In [3]:

```
from __future__ import division
import math
#Variables
d1=0.1; # diameter of sphere, m
T1=303; # environment temp.,K
T2=278; # fridge temp., K
h=6; #convection coefficient, W/(m**2*K)
k=0.603; #thermal conductivity,W/(m*K)
a=997.6; # density of water, kg/m**3
c=4180; #heat capacity, J/(kg*K)
#Calculations
F=(k/(a*c))*3600/(math.pow(d1,2))/4;
a1=math.pow(h*(d1/2)/k,-1) # Biot no.=1/2.01 therefore we read from fig. in upper left hand corner
Tcen=a1*(T1-T2)+T2; # temperature of the center of apple after 1 hour, K
Tc=Tcen-273; # temperature of the center of apple after 1 hour, °C
F1=1.29 #Bi is still 1/2.01, by looking at the graph we can find time.
t=F1*a*c*math.pow(d1/2,2)/k-2; #Time to bring the temp equal to 283k , seconds.
#finally we look up at Bi=1/2.01 and fouling factor is 1.29, for spheres heta removal is 43.67 kJ per apple.
x=43.67; #heat removal for an apple, kJ
X=12*x; #total heat removal,kJ
#Results
print "Temperature after an hour is :",Tc,"C\n"
print "Time to bring the temp equal to 283k is :",round(t,3),"s or 6 hr 12 min\n"
print "Total energy removal is :",X,"kJ\n"
#end
```

In [6]:

```
from __future__ import division
import math
#Variables
d1=0.001; #diameter of nichrome, m
h=30000; #convection coefficient , W/(m**2*K)
T1=373; # wire temperature, K
k=13.8; #thermal conductivity,W/(m*K)
a=3.43*10**-6; # Thermal Diffusivity, m^2/s
w = 2*math.pi*60; # Frequency of current in rad/s
#heat is being generated in proportion to product of voltage and current, if the boiling action removes heat rapidly enough in comparison with the heat capacity of the wire,the surface temperature may well vary.
#Calculations
Bi = h*d1/2/k; # biot number comes ot to be 1.09 by looking at the chart of cylinders, we find that, (Tmax-Tav)/(Tav-To)=0.04
phi = w*d1**2/4/a; # value of a= w*d1**(2)/4/a1 comes out to be 27.5.
TF=0.04; # (from the charts for cylinders, Fig. 5.12)temperature fluctuation of 4 percent is not serious and experiment is valid.
#Results
print "Biot number: ",round(Bi,1);
print "Psi(Ψ): ",round(phi,1);
print "The temperature fluctuation is : ", TF,"this fluctuation is probably not serious.It therefore appears that the experiment is valid.\n"
```

In [5]:

```
from __future__ import division
import math
#Variables
t=0.003; #half thickness of sword, m
a=1.5*math.pow(10,-5); #Thermal diffusivity, m^2/s
#Calculations
Tmax=math.pow(t,2)/(math.pow(3.64,2)*a); #Maximum time for sword to be in semi infinite region, seconds
#Results
print "Maximum time for sword to be in semi infinite region is :",round(Tmax,4),"s\n"
print "Thus the quench would be felt at the centerline of the sword within only 1/20 s. the thermal diffusivity of clay is smaller than that of steel by a factor of about 30, so the quench time of coated steel must continue for over 1s before the temperature of the steel is affected at all, if the clay and sword thickness are comparable"
```

In [10]:

```
from __future__ import division
import math
#Variables
Tburn=65; #Skin threshold, C
Tbody=37; #Body temp, C
Tflame=800; #Flame temp, C
h=100; #convective heat transfer coefficient, W/(m**2*K)
k=0.63; # thermal conductivity,W/(m*K)
#the short exposure to the flame causes only a very superficial heating,so we consider the finger to be semi-infinite region.it turns out that the burn threshold of human skin,Tburn is about 65 C. h=100 W/(m**2*K), we shall assume that the thermal conductivity of human flesh equals that of its major component - water and that the thermal diffusivity is equal to the known value for beef.
# a=0.963, BE=h*x/k=0(since x=0 at the surface)
# b**2=(h**2)*(0.135*10**-6)*t/(k**2)=0.0034*t. On solving error function by trial and error method, we get the value of t=0.33 sec.
#Calculations
a=(Tburn-Tflame)/(Tbody-Tflame);
beta=(1-a)*(math.sqrt(math.pi))/2;
# from fig. 5.16, it would require about 1/3 se to bring the skin to burn point.
#Results
print "It would require about 1/3 sec to bring the skin to burn point"
```

