Chapter 3 : Conduction of heat in the steady state¶

Example 3.1 page no : 28¶

In [1]:
# Variables
deltax = 9./12;                 # thickness of wall in ft
k = 0.18;                     # thermal conductivity of wall in B/hr-ft-degF
t1 = 1500;                     # inside temperature of oven wall in degF
t2 = 400;                      # outside temperature of oven wall in degF

# Calculations
q = k*(t1-t2)/deltax;          # heat loss in Btu/hr

# Results
print " The heat loss for each square foot of wall surface is %d Btu/hr-ft**2"%(q);

 The heat loss for each square foot of wall surface is 264 Btu/hr-ft**2


Example 3.2 page : 29¶

In [2]:
# Variables
x1 = 9./12;             # thickness of firebrick in ft
k1 = 0.72;             # thermal conductivity of firebrick in Btu/hr-ft-degF
x2 = 5./12;             # thickness of insulating brick in ft
k2 = 0.08;             # thermal conductivity of insulating brick in Btu/hr-ft-degF
x3 = 7.5/12;           # thickness of redbrick in ft
k3 = 0.5;              # thermal conductivity of firebrick in Btu/hr-ft-degF
t1 = 1500.;             # inner temperature of wall in degF
t2 = 150.;              # outer temperature of wall in degF

# Calculations and Results
# resistances of mortar joints are neglected
q = (t1-t2)/(x1/k1+x2/k2+x3/k3);   # heat flow per square ft in Btu/hr
t2 = t1-(q*x1/k1);                 # first contact temperature in degF
print " The temperature at the contact of firebrick and insulating brick is %d degF"%(t2);

t3 = t2-(q*x2/k2);                 # second contact temperature in degF
print " The temperature at the contact of insulating brick and red brick is %d degF"%(t3);

 The temperature at the contact of firebrick and insulating brick is 1312 degF
The temperature at the contact of insulating brick and red brick is 375 degF


Example 3.3 page : 31¶

In [4]:
import math

# Variables
d1 = 2.375/12;                   # internal diameter of pipe in ft
t = 1./12;                        # thickness of insulating material in ft
d2 = d1+2*t;                     # external (insulation)diameter of pipe in ft
k = 0.0375;                      # thermal conductivity of insulating material in Btu/hr-ft-F
l = 30.;                          # length of pipe in ft
t1 = 380.;                        # inner surface temperature of insulation
t2 = 80.;                         # outer surface temperature of insulation

# Calculations and Results
q = 2*math.pi*k*(t1-t2)/math.log(d2/d1);  # heat loss per unit length
print " Heat loss per linear foot is %.f Btu/hr"%(q)

qtot = round(q)*l;               # heat loss for 30 ft pipe
print " Total heat loss through 30 ft of pipe is %d Btu/hr"%(qtot)

 Heat loss per linear foot is 116 Btu/hr
Total heat loss through 30 ft of pipe is 3480 Btu/hr


Example 3.4 page :33¶

In [5]:
import math

# Variables
d1 = 10.75/12;            # outer diameter of pipe in ft
x1 = 1.5/12;              # thickness of insulation 1 in ft
x2 = 2./12;                # thickness of insulation 2 in ft
d2 =  d1+2*x1;            # diameter of insulation 1 in ft
d3 = d2+2*x2;             # diameter of insulation 1 in ft
t1 = 700.;                 # inner surface temperature of composite insulation in degF
t2 = 110.;                 # outer surface temperature of composite insulation in degF
k1 = 0.05;                #thermal conductivity of material 1 in Btu/hr-ft-degF
k2 = 0.039;               # thermal conductivity of material 2 in Btu/hr-ft-degF

# Calculations
q = 2*math.pi*(t1-t2)/(math.log(d2/d1)/k1+math.log(d3/d2)/k2);                                     # heat loss per linear foot in Btu/hr

