# Chapter 11 : Heat Transfer¶

## Example 11.1 Page No : 553¶

In [1]:
# given data
deltaX = 6./12; 			#6 inch  =  6/12 feet 			#deltaX = length 			#unit:feet
k = 0.40; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
T1 = 150; 			#temperature maintained at one face 			#fahrenheit
T2 = 80.; 			#tempetature maintained at other face 			#fahrenheit

# calculations
deltaT = T2-T1; 			#fahrenheit 			#Change in temperature
Q = (-k*deltaT)/deltaX; 			#Heat transfer per square foot of wall 			#Unit:Btu/hr*ft**2

# results
print "Heat transfer per square foot of wall is %.2f Btu/hr*ft**2"%(Q);

Heat transfer per square foot of wall is 56.00 Btu/hr*ft**2


## Example 11.2 Page No : 553¶

In [2]:
# given data
deltaX = 0.150; 			#Given,150 mm  = 0.150 meter 			# 			#deltaX = length 			#Unit:meter
k = 0.692; 			#Unit:W/(m*celcius) 			#k = proportionality constant 			#k = thermal conductivity
T1 = 70; 			#temperature maintained at one face 			#celcius
T2 = 30; 			#tempetature maintained at other face 			#celcius

# calculations
deltaT = T2-T1; 			#celcius 			#change in temperature
Q = (-k*deltaT)/deltaX; 			#Heat transfer per square foot of wall 			#unit:W/m**2

# results
print "Heat transfer per square foot of wall is %.2f W/m**2"%(Q);

Heat transfer per square foot of wall is 184.53 W/m**2


## Example 11.3 Page No : 556¶

In [5]:
#From example 11.1,
deltaX = 6./12; 			#6 inch  =  6/12 feet 			#deltaX = length 			#unit:feet
A = 1; 			#area  			#ft**2
k = 0.40; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table

Rt = deltaX/(k*A); 			#Thermal resistance 			#Unit:(hr*f)/Btu

#Q = deltaT/Rt 			#Q = heat transfer 			#ohm's law (fourier's equation)
#i = deltaE/Re 			#i = current in amperes 			#deltaE = The potential difference 			#Re = the electrical resistance 			#ohm's law
# Q/i  =  (deltaT/Rt)*(deltaE/Re)
#Q/i = 100; 			#Given 			# 1 A correspond to 100 Btu/(hr*ft**2)
deltaE = 9.; 			#Unit:Volt 			#potential difference
T1 = 150.; 			#temperature maintained at one face 			#fahrenheit
T2 = 80.; 			#tempetature maintained at other face 			#fahrenheit
deltaT = T2-T1; 			#fahrenheit 			#Change in temperature
Re = (100*deltaE*Rt)/deltaT; 			#Unit:Ohms 			#The electrical resistance needed
print "The electrical resistance needed is %.2f ohms"%(abs(Re));
i = deltaE/Re; 			#current 			#Unit:amperes
Q = 100*i; 			#Heat transfer per square foot of wall 			#Unit:Btu/hr*ft**2
print "Heat transfer per square foot of wall is %.2f Btu/hr*ft**2"%(abs(Q));

The electrical resistance needed is 16.07 ohms
Heat transfer per square foot of wall is 56.00 Btu/hr*ft**2


## Example 11.4 Page No : 558¶

In [6]:
#For Brick,
deltaX = 6./12; 			#6 inch  =  6/12 feet 			#deltaX = length 			#unit:ft
A = 1.; 			#area  			#unit:ft**2
k = 0.40; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:(hr*f)/Btu
print "For brick"
print "The resistance is %.2f hr*F)/Btu"%(R);
R1 = R;

#For Concrete,
deltaX = (1./2)/12; 			#(1/2) inch  =  (1/2)/12 feet 			#deltaX = length 			#unit:ft
A = 1; 			#area 			#ft**2
k = 0.80; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:(hr*f)/Btu
print "For Concrete"
print "The resistance is %.2f hr*F)/Btu"%(R);
R2 = R;

#For plaster,
deltaX = (1./2)/12; 			# (1/2) inch  =  6/12 feet 			#deltaX = length 			#unit:ft
A = 1; 			#area 			#ft**2
k = 0.30; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:(hr*f)/Btu
print "For plaster"
print "The resistance is %.2f hr*F)/Btu"%(R);
R3 = R;

Rot = R1+R2+R3; 			#Rot = The overall resistance 			#unit:(hr*F)/Btu
print "The overall resistance is %.2f hr*F)/Btu"%(Rot);
T1 = 70.; 			#temperature maintained at one face 			#fahrenheit
T2 = 30; 			#tempetature maintained at other face 			#fahrenheit
deltaT = T2-T1; 			#fahrenheit 			#Change in temperature
Q = deltaT/Rot; 			#Q = Heat transfer 			#Unit:Btu/(hr*ft**2); 			#ohm's law (fourier's equation)
print "Heat transfer per square foot of wall is %.2f Btu/hr*ft**2"%(abs(Q));

For brick
The resistance is 1.25 hr*F)/Btu
For Concrete
The resistance is 0.05 hr*F)/Btu
For plaster
The resistance is 0.14 hr*F)/Btu
The overall resistance is 1.44 hr*F)/Btu
Heat transfer per square foot of wall is 27.76 Btu/hr*ft**2


## Example 11.5 Page No : 558¶

In [9]:
print "In problem 11.4"
#From example 11.4,,,
#For Brick,
deltaX = 6./12; 			#6 inch  =  6/12 feet 			#deltaX = length 			#unit:ft
A = 1; 			#area  			#unit:ft**2
k = 0.40; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:(hr*f)/Btu
print "For brick"
print "The resistance is %.2f hr*F)/Btu"%(R);
R1 = R;

