# Chapter 8 Radiation Heat Transfer¶

## Exa 8.1¶

In [1]:
#Example Number 8.1
# transmission and absorption in a gas plate

# Variable declaration

T = 2000+273 				# [K] furnace temperature
L = 0.3 				# [m] side length of glass plate
t1 = 0.5 				# transmissivity of glass betweenb/n lambda1 					to lambda2
lambda1 = 0.2 				# [micro m]
lambda2 = 3.5 				# [micro m]
E1 = 0.3 				# emissivity of glass upto lambda2
E2 = 0.9 				# emissivity of glass above lambda2
t2 = 0 					# transmissivity of glass except in the range 					of lambda1 to lambda2

#Calculation

sigma = 5.669*10**(-8) 			# [W/square meter K**(4)]
A = L**(2) 				# [square meter] area of glass plate
# calculating constants to use table 8-1(page no.-379-380)
K1 = lambda1*T 				# [micro m K]
K2 = lambda2*T 				# [micro m K]
# from table 8-1
Eb_0_lam1_by_sigmaT4 = 0
Eb_0_lam2_by_sigmaT4 = 0.85443
Eb = sigma*T**(4) 			# [W/square meter]
# for 0.2 micro m to 3.5 micro m
TIR = Eb*(Eb_0_lam2_by_sigmaT4-Eb_0_lam1_by_sigmaT4)*A 	# [W]
TRT = t1*TIR 				# [W]
RA1 = E1*TIR 				# [W] for 0<lambda<3.5 micro m
RA2 = E2*(1-Eb_0_lam2_by_sigmaT4)*Eb*A # [W] for 3.5 micro m <lambda< infinity
TRA = RA1+RA2 				# [W]

#Result

print "Total energy absorbed =",round(TRA/1000,2),"kW"
print "total energy transmitted=",round(TRT/1000,1),"kW"

Total energy absorbed = 52.75 kW
total energy transmitted= 58.2 kW


## Exa 8.2¶

In [2]:
#Example Number 8.2
# heat transfer between black surfaces

# Variable declaration

L = 1  					# [m] length of black plate
W = 0.5  				# [m] width of black plate
T1 = 1000+273  				# [K] first plate temperature
T2 = 500+273  				# [K] second plate temperature
sigma = 5.669*10**(-8)  		# [W/square meter K**(4)]
# the ratios for use with figure 8-12(page no.-386) are
Y_by_D = W/W
X_by_D = L/W
# so that
F12 = 0.285  				# radiation shape factor
# the heat transfer is calculated from
q = sigma*L*W*F12*(T1**(4)-T2**(4))
print "net radiant heat exchange between the two plates is",round(q/1000,2),"kW"

net radiant heat exchange between the two plates is 18.33 kW


## Exa 8.3¶

In [3]:
#Example Number 8.3
# shape-factor algebra for open ends of cylinder

# Variable declaration

d1 = 0.1 				# [m] diameter of first cylinder
d2 = 0.2 				# [m] diameter of second cylinder
L = 0.2 				# [m] length of cylinder
# we use the nomenclature of figure 8-15(page no.-388) for this 		problem and designate the open ends as surfaces 3 and 4.
# we have
L_by_r2 = L/(d2/2)
r1_by_r2 = 0.5
# so from figure 8-15 or table 8-2(page no.-389) we obtain
F21 = 0.4126
F22 = 0.3286
# using the reciprocity relation (equation 8-18) we have

#Calculation
F12 = (d2/d1)*F21
# for surface 2 we have F12+F22+F23+F24 = 1.0
# and from symmetry F23 = F24 so that
F23 = (1-F21-F22)/2
F24 = F23
# using reciprocity again,
import math

A2 = math.pi*d2*L 			# [m**2]
A3 = math.pi*(d2**2-d1**2)/4 		# [m**2]
F32 = A2*F23/A3
# we observe that F11 = F33 = F44 = 0 & for surface 3 F31+F32+F34 =1.0
# so, if F31 can be determined, we can calculate the desired quantity 		F34. for surface 1 F12+F13+F14 = 1.0
# and from symmetry F13 = F14 so that
F13 = (1-F12)/2
F14 = F13
# using reciprocity gives
A1 = math.pi*d1*L 			# [square meter]
F31 = (A1/A3)*F13
# then
F34 = 1-F31-F32

