# Chapter 2 :Steady State conduction In one dimension¶

## Example 2.1 Page No : 14¶

In [1]:
# Variables
A = 1.           			#m**2, area
#for inner layer (cement)
ti = 0.06       			#m, thickness
ki = 0.72       			#W/m C,  thermal conductivity
Ti = -15.        			#C, temprature
#for middle layer (cork)
tm = 0.1        			#m, thickness
km = 0.043      			#W/m C,  thermal conductivity
#for outer layer(brick)
to = 0.25       			#m, thickness
ko = 0.7        			#W/m C,  thermal conductivity
To = 30.         			#C, temprature

# Calculation and Results
#Thermal resistance of outer layer  			#C/W
Ro = to/(ko*A)
#Thermal resistance of middle layer   			#C/W
Rm = tm/(km*A)
#Thermal resistance of inner layer    			#C/W
Ri = ti/(ki*A)
Rt = Ro+Rm+Ri
tdf = To-Ti        			#temp driving force
#(a)
Q = tdf/Rt      			#rate of heat gain
print "the rate of heat gain is %.2f W"%(Q)

#(b)
#from fig. 2.4
td1 = Q*to/(ko*A)   			#C temp. drop across the brick layer
T1 = To-td1         			#interface temp. between brick and cork
#similarly
td2 = Q*tm/(km*A)   			#C temp. drop across the cork layer
T2 = T1-td2         			#C, interface temp. between cement and cork
print "interface temp. between brick and cork is %.1f C"%(T1)
print "interface temp. between cement and cork is %.1f C"%(T2)

#(c)
Rpo = Ro/Rt         			#thermal resistance offered by brick layer
Rpm = Rm/Rt         			#thermal resistance offered by cork layer
Rpi = Ri/Rt         			#thermal resistance offered by cement layer
print "thermal resistance offered by brick layer is %.1f percent"%(Rpo*100)
print "thermal resistance offered by cork layer is %.1f percent"%(Rpm*100)
print "thermal resistance offered by cement layer is %.1f percent"%(Rpi*100)

#second part
x = 30.              			#percentage dec in heat transfer
Q1 = Q*(1-x/100)    			#W, desired rate of heat flow
Rth = tdf/Q1        			#C/W, required thermal resistance

the rate of heat gain is 16.27 W
interface temp. between brick and cork is 24.2 C
interface temp. between cement and cork is -13.6 C
thermal resistance offered by brick layer is 12.9 percent
thermal resistance offered by cork layer is 84.1 percent
thermal resistance offered by cement layer is 3.0 percent
Additional thickness of cork to be provided  = 5.1 cm


## Example 2.2 Page No : 15¶

In [2]:
# Variables
#outer thickness of brickwork (to) & inner thickness (ti)
to = 0.15    			#m thickness
ti = 0.012   			#m thickness
#thickness of intermediate layer(til)
til = 0.07   			#m  thick
#thermal conductivities of brick  and wood
kb = 0.70    			#W/m celcius
kw = 0.18    			#W/m celcius
#temp. of outside and inside wall
To = -15    			#celcius
Ti = 21     			#celcius
#area
A = 1       			#m**2

# Calculations and Results
#(a)
#Thermal resistance of brick  , wood and insulating layer
TRb = to/(kb*A)   			#C/W
TRw = ti/(kw*A)   			#C/W
TRi = 2*TRb       			#C/W
#Total thermal resistance
TR = TRb+TRw+TRi  			#C/W
#Temp. driving force
T = Ti-To         			#C
#Rate of heat loss
Q = T/TR
print "Rate of heat loss is %.1f W"%(Q)
#(b)thermal conductivities of insulating layer
k = til/(A*TRi)
print "thermal conductivities of insulating layer is %.4f W/m C"%(k)

Rate of heat loss is 50.7 W
thermal conductivities of insulating layer is 0.1633 W/m C


## Example 2.3 Page No : 19¶

In [5]:
import math

# Variables
#Length & Inside rdius of gas duct
L = 1.       			#m
#Properties of inner and outer layer
ki = 1.3    			#W/m C, thermal conductivity of inner bricks
ti = 0.27   			#m, inner layer thickness
ko = 0.92   			#W/m C, thermal conductivity of special bricks
to = 0.14   			#m, outer layer thickness
Ti = 400.    			#C, inner layer temp.
To = 65.     			#C, outer layer temp.