#For Concrete,
deltaX = (1./2)/12; 			#(1/2) inch  =  (1/2)/12 feet 			#deltaX = length 			#unit:ft
A = 1; 			#area 			#ft**2
k = 0.80; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:(hr*f)/Btu
print "For Concrete"
print "The resistance is %.2f hr*F)/Btu"%(R);
R2 = R;

#For plaster,
deltaX = (1./2)/12; 			# (1/2) inch  =  6/12 feet 			#deltaX = length 			#unit:ft
A = 1; 			#area 			#ft**2
k = 0.30; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:(hr*f)/Btu
print "For plaster"
print "The resistance is %.2f hr*F)/Btu"%(R);
R3 = R;

Rot = R1+R2+R3; 			#Rot = The overall resistance 			#unit:(hr*F)/Btu
print "The overall resistance is %.2f hr*F)/Btu"%(Rot);
T1 = 70; 			#temperature maintained at one face 			#fahrenheit
T2 = 30; 			#tempetature maintained at other face 			#fahrenheit
deltaT = T2-T1; 			#fahrenheit 			#Change in temperature
Q = deltaT/Rot; 			#Q = Heat transfer 			#Unit:Btu/(hr*ft**2);
print "Heat transfer per square foot of wall is %.2f Btu/hr*ft**2"%(abs(Q));

print "Now in problem 11.5"
deltaT = R*Q 			#ohm's law (fourier's equation) 			#Change in temperature 			#fahrenheit
#For Brick,
deltaT = Q*R1; 			#Unit:fahrenheit 			#ohm's law (fourier's equation) 			#Change in temperature
t1 = deltaT;
#For Concrete,
deltaT = Q*R2; 			#Unit:fahrenheit 			#ohm's law (fourier's equation) 			#Change in temperature
t2 = deltaT;
#For plaster,
deltaT = Q*R3; 			#Unit:fahrenheit 			#ohm's law (fourier's equation) 			#Change in temperature
t3 = deltaT;

deltaTo = t1+t2+t3; 			#Overall Change in temperature 			#fahrenheit
print "The overall change in temperature is %.2f F"%(abs(deltaTo));
#The interface temperature are:
print "The interface temperature are:";
print "For brick-concrete : %.2f fahrenheit"%(abs(T2)+abs(t1));
print "For concrete-plaster : %.2f fahrenheit"%(abs(T2)+abs(t1)+abs(t2));

In problem 11.4
For brick
The resistance is 1.25 hr*F)/Btu
For Concrete
The resistance is 0.05 hr*F)/Btu
For plaster
The resistance is 0.14 hr*F)/Btu
The overall resistance is 1.44 hr*F)/Btu
Heat transfer per square foot of wall is 27.76 Btu/hr*ft**2
Now in problem 11.5
The overall change in temperature is 40.00 F
The interface temperature are:
For brick-concrete : 64.70 fahrenheit
For concrete-plaster : 66.14 fahrenheit


## Example 11.6 Page No : 559¶

In [10]:
#For Brick,
deltaX = 0.150; 			#Unit:m 			#150 mm  =  0.150 m 			#deltaX = length 			#unit:meter
A = 1; 			#area  			#unit:m**2
k = 0.692; 			#Unit:W/(m*C) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:C/W
print "For brick"
print "The resistance is %.2f Celcius/W"%(R);
R1 = R;

#For Concrete,
deltaX = 0.012; 			#Unit:m 			#12 mm  =  0.0120 m 			#deltaX = length 			#unit:meter
A = 1; 			#area  			#unit:m**2
k = 1.385; 			#Unit:W/(m*C) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:C/W
print "For Concrete"
print "The resistance is %.2f Celcius/W"%(R);
R2 = R;

#For plaster,
deltaX = 0.0120; 			#Unit:m 			#12 mm  =  0.0120 m 			#deltaX = length 			#unit:meter
A = 1; 			#area  			#unit:m**2
k = 0.519; 			#Unit:W/(m*C) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:C/W
print "For plaster"
print "The resistance is %.2f Celcius/W"%(R);
R3 = R;

Ro = R1+R2+R3; 			#Rot = The overall resistance 			#unit:C/W
print "The overall resistance is %.2f Celcius/W"%(Ro);
T1 = 0; 			#temperature maintained at one face 			#Celcius
T2 = 20; 			#tempetature maintained at other face 			#Celcius
deltaT = T2-T1; 			#Change in temperature  			#Celcius
Q = deltaT/Ro; 			#Q = Heat transfer 			#Unit:W/m**2; 			#ohm's law (fourier's equation)
print "Heat transfer per square meter of wall is %.2f W/m**2"%(abs(Q));

For brick
The resistance is 0.22 Celcius/W
For Concrete
The resistance is 0.01 Celcius/W
For plaster
The resistance is 0.02 Celcius/W
The overall resistance is 0.25 Celcius/W
Heat transfer per square meter of wall is 80.47 W/m**2


## Example 11.7 Page No : 560¶

In [11]:
print "In problem 11.6"
#For Brick,
deltaX = 0.150; 			#Unit:m 			#150 mm  =  0.150 m 			#deltaX = length 			#unit:meter
A = 1; 			#area 			#unit:meter**2
k = 0.692; 			#Unit:W/(m*C) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:Celcius/W
print "For brick"
print "The resistance is %.2f Celcius/W"%(R);
R1 = R;

#For Concrete,
deltaX = 0.012; 			#Unit:m 			#12 mm  =  0.0120 m 			#deltaX = length 			#unit:meter
A = 1; 			#area  			#unit:meter**2
k = 1.385; 			#Unit:W/(m*C) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:Celcius/W
print "For Concrete"
print "The resistance is %.2f Celcius/W"%(R);
R2 = R;

