Chapter 06:Second Law of Thermodynamics

Ex6.1:pg-138

In [1]:
import math
T1 = 800 # Source temperature in degree Celsius

T2 = 30 # Sink temperature in degree Celsius

e_max = 1-((T2+273)/(T1+273)) # maximum possible efficiency 

Wnet = 1  # in kW

Q1 = Wnet/e_max # Least rate of heat required in kJ/s

Q2 = Q1-Wnet  # Least rate of heat rejection kJ/s



print "\n Example 6.1"

print "\n Least rate of heat rejection is ",Q2," kW"

#The answers vary due to round off error
 Example 6.1

 Least rate of heat rejection is  0  kW

Ex6.2:pg-139

In [2]:
import math
T1 = -15 # Source temperature in degree Celsius

T2 = 30 # Sink temperature in degree Celsius

Q2 = 1.75  # in kJ/sec

print "\n Example 6.2"

W= Q2*((T2+273)-(T1+273))/(T1+273) # Least Power necessary to pump the heat out

print "\n Least Power necessary to pump the heat out is ",round(W,2),"kW"
    
    #The answers vary due to round off error
    
    
 Example 6.2

 Least Power necessary to pump the heat out is  0.31 kW

Ex6.3:pg-140

In [3]:
import math
#Given 

T1 = 600 # Source temperature of heat engine in degree Celsius

T2 = 40 # Sink temperature of heat engine in degree Celsius 

T3 = -20 # Source temperature of refrigerator in degree Celsius

Q1 = 2000  # Heat transfer to heat engine in kJ

W = 360  # Net work output of plant in kJ

# Part (a)

e_max = 1.0-((T2+273)/(T1+273)) # maximum efficiency 

W1 = e_max*Q1 # maximum work output 

COP = (T3+273)/((T2-273)-(T3-273)) # coefficient of performance of refrigerator

W2 = W1-W # work done to drive refrigerator 

Q4 = COP*W2 # Heat extracted by refrigerator

Q3 = Q4+W2 # Heat rejected by refrigerator

Q2 = Q1-W1 # Heat rejected by heat engine

Qt = Q2+Q3 # combined heat rejection by heat engine and refrigerator 

print "\n Example 6.3"

print "\n\n Part A:"

print "\n The heat transfer to refrigerant is ",round(Q2,3) ," kJ"

print "\n The heat rejection to the 40 degree reservoir is ",round(Qt,3) ," kJ"



# Part (b)

print "\n\n Part B:"

e_max_ = 0.4*e_max # maximum efficiency

W1_ = e_max_*Q1 # maximum work output 

W2_ = W1_-W # work done to drive refrigerator 

COP_ = 0.4*COP # coefficient of performance of refrigerator

Q4_ = COP_*W2_  # Heat extracted by refrigerator

Q3_ = Q4_+W2_ # Heat rejected by refrigerator

Q2_ = Q1-W1_ # Heat rejected by heat engine

QT = Q2_+Q3_# combined heat rejection by heat engine and refrigerator 

print "\n The heat transfer to refrigerant is ",round(Q2_,3) ," kJ"

print "\n The heat rejection to the 40 degree reservoir is ",round(QT,3) ," kJ"

#The answers vary due to round off error
 Example 6.3


 Part A:

 The heat transfer to refrigerant is  0.0  kJ

 The heat rejection to the 40 degree reservoir is  8200.0  kJ


 Part B:

 The heat transfer to refrigerant is  1200.0  kJ

 The heat rejection to the 40 degree reservoir is  2344.0  kJ

Ex6.5:pg-142

In [4]:
import math
T1 = 473 # Boiler temperature in K

T2 = 293 # Home temperature in K

T3 = 273 # Outside temperature in K

print "\n Example 6.5"

MF = (T2*(T1-T3))/(T1*(T2-T3)) 

print "\n  The multiplication factor is ",MF 

#The answers vary due to round off error
 Example 6.5

  The multiplication factor is  6

Ex6.6:pg-144

In [5]:
import math
T1 = 90.0 # Operating temperature of power plant in degree Celsius 

T2 = 20.0 # Atmospheric temperature in degree Celsius

W = 1.0  # Power production from power plant in kW

E = 1880  # Capability of energy collection in kJ/m**2 h



print "\n Example 6.6"

e_max = 1.0-((T2+273.0)/(T1+273.0))  # maximum efficiency

Qmin = W/e_max  # Minimum heat requirement per second

Qmin_ = Qmin*3600.0 # Minimum heat requirement per hour

Amin = Qmin_/E  # Minimum area requirement

print "\n Minimum area required for the collector plate is ",math. ceil(Amin) ," m**2"
 Example 6.6

 Minimum area required for the collector plate is  10.0  m**2

Ex6.7:pg-144

In [6]:
import math
T1 = 1000 # Temperature of hot reservoir in K

W = 1000 # Power requirement in kW

K = 5.67e-08 # constant 

print "\n Example 6.7"

Amin = (256*W)/(27*K*T1**4) # minimum area required

print "\n Area of the panel ",Amin ," m**2"
 Example 6.7

 Area of the panel  0.167221895617  m**2