothermal;
print ("Isothermal compression"), (Q3)

Q4 = 0;
print ("Adiabatic reversible compression"), (Q4)

(i) The maximum and minimum temperatures
Maximun temperature  = 585.366 K
Minimum temperature  = 292.683 K
(ii) The volume at the end of isothermal expansion  = 0.193 m**3
(iii) The heat transfer for each of the four processes
Isothermal expansion 40.000 kJ
Adiabatic reversible expansion 0
Isothermal compression -40.0
Adiabatic reversible compression 0

Example 13.3 page no : 608

In [5]:
import math

# Variables
p1 = 18.*10**5; 	#Pa
T1 = 683.; 			#K
T2 = T1;
r1 = 6.; 			#ratio V4/V1; Isentropic compression
r2 = 1.5; 			#ratio V2/V1; Isothermal expansion
y = 1.4;
V1 = 0.18; 			#m**3

# Calculations and Results
print ("(i) Temperatures and pressures at the main points in the cycle")
T4 = T1/(r1)**(y-1);
print ("T4 = %.1f")% (T4), ("K")

T3 = T4;
print ("T3 = %.3f")% (T3), ("K")

p2 = p1/r2;
print ("p2 = %.3f")% (p2/10**5), ("bar")

p3 = p2/(r1)**y;
print ("p3 = %.3f")% (p3/10**5), ("bar")

p4 = p1/(r1)**y;
print ("p4 = %.3f")% (p4/10**5), ("bar")


dS = p1*V1/T1/10**3*math.log(r2);
print ("(ii) Change in entropy  = %.3f")% (dS), ("kJ/K")

print ("(iii) Mean thermal efficiency of the cycle")
Qs = T1*(dS);
Qr = T4*(dS);

n = 1-Qr/Qs;
print ("n = %.3f")% (n)


pm = (Qs-Qr)/8/V1/100; 			#bar
print ("(iv) Mean effective pressure of the cycle  = %.3f")% (pm), ("bar")


n = 210.; 			#cycles per minute
P = (Qs-Qr)*n/60; 			#kW
print ("(v) Power of the engine  = %.3f")% (P), ("kW")

# answers are slightly different because of rounding error.

(i) Temperatures and pressures at the main points in the cycle
T4 = 333.5 K
T3 = 333.549 K
p2 = 12.000 bar
p3 = 0.977 bar
p4 = 1.465 bar
(ii) Change in entropy  = 0.192 kJ/K
(iii) Mean thermal efficiency of the cycle
n = 0.512
(iv) Mean effective pressure of the cycle  = 0.467 bar
(v) Power of the engine  = 235.251 kW

Example 13.4 page no : 611

In [4]:
# Calculations
T2 = 1029/0.6;
T1 = 1.2*T2;

# Results
print ("Temperature of the source  = "), (T1), ("K")

print ("Temperature of the math.sink = "), (T2), ("K")

Temperature of the source  =  2058.0 K
Temperature of the math.sink =  1715.0 K

Example 13.5 page no : 611

In [5]:
# Variables
T1 = 1990.; 			#K
T2 = 850.; 			#K
Q = 32.5/60; 			#kJ/s
P = 0.4; 			#kW

# Calculations
n_carnot = (T1-T2)/T1;
n_th = P/Q;

# Results
print ("most efficient engine is one that works on Carnot cycle %.3f")% (n_carnot)

print ("n_thermal  = %.3f")% (n_th)

print ("which is not feasible as no engine can be more efficient than that working on Carnot")
print ("Hence claims of the inventor is not true.")

most efficient engine is one that works on Carnot cycle 0.573
n_thermal  = 0.738
which is not feasible as no engine can be more efficient than that working on Carnot
Hence claims of the inventor is not true.

Example 13.7 page no : 615

In [6]:
# Variables
n = 0.6; 
y = 1.5;

# Calculations
r = (1./(1-n))**(1./(y-1));

# Results
print ("Compression ratio  = "), (r)

Compression ratio  =  6.25

Example 13.8 page no : 615

In [6]:
import math

# Variables
D = 0.25; 			#m
L = 0.375; 			#m
Vc = 0.00263; 		#m**3
p1 = 1.; 			#bar
T1 = 323.; 			#K
p3 = 25.; 			#bar

# Calculations and Results
Vs = math.pi/4*D**2*L;
r = (Vs+Vc)/Vc;
y = 1.4;

n_otto = 1-1/(r**(y-1));
print ("(i) Air standard efficiency = %.3f")% (n_otto)



p2 = p1*(r)**(y);
r_p = p3/p2;

p_m = p1*r*(r**(y-1) - 1)*(r_p - 1)/(y-1)/(r-1);
print ("(ii)Mean effective pressure  = %.3f")%(p_m), ("bar")

(i) Air standard efficiency = 0.565
(ii)Mean effective pressure  = 1.336 bar

Example 13.9 page no : 617

In [8]:
# Variables
cv = 0.72; 			#kJ/kg K
y = 1.4;
p1 = 1.; 			#bar
T1 = 300.; 			#K
Q = 1500.; 			#kJ/kg
r = 8.;
y = 1.4;

# Calculations and Results
print ("(i) Pressures and temperatures at all points")
T2 = T1*(r)**(y-1);
print ("T2 = %.3f")% (T2), ("K")

p2 = p1*(r)**y;
print ("p2 = %.3f")%(p2), ("bar")

T3 = Q/cv + T2;
print ("T3 = %.3f")% (T3), ("K")

p3 = p2*T3/T2;
print ("p3 = %.3f")% (p3), ("bar")

T4 = T3/r**(y-1);
print ("T4 = %.3f")% (T4), ("K")

p4 = p3/r**(y);
print ("p4 = %.3f")% (p4), ("bar")


print ("(ii) Specific work and thermal efficiency")
SW = cv*((T3-T2) - (T4-T1));
print ("Specific work  = %.3f")% (SW), ("kJ/kg")

n_th = 1-1./r**(y-1);
print ("Thermal efficiency  = %.3f")% (n_th)

(i) Pressures and temperatures at all points
T2 = 689.219 K
p2 = 18.379 bar
T3 = 2772.552 K
p3 = 73.935 bar
T4 = 1206.824 K
p4 = 4.023 bar
(ii) Specific work and thermal efficiency
Specific work  = 847.087 kJ/kg
Thermal efficiency  = 0.565

Example 13.10 page no : 618

In [7]:
# Variables
r = 6.; 			#v1/v2 = v4/v3 = r
p1 = 1.; 			#bar
T1 = 300.; 			#K
T3 = 1842.; 		#K
y = 1.4;