#Result
print "Shape factor between the open ends of the cylinder is",F34

Shape factor between the open ends of the cylinder is 0.0768


## Exa 8.4¶

In [4]:
#Example Number 8.4
# shape-factor algebra for truncated cone

#Variable declaration

d1 = 0.1 			# [m] diameter of top of cone
d2 = 0.2 			# [m] diameter of bottom of cone
L = 0.1 			# [m] height of cone
#we employ figure 8-16(page no.-390) for solution of this problem and 		take the nomenclature as shown, designating the top as surface 2,
# the bottom as surface 1, and the side as surface 3. thus the desired 		quantities are F23 and F33. we have
Z = L/(d2/2)
Y = (d1/2)/L
# thus from figure 8-16(page no.-390)
F12 = 0.12
# from reciprcity(equatin 8-18)

import math
A1 = math.pi*d2**(2)/4 		# [square meter]
A2 = math.pi*d1**(2)/4 		# [square meter]
F21 = A1*F12/A2
#and
F22 = 0
# so that
F23 = 1-F21
# for surface 3 F31+F32+F33 = 1,  so we must find F31 and F32 in order 		to evaluate F33. since F11 = 0 we have
F13 = 1-F12
# and from reciprocity
A3 = math.pi*((d1+d2)/2)*((d1/2-d2/2)**(2)+L**(2))**(1.0/2.0) 	# [square meter]
# so from above equation
F31 = A1*F13/A3
# a similar procedure is applies with surface 2 so that
F32 = A2*F23/A3
# finally from above equation
F33 = 1-F32-F31
print "Shape factor between the top surface and the side is",F23
print "Shape factor between the side and itself is",round(F33,3)

Shape factor between the top surface and the side is 0.52
Shape factor between the side and itself is 0.398


## Exa 8.5¶

In [5]:
#Example Number 8.5
# shape-factor algebra for cylindrical reflactor

# Variable declaration

d = 0.6 				# [m] diameter of long half-circular cylinder
L = 0.2 				# [m] length of square rod
# we have given figure example 8-5(page no.-397) for solution of this 			problem and take the nomenclature as shown,
# from symmetry we have
F21 = 0.5
F23 = F21
# in general, F11+F12+F13 = 1. to aid in the analysis we create the 			fictious surface 4 shown in figure example 8-5 as dashed line.
# for this surface
F41 = 1.0
# now, all radiation leaving surface 1 will arrive either at 2 or at 			3. likewise,this radiation will arrive at the imaginary surface 4, so 			that F41 = F12+F13 say eqn a
# from reciprocity
#Calculation

import math

A1 =math.pi*d/2 			# [square meter]
A4 = L+2*math.sqrt(0.1**(2)+L**(2)) 	# [square meter]
A2 = 4*L 				# [square meter]
# so that
F14 = A4*F41/A1 			# say eqn b
# we also have from reciprocity
F12 = A2*F21/A1 			# say eqn c
# combining a,b,c, gives
F13 = F14-F12
# finally
F11 = 1-F12-F13

#Result

print "Value of F12 is ",round(F12,3)
print "Value of F13 is ",round(F13,3)
print "Value of F11 is ",round(F11,3)

Value of F12 is  0.424
Value of F13 is  0.262
Value of F11 is  0.313


## Exa 8.7¶

In [7]:
#xample Number 8.7

# Variable declaration

w = 0.5 			# [m] width of plate
L = 0.5 			# [m] length of plate
sigma = 5.669*10**(-8) 		# [W/square meter K**(4)]
# from the data of the problem
T1 = 1000 			# [K] temperature of first surface
T2 = 27+273 			# [K] temperature of room
A1 = w*L 			# [square meter] area of rectangle
A2 = A1 			# [square meter] area of rectangle
E1 = 0.6 			# emissivity of surface 1
#eq(8-41) may not be used for the calculation because 	one of the heat-exchanging surfaces is not convex. The radiation 			network is shown in figure example 8-7(page no.-404) where surface 3 is the 		room and surface 2 is the insulated surface. n	J2 "floats" in the network and is determined from the overall radiant balance.