#For plaster,
deltaX = 0.0120; 			#Unit:m 			#12 mm  =  0.0120 m 			#deltaX = length 			#unit:meter
A = 1; 			#area  			#unit:meter**2
k = 0.519; 			#Unit:W/(m*C) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
R = deltaX/(k*A); 			#Thermal resistance 			#Unit:Celcius/W
print "For plaster"
print "The resistance is %.2f Celcius/W"%(R);
R3 = R;

Ro = R1+R2+R3; 			#Rot = The overall resistance Celcius/W
print "The overall resistance is %.2f Celcius/W"%(Ro);
T1 = 0; 			#temperature maintained at one face 			#Celcius
T2 = 20; 			#tempetature maintained at other face 			#Celcius
deltaT = T2-T1; 			#Change in temperature 			#Celcius
Q = deltaT/Ro; 			#Q = Heat transfer 			#Unit:W/m**2;
print "Heat transfer per square meter of wall is %.2f W/m**2"%(abs(Q));

print "Now in problem 11.5"
#deltaT = R*Q 			#ohm's law (fourier's equation)
#For Brick,
deltaT = Q*R1; 			#Unit:Celcius 			#Change in temperature
t1 = deltaT;
#For Concrete,
deltaT = Q*R2; 			#Unit:Celcius 			#Change in temperature
t2 = deltaT;
#For plaster,
deltaT = Q*R3; 			#Unit:Celcius 			#Change in temperature
t3 = deltaT;

deltaTo = t1+t2+t3; 			#The overall Change in temperature 			#Celcius
print "The overall change in temperature is %.2f celcius"%(abs(deltaTo));
#The interface temperature are:
print "The interface temperature are:";
print "%.2f Celcius"%(abs(deltaTo)-abs(t1));
print "%.2f Celcius"%(abs(deltaTo)-abs(t1)-abs(t2));
print "%.2f Celcius"%(abs(deltaTo)-abs(t1)-abs(t2)-abs(t3));

In problem 11.6
For brick
The resistance is 0.22 Celcius/W
For Concrete
The resistance is 0.01 Celcius/W
For plaster
The resistance is 0.02 Celcius/W
The overall resistance is 0.25 Celcius/W
Heat transfer per square meter of wall is 80.47 W/m**2
Now in problem 11.5
The overall change in temperature is 20.00 celcius
The interface temperature are:
2.56 Celcius
1.86 Celcius
-0.00 Celcius


## Example 11.8 Page No : 561¶

In [6]:
from numpy.linalg import inv

deltaX = 4./12; 			#4 inch  =  6/12 feet 			#deltaX = length 			#unit:ft
A = 7*2.; 			#area 			#area = hight*width 			#unit:ft**2
k = 0.090; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity for fir 			#From the table
Rfir = deltaX/(k*A); 			#Resistance of fir 			#Unit:(hr*F)/Btu
print "For fir"
print "The resistance is %.2f hr*F)/Btu"%(Rfir);

deltaX = 4./12; 			#4 inch  =  6/12 feet 			#deltaX = length 			#unit:ft
A = 7*2; 			#area 			#area = hight*width  			#unit:ft**2
k = 0.065; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity for pine 			#From the table
Rpine = deltaX/(k*A); 			#Resistance of pine 			#Unit:(hr*F)/Btu
print "For pine"
print "The resistance is %.2f hr*F)/Btu"%(Rpine);

deltaX = 4./12; 			#4 inch  =  6/12 feet 			#deltaX = length 			#unit:ft
A = 7*2; 			#area 			#area = hight*width 			#unit:ft**2
k = 0.025; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity for corkboard 			#From the table
Rcorkboard = deltaX/(k*A); 			#Resistance of corkboard 			#Unit:(hr*F)/Btu
print "For corkboard"
print "The resistance is %.2f hr*F)/Btu"%(Rcorkboard);

Roverall = inv([[inv([[Rfir]])+inv([[Rpine]])+inv([[Rcorkboard]])]]);
print "The overall resistance is %.2f hr*F)/Btu"%(Roverall);

T1 = 60.; 			#temperature maintained at one face 			#unit:fahrenheit
T2 = 80.; 			#tempetature maintained at other face 			#unit:fahrenheit
deltaT = T2-T1; 			#Change in temperature 			#unit:fahrenheit
Qtotal = deltaT/Roverall; 			#Q = Total Heat loss 			#Unit:Btu/hr; 			#ohm's law (fourier's equation)
print "Total Heat loss from the wall is %.2f Btu/hr"%(abs(Qtotal));

#As a check,
Qfir = deltaT/Rfir; 			#Q = Fir Heat loss 			#Unit:Btu/hr; 			#ohm's law (fourier's equation)
print "Heat loss from the wall made of fir is %.2f Btu/hr"%(abs(Qfir));
Qpine = deltaT/Rpine; 			#Q = Pine Heat loss 			#Unit:Btu/hr; 			#ohm's law (fourier's equation)
print "Heat loss from the wall made of pine is %.2f Btu/hr"%(abs(Qpine));
Qcorkboard = deltaT/Rcorkboard; 			#Q = corkboard Heat loss 			#Unit:Btu/hr; 			#ohm's law (fourier's equation)
print "Heat loss from the wall made of corkboard is %.2f Btu/hr"%(abs(Qcorkboard));
Qtotal = Qfir+Qpine+Qcorkboard; 			#Total Heat loss from the wall 			#unit:Btu/hr
print "Total Heat loss from the wall is %.2f Btu/hr"%(abs(Qtotal));

For fir
The resistance is 0.26 hr*F)/Btu
For pine
The resistance is 0.37 hr*F)/Btu
For corkboard
The resistance is 0.95 hr*F)/Btu
The overall resistance is 0.13 hr*F)/Btu
Total Heat loss from the wall is 151.20 Btu/hr
Heat loss from the wall made of fir is 75.60 Btu/hr
Heat loss from the wall made of pine is 54.60 Btu/hr
Heat loss from the wall made of corkboard is 21.00 Btu/hr
Total Heat loss from the wall is 151.20 Btu/hr


