Chapter 8 : Available Energy, Availability and Irreversibility

Example 8.1 Page No : 243

In [1]:
import math 

# Variables
T0 = 308.;
T1 = 693.; 
T1_ = 523.; 			# T1_ = T1'
T1_ = 523.; 			# ""

# Calculation
f = (T0*(T1-T1_))/(T1_*(T1-T0));

# Results
print "The fraction of energy that becomes unavailable due to irreversible heat transfer is %.2f"%f
The fraction of energy that becomes unavailable due to irreversible heat transfer is 0.26

Example 8.2 Page No : 244

In [1]:
import math 
from scipy.integrate import quad 

# Variables
lhw = 1858.5; 			# Latent heat of water
Tew = 220+273.;
Sw = lhw/Tew; 
Tig = 1100.; 			# Initial temperature of the gas
Tfg = 550.; 			# Final ""

# Calculation
k = 1*lhw/(Tig-Tfg); 			# k = mg_dot*cpg
Tg2 = 823.
Tg1 = 1373.
T0 = 303.;

def f2(T): 
	 return k/T

Sg =  quad(f2,Tg1,Tg2)[0]

St = Sg+Sw; 

print "Total change in entropy is %.3f kJ/K"%St
print "Increase in unavailable energy is %.0f Kj"%(round(T0*St,-1))
Total change in entropy is 2.040 kJ/K
Increase in unavailable energy is 620 Kj

Example 8.3 Page No : 245

In [4]:
import math 
from scipy.integrate import quad 

# Variables
Tw = 75. + 273;
Ts = 5. + 273; 			# Ts = T0
m = 40.;
cp = 4.2;

# Calculation
def f9(T): 
	 return m*cp*(1-(Tw/T))

W =  -quad(f9,Ts,Tw)[0]
Q1 = m*cp*(Tw-Ts);
UE = Q1-W;

# Results
print "Total work %.0f kJ"%W
print "Heat released",Q1,"kJ"
print "Internal energy change %.0f kJ"%UE

# quad gives slightly different answer than book has. pLease check.
Total work 1370 kJ
Heat released 11760.0 kJ
Internal energy change 10390 kJ

Example 8.4 Page No : 246

In [5]:
import math 
from scipy.integrate import quad 

# Variables
Ts = 273+15.         #temperature 
Tw1 = 95+273.        #Water
Tw2 = 35+273.        #water
m1 = 25.             #energy
m2 = 35.             
cp = 4.2             #Water

# Calculation
def f3(T): 
	 return m1*cp*(1-(Ts/T))

AE25 =  quad(f3,Ts,Tw1)[0]


def f4(T): 
	 return m2*cp*(1-(Ts/T))
AE35 =  quad(f4,Ts,Tw2)[0]
AEt = AE25 + AE35;
Tm = (m1*Tw1+m2*Tw2)/(m1+m2); 			# Temperature after mixing

def f5(T): 
	 return (m1+m2)*cp*(1-(Ts/T))

AE60 =  quad(f5,Ts,Tm)[0]

AE = AEt - AE60;
print "The decrease in the totla energy is %.2f kJ"%AE
The decrease in the totla energy is 281.82 kJ

Example 8.5 Page No : 247

In [6]:
import math 
from scipy.integrate import quad 

# Variables
N1 = 3000.;                   #RPM
w1 = (2*math.pi*N1)/60.;      
I = 0.54;
Ei = 0.5*I*w1**2;
ti = 15.+273;                #temperature
m = 2.;                      #water equivalent

# Calculation
dt = Ei/(1000*2*4.187);
tf = ti+dt;

def f6(T): 
	 return m*4.187*(1-(ti/T))

AE =  quad(f6,ti,tf)[0]

UE = Ei/1000 - AE;
w2 = math.sqrt(AE*1000*2/I);
N2 = (w2*60)/(2*math.pi);

# Results
print "The final RPM of the flywheel would be %.0f"%N2
The final RPM of the flywheel would be 222

Example 8.6 Page No : 248

In [5]:
import math 
from scipy.integrate import quad 

# Variables
T1 = 353.
T2 = 278.
V2 = 2.
V1 = 1.
P0 = 100.             #tempareture
P1 = 500.             #air
R = 0.287
cv = 0.718;           #air 
m = 2.                