# Results
print " The heat loss is found to be %d Btu/hr-ft"%( q);

 The heat loss is found to be 323 Btu/hr-ft


Example 3.5 page : 34¶

In [6]:
# Variables
km = 0.0377;               # Mean thermal conductivity at 220degF
t1 = 260.;                  # Inner surface temperature of slab in degF
t2 = 180.;                  # Outer surface temperature of slab in degF
A = 1.;                     # Area of slab in ft
x = 2./12;                  # Thickness of insulation in ft

# Calculations
q = km*A*(t1-t2)/x;        # Heat loss through slab in Btu/hr

# Results
print " Heat loss through flat slab is %.1f Btu/hr"%(q);

 Heat loss through flat slab is 18.1 Btu/hr


Example 3.6 page : 37¶

In [11]:
# Variables
k = 0.8                     # Avg. thermal conductivity in Btu/hr-ft-degF
T1 = 400.                    # Inner surface temperature of furnace in degF
T2 = 100.                    # Outer surface temperature of furnace in degF
a = 3.                       # Length of furnace in ft
b = 4.                       # Breadth of furnace in ft
c = 2.5                     # Height of furnace in ft
Aa = 2*a*b                  # Area of surface A in ft**2
Ab = 2*b*c                  # Area of surface A in ft**2
Ac = 2*a*c                  # Area of surface A in ft**2
x = 4.5/12                  # Thickness of insulation in ft
t = 24.                      # Time elapsed in hr
M = 4.                       # Number of edges
N = 8.                       # Number of corners

# Calculations
S = Aa/x+Ab/x+Ac/x+0.54*(a+b+c)*M+0.15*x*N       # Shape factor
qo = round(S*k*(T1-T2),-2)                                 # Heat flow per hour
q = qo*t                                         # Heat loss in 24 hr

# Results
print "The heat loss in 24 hr is %d Btu"%(q)

# note : book answer is wrong. Kindly check.

The heat loss in 24 hr is 1027200 Btu


Example 3.7 page : 40¶

In [12]:
# Variables
M = 8*9;                  # number of flow channels for the entire section
N = 8.37;                 # number of equal channel intervals
# the fractional part arises due to the fractional part of temperature close to border EG

# Calculations
k  =  M/N;               # Ratio of shape factor to wall length

# Results
print " Shape factor for the special section where the ratio of radius of\
circle to half side length is 0.5,S is %.2fL"%( k );

 Shape factor for the special section where the ratio of radius of circle to half side length is 0.5,S is 8.60L


Example 3.8 page : 43¶

In [13]:
from numpy import zeros
# Variables
t1 = 800.;            # inner surface temperature of wall in degF
t4 = 200.;           #  outer surface temperature of wall in degF

#Grids are square in shape so delx  = dely where delx,y sre dimensions of square grid
t2 = [700, 550 ,550, 587.5, 587.5, 596.9, 596.9, 599.3, 599.3, 599.8];             # Assumed temperature of grid point 1
t3 = [300, 300 ,375 ,375, 393.8, 393.8, 398.5, 398.5, 399.6, 399.6];               # Assumed temperature of grid point 2

th2 = zeros(9)
th3 = zeros(9)

# Calculations and Results
for i in range(9):
th2[i] = t1+t3[i]-2*t2[i]; # th1 =  q/kz at grid pt1
th3[i] = t2[i]+t4-2*t3[i];# th2 =  q/kz at grid pt2
print " Assuming t2 = %.1f  degF and t2 = %.1f degF  th1[%d] = %.1f degF and th2[%d] = %.1f degF "%(t2[i],t3[i],i,th2[i],i,th3[i]);
print " Since th2[%d] is not equal to th3[%d], hence other values of t2 and t3 are to be assumed"%(i,i);

print "Assuming t2 = 600 degF and t3 = 400 degF, th2 = th3."
print "Hence Steady state condition is satisfied at grid temperatures of 400 degF and 600 degF";