# from figure 8-14(page no.-387) the shape factors are
F12 = 0.2
F21 = F12
# because
F11 = 0
F22 = 0
F13 = 1-F12
F23 = F13
# the resistances are
R1 = (1-E1)/(E1*A1)
R2 = 1/(A1*F13)
R3 = 1/(A2*F23)
R4 = 1/(A1*F12)
# we also have
Eb1 = sigma*T1**(4) 		# [W/square meter]
Eb3 = sigma*T2**(4) 		# [W/square meter]
J3 = Eb3 			# [W/square meter]
# the overall ckt is a series parallel arrangement and the heat transfer is
R_equiv = R1+(1/((1/R2)+1/(R3+R4)))
q = (Eb1-Eb3)/R_equiv 		 # [W]
# this heat transfer can also be written as q = (Eb1-J1)/((1-E1)/(E1*A1))
# inserting the values
J1 = Eb1-q*((1-E1)/(E1*A1)) 	 # [W/square meter]
# the value of J2 is determined from proportioning the resistances between J1 		and J3, so that
# (J1-J2)/R4 = (J1-J3)/(R4+R2)
J2 = J1-((J1-J3)/(R4+R2))*R4 	 # [W/square meter]
Eb2 = J2			 # [W/square meter]
# finally, we obtain the temperature of the insulated surface as
T2 = (Eb2/sigma)**(1.0/4.0) 	 # [K]

print "Temperature of the insulated surface is",round(T2,1),"K"
print "Heat lost by the surface at 1000K is",round(q/1000,3),"kW"

Temperature of the insulated surface is 599.4 K
Heat lost by the surface at 1000K is 8.229 kW


## Exa 8.8¶

In [9]:
#Example Number 8.8
# open hemisphere in large room

# Variable declaration

d = 0.3 			# [m] diameter of hemisphere
T1 = 500+273 			# [degree celsius] temperature of hemisphere
T2 = 30+273 			# [degree celsius] temperature of enclosure
E = 0.4 			# surface emissivity of hemisphere
sigma = 5.669*10**(-8) 		# [W/square meter K**(4)] constant

# in the given figure example 8-8(page no.-407) we take the inside of the 		sphere as surface 1 and the enclosure as surface 2.
# we also create an imaginary surface 3 covering the opening.
# then the heat transfer is given by
#Calculation
Eb1 = sigma*T1**(4) 		# [W/square meter]
Eb2 = sigma*T2**(4) 		# [W/square meter]

import math
A1 = 2*math.pi*(d/2)**(2) 	# [square meter] area of surface 1
# calculating the surface resistance
R1 = (1-E)/(E*A1)
# since A2 tends to 0 so R2 also tends to 0
R2 = 0
# recognize that all of the radiation leaving surface 1 which 	will 	eventually arrive at enclosure 2 will also hit the imaginary   surface 			3(F12=F13). we also recognize that A1*F13 = A3*F31. but
F31 = 1.0
A3 = math.pi*(d/2)**(2) 	# [square meter]
F13 = (A3/A1)*F31
F12 = F13
# then calculating space resistance
R3 = 1/(A1*F12)
# we can claculate heat transfer by inserting the quantities in eq (8-40):
q = (Eb1-Eb2)/(R1+R2+R3) 	# [W]

#Result


Net radiant exchange is 798.0 W


## Exa 8.9¶

In [10]:
#Example Number 8.9
# effective emissivity of finned surface

# Variable declaration

# for unit depth in the z-dimension we have
A1 = 10 			# [square meter]
A2 = 5 				# [square meter]
A3 = 60 			# [square meter]
# the apparent emissivity of the open cavity area A1 is given by 	equation(8-47) as
# Ea1 = E*A3/[A1+E*(A3-A1)]
# for const surface emissivity the emitted energy from the total area A1+A2 is
# e1 = Ea1*A1+E*A2*Eb
# and the energy emitted per unit area for that total area is
# e_t = [(Ea1*A1+E*A2)/(A1+A2)]*Eb
# the coeff of Eb is the effective emissivity, E_eff of the combination of the 	surface and open cavity. inserting
# above equations gives the following values

#Calculation & Results

# for E = 0.2

E = 0.2
Ea1 = E*A3/(A1+E*(A3-A1))
E_eff = ((Ea1*A1+E*A2)/(A1+A2))

print "For emissivity of 0.2 the value of effective emissivity is",round(E_eff,3)

# for E = 0.5

E = 0.5
Ea1 = E*A3/(A1+E*(A3-A1))
E_eff = ((Ea1*A1+E*A2)/(A1+A2))

print " For emissivity of 0.5 the value of effective emissivity is ",round(E_eff,3)

# for E = 0.8

E = 0.8
Ea1 = E*A3/(A1+E*(A3-A1))
E_eff = ((Ea1*A1+E*A2)/(A1+A2))

print "For emissivity of 0.8 the value of effective emissivity is ",round(E_eff,3)

For emissivity of 0.2 the value of effective emissivity is 0.467
For emissivity of 0.5 the value of effective emissivity is  0.738
For emissivity of 0.8 the value of effective emissivity is  0.907