## Example 11.9 Page No : 565¶

In [13]:
import math

#A bare steel pipe
ro = 3.50; 			#Outside diameter 			#Unit:in.
ri = 3.00; 			#inside diameter 			#Unit:in.
Ti = 240; 			#Inside temperature 			#unit:fahrenheit
To = 120; 			#Outside temperature 			#unit:fahrenheit
L = 5; 			#Length 			#Unit:ft
deltaT = Ti-To; 			#Change in temperature 			#unit:fahrenheit
k = 26 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity

# Calculations
Q = (2*math.pi*k*L*deltaT)/math.log(ro/ri); 			#The heat loss from the pipe 			#unit:Btu/hr

# Results
print "The heat loss from the pipe is %.2f Btu/hr"%(Q);

The heat loss from the pipe is 635856.36 Btu/hr


## Example 11.10 Page No : 566¶

In [14]:
import math

#A bare steel pipe
ro = 90.; 			#Outside diameter 			#Unit:mm
ri = 75; 			#inside diameter 			#Unit:mm
Ti = 110; 			#Inside temperature 			#Unit:Celcius
To = 40; 			#Outside temperature 			#Unit:Celcius
L = 2; 			#Length 			#Unit:m

# Calculations
deltaT = Ti-To; 			#Change in temperature 			#Unit:Celcius
k = 45 			#Unit:W/(m*C) 			#k = proportionality constant 			#k = thermal conductivity
Q = (2*math.pi*k*L*deltaT)/math.log(ro/ri); 			#The heat loss from the pipe  			#unit:W

# Results
print "The heat loss from the pipe is %.2f W"%(Q);

The heat loss from the pipe is 217111.28 W


## Example 11.11 Page No : 567¶

In [7]:
import math
from numpy.linalg import inv

#From problem 11.9,
#A bare steel pipe
r2 = 3.50; 			#Outside diameter 			#Unit:in.
r1 = 3.00; 			#inside diameter 			#Unit:in.
Ti = 240; 			#Inside temperature 			#unit:fahrenheit
L = 5; 			#Length 			#Unit:ft
k1 = [[26]]; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity
ans1 = (inv(k1)*math.log(r2/r1));

#Now,in problem 11.11,
#Mineral wool
r3 = 5.50; 			#inside diameter 			#Unit:in.
r2 = 3.50; 			#outside diameter 			#Unit:in.
To = 85; 			#Outside temperature 			#unit:fahrenheit
deltaT = Ti-To; 			#Change in temperature 			#unit:fahrenheit
k2 = [[0.026]] 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity
ans2 = (inv(k2)*math.log(r3/r2));

Q = (2*math.pi*L*deltaT)/(ans1+ans2); 			#The heat loss from the pipe  			#unit:Btu/hr

# Results
print "The heat loss from the pipe is %.2f Btu/hr"%(Q);

The heat loss from the pipe is 280.02 Btu/hr


## Example 11.12 Page No : 569¶

In [13]:
import math
from numpy.linalg import inv

#From problem 11.9,
#The bare pipe
r2 = 3.50; 			#Outside diameter 			#Unit:in.
r1 = 3.00; 			#inside diameter 			#Unit:in.
Ti = 240.; 			#Inside temperature 			#unit:fahrenheit
L = 5.; 			#Length 			#Unit:ft
k = 26.; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity
Rpipe = math.log(r2/r1)/(2*math.pi*k*L); 			#the resistance of pipe 			#Unit:(hr*F)/Btu
print "The resistance of pipe is %.5f hr*F)/Btu"%(Rpipe);

#Now,in problem 11.12,
To = 70; 			#Outside temperature 			#unit:fahrenheit
deltaT = Ti-To; 			#Change in temperature 			#unit:fahrenheit
h = 0.9; 			#Coefficient of heat transfer 			#Unit:Btu/(hr*ft**2*F)
A = (math.pi*r2)/12*L;  			#Area 			#Unit:ft**2 			#1 inch  =  1/12 feet 			#unit:ft**2
Rconvection = inv([[h*A]]); 			#The resistance due to natural convection to the surrounding air 			#Unit:(hr*F)/Btu
print "The resistance due to natural convection to the surrounding air is %.2f hr*F)/Btu"%(Rconvection);

Rtotal = Rpipe+Rconvection; 			#The total resistance   			#unit:(hr*F)/Btu
print "The total resistance is %.2f hr*F)/Btu"%(Rtotal);
Q = deltaT/Rtotal; 			#ohm's law (fourier's equation) 			#The heat transfer from the pipe to the surrounding air   			#unit:Btu/hr
print "The heat transfer from the pipe to the surrounding air is %.2f Btu/hr"%(Q);

The resistance of pipe is 0.00019 hr*F)/Btu
The resistance due to natural convection to the surrounding air is 0.24 hr*F)/Btu
The total resistance is 0.24 hr*F)/Btu
The heat transfer from the pipe to the surrounding air is 700.42 Btu/hr


## Example 11.15 Page No : 574¶

In [17]:
import math

# given data
D = 3.5/12; 			#3.5 inch  =  3.5/12 feet			#Unit:ft 			#Outside diameter
Ti = 120; 			#Inside temperature 			#unit:fahrenheit
To = 70; 			#Outside temperature 			#unit:fahrenheit
deltaT = Ti-To; 			#unit:fahrenheit 			#Change in temperature
h = 0.9; 			#Coefficient of heat transfer 			#Unit:Btu/(hr*ft**2*F)
L = 5; 			#Length 			#Unit:ft 			#From problem 11.10
A = (math.pi*D)*L;  			#Area 			#Unit:ft**2
Q = h*A*deltaT; 			#The heat loss due to convection 			#Unit:Btu/hr 			#Newton's law of cooling

