# Chapter 6 : Availability and Irreversibility¶

### Example 6.1 Page no : 313¶

In [1]:
import math

# Variables
T0 = 293.; 			#K
T1 = 300.; 			#K
T2 = 370.; 			#K
cv = 0.716;
cp = 1.005;
R = 0.287;
p1 = 1.; 			#bar
p2 = 6.8; 			#bar
m = 1.; 			#kg

# Calculations
Wmax = -(cv*(T2-T1) - T0*(cp*math.log(T2/T1)-R*math.log(p2/p1)));
n = 1/(1-(math.log(T2/T1)/math.log(p2/p1)));
Wact = m*R*(T1-T2)/(n-1);
I = Wmax - Wact;

# Results
print ("Irreversibility  =  %.3f")%(I),("kJ/kg")

Irreversibility  =  13.979 kJ/kg


### Example 6.2 Page no : 313¶

In [2]:
# Variables
T1 = 1000.; 			#K
T2 = 500.; 			#K
T0 = 300.; 			#K
Q = 7200.; 			#kJ/min

# Calculations and Results
print ("(i) Net change of entropy :")
dS_source = -Q/T1;
dS_system = Q/T2;
dS_net = dS_source+dS_system;
print ("dS_net = "), (dS_net), ("kJ/min.K")

print ("(ii) Decrease in available energy :")
AE_source = (T1-T0)*(-dS_source); 			#Available energy with the source
AE_system = (T2-T0)*dS_system; 			#Available energy with the system
dAE = AE_source - AE_system; 			#Decrease in available energy
print ("dAE = "), (dAE), ("kJ")

(i) Net change of entropy :
dS_net =  7.2 kJ/min.K
(ii) Decrease in available energy :
dAE =  2160.0 kJ


### Example 6.3 Page no : 315¶

In [3]:
import math

# Variables
m = 8.; 			#kg
T1 = 650.; 			#K
p1 = 5.5*10**5; 			#Pa
p0 = 1*10.**5; 			#Pa
T0 = 300.; 			#K
cp = 1.005; 			#kJ/kg.K
cv = 0.718;
R = 0.287;
#p1*v1/T1 = p0*v0/T0
#Let r = v1/v0 = 1/2.54
r = 1/2.54;

# Calculations and Results
print ("(i) Change in available energy(for bringing the system to dead state) = ")
ds = cv*math.log(T1/T0) + R*math.log(r);
dAE = m*(cv*(T1-T0) - T0*ds);
#dAE is the change in available energy in kJ
V1 = m*R*10**3*T1/p1;
V0 = V1/r;
L = p0*(V0 - V1)/10**3;
print ("Loss of availability, L = "), (L), ("kJ")

print ("(ii) Available Energy and Effectiveness")
Q = m*cp*(T1-T0);
ds = m*cp*math.log(T1/T0);
Unavailable_energy = T0*ds;
Available_energy  =  Q - Unavailable_energy;
print ("Available energy  =  %.3f")% (Available_energy), ("kJ")

Effectiveness = Available_energy/dAE;
print ("Effectiveness  =  %.3f")% (Effectiveness)

(i) Change in available energy(for bringing the system to dead state) =
Loss of availability, L =  417.872 kJ
(ii) Available Energy and Effectiveness
Available energy  =  949.066 kJ
Effectiveness  =  0.719


### Example 6.4 Page no : 316¶

In [1]:
import math

# Variables
c_pg = 1.;  			#kJ/kgK
h_fg = 1940.7; 			#kJ/kg
Ts = 473.; 	    		#K ; Temperature of saturation of steam
T1 = 1273.; 			#K ; Initial temperature of gases
T2 = 773.; 		    	#K ; Final temperature of gases
T0 = 293.; 			    #K ; atmospheric temperature

# Calculations
#Heat lost by gases = Heat gained by 1 kg saturated water when it is converted to steam at 200 0C
m_g = h_fg/c_pg/(T1-T2);
dS_g = m_g*c_pg*math.log(T2/T1);
dS_w = h_fg/Ts;
dS_net = dS_g + dS_w;

# Results
print ("Net change in entropy  =  %.3f")% (dS_net), ("kJ/K")

E = T0*dS_net; 			#Increase in unavailable energy due to hea transfer
print ("Increase in unavailable energy  = %.3f")%(E), ("kJ per kg of steam formed")

Net change in entropy  =  2.167 kJ/K
Increase in unavailable energy  = 634.847 kJ per kg of steam formed