## Exa 8.10¶

In [11]:
#Example Number 8.10
# heat transfer reduction with parallel plate shield

# Variable declaration

E1 = 0.3 			# emissivity of first plane
E2 = 0.8 			# emissivity of second plane
E3 = 0.04 			# emissivity of shield

#Calculation
sigma = 5.669*10**(-8) 		# [W/square meter K**(4)]
# the heat transfer without the shield is given by
# q_by_A = sigma*(T1**4-T2**4)/((1/E1)+(1/E2)-1) = 		0.279*sigma*(T1**4-T2**4)
# T1 is temp of 1st plane & T2 is temperature of second plane
# the radiation network for the problem with the shield in place is 			shown in figure (8-32) (page no.-410).
# the resistances are
R1 = (1-E1)/E1
R2 = (1-E2)/E2
R3 = (1-E3)/E3
# the total resistance with the shield is
R = R1+R2+R3
# and the heat transfer is
# q_by_A = sigma*(T1**4-T2**4)/R = 0.01902*sigma*(T1**4-T2**4)

#Result

print "Heat tranfer is reduced by",round((((0.279-0.01902)/0.279)*100),1),"percent"

Heat tranfer is reduced by 93.2 percent


## Exa 8.11¶

In [13]:
#Example Number 8.11
# open cylindrical shield in large room

# Variable declaration

# two concentric cylinders of example(8.3) have
T1 = 1000 			# [K]
E1 = 0.8
E2 = 0.2
T3 = 300 			# [K] room temperature
sigma = 5.669*10**(-8) 		# [W/square meter K**(4)]
# refer to figure example 8-11(page no.-413) for radiation network
# the room is designed as surface 3 and J3 = Eb3, because the room is very 		large,(i.e.its surface is very small)
# in this problem we must consider the inside and outside of surface 2 and 		thus have subscripts i and o to designate the respective quantities.
# the shape factor can be obtained from example 8-3 as
F12 = 0.8253
F13 = 0.1747
F23i = 0.2588
F23o = 1.0
# also

#Calculations
import math
A1 = math.pi*0.1*0.2 		# [square meter] area of first cylinder
A2 = math.pi*0.2*0.2 		# [square meter] area of second cylinder
Eb1 = sigma*T1**4 		# [W/square meter]
Eb3 = sigma*T3**4 		# [W/square meter]
# the resistances may be calculated as
R1 = (1-E1)/(E1*A1)
R2 = (1-E2)/(E2*A2)
R3 = 1/(A1*F12)
R4 = 1/(A2*F23i)
R5 = 1/(A2*F23o)
R6 = 1/(A1*F13)
# the network could be solved as a series-parallel circuit to obtain the heat 		transfer, butwe will need the radiosities anyway, so we setup three nodal		equations to solve for J1,J2i, and J2o.
# we sum the currents into each node and set them equal to zero:
# node J1: (Eb1-J1)/R1+(Eb3-J3)/R6+(J2i-J1)/R3 = 0
# node J2i: (J1-J2i)/R3+(Eb3-J2i)/R4+(J2o-J2i)/(2*R2) = 0
# node J2o: (Eb3-J2o)/R5+(J2i-J2o)/(2*R2) = 0
# these equations can be solved by matrix method and the solution is
J1 = 49732 			# [W/square meter]
J2i = 26444 			# [W/square meter]
J2o = 3346 			# [W/square meter]
# the heat transfer is then calculated from
q = (Eb1-J1)/((1-E1)/(E1*A1)) 	# [W]
# from the network we see that
Eb2 = (J2i+J2o)/2		 # [W/square meter]
# and
T2 = (Eb2/sigma)**(1.0/4.0) 	# [K]
# if the outer cylinder had not been in place acting as a "shield" the heat 		loss from cylinder 1 could have been calculated from equation(8-43a) as
q1 = E1*A1*(Eb1-Eb3)		 # [W]

#Result

print "Temperature of the outer cylinder is",round(T2),"K"
print "Total heat lost by inner cylinder is",round(q1),"W"

Temperature of the outer cylinder is 716.0 K
Total heat lost by inner cylinder is 2826.0 W


## Exa 8.12¶

In [14]:
#Example Number 8.12
# network for gas radiation between parallel plates

