# Chapter 01:Fundamental concepts¶

## Ex1.1:pg- 4¶

In :
print"Introduction to heat transfer by S.K.Som, Chapter 1, Example 1"
#The temprature of two faces of the slabs are T1=40Â°C & T2=20Â°C
#The thickness of the slab(L) is 80mm or .08m
#The thermal conductivity(k)of the material is .20 W/(m*K)
T1=40;
T2=20;
L=.08;
k=.20;
#The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L
print"The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m**2 "
q=k*(T1-T2)/L
print"q=",q

Introduction to heat transfer by S.K.Som, Chapter 1, Example 1
The steady state heat transfer rate per unit area through the thick slab is given by q=k(T1-T2)/L in W/m**2
q= 50.0


## Ex1.2:pg- 4¶

In :
print"Introduction to heat transfer by S.K.Som, Chapter 1, Example 2"
#The thermal conductivity(km)of masonry wall is .8 W/(mK)
#The thermal conductivity(kc)of composite wall is .2 W/(mK)
#The thickness of composite wall(Lc) is 100 mm or .1 m
km=.8;
kc=.2;
Lc=.1;
#The thickness of masonry wall(Lm) is to be found.
#The steady state heat flow(qm)through masonry wall is km(T1-T2)/L
# The steady state heat flow(qc)through composite wall is kc(T1-T2)/L
#As the steady rate of heat flow through masonry wall is 80% that through composite wall and both the wall have same surface area and same temp. difference so qm/qc=0.8=(km/kc)*(Lc/Lm)
#The thickness of masonry wall is Lm.
print"The thickness of masonry wall is Lm in m"
Lm=(km/kc)*(Lc/(0.8))
print"Lm=",Lm

Introduction to heat transfer by S.K.Som, Chapter 1, Example 2
The thickness of masonry wall is Lm in m
Lm= 0.5


## Ex1.4:pg-8¶

In :
print"Introduction to heat transfer by S.K.Som, Chapter 1, Example 4"
#The average forced convective heat transfer coefficient(hbr) is 200 W/( m**2 Â°C)
#The fluid temprature(Tinf) upstream of the cold surface is 100Â°C
#The surface temprature(Ts) is 20Â°C
hbr=200;
Tinf=100;
Ts=20;
#The rate of heat transfer per unit area is q
print"The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m**2"
q=hbr*(Tinf-Ts)
print"q=",q

Introduction to heat transfer by S.K.Som, Chapter 1, Example 4
The rate of heat transfer per unit area q=hbr*(Tinf-Ts) in W/m**2
q= 16000


## Ex1.5:pg-9¶

In :
print"Introduction to heat transfer by S.K.Som, Chapter 1, Example 5"
#The average heat transfer coefficient(hbr) is 800 W/(m**2Â°C)
#The surface temprature of heat exchanger is 75Â°C and air temprature is 25Â°C so deltaT=(75-25)
#The amount of heat exchanged(Q) is 20 MJ/h
#The heat exchanger surface area(A) is given by A=Q/(hbr*âˆ†T)
hbr=800;
deltaT=(75-25);
Q=20;
print"The heat exchanger surface area(A)in m**2 required for 20 MJ/h of heating is "
A = (Q*10**6)/(3600*hbr*deltaT)
print"A=",A

Introduction to heat transfer by S.K.Som, Chapter 1, Example 5
The heat exchanger surface area(A)in m**2 required for 20 MJ/h of heating is
A= 0


## Ex1.6:pg-9¶

In :
print"Introduction to heat transfer by S.K.Som, Chapter 1, Example 6"
#The temprature of the plate(Ts) is 225Â°C
#The ambient temprature (Tinf) is 25Â°C
#The change in plate temprature with time is dT/dt=-.02K/s
#The plate area (A)=.1m**2 , mass(m)= 4Kg and specific heat(cp)=2.8KJ/(Kg*K)
#The average free convective heat coefficient(hbr) is to be found
Ts=225;
Tinf=25;
#|dT/dt|=0.2,because it is modulus function and it converts negative values to positive value.
#Let |dT/dt|=X
X=0.02;
A=.1;
m=4;
cp=2.8;
print"The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)"
Q=hbr*A*(Ts-Tinf)
print"Q=",Q
print"The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance."
print"Q=",Q
print"Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2Â°C)"
hbr=(m*cp*10**3*X)/(A*(Ts-Tinf))
print"hbr=",hbr

