# Chapter 5 : Transient Heat Conduction¶

## Example 5.1 Page No : 159¶

In :
import math

# Variables
t = 0.5;			#Thickness of slab in m
A = 5;			#Area of slab in m**2
k = 1.2;			#Thermal conductivity in W/m.K
a = 0.00177;			#Thermal diffusivity in m**2/h

# Calculations
x = 0;
y = -50+(24*x)+(60*x**2)-(60*x**3);			#Temperature when x = 0
Qo = (-k*A*y);			#Heat entering the slab in W
x = 0.5;
y = -50+(24*x)+(60*x**2)-(60*x**3);			#Temperature when x = 0.5
QL = (-k*A*y);			#Heat leaving the slab in W
R = (Qo-QL);			#Rate of heat storage in W
x = 0;
z1 = 24+(120*x)-(180*x**2);			#T' when x = 0
p1 = (a*z1);			#Rate of temperature change at one side of slab in degree C/h
x = 0.5;
z2 = 24+(120*x)-(180*x**2);			#T' when x = 0.5
p2 = (a*z2);			#Rate of temperature change at one side of slab in degree C/h
#For the rate of heating or cooling to be maximum, T''' = 0
x = (120./360);

# Results
print 'a)\n \
i)Heat entering the slab is %i W \n \
ii)Heat leaving the slab is %i W \n\n\
b)Rate of heat storage is %i Wc \n \
i)Rate of temperature change at one side of slab is %3.4f degree C/h \n \
ii)Rate of temperature change at other side of slab is %3.4f degree C/h \
\n\nd)For the rate of heating or cooling to be maximum x = %3.2f'%(Qo,QL,R,p1,p2,x)

a)
i)Heat entering the slab is 300 W
ii)Heat leaving the slab is 183 W

b)Rate of heat storage is 117 Wc
i)Rate of temperature change at one side of slab is 0.0425 degree C/h
ii)Rate of temperature change at other side of slab is 0.0690 degree C/h

d)For the rate of heating or cooling to be maximum x = 0.33


## Example 5.2 Page No : 164¶

In :
import math

# Variables
A = (0.4*0.4);			#Area of copper slab in m**2
t = 0.005;			#Thickness of copper slab in m
T = 250.;			#Uniform teperature  in degree c
Ts = 30.;			#Surface temperature in degree C
Tsl = 90.;			#Slab temperature in degree C
p = 9000.;			#Density in kg/m**3
c = 380.;			#Specific heat in J/kg.K
k = 370.;			#Thermal conductivity in W/m.K
h = 90.;			#Heat transfer coefficient in W/m**2.K

# Calculations
A1 = (2*A);			#Area of two sides in m**2
V = (A*t);			#Volume of the slab in m**3
Lc = (V/A1);			#Corrected length in m
Bi = ((h*Lc)/k);			#Biot number
t = -math.log((Tsl-Ts)/(T-Ts))/((h*A1)/(p*c*V));			#Time at which slab temperature becomes 90 degree C in s
y = (h*A1)/(p*c*V);

# Results
print 'Time at which slab temperature becomes 90 degree C is %3.2f s'%(t)

Time at which slab temperature becomes 90 degree C is 123.43 s


## Example 5.3 Page No : 164¶

In :
import math

# Variables
D = 0.01;			#Outer diameter of the rod in m
T = 320.;			#Original temperature in degree C
Tl = 120.;			#Temperature of liquid in degree C
h = 100.;			#Heat transfer coefficient in W/m**2.K
Tf = 200.;			#Final temperature of rod in degree C
k = 40.;			#Thermal conductivity in W/m.K
c = 460.;			#Specific heat in J/kg.K
p = 7800.;			#Density in kg/m**3

# Calculations
V = (3.14*D**2*1)/4;		#Volume of rod in m**3 taking 1m length
A = (3.14*D*1);		    	#Surface area of rod in m**2 taking 1m length
Lc = (D/4);			        #Corrected length in m
Bi = ((h*Lc)/k);			#Biot number
t = -math.log((Tf-Tl)/(T-Tl))/((h*4)/(p*c*D));			#Time at which rod temperature becomes 200 degree C in s