# Calculation
def f7(T): 
	 return (m*cv)/T

def f8(V):
    return m*R/V

S =  quad(f7,T1,T2)[0] + quad(f8,V1,V2)[0]
U = m*cv*(T1-T2);
Wmax = U-(T2*(-S));
V1_ = (m*R*T1)/P1;
CA = Wmax-P0*(V1_); 			# Change in availability
I = T2*S; 

# Results
print "The maximum work is %.2f kJ"%Wmax
print "Change in availability is %.2f kJ"%CA
print "Irreversibility is %.1f kJ"%I

# rounding off error. please check.
The maximum work is 122.96 kJ
Change in availability is 82.43 kJ
Irreversibility is 15.3 kJ

Example 8.7 Page No : 250

In [6]:
import math 

# Variables
P1 = 500.          # air
P2 = 100.;         # kPa    
T1 = 793.
T2 = 573.
cp = 1.005         # air
T0 = 293.
R = 0.287;

# Calculation and Results
S21 = (R*math.log(P2/P1))-(cp*math.log(T2/T1))
CA = cp*(T1-T2)-T0*S21; 			# Change in v=availability
print "The decrease in availability is %.1f kJ/Kg"%CA

Wmax = CA;
print "The maximum work is %.1f kJ/Kg"%Wmax

Q = -10.;
W = cp*(T1-T2)+Q ;
I = Wmax-W;
print "The irreversibility is %.1f kJ/Kg"%I

# rounding off error.
The decrease in availability is 260.8 kJ/Kg
The maximum work is 260.8 kJ/Kg
The irreversibility is 49.7 kJ/Kg

Example 8.8 Page No : 251

In [7]:
import math 

# Variables
T0 = 300.        # cooled
Tg1 = 573.
Tg2 = 473.;
Ta1 = 313.; 
cpg = 1.09      # tempareture
cpa = 1.005;    # air
mg = 12.5
ma = 11.15;

# Calculation and Results
f1 = cpg*(Tg1-T0)-T0*cpg*(math.log(Tg1/T0));
f2 = cpg*(Tg2-T0)-T0*cpg*(math.log(Tg2/T0));
print "The initial and final availbility of the products are %.2f and %.2f kJ/Kg respectively"%(f1,f2)

# Part (b)
Dfg = f1-f2;
Ta2 = Ta1 + (mg/ma)*(cpg/cpa)*(Tg1-Tg2);
Ifa = cpa*(Ta2-Ta1)-T0*cpa*(math.log(Ta2/Ta1));
I = mg*Dfg-ma*Ifa;
print "The irreversibility of the process is %.2f kW"%I

# Part (c)
Ta2_ = Ta1*(math.e**(-(mg/ma)*(cpg/cpa)*math.log(Tg2/Tg1)));
Q1 = mg*cpg*(Tg1-Tg2);
Q2 = ma*cpa*(Ta2_-Ta1);
W = Q1-Q2;
print "Tota power generated by the heat engine %.2f kW"%W

# rounding off error. please check.
The initial and final availbility of the products are 85.97 and 39.68 kJ/Kg respectively
The irreversibility of the process is 319.37 kW
Tota power generated by the heat engine 441.35 kW

Example 8.9 Page No : 253

In [14]:
import math 

# Variables
T2 = 1063.;
T1 = 1073.;
m = 2.;       # pipe
cp = 1.1;

# Calculation and Results
I = m*cp*((T1-T2)-T0*(math.log(T1/T2)));
print "The irrevesibility rate is %.3f kW"%I

# At lower temperature
T1_ = 353.
T2_ = 343.
I_ = m*cp*((T1_-T2_)-T0*(math.log(T1_/T2_)));
print "The irrevesibility rate at lower temperature is %.3f kW"%I_
The irrevesibility rate is 15.820 kW
The irrevesibility rate at lower temperature is 3.033 kW

Example 8.10 Page No : 254

In [11]:
# Variables
m = 3.;       # insulated pipe
R = 0.287; 
T0 = 300.
k = 0.10; 			# k = dP/P1

# Calculation
Sgen = m*R*k;
I = Sgen*T0;

# Results
print "The rate of energy loss because of the pressure drop due to friction",I,"kW"
The rate of energy loss because of the pressure drop due to friction 25.83 kW