#calculation and Results
r_ = ri+ti   			#m, outer radius of fireclay  brick layer
ro = r_+to   			#m, outer radius of special brick layer
#Heat transfer resistance
#Heat transfer resistance of fireclay brick
R1 = (math.log(r_/ri))/(2*math.pi*L*ki)
#Heat transfer resistance of special brick
R2 = (math.log(ro/r_))/(2*math.pi*L*ko)
#Total resistance
R = round(R1+R2,4)
#Driving force
T = Ti-To
#Rate of heat loss
Q = T/(R)
print "Rate of heat loss is %d W"%(Q)
#interface temp.
Tif = Ti-(Q*R1)
print "interface temp.is  %d C"%(Tif)
#Fractional resistance offered  by the special brick layer
FR = R2/(R1+R2)
print "Fractional resistance offered  by the special brick layer is %.3f "%(FR)

Rate of heat loss is 4095 W
interface temp.is  183 C
Fractional resistance offered  by the special brick layer is 0.353


## Example 2.4 Page No : 20¶

In [6]:
import math

# Variables
d1 = 0.06       			#m, one end diameter of steel rod
d2 = 0.12       			#m,other end diameter of steel rod
l = 0.2         			#m length of rod
T2 = 30.         			#C, temp. at end 2
Q = 50.          			#W, heat loss
k = 15.           			#W/m c, thermal conductivity of rod

# Calculation and Results
#T = 265.8-(7.07/(0.06-0.15*x))........(a)
#(a)
x1 = 0
#from eq. (a)
T1 = 265.8-(7.07/(0.06-0.15*x1))
print "The hot  end temp. is %.0f C"%(T1)
#(b)  from eq. (i)
C = 50                 			#integration consmath.tant
#from eq. (i)
D1 = -C/(math.pi*d1**2*k)   			#D = dT/dx, temprature gradient
print "The temprature gradient at hot end is %.1f C/m"%(D1)
#similarly
D2 = -1179             			#at x =  0.2m
print "The temprature gradient at cold end is %.0f C/m"%(D2)

#(c)
x2 = 0.15              			#m, given,
x3 = l-x2              			#m, section away from the cold end
#from eq. (a)
T2 = 265.8-(7.07/(0.06-0.15*x3))
print "the temprature at 0.15m away from the cold end is %.0f C"%(T2)

The hot  end temp. is 148 C
The temprature gradient at hot end is -294.7 C/m
The temprature gradient at cold end is -1179 C/m
the temprature at 0.15m away from the cold end is 131 C


## Example 2.5 Page No : 24¶

In [7]:
import math

# Variables
#inside and outside diameter and Temp.  of sphorical vessel
do = 16.         			#m, diameter
t = 0.1          			#m, thick
Ri = do/2        			#m, inside radius
Ro = Ri+t        			#m. outside radius
To = 27.          			#C, temperature
Ti = 4.           			#C  ammonia
k = 0.02         			#W/m C, thermal conductivity of foam layer

# Calculations and Results
#from eq. 2.23 the rate of heat transfer
Q = (Ti-To)*(4*math.pi*k*Ro*Ri)/(Ro-Ri)
print "the rate of heat transfer is %.0f W"%(Q)
#Refrigeration capacity(RC)
#3516 Watt =  1 ton
RC = -Q/3516
print "Refrigeration capacity is %.2f tons"%(RC)

the rate of heat transfer is -3746 W
Refrigeration capacity is 1.07 tons


## Example 2.6 Page No : 28¶

In [8]:
import math
# Variables
d = 0.05        			#m, diameter of rod
l = 0.5         			#m, length of rod
T1 = 30.         			#CTemp. at  one end (1)
T2 = 300.        			#C, temp  at other end (2)

# Calculations and Results
x1 = l/2         			#m, at mid plane
#temprature distribution ,
#and its solution as x = (-b+math.sqrt(b**2-4*a*c))/2*a
a = 1.35*10**-4
b = 1
c = -(564*x1+30.1)
T = (-b+math.sqrt(b**2-4*a*c))/(2*a)
print "the temprature midway in the rod at steady state is %.1f C"%(T)

#Temprature gradient at the ends of the rod
x2 = 0               			#m, at one end
a1 = 1.35*10**-4
b1 = 1
c1 = -(564*x2+30.1)
T1 = (-b1+math.sqrt(b1**2-4*a1*c1))/(2*a1)
k1 = 202+0.0545*T1
C1 = 113930           			#integration consmath.tant from eq. (1)
TG1 = C1/k1           			#C/W, temprature gradient, dT/dx
#similarly
x3 = 0.5
a2 = 1.35*10**-4
b2 = 1
c2 = -(564*x3+30.1)
T2 = (-b2+math.sqrt(b2**2-4*a2*c2))/(2*a2)
k2 = 202+0.0545*T2
TG2 = C1/k2
print "Temprature gradient at one end of the rod is %.0f C/W"%(TG1)
print "Temprature gradient at other end of the rod is %.1f C/W"%(TG2)

the temprature midway in the rod at steady state is 167.3 C
Temprature gradient at one end of the rod is 559 C/W
Temprature gradient at other end of the rod is 521.8 C/W