# Results
print 'Time at which rod temperature becomes 200 degree C is %3.2f s'%(t)

Time at which rod temperature becomes 200 degree C is 82.19 s


## Example 5.4 Page No : 165¶

In :
import math

# Variables
w = 5.5;			#Weight of the sphere in kg
Ti = 290.;			#Initial temperature in degree C
Tl = 15.;			#Temperature of liquid in degree C
h = 58.;			#Heat transfer coefficient in W/m**2.K
Tf = 95.;			#Final temperature in degree C
k = 205.;			#Thermal conductivity in W/m.K
c = 900.;			#Specific heat in J/kg.K
p = 2700.;			#Density in kg/m**3

# Calculations
V = (w/p);			#Volume of the sphere in m**3
R = ((3*V)/(4*3.14))**(1./3);			#Radius of sphere in m
Lc = (R/3.);			#Corrected length in m
t = -math.log((Tf-Tl)/(Ti-Tl))/((h*3)/(p*c*R));			#Time at which rod temperature becomes 95 degree C in s

# Results
print 'Time at which rod temperature becomes 95 degree C is %3.0f s'%(t)

# note : answer is slightly different because of rouding off error.

Time at which rod temperature becomes 95 degree C is 1356 s


## Example 5.5 Page No : 166¶

In :
import math

# Variables
Ti = 100.;			#Temperature of air in degree C
t = 0.03;			#Thickness of slab in m
To = 210.;			#Initial temperature of the plate in degree C
t = 300.;			#Time for attaining temperature in s
T = 170.;			#Temperature decreased in degree C
c = 380.;			#Specific heat in J/kg.K
p = 9000.;			#Density in kg/m**3

# Calculations
Lc = (t/2);			#Corrected length in m
h = -math.log((T-Ti)/(To-Ti))/((t*10**4)/(p*c*Lc));			#Heat transfer coefficient in W/m**2.K

# Results
print 'Heat transfer coefficient is %3.2f W/m**2.K'%(h)

Heat transfer coefficient is 77.29 W/m**2.K


## Example 5.6 Page No : 167¶

In :
import math

# Variables
D = 0.00071;			#Diameter of thermocouple in m
h = 600.;			#Heat transfer coefficient in W/m**2.K
c = 420.;			#Specific heat in J/kg.K
p = 8600.;			#Density in kg/m**3

# Calculations
t = (p*c*D)/(4*h);			#Time period in s
T = math.exp(-1);			#Temperture distribution ratio
t1 = (4*t);			#Total time in s

# Results
print 'At the end of time period t* = %3.3f s the temperature difference \
between the body and the source would be %3.3f of the initial temperature differnce.\n\
To get a true reading of gas temperature, it should be recorded after 4t*  =  %i seconds after\
the thermocouple has been \nintroduced into the stream'%(t,T,t1)

At the end of time period t* = 1.069 s the temperature difference  between the body and the source would be 0.368 of the initial temperature differnce.
To get a true reading of gas temperature, it should be recorded after 4t*  =  4 seconds after the thermocouple has been
introduced into the stream


## Example 5.8 Page No : 177¶

In :
import math

# Variables
x = 0.2;			#Distance of plane from the wall in m
t = 10;			#Time for heat flow in h
T = [25,800];			#Initial and final tempertaure in degree C
k = 0.8;			#Thermal conductivity in W/m.K
a = 0.003;			#Thermal diffusivity in m**2/h

# Calculations
X = (x*(2*math.sqrt(a*t)));			#Calculation of X for erf function
Y = 0.585                           #erf(X) from table 5.1
Ti = T-((T-T))*Y;			#Temperarture of the plane in degree C
Qi = ((-k*(T-T)*math.exp(-x**2/(4*a*t)))/(math.sqrt(3.14*a*t)));			#Instanteneous heat flow rate per unit area in W/m**2
Q = ((2*k*(T-T)*3600)/(math.sqrt((3.14*a)/t)))/10**8;			            #Total heat energy taken up by the wall in 10 hours in J/m**2
print
# Results
print 'Temperarture of the plane is %3.2f degree C\nInstanteneous heat flow rate per \
unit area is %i W/m**2 \nTotal heat energy taken up by the wall in 10 hours is %3.3f*10**8 J/m**2'%(Ti,Qi,Q)

