In [1]:

```
print"\t approximate values are mentioned in the book \n";
#given
T1=1000+460; # R
T2=800+460; # R
#solution
Q=((0.173)*((14.6)**4-(12.6)**4)); # using eq.4.24,Btu/(hr)*(ft**2)
#results
print"\t heat removed from colder wall per unit area is : Btu/(hr)*(ft**2) ",round(Q,0)
# end
```

In [2]:

```
print"\t approximate values are mentioned in the book \n"
#given
T1=1000+460; # R
T2=800+460; # R
e1=0.6; # emissivity of hotter wall
e2=0.8; # emissivity of colder wall
#solution
Q=(((0.173)/((1/0.6)+(1/0.8)-1))*((14.6)**4-(12.6)**4)); # using eq.4.26,heat loss per unit area,Btu/(hr)*(ft**2)
#results
print"\t heat removed from colder wall per unit area is : Btu/(hr)*(ft**2) ",round(Q,0)
print"\t For perfect black bodies the value was 3500 Btu/(hr)(ft**2) "
# end
```

In [5]:

```
print"\t approximate values are mentioned in the book \n"
#given
T1=125.+460.; # R
T2=70.+460.; # R
e=0.9; # emissivity,using table 4.1B
#solution
from math import pi
A=(pi)*(3.375/12)*(1); # area,ft**2/lin ft
print"\t area is : ft**2/lin ft ",round(A,2)
Q=(0.9)*(0.88)*(0.173)*((T1/100)**4-(T2/100)**4); # heat loss using eq.4.32,Btu/(hr)*(lin ft)
print"\t heat loss is : Btu/(hr)*(lin ft) ",round(Q,1)
hr=(Q)/((A)*(T1-T2)); # fictitious film coefficient,using eq 4.33,Btu/(hr)(ft**2)(F)
print"\t fictitious film coefficient is : Btu/(hr)(ft**2)(F) ",round(hr,2)
#end
```

In [4]:

```
print"\t approximate values are mentioned in the book \n"
#given
T1=300+460; # R
T2=75+460; #R
A1=0.622; # area from table 11 in the appendix A,ft**2/lin ft
A2=4*(1*1); # surface area of duct,ft**2/lin ft
e1=0.79; # emissivity of oxidized steel from table 4.1
e2=0.276; # emissivity of oxidized zinc from table 4.1
print"\t surface area of pipe is : ft**2/lin ft ",round(A1,2)
print"\t surface area of duct is : ft**2/lin ft ",A2
print"\t The surface of the pipe is not negligible by comparison with that of the duct, and(f) of Table 4.2 applies most nearly"
Fa=1; # from table 4.2
Fe=((1)/((1/e1)+((A1/A2)*((1/e2)-1)))); # from table 4.2
print"\t Fe is : ",round(Fe,2)
Q=(0.173*10**-8)*(Fa)*(Fe)*(A1)*((T1)**4-(T2)**4); # heat loss due to radiation,Btu/(hr)*(lin ft)
print"\t heat loss due to radiation is : Btu/(hr)*(lin ft) ",round(Q,0)
# end
```