# Results
print "The heat loss due to convection is %.2f Btu/hr"%(Q);

The heat loss due to convection is 206.17 Btu/hr


## Example 11.16 Page No : 575¶

In [16]:
from numpy.linalg import inv

#This problem can not be solved directly,because the individual film resistances aree functions of unknown temperature differences.Therefore,
#From the first approximation,
h = 1./2; 			#Coefficient of heat transfer 			#unit:Btu/(hr*ft**2*F)
#For area 1 ft**2,
R = (3./12)/0.07; 			#The wall resistance is deltax/(k*A) 			#k = 0.07 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity
Roverall = inv([[1./2]])+inv([[1./2]])+R; 			#the overall series resistance 			#Unit:Btu/(hr*ft*F)
print "For h = 0.5, the overall series resistance is %.2f Btu/hr*ft*F)"%(Roverall);
#Using the value of Roverall,we can now obtain Q and individual temperature differences,
Ti = 80.; 			#warm air temperature 			#unit:fahrenheit
To = 50; 			#cold air temperature 			#unit:fahrenheit
deltaT = Ti-To; 			#unit:fahrenheit  			#Change in temperature
Q = deltaT/Roverall; 			#Unit:Btu/(hr*ft**2) 			#heat transfer 			#ohm's law (fourier's equation)
print "For h = 0.5, heat transfer is %.2f Btu/hr*ft**2)"%(Q);
print "For h = 0.5"
#deltaT through the hot air film is Q/(1/2)
print "Temperaure difference through the hot air film is %.2f F"%(Q/1./2);
#Throught the wall deltaT is R*Q
print "Temperaure difference through the wall is %.2f F"%(Q*R);
#deltaT through the cold air film is Q/(1/2)
print "Temperaure difference through the cold air film is %.2f F"%(Q/1./2);

#With these temperature differences,we can now enter figures 11.12 and 11.14 to verify our approximation.From figure 11.14,we find h = 0.42 Btu/(hr*ft*2*F)
#Using h = 0.42,we have for the overall resistance (1/0.42)+(1/0.42)+R
h = [[0.42]]; 			#Coefficient of heat transfer  			#unit:Btu/(hr*ft**2*F)
Roverall = inv(h)+inv(h)+R; 			#the overall series resistance 			#Unit:Btu/(hr*ft*F)
print "For h = 0.42, the overall series resistance is %.2f Btu/hr*ft*F)"%(Roverall);
Q = deltaT/Roverall; 			#Unit:Btu/(hr*ft**2) 			#heat transfer 			#ohm's law (fourier's equation)
print "For h = 0.42, heat transfer is %.2f Btu/hr*ft**2)"%(Q);
print "For h = 0.42"
# deltat through both air films is Q/h
print "Temperaure difference through the hot and cold air film is %.2f F"%(Q/h);
#and through the wall,deltat is Q*R
print "Temperaure difference through the wall is %.2f F"%(Q*R);

#Entering figure 11.14,we find that h stays essentially 0.42,and our solution is that the heat flow is Q,the "hot" side of the wall is at Ti-(Q/h),the "cold" side  is at To+(Q/h) ,and temperature drop in the wall is Ti-(Q/h)-(To+(Q/h)).
print "The temperature drop on the hot side of the wall is %.2f F"%(Ti-Q/h)
print "The temperature drop on the cold side of the wall is %.2f F"%(To+Q/h)
print "The temperature drop in the wall is %.2f F"%(((Ti-Q/h)-To+Q/h));
#Which checks our wall deltat calculation.

For h = 0.5, the overall series resistance is 7.57 Btu/hr*ft*F)
For h = 0.5, heat transfer is 3.96 Btu/hr*ft**2)
For h = 0.5
Temperaure difference through the hot air film is 1.98 F
Temperaure difference through the wall is 14.15 F
Temperaure difference through the cold air film is 1.98 F
For h = 0.42, the overall series resistance is 8.33 Btu/hr*ft*F)
For h = 0.42, heat transfer is 3.60 Btu/hr*ft**2)
For h = 0.42
Temperaure difference through the hot and cold air film is 8.57 F
Temperaure difference through the wall is 12.86 F
The temperature drop on the hot side of the wall is 71.43 F
The temperature drop on the cold side of the wall is 58.57 F
The temperature drop in the wall is 30.00 F


## Example 11.17 Page No : 578¶

In [22]:
#The first step is to check Reynolds number.It will be recalled that the Reynolds number is given by (D*V*rho)/mu and is dimensionless.Therefore,we can use D,      diameter in feet;V velocity in ft/hr;rho density in lbm/ft**3 and mu vismath.cosity in lbm/(ft*hr).
#Alternatively,the Reynolds number is given by (D*G)/mu,where G is the mass flow rate per unit area (lbm/(hr*ft**2)).
G = ((20*60)*(4*144)/(math.pi*0.87**2)); 			#Unit:lbm/(hr*ft**2)  			#Inside diameter = 0.87 inch 			#			#1 in.**2 = 144 ft**2 			#20 lbm/min of water(min converted to second)
#the vismath.cosity of air at these conditions is obtained from figure 11.17 as 0.062 lbm/(ft*hr).So,
mu = 0.33; 			#the vismath.cosity of air 			#unit:lbm/(ft*hr)
D = 0.87/12; 			#Inside diameter 			#1 in**2 = 144 ft**2
#Therefore Reynolds number is
Re = (D*G)/mu; 			#Reynolds number
#which is well into the turbulent flow regime.
print "The Reynolds number is %.2f"%(Re);
#The next step is to enter Figure 11.18 at W/1000 of 20*(60/1000) = 1.2 and 400F to obtain h1 = 630.
#From the figure 11.20,we obtain F = 1.25 for an inside diameter of 0.87 inch.So,
h1 = 630; 			#basic heat transfer coefficient 			#unit:Btu/(hr*ft**2*F)
F = 1.25; 			#correction factor
h = h1*F; 			#heat transfer coefficient 			#the inside film coefficient 			#unit:Btu/(hr*ft**2*F)
print "The heat-transfer coefficient is %.2f Btu/hr*ft**2*F)"%(h);