Temperarture of the plane is 346.62 degree C
Instanteneous heat flow rate per  unit area is 1447 W/m**2
Total heat energy taken up by the wall in 10 hours is 1.454*10**8 J/m**2


## Example 5.9 Page No : 177¶

In :
# Variables
Tc = 55;			#Tempertaure of the concrete in degree C
Ts = 35;			#Temperature lowered in degree C
Tf = 45;			#Final temperature in degree C
x = 0.05;			#Depth of the slab in m
k = 1.279;			#Thermal conductivity in W/m.K
a = 0.00177;			#Thermal diffusivity in m**2/h

# Calculations
T = (Tf-Ts)/(Tc-Ts);			#Temperature distribution
X = 0.485;			#Taking 0.5 = erf(0.482) from table 5.1 on page no. 175
t = (x**2)/(4*X**2*a);			#Time taken to cool the concrete to 45 degree C in h

# Results
print 'Time taken to cool the concrete to 45 degree C is %3.2f h'%(t)

Time taken to cool the concrete to 45 degree C is 1.50 h


## Example 5.10 Page No : 178¶

In :
import math

# Variables
q = (0.3*10**6);			#Heat flux in W/m**2
t = (10./60);			#Time taken for heat transfer in s
Ti = 30.;			#Initial temperature of the slab in degree C
x = 0.2;			#Distance of the plane from the surface in m
k = 386.;			#Thermal conductivity in W/m.K
a = 0.404;			#Thermal diffusivity in m**2/h

# Calculations
Ts = ((q*math.sqrt(3.14*a*t))/k)+Ti;			#Surface temperature in degree C
X = (x/(2*math.sqrt(a*t)));			#X for calculating erf function
Y = 0.4134;			#Taking ref(0.385) = 0.4134 from table 5.1 on page no. 175
T = Ts-(Y*(Ts-Ti));			#Tempertaure at a distance of 20 cm from the surface after 10 min in degree C

# Results
print 'Tempertaure at a distance of 20 cm from the surface after 10 min is %3.2f degree C'%(T)

Tempertaure at a distance of 20 cm from the surface after 10 min is 239.63 degree C


## Example 5.11 Page No : 178¶

In :
import math

# Variables
a = 0.405;			#Thermal diffusivity in m**2/h
Ti = 100;			#Initial temperture in degree C
Tf = 0;			#Final tempertaure in degree C
Tg = (4*100);	#Temperature gradient in degree C/m
t1 = 1;			#Time taken in m

# Calculations
t = (Ti-Tf)**2/(Tg**2*3.14*a);			#Time required for the temperature gradient at the surface to reach 4 degree/cm in h
x = math.sqrt(2*a*(t1/60.));			#The depth at which the rate of cooling is maximum after 1 minute in m

# Results
print 'Time required for the temperature gradient at the surface to reach 4 degree/cm is %3.3f h \
\nThe depth at which the rate of cooling is maximum after 1 minute is %3.4f m'%(t,x)

Time required for the temperature gradient at the surface to reach 4 degree/cm is 0.049 h
The depth at which the rate of cooling is maximum after 1 minute is 0.1162 m


## Example 5.12 Page No : 185¶

In :
# Variables
x = 0.1;			#Thickness of the slab in m
Ti = 500;			#Initial temperature in degree C
Tl = 100;			#Liquid temperature in degree C
h = 1200;			#Heat transfer coefficient in W/m**2.K
t = (1*60);			#Time for immersion in s
k = 215;			#Thermal conductivity in W/m.K
a = (8.4*10**-5);			#Thermal diffusivity in m**2/h
c = 900;			#Specific heat in J/kg/K
p = 2700;			#Density in kg/m**3

