# Chapter 5 :- The Second Law of Thermodynamics¶

## Example 5.1 Page no- 194

In [1]:
# Given :-
W = 410.00                             # net work output in kj claimed
Q = 1000.00                            # energy input by heat transfer in kj
Tc = 300.00                            # temperature of cold reservoir in kelvin
TH = 500.00                            # temperature of hot reservoir in kelvin

# Calculations
eta = W/Q                              # thermal efficiency
etamax = 1-(Tc/TH)

# Results
print '-> Eta = ',round(eta,4)
print '-> Etamax  = ',round(etamax,4)
print '-> Since eta is more than etamax, the claim is not authentic'

-> Eta =  0.41
-> Etamax  =  0.4
-> Since eta is more than etamax, the claim is not authentic

## Example 5.2 Page no- 195

In [2]:
# Given :-
Qcdot = 8000.00                                # in kj/h
Wcycledot = 3200.00                            # in kj/h
Tc = 268.00                                    # temperature of compartment in kelvin
TH = 295.00                                    # temperature of the surrounding air in kelvin

# Calculations
beta = Qcdot/Wcycledot                         # coefficient of performance
betamax = Tc/(TH-Tc)                           # reversible coefficient of performance

# Results
print '-> Coefficient of performance is ',round(beta,3)
print '-> Coefficient of performance of a reversible cycle is ',round(betamax,3)

-> Coefficient of performance is  2.5
-> Coefficient of performance of a reversible cycle is  9.926

## Eample 5.3 Page no- 196

In [3]:
# Given :-
Tc = 283.0                         # in kelvin
TH = 295.0                         # in kelvin
QH = 5*(10**5)                     # in kj per day

# Calculations
Wcyclemin = (1-(Tc/TH))*QH

# Results
print '-> Minimum theoretical work input for one day of operation in kJ is: ',round(Wcyclemin,2)

-> Minimum theoretical work input for one day of operation in kJ is:  20338.98