Chapter 12: Thermal Properties¶

Example 12.1, page no-350¶

In [1]:
# Change in length due to heating

import math
#Variable Declaration
alfe=8.8*10**-6                    # linear coefficient of thermal expansion for alumina
lo=0.1                             # length of the alumina rod
delT=973.0                         # difference in temperature

#Calculation
delL=alfe*lo*delT

#Result
print('The change in length produced by heating is %.3f mm'%(delL*10**3))

The change in length produced by heating is 0.856 mm


Example 12.2, page no-350¶

In [2]:
# Change in length due to heating

import math
#Variable Declaration
alfe=5.3*10**-6                  # linear coefficient of thermal expansion or alumina
lo=0.1                           # length of the alumina rod
delT=973.0                       # difference in temperature

#Calculation
delL=alfe*lo*delT

#Result
print('The change in length produced by heating is %.3f mm'%(delL*10**3))

The change in length produced by heating is 0.516 mm


Example 12.3, page no-351¶

In [3]:
# Steady state heat Transfer

import math
#Variable Declaration
k=371.0                    # Thermal conductivity of copper in J/msk
delT=50.0                  # change in temperature
delx=10*10**-3             # change in thickness of the copper's sheet

#Calculation
ht=k*delT/delx

#Result
print('The steady state heat transfer of 10 mm copper sheet is %.3f *10^6 J.m^-2.s^-1'%(ht*10**-6))

The steady state heat transfer of 10 mm copper sheet is 1.855 *10^6 J.m^-2.s^-1


Example 12.4, page no-351¶

In [4]:
# Compression Stress due to Heating

import math
#Variable Declaration
alfe=8.8*10**-6                           # linear coefficient of thermal expansion for alumina
t1=1300.0                                 # Temperature 1
t2=327.0                                  # Temperature 2
E=370.0                                   # modulus of elasticity

#Calculation
delT=t1-t2
ep=alfe*delT
sig=ep*E

#Result
print('\nThe unconstrained thermal expansion produced by the heating is %.4f *10^-3'%(ep*10**3))
print('\nthe compression stress produced by heating is %.3f GPa'%(math.ceil(sig*1000)/1000))

The unconstrained thermal expansion produced by the heating is 8.5624 *10^-3

the compression stress produced by heating is 3.169 GPa


Example 12.5, page no-352¶

In [5]:
# Heat flux transmitted

import math
#Variable Declaration
K=120.0                       # thermal conductivity of brass
t2=423.0                      # Temperature 2
t1=323.0                      # Temperature 1
delT=t2-t1                    # temperature difference
delx=7.5*10**-3               # change in thickness of the brass's sheet
A=0.5                         # Area of the sheet

#Calculation
Q=K*A*(delT/delx)
hph=Q*3600

#Result
print('The heat flux transmitted through a sheet per hour is %.2f *10^9 J.h^-1'%(hph*10**-9))

The heat flux transmitted through a sheet per hour is 2.88 *10^9 J.h^-1


Example 12.6, page no-353¶

In [6]:
# Young's Modulus

import math
#Variable Declaration
alfe=17*10**-6               # linear coefficient of thermal expansion for copper
t2=293.0                     # Temperature 2
t1=233.0                     # Temperature 1
delT=t2-t1                   # temperature difference
st=119.0                     # Maximum thermally induced stress

#Calculation
k=alfe*delT
E=(st*10**6)/k

#Result
print('\nThe strain produced in te rod is %.2f * 10^-3'%(k*10**3))
print('\nThe Youngs Modulus of the rod is %.1f GPa'%(E*10**-9))

The strain produced in te rod is 1.02 * 10^-3

The Youngs Modulus of the rod is 116.7 GPa


Example 12.7, page no-353¶

In [7]:
# Temperature Change

import math
#Variable Declaration
lo=11.6                       # length of the steel rod
delx=5.4*10**-3               # difference in length
alfL=12*10**-6                # Linear coefficient of thermal expansion for steel

#Calculation
delT=delx/(lo*alfL)

#Result
print('The maximum temperature cange can withstand without any thermal stress is %.2f K'%delT)

The maximum temperature cange can withstand without any thermal stress is 38.79 K


Example 12.8, page no-354¶

In [8]:
#compressive Sress

import math
#Variable Declaration
lo=0.35                         # length of the Al rod
alfe=23.6*10**-6                # Linear coefficient of thermal expansion for Al
t2=358.0                        # temperature 2
t1=288.0                        # temperature 1
delT=t2-t1                      # temperature difference
ym=69.0                         # Young's modulus

#Calculation
k=alfe*delT
E=ym*k*10**9

#Result
print('\nThe strain produced in te rod is %.3f * 10^-3'%(k*10**3))
print('\nThe compressive stress produced in Al rod is %.3f GPa'%(E*10**-9))

The strain produced in te rod is 1.652 * 10^-3

The compressive stress produced in Al rod is 0.114 GPa


Example 12.9, page no-355¶

In [9]:
# limit to compression stress

import math
#Variable Declaration
alfe=20*10**-6               # Linear coefficient of thermal expansion for alumina
t1=293.0                     # temperature
sig=172.0                    # compressive stress
E=100.0                      # modulus of elasticity

#Calculation
delT=(sig*10**6)/(E*alfe*10**9)

#Result
print('\nTf-Ti=%.0f'%delT)
print('\n\nThe maximum temperature at which the rod may be heated without\nexceeding a compresssive stress of %.0f MPa is %.0f K'%(sig,delT+t1))

Tf-Ti=86

The maximum temperature at which the rod may be heated without
exceeding a compresssive stress of 172 MPa is 379 K


Example 12.10, page no-356¶

In [10]:
# Heat energy Requirement

import math
#Variable Declaration
h_ir=444.0                   # specific heat capacity of iron in J.kg^-1.K^-1
h_gr=711.0                   # specific heat capacity of graphite in J.kg^-1.K^-1
h_pl=1880.0                  # specific heat capacity of polypropylene in J.kg^-1.K^-1
t2=373.0                     # Temperature 2
t1=300.0                     # Temperature 1
delT=t2-t1                   # difference in temperature
W=2.0                        # weight

#Calculation

#(a) For Iron
q=W*h_ir*delT

#(b)for Graphite
q1=W*h_gr*delT

#(b)for polypropylene
q2=W*h_pl*delT

#Result
print('The heat energy required to raise temperature %.0f K from its temperature of \niron, graphite and polypropylene is %.0f,%.0f,%.0f J respectively'%(delT,q,q1,q2))

The heat energy required to raise temperature 73 K from its temperature of
iron, graphite and polypropylene is 64824,103806,274480 J respectively