# Calculations
X = (a*t)/(x/2)**2;			#Calculation for input in Heisler charts
B = (k/(h*(x/2)));			#Calculation for input in Heisler charts
T1 = 0.68;			#T value taken from Fig. 5.7 on page no. 183
Tc1 = (T1*(Ti-Tl));			#Temperature in degree C
To = Tc1+Tl;			#Temperature in degree C
T2 = 0.880;			#From Fig 5.8 on page no. 184 at x/L = 1.0 and for k/hL = 3.583, tempertaure in degree C
Tc2 = (T2*(To-Tl))+Tl;			#Temperature in degree C
Y = (h**2*a*t)/(k**2);			#Y to calculate the energy losses
Bi = (h*(x/2))/k;			#Biot number
U = 0.32;			#U/Uo value from Fig. 5.9 on page no.185
Uo = (p*c*x*(Ti-Tl));			#For unit area in J/m**2
U1 = (U*Uo)/(10**6);			#Heat removed per unit surface area in MJ/m**2

# Results
print 'Temperature at the centreline and the surface 1 minute after the immersion is %3.2f degree C  \n \
Heat removed per unit surface area is %3.1f*10**6 J/m**2'%(Tc2,U1)

Temperature at the centreline and the surface 1 minute after the immersion is 339.36 degree C
Heat removed per unit surface area is 31.1*10**6 J/m**2


## Example 5.13 Page No : 186¶

In :
# Variables
D = 0.12;			#Diameter of cylinder in m
Ti = 20;			#Initial temperature in degree C
Tf = 820;			#Temperature of furnace in degree C
h = 140;			#Heat transfer coefficient in W/m**2.K
Ta = 800;			#Axis temperature in degree C
r = 0.054;			#Radius in m
k = 21;			#Thermal conductivity in W/m.K
a = (6.11*10**-6);			#Thermal diffusivity in m**2/h

# Calculations
Bi = (h*(D/2))/(2*k);			#Biot number
T = (Ta-Tf)/(Ti-Tf);			#Temperature distribution
Fo = 5.2;			#Umath.sing Fig.5.10, on page no.187 for 1./(2Bi) = 2.5
t = (Fo*(D/2)**2)/a;			#Time required for the axis temperature to reach 800 degree C in s
r1 = (r/(D/2));			#Ratio at a radius of 5.4 cm
X = 0.85;			#From Fig.5.11 on page no. 188 the temperature at r = 5.4 i sgiven by X
T1 = X*(Ta-Tf)+Tf;			#Temperature at a radius of 5.4 cm at that tim ein degree C

# Results
print 'Time required for the axis temperature to reach 800 degree C is %3.0f s \n \
Temperature at a radius of 5.4 cm at that time is %i degree C'%(t,T1)

Time required for the axis temperature to reach 800 degree C is 3064 s
Temperature at a radius of 5.4 cm at that time is 803 degree C


## Example 5.14 Page No : 189¶

In :
import math
import math
# Variables
r = 0.01;			#Radius of the mettalic sphere in m
Ti = 400.;			#Initial temperature in degree C
h1 = 10.;			#Heat transfer coefficient in W/m**2.K
Ta = 20.;			#Temperature of air in degree C
Tc = 335.;			#Central temperature in degree C
Tw = 20.;			#Temperature of water bath in degree C
h2 = 6000.;			#Heat transfer coefficient in W/m**2.K
Tf = 50.;			#Final temperature of the sphere in degree C
k = 20.;			#Thermal conductivity in W/m.K
a = (6.66*10**-6);			#Thermal diffusivity in m**2/h
c = 1000.;			#Specific heat in J/kg/K
p = 3000.;			#Density in kg/m**3

# Calculations
Bi1 = (h1*r)/(3*k);			#Biot number
t = ((p*r*c)/(3*h1)*math.log((Ti-Ta)/(Tc-Ta)))			#Time required for cooling in air in s
Bi2 = (h2*r)/(3*k);			#Biot number
X = 1./(3*Bi2);			#X value for lumped capacity method
T = (Tf-Ta)/(Tc-Ta);			#Temperature distribution
Fo = 0.5;			#Umath.sing Fig.5.13, on page no.190
t1 = (Fo*r**2)/a;			#Time required for cooling in water in s
Z = 0.33;			#Umath.sing Fig.5.14, on page no.191
Tr = Z*(Tf-Ta)+Ta;			#Surface temperature at the end of cooling in degree C

