Chapter 19: Insulation and Refractory¶

ILLUSTRATIVE EXAMPLE 19.1, Page number: 413¶

In :
from __future__ import division

#Variable declaration:
H = 2.5                         #Height of wall (m)
W = 4                           #Width of wall (m)
h = 11                          #Convective heat transfer coefficient (W/m^2.K)
T1 = 24                         #Outside surface temperature (°C)
T3 = -15                        #Outside air temperature (°C)
L = 7.62/10**3                  #Insulation thickness (m)
k = 0.04                        #Thermal conductivity of wool (W/m.K)

#Calculation:
A = H*W                         #Heat transfer area (m^2)
Q = h*A*(T1-T3)                 #Heat transfer rate (W)
Ri = L/(k*A)                    #Insuation resistance (K/W)
Rc = 1/(h*A)                    #Convective resitance (K/W)
R = Ri+Rc                       #Total resistance (K/W)
Qt = (T1-T3)/R                  #Revised heat transfer rate (Btu/h)

#Result:
print "1. The heat transfer rate without insulation is :",round(Q)," W ."
print "Or, the heat transfer rate without insulation is :",round(Q*3.412)," Btu/h ."
print "2. The revised heat transfer rate with insulation is :",round(Qt)," W ."
print "Or, the revised heat transfer rate with insulation is :",round(Qt*3.412)," Btu/h ."
print "There is a calculation mistake in book."
1. The heat transfer rate without insulation is : 4290.0  W .
Or, the heat transfer rate without insulation is : 14637.0  Btu/h .
2. The revised heat transfer rate with insulation is : 1386.0  W .
Or, the revised heat transfer rate with insulation is : 4729.0  Btu/h .
There is a calculation mistake in book.

ILLUSTRATIVE EXAMPLE 19.2, Page number: 414¶

In :
#Variable declaration:
#From example 19.1:
T1 = 24                         #Outside surface temperature (°C)
Ri = 0.0191                     #Insulation resistance (K/W)
Q = 1383                        #Revised heat transfer rate (Btu/h)

#Calculation:
T2 = T1-Q*Ri                    #Temperature at outer surface of insulation (°C)

#Result:
print "The temperature at the outer surface of the insulation is :",round(T2,1)," °C ."
The temperature at the outer surface of the insulation is : -2.4  °C .

ILLUSTRATIVE EXAMPLE 19.3, Page number: 415¶

In :
#Variable declaration:
#From example 19.1:
h = 11                          #Convective heat transfer coefficient (W/m^2.K)
L = 7.62/10**3                  #Insulation thickness (m)
k = 0.04                        #Thermal conductivity of wool (W/m.K)

#Calculation:
Bi = h*L/k                      #Biot number

#Result:
print "The Biot nmuber is :",round(Bi,1)," ."
The Biot nmuber is : 2.1  .

ILLUSTRATIVE EXAMPLE 19.4, Page number: 415¶

In :
from __future__ import division

#Variable declaration:
k = 0.022                       #Thermal conductivity of glass wool (Btu/h.ft.°F)
T1 = 400                        #Inside wall temperature (°F)
T2 = 25                         #Outside wall temperature (°C)
L = 3/12                        #Length of insulation cover (ft)

#Calculation:
T_2 = T2*(9/5)+32               #Outside wall temperature in fahrenheit scale (°F)
QbyA = k*(T1-T_2)/L              #Heat flux across the wall (Btu/h.ft^2)

#Result:
print "The heat flux across the wall is :",round(QbyA,1)," Btu/h.ft^2 ."
The heat flux across the wall is : 28.4  Btu/h.ft^2 .

ILLUSTRATIVE EXAMPLE 19.5, Page number: 415¶

In :
from __future__ import division

#Variable declaration:
w = 8                           #Width of wall (m)
H = 3                           #Height of wall (m)
h = 21                          #Convective heat transfer coefficient between the air and the surface (W/m^2.K)
T1 = -18                        #Outside surace of wall temperature (°C)
T3 = 26                         #Surrounding air temperature (°C)
l1 = 80/100                     #Reduction in cooling load
k = 0.0433                      #Thermal conductivity of cork board insulation (W/m.K)
T = 12000                       #Units Btu/h in 1 ton of refrigeration

