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);
```

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);
```

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));
```

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));
```

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 [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));
```

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 [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));
```

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);
```

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);
```

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);
```

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);
```

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);
```

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.
```

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);
```

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
```

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);
```

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);
```

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);
```

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.
```

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
r3 = 5.50/2; #radius #Unit:inch
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);
```

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);
```

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);
```

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);
```

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);
```