Chapter 7: Steady-State Heat Conduction

ILLUSTRATIVE EXAMPLE 7.1, Page number: 93

In [1]:
#Variable declaration:
Q = 3000.0              #The rate of heat flow through the glass window (W)
L = 0.01                #Thickness of glass window (m)
A = 3.0                 #Area of heat transfer (m^2)
TC = 10+273             #Temperature at the outside surface (K)
k = 1.4                 #Thermal onductivity of glass (W/m.K)

#Calculation:
TH = TC+Q*L/k/A         #Temperature at the inner surface (K)

#Result:
print "The temperature at the inner surface is :",round(TH,1)," K"
print "The temperature at the inner surface is :",round(TH-273,1)," °C"

The temperature at the inner surface is : 290.1  K
The temperature at the inner surface is : 17.1  °C

ILLUSTRATIVE EXAMPLE 7.2, Page number: 94

In [2]:
#Variable declaration:
k = 0.026               #Thermal conductivity of insulating material (Btu/ft.h.°F)
L = 1.0                 #Thickness of insulating material (ft)
TC = 70.0               #Temperature on the cold side surface (°F)
TH = 210.0              #Temperature on the hot side surface (°F)
c = 0.252               #Kilocalorie per hour in a Btu per hour
m = 0.093               #meter square in a feet square

#Calculation:
DT = TH-TC              #Change in temperature (°F)
Q1 = k*DT/L             #Rate of heat flux throughthe wall (Btu/f^t2.h.)
Q2 = Q1*c/m             #Rate of heat flux throughthe wall in SI units (kcal/m^2.h)

#Result:
print "The rate of heat flux in Btu/ft^2.h is :",round(Q1,3)," Btu/ft^2.h ."
print "The rate of heat flux in SI units is :",round(Q2,3)," kcal/m^2.h ."

The rate of heat flux in Btu/ft^2.h is : 3.64  Btu/ft^2.h .
The rate of heat flux in SI units is : 9.863  kcal/m^2.h .

ILLUSTRATIVE EXAMPLE 7.3, Page number: 94

In [3]:
#Variable declaration:
TH = 1592.0                     #Temperature of inside surface (K)
TC = 1364.0                     #Temperature of outside surface (K)
H = 3.0                         #Height of furnace wall (m)
W = 1.2                         #Width of furnace wall (m)
L = 0.17                        #Thickness furnace wall (m)
m = 0.0929                      #Meter square per second in a feet square per second
Btu = 3.412                     #Btu per hour in a Watt
Btu2 = 0.3171                   #Btu per feet square hour in a watt per meter square

#Calculation:
Tav = (TH+TC)/2                 #Average wall temperature (K)
#From Table in Appendix:
p = 2645.0                      #Density of material (kg/m^3)
k = 1.8                         #Thermal conductivity (W/m.K)
Cp = 960.0                      #Heat capacity of material (J/kg.K)
a = k/(p*Cp)/m                  #Thermal diffusivity (ft^2/s)
t = (TC-TH)/L                   #Temperature gradient (°C/m)
A = H*W                         #Heat transfer area (m^2)
Q1 = k*A*(TH-TC)/L*Btu          #Heat transfer rate (Btu/h)
Q2 = k*(TH-TC)/L*Btu2           #Heat transfer flux (Btu/h.ft^2)
R = L/(k*A)                     #Thermal resistance (°C/W)

#Result:
print "The temperature gradient is :",round(t)," °C/m."
print "The heat transfer rate is :",round(Q1)," Btu/h."
print "The heat transfer flux is :",round(Q2,1)," Btu/h.ft^2."

The temperature gradient is : -1341.0  °C/m.
The heat transfer rate is : 29653.0  Btu/h.
The heat transfer flux is : 765.5  Btu/h.ft^2.

ILLUSTRATIVE EXAMPLE 7.4, Page number: 96

In [4]:
#Variable declaration:
TH = 25.0                   #Temperature at inner suface of wall (°C)
TC = -15.0                  #Temperature at outer suface of wall (°C)
L = 0.3                     #Thickness of wall (m)
k = 1.0                     #Thermal conductivity of concrete (W/m)
A = 30.0                    #Sueface area of wall (m^2)

#Calculation:
DT = TH-TC                  #Driving force for heat transfer (°C) (part 2)
R = L/(k*A)                 #Thermal resistance (°C/W) (part 3)
Q = DT/R/10**3              #Heat loss through the wall (kW)

#Result:
print "1. Theoretical part."
print "2. The driving force for heat transfer is :",DT," °C."
print "3. The heat loss through the wall is :",Q," kW."

1. Theoretical part.
2. The driving force for heat transfer is : 40.0  °C.
3. The heat loss through the wall is : 4.0  kW.