Example 8.11 Page No : 254

In [8]:
import math 

# Variables
m1 = 2.; 			# m1_dot
m2 = 1.;
T1 = 90.+273; 
T2 = 30.+273;
T0 =300.;

# Calculation
m = m1+m2;
x = m1/m;
t = T2/T1; 			# Tau
cp = 4.187;
Sgen = m*cp*math.log((x+t*(1-x))/(t**(1-x)));
I = T0*Sgen;


print "The rate of entropy generation is %.4f kW/K"%Sgen
print "The rate of energy loss due to mixing is %.1f kW"%I

# rounding off error.
The rate of entropy generation is 0.0446 kW/K
The rate of energy loss due to mixing is 13.4 kW

Example 8.12 Page No : 255

In [19]:
# Variables
Qr = 500.; 			# Heat release in kW
Tr = 2000.;
T0 = 300.;          # chemical process     

# Calculation and Results
# Part (a)
Qa = 480.
Ta = 1000.;
n1a = (Qa/Qr);
n2a = n1a*(1-(T0/Ta))/(1-(T0/Tr));
print ("PART (A)")
print "The first law efficiency is",n1a*100,"%"
print "The Second law efficiency is %.0f %%"%(n2a*100)

# Part (b)
Qb = 450.
Tb = 500.
n1b = (Qb/Qr);
n2b = n1b*(1-(T0/Tb))/(1-(T0/Tr));
print ("PART (B)")
print "The first law efficiency is",n1b*100,"%"
print "The Second law efficiency is %.1f %%"%(n2b*100)

# Part (c)
Qc = 300.
Tc = 320.
n1c = (Qc/Qr);
n2c = n1c*(1-(T0/Tc))/(1-(T0/Tr));
print ("PART (C)")
print "The first law efficiency is",n1c*100,"%"
print "The Second law efficiency is %.2f %%"%(n2c*100)

# Part (d)
Qd = 450.; 
n1d = (Qd/Qr);
n2a_= n1d*(1-(T0/Ta))/(1-(T0/Tr));
n2b_= n1d*(1-(T0/Tb))/(1-(T0/Tr));
n2c_= n1d*(1-(T0/Tc))/(1-(T0/Tr));
print ("Part (D)")
print "The first law efficiency is",n1d
print "The Second law efficiency of part (a) is %.2f %%"%(n2a_*100)
print "The Second law efficiency of part (b) is %.1f %%"%(n2b_*100)
print "The Second law efficiency of part (c) is %.2f %%"%(n2c_*100)
PART (A)
The first law efficiency is 96.0 %
The Second law efficiency is 79 %
PART (B)
The first law efficiency is 90.0 %
The Second law efficiency is 42.4 %
PART (C)
The first law efficiency is 60.0 %
The Second law efficiency is 4.41 %
Part (D)
The first law efficiency is 0.9
The Second law efficiency of part (a) is 74.12 %
The Second law efficiency of part (b) is 42.4 %
The Second law efficiency of part (c) is 6.62 %

Example 8.14 Page No : 258

In [21]:
import math 

# Variables
cp = 1.005
T2 = 433.
T1 = 298.;
T0 = 298.
R = 0.287
P2 = 8.
P1 = 1.
Q = -100.
m = 1.

# Calculation
W = Q + m*cp*(T1-T2);
AF = cp*(T2-T1)-T0*((cp*math.log(T2/T1))-(R*math.log(P2/P1))) ; 			# AF = af2-af1
e = AF/-W 			# efficiency 

# Results
print "The power input is %.1f kW"%W
print "The second law efficiency of the compressor is %.3f"%e
The power input is -235.7 kW
The second law efficiency of the compressor is 0.855

Example 8.15 Page No : 259

In [16]:
import math 

# Variables
U = 0
H0 = 0
S = 0;
# If the vaccume ha reduced to dead state
U0 = 0
H0 = 0
S0 = H0
V0 = 0;
P0 = 100
V = 1;

# Calculation
fi = P0*V;

# Results
print "The energy of the complete vaccume is",fi,"kJ"
The energy of the complete vaccume is 100 kJ

Example 8.16 Page No : 259

In [10]:
import math 

# Variables
m = 1000.    # mass in kg
T0 = 300.
P0 = 1. 
T1 = 300.;
T2 = 273-20.
Tf = 273-2.2; # freezing point in C
Cb = 1.7 
Ca = 3.2;
Lh = 235.     # latent heat