# Calculations and Results
print ("(i) Temperature and pressure after the isentropic expansion")
p2 = p1*(r)**y;
T2 = T1*r**(y-1);
p3 = p2*(T3/T2);
T4 = T3/r**(y-1);
print ("T4 = %.3f")% (T4), ("K")

p4 = p3/(r)**(y);
print ("p4  = %.3f")% (p4), ("bar")

print ("(ii)Process required to complete the cycle is the consmath.tant pressure scavenging. The cycle is called Atkinson cycle")
print ("(iii) Percentage improvement/increase in efficiency")
p5 = 1.; 			#bar
T5 = T3*(p5/p3)**((y-1)/y);
n_otto = (1-1./r**(y-1))*100;
print ("n_otto  =  %.3f")% (n_otto), ("%")

n_atkinson = (1.-y*(T5-T1)/(T3-T2))*100;
print ("n_atkinson = %.3f")% (n_atkinson), ("%")

dn = n_atkinson - n_otto; 			#Improvement in efficiency
print ("Improvement in efficiency  = %.3f")% (dn), ("%")

(i) Temperature and pressure after the isentropic expansion
T4 = 899.558 K
p4  = 2.999 bar
(ii)Process required to complete the cycle is the consmath.tant pressure scavenging. The cycle is called Atkinson cycle
(iii) Percentage improvement/increase in efficiency
n_otto  =  51.164 %
n_atkinson = 59.254 %
Improvement in efficiency  = 8.090 %

Example 13.11 page no : 620

In [10]:
# Variables
p1 = 1.;     			#bar
T1 = 343.; 	    		#K
p2 = 7.; 		    	#bar
Qs = 465.; 			    #kJ/kg of air
cp = 1.; 			    #kJ/kg K
cv = 0.706; 			#kJ/kg K
y = 1.41;

# Calculations and Results
r = (p2/p1)**(1./y);
print ("(i) Compression ratio of engine  = %.3f")% (r)

T2 = T1*(r)**(y-1);
t2 = T2-273;
print ("(ii) Temperature at the end of compression  = %.3f")% (t2), ("0C")

T3 = Qs/cv+T2;
t3 = T3-273;
print ("(iii) Temperature at the end of heat addition  = %.3f")% (t3), ("0C")

(i) Compression ratio of engine  = 3.975
(ii) Temperature at the end of compression  = 330.993 0C
(iii) Temperature at the end of heat addition  = 989.633 0C

Example 13.12 page no : 621

In [8]:
# Variables
y = 1.4;
R = 0.287; 	    		#kJ/kg K
T1 = 311.;   			#K
T3 = 2223.; 			#K

#p2/p1 = 15
r = 15**(1/1.4);
print ("(i) Compression ratio  = %.3f")% (r)

n_th = 1-1./r**(y-1);
print ("(ii) Thermal efficiency  = %.3f")% (n_th)

T2 = T1*(r)**(y-1);
T4 = T3/r**(y-1);
cv = R/(y-1);

Q_supplied = cv*(T3-T2);
Q_rejected = cv*(T4-T1);

W = Q_supplied-Q_rejected;
print ("(iii)Work done = %.3f")% (W), ("kJ")

(i) Compression ratio  = 6.919
(ii) Thermal efficiency  = 0.539
(iii)Work done = 598.651 kJ

Example 13.13 page no : 623

In [9]:
# Variables
V1 = 0.45; 			#m**3
p1 = 1.; 			#bar
T1 = 303.; 			#K
p2 = 11.; 			#bar
Qs = 210.; 			#kJ
n = 210.; 			#number of working cycles/min
R = 287.; 			#J/kg K
cv = 0.71; 			#kJ/kg K
y = 1.4;

# Calculations and Results
print ("(i) Pressures, temperatures and volumes at salient points")
r = (p2/p1)**(1./y);

T2 = T1*(r)**(y-1);
print ("T2 = %.3f")% (T2), ("K")

V2 = T2/T1*p1/p2*V1;
print ("V2 = %.3f")% (V2), ("m**3")

m = p1*10**5*V1/R/T1;
T3 = Qs/m/cv+T2;
print ("T3 = %.3f")% (T3), ("K")

p3 = T3/T2*p2;
print ("p3 = %.3f")% (p3), ("bar")

V3 = V2;
print ("V3 = %.3f")% (V3), ("m**3")

p4 = p3/r**y;
print ("p4 = %.3f")%(p4), ("bar")

T4 = T3/r**(y-1);
print ("T4 = %.3f")% (T4), ("K")

V4 = V1;
print ("V4 = %.3f")%(V4), ("m**3")

Qr = m*cv*(T4-T1);
n_otto = (Qs-Qr)/Qs;
print ("(iii) Efficiency  = %.3f")% (n_otto)


p_m = (Qs-Qr)/(V1-V2)/100; 			#bar
print ("(iv) Mean effective pressure  = %.3f")% (p_m), ("bar")

P = (Qs-Qr)*n/60;
print ("(v) Power developed  = %.3f")% (P), ("kW")

(i) Pressures, temperatures and volumes at salient points
T2 = 601.151 K
V2 = 0.081 m**3
T3 = 1172.725 K
p3 = 21.459 bar
V3 = 0.081 m**3
p4 = 1.951 bar
T4 = 591.093 K
V4 = 0.450 m**3
(iii) Efficiency  = 0.496
(iv) Mean effective pressure  = 2.824 bar
(v) Power developed  = 364.536 kW

Example 13.14 page no : 625

In [14]:
# Variables
print ("r = (T3/T1)**(1/2/(y-1))")
print ("(b)Change in efficiency")
T3 = 1220.; 			#K
T1 = 310.    			#K

# Calculations and Results
# For air
y = 1.4;
r1 = (T3/T1)**(1./2./(y-1));
n1 = 1-1/r1**(y-1); 			#air smath.radians(numpy.arcmath.tan(ard Efficiency
print ("Air standard Efficiency  = %.3f")% (n1)

#For helium
cp = 5.22; 			#kJ/kg K
cv = 3.13; 			#kJ/kg K
y = cp/cv;
r2 = (T3/T1)**(1./2/(y-1));

n2 = 1-1/r2**(y-1);
print ("Air standard efficiency for helium  = %.3f")% (n2)

change = n1-n2;
print ("Change in efficiency = %.3f")% (change)

print ("Hence change in efficiency is nil")

r = (T3/T1)**(1/2/(y-1))
(b)Change in efficiency
Air standard Efficiency  = 0.496
Air standard efficiency for helium  = 0.496
Change in efficiency = -0.000
Hence change in efficiency is nil

Example 13.15 page no : 627

In [10]:
print ("(b) Power developed ")