# Results
print 'Time required for cooling in air is %3.0f s \n \
Time required for cooling in water is %3.1f s \n \
Surface temperature at the end of cooling is %3.0f degree C'%(t,t1,Tr)

Time required for cooling in air is 188 s
Time required for cooling in water is 7.5 s
Surface temperature at the end of cooling is  30 degree C


## Example 5.15 Page No : 192¶

In :
import math
# Variables
Ti = 250.;			#Temperature of aluminium slab in degree C
Tc = 50.;			#Convective environment temperature in degree C
h = 500.;			#Heat transfer coefficient in W/m**2.K
x = 0.05;			#Depth of the plane in m
t = (1.*3600);			#Time in s
k = 215.;			#Thermal conductivity in W/m.K
a = (8.4*10**-5);			#Thermal diffusivity in m**2/h

# Calculations
X = (h*math.sqrt(a*t))/k;			#X for calculating Temperature
Y = (x/(2*math.sqrt(a*t)));			#Y for calculating Temperature
Z = 0.62;			#From Fig. 5.16 on page no.193
T = (Z*(Tc-Ti)+Ti);			#Temperature at a depth of 5 cm after 1 hour in degree C

# Results
print 'Temperature at a depth of 5 cm after 1 hour is %3.0f degree C'%(T)

Temperature at a depth of 5 cm after 1 hour is 126 degree C


## Example 5.16 Page No : 196¶

In :
# Variables
D = 0.08;			#Diameter of the cylinder in m
L = 0.16;			#Length of the cylinder in m
Ti = 800;			#Initial tempertaure in degree C
Tm = 30;			#Temperature of the medium in degree C
h = 120;			#Heat transfer coefficient in W/m**2.K
t = (30*60);			#Time for cooling in s
k = 23.5;			#Thermal conductivity in W/m.K
a = 0.022;			#Thermal diffusivity in m**2/h

# Calculations
Bi2 = (h*(D/2))/k;			#2 times the Biot number
X = (a*t)/(D/2)**2;			#X for calculating C(R)
CR = 0.068;			#From Fig.5.10 on page no.187
Bi1 = (k/(h*L));			#Biot number
Y = (a*t)/L**2;			#Y for calculating P(X)
PX = 0.54;			#From Fig.5.7 on page no.183
T = CR*PX;			#Temperature at the centre of the cylinder in degree C
T30 = T*(Ti-Tm)+Tm;			#Temperature at the centre of cylinder 30 minutes after cooling is initiated in degree C

# Results
print 'Temperature at the centre of cylinder 30 minutes after cooling is initiated is %3.2f degree C'%(T30)

Temperature at the centre of cylinder 30 minutes after cooling is initiated is 58.27 degree C


## Example 5.17 Page No : 197¶

In :
import math
from numpy import *

# Variables
L = array([0.5,0.4,0.2]);			#Lengths of sides of a recmath.tangular steel billet in m
Ti = 30;			#Initial temperature in degree C
Tf = 1000;			#Final temperature in degree C
t = (90*60);			#Time for heating in s
h = 185;			#Heat transfer coefficient in W/m**2.K
k = 37;			#Thermal conductivity in W/m.K
a = 0.025;			#Thermal diffusivity in m**2/h

# Calculations
L1 = L/2;			#L values of the parallelepiped in m
Bi1 = (h*L)/k;			#Biot number
X1 = (a*t)/L**2;			#X1 for calculating P(X1)
PX1 = 0.68;			#P(X1) value from From Fig.5.7 on page no.183
Bi2 = (h*L)/k;			#Biot number
X1 = (a*t)/L**2;			#X1 for calculating P(X2)
PX2 = 0.57;			#P(X2) value from From Fig.5.7 on page no.183
Bi3 = (h*L)/k;			#Biot number
Y = (1./Bi3);			#Inverse of Biot number
X1 = (a*t)/L**2;			#X1 for calculating P(X3)
PX3 = 0.22;			#P(X3) value from From Fig.5.7 on page no.183
T = PX1*PX2*PX3;			#Temperature at the centre of billet in degree C
T1 = T*(Ti-Tf)+Tf;			#Temperature at the centre of cylinder 90 minutes after heating is initiated in degree C