Introduction to heat transfer by S.K.Som, Chapter 1, Example 6
The rate of heat transfer from the plate is given by Q=hbr*A*(Ts-Tinf)
Q= 224.0
The rate of heat transfer can also be written in the form of Q=m*cp*|dT/dt| from an energy balance.
Q= 224.0
Equating the above two equations we get hbr=(m*cp*|dT/dt|)/(A*(Ts-Tinf)) in W/(m**2Â°C)
hbr= 11.2


## Ex1.7:pg-10¶

In :
print"Introduction to heat transfer by S.K.Som, Chapter 1, Example 7"
#The temprature(T) of brick wall after sunset is 50Â°C
#The emissity value(emi)=0.9
#The radiant heat flux per square meter =E/A Where E is radiant heat energy and A is area of brick wall.
#The stefan-Boltzman constant(sigma)=5.6697*10**-8 W/(m**2*K**4).
T=50;
emi=.9;
sigma=5.6697*10**-8;
print"The heat flux per square meter is given by E/A=emi*sigma*T**4 in W/m**2"
#Let E/A=F
F=emi*sigma*(T+273.15)**4
print"F=",F

Introduction to heat transfer by S.K.Som, Chapter 1, Example 7
The heat flux per square meter is given by E/A=emi*sigma*T**4 in W/m**2
F= 556.4411381


## Ex1.8:pg-11¶

In :
print"Introduction to heat transfer by S.K.Som, Chapter 1, Example 8"
#The temprature(T) of asphalt pavement = 50Â°C
#The stefan-Boltzman constant(sigma)=5.6697*10**-8 W/(m**2*K**4).
T=50;
sigma=5.6697*10**-8;
#The emitted radiant energy per unit surface area is given by (Eb/A)=sigma*T**4
print"The emitted radiant energy per unit surface area is given by Eb/A=sigma*T**4 in W/m**2"
#Let Eb/A=F
F=sigma*(50+273.15)**4
print"F=",F

Introduction to heat transfer by S.K.Som, Chapter 1, Example 8
The emitted radiant energy per unit surface area is given by Eb/A=sigma*T**4 in W/m**2
F= 618.267931222


## Ex1.9:pg-12¶

In :
print"Introduction to heat transfer by S.K.Som, Chapter 1, Example 9"
#The Thickness(L) of wall= 150 mm or 0.15 m.
#The wall on one side is exposed to air at temprature(Ta)= 60Â°C and on the other side to air at temprature(Tb) = 20Â°C
#The average convective heat transfer coefficients are hbr1=40 W/(m**2Â°C) on the 60Â°C and hbr2= 10 W/(m**2Â°C) on 20Â°C side.
#The thermal conductivity(k)=.8 W/(mÂ°C)
L=0.15;
Ta=60;
Tb=20;
hbr1=40;
hbr2=10;
k=0.8;
#Area(A=1 m**2 )since unit surface area is required.
A=1;
#The rate of heat transfer per unit surface area of wall is given by (Q/A)=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))
print"The rate of heat transfer per unit surface area of wall is given by Q/A=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))in W/m**2"
#Let Q/A=F
F=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))
print"F=",F
#The surface tempratures of wall on 60Â°C side is T1 and on 20Â°C side is T2
print"The surface tempratures of wall on 60Â°C side is T1 =Ta-(Q/(A*hbr1)) in Â°C"
T1 =Ta-(F/hbr1)
print"T1=",T1
print"The surface tempratures of wall on 20Â°C side is T2 =Tb+(Q/(A*hbr2)) in Â°C"
T2 =Tb+(F/hbr2)
print"T2=",T2

Introduction to heat transfer by S.K.Som, Chapter 1, Example 9
The rate of heat transfer per unit surface area of wall is given by Q/A=(Ta-Tb)/((1/hbr1*A)+(L/(k*A))+(1/hbr2*A))in W/m**2
F= 213.333333333
The surface tempratures of wall on 60Â°C side is T1 =Ta-(Q/(A*hbr1)) in Â°C
T1= 54.6666666667
The surface tempratures of wall on 20Â°C side is T2 =Tb+(Q/(A*hbr2)) in Â°C
T2= 41.3333333333