# Variables
T1 = 310.; 			#K
T3 = 1450.; 		#K
m = 0.38; 			#kg
cv = 0.71; 			#kJ/kg K

# Calculations
T2 = math.sqrt(T1*T3);
T4 = T2;

W1 = cv*((T3-T2) - (T4-T1)); 			#Work done
W = m/60*W1; 			#Work done per second

# Results
print ("Power  = %.3f")%(W), ("kW")

(b) Power developed 
Power  = 1.885 kW

Example 13.17 page no : 632

In [16]:
# Variables
r = 15.;
y = 1.4;
#V3-V2 = 0.06*(V1-V2)
rho = 1.84; 			#cut off ratio rho = V3/V2

# Calculations
n_diesel = 1-1/y/r**(y-1)*((rho**y-1)/(rho-1));

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

efficiency  = 0.612

Example 13.18 page no : 633

In [17]:
# Variables
L = 0.25; 			#m
D = 0.15; 			#m
V2 = 0.0004; 		#m**3

# Calculations
Vs = math.pi/4*D**2*L;
V_total = Vs+V2;
y = 1.4;
V3 = V2+5./100*Vs;
rho = V3/V2;
r = (Vs+V2)/V2; 			#V1 = Vs+V2
n_diesel = 1-1/y/r**(y-1)*((rho**y-1)/(rho-1));

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

efficiency  = 0.593

Example 13.19 page no : 633

In [12]:
# Variables
y = 1.4;
r = 14

# Calculations
#When the fuel is cut-off at 5%
rho1 = 5./100*(r-1)+1;
n_diesel1 = 1-1./y/r**(y-1)*((rho1**y-1)/(rho1-1));

#When the fuel is cut-off at 8%
rho2 = 8./100*(r-1)+1;
n_diesel2 = 1-1./y/r**(y-1)*((rho2**y-1)/(rho2-1));
loss = (n_diesel1-n_diesel2)*100;

# Results
print ("percentage loss in efficiency due to delay in fuel cut off  = %.1f")% (loss), ("%")

percentage loss in efficiency due to delay in fuel cut off  = 2.1 %

Example 13.20 page no : 634

In [19]:
# Variables
pm = 7.5; 			#bar
r = 12.5;
p1 = 1; 			#bar
y = 1.4;
rho = 2.24;

# Calculations
cutoff = (rho-1)/(r-1)*100;

# Results
print ("cutoff = %.3f")% (cutoff), ("%")

cutoff = 10.783 %

Example 13.21 page no : 634

In [12]:
# Variables
D = 0.2; 			#m
L = 0.3; 			#m
p1 = 1.; 			#bar
T1 = 300.; 			#K
R = 287.;
r = 15.;
y = 1.4;

# Calculations and Results
print ("(i) Pressures and temperatures at salient points")
Vs = math.pi/4*D**2*L;

V1 = r/(r-1)*Vs;
print ("V1 = %.3f")% (V1), ("m**3")

m = p1*10**5*V1/R/T1;

p2 = p1*r**y;
print ("p2 = %.3f")% (p2), ("bar")

T2 = T1*r**(y-1);
print ("T2 = %.3f")% (T2), ("K")

V2 = Vs/(r-1);
print ("V2 = %.5f")% (V2), ("m**3")

rho = 8./100*(r-1) + 1;
V3 = rho*V2;
print ("V3 = %.5f")% (V3), ("m**3")

T3 = T2*V3/V2;
print ("T3 = %.3f")% (T3), ("K")

p3 = p2;
print ("p3 = %.3f")% (p3), ("bar")

p4 = p3*(rho/r)**y;
print ("p4 = %.3f")% (p4), ("bar")

T4 = T3*(rho/r)**(y-1);
print ("T4 = %.3f")% (T4), ("K")

V4 = V1;
print ("V4 = %.3f")% (V4), ("m**3")

print ("(ii) Theoretical air standard efficiency  = "),
n_diesel = 1-1/y/r**(y-1)*((rho**y-1)/(rho-1));
print ("efficiency  = %.3f")% (n_diesel)


pm = (p1*r**y*(y*(rho-1) - r**(1-y)*(rho**y-1)))/(y-1)/(r-1);
print ("(iii) Mean effective pressure  = %.3f")% (pm), ("bar")

n = 380; 			#number of cycles per min
P = n/60.*pm*Vs*100; 			#kW
print ("(iv) Power of the engine  = %.3f")% (P), ("kW")

(i) Pressures and temperatures at salient points
V1 = 0.010 m**3
p2 = 44.313 bar
T2 = 886.253 K
V2 = 0.00067 m**3
V3 = 0.00143 m**3
T3 = 1878.857 K
p3 = 44.313 bar
p4 = 2.863 bar
T4 = 858.997 K
V4 = 0.010 m**3
(ii) Theoretical air standard efficiency  =  efficiency  = 0.598
(iii) Mean effective pressure  = 7.417 bar
(iv) Power of the engine  = 44.269 kW

Example 13.22 page no : 637

In [21]:
# Variables
r1 = 15.3; 			#V1/V2
r2 = 7.5;  			#V4/V3
p1 = 1.; 			#bar
T1 = 300.; 			#K
n_mech = 0.8;
C = 42000.; 		#kJ/kg
y = 1.4;
R = 287.;
cp = 1.005;
cv = 0.718;
V2 = 1.; 					#Assuming V2 = 1 m**3

# Calculations
T2 = T1*r1**(y-1);
p2 = p1*r1**y;
T3 = r1/r2*T2;
m = p2*10**5*V2/R/T2;
T4 = T3/r2**(y-1);

Q_added = m*cp*(T3-T2);
Q_rejected = m*cv*(T4-T1);
W = Q_added-Q_rejected;

pm = W/(r1-1)/V2/100;

# Results
print ("Mean effective pressure  = %.3f")% (pm), ("bar")

ratio = p2/pm;
print ("Ratio of maximum pressure to mean effective pressure  = %.3f")% (ratio)

n_cycle = W/Q_added;
print ("Cycle efficiency  = %.3f")% (n_cycle)

n_thI = 0.5;
n_cycle1 = n_thI*n_cycle;

n_thB = n_mech*n_cycle1;

BP = 1;
mf = BP/C/n_thB*3600;
print ("Fuel consumption per kWh  = %.3f ")% (mf), ("kg/kWh")

Mean effective pressure  = 7.017 bar
Ratio of maximum pressure to mean effective pressure  = 6.493
Cycle efficiency  = 0.605
Fuel consumption per kWh  = 0.354  kg/kWh