#Calculation:
A = w*H                         #Heat transfer area (m^2) (part 1)
Q1 = h*A*(T1-T3)                #Rate of heat flow in the absence of insulation (W)
Q2 = Q1*3.4123/T                #Rate of heat flow in the absence of insulation (ton of refrigeration)
l2 = 1-l1                       #Reduced cooling load (part 2)
Q3 = l2*Q1                      #Heat rate with insulation (W)
Rt = (T1-T3)/Q3                 #Total thermal resistance (°C/W)
R2 = 1/(h*A)                    #Convection thermal resistance (°C/W)
R1 = Rt-R2                      #Insulation conduction resistance (°C/W)
L = R1*k*A                      #Required insulation thickness (m)

#Result:
print "1. The rate of heat flow through the rectangular wall without insulation is :",round(Q1/10**3,2)," kW ."
print "Or, the rate of heat flow through the rectangular wall without insulation in tons of refrigeration is :",round(Q2,1)," ton of refrigeration ."
if (Q1<0):
print "   The negative sign indicates heat flow from the surrounding air into the cold room."
else :
print "   The positive sign indicates heat flow from the surrounding air into the cold room."
print "2. The required thickness of the insulation board is :",round(L*10**3,2)," mm ."
1. The rate of heat flow through the rectangular wall without insulation is : -22.18  kW .
Or, the rate of heat flow through the rectangular wall without insulation in tons of refrigeration is : -6.3  ton of refrigeration .
The negative sign indicates heat flow from the surrounding air into the cold room.
2. The required thickness of the insulation board is : 8.25  mm .

ILLUSTRATIVE EXAMPLE 19.6, Page number: 417¶

In :
#Variable declaration:
#From example 19.5:
Q = -4435.2                     #Heat rate with insulation (W)
R2 = 0.00198                    #Convection thermal resistance (°C/W)
T3 = 26                         #Surrounding air temperature (°C)
h = 21                          #Convective heat transfer coefficient between the air and the surface (W/m^2.K)
k = 0.0433                      #Thermal conductivity of cork board insulation (W/m.K)
L = 0.00825                     #Required insulation thickness (m)

#Calculation:
T2 = T3+Q*R2                    #Interface temperature (°C) (part 1)
Bi = h*L/k                      #Biot number (part 2)

#Result:
print "1. The interface temperature is :",round(T2,2)," °C ."
print "2. The Biot number is :",round(Bi)," ."
print "3. Theoretical part."
1. The interface temperature is : 17.22  °C .
2. The Biot number is : 4.0  .
3. Theoretical part.

ILLUSTRATIVE EXAMPLE 19.7, Page number: 417¶

In :
from __future__ import division
from math import pi,log

#Variable declaration:
D2 = 0.5/10**3                  #External diameter of needle (m)
h3 = 12                         #Heat transfer coefficient (W/m^2.K)
L = 1                           #Insulation thickness (m)
T1 = 95                         #Reactant temperature (°C)
T3 = 20                         #Ambient air temperature (°C)
k1 = 16                         #Thermal conductivity of needle (W/m.K)
k3 = 0.0242                     #Thermal conductivity of air (W/m.K)
D3 = 2/10**3                    #Diameter of rubber tube (m)

#Calculation:
r2 = D2/2                       #External radius of needle (m)
r3 = D3/2                       #Radius of rubber tube (m)
Rt1 = 1/(h3*(2*pi*r2*L))        #Thermal resistance (°C/W)
Q1 = (T1-T3)/Rt1                #Rate of heat flow in the absence of insulation (W)
Bi = h3*D2/k1                   #Biot number
Nu = h3*D2/k3                   #Nusselt number
R2 = log(r3/r2)                 #Thermal resistance of needle (°C/W)
R3 = 1/(h3*(2*pi*r3*L))         #Thermal resistance of rubber tube (°C/W)
Rt2 = R2+R3                     #Total thermal resistance (°C/W)
Q2 = (T1-T3)/Rt2                #Rate of heat loss (W)

#Result:
print "1. The rate of the heat loss from the hypodermic needle with the rubber insulation is :",round(Q1,2)," W ."
print "   The rate of the heat loss from the hypodermic needle without the rubber insulation is :",round(Q2,2)," W ."
print "2. The Biot number is :",round(Bi,6)," ."
print "   The nusselt number is :",round(Nu,3)," ."
1. The rate of the heat loss from the hypodermic needle with the rubber insulation is : 1.41  W .
The rate of the heat loss from the hypodermic needle without the rubber insulation is : 5.12  W .
2. The Biot number is : 0.000375  .
The nusselt number is : 0.248  .