## Ex1.10:pg-13¶

In :
print"Introduction to heat transfer by S.K.Som, Chapter 1, Example 10"
#The spacecraft panel has thickness(L)=.01 m
#The spacecraft has inner temprature (Ti)=298 K
#The spacecraft has outer temprature(T2)
#The panel is exposed to deep space where temprature(To)= 0K
#The material has Thermal conductivity(k)= 5.0 W/(m*K)
#The emissivity(emi)=0.8
#The inner surface of the panel is exposed to airflow resulting in an average heat transfer coefficient(hbri)=70 W/(m**2*K)
L=0.01;
Ti=298.0;
To=0.0;
k=5.0;
emi=0.8;
hbri=70.0;
#The stefan Boltzman constant(sigma)= 5.67*10**-8 W/(m**2/K**4)
sigma=5.67*10**(-8);
#Heat transfer from the outer surface takes place only by radiation is given by  Q/A=emi*sigma*(T2**4-T0**4)in W/m**2=F1
#heat transfer from the outer surface can also be written as Q/A=(Ti-To)/((1/hbri)+(L/k)+(1/hr))=F2
#Radiation heat transfer coefficient(hr) is defined as Q/A=hr(T2-To)
#so hr=4.536*10**-8*T2**3
print"Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K"
print"F1=",F1
print"heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K"
print"F2=",F2
print"The values of temprature that are considered are <298 K"
for i in range(285,292):
T2=i
hr=4.536*10**(-8)*i**3
F1=emi*sigma*(T2**4-To**4)
F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))
if F1==F2:
T2=i
else:
T2=292.5
hr=4.536*10**(-8)*T2**3
F1=emi*sigma*(T2**4-To**4)
F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr))
print"Satisfactory solutions for Temprature in K is"
print"T2=",T2
print"Approximate Rate of Heat Transfer in W/m**2 is"
print"F1=",F1
print"F2=",F2

Introduction to heat transfer by S.K.Som, Chapter 1, Example 10
Heat transfer from the outer surface takes place only by radiation is given by Q/A=F1=emi*sigma*(T2**4-T0**4)in W/m**2 for different values of tempratures in K
F1= 332.029390022
heat transfer from the outer surface can also be written as Q/A=F2=(Ti-To)/((1/hbri)+(L/k)+(1/hr)) in W/m**2 at different tempratures in K
F2= 332.132667923
The values of temprature that are considered are <298 K
Satisfactory solutions for Temprature in K is
T2= 292.5
Approximate Rate of Heat Transfer in W/m**2 is
F1= 332.029390022
F2= 332.132667923


## Ex1.11:pg-15¶

In :
import math

print"Introduction to heat transfer by S.K.Som, Chapter 1, Example 11"
#The horizontal steel pipe has outer diameter(D)=80 mm or.08 m
#The pipe is maintained at a temprature(T1)=60Â°C where the air and wall temprature(T2)=20 Â°C
#The average free convective heat transfer coefficient(hbr)=6.5 W/(m**2/K) b/w the outer surface of the pipe and air
D=.08;
T1=60;
T2=20;
hbr=6.5;
#Length(L=1) since per unit length is considered
L=1;
#The surface area of pipe is given by A=(math.pi*D*L)
print"L=",L
A=(math.pi*D*L);
#The surface emissivity(emi) of steel = 0.8
#The stefan -Boltzman constant(sigma)= 5.7*10**-8 W/(m**2*K**4)
print"A=",A
sigma=5.67*10**-8;
emi=.8;
#The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4)
print"The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4) in W/m"
#Let Q/L=F
F=hbr*A*((T1+273.15)-(T2+273.15))+sigma*emi*A*((T1+273.15)**4-(T2+273.15)**4)
print"F=",F

Introduction to heat transfer by S.K.Som, Chapter 1, Example 11
L= 1
A= 0.251327412287
The total heat loss by The pipe per unit length is given by Q/L=hbr*A*(T1-T2)+sigma*emi*A*(T1**4-T2**4) in W/m
F= 121.586773684