In [1]:

```
import math
# View Factors Associated with two Concentric Spheres
# Calculations and Results
#The outer surface of the smaller sphere and inner surface of the larger sphere form a two surface enclosure
N = 2;
print "This enclosure involves ",N**2,"View Factors"
x = (1./2)*N*(N-1)
print "W need to determine only",x,"view factor directly"
F11 = 0
F12 = 1
print ("The Two view Factors")
print "Since no radiation leaving surface 1 strikes itself..... F11 = ",F11
print "Since all radiation leaving surface 1 strikes surface 2 F12 = ",F12
print ("F12 = ((r1/r2)**2)")
print ("F22 = 1-((r1/r2)**2)")
print ("where r1 and r2 are radius of surface 1 and surface 2")
```

In [2]:

```
import math
# Fraction of Radiation Leaving through an Opening
# Variables
r1 = 0.1 #Radius of enclosure[m]
L = 0.1 #Length of Enclosure[m]
r2 = 0.05
r3 = 0.08 #Inner and outer radii of the ring[m]
# Calculations
#Using Chart in Fig 13.7
F12 = 0.11;
F13 = 0.28;
F1_ring = F13-F12;
# Results
print "The fraction of the radiation leaving the base cyllinder enclosure that escapes \
through coaxial ring opening at its top surface is",F1_ring
```

In [4]:

```
import math
# View Factors Associated with a Tetragon
# Variables
#A pyramid with square base and it's sides being isoceles triangle
F11 = 0 #Since base is a flat surface
# Calculations
#F12 = F13 = F14 = F15 = x
x = (1-F11)/4.;
# Results
print "Each side pf the four surfaces of the pyramid recieves",x,"of total radiation"
```

In [5]:

```
import math
# Variables
# The Crossed-Strings Method for View Factors
a = 12.
b = 5. #With od long parallel plates[cm]
c = 6. #Distance between the plates
L1 = a
L2 = b
L3 = c;
# Calculations
L4 = math.sqrt((7**2)+(6**2));
L5 = math.sqrt((5**2)+(6**2));
L6 = math.sqrt((12**2)+(6**2));
F12_1 = ((L5+L6)-(L3+L4))/(2*L1);
F13 = (L1+L3-L6)/(2*L1);
F14 = (L1+L4+L5)/(2*L1);
F12_2 = 1-F13-F14;
# Results
print "F13 = ",F13,"F14 = ",F14,"Therefore from two different methods F12_1 = F12_2 = ",F12_1,
```

In [6]:

```
import math
# Radiation Heat Transfer in a Black Furnace
# Variables
F12 = 0.2;
A = 5.*5 #Area of 1 surface of cube[m**2]
Tb = 800
Tt = 1500
Ts = 500 #Temperature of base top and the side surfaces of the furbace[K]
# Calculations and Results
F11 = 0;
Q11 = 0;
F13 = 1-F11-F12;
Q13 = A*F13*(5.67*10**(-8))*((Tb**4)-(Ts**4)) #[kW]
print "The net rate of heat transfer from surface1 to surface3 is",round(Q13/1000),"kW"
Q12 = A*F12*(5.67*10**(-8))*((Tb**4)-(Tt**4)) #[kW]
print "The net rate of radiation heat transfer from siurface1 to surface2 is",round(Q12/1000),"kW"
Q1 = Q11+Q12+Q13 #[kW]
print "Rhe net radiation heat transfer from the base surface is",round(Q1/1000),"kW"
```

In [8]:

```
import math
# Radiation Heat Transfer between Parallel Plates
# Variables
T1 = 800
T2 = 500 #Temp of parallel plates[K]
e1 = 0.2
e2 = 0.7 #Emissivities
# Calculations
q12 = (5.67*10**(-8))*((T1**4)-(T2**4))/((1/e1)+(1/e2)-1);
# Results
print "The net heat at the rate of" \
,round(q12),"W","is transferred from plate 1 to plate 2 by radiation per unit surface area of either plate"
```