Example 13.23 page no : 642

In [22]:
# Variables
Vs = 0.0053; 			#m**3
Vc = 0.00035; 			#m**3
V3 = Vc;
V2 = V3;
p3 = 65.; 			#bar
p4 = 65.; 			#bar
T1 = 353.; 			#K
p1 = 0.9; 			#bar
y = 1.4;

# Calculations
r = (Vs+Vc)/Vc;
rho = (5/100*Vs+V3)/V3;
p2 = p1*(r)**y;
B = p3/p2;
n_dual = 1-1/r**(y-1)*((B*rho**y-1)/((B-1)+B*y*(rho-1)));

# Results
print ("Efficiency of the cycle  = %.3f")% (n_dual)

Efficiency of the cycle  = 0.671

Example 13.24 page no : 643

In [23]:
# Variables
r = 14.;
B = 1.4;
rho = 6./100*(r-1) + 1;
y = 1.4;

# Calculations
n_dual = 1-1./r**(y-1)*((B*rho**y-1)/((B-1)+B*y*(rho-1)))

# Results
print ("Efficiency of the cycle  = %.3f")% (n_dual)

Efficiency of the cycle  = 0.614

Example 13.25 page no : 643

In [24]:
# Variables
D = 0.25; 			#m
r = 9.; 
L = 0.3; 			#m
cv = 0.71; 			#kJ/kg K
cp = 1.; 			#kJ/kg K
p1 = 1.; 			#bar
T1 = 303.; 			#K
p3 = 60.; 			#bar
p4 = p3;
n = 3.; 			#number of working cycles/ sec
y = 1.4;
R = 287.;

# Calculations and Results
print ("(i) Air standard efficiency")
Vs = math.pi/4*D**2*L;
Vc = Vs/(r-1);
V1 = Vs+Vc;
p2 = p1*(r)**y;
T2 = T1*r**(y-1);
T3 = T2*p3/p2;
rho = 4./100*(r-1)+1;
T4 = T3*rho;
T5 = T4*(rho/r)**(y-1);
p5 = p4*(r/rho)**(y);
Qs = cv*(T3-T2)+cp*(T4-T3)
Qr = cv*(T5-T1);

n_airsard = (Qs-Qr)/Qs;
print ("efficiency  = %.3f")% (n_airsard)

print ("(ii) Power developed by the engine")
m = p1*10**5*V1/R/T1;

W = m*(Qs-Qr);

P = W*n;
print ("P = %.3f")% (P), ("kW")

(i) Air standard efficiency
efficiency  = 0.575
(ii) Power developed by the engine
P = 51.392 kW

Example 13.26 page no : 646

In [25]:
# Variables
p1 = 1.; 			#bar
T1 = 363.; 			#K
r = 9.; 
p3 = 68.; 			#bar
p4 = 68.; 			#bar
Q = 1750.; 			#kJ/kg
y = 1.4;
cv = 0.71;
cp = 1.0;

# Calculations and Results
print ("(i) Pressures and temperatures at salient points")
p2 = p1*(r)**y;
print ("p2 = %.3f")% (p2), ("bar")

T2 = T1*r**(y-1);
print ("T2 = %.3f")% (T2), ("K")

print ("p3 = %.3f")% (p3), ("bar")

print ("p4 = %.3f")% (p4), ("bar")

T3 = T2*(p3/p2);
print ("T3 = %.3f")% (T3), ("K")

Q1 = cv*(T3-T2); 			#heat added at consmath.tant volume
Q2 = Q-Q1; 	        		#heat added at consmath.tant pressure

T4 = Q2/cp+T3;
print ("T4 = %.3f")% (T4), ("K")

rho = T4/T3; 			#V4/V3 = T4/T3

p5 = p4*(rho/r)**y;
print ("p5 = %.3f")% (p5), ("bar")

T5 = T4*(rho/r)**(y-1);
print ("T5 = %.3f")% (T5), ("K")


Qr = cv*(T5-T1);
n_airard = (Q-Qr)/Q;
print ("(ii) Air standard efficiency  = %.3f")% (n_airard)


pm = 1./(r-1)*(p3*(rho-1) + (p4*rho-p5*r)/(y-1) - (p2-p1*r)/(y-1));
print ("(iii) Mean effective pressure  = %.3f")% (pm), ("bar")

(i) Pressures and temperatures at salient points
p2 = 21.674 bar
T2 = 874.186 K
p3 = 68.000 bar
p4 = 68.000 bar
T3 = 2742.667 K
T4 = 3166.045 K
p5 = 3.836 bar
T5 = 1392.380 K
(ii) Air standard efficiency  = 0.582
(iii) Mean effective pressure  = 11.094 bar

Example 13.27 page no : 648

In [13]:
# Variables
T1 = 300.; 			#K
r = 15.;
y = 1.4;

# Calculations
#p3/p1 = 70
T2 = T1*(r)**(y-1);
#p2/p1 = r**y
#p2 = 44.3*p1
T3 = 1400.; 			#K; T3 = T2*p3/p2
T4 = T3 + (T3-T2)/y;
T5 = 656.9; 			#K
n_airard = 1-(T5-T1)/((T3-T2) + y*(T4-T3));

# Results
print ("Efficiency  = %.3f")% (n_airard)

Efficiency  = 0.653

Example 13.28 page no : 650

In [27]:
# Variables
T1 = 373.; 			#K
p1 = 1.; 			#bar
p3 = 65.; 			#bar
p4 = p3;
Vs = 0.0085; 			#m**3
ratio = 21.; 			#Air fuel ratio
r = 15.;
C = 43890.; 			#kJ/kg
cp = 1.;
cv = 0.71;
V2 = 0.0006; 			#m**3
V1 = 0.009; 			#m**3
y = 1.41;
V5 = V1;
V3 = V2;
R = 287.;

# Calculations
p2 = p1*(r)**y;
T2 = T1*r**(y-1);
T3 = T2*p3/p2;
m = p1*10**5*V1/R/T1;
Q1 = m*cv*(T3-T2); 			#Heat added during consmath.tant volume process 2-3
amt = Q1/C; 			    #Amount of fuel added during the consmath.tant volume process 2-3
total = m/ratio; 			#Total amount of fuel added
quantity = total-amt; 		#Quantity of fuel added during the process 3-4
Q2 = quantity*C; 			#Heat added during consmath.tant pressure process
T4 = Q2/(m+total)/cp+T3;
V4 = V3*T4/T3;
T5 = T4*(V4/V5)**(y-1);
Q3 = (m+total)*cv*(T5-T1); 	#Heat rejected during consmath.tant volume process 5-1
W = (Q1+Q2) - Q3;
n_th = W/(Q1+Q2);

