Chapter 6 : Second Law of Thermodynamics

Example 6.1 Page No : 141

In [1]:
# Variables
T1 = 800.;
T2 = 30.;

# Calculation
e_max = 1.-((T2+273)/(T1+273));
Wnet = 1.    			# in kW
Q1 = Wnet/e_max;
Q2 = Q1-Wnet;

# Results
print "Least rate of heat rejection is %.3f KW"%Q2

Least rate of heat rejection is 0.394 KW

Example 6.2 Page No : 142

In [10]:
# Variables
T1 = -15.+273;
T2 = 30.+273;
Q2 = 1.75; 			# in kJ/sec

# Calculation
Q1 = Q2*T2/T1
W = Q1-Q2;

# Results
print "Least Power necessary to pump the heat out is",round(W,2),"kW"

Least Power necessary to pump the heat out is 0.31 kW

Example 6.4 Page No : 144

In [2]:
# Variables
Q1 = 200.;
T1 = 373.15;
T2 = 273.16;

# Calculation
Q2 = Q1*(T2/T1);
W = Q1-Q2;
e = W/Q1;

# Results
print "The heat rejected, the work done and the thermal effiency of the engine is %.3f J, %.1f J, %.1f respectively"%(e,W,Q2)

The heat rejected, the work done and the thermal effiency of the engine is 0.268 J, 53.6 J, 146.4 respectively

Example 6.5 Page No : 144

In [6]:
# Variables
T1 = 873.;
T2 = 313.;
T3 = 253.;
Q1 = 2000.; 			# In joule
W = 360.; 			# in joule

# Calculation
# Part (a)
e_max = round(1-(T2/T1),3);
W1 = e_max*Q1;
COP = T3/(T2-T3);
W2 = W1-W;
Q4 = COP*W2;
Q3 = Q4+W2;
Q2 = Q1-W1;

# Results
print "The heat rejection to the 40 degree reservior is",round((Q2+Q3)),"kJ"

# Part (b)
e_max_ = 0.4*e_max;
W1_ = e_max_*Q1;
W2_ = W1_-W;
COP_ = 0.4*COP;
Q4_ = COP_*W2_;
Q3_ = Q4_+W2_;
Q2_ = Q1-W1_;

print "Q4 = %.1f kJ"%round(Q4_,1)
print "Q3 = %.1f kJ"%round(Q3_,1)
print "Q2 = %.1f kJ"%round(Q2_,1)
print "The heat rejection to the 40 degree reservior is",round((Q2_+Q3_)),"kJ"

# note:  answers are slightly different because of rounding off error. please check.

The heat rejection to the 40 degree reservior is 5528.0 kJ
Q4 = 257.7 kJ
Q3 = 410.5 kJ
Q2 = 1487.2 kJ
The heat rejection to the 40 degree reservior is 1898.0 kJ

Example 6.7 Page No : 147

In [22]:
# Variables
T1 = 473.;
T2 = 293.;
T3 = 273.;

# Calculation
MF = (T2*(T1-T3))/(T1*(T2-T3));

# Results
print " The multiplication factor is %.1f"%MF

# rounding off error.

 The multiplication factor is 6.2

Example 6.8 Page No : 148

In [6]:
# Variables
T1 = 363.;
T2 = 293.;
W = 1.; 			# Kj/s

# Calculation
e_max = 1.-(T2/T1);
Qmin = W/e_max ;
Qmin_ = Qmin*3600.;
E = 1880.; 			# in kJ/m2 h
Amin = Qmin_/E ;

# Results
print "Minimum area required for the collector plate %.0f m2"%Amin

Minimum area required for the collector plate 10 m2

Example 6.9 Page No : 149

In [8]:
# Variables
T1 = 1000.;
W = 1000.; 			# in W
K = 5.67e-08;

# Calculation
Amin = (256.*W)/(27.*K*T1**4);

# Results
print "Area of the panel %.4f m^2"%Amin

Area of the panel 0.1672 m^2