ILLUSTRATIVE EXAMPLE 19.9, Page number: 420¶

In :
from __future__ import division
from math import log, pi

#Variable declaration:
h = 140                             #Convention heat transfer coefficient (W/m^2.K)
D1 = 10/10**3                       #Rod diameter (m)
L = 2.5                             #Rod length (m)
T1 = 200                            #Surface temperature of rod (°C)
T2 = 25                             #Fluid temperature (°C)
k = 1.4                             #Thermal conductivity of bakellite (W/m.K)
l = 55/10**3                        #Insulation thickness (m)

#Calculation:
Q1 = h*pi*D1*L*(T1-T2)              #Rate of heat transfer for the bare rod (W) (part 1)
Bi = 2                              #Critical Biot number (part 2)
D2 = Bi*k/h                         #Critical diameter associated with the bakelite coating (m)
r2 = D2/2                           #Critical radius associated with the bakelite coating (m)
r1 = D1/2                           #Rod radius (m)
R1 = log(r2/r1)/(2*pi*k*L)          #Insulation conduction resistance (°C/W)
R2 = 1/(h*(2*pi*r2*L))              #Convection thermal resistance (°C/W)
Rt1 = R1+R2                         #Total thermal resistance (°C/W)
Qc = (T1-T2)/Rt1                    #Heat transfer rate at the critical radius (W)
r3 = r1+l                           #New radius associated with the bakelite coating after insulation (m) (part 3)
R3 = log(r3/r1)/(2*pi*k*L)          #Insulation conduction bakelite resistance (°C/W)
R4 = 1/(h*(2*pi*r3*L))              #Convection bakelite thermal resistance (°C/W)
Rt2 = R3+R4                         #Total bakelite thermal resistance (°C/W)
Q2 = (T1-T2)/Rt2                    #Heat transfer rate at the bakelite critical radius (W)
Re = ((Q1-Q2)/Q1)*100               #Percent reduction in heat transfer rate relative to the case of a bare rod (%)

#Result:
print "1. The rate of heat transfer for the bare rod is :",round(Q1)," W ."
print "2. The critical radius associated with the bakelite coating is :",round(r2*10**3)," mm."
print " & the heat transfer rate at the critical radius is :",round(Qc)," W ."
print "3. The fractional reduction in heat transfer rate relative to the case of a bare rod is :",round(Re,1)," % ."
1. The rate of heat transfer for the bare rod is : 1924.0  W .
2. The critical radius associated with the bakelite coating is : 10.0  mm.
& the heat transfer rate at the critical radius is : 2273.0  W .
3. The fractional reduction in heat transfer rate relative to the case of a bare rod is : 24.6  % .

ILLUSTRATIVE EXAMPLE 19.10, Page number: 421¶

In :
from __future__ import division
from math import pi, log

#Variable declaration:
r1 = 1.1/100                    #Inside radius of pipe (m)
r2 = 1.3/100                    #Outside radius of pipe (m)
r3 = 3.8/100                    #Outside radius of asbestos insulation (m)
L = 1                           #Length of tube (m)
h1 = 190                        #Heat transfer coefficient from ethylene glycol to the stainless steel pipe (W/m^2.K)
k2 = 19                         #Thermal conductivity of pipe (W/m.K)
h2 = 14                         #Outside heat transfer coefficient from the air to the surface of the insulation (W/m^2.K)
k3 = 0.2                        #Thermal conductivity of asbestos (W/m.K)
T1 = 124                        #Hot ethylene glycol temperature (°C)
T5 = 2                          #Surrounding air temperature (°C)
k4 = 0.0242                     #Thermal conductivity of air (W/m.K)

#Calculation:
A1 = 2*pi*r1*L                  #Inside surface area of pipe (m^2) (part1)
A2 = 2*pi*r2*L                  #Outside surface area of pipe (m^2)
A3 = 2*pi*r3*L                  #Outside surface area of asbestos insulation (m^2)
R1 = 1/(h1*A1)                  #Inside convection resistance (°C/W)
R2 = log(r2/r1)/(2*pi*k2*L)     #Conduction resistance through the tube (°C/W)
R3 = 1/(h2*A2)                  #Outside convection resistance (°C/W)
Rt1 = R1+R2+R3                  #Total resistance without insulation (°C/W)
Q1 = (T1 - T5)/Rt1              #Heat transfer rate without insulation (W)
R4 = log(r3/r2)/(2*pi*k3*L)     #Conduction resistance associated with the insulation (°C/W) (part 2)
R5 = 1/(h2*A3)                  #Outside convection resistance (°C/W)
Rt2 = R1+R2+R4+R5               #Total rsistance with the insulation (°C/W)
Q2 = (T1-T5)/Rt2                #Heat transfer rate with the insulation (W)
U1 = 1/(Rt2*A1)                 #Overall heat transfer coefficient based on the inside area (W/m^2.K) (part 3)
U3 = 1/(Rt2*A3)                 #Overall heat transfer coefficient based on the outside area (W/m^2.K) (part 4)
T3 = T1-(R1+R2)*Q2              #Temperature at the steel–insulation interface (°C) (part 5)
Bi1 = h2*(2*r3)/k3              #Outside Biot number (part 6)
Bi2 = h1*(2*r1)/k2              #Inside Biot number
Nu = h1*(2*r1)/k4               #Nusselt number of the air
rlm = (r3-r2)/log(r3/r2)        #Log mean radius of the insulation (m) (part 7)