# Results
print ("Thermal efficiency  = %.3f")% (n_th)

Thermal efficiency  = 0.618

Example 13.29 page no : 652

In [15]:
# Variables
T1 = 303.; 			#K
p1 = 1.; 			#bar
rc = 9.;
re = 5.;
n = 1.25;
D = 0.25; 			#m
L = 0.4; 			#m
R = 287.;
cv = 0.71;
cp = 1.;
num = 8.; 			#no. 0f cycles/sec

# Calculations and Results
print ("(i) Pressure and temperatures at all salient points  = "),
p2 = p1*(rc)**n;
print ("p2 = %.3f")% (p2), ("bar")

T2 = T1*(rc)**(n-1);
print ("T2 = "), (T2), ("K")

rho = rc/re;
T3 = 1201.9; 			#K
print ("T3 = "), (T3), ("K")

p3 = p2*T3/T2;
print ("p3 = "), (p3), ("bar")

p4 = p3;
print ("p4 = "),(p4), ("bar")

T4 = 1.8*T3;
print ("T4 = "), (T4), ("K")

p5 = p4*(1./re)**(n);
print ("p5 = %.3f")%(p5), ("bar")

T5 = T4*(1./re)**(n-1)
print ("T5 = %.3f")%(T5),("K")


pm = 1./(rc-1)*(p3*(rho-1)+(p4*rho-p5*rc)/(n-1)-(p2-p1*rc)/(n-1));
print ("(ii) Mean effective pressure  =  %.3f")% (pm), ("bar")

print ("(iii) Efficiency of the cycle")
Vs = math.pi/4*D**2*L;
W = pm*10**5*Vs/1000;

V1 = rc/(rc-1)*Vs
m = p1*10**5*V1/R/T1;
Q = m*(cv*(T3-T2) + cp*(T4-T3));

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

P = W*num;
print ("(iv) Power of the engine  = %.3f")% (P), ("kW")

(i) Pressure and temperatures at all salient points  =  p2 = 15.588 bar
T2 =  524.811394693 K
T3 =  1201.9 K
p3 =  35.7 bar
p4 =  35.7 bar
T4 =  2163.42 K
p5 = 4.775 bar
T5 = 1446.766 K
(ii) Mean effective pressure  =  10.919 bar
(iii) Efficiency of the cycle
Efficiency  = 0.585
(iv) Power of the engine  = 171.518 kW

Example 13.30 page no : 657

In [1]:
%matplotlib inline

from numpy import *
from matplotlib.pyplot import *

v = linspace(10,100,90)
def f(v):
    return 1./v**1.4;

f1 = f(v)
plot(v,f1)
v = [10, 20]
p = [f(10), f(10)]
plot(v,p,'r')

v = linspace(20,100,80)
def fa(v):
    return 2.6515/v**1.4;
f1 = fa(v)
plot(v,f1,'g')

v = [100, 100]
p = [f(100), fa(100)]
plot(v,p,'--p')

v = [15, 15]
p = [f(15), 0.040]
plot(v,p,'--')

v = [20 ,20]
p = [f(20), 0.040]
plot(v,p,'--r')

s = linspace(10,50,40);

def fb(s):
    return s**2
f1 = fb(s)
plot(s,f1)

s = linspace(10,50,40)

def fc(s):
    return (s+30)**2
f1 = fc(s)
plot(s,f1,'r')

s = [12, 12];
T = [fb(12), fc(12)];
plot(s,T,'--p')

s = [45, 45];
T = [fb(45) ,fc(45)]
plot(s,T,'m')

s = linspace(10,27,17);
T = 5*(s)**2;
plot(s,T,'g')

s = linspace(10,20,10);
T = 7*s**2;
plot(s,T,'--r')


print ("Thus, ηdiesel > ηdual > ηotto")

Populating the interactive namespace from numpy and matplotlib
Thus, ηdiesel > ηdual > ηotto

Example 13.31 page no : 659

In [32]:
# Variables
cp = 0.92;
cv = 0.75;
y = 1.22; 			#y = cp/cv
p1 = 1.; 			#bar
p2 = p1;
p3 = 4.; 			#bar
p4 = 16.; 			#bar
T2 = 300.; 			#K

# Calculations and Results
T3 = T2*(p3/p2)**((y-1)/y);
T4 = p4/p3*T3;
T1 = T4/(p4/p1)**((y-1)/y);

print ("(i) Work done per kg of gas ")
Q_supplied = cv*(T4-T3);
Q_rejected = cp*(T1-T2);

W = Q_supplied-Q_rejected;
print ("W = %.3f")% (W), ("kJ/kg")

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

(i) Work done per kg of gas 
W = 282.900 kJ/kg
(ii) Efficiency of the cycle  = 0.326

Example 13.32 page no : 680

In [16]:
# Variables
p1 = 101.325; 		#kPa
T1 = 300.; 			#K
rp = 6.;
y = 1.4;

# Calculations
T2 = T1*rp**((y-1)/y);
T3 = 2.5*(T2-T1)/(1-1/1.668);

# Results
print ("(i) Maximum temperature in the cycle  = %.3f")% (T3), ("K")


T4 = T3/1.668;

n_cycle = ((T3-T4) - (T2-T1))/(T3-T2);
print ("(ii)Cycle efficiency  = %.3f")% (n_cycle)

(i) Maximum temperature in the cycle  = 1251.956 K
(ii)Cycle efficiency  = 0.400

Example 13.33 page no : 681

In [36]:
%matplotlib inline
from numpy import *
from matplotlib.pyplot import *

# Variables
p1 = 1.; 			#bar
p2 = 5.; 			#bar
T3 = 1000.; 			#K
cp = 1.0425; 			#kJ/kg K
cv = 0.7662; 			#kJ/kg K
y = cp/cv;

print ("(i)Temperature entropy diagram")

# Calculations
s = linspace(10,50,40);
def fb(s):
    return s**2

plot(s,fb(s),'--')

s = linspace(10,50,40);

def fc(s):
    return (s+30)**2

plot(s,fc(s),'r')

s = [12, 12];
T = [fb(12), fc(12)];
plot(s,T,'m')

s = [45 ,45];
T = [fb(45), fc(45)]
plot(s,T,'g')


# Results
print ("(ii) Power required  = ")
T4 = T3*(p1/p2)**((y-1)/y);
P = cp*(T3-T4);
print ("P = %.3f")% (P), ("kW")
show()

Populating the interactive namespace from numpy and matplotlib
(i)Temperature entropy diagram
(ii) Power required  = 
P = 362.007 kW

