In [1]:

```
#given
Tavg=900; # average temperature of the wall,F
k=0.15; # Thermal conductivity at 932 F,Btu/(hr)(ft^2)(F/ft)
T1=1500; # hot side temperature,F
T2=300; # cold side temperature,F
A=192; # surface area,ft^2
L=0.5; # thickness,ft
#solution
Q=(k)*(A)*(T1-T2)/L; # formula for heat,Btu/hr
print " heat is Btu/hr ",Q
print"\t approximate values are mentioned in the book \n"
#end
```

In [3]:

```
#given
La=0.66; # Thickness of firebrick wall,ft
Lb=0.33; # Thickness of insulating brick wall,ft
Lc=0.5; # Thickness of building brick wall,ft
Ka=0.68; # themal conductivity of firebrick,Btu/(hr)*(ft^2)*(F/ft)
Kb=0.15; # themal conductivity of insulating brick,Btu/(hr)*(ft^2)*(F/ft)
Kc=0.40; # themal conductivity of building brick,Btu/(hr)*(ft^2)*(F/ft)
A=1.; # surface area,ft^2
Ta=1600.; # temperature of inner wall,F
Tb=125.; # temperature of outer wall.F
#solution
Ra=La/(Ka)*(A); # formula for resistance,(hr)*(F)/Btu
print"\t resistance offered by firebrick : (hr)*(F)/Btu ",round(Ra,2)
Rb=Lb/(Kb)*(A); # formula for resistance,(hr)*(F)/Btu
print"\t resistance offered by insulating brick : (hr)*(F)/Btu ",round(Rb,2)
Rc=Lc/(Kc)*(A); # formula for resistance,(hr)*(F)/Btu
print"\t resistance offered by buildingbrick : (hr)*(F)/Btu ",round(Rc,2)
R=Ra+Rb+Rc; # total resistance offered by three walls,(hr)*(F)/Btu
print"\t total resistance offered by three walls : (hr)*(F)/Btu ",round(R,2)
Q=(1600-125)/4.45; # using formula for heat loss/ft^2,Btu/hr
print"\t heat loss/ft^2 : Btu/hr ",round(Q,0)
# T1,T2 are temperatures at interface of firebrick and insulating brick, and insulating brick and building brick respectively,F
delta=(Q)*(Ra); # formula for temperature difference,F
print"\t delta is : F ",round(delta,0)
T1=Ta-((Q)*(Ra)); # temperature at interface of firebrick and insulating brick,F
print"\t temperature at interface of firebrick and insulating brick F ",round(T1,0)
deltb=Q*(Rb);
print"\t deltb is : F ",round(deltb,0)
T2=T1-((Q)*(Rb)); #temperature at interface of insulating brick and building brick,F
print"\t temperature at interface of insulating brick and building brick F ",round(T2,0)
print"\t approximate values are mentioned in the book \n"
#end
```

In [4]:

```
print"\t approximate values are mentioned in the book \n"
#given
Lair=0.25/12; # thickness of air film,ft
Kair=0.0265; # thermal conductivity of air at 572F,Btu/(hr)*(ft^2)(F/ft)
A=1; # surface area,ft^2
#solution
Rair=Lair/(Kair*(A)); # resistance offered by air film, (hr)(F)/Btu
print"\t resistance offered by air film (hr)(F)/Btu ",round(Rair,2)
R=4.45; # resistance from previous example 2.2,(hr)(F)/Btu
Rt=(R)+Rair; # total resistance,(hr)(F)/Btu
print"\t total resistance (hr)(F)/Btu ",round(Rt,2)
Ta=1600; # temperature of inner wall,F
Tb=125; # temperature of outer wall,F
Q=(1600-125)/Rt; # heat loss, Btu/hr
print"\t heat loss Btu/hr ",round(Q,0)
```

In [1]:

```
#given
k=0.63; # thermal conductivity of pipe, Btu/(hr)*(ft^2)*(F/ft)
Do=6. # in
Di=5. # in
Ti=200.;# inner side temperature,F
To=175.; # outer side temperature,F
#solution
import math
from math import log
q=(2*(3.14)*(k)*(Ti-To))/(log (Do/Di)); # formula for heat flow,Btu/(hr)*(ft)
print"\t heat flow is : Btu/(hr)*(ft) ",round(q,0)
print"\t approximate values are mentioned in the book \n"
# caculation mistake in book
# end
```

In [1]:

```
print"\t approximate values are mentioned in the book \n"
#given
t1=150; # assume temperature of outer surface of rockwool,F
ta=70; # temperature of surrounding air,F
ha=2.23; # surface coefficient,Btu/(hr)*(ft^2)*(F)
#solution
import math
from math import log
q=(3.14)*(300-70)/(((1/(2*0.033))*log(3.375/2.375))+(1/((2.23)*(3.375/12)))); # using formula for heat loss,Btu/(hr)*(lin ft), calculation mistake
print"\t heat loss for linear foot is : Btu/(hr)*(lin ft) ",round(q,1)
print"\t Check between ts and t1, since delt/R = deltc/Rc "
t1=300-(((104.8)*((1)*(log(3.375/2.375))))/((2)*(3.14)*(.033))); # using eq 2.31,F
print"\t t1 is : F ",t1
t1=125; # assume temperature of outer surface of rockwool,F
ha=2.10; # surface coefficient,Btu/(hr)*(ft^2)*(F)
q=((3.14)*(300-70))/(((1/(2*0.033))*log(3.375/2.375))+(1/((2.10)*(3.375/12)))); # using formula for heat loss,Btu/(hr)*(lin ft)
print"\t heat loss for linear foot is : Btu/(hr)*(lin ft)",round(q,1)
print"\t Check between ts and t1, since delt/R = deltc/Rc "
t1=300-(((103)*((1)*(log(3.375/2.375))))/((2)*(3.14)*(.033))); # using eq 2.31,F
print"\t t1 is : F ",round(t1,1)
# end
```