#Result:
print "1. The rate of heat transfer without insulation is :",round(Q1,1)," W."
print "2. The rate of heat transfer with insulation is :",round(Q2,1)," W."
print "3. The overall heat transfer coefficient based on the inside area of the tube is :",round(U1,2)," W/m^2.K ."
print "4. The overall heat transfer coefficient based on the outside area of the insulation is :",round(U3,1)," W/m^2.K ."
print "5. The temperature, T3, at the steel–insulation interface is :",round(T3,1)," °C."
print "6. The inside Biot numbers is :",round(Bi2,2)," ."
print "   The outside Biot numbers is :",round(Bi1,2)," ."
print "   The Nusselt number is :",round(Nu,1)," ."
print "7. The log mean radius of insulation is :",round(rlm*100,2)," cm."
print "There is a printing mistake in book for unit in part 7."
1. The rate of heat transfer without insulation is : 128.1  W.
2. The rate of heat transfer with insulation is : 99.2  W.
3. The overall heat transfer coefficient based on the inside area of the tube is : 11.76  W/m^2.K .
4. The overall heat transfer coefficient based on the outside area of the insulation is : 3.4  W/m^2.K .
5. The temperature, T3, at the steel–insulation interface is : 116.3  °C.
6. The inside Biot numbers is : 0.22  .
The outside Biot numbers is : 5.32  .
The Nusselt number is : 172.7  .
7. The log mean radius of insulation is : 2.33  cm.
There is a printing mistake in book for unit in part 7.

ILLUSTRATIVE EXAMPLE 19.11, Page number: 424¶

In :
from __future__ import division
from math import pi

#Variable declaration:
h1 = 800                        #Heat transfer coefficient for steam condensing inside coil (Btu/h.ft^2.°F)
h2 = 40                         #Heat transfer coefficient for oil outside coil (Btu/h.ft^2.°F)
h3 = 40                         #Heat transfer coefficient for oil inside tank wal (Btu/h.ft^2.°F)
h4 = 2                          #Heat transfer coefficient for outer tank wall to ambient air (Btu/h.ft^2.°F)
k1 = 0.039                      #Thermal conductivity of insulation layer (Btu/h.ft.°F)
l1 = 2/12                       #Thickness of insulation layer (ft)
D = 10                          #Diameter of tank (ft)
H = 30                          #Height of tank (ft)
k2 = 224                        #Thermal conductivity of copper tube (Btu/h.ft.°F)
l2 = (3/4)/12                   #Thickness of insulation layer (ft)
T1 = 120                        #Temperature of tank (°F)
T2 = 5                          #Outdoor temperature (°F)

#Calculation:
Uo1 = 1/(1/h3+(l1/k1)+1/h4)     #Overall heat transfer coefficient for tank (Btu/h.ft^2.°F)
At = pi*(D+2*l1)*H              #Surface area of tank (ft^2)
Q = Uo1*At*(T1-T2)              #Heat transfer rate lost from the tank (Btu/h)
#From table 6.3:
l2 = 0.049/12                   #Thickness of coil (ft)
A = 0.1963                      #Area of 18 guage, 3/4-inch copper tube (ft^2/ft)
Uo2 = 1/(1/h2+(l2/k2)+1/h1)     #Overall heat transfer coefficient for coil (Btu/h.ft^2.°F)
#From steam tables:
Tst = 240                       #Temperature for 10 psia (24.7 psia) steam (°F)
Ac = Q/(Uo2*(Tst-T1))           #Area of tube (ft^2)
L = Ac/A                        #Lengt of tube (ft)

#Result:
print "The length ofcopper tubing required is :",round(L,1)," ft ."
The length ofcopper tubing required is : 26.0  ft .