### Example 6.5 Page no : 317¶

In [5]:
import math

# Variables
m_g = 3.; 			#kg
p1 = 2.5; 			#bar
T1 = 1200.; 			#K; Temperature of infinite source
T1a = 400.; 			#K; Initial temperature
Q = 600.; 			#kJ
cv = 0.81; 			#kJ/kg.K
T0 = 290.; 			#K; Surrounding Temperature

# Calculations
#final temperature  =  T2a
T2a = Q/m_g/cv + T1a;
AE = (T1-T0)*Q/T1; 			#Available energy with the source
dS = m_g*cv*math.log(T2a/T1a); 			#Change in entropy of the gas
UAE = T0*dS; 			#Unavailability of the gas
A = Q-UAE; 			#Available energy with the gas
loss = AE-A;

# Results
print ("Loss in available energy due to heat transfer  = %.3f")%(loss),("kJ")

Loss in available energy due to heat transfer  = 193.783 kJ


### Example 6.6 Page no : 318¶

In [7]:
import math

# Variables
m = 60.; 			#kg
T1 = 333.; 			#K
T0 = 279.; 			#K
p = 1.; 			#atm
cp = 4.187;

# Calculations
def f16( T):
return m*cp*(1-T0/T)

Q1 = m*cp*(T1-T0);
#Let unavailable energy = E
E = Q1-Wmax;

# Results
print ("unavailable energy  =  %.3f")%(E), ("kJ")

unavailable energy  =  12401.141 kJ


### Example 6.7 Page no : 319¶

In [8]:
import math

# Variables
m = 15.; 			#kg
T1 = 340.; 			#K
T0 = 300.; 			#K
cp = 4.187; 			#kJ/kgK

# Calculations
#Work added during churning  =  Increase in enthalpy of water
W = m*cp*(T1-T0);
ds = cp*math.log(T1/T0);
AE = m*(cp*(T1-T0)-T0*ds);
AE_loss = W-AE; 			#Loss in availability

# Results
print ("Loss in availability %.3f")% (AE_loss), ("kJ")

Loss in availability 2358.261 kJ


### Example 6.8 Page no : 320¶

In [9]:
import math

# Variables
m = 5.; 			#kg
T1 = 550.; 			#K
p1 = 4*10.**5; 			#Pa
T2 = 290.; 			#K
T0 = T2;
p2 = 1.*10**5; 			#Pa
p0 = p2;
cp = 1.005; 			#kJ/kg K
cv = 0.718; 			#kJ/kg K
R = 0.287; 			#kJ/kg K

# Calculations and Results
print ("(i) Availability of the system :")
ds = cp*math.log(T1/T0) - R*math.log(p1/p0);
Availability = m*(cv*(T1-T0) - T0*ds);
print ("Availability of the system  = %.3f")% (Availability), ("kJ")

print ("(ii) Available energy and Effectiveness")
Q = m*cp*(T1-T0);
dS = m*cp*math.log(T1/T0);
E = T0*dS; 			#Unavailable energy
AE = Q-E;
print ("Available Energy  =  %.3f")%(AE), ("kJ")

Effectiveness = AE/Availability;
print ("Effectiveness =  %.3f")%(Effectiveness)

(i) Availability of the system :
Availability of the system  = 577.612 kJ
(ii) Available energy and Effectiveness
Available Energy  =  373.806 kJ
Effectiveness =  0.647


### Example 6.9 Page no : 321¶

In [10]:
import math

# Variables
R = 0.287; 			#kJ/kgK
cp = 1.005; 			#kJ/kgK
m = 25./60; 			#kg/s
p1 = 1.; 			#bar
p2 = 2.; 			#bar
T1 = 288.; 			#K
T0 = T1;
T2 = 373.; 			#K

# Calculations and Results
W_act = cp*(T2-T1); 			#W_actual
W_total =  m*W_act;

print ("Total actual power required  = %.3f")%(W_total), ("kW")

ds = cp*math.log(T2/T1) - R*math.log(p2/p1);
Wmin = cp*(T2-T1) - T0*(ds);

W = m*Wmin;
print ("Minimuumm work required  =  %.3f")%(W), ("kW")

Total actual power required  = 35.594 kW
Minimuumm work required  =  28.276 kW


### Example 6.10 Page no : 322¶

In [11]:
import math

# Variables
m_O2 = 1.; 			#kg
m_H2 = 1.; 			#kg
p = 1*10.**5; 			#Pa
T_O2 = 450.; 			#K
T_H2 = 450.; 			#K
T0 = 290.; 			#K
R0 = 8.314;
M_O2 = 32.;
M_H2 = 2.;