In [1]:

```
import math
# Radiation Heat Transfer in Cylindrical Furnace
# Variablesa
ro = 1
H = 1 #Radius amd height of cylinder[m]
e1 = 0.8
e2 = 0.4 #Emissivities
T1 = 700
T2 = 500 #Top and base temperatures of furnace[K]
T3 = 400 #Side durface temperature[K]
F11 = 0
F12 = 0.38;
# Calculations and Results
A1 = math.pi*(ro**2) #[m**2]
A2 = A1 #[m**2]
A3 = 2*math.pi*ro*H #[m**2]
F13 = 1-F11-F12;
F21 = F12 #Top and Bottom are symmetric
F31 = F13*(A1/A3);
F23 = F13;
F32 = F31;
def rad(J):
i = [0,0,0]
i[0] = J[0]+(((1-e1)/e1)*((F12*(J[0]-(J[1])))+(F13*((J[0])-(J[2])))))-((T1**4)*(5.67*10**(-8)));
i[1] = J[1]+(((1-e2)/e2)*((F21*(J[1]-J[0]))+(F23*(J[1]-J[2]))))-((T2**4)*(5.67*10**(-8)));
i[2] = J[2]-((T3**4)*(5.67*10**(-8)));
#print (J[2],J[1],J[0])
#Q1 = A1*((F12*(J[0]-J[1]))+(F13*(J[0]-J[2]))) #[kW]
#Q2 = A2*((F21*(J[1]-J[0]))+(F13*(J[1]-J[2]))) #[kW]
#Q3 = A3*((F31*(J[2]-J[0]))+(F32*(J[2]-J[1]))) #[kW]
#print "The net rates of radiation heat transfer at the three surfaces are", \
#Q1/1000,Q2/1000,Q3/1000,"kW"
```

In [11]:

```
import math
# Radiation Heat Transfer in a Triangular Furnace
# Variables
A1 = 1.
A2 = 1.
A3 = 1. #Area of each side[m**2]
T1 = 600.
T2 = 1000. #[K]
e = 0.7;
F12 = 0.5
F13 = 0.5
F23 = 0.5 #Symmetry
# Calculations
Eb1 = 5.67*10**(-8)*(T1**4) #[W/m**2]
Eb2 = 5.67*10**(-8)*(T2**4) #[W/m**2]
Q = (Eb2-Eb1)/(((1-e)/(A1*e))+((((A1*F12)+(1/((1/(A1*F13))+(1/(A2*F23))))))**(-1))) #[kW]
# Results
print "Heat at the rate of",round(Q/1000),"kW"
print "must be supplied to the heated surface per unit lemgth of the duct \
to maintain steady operation in the furnace"
```

In [13]:

```
import math
# Heat Transfer through a Tubular Solar Collector
# Variables
k = 0.02588 #[W/m.degree Celcius]
Pr1 = 0.7282
Pr2 = 0.7255 #Prandtl no
nu1 = 1.608*(10**(-5))
nu2 = 1.702*10**(-5) #[m**2/s]
T1 = 20
T2 = 40 #[degree Celcius]
Tavg = ((T1+T2)/2)+273 #[K]
Do = 0.1
L = 1 #Dimensions of glass tube[m]
Di = 0.05 #Inner diameter of tube[m]
Q_glass = 30 #Rate of heat transfer from the outer surface of the glass cover[W]
g = 9.81 #[m**2/s]
eo = 0.9
ei = 0.95 #Emissivity
# Calculations and Results
Ao = math.pi*Do*L #Heat transfer surface area of the glass cover[m**2]
print (Ao,Tavg)
Ra_Do = g*Tavg*(T2-T1)*(Do**3)*Pr1/(nu1);
print "The Rayleigh number is",Ra_Do
Nu = ((0.6+((0.387*(Ra_Do**(1./6)))/((1+((0.559/Pr1)**(9./16)))**(8./27))))**2);
print "The nusselt number is",Nu
ho = k*Nu/Do #[W/m**2.degree Celcius]
Qo_conv = ho*Ao*(T2-T1) #[W]
Qo_rad = eo*5.67*10**(-8)*Ao*(((T2+273)**4)-((T1+273)**4)) #[W]
Qo_total = Qo_conv+Qo_rad #[W]
print "The total rate of heat loss from the glass cover ",Qo_total,"W"
Lc = (Do-Di)/2 #The characteristic length
Ai = math.pi*Di*L #[m**2]
#Assuming
T_tube = 54
T_cover = 26 #Temperature of tube and glass cover[degree Celcius]
T_avg = ((T_tube+T_cover)/2)+273 #[K]
Ra_L = g*T_avg*(T_tube-T_cover)*(Lc**3)*Pr2/(nu2);
print "The Rayleigh number in this case is",Ra_L
F_cyl = ((math.log(Do/Di))**4)/((Lc**3)*(((Di**(-3./5))+(Do**(-3./5)))**5));
k_eff = 0.386*k*((Pr2/(0.861+Pr2))**(1./4))*((F_cyl*Ra_L)**(1./4));
print "The effective thermal conductivity is",k_eff,"W/m.degree Celcius"
QL_conv = 2*math.pi*k_eff*(T_tube-T_cover)/(math.log(Do/Di));
print "The rate of heat transfer between the cylinders by convection is",QL_conv,"W"
QL_rad = ((5.67*10**(-8))*Ai*(((T_tube+273)**4)-((T_cover+273)**4)))/((1/ei)+(((1-eo)/eo)*(Di/Do)));
print "The radiation rate of heat transfer is",QL_rad,"W"
QL_total = QL_conv+QL_rad #[W]
print "The total rate of heat loss from the glass cover is",QL_total,"W"
```