ILLUSTRATIVE EXAMPLE 7.5, Page number: 97

In [5]:
#Variable declaration:
TC = 27.0                   #Inside temperature of walls (°C)
TH = 68.7                   #Outside temperature of walls (°C)
LC = 6*0.0254               #Thickness of concrete (m)
LB = 8*0.0254               #Thickness of cork-board (m)
LW = 1*0.0254               #Thickness of wood (m)
kC = 0.762                  #Thermal conductivity of concrete (W/m.K)
kB = 0.0433                 #Thermal conductivity of cork-board (W/m.K)
kW = 0.151                  #Thermal conductivity of wood (W/m.K)

#Calculation:
RC = LC/kC                  #Thermal resistance of concrete (K/W)
RB = LB/kB                  #Thermal resistance of cork-board (K/W)
RW = LW/kW                  #Thermal resistance of wood (K/W)
Q = (TC-TH)/(RC+RB+RW)      #Heat transfer rate across the wall (W)
T = -(Q*RW-TC)              #Interface temperature between wood and cork-board (K)

#Result:
print "The heat transfer rate across the wall is :",round(Q,3)," W."
print "The interface temperature between wood and cork-board is :",round(T,1)," °C."
print "The interface temperature between wood and cork-board is :",round(T+273,1)," K."

The heat transfer rate across the wall is : -8.239  W.
The interface temperature between wood and cork-board is : 28.4  °C.
The interface temperature between wood and cork-board is : 301.4  K.

ILLUSTRATIVE EXAMPLE 7.6, Page number: 98

In [6]:
from math import pi, log
from sympy import symbols

#Variable declaration:
Z = symbols ('z')                       #Length of pipe
D1s = 4.0                               #Glass wool inside diameter (in)
D2s = 8.0                               #Glass wool outside diameter (in)
D1a = 3.0                               #Asbestos inside diameter (in)
D2a = 4.0                               #Asbestos outside diameter (in)
TH = 500.0                              #Outer surface temperature of pipe (°F)
TC = 100.0                              #Outer surface temperature of glass wool (°F)
La = 0.5/12.0                           #Thickness of asbestos (ft)
Lb = 2.0/12.0                           #Thickness of glss wool (ft)
ka = 0.120                              #Thermal conductivity of asbestos (Btu/h.ft.°F)
kb = 0.0317                             #Thermal conductivity of asbestos (Btu/h.ft.°F)

#Calculation:
Aa = (pi*Z*(D2a-D1a)/12.0)/log(D2a/D1a) #Area of asbestos (ft^2)
Ab = (pi*Z*(D2s-D1s)/12.0)/log(D2s/D1s) #Area of glass wool (ft^2)
Q1 = (TH-TC)/(La/(ka*Aa)+Lb/(kb*Ab))    #Steady-state heat transfer per foot of pipe (Btu/h.)
Q2 = Q1/Z                               #Factorization of Q/Z (Btu/h.ft)

#Result:
print "The steady-state heat transfer per foot of pipe, Z, is :",round(Q1/Z,1)," x z Btu/h."
print "The steady-state heat transfer factorizating out Z is :",round(Q2,1)," Btu/h.ft."

The steady-state heat transfer per foot of pipe, Z, is : 103.6  x z Btu/h.
The steady-state heat transfer factorizating out Z is : 103.6  Btu/h.ft.

ILLUSTRATIVE EXAMPLE 7.7, Page number: 99

In [7]:
#Variable declaration:
#From example 7.6:
TH = 500                                #Outer surface temperature of pipe (°F)
Lb = 2.0/12.0                           #Thickness of glss wool (ft)
kb = 0.0317                             #Thermal conductivity of asbestos (Btu/h.ft.°F)
Ab = 1.51                               #Area of glass wool (ft^2)
Q = 103.5                               #Steady-state heat transfer per foot of pipe (Btu/h.)
La = 0.5/12.0                           #Thickness of asbestos (ft)
ka = 0.120                              #Thermal conductivity of asbestos (Btu/h.ft.°F)
Aa = 0.91                               #Area of asbestos (ft^2)
TC = 100                                #Outer surface temperature of glass wool (°F)

#Calculation:
Ti_b = -((Lb*Q)/(kb*Ab)-TH)             #Interfacial temperature of glass wool layer (°F)
Ti_a = (Q*La)/(ka*Aa)+TC                #Interfacial temperature of asbestos layer (°F)

#Result:
print "The interfacial temperature of glass wool layer is :",round(Ti_b)," °F."
print "The interfacial temperature of asbestos layer is :",round(Ti_a,1)," °F."

The interfacial temperature of glass wool layer is : 140.0  °F.
The interfacial temperature of asbestos layer is : 139.5  °F.

ILLUSTRATIVE EXAMPLE 7.8, Page number: 100

In [8]:
from sympy import cos,symbols,diff,pi

#Variable declaration:
z,h,k = symbols('z, h, k')      #Length, height, thermal conductivity
T = 100*cos((pi*z)/(2*h))       #Temperature of solid slab

#Calculation:
DT = diff(T,z)                  #Temperature at z
Q = -k*(DT)                     #Heat flux in slab (Btu/s.ft^2)
Q1 = Q.subs(z,0)                #Heat flux in slab at z = 0 (Btu/s.ft^2)
Q2 = Q.subs(z,h)                #Heat flux in slab at z = h (Btu/s.ft^2)

#Result:
print "The heat flux in slab is :",Q," Btu/s.ft^2 ."
print "The heat flux in slab at z = 0 is :",Q1," Btu/s.ft^2 ."
print "The heat flux in slab at z = h is :5",Q2," Btu/s.ft^2 ."

The heat flux in slab is : 50*pi*k*sin(pi*z/(2*h))/h  Btu/s.ft^2 .
The heat flux in slab at z = 0 is : 0  Btu/s.ft^2 .
The heat flux in slab at z = h is :5 50*pi*k/h  Btu/s.ft^2 .