# Calculations
R_O2 = R0/M_O2;
v_O2 = m_O2*R_O2*T_O2/p;
R_H2 = R0/M_H2;
v_H2 = m_H2*R_H2*T_H2/p;
v_f = v_O2 + v_H2; 			#total volume after mixing
dS_O2 = R_O2*math.log(v_f/v_O2);
dS_H2 = R_H2*math.log(v_f/v_H2);
dS_net = dS_O2 + dS_H2;
#Let E be the loss in availability
E = T0*dS_net;

# Results
print ("Loss in availability = %.3f")% (E), ("kJ")

Loss in availability = 286.555 kJ


### Example 6.11 Page no : 323¶

In [2]:
import math

# Variables
T0 = 283.; 			#K
cp = 4.18; 			#kJ/kgK
m1 = 20.; 			#kg
T1 = 363.; 			#K
m2 = 30.; 			#kg
T2 = 303.; 			#K
T3 = 327.; 			#K

# Calculations
def f13( T):
return m1*cp*(1-T0/T)

def f14( T):
return m2*cp*(1-T0/T)

AE_total = AE1 + AE2; 			#before mixing
#If T K is the final temperature after mixing
T = (m1*T1+m2*T2)/(m1+m2);
m_total = m1+m2;

#Available energy of 50kg of water at 54 0C
AE3 = m_total*cp*((T3-T0) - T0*math.log(T3/T0));

#Decrease in available energy due to mixing dAE
dAE = AE_total - AE3;

# Results
print ("Decrease in avialble energy = %.3f")% (dAE), ("kJ")

Decrease in avialble energy = 234.184 kJ


### Example 6.12 Page no : 324¶

In [13]:
import math

# Variables
T_w1 = 323.; 			#K
T_w2 = 343.; 			#K
T_o1 = 513.; 			#K
T_o2 = 363.; 			#K
SG_oil = 0.82;
c_po = 2.6; 			#kJ/kg K
c_pw = 4.18; 			#kJ/kg K
T0 = 300.; 			#K
m_o = 1.; 			#kg

# Calculations
#Heat lost by oil = Heat gained by water
m_w = (m_o*c_po*(T_o1-T_o2))/(c_pw*(T_w2-T_w1));
dS_w = m_w*c_pw*math.log(T_w2/T_w1);
dS_o = m_o*c_po*math.log(T_o2/T_o1);
dAE_w = m_w*(c_pw*(T_w2-T_w1))-T0*dS_w;
dAE_o = m_o*(c_po*(T_o2-T_o1))-T0*dS_o;

# Loss in availability E =
E = dAE_w+dAE_o;

# Results
print ("Loss in availability  = %.3f")%(E),("kJ")

Loss in availability  = -81.676 kJ


### Example 6.13 Page no : 325¶

In [14]:
import math

# Variables
m_i = 1.; 			#kg
T_i = 273.; 			#K
m_w = 12.; 			#kg
T_w = 300.; 			#K
T0 = 288.; 			#K
c_pw = 4.18; 			#kJ/kg K
c_pi = 2.1; 			#kJ/kg K
L_i = 333.5; 			#kJ/kg

# Calculations
Tc = (m_w*c_pw*T_w + m_i*c_pw*T_i - L_i)/(m_w*c_pw + m_i*c_pw);
dS_w = m_w*c_pw*math.log(Tc/T_w);
dS_i = m_i*c_pw*math.log(Tc/T_i) + L_i/T_i;
dS_net = dS_w+dS_i;

# Results
print ("Increase in entropy  = %.3f")% (dS_net), ("kJ/K")

dAE = T0*dS_net;
print ("Increase in unavailable energy  =  %.3f")% (dAE),("kJ")

Increase in entropy  = 0.107 kJ/K
Increase in unavailable energy  =  30.878 kJ


### Example 6.14 Page no : 326¶

In [15]:
# Variables
T1 = 673.; 			#K
T2 = 473.; 			#K
T0 = 303.; 			#K
T1a = T2;

# Calculations
UAE = T0*(T1-T1a)/T1a/(T1-T0);

# Results
print ("the fraction of energy that becomes unavailable  = %.3f")%(UAE)

the fraction of energy that becomes unavailable  = 0.346


### Example 6.15 Page no : 327¶

In [16]:
import math

# Variables
T1 = 293.; 			#K
T2 = 353.; 			#K
Tf = 1773.; 			#K
T0 = 288.; 			#K
c_pl = 6.3; 			#kJ/kg K

# Calculations
dAE = c_pl*(T2-T1) - T0*c_pl*math.log(T2/T1);
n = (1-T0/Tf); 			#efficiency
E = c_pl*(T2-T1)*n;
Effectiveness = dAE/E;