WARNING: pylab import has clobbered these variables: ['draw_if_interactive', 'new_figure_manager']
`%pylab --no-import-all` prevents importing * from pylab and numpy

Example 13.34 page no : 682

In [37]:
# Variables
m = 0.1; 			#kg/s
p1 = 1.; 			#bar
T4 = 285.; 			#K
p2 = 4.; 			#bar
cp = 1.; 			#kJ/kg K
y = 1.4;

# Calculations and Results
T3 = T4*(p2/p1)**((y-1)/y);
print ("Temperature at turbine inlet  = %.3f")% (T3), ("K")

P = m*cp*(T3-T4);
print ("Power developed  = %.3f")% (P), ("kW")

Temperature at turbine inlet  = 423.508 K
Power developed  = 13.851 kW

Example 13.35 page no : 682

In [38]:
# Variables
y = 1.4;
cp = 1.005; 			#kJ/kg K
p1 = 1.;    			#bar
T1 = 293.; 	    		#K
p2 = 3.5; 		    	#bar
T3 = 873.; 			    #K
rp = p2/p1;

# Calculations and Results
n_cycle = 1-1/rp**((y-1)/y);
print ("(i) Efficiency of the cycle  = %.3f")% (n_cycle)

T2 = T1*(p2/p1)**((y-1)/y);
Q1 = cp*(T3-T2);
print ("(ii) Heat supplied to air  = %.3f")%(Q1),("kJ/kg")

W = n_cycle*Q1;
print ("(iii) Work available at the shaft  = %.3f")%(W),("kJ/kg")

Q2 = Q1-W;
print ("(iv) Heat rejected in the cooler  = %.3f")%(Q2),("kJ/kg")

T4 = T3/rp**((y-1)/y);
print ("(v) Temperature of air leaving the turbine  = %.3f")%(T4), ("K")

(i) Efficiency of the cycle  = 0.301
(ii) Heat supplied to air  = 456.171 kJ/kg
(iii) Work available at the shaft  = 137.253 kJ/kg
(iv) Heat rejected in the cooler  = 318.919 kJ/kg
(v) Temperature of air leaving the turbine  = 610.332 K

Example 13.36 page no : 683

In [17]:
import math

# Variables
T1 = 303.; 			#K
T3 = 1073.; 			#K
C = 45000.; 			#kJ/kg
cp = 1.; 			#kJ/kg K
y = 1.4;

# Calculations
T2 = math.sqrt(T1*T3);
T4 = T2;

#W_turbine-W_compressor = m_f*C*n = 100;

m_f = 100./C/(1-(T4-T1)/(T3-T2));

# Results
print ("m_f = %.6f")%(m_f), ("kg/s")

m_a = (100-m_f*(T3-T4))/(T3-T4-T2+T1);
print ("m_a = %.3f")% (m_a), ("kg/s")

m_f = 0.004742 kg/s
m_a = 0.414 kg/s

Example 13.37 page no : 684

In [40]:
import math 

# Variables
T1 = 300.; 			#K
p1 = 1.; 			#bar
rp = 6.25;
T3 = 1073.; 		#K
n_comp = 0.8;
n_turbine = 0.8;
cp = 1.005; 		#kJ/kg K
y = 1.4;

# Calculations and Results
T2 = T1*(rp)**((y-1)/y);

#Let T2' = T2a
T2a = (T2-T1)/n_comp + T1;

W_comp = cp*(T2a-T1);
print ("Compressor work  = %.3f")% (W_comp), ("kJ/kg")

T4 = T3/rp**((y-1)/y);
T4a = T3-n_turbine*(T3-T4);

W_turbine = cp*(T3-T4a);
print ("Turbine work  = %.3f")% (W_turbine), ("kJ/kg")

Q_s = cp*(T3-T2a);
print ("Heat supplied  = %.3f")% (Q_s), ("kJ/kg")

W_net = W_turbine - W_comp;

n_cycle = W_net/Q_s*100;
print ("n_cycle %.3f")% (n_cycle), ("%")

t4a = T4a-273;
print ("Turbine exhaust temperature  = %.3f")% (t4a), ("0C")

Compressor work  = 259.322 kJ/kg
Turbine work  = 351.644 kJ/kg
Heat supplied  = 517.543 kJ/kg
n_cycle 17.839 %
Turbine exhaust temperature  = 450.105 0C

Example 13.38 page no : 685

In [41]:
# Variables
n_turbine = 0.85;
n_compressor = 0.80;
T3 = 1148.; 		#K
T1 = 300.; 			#K
cp = 1.; 			#kJ/kg K
y = 1.4;
p1 = 1.; 			#bar
p2 = 4.; 			#bar
C = 42000.; 		#kJ/kg K
n_cc = 0.90;

# Calculations
T2 = T1*(p2/p1)**((y-1)/y);
T2a = (T2-T1)/n_compressor + T1;
ratio = 0.9*C/cp/(T3-T2a) - 1; 			#ratio = ma/mf

# Results
print ("A/F ratio  = %.3f")% (ratio)

A/F ratio  = 55.778

Example 13.39 page no : 686

In [42]:
# Variables
cp = 1.005; 		#kJ/kg K
y1 = 1.4;
y2 = 1.333;
p1 = 1.; 			#bar
p4 = p1;
T1 = 300.; 			#K
p2 = 6.2; 			#bar
p3 = p2;
n_compressor = 0.88;
C = 44186.; 			#kJ/kg
ratio = 0.017; 			#Fuel-air ratio; kJ/kg of air
n_turbine = 0.9; 		
cpg = 1.147;

# Calculations
T2 = T1*(p2/p1)**((y1-1)/y1);
T2a = (T2-T1)/n_compressor + T1; 		#T2'
T3 = ratio*C/(1+ratio)/cp + T2a;
T4 = T3*(p4/p3)**((y2-1)/y2);
T4a = T3-n_turbine*(T3-T4);
W_compressor = cp*(T2a-T1);
W_turbine = cpg*(T3-T4a);
W_net = W_turbine-W_compressor;
Qs = ratio*C;
n_th = W_net/Qs*100;

# Results
print ("Thermal efficiency  = %.3f")% (n_th), ("%")

Thermal efficiency  = 32.590 %

Example 13.40 page no : 688

In [43]:
# Variables
cp = 1.; 			#kJ/kg K
y = 1.4;
C = 41800.; 		#kJ/kg
p1 = 1.; 			#bar
T1 = 293.; 			#K
p2 = 4.; 			#bar
p4 = p1;
p3 = p2;
n_compressor = 0.80;
n_turbine = 0.85; 
ratio = 90.; 			#Air-Fuel ratio
m_a = 3.;    			#kg/s