# Variable declaration

T1 = 800 				# [K] temperature of first plate
E1 = 0.3 				# emissivity
T2 = 400 				# [K] temperature of second plate
E2 = 0.7 				# emissivity
Eg = 0.2 				# emissivity of gray gas
tg = 0.8 				# transmissivity of gray gas
sigma = 5.669*10**(-8) 			# [W/square meter K**(4)]
# the network shown in figure 8-39(page no.-419) applies to this problem. all 		the shape factors are unity for large planes and the various resistors can be 		computed on a unit area basis as

#Calculation
F12 = 1
F1g = 1
F2g = F1g
R1 = (1-E1)/E1
R2 = (1-E2)/E2
R3 = 1/(F12*(1-Eg))
R4 = 1/(F1g*Eg)
R5 = 1/(F2g*Eg)
Eb1 = sigma*T1**(4) 			# [W/square meter]
Eb2 = sigma*T2**(4) 			# [W/square meter]

# the equivalent resistance of the center "triangle" is

R = 1/((1/R3)+(1/(R4+R5)))

# the total heat transfer is then

q_by_A = (Eb1-Eb2)/(R1+R2+R) 		# [W/square meter]

# heat transfer would be given by equation (8-42):

q_by_A1 = (Eb1-Eb2)/((1/E1)+(1/E2)-1) 	# [W/square meter]

# the radiosities may be computed from q_by_A = (Eb1-J1)*(E1/(1-E1)) = 	(J2-Eb2)*(E2/(1-E2))

J1 = Eb1-q_by_A*((1-E1)/E1) 		# [W/square meter]
J2 = Eb2+q_by_A*((1-E2)/E2) 		# [W/square meter]

# for the network Ebg is just the mean of these values

Ebg = (J1+J2)/2 			# [W/square meter]

# so that the temperature of the gas is
Tg = (Ebg/sigma)**(1.0/4.0) 		# [K]

#Result

print "Heat-transfer rate between the two planes is",round(q_by_A),"W/square meter"
print "Temperature of the gas is",round(Tg,1),"K"
print "Ratio of heat-transfer with presence of gas to without presence of gas is",round(q_by_A/q_by_A1,2)

Heat-transfer rate between the two planes is 5621.0 W/square meter
Temperature of the gas is 592.4 K
Ratio of heat-transfer with presence of gas to without presence of gas is 0.97


## Exa 8.13¶

In [15]:
#Example Number 8.13
# cavity with transparent cover

#Variable declaration

E1 = 0.5 			# emissivity of rectangular cavity
t2 = 0.5 			# transmissivity
rho2 = 0.1 			# reflectivity
E2 = 0.4 			# emissivity
# from example 8-9 we have
# per unit depth in the z direction we have

#Calculation
A1 = 60
A2 = 10.0
# we may evaluate K from equation(8-96a)
K = E1/(t2+(E2/2))

# the value of Ea is then computed from equation (8-96) as
Ea = (t2+(E2/2))*K/((A2/A1)*(1-E1)+K)
print "Apparent emissivity of covered opening is ",round(Ea,4)
# when no cover present, the value of Ea would be given by eq (8-47) as
Ea1 = E1*A1/(A2+E1*(A1-A2))

#Result
print "If there were no cover present, the value of Ea(apparent emissivity) would be ",round(Ea1,4)

Apparent emissivity of covered opening is  0.6269
If there were no cover present, the value of Ea(apparent emissivity) would be  0.8571


## Exa 8.14¶

In [16]:
#Example Number 8.14
# Transmitting and reflecting system for furnace opening

# Variable declaration

T1 = 1000+273 			# [K] temperature of furnace
lamda = 4.0 			# [micro meter]
#for  0 < lamda < 4 micro meter
t1 = 0.9
E1 = 0.1
rho1 = 0
#for  4 micro meter < lamda < infinity
t2 = 0
E2 = 0.8
rho2 = 0.2
sigma = 5.669*10**(-8) 		# [W/square meter K**(4)]
T3 = 30+273		 	# [K] room temperature
# because the room is large it may be treated as a blackbody.
# analyze the problem by calculating the heat transfer for each 			wavelength band and then adding them together to obtain the total. the 		network for each band is a modification of figure 8-57
A1 = 1.0 			# [square meter]
A2 = 1.0 			# [square meter]
A3 = 1.0 			# [square meter]
F12 = 1.0
F13 = 1.0
F32 = 1.0
# the total emissive powers are

#Calculation

Eb1 = sigma*T1**(4) 		# [W/square meter]
Eb3 = sigma*T3**(4) 		# [W/square meter]
# to determine the fraction of radiation in each wavelength band
lamba_into_T1 = lamda*T1 	# [micro meter K]
lamba_into_T3 = lamda*T3 	# [micro meter K]
# consulting table 8-1, we find
Eb1_0_to_4 = 0.6450*Eb1 	# [W/square meter]
Eb3_0_to_4 = 0.00235*Eb3 	# [W/square meter]
Eb1_4_to_inf = (1-0.6450)*Eb1 	# [W/square meter]
Eb3_4_to_inf = (1-0.00235)*Eb3 	# [W/square meter]
#  apply these numbers to the network for the two wavelengths bands, 			with unit areas.

