# Chapter 4: Analysis of heat conduction and some steady one-dimensional problems¶

### Example 4.8, Page number: 172¶

In :
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."

The exact tip temperature is : 111.428 C

Approximate tip temperature is :  114.659  C

Thus the insulated tip approximation is adequate for the computation in this case.


### Example 4.9, Page number: 174¶

In :
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"

Actual temperature of root is : 143.8 C , the correction is modest in this



### Example 4.10, Page number: 178¶

In :
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"

Resistor temperature is : 62.68 C or about 10 C lower than before.



### Example 4.11, Page number: 181¶

In :
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"

Heat trnsferred without fins is : 199.0 W/m

Heat transferred with fins is : 478.0 W/m or 4.02 times heat loss without fins.