# Results
print ("Effectiveness of the heating process  = %.3f")%(Effectiveness)

Effectiveness of the heating process  = 0.126


### Example 6.16 Page no : 328¶

In [3]:
import math

# Variables
T0 = 293.; 			#K
T1 = 293.; 			#K
T2 = 373.; 			#K
T3 = 323.; 			#K
cp = 1.005;

# Calculations and Results
print ("(i) The ratio of mass flow")
x = (T3-T1)/(T2-T3);
print ("x = "), (x)

ds_13 = cp*math.log(T3/T1);
ds_32 = cp*math.log(T2/T3);
A = cp*(T3-T1) - T1*ds_13; 	    		#Increase of availability of system
B = x*(cp*(T2-T3)-T0*(ds_32));			# Loss of availability of surroundings
Effectiveness = A/B;
print ("Effectiveness of heating process = %.3f")%(Effectiveness)

(i) The ratio of mass flow
x =  0.6
Effectiveness of heating process = 0.306


### Example 6.17 Page no : 329¶

In [6]:
import math

# Variables
m = 2.5; 			#kg
p1 = 6.*10**5; 			#Pa
r = 2.; 			#r = V2/V1
cv = 0.718; 			#kJ/kg K
R = 0.287; 			#kJ/kg K
T1 = 363.; 			#K
p2 = 1.*10**5; 			#Pa
T2 = 278.; 			#K
V1 = m*R*T1/p1;
V2 = 2*V1;
T0 = 278.; 			#K
p0 = 1.*10**5; 			#Pa

# Calculations and Results
dS = m*cv*math.log(T2/T1) + m*R*math.log(V2/V1);
Wmax = m*(cv*(T1-T2)) + T0*(cv*math.log(T2/T1) + R*math.log(V2/V1));
print ("(i)The maximum work"),("Wmax = %.3f")% (Wmax), ("kJ")

dA = Wmax+p0*(V1-V2);
print ("(ii)Change in availability  = %.3f")%(dA), ("kJ")

I = T0*m*(cv*math.log(T2/T1)+R*math.log(V2/V1));
print ("(iii)Irreversibility  = %.3f")% (I),("kJ")

(i)The maximum work Wmax = 154.628 kJ
(ii)Change in availability  = 111.219 kJ
(iii)Irreversibility  = 5.132 kJ


### Example 6.18 Page no : 330¶

In [8]:
import math

# Variables
m = 1.; 			#kg
p1 = 7.*10**5; 			#Pa
T1 = 873.; 			#K
p2 = 1.*10**5; 			#Pa
T2 = 523.; 			#K
T0 = 288.; 			#K
Q = -9.; 			#kJ/kg
cp = 1.005; 			#kJ/kg K
R = 0.287; 			#kJ/kg K

# Calculations and Results
print ("(i) The decrease in availability ")
dA = cp*(T1-T2) - T0*(R*math.log(p2/p1) - cp*math.log(T2/T1));
print ("dA = %.3f")%(dA), ("kJ/kg")

print ("(ii) The maximum work")
Wmax = dA; 			#change in availability
print ("Wmax %.3f")% (Wmax), ("kJ/kg")

W = cp*(T1-T2) + Q;
I = Wmax - W;
print ("(iii)Irreversibility  = %.3f")%(I), ("kJ/kg")

(i) The decrease in availability
dA = 364.295 kJ/kg
(ii) The maximum work
Wmax 364.295 kJ/kg
(iii)Irreversibility  = 21.545 kJ/kg


### Example 6.19 Page no : 331¶

In [9]:
import math

# Variables
cp = 1.005; 			#kJ/kg K
cv = 0.718; 			#kJ/kg K
R = 0.287; 			#kJ/kg K
m = 1.; 			#kg
T1 = 290.; 			#K
T0 = 290.; 			#K
T2 = 400.; 			#K
p1 = 1.; 			#bar
p0 = 1.; 			#bar
p2 = 6.; 			#bar

# Calculations and Results
#Wrev = change in internal energy - T0*change in entropy
Wrev = -(cv*(T2-T1) - T0*(cp*math.log(T2/T1) - R*math.log(p2/p1)));
n = (1./(1-math.log(T2/T1)/math.log(p2/p1)));
Wact = m*R*(T1-T2)/(n-1);

I = Wrev-Wact;
print ("(i)Irreversibility = %.3f")% (I), ("kJ")

effectiveness = Wrev/Wact*100;
print ("(ii)The effectiveness  =  %.3f")%(effectiveness), ("%")