ILLUSTRATIVE EXAMPLE 19.12, Page number: 426¶

In :
from __future__ import division
from math import pi, log
from numpy import array,log as LOG

#Variable declaration:
#For 1-inch pipe schedule 40:
Di = 1.049/12                   #Inside diameter (ft)
Do = 1.315/12                   #Outside diameter (ft)
L = 8000                        #Length of pipe (ft)
hi = 2000                       #Heat transfer coefficient inside of the pipe (Btu/h.ft^2.°F)
ho = 100                        #Outside heat transfer coefficient (Btu/h.ft.°F)
kl = 0.01                       #Thermal conductivity of insulation (Btu/h.ft.°F)
T1 = 240                        #Steam temperature (°F)
T2 = 20                         #Air temperature (°F)
k = 24.8                        #Thermal conductivity for steel (Btu/h.ft.°F)
Dxl = array([3/8,1/2,3/4,1])/12 #thickness(ft)
amt = array([1.51,3.54,5.54,8.36])/6 #Cost per feet($) #Calculation: D_ = (Do-Di)/log(Do/Di) #Log-mean diameter of the pipe (ft) Dl = Do+2*(Dxl) #Insulation thickness (ft) D_l = (Dl-Do)/LOG(Dl/Do) #Log mean diameter of pipe (ft) Dxw = (Do-Di)/2 #Pipe thickness (ft) Rw = Dxw/(k*pi*D_*L) #Wall resistance ((Btu/h.°F)^-1) Ri = 1/(hi*pi*Di*L) #Inside steam convection resistance ((Btu/h.°F)^-1) Rl = Dxl/(kl*pi*D_l*L) #Insulation resistance ((Btu/h.°F)^-1) Ro = 1/(ho*pi*Dl*L) #Outside air convection resistance ((Btu/h.°F)^-1) R = Ri+Rw+Rl+Ro #Total resistance ((Btu/h.°F)^-1) Uo = 1/(R*pi*Dl*L) #Overall outside heat transfer coefficient (Btu/h.ft^2.°F) Ui = 1/(R*pi*Di*L) #Overall inside heat transfer coefficient (Btu/h.ft^2.°F) dT = T1-T2 Ai = pi*Di*L #Inside area (ft^2) Q = Ui*Ai*dT #Energy loss (Btu/h) def energyPerDollar(Q1,Q2,amt1,amt2): return round((Q1-Q2)/(8000*(amt2-amt1)),1) #Results: print "Energy saved per dollar ingoing from 3/8 to 1/2 inch is :",energyPerDollar(Q,Q,amt,amt),' Btu/h.$'
print "Energy saved per dollar ingoing from 1/2 to 3/4 inch is :",energyPerDollar(Q,Q,amt,amt),' Btu/h.$' print "Energy saved per dollar ingoing from 3/4 to 1 inch is :",energyPerDollar(Q,Q,amt,amt),' Btu/h.$'
Energy saved per dollar ingoing from 3/8 to 1/2 inch is : 18.2  Btu/h.$Energy saved per dollar ingoing from 1/2 to 3/4 inch is : 18.8 Btu/h.$
Energy saved per dollar ingoing from 3/4 to 1 inch is : 6.8  Btu/h.\$

ILLUSTRATIVE EXAMPLE 19.16, Page number: 434¶

In :
from __future__ import division

#Variable declaration:
ki = 0.44                       #Thermal conductivity of insulation (Btu/h.ft.°F)
ho = 1.32                       #Air flow coefficient (Btu/h.ft^2.°F)
OD = 2                          #Outside diameter of pipe (in)

#Calculation:
rc = (ki/ho)*12                 #Outer critical radius of insulation (in)
ro = OD/2                       #Outside radius of pipe (in)
L = rc-ro                       #Critical insulation thickness (in)

#Result:
print "The outer critical radius of insulation is :",round(rc)," in ."
if ro<rc:
print "Since, ro<rc, the heat loss will increase as insulation is added."
else :
print "Sice, ro>rc, the heat loss will decrease as insulation is added."
The outer critical radius of insulation is : 4.0  in .
Since, ro<rc, the heat loss will increase as insulation is added.