The Reynolds number is 63861.54
The heat-transfer coefficient is 787.50 Btu/hr*ft**2*F)


## Example 11.18 Page No : 579¶

In [23]:
#We first check the Reynolds number and note that G is same as for problem 11.17.So,
#G is the mass flow rate per unit area (lbm/(hr*ft**2)).
G = ((20*60)*(4*144))/(math.pi*(0.87**2)); 			#Unit:lbm/(hr*ft**2)  			#Inside diameter = 0.87 inch 			#			#1 in.**2 = 144 ft**2 			#20 lbm/min of water(min converted to second)
#the vismath.cosity of air at these conditions is obtained from figure 11.17 as 0.062 lbm/(ft*hr).So,
mu = 0.062; 			#the vismath.cosity of air 			#unit:lbm/(ft*hr)
D = 0.87/12; 			#Inside diameter 			#1 in**2 = 144 ft**2
#Reynolds number is DG/mu,therefore
Re = (D*G)/mu; 			#Reynolds number
print "The Reynolds number is %.2f"%(Re);
#which places the flow in the turbulent regime.Because W/1000(W = weight flow) is same as for problem 11.17 and equals 1.2,we now enter figure 11.19 at 1.2 and 400F   to obtain h1 = 135.Because the inside tube diameter is same as before,F = 1.25.Therefore,
h1 = 135; 			#basic heat transfer coefficient 			#unit:Btu/(hr*ft**2*F)
F = 1.25; 			#correction factor
h = h1*F; 			#heat transfer coefficient 			#the inside film coefficient 			#unit:Btu/(hr*ft**2*F)
print "The inside film coefficient is %.2f Btu/hr*ft**2*F)"%(h);
#It is interesting that for equal mass flow rates,water yields a heat-transfer coefficient almost five times greater than air

The Reynolds number is 339908.22
The inside film coefficient is 168.75 Btu/hr*ft**2*F)


## Example 11.19 Page No : 586¶

In [24]:
#A bare steel pipe
#From the Table 11.5,case 2,
Fe = 0.79; 			#Emissivity factor to allow for the departure of the surfaces interchanging heat from complete blackness;Fe is a function of the surface emissivities and       configurations
FA = 1; 			#geometric factor to allow for the average solid angle through which one surface "sees" the other
sigma = 0.173*10**-8; 			#Stefan-Boltzmann constant 			#Unit:Btu/(hr*ft**2*R**4)
T1 = 120+460; 			#outside temperature 			#Unit:R 			#fahrenheit converted to absolute temperature
T2 = 70+460; 			#inside temperature 			#Unit:R 			#fahrenheit converted to absolute temperature
D = 3.5/12; 			#3.5 inch  =  3.5/12 feet			#Unit:ft 			#Outside diameter
L = 5; 			#Length 			#Unit:ft 			#From problem 11.10
A = (math.pi*D)*L;  			#Area 			#Unit:ft**2
Q = sigma*Fe*FA*A*(T1**4-T2**4); 			#The net interchange of heat by radiation between two bodies at different temperatures 			#Unit:Btu/hr 			#			#Stefan-Boltzmann law

# Results
print "The heat loss by radiation is %.2f Btu/hr"%(Q);

The heat loss by radiation is 214.52 Btu/hr


## Example 11.20 Page No : 588¶

In [25]:
#The upper temperature is given as 120 F and the temperature difference is
Ti = 120; 			#Inside temperature 			#unit:fahrenheit
To = 70; 			#Outside temperature 			#unit:fahrenheit
deltaT = 120-70; 			#unit:fahrenheit 			#Change in temperature
#Using figure 11.28,
hrdash = 1.18; 			#factor for radiation coefficient 			#Unit:Btu/(hr*ft**2*F)
Fe = 1; 			#Emissivity factor to allow for the departure of the surfaces interchanging heat from complete blackness;Fe is a function of the surface emissivities and       configurations
FA = 0.79; 			#geometric factor to allow for the average solid angle through which one surface "sees" the other
hr = Fe*FA*hrdash; 			#The radiation heat-transfer coefficient for the pipe 			#Unit:Btu/(hr*ft**2*F)
print "The radiation heat-transfer coefficient for the pipe is %.2f Btu/hr*ft**2*F)"%(hr);

#As a check,Using the results of problem 11.17,
print "As a check, Using the results of problem 11.17"
D = 3.5/12; 			#3.5 inch  =  3.5/12 feet			#Unit:ft 			#Outside diameter
L = 5; 			#Length 			#Unit:ft 			#From problem 11.10
A = (math.pi*D)*L;  			#Area 			#Unit:ft**2
Q = 214.5; 			#heat loss 			#Unit:Btu/hr
hr = Q/(A*deltaT); 			#The radiation heat-transfer coefficient for the pipe 			#Unit:Btu/(hr*ft**2*F) 			#Newton's law of cooling
print "The radiation heat-transfer coefficient for the pipe is %.2f Btu/hr*ft**2*F)"%(hr);

The radiation heat-transfer coefficient for the pipe is 0.93 Btu/hr*ft**2*F)
As a check, Using the results of problem 11.17
The radiation heat-transfer coefficient for the pipe is 0.94 Btu/hr*ft**2*F)


## Example 11.21 Page No : 589¶

In [26]:
#Because the conditions of illustrative problem 11.15 are the same as for problem 11.19 and 11.20,we can solve this problem in two ways to obtain a check.
#Thus,adding the results of these problems yields,
print "Adding the results of the problems yields"
Qtotal = 206.2+214.5; 			#Unit:Btu/hr 			#total heat loss
print "The heat loss due to convection is %.2f Btu/hr"%(Qtotal);

