# Chapter 11 : Heat transfer by radiation¶

## Example 11.1 page : 222¶

In :
# 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.

The net radiant interchange between two bodies of unit area is 8826 Btu/hr-ft**2


## Example 11.2 page : 224¶

In :
# 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.

The net radiant interchange between two bodies of unit area is 4291391 Btu/hr-ft**2


## Example 11.3 page : 225¶

In :
# 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.

The net radiant interchange between two bodies of unit area is 76367 Btu/hr-ft**2


## Example 11.4 page : 227¶

In :
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

The net radiant interchange between two parallel black discs is 6484 Btu/hr


## Example 11.5 page : 228¶

In :
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.

The net radiant interchange between two parallel very large planes per square foot is 4019 Btu/hr


## Example 11.6 page : 230¶

In :
# 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));

The net radiant interchange between two bodies of unit area is 3710 Btu/hr-ft**2


## Example 11.7 page : 231¶

In :
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);

The net radiant interchange per foot length of pipe is 215.2 Btu/hr-ft