Chapter 8 : Availability and Irreversibility

Example 8.1 Page No : 185

In [1]:
# Variables
Q = 16.;			# in MJ
Q = Q * 10**3;			# in kJ
T_H = 227.;			# in °C
T_H = T_H + 273;			# in K
T_L = 15.;			# in °C
T_L = T_L + 273;			# in K

# Calculations and Results
del_S = Q/T_H;			# in kJ/K
A = Q - (T_L * del_S);			# in kJ
print "The available part of heat in kJ is ",A

U_P_ofHeat = T_L * del_S;			# unavailable part of heat in kJ
print "The unavailable part of heat in kJ is :",U_P_ofHeat

The available part of heat in kJ is  6784.0
The unavailable part of heat in kJ is : 9216.0

Example 8.2 Page No : 185

In [1]:
# Variables
Q = 12000.;			# in kJ
T_H = 600.;			# in K
T_L = 300.;			# in K

# Calculations and Results
dS = Q / T_H;			#in kJ/K
A = Q - (T_L * dS);			#available work in kJ
print "Available work is in kJ",A

UA = T_L * dS;			#unavailable work in kJ
print "Unavailable work is in kJ",UA

Available work is in kJ 6000.0
Unavailable work is in kJ 6000.0

Example 8.3 Page No : 185

In [4]:
import math 

# Variables
m = 800.;			# in kg
C_p = 0.5;			# in kJ/kg K
T2 = 500.;			# in K
T1 = 1250.;			# in K
T_o = 300.;			# in K

# Calculations and Results
del_t = T1 - T2;                    			# in K
Q = m * C_p * del_t;            	    		# in kJ
dS = abs(m * C_p * math.log(T2/T1));			# in kJ/K
availableEnergy = Q - (T_o * dS);			#in kJ
print "Available energy in MJ is :",round(availableEnergy*10**-3)

unavailableEnergy = T_o * dS;			# UA  for unavailable energy in kJ
print "Unavailable energy in MJ is :",(round(unavailableEnergy*10**-3))

Available energy in MJ is : 190.0
Unavailable energy in MJ is : 110.0

Example 8.4 Page No : 197

In [2]:
# Variables
h_i = 726.1;
h_o = 25.03;
T_o = 298;			# in K
s_i = 1.582;
s_o = 0.087;
h2 = 669.;
s2 = 1.677;

# Calculations and Results
h3 = 52.17 + (0.9 * 567.7);
s3 = 0.1748 + (0.9 * 1.7448);
sai_i = (h_i - h_o) - (T_o * (s_i - s_o));			# in kcl/kg
print "The availablibity per kg of steam entering in kcl/kg is : %.2f"%sai_i

sai_e = (0.25 * ((h2 - h_o) - (T_o * (s2 - s_o)))) + (0.75 * ((h3 - h_o) - (T_o * (s3 - s_o)))) ;			# in kcl/kg
print "The availablibity per kg of steam leaving in kcl/kg is : %.2f"%sai_e

w_rev = sai_i - sai_e;			# in kcl/kg
print "reversible work per kg of steam in kcl/kg is : %.4f"%w_rev

# Note: There is calculation error in evaluating the value of availability per kg of 
# steam leaving in kcl/kg . so the answer in the book is wrong and coding is right.

The availablibity per kg of steam entering in kcl/kg is : 255.56
The availablibity per kg of steam leaving in kcl/kg is : 75.50
reversible work per kg of steam in kcl/kg is : 180.0598

Example 8.5 Page No : 198

In [1]:
# Variables
T_o = 298;			# in K
m2 = 25000;
s2 = 16775;
m3 = 75000;
s3 = 17448;
m1 = 1000000;
s1 = 1582;
Q = -16;			# in MJ

# Calculations
Q = Q * 10**3;			# in kJ
I = (T_o * ((m2 * s2) + (m3 * s3) - (m1 * s1))) - Q;			# in cal/hr
I=I*10**-3;			# in kcal/hr

# Results
print "The irreversiblity in kcal/hr",I

# Note: There is calculation error in evaluating the value of the irreversibility
# in kcal/hr. so the answer in the book is wrong and coding is right.

The irreversiblity in kcal/hr 43500566.0

Example 8.6 Page No : 198

In [2]:
# Variables
h_i = 749.2;
h_o = 25.03;
T_o = 298;			# in K
s_i = 1.6202;
s_o = 0.0877;

# Calculations and Results
phi_i = (h_i - h_o)- (T_o * (s_i - s_o));			# kcal/kg
print "The availablibity before adiabatic throttling in kcal/kg is : %.2f"%phi_i

h_e = 749.2;
s_e = 1.6936;
phi_e = (h_e - h_o) - (T_o * (s_e - s_o));			# in kcal/kg
print "The availablibity before adiabatic throttling in kcal/kg is : %.2f"%phi_e

Wrev = phi_i - phi_e;			# in kcal/kg
print "Reversible work in kcal/kg is : %.2f"%Wrev

Wactual = 0;
i = Wrev-Wactual;			# in kcal/kg
print "Irreversibility per kg of steam in kcal/kg is : %.2f"%i

The availablibity before adiabatic throttling in kcal/kg is : 267.49
The availablibity before adiabatic throttling in kcal/kg is : 245.61
Reversible work in kcal/kg is : 21.87
Irreversibility per kg of steam in kcal/kg is : 21.87

Example 8.7 Page No : 199

In [3]:
import math 

# Variables
# del_W = T * ds - del_Q
T = 600;			# in K
p_i = 7;			#kgf/cm**2
p_e = 1.5;			#kgf/cm**2
T_o = 298;			# in K

# Calculations and Results
R = 29.27/427;
del_W_lost = T * ( R *math.log(p_i/p_e));			# in kcal/kg
print "Lost work in kcal/kg is",round(del_W_lost,4)

i = T_o * (R * (math.log(p_i/p_e)));			# in kcal/kg
print "Irreversebility per kg of air flow in kcal/kg is : %.3f"%i

Lost work in kcal/kg is 63.3567
Irreversebility per kg of air flow in kcal/kg is : 31.467