#We can also approach this solution by obtaining radiation and convection heat-transfer co-efficcient.Thus,
hcombined = 0.9+0.94; 			#Coefficient of heat transfer 			#Unit:Btu/(hr*ft**2*F)
D = 3.5/12; 			#3.5 inch  =  3.5/12 feet 			#Unit:ft 			#Outside diameter
Ti = 120; 			#Inside temperature 			#unit:fahrenheit
To = 70; 			#Outside temperature 			#unit:fahrenheit
deltaT = Ti-To; 			#unit:fahrenheit 			#Change in temperature
L = 5; 			#Length 			#Unit:ft 			#From problem 11.10
A = (math.pi*D)*L;  			#Area 			#Unit:ft**2
Qtotal = hcombined*A*deltaT; 			#Unit:Btu/hr 			#total heat loss due to convection 			#Newton's law of cooling
print "By obtaining radiation and convection heat-transfer co-efficcient"
print "The heat loss due to convection is %.2f Btu/hr"%(Qtotal);

Adding the results of the problems yields
The heat loss due to convection is 420.70 Btu/hr
By obtaining radiation and convection heat-transfer co-efficcient
The heat loss due to convection is 421.50 Btu/hr


## Example 11.22 Page No : 595¶

In [18]:
from numpy.linalg import inv

#For brick,concrete,plaster,hot film and cold film,
A = 1.; 			#area 			#Unit:ft**2
#For a plane wall,the areas are all the same,and if we use 1 ft**2 of wall surface as the reference area,
#For Brick,
deltax = 6./12; 			#6 inch  =  6/12 feet 			#deltax = length 			#unit:ft
k = 0.40; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
brickResistance = deltax/(k*A); 			#Thermal resistance 			#Unit:(hr*f)/Btu
print "For brick";
print "The resistance is %.2f hr*F)/Btu"%(brickResistance);

#For Concrete,
deltax = (1./2)/12; 			#(1/2) inch  =  (1/2)/12 feet 			#deltax = length 			#unit:ft
k = 0.80; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
concreteResistance = deltax/(k*A); 			#Thermal resistance 			#Unit:(hr*f)/Btu
print "For Concrete"
print "The resistance is %.2f hr*F)/Btu"%(concreteResistance);

#For plaster,
deltax = (1./2)/12; 			# (1/2) inch  =  6/12 feet 			#deltax = length 			#unit:ft
k = 0.30; 			#Unit:Btu/(hr*ft*F) 			#k = proportionality constant 			#k = thermal conductivity 			#From the table
plasterResistance = deltax/(k*A); 			#Thermal resistance 			#Unit:(hr*f)/Btu
print "For plaster";
print "The resistance is %.2f hr*F)/Btu"%(plasterResistance);

#For "hot film",
h = 0.9; 			#Coefficient of heat transfer 			#Unit:Btu/(hr*ft**2*F)
hotfilmResistance = inv([[h*A]]); 			#Thermal resistance 			#Unit:(hr*f)/Btu
print "For hot film"
print "The resistance is %.2f hr*F)/Btu"%(hotfilmResistance);

#For "cold film",
h = 1.5; 			#Coefficient of heat transfer 			#Unit:Btu/(hr*ft**2*F)
coldfilmResistance = inv([[h*A]]); 			#Thermal resistance 			#Unit:(hr*f)/Btu
print "For cold film"
print "The resistance is %.2f hr*F)/Btu"%(coldfilmResistance);

totalResistance = brickResistance+concreteResistance+plasterResistance+hotfilmResistance+coldfilmResistance; 			#the overall resistance  			#Unit:(hr*f)/Btu
print "The overall resistance is %.2f hr*F)/Btu"%(totalResistance);

U = inv([[totalResistance]]); 			#Unit:Btu/(hr*ft**2) 			#The overall conducmath.tance(or overall heat-transfer coefficient)
print "The overall conductanceor overall heat-transfer coefficient) is %.2f Btu/hr/ft**2)"%(U);
#In problem 11.21,the solution is straightforward,because the heat-transfer area is constant for all series resistances.

For brick
The resistance is 1.25 hr*F)/Btu
For Concrete
The resistance is 0.05 hr*F)/Btu
For plaster
The resistance is 0.14 hr*F)/Btu
For hot film
The resistance is 1.11 hr*F)/Btu
For cold film
The resistance is 0.67 hr*F)/Btu
The overall resistance is 3.22 hr*F)/Btu
The overall conductanceor overall heat-transfer coefficient) is 0.31 Btu/hr/ft**2)


## Example 11.23 Page No : 596¶

In [29]:
import math

# given data
hi = 45; 			#Film coefficient on the inside of the pipe 			#Unit:Btu/(hr*ft**2*F)
r1 = 3.0/2; 			#Inside radius 			#Unit:inch
k1 = 26; 			#Unit:Btu/(hr*ft**2*F) 			#k = proportionality constant for steel pipe 			#k = thermal conductivity for fir 			#From the table
r2 = 3.5/2; 			#outide radius 			#Unit:inch
k2 = 0.026; 			#Unit:Btu/(hr*ft**2*F) 			#k = proportionality constant for mineral wool 			#k = thermal conductivity for fir 			#From the table
ho = 0.9; 			#Film coefficient on the outside of the pipe 			#Unit:Btu/(hr*ft**2*F)
#Results of problem 11.23,
Ui = 1/((1/hi)+((r1/(k1*12))*math.log(r2/r1))+((r1/(k2*12))*math.log(r3/r2))+(1/(ho*(r3/r1)))); 			#Unit:Btu/(hr*ft**2*F) 			#1 in. = 12 ft 			#Heat transfer coefficient based on inside surface
print "Heat transfer coefficient based on inside surface is %.2f Btu/hr*ft**2*F)"%(Ui);
#Because Uo*Ao = Ui*Ai
Uo = Ui*(r1/r3); 			#Heat transfer coefficient based on outside surface 			#Unit:Btu/(hr*ft**2*F)
print "Heat transfer coefficient based on outside surface is %.2f Btu/hr*ft**2*F)"%(Uo);