# Calculations and Results
print ("(i)Power developed ")
T2 = T1*(p2/p1)**((y-1)/y);
T2a = (T2-T1)/n_compressor + T1;
T3 = C/(1+ratio)/cp + T2a;
T4 = T3*(p4/p3)**((y-1)/y);
T4a = T3-n_turbine*(T3-T4);

W_turbine = (ratio+1)/ratio*cp*(T3-T4a);
W_compressor = cp*(T2a-T1);
W_net = W_turbine-W_compressor;
Qs = 1/ratio*C;

P = m_a*W_net;
print ("Power = %.3f")% (P), ("kW/kg of air")

n_thermal = W_net/Qs;
print ("(ii) Thermal efficiency of cycle  = %.3f")%(n_thermal), ("%")

(i)Power developed 
Power = 250.514 kW/kg of air
(ii) Thermal efficiency of cycle  = 0.180 %

Example 13.41 page no : 689

In [44]:
# Variables
T1 = 288.; 			#K
T3 = 883.; 			#K
rp = 6.; 			#rp = p2/p1
n_compressor = 0.80;
n_turbine = 0.82;
m_a = 16.; 			#kg/s
cp1 = 1.005; 		#kJ/kg K, For compression process
y1 = 1.4; 			# For compression process
cp2 = 1.11; 		#kJ/kg K
y2 = 1.333;

# Calculations
T2 = T1*(rp)**((y1-1)/y1);
T2a = (T2-T1)/n_compressor + T1;
T4 = T3/rp**((y2-1)/y2);
T4a = T3-n_turbine*(T3-T4);

W_compressor = cp1*(T2a-T1);
W_turbine = cp2*(T3-T4a);
W_net = W_turbine-W_compressor;
Power = m_a*W_net;

# Results
print ("Power  = %.3f")%(Power), ("kW")

Power  = 770.306 kW

Example 13.42 page no : 691

In [45]:
# Variables
cp = 1.11;
T3 = 883.; 			#K
T2a = 529.; 		#K
W_turbine = 290.4; 	#kJ/kg
W_net = 48.2; 		#kJ/kg

# Calculations
Qs = cp*(T3-T2a);
n_thermal = W_net/Qs*100;
W_ratio = W_net/W_turbine; 			#Work ratio = net work output/Gross work output

# Results
print ("Thermal efficiency  = %.3f")%(n_thermal),("%")

print ("Work ratio  = %.3f")% (W_ratio)

Thermal efficiency  = 12.267 %
Work ratio  = 0.166

Example 13.43 page no : 691

In [46]:
# Variables
p1 = 1.; 			#bar
p2 = 5.; 			#bar
p3 = 4.9; 			#bar
p4 = 1.; 			#bar
T1 = 293.; 			#K
T3 = 953.; 			#K
n_compressor = 0.85;
n_turbine = 0.80;
n_combustion = 0.85;
y = 1.4;
cp = 1.024; 			#kJ/kg K
P = 1065.;   			#kW

# Calculations and Results
print ("(i) The quantity of air circulation")
T2 = T1*(p2/p1)**((y-1)/y);
T2a = (T2-T1)/n_compressor + T1;
T4 = T3*(p4/p3)**((y-1)/y);
T4a = T3-n_turbine*(T3-T4);

W_compressor = cp*(T2a-T1);
W_turbine = cp*(T3-T4a);
W_net = W_turbine-W_compressor;

m_a = P/W_net;
print ("m_a  = %.3f")% (m_a), ("kg")


Qs = cp*(T3-T2a)/n_combustion;
print ("(ii) Heat supplied per kg of air circulation  = %.3f")%(Qs), ("kJ/kg")

n_thermal = W_net/Qs*100;
print ("(iii) Thermal efficiency of the cycle  = %.3f")% (n_thermal), ("%")

(i) The quantity of air circulation
m_a  = 13.507 kg
(ii) Heat supplied per kg of air circulation  = 552.664 kJ/kg
(iii) Thermal efficiency of the cycle  = 14.267 %

Example 13.44 page no : 693

In [47]:
# Variables
m_a = 20.; 			#kg/s
T1 = 300.; 			#K
T3 = 1000.; 		#K
rp = 4.; 			#rp = p2/p1
cp = 1.; 			#kJ/kg K
y = 1.4;

# Calculations
T2 = T1*(rp)**((y-1)/y);
T4 = T3-T2+T1;
r1 = (T3/T4)**(y/(y-1));

r2 = 1./4*r1;
P_ratio = 1./r2; 			#Pressure ratio of low pressure turbine

# Results
print ("Pressure ratio of low pressure turbine  = %.3f")% (P_ratio)

T5 = T4/(P_ratio)**((y-1)/y);
print ("Temperature of the exhaust from the unit  = %.3f")%(T5), ("K")

Pressure ratio of low pressure turbine  = 2.304
Temperature of the exhaust from the unit  = 672.950 K

Example 13.45 page no : 694

In [18]:
# Variables
T1 = 288.; 			#K
p1 = 1.01; 			#bar
rp = 7.;
p2 = rp*p1;
p3 = p2;
p5 = p1;
n_compressor = 0.82;
n_turbine = 0.85;
n_turbine = 0.85;
T3 = 883.; 			#K
cpa = 1.005;
cpg = 1.15;
y1 = 1.4;
y2 = 1.33;

# Calculations and Results
print ("(i) Pressure and temperature of the gases entering the power turbine   ")
T2 = T1*rp**((y1-1)/y1);
T2a = (T2-T1)/n_compressor + T1;
W_compressor = cpa*(T2a-T1);

T4a = (cpg*T3-W_compressor)/cpg;
print ("Temperature of gases entering the power turbine  = %.3f")% (T4a), ("K")

T4 = T3-(T3-T4a)/n_turbine;

p4 = p3/(T3/T4)**(y2/(y2-1));
print ("Pressure of gases entering the power turbine  = %.3f")%(p4),("bar")

print ("(ii) Net power developed per kg/s mass flow")
T5 = T4a/(p4/p5)**((y2-1)/y2);
T5a = T4a-n_turbine*(T4a-T5);

W_turbine = cpg*(T4a-T5a);
print (" Net power developed per kg/s mass flow  = %.3f")%(W_turbine), ("kW")


W_ratio = W_turbine/(W_turbine+W_compressor);
print ("(iii) Work ratio  = %.3f")% (W_ratio)


print ("(iv) Thermal efficiency of the unit")
Qs = cpg*(T3-T2a);
n_thermal = W_turbine/Qs*100;
print ("n_thermal  = %.3f")% (n_thermal), ("%")