In [14]:

```
from __future__ import division
import math
#Variables
b=0.139*math.pow(10,-6); #thermal diffusivity, m**2/s
t=365*24*3600; #seconds in a year
min=2.356; #first minima from fig 5.19
#w=2*3.14 rad/yr , a=w*t=0 at present.first we find the depths at which a=0 curve reaches its local extrema.(we pick the a=0 curve because it) gives the highest temperature at t=o.).tan(o-e)=1 so e=3%pi/4, 7%pi/4....and the first minima occurs where e=3%pi/4=2.356.
#Calculations
x=min/math.sqrt((2*math.pi/(2*b*t))); #depth of digging of earth to find the temperature wave, m
#Results
print "Depth of digging of earth is :",round(x,3)," m, if we dug in the earth, we would find it growing older until it reached a maximum coldness at a depth of about 2.8 m.Farther down, it would begin to warm up again, but nt much. in midwinter, the reverse would be true \n"
```

In [15]:

```
from __future__ import division
import math
#Variables
l=0.08; #distance between metal walls,m
k=0.12; #thermal conductivity of insulating material, w/(m*K)
l1=0.04; #length of ribs,m
l2=0.14; #projected legth of wall,m
T1=40; #temperature of 1st wall,C
T2=0; #temperature of wall, C
hc=6.15; #Heat flow channels
ii=5.6; #Isothermal increments
#Calculations
#by looking at the configuration plot, there are approximately 5.6 isothermal increments and 6.15 flow channels.
Q=2*(hc/ii)*k*(T1-T2); #factor of 2 accounts for the fact that there are two halves in the section.
T=2.1/ii*(T1-T2); #(using proportionality and fig 5.23)Temperature in the middle of of wall, 2 cm from a rib, °C
#Results
print "Temperature in the middle of of wall, 2 cm from a rib is :",T,"C\n"
```

In [4]:

```
from __future__ import division
import math
#Variables
r=3; # radius ratio of one-quarter section of cylinder
#Calculations
S=math.pi/(2*math.log(r)); # shape factor
#Results
print "Shape factor is :",round(S,3),"\nThe quarter cylinder will be pictured for the radius ratio of 3, but for the different sizes, in both the cases it will be 1.43.\n"
```

In [3]:

```
from __future__ import division
import math
#Variables
Q=14; #steady heat transfer,W
D=0.06; #diameter of heat source,m
l=0.3; # length of source below surface ,m
T=308; #temperature of heat source,K
T1=294; #temperature of surface,K
#Calculations
k=(Q/(T-T1))*(1-(D/2)/(D*10))/(4*3.14*D/2)+0.025;# thermal conductivity of soil, W/m.K
#Results
print "Thermal conductivity is :",round(k,3),"W/(m*K)\n"
```

In [2]:

```
from __future__ import division
import math
#Variables
l=0.04; # length of square rod, m
T1=373; # temerature of rod, K
T2=293; # temperature of coolant,K
h=800; #convective heat transfer coefficient, W/(m**2*K)
a1=0.93; # ratio of temperature difference for Fo1=0.565, Bi1=0.2105, (x/l)1=0
a2=0.91; # ratio of temperature difference for Fo2=0.565, Bi2=0.2105, (x/l)2=0.5
#Calculations
a=a1*a2; #ratio of temperature difference at the axial line of interest
T=(T1-T2)*a+T2; #temperature on a line 1 cm. from one side and 2 cm. from the adjoining side after 10 sec. in K
Ta=T-273; #in °C
#Results
print "Temperature is : ",Ta,"C\n"
```

In [1]:

```
from __future__ import division
import math
#Variables
T1=373; # temperature of iron rod,K
T2=293; # temperature of coolant,K
#Biot no., Bi1=Bi2=0.2105,Fo1=Fo2=0.565
a1=0.10; #Fin effectiveness
a2=0.10; #Fin effectiveness
#Calculations
a=a1+a2*(1-a1);
T=(T1-T2)*(1-a)+T2; #mean temperature,K
Ta=T-273; #mean temperature in °C
#Results
print "Mean temperature is :",Ta,"C\n"
```