# Results
print 'Temperature at the centre of cylinder 90 minutes after heating is initiated is %3.2f degree C'%(T1)

Temperature at the centre of cylinder 90 minutes after heating is initiated is 917.29 degree C


## Example 5.18 Page No : 202¶

In :
# Variables
Ti = 30;			#Initial temperature of the slab in degree C
q = (2*10**5);			#Constant heat flux in W/m**2
k = 400;			#Thermal conductivity in W/m.K
a = (117*10**-6);			#Thermal diffusivity in m**2/h
n = 0.075;			#Nodal spacing in m
x = 0.15;			#Depth in m
t = (4*60);			#Time elapsed in s

#CALCULATION
R = (x**2/(a*t));			#R value for t1
t1 = (n**2/(R*a));			#Value of t1 in s
To = 121.9;			#The surface temperature after 4 min in degree C from the table on page no. 203
T2 = 64;			#Temperature at 0.15 m from the surface after 4 minutes in degree C from the table on page no. 203

# Results
print 'The surface temperature after 4 min is %3.1f degree C  \n \
Temperature at 0.15 m from the surface after 4 minutes is %i degree C'%(To,T2)

The surface temperature after 4 min is 121.9 degree C
Temperature at 0.15 m from the surface after 4 minutes is 64 degree C


## Example 5.19 Page No : 205¶

In :
# Variables
t = 0.6;			#Thickness of the wall in m
x = 0.1;			#x value taken from Fig.Ex. 5.19 on page no. 205
Ti = 20;			#Initial temperature in degree C
T = [150,300];			#Temperatures of the sides of the wall in degree C
Tf = 150;			#Final temperature of the wall in degree C
a = (1.66*10**-3);			#Thermal diffusivity in m**2/h

# Calculations
t = (x**2/(2*a));			#Length of one time increment in h
t1 = (9*t);	        		#Elapsed time in h

# Results
print 'Elasped time before the centre of the wall attains a temperature of 150 degree C is %3.0f h'%(t1)

Elasped time before the centre of the wall attains a temperature of 150 degree C is  27 h


## Example 5.20 Page No : 206¶

In :
# Variables
k = 0.175;			#Thermal conductivity in W/m.K
a = (0.833*10**-7);			#Thermal diffusivity in m**2/h
Th = 144;			#Heated temeparture in degree C
Tc = 15;			#Cooled temperature in degree C
x = 0.02;			#Thickness of the plate in m
h = 65;			#Heat transfer coefficient in W/m**2.K
t = (4*60);			#Tiem elapsed in s

# Calculations
s = 0.002;			#Space increment in m from FIg. Ex. 5.20 on page no. 207
t1 = (s**2/(2*a));			#Time increment for the space increment in s
x1 = (k/h);			#Convective film thickness in mm
Tn = 114;			#Temperature at the centre in degree C from Fig. Ex.5.20 on page no. 207
Ts = 50;			#Surface temperature in degree C from Fig. Ex.5.20 on page no. 207

# Results
print 'Temperature at the centre is %i degree C \n \
Surface temperature is %i degree C'%(Tn,Ts)

Temperature at the centre is 114 degree C
Surface temperature is 50 degree C


## Example 5.21 Page No : 213¶

In :
import math
# Variables
t = 24;			#Time period in h
T = [-10,10];			#Range of temperatures in degree C
x = 0.1;			#Depth in m
c = 1970;			#Specific heat in J/kg/K
p = 1000;			#Density in kg/m**3
k = 0.349;			#Thermal conductivity in W/m.K
ta = 5;			#Time in h

