In [2]:

```
# Variables
T1 = 1660./100; # Temperature of first black plane in degR
T2 = 1260./100; # Temperature of second black plane in degR
s = 0.174; # Stephan Boltzman's consmath.tant
# Calculations
q = s*(T1**4-T2**4);
# Results
print "The net radiant interchange between two bodies of unit area is %d Btu/hr-ft**2"%(q);
# book answer is rounded off.
```

In [5]:

```
# Variables
A1 = 15.*15; # Area of floor in ft**2
A2 = A1; # Area of roof in ft**2
T1 = 2460./100; # Temperature of floor in degR
T2 = 1060./100; # temperature of roof in degR
s = 0.174; # Stephan Boltzman's consmath.tant
# S/L = 1.5, So considering graph F12 = 0.31
# Calculations
F12 = 0.31;
q = s*F12*A1*(T1**4-T2**4);
# Results
print "The net radiant interchange between two bodies of unit area is %d Btu/hr-ft**2"%(q);
# note : book answer is rounded off.
```

In [7]:

```
# Variables
x = 6.; # length of wall in ft
y = 12.; # breadth of wall in ft
z = 18.; # height of wall in ft
A1 = x*y;
s = 0.174; # Stephan Boltzman's consmath.tant
T1 = 1000.; # Temperature of floor in degF
T2 = 500.; # Temperature of wall in degF
Y = y/x; # Ratios
Z = z/x;
# Calculations
# Seeing the graph, F12 could be found out
F12 = 0.165;
q12 = s*F12*A1*((((T1+460)/100)**4)-((T2+460)/100)**4); # Radiant interchange
# Results
print "The net radiant interchange between two bodies of unit area is %d Btu/hr-ft**2"%(q12);
# note : book answer is rounded off.
```

In [9]:

```
import math
# Variables
D = 10./12; # Diameter of black disc
L = 5./12; # distance between two discs
T1 = (1500.+460)/100; # Temperature of disc 1 in degR
T2 = (1000.+460)/100; # Temperature of disc 2 in degR
# From the ratio of S/L, the value of F1r2 can be found out
F1r2 = 0.669; # Shape factor
# Calculations
A1 = math.pi*D*D/4; # Area of disc 1 in ft**2
A2 = math.pi*D*D/4; # Area of disc 2 in ft**2
s = 0.174; # Stephan Boltzman's consmath.tant
q12 = s*F1r2*A1*((T1**4)-(T2**4)); # Radiant interchange in Btu/hr
# Results
print "The net radiant interchange between two parallel black discs is %d Btu/hr"%(q12);
# note : book answer is rounded off
```

In [11]:

```
import math
# Variables
T1 = (1500.+460)/100; # Temperature of plane 1 in degR
T2 = (1000.+460)/100; # Temperature of plane 2 in degR
e1 = 0.8; # Emmisivity for higher temperature
e2 = 0.6; # Emmisivity for lower temperature
s = 0.174; # Stephan Boltzman's consmath.tant
D = 10./12; # Diameter of disc in ft
# Calculations
A = math.pi/4*D**2; # Area of disc in ft**2
F1r2 = 0.669;
F1r2g = 1/((1/F1r2)+(1/e1)+(1/e2)-2); # Shape factor
q12 = s*F1r2g*A*((T1**4)-(T2**4)); # Radiant interchange in Btu/hr
# Results
print "The net radiant interchange between two parallel very large planes per square foot is %d Btu/hr"%(q12);
# book answer is rounded off.
```

In [13]:

```
# Variables
T1 = 1460./100; # Temperature of first black plane in degK
T2 = 1060./100; # temperature of second black plane in degK
s = 0.174; # Stephan Boltzman's consmath.tant
e1 = 0.9; # Emmisivity for higher temperature
e2 = 0.7; # Emmisivity for higher temperature
# Calculations
F1r2 = 1./((1/e1)+(1/e2)-1); # Shape factor
q = s*F1r2*(T1**4-T2**4);
# Results
print "The net radiant interchange between two bodies of unit area is %d Btu/hr-ft**2"%(round(q,-1));
```

In [14]:

```
import math
# Variables
e = 0.8; # emmisivity of pipe metal
D = 2.375/12; # Diameter of pipe in ft
s = 0.174; # Stephans Boltzman's consmath.tant
T1 = (300.+460)/100; # Temperature of disc 1 in degF
T2 = (80.+460)/100; # Temperature of disc 2 in degF
# Calculations
A1 = math.pi*D; # Area of one foot of pipe in ft**2
q12 = s*e*A1*((T1**4)-(T2**4)); # Radiant interchange in Btu/hr
# Results
print "The net radiant interchange per foot length of pipe is %.1f Btu/hr-ft"%(q12);
```