(i) Pressure and temperature of the gases entering the power turbine   
Temperature of gases entering the power turbine  = 654.751 K
Pressure of gases entering the power turbine  = 1.640 bar
(ii) Net power developed per kg/s mass flow
 Net power developed per kg/s mass flow  = 72.519 kW
(iii) Work ratio  = 0.216
(iv) Thermal efficiency of the unit
n_thermal  = 18.890 %

Example 13.46 page no : 696

In [50]:
# Variables
T1 = 288.; 			#K
rp = 4.; 			#rp = p2/p1 = p3/p4
n_compressor = 0.82;
e = 0.78; 			#Effectiveness of the heat exchanger
n_turbine = 0.70;
T3 = 873.; 			#K
y = 1.4;
R = 0.287;

# Calculations
T2 = T1*(rp)**((y-1)/y);
T2a = (T2-T1)/n_compressor + T1;
T4 = T3/rp**((y-1)/y);
T4a = T3-n_turbine*(T3-T4);

cp = R*y/(y-1);
W_compressor = cp*(T2a-T1);
W_turbine = cp*(T3-T4a);
W_net = W_turbine-W_compressor;

T5 = e*(T4a-T2a) + T2a;
Qs = cp*(T3-T5);

n_cycle = W_net/Qs*100;

# Results
print ("Efficiency  = %.3f")% (n_cycle), ("%")

Efficiency  = 11.808 %

Example 13.47 page no : 698

In [51]:
# Variables
p2 = 4.; 			#bar
p1 = 1.; 			#bar
T1 = 293.;
n_compressor = 0.8;
n_turbine = 0.85;
ratio = 90.; 			#Air Fuel ratio
C = 41800.; 			#kJ/kg
cp = 1.024;
p4 = 1.01; 			#bar
p3 = 3.9; 			#bar
y = 1.4;
e = 0.72; 			#thermal ratio

# Calculations and Results
T2 = T1*(p2/p1)**((y-1)/y);
T2a = (T2-T1)/n_compressor + T1;
T3 = C/cp/(ratio+1)+471;
T4 = T3*(p4/p3)**((y-1)/y);
T4a = T3-n_turbine*(T3-T4);
n_thermal1 = ((T3-T4a)-(T2a-T1))/(T3-T2a)*100;

print ("Thermal efficiency of simple cycle = %.3f")% (n_thermal1), ("%")


T2a = 471.; 			# K (as for simple cycle)
T3 = 919.5; 			# K (as for simple cycle)
p3 = 4.04-0.14-0.05; 			#bar
p4 = 1.01+0.05; 			#bar

T4 = T3*(p4/p3)**((y-1)/y);
T4a = T3-n_turbine*(T3-T4);
T5 = e*(T4a-T2a) + T2a;
n_thermal2 = ((T3-T4a) - (T2a-T1))/(T3-T5)*100;

print ("Thermal efficiency of heat exchanger cycle  = %.3f")%(n_thermal2), ("%")

dn = n_thermal2-n_thermal1;
print ("Increase in thermal efficiency  = %.3f")% (dn), ("%")

Thermal efficiency of simple cycle = 16.120 %
Thermal efficiency of heat exchanger cycle  = 21.038 %
Increase in thermal efficiency  = 4.918 %

Example 13.48 page no : 700

In [14]:
# Variables
T1 = 293.; 			#K
T6 = 898.; 			#K
T8 = T6;
n_c = 0.8; 			#Efficiency of each compressor stage
n_t = 0.85; 			#Efficiency of each turbine stage
n_mech = 0.95;
e = 0.8;
cpa = 1.005; 			#kJ/kg K
cpg = 1.15; 			#kJ/kg K
y1 = 1.4;
y2 = 1.333;

# Calculations and Results
print ("(i) Thermal efficiency")
T3 = T1;

# p2/p1 = math.sqrt(9) = 3
T2 = T1*(3)**((y1-1)/y1);
T2a = (T2-T1)/n_c + T1;
T4a = T2a;
W_c = cpa*(T2a-T1); 			#Work input per compressor stage
W_t = 2*W_c/n_mech; 			#Work output of H.P. turbine
T7a = T6-W_t/cpg;
T7 = T6-(T6-T7a)/n_t;

T9 = T8/(1.86)**((y2-1)/y2);
T9a = T8-n_t*(T8-T9);

W = cpg*(T8-T9a)*n_mech; 			#Net work output
T5 = e*(T9a-T4a)+T4a;

Q = cpg*(T6-T5)+cpg*(T8-T7a); 			#Heat supplied
n_thermal = W/Q*100;

print ("n_thermal  = %.3f")% (n_thermal), ("%")

print ("(ii) Work ratio")
Gross_work = W_t+W/n_mech;
W_ratio = W/Gross_work;
print ("Work ratio = %.3f")% (W_ratio)

m = 4500./W;
print ("(iii) Mass flow rate  = %.3f")%(m), ("kg/s")

# Note : answers are different because of rounding error.

(i) Thermal efficiency
n_thermal  = 24.209 %
(ii) Work ratio
Work ratio = 0.291
(iii) Mass flow rate  = 37.576 kg/s

Example 13.49 page no : 704

In [53]:
import math 

# Variables
T1 = 293.; 			#K
T5 = 1023.; 		#K
T7 = T5;
p1 = 1.5; 			#bar
p2 = 6.; 			#bar
n_c = 0.82;
n_t = 0.82;
e = 0.70;
P = 350.; 			#kW
cp = 1.005; 		#kJ/kg K
y = 1.4;

# Calculations
T3 = T1;
px = math.sqrt(p1*p2);
T2 = T1*(px/p1)**((y-1)/y);
T2a = T1+(T2-T1)/n_c;
T4a = T2a;
p5 = p2;
T6 = T5/(p5/px)**((y-1)/y);
T6a = T5-n_t*(T5-T6);
T8a = T6a;
Ta = T4a+e*(T8a-T4a);
W_net = 2*cp*((T5-T6a)-(T2a-T1));
Q1 = cp*(T5-T4a)+cp*(T7-T6a); 			#Without regenerator
Q2 = cp*(T5-Ta)+cp*(T7-T6a);

# Results
n1 = W_net/Q1*100;
print ("n_thermal without regenerator  = %.3f")%(n1), ("%")

n2 = W_net/Q2*100;
print ("n_thermal woth regenerator  = %.3f")% (n2), ("%")

m = P/W_net;
print ("(iii) Mass of fluid circulated  = %.3f")% (m), ("kg/s")

n_thermal without regenerator  = 18.059 %
n_thermal woth regenerator  = 32.079 %
(iii) Mass of fluid circulated  = 2.403 kg/s