# Calculations
w = (2*3.14)/t;			#Angular velocity in rad/h
Tm = (T+T)/2;			#Mean teperature in degree C
Tmax = T-Tm;			#Maximum temperature in degree C
a = ((k*3600)/(p*c));			#Thermal diffusivity in m**2/h
Txmax = Tmax*exp(-math.sqrt(w/(2*a))*x);			#Amplitude of temperature variation in degree C
t1 = math.sqrt(1./(2*a*w))*x;			#Time lag of temperature wave at a depth of 0.1 m in h
t2 = (3.14/w);			#Time for surface temperature is minimum in h
t3 = t2+ta;			#Time in h
Tx = Tmax*exp(-math.sqrt(w/(2*a))*x)*math.cos((w*t3)-(x*x*math.sqrt(w/(2*a))));			#Temperature at 0.1m 5 hours after the surface temperature reaches the minimum in degree C

# Results
print 'Amplitude of temperature variation at a depth of 0.1m is %3.2f degree C  \n \
Time lag of temperature wave at a depth of 0.1 m is %3.2f h  \n \
Temperature at 0.1m 5 hours after the surface temperature reaches the minimum is %3.3f degree C'%(Txmax,t1,Tx)

Amplitude of temperature variation at a depth of 0.1m is 2.39 degree C
Time lag of temperature wave at a depth of 0.1 m is 5.47 h
Temperature at 0.1m 5 hours after the surface temperature reaches the minimum is -0.946 degree C


## Example 5.22 Page No : 214¶

In :
import math

# Variables
T = [800.,200.];			#Limits in which temperature varies in degree C
t = 12.;			#Cycle time in h
x = 0.1;			#Depth of penetration in m
k = 1.8;			#Thermal conductivity in W/m.K
a = 0.02;			#Thermal diffusivity in m**2/h

# Calculations
w = (2*3.14)/t;			#Angular velocity in rad/h
t1 = math.sqrt(1./(2*a*w))*x;			#Time lag in h
Tmax = (T-T)/2;			#Range of maximum temperature in degree C
q = ((2*k*Tmax)/math.sqrt(math.pi/6*a))*(3600./1000);			#Heat flow through the surface in kJ/m**2

# Results
print 'i)Time lag of the temperature wave at a depth of 10 cm from the inner surface is %3.2f h \n \
ii)The flow through a surface located at a distance of 10 cm from the surface during the first\
six hours interval while the temperature is above the mean value is %i kJ/m**2'%(t1,q)

i)Time lag of the temperature wave at a depth of 10 cm from the inner surface is 0.69 h
ii)The flow through a surface located at a distance of 10 cm from the surface during the first six hours interval while the temperature is above the mean value is 37993 kJ/m**2


## Example 5.23 Page No : 215¶

In :
# Variables
N = 2000;			#Speed of the engine
a = 0.06;			#Thermal diffusivity in m**2/h

# Calculations
t = 1./(60*N);			#Period of on oscillation in h
x = (1.6*math.sqrt(3.14*a*t))*1000;			#Depth of penetration in mm

# Results
print 'Depth of penetration of the temperature oscillation into the cylinder wall of a \
single acting cylinder two stroke IC engine is%3.0f mm'%(x)

Depth of penetration of the temperature oscillation into the cylinder wall of a single acting cylinder two stroke IC engine is  2 mm


## Example 5.24 Page No : 218¶

In :
import math
# Variables
Tc = 55;			#Tempaerature of concrete hyway in degree C
Tl = 35;			#Temperature lowered in degree C
Tf = 45;			#Final temperature in degree C
x = 0.05;			#Depth in m
k = 1.279;			#Thermal conductivity in W/m.K
a = (1.77*10**-3);			#Thermal diffusivity in m**2/h

# Calculations
t = 1.4;			#Time taken from page no. 219 in h
q = 2*(k*(Tl-Tf))/(math.sqrt(3*a*t));			#Instantaneous heat removal rate in W/m**2

# Results
print 'Instantaneous heat removal rate is %3.1f W/m**2'%(q)

Instantaneous heat removal rate is -296.7 W/m**2