# Chapter 11 : Conduction of heat¶

## Example 11.1 pageno : 375¶

In [1]:
# Variables
k = 0.12;			#thermal conductivity in cgs unit
t1 = 200;			#temperature at one side in deg.C
t2 = 50;			#temperature at other side in deg.C
t = 3600;			#time in sec
a = 1;			#area in sq.cm
t3 = 3;			#thickness of the plate in cm

# Calculations
q = k*a*(t1-t2)*t/t3;			#amount of heat conducted in cal

# Result
print 'the amount of heat conducted is %3.2f cal'%(q)

the amount of heat conducted is 21600.00 cal


## Example 11.2 pageno : 375¶

In [2]:
# Variables
k = 0.9;			#thermal conductivity in cgs unit
a = 10;			#area of the copper bar in sq.cm
t1 = 100;			#hot side temperature in deg.C
t2 = 20;			#cool side temperature in deg.C
d = 25;			#thickness of the bar in cm
t3 = 14;			#temperature of water when entering in deg.C

# Calculations
m = k*a*(t1-t2)/(d*(t2-t3));			#rate flow of water in gm/sec

# Result
print 'rate flow of water is %3.2f gm/sec'%(m)

rate flow of water is 4.80 gm/sec


## Example 11.3 pageno : 375¶

In [3]:
# Variables
i = 1.18;			#current in amperes
e = 20;			#potential difference across its ends in volts
j = 4.2;			#joules constant in joule/cal
a = 2*10**4;			#area of the slab in sq.cm
t = 5;			#thickness of the plate in cm
t1 = 12.5;			#temperature at hot side in K
t2 = 0;			#temperature at cold side in k

# Calculations
k = e*i*t/(j*a*(t1-t2));			#thermal conductivity in cgs unit

# Result
print 'thermal conductivity of slab is %3.5f cgs unit'%(k)

thermal conductivity of slab is 0.00011 cgs unit


## Example 11.4 pageno : 375¶

In [4]:
import math

# Variables
l = 30.;			#length of the tube in cm
t = 100.;			#temperature at outside in deg.C
t1 = 40.;			#tempertaure of water when leaving tube in deg.C
t2 = 20.;			#temperature of water when entering tube in deg.C
m = 165./60			#mass flow rete of water in cc/sec
r1 = 6.;			#internal radii in mm
r2 = 8.;			#external radii in mm

# Calculations
k = m*(t1-t2)*math.log(r2/r1)/(2*3.14*l*(t-((t1+t2)/2)));			#thermal conductivity in cgs unit

# Result
print 'thermal conductivity of the tube is %3.4f cgs unit'%(k)

thermal conductivity of the tube is 0.0012 cgs unit


## Example 11.5 pageno : 376¶

In [5]:
# Variables
l1 = 1.9;			#length of the first bar in cm
l2 = 5; 			#length of the second bar in cm
k2 = 0.92;			#thermal conductivity in cgs unit

# Calculations
k1 = k2*(l1/l2)**2;			#thermal conductivity if first bar in cgs unit

# Result
print 'thermal conductivity of first bar is %3.3f cgs unit'%(k1)

thermal conductivity of first bar is 0.133 cgs unit


## Example 11.6 pageno : 376¶

In [6]:
# Variables
k1 = 0.92;			#thermal conductivity of copper in cgs unit
k2 = 0.5;			#thermal conductivity of alluminium in cgs unit
t1 = 100;			#temperature of copper in deg.C
t2 = 0;			#temperature of alluminium in deg.C

# Calculations
t = k1*t1/(k1+k2);			#welded teperature in deg.C

# Result
print 'welded temperature is %3.1f deg.C'%(t)

welded temperature is 64.8 deg.C


## Example 11.7 pageno : 376¶

In [9]:
# Variables
w = 23;			#thermal capacity of calorimeter in cal
m = 440;			#mass of water in gm
l = 14.6;			#lenght of the rubber tube in cm
dt = 0.019;			#rate of change in temperature in deg.C/sec
t = 100;			#temperature of steam in deg.C
t1 = 22;			#temperature of the water in deg.C
t2 = t1;			#temperature of calorimeter in deg.C
r1 = 1;			#external radii in cm
r2 = 0.75;			#internal radii in cm

# Calculations
k = (w+m)*dt*math.log(r1/r2)/(2*3.14*l*(t-((t1+t2)/2)));			#thermal conductivity in cgs unit

# Result
print 'thermal cnductivity of rubber tube is %3.6f cgs unit'%(k)

thermal cnductivity of rubber tube is 0.000354 cgs unit


## Example 11.8 pageno : 377¶

In [10]:
# Variables
ti = 18;			#inside temperature in deg.C
to = 4;			#outside temperature in deg.C
k1 = 0.008;			#thermal conductivity of stone in cgs unit
k2 = 0.12;			#thermal conductivity of steel in cgs unit
t = 3600;			#time in sec
t1 = 25;			#thickness of the stone in cm
t2 = 2;			#thickness of the steel in cm
a = 10**4;			#area of the cottage in sq.cm

# Calculations
q1 = k1*a*(ti-to)*t/(t1);			#heat lost by stone per hour in cal
q2 = k2*a*(ti-to)*t/t2;			#heat lost by steel per hour in cal

# Result
print 'heat lost by stone is %3.4e cal  \
\nheat lost by steel is %.3e cal'%(q1,q2)

heat lost by stone is 1.6128e+05 cal
heat lost by steel is 3.024e+07 cal


## Example 11.9 pageno: 377¶

In [11]:
# Variables
l1 = 4;			#length of the slab1 in cm
l2 = 2;			#length of the slab2 in cm
k1 = 0.5;			#thermal conductivity in cgs unit
k2 = 0.36;			#thermal conductivity in cgs unit
t1 = 100;			#temperature of the slab1 in deg.C
t2 = 0;			#temperature of the slab2 in deg.C

# Calculations
t = k1*l2*t1/((k2*l1)+(k1*l2));			#temperature of the commaon surface in deg.C

# Result
print 'the temperature of the common surface is %3.0f deg.C'%(t)

the temperature of the common surface is  41 deg.C


## Example 11.10 pageno : 378¶

In [14]:
# The distance

# Variables
t1 = 15.;			#temperature of the one end of the slab in deg.C
t2 = 45.;			#temperature of the other end of the slab in deg.C
k = 0.3;			#thermal conductivity in cgs unit
d = 7.;			#density of the material in gm/cc
cp = 1.;			#specific heat of the material in kj/kg.K
t = 5.*3600;			#time in sec
dt = 1./10;			#thermometer reading in deg.C

# Calculations
b = (3.14*d*cp/(t*k))**(0.5);
x = (math.log((t2-t1)/dt))/b;			#distance from which temparature variation can be detected in cm

# Result
print 'the distance from which temparature variation can be detected is %3.1f cm'%(x)

the distance from which temparature variation can be detected is 89.4 cm