ILLUSTRATIVE EXAMPLE 19.18, Page number: 439¶

In :
from __future__ import division

#Variable declaration:
Lf = 6/12                           #Length of firebrick (ft)
kf = 0.61                           #Thermal conductivity of firebrick (Btu/h.ft.°F)
A = 480                             #Surface area of wall (ft^2)
Lw = 8/12                           #Length of rock wool (ft)
kw = 0.023                          #Thermal conductivity of rock wool (Btu/h.ft.°F)
T1 = 1900                           #Temperature of insulation of firebrick (°F)
T2 = 140                            #Temperature of insulation of rock wool (°F)

#Calculation:
Rf = Lf/(kf*A)                      #Resistance of firebrick (h.°F/Btu)
Rw = Lw/(kw*A)                      #Resistance of rock wool (h.°F/Btu)
R = Rf+Rw                           #Total resitance (h.°F/Btu)
Q = (T1-T2)/R                       #Heat loss through the wall (Btu/h)

#Result:
print "The heat loss through the wall is :",round(Q)," Btu/h ."

The heat loss through the wall is : 28344.0  Btu/h .

ILLUSTRATIVE EXAMPLE 19.19, Page number: 440¶

In :
from __future__ import division

#Variable declaration:
h1 = 1700                       #Steam heat-transfer coefficient (Btu/h.ft^2.°F)
h2 = 2                          #Air heat-transfer coefficient (Btu/h.ft^2.°F)
A = 1                           #Area of base (ft^2) (assumption)
k1 = 26                         #Thermal conductivity of steel (Btu/h.ft.°F)
k2 = 218                        #Thermal conductivity of copper (Btu/h.ft.°F)
t = 0.375                       #Thickness of steel sheet (in)
h3 = 2500                       #Increased steam heat-transfer coefficient (Btu/h.ft^2.°F)
h4 = 12                         #Increased air heat-transfer coefficient (Btu/h.ft^2.°F)

#Calculation:
R1 = 1/(h1*A)                   #Steam resistance (h.°F/Btu)
R2 = 1/(h2*A)                   #Air resistance (h.°F/Btu)
R3 = (t/12)/(k1*A)              #Steel resistance (h.°F/Btu)
Rt1 = R1+R2+R3                  #Total resistance (with steel) (h.°F/Btu)
R4 = (t/12)/(k2*A)              #Copper resistance (h.°F/Btu) (part 1)
Rt2 = R1+R2+R4                  #Total resistance (with copper) (h.°F/Btu)
R5 = 1/(h1*A)                   #New steam resistance (h.°F/Btu)
Rt3 = R5+R2+R3                  #Total resistance after increasing the steam coefficient (h.°F/Btu)
R6 = 1/(h4*A)                   #Air resistance (h.°F/Btu)
Rt4 = R1+R6+R3                  #Total resistance after increasing the air coefficient (h.°F/Btu)

#Result:
if (Rt1==Rt2):
print "1.The rate of heat transfer is essentially unaffected."
else :
print "1. The rate of heat transfer is essentially affected."
if (Rt1==Rt3):
print "2. The rate is again unaffected."
else :
print "2. The rate is again affected."
if (Rt1==Rt4):
print "3. The rate is unaffected for this case."
else :
print "3. The rate is affected for this case."
1. The rate of heat transfer is essentially affected.
2. The rate is again unaffected.
3. The rate is affected for this case.

ILLUSTRATIVE EXAMPLE 19.20, Page number: 441¶

In :
from __future__ import division
from math import pi, log, exp

#Variable declaration:
rfo = 12/2                      #Outside radius of firebrick (ft)
rfi = 5.167                     #Inside radius of firebrick (ft)
rso = 6.479                     #Outside radius of sil-o-cel (ft)
rsi = 6.063                     #Inside radius of fsil-o-cel (ft)
L = 30                          #Length of incinerator (ft)
kf = 0.608                      #Thermal conductivity of firebrick (Btu/h.ft.°F)
ks = 0.035                      #Thermal conductivity of sil-o-cel (Btu/h.ft.°F)

#Calculation:
Rf= log(rfo/rfi)/(2*pi*L*kf)    #Resistance of firebrick (h.ft.°F/Btu)
Rs= log(rso/rsi)/(2*pi*L*ks)    #Resistance of sil-o-cel (h.ft.°F/Btu)
R = Rf+Rs                       #Total resistance (h.ft.°F/Btu)
ro = exp(R*(2*pi*L*ks))*rso     #New outside radius of sil-o-cel (ft)
r= ro-rso                       #Extra thickness (ft)

#Result:
print "The extra thickness is :",round(r,3)," ft."
print "Or, the extra thickness is :",round(r*12,2)," in."
The extra thickness is : 0.504  ft.
Or, the extra thickness is : 6.05  in.