In [1]:

```
from __future__ import division
import math
#Variables
d=0.02; #diameter of alluminium rod,m
k=205; #thermal conductivity of rod,W/(m.K)
l=0.08; #length of rod, m
T1=423; #wall temperature, K
T2=299; #air temperatutre, K
h=120; #convective coefficient, W/(m**2*K)
#Calculations
mL=math.sqrt(h*(math.pow(l,2))/(k*d/4)); # formula for mL=((h*Perimeter*l**2)/(k*Area))**0.5
Bi=h*l/k #Biot no.
a1=(math.cosh(0)+(Bi/mL)*math.sinh(0))/(math.cosh(mL)+(Bi/mL)*math.sinh(mL));#formula for temperature difference T-Ttip
Ttip1=T2+a1*(T1-T2); # exact tip temperature, C
Tt1=Ttip1-273; #Exact tip temp., K
a2=(math.cosh(0)+(Bi/mL)*math.sinh(0))/(math.cosh(mL));#dimensionless temp. at tip if heat transfer from the tip is not considered
Ttip2=T2+a2*(T1-T2); #Approximate tip temp., K
Tt2=Ttip2-273; #Approximate tip temp., C
#Results
print "The exact tip temperature is :",round(Tt1,3),"C\n"
print "Approximate tip temperature is : ",round(Tt2,3)," C\n"
print "Thus the insulated tip approximation is adequate for the computation in this case."
```

In [2]:

```
from __future__ import division
import math
#Variables
T1=423; #wall temperature, K
d=0.02; #diameter of alluminium rod,m
k=205; #thermal conductivity of rod,W/(m.K)
l=0.08; #length of rod, m
T2=299; #air temperatutre, K
h=120; #convective coefficient, W/(m**2*K)
mL=0.8656;
#Calculations
mr=mL*(d/(2*l)); # by looking at graph of 1-Qact/Q(no temp.depression) vs. mr*math.tanh(mL), we can find out the value of Troot. 1-Qact./Q(no temp. depression) = 0.05 so heat flow is reduced by 5 percent
Troot=T1-(T1-T2)*0.05; #Actual temperature of root, K (0.05 is from graph)
Tr=Troot-273; #Actual temperature of root, °C
#Results
print "Actual temperature of root is :",Tr,"C , the correction is modest in this \n"
```

In [11]:

```
from __future__ import division
import math
#Variables
T1=308; #air temperature, K
Q=0.1; # heat transferred,W
k=16; #thermal conductivity of wires, W/(m*K)
d=0.00062; #diameter of wire,m
Heff=23; #convection coefficient, W/(m**2*K)
A=1.33*math.pow(10,-4); #Aera of resistor surface, m^2 (from example 2.8)
#the wires act actn as very long fins connected to ressistor hence math.tanh(mL)=1
#Calculations
R1=1/math.sqrt(k*Heff*math.pow(math.pi,2)*math.pow(d,3)/4); #Fin resistance, K/W
Req=math.pow((1/R1+1/R1+7.17*A+13*A),-1); #the 2 thermal ressistances are in parallel to the thermal ressistance for natural...
#convection and thermal radiation from the ressistor's surface found in previous eg.
Tres=T1+Q*Req; #Resistor temperature, K
Trs=Tres-273; #Resistor temperature, °C
#Results
print "Resistor temperature is :",round(Trs,2),"C or about 10 C lower than before.\n"
```

In [9]:

```
from __future__ import division
import math
#Variables
D1=0.03; # outer diameter, m
T1=358; #hot water temperature, K
t1=0.0008; #thickness of fins, m
D2=0.08; # diameter of fins, m
t2=0.02; # spacing between fins, m
h1=20; # convection coefficient, W/(m**2*K)
h2=15; #convection coefficient with fins, W/(m**2*K)
To=295; #surrounding temperature, K
#Calculations
Q=math.pi*D1*h1*(T1-To); # if fins are not added.
Q1=math.ceil(Q1); #heat loss without fins,W/m
# we set wall temp.=water temp..since the wall is constantly heated by water, we should not have a root temp. depression problem after the fins are added.hence by looking at the graph, ml(l/Perimeter)**0.5=(h*(D2/2-D1/2)/(125*0.025*t1)) = 0.306, we obtain n(efficiency)=89 percent
Qfin=math.ceil(Q*(t2-t1)/t2 + 0.89*(2*3.14*(math.pow(D2,2)/4-math.pow(D1,2)/4))*50*h2*(T1-To)) #Heat transferred with fins, K/W
#Results
print "Heat trnsferred without fins is :",Q1,"W/m\n"
print "Heat transferred with fins is :", round(Qfin,3),"W/m or 4.02 times heat loss without fins.\n"
```