Heat transfer coefficient based on inside surface is 0.36 Btu/hr*ft**2*F)
Heat transfer coefficient based on outside surface is 0.20 Btu/hr*ft**2*F)


## Example 11.24 Page No : 601¶

In [30]:
import math

#A COUNTERFLOW HEAT EXCHANGER
#Hot oil enters at 215 F and leaves at 125 F
#Water enters the unit at 60 F and leaves at 90 F
#Therefore,From figure 11.34,
thetaA = 215.-90; 			#the greatest temperature difference between the fluids(at either inlet or outlet) 			#Unit:fahrenheit
thetaB = 125.-60; 			#the least temperature difference between the fluids(at either inlet or outlet) 			#Unit:fahrenheit
deltaTm = (thetaA-thetaB)/math.log(thetaA/thetaB); 			#math.logarithmic mean temperature difference 			#Unit:fahrenheit
#From the oil data,
m = 400*60;  			#mass 			#Unit:lb/sec 			#1 min = 60 sec
Cp = 0.85; 			#Specific heat of the oil 			#Unit:Btu/(lb*F)
deltaT = 215.-125;  			#Change in temperature 			#Unit:fahrenheit
Q = m*Cp*deltaT 			#The heat transfer 			#Unit:Btu/hr
#Q = U*A*deltaTm
U = 40.;			#The overall coefficient of heat transfer of the unit 			#Unit:Btu/(hr*ft**2*F)
A = Q/(U*deltaTm); 			#Umit:ft**2 			#The outside surface area
print "The outside surface area required is %.2f ft**2"%(A);

The outside surface area required is 500.25 ft**2


## Example 11.25 Page No : 602¶

In [32]:
import math

#In problem 11.24, A COUNTERFLOW HEAT EXCHANGER is operated in the parallel flow
#Hot oil enters at 215 F and leaves at 125 F
#Water enters the unit at 60 F and leaves at 90 F
#Therefore,From figure 11.35,
thetaA = 215.-60; 			#the greatest temperature difference between the fluids(at either inlet or outlet) 			#Unit:fahrenheit
thetaB = 125.-90; 			#the least temperature difference between the fluids(at either inlet or outlet) 			#Unit:fahrenheit
deltaTm = (thetaA-thetaB)/math.log(thetaA/thetaB); 			#math.logarithmic mean temperature difference 			#Unit:fahrenheit
#From the oil data,
m = 400*60.;  			#mass 			#Unit:lb/sec 			#1 min = 60 sec
Cp = 0.85; 			#Specific heat of the oil 			#Unit:Btu/(lb*F)
deltaT = 215.-125;  			#Change in temperature  			#Unit:fahrenheit
Q = m*Cp*deltaT 			#The heat transfer 			#Unit:Btu/hr
#Q = U*A*deltaTm
U = 40.;			#The overall coefficient of heat transfer of the unit 			#Unit:Btu/(hr*ft**2*F)
A = Q/(U*deltaTm); 			#Umit:ft**2 			#The outside surface area
print "The outside surface area required is %.2f ft**2"%(A);

The outside surface area required is 569.19 ft**2


## Example 11.26 Page No : 603¶

In [3]:
#From the table 11.7,
#For the oil side,a resistance(fouling factor) of 0.005 (hr*F*ft**2)/Btu can be used
#and for the water side,a fouling factor of 0.001 (hr*F*ft**2)/Btu can be used
#From problem 11.25,
U = 40;			#The coefficient of heat transfer of the unit 			#Unit:Btu/(hr*ft**2*F)
#therefore,
Roil = 0.005; 			#unit:(hr*ft**2*F)/Btu 			#resistance at oil side
Rwater = 0.001; 			#unit:(hr*ft**2*F)/Btu 			#resistance for water side
invU = 0.025             # Inv of U
Rcleanunit = invU; 			#unit:(hr*ft**2*F)/Btu 			#resistance at clean unit
Roverall = Roil+Rwater+Rcleanunit; 			#unit:(hr*ft**2*F)/Btu 			#overall resistance
invRoverall = 32.258065;
Uoverall = invRoverall; 			#Unit:Btu/(hr*ft**2*F) 			#The overall coefficient of heat transfer of the unit
#Because all the parameters are the same,the surface area required will vary inversely as U
A = 569*(U/Uoverall); 			#A = 569 ft**2 in the problem 11.25 			#unit:ft**2  			#The outside surface area
print "The outside surface area required is %.2f ft**2"%(A);

The outside surface area required is 705.56 ft**2


## Example 11.27 Page No : 605¶

In [34]:
#HEAT EXCHANGER
#Oil flows in the tube side and is cooled from  280 F to 140 F
#Therefore,
t2 = 140.; 			#Unit:fahrenheit
t1 = 280.; 			#Unit:fahrenheit
#On the shell side,water is heated from 85 F to 115 F
T1 = 85.; 			#Unit:fahrenheit
T2 = 115.; 			#Unit:fahrenheit
P = (t2-t1)/(T1-t1);
R = (T1-T2)/(t2-t1);
#From the figure,
F = 0.91;			#Correction factor
LMTD = ((t1-T2)-(t2-T1))/math.log((t1-T2)/(t2-T1)); 			#LMTD = Log mean temperature difference 			#Unit:fahrenheit
TMTD = F*LMTD; 			#TMTD = True mean temperature difference 			#Unit:fahrenheit

# Results
print "The true mean temperature is %.2f fahrenheit"%(TMTD);

The true mean temperature is 91.11 fahrenheit