#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]
# total incident radiation is
# 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"
#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"
#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
#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)
#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)
#xample Number 8.7
# surface in radiant balance
# 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"
#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
print "Net radiant exchange is",round(q),"W"
#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)
#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"
#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"
#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)
#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)
#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"
#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"
#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"