# 0 < lamda < 4 micro meter band:
R1 = 1/(F13*t1)
R2 = 1/(F32*(1-t1))
R3 = 1/(F12*(1-t1))
R4 = rho1/(E1*(1-t1))
# the net heat transfer from the network is then
R_equiv_1 = 1/(1/R1+1/(R2+R3+R4))
q1 = (Eb1_0_to_4-Eb3_0_to_4)/R_equiv_1 # [W/square meter]

# 4 micro meter < lamda < infinity band:
R2 = 1/(F32*(1-t2))
R3 = 1/(F12*(1-t2))
R4 = rho2/(E2*(1-t2))
# the net heat transfer from the network is then
# R1 is infinity
R_equiv_2 = R2+R3+R4*2
q2 = (Eb1_4_to_inf-Eb3_4_to_inf)/R_equiv_2 # [W/square meter]

# the total heat loss is then
q_total = q1+q2 			# [W/square meter]
# with no windows at all, the heat transfer would have been the 			difference in blackbody emissive powers,
Q = Eb1-Eb3 				# [W/square meter]

#Result

print "Radiation lost through the quartz window to a room temperature of 30 degree celsius is",round(q_total),"W/square meter"

print "With no windows at all, the heat transfer would be",round(Q),"W/square meter"

Radiation lost through the quartz window to a room temperature of 30 degree celsius is 112171.0 W/square meter
With no windows at all, the heat transfer would be 148397.0 W/square meter


## Exa 8.20¶

In [17]:
#Example Number 8.20
# solar-environment equilibrium  temperatures

# Variable declaration

q_by_A_sun = 700 			# [W/m**(2)] solar flux
T_surr = 25+273 			# [K] surrounding temperature
sigma = 5.669*10**(-8) 			# [W/square meter K**(4)]
# at radiation equilibrium the netenergy absorbed from sun must equal the 	long-wavelength radiation exchange with the surroundings,or
# (q_by_A_sun)*alpha_sun = alpha_low_temp*sigma*(T**4-T_surr**4)         (a)

# case (a) for white paint

# for white paint we obtain from table 8-4

#Calculation

alpha_sun = 0.12
alpha_low_temp = 0.9
# so that equation (a) becomes
T = ((q_by_A_sun)*alpha_sun/(alpha_low_temp*sigma)+T_surr**(4))**(1.0/4.0)	 # [K]

#Result

print"Radiation equilibrium temperature for the plate exposed to solar flux if the surface is coated with :"
print "White paint is",round(T-273,1),"degree celsius"

# case (b) for flat black lacquer we obtain

alpha_sun = 0.96
alpha_low_temp = 0.95
# so that equation (a) becomes
T = ((q_by_A_sun)*alpha_sun/(alpha_low_temp*sigma)+T_surr**(4))**(1.0/4.0) 	# [K]

print "Flat black lacquer is",round(T-273,1),"degree celsius"

Radiation equilibrium temperature for the plate exposed to solar flux if the surface is coated with :
White paint is 39.5 degree celsius
Flat black lacquer is 104.8 degree celsius


## Exa 8.23¶

In [18]:
#Example Number 8.23
# temperature measurement error caused by radiation

# Variable declaration

E = 0.9 				# emissivity of mercury-in-glass thermometer
Tt = 20+273 				# [K] temperature indicated by thermometer
Ts = 5+273 				# [K] temperature of walls
sigma = 5.669*10**(-8) 			# [W/square meter K^(4)]
h = 8.3 				# [W/sq m] heat transfer coefficient for 					thermometer
# we employ equation(8-113) for the solution: h*(Tinf-Tt) =sigma*E*(Tt^4-Ts^4)
# inserting the values in above equation

Tinf = sigma*E*(Tt**4-Ts**4)/h+Tt 	# [K]
print"the true air temperature is",round(Tinf-273,1),"degree celsius"

the true air temperature is 28.6 degree celsius