# Chapter 12: Radient Heat Transfer between Surfaces¶

### Example 12.3 Page No 598¶

In [3]:
a_13=100.0
b_13=250.0
c_13=100.0
X_13=a_13/c_13
Y_13=b_13/c_13
Fd1_3=0.17      # value for Fd1_3 corresponding to above calculated values of a/c and b/c
a_14=300
b_14=50
c_14=100

X_14=a_14/c_14
Y_14=b_14/c_14
Fd1_4=0.11       #value for Fd1_4 corresponding to above calculated values of a/c and b/c
a_15=100
b_15=50
c_15=100
X_15=a_15/c_15
Y_15=b_15/c_15
Fd1_5=0.09       #value for Fd1_3 corresponding to above calculated values of a/c and b/c
Fd1_2=Fd1_3+Fd1_4-Fd1_5
sigma=0.1714e-8  # Stefan-Boltzmann constant
T1=660
T2=560
q12_A1=sigma*Fd1_2*(T1**4-T2**4)

print"The net heat transferred is ",round(q12_A1,1),"BTU/(hr.sq.ft)"

The net heat transferred is  29.8 BTU/(hr.sq.ft)


### Example 12.4 Page No 601¶

In [6]:
import math
L1=1
angle=math.pi*45/180.0
L2=L1*math.sin(angle)
L3=L2
T1=303
T2=473

sigma=5.67e-8      # Stefan-Boltzmann constant
q21_A2=sigma*(T2**4-T1**4)*((L1/L2)+1-(L3/L2))/2.0
q31_A3=sigma*(T2**4-T1**4)*((L1/L2)-1+(L3/L2))/2.0

print"The heat transferred from A3 to A1 is ",round(q31_A3,0)," W/sq.m"

The heat transferred from A3 to A1 is  1669.0  W/sq.m


### Example 12.5 Page No 605¶

In [8]:
ac=1
bd=1

uncrossed_strings=ac+bd
L1_F12=(1/2.0)*(crossed_strings-uncrossed_strings)
L1=3
F12=L1_F12/L1
sigma=5.67e-8 # Stefan-Boltzmann constant
T1=560
T2=460
q12_A1=sigma*(T1**4-T2**4)*F12

print"The heat transfer rate is ",round(q12_A1,0),"W/sq m"

The heat transfer rate is  2189.0 W/sq m


### Example 12.6 Page No 608¶

In [16]:
T1=1000.0
T3=500.0
q2=0
F12=1/2.0
F13=1/2.0
F21=1/2.0
F23=1/2.0
F31=1/2.0
F32=1/2.0

T2=((T1**4+T3**4)/2.0)**(1/4.0) # using equation (2)
sigma=0.1714e-8                 # Stefan-Boltzmann constant
q1_A1=sigma*((T1**4-T2**4)*F12+(T1**4-T3**4)*F13) # using equation (1)
q3_A3=sigma*((T3**4-T1**4)*F31+(T3**4-T2**4)*F32) # using equation (3)

print"The temperature is ",round(T2,1),"R"
print"The heat flux through area A1 is",round(q1_A1,0),"BTU/(hr.sq.ft)"
print"The heat flux through area A3 is",round(q3_A3,0),"BTU/(hr.sq.ft)"
print"In the book there is calculation mistake"

The temperature is  853.7 R
The heat flux through area A1 is 1205.0 BTU/(hr.sq.ft)
The heat flux through area A3 is -1205.0 BTU/(hr.sq.ft)
In the book there is calculation mistake


### Example 12.7 Page No 613¶

In [21]:
F12=1
F21=1
F11=0             # the surfaces are flat
F22=0
emissivity1=0.94  # for oxidized steel from appendix table E1
emissivity2=0.94
T1=533
T2=323
sigma=5.67e-8      # Stefan-Boltzmann constant

q1=(sigma*(T1**4-T2**4))/((1/emissivity1)+(1/emissivity2)-1)
q2=-q1

print"The heat lost through bottom surface is ",round(q1,1),"W/sq m"
print"The heat lost through top surface is ",round(q2,1),"W/sq m"

The heat lost through bottom surface is  3510.7 W/sq m
The heat lost through top surface is  -3510.7 W/sq m


### Example 12.8 Page No 616¶

In [25]:
D=1.0           # diameter in ft
L=6/12.0        # length in ft

A=2*math.pi*D**2/4+math.pi*D*L
F12=1           # the view factor between the dish and the surroundings is unity
T1=810
T2=530
sigma=0.1714e-8 # Stefan-Boltzmann constant
q1=sigma*A*(T1**4-T2**4)*F12

F11=0
e1=0.82
e2=0.93
q1_=A*e1*(sigma*T1**4-F12*sigma*T2**4)

print"(a)The heat exchanged between the dish and the surroundings is",round(q1,0),"BTU/hr"
print"(b)The heat exchanged between the dish and the surroundings for the second case is ",round(q1_,0),"BTU/hr"

(a)The heat exchanged between the dish and the surroundings is 1893.0 BTU/hr
(b)The heat exchanged between the dish and the surroundings for the second case is  1552.0 BTU/hr

In [ ]: