# Chapter4-Heat transfer theory¶

## Ex1-pg88¶

In :
import math
## Example 4.1
print('Example 4.1\n\n');
print('Page No. 88\n\n');

## given
K = 45.## Thermal Conductivity in W/m-K
L = 5.*10**-3;## thickness in metre
T1 = 100.;## in degree celcius
T2 = 99.9;## in degree celcius
A = 1.;## Area in m^2

##By Fourier law of conduction
Q = ((K*A*(T1-T2))/L);## in Watts
print'%s %.2f %s'%('The rate of conductive heat transfer is ',Q,' W \n')

Example 4.1

Page No. 88

The rate of conductive heat transfer is  900.00  W



## Ex2-pg89¶

In :
import math
## Example 4.2
print('Example 4.2\n\n');
print('Page No. 89\n\n');
## given
K1 = 45.## Thermal Conductivity  of mild steel in W/m-K
K2 = 0.040## Thermal Conductivity  of insulaton in W/m-K
L1 = 5.*10**-3;## thickness of mild steel in metre
L2 = 50.*10**-3;## thickness of insulation in metre
T1 = 100.;## in degree celcius
T2 = 25.;## in degree celcius
A = 1.;## Area in m^2

##By Fourier law of conduction
Q = (((T1-T2)/((L1/(K1*A))+(L2/(K2*A)))))## in Watts
print'%s %.2f %s'%('The rate of conductive heat transfer is ',Q,' W \n')

Example 4.2

Page No. 89

The rate of conductive heat transfer is  59.99  W



## Ex3-pg90¶

In :
import math
## Example 4.3
print('Example 4.3\n\n');
print('Page No. 90\n\n');

## given
K1 = 26.;## Thermal Conductivity  of stainless steel in W/m-K
K2 = 0.038;## Thermal Conductivity  of insulaton in W/m-K
L1 = 3.*10**-3;## thickness of stainless steel in metre
L2 = 40.*10**-3;## thickness of insulation in metre
T1 = 105.;## in degree celcius
T2 = 25.;## in degree celcius
L = 15.;## Length of pipe in metre
d1 = 50.*10**-3;## Internal diameter of pipe in metre
d2 = 56.*10**-3;## External diameter of pipe in metre

r1 = d1/2.;## in metre
r2 = d2/2.;## in metre

rm_p = ((r2-r1)/math.log(r2/r1));## logarithmic mean radius of pipe in m
rm_i = (((r2+L2)-r2)/math.log((r2+L2)/r2));## logarithmic mean radius of insulation in m

##By Fourier law of conduction
Q = (((T1-T2)/((L1/(K1*2.*math.pi*rm_p))+(L2/(K2*2.*math.pi*rm_i)))));## in W/m
Q_L = Q*L;
print'%s %.2f %s'%('The rate of conductive heat transfer per 15 m length of pie is ',Q_L,'W')## Deviation in answer due to direct substitution

Example 4.3

Page No. 90

The rate of conductive heat transfer per 15 m length of pie is  322.84 W


## Ex4-pg93¶

In :
import math
## Example 4.4
print('Example 4.4\n\n');
print('Page No. 93\n\n');

## given
dH = 12.*10**-3;## Outer diameter of pipe in m
dC = 10.*10**-3;## Inner diameter of pipe in m
L = 1.*10**-3;## im m
h_H = 10.*10**3;## Heat Transfer Coefficient  on vapour side in W/m^2-K
h_C = 4.5*10**3;## Heat Transfer Coefficient  on vapour side in W/m^2-K
K = 26.;## Thermal Conductivity of metal in W/m-K
dM = (dH + dC)/2.;## mean diameter in m
h_Hf = 6.*10**3;## Fouling factor for hot side
h_Cf = 6.*10**3;## Fouling factor for cold side

U = (1./h_H)+((L*dH)/(K*dM))+(dH/(dC*h_C));
Uh = (1./U);## in W/m^2-K
print'%s %.2f %s'%('The original heat transfer coefficient is ',Uh,' W/sq.m K \n' )## Deviation in answer due to direct substitution

u = (1./h_H)+(1./h_Hf)+((L*dH)/(K*dM))+(dH/(dC*h_C))+(dH/(dC*h_Cf));
Uf = (1./u);## in W/m^2-K
print'%s %.2f %s'%('The final heat transfer coefficient due to fouling is ',math.ceil(Uf),' W/m^2-K \n')

Example 4.4

Page No. 93

The original heat transfer coefficient is  2447.23  W/sq.m K

The final heat transfer coefficient due to fouling is  1290.00  W/m^2-K



## Ex5-pg95¶

In :
import math
## Example 4.5
print('Example 4.5\n\n');
print('Page No. 95\n\n');

## given
m_h = 1.05;## Mass flow rate of hot liquid in kg/s
Thi = 130.;## Inlet Temperature of hot liquid in degree celcius
Tho = 30.;## Outlet Temperature of hot fluid in degree celcius
Cph =  2.45*10**3;## Specific heat capacity of hot liquid in J/kg-K

m_c = 4.10;## Mass flow rate of cold liquid in kg/s
Tci =  20.;## Inlet Temperature of cold liquid in degree celcius
Cpc =  4.18*10**3;## Specific heat capacity of cold liquid in J/kg-K

A = 6.8;## Area of heat exchanger in m^2
Q = m_h*Cph*(Thi-Tho);## in Watts

##From heat balance
## m_c*Cpc*(Tci-Tco)= m_h*Cph*(Thi-Tho)= UAlTm = Q
Tco = ((Q/(m_c*Cpc))+Tci);
print'%s %.2f %s'%(' The Outlet Temperature of cold fluid is ',Tco,' degree celcius\n')
## As counter flow  heat exchanger
T1 = Thi-Tco;
T2 = Tho-Tci;
Tm = ((T1-T2)/math.log(T1/T2));

U = (Q/(A*Tm));
print'%s %.2f %s'%('The overall heat transfer coefficient is ',U,' W/sq.m K \n')## Deviation in answer due to direct substitution

Example 4.5

Page No. 95

The Outlet Temperature of cold fluid is  35.01  degree celcius

The overall heat transfer coefficient is  1002.06  W/sq.m K



## Ex6-pg98¶

In :
import math
## Example 4.6
print('Example 4.6\n\n');
print('Page No. 98\n\n');

## given
v = 1.23;## velocity in m/s
d = 25.*10**-3;## diameter in m
p = 980.;## density in kg/m^3
u = 0.502*10**-3;## viscosity in Ns/m^2
Cp = 3.76*10**3;## Specific heat capacity in J/kg-K
K = 0.532;## Thermal conductivity in W/m-K

Re = (d*v*p)/u;##Reynolds Number
Pr = (Cp*u)/K;## Prandtl Number
Re_d = (Re)**0.8;
Pr_d = (Pr)**0.4;

## By Dittus-Boelter Equation
##Nu = 0.0232 * Re^0.8 Pr^0.4 = (hd)/K
Nu = 0.0232 * Re_d * Pr_d;## Nusselt Number
h = (Nu*K)/d;##W/m^2-K
print'%s %.2f %s'%('The film heat transfer coefficient is ',h,' W/sq.m K\n')## Deviation in answer due to direct substitution

Example 4.6

Page No. 98

The film heat transfer coefficient is  5446.85  W/sq.m K



## Ex7-pg99¶

In :
import math
## Example 4.7
print('Example 4.7\n\n');
print('Page No. 99\n\n');

## (a) without insulation
## given
d_a = 0.150;## Diameter of pipe in m
T1_a = 60.;## Surface temperature in degree celcius
T2_a = 10.;## Ambient temperature in degree celcius

##For laminar flow in pipe,h= 1.41*((T1-T2)/d)^0.25
h_a = 1.41*((T1_a-T2_a)/d_a)**0.25;##W/m^2-K
A_a = math.pi * d_a;## Surface Area per unit length in m^2/m
Q_a = h_a*A_a*(T1_a - T2_a);## in W/m
print'%s %.2f %s'%('The heat loss per unit length without insulation is ',math.ceil(Q_a),' W/m \n')

## (b) with insulation
## given
d_b = 0.200;## Diameter of pipe in m
T1_b = 20.;## Surface temperature in degree celcius
T2_b = 10.;## Ambient temperature in degree celcius

##For laminar flow in pipe,h= 1.41*((T1-T2)/d)^0.25
h_b = 1.41*((T1_b-T2_b)/d_b)**0.25;##W/m^2-K
A_b = math.pi * d_b;## Surface Area per unit length in m^2/m
Q_b = h_b*A_b*(T1_b - T2_b);## in W/m
print'%s %.2f %s'%('the heat loss per unit length with insulation is ',Q_b,' W/m')
## Deviation in answer due to direct substitution

Example 4.7

Page No. 99

The heat loss per unit length without insulation is  142.00  W/m

the heat loss per unit length with insulation is  23.56  W/m


## Ex8-pg103¶

In :
import math
## Example 4.8
print('Example 4.8\n\n');
print('Page No. 103\n\n');

## given
d = 0.100;## Diameter of pipe in m
T1 = 383.;## Surface temperature in Kelvin
T2 = 288.;## Surrounding air temperature in Kelvin
e = 0.9;## Emissivity of pipe
A = math.pi * d;## Surface Area per unit length in m^2/m

## By Stefan-Blotzmann law, the radiative heat transfer rate is   Q = 5.669*e*A*((T1/100)^4-(T2/100)^4)
Q = 5.669*e*A*((T1/100.)**4-(T2/100.)**4);## in W/m
print'%s %.2f %s'%('The radiative heat loss per unit length is ',math.ceil(Q),' W/sq.m')

Example 4.8

Page No. 103

The radiative heat loss per unit length is  235.00  W/sq.m


## Ex9-pg103¶

In :
import math
## Example 4.9
print('Example 4.9\n\n');
print('Page No. 103\n\n');

## given
A = 1.;## Area in m^2
T1 = 423.;## Surface temperature in Kelvin
T2 = 293.;## Surrounding air temperature in Kelvin
T1_c = 150.;## Surface temperature in degree celcius
T2_c = 20.;## Ambient temperature in degree celcius
e = 0.9;## Emissivity of pipe

##(a) Horizontal Pipe
d = 0.100;## Diameter of pipe in m
##For laminar flow in pipe,Q= (1.41*((T1-T2)/d)^0.25)*(T1-T2)
Q_Ca = (1.41*((T1_c-T2_c)/d)**0.25)*(T1_c-T2_c);## Convective heat transfer rate in W/m^2
## By Stefan-Blotzmann law, the radiative heat transfer rate is   Q = 5.669*e*((T1/100)^4-(T2/100)^4)
Q_Ra = 5.669*e*((T1/100.)**4-(T2/100.)**4);## in W/m^2
Q_Ta = Q_Ra + Q_Ca;## IN W/m^2
print'%s %.2f %s'%('The total heat loss from per square meter area is ',Q_Ta,' W/sq.m\n')## Deviation in answer due to direct substitution

##(b) Vertical Pipe
##For turbulent flow in pipe,Q= (1.24*(T1-T2)^1.33)
Q_Cb = (1.24*(T1-T2)**1.33);## Convective heat transfer rate in W/m^2
## By Stefan-Blotzmann law, the radiative heat transfer rate is   Q = 5.669*e*((T1/100)^4-(T2/100)^4)
Q_Rb = 5.669*e*((T1/100.)**4-(T2/100.)**4);## in W/m^2
Q_Tb = Q_Rb + Q_Cb;## IN W/m^2
print'%s %.2f %s'%('The total heat loss from per square meter area is ',math.floor(Q_Tb),' W/sq.m\n')

Example 4.9

Page No. 103

The total heat loss from per square meter area is  2358.09  W/sq.m

The total heat loss from per square meter area is  2060.00  W/sq.m



## Ex10-pg106¶

In :
import math
## Example 4.10
print('Example 4.10\n\n');
print('Page No. 106\n\n');

## given
T1 = 150.;## Surface temperature in degree celcius
T2 = 20.;## Ambient temperature in degree celcius
d = 0.100; ##Outside diametr of pipe in m
h = 10.;## Outside film coefficient in W/m^2-K
t = 25.*10**-3;## thickness of insulation in m
K = 0.040;## Thermal conductivity of insulation in W/m-K

r2 = d/2.;##in m
r1 = r2+t;## in m
Q = ((T1-T2)/((1./(2.*math.pi*r1*h))+(math.log(r1/r2)/(2.*math.pi*K))));## in W/m
print'%s %.2f %s'%('The heat loss per unit length is ',Q,' W/m')

Example 4.10

Page No. 106

The heat loss per unit length is  71.21  W/m