## Example 2.7 Page No : 29¶

In [3]:
import math
# Variables
#temprature distribution in wall

t = 0.3             			#m, thickness of wall
k = 23.5            			#W/m c  thermal conductivity of wall

# Calculations and Results
x1 = 0
T1 = 600+2500*x1-12000*x1**2    			#C, at surface
x2 = 0.3
T2 = 600+2500*x2-12000*x2**2    			#C, at x = 0.3

def f3(x):
return 600+2500*x-12000*x**2

print "At the surface x = 0, the temp. is %.0f C"%(T1)
print "At the surface x = 0.3m, the temp. is %.0f C"%(T2)
print "Rhe average temprature of the wall is %.0f C"%(Tav)

#(b)

#for maximum temprature D = 0
x3 = 2500/24000.
print "The maximum temprature occurs at %.3f m"%(x3)
Tmax = 600+2500*x3-12000*x3**2
print "The maximum temp. is %.0f C"%(Tmax)

#(c)
D1 = 2500-24000*x1         			#at x = 0, temprature gradient
Hf1 = -k*D1                			#W/m**2, heat flux at left surface(x = 0)
D2 = 2500-24000*x2         			#at x = 0.3, temprature gradient
Hf2 = -k*D2                			#W/m**2, heat flux at right surface(x = 0.3)
print "heat flux at left surface is %.0f W/m**2"%(Hf1)
print "heat flux at right surface is %.0f W/m**2"%(Hf2)

#(d)
Qt = Hf2-Hf1              			#W/m**2, total rate of heat loss
Vw = 0.3                  			#m**3/m**2, volume of wall per unit surface area
Hav = Qt/Vw               			#W/m**3, average volumetric rate
print "The average volumetric rate if heat genaration is %.0f W/m**3 "%(Hav)

At the surface x = 0, the temp. is 600 C
At the surface x = 0.3m, the temp. is 270 C
Rhe average temprature of the wall is 615 C
The maximum temprature occurs at 0.104 m
The maximum temp. is 730 C
heat flux at left surface is -58750 W/m**2
heat flux at right surface is 110450 W/m**2
The average volumetric rate if heat genaration is 564000 W/m**3


## Example 2.8 Page No : 30¶

In [10]:
import math
# Variables
ka = 24         			#W/mC thermal conductivitiy of material A
tA = 0.1        			#m, thickness of A material
kB = 230        			#W/mC thermal conductivity of metl B
kC = 200        			#W/mC thermal conductivity of metal C
tB = 0.1        			#m, thickness of B metal
tC = 0.1        			#m, thickness of C metal
TBo = 100       			#C, outer surface temp. of B wall
TCo = 100       			#C, outer surface temp. of C wall
Q = 2.5*10**5    			#W/m**3, heat generated

# Calculations and Results
#At D = 0
x = 2175./10416
print "The maximum temp. will occur at a position %.3f m"%(x)
x1 = x
TA = -5208*x1**2+2175*x1-74.5
print "The maximum temprature is %.1f C"%(TA)

The maximum temp. will occur at a position 0.209 m
The maximum temprature is 152.6 C


## Example 2.9 Page No : 36¶

In [11]:
import math

# Variables
di = 0.15         			#m, inner diameter
do = 0.3          			#m, outer diameter
Q1 = 100.*10**3  			#W/,m**3,inner  rate of heat generation
Q2 = 40.*10**3   			#W/m**3, outer rate of heat generation
Ti = 100.       			#C, temp.at inside surface
To = 200.       			#C, temp. at outside surface
k1 = 30.        			#W/m C, thermal conductivity of material for inner layer
k2 = 10.        			#W/m C, thermal conductivity of material for outer layer

# Calculations and Results
#T1 = 364+100*math.log(r)-833.3*r**2          (1)
#T2 = 718+216*math.log(r)-1000*r**2           (2)
#(b)from eq. 1
r = math.sqrt(100./2*833.3)
print "This radial position does not fall within layer 1. Therefore no temprature maximum occurs in this layer."
#similarly
print " Similarly no temprature maximum occurs in  layer 2."
ro = di        		        # m, outer boundary
Tmax = To
print "The maximum temprature at the outer boundary is %.0f C"%(Tmax)

This radial position does not fall within layer 1. Therefore no temprature maximum occurs in this layer.
Similarly no temprature maximum occurs in  layer 2.
The maximum temprature at the outer boundary is 200 C