# Chapter 5: The Second Law of Thermodynamics¶

## Example 5.1, page no. 87¶

In [3]:
#initilisation
Tr = 540.0                      #R
Te = 2000.0                     #R
m = 200.0                       #B/lbm

#calculation
eta = 1-(Tr/Te)
Qr = m*(1-eta)

#result
print "Thermal efficiency is ", eta*100, "%"
print "Heat rejected = %d B/lbm" %Qr

Thermal efficiency is  73.0 %
Heat rejected = 54 B/lbm


## Example 5.2, page no. 90¶

In [1]:
import scipy.integrate

#initilisation
cv=0.171                    #B/lbm F
T2=580                      #F
T1=520                      #F

#calculation
def fun(T):
cp=cv/T
return cp

ds = scipy.integrate.quadrature(fun, T1, T2)[0]

#result
print "Change in entropy = %.4f B/lbm R" %ds

Change in entropy = 0.0187 B/lbm R


## Example 5.3, page no. 95¶

In [6]:
import scipy.integrate

#initilisation

Q = 100.0                         #B/lbm
Cp = 0.24                         #B/lbm F
T1 = 70.0+460.0                   #R
T2 = 550.0+460.0                  #R
Ts = 50.0+460.0                   #R

#calculation
def fun(T):
cp = Cp/T
return cp

ds1 = scipy.integrate.quadrature(fun, T1, T2)[0]
Tf  = Q/Cp + T1
ds2 = scipy.integrate.quadrature(fun, T1, Tf)[0]
Qr = Ts*(ds2)
Qa = Q-Qr
Qun = Ts*(ds1)
Qa2 = Q-Qun

#result
print "Case 1"
print "Change in entropy = %.4f B/lbm R" %ds1
print "case 2"
print "Entropy change = %.4f B/lbm R" %ds2
print "Available energy = %.1f B/lbm" %Qa
print "case 3"
print "Available energy = %.1f B/lbm" %Qa2

Case 1
Change in entropy = 0.1548 B/lbm R
case 2
Entropy change = 0.1392 B/lbm R
Available energy = 29.0 B/lbm
case 3
Available energy = 21.1 B/lbm