(i)Irreversibility = 9.945 kJ
(ii)The effectiveness  =  93.109 %


### Example 6.20 Page no : 333¶

In [21]:
import math

# Variables
I = 0.62; 			#kg/m**2
N1 = 2500.; 			#rpm
m = 1.9; 			#kg;  Water equivalent of shaft bearings
cp = 4.18;
T0 = 293.; 			#K
t0 = 20.; 			#0C

# Calculations and Results
print ("(i)Rise in temperature of bearings")
KE = 1./2*I*w1**2/1000; 			#kJ
dT = KE/(m*cp); 			#rise in temperature of bearings
print ("dT = %.3f")% (dT), ("0C")

t2 = t0+dT;
print ("Final temperature of the bearings  = %.3f")% (t2), ("0C")

T2 = t2+273;
print ("(ii)Final r.p.m. of the flywheel")
def f15( T):
return m*cp*(1-T0/T)

UE = KE - AE;

print ("Available energy  = %.3f")% (AE), ("kJ")

UAE = KE-AE;
print ("Unavailable energy  = %.3f")%(UAE), ("kJ")

w2 = math.sqrt(AE*10**3*2/I);
N2 = w2*60/2/math.pi;
print ("Final rpm of the flywheel  = %.3f")%(N2), ("rpm")

(i)Rise in temperature of bearings
dT = 2.675 0C
Final temperature of the bearings  = 22.675 0C
(ii)Final r.p.m. of the flywheel
Available energy  = 0.096 kJ
Unavailable energy  = 21.151 kJ
Final rpm of the flywheel  = 168.407 rpm


### Example 6.21 Page no : 334¶

In [22]:
import math

# Variables
p1 = 8.; 			#bar
T1 = 453.; 			#K
p2 = 1.4; 			#bar
T2 = 293.; 			#K
T0 = T2;
p0 = 1.; 			#bar
m = 1.; 			#kg
C1 = 80.; 			#m/s
C2 = 40.; 			#m/s
cp = 1.005; 			#kJ/kg K
R = 0.287; 			#kJ/kg K

# Calculations and Results
print ("(i) Reversible work and actual work ")
A1 = cp*(T1-T0)-T0*(cp*math.log(T1/T0)-R*math.log(p1/p0))+C1**2/2/10**3; 			#Availability at the inlet
A2 = cp*(T2-T0)-T0*(cp*math.log(T2/T0)-R*math.log(p2/p0))+C2**2/2/10**3; 			#Availability at the exit

W_rev = A1-A2;
print ("W_rev  = %.3f")%(W_rev), ("kJ/kg")

W_act = cp*(T1-T2) + (C1**2-C2**2)/2/10**3;
print ("W_act  = %.3f")%(W_act),("kJ/kg")

print ("(ii) Irreversibilty and effectiveness  = ")

I = W_rev-W_act;
print ("Irreversibilty  = %.3f")% (I), ("kJ/kg")

Effectiveness = W_act/W_rev*100;
print ("Effectiveness  = %.3f")%(Effectiveness),("%")

(i) Reversible work and actual work
W_rev  = 181.464 kJ/kg
W_act  = 163.200 kJ/kg
(ii) Irreversibilty and effectiveness  =
Irreversibilty  = 18.264 kJ/kg
Effectiveness  = 89.935 %


### Example 6.22 Page no : 335¶

In [10]:
# Variables
p1 = 20.; 			#bar
t1 = 400.; 			#0C
p2 = 4.; 			#bar
t2 = 250.; 			#0C
t0 = 20.; 			#0C
T0 = t0+273;
h1 = 3247.6; 			#kJ/kg
s1 = 7.127; 			#kJ/kg K
#let h2' = h2a and s2' = s2a
h2a = 2964.3; 			#kJ/kg
s2a = 7.379; 			#kJ/kg K
s2 = s1;
s1a = s1;
#By interpolation, we get
h2 = 2840.8; 			#kJ/kg

# Calculations and Results
n_isen = (h1-h2a)/(h1-h2);
print ("(i)Isentropic efficiency  = %.3f")%(n_isen)

A = h1-h2a + T0*(s2a-s1a);
print ("(ii)Loss of availability = %.3f")%(A), ("kJ/kg")

Effectiveness = (h1-h2a)/A;
print ("(iii)Effectiveness  = %.3f")% (Effectiveness)

(i)Isentropic efficiency  = 0.696
(ii)Loss of availability = 357.136 kJ/kg
(iii)Effectiveness  = 0.793