In [14]:

```
import math
# Radiation Shields
# Variables
e = 0.1 #Emissivity of aluminium sheet
T1 = 800
T2 = 500 #Temperatures of two parallel plates[K]
e1 = 0.2
e2 = 0.7 #Emissivities of plates
# Calculations
q12 = ((5.67*10**(-8))*((T1**4)-(T2**4)))/((1/e1)+(1/e2)-1+(1/e)+(1/e)-1) #[W/m**2]
# Results
print "Radiation Heat Transfer",round(q12),"W/m**2"
```

In [15]:

```
import math
# Radiation Effect on Temperature Measurements
# Variables
Tw = 400
Tth = 650 #Temperature of duct wall and hota air flowing in it[K]
e = 0.6 #emissivity
h = 80 #Heat transfer coefficient[W/m**2.K]
# Calculations
Tf = Tth+((e*5.67*10**(-8)*((Tth**4)-(Tw**4)))/h) #[K]
# Results
print "The temperature of actual air is",round(Tf),"K"
```

In [18]:

```
import math
# Effective Emissivity of Combustion Gases
d = 5
H = 5 #Diameter and height of cylindrical furnace[m]
T = 1200 #Temp of gases[K]
P = 2 #Pressure[atm]
yN2 = 0.8
yH2O = 0.08
yO2 = 0.07
yCO2 = 0.05 #Volumetric Composition
# Calculations and Results
Pc = yCO2*P #[atm]
Pw = yH2O*P #[atm]
print "The partial pressures of CO2 and H2O are",Pc,"atm","and",Pw,"atm"
L = 0.6*d #[m]
x = Pc*L
y = Pw*L #[m.atm]
ec_1 = 0.16
ew_1 = 0.23 #Emissivity of CO2 and H2O at 1 atm pressure
Cc = 1.1
Cw = 1.4 #Pressure Correction Factors are
del_e = 0.048 #Emissivity correction factor at T = 1200K
e_g = Cc*ec_1+Cw*ew_1-del_e;
print "The effectivity of the combustion gases is",e_g
```

In [19]:

```
import math
# Radiation Heat Transfer in a Cylindrical Furnace
# Variables
Ts = 600 #Wall Temperature[K]
d = 5
H = 5 #Diameter and Height of cylindrical furnace
Tg = 1200
eg = 0.45 #Average gas temperature and average emissivity of the combustion gases
Pc = 0.10
L = 3
Pw = 0.16 #From Previous s
# Calculations and Results
x = Pc*L*Ts/Tg #[m.atm]
y = Pw*L*Ts/Tg #[m.atm]
ec_1 = 0.11
ew_1 = 0.25 #Emissivities of CO2 and H2O corresponding to 600K and 1atm
Cc = 1.1
Cw = 1.4 #Correction Factors
a_c = Cc*((Tg/Ts)**(0.65))*(ec_1);
a_w = Cw*((Tg/Ts)**(0.45))*ew_1;
print "The absorptivities of CO2 and H2O are",a_c,"and",a_w
del_a = 0.027;
a_g = a_c+a_w-del_a;
print "The absorptivity of the combustion gases is",a_g
As = (math.pi*d*H)+(math.pi*(d**2)/2) #[m**2]
print "the surface area of the cylindrical surface is",round(As),"m**2"
Q_net = round(As)*(5.67*10**(-8))*((eg*(Tg**4))-(a_g*(Ts**4)));
print "The net rate of radiation heat transfer from the combustion gases to walls of the furnace is", \
Q_net,"W"
```

In [20]:

```
import math
# Effect of Clothing on Thermal Comfort
# Variables
h_rad = 4.7
h_conv = 4.0 #The radiation and convection heat transfer coefficient[W/m**2.degree Celcius]
R_clo = 0.6*0.155 #Thermal resismath.tance of clothing[m**2.degree Celcius/W]
T_skin = 33
T_amb = 22 #Skin and Ambient temperature[degree Celcius]
As = 1.8 #Surface area of an average man
# Calculations and Results
h_comb = h_conv+h_rad #combined heat transfer coefficient[W/m**2.degree Celcius]
Q_sen_clo = As*(T_skin-T_amb)/(R_clo+(1/h_comb)) #[W]
print "The sensible heat loss from this person when clothed is",Q_sen_clo,"W"
#On removing the clothes
#R_clo = 0 Clothing resismath.tance on removing clothes
#Setting both heat transfer rates equal to determine new ambient air temperature
T_amb_new = T_skin-(Q_sen_clo*(1/h_comb)/As) #[degree Celcius]
print "The ambient temperature now is",T_amb_new,"degree Celcius"
```