 Assuming t2 = 700.0  degF and t2 = 300.0 degF  th1[0] = -300.0 degF and th2[0] = 300.0 degF
Since th2[0] is not equal to th3[0], hence other values of t2 and t3 are to be assumed
Assuming t2 = 550.0  degF and t2 = 300.0 degF  th1[1] = 0.0 degF and th2[1] = 150.0 degF
Since th2[1] is not equal to th3[1], hence other values of t2 and t3 are to be assumed
Assuming t2 = 550.0  degF and t2 = 375.0 degF  th1[2] = 75.0 degF and th2[2] = 0.0 degF
Since th2[2] is not equal to th3[2], hence other values of t2 and t3 are to be assumed
Assuming t2 = 587.5  degF and t2 = 375.0 degF  th1[3] = 0.0 degF and th2[3] = 37.5 degF
Since th2[3] is not equal to th3[3], hence other values of t2 and t3 are to be assumed
Assuming t2 = 587.5  degF and t2 = 393.8 degF  th1[4] = 18.8 degF and th2[4] = -0.1 degF
Since th2[4] is not equal to th3[4], hence other values of t2 and t3 are to be assumed
Assuming t2 = 596.9  degF and t2 = 393.8 degF  th1[5] = 0.0 degF and th2[5] = 9.3 degF
Since th2[5] is not equal to th3[5], hence other values of t2 and t3 are to be assumed
Assuming t2 = 596.9  degF and t2 = 398.5 degF  th1[6] = 4.7 degF and th2[6] = -0.1 degF
Since th2[6] is not equal to th3[6], hence other values of t2 and t3 are to be assumed
Assuming t2 = 599.3  degF and t2 = 398.5 degF  th1[7] = -0.1 degF and th2[7] = 2.3 degF
Since th2[7] is not equal to th3[7], hence other values of t2 and t3 are to be assumed
Assuming t2 = 599.3  degF and t2 = 399.6 degF  th1[8] = 1.0 degF and th2[8] = 0.1 degF
Since th2[8] is not equal to th3[8], hence other values of t2 and t3 are to be assumed
Assuming t2 = 600 degF and t3 = 400 degF, th2 = th3.
Hence Steady state condition is satisfied at grid temperatures of 400 degF and 600 degF


Example 3.10 page : 46¶

In [14]:
import math

# Variables
k = 0.9;                        # thermal conductivity of section material in Btu/hr-ft-degF

# Heat is considered to flow through fictitious rods and only half of the heat flows through symmetry axes
Ta = 300.;
Tb = 441.;
Tc = 600.;
Td = 300.;
Te = 432.;
Tf = 600.;
Tg = 600.;
Th = 600.;
Ti = 300.;
Tj = 384.;
Tk = 461.;
Tl = 485.;
Tm = 490.;
Tn = 300.;
To = 340.;
Tp = 372.;
Tq = 387.;
Tr = 391.;
Ts = 300.;
Tt = 300.;
Tu = 300.;
Tv = 300.;
Tw = 300.;

# Calculations and Results
# Above grid point temperatures are given in the question for the quarter section considered in degF(a,b,c...w are grid points)
q1 = 4*k*((Tc-Tb)/2+(Tf-Te)+(Tf-Tk)+(Tg-Tl)+(Th-Tm)/2);                           # Amount of heat coming from inside in Btu/hr
q2 = 4*k*((Tb-Ta)/2+(Te-Td)+(Tj-Ti)+(To-Tn)+(To-Tt)+(Tp-Tu)+(Tq-Tu)+(Tr-Tw)/2);   # Amount of heat going outside in Btu/hr
q = (q1+q2)/2;                  # average of heat going in and heat coming out
print " Total heat flow per unit depth is %.1fBtu/hr"%(q);

S = q/(k*(Tc-Ta));              # shape factor in ft
print " Shape factor is %.2fft"%(S)

 Total heat flow per unit depth is 2029.5Btu/hr
Shape factor is 7.52ft