# Calculation
H12 = m*((Cb*(Tf-T2))+Lh+(Ca*(T1-Tf)));
H21 = -H12;
S12 = m*((Cb*math.log(Tf/T2))+(Lh/Tf)+(Ca*math.log(T1/Tf)));
S21 = -S12;
E = H21-T0*S21;

# Results
print "Energy produced is %.1f MJ"%(E/1000)
Energy produced is 34.6 MJ

Example 8.17 Page No : 260

In [11]:
import math 

# Variables
cv = 0.718
T2 = 500.    # tempareture
T1 = 300.
m = 1.
T0 = 300.

# Case (a)
Sua = cv*math.log(T2/T1);
Ia = T0*Sua;
print "The irreversibility in case a is %.1f kJ/Kg"%Ia

# Case (b)
Q = m*cv*(T2-T1);
T = 600.
Sub = Sua-(Q/T);
Ib = T0*Sub;
print "The irreversibility in case b is %.2f kJ/Kg"%Ib
 
# rounding off error
The irreversibility in case a is 110.0 kJ/Kg
The irreversibility in case b is 38.23 kJ/Kg

Example 8.18 Page No : 260

In [28]:
import math 

# Variables
h1 = 3230.9
s1 = 6.69212
V1 = 160.         # velocity
T1 = 273.+400;    # Steam
h2 = 2676.1
s2 = 7.3549
V2 = 100.
T2 = 273. + 100;
T0 = 298.
W = 540.         # water
Tb = 500.        # tempareture

# Calculation
Q = (h1-h2)+((V1**2-V2**2)/2)*1e-03-W;
I = 151.84-Q*(0.404);
AF = W + Q*(1-(T0/Tb)) + I; 			# AF = af1-af2
n2 = W/AF;

# Results
print "Irreversibility per unit mass is %.2f kJ/Kg"%I
print "The second law effiency of the turbine is %.2f"%n2
Irreversibility per unit mass is 142.71 kJ/Kg
The second law effiency of the turbine is 0.78

Example 8.19 Page No : 262

In [21]:
# Variables
T0 = 300.
T = 1500.   # resistor
Q = -8.5
W = 8.5;

# Calculation and Results
# Case (a)
I = Q*(1-T0/T) + W;
R = Q*(1-T0/T);
print "Rate of availability transfer with heat and the irreversibility rate are",R,"and",I,"kW"

# Case (b)
T1 = 500.;
Ib = - Q*(1-T0/T) + Q*(1-T0/T1);
print "Rate of availability in case b is",Ib,"kW"
Rate of availability transfer with heat and the irreversibility rate are -6.8 and 1.7 kW
Rate of availability in case b is 3.4 kW

Example 8.20 Page No : 263

In [12]:
import math 

# Variables
P1 = 1.
T1 = 273.+30; 
P2 = 3.5
T2 = 141.+273 ; 
V = 90.;
T0 = 303.;

# Calculation and Results
# Part (a)
g = 1.4;
T2s = T1*((P2/P1)**((g-1)/g));
print "As T2s> T2 so the process must be polytropic"

# Part (b)
p = math.log(P2/P1); q = math.log(T2/T1);
n = p/(p-q);
print "The polytropic index is %.2f"%n

# Part (c)
cp = 1.0035; R = 0.287;
Wa = cp*(T1-T2)-(V2**2/2)*1e-03 ;
Wt = -R*T0*math.log(P2/P1)-(V2**2/2)*1e-03;
Nt = Wt/Wa;
print "The isothermal effiency is %.3f"%Nt

# Part (d)
f12 = cp*(T1-T2) + T0*((R*math.log(P2/P1))-(cp*math.log(T2/T1))) - (V2**2/2)*1e-03 ;
I = f12-Wa ; 
print "The minimum work input and irreversibility are %.1f and %.1f kJ/Kg "%(f12,I)

# Part (e)
n2 = (f12/Wa);
print "Second law efficiency is %.2f"%n2

# note : answer are slightly different because of rounding off error. please check.
As T2s> T2 so the process must be polytropic
The polytropic index is 1.33
The isothermal effiency is 0.978
The minimum work input and irreversibility are -97.4 and 14.0 kJ/Kg 
Second law efficiency is 0.87